2
ARTES
LIBRARY
1837
VERITAS
SCIENTIA
OF THE
UNIVERSITY OF MICHIGAN
L PLURIBUS UNUM
TUEBOR
SI QUERIS PENINSULAM-AMŒNAM
CIRCUMSPICE
FROM THE LIBRARY OF
PROFESSOR W. W. BEMAN
AB.1870: AM,1873
TEACHER OF MATHEMATICS
1871 - 1922
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ELEMENTS
OF
GEOMETRY.
1
.
EDINBURGH :
Printed by Abernethy & Walker.
ELEMENTS
OF

GEOMETRY,
GEOMETRICAL ANALYSIS,
AND
PLANE TRIGONOMETRY.
WITH AN
APPENDIX,
AND COPIOUS NOTES AND ILLUSTRATIONS.
BY
JOHN LESLIE, F. R.S. E.
PROFESSOR OF MATHEMATICS IN THE UNIVERSITY
OF EDINBURGH.
SECOND EDITION,
IMPROVED AND ENLARGED.
EDINBURGH:
PRINTED FOR JOHN BALLANTYNE AND COMPANY;
AND FOR
LONGMAN, HURST, REES, ORME AND BROWN,
LONDON.
1811.
1
gift
Thof. W. W. Bernie vý
4-12-33
$
PREFACE.
THE Volume now laid before the public, is the first
of a projected Course of Mathematical Science.
Many compendiums or elementary treatises have
appeared at different times, and of various merit;
but there seemed still wanting in our language, a work
that should embrace the subject in its full extent,—
that should unite theory with practice, and con-
nect the ancient with the modern discoveries. The
magnitude and difficulty of such a task might deter
an individual from the attempt, if he were not deep-
ly impressed with the importance of the underta-
king, and felt his exertions to accomplish it anima-
ted by zeal, and supported by active perseverance.
The study of Mathematics holds forth two capi-
tal objects:-While it traces the beautiful relations
of figure and quantity, it likewise accustoms the
mind to the invaluable exercise of patient attention
and accurate reasoning. Of these distinct objects,
the last is perhaps the most important in a course of
vi
PREFACE.
liberal education. For this purpose, the geometry
of the Greeks is the most powerfully recommended,
as bearing the stamp of that acute people, and dis-
playing the finest specimens of logical deduction.
Some of the conclusions, indeed, might be reached
by a sort of calculation; but such an artificial
mode of procedure gives only an apparent facility,
and leaves no clear or permanent impression on the
mind.
We should form a wrong estimate, however, did
we consider the Elements of Euclid, with all its
merits, as a finished production. That admirable
work was composed at the period when geometry
was making its most rapid advances, and new pro-
spects were opening on every side. No wonder
that its structure should now appear loose and defec-
tive. In adapting it to the actual state of the sci-
ence, I have therefore endeavoured carefully to re-
tain the spirit of the original, but have sought to
enlarge the basis, and to dispose the accumulated
materials into a regular and more compact system.
By simplifying the order of arrangement, I presume
that I have materially abridged the labour of the
student. The numerous additions which are incor-
porated in the text, so far from retarding, will ra-
ther facilitate his progress, by rendering more con-
tinous the chain of demonstration.
The view which I have given of the nature of
Proportion, in the fifth Book, will, I hope, contribute
to remove the chief difficulties attending that im-
PREFACE.
vii
portant subject. The sixth Book, which exhibits
the application of the doctrine of ratios, contains 'a
copious selection of propositions, not only beautiful
in themselves, but which pave the way to the higher
branches of Geometry, or lead immediately to va-
luable practical results. The Appendix, without
claiming the same degree of utility, will not per-
haps be deemed the least interesting portion of the
volume, since the ingenious resources which it dis-
closes for the construction of certain problems are
calculated to afford a very pleasing and instructive
exercise.
The part which has cost me the greatest pains,
is that devoted to Geometrical Analysis. The first
Book consists of a series of the choicest problems,
rising above each other in gradual succession. The
second and third Books are almost wholly occupied
with the researches of the Ancient Analysis. In
framing them, I have consulted a great diversity of
authors, some of whom are of difficult access. The
labour of condensing the scattered materials, will be
duly estimated by those, who, taking delight in such
fine speculations, are thus admitted at once to a rich
and varied repast. The analytical investigations of
the Greek geometers are indeed models of simplici-
ty, clearness, and unrivalled elegance; and though
miserably defaced by the riot of time and barbarism,
they will yet be regarded by every person capable
of appreciating their beauties, as among the noblest
monuments of human genius. It is matter of deep
regret, that Algebra, or the Modern Analysis, from
.1
viii
PREFACE.
[
the mechanical facility of its operations, has contri-
buted, especially on the Continent, to vitiate the
taste and destroy the proper relish for the strictness
and purity so conspicuous in the ancient method of
demonstration. The study of geometrical analysis
appears admirably fitted to improve the intellect, by
training it to habits of precision, arrangement, and
close application. If the taste thus acquired be not
allowed to obtain undue ascendency, it may be
transferred with eminent utility to Algebra, which,
having shot up prematurely, wants reform in almost
every department.
The Elements of Trigonometry are as ample as
my plan would allow. I have explained fully the
properties of the lines about the circle, and the cal-
culation of the trigonometrical tables; nor have I
omitted any proposition which has a distinct refe-
rence to practice. Some of the problems annexed
are of essential consequence in marine surveying.
In the improvement of this edition, I have spared
no trouble or expence. The whole has undergone a
careful and minute revision; and by adopting a
denser page, I have been able, without adding much
to the size of the volume, yet greatly to augment
the work. The text has been simplified and reduced
to a shorter compass, by throwing such propositions
as were less elementary to the Notes. Other Notes
of a simpler kind, and marked by the reference in
italics, are intended chiefly to engage the attention
of the young student. In various parts, the demon-
PREFACE.
ix
strations are occasionally abridged. The books of
Geometrical Analysis have been rendered more
complete by numerous insertions, and particularly
an abstract of all the different investigations left
us by the ancients, relating to the trisection of an
angle and the duplication of the cube. The Ele-
ments of Trigonometry are much expanded, and
brought to include whatever appears most valuable
in recent practice. But the greatest additions have
been made in the Notes and Illustrations, which
will be found, I should hope, to contain various and
useful information. The more advanced student
may peruse with advantage the historical and criti-
cal remarks; and some of the disquisitions, and the
solutions of certain more difficult problems relative
to trigonometry and geodesiacal operations, in which
the modern analysis is sparingly introduced, are of
a nature sufficiently interesting, perhaps, to claim
the notice of proficients in science. I have sim-
plified, and materially enlarged the formulæ con-
nected with trigonometrical computation; explain-
ed the art of surveying, in its different branches;
and given reduced plans, blended with the narrative
of the great operations lately carried on both in
England and France. I have likewise shown a very
simple method for calculating heights from barome-
trical observations, accompanied by illustrative sec-
tions;
and I have been thence led to state the law
of climate, as it is modified by elevation. On this
attractive subject, I should have dwelt with pleasure,
had the limits of the volume permitted.
.
X
PREFACE.
.
The plan now in contemplation might perhaps
be comprised in five volumes. The next volume
is intended to treat of the Geometry of curve lines,
the intersections of planes, and the properties of
solids, including the doctrine of the sphere and
the calculation of spherical triangles, with the
elements of perspective and projection. The third
volume will be devoted to Algebra,-a wide and rank
field, which still needs the most sedulous cultivation.
As an introduction to that difficult task, I design to
prepare, with all convenient speed, a short tract on
the Principles of Arithmetic; a work much wanted
in our ordinary course of education, and which,
were it executed with taste, and in the spirit of phi-
losophy, would be admirably fitted for opening the
mind of the pupil. The fourth and fifth volumes
will embrace the differential and integral calculus,
with their principal applications. But to accomplish
this vast undertaking would require years of health
and inflexible resolution; and under all the dis-
couragements which attend the publication of sci-
entific works, I cannot look forward to its comple-
tion without feeling extreme solicitude.
It is the nature of mathematical science to ad-
vance in continual progression. Each step carries
it to others still higher; new relations are descried;
and the most distant objects seem graduaily to ap-
proximate. But, while science thus enlarges its
bounds, it likewise tends uniformly to simplicity and
concentration. The discoveries of one age are, per
haps in the next, melted down into the mass of ele-
PREFACE.
採
​mentary truths. What are deemed at first merely
objects of enlightened curiosity, become, in due
time, subservient to the most important interests.
Theory soon descends to guide and assist the opera-
tions of practice. To the geometrical speculations
of the Greeks, we may distinctly trace whatever
progress the moderns have been enabled to achieve
in mechanics, navigation, and the various complicat-
ed arts of life. A refined analysis has unfolded the
harmony of the celestial motions, and conducted the
philosopher, through a maze of intricate phenome-
na, to the great laws appointed for the government
of the Universe.
COLLEGE OF EDINBURGH,
September 2. 1811.
!
:
}
1
ELEMENTS
OF
GEOMETRY.
GEOMETRY is that branch of natural science which
treats of figured space.
Our knowledge concerning external objects is
grounded entirely on the information received
through the medium of the senses. The science of
physics considers bodies as they actually exist, in-
vested at once with all their various qualities, and
endued with their peculiar affections. Its researches
are hence directed by that refined species of obser-
vation which is termed Experiment. Geometry takes
a more limited view, and, selecting only the generic
property of magnitude, it can, from the extreme sim-
plicity of its basis, safely pursue the most lengthened
train of investigation, and arrive with perfect certain-
ty at the remotest conclusion. It contemplates mere-
ly the forms which bodies present, and the spaces
which they occupy. Geometry is thus likewise found-
ed on external observation; but such observation is
A
$
ELEMENTS OF GEOMETRY.
so familiar and obvious, that the primary notions which
it furnishes might seem intuitive, and have often been
regarded as innate. This science is therefore super-
eminently distinguished by the luminous evidence
which constantly attends every step of its progress.
PRINCIPLES.
IN contemplating an external object, we can, by
successive acts of abstraction, reduce the complex
idea which arises in the mind into others that are pro-
gressively simpler. Body, divested of its essential cha-
racters, presents the mere idea of surface; a surface,
considered apart from its peculiar qualities, exhibits
-only linear boundaries; and a line, abstracting its
continuity, leaves nothing in the imagination but the
points which form its extremities. A solid is bounded
by surfaces; a surface is circumscribed by lines; and
a line is terminated by points. A point marks posi-
tion; a line measures distance; and a surface represents
extension. A line has only length; a surface has both
length and breadth; and a solid combines all the three
dimensions of length, breadth, and thickness.
The uniform description of a line which through its
whole extent stretches in the same direction, gives
the idea of a straight line. No more than one straight
line can therefore join two points.
From our idea of the straight line is derived that of
a plane surface, which, though more complex, has a
ELEMENTS OF GEOMETRY.
3
like uniformity of character. A straight line joining
any two points situate in a plane, lies wholly on the
surface; and consequently planes admit, in every
way, a mutual and perfect application.
Two points ascertain the position of a straight line;
for the line may continue to turn about one of the
points till it falls upon the other. But to determine.
the position of a plane, it requires three points; be-
cause a plane touching the straight line which joins
two of the points, may be made to revolve, till it
meets the third point.
The separation or opening of two straight lines at
their point of intersection, constitutes an angle. If
we obtain the idea of distance, or linear extent, from
progressive motion, we derive that of divergence, or
angular magnitude, from revolving motion *.
GEOMETRY is divided into Plane and Solid; the
former confining its views to the properties of space
delineated on the same plane; the latter embracing
the relations of different planes or surfaces, and of
the solids which these describe or terminate. In the
following definitions, therefore, the points and lines
are all considered as existing in the same plane.
*See Note I.
גי
A 2
1
Z
1
[ 5 ]
BOOK I.
DEFINITIONS.
1. A crooked line is that which consists of
straight lines not continued in the same di-
rection.
2. A curved line is that of which no por-
tion is a straight line.
о
3. The straight lines which contain an angle are termed its
sides, and their point of origin or intersection, its vertex.
To abridge the reference, it is usual to denote an angle by tracing
over its sides; the letter at the vertex, which is common to them
both, being placed in the middle.
Thus, the angle contained by the straight lines.
AB and BC, or the opening formed by turning
BA about the point B into the position BC, is A
named ABC or CBA,
4. A right angle is the fourth part of an entire
circuit or revolution.
B
If a straight line CB stand at equal angles CBA and CBD on ano`
ther straight line AD, and if the surface ACD be laid over towards
the opposite part, the point B and the line
AD remaining the same; CB will, in this
new position EB, make angles EBA and
EBD equal to the former, and therefore all
of them equal to each other, But the four
angles ABC, CBD, DBE and EBA consti-

A
C
I
D
B
tute about the point B a complete revolution; or the line BA in
forming them, by its successive openings, would return into its origi-
nal place, and consequently each of those angles is a right angle.
6
ELEMENTS OF GEOMETRY.
The angle contained by the opposite portions DA and DB of a
straight line is hence equal to two right
angles; and, for the same reason, all the
angles ADC, CDE, EDF and FDB, formed
at the point D and on the same side of the
straight line AB, are together equal to two
right angles.
E
A-
-B
D
It is manifest that all right angles, being derived from the same
measure, must be equal to each other.
5. The sides of a right angle are said to be perpendicular to
each other.
6. An acute angle is less than a right an-
gle.
7. An obtuse angle is greater than a right
angle.
8. One side of an angle forms with the
other produced a supplemental or exterior
angle.
9. A vertical angle is formed by the pro-
duction of both its sides.
10. The retroflected divergence of the two sides, or the
defect of the angle from four right angles, is named a reverse
angle.
The angle DBE is vertical to ABC, ABD is the supplemental or
exterior angle, and the angle made up of ABD, A
DBE, and EBC, or the opening formed by the
regression of AB through the points D and E
-D
B
into the position BC, is the reverse angle.
E
It is apparent that vertical angles, or those formed by the same
1




BOOK I.
7
lines in opposite directions, must be equal; for the angles CBA and
ABD which stand on the straight line CD, being equal to two right
angles, are equal to ABD and DBE, and, omitting the common angle
ABD, there remains CBA equal to DBE.
11. Two straight lines are said to be
inclined to each other, if they meet when
produced; and the angle so formed is
called their inclination.
12. Straight lines which have no inclina-
tion, are termed parallel.
13. A figure is a plane surface included by a linear boundary
called its perimeter.
14. Of rectilineal figures, the triangle is contained by three
straight lines.
15. An isosceles triangle is that which has two
of its sides equal.
16. An equilateral triangle is that which has all
its sides equal.

17. A triangle whose sides are unequal, is
named scalene.
It will be shown (I. 10.) that every triangle has at least two acute
angles. The third angle may therefore, by its character, serve to
discriminate a triangle.

18. A right-angled triangle is that which has
a right angle.
19. An obtuse angled triangle is that
which has an obtuse angle.
ELEMENTS OF GEOMETRY.

ހ
20. An acute angled triangle is that which has
all its angles acute.
21. Two triangles which are both of them right angled, or
obtuse, or acute, are said to have the same affection.
22. Any side of a triangle may be called its base, and the
opposite angular point its vertex.
23. A quadrilateral figure is contained by four straight lines.
24. Of quadrilateral figures, a trapezoid
(1) has two parallel sides :
25. A trapezium (2) has two of its sides pa-
rallel and the other two equal, though not pa-
rallel, to each other:


26. A rhomboid (3) has its opposite
sides equal:

27. A rhombus (4) has all its sides equal:

28. An oblong, or rectangle, (5) has a right
angle, and its opposite sides equal :

29. A square (6) has a right angle, and all its
sides equal.
30. The straight line which joins obli-
quely the opposite angular points of a qua-
drilateral figure, is named a diagonal.

BOOK I.
9
31. A rectilineal figure having more than four sides, bears
the general name of a polygon.
32. If an angle of a polygon be less than two right angles,
it protrudes and is called salient; if it be greater than two
right angles, it makes a sinuosity and is termed re-entrant.
C
D
Thus the angle ABC is re-entrant, and the rest
of the angles of the polygon ABCDEF are sali-
ent at A, C, D, E and F.
E
B
A
F
}

33. A circle is a figure described by the revolution of a
straight line about one of its extremities.
34. The fixed point is called the centre of the
circle, the describing line its radius, and the
boundary traced by the remote end of that line
its circumference.

35. The diameter of a circle is a straight line
drawn through the centre, and terminated both ways by the
circumference.
It is obvious that all radii of the same circle are equal to each
other and to a semidiameter. It likewise appears from the slightest
inspection, that a circle can only have one centre, and that circles are
equal which have equal diameters.
36. Figures are said to be equal, when, applied to each
other, they wholly coincide; they are equivalent, if, without
coinciding, they yet contain the same space *.
* See Note II.
10
ELEMENTS OF GEOMETRY.
A PROPOSITION is a distinct portion of abstract science. It
is either a problem or a theorem.
A PROBLEM proposes to effect some combination
A THEOREM advances some truth, which is to be establish-
ed.
A problem requires solution, a theorem wants demonstration;
the former implies an operation, and the latter generally needs
a previous construction.
A direct demonstration proceeds from the premises, by a re-
gular deduction.
An indirect demonstration attains its object, by showing that,
any other hypothesis than the one advanced would involve a
contradiction, or lead to an absurd conclusion.
A subordinate property, included in a demonstration, is
sometimes, for the sake of unity, detached, and then it forms
a LEMMA.
A COROLLARY is an obvious consequence that results from
à proposition.
A SCHOLIUM is an excursive remark on the nature and ap-
plication of a train of reasoning.
The operations in Geometry suppose the drawing of straight
tines and the description of circles, or they require in practice
the use of the rule and compasses.
BOOK I.
11
PROPOSITION I. PROBLEM.
To construct a triangle, of which the three sides
are given.
Let AB represent the base, and G, H two sides of the tri-
angle, which it is required to construct.
From the centre A, with the distance G, describe a circle,
and, from the centre B with the distance H, describe another
circle, meeting the former in the
point C: ACB is the triangle re-
quired.
Because all the radii of the same
circle are equal, AC is equal to G;
and, for the same reason, BC is
A.
G

H
C
B
equal to H. Consequently the triangle ACB answers the con-
ditions of the problem.
Corollary. If the radii G and H be equal to each other,
the triangle will evidently be isosceles; and if those lines be
likewise equal to the base AB, the triangle must be equilateral.
—The limiting circles, after intersecting, must obviously diverge
from each other, till, crossing the extension of the base AB,
they return again and meet below it; thus marking two posi-
tions for the required triangle.
PROP. II. THEOREM.
Two triangles are equal, which have all the sides of
the one equal to the corresponding sides of the other.
Let the two triangles ABC and DFE have the side AB
equal to DF, AC to DE, and BC to FE: These triangles are
equal.
12
ELEMENTS OF GEOMETRY.
For let the triangle ACB be applied to DEF, in the same
position. The point A being laid on D, and the side AC on
DE, their other extremities C and E must coincide, since AC
is equal to DE. And because AB is equal to DF, the point
B must be found in the circumference
of a circle described from D with the
distance DF; and, for the same rea-
son, B must also be found in the cir-
cumference of a circle described from D

B
AA
EA
E with the distance EF: The vertex of the triangle ACB
must, therefore, occur in a point which is common to both
those circles, or, by the first proposition, in F the vertex of
the triangle DFE. Consequently these two triangles, being
rectilineal, must entirely coincide. The angle CAB is equal
to EDF, ACB to DEF, and CBA to EFD; the equal an-
gles being thus always opposite to the equal sides.
PROP. III. THEOR.
Two triangles are equal, if two sides and the angle
contained by these in the one be respectively equal
to two sides and the contained angle in the other.
Let ABC and DEF be two triangles, of which the side AB
is equal to DE, the side BC to EF, and the angle ABC con-
tained by the former equal to DEF which is contained by the
latter: These triangles are equal.
For let the triangle ABC be applied to DEF: The vertex
B being placed on E, and the side BA on ED, the extremity
A must fall upon D, since AB is
+
B
I


equal to DE. And because the
angle or divergence ABC is e-
qual to DEF, and the side AB A
CD
coincides with DE, the other side BC must lie in the same di-
rection with EF, and being of the same length, must entirely
coincide with it; and consequently, the points A and C rest-
BOOK I
13
ing on D and F, the straight lines AC and DF will also coin-
cide. Wherefore, the one triangle being thus perfectly a-
dapted to the other, a general equality must obtain between
them: The third sides AC and DF are hence equal, and the
angles BAC, BCA opposite to BC and BA are equal respec-
tively to EDF and EFD, which the corresponding sides EF
and ED subtend.
PROP. IV. PROB.
At a point in a straight line, to make an angle
equal to a given angle.
At the point D in the given straight line DE, to form an
angle equal to the given angle BAC.
In the sides AB and AC of the given angle, assume the
points G and H, join GH, from
DE cut off DI equal to AG, and
on DI constitute (I. 1.) a trian-
gle DKI, having the sides DK
and IK equal to AH and GH: B
АД

G
EDK or EDF is the angle required.
CE
K
For all the sides of the triangles GAH and IDK being re-
spectively equal, the angles opposite to the equal sides must be
likewise equal (I. 2.), and consequently IDK is equal to GAH.
F
Scholium. If the segments AG, AH be taken equal, the con-
struction will be rendered simpler and more commodious.-By
the successive application of this problem, an angle may be con-
tinually multiplied. Two circles CEG and ADF being descri-
bed from the vertex B of the given angle
with radii BC and BA equal to its sides,
and the base AC being repeated between
those circumferences; a multitude of tri-
angles are thus formed, all of them equal
to the original triangle ABC. Conse-

G
E
quently the angle ABD is double of ABC, ABE triple, ABF
quadruple, ABG quintuple, &c.
14
ELEMENTS OF GEOMETRY.
H
B
If the sides AB and BC of the given angle be supposed
equal, only one circle will be required, a
series of equal isosceles triangles being con-
stituted about its centre. It is evident that
this addition is without limit, and that the
angle so produced may continue to spread
out, and its expanding side even make re-
peated revolutions.

PROP. V. PROB.
To bisect a given angle.
Let ABC be an angle which it is required to bisect.
B
In the side AB take any point D, and from BC cut off BE
equal to BD; join DE, on which construct the isosceles triangle
DFE (I. 1.), and draw the straight line
BF: The angle ABC is bisected by BF.
For the two triangles DBF and EBF,
having the side DB equal to EB, the side
DF to EF, and BF common to both, are Á
(I. 2.) equal, and consequently the angle

DBF is equal to EBF.
D
E
C
F
Cor. Hence the mode of drawing a perpendicular from a
given point B in the straight line AC; for the angle ABC,
which the opposite segments BA and
BC make with each other, being c-
F

qual to two right angles, the straight
line that bisects it must be the per-
pendicular required. Taking BD,
B
E C
therefore, equal to BE, and con- A D
structing the isosceles triangle DFE; the straight line BF
which joins the vertex of the triangle, is perpendicular to AC.
BOOK I.
15
1
PROP. VI. PROB.
To let fall a perpendicular upon a straight line,
from a given point without it.
From the point C, to let fall a perpendicular upon a given
straight line AB.
In AB take the point D, and with the distance DC describe
a circle; and in the same line take another point E, and with
distance EC describe a second circle intersecting the former in
F; join CF, crossing the given line in
G: CG is perpendicular to AB.

B
D
G
F
For the triangles DCE and DFE have
the side DC equal to DF, EC to EF, and
DE common to them both; whence (I. 2.)
the angle CDE or CDG is equal to F'DE
or FDG. And because in the triangles DCG and DFG, the
side DC is equal to DF, DG common, and the contained angles
CDG and FDG are proved to be equal; these triangles are
(I.3.) equal, and consequently the angle DGC is equal to DGF,
and each of them a right angle, or CG is perpendicular to AB.
PROP. VII. PROB.
To bisect a given finite straight line.
On the given straight line AB, construct two isosceles trian-
gles (I. 1.) ACB and ADB, and join their vertices C and D by
a straight line cutting AB in the point E: AB is bisected in E.
For the sides AC and AD of the triangle CAD
being respectively equal to BC and BD of the
triangle CBD, and the side CD common to them
C
both; these triangles (I. 2.) are equal, and the A
angle ACD or ACE is equal to BCD or BCE.
B
E
Again, the triangles ACE and BCE, having the
side AC equal to BC, CE common, and the contained angle
ACE equal to BCE, are (I. 3.) equal, and consequently the
base AE is equal to BE.
1

16
ELEMENTS OF GEOMETRY.
PROP. VIII. THEOR.
The exterior angle of a triangle is greater than ei-
ther of its interior opposite angles.
1
The exterior angle BCF, formed by producing a side AC
of the triangle ABC, is greater than either of the opposite
and interior angles CAB and CBA.
For bisect the side BC in D (I. 7.),
draw AD, and produce it until DE
be equal to AD, and join EC.
The triangles ADB and EDChave,
by construction, the side DA equal
to DE, the side DB to DC, and the
vertical angle BDA equal to CDE;

A
B
E
C
F
G
these triangles are, therefore, equal (I. 3.), and the angle DCE
is equal to DBA. But the angle BCF is evidently greater
than DCE; it is consequently greater than DBA or CBA.
In like manner, it may be shown, that if BC be produced,
the exterior angle ACG is greater than CAB. But ACG is
equal to the vertical angle BCF, and hence BCF must be
greater than either the angle CBA or CAB. ·
PROP. IX. THEOR.
Any two angles of a triangle are together less than
two right angles.
The two angles BAC and BCA of the triangle ABC are
together less than two right angles.

For produce the common side AC.
And, by the last proposition, the exte-
B
rior angle BCD is greater than BAC,
add BCA to cach, and the two angles A
C
D
BCD and BCA are greater than BAC and BCA, or BAC
and BCA are together less than BCD and BCA, that is, less
than two right angles (Def. 4).
BOOK I.
17
PROP. X. THEOR.
Every triangle has two acute angles.
The triangle ABC obviously may have one angle right, or
obtuse or acute. And first, let it be right
angled at C. By the last proposition, the an-
A
B
C
gles ACB and CAB are less than two right
angles, and so are the angles ACB and ABC.
Consequently the angles CAB and CBA are each of them
less than one right angle, or they are both acute. '
Next, let the triangle have an obtuse angle ACB. The an-
gles ACB and CAB, being less than two right angles, and
ACB being greater than one right angle,
CAB must be still less than a right angle.
B

And the angles ACB and ABC being less
than two right angles, ABC must also be á
still less than one right angle. Consequently the angles
CAB and CBA are both of them acute.
Lastly, let the triangle have the angle at C acute.
of the remaining angles, such as BAC, be like-
If one
B

wise acute, the two angles ACB and BAC
are both of them acute. But if the angle
BAC be either obtuse or a right angle, it A
comes under the two former cases, and the other angles ABC
and ACB are, therefore, acute.
PROP. XI. THEOR.
The angles at the base of an isosceles triangle are
equal.
The angles BAC and BCA at the base of the isosceles
triangle ABC are equal.
B
18
ELEMENTS OF GEOMETRY.
B
For draw (I. 5.) BD bisecting the vertical angle ABC.
Because AB is equal to BC, the side BD
common to the two triangles BDA and BDC,
and the angles ABD and CBD contained by
them are equal; these triangles are equal (I. 3.)
and consequently the angle BAD is equal to Á
BCD.
Cor. Every equilateral triangle is also equiangular *.

PROP. XII. THEOR.
D
If two angles of a triangle be equal, the sides op-
posite to them are likewise equal.
Let the triangle ABC have two equal angles BCA and
BAC; the opposite sides AB and BC are also equal.
For if AB be not equal to CB, let it be equal to CD, and
join AD.
B
D
Comparing now the triangles BAC and DCA, the side AB
is by supposition equal to CD, AC is common
to both, and the contained angle BAC is equal
to DCA; the two triangles (I. 3.) are, therefore,
equal. But this conclusion is manifestly absurd.

C
To suppose then the inequality of AB and BC A
involves a contradiction; and consequently those sides must
be equal.
Cor. Every equiangular triangle is also equilateral.
PROP. XIII. THEOR.
In a triangle, that angle is the greater which lies
opposite to a greater side.
If a side BC of the triangle ABC be greater than BA; the
opposite angle BAC is greater than BCA.
See Note III.
BOOK I.
19
For make BD equal to BA, and join AD. The angle CAB
B
D
C
is greater than DAB; but since BA is equal
to BD, the angle DAB (I. 11.) is equal to
ADB, and consequently CAB is greater than
ADB. Again, the angle ADB, being an ex-
terior angle of the triangle CAD, is (I. 8.) greater than ACD
or ACB; wherefore the angle CAB is much greater than ACB.

A
PROP. XIV. THEOR.
That side of a triangle is the greater which sub-
tends a greater angle.
If, in the triangle ABC, the angle CAB be greater than
ACB; its opposite side BC is greater than AB.
For if BC be not greater than AB, it must be either equal
or less. But it cannot be equal, because the
angle CAB would then be equal to ACB
(I. 11.); nor can BC be less than AB, for then
AB would be greater than BC, and conse- A
B

quently (I. 13.) the angle ACB would be greater than CAB,
or CAB less than ACB, which is absurd. The side BC being
thus neither equal to AB, nor less than it, must therefore
be greater than AB.
PROP. XV. THEOR.
Two sides of a triangle are together greater than
the third side.
The two sides AB and BC of the triangle ABC are togc-
ther greater than the third side AC.
For produce AB until DB be equal to the side BC, and
join CD.
B 2
20
ELEMENTS OF GEOMETRY.
*
Because BC is equal to BD, the angle BCD is equal to
BDC (I. 11.); but the angle ACD is
greater than BCD, and therefore greater
than BDC, or ADC; consequently the
opposite side AD is greater than AC
(I. 14.); and since AD is equal to AB
and BD, or to AB and BC, the two sides A
D

B
C
AB and BC are together greater than the third AC*.
Cor. By an extension of this proposition, it may be shown
that a straight line is the shortest line which will connect two
points +.
PROP. XVI.
THEOR.
The difference between two sides of a triangle is
less than the third side.
Let the side AC be greater than AB, and from it cut off
a part AE equal to AB; the remainder EC
is less than the third side BC.
For the two sides AB and BC are toge-
ther greater than AC (I. 15.); take away the A
B
E
equal lines AB and AE, and there remains BC greater than
EC, or EC is less than BC ‡.
PROP. XVII. THEOR.
Two straight lines drawn to a point within a triangle
from the extremities of its base, are together less than
the sides of the triangle, but contain a greater angle.
The straight lines AD and CD, projected to a point D
within the triangle ABC from the extremities of the base AC,
are together less than the sides AB and CB of the triangle,
but contain a greater angle.
For produce AD to meet CB in E. The two sides AB
and BE of the triangle ABE are greater than the third side AE
(I. 15.); add EC to each, and AB, BE, EC, or AB and BC,
* Sce Note IV.
+ See Note V.
‡ See Note VI,
BOOK I.
21
D
B
E
are greater than AE and EC. But the sides CE and ED
of the triangle DEC are (I. 15.) greater than
DC, and consequently CE, ED, together with
DA, or CE and EA, are greater than CD
and DA. Wherefore the sides AB and BC,
being greater than AE and EC, which are
themselves greater than AD and DC, must be still great-
er than AD and DC, or the lines AD and DC are less than
AB and BC, the sides of the triangle.

A
Again, the angle ADC, being the exterior angle of the
triangle DCE, is greater than DEC (I. 8.); and, for the same
reason, DEC is greater than ABE, the opposite interior angle
of the triangle EAB. Consequently ADC is still greater than
ABE or ABC.
PROP. XVIII. THEOR.
If straight lines be drawn from the same point to
another straight line, the perpendicular is the short-
est of them all; the lines equidistant from it on both
sides are equal; and those more remote are greater
than such as are nearer.
Of the straight lines CG, CE, CD, and CF drawn from a
given point C to the straight line AB, the perpendicular CD
is the least, the equidistant lines CE and CF are equal, but
the remoter line CG is greater than either of these two.
For the right angle CDE, which is equal to CDF, is (I. 8.)
greater than the interior angle CFD of the triangle DCF, and
consequently the opposite side CF
is (I. 14.) greater than CD, or
CD is less than CF.
AGE
C

D F
B
But a straight line drawn of a
determinate length from C to AB,
may have two positions; for, if CE be supposed to turn about
the point C, the angle CEA will (I. 8.) continually decrease,
till, passing from obtuse to acute, it becomes equal to CEF,
22
ELEMENTS OF GEOMETRY.
and then forms (I. 12.) the isosceles triangle ECF.-Because
ED then is by hypothesis equal to FD, CD common to the two
triangles ECD and FCD, and the contained angles. CDE and
CDF equal; these triangles (I. 3.) are equal, and consequent-
ly their bases CE and CF are equal.
Again, because GCD is a right angled triangle, the angle
CGD or CGE is acute (I. 10.), and, for the same reason, the
angle CED of the triangle CDE is acute, and consequently its
adjacent angle CEG is obtuse. Wherefore CEG is still great-
er than CGE, and the opposite side CG greater (I. 14.) than
CE.
Cor. Hence only a single perpendicular CD can be let fall
from the same point C upon a given straight line AB; and
hence also a pair only of equal straight lines greater than CD
can at once be extended from C to AB, making on the same
side, the one an obtuse angle CEA, and the other an acute
angle CFA.—As the term distance signifies the shortest road,
the distance between two points is the straight line which joins
them; and the distance from a point to a straight line, is the
perpendicular let fall upon it.
PROP. XIX. THEOR.
If two sides of one triangle be respectively equal to
those of another, but contain a greater angle; the base
also of the former will be greater than that of the lat-
ter.
In the triangles ABC and DEF, let the sides AB and BC
be equal to DE and EF, but the angle ABC greater than
DEF; then is the base AC greater than DF.
For, suppose AB one of the sides to be not greater than
BC or EF, and (I. 4.) draw BG equal to EF making an
angle ABG equal to DEF, join AG and GC.
Because AB and BG are equal to DE and EF, and the
BOOK I.
23
1
contained angle ABG is equal to DEF; the triangles ABG
and DEF (I. 3.) are equal, and have equal bases AG and DF.
First, let the triangles, ABC and DEF be isosceles. Since
the side AB is equal to BC, the
angle BAC (I. 11.) is equal to
BCA; but (I.8.), the angle BHC
B
E


is greater than BAH or BCH,
and consequently (I. 14.) the side
D
H.
G
BC or BG is greater than BH, or the point G lies beyond H.
Next, suppose the side BC
or EF to be greater than AB
or DE. Wherefore (I. 13.)
the angle BAC is greater than
BCA; but (I. 8.) the exterior
angle BHC of the triangle
ABH is greater than BAH or
A
B


D
H
G
C
F
Ε
BAC, and hence still greater than BCA or BCH; consequent-
ly the side BC or EF is (I. 14.) greater than BH.
In every case; therefore, the point G must lie below the
base AC. But the triangle GBC being evidently isosceles, its
angles BGC and BCG (I. 11.) are equal. Whence the angle
AGC, being greater than BGC or BCG, which again is great-
er than ACG, must be still greater than ACG; and therefore
the opposite side AC is (I. 14.) greater than AG or DF*.
PROP. XX. THEOR.
If two sides of one triangle be respectively equal
to those of another, but stand on a greater base; the
angle contained by the former will be likewise
er than what is contained by the latter.
great-
Let the triangles ABC and DEF have the sides AB and
BC equal to DE and EF, but the base AC greater than DF;
the vertical angle ABC is greater than DEF.
See Note VII.
24
ELEMENTS OF GEOMETRY.
For if ABC be not greater than the angle DEF, it must ei-
ther be equal or less. But it cannot
B
be equal to DEF, for the sides AB,
E

A
C
D
F
BC being then equal to DE, EF, and
containing equal angles, the base
AC would (I. 3.) be equal to DF,
which is contrary to the hypothesis. Still more absurd it would
be to suppose the angle ABC less than DEF, since the triangles
BAC and EDF, having their sides AB, BC equal to DE, EF, -
but the contained angle ABC less than DEF, or DEF greater
than ABC, the base DF would, from the preceding proposi-
tion, be greater than AC, or AC would be less than DF *.
PROP. XXI. THEOR.
Two triangles are equal, which have two angles
and a corresponding side in the one respectively
equal to those in the other.
Let the triangles ABC and DEF have the angle BAC equal
to EDF, the angle BCA to EFD, and a side of the one equal
to a side of the other, whether it be interjacent or opposite to
those equal angles; the triangles will be equal.
First, let the equal sides be AC and DF, which are interja-
cent to the equal angles in both triangles.—Apply the triangle
ABC to DEF; the point A being laid on D, and the straight
line AC on DF, the other extremities C and F must coincide,
since those lines are equal. And be-

cause the angle BAC is equal to EDF,
B
E
and the side AC is applied to DF, the
other side AB must lie along DE; and
for the same reason, the angles BCA A
CD
and EFD being equal, the side CB must lie along FE. Where-
fore the point B, which is common to both the lines AB and
CB, will be found likewise in both DE and FE; that is, it
* See Note VIII.
¿
BOOK I.
25
AC of the angle
B
E
must fall upon the corresponding vertex E. The two triangles
ABC and DEF, thus adapting, are hence entirely equal.
Next, let the equal sides be AB and DE, which are oppo-
site to the equal angles BCA and EFD. The triangle ABC
being laid on DEF, the sides AB and
BAC will apply to DE and DF, the
sides of the equal angle EDF; and
since AB is equal to DE, the points
B and E must coincide; but, by hy-
pothesis, the angles BCA and EFD
being equal, BC must adapt itself to
EF, for otherwise one of those angles becoming exterior,
would (I. 8.) be greater than the other. Whence the triangles
ABC, DEF are entirely coincident, and have those sides equal
which subtend equal angles.
44
A
PROP. XXII. THEOR.
C D
Two triangles are equal, which, being of the same
affection, have two sides and an opposite angle in
the one equal to those in the other.
Let the triangles ABC and DEF have the side AB equal to
DE, BC to EF, and the angles BAC, EDF, opposite to BC,
EF, also equal; the triangles themselves are equal, if both the
angles BCA and EFD be right, or acute, or obtuse.
For, the triangle ABC being applied to DEF, the angle BAC
will adapt itself to EDF, since they are equal; and the point
B must coincide with E, because the side AB is equal to DE.
But the other equal sides BC and EF, now stretching from the
same point E towards DF, must likewise
coincide; for if the angle at C or F be
right, there can exist no more than one

AA
perpendicular EF (I. 18. cor.) and, in Ấ
like manner, if this angle at F be either obtuse or acute, the
line EF, which forms it, can have only one corresponding po-
26
ELEMENTS OF GEOMETRY.
sition. Whence, in each of these three cases, the triangle
ABC admits of a perfect adaptation with DEF *.
PROP. XXIII. THEOR.
If a straight line fall upon two parallel straight
lines, it will make the alternate angles equal, the ex-
terior angle equal to the interior opposite one, and
the two interior angles on the same side together
equal to two right angles.
Let the straight line EFG fall upon the parallels AB and
CD; the alternate angles AGF and DFG are equal, the ex-
terior angle EFC is equal to the interior angle EGA, and the
interior angles CFG and AGF, or FGB and GFD, are toge-
ther equal to two right angles.
1
For suppose the straight line EFG, produced both ways
from F, to turn about that point in the direction BA; it
will first cut the extended line AB towards A, and will in its
progress afterwards meet the same line on the other side to-
wards B. In the position IFH, the angle EFH is the exte-
rior angle of the triangle FHG, and therefore greater than
FGH or EGA (I. 8.) But in the last position LFK, the
exterior angle EFL is equal to its vertical angle GFK in the
triangle FKG, and to which the angle
FGA is exterior; consequently (I. 8.)
FGA is greater than EFL, or the angle
EFL is less than FGA or EGA. When
the incident line EFG, therefore, meets
AB above the point G, it makes an
angle EFH greater than EGA; and
when it meets AB below that point, it
makes an angle EFL, which is less than B

K
D
L
H
F
E
the same angle. But in passing through all the degrees front
* See Note IX.
BOOK I.
27
greater to less, a varying magnitude must evidently rencoun-
ter, as it proceeds, the single intermediate limit of equality.
Wherefore, there is a certain position, CD, in which the line
revolving about the point F makes the exterior angle EFC
equal to the interior EGA, and at the same time meets AB
neither towards the one part nor the other, or is parallel to it.
And now, since EFC is proved to be equal to EGA, and
is also equal to the vertical angle GFD; the alternate angles
FGA and GFD are equal. Again, because GFD and FGA
are equal, add the angle FGB to each, and the two angles
GFD and FGB are equal to FGA and FGB; but the angles
FGA and FGB, on the same side of AB, are equal to two
right angles, and consequently the interior angles GFD and
FGB are likewise equal to two right angles.
Cor. Since the position CD is individual, or that only one
straight line can be drawn through the point F parallel to
AB, it follows that the converse of the proposition is likewise
true, and that those three properties of parallel lines are criteria
for distinguishing parallels *.
PROP. XXIV. PROB.
Through a given point, to draw a straight line pa-
rallel to a given straight line.
A
C
E
D
B
To draw, through the point C, a straight line parallel to AB.
In AB take any point D, join CD, and
at the point C make (I. 4.) an angle DCE
equal to CDA; CE is parallel to AB.
For the angles CDA and DCE, thus formed equal, are the
alternate angles which CD makes with the straight lines CE
and AB, and, therefore, by the corollary to the last proposi-
tion, these lines are parallel.

* See Note X.
28
ELEMENts of GEOMETRY.
*
PROP. XXV. THEOR.
Parallel lines are equidistant, and equidistant
straight lines are parallel.
The perpendiculars EG, FH, let fall from any points E, F
in the straight line AB upon its parallel CD, are equal; and
if these perpendiculars be equal, the straight lines AB and CD
are parallel.
A
E
F
B
For join EH: and because each of the interior angles EGH
and FHG is a right angle, they are together equal to two
right angles, and consequently the perpendiculars EG and FH
are (I.23. cor.) parallel to each other;
wherefore (I.23.) the alternate angles
HEG and EHF are equal. But,
EF being parallel to GH, the alter- C
nate angles EHG and HEF are likewise equal; and thus the
two triangles HGE and HFE, having the angles HEG and
EHG respectively equal to EHF and HEF, and the side EH
common to both, are (I. 21.) equal, and hence the side EG is
equal to FH.

G
H
D
Again, if the perpendiculars EG and FH be equal, the two
triangles EGH and EFH, having the side EG equal to FH,
EH common, and the contained angle HEG equal to EHF,
are (I. 3.) equal, and therefore the angle EHG equal to HEF,
and (I. 23.) the straight line AB parallel to CD.
PROP. XXVI. THEOR.
The opposite sides of a rhomboid are parallel.
If the opposite sides AB, DC, and AD, BC of the quadri-
lateral figure ABCD be equal, they are also parallel.
BOOK I.
29
For join AC. And because AB is equal to DC, BC to AD,
A
B
and AC is common; the two triangles
ABC and ADC are (I. 2.) equal. Con-
sequently the angle ACD is equal to
CAB, and the side AB (I. 23. cor.)
parallel to CD; and, for the same reason, the angle CAD is
equal to ACB, whence the side AD is parallel to BC.

D
C
Cor. Hence the angles of a square or rectangle are all of
them right angles; for the opposite sides being equal, are pa-
rallel; and if the angle at A be right, the other interior one
at B is also a right angle (I. 25.), and consequently the angles
at C and D, opposite to these, are right.
PROP. XXVII. THEOR.
The opposite sides and angles of a parallelogram
are equal.
Let the quadrilateral figure ABCD have the sides AB, BC
parallel to CD, AD; these are respectively equal, and so are
the opposite angles at A and C, and at B and D.
For join AC. Because AB is parallel to CD, the alternate
angles BAC and ACD are (I. 25.) equal; and since AD is
parallel to BC, the alternate angles ACB and CAD are like-
wise equal. Wherefore the triangles
A
B

ABC and ADC, having the angles
D
C
CAB and ACB equal to ACD and
CAD, and the interjacent side AC
common to both, are (I. 21.) equal. Consequently, the side
AB is equal to CD, and the side BC to AD; and these op-
posite sides being thus equal, the opposite angles (I. 26.) must
also be equal.
Cor. Hence the diagonal divides a rhomboid or parallelo-
gram into two equal triangles. Hence also an oblong is a
rectangular parallelogram; for if the angle at A be right, the
30
ELEMENTS OF GEOMETRY.
opposite angle at C is right, and the remaining angles at B
and D, being equal to each other and to two right angles,
must be right angled.
PROP. XXVIII. THEOR.
If the parallel sides of a trapezoid be equal, the
other sides are likewise equal and parallel.
Let the sides AB and DC be equal and parallel; the sides
AD and BC are themselves equal and parallel.
For join AC. Because AB is parallel to CD, the alternate
angles CAB and ACD are (I. 23.) equal; and the triangles
ABC and ADC, having the side AB equal to CD, AC com-
mon to both, and the contained angle
CAB equal to ACD, are, therefore, equal
(I. 3.) Whence the side BC is equal to
D
AD, and the angle ACB equal to CAD;
B

but these angles being alternate, BC must also be parallel to
AD (I. 23. cor.)
PROP. XXIX. THEOR.
The diagonals of a rhomboid mutually bisect each
other.
If the diagonals of the rhomboid ABCD intersect each other
in E; the part AE is equal to CE, and DE to BE.
B
For because a rhomboid is also a parallelogram (I. 26.), the
alternate angles BAC and ACD are equal (I. 23.) and like-
wise ABD and BDC. The triangles A
AEB and CED, having thus the angles
BAE, ABE respectively equal to DCE
and CDE, and the interjacent sides AB
and CD equal, are (I. 21.) wholly equal.
equal to the corresponding side CE, and BE to DE.

D
E
Wherefore AE is
BOOK I.
31
Cor. Hence the diagonals of a rectangle are equal to each
other; for if the angles at A and B were right angles, the
triangles DAB and CBA would be equał (I. 3.) and conse-
quently the base DB equal to AC.
PROP. XXX. THEOR.
Lines parallel to the same straight line, are paral-
lel to each other.
If the straight line AB be parallel to CD, and CD parallel
to EF; then is AB parallel to EF.
For let a straight line GH cut these lines.
And because AB is parallel to CD, the
exterior angle GIA is equal (I. 25.) to the
interior GKC; and since CD is parallel to
EF, this angle GKC is, for the same rea-
son, equal to GLE. Therefore the angle

A
C
G
I
B
K
D
E
L F
T-J
GIA is equal to GLE, and consequently AB is parallel to EF
(I. 23. cor.)
PROP. XXXI. THEOR.
Straight lines drawn parallel to the sides of an
angle, contain an equal angle.
If the straight lines AB, AC be parallel
to DE, DF; the angle BAC is equal to
EDF.
For draw the straight line GAD through
the vertices. And since AC is parallel to
DF, the exterior angle GAC is (I. 23.)
equal to GDF; and, for the same reason,

B
T
C
D
F
GAB is equal to GDE; there consequently remains the angle
BAC equal to EDF.
32
ELEMENTS OF GEOMETRY.
PROP. XXXII. THEOR.
An exterior angle of a triangle is equal to both
its opposite interior angles, and all the interior angles
of a triangle are together equal to two right angles.
The exterior angle BCD, formed by the production of the side
AC of the triangle ABC, is equal to the two opposite interior
angles CAB and CBA, and all the interior angles CAB, CBA
and BCA of the triangle are together equal to two right angles.
For, through the point C, draw (I. 24.) the straight line
CE parallel to AB. And, AB being parallel to CE, the inte-
rior angle BAC is (I. 23.) equal to the exterior one ECD;
and, for the same reason, the alternate angle ABC is equal to
BCE. Wherefore the two angles CAB and

B
E
ABC are equal to DCE and ECB, or to
the whole exterior angle BCD. Add to
A
C
D
each the adjacent angle BCA; and all
the interior angles of the triangle ABC are
together equal to the angles BCD and BCA on the same side
of the straight line AD, that is, to two right angles.
Cor. 1. Hence the two acute angles of a right angled tri-
angle are together equal to one right angle; and hence each
angle of an equilateral triangle is two third parts of a right
angle.
Cor. 2. Hence if a triangle have its exterior angle, and one
of its opposite interior angles, double of those in another tri-
angle; its remaining opposite interior angle will also be double
of the corresponding angle in the other *.
* See Note XI.
BOOK I.
33
PROP. XXXIII. THEOR.
The angles round any rectilineal figure are toge-
ther equal to twice as many right angles (abating four
from the result) as the figure has sides.
For assume a point O within the figure, and draw straight
lines OA, OB, OC, OD, and OE, to the several corners. It
is obvious, that the figure is thus resolved into as many tri-
angles as it has sides, and whose collected

B
angles must be therefore equal to twice as
many right angles. But the angles at the
bases of these triangles constitute the in-
ternal angles of the figure. Consequently,
E
A
D
from the whole amount there is to be deducted the vertical
angles about the point O, and which are (Def. 4.) equal to
four right angles.
Cor. Hence all the angles of a quadrilateral figure are equal
to four right angles, those of a pentelateral figure equal to six
right angles, and so forth; increasing the amount by two
right angles, for each additional side.
PROP. XXXIV. THEOR.
The exterior angles of a rectilineal figure are to-
gether equal to four right angles.
The exterior angles DEF, CDG, BCH, ABI, and EAK of
the rectilineal figure ABCDE are taken together equal to four
right angles.
C
i
}
A
34
ELEMENTS OF GEOMETRY.
H
For each exterior angle DEF, with its adjacent interior one
AED, is equal to two right angles.
All the exterior angles, therefore,
added to the interior angles, are
equal to twice as many right
angles as the figure has sides.
Consequently the exterior angles
are equal to the four right angles
which, by the last Proposition,
were abated, to form the aggre-

gate of the interior angles.
I
B
K
D
E
F
Cor. If the figure has a re-entrant angle BCD, the angle
BCK which occurs in place of
an exterior angle, must be taken
away in forming the amount; for
the corresponding interior angle
BCD, in this case, exceeds two Τ.
right angles, by BCK. Hence
the angles EFG, DEH, CDI,
ABL, FAM, diminished by BCK,
are equal to four right angles.
B
I

K
M
PROP. XXXV. THEOR.
I-I
D
E
F
G
If the opposite angles of a quadrilateral figure be
equal, its opposite sides will be likewise equal and
parallel.
In the quadrilateral figure ABCD, let the angle at B be
equal to the opposite one at D, and the angle at A equal to
that at C; the sides AB and BC are equal and parallel to
DC and DA.
}
i
;
BOOK I.
35
?
For all the angles of the figure being equal to four right
angles (I. 33. cor.), and the opposite
B
angles being mutually equal, each

pair of adjacent angles must be equal
to two right angles. Wherefore ABC A
and BCD are equal to two right
D
angles, and the lines AB and DC (I. 23. cor.) parallel; for
the same reason, ABC and BAD being together equal to two
right angles, the sides BC and AD, which limit them, are pa-
rallel. But (I. 27.) the parallel sides of the figure are also equal.
Cor. Hence a rectangle has its opposite sides equal and pa-
rallel.
PROP. XXXVI. PROB.
To draw a perpendicular from the extremity of a
given straight line.
From the point B, to draw a perpendicular to AB, without
producing that line.
In AB take any point C, and on BC (I. 1. cor.) describe
an isosceles triangle BDC, produce CD till DF equal it; and
BF being joined, is the perpendicular required.
A C
D
B
For, since by construction DF is equal to
CD or BD, the triangle BDF is isosceles,
and (I. 11.) the angle DBF equal to DFB;
whence the angle CDB, being equal (I. 8.)
to the interior angles DBF and DFB, is
double of DBF, or the angle DBF is half of CDB. But the
triangle BDC being isosceles, the angle CBD is equal to BCD;
consequently the angles DBF and DBC are the halves of the
vertical and base angles of BDC, and therefore (I. 32.Jthe
whole angle CBF is the half of two right angles, or it is equal
to one right angle

*
* See Note XII.
C 2
36
ELEMENTS OF GEOMETRY.
PROP. XXXVII. PROB.
On a given finite straight line, to construct a
square.
Let AB be the side of the square which it is required to
construct.
From the extremity B draw (I. 36.)
BC perpendicular to BA and equal to
it, and, from the points A and C with
the distance BA or BC describe two
circles intersecting each other in the
point D, join AD and CD; the quadri-
lateral figure ABCD is the square requi-
red.

D
C
A
B
For, by this construction, the figure has all its sides equal,
and one of its angles ABC a right angle; which comprehends
the whole of the definition of a square.
PROP. XXXVIII. PROB.
To divide a given straight line into any number
of equal parts.
Let it be required to divide the straight line AB into a gi-
ven number of equal parts, suppose five.
From the point A and at any oblique
angle with AB, draw a straight line AC,
in which take the portion AD, and re-
peat it five times from A to C, join CB,
and from the several points of section
D, E, F, and G draw the parallels DH,
EI, FK, and GL, (I. 24.), cutting AB
in H, I, K, and L: AB is divided at
these points into five equal parts.
C

G
P
T
E
N
D
M
A HIK L B
BOOK I.
37
For (I. 24.) draw DM, EN, FO, and GP parallel to AB.
And because DH is parallel to EM, the exterior angle ADH is
equal to DEM (I. 23.); and, for the same reason, since AH
is parallel to DM, the angle DAH is equal to EDM. Where-
fore the triangles ADH and DEM, having two angles respec-
tively equal and the interjacent sides AD, DE—are (I. 21.)
equal, and consequently AH is equal to DM. In the same
manner, the triangle ADH is proved to be equal to EFN, to
FGO, and GCP, and therefore their bases EN, FO, and GP
are all equal to AH. But these lines are equal to HI, IK,
KL, and LB, for the opposite sides of parallelograms are
equal (I. 29.). Wherefore the several segments AH, HI, IK,
KL, and LB, into which the straight line AB is divided, are
all equal to each other.
Scholium. The construction of this problem may be facilita-
ted in practice, by drawing from B in the opposite direction
a straight line parallel to AC, and repeating on both of them
portions equal to the assumed segment AD, but only four
times, or one fewer than the number of divisions required;
then joining D, the first section of AC, with the last of its pa-
rallel, E with the next, and so on till G, which connecting
lines are (I. 28.) all parallel, and consequently the former de-
monstration still holds.
ELEMENTS
OF
GEOMETRY.
BOOK II.
DEFINITIONS.
1. In a right-angled triangle, the side that subtends the
right angle is termed the hypotenuse; either of the sides which
contain it, the base; and the other side, the perpendicular.
2. The altitude of a triangle is a
perpendicular let fall from the vertex
upon the base or its extension.
3. The altitude of a trapezoid is the per-
pendicular drawn from one of its parallel
sides to the other.
4. The complements of rhomboids about the diagonal of a
rhomboid, are the spaces required to complete
the rhomboid; and the defect of each rhom-
boid from the whole figure, is termed a
gnomon.
5. A rhomboid or rectangle is said to be contained by
any two adjacent sides.
A rhomboid is often indicated merely by the two letters placed
at opposite corners.
40
ELEMENTS OF GEOMETRY.
PROP. I. THEOR.
Triangles which have the same altitude, and stand
on the same base, are equivalent.
The triangles ABC and ADC which stand on the same base
AC and have the same altitude, contain equal spaces.
For join the vertices B, D by a straight line, which pro-
duce both ways; and from A draw AE (I. 24.) parallel to
CB, and from C draw CF parallel to AD.
F
Because the triangles ABC, ADC have the same altitude,
the straight line EF is parallel to AC (I. 25.), and consequent-
ly the figures CE and AF are parallelo-
grams. Wherefore EB, being equal
to AC (I. 27.), which is equal to DF, is

A
C
itself equal to DF. Add BD to each,
and ED is equal to BF; but EA is equal to BC (I. 27.), and
the interior angle AED is equal to the exterior angle CBF
(I. 23.). Thus the two triangles EDA, BFC have the sides
ED, EA equal to BF, BC, and the contained angle AED equal
to CBF, and are therefore equal (I. 3.). Take these equal tri-
angles CBF and EDA from the whole quadrilateral space
AEFC, and there remains the rhomboid AEBC equivalent to
ADFC. Whence the triangles ABC and ADC, which are the
halves of these rhomboids (I. 27. cor.), are likewise equivalent.
Cor. Hence the rhomboids on the same base and between
the same parallels, are equivalent.
PROP. II. THEOR.
Triangles which have the same altitude and stand
on equal bases, are equivalent.
The triangles ABC, DEF, standing on equal bases AC and
DF and having the same altitude, contain equal spaces.
BOOK II.
41
**
For let the bases AC, DF be placed in the same straight
line, join BE, and produce it both ways, draw AG and DH
parallel to CB and FE (I. 24.), and join AH, CE.
G
B
II
E
A
C D
F
Because the triangles ABC, DEF are of equal altitude, GE
is parallel to AF (I. 25.), and GC, HF are parallelograms.
But AC, being equal to DF, and DF
equal (I. 27.) to HE, must also be
equal to HE, and therefore (I. 28.)
AE is a rhomboid or parallelogram.
Whence the rhomboid GC is equivalent to AE (II. 1. cor.), and
this again is, for the same reason, equivalent to HF; conse-
quently GC is equal to HF, and therefore their halves or
(I. 27. cor.) the triangles ABC and DEF are equivalent.
Cor. Hence rhomboids on equal bases and between the
same parallels, are equivalent.

PROP. III. THEOR.
Equivalent triangles on the same or equal bases,
have the same altitude.
If the triangles ABC and ADC, standing on the same base
AC, contain equal spaces, they have the same altitude, or the
straight line which joins their vertices is parallel to AC.
For if BD be not parallel to AC, draw the parallel BE meet-
ing AD or that side produced, in E, and join CE.
B
רן
Because BE is made paralel to AC, the triangle ABC is
(II. 1.) equivalent to AEC; but ABC is
by hypothesis equivalent to ADC, and there-
fore AEC is equivalent to ADC, which is
absurd. The supposition then that BD is
not parallel to AC, involves a contradiction.

A
The same mode of demonstration, it is obvious, will apply in
the case where the equivalent triangles stand on equal bases.
Cor. Hence equivalent rhomboids on the same or equal
bases, have the same altitude.
42
ELEMENTS OF GEOMETRY.
PROP. IV. THEOR.
A straight line bisecting two sides of a triangle,
is parallel to the base.
The straight line DE which joins the middle points of the
sides AB and BC, is parallel to the base AC of the triangle
ABC.
For join AE and CD. Because the triangles ADC, BCD
stand on equal bases AD, DB, and have the same vertex or
altitude, they are (II. 2.) equivalent, and therefore ADC is
half of the whole triangle ABC. For the
same reason, since CE is equal to EB, the
B

D
E
A
G
C
triangle AEC is equivalent to AEB, and is
consequently half of the whole triangle
ABC. Whence the triangles ADC and
AEC are equivalent; and they stand on the same base AC,
and have therefore the same altitude (II. 3.), or DE is paral-
lel to AC.
Cor. Hence the triangle DBE cut off by the line DE, is the
fourth part of the original triangle. For bisect AC in G, and
join DG, which is therefore parallel to BC. The triangle ADG
is equivalent to GDC (II. 2.), and GDC, being the half of the
rhomboid GE, is equivalent to DEC, which again is (II. 2.)
equivalent to DEB. The triangle ABC is thus divided into
four equivalent triangles, of which DBE is one.
the rhomboid GDEC is half of the original triangle*.
PROP. V. PROB.
Hence also
To find a triangle equivalent to any rectilineal
figure.
* Sce Note XIII.
BOOK II.
43
Let it be required to reduce the five-sided figure ABCDE to
a triangle, or to find a triangle that shall contain an equal space.
C
Join any two alternate points A, C, and, through the inter-
mediate point B, draw BF parallel to AC, meeting either
of the adjoining sides AE or CD in F; which point, when the
angle ABC is re-entrant will lie within the figure: Join CF.
Again, join the alternate points C, E,
and through the intermediate point
D draw the parallel DG to meet in G
either of the adjoining sides AE or
BC, and which, since the angle CDE
is salient, must for that effect be
duced; and join CG. The triangle FCG is equivalent to the
five-sided figure ABCDE.

pro- A F
B
E
D
Because the triangles CFA and CBA have by construction
the same altitude and stand on the same base AC, they are
(II. 1.) equivalent; take each away from the space ACDE,
and there remains the quadrilateral figure FCDE equivalent
to the five-sided figure ABCDE. Again, because the triangles
CDE and CGE are equal, having the same altitude and the
same base; add the triangle FCE to each, and the triangle
FCG is equivalent to the quadrilateral figure FCDE, and is
consequently equivalent to the original figure ABCDE.
In this manner, any polygon may, by successive steps, be
reduced to a triangle; for an exterior triangle is always ex-
changed for another equivalent one, which, attaching itself to
either of the adjoining sides, coalesces with the rest of the
figure*.
PROP. VI. PROB.
A triangle is equivalent to a rhomboid which has
the same altitude and stands on half the base.
The triangle ABC is equivalent to the rhomboid DEFC,
which stands on half the base DC, but has the same altitude.
**See Note XIV.
44
ELEMENTS OF GEOMETRY.
For join BD and EC. The triangles ABD and DBC ha-
ving the same vertex and equal bases, are
(II. 2.) equivalent. But the diagonal EC
bisects the rhomboid DEFC (I. 27. cor.),
and the triangles DBC and DEC, ha-

A D
ving the same altitude, are equivalent (II. 1.); consequently
their doubles, or the triangle ABC and the rhomboid
DEFC, are equivalent.
PROP. VII. PROB.
To construct a rhomboid equivalent to a given rec-
tilineal figure, and having its angle equal to a given
angle.
Let it be required to construct a rhomboid which shall be
equivalent to a given rectilineal figure and contain an angle
equal to G.
Reduce the rectilineal figure to an equivalent triangle
ABC (II. 5.), bisect the base AC in the point D (I. 7.), and
draw DE making an angle CDE equal B

E
F
A
D
C
to the given angle G (I. 4.), through
B draw BF parallel to AC (I. 24.),
and through C the straight line CF
parallel to DE: DEFC is the rhomboid that was required.
For the figure DF is, by construction, a rhomboid, contains
an angle CDE equal to G, and is equivalent to the triangle
ABC (II. 6.), and consequently to the given rectilineal figure.
PROP. VIII. THEOR.
The complements of the rhomboids about the dia-
gonal of a rhomboid, are equivalent.
Let EI and HG be rhomboids about the diagonal of the
rhomboid BD; their complements BF and FD contain equal
spaces.
BOOK II.
45
E
B
H C
F
I
D
Since the diagonal AF bisects the rhomboid EI (I. 27. cor.),
the triangle AEF is equivalent to AIF; and, for the same
reason, the triangle FHC is equivalent to
FGC. From the whole triangle ABC on
the one side of the diagonal, take away
the two triangles AEF and FHC; and
from the triangle ADC, which is equal to it, take away, on the
other side, the two triangles AIF and FGC, and there re-
mains the rhomboid BF equivalent to FD.
PROP. IX. PROB.
A
With a given straight line to construct a rhom-
boid equivalent to a given rectilineal figure, and ha-
ving an angle equal to a given angle.
Let it be required to construct, with the straight line L, a
rhomboid, containing a given space, and having an angle
equal to K.
E
B
]
C
Construct (II. 7.) the rhomboid BF equivalent to the given
rectilineal figure, and having an angle BEF equal to K; produce
EF until FG be equal to L, through
G draw DGC parallel to EB and
meeting the extension of BH in C,
join CF and produce it to meet the
extension of BE in A; draw AD parallel to EF, meeting CG
in D, and produce HF to I: FD is the rhomboid required.

A
K
D
L
For FD and FB are evidently complementary rhomboids,
and therefore (II. 8.) equivalent; and, by reason of the paral-
lels AE, IF, the angle FID is equal to EAI (I. 23.), which
again is equal to BEF or the given angle K.
PROP. X. THEOR.
A trapezoid is equivalent to the rectangle contain-
ed by its altitude and half the sum of its parallel sides.
46
ELEMENTS OF GEOMETRY.
D
The trapezoid ABCD is equivalent to the rectangle con-
tained by its altitude and half the sum of the parallel sides BC
and AD.
For draw CE parallel to AB (I. 24.), bisect ED (1. 7.) in
F, and draw FG parallel to AB, meeting the production of
BC in G.
Because BC is equal to AE (I. 27.), BC and AD are toge
ther equal to AE and AD, or to twice AE with ED, or to
twice AE and twice EF, that is, to twice AF; conséquently
AF is half the sum of BC and AD.
Wherefore the rectangle contained by
1}
C G

E F D
the altitude of the trapezoid and half A
the sum of its parallel sides, is equivalent to the rhomboid BF
(II. 1. cor.); but the rhomboid EG is equivalent to the tri-
angle ECD (II. 6.), add to each the rhomboid BE, and the
rhomboid BF is equivalent to the trapezoid ABCD *.
PROP. XI. THEOR.
The square described on the hypotenuse of a right-
angled triangle, is equivalent to the squares of the
two sides.
-
Let ACB be a triangle which is right-angled at B; the
square of the hypotenuse AC is equivalent to the two squares
of AB and BC.
For produce the base BA until AD be equal to the per-
pendicular BC, and on DB describe (I. 37.) the square DEFB,
make EG and FH equal to AD or BC, join AG, GH, and
HC, and through the points A and C (I. 24.) draw AL and
CI parallel to BF and BD.
Because the whole line BD is equal to DE, and the part of
it AD equal to GE, the remainder AB is equal to DG;
*See Note XV.
BOOK II.
47
E
L H
F
wherefore (I. 3.) the triangles ACB and AGD are equal, since
they have the sides AB, BC equal to DG, DA, and the con-
tained angle ABC equal to ADG, both of these being right.
angles. In the same manner, it is
proved, that the triangle ACB is equal
to GEH, and to HFC. Consequent-
ly the sides AC, AG, GH, and HC
are all equal. But the angle CAB,
being equal to AGD, is equal to the
alternate angle GAL (I. 23.); add
LAC to each, and the whole angle LAB or (I. 27.) EDB is
equal to GAC, which is therefore a right angle. Hence the
figure AGHC, having all its sides equal and one of its angles
right, is a square.

I
D
C
K
A
B
Again, the rhomboids KB and KE are evidently rectan-
gular; they are also equal, being contained by equal sides;
and each of them being double of the original triangle ACB,
they are together equal to the four triangles ACB, AGD,
EHG, and HCF. The other inscribed figures LC and IA
are obviously the squares of KC and AD, which are equal to
the base and perpendicular of the triangle ABC. From the
whole square DEFB, therefore, take away separately those
four encompassing triangles with the two interjacent rectan-
gles KB and KE, and the remainders must be equal; that is,
the square AGHC is equal in space to both the squares
ADIK and KLFC.
Otherwise thus.
Let the triangle ABC be right-angled at B; the square
described on the hypotenuse AC is equivalent to BF and BI
the squares of the sides AB and BC.
For produce DA to K, and through B draw MBL paral
lel to DA (I. 24.) and meeting FG produced in L.
48
ELEMENTS OF GEOMETRY.
Because the angle CAK, adjacent to CAD, is a right
angle, it is equal to BAF: from each take away the angle
BAK, and there remains the
angle BAC equal to FAK. But
the angle ABC is equal to
AFK, both being right angles.
Wherefore the triangles ABC
and AFK, having thus two
angles of the one respectively
equal to those of the other, and
the interjacent side AF equal
to AB, are equal (I. 21.), and
consequently the side AC is
equal to AK. Hence the rect-
Ε
K
L

H
B
I
N
C
D
M
E
angle or rhomboid AM is equivalent to ABLK (II. 2. cor.),
since they stand on equal bases AD and AK, and between the
same parallels DK and ML. But ABLK is (II. 1. cor:) equiva-
lent to the rhomboid or square BF, for it stands on the same
base AB and between the same parallels FL and AH.
Wherefore the rectangle AM is equivalent to the square of
AB. And in like manner, by drawing MB to meet the pro-
duction of HI, it may be proved, that the rectangle CM is
equivalent to the square of BC. Consequently the whole
square, ADEC, of the hypotenuse, contains the same space
as both together of the squares described on the two sides AB
and BC *
PROP. XII. THEOR.
If the square of a side of a triangle be equivalent
to the squares of both the other sides, that side sub-
tends a right angle.
Let the square described on AC be equivalent to the two
squares of AB and BC; the triangle ABC is right-angled at B.
* Sce Note XVI.
!
BOOK II.
49
For draw BD perpendicular to AB (I. 36.) and equal to
BC, and join AD.
B
Because BC is equal to BD, the square of BC is equal to
the square of BD, and consequently the squares of AB and
BC are equal to the squares of AB and BD. But the squares
of AB and BC are, by hypothesis, equivalent
to the square of AC; and since ABD is, by
construction, a right angle, the squares of AB
and BD are (II. 11.) equivalent to the square
of AD. Whence the square of AC is equiva-
lent to that of AD, and the straight line AC
equal to AD. The two triangles ACB and ADB, having all
the sides in the one respectively equal to those in the other,
are therefore equal (I. 2.), and consequently the angle ABC
is equal to the corresponding angle ABD, that is, to a right
angle *.

PROP. XIII. PROB.
D
To find the side of a square equivalent to any
number of given squares.
Let A, B, and C be the sides of the squares, to which it is
required to find an equivalent square.
Draw DE equal to A, and from its extremity E erect (I. 36.)
the perpendicular EF equal to B, join DF, and again perpen-
dicular to this draw FG equal to C, and join DG: DG is
the side of the square which was required.
D
G
F
For because DEF is a right-angled triangle,
the square of DF is equivalent to the squares
of DE and EF (II. 11.), or of A and B. Add
the square of FG or C, and the squares of DF
and FG, which are equivalent to the square of
DG (II. 11.), are equivalent to the aggregate
squares of A, B, and C. And by thus repeat-
ing the process, it may be extended to any number of squares.
* See Note XVII.
Ꭰ D
E
A
B
C
50
ELEMENTS OF GEOMETRY.
PROP. XIV. PROB.
To find the side of a square equivalent to the dif
ference between two given squares.
Let A and B be the sides of two squares; it is required to
find a square equivalent to their difference.
Draw CD equal to the smaller line B, from its extremity
erect (I. 36.) the indefinite perpendicu-
lar DE, and about the centre C with a
distance equal to the greater line A de-
scribe a circle cutting DE in F: DF is
the side of the square required.
For join CF. The triangle CDF be-
ing right-angled, the square of the hy-
potenuse CF is equivalent to the squares
of CD and DF (II. 11.), and consequent-
A
B

IE
F
C
D
ly taking the square of CD from both, the excess of the square
of CF above that of CD is equivalent to the square of DF,
or the square of DF is equivalent to the excess of the square
of A above that of B.
PROP. XV. THEOR.
In any triangle, the rhomboids described on two
sides, are together equivalent to a rhomboid descri-
bed on the base, and limited by these and by paral-
lels to the line which joins the vertex with their point
of concourse.
Let ADEB and BGFC be rhomboids described on the two
sides AB and BC of the triangle ABC; produce the summits
DE and FG to meet in H, join this point with the vertex B,
to BH draw the parallels AK, CL, and join KL. It is ob-
vious that AK and CL, being equal and parallel to BH, are
BOOK II.
51
likewise equal and parallel to each other, and that the figure
AKLC is a parallelogram or rhomboid.-This rhomboid is
equivalent to the two rhomboids BD and BF.
For produce HB to meet the base AC in I. And because
the rhomboids KI and AH
H

stand on the same base AK
and between the same paral-
lels, they are equivalent (II.
1. cor.); but the rhomboids
AH and BD, standing on
the same base AB and be-
D
A
B
G
E
I
L
F
tween the same parallels, are also equivalent. Whence KI is
equivalent to BD. And in the same manner, it may be proved
that LI is equivalent to BF. Consequently the whole rhom-
boid KC is equivalent to the two rhomboids BD and BF *.
PROP. XVI. THEOR.
The rectangle contained by two straight lines, is
equivalent to the rectangles contained under one of
them and the several segments into which the other
is divided.
The rectangle under AC and AB, is equivalent to the rect-
angles contained by AC and the segments AD, DE, and EB.
For, through the points D and E, draw DF and EG pa-
rallel and equal to AC (I. 24.).
D E
B
The figures AF, DG, and EH are evidently rhomboidal;
they are also rectangular, for the angles
ADF, AEG, and ABH are each equal to the A
opposite angle ACF (I. 21.). And the op-
posite sides DF, EG, and BH, being equal
to AC,--the spaces into which the rectangle

C
F G
II
BC is resolved, are equal to the rectangles contained by AC
and AD, DE and EB +.
*See Note XVIII.
† See Note XIX.
D 2
52
1
ELEMENTS OF GEOMETRY.
PROP. XVII. THEOR.
The square described on the sum of two straight
lines, is equivalent to the squares of those lines, to-
gether with twice their rectangle.
If AB and BC be two straight lines placed continuous; the
square described on their sum AC, is equivalent to the two
squares of AB, BC, and twice the rectangle contained by them.
For through B draw BI (I. 24.) parallel to AD, make AF
equal to AB, and through F draw FH parallel to DE.
D
I
E
It is manifest that the spaces AG, GE, DG and CG,
into which the square of AC is divided,
are all rhomboidal and rectangular. And
because AB is equal to AF, and the oppo-
site sides equal, the figure AG is equila-
teral, and having a right angle at A, is
hence a square. Again, AD being equal
to AC, take away the equals AF and AB,

F
H
G
B C
and there remains DF equal to BC, and consequently IG
equal to GH (I. 27.); wherefore IH is likewise a square.
The rectangle DG is contained by the sides FG and DF,
which are equal to AB and BC; and the rectangle CG is
contained by the sides GB and GH, which are likewise equal
to AB and BC. Consequently the whole square of AC is
composed of the two squares of AB and BC, together with
twice the rectangle contained by these lines.
PROP. XVIII. THEOR.
The square described on the difference of two
straight lines, is equivalent to the squares of those
lines, diminished by twice their rectangle.
Let AC be the difference of two straight lines AB and BC;
the square of AC is equivalent to the excess of the two squares
of AB and BC above twice their rectangle.
BOOK II.
53
For make AD equal to AC, draw CH and DI (1. 24.) pa-
rallel to AF and AB, produce FG until GL be equal to BC,
and complete the figure GK.
It is evident, from the demon-
stration of the last Proposition,
that DC is the square of AC, and
GK the square of BC. From the
compound surface AFLKIB,
which is made up of the squares
of AB and BC, take away twice
H
G
L

D
E
I
K
A
C
B
the rectangle AB, BC, or the two rectangles FI and CG, or
the rectangle FI with the rectangle CI and the square IL,-
and there remains ADEC, or the square of the difference AC
of the two lines AB and BC.
PROP. XIX. THEOR.
The rectangle contained by the sum and dif-
ference of two straight lines, is equivalent to the
difference of their squares.
Let AB and BD be two continuous straight lines, of which
AD is the sum and AC the difference; the rectangle under
AD and AC, is equivalent to the excess of the square of AB
above that of BC.
For, having made AG equal to AC, draw GH parallel to
AD (I. 24.), and CI, DH parallel to AE.
E
T
F
Because GK is equal to KC or HD, and EG is equal to
CB or BD, the rectangle EK is equal
to LD (II. 2. cor.); and consequent-
ly, adding the rectangle BG to each,
the space AEIKLB is equivalent to
the rectangle AH.
But this space
AEIKLB is the excess of the square

C
L
I
A
C
B D
of AB above IL or the square of BC;
and the rectangle AH is contained by AD and DH or AC.
54
ELEMENTS OF GEOMETRY.
Wherefore the rectangle under AD and AC is equivalent to
the difference of the squares of AB and BC..
Cor. 1. Hence if a straight line AB be bisected in C and
cut unequally in D, the rectangle under the unequal segments
AD, DB, together with the square of CD, the interval be-
tween the points of section, is equivalent to
the square of AC, the half line.
For AD
is the sum of AC, CD, and DB is evidently
A
CD B
their difference; whence, by the Proposition, the rectangle
AD, DB is equivalent to the excess of the square of AC above
that of CD, and consequently the rectangle AD, DB, with
the square of CD, is equal to the square of AC.
Cor. 2. If a straight line AB be bisected in C and produ-
ced to D, the rectangle contained by AD the whole line thus
produced, and the produced part DB, together with the square
of the half line AC, is equivalent to the square of CD, which
is made up of the half line and the pro-
duced part.
For AD is the sum of AC,
A
C
B D
CD, and DB is their difference; whence
the rectangle AD, DB is equivalent to the excess of the square
of CD above AC, or the rectangle AD, DB, with the square
of AC, is equivalent to the square of CD.
Scholium. If we consider the distances DA, DB of the point
D from the extremities of AB as segments of this line, whe-
ther formed by internal or external section; both corollaries
may be comprehended under the same enunciation, namely,
that if a straight line be divided equally and unequally, the
rectangle contained by the unequal segments is equivalent to
the difference of the squares of the half line and of the inter-
val between the points of section.
PROP. XX. THEOR.
The square described on a straight line, is equiva-
lent to the squares of the segments into which it is
BOOK II.
55
*
divided, and twice the rectangles contained by each
pair of these segments.
The square of AB is equivalent to the squares of AC, of
CD and of DB, with twice the rectangles of AC, CD, of AC,
DB, and of CD, DB.
T
G
H
ΙΚ
L
For make AE and EF equal to AC and CD, draw EM,
FL parallel to AB, and CH, DI parallel to AG.
It is manifest that AO is the square of
AC, OQ the square of CD, and QK the
square of DB. Nor is it less obvious
that the two rectangles CN and EP are
contained by AC, CD, that the two rect-
angles NL and PI are contained by CD,
DB, and that the two rectangles DM
and FH are contained by AC, DB. But
those squares and those double rectangles complete the whole
square of AB. Wherefore the truth of the Proposition is
established.

E
A
DE
M
Cor. Hence if a straight line be divided into three portions,
the squares of the double segments AD, BC, together with
twice the rectangle under the extreme segments AC, BD, are
equivalent to the squares of the whole line AB and of the in-
termediate segment CD. For the squares FD, HM, together
with the equal rectangles GP, NB, evidently fill up the whole
square AB, with the repetition of the internal square OQ;
that is, the squares of AD and BC, with twice the rectangle
AC, DB, are equivalent to the squares of AB and CD.
PROP. XXI. THEOR.
The sum of the squares of two straight lines, is
equivalent to twice the squares of half their sum and
of half their difference.
56
ELEMENTS OF GEOMETRY.
Let AB, BC be two continuous straight lines, D the middle
point of AC, and consequently AD half the sum of these lines
and DB half their difference; the squares of AB and BC
are together equivalent to twice the square of AD with twice
the square of DB.
A D B C
For (II. 17.) the square of AB, or the square of the sum of
AD and DB, is equivalent to the squares of these segments,
with double their rectangle; and (II. 18.)
the square of BC, or that of the difference
of AD and DB, is equal to the squares of
AD and DB, diminished by double the rectangle contained
by the same lines AD, DB. Wherefore the squares of AB
and BC taken together, are equivalent simply to twice the
squares of AD and DB.
Otherwise thus.
Bisect AC in D (I. 7.), and erect (I. 5. cor.) the perpendi-
cular DE equal to AD or DC, join AE and EC, through B
and F draw (I. 24.) BF and FG parallel to DE and AC, and
join AF.
Because AD is equal to DE, the angle DAE (I. 11.) is
equal to DEA, and since (I. 32. cor.) they make up together one
right angle, each of them must be half a right angle. In the
same manner, the angles DEC and DCE of the triangle EDC
are proved to be each half a right angle; consequently the
angle AEC, composed of AED and CED, is equal to a whole
right angle. And in the triangle FBC, the angle CBF being
equal to CDE (I. 23.) which is a right angle, and the angle
BCF being half a right angle-the re-
maining angle BFC is also half a right
angle (I. 32.), and therefore equal to
the angle BCF; whence (I. 1.) the side
BF is equal to BC. By the same rea-
soning, it may be shown, that the right-
E

F
A
D B C
angled triangle GEF is likewise isosceles. The square of the
hypotenuse EF, which is equivalent to the squares of EG and
BOOK II.
57
GF (II. 11.) is therefore equivalent to twice the square of
GF or of DB; and the square of AE, in the right-angled
triangle ADE, is equivalent to the squares of AD and DE,
or twice the square of AD. But since ABF is a right angle,
the square of AF is equivalent to the squares of AB and BF,
or AB and BC; and because AEF is also a right angle, the
square of the same line AF is equivalent to the squares of AE
and EF, that is, to twice the squares of AD and DB. Where-
fore the squares of AB, BC are together equivalent to twice
the squares of AD and DB.
A
CD B
A
C
B D
Cor. Hence if a straight line AB be bisected in C and cut
unequally in D, whether by internal or
external section, the squares of the un-
equal segments AD and DB are toge-
ther equivalent to twice the square of
the half line AC, and twice the square of CD the interval be-
tween the points of division.
PROP. XXII. PROB.
To cut a given straight line, such that the square
of one part shall be equivalent to the rectangle con-
tained by the whole line and the remaining part.
Let AB be the straight line which it is required to divide
into two segments, such that the square of the one shall be
equivalent to the rectangle contained by the whole line and
the other.
Produce AB till BC be equal to it,
erect (I. 5. cor.) the perpendicular BD
equal to AB or BC, bisect BC in E
(I. 7.), join ED and make EF equal
to it; the square of the segment BF

D
G
K
I
H
F
B E
C
is equivalent to the rectangle contained by the whole line BA
and its remaining segment AF.
58
ELEMENTS OF GEOMETRY.
For on BC construct the square BG (I. 37.), make BH
equal to BF, and draw IHK and FI parallel to AC and BD
(I. 24.). Since AB is equal to BD, and BF to BH; the re-
mainder AF is equal to HD: and it is farther evident, that
FH is a square, and that IC and DK are rectangles. But
BC being bisected in E and produced to F, the rectangle un-
der CF, FB, or the rectangle IC, together with the square of
BE, is equivalent to the square of EF or of DE (II. 19. cor.
2.). But the square of DE is equivalent to the squares of
DB and BE (II. 11.); whence the rectangle IC, with the
square of BE, is equivalent to the squares of DB and BE;
or, omitting the common square of BE, the rectangle IC is
equivalent to the square of DB. Take away from both the
rectangle BK, and there remains the square BI, or the square
of BF, equivalent to the rectangle HG, or the rectangle con-
tained by BA and AF.
Cor. 1. Since the rectangle under CF and FB is equiva-
lent to the square of BC, it is evident that the line CF is like-
wise divided at B in a manner similar to the original line AB.
But this line CF is made up, by joining the whole line AB,
now become only the larger portion, to its greater segment
BF, which next forms the smaller portion in the new com-
pound. Hence this division of a line being once obtained, a
series of other lines possessing the same property may readily
be found, by repeated additions. Thus, let AB be so cut,
that the square of BC is equivalent to the rectangle BA,
AC:
Make successively BD equal to BA, DE equal to DC, EF
AC B D
E
F
G
equal to EB, and FG equal to FD; the lines CD, BE, DF,
and EG are divided at the points B, D, E, and F, such that,
in each of them, the square of the larger part is equivalent to
the rectangle contained by the whole and the smaller part.-
It is obvious, that this procedure might likewise be reversed.
BOOK II.
54
If FD, EB, and DC be made successively equal to FG, EF
and DE, the lines DF, BE, and CD will be divided in the
same manner at the points E, D and B.
Cor. 2. Hence also the construction of another problem of
the same nature; in which it is required to produce a straight
line AB, such that the rectangle contained by the whole line
thus produced and the part produced, shall be equivalent to
the square of the line AB itself. Divide AB in C, so that
the rectangle BA, AC is equivalent to
the square of BC, and produce AB un-
til BD be equal to BC: Then, from
what has been demonstrated, it follows that the rectangle un-
der AD and DB is equivalent to the square of AB *.
A C
B
It will be convenient, for the sake of conciseness, to designate
in future this remarkable division of a line, where the rectangle
under the whole and one part is equivalent to the square of the
other, by the term Medial Section.
PROP. XXIII. THEOR.
The square of the side of an isosceles triangle is
equivalent to the square of a straight line drawn from
the vertex to the base, together with the rectangle
contained by the segments thus formed.
If BD be drawn from the vertex of the iscsceles triangle
ABC to a point D in the base; the square of AB is equiva-
lent to the square of BD, together with the rectangle under
the segments AD, DC.
For (I. 7.) bisect the base AC in E, and join BE. Because
the triangles ABE and CBE have the sides AB, AE equal to
BC, CE, and the side BE common, they are equal (I. 2.),
* See Note XX.
60
ELEMENTS OF GEOMETRY.
and consequently the corresponding angles BEA, BEC are
equal, and each of them (Def. 4.) a right an-
gle. Wherefore the square of AB is equiva-
B

AD E
C
lent to the squares of AE and BE (II. 11.);
and since AC is cut equally in E and unequal-
ly in D, the square of AE is equivalent to
the square of DE, together with the rectan-
gles AD, DC (II. 19. cor. 1.); and conse-
quently the square of AB is equivalent to the
squares of BE and DE, together with the rectangle AD, DC.
But the square of BD is equivalent to the squares of BE and
DE (II. 11.); whence the square of AB is equivalent to the
square of BD, together with the rectangle AD, DC.
Cor. The square of a straight line BD drawn from the
vertex of an isosceles triangle to any point in the base produ-
ced, is equivalent to the square of BA the side of the trian-
gle, together with the rectangle con-
tained by AD and DC, the external
segments of the base.
B

A E C
For draw BE, as before, to bisect
the base AC. The square of DE is
equivalent to the square of AE, to- D
gether with the rectangle AD, DC (II. 19. cor. 2.); to each
of these, add the square of BE, and the squares of DE and
BE,—that is, the square of BD (II. 11.)—are equal to the
squares of AE and BE, or the square of BA, together with
the rectangle AD, DC *.
PROP. XXIV. THEOR.
In a scalene triangle, the difference between the
squares of the sides, is equivalent to twice the rect-
angle contained by the base and the distance of its
middle point from the perpendicular.
* See Note XXI.
BOOK II.
61
Let the side AB of the triangle ABC be greater than BC;
and, having let fall the perpendicular BE, and bisected AC
in D, the excess of the square of AB above that of BC is equi-
valent to twice the rectangle contained by AC and DE.
For the square of AB is equivalent to the squares of AE
and BE (II. 11.), and the square of BC is equivalent to the
squares CE and BE; wherefore the excess of the
AB above that of BC is equivalent to the
excess of the square of AE above that of
CE. But the excess of the square of AE
above that of CE, is (II. 19.) equivalent
to the rectangle contained by their sum AC А
and their difference, which is evidently the
square of
B

DE
double of DE; and consequently the difference between the
squares of AE and CE, being equivalent to the rectangle
contained by AC and the double of DE, is equivalent to twice
the rectangle under AC and DE.
Cor. The difference between the squares of the sides of a
triangle, is equivalent to the difference between the squares of
the segments of the base made by a perpendicular *.
PROP. XXV. THEOR.
In any triangle, the sum of the squares of the sides,
is equivalent to twice the square of half the base and
twice the square of the straight line which joins the
point of bisection with the vertex.
Let BD be drawn from the vertex B of the triangle ABC
to bisect the base; the squares of the sides AB and BC are
together equivalent to twice the squares of AD and DB.
B
For let fall the perpendicular BE (I. 6.); and if the point
D coincide with E, the triangle ABC being
evidently isosceles, the squares of AB and
BC are the same with twice the square of AB,
or twice the squares of AE and EB, or of AD
and DB (II. 11.)
* See Note XXII.
A
DE
62
ELEMENTS OF GEOMETRY.
B
But if the perpendicular fall upon C, the triangle is right-
angled, and the squares of AB and BC
are then equivalent to the square of AC,
and twice the square of BC, or to twice
the squares of AD, DC and BC; but
(II. 11.) twice the squares of DC and BC
A
D
C
are equivalent to twice the square of DB,
and consequently the squares of AB and BC are equivalent
to twice the squares of AD and DB.
B
In every other case, whether the perpendicular BE fall
within or without the base AC, the squares
of AE, EC, the unequal segments of AC,
are (II. 21. cor.) equivalent to twice the
square of AD and twice the square of DE;
add twice the square of EB to both, and
the squares of AE, EB and of CE, EB-
or the squares of the hypotenuses AB, BC
--are equivalent to twice the square of AD,
and twice the squares of DE, EB, that is
(II. 11.) to twice the square of DB.
PROP. XXVI. THEOR.
A
DE C

B
A
D
C E
The square of the side of a triangle is greater or
less than the squares of the base and the other side,
according as the opposite angle is obtuse or acute,
-by twice the rectangle contained by the base and
the distance intercepted between the vertex of that
angle and the perpendicular.
In the oblique-angled triangle ABC, where the perpendi-
cular BD falls without the base; the square of the side AB
which subtends the oblique angle exceeds the squares of the
sides AC and BC which contain it, by twice the rectangle
under AC and CD.
BOOK II.
63
B
For the square of AD, or of the sum of AC and CD, is
(II. 17.) equivalent to the squares of these lines AC, CD, to-
gether with twice their rectangle. Add the square of DB to
each side, and the squares of AD, DB, or
(II. 11.) the square of AB is equivalent to the
square of AC, and the squares of CD, DB,
together with twice the rectangle AC, CD;
but the squares of CD, DB are (II. 11.)
equivalent to the square of CB; whence the
square of AB exceeds the squares of AC, BC, by twice the
rectangle under AC and CD.
A
Again, in the acute-angled triangle ABC, where the
pendicular BD falls within the triangle; the
square of the side AB that subtends the acute
angle, is less than the squares of the contain-
ing sides AC, BC, by twice the rectangle
under the base AC and its intercepted por-
tion CD.
C D
per-
For the square of AD, or of the difference between AC
and CD, is (II. 18.) equivalent to the squares of AC and CD,
diminished by twice their rectangle. Add to each the square
of DB, and the squares of AD and DB-or the square of
AB-are equivalent to the square of AC, with the squares of
CD and DB, or the square of BC diminished by twice the
rectangle under AC and CD. Consequently the square of
AB is less than the squares of AC and BC, by twice the rect-
angle under AC and CD.
Cor. If the side BC be equal to the base AC, the square
of the other side AB is equivalent to twice the rectangle un-
der AC and AD, whether the perpendicular BD fall without
or within the triangle *.
* See Note XXIII.
64
ELEMENTS OF GEOMETRY.
PROP. XXVII. THEOR.
The squares
squares of lines drawn from any point to the op-
posite corners of a rectangle are together equivalent.
If from a point E, either within or without the rectangle
ABCD, straight lines be drawn to the four corners, the
squares of AE, EC are together equivalent to the squares of
BE, ED.
B
C
E
F
E
For join E with F, the intersection of the diagonals AC, BD.
Because (I. 29. and cor.) these diago-
nals are equal, and bisect each other,
the lines AF, BF, CF, and DF are all
equal. Wherefore the squares of AE,
EC are equivalent to twice the square
of EF (II. 25.), and the squares of BE,
ED are likewise equivalent to twice the
square of BF and twice the same square
of EF; consequently, the squares of
AF and BF being equal, the squares
of AE, EC, are together equivalent
to the squares of BE, ED *.


B
A
PROP. XXVIII. THEOR.
The squares of the sides of a rhomboid, are toge-
ther equivalent to the squares of its diagonals.
Let ABCD be a rhomboid: The squares of all the sides
AB, BC, CD, and AD, are together equivalent to the squares
of the diagonals AC, BD.
For the diagonals bisect each other
(I. 31.), and consequently the squares
of AB, BC, are equivalent to twice the
square of AE and twice the square of
BE (II. 30.); wherefore twice the
B
E
A
D
C

squares of AB, BC, or the squares of all the sides of the
Sce Note XXIV.
BOOK II.
65
rhomboid, are equivalent to four times the square of AE and
four times the square of BE, that is, to the squares of AC
and BD.
PROP. XXIX. THEOR.
The squares of the sides of a quadrilateral figure
are together equivalent to the squares of its diago-
nals, together with four times the square of the
straight line joining their middle points.
Let ABCD be a quadrilateral figure, in which the straight
lines AC, BD, drawn to the opposite corners, arc bisected at
the points E, F; the squares of AB, BC, CD, and DA, are
together equivalent to the squares of AC, BD, together with
four times the square of EF.
For join EF. And because AC is bisected in F, the
squares of AB and BC are equivalent to twice the square of AF
and twice the square of BF (II. 25.);
and, for the same reason, the squares
B

D
E
F
of CD and DA are equivalent to twice
the square of AF and twice the square
of DF. Consequently the squares of
all the sides AB, BC, CD, and DA,
are equivalent to four times the square
of AF-or the square of AC-with
twice the square of BF and of DF. But twice these squares
of BF and DF is equivalent (II. 21.) to four times the square
of BE, or the square of BD, with four times the square of
EF; whence the squares of all the sides of the quadrilateral
figure are together equivalent to the squares of its diagonals
AC, BD, with four times the square of the straight line EF
which joins their points of equal section *.
* See Note XXV.
E
"
ELEMENTS
OF

GEOMETRY.
BOOK III.
DEFINITIONS.

1. ANY portion of the circumfe-
rence of a circle is called an arc,.
and the straight line which joins
the two extremities, a chord.
2. The space included between an arc and its chord, is
named a segment.

3. A sector is the portion of a circle con-
tained by two radii and the arc between
them.
4. The tangent to a circle is a straight
line which touches the circumference, or
meets it only in a single point.

£ 2
68
ELEMENTs of geometry.

5. Circles are said to touch
mutually, if they meet, but do not
cut each other.
6. The point where a straight line touches a circle, or one
circle touches another, is called the point of contact.
7. A straight line is said to be inflected
from a point, when it terminates in another
straight line, or at the circumference of a
circle.

BOOK III
69
PROP. Į. THEOR.
A circle is bisected by its diameter.
The circle ADBE is divided into two equal portions by the
diameter AB.
For let the portion ADB be reversed and applied to AEB,
the straight line AB and its middle
point, or the centre C, remaining the
same.
And since the radii of the cir-
cle are all equal, or the distance of C
from any point in the boundary ADB
is equal to its distance from any point
of the opposite boundary AEB, every
point D of the former must meet with
A
E
D

B
a corresponding point of the latter, and consequently the two
portions ADB and AEB will entirely coincide.
Cor. The portion ADB limited by a diameter, is thus a
semicircle, and the arc ADB is a semicircumference.
PROP. II. THEOR.
A straight line cuts the circumference of a circle
only in two points.
If the straight line AB cut the
circumference of a circle in D, it can
only meet it again in another point
E.
For join D and the centre C; and
because from the point C only two
equal straight lines, such as CD and

A D
E B
CE, can be drawn to AB (I. 18. cor.), the circle described
from C through the point D will cross AB again only at E.
70
ELEMENTS OF GEOMETRY.
PROP. III. THEOR.
The chord of an arc lies wholly within the circle.
The straight line AB which joins two points A, B in the
circumference of a circle, lies wholly within the figure.
For, from the centre C, draw CD to
any point in AB, and join CA and CB.
Because CDA is the exterior angle of
the triangle CDB, it is greater (I. 8.)
than the interior CBD or CBA; but
CBA, being (I. 11.) equal to CAB or
CAD, the angle CDA is consequently
greater than CAD, and its opposite side

D B
CA (I. 14.) greater than CD, or CD is less than CA, and
therefore the point D must lie within the circle.
Cor. Hence a circle is concave towards its centre.
PROP. IV. THEOR.
A straight line drawn from the centre of a circle
at right angles to a chord, likewise bisects it; and,
conversely, the straight line which joins the centre
with the middle of a chord, is perpendicular to it.
The perpendicular let fall from the centre C upon the chord
AB, cuts it into two equal parts AD, DB.
For join CA, CB: And, in the triangles ACD, BCD, the
side AC is equal to CB, CD is common to
both, and the right angle ADC is equal
to BDC; these triangles, being of the
same affection, are equal (I. 22.) and conse-
quently the corresponding side AD is equal
to BD.
Again, let AD be equal to BD; the bi-
secting line CD is at right angles to AB.

C
D
B
BOOK III.
71
For join CA, CB. The triangles ACD and BCD, having
the sides AC, AD equal to CB, BD, and the remaining side
CD common to both, are equal (I. 2.), and consequently the
angle CDA is equal to CDB, and each of them a right angle.
Cor. Hence a straight line cutting two concentric circles
has equal portions intercepted.
PROP. V. THEOR.
A straight line which bisects a chord at right
angles, passes through the centre of the circle.
If the perpendicular FE bisect a chord AB, it will pass
through G the centre of the circle.
For in FE take any point D, and join
DA and DB. The triangles ADC and
BDC, having the side AC equal to BC,
CD common, and the right angle ACD
equal to BCD, are equal (I. 3.), and
consequently the base AD is equal to BD.
The point D is, therefore, the centre of
E

D
G
C
B
F
a circle described through A and B; and thus the centres of
the circles that can pass through A and B are all found in the
straight line EF. The centre G of the circle AEBF must
hence occur in that perpendicular.
Cor. The centre of a circle may hence be found by bisect-
ing the chord AB by the diameter EF (I. 7.), and bisecting
this again in G.
PROP. VI. THEOR.
The greatest line that can be inflected within a
circle, is the diameter.
72
ELEMENTS OF GEOMETRY.
The diameter AB is greater than any
chord DE.
For join CD and CE. The two sides
DC and EC of the triangle DCE are
together greater than the third side DE
(I. 15.); but DC and CE are equal to
AC and CB, or to the whole diameter
AB. Wherefore AB is greater than DE.
PROP. VII. THEOR.

E
B
If from any eccentric point, two straight lines be
drawn to the circumference of a circle; the one
which passes nearer the centre, is greater than that
which lies more remote.
Let C be the centre of a circle,
and A a different point, from
which two straight lines AD and
AE are drawn to the circumfe-
rence; of these lines, AD, which
lies nearer to B the opposite extre-
mity of the diameter, is greater
than AE.
For the triangles ADC and
AEC have the side CD equal to
CE, the side CA common to both,
but the contained angle DCA
greater than ECA; wherefore (I.
19.) the base AD is likewise great-
er than the base AE.
Cor. 1. Hence the straight line
ACB, which passes through the
centre, is the greatest of all those

A
F
A H
A
E
E
B
A
E

D
B

B
that can be drawn to the circumference of the circle from the
eccentric point A. For it is evident from the Proposition,
BOOK III.
73
that the nearer the point D approaches to B, the greater is
AD; consequently the point B forms the extreme limit of
majority, or AB is the greatest line that can be drawn from A
to the circumference.
Cor. 2. Hence also, whether the eccentric point be within or
without the circle, the straight line AH is the shortest that
can be drawn from A to the circumference. For AE is less
than AD, and AG less than AF; and the nearer the termi-
nating point approaches to H, which is obviously the most
remote from B, the shorter must be its distance from A.
Wherefore the point H marks the limit of minority, and AH
is the shortest line that can be drawn from A to the circum-
ference of the circle.
PROP. VIII. THEOR.
From any eccentric point, not more than two equal
straight lines can be drawn to the circumference, one
on each side of the diameter.
Let A be a point which is not the
centre of the circle, and AD a straight
line drawn from it to the circumfe-
rence.
Find the centre C (III. 5. cor.) join
CA and CD, draw (I. 4.) CE making
an angle ACE equal to ACD and cut-
ting the circumference in E, and join
AE: The straight lines AE, AD are
equal.

H
A
C
B
G

B
For the triangles ADC, AEC, ha-
ving the side CD equal to CE, the
AH
side AC common, and the contained
E
C
angle ACD equal to ACE, are equal (I. 3.), and consequent-
ly the base AD is equal to AE.
74
ELEMENTS OF GEOMETRY.
But, except AE, no straight line can be drawn from A on
the same side of the diameter HB, that shall be equal to AD:
For if the line terminate in a point F between E and B, it
will be greater than AE (III. 7.); and if the line terminate in
G between E and H, it will, for the same reason, be less than
AE.
Cor. That point from which more than two equal straight
lines can be drawn to the circumference, is the centre of the
circle.
PROP. IX. THEOR.
One circle will not cut another in more than two
points.
Let DCF and DCE be two cir-
cles, of which A and B are the cen-
tres; join B with the intersections C
and D.
And because B is a point different
from the centre A of the circle DCF,
not more than two equal straight
lines BC and BD can be drawn from
it to the circumference of that circle

I
E
F
BA
D

B
Α.
(III. 8.); consequently the circle, described from B as a cen-
tre and through the points C and D, will not again meet the
circumference DCF.
PROP. X. THEOR.
A circle may be described through three points
which are not in the same straight line.
Let A, B, C, be three points not lying in the same direc-
BOOK III.
75
tion; the circumference of a circle may be made to pass
through these points.
For (I. 7.) bisect AB by the perpen-
dicular DF, and BC by the perpendi-
cular EF. These straight lines DF,
EF will meet; because, DE being join-
ed, the angles EDF, DEF are less than
BDF, BEF, and consequently are to-
F
B

gether less than two right angles, and DF, EF are not paral-
lel (I. 23.) but concur to form a triangle whose vertex is F.
Again, every circle that passes through the two points A
and B, has its centre in the perpendicular DF (III. 5.); and,
for the same reason, every circle that passes through B and
C has its centre in EF; consequently the circle which would
pass through all the three points, must have its centre in F,
the point common to both perpendiculars DF and EF.
It is manifest, that there is only one circle which can be´
made to pass through the three points A, B, C; for the in-
tersection of the straight lines DF and EF, which marks the
centre, is a single point.
Cor. Hence the mode of describing a circle about a given
triangle ABC.
PROP. XI. THEOR.
Equal chords are equidistant from the centre of a
circle; and chords which are equidistant from the
centre, are likewise equal.
Let AB, DE be equal chords inflected within the same cir-
cle; their distances from the centre, or the perpendiculars
CF, CG, let fall upon them, are equal.
For the perpendiculars CF and CG bisect the chords AB
and DE (III. 4.), and consequently BF, DG, the halves of
*76
ELEMENTS OF GEOMETRY.
A
E
these, are likewise equal. The right-angled triangles CBF
and CDG, which are thus of the same
affection, having the two sides BC, BF
equal respectively to DC, DG, and the
corresponding angle BFC equal to DGC,
are equal (I. 22.), and consequently the
side FC is equal to GC.
Again, if the chords AB, DE be equal-

F
B
ly distant from the centre, they are themselves equal.
D
For the same construction remaining: The triangles CBF
and CDG are still of the same affection; and have now the
two sides CB, CF equal to CD, CG, and the angle BFC
equal to DGC; consequently they are equal, and the side
BF equal to DG; the doubles of these, therefore, or the
whole chords AB, DE, are equal.
PROP. XII. THEOR.
The greater chord is nearer the centre of the cir-
cle; and the chord which is nearer the centre is like-
wise the greater.
Let the chord DE be greater than AB; its distance from
the centre, or the perpendicular CG let fall upon it, is less
than the distance CF.
For in the right-angled triangle BCF,
the square of the hypotenuse BC is equi-
valent to the squares of BF and FC
(II. 11.); and, for the same reason, the
square of the hypotenuse DC of the right-
angled triangle DCG is equivalent to the
squares of DG and GC. But BC and

A
F
B D
E
DC are equal, and consequently their squares; wherefore
the squares of DG and GC are equivalent to the squares
of BF and FC. And since DE is greater than AB, its
half DG is greater than BF, and consequently the square
BOOK III.
77
of DG is greater than the square of BF; the square of GC is,
therefore, less than the square of FC, because, when joined
to the squares of DG and BF, they produce the same amount,
or the square of the radius of the circle. Hence the perpen-
dicular GC itself is less than FC.
Again, if the chord DE be nearer the centre than AB, it
is also greater.
For the same construction remaining: It is proved that
the squares of BF and FC are together equal to the squares.
of DG and GC; but GC being less than FC, the square of
GC is less than the square of FC, and consequently the square
of DG is greater than the square of BF; whence the side DG
is greater than BF, and its double, or the chord DE, greater
than AB.
PROP. XIII. THEOR.
In the same or equal circles, equal angles at the
centre are subtended by equal chords, and termi-
nated by equal arcs.
If the angle ACB at the centre C be equal to DCE, the
chord AB is equal to DE, and the arc AFB is equal to DGE.
For let the sector ACB be applied to DCE. The centre
remaining in its place, the radius CA will lie on CD; and
the angle ACB being equal to DCE, the radius CB will adapt
itself to CE. And because all the radii
are equal, their extreme points A and
B must coincide with D and E; where-
fore the straight lines which join those
points, or the chords AB and DE,
must coincide. But the arcs AFB
and DGE that connect the same
points, will also coincide; for any in-

A
F
I
D
E
termediate point F in the one, being at the same distance
78
ELEMENTS OF GEOMETRY.
5
from the centre as every point of the other, must, on its ap-
plication, find always a corresponding point G.
The same mode of reasoning is applicable to the case of
equal circles.
Cor. Hence, in the same or equal circles, equal arcs are
subtended by equal chords, and terminate equal angles at the
centre.
PROP. XIV. THEOR.
In the same or equal circles, equal chords subtend
equal arcs of a like kind.
If the chord AB be different from the diameter, it will evi-
dently subtend at the same time two unequal portions of the
circumference of a circle, the one terminating the angle ACB
at the centre and less than the semicircumference, the other
greater than this and terminating the reversed angle.
In the equal circles GAHB
and IDKE the chord AB sub-
tends the arcs AGB and AHB,
which are respectively equal to
DIE and DKE subtended by
the equal chord DE.
For join CA, CB, and FD,
H
K.

C
F
BD
E
I
FE. The two triangles CAB and FDE, having all the sides
of the one equal to those of the other, are equal (I. 2.); and
consequently the angle ACB is equal to DFE. Wherefore
the arcs AGB and DIE, which terminate these equal angles,
are (III. 13.) themselves equal; and hence the remaining por-
tions AHB and DKE of the equal circumferences are like-
wise equal,
This demonstration, it is evident, will likewise apply in the
case of the same circle.
BOOK III.
79
PROP. XV. PROB.
To bisect an arc of a circle.
Let it be required to divide the arc AEB into two equal
portions.
Draw the chord AB, and bisect it (I. 7.) by the perpendi-
cular EF cutting the arc AB in E: The arc AE is equal to
EB.
For the triangles ADE, BDE, have
the side AD equal to BD, the side DE
common, and the containing right angle
ADE equal to BDE; they are (I. 3.)
consequently equal, and the base AE
equal to BE. But these equal chords
AE, BE must subtend equal arcs of a
like kind (III. 14.), and the arcs AE,
A
E

B
D
-C
F
BE are evidently each of them less than a semicircumference.
Cor. The correlative arc AFB is also bisected by the
pendicular EF.
per-
PROP. XVI. PROB.
Given an arc, to complete its circle.
Let ADB be an arc; it is required to trace the circle to
which it belongs.
Draw the chord AB, and bisect it by
the perpendicular CD (I. 7.) cutting the
arc in D, join AD, and from A draw A
AC making an angle DAC equal to
ADC (I. 4.): The intersection C of this
straight line with the perpendicular, is
the centre of the circle required.
D

B
E
C
80
ELEMENTS OF GEOMETRY.
For join CB. The triangles ACE and
BCE, having the side EA equal to EB,
the side EC common, and the contained
angle AEC equal to BEC, are equal (I.
3.), and consequently AC is equal to BC.
But (I. 12.) AC is also equal to CD, be-
D

C
A
B
E
cause the angle DAC was made equal to ADC. Wherefore
(III. 8. cor.) the three straight lines CA, CD, and CB being
all equal, the point C is the centre of the circle.
PROP. XVII. THEOR.
The angle at the centre of a circle is double of
the angle which, standing on the same arc, has its
vertex in the circumference.
Let AB be an arc of a circle; the angle which it terminates
at the centre, is double of ADB the corresponding angle at
the circumference.
For join DC and produce it to the opposite circumference.
This diameter DCE, if it lie not on one of the sides of the angle
ADB, must either fall within that angle or without it.
First, let DC coincide with DB. And
because AC is equal to DC, the angle
ADC is equal to DAC (I. 11.); but the
exterior angle ACB is equal to both of
these (I. 32.) and therefore equal to double
of either, or the angle ACB at the centre
is double of the angle ADB at the cir-
cumference.
Next, let the straight line DCE lie
within the angle ADB. From what has
been demonstrated, it is apparent, that
the angle ACE is double of ADE, and
the angle BCE double of BDE; where-
fore the angles ACE, BCE taken toge-

D
A
B
E
D

B
ther, or the whole angle ACB, are double of the collected
BOOK LII.
81
angles ADE, BDE, or the angle ADB at the circumfe-
rence.
Lastly, let DCE fall without the angle
ADB. Because the angle BCE is double
of BDE, and the angle ACE is double of
ADE; the excess of BCE above ACE,
or the angle ACB at the centre, is double
of the excess of BDE above ADE, that
is, of the angle ADB at the circumference*.

E
PROP. XVIII. THEOR.
B
A
D
The angles in the same segment of a circle are
equal.
Let ADB be the segment of a circle;
the angles AFB, AGB contained in it,
or which stand on the opposite portion
AEB of the circumference, are equal to
each other.
For join CA, CB. The angle ACB
at the centre is double of the angle AFB
or AGB at the circumference (III. 17.);
these angles AFB, AGB, which stand
on the same arc AEB, are, therefore,
the halves of the same central angle
ACB, and are consequently equal to
each other.
D
F

G
E
F D
E
B
G
B
Cor. Hence equal angles at the circumference must stand
on equal arcs; for their doubles or the central angles, being
equal, are terminated by equal arcs (III. 13.) Hence also
equal angles that stand on the same base, have their vertices
in the same segment of a circle.
* See Note XXVI.
F
82
ELEMENTS OF GEOMETRY.
PROP. XIX.
THEOR.
The opposite angles of a quadrilateral figure con-
tained within a circle, are together equal to two
right angles.
Let ABCD be a quadrilateral figure described in a circle;
the angles A and C are together equal to two right angles,
and so are those at B and D.
B
C
E
For join EB and ED. The angle BED at the centre is
double of the angle BCD at the circum-
ference (III. 17.); and for the same rea-
son, the reversed angle BED is double
of BAD. Consequently the angles BCD
and BAD are the halves of angles about
the point E, and which make up four
right angles; wherefore the angles BCD
and BAD are together equal to two
right angles.
A
In the same manner, by joining EA and EC, it may be
proved, that the angles ABC and ADC are together equal to
two right angles.
Cor. Hence a circle may be described about a quadrilate-
ral figure which has its opposite angles equal to two right
angles; for if a circle be made to circumscribe the triangle
BCD (III. 10. cor.), the angles opposite to the base BD are
equal to two right angles, and therefore equal to the angles
BCD' and BAD; consequently the angle BAD is equal to an
angle in the segment BAD, and hence (III. 18. cor.) they
are contained in the same segment, or the circumference of
the circle passes through all the four points A, B, C, and D.
PROP. XX. THEOR.
Parallel chords intercept equal arcs of a circle.
Let the chord AB be parallel to CD; the intercepted arc
AC is equal to BD.
+

BOOK III.
83
For join AD. And because the
straight lines AB and CD are parallel,
the alternate angles BAD and ADC
are equal (I. 23.); wherefore these
angles, having their vertices in the cir-
cumference of the circle, must stand on
equal arcs (III. 18. cor.), and conse-
quently the arcs AC and BD are equal to each other.

B
D
Cor. Hence, conversely, the straight lines which intercept
equal arcs of a circle are parallel; and hence another mode of
drawing a parallel through a given point to a given straight
line*.
PROP. XXI. THEOR.
The inclination of two straight lines is equal to the
angle terminated at the circumference by the sum
or difference of the arcs which they intercept, ac-
cording as their vertex is within or without the circle.
If the two straight lines AB and CD intersect each other
in the point E within a circle; the angle AED which they
form, is equal to an angle at the circumference and standing
on the sum of the intercepted arcs AD and BC.
For draw the chord BF parallel to CD (III. 20. cor.).
Because ED and BF are parallel, the
angle AED (I. 23.) is equal to the inte-
rior angle ABF, which stands on the
arc AF; but since the chords BF and
CD are parallel, the arc BC is equal to
DF (III. 20.) and consequently the arc
AF, which terminates at the circumfe-
C
B

E
F
D
rence an angle equal to AED, is the sum of the two inter-
cepted arcs AD and BC.
* See Note XXVII.
F 2
84
ELEMENTS OF GEOMETRY.
Again, if the straight lines AB and
CD meet at E, without the circle,
their inclination AED is equal to an
angle at the circumference, having for
its base the excess of the arc AD a-
bove BC.
For BF being drawn parallel to CD,
the arc BC is equal to FD, and conse-
quently the arc AF is the excess of AD
13
E

C
D
F
།
above BC; but the angle ABF which stands on AF, is equal
to the interior angle AED.
Cor. Hence if two chords intersect each other at right
angles within a circle, the opposite intercepted arcs are equal
to the semicircumference *.
PROP. XXII. THEOR.
The angle in a semicircle is a right angle, the
angle in a greater segment is acute, and the angle
in a smaller segment is obtuse.
Let ABD be an angle in a semicircle, or that stands on
the semicircumference AED; it is a right angle.
For ABD, being an angle at the
circumference, is half of the angle at
the centre on the same base AED
(III. 17.); it is, therefore, half of the
angle ACD formed by the opposite
portions CA, CD of the diameter, or
half of two right angles, and is con-
sequently equal to one right angle.

B
D
Again, let ABD be an angle in a segment greater than a
semicircle, or which stands on a less arc AED than the semi-
circumference; it is an acute angle.
* See Note XXVIII.
BOOK III.
85
For join CA, CD. The angle ABD
is half of the central angle ACD, which
is evidently less than two right angles;
wherefore ABD is less than one right
angle, or it is acute.
But the angle AED, in the smaller
segment, is obtuse. For AED stands

on the arc ABD, which is greater than a semicircumference,
and is the base of an angle at the centre, the reverse of
ACD, and greater, therefore, than two right angles; AED
is hence an obtuse angle.
Scholium. From the remarkable property, that the angle in
a semicircle is a right angle, may be derived an elegant me-
thod of drawing perpendiculars *.
PROP. XXIII. THEOR.
If a circle be described on the radius of another
circle, any straight line drawn from the point where
they meet to the outer circumference, is bisected by
the interior one.
Let AEC be a circle described on
the radius AC of the circle ADB, and
AD a straight line drawn from A to
terminate in the exterior circumfe-
rence; the part AE in the smaller cir-
cle is equal to the part ED intercepted
between the two circumferences.

A
D
B
For join CE. And because AEC is a semicircle, the
angle contained in it is a right angle (III. 22.); consequently
the straight line CE, drawn from the centre C, is perpendi-
cular to the chord AD, and therefore (III. 4.) bisects it.
* See Note XXZX,
86
ELEMENTS OF GEOMETRY.
}
PROP. XXIV. THEOR.
:
The perpendicular at the extremity of a diameter
is a tangent to the circle, and the only tangent which
can be applied at that point.
Let ACB be the diameter of a circle, to which the straight
line EBD is drawn at right angles from the extremity B; it
will touch the circumference at that point.
For CB, being perpendicular,
is the shortest distance of the cen-
tre C from the straight line EBD
(I. 18.); wherefore every 6ther
point in this line is farther from
the centre than B, and conse-
quently falls without the circle.
But the perpendicular EBD is

A
F
D
G
B
E
H
the only straight line which can be drawn through the point
B that will not cut the circle. For if HBF were such a line,
the perpendicular CG, let fall upon it from the centre, would
be less than CB (I. 18.), and would therefore lie within the
circle; consequently HBG, being extended, would again
meet the circumference, before it effected its escape.
Cor. Hence a straight line drawn from the point of contact
at right angles to a tangent, must be a diameter, or pass
through the centre of the circle.
ம் D D
FI
Scholium. The nature of a tangent to the circle is easily
discovered from the consideration of limits. For suppose the
straight line DE, extending both ways,
to turn about the extremity B of the
diameter AB; it will cut the circle first
on the one side of AB, and afterwards
on the other. But the arc AH being
less than a semicircumference, the
angle HBA which the line D'E'
makes with the diameter is acute

A
K
B
(III. 22.); and, for the same reason, the angle KBA is acute,
་
BOOK III.
87
and consequently its adjacent angle D'BA is obtuse. Thus
the revolving line DE, when it meets the semicircumference
AHB, makes an acute angle with the diameter; but when it
comes to meet the opposite semicircumference, it makes an ob-
tuse angle. In passing, therefore, through all the intermediate
gradations from minority to majority, the line DE must find
a certain individual position in which it is at right angles to
the diameter, and cuts the circle neither on the one side nor
the other.
A similar inference might be derived from Prop. 20. of
this Book; one of the parallel chords being supposed to con-
tract, until its extreme points are about to coalesce in the
position of the tangent.
PROP. XXV. THEOR.
If, from the point of contact, a straight line be
drawn to cut the circumference, the angles which it
makes with the tangent are equal to those in the al-
ternate segments of the circle.
Let CD be a tangent, and BE a straight line drawn from
the point of contact, cutting the circle into two segments.
BAE and BFE; the angle EBD is equal to EAB, and the
angle EBC to EFB.
A
For draw BA perpendicular to CD (I. 5. cor.), join AE,
and, from any point F in the opposite arc, draw FB and FE.
Because BA is perpendicular to the tangent at B, it is a
diameter (III.24. cor.), and consequent-
ly AEFB is a semicircle; wherefore
AEB is a right angle (III. 22.), and the
remaining acute angles BAE, ABE of
the triangle, being together equal to
another right angle, are equal to ABE
and EBD, which compose the right
angle ABD. Take the angle ABE

E
C
B
D
away from both, and the angle BAE remains equal to EBD
88
ELEMENTS OF GEOMETRY.
Again, the opposite angles BAE and BFE of the quadri-.
lateral figure BAEF, being equal to two right angles (III. 19.),
are equal to the angle EBD with its adjacent angle EBC;
and taking away the equals BAE and EBD, there remains
the angle BFE equal to EBC.
Cor. If a straight line meet the circumference of a circle,
and make an angle with an inflected line equal to that in the
alternate segment, it touches the circle.
PROP. XXVI. PROB.
To draw a tangent to a circle, from a given point
without it.
Let A be a given point, from which it is required to draw
a straight line that shall touch the circle DGH.
Find the centre C (III. 5. cor.), join CA and draw DE
(I. 5. cor.), perpendicular to CA, from
C with the distance CA describe a cir-
cle meeting DE in F, join CF cutting
the interior circumference in G; AG
being joined, is the tangent which was
required.
For the triangles ACG and FCD have
the sides CA, CG equal to CF, CD, and
the containing angle ACF common to

E
F
G
A
D
H
both; they are, therefore, equal (I. 3.), and consequently the
angle CGA is equal to CDF. But CDF is a right angle;
whence CGA is likewise a right angle, and AG a tangent to
the circle (III. 24.)
Or thus.

On AC as a diameter describe the
circle AGCK, cutting the given cir-
cle in the points G, K: Join AG,
AK; either of these lines is the tan-
gent required.
For join CG, CK. And the angles
I
G
K
ப
CGA, CKA, being each in a semicircle, are right angles
BOOK III.
89
(III.22.), and consequently AG, AK, touch the circle DGHK
at the points G, K (III. 24.).
Cor. Hence tangents drawn from the same point to a circle
are equal; for the triangles ACG and ACK having the side
CG equal to CK, CA common, and the angles at G and K
right, are equal (I. 22.), and consequently AG is equal to AK.
PROP. XXVII. PROB.
On a given straight line, to describe a segment of
a circle, that shall contain an angle equal to a given
angle.
Let AB be a straight line, on which it is required to de-
scribe a segment containing an angle equal to C.
If C be a right angle, it is evident that the problem will be
performed, by describing a semicircle on AB. But if the angle
C be either acute or obtuse; draw
AD (I. 4.) making an angle BAD
equal to C (I. 36.), erect AE per-
pendicular to AD, draw EF (I. 5.
cor.) to bisect AB at right angles
and meeting AE in E, and, from
this point as a centre and with the
distance EA describe the required
segment AGB.
Because EF bisects AB at right D
G

E
F
B
C
D
G

F
B
C
E
angles, the circle described through A must also pass through
(III. 5.) the point B; and since EAD is a right angle, AD
touches the circle at A (III. 24.), and the angle BAD, which
was made equal to C, is equal (III. 25.) to the angle in the al-
ternate segment AGB.
PROP. XXVIII. THEOR.
Two circles which meet in the straight line join-
ing their centres or in its continuation, touch each
other.
90
ELEMENTS OF GEOMETRY.
Let the circles DCE, FCG meet at
C, in the direction of the straight line
which joins their centres A, B; they
touch each other at that point.
For draw BH to another point H in
the circumference DCE. And because
B is distinct from the centre A, the line
BH is greater than BC (III. 7. cor. 2.),
and consequently the point H lies with-

1
IL
F
B
C
A
E

I
C
BA
G
E
out the circle FCG. Except, therefore, at the single point
C, thé circumference DCE does not meet FCG.
Cor. Hence a straight line extending through the centres
of two circles, will pass through their points of contact.
PROP. XXIX. THEOR.
Two straight lines drawn through the point of con-
tact of two circles, intercept arcs of which the chords
are parallel.
Let the circles ACE and ABD touch mutually in A, and
from this point the straight lines AC, AE be drawn to cut the
circumferences; the chords CE and BD are parallel.
For draw the tangent FAG, (III. 26.), which must touch
both circles.
In the case of internal contact, the angle
GAE is equal to ACE in the alternate
segment, (III. 25.); and, for the same
reason, GAE or GAD is equal to ABD;
consequently the angles ACE and ABD
are equal, and therefore (I. 23.) the straight
lines CE and BD are parallel.

F
A
B
E
BOOK III.
When the contact is external,
the angle GAE is still equal to
ACE, and its vertical angle FAD
is, for the same reason, equal to
ABD; whence ACE is equal to
ABD; and these being alternate
D
B
angles, the straight line CE (I. 23.) is parallel to BD.
PROP. XXX. THEOR.
91

E
If from any point in the diameter of a circle or its
extension, straight lines be drawn to the ends of a
- parallel chord; the squares of these lines are toge-
ther equivalent to the squares of the segments into
which the diameter is divided.
Let BEFD be a circle, and from the point A in its extend-
ed diameter the straight lines AE and AF be drawn to the
ends of the parallel chord EF; the squares of AE and AF
are together equivalent to the squares of AB and AD.
For from the centre C, let fall the perpendicular CG upon
EF (I. 6.), and join AG and CE.
Because CG cuts the chord EF at
right angles, GE is equal to GF (III.
4.); wherefore the squares of AE and
AF are equivalent to twice the squares
of AG and GE (II. 30.) But ACG
being a right-angled triangle, the
square of AG is equivalent to the
squares of AC and CG (II. 11.), or
twice the square of AG is equivalent
to twice the squares of AC and CG.
Wherefore the squares of AE and
AF are equivalent to twice the three

E
B A
G
F
D

E
G
L
D
AB
C
squares of AC, CG, and GE. Of these, the two squares of
CG and GE are equivalent to the square of CE or CB, for
92
ELEMENTS OF GEOMETRY.
the triangle CGE is right-angled. Consequently the squares
of AE and AF are equivalent to twice the squares of AC and
CB. But the straight line BD being cut equally at C and
unequally at A, the squares of the unequal segments AB and
AD are together equivalent to twice the squares of AC and
CB (II. 21. cor.); whence the squares of AE and AF are to-
gether equivalent to the squares of AB and AD.
PROP. XXXI. THEOR.
If through a point, within or without a circle, two
perpendicular lines be drawn to meet the circumfe-
rence, the squares of all the intercepted distances
are together equivalent to the square of the diameter.
Let E be a point within or without the circle, and AB, CD
two straight lines drawn through it at right angles to the cir-
cumference; the squares of the four segments EA, EB, ED,
and EC, are together equivalent to the square of the diame-
ter of the circle.
For draw BF parallel to CD, and join AF, AD, CB, and
DF.
Because BF is parallel to CD, the arc
BC is equal to the arc FD (III. 20.), and
consequently the chord BC is also equal
to the chord FD (III. 15. cor.); but BC
being the hypotenuse of the right-angled
triangle BEC, its square, or that of FD
is equivalent to the squares of EB and EC
(II. 11.), and AED being likewise right-
angled, the square of AD is equivalent to
the squares of EA and ED. Whence
the squares of AD and FD are equiva-
lent to the four squares of EA, EB, ED,
C

В
E
D

BE
C
מ.
and EC. But since ED is parallel to BF,
17
the interior angle ABF is equal to AED
(I. 23.), and therefore a right angle; consequently ACBF is
BOOK III.
93
a semicircle (III. 23. cor.) and AF the diameter. The angle
ADF in the opposite semicircle is hence a right angle (III. 22.),
and the square of the diameter AF is equal to the squares of
AD and FD, or to the sum of the squares of the four seg-
ments EA, EB, ED, and EC intercepted between the circum-
ference and the point Ę.
PROP. XXXII. THEOR.
If through a point, within or without a circle, two
straight lines be drawn to cut the circumference;
the rectangle under the segments of the one, is equi-
valent to that contained by the segments of the
other.
Let the two straight lines AD and AF be extended through
the point A, to cut the circumference BFD of a circle; the
rectangle contained by the segments AE and AF of the one,
is equivalent to the rectangle under AB and AD, the distan-
ces intercepted from A in the other.
For draw AC to the centre, and produce it both ways to
terminate in the circumference at G and H; let fall the per-
pendicular CI upon BD (I. 6.), and join CD.
T
Because CI is perpendicular to AD, the difference between
the squares of CA and CD, the sides of the triangle ACD is
equivalent to the difference between
the squares of the segments AI and
ID the segments of the base (II. 21.
cor.); and the difference between
the squares of two straight lines be-
ing equivalent to the rectangle under
their sum and their difference (II.
19.), the rectangle contained by the

B
C
G
H
E
1
94
ELEMENTS OF GEOMETRY.
!
sum and difference of AC, CD is
equivalent to the rectangle contain-
ed by the sum and difference of AI,
ID. But since the radius CG is
equal to CH, the sum of AC and
CD is AH, and their difference is
AG; and because the perpendicular

G
E
C
H
B
D
CI bisects the chord BD (III. 4.), the sum of AI and ID is
AD, and their difference AB. Wherefore the rectangle AH,
AG is equivalent to the rectangle AB, AD. In the same way
it is proved, that the rectangle AH, AG is equivalent to the
rectangle AE, AF; and consequently the rectangle AE, AF
is equivalent to the rectangle AB, AD.
Or thus.
Draw the diameter GAH, and join CB and CD. And be-
cause BCD is an isosceles triangle and CA is drawn from the
vertex C to a point in the direction of its base, the difference
between the square of CA and CD or CG is equivalent to the
rectangle contained by the segments AB, AD of the base (II.
24. cor.). In like manner, it is proved that the same difference
between the square of CA and CG is equivalent to the rect-
angle contained by the segments AE, AF; whence the rect-
angle under AB, AD is equivalent to the rectangle under
AE, AF.
G
B
II
AC
Cor. 1. If the vertex A of the straight lines
lie within the circle and the point I coincide
with it, BD, being then at right angles to
CA, is bisected at A (III. 4.), and the rect-
angle AB, AD is the same as the square of
AB. Consequently the square of a perpendicular AB limit-
ed by the circumference, is equivalent to the rectangle under
the segments AG, AH of the diameter.

D
BOOK III.
95
Cor. 2. If the vertex A lie without the
circle and the point I coincide with B or
D, the angle ABC being then a right A
angle, the incident line AB must be a
tangent (III. 24.), and consequently the
two points of section B and D must coa-

H
D
B
lesce in a single point of contact. Wherefore the rectangle
under the distances AB, AD becomes the same as the square
of AB; and consequently the rectangle contained by the seg-
ments AG, AH of the diameter, is equivalent to the square of
the tangent AB.
PROP. XXXIII. PROB.
To construct a square equivalent to a given recti-
lineal figure.
A
B
F
Let the rectilineal figure be reduced by Proposition 7.
Book II. to an equivalent rectangle, of
which A and B are the two containing
sides; draw an indefinite straight line
CE, in which take the part CD equal to
A and DE to B, on CE describe a semi-
circle, and erect the perpendicular DF
from the diameter to meet the circumfe-
rence: DF is the side of the square equivalent to the given
rectilineal figure.

C
D
E
For, by Cor. 1. to the last Proposition, the square of the
perpendicular DF is equivalent to the rectangle under the
segments CD, DE of the diameter, and is consequently equi-
valent to the rectangle contained by the sides A and B of a
rectangle that was made equivalent to the rectilineal figure.
PROP. XXXIV. THEOR.
A quadrilateral figure may have a circle described
about it, if the rectangles under the segments made
96
ELEMENTS OF GEOMETRY.
by the intersection of its diagonals be equivalent, or
if those rectangles are equivalent which are contain-
ed by the external segments formed by producing
its opposite sides.
Let ABCD be a quadrilateral figure, of which AC and BD
are the diagonals, and such that the rectangle AE, EC is
equivalent to the rectangle BE, ED; a circle may be made
to pass through the four points A, B, C, and D.
For describe a circle through the
three points A, B, C (III. 10. cor.),
and let it cut BD in G. Because AC
and BG intersect each other within
a circle, the rectangle AE, EC is
equivalent to the rectangle BE, EG
(III. 31.); but the rectangle AE, EC
is by hypothesis equivalent to the

B
F
E
C
rectangle BE, ED. Wherefore BE, EG is equivalent to
BE, ED; and these rectangles have a common base BE,
consequently (II. 3. cor.) their altitudes EG and ED are
equal, and hence the point G is the same as D, or the circle
passes through all the four points A, B, C, and D.
Again, if the opposite sides CB and DA be produced to
meet at F, and the rectangle CF, FB be equal to DF, FA,
a circle may be described about the figure.
For, as before, let a circle pass through the three points
A, B, C, but cut AD in H. And from the property of the
circle, the rectangle CF, FB is equivalent to HF, FA; but
the rectangle CF, FB is also equivalent to DF, FA; whence
the rectangle HF, FA is equivalent to DF, FA, and the base
HF equal to DF, or the point H is the same as D *.
* See Note XXX.
ELEMENTS
OF
GEOMETRY.
BOOK IV.
DEFINITIONS.

1. A rectilineal figure is said to be inscribed
in a circle, when all its angular points lie on
the circumference.

2. A rectilineal figure circumscribes a
circle, when each of its sides is a tan-
gent.

3. A circle is inscribed in a rectilineal fi-
gure, when it touches all the sides.
G
98
ELEMENTS OF GEOMETRY.
4. A circle is described about a rectilineal
figure or circumscribes it, when the circumfe-
rence passes through all the angular points of
the figure.

5. Polygons are equilateral, when their sides, in the same
order, are respectively equal: They are equiangular, if an
equality obtains between their corresponding angles.
6. Polygons are said to be regular, when all their sides and
their angles are equal.
BOOK IV.
99
PROP. I. PROB.
Given an isosceles triangle, to construct another
on the same base, but with half the vertical angle.
Let ABC be an isosceles triangle standing on AC; it is re-
quired, on the same base, to construct another isosceles tri-
angle, that shall have its vertical angle half of the angle ABC.
Bisect AC in D (I. 7.), join DB, which
produce till BE be equal to BA or BC,
and join AE, CE: AEC is the isosceles
triangle required.
For, the straight line BE being equal to
BA and BC, the point B is the centre of
a circle which passes through A, E, and C;
and consequently the angle ABC is double
E

A.
D
B
of AEC at the circumference (III. 17.), or the vertical angle
AEC is half of ABC. But the triangles AED and CED,
having the side DA equal to DC, the side DE common to
both, and the right angle ADE (III. 4.) equal to CDE are
(I. 3.) equal, and consequently AE is equal to CE. Where-
fore the triangle AEC is likewise isosceles.
PROP. II. PROB.
Given an acute-angled isosceles triangle, to con-
struct another on the same base, which shall have
double the vertical angle.
Let ABC be an acute-angled isosceles
triangle; it is required, on the base AC,
to construct another isosceles triangle,
having its vertical angle double of the
angle ABC.
Describe a circle through the three
B

D
C
points A, B, and C (III. 10. cor.), and draw AD, CD to
the centre D; the triangle ADC is the isosceles triangle re-
G 2
ELEMENTS OF GEOMETRY.
1
100
quired. For the angle ADC, being at the centre of the cir-
cle, is (III. 17.) double of ABC, the angle at the circumfe-
rence.
PROP. III. THEOR.
If an isosceles triangle have each angle at the base
double of the vertical angle, its base will be equal to
the greater segment of one of its sides divided by a
medial section.
Let ABC be an isosceles triangle which has each of the
angles BAC, BCA double of the vertical angle ABC; the
base AC is equal to the greater segment of the side BA form-
ed by a medial section.
For draw CD to bisect the angle BCA (I. 5.), and about
the triangle BDC describe a circle (III. 10. cor.).
D
A
B
C
Because the angle BCA is double of
ABC and has been bisected by CD, the
angles ACD, BCD are each of them
equal to CBD, and consequently the
side BD is equal to CD (I. 12.). But
the triangles BAC and DAC, having
the angle ACD equal to ABC, and the
angle at A common to both, must have also (I. 32.) the re-
maining angle CDA equal to BCA or CAD; whence (I. 12.)
the triangle DAC is likewise isosccles, and the side AC equal
to CD; and CD being equal to BD, therefore AC is equal to
BD. And since the angle ACD is equal to CBD in the al-
ternate segment of the circle, the straight line AC touches the
circumference at C (III. 25. cor.); wherefore the rectangle
contained by AB and AD (III. 31. cor. 2.) is equivalent to the
square of AC, or the square of BD. Consequently the base
AC of this isosceles triangle is equal to the greater segment
BD of the side AB cut by a medial section.

BOOK IV.
101
Cor. Hence the interior triangle ACD is likewise isosceles
and of the same nature with ABC, having the greater seg-
ment of AB for its side, and the smaller segment for its base.
PROP. IV. PROB.
Given either one of the sides or the base, to con-
struct an isosceles triangle, so that each of the angles
at the base may be double of its vertical angle.
First, let one of the sides AB be given, to construct such
an isosceles triangle.
Divide AB by a medial section at C (II. 22.), and on CB,
as a base with the distance AB for each of the sides, describe
an isosceles triangle (I. 1.)
Next, let the base AB be given, to
construct an isosceles triangle of this
nature.
C
B
A
B
Produce AB to C, such that the rectangle AC, CB be
equal to the square of AB (II. 22. cor. 2.), and on the base
AB, with the distance AC for each of the sides, describe an
isosceles triangle.
These isosceles triangles will fulfil the conditions required.
For it is evident, from the last Proposition, that isosceles
triangles constituted on CB and AB, with each of the angles
at the base double the vertical angle, would have AB and AC
for their sides, and consequently (I. 2.) must coincide with the
triangles now described.
Cor. Hence an isosceles triangle of this kind has its verti-
cal angle equal to the fifth part of two right angles; for each
of the angles at the base being double of the vertical angle,
they are both equal to four times it, and consequently this
vertical angle is the fifth part of all the angles of the triangle,
or of two right angles.
-102
ELEMENTS OF GEOMETRY.
PROP. V. PROB.
On a given finite straight line, to describe a regu-
lar pentagon.
Let AB be the straight line, on which it is required to de-
scribe a regular pentagon.
On AB erect (IV. 4.) the isosceles triangle ACB having
each of the angles at the base double of its vertical angle, on
AB again construct (IV. 2.) another isosceles triangle whose
vertical angle AOB is double of
ACB, and about the vertex O place
(I. 1.) the isosceles triangles AOD,
DOC, COE, and EOB: These tri-
angles, with AOB, will compose a
regular pentagon.

D
E
A
B
For the angle AOB, being the
double of ACB, which is the fifth
part of two right angles (IV. 4. cor.),
must be equal to the fifth part of four right angles; and con-
sequently five angles, each of them equal to AOB, will adapt
themselves about the point O.
But the bases of those central
sides of the pentagon, are all
triangles, and which form the
equal; and the angles at their bases being likewise equal,
they are equal in the collective pairs which constitute the in-
ternal angles of the figure: It is therefore a regular penta-
gon.
Or thus.
Having erected the isosceles triangle ACB, from the cen-
tre A with the distance AC describe an arc of a circle, and
from the centre B with the same distance describe another
arc, and from C inflect the straight lines CE, CD equal to
BOOK IV.
103
AB: The points D, E mark out the pentagon. For it is ap-
parent, that, the three straight lines AO, BO, and CO being
equal (IV. 2.), and the triangles ACB, CAE, and CBD be-
ing likewise equal, the point O must have the same relation
to all of them, and consequently the central triangles COD
and COE are equal to AOB.
PROP. VI. PROB.
On a given finite straight line, to describe a regu-
lar hexagon.
Let AB be the given straight line, on which it is required
to describe a regular hexagon.
On AB construct (I. 1.) the equilateral triangle AOB, and
repeat equal triangles about the vertex O; these triangles will
together compose the hexagon required.
Because AOB is an equilateral triangle, each of its angles
is equal to the third part of two
right angles (I. 32. cor. 1.);
wherefore the vertical angle AOB
is the sixth part of four right
angles, or six of such angles may
be placed about the point O.
But the bases of the triangles
AOB, AOC, COD, DOE, EOF,
and BOF are all equal; and so
D
E

A
B
F
are the angles at the bases, and which, taken by pairs, form
the internal angles of the figure BACDEF. This figure is,
therefore, a regular hexagon.
PROP. VII. PROB.
On a given finite straight line, to describe a regu-
lar octagon.
Let AB be the given straight line, on which it is required
to describe a regular octagon.
104
ELEMENTS OF GEOMETRY.
1
Bisect AB (I. 5.) by the perpendicular CD, which make
equal to CA or CB, join DA and DB, produce CD until
DO be equal to DA or DB, draw •
AO and BO, thus forming (IV. 1.)
an angle equal to the half of ADB,.
and, about the vertex O, repeat the
equal triangles AOB, AOE, EOF,
FOG, GOH, HOI, IOK, and
KOB to compose the octagon.
F
G
H

E
K
D
A C B
For the distances AD, BD are
evidently equal; and because CA, CD, and CB are all equal
the angle ADB is contained in a semicircle, and is therefore
a right anglẹ (III. 17.). Consequently AOB is equal to the
half of a right angle, and eight such angles will adapt them-
selves about the point O. Whence the figure BAEFGHIK,
having eight equal sides and equal angles, is a regular octa-
gon.
PROP. VIII. PROB.
On a given finite straight line, to describe a regu-
lar decagon.
Let AB be the straight line, on which it is required to de-
scribe a regular decagon.
On AB construct (IV. 4.) an isosceles triangle having each
of the angles at its base double of the vertical angle, and, a-
bout the point O, place a series of
triangles all equal to AOB: A re-
gular decagon will result from
this composition.
For the vertical angle AOB of
the isosceles triangle is equal to
the fifth part of two right angles
(IV. 4. cor.), or to the tenth part
D
F G

E
IL
K
A
B
I
of four right angles; whence ten such angles may be formed
BOOK IV.
105
about the point O. The figure BACDEFGHIK, having
therefore ten equal sides and equal angles, is a regular decagon.
PROP. IX. PROB.
On a given finite straight line, to describe a regu-
lar dodecagon.
Let AB be the straight line, on which it is required to de-
scribe a regular twelve-sided figure.
On AB construct (I. 1.) the equilateral triangle ACB, and
again (IV. 1.) the isosceles triangle AOB, having its vertical
angle equal to the half of ACB, and repeat this triangle
AOB about the point O; a regular
dodecagon will be thus formed.
F
E
H I

K
L
For ACB being an equilateral
triangle, each of its angles is the
third part of two right angles (I.
32. cor. 1.); consequently the angle
AOB is the sixth part of two right
angles or the twelfth part of four
right angles, and twelve such angles can, therefore, be placed
about the vertex O.
D
M
N
A
B
Scholium. Hence a regular twenty-sided figure may be de-
scribed on a given straight line, by first constructing on it
an isosceles triangle having each of the angles at the base dou-
ble of the vertical angle, and then erecting another isosceles
triangle with its vertical angle equal to the half of this. And,
by thus changing the elementary triangle, a regular polygon
may be always described, with twice the number of sides.
PROP. X. PROB.
In a given triangle, to inscribe a circle.
Let ABC be a triangle, in which it is required to inscribe
a circle.
106
ELEMENTS OF GEOMETRY.
Draw AD and CD (I. 5.) to bisect the angles CAB and
ACB, and from their point of concourse D, with its distance.
DE from the base, describe the circle EFG: This circle will
touch the triangle internally.
For let fall the perpendiculars DG and DF upon the sides
AB and BC (I. 6.). The triangles
ADE, ADG, having the angle DAE
B

equal to DAG, the right angle DEA
A
D
E
equal to DGA, and the interjacent
side AD common, are equal (I. 21.),
and therefore the side DE is equal to
DG. In the same manner, it is
proved, from the equality of the triangles CDE, CDF, that
DE is equal to DF; consequently DG is equal to DF, and
the circle passes through the three points E, G, and F. But
it also touches (III. 24.) the sides of the triangle in those
points, for the angles DEA, DGA, and DFC are all of them
right angles.
PROP. XI. PROB.
In a given circle, to inscribe a triangle equiangu-
lar to a given triangle.
Let GDH be a circle, in which it is required to inscribe a
triangle that shall have its angles equal to those of the tri-
angle ABC.
Assuming any point D in the circumference of the circle,
draw (III. 24.) the tangent
E

EDF, and make the angles
D
EDG, FDH equal to BCA,
T
B
BAC, and join GH: The tri-
angle GDH is equiangular to
ABC.
A
C
For EF being a tangent, and
DG drawn from the point of contact, the angle EDG, which
BOOK IV. -
107
was made equal to BCA, is equal to the angle DHG in the
alternate segment (III. 25.); consequently DHG is equal to
BCA. And for the same reason, the angle DGH is equal to
BAC; wherefore (I. 32.) the remaining angle GDH of the
triangle GHD is equal to the remaining angle ABC of the
triangle ACB, and these triangles are mutually equiangular.
PROP. XII. PROB.
About a given circle, to describe a triangle equi-
angular to a given triangle.
Let GIH be a circle, about which it is required to describe
a triangle, having its angles equal to those of the triangle
ABC.
Draw any radius FG, and with it make (I. 4.) the angles
GFI, GFH equal to the exterior angles BAE, BCD of the
triangle ABC, and, from the points G, I, and H draw the
tangents KM, KL, and LM to form the triangle KLM:
This triangle is equiangular to ABC.
For all the angles of the quadrilateral figure KIFG being
equal to four right angles,
L

and the angles KIF and
KGF being each a right
I
angle (III. 24.), the remain-
F
G
ing angles GKI and GFI
are together equal to two K
B
H
EA.
C D
M
right angles, and are consequently equal to the angles BAC
and BAE on the same side of the straight line ED. But the
angle GFI was made equal to BAE; whence GKI is equal
to CAB. In like manner, it is proved that the angle GMH
is equal to ACB; and the angles at K and M being thus
equal to BAC and BCA, the remaining angle at L is (I.
32.) equal to that at B, and the two triangles are, therefore,
equiangular.
108
ELEMENTS OF GEOMETRY.
PROP. XIII. PROB.
In and about a given circle, to inscribe and cir-
cumscribe an equilateral triangle.
Let AEB be a circle, in which it is required to inscribe an
isosceles triangle.
Draw the diameter AB, describe (I. 1.) the equilateral tri-
angle ADB, join CD meeting the circumference in E, draw
(I. 24.) EF, EG parallel to AD, BD, and join FG: The
triangle EFG is equilateral.
For the triangles ADC, BDC having the two sides DA,
AC equal to DB, BC, and the third side DC common to
both, are (I. 2.) equal, and the angle DCA is equal to DCB;
whence the arc AE is (III. 14.) equal to BE. And the tri-
angle ADB (I. 11. cor.) being
likewise equiangular, the angle
DBA is equal to DAB, and the
arc AEM equal to BEL, and the
remaining arc ME equal to LE.
But EF and EG being parallel to
LA and MB, the arcs AF and
BG are (III. 20.) equal to LE and
ME, and to each other; hence
(III. 20. cor.) FG is parallel to
D

H
E
K
L
M
B
F
G.
AB, and the inscribed triangle FEG is (I. 31.) equiangular,
and consequently equilateral.
Again, let it be required to describe an equilateral triangle
about the circle AEB.
The same construction remaining; at the points F, E, and
G, apply the tangents HI, HK, and KI, to form the circum-
scribing triangle IHK: This triangle is equilateral.
For because IH is a tangent and FG is inflected from the
point of contact, the angle IFG is equal to the angle FEG in
the alternate segment (III. 25.), and therefore IH is parallel
BOOK IV.
109
to EG (I. 23. cor.). In like manner it is proved, that HK,
KI are parallel to GF, FE, and consequently (I. 31.) the
angles of the triangle IHK are equal to those of FEG, and
therefore equal to each other.
Cor. Hence the circumscribing equilateral triangle contains
four times that which is inscribed; for the figures EFIG,
EHFG, and EFGK are evidently equal rhombuses, and con-
tain equilateral triangles which are all equal. Hence also the
side of the circumscribing, is double of that of the inscribed,
equilateral triangle.
PROP. XIV. THEOR.
A straight line drawn from the vertex of an equi-
lateral triangle inscribed in a circle to any point in
the opposite circumference, is equal to the two
chords inflected from the same point to the extre-
mities of the base.
Let ABC be an equilateral triangle inscribed in a circle,
and BD, AD, and CD chords drawn from it to a point D in
the circumference; BD is equal to AD and CD taken toge-
ther.
For, make DE equal to DA, and join
AE. The angle ADB, being (III. 18.) e
qual to ACB in the same segment is equal
B

E
C
(I. 32. cor.) to the third part of two right
angles. But the triangle ADE being isos-
celes by construction, the angles DAE, DEA
at its base are equal (I. 11.), and each of
them is, therefore, equal to half of the remaining two-thirds
of two right angles, or to one-third part. Consequently ADE
is an equilateral triangle (I. 12. cor.), and the angle DAE
equal to CAB; take CAE from both, and there remains the
angle DAC equal to EAB; but the angle ABD is equal to
110
ELEMENTS OF GEOMETRY.
ACD in the same segment. And thus the triangles ADC and
AEB have the angles DAC, DCA equal to EAB, EBA, and
the interjacent side AC equal to AB; they are consequently
equal (I. 21.), and the side DC is equal to EB. But DE was
made equal to DA; wherefore DA and DC are together equal
to DE and EB, or to DB.
PROP. XV. PROB.
About and in a given square, to circumscribe and
inscribe a circle.
Let ABCD be a figure, about which it is required to cir-
cumscribe a circle.
Draw the diagonals AC, DB, intersecting each other in O,
and, from that point with the distance AO, describe the circle
ABCD: This circle will circumscribe the square.
Because the diagonals of the square ABCD are equal and
bisect each other (I. 29. and cor.),
the straight lines OA, OB, OC, and
OD are all equal, and consequent-
ly the circle described through A
passes through the other points
B, C, and D.
Again, let it be required to in-
scribe a circle in the square ABCD.
From the intersection of the

B
SHT
R
W
F
Y
A
PE Z
D
diagonals and with its distance from the side AD, describe
the circle EGHF: This circle will touch the square inter-
nally.
For let fall the perpendiculars OG, OH, and OF (I. 6.).
And because the straight lines AB, BC, CD, and DA are
equal, they are equally distant from the centre O of the exte-
rior circle (III. 11.); wherefore the perpendiculars OE, OG,
OH, and OF are all equal, and the interior circle passes
BOOK IV.
111
through the points G, H, and F; but (I. 24.) it likewise
touches the sides of the square, since they are perpendicular
to the radii drawn from O.
Cor. Hence an octagon may be inscribed within a square.
For let tangents be applied at the points I, K, L, and M,
where the diagonals cut the interior circle. It is evident, that
the triangle AOE is equal to DOE, IOP to EOP, and EOZ
to MOZ; whence the angles POE and ZOE are equal, be-
ing the halves of EOA and EOD, and consequently the tri-
angles PEO and ZEO are equal. Wherefore PZ, the double.
of PE, is equal to PQ, the double of PI; and the angle
EZM is, for a like reason, equal to EPI. And, in this man-
ner, all the sides and all the angles about the eight-sided figure
PQRSTWYZ are proved to be equal.
PROP. XVI. PROB.
In and about a given circle, to inscribe and cir-
cumscribe a square.
Let EADB be a circle in which it is required to inscribe a
square.
Draw the diameter AB, the perpendicular ED, and join
AD, DB, BE, and EA: The inscribed figure ADBE is a
square.
The angles about the centre C, being right angles, are
equal to each other, and are, therefore, subtended by equal
chords AD, DB, BE, and AE, but one
of the angles ADB, being in a semicircle,
is (I. 22.) a right angle, and consequent-
ly ADBE is a square.
Next, let it be required to circumscribe
a square about the circle.
Apply tangents FG, GH, HI, and FI
G D
H

A
C
E
I
at the extremities of the perpendicular diameters: These will
form a square.
112
ELEMENTS OF GEOMETRY.
For all the angles of the quadrilateral figure CG being to-
gether equal to four right angles, and those at C, A, and D
being each a right angle, the remaining angle at G is also a
right angle, CG is a rectangle; and AC being equal to CD,
it is likewise a square. In the same manner, CH, CI, and
CF are proved to be squares; the sides FG, GH, HI, and
IF of the exterior figure, being therefore the doubles of equal
lines, are mutually equal, and the angle at G being a right
angle, FH is consequently a square.
Cor. Hence the circumscribing square is double of the in-
scribed square, and this again is double of the square de-
scribed on the radius of the circle.
PROP. XVII. PROB.
To inscribe and circumscribe a circle in and a-
bout a given regular pentagon.
Let ABCDE be a regular pentagon, in which it is requi-
red to inscribe a circle.
Draw AO and EO to bisect the angles at A and E, join C
with the point of concourse O and produce it to meet AE in
F, and from O as a centre, with the distance OF, describe a
circle FGHIK: This circle will touch the pentagon inter-
nally.
B
H
C
I
D
For, from the point O, let fall perpendiculars on the op-
posite sides of the figure. The
angles EAO and AEO, being
the halves of the angles of the
pentagon, are equal, and con-
sequently the triangle AOE is
isosceles, and the perpendicular
OF bisects the base. And the
triangles AOG and BOG, ha-
ving the angles OAG and
OGA equal to OBG and OGB

E
K
and the common side OG, are (I. 3.) equal. Again the tri-
BOOK IV.
113
angles BOG and BOH have now the angles OBG and OGB
equal to OBH and OHB, with the side BO common to both,
and are therefore equal. In like manner, all the triangles
about the centre O are proved to be equal; consequently the
perpendiculars OF, OG, OH, OI, and OK are equal, and the
circle touches the pentagon in the points F, G, H, I, and K.
Next, let it be required to describe a circle about the pen-
tagon.
From the same centre O, with the distance OA, describe
a circle: It will pass through the points B, C, D, E, for the
triangles about O being all equal, the straight lines OA, OB,
OC, OD, and OE must be likewise equal.
PROP. XVIII. PROB.
In and about a given circle, to inscribe and cir-
cumscribe a regular pentagon
Let ABCDE be a circle in which it is required to inscribe
a regular pentagon.
Construct an isosceles triangle having each of its angles at
the base double of its vertical angle (IV. 4.), and equiangular
to this, inscribe the triangle ACE within the circle (IV. 11.),
draw AD, EB bisecting the angles CAE, CEA (I. 5.), and
join AB, BC, CD, and DE: The figure ABCDE is a regu-
lar pentagon.
For the angles AEB, BEC are
each the half of CEA, and there-
foré equal to ACE; but the an-
gles EAD, DAC are likewise e-
qual to ACE. Hence these an-
gles, being all equal, must stand
on equal arcs (III. 18. cor.); and
the chords of these arcs, or the
F
B
G
7.

E
K
sides AB, BC, CD, DE, and AE are equal (III. 13. cor.).
And because the segments EAB, ABC, BCD, CDE, and
H
114
ELEMENTS OF geometry.
DEA are evidently equal, the interior angles of the figure
are all equal (III. 18.), and it is, therefore, a regular pen-
tagon.
Next, let it be required to circumscribe a regular penta-
gon about the circle.
At the points A, B, C, D, and E apply tangents; these
will form a regular pentagon.
For FAK being a tangent, the angle KAE is equal to
ACE (III. 25.); and in like manner it is shown that the
angles AEK, DEI, EDI, CDH, DCH, BCG, CBG, ABF,
BAF are all equal to ACE. The isosceles triangles AKE,
BFA, having, therefore, the angles at the base equal and the
bases themselves AE, AB,-are equal (I. 21.); for the same
reason, the triangles BGC, CHD, DIE, EKA, are equal.
Whence the internal angles of the figure are equal, and its
sides, being double of those of the annexed triangles, are like-
wise equal: The figure is, therefore, a regular pentagon.
PROP. XIX. PROB.
In and about a regular hexagon, to inscribe and
circumscribe a circle.
Let ABCDEF be a regular hexagon, in which it is requi-
red to inscribe a circle.
Draw AO and FO, bisecting the angles BAF and AFE
(I. 5.); and from the point of intersection O, with its distance
from the side AF, describe a circle: This circle will touch the
hexagon internally.
For let fall perpendiculars from O upon the sides of the
figure. It may be demonstrated, as in Prop. XVII. that the
triangles AOB, BOC, COD, DOE, and EOF are all equal
BOOK IY.
115
to AOF; and, in like manner, it will appear that the inter-
mediate bisected triangles are

equal. Hence the perpendicu-
lars OG, OH, OI, OK, OL,
and OM, are all equal, and a
circle must touch these at the
points G, H, I, K, L, and M.
Again, let it be required to
describe a circle about the
hexagon.
From the same point O, as
B
H
C
K
D
E
a centre, with the distance OA, describe a circle, which must
pass through the points B, C, D, E, and F; for the straight
lines OA, OB, OC, OD, OE, and OF were proved to be equal.
Cor. Hence, in any regular polygon, the centre of the in-
scribing and circumscribing circle is the same, and may be
determined in general, by drawing lines to bisect the adjacent
angles of the figure.
PROP. XX. PROB.
To inscribe a regular hexagon in a given circle.
Let it be required, in the circle FBD, to inscribe a hexagon.
Draw the radius OA, on which construct the equilateral
triangle ABO (I. 1. cor.), and repeat the equal triangles about
the vertex O: These triangles will compose a hexagon.
B
For the triangle ABO, being equilateral, each of its angles,
AOB, is the third part of two right angles; and consequently
six of such angles may be placed
about the centre O. But the bases
of the triangles AOB, BOC, COD,
DOE, and EOF form the sides
of the figure, and the angles at
those bases its internal angles ;

wherefore it is a regular hexagon.
F
E
D
Cor. 1. Tangents applied at the points A, B, C, D, E, and
H 2
116
ELEMENTS OF GEOMETRY.
F, would evidently form a regular circumscribing hexagon.-
An equilateral triangle might be inscribed by joining the al-
ternate points; and, by applying tangents at those points, an
equilateral triangle would be made to circumscribe the circle.
Cor. 2. The side AB of the inscribed hexagon is equal to
the radius; and since ABD is a right-angled triangle, and the
squares of AB and BD are equal to the square of AD or to
four times the square of AO, the square of BD the side of an
inscribed equilateral triangle is triple the square of the radius.
Cor. 3. The perimeter of the inscribed hexagon is equal to
six times the radius, or three times the diameter, of the circle.
Hence the circumference of a circle being, from its perpetual
curvature, greater than any intermediate system of straight
lines, is more than triple its diameter.
PROP. XXI. PROB.
To inscribe a regular decagon in a given circle.
Let ADH be a circle, in which it is required to inscribe a
regular decagon.
Draw the radius OA, and with OA as its side describe the
isosceles triangle AOB, having each of its angles at the base
double of its vertical angle (IV. 4.), repeat the equal triangles
about the centre O: These triangles will compose a decagon.
For the vertical angle AOB of
the component isosceles triangle, is
the fifth part of two right angles
(IV. 4. cor.), and consequently ten
such angles can be placed about
the point O. But the sides and
angles of the resulting figure are
all evidently equal; it is, therefore,
a regular decagon.
A
E
K
D

E
G
I
H
Cor. Hence a regular pentagon will be formed, by joining
BOOK IV.
117
the alternate points A, C, E, G, I, and A. It is also mani-
fest, that a decagon and a pentagon may be circumscribed
about the circle, by applying tangents at their several angu-
lar points.
PROP. XXII. THEOR.
The square of the side of a pentagon inscribed in
a circle, is equivalent to the squares of the sides of
the inscribed hexagon and decagon.
Let ABCDEF be half of a decagon inscribed in a circle,
whose diameter is AF; the square of AC the side of the in-
cribed pentagon, is equivalent to the square of AB the side
of the inscribed decagon, and the square of the radius AO
which is equal to a side of the inscribed hexagon.
For join AD, AE, and draw OB, OC, OD, and OE. The
angle FAD at the circumference, being half of the angle FOD
at the centre (III. 17.), is equal to the angle AOB; and, for
the same reason, the angle FAB, being half of FOB, is equal
to FOD or COA. The triangles ABO and AGO, having,
therefore, the angles AOB, OAB equal to OAG, AOG, and
the side AO common to both, are equal (I. 21.) and isosceles,
and consequently the base AB is equal to OG. But the angles
FAE and EAD, standing on
A
B
D

E
F
equal arcs, are equal (III. 18.
cor.); wherefore the triangles
OAH and GAH, having the
side AG equal to AO, the
side AH common, and the
contained angle OAH equal to GAH, are equal (I. 3.), and
hence OH is equal to GH, and the angles AHO and AHG
are equal and right angles. And because AO is equal to CO
and AH perpendicular to it, the square of AC is equivalent
to twice the rectangle under OC and CH (II. 26. cor.), or the
rectangle under OC and twice CH, which is evidently the
118
ELEMENTS OF GEOMETRY.
sum of OC and CG. The square of AC is, therefore, equi-
valent to the square of OC, with the rectangle under OC and
CG; but OG being equal to AB, the radius OC is divided
by a medial section at G, and consequently the rectangle un-
der OC and CG is equivalent to the square of OG or AB.
Whence the square of AC is equivalent to the two squares of
AO and AB.
Cor. 1. The triple chord AD of the decagon, is equal to
the sides AO and AB of the inscribed hexagon and decagon.
For AO being equal to DO, the angle OAD is equal to ODA
(L. 11.); but OAD, or FAD, is equal to the angle DOC
(III. 17.), and consequently the angle DOG is equal to ODG,
and the side OG equal to DG (I. 12.) Wherefore AD being
equal to AG and GD, is equal to AO with OG or AB.
Cor. 2. Hence the sides of the inscribed decagon and pen-
tagon may be found by a single construction. For draw the
perpendicular diameters AC and
EF, bisect OC in D, join DE,
make DG equal to it, and join
GE. It is evident, that AO is cut
medially in G (II. 22.), and conse-
quently that OG is equal to a side
of the inscribed decagon.

But
GOE being a right-angled triangle,
EK
FIL
A
G
D
M
PFN
the square of GE is equivalent to the squares of GO and OE
(II. 11.), or the squares of the sides of the decagon and hex-
agon; whence GE is equal to the side of the inscribed pen-
tagon. It also follows, that CG is equal to CI or CP, the
triple chords of the inscribed decagon *.
PROP. XXIII. PROB.
In a given circle, to inscribe regular polygons of
fifteen and of thirty sides.
See Note XXXI.
BOOK IV.
119
Let AB and BC be the sides of an inscribed decagon, and
AD the side of a hexagon inscribed; the arc BD will be the
fifteenth part of the circumference of the circle, and DC the
thirtieth part.
A
B
C
D
I
For; if the circumference were di-
vided into thirty equal portions, the
arc AB would be equal to three of
these, and the arc AD to five; con-
sequently the excess BD is equal to
two of these portions, or it is the fifteenth part of the whole
circumference. Again, the double arc ABC being equal to
six portions, and ABD to five, the defect DC is equal to one
portion, or to the thirtieth part of the circumference.
Scholium. From the inscription of the square, the penta-
gon, and the hexagon,—may be derived that of a variety of
other regular polygons: For, by continually bisecting the in-
tercepted arcs and inserting new chords, the inscribed figure
will, at each successive operation, have the number of its sides
doubled. Hence polygons will arise of 6, 8, and 10 sides;
then, of 12, 16, and 20; next of 24, 32, and 40; again, of
48, 64, and 80; and so forth repeatedly. The excess of the
arc of the hexagon and above that of the decagon, gives the
arc of a fifteen-sided figure; and the continued bisection of
this arc will mark out polygons with 30, 60, or 120 equal
sides, in perpetual succession. The same results might also
be obtained from the differences of the preceding arcs *.
Of the regular polygons, three only are susceptible of per-
fect adaptation, and capable therefore of covering, by their
repeated addition, a plane surface. These are the equilateral
triangle, the square, and the hexagon. The angles of an
equilateral triangle are each two-thirds of a right angle, those
of a square are right angles, and the angles of a hexagon are

* See Note XXXII.
120
ELEMENTS OF GEOMETRY.
each equal to four-third parts of a right angle. Hence there
may be constituted about a point, six equilateral triangles,
four squares, and three hexagons. But no other regular
polygon can admit of a like disposition. The pentagon, for
instance, having each of its angles equal to six-fifths of a
right angle, would not fill up the whole space about a point,
on being repeated three times; yet it would do more than
cover that space, if added four times. On the other hand,
since each angle of a polygon which has more than six sides
must exceed four third parts of a right angle, three such
polygons cannot stand round a point. Nor can the space
about a point ever be bisected by the application of any re-
gular polygons, of whatever number of sides; for their angles
are always necessarily each less than two right angles *.
* See Note XXXIII.
ELEMENTS
OF
GEOMETRY.
BOOK V.
OF PROPORTION.
THE preceding Books treat of magnitude as con-
crete, or having mere extension; and the simpler
propertics of lines, of angles, and of surfaces, were
deduced, by a continuous process of reasoning,
grounded originally on superposition. But this
mode of investigation, however satisfactory to the
mind, is, from its nature, very limited and laborious.
By introducing the idea of Number into geometry,
a new scene is opened, and a far wider prospect
rises into view. Magnitude, being considered as
discrete, or composed of integrant parts, becomes
assimilated to multitude; and under that aspect, it
presents a vast system of relations, which may be
traced out with the utmost facility.
122
ÉLEMENTS OF GEOMETRY.
、
Numbers were first employed, to denote the col-
lection of distinct, though kindred, objects; but the
subdivision of extent, whether actually effected or
only conceived to exist, bestowing a sort of indi-
viduality, they came afterwards to acquire a more
comprehensive application. In comparing together
two quantities of the same kind, the one may contain
the other, or be contained by it; that is, the one may
result from the repeated addition of the other, or it
may in its turn produce this other by a successive
composition. The one quantity is, therefore, equal,
either to so many times the other, or to a certain
aliquot part of it.
Such seems to be the simplest of numerical rela-
tions. It is very confined, however, in its applica-
tion, and is evidently, in that shape, insufficient al-
together for the purpose of general comparison. But
this object is attained, by adopting some interme-
diate reference. Though a quantity neither contain
another exactly, nor be contained by it; there may
yet exist a third and smaller quantity, which is at
once capable of measuring them both. This mcasure
corresponds to the arithmetical unit; and as number
denotes the collection of units, so quantity may be
viewed as the aggregate of its component measures.
But mathematical quantities are not all suscep-
tible of such perfect mensuration. Two quantities
may be conceived to be so constituted, as not to ad-
mit of any other that will measure them completely,
or be contained in both without leaving a remain-
der. Yet this apparent imperfection, which pro-
ceeds entirely from the infinite variety ascribed to
BOOK V.
123
possible magnitude, creates no real obstacle to the
progress of accurate science. The measure or pri-
mary element, being assumed successively still
smaller and smaller, its corresponding remainder
must be perpetually diminished. This continued
exhaustion will hence approach to its absolute term,
nearer than by any assignable difference.
Quantities in general can, therefore, either exact-
ly or to any required degree of precision, be repre-
sented abstractly by numbers; and thus the science
of Geometry is at last brought under the dominion
of Arithmetic.
It is obvious, that quantities of any kind must
have the same composition, when each contains its
measure the same number of times. But quantities,
viewed in pairs, may be considered as having a simi-
lar composition, if the corresponding terms of each
pair contain its measure equally. Two pairs of
quantities of a similar composition, being thus form-
ed by the same distinct aggregations of their ele-
mentary parts, constitute a proportion*.
* See Note XXXIV.
124
ELEMENTS OF GEOMETRY.
7
DEFINITIONS.
1. Quantities are homogeneous, which can be added toge-
ther.
2. One quantity is said to contain another, when the sub-
traction of this-continued if necessary-leaves no remain-
der.
3. A quantity which is contained in another, is said to
measure it.
4. The quantity which is measured by another, is called
its multiple; and that which measures the other, its sub-
multiple.
5. Like multiples and submultiples are those which contain
their measures equally, or which equally measure their corre-
sponding compounds.
6. Quantities are commensurable, which have a finite com-
mon measure; they are incommensurable, if they will admit of
no such measure.
7. That relation which one quantity is conceived to bear to
another in regard to their composition, is named a ratio.
8. When both terms of comparison are equal, it is called
a ratio of equality; if the first of these be greater than the
second, it is a ratio of majority; and if the first be less than
the second, it is a ratio of minority.
9. A proportion or analogy consists in the identity of ratios.
BOOK V.
125
10. Four quantities are said to be proportional, when a sub-
multiple of the first is contained in the second as often as a
like submultiple of the third is contained in the fourth.
11. Of proportional quantities, the first of each pair is na-
med the antecedent, and the second the consequent.
12. The antecedents are homologous terms; and so are the
consequents.
13. One antecedent is said to be to its consequent, as ano-
ther antecedent to its consequent.
14. The first and last terms of a proportion are called the
extremes, and the intermediate ones, the means.
15. A ratio is direct, if it follows the order of the terms
compared ; it is inverse or reciprocal, when it holds a reversed
order.
Thus, if the ratio of A to B be direct, that of B to A is the inverse
or reciprocal ratio.
16. Quantities form a continued proportion, when the inter-
vening terms stand in the double relation of consequents and
antecedents.
17. When a proportion consists of three terms, the middle
one is said to be a mean proportional between the two ex-
tremes.
18. The ratio which one quantity has to another may be
considered as compounded of all the connecting ratios among
any interposed quantities.
Thus, the ratio of A to D is viewed as compounded of that of A to
B, that of B to C, and that of C to D.
19. Of quantities in a continued proportion, the first is
126
ELEMENTS OF GEOMETRY.
said to have to the third, a ratio the duplicate of what it has
to the second; to have to the fourth, a triplicate ratio; to the
fifth, a quadruplicate ratio; and so forth, according to the
number of equal ratios inserted between the extreme terms.
20. If quantities be continually proportional, the ratio of
the first to the second is called the subduplicate of the ratio of
the first to the third, the subtriplicate of the ratio of the
first to the fourth, &c.
21, A straight line is said to be cut in extreme and mean
ratio, when the one segment is a mean proportional between
the other segment and the whole line.
To facilitate the language of demonstration relative to num-
bers or abstract quantities, it is expedient to adopt a clear and
concise mode of notation.
=
<
1. The sign expresses equality, majority, and mi-
nority: Thus A=B denotes that A is equal to B, AB signi-
fies that A is greater than B, and AB imports that A is
less than B.
2. The signs and mark the addition and subtraction
of the quantities to which they are prefixed: Thus, A+B de-
notes that B is to be joined to A, and A—B signifies that B
is to be taken away from A. Sometimes these two symbols
are combined together: Thus, AB represents either the
sum of A and B, or the excess of A above B.
3. To express multiplication, the quantities are placed close
together; or they may be connected by the point (.), or the
cross: Thus, AB, or A.B, or A x B, denotes the product of
A by B; and ABC indicates the result of the continued mul-
tiplication of A by B, and of this product again by C.
BOOK V.
127
4. When the same number is repeatedly multiplied, the
product is termed its power; and the number itself, in refe
rence to that power, is called the root. The notation is here
still farther abridged, by retaining only a single letter with a
small figure over it, to mark how often it is understood to be
repeated: This figure serves also to distinguish the order of
the power. Thus AA, or A², signifies that A is multiplied
by A, and that the product is the second power of A; and
AAA, or A³, in like manner, imports that AA is again mul-
tiplied by A, and that the result is the third power of A.
5. The roots are denoted, by prefixing a contracted r, or
the symbol ✔. Thus A or A marks the second root of
2
3
A, or that number of which A is the second power; ✔A sig-
nifies the third root of A, or the number which has A for its
third power.
6. To represent the multiplication of complex quantities,
they are included by a parenthesis. Thus, A(B+C—D) de-
notes that the amount of B+C-D, considered as a single
quantity, is multiplied into A.
7. Ratios and analogies are expressed, by inserting points
in pairs between the terms. Thus A: B denotes the ratio of
A to B, and the compound symbols A: B:: C: D, signify
that the ratio of A to B is the same as that of C to D, or that
A is to B as C to D.
128
ELEMENTS OF GEOMETRY.
PROP. I. THEOR.
The product of a number into the sum or difference
of two numbers, is equal to the sum or difference of
its products by those numbers.
Let A, B, and C be three numbers; the product of the sum
or difference of B and C by the number A, is equal to the
sum or difference of the products AB and AC.
For the product AB is the same as each unit contained in
B repeated A times, and the product AC is the same as the
units in C likewise repeated A times; whence the sum of the
products. AB and AC, is equal to the units contained in both
B and C, all repeated A times, or it is equal to the sum of the
numbers B and C multiplied by A.
Again, for the same reason, the difference between the pro-
ducts AB and AC must be equal to the difference between
the units contained in B and in C, repeated A times; that is,
it must be equal to the difference between the numbers B and
C multiplied by A.
Cor. 1. Hence a number which measures any two numbers,
will measure also their sum and their difference.
Cor. 2. It is hence manifest, that the first part of the pro-
position may be extended to more numbers than two; or that
AB+AC+AD+, &c.=A(B+C+D+, &c.)
PROP. II. THEOR.
The product which arises from the continued mul-
tiplication of any numbers, is the same, in whatever
order that operation be performed.
Let A and B be two numbers; the product AB is equal to
BA.
For the product AB is the same as each unit in B added
BOOK V.
129
together A times, that is, the same as A itself repeated B
times, or BA.
Next, let there be three numbers A, B, and C; the pro-
ducts ABC, ACB, BAC, BCA, CAB, and CBA are all
equal.
For put D=AB or BA; then DC=CD, that is, ABC=
CAB, and BAC=CBA.
Again, put E=AC or CA; then EB=BE, that is, ACB
BAC, and CAB=BCA.
Lastly, put FBC or CB; then FA=AF, that is, BCA
=ABC, and CBA ACB.
And thus the several products are all mutually equal.
It is also manifest, that the same mode of reasoning might
be extended to the products of any multitude of numbers.
PROP. III. THEOR.
Homogeneous quantities are proportional to their
like multiples or submultiples.
Let A, B be two quantities of the same kind, and pA, pB
their like multiples; A: B::pA :pB.
For, since A and B are capable of being measured to any
required degree of precision, suppose A=m.a and B=n.a;
then pA=p.ma, and pB=p.na. But (V. 2.) p.mam.pa, and
p.nan.pa. Wherefore a and pa are like submultiples of A
and of pA, which contain them respectively m times; and these
like submultiples are both contained equally, or n times, in B
and in pB. Consequently (V. def. 10.) the quantities A, B, and
pA, pB are proportional; and A, pA are the antecedents, and
B, pB, the consequents, of the analogy.
Again, because the ratio of pA to pB is thus the same as
that of A to B, which, in reference to pA and pB, are only
like submultiples, it follows that homogeneous quantities are
also proportional to their like submultiples.
I
130
ELEMENTS OF GEOMETRY.
!
PROP. IV. THEOR.
In proportional quantities, according as the first
term is greater, equal, or less than the second, the
third term is greater, equal, or less than the fourth.
Let A B C: D; then if AB, CD; if A=B,
C=D; and if A≤B, CZD.
For, if A be greater than B, A: B is a ratio of majority;
whence CD, being the same with it, is likewise a ratio of
majority, and consequently C is greater than D.
If A be equal to B, A: B must be a ratio of equality, and
hence CD is also a ratio of equality, or C is equal to D.
But, if A be less than B, A : B is a ratio of minority, and
so is, therefore, C: D, or C is less than D.
PROP. V. THEOR.
Of four proportionals, if the first be a multiple or
submultiple of the second, the third is a like mul-
tiple or submultiple of the fourth.
Let A B C: D; if A=pB, then C-pD.
:
For, suppose the approximate measures of A and C to be
a and c, and let A=mp.a, and C-mp.c. It is evident, from
the hypothesis, that, A=pB=mp.a, or B=m.a; but the con-
sequents B and D must contain their measures equally (V.
def. 10.), and therefore D=m.c. Whence C=mp.c=(V. 2.)
p.mc=pD.
Again, if qA=B; then will qC=D.
For, let A=na, and C=nc; therefore B=qA=qna= (V. 2.)
nq.a, and, from the definition of proportion, D=nq.c=
(V. 2.) q.ne=qC.
BOOK V.
131
i
PROP. VI. THEOR.
If four numbers be proportional, the product of
the extremes is equal to that of the means; and of
two equal products, the factors are convertible into
an analogy, of which these form severally the ex-
treme and the mean terms.
Let A: B::C:D; then AD=BC.
For (V. 3.) A.D: B.D:: B.C: B.D; and the second
term of this analogy being equal to the fourth, therefore
(V. 4.) AD=BC.
Again, let AD=BC; then A : B : : C : D.
For, by identity of ratios, AD : BD : : BC : BD, and hence
(V. 3.) A: B:: C: D.
Cor. 1. Hence the greatest and least terms of a proportion,
are either extremes or means.
Cor. 2. Hence also a proportion is not affected, by trans-
posing or interchanging its extreme and mean terms.—On
this principle, is founded the two following theorems.
PROP. VII. THEOR.
The terms of an analogy are proportional by in-
version, or the second is to the first, as the fourth to
the third.
::
Let A B C D; then inversely B: A:: D: C.
For the extreme and mean terms are thus only mutually
interchanged, and consequently the same equality of products
AD and BC still obtains.
PROP. VIII. THEOR.
Numbers are likewise proportional by alternation,
or the first is to the third, as the second to the fourth.
I 2
132
`ELEMENTS OF GEOMETRY.
Let A B C D; then alternately A: C:: B : D.
:
For the extreme terms are still retained, and the mean
terms are merely transposed with respect to each other; the
same equality, therefore, of products here also subsists.
PROP. IX. THEOR.
The terms of an analogy are proportional by com-
position; or the sum of the first and second is to the
second, as the sum of the third and fourth to the
fourth.
Let A B C: D; then by composition A+B: B::
C+D: D.
Because A: B:: C: D, the product AD=BC (V. 6.);
add to each of these the product BD, and AD+BD=BC+
BD. But (V. 1.) AD+BD=D(A+B), and BC+BD=
B(C+D); wherefore (V. 6.) assuming the factors of these
equal products for the extreme and mean terms, A+B:B::
C+D: D.
PROP. X. THEOR.
The terms of an analogy are proportional by divi-
sion; or the difference of the first and second is to
the second, as the difference of the third and fourth
to the fourth.
:
Let A B :: C: D; suppose A to be greater than B, then
will C be greater than D (V. 4.): It is to be proved that
A-B: B:: C-D: D.
For, since A: B :: C: D, the product AD=BC (V. 6.),
and, taking BD from both, the compound product AD—BD
is equal to BC-BD; wherefore, by resolution, (A—B) D=
B(C—D), and consequently A-B: B :: C-D: D.
BOOK V.
133
f
If B be greater than A, then BD¬AD÷BD—BC, and,
by resolution, (B—A) D=B (D—C); whence B-—A : B : :
D-C: D.
PROP. XI. THEOR.
The terms of an analogy are proportional by con-
version; that is, the first is to the sum or difference
of the first and second, as the third to the sum or
difference of the third and fourth.
:
Let A B C : D, and suppose A B; then A: AB
:: C: C±D.
For, since (V. 6.) the product AD=BC, add or subtract
these to or from the product AC; and AC÷AD=AC÷BC.
Wherefore, by resolution, A(CD)=C(A+B), and conse-
quently A: A+B :: C: C÷D.
If AB, then AD-AC-BC-AC, and, by resolution,
A(D—C)=C(B-A), whence A: B-A :: C : D—C.
Cor. Hence, by inversion, A÷B: A :: C÷D : C, or
B-A: A :: D-C : C.
PROP. XII. THEOR.
The terms of an analogy are proportional by mix-
ing; or the sum of the first and second is to the dif
ference, as the sum of the third and fourth to their
difference.
Let A B :: C: D, and suppose AB; then A+B: A—B
::C+D: C—D.
For, by conversion, A: A+B :: C: C+D, and alternate-
ly A : C : : A+B : C+D.
Again, by conversion, A : A-B : : C : C—D, and alter-
nately AC A-B: C-D. Whence, by identity of ra-
134
ELEMENTS OF GEOMETRY.
1
:
tios, A+B C+D:: A-B: C-D, and alternately A+B :
A-BC+D: C-D.
The same reasoning will hold if A be less than B, the
order of these terms being only changed.
PROP. XIII. THEOR.
A proportion will subsist, if the homologous terms
be multiplied by the same numbers.
Let A B C : D; then pA : qB :: pC : qD.
:
For, since A: B:: C: D, alternately A: C:: B : D; but
the ratio of A to C is the same as pA : pC (V. 3.), and the
ratio of B to D is the same as qB: qD. Wherefore pA : pC
:: qB: qD, and, by alternation, pA: qB:: pC: qD.
Cor. The Proposition may be extended likewise to the di-
vision of homologous terms, by employing submultiples.
PROP. XIV. THEOR.
The greatest and least terms of a proportion, are
together greater than the intermediate ones.
:
:
Let A B C D; and A being supposed to be the
greatest term, the other extreme D is the least (V. 6. cor. 1.):
The sum of A and D is greater than the sum of B and C.
B::
Because A: B::C: D, by conversion A: A
C: C-D, and alternately A: C:: A-B: C-D; but A, be-
ing the greatest term, is therefore greater than C, and conse-
quently (V. 4.) A—B is greater than C-D; to each add
B+D, and (A+D)⇒(B+C.).
The same mode of reasoning is applicable, should any
other term of the analogy be supposed to be the greatest.
Cor. Hence the mean term of three proportionals, is less
than half the sum of both extremes *
* See Note XXXV..
BOOK V.
135
PROP. XV. THEOR.
If two analogies have the same antecedents, ano-
ther analogy may be formed, having the consequents
of the one as antecedents, and those of the other as
consequents.
:
:
:
Let A B C D and A: E:: C: F; then B:E::
D: F.
For, alternating the first analogy, A: C : : B : D, and al-
ternating the second, A: C:: E: F; whence, by identity of
ratios, B: D:: E: F,-which inference is named a direct
equality.
PROP. XVI. THEOR.
If the consequents of one analogy be antecedents
in another, a third analogy will obtain, having the
same antecedents as the former, and the same con-
sequents as the latter.
:
Let A B :: C: D, and B: E::D: F; then A:E::
C: F.
For, alternating both analogies, A: C:: B: D, and
B:D ::E: F; whence, by identity of ratios, A: C:: E: F,
--which conclusion is also named a direct equality.
PROP. XVII. THEOR.
If two analogies have the same means, the ex-
tremes of the one, with those of the other as mean
terms, will form a third analogy.
Let A B C D, and E: B:: C: F; then A: E::
F: D.
136
ELEMENTS OF GEOMETRY.
1
For, since A: B:: C: D, AD=BC (V. 6.); and because
E: B::C: F, EF-BC. Whence AD EF, and A: E::
F: D.
Cor. Hence the extreme and mean terms being inter-
changeable, it likewise follows, that, if A: B::C: D, and
A: E:: F: D, then B: E:: F: C.
PROP. XVIII. THEOR.
If the extremes of one analogy are the mean
terms in another, a third analogy will subsist, ha-
ving the means of the former as its extremes, and
the extremes of the latter as its means.
:
Let A B C : D, and E: A:: D: F; then B: E::
F: C.
For, from the first analogy AD=BC, and, from the se-
cond, EF=AD; whence BC=EF, and consequently B: E
:: F: C.
Cor. Hence also, if A: B:: C: D and B: E:: F: C;
then E: A :: D: F. The principle of this and the prece-
ding Proposition, is named inverse, or perturbate, equality.
PROP. XIX. THEOR.
If there be any number of proportionals, as one
antecedent is to its consequent, so is the sum of all
the antecedents to the sum of all the consequents.
Let A: B:: C: D::E:F::G: H; then A:B::A+C+
E+G: B+D+F+H.
Because A: B:: C: D, AD=BC; and since A: B::
E: F, AF-BE, and, for the same reason, AH=BG. Con-
sequently, the aggregate products, AB+ AD+AF+AH=
BA+BC÷BE+BG, and, by resolution, A(B+D+F+H)
·B(A+C+E+G), whence AB:: A+ C+E+G:
B+D+F+H.
BOOK V.
137
Cor. 1. It is obvious, that the Proposition will extend like-
wise to the difference of the homologous terms, and may,
therefore, be more generally expressed thus: A: B:: A±
C÷E÷G: B÷D÷F÷H.
Cor. 2. Hence in continued proportionals, as one antece-
dent is to its consequent, so is the sum or difference of the
several antecedents to the corresponding sum or difference of
the consequents. For, if A; B:: B:C::C: D; then
A:B::A±B÷C: B÷±C÷D; or, omitting B and C which
stand in the relation of antecedent and consequent, A : B, or
B:C::A±C: B±D.
PROP. XX. THEOR.
If two analogies have the same antecedents, ano-
ther analogy may be formed of these antecedents,
and the sum or difference of the consequents.
:
Let A B C : D, and A : E::C: F; then A : B÷E: :
C: D±F. For, by alternation, these analogies become A : C
:: B: D, and A: C:: E: F; whence (V. 19.) A: C::
B±E : D±F, and alternately A: B÷E :: C : D±F.
Cor. If A : B :: C: D, and E : B :: F: D; then A±E: B
:: C±F: D. For, by alternating the analogies, A: C::
B: D, and EF:: B: D; whence B: D:: A+E: C÷F,
and, by alternation and inversion, AE: B:: C÷F: D.
PROP. XXI. THEOR.
In continued proportionals, the difference between
the first and second is to the first, as the difference
between the first and last terms to the sum of all the
terms, excepting the last.
:
::
Let A B B: C::C:D:D: E; then if AB,
A-B: A: A-E: A+B+C+D.
138
ELEMENTS OF GEOMETRY.
For (V. 19.), A : B :: A+B+C+D:¡B+C+D+E,
and consequently (V. 11. cor.), A-B: A:: (A+B+C+D)—
(B+C+D+E): A+B+C+D; that is, omitting B+C+D
in the third term, A-B: A:: A—E: A+B+C+D.
If AB, then B-A: A:: (B+C+D+E)—(A+B+
C+D): A+B+C+D, that is, B-A: A::E—A: A+
B+C+D.
The same reasoning, it is evident, will hold for any num-
ber of terms.
PROP. XXII. THEOR.
The products of the like terms of any numeri
cal proportions, are themselves proportional.
Let A: B::C:D
EFG: H
IK:: L: M;
then AEI: BFK :: CGL : DHM.
For (V. 6.), from the first analogy AD=BC, from the se-
cond analogy EH = FG, and from the third analogy IM=KL;
whence the compound product AD.EH.IM=BC.FG.KL.
But AD.EH.IM=AEI.DHM (V. 2.), and BC.FG.KL=
BFK.CGL; wherefore AEI.DHM=BFK.CGL, and con-
sequently (V. 6.) AEI : BFK :: CGL : DHM.
The same reasoning, it is obvious, applies to any number
of proportionals.
Cor. 1. Hence the powers of the successive terms of nume-
rical proportions, are likewise proportional. For, if A : B::
C: D, and, repeating the analogy, A: B :: C: D; then, by
multiplication, AA: BB :: CC: DD, or A² : B² : : C² : D².
Again, let A: B :: C: D, and, repeating the analogy,
A: B:: C: D,
:
and A B C: D; whence, by multiplying the
corresponding terms,
A³: B³ :: C³: D³.
And so the induction may be pursued generally.
BOOK V.
139
Cor. 2. Hence also the roots of the terms of a numerical
proportion, are proportional. If A: B:: C: D, then √ A:√B
:: VC: VD. For let A: VB:: C: E, and, by
the last corollary, A: B:: C: E; but A: B:: C: D, whence
CE::C: D, and consequently E-D, or VA: ✔B::
✓C: D.-In the same manner, it may be shown in general
that, if A : B :: C: D, ✔A: √B:: √C:√D.
12
PROP. XXIII. THEOR.
The ratio which is conceived to be compounded
of other ratios, is the same as that of the products of
their corresponding numerical expressions.
Suppose the ratio of A: D is compounded of A: B, of B : C,
and of CD, and let A: B:: K:L, B: C:: M: N, and
C: D::O: P; then will A: D:: KMO : LNP.
For, since AB::K: L,
B: CM: N,
and CDO: P,
the products of the similar terms are proportional (V. 22.),
or ABC BCD:: KMO: LNP. But A: D:: ABC: BCD
(V. 3.), and consequently A: D :: KMO : LNP.
The same mode of reasoning is applicable to any number
of component ratios.
PROP. XXIV. THEOR.
A duplicate ratio is the same as the ratio of the
second powers of the terms of its numerical expres-
sion, and a triplicate ratio is the same as the third
powers of those terms.
The duplicate ratio of A: B is denoted by A: B², and
the triplicate ratio by A³: B³.
3
For the duplicate ratio of A: B, being the double com-
140
ELEMENTS OF GEOMETRY.
0
pound of A: B and of A: B, is (V. 22.) the same as that of
the corresponding products A.A: B.B, or A2: B².
Again, the triplicate ratio of A: B, being the triple com-
pound of A: B, of A: B, of A: B, is the same as that of the
corresponding products AAA : BBB, or A³: B³.
3
Cor. Hence the subduplicate ratio of A: B, is √A: √B,
and the subtriplicate ratio of A: B, is A: VB.
3
PROP. XXV. THEOR.
3
The product of the numbers expressing the sides
of a rectangle, will represent its quantity of surface,
as measured by a square described on the linear unit.
D
с
P
Let ABCD be a rectangle and OP the linear measure;
and suppose the side AB to contain OP, m times, and the
side BC to contain it, n times.
Divide these sides accordingly
(I. 38.), and, through the points
of section, draw straight lines (I.
24.) parallel to AD and DC : the
whole rectangle will thus be divi-
ded into cells, each of them equal to the square of OP. It is
evident, that there stand on BC, n columns, and that each of
these columns contains, m cells; consequently the entire space
includes, m.n cells, or is equal to the square of OP repeated
mn times.

B
C
Cor. 1. If m=n, then AB=BC, and the rectangle becomes
a square; but mn is in that case cqual to nn, or n². Whence
the surface of a square is equal to the second power of the
number denoting its side.
Cor. 2. Rectangles which have the same altitude m are as
their bases n and p; for (V. 3. mn: mp :: n: p. And tri-
angles having the same altitude, being (I. 27. cor.) the halves
of these rectangles, must likewise be as their bases.
BOOK V.
141
Cor. 3. If two rectangles be equal, their respective sides are
reciprocally proportional, or form the extremes and means of
an analogy. For if mn=pq, then (V. 6.) m :p :: :9 q: n.
PROP. XXVI. THEOR.
If three straight lines be in continued proportion,
the first is to the third, as the square of the sum or
difference of the first and second to the square of the
sum or difference of the second and third.
Let A: B:: B: C; then A: C::(A+B)² : (B+C)², and
A: C : : (A—B)² : (B—C)², or (B—A)² : (C—B)².
For (V. 19. cor. 2.) A : B :: A÷B : B÷C, and conse-
quently (V. 22. cor. 1. and V. 25. cor. 1.) A : B ::
(A+B)² : (B÷C)². But (V. 24.) A : C : : A² : B²; where-
fore A: C:: (A±B)² : (B÷C)².
Cor. The converse of this proposition is likewise true.
PROP. XXVII. PROB.
Given two homogeneous quantities, to find, if pos-
sible, their greatest common measure.
Let it be required to find the greatest common measure,
that two quantities A and B, of the same kind, will admit.
Supposing A to be greater than B, take B out of A, till
the remainder C be less than it; again, take C out of B, till
there remain only D; and continue this alternate operation,
till the last divisor, suppose E, leave no remainder whatever;
E is the greatest common measure of the quantities proposed.
For, that which measures B will measure its multiple; and
being a common measure, it also measures A, and measures,
therefore, the difference between the multiple of B and A
(V. 1. cor. 1,), that is, C; the required measure, hence, mea-
sures the multiple of C, and conscquently the difference of this
multiple and B, which it measured,—that is D: And lastly,
}
!
142
ELEMENTS OF GEOMETRY.
this measure, as it measures the multiple of D, must conse-
quently measure the difference of this from C, or it must mea-
sure E. Here the decomposition is presumed to terminate.
Wherefore, the common measure of A and B, since it mea-
sures E, may be E itself; and it is also the greatest possible
measure, for nothing greater than E can be contained in this
quantity.
By retracing the steps likewise, it might be shown, that E
measures, in succession, all the preceding terms D, C, B,
and A.
If the process of decomposition should never come to a
close, the quantities A and B do not admit a common mea-
sure, or they are incommensurable. But, as the residue of
the subdivision is necessarily diminished at each step of this
operation, it is evident that an element may be always disco-
vered, which will measure A and B nearer than any assigna-
ble difference whatever.
PROP. XXVIII. PROB.
To express by numbers, either exactly or approxi-
mately, the ratio of two given homogeneous quanti-
ties.
Let A and B be two quantities of the same kind, whose nu-
merical ratio it is required to discover.
Find, by the last Proposition, the greatest common mea-
sure E of the two quantities; and let A contain this measure
K times, and B contain it L times: Then will the ratio
K: L express the ratio of A : B.
For the numbers K and L severally consist of as many
units, as the quantities A and B contain their measure E. It
is also manifest, since E is the greatest possible divisor, that
K and L are the smallest numbers capable of expressing the
ratio of A to B.
BOOK V.
143
If A and B be incommensurable quantities, their decom-
position is capable at least of being pushed to an unlimited
extent; and, consequently, a divisor can always be found so
extremely minute, as to measure them both to any degree of
precision.
Otherwise thus.
;
But the numerical expression of the ratio A: B, may be
deduced indirectly, from the series of quotients obtained in
the operation for discovering their common measure.
Let A contain B, m times, with a remainder C; B contain
C, n times, with a remainder D; and, lastly, suppose C to
contain D, p times, with a remainder E, and which is con-
tained in D, q times exactly. Then D=qE, C=pD+E,
B=nC+D, and A=mB+C; whence the terms D, C, B,
and A, are successively computed, as multiples of E;—A and
B will, therefore, be found to contain E their common mea-
sure K and L times, or the numerical expression for the ratio
of those quantities, is K : L*.
PROP. XXIX. THEOR.
A straight line is incommensurable with its seg-
ments formed by medial section.
If the straight line AB be cut in C, such that the rectangle
AB, BC is equivalent to the square of AC; no part of AB,
however small, will measure the segments AC, BC.
For (V. 27.) take AC out of
AB, and again the remainder
BC out of AC. But AD, be-
A FED
C
ing made equal to BC, the straight line AC is likewise divided
in D, by a medial section (II. 22. cor. 1.); and, for the same
reason, taking away the successive remainders CD, or AE, from
AD, and DE or AF from AE, the subordinate lines AD and
* See Note XXXFI.
144
ELEMENTS OF GEOMETRY.
AE are also divided medially in the points E and F. This ope-
ration produces, therefore, a series of decreasing lines, all of
them divided by medial section: Nor can the process of decom-
position ever terminate; for though the remainders BC, CD,
DE, and EF thus continually diminish, they still must consti-
tute the segments of a similar division. Consequently there ex-
ists no final quantity which would measure both AB and AC.
PROP. XXX. THEOR.
The side of a square is incommensurable with its
diagonal.
Let ABCD be a square and AC its diagonal; AC and AB
are incommensurable.
For make CE equal to AB or BC, draw (I. 5. cor.) the
perpendicular EF, and join BE.
G F
B
Because CE is equal to BC, the angle CEB (I. 11.) is equal
to CRE; and since CEF and CBF are right angles, the re-
maining angle BEF is equal to
EBF, and the side EF (I. 12.)
equal to BF; but EF is also
equal to AE, for the angles EAF
and EFA of the triangle AEF are
evidently each half a right angle.
Whence, making FH equal to
FB, FE or AE,-the excess AE

I
KL H
A
C
E
of the diagonal AC above the side AB, is contained twice in
AB, with a remainder AH; and AH again, being the excess
of the diagonal AF of the square GE above the side AE,
must, for the same reason, be contained twice in AG, with a-
new remainder AL; and this remainder will likewise be con-
tained twice in AH, the side of the square KH. This pro-
cess of subdivision is, therefore, interminable, and the same
relations are continually reproduced *.
* See Note AXXVII.
ELEMENTS
OF

GEOMETRY.
BOOK VI.
THE doctrine of Proportion, grounded on the sim-
plest theory of numbers, furnishes a most powerful
instrument, for abridging and extending mathemati
cal investigations. It easily unfolds the primary re-
lations of figures, and the sections of lines and cir-
cles; but it also discloses with admirable felicity
that vast concatenation of general properties, not
less important than remote, which, without such aid,
might for ever have escaped the penetration of the
geometer. The application of Arithmetic to Geo-
metry forms, therefore, one of those grand epochs
which occur, in the lapse of ages, to mark and acce-
lerate the progress of scientific discovery.
K
146
ELEMENTS OF GEOMETRY.
t
!
DEFINITIONS.
1. Straight lines which proceed from the same point, are
termed diverging lines.
2. Straight lines are divided similarly, when their corre-
sponding segments have the same ratio.
3. A straight line is said to be cut harmonically, if it con-
sist of three segments, such that the whole line is to one ex-
treme, as the other extreme to the middle part.
4. The area of a figure is its surface, or the quantity of
space which it occupies.
5. Similar figures are such as have their angles respective-
ly equal, and the containing sides proportional.
6. If two sides of a rectilineal figure be the extremes of an
analogy, of which the means are two sides containing an equal
angle in another rectilineal figure; these sides are said to be
reciprocally proportional.
5
BOOK VI.
147
C
#
PROP. I. THEOR.
Parallels cut diverging lines proportionally.
The parallels DE and BC cut the diverging lines AB and
AC into proportional segments.
Those parallels may lie on the same side of the vertex, or
on opposite sides; and they may consist of two, or of more
lines.
1. Let the two parallels DE and BC intersect the diver-
ging lines AB and AC, on the same side of the vertex A;
then are AB and AC cut proportionally, in the points D and
E,-or AD: AB:: AE: AC.
E
C
For if AD be commensurable with AB, find (V. 27.) their
common measure M, and, from the corresponding points of
section in AD and AB, draw (I. 24.) the parallels FI, GK,
and HL. It is evident, from Book I.
Prop. 38. that these parallels will also
divide the straight lines AE and AC
equally. Wherefore the measure M,
or AF the submultiple of AD, is con-
tained in AB, as often as AI, the like
submultiple of AE, is contained in AC; consequently (V. def.
10.) the ratio of AD to AB is the same with that of AE to
AC.

..
K
T
AF G. DH B
But, should the segments AD and AB be incommensura-
ble, they may still be expressed numerically, and this to any
required degree of precision. AD being divided (I. 38.) into
equal parts, these parts, continued towards B, will, together
with a residuary portion, compose the whole of AB. Let this
division of AD extend in DB to b, and draw the parallel be.
If the parts of AD and AB be again subdivided, the corre-
sponding residue will evidently be diminished; and thus, ap
K 2
148
ELEMENTS OF GEOMETRY.
A
E
D
B
each successive subdivision, the termi-
nating parallel be must approximate
perpetually to BC. Wherefore, by
continuing this process of exhaustion,
the divided lines Ab and Ac will ap-
proach the limits AB and AC, nearer
than any finite or assignable interval.
the preceding demonstration, AD: AB :: AE: AC.
And since AD:AB:: AE: AC, it follows, by conversion
(V. 11.), that AD: DB:: AE: EC, and again, by composi-
tion (V. 9.), that AB : DB:: AC: EC.

Consequently, from
2. Let the two parallels DE and BC cut the diverging
lines DB and EC, on opposite sides of A; the segments AB,
AD have the same ratio with AC, AE,-or AB: AD::
AC: AE.
i
For, make AO equal to AD, AP to AE, and join OP.
The triangles APO and AED,
P
A
B
having the sides AO, AP equal
to AD, AE, and the contained
vertical angle OAP equal to
DAE, are equal (I. 3.), and D
consequently the angle AOP
is equal to ADE; but these
being alternate angles, the
straight line OP (I. 23.) is parallel to DE or BC, and hence,
from what was already demonstrated, AB : AO or AD ::
AC: AP or AE.

E
And since AB : AD: : AC : AE, by composition BD :
AD: CE AE, and, by conversion, BD: AB:: CE:
AC.
3. Lastly, let more than two parallels, BC, DE, FH, and
GI, intersect the diverging lines AB and AC; the segments
DA, AF, FG, and GB, in DB, are proportional respective-
ly to EA, AH, HI, and IC, the corresponding segments in
EC.
BOOK VI.
149
For, from the second case, AD:
AF :: AE: AH; and, from the
first case, AF: FG:: AH: HI.
But from the same case, AG : FG
:: AI: HI, and AG: GB:: AI
:IC; whence (V. 15.) FG: GB
:: HI: IC.
D
E
C

H
A
F G B
Cor. 1. Hence the converse of the proposition is also true,
or that straight lines which cut diverging lines proportionally
are parallel; for it would otherwise follow, that a new divi-
sion of the same line would not alter the relation among the
segments, which is evidently absurd.
Cor. 2. Hence, if the segments of one diverging line be
equal to those of another, the straight lines which join them
are parallel.
PROP. II. THEOR.
Diverging lines are proportional to the correspond-
ing segments into which they divide parallels.
Let two diverging lines AB
and AC cut the parallels BC and
DE; then AB : AD :: BC :
DE.
For draw DF parallel to AC.
And, by the last Proposition,
the parallels AC and DF must
cut the straight lines AB and
BC proportionally, or AB: AD
:: BC: CF. But CF is equal
(I. 27.) to the opposite side DE
of the parallelogram DECF;
and consequently AB : AD ::
BC: DE.

D
E
E
LA.
זן
D
B

150
ELEMENTS OF GEOMETRY.
Next, let more than two diverging lines AB, AF, and AC
intersect the parallels BC and DE; the segments BF and FC
have respectively to DG and GE
the same ratio as AB has to AD.
From what has been already
demonstrated, it appears, that
AB: AD: BF: DG, and al-
so that AF: AG :: FC: GE.
But by the last Proposition, AB A
: AD AF AG; wherefore

::
E
طرح
G
F
D
B
AB: AD :: FC: GE. The same mode of reasoning, it is
obvious, might be extended to any number of sections. Whence
AB: AD:: BF: DG:: FC: GE.
Cor. 1. Hence the straight lines which cut diverging lines
equally, being parallel (VI. 1. cor. 2.), are themselves propor-
tional to the segments intercepted from the vertex.
Cor. 2. Hence parallels are cut proportionally by diverging
lines *.
PROP. III. PROB.
To find a fourth proportional to three given straight
lines.
Let A, B, and C be three straight lines, to which it is re-
quired to find a fourth proportional.
Draw the diverging lines DG
and DH, make DE equal to A,
DF to B, and DG to C, join EF,
and through G draw (I. 24.) GH
parallel to EF and meeting DH
in H; DH is a fourth proportion-
al to the straight lines A, B, and C.
AF
Br

H
F
D
E
See Note XXXVIII.
BOOK VI.
151
For the diverging lines DG and DH are cut proportionally
by the parallels EF and GH (VI. 1.), or DE : DF : : DG :
DH, that is, A: B:: C: DH.
Cor. If the mean terms B and C be equal, it is obvious that
DG will become equal to DF, and that DH will be found a
third proportional to the two given terms A and B.
PROP. IV. PROB.
To cut a given straight line into segments, which
shall be proportional to those of a divided straight
line.
Let AB be a straight line, which it is required to cut into
segments proportional to those of a given divided straight
line.
D
E
Draw the diverging line AC, and
make AD, DE, and EC, equal re-
spectively to the segments of the
divided line, join CB, and draw EG
and DF parallel to it (I. 24.) and
meeting AB in F and G; AB is
cut in those points proportionally to the segments of AC.
For the parallels DF, EG, and CB must cut the di-
verging lines AB and AC proportionally (VI. 1.), or AF:
FG:: AD: DE, and FG: GB:: DE: EC.

A
PROP. V. PROB.
B
To cut off the successive parts of a given straight
line.
Let AB be a straight line from which it is required to cut
off successively the half, the third, the fourth, the fifth, &c.
Through B draw the inclined straight line CBG extended
both ways, in this take any point C, and make BD, DE, EF,
152
ELEMENTS OF GEOMETRY.
FG, &c. each equal to BC, complete the parallelogram ABCI,
and join ID, IE, IF, IG, &c. cutting AB in the points
K, L, M, N, &c.; then is the segment AK the half of AB,
AL the third, AM the fourth,
and AN the fifth part, of the
same given line.
For the segments of the
straight line AB must be pro-
portional to the segments of
the parallels AI and BG, in-
tercepted by the diverging
lines ID, IE, IF, IG, &c.

:
Thus, AK KB:: AI: BD;
but, by construction, BC or
Ar
B
K
I
C
D
F
E
G
AI BD, whence (V. 4.) AK-KB, and therefore AK is the
half of AB. Again, AL: LB:: AI: BE; and since BE=
2AI, it follows, that LB-2AL, or AL is the third part of
AB. In the same manner, AM: MB:: AI: BF; but BF-
3AI, whence MB=3AM, or AM is the fourth part of AB.
And, by a like process, it may be shown that AN is the fifth
part of AB.
Otherwise thus.
On AB describe the rhomboid ABCD, and through E, the
intersection of its diagonals AC and BD, draw EF parallel
to AD (I. 24.), join DF, and through G, where it cuts AC,
draw GH likewise parallel to AD, again join DH and draw
the parallel IK, and so repeat the operation: Then will AF
be the half of AB, AH the third, AK the fourth, and AM
the fifth part of it.
Because AD and EF are parallel, DE: EB :: AF : FB
(VI. 1.); but DE=EB (I. 29.), wherefore AF=FB, or AF
is the half of AB. And AD and EF being intercepted pa-
rallels, AD: EF ; : AB ; BF (VI. 2.); consequently since
BOOK VI.
153
AB is double of BF, AD is
likewise double of EF (V. 5.).
-Again, the diverging lines
AGE and DGF are propor-
tional to the intercepted paral-
lels AD and EF (VI. 2.), or
AD: EF :: AG: GE; and
D
G
E
A MKH F
B
C

GH being parallel to EF, AG: GE :: AH: HF (VI. 1.),
whence AD: EF :: AH : HF; but AD was shown to be
double of EF, wherefore AH is double of HF (V. 5.), or
AH is two-thirds of AF, or of the half of AB, and is conse,
quently the third part of the whole AB. And, since AF:
HF:: AD: GH and AF is triple of HF, it is evident that
AD is triple of GH; but AD: GH:: AI: IG :: AK: KH,
and, AD being triple of GH. AK must also be triple of KH;
or AK is three-fourths of AH, which was proved to be the
third of AB, whence the segment AK is the fourth part of
the whole line AB. By a like process, it is shown that AM
is the fifth part of AB*.
1
PROP. VI. PROB.
To divide a straight line harmonically, in a given
ratio.
Let AB be a straight line, which it is required to cut har-
monically, in the ratio of K to L.
Through A draw the diverging line AC, and produce it
both ways till AC and AD be each equal to K, make AE
* See Note XXXIX.
154
ELEMENTS OF GEOMETRY.
equal to L, join CB, draw
EF parallel to AB, and
FG parallel to CA, and
join DF; the straight line
AB is divided harmoni-
cally in the points H and
G, such that K: L:: AB
: BG:: AH: HG.
For the parallels AC
and GF, being intercept-
ed by the diverging lines
D
I
K-

I
T
A
H G
B
AB and CB, AC: GF:: AB: BG (VI. 2.). Again, the di-
verging lines AG and DF are cut by the parallels AD and
FG, whence (VI. 1.) AD : GF:: AH: HG. Wherefore,
AB: BG:: AH: HG; and each of these ratios is the same
as that of AC or AD to GF, or that of K to L.
Cor. Hence AG is divided, internally in H and externally
in B, in the same ratio. In like manner, BH is divided pro-
portionally, by an external and internal section in A and G;
for AB: BG:: AH : HG, and alternately AB : AH : :
BG: HG.
PROP. VII. THEOR.
If a straight line be divided internally and exter-
nally in the same ratio, half the line is a mean pro-
portional between the distances of the middle from
the two points of unequal section.
Let the straight line AB be divided in the same ratio, in-
ternally and externally in C and D, and also be bisected in
BOOK VI.
155
E; the half EB is a mean pro-
portional between EC and ED,
or EC EB :: EB: ED.
I C B
D
For since AC: CB :: AD: DB, by mixing and inversion
AC-CB: AC+CB:: AD-DB: AD+DB, that is, 2EC:
AB:: AB: 2ED, and, halving all the terms of the analogy,
(V. 3.) EC : EB : : EB : ED.
Cor. Hence if a straight line be cut internally and exter-
nally in the same ratio, the square of the interval between the
points of section is equivalent to the difference between the
rectangles under the internal and external segments. For
(II. 19. cors.) AD.DB=ED²—EB², and AC.CB=EB*—
EC²; consequently AD.DB-AC.CB — ED² —2EB² +
EC², or (V. 6.) ED-2ED.EC+EC, which (II. 18.) is the
square of ED-EC or of CD.-By a similar procedure, the
converse of the proposition and its corollary may be establish-
ed.
=
PROP. VIII. THEOR.
If diverging lines divide a straight line harmoni-
cally, they will cut every intercepted straight line
also in harmonic proportion.
Let the diverging lines EA, EC, EB, and ED terminate
in the harmonic section of the straight line AD; any inter-
cepted straight line FG will be likewise cut by them harmo-
nically, or FG: GI:: FH: HI.
For, through the points B and I, draw (I. 24.) KL and
MN parallel to AE.
Because the parallels AE and BL are intercepted by the
diverging lines DA and DE, AD: DB:: AE: BL (VI. 2.);
and for the same reason, the parallels AE and BK being in-
156
ELEMENTS OF GEOMETRY.
tercepted by the diverging
lines AB and EK, AC: CB ::
AE: BK. And since AD
is divided harmonically, AD;
DB:: AC: CB; wherefore
AE: BL AE: BK, and
consequently BL= BK. But,
KL being parallel to MN,
E

N
G
H
I
M
I
F
A
C
B
D
BL: BK:: IN: IM (VI. 2.
cor. 2.); consequently, BL
K
being equal to BK, IN must
also be equal to IM (V. 4.); whence FE: IN: : FE: IM.
Again, FE : IN :: FG: GI, for the parallels FE and IN are
cut by the diverging lines GF and GE; and FE : IM ::
FH: HI, since the parallels FE and IM are cut by the
diverging lines FI and EM. Wherefore, by identity of ra-
tios, FG: GI :: FH: HI; or the intercepted straight line
FG is cut harmonically in the points H and I.
PROP. IX. THEOR.
If from any point in the circumference of a circle,
straight lines be drawn to the extremities of a chord
and meeting the perpendicular diameter, they will
divide that diameter, internally and externally, in
the same ratio.
Let the chord EF be perpendicular to the diameter AB of
a circle, and from its extremities F and E straight lines FG
and EG be inflected to a point G in the circumference, and
cutting the diameter internally and externally in C and D ;
then will AC: CB:: AD: DB.
For join AG and BG, and draw HBI parallel to AG.
Because AEGB is a semicircle, the angle AGB is a right
BOOK VI.
157
angle (III. 22.); wherefore AG and HI being parallel, the alter-
nate angle GBI is right (I. 23.), and likewise its adjacent angle
GBH. But the diameter AB, being perpendicular to the chord
EF, must (III. 4. and 15.) bisect the arc FAE, and therefore
the angle EGA is equal to AGF (III. 13. cor.) or (III. 19.),
its supplement. And since
AG is parallel to HI, the
angle EGA is equal to the
angle GIB or its supple-
ment (I. 23.); and, for the
same reason, the angle
AGF is equal to the alter-
nate angle GHB. Whence
the angle GIB is equal to
GHB; but the angles GBI
and GBH being both right
angles, are equal, and the
side GB is common to the
two triangles BIG and
E

I
A
C
IT
B
D
G

K I
H
B
BHG, which are, therefore, equal (I. 21.), and consequently
BH is equal to BI, and AG: BH :: AG: BI. Now, be-
cause the parallels AG and BH are intercepted by the diver-
ging lines AB and GH, AG: BH:: AC: CB (VI. 2.); and
since the parallels AG and BI are intercepted by the diverging
lines GD and AD, AG: BI:: AD: DB. Wherefore, by iden-
tity of ratios, AC: CB :: AD: DB, that is, the straight line
AB is cut in the same ratio, internally and externally, or the
whole line AD is divided harmonically in the points C and B.
Cor. 1. As the points E and G come nearer each other, it
is obvious that the straight line EGD will approach, conti-
nually to the position of the tangent, which is its ultimate li-
mit. Hence the tangent and the perpendicular, from the
point of contact or mutual coincidence, cut the diameter pro-
158
ELEMENTS OF GEOMETRY.
portionally, or AC: CB :: AD :
DB. It is, therefore, evident
(VI. 7.) that, O being the centre,
OC: OB:: OB: OD.

E
C
B
D
Cor. 2. Since OC: OB:: OB:
OD, it follows (V. 19. cor. 2.) that OC: OD :: OB-OC²
or AC.CB: OD²-OB or AD.DB; whence, by division,
CD: OD :: AD.DB-AC.CB, or (VI. 7. cor.) CD² :
AD.DB *.
PROP. X. THEOR.
A straight line drawn from the concourse of two
tangents to the concave circumference of a circle,
is divided harmonically, by the convex circumfe-
rence and the chord which joins the points of con-
tact.
Let ED and FD be two tangents applied to the circle
AEBF; the secant DA, drawn from their point of concourse,
will be cut in harmonic proportion, by the convex circumfe-
rence EBF and the chord EF which joins the points of con-
tact,—or AD : DB : : AC : CB.
E
For the tangents ED and FD are equal (III, 26. cor.),
and EDF being thus an isos-
celes triangle, DE² = DC²+
EC.CF (II. 23.); (but III. 32.)
DE is also equal to AD.DB,
and the chords AB and EF,
by their mutual intersection,
make the rectangle EC, CF e-

B
Ꭰ
qual to AC, CB. Whence DC=AD.DB-AC.CB, and
therefore (VI. 7. cor.) AC: CB:: AD: DB.
* See Note XL,
BOOK VL
159
PROP. XI. THEOR.
A straight line which bisects, either internally or
externally, the vertical angle of a triangle, will di-
vide its base into segments, internal or external,
that are proportional to the adjacent sides of the
triangle.
Let the straight line BD bisect the vertical angle of the
triangle ABC; it will cut the base AC into segments which
have the same ratio as the adjacent sides, or AD: DC : :
AB: BC.
For through C draw CE parallel to DB (I. 24.), and meet-
ing the production of AB in E.
B
E
Because DB and CE are parallel, the exterior angle ABD
is equal to BEC, and the alter-
nate angle DBC equal to BCE
(I. 23.); wherefore the angle ABD
being equal by hypothesis to DBC,
the angle BEC is equal to BCE,
and consequently (I. 12.) the tri-
angle CBE is isosceles, or BE is
equal to BC. But the parallels
DB and CE cut the diverging lines
AC and AE proportionally (VI. 1.),
or AD: DC:: AB: BE; that is, since BE-BC, AD:
DC:: AB: BC.
A
D
C
Again, let the vertical line BD bisect the exterior angle
CBG of the triangle; it will divide the base into external
segments AD and DC, which are also proportional to the
adjacent sides AB and BC.
For through C draw CE parallel to DB, and meeting AB
in E.
1
1

160
ELEMENTS OF GEOMETRY.
The equal angles
GBD and DBC are,
from the properties of
parallel straight lines,
respectively equal to
E
BEC and BCE, and
!
B
G

consequently the tri-
angle CBE is isosceles,
A.
or the side BC is equal to BE. And since the diverging
lines AD and AB are cut by the parallels DB and CE pro-
portionally, AD: DC :: AB : BE or BC.
Cor. Hence the converse of the Proposition is likewise true,
or if a straight line be drawn from the vertex of a triangle to
cut the base in the ratio of the adjacent sides, it will bisect
the vertical angle; for it is evident, from VI. 6. cor., that a
straight line is only capable of a single section, whether in-
ternal or external, in a given proportion.
Scholium. The vertical line BD must bisect the base AC
of the triangle, when the sides AB and BC are equal. In
the case where BD bisects the exterior angle CBG, if AB be
supposed to approach to an equality with BC, the straight
line EC will come nearer to AC, and consequently the inci-
dence D of the parallel BD with AC will be thrown conti-
nually more remote. But when the side AB is equal to BC,
the straight line BD, being now parallel to AC, will never
meet it, or there can be no equality of external section; for
though the ratio of AD to CD tends towards the ratio of
equality as the point D retires, yet the constant difference AC
between those distances must always bear a sensible relation
to them. After BD, in turning about the point B, has pass-
ed the limits of distance beyond C, it re-appears in an oppo-
site direction beyond A, when AB, receding from equality,
has become less than BC*.
* See Note XLI.
BOOK VI.
161
PROP. XII. THEOR.
Triangles are similar, which have their correspond-
ing angles equal.
Let the triangles ABC and DEF, have the angle CAB
equal to FDE, CBA to FED, and consequently (I. 32.) the
remaining angle BCA equal to EFD; these triangles are si-
milar, or the sides in both which contain equal angles are pro-
portional.
B
For make BG equal to ED, and draw GH parallel to AC.
Because GH is parallel to
AC, the exterior angle BGH
is equal (I. 23.) to BAC, that
is, to EDF; and the angle at
Bis, by hypothesis, equal to
that at E, and the interjacent.
side BG was made equal to
ED; wherefore (I. 21.) the

A
GA
I
F
triangle GBH is equal to DEF. But, the diverging lines BA
and BC being cut proportionally by the parallels AC and GH
(VI, 1.), AB is to BC as BG to BH, or as ED to EF.
Again, those diverging lines being proportional to the inter-
cepted segments AC and GH of the parallels (VI. 2.), AB is
to BG as AC is to GH, and alternately AB is to AC as BG
is to GH, or as ED to DF. In the same manner, as BC is to
BH so is AC to GH, and alternately, as BC is to AC so is
BH or EF to GH or DF. And thus, the sides opposite to
equal angles in the triangles ABC and DEF, are the homo-
logous terms of a proportion.
Cor. Isosceles triangles are similar which have their verti-
cal angles equal. For (I. 32.) the supplementary angles at
the base must be together equal, and consequently they are
equal to each other.
L
162
ELEMENTS OF GEOMETRY.
:
PROP. XIII. THEOR.
Triangles which have the sides about two of their
angles proportional, are similar.
In the triangles ABC and DEF, let AB: AC:: DE: DF
and BC: AC:: EF: DF; then is the angle BAC equal to
EDF, and the angle BCA to EFD.
For (1. 4.) draw DG and FG, making angles FDG and
DFG equal to CAB and ACB.
By the last Proposition, the triangle ABC is similar to
DGF, and consequently AB: AC :: DG : DF; but, by hy-
pothesis, AB: AC:: DE: DF, and hence, from identity of ra-
tios, DG: DF:: DE: DF,
or DG is equal to DE. In
the same manner, BC:AC::
EF: DF, and BC: AC::
GF: DF; whence EF: DF

E

D
F
:: GF: DF, and EF is
A.
C
equal to FG. Wherefore
the triangles DEF and
DGF, having thus the sides DE and EF equal to DG and
FG, and the side DF common to both, are (I. 2.) equal;
consequently the angle EDF is equal to FDG or BAC, and
the angle EFD is equal to DFG or BCA.
Cor. Hence isosceles triangles which have either side pro-
portional to the base, are similar.
PROP. XIV. THEOR.
Triangles are similar, if each have an equal angle
and its containing sides proportional.
In the triangles BAC and EDF, let the angle ABC be
equal to DEF, and the sides which contain the one be pro-
portional to those which contain the other, or AB : BC : :
DE: EF; the triangles BAC and EDF are similar.
BOOK VI.
163
For, from the points E and F, draw EG and FG, making
the angles FEG and EFG equal to CBA and BCA.
The triangles BAC and EGF, having thus their corre-
sponding angles equal, are similar (VI. 12.), and therefore
AB: BC :: EG: EF. But by hypothesis, AB BC :
ED EF; wherefore
EG: EF:: ED: EF,
and consequently EG
is equal to ED. Hence
the triangles GFE and
DFE, having the side
:
Да


C D
EG equal to ED, EF common to both, and the contained
angle GEF equal to ABC or DEF, are equal (I. 3.), and
therefore the angle EFG or BCA is equal to EFD; con-
sequently the remaining angles BAC and EDF of the tri-
angles ABC and DEF, are equal (I. 32.), and these triangles
are (VI. 12.) similar.
PROP. XV. THEOR.
Triangles are similar, which, being of the same af-
fection, have each an equal angle, and the sides con-
taining another angle proportional.
Let the triangles ABC and DEF, which are of the same
affection, have the angle ABC equal to DEF and the sides
that contain the angles at C and F proportional, or BC: AC::
EF: FD; the triangles ABC and DEF are similar.
For, from the
points E and F
drawEG and FG,
making the angles
FEG and EFG
equal to ABC and
BCA.
Да


C D
The triangle ABC is evidently similar to GEF, and BC:
2
#
164
ELEMENTS OF GEOMETRY.
CA:: EF: FG; but, by hypothesis, BC: CA:: EF: FD,
and therefore EF: FG :: EF: FD, and FG is equal to FD.
Whence the triangles EGF and EDF, having the side FG
equal to FD and the side EF common, and being both of
the same affection with CAB, are equal (I. 22.); consequent-
ly the angle GFE is equal to DFE or ACB, and therefore
(VI. 12.) the triangles ABC and DEF are similar.
PROP. XVI. THEOR.
A perpendicular let fall upon the hypotenuse of
a right-angled triangle from the opposite vertex will
divide it into two triangles which are similar to the
whole and to each other.
Let the triangle ABC be right-angled at B, from which
the perpendicular BD falls upon the hypotenuse AC; the
triangles ABD and DBC, thus formed, are similar to each
other, and to the whole triangle ACB.
For the triangles ABD and ACB, having the angle BAC
common, and the right angle ADB equal to ABC, are simi-
lar (VI. 12.). Again, the triangles
B

DBC and ACB are similar, since
they have the angle BCD common,
and the right angle BDC equal to
ABC. The triangles ABD and
A
כ.
1
DBC being, therefore, both similar to the same triangle ABC,
are evidently similar to each other (VI. 12.).
Cor. 1. Hence the side of a right-angled triangle is a mean
proportional between the hypotenuse and the adjacent seg-
ment, formed by a perpendicular let fall upon it from the op-
posite vertex; and the perpendicular itself is a mean propor-
tional between those segments of the hypotenuse. For the
triangles ABC and ADB being similar, AC: AB:: AB: AD;
and the triangles ABC and BDC being similar, AC: BC::
BC: CD; again, the triangles ADB and BDC are similar,
and therefore AD: DB:: DB: DC.
BOOK VI.
165
. Cor. 2. If the hypotenuse and the sides of a right-angled
triangle form a continued proportion, the hypotenuse will be
divided in extreme and mean ratio, by the perpendicular
let fall upon it from the opposite vertex. For, by the last Co-
rollary, AC AB AB: AD, and therefore (V. 6.) AB'=
AC.AD; in like manner, AC: BC: BC: CD. But, by
hypothesis, AC BC BC AB; whence BC: CD: :
BC: AB, and consequently AB=DC, and AB²=AC.AD=
CD. Wherefore (V. 6.) AC : CD : : CD : AD *.
+
:
PROP. XVII. THEOR.
The perpendicular within a circle, is a mean pro-
portional to the segments formed on it by straight
lines, drawn from the extremities of the diameter,
through any point in the circumference.
نا
Let the straight lines AEC and BCG, drawn from the ex-
tremities of the diameter of a circle
through a point C in the circumfe-
rence, cut the perpendicular to AB;
the part DF within the circle is a
mean proportional between the seg-
ments DE and DG.
For the angle ACB, being in a se-
micircle, is a right angle (III. 22.),
and the angle ABG is common to
the two triangles ABC and GBD,
which are, therefore, similar (VI. 12.).
Hence the remaining angle BAC is
equal to BGD, and consequently the
triangles ADE and GDB are simi-
lar; wherefore AD: DE:: DG: DB,
and (V. 6.) AD.DB=DE.DG. But
(III. 32. cor.), the rectangle under
AD and DB is equivalent to the

* See Note XLII.
E
B
D
E

B
D
166
ELEMENTS OF GEOMETRY.
square of DF; whence DE.DG=DF, and (V. 6.) DE: DE::
DF: DG *.
PROP. XVIII. PROB.
To find the mean proportional between two given
straight lines.
Let it be required to find the mean proportional between
the straight lines A and B.
A
Find C (III. 33.) the side of a square
which is equivalent to the rectangle B
contained by A and B; C is the mean
proportional required.
'CH
+
For since CAB, it follows (V. 6.) that A: C:: C: B.
Otherwise thus.
Make DF-A and DE-B, on DE describe the semicircle
DGE, draw FG perpendicular to the diameter DE, and join
DG; the chord DG is the mean proportional required.
For join GE. The triangle DGE,
being contained in a semicircle, is right-
angled, and therefore (VI. 16. cor. 1.)
DG is a mean proportional between
DF and DE, that is, between the given
straightl ines A and B.

Or thus.
Having made DF and DE equal to A
and B, on FE describe the semicircle
FGE; and the tangent DG being drawn,
is the mean proportional required.
D
D
T
I
For
(III. 32. cor. 2.) DF × DE=DG², and consequently (V. 6.)
DF: DG:: DG: DE+.
* See Note XLIII. + See Note XLIV.
BOOK VI.
167
PROP. XIX. PROB.
To divide a straight line, whether internally or
externally, so that the rectangle under its segments
shall be equivalent to a given rectangle.
Let AB be a straight line, which it is required to cut, so
that the rectangle under its segments shall be equivalent to
a given rectangle.
On AB describe the semicircle AFB, at A and B apply
tangents AD and BE equal to the sides of the given rect-
angle, join DE, to which, and from the point F where it
meets the circumference, draw the perpendicular FC; this
will divide AB into the segments required.
D
F
E
For join AF and BF. And because AD is a tangent and
AF a straight line inflected to
the circumference, the exte-
rior angle DAF is equal to
CBF which stands in the al-
ternate segment (III. 25. and
III. 19. cor. 1.); and, for the
same reason, the exterior angle
EBF is equal to CAF. But
the opposite angles DAC and
DFC of the quadrilateral figure
ADFC are, in the first case,
two right angles, and therefore
the angle ADFis (III. 19. cor.1.)
equal to BCF; and, in the
second case, the angles DAC

A
C
B

E
T
B
and DFC being both right angles, the figure DAFC is con-
tained in a semicircle, consequently (III. 18.) the angle ADF
is equal to BCF. In the same manner, it is proved in both
cases, that the angle BEF is equal to ACF; wherefore the
triangles DAF and AFC are similar to BCF and BFE; and
168
ELEMENTS OF GEOMETRY.
hence AD: AF:: CB: BF, and AF: AC:: BF: BE;
consequently (V. 16.) AD: AC: CB BE, and (V. 6.)
AD.BE-AC.CB.
:
Cor. If the sides of the given rectangle be equal, the con-
struction of the problem will become materially simplified.
First, in the case of internal section: The tangents AD,
BE being equal, it is evident that DE
must be parallel to AB and the per-
pendicular FC parallel to EB. Whence,
employing this construction, the rect-
angle under the segments AC and CB

А.
FE
CB
is equivalent to the square of BE; which also follows from
Prop. 32. Book III.
Next, in the case of external section: The opposite tan-
gents AD, BE being equal,
the triangles AGD and BGE
are evidently equal, and there-
fore DE passes through the
centre. Hence the triangles
BGE and FGC are also equal,
and GC equal to GE.-This
construction being effected,

A
G
B
the rectangle AC, CB will be equal to the square of BE;
which is also deduced from Prop. 32. Book III., since CF is
now a tangent and AC.CB-CF2 or BE².
If AB be equal to BE, the construction will exactly cor-
respond with what was before given.
PROP. XX. THEOR.
The rectangle under any two sides of a triangle,
is equivalent to the rectangle under the perpendicu-
lar drawn to the base and the diameter of the cir-
cumscribing circle.
Let ABC be a triangle, about which is described a circle
BOOK VI.
169
having the diameter BE; the rectangle under the sides AB
and BC is equivalent to the rectangle under BE and the per-
pendicular BD let fall from the vertex of the triangle upon
the base AC.
For join CE. And the angle BAD
is equal to BEC (III. 18.), since they
both stand upon the same arc BC;
and the angle ADB, being a right an-
gle, is equal to ECB, which is contain-
ed in a semicircle (III. 22.).

Where-
A
E
B
D
fore the triangles ABD and EBC, being thus similar (VI.
12.), AB : BD : : EB: BC, and consequently (V. 6.) AB.BC
=EB.BD.
PROP. XXI. THEOR.
The square of a straight line that bisects, whether
internally or externally, the vertical angle of a tri-
angle, is equivalent to the difference between the
rectangle under the sides, and the rectangle under
the segments into which it divides the base.
In the triangle ABC, let BE bisect the vertical angle CBA
or its adjacent angle CBF; then BE AB.BC-AE.EC, or
AE.EC-AB.BC.
=
For (III. 10. cor.) about the triangle describe a circle,
produce BE to the circumference, and join CD.
The angles BAE and BDC, stand-
ing upon the same arc BC, are (III.
18.) equal, and the angle ABE is, by
hypothesis, equal to DBC; wherefore
(VI. 12.) the triangles AEB and DCB
are similar, and AB: BE::DB: BC.
Consequently (V. 6.) AB.BC=BE.BD;

D
B
C
1
170
ELEMENTS OF GEOMETRY.
but BE.BD BE.ED + BE,
or BE.ED-BE², and (III. 32.)
BE.ED AE.EC; wherefore

=
AB.BCBE.ED + BE², or
BE.ED-BE2, and consequently
BE² = AB.BC AE.EC or
AE.EC-AB.BC *.
A
PROP. XXII. THEOR.
The rectangles under the opposite sides of a qua-
drilateral figure inscribed in a circle, are together
equivalent to the rectangle under its diagonals.
In the circle ABCD, let a quadrilateral figure be inscribed,
and join the diagonals AC, BD; the rectangles AB, CD and
BC, AD, are together equivalent to the rectangle AC, BD.
For (I. 4.) draw BE, making an angle ABE equal to
CBD.
The triangles AEB and DCB ha-
ving thus the angle ABE equal to
DBC, and the angle BAE or BAC
equal (III. 18.) to BDC, are similar
(VI. 12.), and therefore AB: AE::
BD : CD; whence (V. 6.) AB.CD=
AE.BD. Again, because the angle

B
C
E
D
ABE is equal to DBC, add EBD to each, and the whole
angle ABD is equal to EBC; and the angle ADB is equal
to ECB (III. 18.); wherefore the triangles DAB and CEB
are similar (VI. 12.), and AD: BD :: EC : BC, and conse-
quently BC.AD=EC.BD. Whence the rectangles AB, CD
and BC, AD are together equal to the rectangles AE, BD
and EC, BD, that is, to the whole rectangle AC, BD†.
* See Note XLV. ↑ See Note XLVI.
BOOK VI.
171
PROP. XXIII. THEOR.
Triangles which have a common angle, are to each
other in the compound ratio of the containing sides.
Let ABC and DBE be two triangles, having the same or
an equal angle at B; ABC is to
DBE in the ratio compounded of
that of BA to BD, and of BC to
BE.
For join AE and CD. The ratio
of the triangle ABC to DBE may
be conceived as compounded of that
of ABC to DBC, and of DBC
to DBE. But (V. 25. cor. 2.)
A
D
B

E
the triangle ABC is to DBC, as the base BA to BD; and,
for the same reason, the triangle DBC is to DBE, as the base
BC to BE; consequently the triangle ABC is to DBE in the
ratio compounded of that of BA to BD, and of BC to BE,
or (V. 23.) in the ratio of the rectangle under BA and BC to
the rectangle under BD and BE.
Cor. 1. Hence similar triangles are in the duplicate ratio
of their homologous sides. For, if the angle at B be equal to
that at E, the tri-
angle ABC is to
DEF in the ratio
compounded of
that of ABto DE,
B

E

and of CB to FE;
but, these trian-
A
gles being similar,
C D
F
the ratio of AB to DE is the same as that of CB to FE (VI.
12.), and consequently the triangle ABC is to DEF in the
duplicate ratio of AB to DE, or (V. 24.) as the square of AB
to the square of DE.
172
ELEMENTS OF GEOMETRY.
Cor. 2. Hence triangles which have the sides that contain
an equal angle reciprocally
proportional, are equiva-
lent. For, the angle at B
being equal to that at E,
the triangle ABCisto DEF,
as AB.CB to DE.FE; but
:
AB DE FE: CB, and
(V. 6.) AB.CB=DE.FE;
consequently (V. 4.) the
A
B
F


C
F
third and fourth terms of the analogy being equal, the first
and second must also be equal.
PROP. XXIV. THEOR.
Similar rectilineal figures may be divided into cor-
responding similar triangles.
Let ABCDE and FGHIK be similar rectilineal figures, of
which A and F are corresponding points; these figures may
be resolved into a like number of triangles respectively simi-
lar.
For, from the point A in the one figure, draw the straight
lines AC, AD, and, from F in the other, draw FH, FI; the
triangles BAC, CAD, and DAE are respectively similar to
GFH, HFI, and IFK.
Because the polygon ABCDE is similar to FGHIK, the
angle ABC is equal to
FGH, and AB: BC::
FG: GH; wherefore
C

(VI. 14.) the triangle
BAC is similar to GFH.
B
H

G
Hence the angle BCA
is equal to GHF; and
A
1
I
K
the whole angle BCD
T
being equal to GHI, the remaining angle ACD must be equal
BOOK VL
179
to FHI. But BC: AC:: GH: FH, and BC: CD:: GH: HI,
consequently (V. 15.) AC: CD:: FH: HI, and the tri-
angles CAD and HFI (VI. 14.) are similar. Whence the
angle CDA being equal to HIF and the angle CDE to HIK,
the angle ADE is equal to FIK; and since CD : DA :: HI : IF,
and CD: DE:: HI: IK, therefore (V.15.) DA : DE: : IF : IK,
and the triangles DAE and IFK are similar.
The same train of reasoning, it is obvious, would apply to
polygons of any number of sides.
PROP. XXV. PROB.
On a given straight line, to construct a rectilineal
figure similar to a given rectilineal figure.
Let FK be a straight line, on which it is required to con-
struct a rectilineal figure similar to the figure ABCDE.
Join AC and AD, dividing the given rectilineal figure into
its component triangles: From the points F and K draw FI
and KI, making the angles KFI and FKI equal to EAD and
AED; from F and I draw FH and IH making the angles
IFH and FIH equal to DAC and ADC; and lastly from F
and H draw FG and HG making the angles HFG and FHG
equal to CAB and ACB. The figure FGHIK is similar to
ABCDE.
For the several triangles KFI, IFH, and HFG, which com-
pose the figure FGHIK, are, by the construction, evidently
similar to the triangles EAD, DAC, and CAB, into which
the figure ABCDE
was resolved. Whence
FK: KI:: AE: ED; B
also KI: IF:: ED: DA,
D

I

TT
and IF:IH:: DA:DC,
and consequently (V.
A
15.) KI IH: ED:
E
F
K
:
DC. Again, IH: HF: DC: CA, and HF: HG ::
:
174
ELEMENTS OF GEOMETRY.
CA: CB; and hence IH: HG:: DC: CB. But HG: GF::
CB: BA: and the ratio of GF to FK, being compounded of
that of GF to FH, of FH to FI, and of FI to FK, is the same
with the ratio of BA to AE, which is compounded of the like
ratios of BA to AC, of AC to AD, and of AD to AE.
Wherefore all the sides about the figure FGHIK are propor-
tional to those about ABCDE; but the several angles of the
former, having a like composition, are respectively equal to
those of the latter. Whence the figure FGHIK is similar to
the given figure.
The same reasoning, it is manifest, would extend to poly-
gons of any number of sides.
Scholium. The general solution of this problem is derived
from the principle, that similar triangles, by their composi-
tion, form similar polygons. The mode of construction, how-
ever, admits of some variation. For instance, if the straight
line FK be parallel to AE, or in the same extension with that
'homologous side,—the several triangles FIK, FHI, and FGH
may be more easily constituted in succession, by drawing the
straight lines FI and KI, FH and IH, and FG and GH pa-
rallel to the corresponding sides in the original figure
ABCDE; because (I. 31.) a corresponding equality of angles
will be thus produced.
But, if FK have no determinate position, the construction
may be still farther simplified;
For, having made AK equal to
13
G
A
K E
D
that base and joined AD and
AC, draw KI, IH, and HG pa-
rallel to ED, DC, and CB. The
figure AKIHG is evidently simi-
lar to AEDCB, since its component triangles have the same
vertical angles as those of the original figure, and (I. 23.) the
angles at the base are equal.
If the given base FK be parallel to the corresponding side
AE of the original figure, a more general construction will re-
sult. Join AF, EK and produce them to meet in O; join
1

BOOK VI.
175
OB, OC, and OD, and draw FG, GH, HI, and therefore
IK, parallel to AB, BC, CD, and DE: The figure FGHIK
thus formed is similar to ABCDE. For the triangles KOF,
FOG, GOH, HOI, and IOK are evidently similar to the
triangles EOA, AOB, BOC, COD, and DOE. But these
triangles compose
severally the two
polygons, when
the point O lies
within the origi-
nal figure; and
when that point
of concurrence lies
without thefigure
ABCDE, the si-
milar triangles
B
A
B
I
H
I
KV
A
E
D
I
H
T
K
E
D


IOK and DOE being taken away from the similar compound
polygons FGHIOK and ABCDOE, there remains the figure
FGHIK similar to the original one.
It farther appears, from these investigations, that a rectili-
neal figure may have its sides reduced or enlarged in a given
ratio, by assuming any point O and cutting the diverging
lines OE, OA, OB, OC, and OD in that ratio; the corre-
sponding points of section being joined, will exhibit the fi-
gure required *.
PROP. XXVI. THEOR.
Of similar figures, the perimeters are proportional
to the corresponding sides, and the areas are in the
duplicate ratio of those homologous terms.
Let ABCDE and FGHIK be similar polygons, which have
the corresponding sides AB and FG; the perimeter, or linear
boundary, ABCDE is to the perimeter FGHIK, as AB to
* See Note XLVII.
176
ELEMENTS OF GEOMETRY.
FG, BC to GH, CD to HI, DE to IK, or EA to KF; but
the area of ABCDE, or the contained surface, is to the area
of FGHIK, in the duplicate ratio of AB to FG, of BC to
GH, of CD to HI, of DE to IK, or of EA to KF.
For, by drawing the diagonals AC, AD in the one, and
FH, IF in the other,
these polygons will be

resolved into similar tri-
angles. Whence the se-
veral analogies AB: BC
I

13
G
:: FG: GH, BC: AC
A
E
T
K
::GH: FH, AC: CD
I
:: FH: HI, CD: AD:: HI FI, and AD: DE:: FI: IK;
wherefore, by equality and alternation, AB FG :: BC :
GH:: CD: HI :: DE: IK, and consequently (V. 19.) as one
of the antecedents AB, BC, CD, DE or AE, is to its conse-
quent FG, GH, HI, IK or FK, so is the amount of all those
antecedents, or the perimeter ABCDE, to the amount of all
the consequents, or the perimeter FGHIK.
Again, the triangle CAB is to the triangle HFG (VI. 23.
cor. 1.) in the duplicate ratio of AB to FG,-the triangle
DAC is to the triangle IFH in the duplicate ratio of AC to
FH, or of AB to FG,-and the triangle EAD is to KFI in
the duplicate ratio of AD to FI or of AB to FG; wherefore
(V. 19.) the aggregate of the triangles CAB, DAC, and EAD,
or the area of the polygon ABCDE, is to the aggregate of
the triangles HFG, IFH, and KFI, or the area of the poly-
gon FGHIK, in the duplicate ratio of AB to FG, of BC to
GH, of CD to HI, or of DE to IK.
Cor. Hence also the perimeter ABCDE is to the perimeter
FGHIK, as any diagonal AD to the corresponding diagonal
FI, and the area ABCDE is to the area FGHIK in the du-
plicate ratio of AD to FI.
PROP. XXVII. PROB.
To construct a rectilineal figure that shall be similar
BOOK VI.
177
to one, and equivalent to another, given rectilineal
figure.
Let it be required to describe a rectilineal figure similar to
A, and equivalent to B.
On CD a side of A, describe (II. 9.) equivalent to that
figure, the rectangle CDFE, and on DF describe the rect-
angle DGHF equivalent to the figure B, find (VI. 18.) IK a
mean proportional be-
tween CD and DG,
and on IK construct,
in the same position, a
figure X similar to the
rectilineal figure A 1 ;

this will be likewise
equivalent to B.
For the figures A
B
A
C
I
th
T
H
بھیونڈ
I
X
and X, being similar, must (VI. 26.) be in the duplicate ratio
of their homologous sides CD and IK; and since IK is a
mean proportional between CD and DG, the duplicate ratio
of CD to IK is the same as the ratio of CD to DG (V. 28.);
consequently the figure A is to the figure X as CD to DG,
or (V. 25.) as the rectangle CF to the rectangle DH; but
the figure A is equivalent to the rectangle CF, and therefore
(V. 4.) the figure X is equivalent to the rectangle DH, that
is, to the figure B.
PROP. XXVIII. THEOR.
A rectilineal figure described on the hypotenuse
of a right-angled triangle, is equivalent to similar fi-
gures described on the two sides.
Let ABC be a right-angled triangle; the figure ACFE de-
scribed on the hypotenuse, is equivalent to the similar figures
AGHB and BIKC, described on the sides AB and BC.
M
R
178
ELEMENTS OF GEOMETRY.
|
For draw BD perpendicular to the hypotenuse. And since
(VI.16.cor.1.) AC: AB:: AB: AD,
therefore AC is to AD in the du-
plicate ratio of AC to AB, that
is, (VI. 26.), as the figure on AC
to the figure on AB. For the
same reason, AC is to CD in
the duplicate ratio of AC to BC,
or as the figure on AC to the fi-
gure on BC. Whence (V. 19.
cor. 2.) AC is to the two seg-
H

1
B
G
I
K
D
F
ments AD and CD taken together, as the figure on AC to
both the figures on AB and BC; and the first term of the
analogy being thus equal to the second, the third must be
equal to the fourth (V. 4.), or the figure described on the hy-
potenuse is equivalent to the similar figures described on the
two sides.
PROP. XXIX. THEOR,
The arcs of a circle are proportional to the angles
which they subtend at the centre.
Let the radii CA, CB, and CD intercept arcs AB and BD;
the arc AB is to BD, as the angle ACB to BCD.
For (I. 5.) bisect the angle ACB, bisect again each of its
halves, and repeat the operation indefinitely. An angle ACa
will be thus obtained less than any assignable angle. Let this
angle ACa or BCb (I. 4.) be repeat-
edly applied about the point C, from
BC towards DC'; it must hence, by
its multiplication, fill up the angle
BCD, nearer than any possible dif-
ference. But the elementary angle
ACa being equal to BCb, the corre-
sponding arc Aa is (III. 13.) equal

a
Bb
to Bb. Consequently this arc Aa and its angle ACa, are like
BOOK VI.
179
measures of the arc AB and the angle ACB, and they are
both contained equally in the arc BD and its corresponding
angle BCD. Wherefore AB: BD :: ACB: BCD.
Cor. Hence the arc AB is also to BD, as the sector ACB
to the sector BCD; for these sectors may be viewed as alike
composed of the elementary sector ACa.
PROP. XXX. THEOR.
The circumference of a circle is proportional to
the diameter, and its area to the square of that dia-
meter.
Let AB and CD be the diameters of two circles ;-the cir-
cumference AFG is to the circumference CKL, as AB to CD;
and the area contained by AFG is to the area contained by
CKL, as the square of AB to the square of CD.
For inscribe the regular hexagons AEFBGH and CIKDLM.
Because these polygons are equilateral and equiangular, they
are similar; and consequently (VI. 26. cor.) the diagonal AB is
to the corresponding diagonal CD, as the perimeter AEFBGH
to the perimeter CIKDLM. But this proportion must sub-
sist, whatever be the number of chords inscribed in either se-
micircumference. Insert a dodecagon in each circle between
the hexagon and the circumference, and its perimeter will
evidently approach
nearer to the length
of that circumfe-
rence. Proceeding
thus, by repeated
duplications, the
perimeters of the se-

E
F
H
BC
M
K

D
ries of polygons that
emerge in succession, will continually approximate to the
curvilineal boundary, which forms their ultimate limit. Where-
M 2
180
ELEMENTS OF GEOMETRY.
ENTS
fore this extreme term, or the circumference AEFBGH, is
to the circumference CIKDLM, as the diameter AB to the
diameter CD..
Again, the hexagon AEFBGH (VI. 26. cor.) is to the hexa-
gón CIKDLM in the duplicate ratio of the diagonal AB to
the corresponding diagonal CD, or (V. 24.) as the square of
AB to the square of CD. Wherefore the successive polygons,
which arise from a repeated bisection of the intermediate arcs,
and which approach continually to the areas of their contain-
ing circles, must have still that same ratio. Consequently the
limiting space, or the circle AEFBGH, is to the circle
CIKDLM, as the square of AB to the square of CD.
1 Cor. 1. It hence follows, that if semicircles be described on the
sides AB, BC of a right-angled triangle, and on the hypote-
nuse AC another semicircle be described, passing (III. 22.)
through the vertex B, the crescents AFBD and BGCE are
together equivalent to the triangle ABC. For, by the Pro-
position, the square of AC is to the square of AB, as the circle
on AC to the circle on AB, or (V. 3.) as the semicircle ADBEC
to the semicircle AFB; and, for the same reason, the square
of AC is to the square of BC, as
the semicircle ADBEC to the se-
micircle BGC. Whence (V. 20.)
the square of AC is to the
squares of AB and BC, as the se-
micircle ADBEC to the semicircles
AFB and BGC. But (II. 11.)
A
F

D
B
G
the square of AC is equivalent to the squares of AB and BC,
and therefore (V. 4.) the semicircle ADBEC is equivalent to
the two semicircles AFB and BGC; take away the common
segments ADB and BEC, and there remains the triangle
ABC equivalent to the two crescents or lunes AFBD and
BGCE.
Cor. 2. Hence the method of dividing a circle into equal
portions, by means of concentric circles. Let it be required,
for instance, to trisect the circle of which AB is a diameter.
BOOK VI.
181
Divide the radius AC into three equal parts, from the points
of section draw perpendiculars DF, EG meeting the circum-
ference of a semicircle described
on AC, join CF, CG, and from
C as a centre, with the distances
CF, CG, describe the circles
FHI, GKL: The circle on AB
will be divided into three equal
portions, by those interior circles.
For, join AF and AG: Because
AFC, being in a semicircle, is

G
A
E
D
FORB
L
a right angle (III. 22.), AC is to CD (VI. 16. cor. 1. and
V. 24.), as the square of AC to the square of CF, that is, as
the circle on AB to the circle FHI; but CD is the third part
of AC; wherefore (V. 5.) the circle FHI is the third part of
the circle on AB. In like manner, it is proved, that the circle
GKL is two-third parts of the circle on AB. Consequently,
the intervening annular spaces, and the circle FHI, are all
equal*.
PROP. XXXI. THEOR.
The area of any triangle is a mean proportional
between the rectangle under the semiperimeter and
its excess above the base, and the rectangle under
the separate excesses of that semiperimeter above
the two remaining sides.
The area of the triangle ABC is a mean proportional be-
tween the rectangle under half the sum of all the sides and its
excess above AC, and the rectangle under the excess of that
semiperimeter above AB and its excess above BC.
For produce the sides BA and BC, draw the straight lines
BE, AD, and AE bisecting the angles CBA, BAC, and CAI,
* See Note XLVIII.
182
ELEMENTS OF GEOMETRY.
and let fall the perpendiculars DF, DG, and DH within the
triangle, and the perpendiculars EI, EK, and EL without it.
B
The triangles ADF and ADG, having the angle DAF equal
to DAG, the angles F and G right angles, and the common
side AD,-are (I. 21.) equal; for the same reason, the triangles
BDG and BDH are equal. In like manner, it is proved, that the
triangles AEI and AEK are equal, and the triangles BEI and
BEL. Whence the triangles CDH and CDF, having the
side DH equal to DF, the side DC common, and the right
angle CHD equal to CFD,-are (I. 22.) equal; and, for the
same reason, the triangles CEK and CEL are equal. The
perimeter of the triangle ABC
is therefore equal to twice the
segments AF, FC, and BG ;
consequently BG is the ex-
cess of the semiperimeter
above the base AC, and AG
is the excess of that semipe-
rimeter—or of the segments
BH, HC, and AG,-above
the side BC. But the sides
AB and BC, with the seg-

Ꮐ
H
C
KF
I
E
L
ments AK and CK, or AI and CL, also form the perime-
ter; whence, BI being equal to BL, the part AI is the excess
of the semiperimeter above the side AB.
Now, because DG and EI, being perpendicular to BI, are
parallel, BG: DG :: BI: EI (VI. 2.), and consequently
(V. 25. cor. 2.) BIX BG: BIX DG:: DGX BI: DG × EI.
But since AD and AE bisect the angle BAC and its adjacent
angle CAI, the angles GAD and EAI are together equal to
a right angle, and equal, therefore, to IEA and EAI; whence
the angle GAD is equal to IEA, and the right-angled triangles
DGA and AIE are similar. Wherefore (VI. 12.) DG : AG
:: AI: EI, and (V. 6.) DG × EI=AG× AI; consequently
BIX BG: DGX BI:: DGX BI: AGX AI. But the tri-
BOOK VI.
183
angle ABC is composed of three triangles ADB, BDC, and
CDA, which have the same altitude; and therefore its area
is equal to the rectangle under DG and half their bases
AB, BC, and AC, or the semiperimeter BI. Whence the
area of the triangle ABC is a mean proportional between the
rectangle under BI and its excess above AC, and the rect-
angle under its excess above BC and that above AB.
Cor. If the area of a triangle be expressed by A, its sideș
by a, b, and c, and the semiperimeter by s; then s(s—a) : A
:: A : (s—b)(s—c), and consequently A²=s(s—a)(s—b)(s—c),
and A= √ (s(s—a)(s—b)(s—c)) *.
PROP. XXXII. PROB.
Given the area of an inscribed, and that of a cir-
cumscribed, regular polygon; to find the areas of
inscribed and circumscribed regular polygons, having
double the number of sides.
Let TKNQ and HBDF be given similar inscribed and cir-
cumscribed rectilineal figures; it is required thence to deter-
mine the surfaces of the corresponding inscribed and circum-
scribed polygons AKCNEQGT and VILMOPRS, which
have twice the number of sides.
From the centre of the circle, draw radiating lines to all
the angular points. It is evident that the triangles ZXK and
ZAB are like portions of the given inscribed and circumscri-
bed figures TKNQ and HBDF; and that the triangle ZAK,
and the quadrilateral figure ZAIK are also like portions of
the derivative polygons AKCNEQGT and VILMOPRS.
And since XK is parallel to AB, ZX: ZA :: ZK : ZB
(VI. 2.); but ZX is to ZA as the triangle ZXK is to the tri-
* See Note XLIX.
181
ELEMENTS OF GEOMETRY.
angle ZAK (V. 25. cor. 2.), and, for the same reason, ZK is
to ZB as the triangle ZAK is to the triangle ZAB; whence
ZXK : ZAK :: ZAK: ZAB, and consequently the deriva-
tive inscribed polygon AKCNEQGT is a mean proportional
between the inscribed and cir-
cumscribed figures TKNQ
and HBDF.
Again, because ZI bisects
the angle AZB, ZA is to ZB,
or ZX is to ZK, as AI to IB
(VI. 11.), and consequently
(V. 25. cor. 2.) the triangle
XZK is to the triangle AZK,
as the triangle AZI to the
triangle IZB. Hence the
inscribed figure TKNQ is to
B
V
ITL
K
I

E
T
F
C
R
its derivative inscribed figure AKCNEQGT as the triangle
AZI to the triangle IZB; wherefore (V. 11. and 13.) TKNQ
and AKCNEQGT together are to twice TKNQ, as the tri-
angles AZI and IZB, or AZB, to twice the triangle AZI,
or the space AIKZ,-that is, as HBDF to VILMOPRS.
And thus the two inscribed polygons are to twice the simple
inscribed polygon, as the surface of the circumscribing poly-
gon, to the surface of the derivative circumscribing polygon
with double the number of sides.
rence.
Cor. Hence the area of a circle is equivalent to the rectangle
under its radius and a straight line equal to half its circumfe-
For the surface of any regular circumscribing poly-
gon, such as VILMOPRS, being composed of a number of
triangles AZI, which have all the same altitude ZA, is equi-
valent (II. 6.) to the rectangle under ZA and half the sum of
their bases, or the semiperimeter of the polygon. But the
circle itself, as it forms the ultimate limit of the polygon, must
have its area, therefore, equivalent to the rectangle under the
radius ZA, and the semicircumference ACE.
BOOK VI.
185
octagon is
Scholium. This Proposition furnishes the best elementary
method of approximating to the numerical expression for the
area of a circle. Suppose the radius of a circle to be denoted
by unit: The surface of the circumscribing square will be ex-
pressed by 4, and consequently (IV. 16. cor.) that of its in-
scribed square by 2.
Wherefore the surface of the inscribed
2x4=2,8284271; and the surface of the cir-
cumscribing octagon is found by the analogy, 2+2,8284271:
2×2::4:3,3137085. Again, ✔ (2,8284271 × 3,3137085)=
3,0614674, which expresses the area of the inscribed polygon
of 16 sides; and 2,8284271+3,0614674: 2×2,8284271, or
5,8898945: 5,6568542:: 3,3137085: 3,1825979, which de-
notes the area of the circumscribing polygon of 16 sides.
Pursuing this mode of calculation, by alternately extracting a
square root and finding a fourth proportional, the following
Table will be formed, in which the numbers expressing the
surfaces of the inscribed and circumscribed polygons con-
tinually approach to each other, and consequently to the
measure of their intermediate circle.

Number of
Sides.
Area of the
inscribed
Polygon.
Area of the
circumscribing
Polygon.
48
4
2,0000000
4,0000000
8 2,8284271 3,3137085
16
3,0614674
3,1825979
32
3,1214451
3,1517249
64 3,1365485
3,1441184
128
3,1403311
3,1422236
256 3,1412772
3,1417504
512
3,1415138
3,1416321
1024
3,1415729
3,1416025
2048
3,1415877
3,1415951
4096
3,1415914
3,1415933
8192
3,1415923 3,1415928
16384
3,1415925
3,1415927
32768 3,1415926 3,1415926
186
ELEMENTS OF GEOMETRY.
Hence 3,1415926 is the nearest expression, consisting of se-
ven decimal places, for the area of a circle whose radius is 1. But
the semicircumference in this case denoting also the surface,
the same number must represent the circumference of a circle
whose diameter is 1. Consequently, if D denote the diame-
ter of any circle, the circumference will be expressed approxi-
mately, by 3,1415926 x D; whence the area will be D² x
3,1415926, or D² ×, 78539815 *.
Since the four last decimals 5926 come so near to 6000, it
will, in most cases, be sufficiently accurate to reckon the cir-
cumference equal to D× 3,1416, and its area equal to D² ×
,7854. But other approximations, expressed in lower num-
bers, may be found, by help of Prop. 28. Book V. For
m=3, n=7, p=16, and q=11; whence, remounting suc-
cessively from these conditional equalities, the ratio of the dia-
meter to the circumference of a circle is denoted in progres-
sion, by 1: 3—by 7:22-by 113: 355-and by 1250: 3927.
Hence also the circle is to its circumscribing square nearly—
as 11 to 14, or still more nearly-as 355 to 452.
*See Note L.
APPENDIX.
THE Constructions used in Elementary Geometry,
were effected, by the combination of straight lines
and circles. Many problems, however, can be re-
solved, by the single application of the straight line
or the circle; and such solutions are not only inte-
resting, from the ingenuity and resources which they
display, but may, in a variety of instances, be em-
ployed with manifest advantage. This Appendix is
intended to exhibit a selection of Geometrical Pro-
blems, resolved by either of those methods singly.
It is accordingly divided into Two Parts, corre-
sponding to the rectilineal and the circular con-
structions.
188
APPENDIX.
PART I.
Problems resolved by help of the Ruler,
or by Straight Lines only.
PROP. I. PROB.
To bisect a given angle.
Let BAC be an angle, which it is required to bisect, by
drawing only straight lines.
In AB take any two points D and E, from AC cut off AF
equal to AD and AG to AE, draw EF and DG, crossing in
the point H: AH will bisect the angle BAC.
For the triangles EAF and DAG, having the sides EA
and AF equal by construction to GA and AD, and the con-
tained angle DAG common to both, are equal (I. 3.), and
consequently the angle AEF is
equal to AGD. And since AE
is equal to AG, and the part
AD to AF, the remainder DE
must be equal to FG; where-
fore the triangles DEH and
HGF, having the angle at E
equal to that at G, the vertical
angles at H equal, and also their
opposite sides DE and FG, are

B
D
F
H
E
G
equal (I. 21.); and hence the side DH is equal to FH. A-
gain, the sides AD and DH are equal to AF and FH, and
AH is common to the two triangles AHD and AHF, which
are therefore equal (I. 2.), and consequently the angle DAH
is equal to FAH.
PART I.
189
PROP. II. PROB.
To bisect a given finite straight line.
Let it be required to bisect AB, by a rectilineal construc-
tion.
Draw AK diverging from AB, and make AC=CD=DE,
join EB and continue it beyond B till BF be equal to BE,
and lastly join FC; which will bisect AB in the point G.
For draw BH parallel to AE...
And because BD evidently bisects
the sides EC and EF of the tri-
angle CEF, it is parallel to the
base CF (II. 4.); wherefore BDCH
is a parallelogram, which has (I. 27.)
its opposite sides BH and CD e-
qual. But AC being parallel to
BH, the angles GAC and GCA
are equal to GBH and GHB, and
the side AC, being made equal to
CD, is hence equal to its corre-
sponding interjacent side BH;
D
T

K
:
whence the triangles AGC and BGH are equal (I. 21.), and
therefore AG is equal to BG.
PROP. III. PROB.
Through a given point, to draw a line parallel to
a given straight line.
Let it be required, by a rectilineal construction, to draw
through C a parallel to AB.
*
190
APPENDIX.
C
G
In AB take any two points D and F, .join CD, which
produce till DE be equal to it; a-
gain join E with the point F, anc
continue this till FG be equal to
EF: Then CG, being joined, will
be parallel to AB.
For, since AB or DF evidently
bisects the sides EC and EG of the
triangle CEG, it must be parallel
to the base CG (II. 4.).

A
D
FB
PROP. IV. PROB.
ID
From a point in a given straight line, to erect a
perpendicular.
Let C be a given point, from which it is required, by help
of straight lines merely, to erect a perpendicular to AB.
In AB, having taken any
point D, draw DE equal to
DC and inclined to AB, join
EC and produce it until CG
be equal to CD or DE, make
CF equal to CE, join FG and
produce this till GH be equal
to GC: Then CH will be per-
pendicular to AB.
For the triangles DCE and
GCF, having the sides DC, CE
A D
E
H

G
F B
equal to GC, CF, and the contained angles vertical at C, are
equal (I. 3.); whence FG=CD=CG=GH. The point G
is therefore the centre of a semicircle which would pass
through F, C, H, and consequently the angle FCH is a right
angle (III. 22.), or CH is perpendicular to AB.
PART I.
191
PROP. V. PROB.
To let fall a perpendicular upon a given straight
line, from a point without it.
Let C be a given point, from which it is required, by a
rectilineal construction, to let fall a perpendicular to AB.
In AB take any point D, draw DF obliquely, and make
DE=DF=DG, join FE
and produce it until EH
be equal to EG, make
EI-EF, join HI, and
(Appendix, Part I. Prop.
3.) draw CK parallel to it:
CK is the perpendicular
required.
For the point D being

·
H
AG
D
E K IB
F
obviously the centre of a semicircle passing through G, F,
and E, the angle GFE is a right angle; and the triangles
EGF, EHI, having the sides GE, EF equal to HE, EI, and
their contained angles vertical,-are equal (I. 3.), and conse-
quently the angle HIE is equal to GFE, or is a right angle;
but since CK and HI are parallel, the angle CKA is equal to
HIE (I. 23.), and therefore is also a right angle, or CK is
perpendicular to AB.
192
APPENDIX.
PART II.
"
Geometrical Problems resolved by means of Compasses,
or by the mere description of Circles.
PROP. I. PROB.
!
To repeat a given distance in the same direction.
Let A and B be two given points; it is required to find, by
means of compasses only, a series of equidistant points in the
same extended line.
From B as a centre, with the given distance BA, describe
a portion of a circle, in which inflect that distance three times
to C; from C, with the same
radius, describe another circle,

and insert the triple chords
P
فظ
E
to D; repeat that process from
D, E, &c. The equidistant
points A, B, C, D, E, &c. will all lie in the same straight
line.
For, by this construction, three equilateral triangles are
formed about the point B, and consequently (I. 32. cor. 1.)
the whole angle ABC, made by the opposite distances BA
and BC, is equal to two right angles, or ABC is a straight
line. The same reason applies to the successive points, D, E,
&c.
PROP. II. PROB.
To find the direction of a perpendicular from a
given point to the straight line joining another given
point.
PART II.
193
Given the points A and B; to find a third point, such tha.
the straight line connecting it with B shall be at right angles
to BA.
From A and B, with any convenient distance, describe two
arcs intersecting in C, from which, with
the same radius, describe a portion of a
circle passing through the points A and
B, and insert that radius three times
from A to D: BD is perpendicular to
BA.
For it is evident, from the last Propo-
sition, that the arc ABD is a semicir-
cumference, and consequently that the angle ABD contained
in it is a right angle.
The construction would be somewhat simplified, by taking
the distance AB for the radius.

PROP. III. PROB.
To find the direction of a perpendicular let fall
from a given point upon the straight line which con-
nects two given points.
Let C be a point, from which a perpendicular is to be let
fall upon the straight line joining A and B.
From A as a centre, with the dis-
tance AC, describe an arc, and from
B as a centre, with the distance BC,
describe another arc, intersecting the
former in the point D: CD is per-
pendicular to AB.
For CAD and CBD are evident-
ly isosceles triangles, and consequent-
ly (I. 7.) their vertices must lie in a
A
B
straight line AB which bisects their base CD at right angles.
N
194
APPENDIX.
PROP. IV. PROB.
To bisect a given distance.
Let A and B be two given points; it is required to find the
middle point in the same direction.
From B as a centre, with the radius BA, describe a semi-
circle, by inserting that distance successively from A to C, D,
and E; from A as a centre, with the distance AE, describe a
portion of a circle FEG, in which, and from E, inflect the
chords EF and EG equal to EC; and from the points F and
G, with the same radius EC describe
arcs intersecting in H: This point
bisects the distance AB.
For, by the first Proposition, the
points A, B, and E extend in a
straight line; but the triangles FAG,
FHG, and FEG, being evidently
isosceles, their vertices A, H, and E
(I. 7.) must lie in a straight line;
whence the point H lies in the di-
rection AB. Again, because EFH
is an isosceles triangle, AF-HF

H
C
-1-
E
B
(II. 23. cor.)=EA. AH; that is, AE-EC or (IV. 20. cor. 2.)
AB²=EA.AH. Wherefore, since EA is double of AB, the
segment AH must be its half.
PROP. V. PROB.
To trisect a given distance.
Let it be required to find two intermediate points that are
situate at equal intervals in the line of communication AB.
Repeat (App. II. 1.) the distance AB on both sides to C
and D, from these points, with the radius CD, describe the
PART II.
195
arcs EDF and GCH, from D and C inflect the chords DE
and DF, CG and CH, all
equal to DB, and, with the


E
same distance and from the
points. E and F, G and H,
describe arcs intersecting
in I and K: The distance
AB is trisected by points I
and K.
For it may be demon-
strated, as in the last pro-
position, that the points I
and K lie in the same di-
rection AB. In like man-
ner, it appears (II. 23. cor.)
that DG-KG*= CD.DK,
D
C
AIRB
or 9AB2-4AB or 5AB=3AB.DK; and consequently
5AB=3DK, or 2AB=3AK, and AB=3BK.
same reason, AB=3AI.
PROP. VI. PROB.
But, for the
To cut off any aliquot part of a given distance.
Suppose it were required to cut off the fifth part of the dis-
tance between the points A and B.
Repeat (App. II. 1.) the distance AB four times, to F;
from F, with the radius FA, describe the arc GAH; infiect
the chords AG and AH
equal to AB, and, with
that radius and from the
points G and H, describe
G
AT B
B ċ
arcs intersecting in I: AI
is the fifth part of the line
of communication AB.
For, as before, the point I is situate in AB. But since
N 2
- 196
APPENDIX.
AGI is evidently an isosceles triangle and AF is equal to FG,
it follows (II. 23. cor.) that AG²=AF.AI, and consequently
AB²=5AB.AI; whence AB=5AI.
PROP. VII. PROB.
To divide a given distance by medial section.
Let it be required to cut the distance AB, such that
BHBA.AH.
From B describe a circle with the radius BA, which insert
successively from A to D, E, C, and F; from the extremities
of the diameter AC and with the double chord AE, describe
two arcs intersecting in G; and, from the points E and F
with the distance BG, describe other two arcs intersecting in
H: This is the point of medial section.
For it is evident that this point
H lies in the straight line AB.
And because the triangles AGB,
CGB have their sides respective-
ly equal, the angle ABG (I. 2.)
is a right angle, and consequent-
ly (II. 11.) AG*= AB² + BG²;
but AG=AE, and AE²=3AB²
(IV. 20. cor. 2.); wherefore
3AB2AB+BG2, and BG² =
2AB2.
Now since BE-EC, it
Ai
&

B
E
follows (II. 23. cor.) that HE-BE-CH.HB; but HE-
BE* — BG² — BE² = AB², and therefore ABCH.HB.
Whence CH is cut by a medial section at B, and consequent-
ly (II 22. cor. 1.) its greater segment BC or AB is likewise
divided medially at H by the remaining portion BH.
PROP. VIII. PROB.
To bisect a given arc of a circle.
PART II.
197
Let it be required to bisect the arc AB of a circle whose
centre is C.
From the extremities A and B, with the radius AC, de-
scribe opposite arcs, and from the centre C inflect the chord
AB to D and E; from these points, with the distance DB
describe arcs intersecting in F; and from D or E, with the
distance CF, cut the given arc AB in G: AB is bisected in
that point.
For the figures ABCD and ABEC being evidently rhom-
boids, DC and CE are parallel to AB, and hence constitute one
straight line; conse-
quently the triangles
DFC and EFC ha-
ving their correspond-
ing sides equal, the
angle DCF is a right
angle, and (II. 11.)
*
B
P-

DF² = DC² + CF³.
D
But, in the rhomboid
ABCD, DB+CA*
=2DC²+2CB² (II. 28.), or DB²=2DC² + CB²; and since
DB=DF, 2DC²+CB²=DC² + CF, whence DC²+CB² =
CF², or DC²+CG²=DG², and therefore (II. 12.) DCG is
a right angle. And because CG is perpendicular to DC, it
is likewise (I. 23.) perpendicular to AB, and the triangles
CAP and CBP are equal (I. 22.) and the angle ACG equal
to BCG; whence (III. 13.) the arc AG-BG.
PROP. IX. PROB.
Given two points, to find the intersection of their
connecting line with a given circumference.
1. Let one of the points be the centre of the circle.
198
L
i
APPENDIX.
T
Take any point D within the circle, and from A, with the
distance AD describe an arc cutting the
circumference in E and F, bisect the arc
EGF in G (App. II. 8.), and determine
the semicircle GEH (App. II. 1.): G
and H are the points of intersection of
the straight line AGH.

E
H
For the triangles AEB and AFB
have their sides respectively equal, and consequently the angle
ABE is equal to ABF (I. 2.); wherefore (III. 13.) the arc
EG is equal to GF, or the straight line AH must bisect the
arc EF.
2. Let neither point lie in the centre of
the circle.
From A and B, with the distances AC
and BC, describe arcs intersecting in D,
from which, with the radius CE, cut the
circumference in E and F: The straight
line'AB would extend through these points.
For the triangles CAD and CBD be-
ing isosceles, it appears from Book I.
Prop. 7., that their vertices A and B lie
in a perpendicular passing through the
middle of the common base CD, and con-
sequently the points E and F, which are
A,

Df
F
D
ex
C
B
C

vertices of the isosceles triangles CED and CFD, must like-
wise occur in the same straight line.
PROP. X. PROB.
To find the sum or difference of two given dis-
tances.
Let AB and CD be two distances, of which it is required
to determine the sum and the difference.
PART II.
199
From A with the distance CD
describe a circle, cut the circumfe-
rence in E and F by any are de-
scribed from B, bisect the arc EF
(App. II. 8.) on both sides at G and
H; BG will be the sum of the two
distances, and BH their difference.
G
C
D

E
A
H R
For GB, bisecting the chord EF at right angles, must pass
through the centre A, and consequently the radius AG or
CD is, on either side, added or taken away from AB.
PROP. XI. PROB.
To find the centre of a circle.
Assume an arc AB greater than a quadrant, and from one
extremity B, with the distance BA, describe a semicircle
ADC, cutting the given circumference in D; from the points
B and C, with the distance CD, describe arcs intersecting in
E, and, from that point with the same distance, describe an
arc cutting ADC in F; and lastly, from the points A and B,
with the distance AF, describe arcs intersecting in G: This
point is the centre of the circle ADB.
E
For the isosceles triangles BEC, BEF, being evidently
equal, the angle FBC is equal to both the angles at the base;
but FBC is (I. 32. El.) equal to the interior angles BAF and
BFA of the isosceles triangle
ABF, and hence that triangle
is similar to BEF. Where-
fore BE : BF : : BA : AF, or
CD: BD :: BA: AG; con-
sequently the isosceles triangles
CBD and BGA are similar,
and the angle BCD is equal
to GBA; BG is, therefore,
parallel to CD, and hence

D
B
(I. 32. El.) the angle BDC, or BCD, is equal to GBD. The
20
APPENDIX.
triangles BGA and BGD, having thus the side BA equal to
BD, BG common, and equal contained angles GBA and
GBD, are (I. 3. El.) equal, and therefore the side GA is
equal to GD. The point G, being thus equidistant from
three points, A, D, and B in the circumference, is hence
(III. S. cor.) the centre of the circle.
PROP. XII. PROB.
To divide the circumference of a given circle
successively into four, eight, twelve, and twenty-four
equal parts.
1. Insert the radius AB three times from A to D, E, and
C; from the extremities of the diameter AC, and with a dis-
tance equal to the double chord AE, describe arcs intersect-
ing in the point F; and from A, with the distance BF, cut
the circumference on opposite sides at G and H : AG, GC,
CH, and HA are quadrants.
For, as before, AF-AE-3AB2; and the triangle ABF
being right-angled, 3AB²=AF²=AB²+BF², and therefore
BF2 AG 2AB; whence (II. 12.) ABG is a right angle,
and AG a quadrant.
2. From the point F with the
radius AB, cut the circle in I and
K, and from A and C inflect the
chord AI to L and M; the cir-
cumference is divided into eight
equal portions by the points A, I,
G, K, C, M, H, and L.
For BF, being equal to 2AB²,
is equal to the squares of BI and

A
J
B
C
ΔΙ
I
IF, and consequently BIF is a right angle; but the triangle
BIF is also isosceles, and therefore the angle IBF at the
base is half a right angle; whence the arc IG is an octant.
PART II.
201
3. The arc DG, on being repeated, will form twelve equal
sections of the circumference.
For the arc AD is the sixth or two-twelfth parts of the cir-
cumference, and AG is the fourth or three-twelfths; conse-
quently the difference DG is one-twelfth.
4. The arc ID is the twenty-fourth part of the circumfe-
rence.
For the octant AI is equal to three twenty-fourths, and
the sextant AD is equal to four twenty-fourths; their dif-
ference ID is hence one twenty-fourth part of the circumfe-
rence.
PROP. XIII. PROB.
To divide the circumference of a given circle suc-
cessively into five, ten, and twenty equal parts.
Mark out the semicircumference ADEC, by the triple in-
sertion of the radius, from A and C, with the double chord
AE, describe arcs intersecting in F, from A, with the dis-
tance BF, cut the circle in G and K, inflect the chords GH
and GI equal to the radius AB, and, from the points H and
I, with distance BF or AG,
describe arcs intersecting in L.
It is evident from App. II.
7, that BL is the greater seg-
ment of the radius BH divided
by a medial section; wherefore
(IV. 22. cor. 2. El.) AL is equal
to the side of the inscribed pen-
tagon, and BL, to that of the
decagon inscribed in the given
circle. Hence AL may be in-

A
HX
D
M K
E
C
flected five times in the circumference, and BL ten times;
and consequently the arc MK, or the excess of the fourth
202
APPENDIX.
above the fifth, is equal to the twentieth part of the whole
circumference.
PROP. XIV. PROB.
From a given side to trace out a square.
Let the points A and B terminate the side of a square,
which it is required to trace.
From B as a centre describe
the semicircle ADEC, from A
and C, with the distance AE,
describe arcs intersecting in F,
from A, with the distance BF,
cut the circumference in G, and
from A and G, with the radius.
D
A
B
C
AB, describe arcs intersecting in H: The points H and G
are corners of the required square.
For (App. II. 12.) the angle ABG is a right angle, and
the distances AB, AH, HG, and GB, are, by construction,
all equal.
PROP. XV. PROB.
Given the side of a regular pentagon, to find the
traces of the figure.
From B describe through A the circle ADECF, in which
the radius is inflected four times, from A and C with the
double chord AE describe arcs intersecting in G, from E
and F, with the distance BG, describe arcs intersecting
in H, from A, with the radius AB, describe a portion of a
circle, inflect BH thrice from B to L and from A to O, and
lastly from L and O, with the radius AB, describe arcs in-
tersecting in P: The points A, L, P, O, B mark out the
polygon.
PART II,
203
For, from App. II. 7, it
is evident that BH is the
greater segment of the dis-
tance AB divided by a me-
dial section. Consequently
(IV. 3. El.) the isosceles tri-
angles BAI, IAK, KAL,
※

K
D
M
E
C
H
B
ABM, MBN, and NBO,
have each of the angles at
the base double their verti-
F
cal angle. Wherefore the angles BAL and ABO are each
of them six-fifths of a right angle (IV. 4. cor.), and hence
(I. 33. cor.) the points L and O are corners of the pentagon;
but P is evidently the vertex of the pentagon, since the sides
LP and OP are each equal to AB.
Scholium. The pentagon might also have been traced, as
in Book IV. Prop. 5, by describing arcs from A and B with
the distance HC, and again, from their intersection P, and
with the radius AB, cutting those arcs in L and O. It is
likewise evident, from Book IV. Prop. 8, that the same pre-
vious construction would serve for describing a decagon, P
being made the centre of a circle in which AB is inflected
ten times.
PROP. XVI. PROB.
The side of a regular octagon being given, to
mark out the figure.
Let the side of an octagon terminate in the points A and
B; to find the remaining corners of the figure.
On AB describe the two semicircles AEFC and BEGD;
with the double chord AF, and from A, C and B, D describe
arcs intersecting in H, I; from these points, with the radius
AB, cut the semicircles in K, L: on HI describe the square
HMNI, by making the diagonals HN, IM equal to BH, and
204
APPENDIX.
!
the sides equal to AB;
and, on MH and NI,
describe the rhombusses
MOKH and NPLI: The
points A, B, K, O, M, N,
P, and L, are the several
corners of the octagon.
D
N
M
E
HX

A
B
For (by App. II. Prop.
12.) BH, AI are both of
them perpendicular to BA,
and BKH, ALI are right angled isosceles triangles; HI is
therefore parallel to BA, and HMNI, consisting of tri-
angles equal to BKH, is a square; whence all the sides AB,
BK, KO, OM, MN, NP, PL, and LA of the octagon are
equal: But they likewise contain equal angles; for ABK,
composed of ABH and HBK, is equal to three half right
angles, and BKO, by reason of the parallels BH and KO,
being the supplement of HBK, is also equal to three half right
angles. In the same manner, the other angles of the figure
may be proved to be equal.
PROP. XVII.
PROB.
On a given diagonal to describe a square.
Let the points A and B be the opposite corners of a square
which it is required to trace.
From B as a centre describe the semicircle ADEC, from
A and C with the double chord AE describe arcs intersect-
ing in F, from C with the distance BF describe an arc and
cut this from A with the radius AD
in G, and lastly from B and A with the
distance BG describe arcs intersect-
ing in H and I: ABHI is the requi-
red square.
For, in the triangle AGC, the
straight line GB bisects the base,
and consequently (II. 25.) AG²+
EX

D
E
HXG
B
CG²=2AB²+2BG²; but, (by App. II. Prop. 12,) CG²=
PART II.
205
BF²=2AB²; whence AGAB²=2BG*, and (II. 12.)
AHB is a right angle; and the sides AH, HB, BI, and IA
being all equal, the figure is therefore a square.
PROP. XVIII. PROB.
Two distances being given, to find a third
tional.
propor-
Let it be required to find a third proportional to the dis-
tances AB and CD.
From any point E, and
with the distance AB, de-
scribe a portion of a cir-
cle, in which inflect FG
equal to CD, and from
G, with that distance, de-
scribe the semicircle FHI;
HI is the third propor-
tional required.
For the angles GEH
and IGH are each of them
double the angle GFH or
E
.AL.
Ci.
F
G
AI
CI-
•
F
G
H
-B
H
D


IFH at the circumference (III. 17. El.); whence the triangles
GEH and IGH must also have the angles at the base
equal, and are consequently similar: Wherefore (VI. 12. El.)
EG: GH:: GH : HI.
If the first term AB be less than half the second term CD,
this construction, without some help, would evidently not
succeed. But AB may be previously doubled, or assumed
4, 8, or 16 times greater, so that the circle FGH shall always
cut FHI; and in that case, HI, being likewise doubled, or
taken 4, 8, or 16 times greater, will give the true result.
206
APPENDIX.
1
PROP. XIX. PROB.
To find a fourth proportional to three given dis-
tances.
Let it be required to find a fourth proportional to the dis-
tances AB, CD, and EF.
From any point G, describe
two concentric circles HI and
KL with the distances AB and
EF, in the circumference of the
first inflect HI equal to CD, as-
sume any point K in the second
circumference, and cut this in L
by an arc described from I with
the distance HK; the chord LK
is the fourth proportional requí-
red.
I
Λι
CI-
E
·
B
·
·
ID

G
For the triangles ILG and HKG are equal, since their
corresponding sides are evidently equal; whence the angle
IGL is equal to HGK, and taking away HGL, the angle
IGH remains equal to LGK; consequently the isosceles tri-
angles GIH and GLK are similar, and GI : IH :: GL:LK,
that is, AB CD :: EF: LK.
If the third term EF be more than double the first AB, this
construction, it is obvious, will not answer without some mo-
dification. It may, however, be made to suit all the variety
of cases, by multiplying equally AB and the chord LK, as in
the last proposition.
PROP. XX. PROB.
To find a mean proportional between two given
distances.
Let AB and CD be the two distances. To AB add
PART II.
207
(App. II. 10.) BE equal to CD, bisect (App. II. 4.) AE in F,
make BG equal to FB, from F
describe the semicircumference
AHE, and, with the same ra-
CID
H

dius FE and from G as a centre,
intersect it in H; BH is the mean
proportional required.
A
1
B EG
For (I. 5. cor.) it is evident, that BH is perpendicular to AE,
and (III 32. cor.), that BH²=AB.BE; whence (V. 6.) AB:
BH :: BH: BE, or CD.
PROP. XXI. PROB.
To find the linear expressions for the square roots
of the natural numbers, from one to ten inclusive.
This problem is evidently the same as, to find the sides of
squares which are equivalent to the successive multiples of the
square constructed on the straight line representing the unit.
Let AB, therefore, be that measure: And from B as a centre,
describe a circle, in which inflect the radius four times, from
A to C, D, E, and F; from the opposite points A and E,
with the double chord AD, describe arcs intersecting in G
and H,-with the same distance, and from the points D, F,
describe arcs intersecting in I,—and, with still the same dis-
tance and from E, cut the
circumference in K; and
from A and K, with the
radius AB, describe arcs
intersecting in L: Then
will AK' 2AB², AD²
=
= 3AB², AE² = 4AB²,
IK² = 5AB², IG² =
6AB²,IC²=7AB²,GH²
=8AB', IA² = 9AB',
and IL-10AB².
G
It
K

A
B
E
L
For, in the isosceles triangles ACB and BDE, the perpen-
208
i
APPENDIX.
diculars CO and DP must bisect the bases AB and BE; and
the triangle ADI being likewise isosceles, IPAP, and con-
sequently IBAE=2AB. But, from what has been former-
ly shown, it is evident that AK²=2AB² and AD²=3AB²;
and since AE=2AB, AE²=4AB. In the right angled tri-
angles IBK and IBG, IK²=IB²+BK²=4EB²+BK²
5AB²,IG²=IB²+BG²=4AB²+2AB²=6AB*; but (II. 26.)
IC²=IB²+BC2+IB.2BO=4AB+ AB² +2AB27AB".
Again, GH being double of BG, GH²=4×2AB² =8AB²,
and AI being the triple of AE, AI²=9AB²; and lastly, IAL
being a right angled triangle, IL²=IA²+AL²=9AB²+
AB210AB2.
If AB, therefore, denote the unit of any scale, it will follow,
that AK=√2, AD≈√3, IK=√5, IG=√6, IC=√7,
GH=8, and IL
10.
2
1
1
GEOMETRICAL ANALYSIS.
!
!
!
1
!
1
1
!
量
​5.
:
!
f
4
GEOMETRICAL ANALYSIS.
ANALYSIS is that procedure by which a proposi-
tion is traced up, through a chain of necessary de-
pendence, to some known operation, or some ad-
mitted principle. It is alike applicable to the in-
vestigation of truth contemplated in a theorem, or to
the discovery of the construction required for a pro-
blem. Analysis, as its name indeed imports, is thus
a sort of inverted form of solution. Assuming the
hypothesis advanced, it remounts, step by step, till
it has reached a source already explored. The re-
verse of this process constitutes Synthesis, or Compo-
sition,—which is the mode usually employed for ex-
plaining the elements of science. Analysis, there-
fore, presents the medium of invention; while syn-
thesis naturally directs the course of instruction *.
* See Note LI.
02
•
BOOK I.
DEFINITIONS.
1. Quantities are said to be given, which are either exhibit-
ed, or may be found.
2. A ratio is said to be given, when it is the same as that
of two given quantities.
3. Points, lines, and spaces, are said to be given in position,
if they have always the same situation, and are either actually
exhibited, or may be found.
4. A circle is given in position, when its centre is given;
it is given in magnitude, if its radius be given.
5. Rectilineal figures are said to be given in species, when
figures similar to them are given.
BOOK I.
213
PROP. I. PROB.
From two given points, to draw straight lines,
making equal angles at the same point in a straight
line given in position.
Let A, B be two given points, and CD a straight line
given in position; it is required to draw AG and GB, so that
the angles AGC and BGD shall be equal.
ANALYSIS.
C-
From B, one of the given
points, let fall the perpendicu-
lar BE, and produce it to meet
AG, or its extension in F.
The angle BGE, being equal
to AGC, is equal to the verti
cal angle FGE, the right angle
BEG is equal to FEG, and the
side GE is common to the tri-
angles GBE and GFE, which
(I. 21. El.) are therefore equal,
and hence the side BE is equal
to FE. But the perpendicu- č
lar BE is given, and conse-
quently FE is given both in po-
sition and magnitude; whence

A
F
B
B
D
E
F

D
E
the point F is given, and therefore G the intersection of the
straight line AF with CD.
COMPOSITION.
Let fall the perpendicular BE, and produce it equally on
the opposite side, join AF meeting CD in G; AG and BG
are the straight lines required.
214
GEOMETRICAL ANALYSIS.
For the triangles GBE and GFE, having the side BE e-
qual to FE, GE common, and the contained angle BEG e-
qual to FEG, are (I. 3. El.) equal; and consequently the
angle BGE is equal to FGE or AGC.
PROP. II. PROB.
Through a given point, to draw a straight line at
equal angles with two straight lines given in position.
Let A be the given point, and CB, CD the straight lines
which are given in position.
ANALYSIS.
\TI
B
E
Draw (I. 24. El.) CH parallel to FE, and produce DC. The
exterior angle GCH (I. 32. El.) is
equal to CFE, and ECH is equal
to the alternate angle CEF; but
the angle CFE is equal to CEF,
and consequently GCH is equal
to ECH, and the angle GCE is
thus bisected by the straight line
CH. Wherefore (I. 5. El.) CH

G.
C
F D
is given in position, and hence (I. 24. El.) the parallel EF is
also given.
COMPOSITION.
Bisect (I. 5. El.) the adjacent angle GCB by the straight
line CH, and parallel to this draw EF (I. 24. El.) through
the given point A; the angle CEF is equal to CFE. For
these angles are equal to the exterior and alternate angles
GCH and ECH, and are consequently equal to each other.
BOOK I.
215
PROP. III. PROB.
Through a given point, to draw a straight line,
such that the segments intercepted by perpendicu-
lars let fall upon it from two given points, shall be
equal.
The points A, B, and C being given,—to draw a straight
line FE, so that the parts CF and CE, cut off by the perpen-
diculars AF and BE, shall be equal.
ANALYSIS.
Produce AC to meet the extension of BE in D. The
right angled triangles AFC and

DEC, having the vertical angle
ACF equal to DCE, and the side
:
CF equal to CE, are (I. 21. El.)
equal, and hence the side CA is
equal to CD. But CA is evident-
ly given; wherefore CD and the
point D are given; BD is conse-
quently given, and hence the per-
pendicular CE is given.
COMPOSITION.
IB
F
E
D
Produce AC till CD be equal to it, join BD, and draw-
CE perpendicular, and AF parallel to it: FCE is the line re-
quired. For the triangles FAC and EDC, having the an-
gles ACF, AFC equal to DCE, DEC, and the side AC
equal to CD,—are equal, and consequently CF is equal to
CE.
216
GEOMETRICAL ANALYSIS.
PROP. IV. PROB.
To bisect a given triangle, by a straight line
drawn from a given point in one of its sides.
Let it be required, from the point D, to draw DF, bisect-
ing the triaugle ABC.
ANALYSIS.
B
T
Bisect (I. 7. El.) the side AC in E, and join EB, EF and
BD. The triangle ABE is (II. 2. El.) equal to EBC, and is
consequently the half of ABC; where-
fore ABE is equal to AFD, and, tak-
ing AFE from both, the remaining
triangle EFB is equal to EFD; and
since these triangles stand on the same
base, they must (II. 3. El.) have the
same altitude, or EF is parallel to BD.
But the points B and D being given, the straight line BD is
given in position, and consequently EF is also given in posi-
tion.

COMPOSITION.
A
E
D
Having bisected AC in E and joined BD, draw EF pa-
rallel to it, meeting AB in F; the straight line DF divides
the triangle ABC into two equal portions.
For join BE. Because BD is parallel to EF, the triangle
EFB (II. 1. El.) is equal to EFD; and, adding AFE to
each, the triangle AFD is equal to ABE, that is, to the half
of the triangle ABC.
PROP. V. PROB.
To find a point within a given triangle, from
which straight lines drawn to the several corners
will divide the triangle into three equal portions.
BOOK I.
217
Let F be the required point, from which the lines FA, FB,
and FC trisect the triangle ABC.
ANALYSIS.
B
Draw FD, FE parallel to the sides BA, BC, and join BD,
BE. Since FD is parallel to AB, the triangle ABF (II. 2.)
El.) is equal to ABD, which is hence the third part of ABC;
and, for the same reason, the tri-
angle BFC is equal to BEC, which
is also the third part of ABC.
Wherefore the bases AD and EC
are each the third part of AC, and
consequently the points of section
D and E are given; hence (I. 24.) A
El.) the parallels DF and EF are
given in position, and their point of concourse is therefore
given.

I
H
G
I
D
E
But the point F may be determined otherwise. For pro-
duce AF and CF to G and H. The triangle DFE is evi-
dently (I. 31. El.) similar to ABC, and therefore AC: AB::
DE: DF, but AC=3DE, and consequently (V. 8. and 5. El.)
AB=3DF. Again, because AH and DF are parallel
AC: AH:: DC: DF, and (V. 13. El.) 2AC: 2AH:: 3DC:
3DF; but 2AC=6AD=3DC, and 2AH=3DF AB.
Hence AB is bisected in H; and, for the same reason, BC
is bisected in G. Wherefore the points H and G being thus
given, the intersection F of the straight lines CH and AG is
likewise given.
COMPOSITION.
Bisect AB and BC (I. 7. El.) in H and G, join CH and
AG, and, from their point of intersection, draw FA, FB,
and FC; the triangle ABC will thus be divided into three
equal portions.
:
218
GEOMETRICAL ANALYSIS,
For, from the points A and B let fall the perpendiculars
AI and BL. The triangles HAI and HBL, having the angles
AHI and AIH equal to BHL and BLH, and the side AH
equal to BH, are (I. 21. El.) equal, and consequently AI=
BL. The triangles AFC and BFC, standing on the same
base CF, and having equal altitudes AI and BL, are equal
(II. 2. El.). And, in the same manner, it is shown that the
triangles AFC and AFB are equal. Wherefore the whole
triangle ABC is divided into three equal triangles, having
their common vertex at the point F.
PROP. VI. PROB.
To trisect a given triangle, by straight lines drawn
from a given point within it.
Let ABC be a triangle which it is required to divide into
three equal portions, by the straight lines DB, DG, and DH,
drawn from the point D.
ANALYSIS.
Join BG, draw DE (I. 24. El.) parallel to it, and join BE.
B
The triangle BDG is equal to BEG,
and consequently the compound
space ABDG is equal to the tri-
angle ABE, which is, therefore,
the third part of the triangle ABC.
Hence the base AE is the third
part of AC, and the point E is
consequently given; wherefore the parallel BG is given, and
also the point G and DG. In like manner, joining BH,
drawing DF parallel to it,—and joining DH, it may be
shown that BH is given.

A G E
COMPOSITION.
FH
Trisect (I. 38. El.) the base AC in the points E and F,
join DE, DF, and parallel to these draw BG, BH, and join
BOOK I.
219
DB, DG, DH; the triangle ABC is thus divided into three
equal portions.
For DE being parallel to BG, the triangle BDG is equal
to BEG, and therefore the space ABDG is equal to the tri-
angle ABE. In the same manner, it is shown that the space
BDHC is equal to the triangle BFC; and consequently the
remaining triangles GDH and EBF are equal. But the tri-
angles ABE, EBF, and FBC, standing on equal bases, are
equal; wherefore the spaces ABDG, GDH, and BDHC, are
each of them the third part of the original triangle ABC.
PROP. VII. PROB.
To inscribe a square in a given triangle.
Let ABC be the triangle in which it is required to inscribe
a square IGFH.
ANALYSIS.
Join AF, and produce it to
meet a parallel to AC in E,
and let fall the perpendiculars
BD and EK.
G
B
E

A ID
H
KC
Because EB is parallel to FG
or AC, AF: AE::FG: EB
(VI. 2. El.); and since the per-
pendicular EK is parallel to
FH, AF: AE::FH:EK. Wherefore FG : EB::FH:EK;
but FG-FH, and consequently (V. 8. and 5. El.) EB=EK.
Again, EK, being equal to BD, the altitude of the triangle
ABC is given, and, therefore, EB is given both in position
and magnitude; whence the point E is given, and the inter-
section of AE with BC is given, and consequently the paral-
lel FG and the perpendicular FH are given, and thence the
square IGFH.
GEOMETRICAL ANALYSIS.
220
COMPOSITION.
From B draw BD perpendicular and BE parallel, to AC,
make BE equal to BD, join AE, intersecting BC in F, and
complete the rectangle IGFH.
Because BE and EK are parallel to GF and FH, AE: AF
:: BE: GF, and AE: AF:: EK: FH; wherefore BE: GF
:: EK: FH; but BE=EK, and consequently GF=FH. It
is hence evident that IGFH is a square.
·
PROP. VIII. PROB.
To draw a straight line through a given point, so
that its portions, terminated by two straight lines
given in position, shall have a given ratio.
Let A be a given point, and BC, BD two straight lines
given in position; it is required to draw EAF, such that EA
shall be to AF as M to N.
ANALYSIS.
Draw AG parallel to BC, and meeting BD in the point G,
which is thus given. The diverging lines FE, FB are cut
proportionally by parallels BE, GA, (VI. 1. El.), and conse-
quently EA: AF:: BG:GF;
but the ratio of EA to AF
is given, and therefore the
ratio of BG to GF; and BG
being given, GF is given,
and the point F, and hence
the straight line EAF is
given.

E
B
G
FD
M
NE
COMPOSITION.
Draw AG parallel to BC, make (VI. 3. El.) BG: GF::
M: N, and join FAE.
BOOK I.
221
For, BE and AG being parallel, EA: AF:: BG : GF;
but BG: GF: M: N, and therefore EA: AF:: M: N.
PROP. IX. PROB.
Through a given point, to draw a straight line that
shall be cut in a given ratio, by the circumference
of a given circle.
Let A be the given point, and BDCE the given circle; it
is required to draw BC, so that BA shall be to AC as M to N.
ANALYSIS.
Draw the diameter DAE, join
DB, CE, and draw CF parallel to
DB. Because the point A and the
centre of the circle are given, the
diameter DE is given in position,
and consequently its extremities D
and E. But, DB being parallel to
CF, BA: AC:: DA: AF (VI. 1.
El.), wherefore the ratio of DA to
AF is given, and since DA is given,
AF is also given. Again, BA.AÇ
=AD.AE (III. 32. El.), and con- Á
sequently AE: AC:: BA: DA;
but BA: DA :: AC: AF (VI.1.El.),
whence AE: AC :: AC : AF, or
D
C

E
A
FG
B
M-
NH
C

B
E
D F G
AC is a mean proportional between AF and AE, and is,
therefore, given. The point C is thus given, and consequent-
ly BC.
COMPOSITION.
Having drawn the diameter DE, make DA: AF:: M: N,
find (VI. 18. El.) AG a mean proportional between AF and
222
GEOMETRICAL ANALYSIS.
AE, and inflect AC equal to it; BAC is the straight line re-
quired.
For join DB, CF, and CE. Since the rectangle BA, AC
is equal to the rectangle DA, AE, it follows that AE: AC::
BA: DA; but, by construction, AE: AC :: AC: AF, and
therefore AC:AF:: BA: DA; hence (VI. 1. cor. 1. El.)
CF is parallel to DB, and consequently BA is to AC, as DA
to AF, that is, as M to N.
PROP. X. PROB.
From two given points in the circumference of a
given circle, to inflect, to another point in the cir-
cumference, straight lines that shall have a given
ratio.
From the points A and B, let it be required to inflect AC
and BC in a given ratio.
ANALYSIS.
Draw (I. 5. El.) CE bisecting the vertical angle ACB.
Therefore (VI. 11. El.) AC: CB:: AD: DB, and conse-
quently the ratio of AD to DB is given,
and thence (VI. 4. El.) the point D is
given. But since the angle ACE is
equal to BCE, the arc AE is (III. 18.
cor. El.) equal to the arc EB, and
therefore the point E is given. Whence
the points E and D being given, the
straight line EDC is given in position,
E
C

B
D
and consequently the point C and the chords AC and BC,
are given.
COMPOSITION.
Bisect (III. 15. El.) the arc AEB in E, divide AB (VI. 4.
El.) in the given ratio at D, join ED, and produce it to meet
BOOK I.
228
the opposite circumference in C; the chords AC and CB are
in the given ratio.
For since the arc AE is equal to BE, the angle ACD is
(III. 18. cor. El.) equal to BCD, and consequently (VI. 11. El.)
AC: CB:: AD: DB, that is, in the given ratio.
PROP. XI. PROB.
Through a given point, to draw a straight line to
a circle, so that the rectangle under the part limited
by the circumference and the segment included with-
in the circle, shall be equal to a given space.
Let it be required through the point A to draw ABC, such
that the rectangle AB, BC shall be equal to a given space.
ANALYSIS.
Through the centre O draw
AF, and (II. 9. El.) find AE,
which forms with AD a rect-
angle equal to the given space.
Because (III. 32. El.) AB.AC=
AD.AF, and, by construction,
AB.BC=AD.AE; it follows (V.
6. El.) that AD: AB:: AC :
AF :: BC: AE; whence (V.
19. cor. 1. El.) AD: AB :: AC
-BC or BC-AC, that is AB
: AF-AE or AE-AF, that is
D
C

B
AD
ΕΟ
C.
AO
F
E
B
EF. Wherefore AB is a mean proportional between AD
and EF; but AE being given, EF is also given, and conse-
quently AB is given both in magnitude and position.
COMPOSITION.
Draw AF through the centre of the circle, make (II. 9. El.)
the rectangle AD, AE equal to the given space, find (VI. 18.
224
GEOMETRICAL ANALYSIS.
•.
El.) a mean proportional to AD and EF, and inflect this from
A towards B; the rectangle AB, BC is equal to the given
space.
For (V. 6.) AD: AB:: AB: EF, and (V. 6. and III. 32.
El.) AD: AB :: AC: AF, whence (V. 19. cor. 1. El.) AD:
AB:: ACAB or BC: AF EF or AE, and consequent-
ly AD.AE=AB.BC.
PROP. XII. PROB.
Through two given points, to describe a circle bi-
secting the circumference of a given circle.
Let A and B be two points, through which it is required
to describe a circle ADGEB, that shall bisect the circumfe-
rence of the circle HDFE.
ANALYSIS.
Join D, E, the points of intersection. Because DFE is,
by hypothesis, a semicircumference,

DE is a diameter, and must, there-
fore, pass through the centre C. Join
AC, and produce it to F. Since DC
=CE, it is evident (III. 32. El.) that
AC.CG=DC=HC.CF; but the
rectangle HC, CF is given, and con-
sequently the rectangle AC, CG is al-
so given; and AC being given, CG
is hence given, and the point G.
D
H
F
E
/C
JAK
B
Wherefore the three points A, G, and B being given, the
circle AGB is (III. 10. El.) given.
COMPOSITION.
Through C, the centre of the given circle, draw ACF,
make (VI. 3. El.) AC: HC:: CF or HC: CG, and through
BOOK I.
225
the three points A, G, and B, describe (III. 10. cor. El.) the
circle AGB: This will bisect the circumference HDFE.
For, through one of the points of intersection, draw the
diameter DCI, and produce it to meet the circumference of
the circle AGB in K. Because AC: HC:: HC: CG, the
square of HC is (V. 6. El.) equal to the rectangle AC and
CG; but (III. 32. El.) HC-DC.CI, and AC.CG=DC.CK;
wherefore DC.CI=DC.CK, and (II. 3. cor. El.) CI=CK, or
the points I and K are one, and the circle AGB passes through
both extremities of the diameter of HDFE.
PROP. XIII. PROB.
To cut a given straight line, such that the square
of one part shall be equivalent to the rectangle un-
der the remainder and another given straight line.
Let AB be a straight line, from which it is required to cut
off a segment whose square shall be equivalent to the rectangle
under the remainder and the straight line C.
ANALYSIS.
F
Produce BA till AD be equal to C, on DB describe a se-
micircle and erect the perpendicular AF. Because AG²=
CxGB, it follows (V. 6. El.) that
DA: AG: AG: GB; wherefore
(V. 19. El.) DA : AG : : DG : AB,
and consequently DA.ABAG.DG;
but (III. 32. cor. 1. El.) DA.AB=
AF², and therefore AG.DGAF²;
whence AF is equal to a tangent

D
E
A GB
drawn from G to a semicircle described on DA. Bisect DA
in E, and join EF; and because AG.DGAF², add EA'
P
226
GEOMETRICAL ANALYSIS.
to each, and AG.DG+EA², or (II. 19. cor. 2. El.) EG², is
equivalent to AF²+EA² or (II. 11.) EF²; whence EG is
equal to EF, and is therefore given.
COMPOSITION.
Having produced AD equal to C, and described on BD a
semicircle, erect the perpendicular AF, bisect. AD in E, join
EF and make EG equal to it; the square of the segment
AG thus formed in AB is equivalent to the rectangle under
the remaining part GB and the given line C.
2
For EFA being a right-angled triangle EFEA²+AF²
(II. 11. El.), and consequently AF²-EF”—EA², or EG²—
EA²; and since (II. 19. El.) EG²- EA² = (EG+EA)
(EG—EA), or DG.AG, therefore AFDG.AG. But
(III. 32. cor. 1. El.) AFDA.AB; whence DG.AG=
DA.AB, and AG: AB :: DA : DG (VI. 6. El.); wherefore
(V. 11. and V. 7. El.) AB—AG, or GB : AG :: DG—DA,
or AG: DA, whence (V. 6. El.) AG²=GB.DA.
Cor. If DA, or C, be equal to AB, then AG=AB.BG,
or AB: AG:: AG: BG, and, therefore, the line AB is now di-
vided in extreme and mean ratio, at the point G. The con-
struction also becomes evidently the same with that which was
given in Book II. Prop. 22. of the Elements, for the medial
section of a line, and which is really a simple case of the same
problem.
PROP. XIV. PROB.
To divide a straight line, such that its segments
shall have the subduplicate ratio of those formed by
another section of the same kind.
such
Let it be required to divide the straight line AB in D,
that the segments AD, DB shall be in the subduplicate ratio
of other like segments AC, CB.
1. Let the given section be internal.
BOOK I.
227
1.
ANALYSIS.
On AB describe a semicircle, erect the perpendicular CE,
and join AE, BE and ED or ED'. Because (III. 22, El.) AEB
is a right angle, the ratio of
AEto BE(VI.16. cor.1. El.)
is the subduplicate of that
of AC to BC, and conse-
quently AE: BE::AD: BD,
or AD' BD'; wherefore
:
D'
F

E
ACD
B
(VI. 11. cor. El.) the vertical angle AEB is bisected in-
ternally or externally by ED or ED'. But the perpendicular
and the semicircle being both given,-the vertex E, the
straight line ED or ED', and the point of section D or D',
are likewise given.
COMPOSITION.
Having on AB described a semicircle, erect the perpendi-
cular CE, join EA, EB, and draw ED or ED' bisecting the
angle AEB or its adjacent angle AEF; the internal segments
AD, DB, or the external segments AD', D'B, are in the sub-
duplicate ratio of AC to CB.
For (VI, 11. El.) AE: BE:: AD: DB or AD' : D'B;
but the triangle AEB being right-angled, AE is to BE
(VI. 16. cor. El.) in the subduplicate ratio of AC to CB, and
consequently AD is to BD or AD' to D'B in the same sub-
duplicate ratio.
2. Let the given section be external.
ANALYSIS.
On AB describe a semicircle, draw the tangent CE, and
join AE, BE and ED or ED'. The triangles ACE and ECB
are similar, for (III. 25. El.) the angle CEA is equal to CBE
in the alternate segment, and BCE is common to both trian-
P 2
228
GEOMETRICAL ANALYSIS.
כן:
T
E
CA D'
B
gles; whence AC: CE::
CE: BC, and consequent-
ly (V def. 20. El.) the ra-
tio of CE to BC is the sub-
duplicate of that of AC to
B.C. But in these similar triangles, AE: CE:: BE: BC,
and alternately AE: BE::CE: BC; wherefore AE: BE
:: AD: DB, or AD': D'B, and the vertical angle AEB
(VI. 11. cor. El.) is bisected externally or internally by ED
or ED'.

COMPOSITION.
Having described a semicircle on AB, apply (III. 26. El.)
the tangent CE, join AE, BE, and draw ED or ED' bisect-
ing externally or internally the vertical angle AEB; the ex-
ternal segments AD, DB, or the internal segments AD', D'B
are in the subduplicate ratio of AC to BC.
For the angle CEA being (III. 25. El.) equal to CBE, and
BCE common to the two triangles ACE and ECB, these are
similar, and AC: CE:: CE: BC; whence the ratio of CE
to BC is the subduplicate of that of AC to BC. Again, from
the same similar triangles, AE:CE:: BE: BC, or alternate-
ly AE: BE:: CE : BC, and therefore AE is to BE in the
subduplicate ratio of AC to BC. But (VI. 16. El.) AE: BE
:: AD: DB, or AD': D'B, and consequently the ratio of
AD to DB or of AD' to D'B is the subduplicate of that of
AC to BC.
Cor. In the second case, the angle CD'E (I. 32. El.) being
equal to D'EB and D'BE, which are equal to D'EA and AEC,
is therefore equal to CED', and the triangle D'CE is hence
isosceles. Again the angle DEF, equal by hypothesis to DEA
or CED and AEC, is (I. 32. El ) equal to CDE and DBE
or AEC, and consequently the triangle DCE is likewise iso-
sceles. Wherefore CE=CD=CD', and thus, without bi-
secting the vertical angle, the point D or D' is found from the
BOOK I.
229
tangent CE, which is a mean proportional between the seg-
ments AC and BC.
PROP. XV. PROB.
To find a point in the diameter of a circle, such
that the square of a straight line inflected from it at
a given angle to the circumference, shall have a gi-
ven ratio to the rectangle under the segments of the
diameter.
Let it be required to draw DE at a given angle with DB,
and so that the square of DE shall have a given ratio to the
rectangle AD, DB.
ANALYSIS.
Make EG=FD, join CF, draw the radius CGH, join AH,
and produce it to meet the extension
of CE in I.
Because CE is equal to CF, the
angle CEF is (I. 11. El.) equal to
CFE. Wherefore the triangles CGE
and CDF, having thus the angle CEG
equal to CFD, and the sides CE and
EG equal to CF and FD,—are (I. 3.
El.) equal, and consequently the angle
A
F
I

IT
E
D
C
B
ECG is equal to FCD; whence (III. 13. El.) the arc HE is
equal to AF, and therefore (III. 20. cor. El.) AH is parallel
to DE. But the angle BDE is given, and thence BAH;
wherefore the chord AH is given. Again, the rectangle
AD,DB, being equal to FD,DE (III. 32. El.), is also equal
to DE,EG; and therefore DE² is to DE.EG, or (V. 25. cor.
2. El.) DE is to EG, in the given ratio; but (VI. 2. El.),
DE: EG :: AI: IH, consequently AI is to IH in a given
ratio, and hence AH is to HI in a given ratio. Wherefore
230
GEOMETRICAL ANALYSIS.
since AH is given, IH and the point I are given; and thence
ÌC, the point E, and DE, are all given.
COMPOSITION.
Draw AH at an inclination with AB equal to the given
angle, and produce it to I, so that AI shall be to IH in the
given ratio, join IC, and draw ED parallel to IA; D is the
point required.
Because AI: IH:: DE: EG, DE is to EG in the given
ratio, and consequently DE² is to DE.EG in the same ratio.
But FE being parallel to AH, the arc HE is equal to AF,
and thence the angle HCE is equal to ACF; the triangles
CGE and CDF, having thus the side CE equal to CF, and
the angles ECG and CEG equal to FCD and CFD,-are
(I. 21. El.) equal, and hence the side EG is equal to FD.
Wherefore DE.EG=DE.FD=AD.DB, and consequently
DE is to AD.DB in the given ratio.
PROP. XVI. PROB.
Through two given points, to draw straight lines
to a point in the circumference of a given circle, so
that the chord of the intercepted segment shall be
parallel to the straight line which connects the given
points.
Let it be required, from the points A and B, to inflect AC
and BC cutting the given circumference in D and E, such
that DE shall be parallel to AB.
ANALYSIS.
Draw the tangent DF meeting AB in F. The angle FDE
is equal to the angle ECD or its supplement in the alternate
segment (III. 25. El.); but DE being parallel to AB, the
BOOK I.
231

i
angle FDE or its supplement is
(I. 23. El.) equal to the alternate
angle AFD, which is conse-
quently equal to the angle ECD
or ACB; wherefore the triangles
ADF and ABC, having likewise
a common angle CAB, are simi-
lar, and AD: AF: AB: AC,
and hence AD.AC-AF.AB.
But since the point A and the
F
G
F
B
D
E
B

circle DCE are given, the rect-
angle AD, AC is also given; for
it is equal to the square of the
tangent AG (III. 32. cor. 2. El.), when A lies without the cir-
cumference,—and equal to the square of AG (III. 82. cor. 1.
El.) a perpendicular to the diameter, in the case where that
point lies within the circle. Hence the rectangle AF, AB is
given; and AB being given, AF is likewise given, and con-
sequently the point F. Wherefore the tangent FD is given
in position; and since the point A is given, the straight line
AC is given, and thence BC and the intersection E.
COMPOSITION.
If the point A be without the circle, draw the tangent AG;
or if it lie within the circle, erect AG perpendicular to the
diameter which passes through it. Make (VI. 3. El.) AB :
AG:: AG: AF, from F draw the tangent FD, join AD,
and produce it to meet the opposite circumference in C, join
CB, cutting the circle in E; the straight line DE is parallel
to AB.
For, since AB: AG :: AG: AF, AG²=AB.AF; but
(III. 32. cor. 1. and 2. El.) AG-CA.AD, whence AB.AF
= CA.AD, and consequently (V. 6. El.) AB: AC:: AD: AF.
Wherefore (VI. 14. El.) the triangles BAC and DAF, having
the sides about their common angle proportional, are similar,
232
GEOMETRICAL ANALYSIS.
1
and hence the angle ACB is equal to AFD; but (III. 25. El.)
ACB or DCE is equal to EDF or its supplement, and
consequently the angle AFD is equal to EDF or its supple-
ment, and (I. 23. cor. El.) the chord DE is parallel to AB.
PROP. XVII. PROB.
From two given points, to inflect straight lines to
the circumference of a circle, such that the chord of
their intercepted arc shall tend to a given point in
the direction of the former.
Let it be required, from the points A and B, to inflect AF
and BF, so that the chord DG produced shall meet the ex-
tension of AB in the point C.
ANALYSIS.
Draw DE parallel to AC, join
EG, and produce it to meet AB
in H.
The angle BHG is equal to the
alternate angle GED, which is
equal (III. 18. El.) to GFD, and
consequently the angles BHG
and BFA are equal, and the tri-
angles BGH and BAF are simi-
F

D
E
A
H
13
lar. Wherefore BG: BH :: BA: BF, and BG.BF=BH.BA;
but the rectangle BG, BF is given, since it is equal to the
square of a tangent drawn from B, and hence BH.BA is gi
ven, and the point H. The problem is thus reduced to the
last Proposition, and only requires, from the points C and H,
to inflect CD and HE, such that DE, the chord of their in-
tercepted arc, may be parallel to HC.
1
BOOK I.
233
î
COMPOSITION.
From the point B draw a tangent BI to the circle, make
BA: BI:: BI: BH, and, by the last Proposition, inflect HE
and CD such that DE shall be parallel to HC; then BG, be-
ing produced to F in the circumference, ADF forms one
straight line.
For since BA : BI : : BI : BH, the rectangle BA, BH will
be equivalent to the square of BI or (III. 32. cor. 2. El.) to the
rectangle BG, BF; consequently (V. 6.) BA: BF:: BG: BH,
and (VI. 14. El.) the triangles BAF and BGH are si-
milar; wherefore the angle BFA is equal to BHG which
(I. 23. El.) is equal to GED, and this again (III. 18. El.) is
equal to GFD; whence BFA is equal to GFD, or the straight
lines FA and FD lie in the same direction from F.
PROP. XVIII. PROB.
From two given points in the circumference of a
given circle, to inflect straight lines to another point
in the opposite circumference, such as to intercept,
on either side of the centre, equal segments of a gi-
ven diameter.
Let it be required, from the points A and B, to inflect AC
and BC, so as to intercept, on the diameter DE, equal por-
tions from the centre.
ANALYSIS.
Join BA, and produce it and the diameter ED to meet in
M, draw COL, from O let fall the perpendicular OK upon
AB, join LK, through A draw AHI parallel to DE, and join
HK.
The parallels FG and AI are cut proportionally by the di-
verging lines CA, CH, and CI (VI. 1. El.); but FO is equal
234
GEOMETRICAL ANALYSIS.
to OG, and consequently AH is equal to HI. Wherefore
(II. 4. El.) HK is parallel to IB, and the angle AKH is
equal to ABI (I. 23. El.); and since the angle ABI or ABC
is equal to ALC (III. 18. El.), the angle AKH is equal to
ALC or ALH, and hence (III. 18. cor. El.) the quadrilateral
figure AHKL is contained in a circle. Consequently
(III. 18. El.) the angle HAK is equal to HLK; but HAK
is equal (I. 23. El.)
to OMK, which is

therefore equal to
HLK or OLK,
and thence the
quadrilateral figure
MOKL is also con-
tained in a circle.
C
M
D
F
Ο
G
E
H
I
A
K
B
L
Wherefore (III. 18. El.) the angle MLO is equal to MKO;
but MKO is a right angle, and consequently MLO is like-
wise a right angle, and thence (III. 24. El.) ML is a tangent.
But the point M, being the concourse of ED and BA, is gi-
ven, and therefore the tangent ML to the given circle is given
(III. 26. El.); whence the diameter LC, and the point C, are
given.
COMPOSITION.
Produce ED and BA to meet in M, draw the tangent ML
and the diameter LC; the straight lines AC and BC will cut
off from the centre equal portions, OF and OG, of the given
diameter ED.
For draw AI parallel to DE, and OK perpendicular to
AB, and join LK and KH.
Because ML is a tangent, MLO is a right angle, and,
therefore, equal to MKO; consequently (III. 18. El.) MKL
is equal to MOL, that is, (I. 23. El.) to AHL. Wherefore
the quadrilateral figure AHKL is contained in a circle, and
BOOK 1.
235
hence (III. 18. El.) the angle ALH is equal to AKH; but, for
the same reason, ALH or ALC is equal to ABC or ABI, and
consequently AKH is equal to ABI, and (I. 23. El.) KH
pa-
rallel to BI. Now since AK is equal to KB, it follows that
AH is equal to HI, and hence that FO is equal to OG.
PROP. XIX. PROB.
Through a given point to draw a straight line, so
that the rectangle under its segments, intercepted
by two straight lines given in position, shall be equal
to a given space.
Let AB, AC be two straight lines, and D a point through
which it is required to draw EF, such that the rectangle un-
der its segments ED, DF shall be equal to a given space.
Join AD, from F
draw (I. 4. El.) FG,
making an angle
DFG equalto DAE,
and meeting AD or
its production in G.
The triangles ADE
and FDG,being thus
evidentlysimilar,AD:
ED: DF: DG,
A
ANALYSIS.

E
E
G
F
B
236
GEOMETRICAL ANALYSIS.
and consequently (V. 6 El.)
AD.DG-ED.DF. But the
rectangle ED, DF is given,
and therefore also the rect-
angle AD, DG; and since
AD is given in position and
magnitude, DG and the
point G are given. Again, the
angle DFG, being equal to

A
D
E
DAC, is given, and thence
E
B
(III. 27. El.) the segment of the circle which contains it;
wherefore the contact or intersection of that arc with the
straight line AB is given, and consequently the position of
EF or E'F' is likewise given.
COMPOSITION.
Join AD, make the rectangle AD, DG equal to the given
space, and on DG describe (III. 27. El.) an arc containing
an angle equal to DAC, and meeting AB in F or F'; EDF
or EDF' is the straight line required.
For the triangles ADE and FDG are similar, and conse-
quently (VI. 12. El.) AD : ED : : DF: DG; whence (V. 6. El.)
ED.DF=AD.DG; but the rectangle AD.DG is equal to the
given space, and therefore the rectangle ED.DF is also equal
to that space.
A limitation evidently takes place, when the points F and
F' coincide, and the circle touches the straight line AB.
In
this case, the angle AFD or BFD, being equal to DGF in
the alternate segment, is therefore equal to AED, and conse-
quently AFE is equal to AEF, and (I. 12. El.) AF=AE.
PROP. XX. PROB.
Two straight lines being given, to draw, through
a given point, another straight line, cutting off seg-
BOOK I.
237
ments which are together equal to a given straight
line.
Let AB, AC be two straight lines, and D a given point,
through which it is required to draw a straight line EF, so as
to cut off the segments AE and AF, that are together equal
to ON.
The point D may lie either within or without the angle
formed by the straight lines AB and AC.
1. Let D have an internal position.
ANALYSIS.
Draw DG and DH (I. 24. El.) parallel to AB and AC.
Because the point D is given, and AB,AC are given in po-
sition, the parallelo-

gram AGDH is given.
And since the triangles
EDG and DFH are
evidently similar, EG:
E
GD:: DH: HF, and
therefore EG.HF =
GD.DH. But AG
and AH, or DH and
GD, being given, the
rectangle GD, DH
is given, and therefore
EG,HFisgiven. Make
FK EG, and therect-
H
E
G
D
1.
M
M
P
F
K B
angle HF, FK is hence given; but HK, being equal to HF
and FK or the excess of AF and AE above GD and DH, is
given, and consequently (VI. 19. El.) its segments HF, FK
are given; whence the point H being given, the point of sec-
tion F or F', and the straight line EDF or E’DF', are given.
COMPOSITION.
Draw the parallels DG and DH. From ON, the sum of
the two segments AE and AF, cut off OP AG+AH, and
=
238
GEOMETRICAL ANALYSIS.
1
make HKPN. On HK describe a semicircle, from the ex-
tremities of the diameter erect the perpendiculars HI and KL
equal to AH and AG, join IL, and at right angles to this,
and from the point or points where it meets the circumfe-
rence, draw MF or M'F'; EDF or E'DF' is the straight
line required.
•
For (VI. 19. El.) HI.KL HF.FK, and consequently
AH.AG=HF.FK. But, from the similar triangles EGD
and DHF, EG GD, or AH :: DH, or AG: HF, and
therefore (V. 6. El.) AH.AG=HF.EG; whence HF.FK=
HF.EG, and FK=EG. And since AG+AH=OP, and
HF+EG HK PN, it follows that AG+EG+AH+
HF, or AE+AF=ON.
2. Let the point D have an external position with respect to
the straight lines AB and AC.
ANALYSIS.
Draw DG parallel to AB, and DH parallel to AC and
meeting AB produced. The triangles EDG and DHF be-
ing similar, EG : DG :: DH: HF, and (V. 6. El.) EG.HF=
DG.DH; but DG and DH are both given, and hence the
rectangle under EG and HF is given. Make FK=EG, and

I
M
E
K
B'
H
F
A
K
F B
I
M
E
L
p
N
BOOK I.
239
4
therefore HK=HF-EG=DG+AF-(DH-AE)=AF+
AE—(DH—DG); whence HK and the rectangle HÈ,FK
are given, and consequently (VI. 19. El.) the point F is
given.
If DF'E' intersect the straight lines AB and AC on the
other side of their vertex A, the triangles E'DG and DF'H
are still similar, and E'G: DG :: DH: HF'; wherefore
E'G.HF', being equal to DG.DH, is given. Make F'K'=E'G,
and thence HK' — E'G—HF'= AE'+DH—(DG—AF′)=
AF'+AE'+(DH-DG); consequently HK' and the rect-
angle HF'.F'K' are given, and therefore (VI. 19. El.) the
point F is given.
COMPOSITION.
Make OP or OP' cqual to the difference of the parallels
DH and DG, from H place likewise towards opposite parts
HKPN and HK'P'N, on HK and HK' describe semi-
circles, from H erect the perpendicular HI equal to DG, and,
from K and K', the perpendiculars KL and K'L', each equal
to DH, join IL and IL', and, at right angles to these, from
the points of section M and M', draw MF and M'F'; the
straight lines DEF and DF'E' will cut off segments from AB
and AC, which are together equal to ON.
=
For (VI. 19. El.) HF.FK HI.KL = DG.DH; but
DG.DHHF.EG, and consequently HF.EG=HF.FK,
or EG=FK. Wherefore HK=HF-EG = AF÷AE-
(DH—DG); and since HKPN=ON—(DH-DG), it
follows that AF+AE=ON.
In like manner, it is shown that E'G=F'K', and hence
HK'=E'G—HF' = AF' + AE' + (DH-DG); but HK'=
P'N'=ON+(DH-DG), and consequently AF'+AE'=ON.
PROP. XXI. PROB.
From one of the corners of a given square, to
draw a straight line, such that its portion, intercept-
240
GEOMETRICAL ANALYSIS.
ed between the opposite sides of the figure, shall be
equal to a given straight line.
Let ABCD be a square, and from the point A let it be re-
quired to draw AEF, so that the part EF, intercepted be-
tween CD and BC, or their extension, may be equal to a gi-
ven straight line.
ANALYSIS.
Draw FG perpendicular to AF, meeting AD produced in
G, from G let fall the perpendicular GH upon BC produced,
and join EG.
The angle EFH is (I. 32. El.) equal to ECF and FEC,
and it is also equal to EFG and GFH; consequently, ECF
and EFG being right an-
gles, the remaining angles

FEC and GFH are e-
qual; whence the triangles
E
EAD and FGH, having
the angle AED or CEF
equal to GFH, the angles
at D and H both right
angles, and the side AD
equal to GH or CD,-
I
B
C
H
A
D
E
are (I. 21. El.) equal, and therefore the side AE is equal to
FG. But EFG and EDG being right-angled triangles,
EF²+FG² = EG² - ED²+DG², (II. 11. El.), or EF²+
AE²=ED²+DG²; but AE²=AD²+ED², and hence EF² +
Where-
AD2+ED²=ED²+DG", or EF+AD-DG².
fore, since EF and AD are both given, DG is also given,
and consequently AG; but the right angle AFG being con-
tained in a semicircle described upon AG, the point F or F",
its contact or intersection with BC, is given, and consequent-
ly the straight line AEF.
BOOK I.
241
COMPOSITION.
Make AI equal to the given straight line, join DI, and,
equal to this, produce AD to G, upon AG describe a semi-
circle meeting the extension of BC in F or F', and join AEF
or AF'E'; EF, the external part of that straight line, is equal
to AI.
For join FG, EG, and let fall the perpendicular GH
upon BF. It is evident that EF+FG=ED²+DG²; and
FG being equal to AE, EF² + AE² = ED²+DG². But
AE2AD+ED, and DGDI AD2+AI; whence
EF²+AD²+ED²=ED²+AD²+AI, and therefore EF² =
AI, and EF-AI*.
PROP. XXII. PROB.
Given the base of a triangle, its altitude, and the
rectangle under its two sides,-to determine the
triangle.
ANALYSIS.
About the triangle ABC describe (III. 10. cor. El.) a
circle, and draw the diameter BF and the radii AE and CE.
Β'
B
B
Because the given rectangle AB.BC is (VI. 20. El.) equal
to BD.BF, this rectangle is likewise given; and since the
perpendicular BD is given, the dia-
meter BF, and therefore the radii
AE, CE, are given. But the base
AC being given, the triangle AEC
is hence given, and consequently
the centre E and the circle ABCF
are given. Again, because BD, the
distance of the vertex of the triangle

A
F
E
C
D
from its base, is given, that point must occur in the parallel
*See Note LII.
A
242
*
GEOMETRICAL ANALYSIS.
BB', and, being thus placed in the contact or intersection o
a given straight line with a given circle, is itself given.
COMPOSITION.
On BD construct (II. 9. El.) a rectangle equal to the given
space, also form on AC the triangle AEC, having AE and CE
each equal to half the greater side of that rectangle, from E
with the radius EA describe a circle, on AC erect a perpendi-
cular DB equal to the altitude of the triangle, and through B
draw a parallel meeting the circumference in B or B′; ABC
is the triangle required.
For ABC has evidently the given altitude BD, and the
rectangle AB.BC, being equal (VI. 20. El.) to BF.BD, is
therefore equal to the given space.
PROP. XXIII. PROB.
Given the hypotenuse of a right-angled triangle,
and the sum or difference of the base and perpendi-
cular, to construct the triangle.
ANALYSIS.
In the base AB, or its production, make BD or BE equal
to the perpendicular BC, and join CD or CE.
The triangles CBD and CBE are right-angled and isosce-
les, and therefore the angles at D and E are each of them.
half a right angle. If AD, the
sum of AB and BC, be given, the
point D is given, and consequent-
ly the straight line DC, making a
given angle with DA, is given in
position; or if AE, the difference
between the base and perpendicu-
lar, be given, the point E is given,

A
B
D
and the straight line EC is given in position. But the hypo-
tenuse AC being given, the point C must, therefore, occur in
BOOK I.
243
the contact or intersection of a circle described from A with
that radius and the straight line CD or CE. Consequently
C is given, the perpendicular CB, and thence the right-
angled triangle ABC.
COMPOSITION.
Make AD or AE equal to the sum or difference of AB and
BC, draw (I. 5. and 4. El.) DC or EC at an angle CDE or
CED equal to half a right angle, from A with the radius AC
describe a circle meeting DC or EC in the point C, and
from C (I. 6. El.) let fall the perpendicular CB : ACB is the
triangle required.
For the right-angled triangles CBD and CBE are evident-
ly isosceles, and therefore AD is equal to the sum, and AE
to the difference, of AB and BC.
PROP. XXIV. PROB.
To investigate the construction of a regular pen-
tagon or decagon.
1. Every regular polygon is capable of being inscribed in
a circle, and therefore the angles, formed at the centre by
drawing radii to the several corners of the figure, are each of
them equal to that part of four right angles corresponding to
the number of sides. Consequently the central angles of a
pentagon are each equal to the fifth, and those of a decagon
are each equal to the tenth, part of four right angles; but an
angle at the circumference being half of that at the centre,
the vertical angle of the isosceles triangle, formed in the pen-
tagon by drawing straight lines from any corner to the ex-
tremities of the opposite side, must also be the tenth part of
four right angles. Whence the construction of a regular
pentagon or decagon involves the description of an isosceles
triangle, whose vertical angle is equal to the tenth part of
four right angles, or the fifth part of two right angles.
Q 2
244.
GEOMETRICAL ANALYSIS.
}
2. Since the vertical angle of that isosceles triangle is the
fifth part of two right angles, the angles at its base must be
together equal to the remaining four-fifths, and each of them
is consequently two-fifths of two right angles. Wherefore
each of the angles at the base of that component triangle is
double of its vertical angle.
B
3. Let ABC be such an isosceles triangle, having each of
the angles at A and C double of the angle at B. Draw CD
bisecting the angle ACB. The angle BCD must then be e-
qual to CBD, and consequently the side CD is equal to BD.
But in the triangles BAC and CAD, the angle ABC is equal
to ACD, the angle CAB common to both, and
consequently the remaining angle BCA is equal
to CDA; whence CDA is equal to CAD, and
therefore the side AC is equal to CD. Thus
the three straight lines AC, CD, and BD are
all equal. Again, because CD bisects the angle
ACB, (VI. 11. El.) BC: AC :: AC: AD, that
is, AB: BD :: BD: AD. Hence AB is divided in extreme
and mean ratio at the point D,-or the square of BD or
AC, the base of the isosceles triangle, is equal to the rect-
angle under the side AB and the remaining segment AD.
Whence the construction of a regular pentagon or decagon,
depends on the medial section of a straight line.

D
Λ
4. Now let the straight line AB be divided by a medial
section, or BC² = BA.AC.
Add to each the rectangle
BA.BC, and BC + BA.BC
BC(BA+BC)= BA'. To
AB annex BD equal to it,
and BC.CD=BD². Bi-
sect BD in E, and the
straight lines CD and BC
are the sum and difference
of CE and BE; whence
A
BA.AC + BA.BC, or
F

G
K
L
B
E
the rectangle under CD and BC, or the square of BA, is
BOOK I.
245
།
equal to the excess of the square of CE above the square of
BE, and therefore CEBA+BE. Erect the perpendi-
cular BF BA, and join EF. It is evident that, EF² =
BA²+BE², and consequently EF2-CE2, and EF=CE;
but EF being given, CE and BC are therefore given.
The composition of this general problem forms a series of
the most interesting propositions in elementary geometry.
Art. 4. corresponds to Prop. 22. Book II.; Art. 3. to Prop.
3. and 4. Book IV.; and Art. 2. and 1. coincide with the 5th
and 8th Propositions of the same Book.
PROP. XXV. PROB.
To discover the conditions required for the trisec-
tion of an angle.
Let ABC be an angle, of which ABD is the third part.
About the vertex B describe a circle, draw DF parallel to AB,
join CF, and produce it to meet the extension of AB in G.
ANALYSIS.
Because the chord DF is parallel to AE, the arc EF
(III. 20. El.) is equal to AD, and consequently (III. 13. cor.
El.) the angle EBF is equal to ABD,

or
is half of the remaining
angle DBC; but half this angle is
equal (III. 17. El.) to the angle DFC
at the circumference, and which
(I. 23. El.) is equal to its opposite
T
A
G F
3
angle BGF. Wherefore the angles BGF and GBF are
equal, and (I. 12. El.) the triangle BFG is isosceles; and thus
the solution of the problem would require, to draw CFG,
such that the extreme part FG shall be equal to the radius of
the circle.
246
GEOMETRICAL ANALYSIS.
1
Otherwise thus.
Let the angle ABD be the third part of ABC. Erect the
perpendicular ADC, complete the rectangle BACE, extend
the side EC to meet BD produced in F, and draw CG
making the angle FCG equal to CFG.
ANALYSIS.
E
B
G
D
F
Because the angle FCG is equal to CFG, the side GF
(I. 12. El.) is equal to GC, and the exterior angle CGB
(I. 32. El.) is double of either of those angles. But the angle
CBA being triple of ABD,
the angle CBG is double
of ABD, or of CFG, and
is therefore equal to CGB;
whence the side BC is e-
qual to GC. Again, from
the right angles EBA and FCD, take away the equal angles
ABD and FCG, and the remaining angles EBD and GCD
are equal; but EBD is equal (I. 23. El.) to the alternate angle
BDA, which is equal to the vertical angle CDF; consequent-
ly the angle GCD is equal to GDC, and therefore the side
GD is equal to GC. Thus it appears, that the four straight
and GF, are all equal. Whence DF,
the external segment of the trisecting line BF, is double of
BC the diagonal of the rectangle BACE.

lines BC, GC, GD,
Scholium. Such then are the final conditions on which the
trisection of an angle is made to depend. But to fulfil them in
general, exceeds the powers of elementary geometry. In
some very limited cases indeed, the trisection of an angle can
be effected, merely by the help of straight lines and circles.
Thus, when the proposed angle ABC is half a right angle, it
may be trisected by the application of Prop. 21. For, pro-
BOOK I.
247
duce BE so that BH=
2BC, join AH, produce
BA till AI AH, and on
BI describe a semicircle
meeting the production of
EC in F; the angle ABF
is the third part of ABC.

TIA
+
C
F
E
This result agrees with
B
1
LI
what is derived from sim-
pler views.
For BH2
2
=
IM
4BC28BA², and AI BH2 + BASBA² + BA² =
9BA; whence AI=3BA, the diameter BI=4BA, and con-
sequently the radius OI=2BA. Let fall the perpendicular
FL, and produce it equally on the other side, join OF and
OM. The triangles OFL and MOL are evidently equal,
and therefore OF, OM, and FM, are all equal to 2BA, or
2FL; consequently the triangle FOM is equilateral, and the
angle FOM two-thirds of a right angle; the angle FOL is
hence one-third of a right angle, and the angle ABF at the
circumference, being the half of it, is therefore equal to the
sixth part of a right angle *.
PROP. XXVI. PROB.
To investigate the conditions required in finding
two mean proportionals.
Let the containing sides AB and AC of the rectangle
ABCD be the extremes of a continued proportion, of which
the successive mean terms are DE and AG.
ANALYSIS.
Join CE and CG. Because AB or CD: DE:: AG: AC,
and CDE, being a right angle, is equal to GAC, the tri-
angles DCE and AGC are (VI. 14. El.) similar; whence the
angle DEC is equal to ACG, and the angles ACG and ACE
* See Note LIII.
248
GEOMETRICAL ANALYSIS.
E
èqual to DEC and ACE, or (I. 23. El.) two right angles, and
consequently ECG forms a straight line. Draw the diago-
nals BC, AD, and join their intersection O with the points
E and G. The triangles BOD and
BOA being (I. 29. cor. El.) isosceles,
therefore (II. 23. cor. El.) OE² =
OD²+BE.ED and OG'OA² +
BG.GA; but (VI. 12. El.) BG: BE
:: GA: AC, or DE: GA, and
hence (V. 6. El.) BE.DE=BG.GA.
Wherefore, OD being equal to
OA, the square of OE is equal to
that of OG, and consequently the

D
H
B
A
G
point O is equidistant from E and G. Hence, likewise, if a
circle were described about the given rectangle, the intercept-
ed segment EC (IV. 4. cor. El.) would be equal to GH.
The solution of the problem, then, requires to draw ECG,
such that the distance OE be equal to OG, or that the part
EC without the circle be equal to the opposite part GH.
Otherwise thus.
E
D
The first part of the construction remaining the same, it was
proved that the rectangle BE.ED
is equivalent to BG, GA;
bisect BD in F, and BE.ED+
DF², or (II. 19. cor. 2. El.)
EF²=BG.GA+DF. On AB
construct the isosceles triangle
BKA, having each of its sides
BK and AK equal to DF, and
join GK; then (II. 23. cor. El.)

2
BG.GA+AK GK, and con-
=
sequently EF-GK², or EF=
F
B
G
M
K
GK. But, by hypothesis, AB: DE::DE:: GA :: GA : AC,
BOOK 1.
249
: :
:
:
and (V. 16. El.) AB GA DE AC, or (V. 13. El.)
2AB : GA : : 2DE: AC; join CF and produce it to meet
the extension of AB in L; the triangles CFD and LFB
(I. 21. El.) are evidently equal, and CD or AB=BL. Where-
fore AL is to GA as 2DE to AC or BD, or (V. 3. El.) as
DE to DF the half of BD, and consequently (V. 9. El.)
GL: GA :: EF: DF. Join LK and draw AM parallel to
it; then (VI. 1. El.) GL: GA:: GK: GM, whence EF: DF::
GK: GM; but EF=GK, and therefore DF-GM. Now
the points F, L and K are evidently given, and consequently
the straight line LK and its parallel AM are given in position.
To effect, therefore, the construction of the problem, it
is required from the point K to draw the straight line KMG,
such that the part MG, intercepted between AM and BA
produced, shall be equal to the half of AC.
Or thus.
Let AB and AC, the extreme terms of the continued pro-
portion, stand as before at right angles, and having produced
CA to D, let AB: AD: : AD: AE:: AE: AC. Since, then,
AD: AE:: AE: AC, it follows (V. 6. El.) that AD.AC=
AE²; whence (III. 32. cor. 1. El.) the point E lies in the circum-
ference of a semicircle described upon CD. Join DE, produce
DB to the circumference, and draw the perpendicular radius
IF. Because AB: AD: : AD: AE, and the angle DAE
is common to the two tri-
angles BAD and DAE-
A
F

C
these triangles (VI. 14. El.)
are similar; consequently
the angle ADB is equal to
AED, and (III. 18. cor. El.)
the arc CG is equal to DE;
Z
M
0
B
K
D
A
I
H
C
whence the arc FG is equal to FE, and (III. 13. and 4. El.)
250
GEOMETRICAL ANALYSIS.
the segment IH of the diameter equal to IA, or the oblique
line GL (VI. 1. El.) is equal to LB.
On this condition therefore, that GD shall have its
intercepted portion GL equal to LB, or that the perpen-
diculars EA and GH shall be equidistant from the cen-
tre, the solution of the problem depends. The ratio
of KI to IC is evidently the same as that of AB to
AC. Wherefore a semicircle being described with the
radius IC could a straight line BD be drawn from D, such
that the part BG, intercepted between the circumference and
the straight line CKM drawn from the other extremity of the
diameter, be bisected in L by the perpendicular radius IF-
the problem would be solved: For make AN-AD, and join
CN meeting IF in O; it is manifest, from what has been
shown, that IK, IO, IL, and IC are continued propor-
tionals
PROP. XXVII. THEOR.
If, from the extremity of the diameter of a circle,
a straight line be drawn to a point in the perpendi-
cular radius, such that triple its square be equal to
the square of a perpendicular from the circumfe-
rence and the squares of the segments into which
the diameter is thus divided; the straight line that
joins the points of section and of termination, will
make a given angle with the diameter.
In the semicircle ADB, EF and the radius CD being at
right angles to AB, and AG drawn so that 3AG²=AE²+
EF² + EB²; if EG be joined, the angle CEG is given.
* See Note LIV.
BOOK I.
251
ANALYSIS.
Because
For join CF.
AB is bisected in C, AE2
+ EB² = 2AC² + 2EC²
(II. 21. cor. El.) and con-
sequently 3AG² = 2AC² +
2EC²+EF²; but (II. 11. El.)
D

H
T
G
A. E
C
B
EC²+EF²=CF², or AC², and hence 3AG=3AC²+EC².
Again, AG²=AC²+CG², or 3AG²=3AC²+3CG²; where-
fore EC2=3CG, or EG-4CG and EG=2CG. The
ratio of EG to CG, and the right angle at C being thus
given, the triangle EGC is (VI. 15. El.) given in species, and
consequently the angle CEG is given.
COMPOSITION.
Inflect BH equal to the radius of the circle, join AH,
draw EG parallel to it meeting CD in G, and join AG; then
3AG AE+EF²+EB².
For join CF. The triangles AHB and ECG being evi-
dently similar, AB : BH : : EG : CG; but AB=2BH, and
therefore (V. 5. El.) EG=2CG. Whence EG²=4CG, and
EC²=3CG²; consequently 3 AG²=3AC²+3CG²=3AC²+
EC2AC +2EC+AC-EC. Now 2AC²+2EC²=
AE²+EB², and AC²-EC² = EF²; wherefore 3AG ≈
AE²+EF²+EB².
PROP. XXVIII. THEOR.
If a triangle have a given angle, the excess of the
square of the sum of the containing sides above the
square of the base, has a given ratio to the area of
the triangle.
Let ABC be a triangle, in which AB is produced till BD
be equal to BC; the excess of the square of AD above the
square of AC, has a given ratio to the area of the triangle.
252
GEOMETRICAL ANALYSIS.
ANALYSIS.
Draw AE parallel to BC, and
meeting DC produced in E, from B
let fall the perpendicular BF, and join
BE.
The triangle CBD being isosceles,
the angle CDB (I. 11. El.) is equal to
DCB, but (I. 23. El.) DCB is equal
to CEA; hence the angles EDA
and DEA are equal, and the tri-
angle DAE is isosceles. Wherefore
(II. 23. El.) ADAC + DC.CE, or
AD²—AC² = DC.CE. Again, be-
cause AE is parallel to
BC, the tri-
angle ABC has (II. 1. El.) the same
B
A
C
E
D

F
area as EBC, or (II. 6. El.) is half the rectangle BF.CE.
Consequently the excess of the square of AD above the square
of AC, is to the area of the triangle ABC, as DC.CE to BF.CE,
that is, (V. 23. cor. 2. El.) as DC to BF, or (V. 3. El.) as 4DF
to BF. But the given angle ABC, being (I. 32.) equal to the
two angles CDB and BCD, is double of either, and thus the
angle BDF is given; whence the right-angled triangle DFB
is given in species, and therefore the ratio of DF to BF is
given. It thence follows, that the ratio of 4DF to BF, or
that of the excess of the square of AD above the square of
AC to the area of the triangle ABC, is given.
COMPOSITION.
The same construction remaining, DC.CE: BF.CE ::
DC: BF; but DC.CE=AD²-AC², and BF.CE is double
of the triangle ABC; whence 2DC is to BF, as the excess of
the
square of AD or that of the sum of the sides AB and BC
above the square of the base AC, to the area of the triangle
ABC *.
1
* See Note LV.
GEOMETRICAL ANALYSIS.
BOOK II.
DEFINITION.
A VARIABLE quantity derived from another given or con-
stant quantity, or which depends on it by some relation ac-
cording to a given law, is necessarily confined between certain
extreme limits. When it has acquired the greatest possible
expansion, it is said to have reached its maximum; and when
it has contracted into its lowest dimensions, it occupies the
state of minimum.
PROP. I. PROB.
From a given point, to draw a straight line inter-
cepting, on two given parallels, segments which shall
have a given ratio.
Let AB and CD be two parallels, in which are two given
points, P and O; and let it be required, from another given
point E, to draw EF, such that PG shall be to OF in the ra-
tio of M to N.
254
GEOMETRICAL ANALYSIS.
ANALYSIS.
Join PO, and produce
it to meet EF, or its ex-
tension in I.
Because PG and OF
are parallel, PI : OI ::
PG: OF (VI. 2. El.); but
the ratio of PG to OF is
given, and hence that of
PI to OI, and of PO to
OI, are given. And since
PO is given, OI and the
Mr
N
A K
P
G B

E
F
D
I
E
point I, are given; wherefore IEF, and the segments PG and
OF are given.
COMPOSITION.
Make PKM and OL=N, join KL, PO, and produce
them to meet in I, and draw IEF; PG and OF are the re-
quired segments.
For (VI. 2. El.) the parallels AB and CD being cut pro-
portionally by the diverging lines IK, IP, and IG,—PG is to
OF as KP to OL, that is, as M to N.
If M be equal to N, the point I vanishes, and EF becomes
evidently a parallel to OP.
If the straight lines KL and PO meet in the given point E,
the problem is by its nature indeterminate, or it admits of in-
definite solution; for, in that case, the segments PG and OF,
intercepted by any straight line whatever, drawn through E,
have all the same ratio.
PROP. II. PROB.
Two diverging lines being given in position, to
draw, through a given point, a straight line inter-
cepting segments which shall have a given ratio.
BOOK II.
255
:
Let it be required, through D, to draw EDF, so that AE
shall be to AF in the ratio of M to N.
ANALYSIS.
Through D, (I. 24. El.)
draw DG parallel to AE,
and meeting AC, or its pro-
duction, in G.
The triangles EAF and
DGF are similar, and there-
fore (VI. 12.) AE: AF::
GD: GF; but the ratio of
AE to AF is given, and
consequently that of GD to
D B

E
D
K
G A
G
F C
M
N
GF. And since GD and the point G are evidently given,
GF and the point F are likewise given.
COMPOSITION.
From AB and AC cut off AK=M, and AL=N, join KL,
and parallel to it draw EDF through D; AE and AF are the
segments required.
For (VI. 1. El.) the parallels EF and KL cut the diverging
lines AB and AC proportionally, and therefore AE is to AF,
as AK to AL, that is, as M to N.
PROP. III. PROB.
Two diverging lines being given in position, to
draw, through a given point, a straight line cutting
off segments-on the one from their intersection,
and on the other from a given point that shall
have a given ratio.
Let AB and AC be two diverging lines, it is required,
through the point D, to draw EDF, so that AE shall be to
the part OF, in the ratio of M to N.
256
GEOMETRICAL ANALYSIS.
ANALYSIS.
Draw DG parallel to AE, and meeting AC, or its produc-
tion in G, and make AE: GD :: OF: OH.
By alternation, AE: OF:: GD: OH; but the ratio of
AE to OF is given, and thence that of GD to OH; and
since GD and the point O are
given, OH and the point H are
also given. Again, because
AE: GD :: OF: OH, and (VI.
2. El.) AE: GD:: AF: GF, it
follows that OF: OH:: AF: GF;
whence (V. 10. El.) FH : OH
:: AG: GF, and (V. 6. El.)
GF.FH-AG.OH. But AG and
OH are both given, and conse-
A
MH
N
E
B

D
O G
H
F C
1
quently the rectangle under the segments GF and FH of the
given portion GH is also given, and thence the point of sec-
tion F is given, and the straight line ED.
COMPOSITION.
Make GD to OH, as M to N, and (VI. 19.) divide GH
in F, so that the rectangle GF, FH shall be equal to AG.OH,
and draw EDF; then the segment AE is to OF as M to N.
Since GF.FH-AG.OH, therefore FH: OH: AG: GF,
and (V. 9. El.) OF : OH :: AF: GF: but (VI. 2. El.)
AE:GD::AF: GF, and consequently AE: GD :: OF: OH,
and alternately AE: OF :: GD: OH, that is, in the given
ratio.
PROP. IV. PROB.
Two diverging lines being given in position, to
draw, through a given point, a straight line, cut-
ting off segments from given points in a given ratio.
BOOK II.
257
Let AB and AC be two diverging lines; it is required,
through the point D, to draw EDF, so that PE shall be to
OF in the ratio of M to N.
Join DP cutting AF
in I, and, through I,
draw IK parallel to AB,
and meeting EF in K.
Because the points D
and P are given, the
straight line DP is gi-
ANALYSIS.
B

E
K
D
I O
F C
ven in position, and
M
P
consequently its inter-
N
It
section I with AC is
given, whence IK, being parallel to AB, is likewise given in
position. But (VI. 2. El.) PE : IK :: PD : ID, and since
PD and ID are both given, the ratio of PE to IK is given;
consequently, the ratio of PE to OF being given, the ratio of
IK to OF is given. Wherefore, by the last proposition, the
straight line KDF is given in position.
COMPOSITION.
Join PD and draw IK parallel to AB, make M to L, as
PD to ID, and draw, by the last proposition, KDF, so that
IK shall be to OF, as L to N; then will PE and OF be the
segments required.
For (VI. 2. El.) PE: IK :: PD: ID :: M: L, and
IK: OF:: L: N; whence (V. 16. El.) PE: OF:: M: N.
PROP. V. PROB.
Two parallels being given, from a point in a given
intersecting line, to draw another straight line cut-
ting off segments which shall contain a given rect-
angle.
R
258
GEOMETRICAL ANALYSIS.
1
Let AB, CD be two parallels, and G a given point, through
which it is required to draw. FE intercepting, from given
points O and P in the same direction OPG, segments OE
and PF, that shall contain a given rectangle.
ANALYSIS.
1

Because AB and CD are paral- A
:
:
lel, GO GP: OE: PF (VI. 2.
El.) and consequently (V. 25. cor.
2. El.) GO GP:: OE: OE.PF;
and GO and GP being given, their
ratio is given, and therefore the
ratio of OE² to OE.PF is given;
but the rectangle OE, PF is given,
EB
I
K
C
P
"I'
D
G-
and hence the square of OE and consequently OE itself, are
given.
COMPOSITION.
Find (VI. 18. El.) GI, a mean proportional between GO
and GP, draw IK parallel to AB or CD, and such (III. 33. El.)
that its square shall be equal to the given rectangle, and join
EKFG; this is the straight line required.
For OE, IK, and PF being parallel, OG: IG:: OE: IK,
and PG : IG :: PF : IK (VI. 2. El.); whence compounding
these analogies (V. 22. El.) OG.FG : IG² : : OE PF : IK²;
but OG.PG=IG, and consequently (V. 4.) OE.PF=IK².
PROP. VI. PROB.
Through a given point, to draw a straight line in-
tercepting, from given points on two given parallels,
segments which shall contain a given rectangle.
Let AB and CD be parallels in which the points O and P
are given, and let it be required through G to draw GFE, so
that the segments OE and PF shall contain a given rectangle.
BOOK II.
259
ANALYSIS.
Draw GO and GP, cutting the parallels in I and H. Be-
cause the points O, P, and G are given, the straight lines
GIO and GPH are given in position, and consequently their
intersections I and H with the given parallels. And since AB
is parallel to CD,
GP: GH::PF: HE
(VI. 2. El.) but (V. 25.
cor. 2. El.) PF:HE::
PF.OE HE.OE, and
consequently GP: GH
:
: : PF.OE : HE.OE.
Now, GP and GH be-
ing given, their ratio is
A K
H
E 3

C
I
I
F D
G
given, and hence that of PF.OE to HE.OE; wherefore the
rectangle PF, OE being given, the rectangle under the seg-
ments HE and OE of the given straight line HO is likewise
given; whence (VI. 19. El.) the point E is given, and conse-
quently the straight line GFE.
COMPOSITION.
Draw GO and GP, find (II. 9. El.) HK the side of a rect-
angle GP, HK which is equal to the given space, and (VI.19.El.)
divide HO in the point E, so that the rectangle under its seg-
ments HE and OE shall be equal to the rectangle HG,HK,
and join GFE; this is the straight line required.
:
•
For HE PF: HG: GP, and hence (V. 25. cor. 2. El.)
HE.OE PF.OE:: HG.HK: GP.HK; but, by construction,
the rectangle HE.OE is equal to GH.HK, and consequently
(V. 8. and 4. El.) PF.OE=GP.HK, or the given space,
PROP. VII. PROB.
To draw through a given point a straight line, cut-
ting from two given diverging lines, segments which
shall contain a given rectangle,
R 2
260
GEOMETRICAL ANALYSIS.
Let AB and AC be two diverging lines given in position,
and let it be required from the point D, to draw DFE, so that
the rectangle under the segments AE, AF shall be equal to a
given space.
ANALYSIS.
Draw HD parallel to AB, and make (II. 9. El.) the rect-
angle DH.AI equal to the given space.
B
E
D
Because AE.AF=DH.AI, AE: DH:: AI: AF (V.6 El.),
but AE: DH:: AF : FH (VI. 2. El.), and therefore AF:
FH:: AI: AF; whence (V.11.
cor. El.) AH : AF :: IF : AI,
´and (V.6. El.) AH.AI= AF IF.
Now DH, being parallel to AB,
is given, and consequently AI
is given; wherefore the rect-
angle AH, AI being given,

A H
T
AF.IF is also given; and since AI is given, its internal or
external section is (VI. 19. El.) given.
COMPOSITION.
Draw DH parallel to AB, find (II. 9. El.) AI, which con-
tains with DH a rectangle equal to the given space, and di-
vide AI (VI 19 El) so that the rectangle under its segments
AF, FI shall be equal to the rectangle AI, AH; EDF is the
straight line required For, by construction, AF.IF=AI.AH,
whence (V. 6. El.) AH: AF:: IF: AI, and (V. 10. and 7. El.)
AF: FH:: AI: AF; but AF: FH:: AE: DH, and con-
sequently AE: DH:: AI: AF, and (V. 6. El.) AE.AF=
DH.AI.
PROP. VIII. PROB.
Through a given point to draw a straight line,
which shall, by its intersection with two given diver-
ging lines, form a triangle containing a given space.
BOOK II.
261
Let it be required, through the point D, to draw a straight
line, EDF intercepting, between the diverging lines AB and
AC, a triangle AEF, which shall contain a given space.
ANALYSIS.
Draw DH parallel to AB, upon AC let fall the perpen-
diculars ES and DT, and find (II 9. and 7 El.) AI the base.
of a triangle, having the altitude DT and containing the gi-
ven space.
Because the rectangles
ES AF and DT AI are
(I. 6 El.) each double of
the triangles AEF and
ADI, they are equal, and
consequently (V. 6. El.)
ES DT:: AI : AF.
But the triangles AES
E
B

D
tR
A H ST
F
1
and HDT are evidently similar, and therefore (VI. 12. EL)
AE: ES:: HD : DT, or alternately AE: HD :: ES : DT;
whence AE HD:: AI: AF, and AE AF-HD.AI. Now
HD is given, and consequently AI; wherefore the rectangle
AE. AF is given, and thence, by the last proposition, the
straight line EDF is given in position.
COMPOSITION.
Draw DH parallel to AB, let fall the perpendicular DT,
bisect this in the point R, find (II. 9. El ) the side AI, which
with RT contains a rectangle equal to the given space,
and, by the last proposition, draw EDF, such that the rect-
angle AE AF shall be equal to DH.AI.
Having let fall the perpendicular ES, and bisected it in Q,
the triangles AES and HDT are similar; whence AE : ES::
HD: DT, and alternately AE: HD :: ES: DT, or (V. 3.
El) AE HD:: QS: RT; wherefore AE AF: HD AI ::
QS,AF : RT.AI; but the rectangle AE.AF=HD.AI, and
262
GEOMETRICAL ANALYSIS.
hence (V. 4. El.) QS.AF=RT.AI, or the triangle AEF is
equal to the given space.
This problem will admit of a simpler construction, in the
case where the given point D lies between the diverging lines
AB and AC. For draw DG parallel to AC, and make (II.
9. El.) the rhomboid AGKI equal to the given space.
Because the triangle AEF is equal to the rhomboid AGKI,
take away from both the figure AEDKLF, and the triangles
GED and ILF remain equal to the triangle DLK; but these
supplementary trian-

gles, being formed by
parallel lines, are evi-
dently similar, and
consequently the ho-
mologous sides GD
and IF are (VI. 28.
El.) sides of a right-
angled triangle, of
which DK is the hy-
H
B
E
D
K
L
A
H
I
I
F C
potenuse; wherefore
(II. 11. El.) GD²+
M
Γ.
IF²=DK², or (I. 27. El.) IF²-HI-AH². And since HI
and AH are both given, it follows that IF is given.
COMPOSITION.
Construct the rhomboid AGKI equal to the given space,
draw DH parallel to AB, on HI describe a semicircle, in
which inflect HM equal to AH, join IM, and make IF, or
IF', equal to it; EDF, or E’DF', is the base of the required
triangle.
For (III. 22. El.) HMI being a right angle, IH²=HM²
+IM² (II. 11. El.), or DK²≈GD²+IF²; whence (VI. 23.
cor. 1. El.) the triangle DLK, or DLK', is equal to the tri-
angles GED and ILF, or to GE'D and IL'F; and, adding
BOOK II.
263
to both the excess of the rhomboid AK above the triangle
DLK, or DL'K', the rhomboid AK is equal to the triangle
AEF or AE'F', which is, therefore, equal to the given space.
PROP. IX. PROB.
Through a given point to draw a straight line,
cutting off segments, from two given diverging
lines-on the one from their intersection, and on the
other from a given point—which shall contain a gi-
ven rectangle.
Let it be required to draw EDF, so that the rectangle
AE, OF shall be equal to a given space.
ANALYSIS.
Draw DH parallel to AB, and (II. 9. El.) make the rect-
angle DH.OI equal to the given space; OI and the point I
are, therefore, given. And
=
since AE.OF DH.OI,
it follows that AE: DH::
OI: OF; but (VI. 2. El.)
AE DH AF FH,
and consequently AF:
FH OI: OF. Where-
::
fore (V. 11. El.) AF:
E
D
H F
F
B
E

ІС
AH :: OI: FI, and (V. 6. El.) AF.FI=AH.OI; hence AI
and the rectangle under its segments, AF and FI, are given,
and consequently (VI. 19. El.) the point of section F and the
straight line EDF are given.
COMPOSITION.
Having drawn DH parallel to AB, and made the rect-
angle DH.OI equal to the given space, divide AI (VI. 19. El.)
in F, or F', such that the rectangle under its segments shall
264
GEOMETRICAL ANALYSIS.
also be equal to the rectangle AH.OI; EDF, or E'DF', is the
required straight line. For since AF.FI=AH.OI, AF : AH
:: OI: IF; whence (V. 11. El.) AF: FH: OI: OF; but
(VI. 2. El.) AF: FH:: AE: DH, and, therefore, AE: DH
::OI: OF, and the rectangle AE.OF is equal to DH.OI, or
the given space.
t
PROP. X. PROB.
Through a given point, to draw a straight line,
cutting off segments from given points, on two gi
ven diverging lines, that shall contain a given rect-
angle.
Let it be required to draw EDF, so that the rectangle
OF.PE shall be equal to a given space.
ANALYSIS.
Join DO meeting AE in Q, and draw QR parallel to
AC.
Because (VI. 2. El.) DO: DQ :: OF: QR, it follows
(V. 25. cor. 2. El.) DO : DQ :: OF.PE: QR.PË; but DO
and DQ are evidently gi-
ven, and therefore the
rectangle OF.PE has to
QR.PE a given ratio;
and since OF.PE is
given, the rectangle
QR.PE is likewise given,
and QR, being parallel
B

E
I
H
D
Р
R
Q
A
H
F
to AC, is given in position. Whence, by the last proposi
tion, the intersecting line EDR or EDF, is given in posi-
tion.
វ
COMPOSITION.
Join DQO, draw DH parallel to AC, and produce it meet-
BOOK II.
265
ing in S the parallel to AB, make the rectangle DS.PI equal
to the given space, and divide PI in E, such that the rect-
angle under its segments PE, IE shall be equal to the rect-
angle AH.PI; EFD is the straight line required.
For DQ: DO:: DH: DS:: QR OF, and consequent-
ly (V. 25. cor. 2. El.) DH.PI : DS.PI :: PE.QR : PE.OF;
but, by the last proposition, DH.PI=PE.QR, whence the
rectangle DS.PI, or the given space, is equal to the rectangle
PE.OF.
PROP. XI. PROB.
To divide a given straight line, so that the reet-
angle under one of its segments and a given line,
shall be equal to the square of the other segment.
Let it be required to divide AB in C, such that the rect-
angle under AC and G shall be equal to the square of CB.
ANALYSIS.
Make BD=G, and since G
AC.G=CB², it follows
ト
​A
C
BD
G
}
(V. 6. El.) that AC: CB::
CB: BD; and consequently
(V. 9. and 10. El.) AB: CB::
CD: BD; whence (V. 6.
El.) AB.BD = CB.CD.
But the rectangle AB.BD
is given, and, therefore, the
A
B
D
G
D
C
A
B
rectangle CB.CD is also given; and BD being given, the
point of section C is (VI. 19. El.) thence given.
COMPOSITION.
In the same straight line AB, make BD equal to G, and
(VI. 19. El.) cut BD such that the rectangle CB.CD be equal
to AB.BD; C is the point of section required. For it is
266
GEOMETRICAL ANALYSIS,
evident (V. 6. El.) that AB: CB:: CD: BD,
CD: BD, and conse-
quently (V. 10. El.) AC: CB:: CB: BD; wherefore (V. 6.
El.) AC.BD, or AC.G=CB²,
PROP. XII. PROB.
To divide a given straight line, so that the rect-
angle under one of its segments and a given line
shall have a given ratio to the square of the other
segment.
Let it be required to divide AB in C, such that AC.G:
CB²:: M: N.
ANALYSIS.
Make (VI. 3. El.) G: H:: M: N,
and H is given; but AC.G:: CB²
G: H, and consequently (V.25. cor. 2.El.)
CB²=AC.H; wherefore, by the last
proposition, the section of AB is given.
COMPOSITION.
ㅏ
​A
G
ㅏ
​H
1
C
B
M
N
Having made M:N::G: H, let AB be divided by the
last proposition, so that AC.H=CB2; then AC.G CB::
M: N. For AC.G; AC.H or CB:: G: H, or M: N.
PROP. XIII. PROB.
In the same straight line, three points being gi-
ven, to find a fourth point, such that the rectangle
under its distance from the first and a given line,
shall be equal to the rectangle under its distances
from the second and third points.
BOOK II.
267
Let it be required to find the point D, so that AD.G=
CD.BD.
A
D
B
E
C
ANALYSIS.
Make BE=G, and because
AD.GCD.BD,
Ꮐ
it follows
:
that AD CD: BD BE;
:
:
whence (V. 9. and 10. El.) AC: CD
:: DE : BE, and AC.BE=
CD.DE. But the rectangle
AC.BE being evidently given,
the rectangle under the segments
G-
H
+
A
++
EB D
C
CD, DE of CE, a given straight A
ㅏ
​BE
C D
line, is also given, and conse-
quently (VI. 19. El.) the point of section D is given.
COMPOSITION.
Having made BE=G, divide (VI. 19. El.) CE in D, so
that CD.DE=AC.BE; D is the point required.
For (V. 6. El.) AC: CD :: DE : BE, and (V. 10. El.)
AD CD: BD: BE; whence AD.BE, or AD.G=
CD.BD.
PROP. XIV. PROB.
In the same straight line, three points being gi-
ven, to find a fourth, so that the rectangle under its
distance from the first and a given line, shall have a
given ratio to the rectangle under its distances from
the second and third points.
Let it be required to find a point D, such that AD.G :
CD.BD :: M: N.
268
GEOMETRICAL ANALYSIS.
Make M
ANALYSIS.
NG: H,
G
H
whence H is given; but since
AD.G: CD.BD :: G : H, A
it is evident that AD.H=
CD.BD; wherefore, by the
D B
M
N
last proposition, the point of section D is given.
COMPOSITION.
Having made G : H :: M: N, find, by the last proposition,
the point D, so that CD.BD = AD.H; D is the section
required. For (V. 25. cor. 2. El.) AD.G: AD.H or CD.BD ::
G: H, or M: N.
PROP. XV. PROB.
In the same straight line, three points being gi-
ven, to find a fourth, so that the square of its dis-
tance from the first, shall be equal to the rectangle
under its distances from the second and third points.
Let it be required to find a point D, such that AD²=
CD.BD.
1. When the point D lies between A and B.
ANALYSIS.
Because ADCD.BD, it follows (V. 6. El.) that CD : AD
:: AD: BD; whence (V. 9. El.) AC AD: : AB: BD,
and alternately AC
: AB :: AD : BD. ADB
:
But the ratio of AC: AB being given, the ratio of AD to BD
is given; and since AB is given, the point D (VI. 4. El.) is
given.
BOOK II.
269:
COMPOSITION.
Divide AB (VI. 4. El.) in the ratio of AC to AB, and the
point of section D is that required.
For, because AD: BD :: AC: AB, AD: BD : : AC—AD,
or CD : AB-BD or AD (V. 19. cor. 1. El.); whence
(V. 6. El) AD2BD.CD.
2. When the point D lies between B and C.
ANALYSIS.
Make DE=AD, and since ADCD.BD, CD: AD, or
DE :: AD : BD, and therefore (V. 10. El.) CE: DE ::
AB: BD, and alternately CE:AB:: DE: BD, or (V. 3.
El.) CE: AB:: 2DE, or AE: 2BD; whence (V. 19. El.) CE:
AB:: CE+AE or
AC AB + 2BD,
or BE, and conse-
quently (V. 6. El.)
+
E
A
B
D
CE.BE=AB.AC; but the rectangle AB.AC being given, the
rectangle CE.BE is likewise given, and BC being given, the
point E is given (VI. 19. El.), and therefore D, the bisection
of AE, is given.
COMPOSITION.
Divide BC (VI. 19. El.) in E, such that CE.BE-AB.AC,
and bisect AE in D; then ADCD.BD.
For since CE.BE AB.AC, it is evident that CE: AB::
AC: BE; whence (V. 19. El.) CE : AB :: AE:2BD or
(V. 3. El.) DE : BD; and alternately, CE: DE: : AB : BD,
and (V. 9. El.) CD: DE, or AD: : AD: BD; wherefore
(V. 6. El.) CD.BD=AD².
This last case is evidently subject to limitation; for the
rectangle AB.AC being equal by construction to CE.BE,
must not exceed the square of the half of BC, which (II. 19.
cor. 1. El.) is the greatest rectangle contained under the seg-
ments of BC. Consequently, if E coincide with the middle
GEOMETRICAL ANALYSIS.
270
point O, it limits the problem; but then AB.AC=BO², or
AB.AC+BO²=(II. 23. cor. 2. El.) AO²=2BO², and there-
fore AO is the diagonal of a square described on BO.
Whence AB: BC :: √2—1 : 2, or 1 : 2+√8; that is, the
1:
ratio of AB to BC has attained its maximum, when it is that
of half the side of a square to the sum of the side and the
diagonal.
PROP. XVI. PROB.
In the same straight line, three points being
given, to find a fourth, such that the square of its
distance from the first, shall have a given ratio to
the rectangle under its distances from the second
and third points.
Let it be required to find a point D, such that AD² shall
be to CD.DB in a given ratio.
1. When D lies between the points A and B.
ANALYSIS.
E
On BC describe a semicircle, and draw the tangent DE;
then (III. 32. cor. 2. El.) DE²=CD.DB, and consequently
the square of AD is to the square of DE in the given ratio;
whence the ratio of AD to
DE is given. Draw the ra-
dius EF, and produce ED
to meet AG aperpendicular
to AC. The triangles ADG
and EDF are evidently
similar, and therefore
AD: AG: DE: EF, or

:
alternately AD : DE : :
DB
F
M
N
의
​AG: EF; and since the ratio AD to DE is given, the ratio
of AG to EF is also given, and the radius EF being given,
AG and the point G are thence given; wherefore the tan-
gent GE and its intersection D with AC, are given.
BOOK H.
271
COMPOSITION.
Let M : N be the given ratio, and to these find (VI. 18.
El.) a mean proportional O, on BC describe a semicircle,
make O: M:: BF: AG, a perpendicular erected from A,
and (III. 26. El.) draw the tangent GDE; the intersection
D is the point required.
For, the triangles DAG, and DEF being similar, AD : AG::
DE: EF, and alternately AD: DE :: AG : EF, or M:0;
wherefore (V 22. cor. 1. El.) AD: DE*:: M: O2, that is,
AD² DE²
(V. 24. El.) M: N; but (III. 32. cor. 2. El.) DE²=CD.DB,
and consequently AD : CD.DB :: M : N.
2. When D lies between the points B and C.
ANALYSIS.
On BC describe a semicircle, draw DF perpendicular to
the diameter, and meeting the circumference in F, and join
AF.
Because (III. 32. cor. 2. El.) BD.DC=DF, the ratio of
AD to DF is given, and consequently that of AD to DF;
but the angle ADF, con-
tained by these sides, being
a right angle, is given, and
therefore the triangle AFD
is given in species. Hence
the angle DAF is given,
and the straight line AF
given in position; where-

.A.
E
F
H
E D'
M
D
C
N
fore the intersection For F', the perpendicular FD, or F'D',
and the point D, or D', are all given.
COMPOSITION.
Let M: N express the given ratio, and to these find (VI. 18.
El.) a mean proportional O, make (VI. 3. El.) M to O as AC
to the perperdicular CE, join AE meeting the circumference
of a semicircle described on BC in the point F or F', and let
272
GEOMETRICAL ANALYSIS.
fall the perpendicular FD or F'D'; then M: N:: AD2:
BD.DC or AD' : BD'.D'C.
For the triangle ACE is evidently similar to ADF or
AD'F', and therefore AC: CE:: AD: DF, and AC²: CE² ::
AD: DF2; but (V. 24. El.) M: N:: M2 : O², or as
M²
AC² : CE², and consequently AD² : DF², that is BD.DC,
:: M: N.
This problem evidently requires limitation; for, if AE
should diverge too much from AC, it will not meet the cir-
cumference at all. Hence an extreme case will occur, when
AE touches the circle. But the ratio of AC to CE, or of
AD to DF, will then be the same as that of a tangent from
A is to the radius HB; and consequently the limiting ratio is
the duplicate of this,--or the ratio of M to N can never ap-
proach nearer to the ratio of equality than that of AB.AC,
or AH'-HB2, to HB2.
3. When the point D lies beyond B and C.
ANALYSIS.
On BC describe a semicircle, draw the tangent DE, and
produce it to meet the perpendicular AG, and join E with
the centre F.
Because (III. 32. cor. 2. El.) BD.DC=DE², the ratio of
AD² to DE² is given, and consequently that of AD to DE.
But the angle DEF, being (III. 24. El.) a right angle is equal
to DAG, and the
angle at D is com-
mon to the tri-
angles DGA and
G

DFE, which are
therefore similar,
and hence AD:AG
:: DE: EF, or al-
ternately AD: DE
:: AG: EF. And
B
M
f
E
D
N
ト
​BOOK II.
273
since the ratio of AD to DE is given, that of AG to EF
is also given, and EF, the half of BC, being given, AG and
the point G are thence given. Wherefore the tangent GE
and its intersection D with AC, are given.
COMPOSITION.
Let M : N be the given ratio, and find the mean propor-
tional O; make O: M:: BF: AG, a perpendicular to AC,
and draw (III. 26. El.) the tangent GED; then M: N::
AD: BD.DC.
For join EF. Because the triangles ADG and EDF are
similar, AG: AD :: EF: ED, and alternately AG: EF::
AD: ED; but AG: EF:: M: O, and therefore M: O::
AD: ED, and M² O²:: AD: ED, that is, M: N::
:
AD: ED or BD.DC.
PROP. XVII. PROB.
In the same straight line, four points being given,
to find a fifth, such that the rectangle under its dis-
tances from the first and second points, shall have a
given ratio to the rectangle under its distances from
the third and fourth.
Let it be required to find a point E, so that AE.EB:
DE.EC:: M ; N.
1. Let M: N be a ratio of equality.
ANALYSIS.
Because AE.EB-DE.EC, it is manifest that AE: CE::
DE EB; whence
(V. 9. and 8. El.) A
AC: BD::CE: EB,
and (V. 9. El.)
B
C
E
AC+BD: BD:: BC: EB; but the ratio of AC+BD to
BD is given, whence that of BC to EB, and therefore BE
and the point E, are given.
a
274
GEOMETRICAL ANALYSIS.
1
}
COMPOSITION.
Make AC+ BD: BD : ; BC: EB, and E is the point re-
quired. For (V. 10. El.) AC: BD:: CE: EB, and (V. 19.
cor. 1. El.) AE: ED :: CE : EB, and hence (V. 6. El.)
AE.EB-CE.ED.
2. Let M: N be a ratio of majority or minority.
ANALYSIS.
Find, by the preceding construction, a point F, such that
AF.FB-DF.FC.
Because AE.EB:
M
DE.EÇ :: M: N, it
N
follows that AE.EB:
B
C
D
AE.EB-DE.EC::
FE
M: M-N; but
AE.EB-DE.EC=(AE.EB-AF.FB)+(DF.FC-DE.EC),
that is, EF (AF + BE) + EF (DF + CE), or = EF
(AD+BC.) Wherefore AE.EB: EF (AD+BC) :: M:
M-N; consequently the point E is assigned by Prop. 14. of
this Book.
The composition of the problem is thence easily derived,
by retracing the steps.
PROP. XVIII. PROB.
In the same straight line, four points being given,
to find a fifth, such that the rectangle under its dis-
tances from the extreme points shall have a given
ratio to the rectangle under its distances from the
mean points.
Let it be required to find a point E, so that AE.ED:
BE. EC :: M: N.
1. Let ABCD.
BOOK II.
275
ANALYSIS.
Because AE.ED=(AB÷BE) (AB+EC), it is evident
that AE.ED=AB.AC+BE.EC, whence AE.ED: AB.AC: :
M: M-N.
The ratio
of AE.ED to AB.AC
M
is therefore given, and
the rectangle under AE
and ED, the segments.
N
PH
A
of AD, being thus gi-
ven, the point E is as-
B
C
D
EE
signed by VI. 19. of the Elements.
COMPOSITION.
Make M-N: M:: AB: P, and (VI. 19. El.) cut AD in
E or E', such that AE.ED=P.AC; E is the point required.
For (V. 7. El.) M: M-N:: P: AB, and hence (V. 25. cor.
2. El.) M : M—N :: P.AC, or AE.ED: AB.AC; conse-
quently M: N:: AE.ED: AE.ED-BA.AC, or BE.EC.
2. Let AB and CD be unequal.
ANALYSIS.
Because AE.ED = (BE + AB) (EC + CD)=BE.EC+
BE.CD+AB.ED, consequently AE.ED-BE.EC-BE.CD
+ AB.ED BD.CD-ED.CD + AB.ED = BD.CD +
(AB—CD) ED. Produce AD to F, so that (AB—CD)DF
BD.CD; and
=
since AB, CD
MH
NH
and BD are
all given, DF 4
B
C D
F
E
and the point
F are given. Thus from construction AE.ED-BE.EC=
(AB—CD) (DF+ED)=(AB—CD) EF. Now, since the
$ 2
276
GEOMETRICAL ANALYSIS.
ratio of AE.ED to BE.EC is given, the ratio of AE.ED to
EF (AB—CD) is also given; wherefore AB-CD being gi-
ven, and the points A, C, and F, the point E is given by
Prop. 14.
Applying that proposition, the construction of the problem
is easily obtained.
It yet remains to assign the limitations of this problem.
On AD describe a circle, erect the perpendiculars BI and
GCH, join IOH, and, parallel to this, draw KEL through
the point of section E, join OG, EG, and IE, which produce
to the circumference, and join MG and ML.
The point O is evidently given. But the ratio of AE.ED
to BE.EC may be considered as compounded of the ratio of
AE.ED, or (III. 32. El.) IE.EM, to KE.EL, and of the ra-
tio of KE.EL to BE.EC.
Now, since BK and CL are parallel, KE: EL :: BE: EC,
or alternately KE: BE :: EL: EC, and therefore (V.
22. El.) KE : BE² : KE.EL: BE.EC. Again, KE
and IO being parallel, KE: IO:: BE: BO, or alternately
KE BE 10: BO, and hence KE²: BE² :: 10² : BO².
Wherefore IO: BO:: KE.EL: BE. EC, and consequently,
BO²
BE.EC,
the ratio of these rectangles is given; let it be that of PQ to
ST.
:
The angle MGL, being equal (III. 18. El.) to MIH in the
same segment, is equal (I. 23. El.) to the exterior angle MEL,
and consequently (III. 18. cor. El.) the quadrilateral figure
MGEL being thus contained in a circle, the angle LME is
(III. 18. El.) equal to LGE. Draw LN making the angle
MLN equal to EGO, and (I. 32. El.) the exterior angle
LNE will be equal to CGO. But the triangles GOC and
HOC are obviously equal, and, therefore, the angle CGO=
CHO=CLE=EKI. Whence the triangles IEK and LEN
are similar, and IE: KE: EL: EN, and consequently,
BOOK II.
277

KE.EL-IE.EN. Make,
therefore, PQ to PR, as
EN to EM. The ratio of
AE.ED to BE. ECis hence
compounded of that of PR
to PQ, and of PQ to ST,
or it is the same with the
ratio of PR to ST. But
as the point of section E
approaches to O, the angle
EGO, or MLN, evident-
ly diminishes, and conse-
quently the point N, in a
K
I
G
A
B
OVE
LAN
M
H
Z
B
P
STT
corresponding degree, approximates to M. Hence the ex-
treme term which PR can never pass, is PQ; and there-
fore the limiting ratio of the rectangle AE, ED to BE, EC
is that of PQ to ST, or of IO² to BO².
The point O of ultimate section, is hence easily determi-
ned. Because BI and CH are parallel, BI: CH :: BO:
OC, and BI² or AB.BD : CH², or AC.CD :: BO² : OC².
Wherefore BC must be divided (I. 14. Anal.) into segments
BO and OC, which are in the subduplicate ratio of the rect-
angle AB, BD to AC, CD.
But the limiting ratio may be found in a more direct man-
ner. For join IG, and draw DV perpendicular to it, join
DG, DI, CV, and BV, and, having drawn the diameter IZ,
join GZ. Because the angles DGC and DIV stand upon
equal arcs DH and DG, they are equal (III. 18. El.);
but the quadrilateral figures DCGV and DBIV, being right
angled at B, at C and V, are each contained in a circle
(III. 19. cor. El.); wherefore (III. 18. El.) the angle DGC
is equal to DVC, and the angle DIV to DBV, and conse-
quently the angles DVC and DBV are equal. Hence the
triangles CDV and VDB, having besides a common vertical
278
GEOMETRICAL ANALYSIS.
angle, are similar; and, therefore, BD: DV: : DV : DC,
and (V. 6. El.) BD.DC=DV². But (VI. 16. cor. 1. El.)
DG²=AD.DC, and consequently DG-DV² or (II. 11. El.)
GV² = AD.DC-BD.DC, or AB.DC. In the same man-
ner, it is shown that IV-AC.DB. Whence IG is given,
being the difference between the sides of two squares that
are equal to the rectangles AC, DB, and AB, DC. Again,
the angle BIO, being equal to the alternate angle GHI, is
equal (III. 18. El.) to GZI, and the right angle OBI is equal
to the angle IGZ in a semicircle; wherefore the triangles IOB
and ZIG are similar, and IO: BO :: IZ or AD : IG.
Hence the limiting ratio of AE.ED to BE. EC, or that which
marks the state of minimum, is the duplicate ratio of AD to
the difference of the sides of squares equal respectively to the
rectangle AC, DB and to the rectangle AB, DC.
PROP. XIX. PROB.
Through a given point, to draw a straight line, so
that the part intercepted by the circumference of a
given circle, shall be equal to a given straight line.
Let A be a point, through which it is required to draw a
straight line HI, limited by a given circumference and equal
to B.
ANALYSIS.
Take any point D in the
given circumference, and in-
flect DE equal to B. Be-
cause DE is equal to B, it is
equal to HI, and, therefore,
(III. 11. El.) the chords HI,
DE are equally distant from
the centre of the circle, or
CG=CF. But DE being gi-
I

G
H
C
D
F
E
B
ven, CF is given, and thence the circle described from C
BOOK II.
279
through F and G; wherefore the point A being given, the
tangent AG to that circle is given, and consequently HI is
given in position.
COMPOSITION.
Inflect DE equal to B, from C let fall the perpendicular
CF, with which distance describe a concentric circle, and
draw (III. 26. El.) the tangent HAI.
It is evident that the chords HI and DE, being equidis-
tant from the centre, are both of them equal to B.
PROP. XX. PROB.
Through a given point, to draw a straight line,
such that the part of it intercepted between two
concentric circles shall be equal to a given straight
line.
Let it be required, through the point A, to draw the
straight line ABC, so that the part BC intercepted by the
two concentric circles HECM and IFBL shall be equal to D.
ANALYSIS.
B
From any point H, in one of the circumferences, inflect
HM=EC, and upon these let fall the perpendiculars OK
and OG. The equal chords HM and EC are therefore
equidistant from the centre, and reciprocally IL is equal to
FB; consequently the halves
of these are equal, or HK=
GC, and IK=GB; whence
the difference HI, being equal
to BC, is given. But since
the point H is given, the
point I and the chord HM
are given; and the circle
which touches at K being gi-
ven, the tangent AGC is also given.

A
E
H
Κ
L
N
DH
280
GEOMETRICAL ANALYSIS.
}
:
COMPOSITION.
In the circumference of one of the circles, having assumed
a point H, place HI equal to D, and produce it to M, upon
this let fall the perpendicular OK, with which as a radius de-
scribe a circle, and apply to it the tangent ABC; then will
the intercepted portion BC be equal to D.
For the chords EC and FB are (III. 11. El.) equal to the
equidistant chords HM and IL; consequently their halves
are equal, or GB=IK, and GC=HK, and hence BC=
HI=D.
It is evident, that the interval BC between the concentric
circles will be least when AC passes through the centre, and
greatest when it touches the inner circle. Wherefore D is
limited on both sides; not being less than the difference of
the radii of the circles, nor its square greater than the differ-
ence of their
squares.
PROP. XXI. PROB.
Two circles described upon the same straight line
being given, to draw from a point similarly placed
in it another straight line, so that the part intercept-
ed by the circumferences shall be equal to a given
straight line.
Let D, E be the centres of the two circles, and let AD: AE
:: DI: EK; it is required from A to draw ABC, such that
BC shall be equal to L.
BOOK II.
281
ANALYSIS.
B
Join BD and CE. Because AD: AE :: DI or DB : EK,
or EC, therefore (VI. 1. cor. El.)
DB is parallel to EC; whence
AD: DE :: AB: BC, and since
AD and DE are given, the ratio
of AB to BC is given; but BC is
given, and consequently AB is gi-
ven, both in magnitude and posi-
tion.

GHA
L
M
DE I
K
COMPOSITION.
Make (VI. 3. El.) EK-DI: DI:: L: M, and from A in-
flect AB equal to M; ABC is the straight line required.
For since, by hypothesis, AD: AE :: DI or DB : EK or
EC, DB is parallel to EC; wherefore DB or DI: EC or EK
:: AB: AC, and consequently (V. 11. cor. El.) EK—DI : DI
:: BC: AB; but EK-DI: DI:: L: M or AB, whence
BC: AB:: L: AB, and therefore (V. def. 10. El.) BC=L.
PROP. XXII. PROB.
Two circles described upon the same straight line
being given, to draw, from the extremity of either
diameter, another straight line, so that the part of it
intercepted by the circumferences shall be equal to
a given straight line.
Let it be required to draw ABC, so that the intercepted
portion BC shall be equal to QR.
ANALYSIS.
Join BG, CH, and FP, from E, the centre of the exterior
circle, let fall upon AC the perpendicular EI, cut off IL=IB
and draw LK parallel to BG, in the extension of AH make
282
GEOMETRICAL ANALYSIS.
(VI. 3. El.) AK : AG :: AF : AM, and, from the point M,
draw MN parallel to FP, and meeting the production of AC.
Because LK is parallel to BG and FP to MN, therefore
(VI. 1. El.) AK: AG :: AL: AB, and AF:AM::AP:AN;
but, by construction,
AK: AG: AF: AM,
and, consequently, AK :
AG AL : AB : :

AP
:
AN. Whence
(V. 19. El.) AK: AG ::
AL+AP or PL: AB+
B
I
M
F
H
KDE
G
P
AN or BN. Now, since
(III. 4. El.) IP=IC,
and IL IB, therefore
PL=BC or QR; and
LK, IE, and BG being
S
Q
R
parallel lines, KE=EG (VI. 1. El.) and thence AK is given;
wherefore three terms of the analogy being given, the fourth
term BN is given, and consequently BN+BC, or NC, is given.
But the angle ACH is equal to AFP (III. 18. El.) which again
(I. 23. El.) is equal to AMN, and hence the triangles CAH
and ANM, having also the same vertical angle, are similar,
consequently AH: AC:: AN: AM, and (V. 6. El.) AH.AM
=AC.AN, wherefore NC and the rectangle under its seg-
ments AC, AN being given, AC is given in magnitude (VI.
19. El.) and hence likewise in position.
COMPOSITION.
Having cut off KE=EG, make (VI. 3. El.) AK : AG : :
AF: AM, and AK: AG :: QR: QS, divide (VI. 19. El.)
SR in O such that SO.OR=AH.AM, and inflect AC=
OR; AC is the straight line required.
For join CH, BG, FP, and draw MN parallel to FP,
and EI and KL parallel to CH. Since ILIB and IP=IC,
therefore PL-BC. The triangles CAH and ANM are si-
BOOK II.
283
milar, AH: AC :: AN: AM, and AH.AMAC.AN; but,
by construction, AH.AM = SO.OR, and AC = OR, con-
sequently AN SO. Now, from the property of paral-
lels, AK: AG :: AL: AB, and, by hypothesis, AK : AG : :
AF: AM, or AP: AN; wherefore (V. 19. El.) AK : AG ::
AL+AP or BC: AN + AB or BN. Whence BC: BN
:: QR: QS, and (V. 11. El.) BC: CN:: QR: SR; but CN=
SR, and consequently BC=QR.
PROP. XXIII. PROB.
From the extremity of the diameter of a given
circle, to draw a straight line, so that the part of it
intercepted between a given perpendicular and the
circumference, shall be equal to a given straight
line.
Let it be required from A to draw AC, such that the in-
tercepted portion BC shall be equal to GH.
ANALYSIS.
Join BD. The angle ABD, being in a semicircle, is a
right angle, and therefore
equal to AEC; consequent-
ly the triangles DAB and
CAE, having besides a
common angle at A, are
similar, and AB : AD::
AE : AC, and hence
AB.ACAD.AE. But
the rectangle AD,AE is
given, and thence AB,AC;

B
E
DE
G
H
H'
I'
and since BC is given in magnitude, therefore (VI. 19. El.)
AB is given in magnitude, and consequently in position.
COMPOSITION.
Produce GH (VI. 19 El.) till GI.IH=AD.AE, and in-
flect IH from A to B; AB is the straight line required. For
284
GEOMETRICAL ANALYSIS.
join BD. The triangles ABD and AEC being evidently si-
milar, AB AD :: AE: AC, and consequently AB.AC=
AD.AE=GI.IH; but AB=IH, whence ACGI, and
:
therefore BC=GH.
PROP. XXIV. PROB.
Through a given point in the line bisecting a
given angle, to draw a straight line limited by the
sides, and equal to a given straight line.
Let it be required, through the point D, situate in the
straight line AD which bisects the angle BAC, to draw BC
equal to a given straight line.
ANALYSIS.
About the points B, A and C, describe (III. 10. El.) a
circle, draw the diameter EF,
and join AF. Because BC and
the angle BAC are given, the
circumscribing circle (III. 27.
El.) and consequently the tri-
angle BAC, are given in mag-
nitude: But since the angle
BAE is equal to CAE, the arc
BE is (III. 18. cor. El.) equal
to CE; and hence the chord
BC is bisected at right angles
by the diameter EF. Where-
fore AD being given, AE is,
by the last proposition, given
in magnitude, and thence DB
is given in magnitude, and con-
sequently in position,
F

D
B
E
B

E
A
F
C
B/
D
COMPOSITION.
On the given straight line describe (III. 27. El.) a seg-
ment BAC, containing an angle equal to the given angle,
and complete the circle, bisect the arc BAC in E, and from
BOOK 11.
285
that point draw, by the last proposition, EAD, such that AD
shall be equal to the distance of the given point from the ver-
tex; and DB,DC are the segments of the required line, from
which its position is immediately determined.
For the angle BAC is equal to the given angle, and AD
bisects it, since the arc BE=CE; but AD is besides equal
to the distance of the given point from the vertex, and
BC is equal to the given straight line. Wherefore all the
points and lines retain, by this construction, their relative
position.
Since AE cannot exceed the diameter FE, the limiting
case will occur when these lines coincide; whence BC is the
least possible when at right angles to AD, and therefore
intercepting equal segments AB and AC.
PROP. XXV. PROB.
Between the side of a given rhombus and its ad-
jacent side produced, to insert a straight line of a
given length, and directed to the opposite corner.
Let ABCD be a rhombus, of which the side BC is pro-
duced; it is required, from the opposite corner A, to draw
AEF, such that the exterior portion EF shall be equal to a
given straight line.
ANALYSIS.
G
B
сл
Join AC, and, meeting this produced, draw EG, making
the angle AEG equal to ACF.
The triangles CAF and EAG are
evidently similar, and AC: CF ::
AE: EG; but CE being paral-
lel to AB, BC: CF:: AE: EF
(VI. 1. El.); whence (V. 17. El.)
AC: BC: EF: EG. But AC,
BC, and EF being given, EG
is (VI. 3. El.) also given. Again,
the angle ACD is (I. 2. El.)

ار
K
1.
E
D)
286
GEOMETRICAL ANALYSIS.
equal to ACB, and therefore to FCG; consequently adding
ECF to each, the whole angle ACF, or AEG, is equal to
ECG. Hence the triangles AGE and EGC are similar, and
AG: EG:: EG: GC, or AG.GC=EG². Wherefore the rect-
angle AG, GC is given, and consequently (VI. 19. El.) the
point G, and thence the point E and the straight line AF.
COMPOSITION.
Let the intercepted segment be equal to K, join AC, make
AC: BC:: K: L, divide AC in G (VI. 19. El.) so that
AG.GC=L2, and from G, with the radius L, describe a cir-
cle cutting CD in E; AEF is the straight line required.
For since AG.GC=L²= EG², AG: EG :: EG: GC, and
therefore the triangles AGE and EGC are similar, and the
angle AEG is equal to ECG, or ACF; whence the triangles
AFC and AGE are likewise similar, and AC: CF:: AE: EG;
but (VI. 1. El.) BC: CF :: AE: EF, and consequently
(V. 17. El.) AC: BC:: EF: EG. Now AC: BC:: K: L
or EG; wherefore EF-K.
Otherwise thus.
ANALYSIS.
B
IT
C
F
Draw FG making the angle AFG equal to ADC, cut off
CH=CE, from C inflect CN=CA, and join CG and AH.
The triangle ACN being isosceles, the angle CAN is
(I. 11. El.) equal to CNA;
and since the diagonal AC
bisects the angle BCD of the
rhombus, the triangles ACE
and ACH are (I. 3. El.) like-
wise equal, and hence AE is
equal to AH, and the angle
CAE equal to CAH.
because the triangles ADE

E
Л.
DN
G-
And
K
I
and AFG are similar, AD: AE
AF: AG and AD.AG=AE.AF. But the angle ACD,
BOOK II.
287
::
:
being equal to CAD, is equal to CNA, and consequently the
triangles ADC and ACN are similar; whence AN AC::
AC: AD, and therefore AN.AD=AC². Again, because
AC bisects the vertical angle HAF (VI. 21. El.) FA.AH=
AC²+FC.CH, that is, FA.AE=AC²+FC.CE; wherefore
FC.CE=FA.AE-AC, that is, AG.AD-AN. AD, or
NG.AD. But BA and CE being parallel, FC: EF::
AD: AE AF: AG, and CE: EF:: AB or AD: AF;
consequently (V. 22. El.) FC.CE : EF² :: AD : AG ::
(V.25. cor. 2.El.) NG.AD: NG.AG; since therefore FC.CE=
NG.AD, it follows (V. 8. and 4. El.) that EF²=NG.AG.
Hence (VI. 19. El.) AG and the point G are given, and the
angle AFG, being equal to ADC, is (III. 27. El.) contained
in a given segment of a circle; wherefore the intersection F
and the inflected line AF, are given.
COMPOSITION.
Let K be equal to the intercepted portion of the straight
line which is to be inflected from A, and find (II. 13. El.) L
the side of a square equivalent to the squares of K and of the
diagonal AC, produce AD, and from C place CG equal to
L, upon AG describe (III. 27. El.) a segment of a circle con-
taining an angle equal to ADC, and join A with the point of
intersection F; AF is the straight line required.
For inflect CN=CA, and join GF and AH.
The triangles AHC and AEC are equal; for the angle
AFG, being by construction equal to ADC, is equal
(I. 23. El.) to the alternate angle formed by the produc-
tion of BA with AD, and consequently (III. 25. cor. El.) AB
touches the circle at A; whence the angle BAH=HFA=
DAE, and taking these from the equal angles BAC and DAC,
there remains CAH-CAE, but the angles ACH and ACE
are also equal, and the side AC is common to the two tri-
angles; wherefore AH-AE, and CH-CE. And because
the triangles ADE and AFG are similar, AD: AE ::
AF: AG, and AD.AG=AE.AF. Again, the triangles
288
GEOMETRICAL ANALYSIS.
ANC and ACD being similar, AN: AC:: AC: AD, and
AN.AD AC². But FC: EF:: AD: AE:: AF: AG, and
CE: EF: AB or AD: AF; consequently FC.CE: EF² ::
AD: AG: NG.AD: NG.AG; and since AC bisects the
angle FAH, FC.CH+ AC² = FA.AH=FA.AE=AG.AD
=AN.AD+NG.AD, it follows that FC.CH, or FC.CE=
NG.AD, and hence EFNG.AG. Now KCG²—AC² =
(II. 23. cor.) NG.AG; wherefore EFK, and EF=K*.
PROP. XXVI. PROB.
Through two given points, to describe a circle
touching a straight line given in position.
Let it be required to describe a circle through the points
A, B, and touching the straight line CD.
It is evident that CD must either be parallel or inclined to
the straight line which joins the points A and B.
1. Let CD be parallel to AB.
ANALYSIS.
From the point of contact E, draw (I. 6. El.) EG perpen-
dicular to CD. Hence (III 24. cor. El.)
EG passes through the centre of the cir- C
cle, and since it is also perpendicular to
E
D

AB (I. 23. El.) it bisects that chord at
right angles (III. 4. El.) the point G is
B
G
therefore given, and the perpendicular
GE; consequently the three points A, E, and B being thus
given, the circle AEB is given.
COMPOSITION.
Draw (I. 7. El.) GE bisecting AB at right angles, and
(III. 10. cor. El.) through the points, A, E and B describe
a circle; this will touch the straight line CD.
See Note LVI.
BOOK II,
289
For (III. 6. El.) GE must pass through the centre of the
circle, and (I. 23. El.) it meets the parallels CD and AB aț
right angles; whence (III. 24. El.) CD is a tangent to the
circle.
2. Let CD be inclined to AB.
ANALYSIS.
Produce BA to meet CD in F. Then (III.32. cor. 2. El.) FE*=
AF.FB; but the point
of concourse being gi-
ven, the rectangle AF,
FB is given, and con-
sequently FE and the
point E. Wherefore
since the three points
A, E, and B are given,
the circle AEB is given.

E
E
D
T
B
COMPOSITION.
Produce BA to meet CD in F, find (VI. 18. El.) FE or
FE' a mean proportional to AF and FB, and (III. 10. cor. El.)
through the points A, B, and E, or A, B, and E', describe a
circle; this will touch the straight line CD.
For since AF:FE::FE : FB, therefore (V. 6. El.) FE²
=AF.FB, and consequently (III. 34. El.) FE, or FE', touches
the circle.
PROP. XXVII. PROB.
Through a given point, to describe a circle touch-
ing two straight lines given in position.
Let it be required, through the point E, to describe a circle
touching AB and CD.
1. Suppose AB parallel to CD.
T
{
290
GEOMETRICAL ANALYSIS.
ANALYSIS.
Through the centre O draw the parallel FO and the com-
mon perpendicular KI. It is
evident that the radius OI is
given, and consequently FO is
given in position; but OE, be-
ing equal to OI, is given, and
therefore the centre O is given.
C
F
A
COMPOSITION.
K
D

6
1
B
Draw a parallel FO bisecting the distance between the
straight lines AB and CD, and from E with a radius equal to
half that distance intersect FO in O, or O'; this point is the
centre of the circle required. For OE=O!=OK, and the
circle which passes through E must touch at K and I.
2. Suppose CD inclined to AB.
ANALYSIS.
Produce BA and DC to meet in F, join OI, OK, and OF,
and from E draw EGH perpendicular to OF.
The triangles OKF and OIF, being (III. 24. El.) right-an-
gled, and having the side OK equal to OI and the side OF
common, are (I. 22. El.) equal, and consequently the angle
OFK is equal to OFI;

K
H
wherefore since the point of
concourse F is given, the
straight line OF is given.
But, the point E being gi-
ven, the perpendicular EH
is thence given, and (III. 4.
El) GH being equal to GE,
the opposite point H is given. Two points E, H, and a
straight line AB, are thus given, and therefore, by the last
proposition, the circle EHKI is given.
F
E
A I' L 1
13
BOOK II.
291
1
COMPOSITION.
Produce BA and DC to meet in F, draw (I. 5. El.) FO bi-
secting the angle BFD, from E (I. 6. El.) let fall the perpen-
dicular EG, and extend it both ways, making GH-GE, find
(VI. 18. El.) LI, or LI', a mean proportional to HL and¸LE,
and through the points H, E, I, or H, E, I', describe a circle;
this circle will touch both the straight lines AB and CD.
For the centre of the circle which passes through E and
H, must (III. 5. El.) occur in FO; let it be O, join OI and
draw the perpendicular OK. Because HL.LE=LI², the
circle touches AB at I, and hence OIF is a right angle; con-
sequently the triangles KOF and IOF having the angles OKF
and OFK equal to OIF and OFI, and the side OF common,
are (I. 21. El.) equal, and therefore OI-OK; whence the
circle described from O passes through K, and (III. 24. El.)
must touch CD at that point.
Cor. If the given point E should fall on AB, and thus coin-
cide with the point of contact,-the problem will become much
simpler; for the centre O, lying in the intermediate or bi-
secting line FO, will be determined by the intersection of the
perpendicular IO.
PROP. XXVIII. PROB.
Through two given points, to describe a circle
touching a given circle.
Let it be required, through the points A and B, to describe
a circle, touching another circle whose centre is C.
ANALYSIS.
Through D, the point of contact, draw ADE and BDF,
join EF, at F (I. 5. cor. 2. El.) apply the tangent FG, and
draw BHCI.
Because FG touches the given circle, the angle BFG is
(III. 25. El.) equal to FED, and therefore equal to BAD,
T 2
292
GEOMETRICAL ANALYSIS.
1
:
since (III. 29. El) FE and AB are parallel; but the triangles
BGF and BDA have likewise
a common angle at B, and are
hence similar; wherefore BF:
BG:: BA: BD, and (V. 6. El.)
BA.BG = BF.BD = (III. 32.
El.) BI.BH. But BI and BH
are given, and thence the rect-
angle BA, BG is given, and
consequently II. .. El. the
point G is given. Hence the
tangent GF, and D, the inter-
section of BF, are given;
wherefore the circle that passes
through the three points A, D,
and B, is given.
COMPOSITION.

E
E
C
B
G-
I
H

1
Ε
G/B
Make (VI. 3. El.) BA BI:: BH: BG, draw (III. 26. El.)
the tangent GF, join BF cutting the given circumference in
D, and (III. 10. cor. El.), through the points A, D, and B,
describe a circle; this will touch the circle FDE.
:
For draw ADE, join FE, and draw BHCI. Since BA : BI
:: BH BG, therefore (V. 6. El.) BA.BG = BI.BH=
(III. 32. El.) BF.BD; whence BF: BG:: BA: BD, and
consequently the triangles BGF and BDA, having the same
vertical angle, are (VI. 14. El.) similar, and hence the angle
BFG is equal to BAD. But (III. 25. El.) BFG is equal to
FED, and thus the alternate angles BAE and FEA are
equal, and FE is parallel to AB; whence (III. 29. El.) the
two circles touch at D.
PROP. XXIX. PROB.
Through a given point, to describe a circle,
touching a given circle and a straight line which is
given in position.
BOOK II.
293
Let it be required, through the point A, to describe a cir-
ce touching the straight line CD and the circle whose centre
is B..
ANALYSIS.
From the centre of the given circle let fall the perpendicu-
lar EBG, join EI and extend it to H in the straight line
CD, also draw FIK and join HK.
The angle HIK, being equal to EIF which stands in
a semicircle, is (III. 22. El.) a right angle, and consequently
HK is the diameter of the
circle ILA, and H the point
of contact.
The triangles
HEG and FEI are there-
fore similar, HE : EG ::
EF EI, whence HE.EI
=EG.EF. Join ELA, and
(III. 32. El.) AE.EL=
HE.EI=EG.EF; but the
rectangle EG, EF is given,
and consequently that of
HE, EI, and EH being given,
the point L is hence given.
Wherefore, since the two
points A, L, and the straight
line CD, are all given,—the
circle HIA is given.
E

B
LK
I
A
Fi
C
G
H
D
T
K
I
P
C
H\G
D
E
1

COMPOSITION.
Join EA, draw the perpendicular EG, make (VI. 3. El.)
AE: EG :: EF EL, and by Prop. 26. of this Book, de-
scribe a circle through the points A, L, and touching the
straight line CD; this circle will also touch the given circle.
For draw the diameter HK, join EH cutting the circum-
ference EIF, and draw FIK meeting HK.
The triangles HEG and FEI being evidently similar,
294
GEOMETRICAL ANALYSIS.
ว
HE: EG:: EF: EI, and HE.EI-EG.EF; but AE: EG::
EF: EL, and AE.EL=EG.EF; wherefore HE.EI=AE.EL,
and (III. 34. El.) the point I must lie in the circumference
HIK. But the two circles also touch in I; for EG being
parallel to HK, the angles IEF and IHK are equal, which
are again equal to those made by a tangent with IF and IK.
Cor. The problem will be greatly simplified, if the given
point A should occur in the straight line or in the circle, and
hence, coincide with either of the points of contact H or I;
for EIH and FIK being drawn, the perpendicular HK is the
diameter of the required circle.
PROP. XXX. PROB.
Through a given point, to describe a circle touch-
ing two given circles.
Let it be required, through the point C, to describe a cir-
cle touching two given circles whose centres are A and B.
ANALYSIS.
Join AB, and produce it to meet, in D, the extension of
the straight line which connects E, F, the points of contact;
join OA and OB, AG and BH, draw CEI, and produce IG
and DC to meet in K.
The isosceles triangles EOF, EAG, and FBH, are evi-
dently similar, and conse-

quently AG is parallel to BF
and AE to BH. Whence
(VI. 2. El.) AE: BH ::
AD: BD; and, this ratio
being therefore given, the
point D is given. Again,
AG: BF:: DG : DF, and
I
A
E
B
H
DG: DF :: DK : DC, for
(III. 29. El.) IG is parallel to
K
C
כ.
FC; consequently, DC being given, DK, and the point K,
BOOK II.
295
are given. Wherefore, by Proposition 17. of the first Book
of Analysis, the straight line GE, included by the reflected
lines KI and CI, and directed to the given point D, is given ;
hence AEO is given in position. Join OC, and the angle
ECO, being equal (I. 11. El.) to CEO, is given; and conse-
quently CO, and the centre O, are given.
COMPOSITION.
Make (VI. 3. El.) AE: BH:: AD: BD, join DC, make
BH: AE :: DC: DK; and, from the points K and C, in-
flect KI and CI, by Prop. 17. Book I. such that GE shall tend
to D, produce AE and CO, making the angle ECO equal to
CEO; the intersection O is the centre of the required circle.
For join AG, CF, OB, and BH. Because AE or
AG BH or BF :: AD: BD, and the triangles ADG and
BDF have a common angle at D, they are (VI. 15. El.) si-
milar; consequently AD: BD :: DG : DF :: DK : DC, and
IG is parallel to FC; and therefore the circles touch at E.
But the triangle, BFH, having its sides BF and BH paral-
lel to AG and AE, the sides of the isosceles triangle GAE,
must likewise be isosceles; wherefore the circles meet at F:
And, since BH is parallel to EO, they must touch at that
point. Again, the angle ECO being equal to CEO, the side
OE is equal to OC; and consequently the circle described
from O, and which touches at E and F, must also pass
through C.
Otherwise thus.
ANALYSIS.
Join the centres A, B and O, and produce AB and the
straight line connecting E, F, the points of contact, till they
meet in D; join also BH and DC, and extend this to cut the
circle in L.
*
296
GEOMETRICAL ANALYSIS.
Since EOF and FBH are isosceles triangles, the vertical
angles OFE and BFH are equal
to OEF and BHF, which are
therefore equal, and consequent-
ly (I. 23. El.) EO is parallel to
BH; whence (VI. 2. El.) AE
: BH :: DA: DB, and the
point D is therefore given.
Again (VI. 1. El.) DA: DB::
DE: DH, or (V. 25. cor. 2. El.)
DE.DF: DH.DF; but the

L
1
E
A
B
T
11
D
is
rectangle DH, DF, being equal (III. 32. El.) to the rectangle
under the segments of DB intercepted by the circle B,
given, and hence DE.DF or DC.DL are given rectangles;
wherefore DC being given, DL and the point L are likewise
given. The problem is thus reduced to Proposition 28. of
this Book.
COMPOSITION.
Make (VI. 3. El.) AE: BH :: AD: BD, join DC, and
produce it to L, such that the rectangle DC, DL shall be to
the rectangle formed by a secant drawn from D to the circle
B, in the ratio of AE to BH, and (II. 28. Anal.) describe a
circle through the points C and L, and touching the circle
A; this will also touch the circle B.
:
For join OA, OB, EH, and draw BH parallel to AO.
Because AE BH:: AD: BD, it is evident that EH, being
produced, will meet AD in D; hence AE: BH : : DE : DH,
or (V. 25. cor. 2. El.) DE.DF :: DH.DF; but, by construc-
tion, AE: BH :: DC.DL : DH.DF, and consequently DC,
DL is equal to DE, DF, and the point F lies in the circle
O. Wherefore the triangle EOF is isosceles, and likewise
the similar triangle HBF; hence F belongs also to the circle
B and (III. 28. El.) is the point of mutual contact.
BOOK II.
297
J
If L should coincide with the point C, the construction
will be effected by the corollary to the preceding Proposition *.
PROP. XXXI. PROB.
To describe a circle that shall touch a given cir-
cle and two straight lines given in position.
Let it required to describe a circle touching the straight
lines AB and CD, and another circle whose centre is E.
ANALYSIS.
Join FE, draw FH,FI to the points of contact, from
F, with the radius FE,
describe a circle meet-
ing FH and FI pro-
duced in K and L, and,
at these points, apply
the tangents MN and
OP.
Because FE=FK=
FL and FG = FH=
FI, therefore GE=HK
=IL. But the tangents
CD and OP, being per-
pendicular to FK, are
parallel; and, for the
same reason, the tan-
gents AB and MN are
parallel. Wherefore OP
and MN are given in
position, and conse-
quently, by Prop. 27.

E
F
K
H
A
B
M
I
N

C
H
K
E
D
M
L
N
the circle EKL is gi-
A
ven; and thence the concentric circle GHI.
* See Note LVII.
1
B
298
GEOMETRICAL ANALYSIS.
પ
COMPOSITION.
At a distance equal to the radius of the given circle, draw
MN and OP parallel to AB and CD; and, by Prop. 27. of
this Book, find F the centre of a circle which passes through
E and touches MN and OP; F is likewise the centre of the
required circle.
For join FE, and draw FK and FL to the points of con-
tact. And because GE=HK=IL, it is evident that FG=
FH=FI. But the circle also touches at the points H and
I, since CD and AB are perpendicular to FK and FL.
Scholium. The six preceding propositions are only cases
of a general problem: "Three things being given,—whether
points, or straight lines, or circles, to describe a circle
limited by them all." This problem comprizes ten dis-
tinct cases. Two of these have been already given in the
Elements: To describe a circle through three given points,
forms the 10th Prop. Book III.: To describe a circle that
shall touch three straight lines given in position, is the basis
of Prop. 10. Book IV., and appears complete in the con-
struction of Prop. 31. Book VI. The same principle, it may
be perceived, runs through all the solutions already given; the
conditions of the problem are only repeatedly simplified, each of
the linear or circular data being exchanged in succession for
a point. Two cases still remain: When there are given three
circles or two circles and a straight line, to describe another
circle limited by these data. These are easily reduced, how-
ever, to the cases already solved, as in the concluding propo-
sition, by drawing a parallel, or describing a concentric cir-
cle, at distances, according to the relative position of the data,
equal to the sum or difference of the given radii *
* See Note LVIII.
!
GEOMETRICAL ANALYSIS.
BOOK III.
DEFINITION.
If a point vary its position according to some determined
law, it will trace a line which is termed its Locus.
PROP. I. THEOR,
If a straight line, drawn through a given point
to a straight line given in position, be divided in a
given ratio, the locus of the point of section is a
straight line given in position.
Let the point A and the straight line BD be given in po-
sition, and let AB, limited by these, be cut in a given ratio
at C; this point will lie in a straight line which is given in
position.
ANALYSIS.
From A let fall the perpendicu-
lar AD upon BD, and, through
C, draw CE parallel to BD. It is
evident (VI. 1. El.) that AC: AB
:: AE : AD, and consequently
that the ratio of AE to AD is
given; but AD is given both in
E
B
C
D
I
position and magnitude, and hence AE and the point E are
given, and therefore CE, which stands at right angles to AD,
is given in position.
}

300
GEOMETRICAL ANALYSIS.
COMPOSITION.
Let fall the perpendicular AD, which divide at E in the
given ratio, and erect the perpendicular CE; this straight line
is the locus required. For CE being parallel to BD, AC :
AB:: AE: AD, that is, in the given ratio.
PROP. II. THEOR.
If a straight line, drawn through a given point to
the circumference of a given circle, be divided in a
given ratio, the locus of the point of section will also
be the circumference of a given circle.
Let AB, terminating in a given circumference, be cut in a
given ratio; the segment AC will likewise terminatę in a gi-
ven circumference.
:
ANALYSIS.
B
Join A with D the centre of the given circle, and draw CE
parallel to BD. It is obvious
(VI. 1. El.) that AC AB :: AE:
AD; whence the ratio of AE to
AD being given, AE and the
point E are given. Again, since
(VI. 2. El.) AC: AB::CE: BD,
the ratio of CE to BD is given,
and consequently CE is given
in magnitude. Wherefore the

A
ED
one extremity E being given, the other extremity of CE must
trace the circumference of a given circle.
COMPOSITION.
Join AD, and divide it at E in the given ratio, and in the
same ratio make DB to the radius EC, with which, and from
the centre E, describe a circle.
BOOK III.
301
For draw AB cutting both circumferences, and join CE
and BD. Because CE BD :: AE: AD, alternately CE :
AE: BD: AD; wherefore the triangles CAE and BAD,
having likewise a common angle, are similar, and consequent-
ly AC: CB:: AE : AD, that is in the given ratio.
PROP. III. THEOR.
If, through a given point, two straight lines be
drawn in a given ratio and containing a given angle;
if the one terminate in a straight line given in posi-
tion, the other will also terminate in a straight line
given in position.
Let the ratio of BA to AC, with the angle BAC and its
vertex A, be given; if the extremity B lie in the straight line
BD, the extremity C will have for its locus another straight
line given likewise in position.
ANALYSIS.
Let fall the perpendicular AD upon BD, draw AE form-
ing with AD an angle DAE equal to BAC, and make AB :
AC:: AD: AE; CE being joined, is the locus required.
Because the angle DAE is, by construction, equal to BAC,
it is given; and the perpendicular AD being given, the
straight line AE is, therefore, given in position. But AB :
AC AD AE, and this being a given ratio, AE is hence
given also in magnitude. Again, since the angle BAC is
equal to DAE, the angle BAD is equal to CAE; and be-
cause AB AC :: AD: AE, alter-
:
nately AB AD AC: AE;

:
: :
wherefore the triangles ABD and
ACE, having their vertical angles
equal, and the sides containing
those angles proportional, are (VI.
14. El.) similar, and consequently
B D
C
E
302
GEOMETRICAL ANALYSIS.
the angle CEA is equal to BDA, and therefore a right angle;
consequently the straight line EC is given in position.
COMPOSITION.
Having let fall the perpendicular AD, and made the angle
DAE equal to BAC, make AD to AE in the given ratio,
and, at right angle to AE, draw EC; this is the locus requi-
red. For the triangles BAD and CAE, having their vertical
angles equal, and the angles at D and E right angles, are si-
milar, and consequently AB : AD: : AC: AE, or alternate-
ly AB AC AD AE, that is, in the given ratio.
:
:
PROP. IV. THEOR.
If, through a given point, two straight lines be
drawn in a given ratio, and containing a given an-
gle; if the one terminate in a given circumference,
the other will also terminate in a given circumfe-
rence.
Let the angle BAC, its vertex A, and the ratio of its sides,
be given; if AB be limited by a given circle, the locus of C
will also be a given circle.
ANALYSIS.
Join A with D the centre of the given circle, draw AE at
the given angle with AD, and in the given ratio, and join
DB and EC.
Because the point A and the centre D are given, the
straight line AD is given; and since the angle DAE, being
equal to BAC, is given, AE is given in position. But AD
being to AE in the given ratio, AE must be given also in
BOOK III.
303-

magnitude, and consequently the
point E is given.
B
Again, the whole angle BAC
being equal to DAE, the part
BAD is equal to CAE; and be-
cause AB : AC :: AD : AE, al-
ternately AB: AD:: AC: AE;
wherefore the triangles ADB and
AEC are similar, and hence
D
E
C
AB: BD: AC CE, or alternately AB: AC:: BD: CE;
consequently the fourth term CE is given in magnitude; and
its extremity E being given, the other must lie in a given cir-
cumference.
COMPOSITION.
Having drawn AE at the given angle with AD, make AD
to AE in the given ratio, and in the same ratio let DB be
made to EC; a circle described from the centre E with the
distance EC, is the locus required.
:
For AD: AE: : DB: EC, and alternately AD : DB : :
AE: EC; but the angle BAD is equal to CAE, because the
whole BAC is equal to DAE; consequently the triangles
ABD and ACE are similar, and AB: AD :: AC: AE or
alternately AB AC:: AD: AE, that is, in the given ratio.
Scholium. Since the tangent of a circle is only the extreme
limit of its adjacent arc, which, in proportion as the circle ex-
pands, must continually approach to that ultimate position—
the rectilineal, may be derived from the circular, locus. Thus, in
Prop. 2. of this Book, if the centres E and D be supposed to
retire to a distance indefinitely remote, the arcs which pass
through C and B may be viewed as merging in their tan-
gents or in perpendiculars let fall from those points upon AD,
which is the first proposition. In like manner, if the circles
in Prop. 4. be supposed immeasurably expanded, the arcs in
which the points B and C lie may be conceived to pass into
tangents perpendicular to AD and AE, as in Prop. 3.
304
GEOMETRICAL ANALYSIS.
J
PROP. V. THEOR.
If a straight line, drawn from a given point to a
straight line given in position, contain a given rect-
angle, the locus of its point of section will be a given
eircle.
Let the rectangle AB, AC be given, while the point B and
the straight line BD are given in position; the point C will
lie in the circumference of a given circle.
ANALYSIS.
Draw AD perpendicular to BD, and make the rectangle
AD.AE=AB.AC. Since AD is evidently given both in po-
sition and magnitude, AE and the point
E are given. Join CE. Because
AD.AE=AB.AC, AD: AB:: AC:
AE, and the triangles DAB and CAE,
having the sides about the common
angle at A proportional, are therefore
similar; and consequently the angle

A
B
D
E
ACE is equal to ADB, or a right angle. Whence (III. 22. El.)
the point C must lie in a semicircle, of which AE the diame-
ter is given.
COMPOSITION.
Having drawn the perpendicular AD, make the rectangle
AD, AE equal to the given space, and upon the diameter AE
describe a circle; this is the locus required. For draw AC
and CE. The triangles ABD and AEC are similar, since
they have a common angle at A, and those at D and C right
angles; wherefore AB AD AE: AC, and AB.AC=
AD.AE, that is, equal to the given space.
::
BOOK III.
305
PROP. VI. THEOR.
If a straight line, containing a given rectangle, be
drawn through a given point to the circumference of
a given circle, the locus of its point of section will
be either a straight line given in position or a given
circle, according as it originates or not in the given
circumference.
Let the rectangle AC, AB be equal to a given space, and
the segment AC terminate in a given circumference, the point
of origin A may lie either in that circumference or not.
1. Suppose the given point A lies in the given circumference;
the locus of B is a straight line given in position.
ANALYSIS.
Draw the diameter AE, and make AE.AD AB.AC;
wherefore the point D is given. Join
CE and BD; and because AE.AD
=AB.AC, AC: AE: AD: AB;
whence the triangles CAE and
DAB, having likewise a common
angle at A, are similar. Consequent-
ly the angle ADB being thus equal
to ACE, is a right angle, and the
straight line DB is hence given in position.

COMPOSITION.
E
Having drawn the diameter AE, make the rectangle
AE, AD equal to the given space, and erect the perpendicular
• DB; this is the locus required. For draw ACB, and join CE.
The right-angled triangles ACE and ADB being evidently
306
GEOMETRICAL ANALYSIS,
similar, AC: AE:: AD: AB, and AC,AB=AE.AD, or
the given space.
2. Suppose that the point A does not lie in the given circum-
ference; then the locus of B is a given circle.
ANALYSIS.
C
B
Draw the diameter EAD, and produce CAF to the circum-
ference. The rectangle AC, AF,
being equal to AD, AE, is given,
and has therefore a given ratio to
the rectangle AC, AB; whence
the ratio of AF to AB is given,
and consequently (III. 2. Anal.) F
AB terminates in the circumfe-
rence of a given circle.

E
COMPOSITION.
A
H
G
Having drawn the diameter EAD, make the rectangle AD,
AH equal to the given space, and (III.2. Anal.) describe a circle
EBGF, such that a straight line, passing through it shall be cut
by the circumference in the ratio of AE to AH; this circle
is the locus required. For AE: AH::AF: AB:: AF.AC:
AB.AC; wherefore AF.AC: AB.AC:: AE.AD: AH.AD,
and the first term of this analogy being equal to the third,
the second term is equal to the fourth, or AB.AC=AH.AD,
that is, equal to the given space.
PROP. VII. THEOR.
If two straight lines, containing a given rectangle,
be drawn from a given point at a given angle; should
the one terminate in a straight line given in position,
the other will terminate in the circumference of a •
given circle.
BOOK III.
307
Let the point A, the angle BAC, and the rectangle under
its sides BA, AC be given; if the direction BD be given,
then will the locus of C be a given circle.
ANALYSIS.

From A let fall the perpendicular
AD upon BD. Draw AE, to con-
tain with AD an angle equal to the
given angle, and a rectangle equal to
the given space; and join CE.
Since AD is evidently given in po-
sition and magnitude, AE is likewise
A
2
B
ཇ་
D
given in position and magnitude; and the rectangle AD, AE
being equal to AB,AC, therefore AD: AB:: AC: AE; but
the angle DAE is equal to BAC, and hence DAB is equal to
EAC. Wherefore the triangles ABD and AEC, having each
an equal angle and its containing sides proportional, are simi-
lar; and consequently the angle ACE is equal to the right
angle ADB. Whence the locus of C is a circle, with AE for
its diameter.
COMPOSITION.
Having let fall the perpendicular AD, draw AE, making
the angle DAE equal to the given angle, and the rectangle
DA, AE equal to the given space, and on AE, as a diameter,
describe a circle; this is the locus required.
For join CE; and the triangles DAB and EAC being right-
angled at D and C, and having the vertical angles at A equal,
are evidently similar, and consequently AD: AB:: AC:AE;
and hence the rectangle AB, AC is equal to AD, AE, that
is, to the given space.
!
U 2
308
GEOMETRICAL ANALYSIS
PROP. VIII. THEOR.
If two straight lines, in a given ratio, stand at gi-
ven angles on two diverging lines which are given
in position, the locus of their vertex will be like-
wise a straight line given in position.
Let the straight lines AB, AC, in a given ratio, form gi-
ven angles ABD and ACD with the given diverging lines
DE, DF; then will their vertex A lie in a given direction.
ANALYSIS.
Join DA, and produce BA to meet DF in G. The tri-
angle DBG is given in species; for the angles at D and B are
given, and, consequently, the angle at G. Again, the triangle
ACG is given in species, since
all its angles are given.
Hence the ratio of AC to
E

M
B
IT
N
D
I
I C
GF
AG is given; but the ratio
of AB to AC is given, and
consequently that of AB to
AG and that of BG to AG.
Hence, also, the ratio of BG to DG is given, and therefore
the ratio of AG to DG; and the angle at G being given, the
triangle DAG is (VI. 14. El.) consequently given in species.
Wherefore the angle GDA is given, and hence the straight
line DA is given in position.
COMPOSITION.
In DE take any point H, and draw HI and HL, making
with DE and DF angles equal to the respective inclinations
of the bounded lines, produce IH to M, so that MH shall
have to HL the given ratio, find (VI. 3. cor. El.) IN a third
proportional to IM and IH, and join DNA; this straight
line is the locus required.
BOOK III.
309
Because IM IH :: IH: IN, therefore (V. 10. El.)
MH: IH:: NH: IN; but (VI. 2. El.) AB: AG:: NH: IN,
and the triangles ACG and HLI being evidently similar,
AG: AC:: IH: HL; therefore (V. 16. El.) AB: AC::
MH: HL, that is, in the given ratio.
PROP. IX. THEOR.
Three diverging lines being given in position, if
a straight line cut them at given angles, and such
that the rectangle of its first segment, by a given
line, shall be equal to both the rectangles of its se-
cond and third segments by given lines; the locus of
its point of origin will be a straight line given in po-
sition.
Let ABCD cut the diverging lines EF, EG and EH at
given angles, and let AB.KL-AC.ML+AD.NM; then will
the locus of the point A be a straight line given in position.
ANALYSIS.
Because AC.ML AB.ML + BC.ML, and AD.NM=
AB.NM+BD.NM, thereforeAB.KL-AB.ML+BC.ML+
AB.NM+BD.NM, and consequently AB.KL-AB(ML+
NM) + BC.ML + BD.NM,
and AB.KN = BC.ML +
BD.NM. Make BC: BD::
NM: MO, and BC.MO=
BD.NM; whence AB.KN=
BC(ML+MO)=BC.OL, and
AB: BC:: OL: KN. The
ratio of AB to BC is there-
fore given; but the triangle
BCE being given in species,
the ratio of BE to BC is given,
E
K
73
B
G
D
H
ON M I
and consequently the ratio of AB to BE is given; and since
+

310
GEOMETRICAL ANALYSIS.
J
the contained angle ABE is given, the triangle BEA is like-
wise given in species; and thence the point A, and the
straight line EA, are given in position.
COMPOSITION.
Having assumed in EH any point H, draw HGF in the
given inclination, make FG: FH:: NM: MO, and pro-
duce HF till KN: OL:: FG: IF; EI is the straight line
required. For BC: AB:: FG :IF :: KN: OL, and AB.KN=
BC.OL; but BC: BD:: FG FH:: NM: MO, and
BC.MOBD.NM. Wherefore AB.KN BC.OL =
BC.ML+BD.NM, and AB.KM = AB.NM+ BC.ML +
BD.NM BC.ML + AD.NM, and hence AB.KL =
AB.ML+BC.ML+AD.NM=AC.ML+AD.NM.
PROP. X. THEOR.
Four diverging lines being given in position, if a
straight line cut them at given angles, and such
that the rectangles of its first and second segments
by given lines shall be equal to both the rectangles
of its third and fourth segments by given lines;
the locus of its point of origin will be a straight line
given in position.
Let ABCDE cut the diverging lines FG, FH, FI, and FK
at given angles, and let AB.MN+AC.NO = AD.OP+
AE.PQ; then will the locus of the point A be a straight line
given in position.
ANALYSIS.
Because AB.MN + AC.NO=AD.OP + AE.PQ, it fol-
lows, by decomposition, that AB.MO+BC.NO=AB.OQ+
BOOK III.
311
BD.OP 4 BE.PQ, and conse-
quently AB.MQ+ BC.NO=
BD.OP + BE.PQ.

I
Make
BD BC: NO OR, and
:
:
BD: BE:: PQ: PS; then
BD.OR=BC.NO, and BD.PS
=BE.PQ; whence AB.MQ+
BD.OR = BD.OP + BD.PS,
or AB.MQBD.SR, and,
therefore, AB: BD :: SR:MQ.
But the triangle BDF being
given in species, the ratio of
BD to BF is given; and con-
A
B
H
C
I
D
F
E
K
M
SQ NP
RO
sequently the ratio of AB to BF is given, and the contained
angle ABF being given, the triangle BFA is likewise given
in species; and hence the straight line FA is given in posi-
tion.
COMPOSITION.
Having assumed in FK any point K, draw KIHG at the
given inclination, make GI: GH:: NO: OR, and GI : GK
:: PQ: PS, and produce KG till MQ : SR :: GI : GL; FL
is the straight line required.
For BD: BC:: GI : GH :: NO : OR, and BD.OR=
BC.NO; but BD: BE:: GI: GK :: PQ: PS, and BD.PS=
BE.PQ; again, MQ:SR :: GI: GL :: BD : AB, and
AB.MQ=BD.SR. Whence AB.MQ+BC.NO=BD.SR+
BD.OR = BD.SO = BD.PS+BD.OPBE.PQ+BD.OP,
add to each AB.OQ = AB.NQ + AB.NO, or AB.PQ+
AB.OP, and AB.MN+AC.NO=AD.OP+AE.PQ *.
PROP. XI. THEOR.
If a straight line given in position, be cut at given
angles by two straight lines, which intercept, from
*See Note LVIII.
312
GEOMETRICAL ANALYSIS.
two given points in it, segments that have a given
ratio, the locus of the point of concourse is a straight
line given in position.
Let AB and AC be drawn, such that the angles ABF,
and ACF, with the ratio of DB to EC, are given; the locus
of A, the point of concourse, is a straight line given in
position.
ANALYSIS.
Make FD to FE in the given ratio, and join FA. Since
therefore FD: FE: DB: EC, it follows (V. 19. El.) that
FD: FE :: FB: FC; consequently the ratio of FB to FC,
and thence that of FB to BC, are each given. But the
angles FBA and FCA being given, the triangle BAC is evi-
dently given in species, and
therefore the ratio of AB to
BC is given, and hence the
ratio of FB to AB is also
given. The triangle FBA
having thus two sides con-
taining a given angle and in
H

A
G
D
E B
C I
a given ratio, is (VI. 14. El.)
given in species; and consequently the angle BFA is given,
and the straight line FA given in position.
COMPOSITION.
Having made FD to FE in the given ratio, draw DG and
EG at the given angles with FI, and join F with their point
of concourse; FGH is the locus required.
For, from any point A in FH, draw AB and AC at the
given angles with FI, and consequently parallel to GD and
GE. Because AB is parallel to GD, and AC to GE,
FG: FA :: FD : FB :: FE : FC (VI. 1. El.) and alternately
FD: FE:: FB: FC; wherefore (V. 19. cor. 1. El.) DB: EC
:: FD: FE, that is, in the given ratio.
BOOK III.
313
PROP. XII. THEOR.
If, from two given points, there be inflected two
straight lines in a given ratio, the locus of their point
of concourse is a straight line, or a circle given in
position.
Let AC and BC, drawn from the points A and B, have a
given ratio; then will C, the point of concourse, lie in a
straight line given in position, or in the circumference of a
given circle.
1. When the inflected lines are equal, they terminate in a
straight line given in position.
ANALYSIS.

Bisect AB in E, and join EC. The tri-
angles ACE and BCE, having the sides
AE and AC equal to BE and BC, and
EC common, are equal (I. 2. El.); where-
fore the angle AEC is equal to BEC, and
EC is perpendicular to AB, and conse- A
quently given in position.
COMPOSITION.
E
B
Bisect AB by the perpendicular EC, which is the locus re-
quired. For draw AC and BC to any point in it, and the
triangles AEC and BEC are (I. 3. El.) evidently equal, and
hence AC is equal to BC.
2. When the inflected lines AC and BC have an unequal
ratio, their point of concourse lies in the circumference of a gi-
ven circle.
ANALYSIS.
Draw CD, making the angle BCD equal to BAC, and
meeting AB produced in D. The triangles DAC and DCB,
having the angle at D common, and the angles at A and C
equal, are evidently similar; and hence AD: AC:: DC: BC,
314
GEOMETRICAL ANALYSIS.
and alternately AD : DC ::
AC: BC, that is, in the given
ratio. But AD:DC:: DC: BD,
and consequently AD is to BD
in the duplicate of the given
ratio of AD to DC, and which
is therefore likewise given.
Consequently BD, and the

E
B D
point D, are given; and DC being thence given, its extremi-
ty C must lie in the circumference of a circle described with
that radius.
COMPOSITION.
Divide AB in the given ratio at E, and in the same ratio
make ED to BD; the circle described from the centre D,
and with the radius DE, is the locus required.
For, since AE: BE:: ED: BD, it follows (V. 19. El.)
that AD: ED, or DC :: ED, or DC: BD; hence the tri-
angles DAC and DCB, thus having the sides which contain
their common angle at D proportional, are similar, and
therefore AC: AD :: BC: DC, or alternately AC : BC : :
AD: DC or ED, that is, in the given ratio.
Scholium. Since, in the second case, AC: BC :: AD: ED,
it is obvious, that as the ratio of AC to BC approaches to
equality, the centre D must continually recede from A or E,
and consequently the arc EC may be conceived as ultimately
passing into the tangent which bisects AB at right angles ;
thus comprehending the first case of the proposition.
PROP. XIII. THEOR.
A point and a straight line being given in posi-
tion, the locus of another point, the square of whose
distance from the former, is equal to the rectangle
under its distance from the latter and a given
straight line-is a given circle.
BOOK III.
315
The point A and the straight line DC being given in
position, let the square of BA be equal to the rectangle
under the perpendicular BC and K; the locus of B is a gi-
ven circle.
ANALYSIS.
Draw DFA parallel to CB, make AO equal to the half of
K, and bisect it in G, join BO, and let fall the perpendicu-
lar BF.
Becausé AO is bisected in G, OB-AB2, or AB'-OB",
(II. 24. El.) = 2AO.GF K.GF; but AB²= K.BC, or
K.DF, and hence OB K.DG. E
Since therefore DG is given,
OB is also given; and the one
extremity O being given, the
other extremity B must lie in
the circumference of a given
circle.

C
B
D
HF AGO
F'
K
B
COMPOSITION.
Having drawn DA parallel to CB, make AOK, and
AG=AO, and find OH a mean proportional between K
and DG; a circle described from O with the radius OH, is
the locus required.
For OB-AB2, or AB'2-OB",=2AO.GF=K.GF; and
since, by construction, OH², or OB², K.DG, it follows that
ABK.DF, or K.BC.
Cor. If the given point A lies in DC, or coincides with D,
then DGK and OH=K, or the circle likewise passes
through D; whence AB becomes a chord, and its square
(VI. 16. cor. 1. El.) is equivalent to the rectangle under the
segment DF, and the diameter or K.
316
GEOMETRICAL ANALYSIS.
PROP. XIV. THEOR.
If, from two given points, there be inflected two
straight lines, such that the difference of the square
of the one and a given space, shall have to the
square of the other, a given unequal ratio-their
point of concourse will lie in the circumference of a
given circle.
Let AC and BC be the inflected lines, and the rectangle
AC, AD be made equal to the given space; then if the dif
ference between the square of AC and that rectangle, or the
remaining rectangle AC, CD, have a given unequal ratio to
the square of BC, the locus of the point C will be a given cir-
cle.
ANALYSIS.
Make (VI. 4. El.) AE to BE in the given ratio, join CE
and BD, produce CB to
meet the circumference of a
circle described about the
triangle ADB, and join AF.

Because (III. 32. El.) the
rectangle AC, CD is equal
to FC, BC, it follows that
the rectangle FC, BC is to
A
T
D
GH/B
E
the square of BC, or (V. 25.
cor. 2. El.) FC is to BC, in the given ratio of AE to BE;
wherefore (VI. 1. cor. 1. El.) AF is parallel to CE, and con-
sequently the angle ECB is equal to AFB, which is equal to
CDB the opposite exterior angle of the quadrilateral figure
ADBF. Through the points C, D, B, describe a circle cut-
ting AB in G, and join CG and DG; then (III. 32. El.) the
rectangle BA, AG is equal to CA, AD, or to the given space,
and hence AG, and the point G are given. The angle CDB,
or ECB, is, therefore, equal to CGB, and consequently the
triangles BEC and CEG are similar, and GE: CE:: CE:
BOOK IIL
817
BE; whence CE=GE.BE, which is a given rectangle, and
thus CE is given, and the locus of C a given circlę.
COMPOSITION.
Make the rectangle AB, AG equal to the given space, and
AE to BE in the given ratio, and find EH a mean propor-
tional between GE and BE; the locus required is a circle de-
scribed from E with the radius EH.
For, through the points A, D, B, and through C, B, G,
describe circles, produce CB to F, and join AF, CG, and
DG. Because GE.BE-HE, GE: HE or CE: HE or
CE: BE, and, therefore, the triangles GEC and CEB are si-
milar, and the angle EGC is equal to ECB; but the angle
EGC, or BGC, is equal to CDB, which again is equal to
AFB; consequently the alternate angles ECB and AFB are
equal, and the straight lines CE and AF parallel. Where-
fore AE: BE:: FC : BC :: FC.BC, or AC.CD: BC². But
CA.AD=BA.AG, or the given space; and hence the differ-
ence between the square of AC and that space, or the rect-
angle AC, CD, is to the square of BC, in the given ratio.
Scholium. If this local theorem were extended to the ex-
treme cases, it would include other propositions which are
exhibited in a separate form. Thus, supposing the given ra-
tio to be that of equality, the sum or difference of the squares
of AC and BC will be equivalent to the given space, accord-
ing as this is greater or less than the square of AC. When
the given space exceeds the square of AC, the centre E of the
circle bisects AB, as in the first case of the sixteenth propo-
sition of this Book. But when the square of AC is deficient
by the given space, the ratio of AE to BE being that of
equality, the centre E, lying beyond B, must be thrown to an
infinite distance, and consequently the arc which crosses AB
will merge in a tangent bisecting GB at right angles, as in Pro-
position 15. Again, if the deficient space be supposed to va-
nish, while the ratio of the squares of AC and BC, or that of
318
GEOMETRICAL ANALYSIS.
the inflected lines themselves is given, the point G will coin-
cide with A, and the centre and radius of the circle are hence
determined, after the same manner as in Proposition 12.
PROP. XV. THEOR.
If from two given points there be inflected two
straight lines, of whose squares the difference is
given, the locus of their point of concourse will be a
straight line given in position.
Let AC and BC, drawn from the points A and B,
have
the difference of their squares given; the locus of C, the point
of concourse, is a straight line given in position.
ANALYSIS.
Draw CD perpendicular to
AB, which bisect in E. The
difference between the squares
of AC and BC is (II. 24. El.)
equal to twice the rectangle
under AB and ED; conse-
quently that rectangle, and
its containing side ED, are
given; whence the point of
bisection E being given, the
point D is given, and the per-
pendicular CD is therefore
given in position.

A
E
B
D

COMPOSITION.
A
E D
B
Bisect AB in E, and make (II. 9. El.) the rectangle under
twice AB and ED equal to the given space; the perpendicu-
lar DC is the locus required.
For (II. 24. El.) AC-BC-AB.2ED=2AB.ED, and
consequently the difference of the squares of AC and BC is
equal to the given space.
BOOK ILL
319
LEMMA.
If a straight line AB be cut any how in the point D, but di-
vided at C, so that the segment AC shall be the nth part of BC;
then n.AD²+BD²=AB.BC+ (n+1) CD².
For upon AB describe a semicircle, and erect the perpen-
dicular CE, join AE, BE, draw DF parallel to CE and meet-
ing AE or its extension, and join BF..
F'
The angle AEB in a semicircle being a right angle,
AC: CE :: CE : BC (VI. 16. cor. El.) and consequently
(V. 24. El.) AC: BC: AC: CE; but BC=n.AC, and
therefore CEn.AC. Again, from the same property,
AB: AE; AE: AC, and
AB: AC: AE²: AC²;
and since AB=(n+1)AC;
it follows (V. 5. El.)
that AE(n+1) AC².
Now CE and DF being
parallel, CE DF:: AC: D

AD, and (V. 22. cor. 1. El.)
"
CE² : DF² :: AC² : AD², F
A
A D C
D'
and CE² being equal to n.AC², therefore (V. 8. and 5. El.)
DF²=n.AD². In the same manner, it is shown that EF2 =
(n+1) CD². But (VI. 16. cor. 1. El.) BE²=AB.BC, and
the triangles BDF and BEF being right angled, BD²+
DF2 = BF2 = BE+EF2, and consequently by substitution,
n.AD²+BD²=AB.BC+(n+1) CD².
PROP. XVI. THEOR.
If, from given points, there be inflected straight
lines, whose squares are together equal to a given
space, their point of concourse will terminate in
the circumference of a given circle.
1. When there are only two given points.
320
GEOMETRICAL ANALYSIS.
Let AP and BP, drawn from the points A and B, have
the sum of their squares given; the locus of their point of
concourse is a given circle.
ANALYSIS.
Bisect AB in O, and join OP. The squares of AP and
BP are (II. 25. El.) equal to
twice the squares of AO and
OP. Hence the sum of the
squares of AO and OP is gi-
ven; but AO and its square
being given, the square of OP

A
O F
E B
and OP itself must be given;
wherefore the locus of the extremity P is a circle, of which the
point of bisection is the centre.
COMPOSITION.
Bisect AB in O, find (III. 33. El.) AF the side of a square
equal to half the given space, and make (II. 14. El.) OE =
AF²-AO²; the point O is the centre, and OE the radius,
of the required circle.
For (II. 25. El.) AP²+BP²=2AO +2OP²=2AO²+
20E²=2AF², or the given space.
2. When three points are given.
Let the straight lines AP, BP and CP, inflected from the
points A, B, and C, have the sum of their squares given; the
locus of their point of concourse is a given circle.
ANALYSIS.
Bisect AB in E, and (II. 25. El.) AP²+BP²=2AE²+
2EP²; consequently AP² +
BP² + CP² = 2AE² +
2EP²+CP2. Now2 AE-
AB.BE, and, letting fall
the perpendicular PF,
(II.11.El.)2EP²=2EF²+
2PF², and CP²=PF²+
CF. Wherefore AP²+

P
E
B
BOOK III.
321
BP + CP²
AB.BE + 3PF² + 2EF² + CF". Trisect EC
(I. 38. El.) in the point O, and join PO; and, by the Lemma,
2EF² + CF² = EC.CO + 30F2. Whence AP² + BP² +
CP² = AB.BE + EC.CO + 3PF² + 30F² = AB.BE +
EC.CO+3PO². But the intermediate points of division E
and O, are evidently given, and thence the rectangles AB,BE
and EC, CO, are given; wherefore 3PO is given, and conse-
quently PO itself. Since one extremity of that line then is given,
the other extremity P must lie in the circumference of a given
circle.
COMPOSITION.
Bisect AB in E, trisect EC in O, and find (III. 33. El.)
OP such that its square shall be triple the excess of the gi-
ven space above the rectangles AB, BE and EC, CO; the
locus required is a circle, of which O is the centre, and PO
the radius. For 3PO 3PF²+30F², 3PO² + EC.CO=
3PF+EC.CO+30F 3PF + 2EF + CF22PE +
PF²+CF²=2PE+CP²; consequently the given space, or
3PO+AB.BE+ EC.CO= 2AE + 2PE*+ CP²
BP +CP*.
3. When there are four given points.
AP²+
Let AP, BP, CP and DP drawn from the points A, B, C,
and D, have the sum of their squares given; the locus of their
concourse P is a given circle.
ANALYSIS.
Bisect AB in E, trisect EC in F, and join PE and PF.
It is manifest, from the last case, that AP+BP²+CP² =
X
322
GEOMETRICAL ANALYSIS.
AB.BE + EC.CF +
3PF; add DP² to each,
and AP²+BP²+CP²+
AB.BE+EC.CF
DP²
A
D

P
E
G-
F
B
+3PF+DP. Let fall
the perpendicular PG up-
on DF, and the given
space is equal to AB.BE
+ EC.CF + 3PG² +
3FG²+PG²+DG²; and
hence 4PG + 3FG +
DG² must be equal to a given space.
fourth part of DF, and join PO:
3FG+ DG FD.DO + 4OG. Wherefore FD.DO +
4OG²+4PG², or FD.DO+4PO², is equal to a given space,
and hence 4PO², and PO itself, are given. Now the point O
being given, P must lie in the circumference of a given circle.
COMPOSITION.
1
Let FO be made the
then, by the Lemma,
Bisect AB in E, trisect EC in F, and quadrisect FD in O;
from the given space take away the accumulate rectangles
AB.BE+EC.CF+FD.DO, and find (III. 33. El.) the sidet
of a square equal to this difference: That straight line is the
diameter of a circle, which is the locus required.
For join PE, PF, PO, and let fall the perpendicular PG
upon DF; then FD.DO+4PO² — FD.DO+4OG²+4PG²=
3FG²+DG²+4PG3FG+3PG+DP3PF+ DP².
Wherefore AB.BE + EC.CF + 3PF² + DP², is equal
to the given space. But, from the composition of the last
case, it is manifest that AP² + BP² + CP² = AB.BE +
EC.CF+3PF*; consequently AP²+BP²+CP²+DP² are
together equal to the given space.
By pursuing this mode of investigation, it is obvious that
the proposition will be successively extended to any number
of given points.
Scholium. The property now demonstrated is capable of be-
ing generalized. Thus, if any multiples of the squares of the
T
BOOK III.
823
2
P
I
B
inflected lines, be together equal to a given space, the locus of
their point of concourse is still a given circle: For, conceive
so many points to be clustered together at each centre A, B,
C, &c. of inflection, and the squares of the lines which pro-
ceed from them will evidently receive in effect a correspond-
ing. multiplication.—But the property may be traced out more
clearly, and through all its shadings, by help of a simple ex-
tension of the Lemma. Let AP and BP be two straight lines
inflected from the points A and B, and let the segment OB=
v.OA; then, joining PO and
drawing the perpendicular PL,
it was proved that v.AL +
BL = AB.BO + (+1) OL";
add (v + 1) PL to each, and
(AL² + PL²) + BL² + PL² =
AB.BO+(v+1) (OL*+PL*), or
v.AP²+BP² = AB.BO+ (v+1)OP2. Multiply both by n,
and suppose nv=m, and there results m.AP² + n.BP² =
n.AB.BO + (m+n) OP². By repeated application of this
principle, it may be demonstrated that m.AP+n.BP² +
p.CP²+q.DP², &c. =(m+n+p+q, &c.) OP², together with
certain multiples of given rectangles, and consequently that
their point of concourse has for its locus a circle, whose centre
is O and radius OP. But the property must likewise hold, if
all those multiple squares were divided by the same number,
that is, if instead of the squares of the inflected lines, there
were substituted only similar rectilineal figures constructed
upon them. If the given space should be equal to the rect-
angles, the circle will evidently contract to a point, and be-
yond this limit the problem becomes impossible. It is like-
wise obvious, that the centre O and radius OP will turn out
the same, in whatever order the successive connected sections
take place *.

* See Note LIX.
1
X 2
324
GEOMETRICAL ANALYSIS.
it
DEFINITION.
A Porism proposes to demonstrate that one or more things
may be found, between which and innumerable other objects
assumed after some given law, a certain specified relation is
to be shown to exist.
The nature of a porism consists in affirming the possibility of find-
ing such conditions, as will render a problem indeterminate, or ca-
pable of innumerable solutions.
PROP. XVIII. PORISM.
Three points being given, a fourth may be found,
such that any straight line drawn through it shall
have its distances from two of those equal to its dis-
tance from the third.
Let A, B, and C be given points, another point D may be
found, so that, HDI being drawn through it, the perpendi-
culars AH and BI, let fall on the one side, shall be equal to
CG on the other.
ANALYSIS.
M
Through the point D, draw CDK, and upon this let fall
the perpendiculars AK, BL, and join AB, meeting KC in E.
Since CDK passes through C, its distances KA and LB
on either side, from the two remaining points, must evident-
ly be equal. Hence (I. 21. El.)
the right-angled triangles AEK
and BEL are equal, and con-
sequently the side AE is equal
to BE; wherefore E, being
thus the point of bisection, is
given. Draw the perpendicu-
lar EF; and it is evident (II.
10. El.) that 2EF = AH and
BI.

Now CG and EF being
H
T
N
A
E
B
K
parallel, CD: DE :: CG; EF, and (V. 13. El.) CD: 2DE::
BOOK III.
325
CG: 2FE, or AH+BI; but, by hypothesis, CG=AH+BI,
and therefore (V. 4. El.) CD=2DE. Whence, CE being
given, the point D is given.
COMPOSITION.
Bisect AB in E, join CE and trisect it in D; this is the
point required.
For let fall the perpendicular EF. Because CG and EF
are parallel, CD: DE:: CG: EF; but CD = 2DE, and
therefore (V. 4. El.) CG=2EF, that is, AH+BI.
The porism now demonstrated may be viewed as origina-
ting in the solution of this problem :-To draw, through the
point M, a straight line MN, such that the perpendiculars
AH and BI, let fall upon it from the points A and B, shall
be together equal to the perpendicular CG, from the point C
on the other side. The point D is found as before, and
thence the position of MDN is assigned. But this straight
line, it is evident, will become indeterminate if the point M
should happen to coincide with D; on that supposition, the
problem would admit of innumerable answers, or the diame-
ter MDN might lie in every possible direction *.
PROP. XIX. PORISM.
A circle and a straight line being given in posi-
tion, a point may be found, such that any straight
line, drawn through it and limited by these, shall
contain a given rectangle.
Let the straight line AB, and the circle HDF, be given in
position; it is required to determine a point F, which may
divide any connecting straight line DFE into segments con-
taining a rectangle that will be given.
* Sce Note LX.
326
GEOMETRICAL ANALYSIS.
ANALYSIS.
Through F draw HFG perpendicular to AB. By hypo-
thesis, the rectangle HF.FG is likewise equal to the given
space, and therefore equal to DF.FE; whence (V. 6. El.)
DF: HF:: FG: FE, and the
triangles DFH and GFE, ha-
ving the vertical angles at F
equal, are consequently simi-
lar, and the angle FDH is H
thus equal to FGE, or is a
right angle. Wherefore HDF
is a semicircle, of which HF is
the diameter; but the centre

¡A
D
хах
F
G
E
B
C being given, the perpendicular HCG is thence given, and
consequently the extremity of the diameter, or the point F.
Again, the points H, F, and G being given, the rectangle un-
der the segments HF and FG is given.
COMPOSITION.
From the centre C, let fall upon AB the perpendicular
HCFG, cutting the circumference in F; this point has the
property, that any intersecting line drawn through it will con-
tain a given rectangle. For join DH, and the triangles FGE
and FDH are similar; whence FG: FE:: FD: FH, and
consequently FE.FD=FG.FH, which is manifestly given.
This porism also may be considered as arising out of the
solution of a simple problem:-Through the point M, to draw
a straight line DMFE, so that its segments DF and FE shall
contain a given rectangle. The point F being found as be-
fore, DME is consequently given in position. But when the
point M coalesces with F, the straight line DE can thus have
no determinate position, or it will fulfil the conditions of the
problem in whatever direction it be drawn.
BOOK III.
327
PROP. XX. PORISM.
A circle and a point being given, another point
may be found, such that straight lines drawn from
them to any point in the circumference, shall have a
ratio which will be given.
The point B may be found, so that AC and BC, inflected
to the given circumference ECF, shall have a ratio which
may be likewise assigned.
ANALYSIS.
D
Draw AB, cutting the circle in E and F; join CE, CF,
and produce AC. Because E, F are points in the circum-
ference, AC: BC:: AE: EB, and AC: BC :: AF : FB;
whence (VI. 11. cor. El.) CE bisects the vertical angle ACB,
and CF the adjacent angle BCD; consequently the angle
ECF, being the half of both of these, is a right angle, and
(III. 22. El.) ECF, a semicircle.
Wherefore AF, thus passing
through the centre O, is given in
position. Now, since AF: FB::
AE : EB, alternately AF: AE::
FB: EB; hence EF, being cut
externally and internally in the
same ratio, EO is (VI. 7. El.) a mean proportional between
AO and BO, or EO²=AO.BO. But AO and EO are gi-
ven, and therefore BO and the point B are given. Again,
because AO: EO:: EO: BO, by division and alternation,
AE EB EO: BO; that is, the inflected lines have the
given ratio of EO to BO.
· •
A
COMPOSITION.
E B
F
Draw AF through the centre of the given circle, and
make AO : EO :: EO: BO; B is the point required. For
?

$328
GEOMETRICAL ANALYSIS.
join CO. Because EO is equal to CO, therefore AO: CO::
CO BO; consequently the triangles ACO and CBO,
having besides the common angle at O, are similar, and
AC: AO :: BC: CO, or alternately AC: BC:: AO : CO,
that is, in a given ratio.
The porism now demonstrated is evidently derived from
the local theorem, which forms the 12th Proposition of this
Book.
PROP. XXI. PORISM.
A circle and a straight line being given in posi-
tion, a point may be found, such that any straight
line drawn from it to the given line, shall be a mean
proportional between the segments intercepted by
the given circumference.
Let the straight line AB, and the circle HKF be given in
position; it is possible to assign a point D, through which a
straight line FDC being drawn, CD shall be a mean propor-
tional between the segments CE and CF.
ANALYSIS.
From D let fall upon AB the perpendicular IDG, and join
CI and HK. Because CE: CD:: CD: CF, CD2 =
CE.CF⇒(III. 32. El.) CK.CI; and, since GI passes through
the point D, GH GD: :
GD: GI, and GD²=GH.GI. A
But (II. 11. El.) CD²=CG²+
GD², and consequently CK.CI=
CG²+GI.GH; take these away
from CI² = CG² + GI², and
there remains CI.KI=GI.HI.
Whence CI: GI :: HI : KI,
and consequently the triangles
CIG and HIK, having a com-

K
G
[
I
D ONP
M
B
F
BOOK III.
329
mon vertical angle, are similar. Wherefore the angle HKI,
being thus equal to CGI, stands in a semicircle, of which HI
is the diameter; consequently GI is given in position, and
the points G, H, and I being thence given, the rectangle
under GH and GI, or the square of GD, is given, and there-
fore the point D.
COMPOSITION.
Through the centre O, draw the perpendicular GOI; and ·
find (VI. 18. El.) GD a mean proportional to GH and GI;
D is the point required. For (III. 32. and II. 19. El.)
CE.CF-CO-HOCG + GO-HOCG + GH.GI;
but (V. 6. El.) GD²=GH.GI, and consequently CE.CF=
CG²+GD²= CD².
This porism may be supposed to derive its origin from the
problem :-" Through a given point P, in the diameter of a
circle, to draw a straight line CLPM to the perpendicular
AB, so that the rectangle under the segments CL and CM
shall be equal to the square of GN." Since (III. 32. El.)
CL.CM=CK.CI=CIª—CI.KI; but (II. 11. El.),CI²=
CG²+GI², and CI.KI=GI.HI; whence CL.CM=CG²+
GI.GH, or making GDGI.GH, CL.CM= CG²+GD² or
CD², and consequently CD2GN², or CD=GN. Wherefore
the point D being given, the point C is also given, and
thence the straight line CLPM. The problem then is solved
by finding GD a mean proportional to GH and GI, and de-
scribing, from D with the radius GN, a circle to intersect the
perpendicular in C. It is hence evident, that C is indepen-
dent of the point P. Let CLM, therefore, coincide with
CEF, and CE.CF=GN²=CD³. But this property must
evidently obtain, whatever be the position of the point C*.
PROP. XXII. PORISM.
A point being given in the diameter of a given
circle, another point in the same extension may be
* See Note LXI.
330
GEOMETRICAL ANALYSIS.
found, such that the angle contained by two straight
lines drawn from it to the extremities of a chord
passing through the given point, shall be bisected by
the diameter.
In the diameter FH of a given circle, let A be a given
point through which any chord BAC is drawn; a point D
may be found in the extension of the diameter, so that DC
and DB being joined, the angle ADC shall be equal to ADB.
ANALYSIS.
Join EB, and draw EO and BO to the centre O. The tri-
angles EOD and BOD, having the side EO equal to BO, OD
common, and the angle ODE
equal to ODB, and being like-
wise of the same affection,
since the angles DEO and
DBO are evidently both acute
-are (I. 22. El.) equal, and
consequently the angle EOG
is equal to BOG. Whence
E

F
D
A
H
B
the triangles OEG and OBG are (I. 3. El.) also equal, and
therefore EB is perpendicular to the diameter FH. Where-
fore (VI. 9. El.) FA: AH :: FD: DH; but the ratio of FA
to AH being given, and consequently that of FD to DH, the
point D (VI. 6. El.) is given.
COMPOSITION.
Make (VI. 3. El.) OA:OH:: OH: OD, and then D is
the point required. For join OC and OB. Because OH=
OC, OA: OC :: OC: OD; wherefore the triangles AOC
and COD, having thus the sides about their common angle
DOC proportional, are similar; and hence the angle OCA is
equal to ODC. In the same manner, it is proved that the
angle OBA is equal to ODB. But BOC being an isosceles
triangle, the angle OCA is equal to OBA; whence the angle
ODC is equal to ODB.
BOOK 11I.
331
:
This porism is likewise derived from the local theorem gi-
ven in Prop. 12. For AC, DC, and AB, DB being inflect-
ed in the same ratio, AC: AB :: DC : DB; and conse-
quently (VI. 11. cor. El.) the angle BDC is bisected by DA.
PROP. XXIII.
PROP. XXIII. PORISM.
A point being given in the circumference of a
circle, another point may be found, so that two
straight lines inflected from them to the opposite
circumference, shall cut off, on a given chord, ex-
treme segments, whose alternate rectangles shall
have a given ratio.
Let the circle ADBE, the point A, and the chord DE, bẹ
given in position,—another point C may be found, such that
straight lines AB and CB inflected to the opposite circumfe-
rence, shall form segments containing rectangles DG, FE,
and DF, GE, in the ratio of KM to LM.
ANALYSIS.
Join CA, and produce it to meet the extension of the chord
ED in H.
K
L M
B
Because KM: LM:: DG.FE: DF.GE, by division KL:
LM :: DG. FE-DF.GE: DF.GE; but DG.FE-DF.GE
=(DF+ FG) (GE + FG)—
DF.GE=FG.DE, and conse-
quently KL LM:: FG.DE:
DF.GE. Make KL : LM ::
DE: DH, then KL: LM: :
FG.DE FG.DH; whence
FG.DH=DF.GE, and, add-
ing DF.FG to both, FH.FG=
H
כר
T G
E
DF.FE=(III. 32. El.) AF.FB. Wherefore FH: FB:: AF
: FG, and (VI. 14. El.) the triangles AFH and GFB are si-
-2

332
GEOMETRICAL ANALYSIS.
milar, and consequently the angle AHF is equal to FBG; but
the angle AHF is given, since the points A, H, and D are
given, and, therefore, the chord AC, cutting off from the gi-
ven circumference, a segment that contains a given angle
ABC or FBG is given, and thence the point C.
COMPOSITION.
Produce the chord ED to H in the ratio of KM to LM,
join HA, and, at any point B in the circumference, make the
angle ABC equal to AHF; C is the point required.
:
:
For, the triangles AFH and GFB being evidently similar,
FH FB AF FG, and FH.FG=FB.AF=DF.FE;
whence FH.FG-DF.FG=DF.FE-DF.FG, or FG.DH=
DF.GE. But KL: LM:: DE: DH:: FG.DE: FG.DH,
and therefore KL : LM :: FG.DE: DF.GE; consequently
(V. 9. El.) KM: LM :: FG.DE+DF.GE, or DG.FE:
DF.GE.
The porism now investigated arises naturally out of this
problem :-" From two given points A and C, one of which
lies in a given circumference, to inflect straight lines AB and
CB, so as to intercept on the chord DE segments that con-
tain rectangles DG, FE and DF, GE, which are in a given
ratio." For, the point H being assumed as before, the ana-
lysis requires that the angle ABC should be made equal to
AHF. Whence, if on AC, a segment of a circle were de-
scribed containing that angle, its contact or intersection with
the given circumference, would determine the point of inflec-
tion. Supposing, therefore, the two circles entirely to coin-
cide, the problem will in that case become indeterminate, or
admit of innumerable answers.
BOOK III.
333
PROP. XXIV. PORISM.
Two points and two diverging lines being given
in position, straight lines, inflected from those points
to one of the diverging lines, intercept segments, on
the other, from points that may be found, and con-
taining a rectangle which will be likewise assign-
able.
Let DF and EF be inflected, from the points D and E, to
the diverging line AC; they will cut off segments, on AB,
from points I and K which may be found, so that the rect-
angle IH, GK shall be given.
ANALYSIS.
Join EI and EA, DA and DK, and produce ED to meet
AC in P. Since A, F, and P are so many points of inflection,
it is evident, from the hypothesis, that IA.AK=IH.GK=
IN.NK; whence IH: IA ::

AK: GK, and, by division,
AH IA :: AG: GK, and
:
alternately AH: AG :: IA
: GK. Through E, draw
LEM parallel to AB and
meeting AC and FD pro-
duced; then (VI. 2. El.)
LE: LM :: AH: AG ::
:
:
I
I
O
P
F
A
H//G/N
K B
EM
D
IA GK. Again, because IA.AK-IN.NK, IN: IA : :
AK NK, by division AN: IA :: AN: NK, and conse-
quently IA=NK. Wherefore, by substitution, LE: LM::
NK : GK, and LE: EM :: NK: GN, or alternately LE :
NK :: EM: GN, that is, (VI. 2. El.) ED : DN; hence
(VI. 14. El.) the triangles LDE and KDN are similar, and
LDK forms one single straight line. Join DO. Since IA=
334
GEOMETRICAL ANALYSIS.
NK, LE: IA :: LENK, that is, (VI. 2. El.) EO : OI ::
ED: DN, and therefore (VI. 1. cor. I. El.) DO is parallel to
AB. But the parallels OD and LM being given in position,
the points O and L, and thence I and K, are given, and con-
sequently the rectangle IA, AK is given.
COMPOSITION.
Draw DO, EL parallel to AB and meeting the extension
of AC, join EO, LD, and produce them to meet AB in I and
K; these are the points required. For DF and EF being in-
flected, LE : IA :: OE : OI :: ED: DN:: DM : DG : :
LM: GK, and alternately LE: LM:: IA: GK; but LE:
LM: AH: AG, and therefore IA: GK :: AH: AG;
consequently (V. 8. and 11. El.) IA : IH : : GK: AK, and
IA.AK=IH.GK.
The porism thus investigated follows from this problem:
"Two straight lines AB and AC being given in position,
with the points I and K, E and D, to find a point F, such
that the inflected lines EF and DF shall intercept segments
IH and GK, containing a given space:" For, when the points
I and K have the position before assigned, the construction
becomes indeterminate.
PROP. XXV. PORISM.
Three diverging lines being given in position, a
fourth may be found, such that straight lines can be
drawn intersecting all these and divided by them into
proportional segments.
Let AB, CD, and AE be given diverging lines, and HIKL
any transverse line cut by them in given ratios; a fourth di-
verging line FG may be found limiting the segment KL.
BOOK III.
335
ANALYSIS..
Produce EA and GF to meet in M, through K and P
draw NO and PO parallel to AB and FG, and meeting in O,
join CO; let H'I'K'L' be another transverse line divided
into proportional seg-
ments, draw P'I'O' pa-
rallel to PIO and meet-
ing CO in O', and join
G

I
T
I
N
"A A
K
N
K
O'K' and produce it to
N'.
Because KO is paral-
lel to PH, HI : IK : :
PI: IO; and, since the
parallels PO and P'O
are cut by the diverg- M
ing lines CP, CI, and
Food
PP' H H‘B
CO, PI: IO: P'I': I'O'; consequently H'I': I'K' : : P'T'
I'O', and O'N' is parallel to ON. Again, IK: KL :: OK :
KN and I'K': K'L' : : O'K': K'N'; wherefore OK: KN ::
O'K' : K'N'; and hence the straight lines OC, EA, and GF
all converge to the same point M. Now CA: AF :: OK:
KN :: IK: KL; whence the ratio of CA to AF being gi-
ven, AF and the point F are given; but the point L is gi-
ven, and, therefore, FLG is given in position.
COMPOSITION.
Make CA to AF in the given ratio of the segment IK to
KL, and join FL; this is the diverging line required. For
draw NK and PI parallel to AB and FG, and meeting in O,
join CO, and, assuming in it another point O', draw likewise
the parallels O'K'N' and O'I'P', intersecting AE and AB in
K' and I'; the transverse line H'I'K'L' is cut similarly to
HIKL.
For, since NO, N'O' are parallel to AB, and OP, O'P' pa-
rallel to FG, it follows that HI: IK:: PI: IO:: P'I': I'O' ::
336
GEOMETRICAL ANALYSIS,
|
:
H'I' I'K'. Again, because CA: AF:: IK: KL::OK:
KN; whence OC, EA, and GF converge to the same point,
and consequently IK KL :: OK : KN : : O'K' : K'N' : :
I'K': K'L'.
:
The porism now demonstrated arises out of the indetermi-
nate case of a celebrated problem :-" Four straight lines,
AB, CD, AE and FG, being given in position, to draw a
transverse line, HIKL, that shall
be cut by them into segments in a
given proportion." Suppose it
done; produce GF and EA to
meet in M, draw the parallels
NKO and PIO, and join MTO.
Because TA: AF:: OK : KN
:: IK: KL, the ratio of TA
to AF is given, and hence the
point T and the straight line MO
are given in position. Again,
PI: IO HI IK, and there-

::
:
G
VE
K
N
R
I
F
A/TC QPIL
B
Z
Y X
M
fore the ratio of PI to PO is gi-
ven; but the triangle CPI, being evidently given in species,
the ratio of CP to PI is given; whence the ratio of CP to PR
is given, and the triangle CPO is given in species. The
straight lines MO and CO being, therefore, both given in po-
sition, their intersection O is given; consequently the paral-
lels NO and PO are given in position, and thence are like-
wise given their intersections K and I, and the transverse line
HIKL.
The construction is easily derived: For, having produced
EA and GF to meet in M, make FA : AT : : Z : Y, and
draw MTO. Again, take any point Q in CB, draw QS pa-
rallel to FG, and make QR: RS:: X: Y, join CS and pro-
duce it to meet MO in O, and draw OI and KO parallel to
FG and AB; HIKL, which passes through the points of in-
tersection I and K, is the straight line required. For HI: IK
BOOK III.
337
:: PI: IO:: QR: RS:: X: Y, and IK: KL:: OK: KN
:: TA: AF :: Y: Z.
Now, if the ratio of CA to AF should be the same as that
of Y to Z, the point T will coincide with C, and the straight
line TO with CO. The problem, therefore, becomes, in this
case, porismatic, or every point whatever in CO has the pro-
perty which belonged before to the single point O*.
DEFINITION.
Isoperimetrical figures are such as have equal perimeters, or
the same extent of linear boundary.
PROP. XXVI. PROB.
In a straight line given in position, to find a point,
whose distances from two given points on the same
side shall together be the least possible.
Let it be required, from the points A and B to some point
in CD, to draw AG and BG, forming jointly a minimum.
ANALYSIS.
From B, either of the given points, let fall BE a perpendi-
cular upon CD, and, having produced it equally on the op-
posite side, join GF. It is obvious that the triangles BEG,
FEG are equal, and conse-
quently that BG=GF; whence
AG+GF is a minimum. But
the points A and F are evi-
dently both given, and since
(I. 15. El.) the shortest com-
munication between them is a
straight line, its intersection G

C-
* See Note LXII.
Y
B
-D
E
338
GEOMETRICAL ANALYSIS.
with CD is given, and therefore the inflected lines AG and
BG are given in position.
It hence appears that, when the combined distance of the
points A and B from the straight line CD is the least possi-
ble, the incident angles AGC and BGD are equal.
Cor. Hence also the solution of a similar problem :-" In
a straight line given in position, to find a point the difference
of whose distances from two given points shall be the greatest
possible." If these points lie on the same side of the straight
line CD, it is evident that the
difference between AG and BG
A

F
being (I. 16. El.) less than the
base AB, this must be the ex-
treme limit, or the difference
D
C
E
G
must reach its maximum when
AG and BG coincide with AB,
and consequently the point G
B
occurs where the production of AB meets CD.-But if A and
B lie on opposite sides of CD, let fall the perpendicular BE
which produce till EF be equal to it, and join AF and GF.
The triangles BEG, FEG are evidently equal, and therefore
BG=GF; but, in the triangle AFG, the difference of AG
and GF, being less than AF, must attain its greatest extent,
when that triangle is supposed to flatten into a straight line ;
in which case the angle AGE is equal to BGC.
PROP. XXVII. THEOR.
Straight lines drawn from two given points to the
circumference of a given circle are the least possi-
ble, when they make equal angles with a tangent ap-
plied at the point of inflection.
Of all the straight lines inflected from the points A and B
to the circumference of the circle GDH, AD and BD which
meet the tangent EF at equal angles, form together a mini-
тит.
BOOK III.
339

For, by the last proposition, AD and
BD, falling at an equal incidence, are
jointly shorter than any other lines in-
flected from the points A and B to the
straight line EF; but (I. 17. El.) such
lines drawn to that tangent are less
than the exterior lines which termi-
nate in the circumference; whence,
for both these reasons combined, AD
E
B
D
F
G
and BD must form the minimum of all the straight lines in-
flected to the circumference GDH.
PROP. XXVIII. PROB.
To find a point, whose distances from three given
points are the least possible.
Let it be required, from the points A, B, and C, to draw
AD, BD, and CD, such that their sum shall be a minimum.
ANALYSIS.
If the distance BD were supposed to remain constant, the
position of D, in the circumference of a circle described from
B with the radius BD, must, by the last proposition, be such,
when AD and CD compose a minimum, that the angle ADB
shall be equal to CDB. For the same reason, if AD conti-
nued invariable, BD and CD, completing the minimum, must
form with it equal angles ADB
and ADC. Whence, uniting
these conditions, the straight lines
AD,BD, and CD all make equal
angles about their point of con-
course.
A
D
B

Hence this construction :-
Connect the triangle ABC, and upon each of the sides AC
and BC describe equilateral triangles, and again circumscribe
these by circles, which will intersect in the point D. For, the
Y 2
340
GEOMETRICAL ANALYSIS.
i
angles ADC and CDB, being the supplements of angles of
equilateral triangles, are each equal to two third parts of two
right angles, or to one-third of four right angles; consequent-
ly three such angles will stand about the point D.
PROP. XXIX. PROB.
In a straight line given in position, to find a point,
at which the straight lines, drawn to two given points
on the same side, will contain the greatest angle.
Let it be required to draw AC and BC, so that the angle
ACB shall be a maximum.
ANALYSIS.
B
Describe a circle about the points C, A, and B. Because the
angle ACB is greater than any other
which has its vertex in DE, the
circumference must lie within that
straight line, and therefore DE
touches the circle.
It is hence evident, that GA.GB
=GC*, and, therefore, the point C D G
is assigned.

PROP. XXX. PROB.
To find a triangle with a given perimeter, and
standing on a given base, which shall contain the
greatest area.
Let it be required to find a triangle ABC, constituted on
the base AC, and containing, within a given perimeter, the
greatest possible surface.
ANALYSIS.
Since the base of the triangle ABC is constant while its
area forms a maximum, the corresponding altitude must evi-
BOOK III.
341
D
B
E
dently be the greatest possible, and consequently the vertex B
must lie in a parallel the remotest from AC. Supposing,
therefore, the parallel DE to retain its place, the sum of the
sides AB and CB, and consequently the
whole perimeter of the triangle, will, by
Proposition 26. of this Book, be the
least possible, when the angle ABD is
equal to CBE. Whence, preserving
the same perimeter, the parallel will be
enabled to recede to the greatest distance from AC, if these
incident angles still maintain their equality; but DE being
parallel to AC, the alternating angles BAC and BCA (I.23. El.)
are likewise equal, and consequently their opposite sides CB
and AB. The triangle ABC is thus isosceles; and it is also
given, for its sides are all given.

Λ
Cor. Hence an equilateral polygon is that which, under a
given number of sides, contains, within the same perimeter,
the greatest possible surface: For, the rest of the figure re-
maining constant, suppose any two adjacent sides to vary, and
the accrescent triangle so formed will, by this proposition, be
a maximum, when those sides are equal. The polygon, de-
riving its expansion from the aggregate of the exterior tri-
angles, must therefore be the greatest possible, when such tri-
angles are in every combination isosceles, and consequently
all the sides of the figure equal.
PROP. XXXI. THEOR.
If a polygon have all its sides given, except one,-
it will contain the greatest area, when it can be in-
scribed in a semicircle, of which that indeterminate
side is the diameter.
Let the polygon ABCDEF, having given sides AB, BC,
CD, DE and EF, stand upon a base AF, which is variable;
342
GEOMETRICAL ANALYSIS.
the area will attain its maximum, when AF becomes the dia-
meter of a circumscribing semicircle.
For, AD and FD being inflected to any point D, the
spaces ABCD and DEF will evidently remain the same,
while the angle ADF is enlarged,

A
B
I
F
or the points A and F are dis-
tended. Whence the polygon
must contain the greatest area,
when the included triangle ADF
contained by given sides AD and
DF, is a maximum. Now, this will take place when the alti-
tude of the triangle, or the perpendicular let fall from the
vertex F upon AD, is the greatest possible. Wherefore
(I. 18. El.) ADF is a right angle, and consequently (III. 22. El.).
the point D lies in a semicircumference. But the same rea-
son applies to every other intermediate point B, C, or E, of
the polygon, which consequently, in its state of maximum, is
disposed within a semicircle described on the variable side AF.
Cor. 1. Hence a polygon, whose sides are all given, con-
tains the greatest area, when it can be inscribed in a circle.
For let ABCD be a polygon, which has each of its sides AB,
BC, CD, and AD given. Draw the diameter AF, and
join DF. The polygon ABCDF is thus a maximum; but the
triangle ADF being evidently determinate, the remaining
polygon ABCD is likewise a maximum.
Cor. 2. Hence a regular polygon is that which, with a gi-
ven perimeter, formed by a given number of sides, contains
the greatest arca. For, by the corollary to the last Proposition,
the sides are all equal; but its angles are (III. 14. and 18. El.)
also equal, since it occupies the circumference of a circle.
BOOK III.
343
PROP. XXXII. THEOR.
A circle contains, within a given perimeter, the
greatest possible area.
M
B
P
G
12
pe-
I
From the preceding investigations, it appears, that the
rimeter and number of sides being given, the figure of great-
est capacity is a regular polygon. Let ABCDEF be such a
polygon, bounded by the given perimeter: Bisect the corre-
sponding arcs of the circumscribing circle, and another regu-
lar polygon MBGCHDIEKFLA will arise, having twice the
number of sides. Draw the diameter MI, and join MD and
OD. Both polygons are alike composed of triangles equal
to ODN and ODI, and consequently the area of the poly-
gon ABCDEF is to that of
MBGCHDIEKFLA as ON
to OI, or as 2ON or PN to
201 or MI. But if this exterior
polygon MBGCHDIEK FLA
were contracted to the same
perimeter with ABCDEF, its
area would (VI. 26. El.) be di-
minished in the ratio of DI² to
DN2, that is, (III. 22. and VI.
16. cor. 1. El.) in the ratio of the rectangle MI, NI to MN, NI,
or that of MI to MN. Whence (V. 16. El.) the origi-
nal polygon is to another of equal perimeter and with double
the number of sides, as PN to MN. An isoperimetrical figure
thus has its area always increased, by doubling the number of
its sides. Continuing this duplication, therefore, the regular
polygons which arise in succession will have their capacity per-
petually enlarged. Whence the circle, as it forms the limit
or extreme boundary of all those polygons, must, with a gi-
ven circumference, contain the greatest possible space *.

* See Note LXIII.
L
K
F
1
}
}
:
1
1
ELEMENTS
OF
PLANE TRIGONOMETRY.
i
!
ག
1
ELEMENTS
OF
PLANE TRIGONOMETRY,
TRIGONOMETRY is the science of calculating the
sides or angles of a triangle. It grounds its conclu-
sions on the application of the principles of Geome-
try and Arithmetic.
The sides of a triangle are measured, by referring
them to some definite portion of linear extent,
which is fixed by convention. The mensuration of
angles is effected, by means of that universal stand-
ard derived from the partition of a circuit. Since
angles were shown to be proportional to the inter-
cepted arcs of a circle described from their vertex,
the subdivision of the circumference therefore de-
termines their magnitude. A quadrant, or the fourth
part of the circumference, as it corresponds to a
right angle, hence forms the basis of angular mea-
But these measures depend on the relation
of certain orders of lines connected with the circle,
and which it is necessary previously to investigate.
sures.
T
348
ELEMENTS OF
DEFINITIONS.
1. The complement of an arc is its defect from a quadrant;
and its supplement is its defect from a semicircumference.
2. The sine of an arc is a perpendicular let fall from
one of its extremities upon a diameter passing through the
other.
3. The versed sine of an arc is that portion of a diameter
intercepted between its sine and the circumference.
4. The tangent of an arc is a perpendicular drawn at one
extremity to a diameter, and limited by a diameter extending
through the other.
5. The secant of an arc is a straight line which joins the
centre with the termination of the tangent.
In naming the sine, tangent, or secant of the complement of an arc,
it is usual to employ the abbreviated terms of cosine, cotangent, and
cosecant. A farther contraction is frequently made in noting the
radius and other lines connected with the circle, by retaining only
the first syllable of the word, or even the mere initial letter.
I
I
B
C
Let ACFE be a circle, of which the diameters AF and CE are at
right angles; having taken any arc
AB, produce the radius OB, and draw
BD, AH perpendicular to AF, and
BG, CI perpendicular to CE. Of this
assumed arc AB, the complement is
BC, and the supplement BCF; the
sine is BD, the cosine BG or OD, the
versed sine AÐ, the coversed sine CG,
and the supplementary versed sine
FD; the tangent of AB is AH, and

T
F
D
A
its cotangent CI; and the secant of the same arc is OH, and its co-
secant Ol.
TRIGONOMETRY.
349
Several obvious consequences flow from these defini-
tions :—
1. Since the diameter which bisects an arc bisects also the
chord at right angles, it follows that half the chord of any
arc is equal to the sine of half that arc.
2. In the right-angled triangle ODB, BD²+OD²=OB²;
and hence the squares of the sine and cosine of an arc are to-
gether equal to the square of the radius.
3. The triangle ODB being evidently similar to OAH,
OD : DB :: OA: AH; that is, the cosine of an arc is to the
sine, as the radius to the tangent.
4. From the similar triangles ODB and OAH, OD : OB
:: OA:OH; wherefore the radius is a mean proportional
between the cosine and the secant of an arc.
5. Since BDAD.FD, it is evident that the sine of an
arc is a mean proportional between the versed sine and the
supplementary versed sine, or between the sum and difference
of the radius and the cosine.
6. Hence also the chord of an arc is a mean proportional
between the versed sine and the diameter; for AB² =
AD.AF.
7. The triangles OAH and ICO being similar, AH : OA
:: OC: CI; and hence the radius is a mean proportional be-
tween the tangent of an arc and its cotangent.
8. Since ODBG²=CG.CE, it follows that the cosine
of an arc is a mean proportional between the sum and the
difference of the radius and the sine.
The circumference of the circle is commonly divided
into 360 equal parts, called degrees, each of them being
subdivided into 60 minutes, and these again being each
350
ELEMENTS or
distinguished into 60 seconds. It very seldom is required
to carry this subdivision any farther. Degrees, minutes,
seconds, or thirds, are conveniently noted by these marks,
"/
///
Thus, 23° 27′ 43″ 42″", signifies 23 degrees, 27 minutes,
43 seconds, and 42 thirds *.
H
B
Scholium. To discern more clearly the connexion of the
lines derived from the circle, it will be proper to trace their
successive values, while the corresponding arc is supposed to
increase. Let the arc AB', on the opposite side, be made e-
qual to AB, draw the diameter FOA, extend the diameters
b'OB and bОB', join BB' and bb', and at A apply the double
tangent HAH'. It is evident that BE-be, or that the sine
of the arc AB is equal to the sine of its supplement ABb.
But B'E and b'e, or the sines of ABFb' and ABF'B', which
lie on the opposite side of the diameter, are likewise equal to
BE; that is, the inverted sine of
an arc is equal to the sine of that
arc or of its supplement, augment-
ed, each by a semicircumference.
The arc AB, and its defect ABFB'
from whole circumference,
have both the same cosine OE;
and the supplemental arc ABʊ,
and its defect from a whole cir-
cumference, have likewise the same
cosine, although with an inverted position. AH and OH
are respectively the tangent and secant not only of AB, but
of the arc ABbFb', which is compounded of the original arc
and a semicircumference; and the similar lines AH' and OH',
on the opposite side, are at once the tangent and secant of
the supplementary arc ABb, and of AB6Fb'B', likewise
compounded of that arc and a semicircumference.

a
e
E
A
ட்
As the prolonged diameter b'OBH, therefore, turns about
*See Note LXIV.
TRIGONOMETRY.
351
the centre, the sine and tangent both increase, till the arc
attains 90°, when the sine becomes equal to the radius, and
the tangent vanishes into unlimited extent. Between 90° and
180°, the sine again diminishes, and the tangent, re-appear-
ing in the opposite direction, likewise contracts by successive
diminutions. In the third quadrant, the sine emerges with a
contrary position, and increases till it becomes equal to the
radius; while the tangent, resuming its first position, stretches
out till it vanishes away. Between 270° and 360°, the oppo-
site sine again contracts, and the tangent, re-appearing on the
same side, shrinks also by degrees to a point. In the first and
fourth quadrants, the cosine lies on the same side of the centre,
while the secant stretches from it in the direction of the ex-
tremity of the arc; but, in the second and third quadrants, the
cosine shifts to the opposite side, and the secant shoots from the
centre in a direction opposite to the termination of the arc.
The same phases are thus repeated at each succeeding re-
volution. Hence, if m denote any integral number, the sine
of an arc a is equal to the sine of the arc (2m-1) 180°—a,
and to the opposite sines of (2m—1) 180° + a and of
2m.180°-a; the cosine and secant of an arc a are equal to
the cosine and secant of 2m.180°—a, and to the opposite co-
sines and secants of (2m-1) 180°-a and of (2m—1) 180°+a;
and the tangent or cotangent of an arc a is equal to the tan-
gent or cotangent of the arc (2m-1) 180°+a, and to the
opposite tangents or cotangents of the arcs (2m—1) 180°—a
and 2m.180°—a.
An arc may, by a simple extension of analogy, be con-
ceived to comprehend innumerable other arcs. Thus, the
arc AB, in fact, represents all the arcs which have their ori-
gin at A and their termination at B; it therefore includes not
only the small arc AB, but that arc as augmented by succes-
sive revolutions, or the repeated addition of entire circumfe-
rences. Hence the sine or tangent of an arc a are the same
with the sine or tangent of any arc n.360° +a*.
See Note LXV.
352
ELEMENTS OF
PROP. I. THEOR.
The rectangle under the radius and the sine of the
sum or difference of two arcs, is equal to the sum or
difference of the rectangles under their alternate
sines and cosines.
Let A and B denote two arcs, of which A is the greater;
then, Rx sin(AB)=sinA × cos B±cosA × sinB.
For, having made BC' BC, it is evident that AC and AC
will represent the sum and difference of the arcs AB and BC;
join OB and CC', and draw HFH' parallel, and CE, FG,
BD, and H'CE perpendicular, to the radius OA.
:
H F
B
The triangles COF and C'OF, having the side CO equal
to C'O, OF common, and the contained angles FOC and
FOC' measured by the equal arcs BC
and BC, are equal; wherefore OF bi-
sects CC' at right angles. But the tri-
angles OBD and OFG being similar,
OB: BD OF: FG, or HE, and
consequently OB.HE=BD.OF. The
triangles OBD and CFH are likewise
similar, for the right angle CFO being
equal to HFG, if HFO be taken from both, the remaining
angle CFH is equal to OFG or OBD; whence OB: OD::
CF: CH, and OB.CH=OD.CF. Wherefore OB.HE +
OB.CH, or OB.CE=BD.OF+OD.CF. But BD and OD
are the sine and cosine of the arc AB, CF and OF the sine
and cosine of BC, and CE is the sine of the compound arc AC.
Consequently, Rx sinAC=sinABX cos BC+cosAB x sin BC.

O E GD E'A
Again, taking the difference of the rectangles OB, H’E' and
OB, C'H', and OBX C'E' BDX OF-OD x CF; whence
RX,sinAC'sinABX cos BC-cosABX sinBC.
2
TRIGONOMETRY.
353
Cor. 1. If the two arcs A and B be equal, it is obvious that
Rx sin2A=sinA × 2cosA.
Cor. 2. Let the arc A contain 45°; then RX sin(45°÷B)=
sin45° (cos B sin B)=√ R²(cos B sinB), or sin(45°÷B)=
√(cos B±sinB.)
Cor. 3. Let 2A=C, and, by the first corollary, Rx sinC=
sin Cx 2cos & C.
✪
PROP. II. THEOR.
The rectangle under the radius and the cosine of
the sum or difference of two arcs, is equal to the dif
ference or the sum of the rectangles under their re-
spective cosines and sines.
Let A and B denote two arcs, of which A is the greater;
then Rx cos(A+B)=cos A x cos BsinA x sin B.
For, in the preceding figure, the triangles OBD and OFG
being similar, OB: OD :: OF: OG, and OB.OG=OD.OF,
and the triangles OBD and CFH being likewise similar,
OB: BD :: CF: FH, or GE, and consequently OB.GE=
BD.CF. Wherefore OB.OG-OB.GE=OB.OE=OD.OF-
BD.CF; that is, RX cos AC cos ABX cosBC-sinAB X sin BC.
Again, taking the sum of those rectangles, OB × OG +
OBX GE=OB x OE=OD × OF + BD x CF; whence
RXcos AC' cos ABX cos BC+sinAB x sinBC.
Cor. 1. If A and B represent two equal arcs, it will follow,
that RX cos2A=cos A²—sin A² = (cosA+sinA)(cosA—sinA);
or, since cosA² — R²—sinA², R × cos2A = R²—2sinA² =
2cos A-R2.
Ꭱ
Cor. 2. Hence sinA² = R(R—cos2A), and cos², =
¿R(R+cos2A); wherefore sin A*—sinB² = {R(cos2 B—cos2A).
Z
354
ELEMENTS OF
Cor. 3. Let the arc be equal to 45°, and RX cos(45° ±B)=
sin 45°(cosВ sinB).
Cor. 4. Let 2A=C, and by the first corollary, R×¢os¤=
R²-2sin C² 2cos C²-R².
PROP. III. THEOR.
Of three equidifferent arcs, the rectangle under
the radius and the sum or difference of the sines of
the extremes, is equal to twice the rectangle under
the cosine or sine of the common difference and the
sine or cosine of the mean arc.
=
Let A-B, A, and A+B represent three arcs increasing
by the difference B; then R (sin(A + B) + sin (A—B))
2cosB× sinA, and R(sin(A+B)—sin(A—B))=2sinB × cosA.
C
These properties are easily deduced by combining the pre-
ceding theorems; but
they are more easily
perceived, by referr-
ing immediately tothe
original figure. The
triangles OBD and
OFG being similar,

OB: BD :: OF: FG,
or OB: BD :: 20F
: 2FG or CE+C'E',
andOB(CE+C'E')=
20FX BD; that is,
R(sinAC+sinAC')=
11
E
C
L
JE GD E' A
K
I
2cos BC × sin AB. Again, OB: OD :: CF: CH:: 2CF:
2CH or CE-C'E', and OB(CE—C'E')=2CF ×OD; con-
sequently R(sin AC-sinAC')—– 2sin BC× cosAB.
TRIGONOMETRY.
355
Cor. 1. Hence also R(cos(A-B)+cos(A+B))=2cosB x cosa,
and R(cos(A-B)-cos(A+B))=2sinB × sinA.
For OB: OD:: OF: OG: 20F: 20G or OE'+OE,
and OB(OE'+OE)=2OF × OD; that is,R(cosAC'+cosAC)
=2cos BC x cos AB. Again, OB BD CF: FH::
::
2CF: 2FH, or OE'—OE, and OB(OE'—OE)=2CF× BD;
that is, R(cosAC'—cosAC)=2sinBC × sinAB.
:
Cor. 2. Let the radius be expressed by unit, and arcs B
and A, denoted by a and na; then collectively 2sin a x cos na
=sin(n+1)a sin(n-1)a, 2cos a x sin na sin(n+1)a+
sin(n-1)a, 2sina x sin na = cos(n-1)a—cos(n + 1)a, and
2cos a x cos na = cos(n-1)a+cos(n+1)a.
×
Cor. 3. Since versB=R-cos B, it follows that R(sin(A+B)
+sin(A-B))=2Rx sinA-2vers B x sinA, and consequently
R× sin(A+B) = 2R × sinA—R × sin(A—B)-2versB × sinA,
or R(sin(A+B)—sin A) = R(sin A-sin(A-B))—2vers B x sinA.
In the same way, it may be shown that R(cos(A-B)-cos A)=
R(cos A-cos(A+B))-2vers B x cosa.
Cor. 4. If the mean arc contain 60°; then R(sin(60°+B)
—sin(60°—B))=2sinB × cos60°, or sinB ×2sin30º. But
twice the sine of 30° being (cor. 1. def.) equal to the chord
of 60° or the radius, it is evident that sin(60° + B)—
sin(60°—B)=sinB, or sin(60° +B)= sin(60°—B)+sinB.—
This property also follows from Prop. 14. Book IV. of the
Elements; for BD=AD+CD, and BD=AD+CD, or
sin BAD = sinḥAD + sin¿CD, that is, sin(60° + §AD)=
sin¸AD+sin(60°—↓ AD.)
Cor. 5. Produce CE to the circumference, join C'I meet-
ing the production of FG in K, and join OK. Since FK is
parallel to CI and bisects CC', it likewise bisects IC'; and
hence OK is perpendicular to KC', which is, therefore, the
sine of half the arc IAC', or of half the sum of the arcs AC
and AC', as CF is the sine of half their difference. But
Z 2
356
ELEMENTS OF
(II.24.El.) IC¹²¬CC'²=IC × 2C'E',orC/K"-CF²=CE× C'E';
consequently sin² AB— sin² BC = sinAC × sinAC', or, em
ploying the general notation, sin A*—sin B²=sin(A+B)×
sin(A—B)= (2. cor. 2.) R(cos2B-cos2A) *.
Scholium. By help of this proposition, the sines and cosines
of multiple arcs are easily determined; but the expressions for
them will become simpler, if, as in cor. 2. the radius be supposed
equal to unit. For A, 2A and 3A being three equidifferent arcs,
sinA + sin3 A = 2cosA × sin2A = 2cosA × 2eosA × sinA, or
sin3A = 4cosÀª × sinA—sinA; and cosA + cos3A = 2cosA ×
cos2A=2cos A (2cos A²-1) = 4cos A3-2 cos A, or cos 3 A =
4cos A3-3cosA. Again, since 2A, 3A, and 4A are equidif-
ferent arcs, sin2A+sin4A=2cos A x sin3 A=8cosA³ x sinA-
2cos A x sinA, or sin4A 8cos A3 x sin A-4cosA x sinA;
×
A³ × ×
cos 2A+cos4 A = 2cos A × cos3 A = 2cos A (4cos A³ —3cos▲), or
cos4 A = 8cos A+—8cos A2+1. In like manner, assuming the
-ScosA²+1.
equidifferent arcs 3A, 4A, 5A, and the sine and cosine of 5A
are found; and this mode of procedure may be continually re-
peated. To abridge the notation, however, it will be proper
to express the sine and the cosine of the arc a, by s and c.
The results are thus expressed in a tabular form:
Sin 2a = 2cs.
Sin 3a
4c²s
S.
4cs.
Sin 4a 8c³s
=
(1.) Sin 5a = 16cªs — 12c²s + s.
Sin 6a= 32c's
· 32c³s + 6cs.
Sin 7a = 64c´s
Cos 2a = 2c²
Cos 3a4c3
(2.) Cos 4a = 8c*
1.
3C.
80c's+24c's — s.
&c. &c. &c.
3c21.
Cos 5a
16cs
20c³ +5c.
Cos 6a = 32c°
48ct of 18c² — 1.
&c. &c. &c.
*See Note LXVI.
TRIGONOMETRY.
357
If in these expressions, I-s² be substituted forc², in the sines
of the odd multiples of a, and in the cosines of the even mul-
tiples,—the sines and cosines of such multiple arcs will be re-
presented merely by the powers of the sine a.
Sin 3a 35 4$3.
(3.) Sin 5a = 5s
Sin 7a
-
20s³ + 16s.
= 7s — 56s³ + 112s5
6457.
&c. &c. &c.
Cos 2a = + 1
252.
(4.) Cos 4a =
+ 1
Ss2Ss+.
Cos 6a = = + 1 — 18s² + 4854
&c. &c. &c.
325.
If the terms of the first table be repeatedly multiplied by
2s, and those of the second by 2c, observing the substitu-
tions of cor. 2, there will result expressions for the sines and
cosines. Thus, 2sin a² = 25.5 == cos 2a + 1, 4sin a³ ——
2s.cos 2a+2s = — sin 3a + sin a +2s =— sin 3a + 3s, and
8sin at —— -2s.sin 3a + 2s.35 = + cos 4a—cos 2a-3cos 2a +3=
cos 4a—4cos 2a+3. Again, 2cos a²=2c.c=cos 2a+1, 4cos a³ =
2c.cos 2a + 2c = cos 3a + cos a + 2c = cos 3a+3cos a, and
8cos at 2c.cos 3a + 2c.3cos a = cos 4a+cos 2a+3cos 2a+3=
cos 4a+4cos 2a+3. In this manner, the following tables are
formed.
4
Sin a=s.
2 Sin a² -
cos 2a + 1.
4 Sin a³
(5.) 8 Sin a^ =
s
=
16 Sin as =
32 Sin a
+ sin 5a
sin 3a + 3s.
+ cos 4a
4cos 2a + 3.
5sin 3a + 10s.
64 Sin a' =
cos Ġa + 6cos 4a
sin 7a7sin 5a
15cos 2a + 10.
21sin 3a + 35s.
&c. &c. &c.
358
ELEMENTS OF
Cos a = c.
2 Cos a² = cos 2a + 1.
4 Cos a³ = cos 3a + 3c.
(6.) 8 Cos a² = cos 4a + 4cos 2a + 3.
16 Cos a³ = cos 5a 5cos 3a + 10c.
+
32 Cos a 6 =cos 6a+6cos 4a+15cos 2a + 10.
64 Cos a² = cos 7a +7cos 5a +21cos 3a + 35c:
&c. &c. &c *.
PROP. IV. THEOR.
The sum of the sines of two arcs is to their diffe-
rence, as the tangent of half the sum of the arcs to
the tangent of half the difference.
If A and B denote two arcs; the sinA+sinB : sinA—sinB
:: tan
A+B
2
: tan
A-B
2
For, let AC and AC' be the sum and difference of the arcs
AB and BC, or BC'; draw the perpendiculars CE, and C'E',
extend the chord CC', and
H
C
B
F
O E
E'A
K L
apply at B the parallel tan-
gent HBL, meeting in K
and L the diameter produ-
ced, and draw OCH, OFB
and OC'H'. Because CE
is parallel to C'E', and CK
to HL,CE: C'E':: CK: C/K
(VI. 2. El.) HL: H'L; and consequently CE+C'E' :
CE-C'E':: HL+H'L: HL-HL', that is, 2BL: 2BH, or
BL: BH. But CE and C'E' are the sines of the arcs AC
and AC', and BL and BH are the tangents of AB and BC,
or of half the sum and half the difference of those arcs.
AC+AC':
2
Wherefore sin AC+sinAC' : sinAC—sinAC' :: tan

AC-AC'.
tan
2
* See Note LXVII.
TRIGONOMETRY.
359
Cor. 1. The sines of the sum and difference of two arcs
are proportional to the sum and difference of their tangents.
For CE : C'E' :: HL, or BL+BH: H'L, or BL—BH;
that is, resuming the general notation, sin(A+B): sin(A—B);::
tan A+tan B: tan A-tan B.
:
Cor. 2. Let the greater arc be equal to a quadrant; and
R+sinB R-sinB :: tan(45° +B) : tan(45°-4B), or
cot (45°+B). But, the radius being a mean proportional
between the tangent and cotangent of any arc, and the co-
sine of an arc being a mean proportional between the sum and
difference of the radius and the sine, it follows that R+sinB:
cosB :: R: tan(45°—B), and R-sinB : cosB, or cosB :
R+sinB, :: R : tan(45°+1B).
Or, if instead of B, there be substituted its complement,
these analogies will become R+cosB: sinB :: R : tanžВ, and
R-cosB: sinB:: R cot B.
:
Cor. 3. Since cosB: R :: R-sinB : tan(45°—B), and
cosB: R:: R+sinВ : tan(45+B), therefore (V. 19. El.)
cosB: R2R tan(45°-B)+tan(45° +B); that is, sup-
posing B to be the complement of 2C, sin2C : 2R :
R:tanC+cotC. But (Prop. 1. cor. 1.)R x sin2C=2cos Cx sinC,
and consequently cosC× sinC : R² :: R : tanC+cotC.
Cor. 4. Since (4. cor. def.) cos B R R secB, and
(3. cor. def.) cosB: sinB::R:tan B, therefore cos B: R+sinB::
R:tanB+secB, and consequently (2. cor. def.)tan(45°+{B)=
tanB+sec B.-This also appears clearly from the figure, on sup-
posing OH'H'L', or the angle LOH' equal to OLH', and
consequently the arc AC' equal to the complement of AB.
PROP. V. THEOR.
:
As the difference or sum of the square of the ra-
dius and the rectangle under the tangents of two
arcs, is to the square of the radius,-so is the sum or
difference of their tangents, to the tangent of the
sum or difference of the arcs.
360
ELEMENTS OF
I
Let A and B denote two arcs, of which A is the greater;
then R*tan Ax tanB: R:: tanA÷tanB: tan(A÷B.)
R²
For (3. cor. def.) R: tanA :: cosA: sinA, and R: tanB::
cosB:sinB; whence (V.22. El.) R2: tanAxtanB:: cosAx cosB
sinAxsinB, and (V. 11. and 7. El.) RtanA x tanВ :
R² cosAx cos B sinA x sinB: cosA x cos B, that is,
Ꭱ ::
R²tanA × tanB: R²:: R× cos(A+B): cosAx cos B. But
(3. cor. def.) cos(A÷B) × tan(A±B)=R× sin(A+B), and
cos A × cos B (tan A± tan B) = cos A × tan A × cos B±
cos Ax cos B x tan B=RX sin A x cos BR x cos Ax sinB=
(Prop. 1.) R² × sin(AB); wherefore (V. 6. and 3. El.)
Rx cos(AB): cos A× cosB:: tanA±tanB: tan(A÷B), and
consequently R² tanAxtanB R2 tanAtan B:
—
tan(A+B)*.
::
Cor. 1. Let the two arcs be equal; and R²—tanA² : R²
2tanA : tan2A.
Cor. 2. Let the greater arc contain 45°; aud since
tan45°-R, it follows that RRXtanB: R² :: R±tanB:
tan(45º±B), or RanB : R±tanB :: R : tan(45°±B) †.
Scholium. Assuming the radius equal to unit, expressions
are hence easily derived for the tangents of multiple arcs.
Let t denote the tangent of an arc a; then I—t² : 1 : : 2t :
tan2a=
2t
I—t²
and I-t.
I-
2t
t
2t
3t-ts
:1::t+
: tan3a=
I-3t²
In like manner, it will be found that,
Tan 4a=
(7.) Tan 5a=
Tan Ga=
4t-413
1—6ť² + 1ª
5t-10t31s
1—10t² + 5t4
6t-2013+6ts
1—15t² + 15tª.
&c. &c. &c. I.
* See Note LXVIII.
+ Sce Note LXIX.
† See Note LXX.
TRIGONOMETRY.
361
1
These formula might also be derived from expressions for
the sine and cosine of the multiple arc which involve the
powers of the tangent. Thus, from (1), sin2a=2cs=c²(25)
=c².2t, and sin3a = 4c²s—s=3c²s—(I—c²)s=c³ ( 3
S s3
C C³
)
c³ (3t-t³); again, from (2), cos2a=2c2-1=c² —s² =
sz
c² ( 1 —— ² ) = c²(1—1²), and cos3a= 4c³ —3c = c³ —3c(1—c² )=
e³ (1—3——)=c³ (1—31²). In this way, the following tables
are formed:
Sin 2a = c.2t.
Sin 3a¸ = c³ (3t—13).
(8.) Sin 4a = c4 (4t— 4t³).
Sin 5a =
cs(5t-101³+15).
Sin 6a = c(6t—201³+6t³).
&c. &c. &c.
Cos 2a = e²(1-ť²).
Cos 3a = c³(1—3ť²).
(9.) Cos 4a=c4(1—6t®+t4).
Cos 5a = c(1-10tª+5t4).
Cos 6a =
c°(1—15t³+15tª—t°).
&c. &c. &c.
7
The first set of expressions being divided by the second,
will evidently give the same results for the tangent of the
multiple arc *.
PROP. VI. THEOR.
The supplemental chord of half an arc, is a mean
proportional between the radius, and the sum of the
diameter and the supplemental chord of the whole
arc.
* See Note LXXI.
263
ELEMENTS OF
This property, which is only a modification of cor. 2. to Pr. 2.
will admit of a more direct demonstration. For draw the
chord AB, the semichords AE and BE, and the supplement-
al chords CB and CE,

E
F
KL
and the radius OE.
The isosceles triangles
AEB and COE are si-
milar, for the` angles
OCE and EAB at the
base stand on equal arcs
AE and EB; consequently AE AB CO CE. But,
ACBE being a quadrilateral figure contained in a circle,
CE.AB=AE.CB+EB.CA=AE (CA+CB), or AE : AB
::CE: CA+CB; wherefore CO : CE :: CE : CA+CB,
or
CA+CB
CE²=CA (CA-
2
C
::
:
D IZA
Cor. Hence, in small arcs, the ratio of the sine to the arc
approaches that of equality. For, let the semiarcs AE and
EB be again bisected in the points F and G; and, continuing
their subdivision indefinitely, let the successive intermediate
chords be drawn. The ratio of the sine BD to the arc AB
may be viewed as compounded of the ratio of BD to the
chord AB, of that of AB to the two chords AE and EB, of
that of AE and EB to the four chords AF, FE, EG, and
GB, and so forth. But these ratios, it has been shown, are
the same respectively as those of the supplemental chords
CB, CE, CF, &c. to the diameter CA. And since each of
the ratios CB: CA, CE: CA, CF: CA, &c. approaches to
equality, it is evident that their compounded ratio, or that of
the sine to its corresponding arc, must also approach to equa-
lity.
Scholium. Hence the ratio of the sine BD to the arc AB is
expressed numerically, by the ratio of the continued product
of the series of supplemental chords CB, CE, CF, &c. to the
relative continued power of the diameter CA. The ratio
may, therefore, be determined to any degree of exactness, by
TRIGONOMETRY.
363
the repeated application of the proposition in computing those
derivative chords. But a very convenient approximation is
more readily assigned. Make CD to CI as CB to CA, CI to
CK as CE to CA, CK to CL as CF to CA, and so forth,
tending always towards the limit Z; then the ratio of CD
to CZ, being compounded of these ratios, must express the ra-
tio of the sine BD to its corresponding arc AB. Now CD : CB
:: CB CA; consequently CI=CB, and CD CI::
CA, or the point I nearly bisects DA. Again,
and therefore CE differs from CA,
CI
CE² = CA (
(CA+CB),
by only the fourth part of the difference between CB and CA.
These differences being small in comparison of the quantities
themselves, the series of supplemental chords may be consi-
dered as forming a regular progression, each succeeding term
of which approaches four times nearer to the length of the
diameter. Wherefore IK= 4DI, KL= 4IK, and so conti-
nually. But (V. 21. El.) as the difference between the first
and second term, is to the first, so is the difference between
the first and last term, or DI itself, to the sum of all the terms,
or the extreme limit DZ; that is, 3 : 4 :: DI: DZ; and con-
sequently DZ=4DA. The ratio of the sine BD to the arc
AB is, therefore, nearly that of CD to CD+DA, or of
3CD to CD+2CA.
This approximation may be differently modified. Since
3CD 60A-3DA, and CD + 2AC 60A-DA, it fol-
=
lows that BD is to AB, as 60A-3DA to 60A-DA.
But this ratio, which approaches to equality, will not be
sensibly affected, by annexing or taking away equal small dif-
ferences. Whence the sine is to the arc, as 60A-6DA
to 60A-4DA, or 30D to OA+20D. But OD is to OA,
as the sine of AB is to its tangent; and consequently the
triple of that arc is equal to its tangent together with twice
its sine.
Again, both terms of the ratio increased by the minute
difference DA become 60A-2DA, and 60A; wherefore
A
364
ELEMENTS OF
BD is to AB, as 30A-DA to 30A, or as 20C+OD to 3CO,
Hence, if CP be made equal
to the radius CO, and PBH
bedrawn to meet the tangent,
-the arc AB will be near-

Bj.
DA
P
ly equal to the intercepted
portion AH. For BD: AH :: PD: PA, or 20C+OD: 30C;
that is, as the sine BD is to its arc AB.
Another approximation, of much higher importance, may
be hence derived; for PD : PA :: BD : AH, or as the
sine to its arc nearly. But (V. 3. El.) PD x CD is to
PAX CD in the same ratio, and PA x CD=PD × CD +
AD × CD = (III. 32. cor. 1.) PD × CD + BD²; whence
PD x CD is to PD × CD + BD², as the sine to its arc
nearly. If the arc be small, it is evident that OD will
be very nearly equal to AO, and consequently PD may be
assumed equal to 3AO, and CD equal to 2AO. Wherefore
6AO² : 6AO² + BD² : : BD : AB nearly; or, the radius
being unit, and a and s denoting a small arc and its sine,
53
6:6+s²::s: a, and hence as + nearly. But since
6
a and s are very small, a³ will approach extremely near to s³,
and it may, therefore, be inferred conversely, that sa-
Q3
6
A convenient approximation for the versed sine of an
arc is easily derived from the fundamental property of the
lines themselves; for 2AO,AD AB² = BD+AD", or
=
employing to denote the versed sine, 2v=s²+v², and
Sz
+
2
-. If, therefore, the arc be small, it may be suffi-
s²
ciently near the truth to assume v= ; but should greater
accuracy be required, substitute this value of v in the second
term of the complete expression, and v
form a very close approximation.
s²
+
which will
2
8
J
TRIGONOMETRY.
365
Calculation of the Trigonometrical Lines.
The preceding theorems contain all the principles required
in constructing Trigonometrical Tables. The radius being
denoted by unit, the several lines connected with the circle
are referred to that standard, and are generally computed to
seven decimal places.
The first object is to compute the SINES for every arc of
the quadrant.
Since the semicircumference of a circle whose radius is
unit was found, by the scholium to Prop. 32. Book VI. of
the Elements, to be 3.1415926, the length of the are of one
minute is .0002909, which, in so small an arc, may be as-
sumed as equal to the sine, and consequently the versed
sine of a minute (.0002909) = .000,000,042,308.
Whence, by cor. 3. to Prop. 3. sin(A +1') = 2sinA-
2sinA × .000,000,042,308—sin (A—1′); and therefore, by a
series of repeated operations, the intermediate arc being suc-
cessively 1', 2', 3′, 4', &c. the sines of 2', 3', 4', 5', &c. in
their order will be calculated.
=
The numbers thus obtained will at first scarcely differ from
an uniform progression, the versed sine of 1', which forms
the multiplier of deviation, being so extremely small. It is
hence superfluous to compute rigidly all those minute varia-
tions. The labour may be greatly shortened, by calculating
the sines for each degree only, and employing some abridged
process for filling up the sines, corresponding to the subdivi-
sion in minutes.
The arc of one degree being equal to .0174533, it fol-
lows from the scholium to Prop. 6., that the sine of
1° = .0174533-(.0174533)³ —.0174524, and hence the
366
ELEMENTS OF
=
versed sine of 1° (.0174524)=.0001523.
Wherefore
sin(A+1°)=2sinA-2sinA x .0001523-sin (A-1°); or, if
from twice the sine of an arc, diminished by its 6566th part,
the sine of an arc one degree lower be subtracted, the remainder
will exhibit the sine of an arc, which is one degree higher.
Thus,
Sin2°-2sin1-2sin 1°.0001523.0349048-.0000053
=.0348995
Sin3°—2sin2º-2s n2°X.0001523-sin 1°.0697990.0000106—
.0174524.0523360.
Sin4°2sin3°—2sin3° × .0001523-sin2°.1046720—.0000160—
.0348995.0697565.
After this manner, the sine for each degree is computed in
succession.
But the sines may be found, independently of the previous
quadrature of the circle. Assuming an arc whose chord is
already known, it is easy, from Prop. 6. to determine the
successive chords and supplemental chords of its continued
bisection. Let that are be 60°; its chord is equal to
the radius, and (IV. 20. cor. 2.) its supplemental chord
=√3=1.7320508076. Whence the supplemental chord
of 30° = √(2+1.7320508076) = 1.9318516525. In this
way, by continued extractions, the supplemental chords of
15°, 7° 30', 3° 45', and 1° 52' are successively computed,
the last one being equal to 1.9997322758. Again, the
chords themselves are deduced by a series of analogies;
for 1.9318516525 : 1 :: 1 :.51763809004 chord of 15º,
and so repeatedly, till the chord of 1° 52', which is
.0327234633. Hence, taking the halves of those num
bers, the sine of 56'4 = .0163617317 and the cosine of
56′ 9998661379, and therefore (cor. 3. defin.) the tan-
gent of that arc is .0163639215; consequently the arc itself
(2 x .0163617317+.0163639215) = .0163624616, and
thence the length of the arc of a minute is .0002908882086.
"3
TRIGONOMETRY,
367
Wherefore the sine of 1'=.0002908882-(.0002908882)³
=.00029088826046, and the versed sine of 1'=
(.00029088826046)²=.000000042308.
Employing these data, therefore,
Sin2'-2sin 1'—2sin1' × .000000042308=.0005817763845;
Sin32sin2'-2sin2'X.000000042308-sin i'.0008726645152;
and so forth *.
n
120
But it is very seldom requisite to push the estimation
to such extreme nicety. The sines being calculated for
each degree as before, those corresponding to the subdivision
in minutes, may be found by a more expeditious method,
though founded on ulterior considerations. If the sines in-
creased uniformly, the sine of A°+n' would exceed that of
A by the quantity (sinA +1°—sinA—1°)=B. But the
rate of this augmentation, being continually retarded, occa-
sions a defect, equal to n² x sinA x .000,000,042308 C. A-
gain, since the retardation itself gradually relaxes, it requires
a small compensation, which may be estimated at (60-7)Bx
.0000013=D. The sine of A°+n' is then very nearly=
sinA + B-C+D. Thus, the sines of 31°, 32°, and
33° being respectively .5150381, .5299193, and .5446390,
let it be required to find the sine of 32° 40'. Here
B =
40
120
-
(sin33°—sin31°) = .0098670, C=1600× sin32° ×
.0000000423.0000359, and D=20 ×.0098670 × .0000013
=.0000003. Whence sin32° 40′.5299193+.0098670—
.0000359+.0000003.5397507.
After the sines are calculated up to 60°, the rest are deduced
from cor. 4. Prop. 3. by simple addition. Thus, sin61°=
sin59° +sin1° =.8571673+.0174524.8746197 †.
The VERSED SINES and supplementary versed sines are
only the difference and sum of the radius and the sines.
* See Note LXXII.
+ See Note LXXIII.
368
ELEMENTS of
}
The TANGENTS are easily derived from the sines, by help
of the analogy given in the 3d corollary to the definitions.
Thus, cos 32° : sin 32° :: R: tan 32°, or, .8480481:.5299193
:: 1.6248694-tan32°. Beyond 45°, the calculation is
simplified, the radius being (cor. 7. defin.) a mean propor-
tional between the tangent and cotangent, or the cotangent
is the reciprocal of the tangent.
The SECANTS are deduced by cor. 4. to the definitions,
since they are the reciprocals of the cosines.
From the lower tangents and secants, the tangents of arcs
that exceed 45° are most easily derived; for (cor. 4. Prop. 4.)
tan(45°+a)=sec 2a+tan 2a. Thus, tan46 sec 2° +tan 2°,
or 1.03553031.0006095+.0349208.
*
PROP. VII. THEOR.
In a right angled triangle, the radius is to the sine
of an oblique angle, as the hypotenuse to the oppo-
site side.
Let the triangle ABC be right angled at B; then
R: sinCAB:: AC: CB.
For assume AR equal to the given radius, describe the arc
RD, and draw the perpendicular
:
RS. The triangles ARS and
ACB are evidently similar, and
therefore AR: RS: AC: CB.
But, AR being the radius, RS
is the sine of the arc RD which
measures the angle RAD or
CAB; and consequently R : sinA
:: AC: CB.

A
R
S D
מ.
B
Cor. Hence the radius is to the cosine of an angle, as the
hypotenuse to the adjacent side; for R:sinC or cos A:: AC: AB.
i
TRIGONOMETRY.
369
PROP. VIII. THEOR.
In a right angled triangle, the radius is to the tan-
gent of an oblique angle, as the adjacent side to the
opposite side.
Let the triangle ABC be right angled at B; then
R: tanBAC::AB: BC.
T
C
For, assuming AR equal to the given radius, describe the
arc RD, and draw the perpendi-
cular RT. The triangles ART
and ABC being similar, AR: RT
::AB: BC. But, AR being the
radius, RT is the tangent of the
arc RD which measures the angle
at A; and therefore R: tan▲ : :
AB: BC.

D
A
R
B
Cor. Hence the radius is to the secant of an angle, as the
adjacent side to the hypotenuse. For AT is the secant of
the arc RD, or of the angle at A; and, from similar tri-
angles, AR AT: : AB: AC.
PROP. IX. THEOR.
The sides of any triangle are as the sines of their
opposite angles.
In the triangle ABC, the side AB is to BC, as the sine of
the angle at C to the sine of that at A.
For let a circle be described about the triangle; and the
sides AB and BC, being chords of the intercepted arcs
A 2
370
ELEMENTS OF

or of the angles at the centre, are
f
(cor. def.) equal to twice the
sines of the halves of those angles,
or the angles ACB and CAB at
the circumference. But, of the
same angles, the chords or sines
(VI. 12. cor. El.) are proportional
B
to the radius; and consequently AB: BC:: sinC : sinA.
Cor. Since the straight lines AB and BC are chor s, not
only of the arcs AB and BC, but of the arcs ACB and BAC,
or the defects of the former from the circumference, it fol-
lows that the sides of the triangle are proportional also to
the sines of half these compound arcs, or to the sines of the
supplements of their opposite angles.-A like inference results
from the definition, for the sine of an arc and that of its sup-
plement are the same.
PROP. X. THEOR.
In any triangle, the sum of two sides, is to the
difference, as the tangent of half the sum of the
angles at the base, to the tangent of half their dif-
ference.
tan
C + B
In the triangle ABC, AB+ AC: AB-AC:: tan 2
C-B
2
For, by the last proposition, AB : AC :: sinC : sinB,
and consequently (V. 12. El.) AB+AC: AB-AC :
sinC+sinB ; sin C-sin B. But, by Prop. 4. sinC+sinB:
C-B
:
tan- ; wherefore, by identity
C+B
sinC-sinB:: tan
2
2
of ratios, AB+AC: AB-AC:: tan-
C+B
2
C-B
: tan-
2
:
TRIGONOMETRY.
371
I
Otherwise thus.
From the vertex A, and with a distance equal to the
greater side AB, describe the semicircle FBD, meeting the
other side AC extended both ways to F and D, join BD and
BF, which produce to meet a straight line DE drawn pa-
rallel to CB.

Because the isosceles tri-
angle DAB, has the same
vertical angle with the tri-
angle CAB, each of its re-
maining angles ADB and
ABD is (I. 32. El.) equal to
F
E
B
A
C D
half the sum of the angles ACB and ABC; and therefore
the defect of ABC from that mean, that is the angle
CBD, or its alternate angle BDE, must be equal to half
the difference of those angles. Now FBD being (III. 22. El.)
a right angle, BF and BE are tangents of the angles BDF
and BDE, to the radius DB, and hence are proportional
to the tangents of those angles with any other radius. But
since CB and DE are parallel, CF, or AB+AC : CD, or
AB-AC :: BF BE; consequently AB+ AC: AB-AC
ACB+ABC ACB-ABC
:: tan
2
:
: tan
2
or AB+AC: AB-AC
:: cot &A: cot(B+A), or-cot(C+&A).
Cor. Suppose another triangle abc to have the sides ab
and ac equal to AB and AC, but containing

a right angle: It is obvious that tan
c+b
2
: tan
2
:: tan
: tan
ACB+ABC ACB-ABC
2
2
, or
R:tan (45°-b):: tan
ACB+ABC
2
ACB-ABC
: tan-
a
2
that is, R: tan(45—b): :cot½Â: cot (B+}A), or—cot(C+{A).
Now, in the right angled triangle abc, ab or AB, is to ac,
or AC, as the radius, to the tangent of the angle at b.
A a 2
ELEMENTS OF
372
PROP. XI.. THEOR.
In any triangle, as twice the rectangle under two
sides, is to the difference between their squares and
the square of the base, so is the radius, to the co-
sine of the contained angle.
In the triangle ABC, 2AB × AC: AB²+AC²—BC² ::
R: cosBAC; the angle BAC being acute or obtuse, accord-
ing as BC² is less or greater than
AB²+AC².
B
For let fall the perpendicular
BD. In the right angled triangle
·ADB, AB: AD:: R:sinABD or
cos BAC; consequently2AB × AC:
2AD × AC::R: cosBAC. But
(II. 26. El.) twice the rectangle
under AD and AC is equal to
the difference of the squares AB
and AC from the square of BC. D
T

A
D

A
Whence 2ABX AC: AB + AC²-BC²:: R:cos BAC.
Cor. The radius being denoted by unit, it follows (V. 6. El.)
that AB+ AC-BC-2ABX ACX cos BAC, and conse-
quently BC AB+ AC-2AB × AC x cos BAC, or
BC= √(AB² + AC²-2AB × AC × cos BAC).
PROP. XII. THEOR.
In any triangle, the rectangle under the semipe-
rimeter and its excess above the base, is to the rect-
angle under its excesses above the two sides, as the
square of the radius, to the square of the tangent of
half the contained angle.
In the triangle ABC, the perimeter being denoted by P,
¿P(↓P—AC): (†P—AB) (¿P—BC) : : R²: tan}B².
TRIGONOMETRY.
373
For, employing the same construction as in Prop. 31.,
Book VI. of the Elements; since the triangles BIE and BGD
are right angled, BI: IE :: R : tanIBE, or tan B, and
BG: GDR tanGBD, or tan B; whence
:
(V. 22. El.) BI× BG: IE× GD :: R²: tan B².
But it was proved that
IE X GD = AI X AG;
wherefore BI × BG:AI × AG
:: R2.tan B. Now BI is e-
qual to the semiperimeter,
BG is its excess above the
base AC, and AI, AG are
its excesses above the sides
AB and BC; consequently
the proportion is establish-
B

KF
I
H
L
ed.
E
PROP. XIII. THEOR.
In any triangle, the rectangle under two sides, is
to the rectangle under the semiperimeter, and its
excess above the base, as the square of the radius,
to the square of the cosine of half the contained
angle.
In the triangle ABC, the perimeter being denoted by P,
ABX BC: P(P-AC) :: R2: cos & B².
For, the same construction remaining; in the right angled
triangles BIE and BGD,
BE: BI:: R: sinBEI, or cos B,
così}B,
and BD: BG::R: sinBDG, or costB;
whence BEX BD: BIx BG:: R: cos B².
But the quadrilateral figure EADC being right angled at
A and C, is (III. 19. cor.) contained in a circle, and conse-
quently (III. 18. El.) the angle AED or AEB is equal to
374
ELEMENTS OF
ACD or to DCB; wherefore, since by construction the angle
ABE is equal to DBC, the triangles BAE and BDC are si-
milar, and BE: AB:: BC: BD, or BEX BD=ABX BC.
Hence ABX BC: BIX BG:: R2: cos&B". The proposition
is therefore demonstrated.
PROP. XIV. THEOR.
In any triangle, as the rectangle under two sides
is to the rectangle under the excesses of the semi-
perimeter above those sides, so is the square of the
radius, to the square of the sine of half their con-
tained angle.
In the triangle ABC, the perimeter being still denoted by P,
ABX BC: (P-AB) (P-BC): R2: sin B².
For, the same construction being retained, in the right
angled triangles BIE and BGD, BE: IE :: R: sin}B,
and BD : GD :: R : sin3B;
whence BEX BD : IE × GD :: R² : sin}B².
But it has been proved that BE × BD = AB × BC, and
IE×GD=AIX AG, or the rectangle under the excesses of
the semiperimeter above the sides AB and BC; wherefore
the proposition is established.
Otherwise thus :
Produce the shorter side BC till BD be equal to AB,
join AD, let BE bisect the
vertical angle, and draw CG
and CF parallel to BE and AD.
Since BE is perpendicular
to ED and FC, it follows that
BD, or AB ED: R: sinB,
and BC: FC, or EG:: R: sin{B.
Wherefore ABX BC: EDx EG
F
B

A
E
G D
:: R²: sin! B². Now (II. 24. El.) 2ED × 2EG=AC²—CD²=
(II. 19. El.) (AC+ CD) (AC-CD), and consequently
+
TRIGONOMETRY.
375
ED X EG=
(AC+CD) (AC÷CD); but
AC+AB¬BC P-2BC
2
AC + CD
2
AC -
CD
=
4P-BC, and
2
2
2
AC-(AB-BC) P-2AB
=
=P-AB. Hence, by sub-
2
2
stitution, AB × BC : (¿P—AB) (¿P—BC) : : R² : sin B².
The eight preceding theorems contain the grounds of
trigonometrical calculation. A triangle has only five variable
parts-the three sides and two angles, the remaining angle
being merely supplemental. Now, it is a general principle,
that, three of those parts being given, the rest may be thence
determined. But the right angled triangle has necessarily
one known angle; and, in consequence of this, the opposite
side is deducible from the containing sides. In right angled
triangles, therefore, the number of parts is reduced to four,
any two of which being the assigned, the others may be
found.
PROP. XV. PROB.
Two variable parts of a right angled triangle
being given, to find the rest.
This problem divides itself into four distinct cases, accord-
ing to the different combination of the data.
1. When the hypotenuse and a side are given.
2. When the two sides containing the right angle are given.
3. When the hypotenuse and an angle are given.
4. When either of the sides and an angle are given.
376
ELEMENTS OF
$
The first and third cases are solved by the application of
Proposition 7, and the second and fourth cases receive their
solution from Proposition 8. It may be proper, however, to
exhibit the several analogies in a tabular form.

A
B

SOLUTION.
I
AC,
AB
A, or C,
BC
AC: AB:: R: sinC, or cosA.
R: sinA AC: BC.
AB, A, or C AB : BC :: R: tanA, or cot C
II
BC
AC.
cosA RAB: AC, or
R: secA AB: AC
AC
AB
R: cosA: AC: AB.
III
A
BC
R: sinA
AC: BC.
AB,
BC
:
R tanA
AB: BC.
A
AC
cos AR
IV
AB : AC, or
R: secA :: AB : AC.
In the first and second cases, BC or AC might also be de-
duced, by the mere application of Prop. 11. Book II. of the
Elements For AC²=AB²+BC², or AC=√(AB²+BC²)
and BC² = AC - AB² = (AC + AB) (AC — AB),
or BC=√((AC+AB)(AC—AB)).
Cor. Hence the first case admits of a simple approxi-
mation. For, by the scholium to Proposition 6, it ap-
pears, that, AC being made the radius, 2AC+AB is to 3AC,
as the side BC is to the arc which measures its opposite
TRIGONOMETRY.
377
angle CAB, or alternately 2AC+AB is to BC, as 3AC to
the arc corresponding to BC. But the radius is equal to an
arc of 57° 17 44 48, or 57 nearly; wherefore 3AC is to the
arc which corresponds to BC, as 3 × 57, or 172°, to the
number of degrees contained in the angle CAB, and conse-
quently 2AC+AB: BC:: 172° : the expression of the angle
at A, or AC÷¿AB: BC : : 86° : number of degrees in the
angle at A.
This approximation will be the more correct, when the
side opposite to the required angle becomes small in compa-
rison with the hypotenuse; but the quantity of error can
never amount to 4 minutes.
PROP. XVI. PROB.
Three variable parts of an oblique angled triangle
being given, to find the other two.
This general problem includes three distinct cases, one of
which again is branched into two subordinate divisions.
1. When all the three sides are given.
2. When two sides and an angle are given ; which angle may
either (1.) be contained by these sides, or (2.) subtended by one
of them.
3. When a side and two of the angles are given.
The first case admits of four different solutions, derived
from Propositions 11, 12, 13, and 14, and which have their
several advantages. The second case, consisting of two
branches, is resolved by the application of propositions 9 and
10; and the solution of the third case flows immediately from
the former of these propositions.
378
ELÉMENTS of
1
Case.
Given.
I.'
AB,
BC,
and
AC.
B.
AB,
A,
1
BC,
and
and
AC.
C.
II.
2
and
B.
AB,
BC,
A,
or
C,
and
AC.
Sought.
“
B

SOLUTION.
AB × BC (†P—AB) (†P-—BC) :: R² : sin B². 1
P(P-AC): (}P—AB) (P-BC):: R²: tan B2 2
AB X BC :
P (P-AC) :: R2: cos B. 3
2AB X BC: AB² + BC² -AC²::R: cos &B, 4
×
:
AB: BC: sinC: sinA; whence B, and
sinC: sinB:: AB : AC.
AB+BC: AB--BC : : cot B::cot(A+B),
56
17
or cot(C+B).
AB: BC: R:: tan b; and
{R: tan(45°-b): : cot B: cot(A + 4B,)
or
sinA sinB BC: AC, or
:
:
cot(C+B.)
AC=√(AB² + BC2—2AB× BC × cos B.)
8
9
10

AB,
A,
BC
:
sinC sinA :: AB: BC.
III.
Б.
AC
:
sinC sinB:: AB AC.
"and
thence
C.
For the resolution of the first Case, the analogy set down
first, is on the whole the most convenient, particularly if the
angle sought do not approach to two right angles. The se-
cond analogy may be applied through a wider extent, but is
liable in practice to some irregularity, when the angle sought
becomes very obtuse. The third and fourth analogies, espe-
cially the latter, are not adapted for the calculation of very
11
12
TRIGONOMETRY.
379
acute angles; they will, however, answer the best when the
angle sought is obtuse. It is to be observed, that the cosines
of an angle and of its supplement are the same, only placed in
opposite directions; and hence the second term of the ana-
logy, or the difference of AB+BC2 from AC, is in excess
or defect, according as the angle at B is acute or obtuse.—
These remarks are founded on the unequal variation of the
sine and tangent, corresponding to the uniform increase of an
arc *
The first part of Case II. is ambiguous, for an arc and its
supplement have the same sine. This ambiguity, however, is
removed if the character of the triangle, as acute or obtuse,
be previously known.
For the solution of the second part of Case II. the first ana-
logy is the most usual, but the double analogy is the best
adapted for logarithms. The direct expression for the side
subtending the given angle is very commodious, where loga-
rithms are not employed †.
PROP. XVII. PROB.
Given the horizontal distance of an object and its
angle of elevation, to find its height and absolute dis-
tance.
Let the angle CAB, which an object A makes at the sta-
tion B, with an horizontal line, and also the distance BC of a
perpendicular AC, to find that per-
pendicular, and the hypotenusal or
aërial distance BA.
In the right angled triangle BCA,
the radius is to the tangent of the

B
angle at B, as BC to AC; and the radius is to the secant of the
* See Note LXXIV.
+ See Note LXXV.
380
ELEMENTS OF
angle at B, or the cosine of the angle at B is to the radius,
as BC to AB.
PROP. XVIII. PROB.
Given the acclivity of a line, to find its correspond-
ing vertical and horizontal length.
In the preceding figure, the angle CBA and the hypote-
nusal distance BA being given to find the height and the ho-
rizontal distance of the extremity A.
The triangle BCA being right angled, the radius is to the
sine of the angle CBA as BA to AC, and the radius is to the
cosine of CBA as BA to BC.
Scholium. If the acclivity be small, and A denote the mea-
sure of that angle in minutes; then AC BAX
A
3438
nearly.
But the expression for AC, will be rendered more accurate,
by subtracting from it, as thus found, the quantity
AC3
BA2
In most cases when CBA is a small angle, the horizontal
distance may be computed with sufficient exactness, by de-
AC2
2BA
tenusal distance.
ducting
or BAXA² x .000,000,0423, from the hypo-
PROP. XIX. PROB.
Given the interval between two stations, and the
direction of an object viewed from them, to find its
distance from each.
Let BC be given, with the angles ABC and ACB, to cal-
culate AB and AC.
TRIGONOMETRY.
In the triangle CBA, the angles ABC and
ACB being given, the remaining or supple-
mental angle BAC is thence given; and con-
sequently, sin BAC : sinACB :: BC : AB,
and sinBAC:: sinABC: BC: AC.
Cor. If the observed angles ABC and ACB
be each of them 60°, the triangle will be evi-
dently equilateral; and if the angle at the sta-
B
381
A

tion B be 90° and that at C 45°, the distance AB will be e-
qual to the base BC.
PROP. XX. PROB.
Given the distances of two objects from any sta-
tion and the angle which they subtend, to find their
mutual distance.
Let AC, BC, and the angle ACB be gi- A
ven, to determine AB.
In the triangle ABC, since two sides and
their contained angle are given, therefore,
by cor. Prop. 10. AC+ BC: AC-BC::
cot C: cot(A+C), and then sin A: sinC ::
BC: AB; or (by corollary to Prop. 11.)
AB= √(AC+BC2-2AC.BC cos C).

B
Cor. By combining this with the preceding proposition,
the distance of an object may be found from two stations, be-
tween which the communication is inter-
rupted. Thus let A be visible from B and
C, though the straight line BC cannot be
traced. Assume a third station D, from
which B and C are both seen. Measure
DB and DC, and observe the angles BDC,
ABC and ACB. In the triangle BDC,
the base BC is found as above; and thence,
B
D
A

by the preceding proposition, the sides AB and AC of the
triangle ABC are determined.
382
ELEMENTS OF
...
PROP. XXI. PROB.
Given the interval between two stations, and the
directions of two remote objects viewed from them
in the same plane, to find the mutual distance, and
relative position of those objects.
Let the points A, B represent the two objects, and C, D
the two stations from which these are observed; the interval
or base CD being measured, and also the angles CDA, CDB
at the first station, and DCA, DCB at the second; it is
thence required to determine the transverse distance AB, and
its direction.
It is obvious that each of the points A and B would be as-
signed geometrically by the intersection of two straight lines,
and consequently that the position of the objects will not be
determined, unless each of them appears in a different direc-
tion at the successive stations.
1. Suppose one of the stations C to lie in the direction of the
two objects A and B.
At C observe the angle BCD, and at D
the angles CDA and BDC. Then by
Prop. 19. sinCAD : sinCDA :: CD: CA,
and sin CBD : sinCDB:: CD: CB; the
difference or sum of CA and CB is AB,
the distance sought.
D
B

2. When neither station lies in the direction of the two ob-
jects, and the base CD has a transverse position.
Find by Prop. 19. the distances AC and BC of both ob-
TRIGONOMETRY.
383
A
B
jects from one of the stations C;
then the contained angle ACB, or
the excess of DCA above DCB,
being likewise given, the angles at
the base AB of the triangle BCA,
and the base itself, may be calcula-
ted, from the analogies exhibited
for the solution of the second
branch of Case II. For AC+BC:
AC-BC:: cotACB: cot(ACB+CAB), and thus the angle
CAB is found. Or more conveniently by two successive opera-
tions, AC: BC:: R:tan b, and R : tan(45° —b):: cot‡ACB
::cot (ACB+CAB.) Now, sinCAB:: sinACB: BC: AB,
or AB=√(AC² + BC²—2AC × BC × cos ACB).

D
The inclination of AB to CD in the first case is given by
observation, and in the second case it is evidently the supple-
ment of the interior angles CAB and DCA. A parallel to
AB may hence be drawn from either station.
Cor. Hence the converse of this problem is readily solved.
Suppose two remote objects A and B, of which the mutual dis-
tance is already known, are observed from the stations C and
D, and it were thence required to determine the interval CD.
Assume unit to denote CD, and calculate AB according to the
same scale of measures; the actual distance AB being then
divided by that result, will give CD: For the several triangles
which combine to form the quadrilateral figure CABD, are
evidently given in species.
PROP. XXII. PROB.
Given the directions of two inaccessible objects
viewed in the same plane from two given stations,
to trace the extension of the straight line connect-
ing them.
384
>
ELEMENTS OF
Let the angles ACD, BCD be observed at C, and
ADC, BDC at D, with the base CD; to find a point E in
the straight line ABF produced through A and B.
By the last proposition, find
AD and the angle DAB, and as-
sume any angle ADE. In the
triangle DAE, the angles at the
base AD, and consequently the
vertical angle AED, being known,
it follows, by Prop. 9., that
sinAED: sinEAD:: AD: DE.

C
D
R
E
Measure out DE, there-
fore, on the ground, and its extremity E will mark the ex-
tension of AB.
PROP. XXII. PROB.
Given on the same plane the direction of two re-
mote objects separately seen from two stations, and
their direction as viewed at once from an interme-
diate station, with the distances of those stations,
from the middle station,-to find the mutual dis-
tance of the objects.
Let object A be visible from the station D, and B from E,
and both of them be seen at once from the station C; the com-
pound base DC, CE being measured,
and the angles DCA, ACB and BCE,
with ADC and BEC, observed-
to determine AB.
In the triangles DAC, CBE, the
sides AC and BC are found by Prop.
19. and in the triangle ACB, the base
AB is thence found by the applica-
tion of Prop. 20.
A

D
B
E
It is evident that the mode of investigation will not be
altered if the three stations D, C and E should lie in the
same straight line.
เ
TRIGONOMETRY.
385
PROP. XXIV. PROB.
Given four stations, with the direction of a re-
mote object viewed from the first and second sta-
tions, and the direction of another remote object
viewed from the third and fourth stations, all in the
same plane,-to find the distance between the ob-
jects.
Let the bases EC, CD, and DF be given, with the angles
ECD and CDF, and suppose that at the stations E and C
the angles CEA and ECA are observed, and the angles
BDF and BFD at D and F; to find the transverse distance
AB.
In the triangles EAC and DBF,
find by Prop. 19. the sides AC
and BD; and, in the triangle
CAD, the sides AC, CD, with
their contained angle ACD, being
given, the base DA and the angle
CDA are found by Case II. But
the distances DA, DB being now

E
D
B
given, with their contained angle ADB, the base AB is found
by Prop. 20.
PROP. XXV. PROB.
The mutual distances of three remote objects be-
ing given, with the angles which they subtend at a
station in the same plane, to find the relative place
of that station.
Let the three points A, B, and C, and the angles ADB
and BDC which they form at a fourth point D, be given; to
determine the position of that point.
Bh
386
ELEMENTS OF
!
1. Suppose the station D to be situate in the direction of two
of the objects A, C.
All the sides AB, AC and BC of the triangle ABC being
given, the angle BAC is found by
Case I.; and in the triangle ABD,
the side AB with the angles at A
B
and D being given, the side AD is
found by Case III. and consequent-
ly the position of the point D is D
determined.
A
D
3 C

C
2. Suppose the three objects A, B and C to lie in the same
direction.
Describe a circle about the extreme objects A, C and the
station D, join DA, DB and DC, produce DB to meet the
circumference in E, and join AE and CE.
In the triangle AEC, the side AC is given, and the
angles EAC and ECA, being equal (III. 18. El.) to CDE and
ADE, are consequently given; wherefore the side AE is
found by Case III. The triangle AEB, having thus the sides
AE, AB, and their contained angle EAB or BDC given, the
angle ABE and its supplement
ABD are found by Case II.
Last-
E

ly, in the triangle ABD, the angles
ABD and ADB, with the side AB, A
C
B
are given; whence BD is found by
Case III. But since the angle ABD
and the distance BD are assigned,
D
the position of the station D is evi-
dently determined.
3. Let the three objects form a triangle, and the station D
lie either without or within it.
TRIGONOMETRY.
387
Through D and the extreme points A and C describe a
circle, draw DB cutting the circumference in E, and join AE
and CE.
1. In the triangle AEC, the side AC, and the angles
ACE and CAE, which are equal (III. 18. El.) to ADB and
BDC, being given, the side AE is found by Case III.
2. All the sides of the triangle
ABC being given, the angle CAB
is found by Case I.
3. In the triangle BAE, the sides
AB and AE are given, and their
contained angle EAB, or the dif-
ference of CAE and CAB, are gi-
ven, whence, by Case II., the angle
ABE or ABD is found.
B

E
D
C
پر
4. Lastly, in the triangle DAB,
the side AB and the angles ABD and ADB being given, the
side AD or BD is found by Case III., and consequently the
position of the point D, with respect to A and B is determi-
ned. By a like process, the relative position of D and C is de-
duced; or CD may be calculated by Case II. from the sides
AC, AD, and the angle ADC, which are given in the tri-
angle CAD.
It is obvious that the calculation will fail, if the points B
and E should happen to coincide. In fact, the circle then
passing through B, any point D whatever in the opposite arc
ADC will answer the conditions required, since the angles
ADB, and BDC, being now in the same segment, must re-
main unaltered.
Otherwise thus.
On AB describe (III. 27. El.) a segment of a circle ADB
containing an angle equal to that subtended by the objects A
and B, and on BC describe another segment BDC contain-
ing an angle equal to that subtended by the objects B and C;
Bh q
J
e
388
ELEMENTS OF
the point D, where the two circumferences intersect, will evi-
dently mark the station required.
B
Join AD, BD, CD, draw the diameters BF, BG, and join
AF, CG, DF and DG.
The angles BDF and BDG,
thus occupying semicircles,
are right angles, and there-
fore DGF forms but one
straight line. Hence these
successive calculations.

1. All the sides of the tri-
angle BAC being given, the
angle ABC is found by Case
I.
D
2. In the right-angled triangles BAF, BCG, the sides AB,
BC, and the angles AFB, BGC, which are equal (III. 18. El.)
to ADB, BDC, being given, the hypotenuses BF, BG, or
the diameters of the circles are thence found.
3. In the triangle FBG the two sides BF, BG being now
given, with the angle FBG=CBG~CBF=CBG—ABC÷
ABF=BAC+BCA-ADC, the angle BFG is found by
Case II.
4. Lastly, in the right angled triangle BDF, the hypote-
nuse BF, and the angle BFD or BFG being given, the side
BD is found; and since the angle FBD is also known, the
position of the point D is assigned.
Should the two circles have the same centre, their circum-
ferences must obviously coincide, and therefore every point in
the containing arc will answer the conditions required. When
this porismatic or indeterminate case of the problem occurs,
the distances AB and BC become chords of the correspond-
ing observed angles, and are consequently, by Proposition IX.
proportional to the sines of those angles *.
See Note LXXVI.
TRIGONOMETRY.
389
PROP. XXVI. PROB.
The mutual distances of three remote objects, two
of which only are seen at once from the same sta-
tion, being given, with the angles observed at two
stations in the same plane, and the intermediate di-
rection of these stations,-to find their relative places.
Suppose the three points A, Band
C are given, with the angle AEB
which A and B subtend at E, and
BFC, which B and C subtend at
F, and likewise the angles AEF
and EFC; to find the relative si-
tuation of each of those stations E
and F.
A
E
D
B

C
Produce AE and CF to meet in D, and join BD. The angle
EDF, being equal to AEF+CFE-180°, is given. Now in
the triangle EBF, sinBFE: sinEBF:: EB: EF; and in
the triangle EDF, sinEDF: sinDFE:: EF: ED; where-
fore, (V. 23. El.) sinBFE x sinEDF: sinEBF × sinDFE ::
EB: ED, and consequently the ratio of EB to ED is found.
Again the angle BED, being the supplement of AEB, is given,
(Prop. 10. cor.) sinBFE x sinEDF: sinEBF x sinDFE::
R: tanb, and R:tan(45°-b):: cot BED:-cot (BED+ EBD),
or cot(180°-BED-EBD), whence the angle EDB is gi-
ven. The angles which all the three objects A, B, and C
subtend at the point D are therefore all given, and hence the
position of D is determined by the preceding proposition.
But BD, being found, the several distances BE, ED, and
BF, FD are thence obtained, and consequently the position of
each of the stations E and F is determined.
Scholium. In all the foregoing problems, the angles on the
ground are supposed to be taken by means of a theodolite. If
the sextant be employed for that purpose, such angles, when
390
ELEMENTS OF
oblique, must be rednced by calculation to their projection on
the horizontal plane *.
In surveying an extensive country, a base is first carefully
measured, and the prominent distant objects are all connect-
ed with it, by a series of triangles. To avoid, in practice,
the multiplication of errors, these triangles should be chosen,
as nearly as possible, equilateral.-After a similar method,
large estates are the most accurately planned and measured †.
The vertical angles employed in the mensuration of
heights, being estimated from the varying direction of the
level or the plummet, will evidently, when the stations are dis-
tant, require some correction. Let the points A and B re-
E
B
Z
A
D
present two remote objects, and C
their centre of gravitation; with
the radius CA describe a circle,
draw CB cutting the circumfe-
rence in D and E, and join EA
and AD. The converging lines
AC and BC will indicate the di-
rection of the plummet at A and
B, the intercepted arc AD, will
trace the contour of a quiescent
fluid, and the tangent AZ, being applied at A, will mark the
line of the horizon from that station. Wherefore the verti-
cal angle observed at A is only ZAB, which is less than the
true angle DAB, by the exterior angle DAZ. But (III. 25. El.)
DAZ being equal to the angle AED in the alternate seg-
ment, is (III. 17. El.) equal to half the angle ACD at the cen-
tre. Hence the true vertical angle at any station will be found,
by adding to the observed angle half the measure of the in-
tercepted arc; and this measure depending on the curvature
of the earth, which is neither uniform nor quite regular, must
be deduced, for each particular place, from the length of the
corresponding degree of latitude.

See Note LXXVII.
+ See Note LXXVIII.
TRIGONOMETRY.
391
Such nicety, however, is very seldom required. It will be
sufficiently accurate in practice to assume the mean quantities,
and to consider the earth as a globe, whose circumference is
24,856 miles, and diameter 7,912. The arc of a minute on the
meridian being, therefore, equal to 6076 feet, the correction
to be added to the observed vertical angle must amount to one
second, for every 69 yards contained in the intervening dis-
tance.
=
The quantity of depression ZD below the horizon is hence
easily computed; for (III. 32. El.) AZ EZ.ZD, or very
nearly ED.ZD; and consequently the depression of an object
is proportional to the square of its distance AZ. In the space
of one mile, this depression will amount to 3
parts of a
foot; and generally, therefore, it may be expressed in feet, by
two-thirds of the square of the distance in miles.
But the effect of the earth's curvature is modified by an-
other cause, arising from optical deception. An object is ne-
direction of the
ver seen by us in its true position, but in the
ray of light which conveys the impression. Now the lumi-
nous particles, in traversing the atmosphere, are, by the force
of superior attraction, refracted or bent continually towards.
the perpendicular, as they penetrate the lower and denser
strata; and consequently they describe a curved track, of
which the last portion, or its tangent, indicates the apparent
elevated situation of a remote point. This trajectory, suffer-
ing almost a regular inflexure, may be considered as very
nearly an arc of a circle, which has for its radius six times the
radius of our globe. Hence, to correct the error occasioned
by refraction, it will only be requisite to diminish the effects
of the earth's curvature by one-sixth part, or to deduct, from
the vertical angles, the twelfth part of the measure of the in-
tervening terrestrial arc. The quantity of horizontal refrac-
tion, however, as it depends on the density of the air at the
surface, is extremely variable, especially in our unsteady cli-
mate.
:
་
}
5.
!
1
1
NOTES
•
AND
ILLUSTRATIONS.
}
NOTES
AND
ILLUSTRATIONS.
!
Note I.-Page 3.
THE primary objects which Geometry contemplates are, from their
nature, incapable of decomposition. No wonder that ingenuity has
only wasted its efforts to define such elementary notions. It appears
more philosophical to invert the usual procedure, and endeavour to
trace the successive steps by which the mind arrives at the princi-
ples of the science. Though no words can paint a simple sound, this
may yet be rendered intelligible, by describing the mode of its arti-
culation.
The founders of mathematical learning among the Greeks were in
general tinctured with a portion of mysticism, transmitted from Py-
thagoras, and cherished in the school of Plato. Geometry became
thus infected at its source. By the later Platonists, who flourished
in the Museum of Alexandria, it was regarded as a pure intellectual
science, far sublimed above the grossness of material contact. Such
visionary metaphysics could not impair the solidity of the superstruc-
ture, but did contribute to perpetuate some misconceptions, and to
give a wrong turn to philosophical speculation. It is full time to re-
store the sobriety of reason. Geometry, like the other sciences
which are not concerned about the operations of mind, must ultimate-
ly rest on external observation. But those ultimate facts are so few, SO
distinct, and obvious, that the subsequent train of reasoning is safely
396
NOTES AND ILLUSTRATIONS.
pursued to unlimited extent, without ever appealing again to the evi-
dence of the senses. The science of Geometry, therefore, owes its
perfection to the extreme simplicity of its basis, and derives no visi-
ble advantage from the artificial mode of its construction. The
axioms are rejected, as being totally useless and rather apt to pro-
duce obscurity.
The term surface, in Latin superficies, and in Greek sQuvad, con-
veys a very just idea, as marking the mere expansion which a body
presents to our sense of sight. Line, or reapua, signifies a stroke ;
and, in reference to the operation of writing, it expresses the boun-
dary or contour of a figure. A straight line has two radical proper-
ties, which are distinctly marked in different languages. It holds the
same undeviating course, and it traces the shortest distance be-
tween its extreme points. The first property is expressed by the epi-
thet recta in Latin, and droite in French; and the last seems intima-
ted by the English term straight, which is evidently derived from
the verb to stretch. Accordingly Proclus defines a straight line as
stretched between its extremities— saxgov Telaμevn.
The word point in every language signifies a mark, thus indicating
its essential character, of denoting position. In Greek, the term
Type was first used; but, this being degraded in its application, the
diminutive σημείον, formed from onpa, a signal, came afterwards to be
preferred. The neatest and most comprehensive description of a point
was given by Pythagoras, who defined it "a monad having posi-
tion." Plato represents the hypostasis, or constitution of a point, as
adamantine; finely alluding to the opinion which then prevailed,
that the diamond is absolutely indivisible, the art of cutting this re-
fractory substance being the discovery of modern ages.
The conception of an angle is one of the most difficult perhaps in
the whole compass of Geometry. The term corresponds, in most
languages, to corner, and therefore exhibits a most imperfect picture
of the object. Apollonius defined it to be" the collection of space
about a point." Euclid makes an angle to consist in “ the mutual in-
clination, or x15, of its containing lines,”—a definition which is ob-
scure and altogether defective. In strictness, this can apply only to
acute angles, nor does it give any idea of angular magnitude; though
NOTES AND ILLUSTRATIONS.
397
this really is as capable of augmentation as the magnitude of lines
themselves. It is curious to observe the shifts to which the author
of the Elements is hence obliged to have recourse. This remark is
particularly exemplified in the 20th and 21st Propositions of his
Third Book. Had Euclid been acquainted with Trigonometry, which
was only begun to be cultivated in his time, he would certainly have
taken a more enlarged view of the nature of an angle.
In the definition of reverse angle, I find that I have been anticipa-
ted by the famous Stevin of Bruges, who flourished about the end of
the sixteenth century. It is satisfactory to have the countenance of
such respectable authority.
Note II.-Page 9.
A square is commonly described as having all its angles right.
This definition errs however by excess, for it contains more than
what is necessary. The original Greek, and even the Latin version,
by employing the general terms ogboywviov, and rectanglum, dexterous-
ly, avoided that objection. The word rhombus comes from psußer,
to sling, as the figure represents only a quadrangular frame disjoint-
ed.
It scarcely deserves notice, but I will anticipate the objection
which may be brought against me, for having changed the definition
of trapezium. The fact is, that I have only restricted the word to its
appropriate meaning, from which Euclid had, according to Proclus,
taken the liberty to depart. In the original, it signifies a table; and
hence we learn the prevailing form of the tables used among the
Greeks. Indeed the ancients would appear to have had some predi-
lection for the figure of the trapezium, since the doors now seen in
the ruins of the temples at Athens are not exactly oblong, but wider
below than above.
Language is capable of more precision, in proportion as it becomes
copious. As I have confined the epithet right to angles, and straight
to lines, I have likewise appropriated the word diagonal to rectilineal
figures, and diameter to the circle. In like manner, I have restricted
the term arc to a portion of the circumference, its synonym arch
being assigned to architecture. For the same reason, I bave adopted
398
NOTES AND ILLUSTRATIONS.
the term equivalent, from the celebrated Legendre, whose Elemens
de Geometrie is one of the ablest works that has appeared in our
timės. These distinctions evidently tend to promote perspicuity,
which is the great object of an elementary treatise.-Euclid and all
his successors define an isosceles triangle to have only two equal
sides, which would absolutely exclude the equilateral triangle. Yet the
equilateral triangle is afterwards assumed by them to be a species of
isosceles triangle, since the equality of its angles is at once inferred as
a corollary from that of the angles at the base of an isosceles triangle.
This inadvertency, slight as it may appear, is now avoided.
Note III.-Page 18.
This proposition may be very simply demonstrated, in the same
manner as its converse, by a direct appeal to superposition or mental
experiment. For suppose a copy of the triangle ABC were inverted
and applied to it, the side BA being laid on BC, the side BC again
will evidently lie on BA, and the base AC coincide with CA. Con-
sequently the angle BAC, occupying now the place of BCA, must
be equal to this angle.
It may be worth while to remark, that Euclid's demonstration,
which, being placed near the commencement of the Elements, has
from its intricacy been styled the Pons Asinorum, is in fact essen-
tially the same. This will readily appear on a review of the several
steps of the reasoning :--
B
The sides BA and BC of the isosceles triangle being produced,
the equal segments AD and CE are assumed, and AE, CD joined.
1. The complex triangles ABE and CBD are com-
pared: The sides AB and BC are equal, and like-
wise BE and BD, which consist of equal parts,
and the contained angles EBA and DBC are the
same with DBE; whence (I. 3.) these triangles
are equivalent, and the base AE equal to CD, the
angle BAE equal to BCD, and the angle BEA to
BDC. 2. The additive triangles CAE and ACD

F
1
C
E
G
江流
​are next compared: The sides EC and EA being equal to DA and
DC, and the contained angle CEA equal to ADC, the triangles are
NOTES AND ILLUSTRATIONS.
399
(I. 3.) equivalent, and therefore the angle CAE is equal to ACD.
3. Lastly, since the whole angle BAE is equal to BCD, and the
part CAE to ACD, the remainder BAC must be equal to BCA.
Now this process of reasoning is at best involved and circuitous.
The compound triangles ABE and CBD consist of the isosceles tri-
angle ABC joined to each of the appended triangles ACE and
CAD; when therefore, as the demonstration implies, ABE is laid on
CBD, the common part ABC is reversed, or it is applied to CBA, and
the other part ACE is laid on CAD. But the superposition of ABC or
CBA is easily perceived by itself; nor is the conception of that invert-
ed application anywise aided by having recourse to the superposition,
first of the enlarged triangles ABE and CBD, and then contracting
these by the superposition of the subsidiary triangles ACE and CAD.
In this, as in some other instances, Euclid has deceived himself, in
attempting a greater than usual strictness of reasoning,
Note IV.-Page 20.
This proposition may be demonstrated otherwise.
Draw (I. 5. El.) BE bisecting the angle ABC. The angle BEA
Б
(I. S. El.) is greater than the interior angle EBC
or EBA, and therefore (I. 14. El.) the side AB
is greater than AE. In like manner, the angle
BEC is greater than the interior angle EBA or
EBC, and consequently (I. 14. El.) the side CB
is greater than CE. Wherefore the two sides AB and CB, being
each of them greater than the adjacent segments AE and CE, are
together greater than the whole base AC.

Note V.-Page 20.
A
E
C
From this property of the sides of a triangle, may be derived the
generic character of a straight line:
The shortest line that can be drawn between two points, is a straight
line.
Let the points A and B be connected by straight lines joining an
intermediate point C; and the two sides AC and BC of the triangle
ACB are greater than AB (I. 15.). Now let a third point D be in-
400
NOTES AND ILLUSTRATIONS.
G
D
H
E
B
terposed between A and C; and because AD and DC are together
greater than AC, add BC to both, and the three lines AD, DC,
and CB are greater than AC and BC, and consequently still greater
than AB. Again, suppose a fourth
point E to connect B with C; and the
sides BE and CE of the triangle BCE
being greater than BC, the four straight
lines AD, DC, CE, and EB are toge-
ther, by a still farther access, greater
than AB. By thus repeatedly multi-
plying the interjacent points, two sides of a triangle will at each
successive step come in place of a third side, and consequently the
aggregate polygonal or crooked line AFDGCHEIB will acquire
continually some farther extension. Nay, since there is no limit to
the possible number of those connecting points, they may approach
each other nearer than any assignable interval; and consequently the
proposition is also true in that extreme case where the boundary is
a curve line, or of which no portion can be deemed rectilineal.
The proposition now demonstrated is commonly assumed as an
axiom. It is indeed forced upon our earliest observation, being sug-
gested by the stretching of a cord, and other familiar occurrences in
life. But thus to multiply principles, appears quite unphilosophical.
The two radical properties of a straight line-the congruity of its
parts-and its shortness of trace-are distinct, though connected.
The latter is shown to be the necessary consequence of the former;
but it would be impossible, by any direct process, to infer the uni-
formity of straight lines, from their marking out the nearest routes.

Note VI.-Page 20.
This proposition may be otherwise demonstrated.
Join BE. The angle BEC (I. 8. El.) is greater than ABE or
(I. 11. El.) AEB, which again (I. 8. El.) is
greater than CBE; wherefore (I. 14. El.) the
side BC is greater than CE, or the difference
between AB and AC.
A
E C
NOTES AND ILLUSTRATIONS.
401
In the demonstration, I could not avoid introducing the considera-
tion of limits. This will occasion, I presume, no material difficulty,
since the reasoning is actually the same as that by which our most
familiar conceptions are gradually expanded.
Note VII.-Page 23.
The ingenious Mr Thomas Simpson has very justly remarked, in
his Elements of Geometry, that the demonstration which Euclid
gives of this proposition is defective, since it assumes that the point
G must lie below the base AC. He has therefore legitimately sup-
plied the deficiency of the proof; and it is surprising that so rigorous
a geometer as Dr Robert Simson, should have so far yielded to his
prejudices, as to resist such a decided improvement. The demon-
stration inserted in the text appears to be rather simpler and more
natural than that of Mr T. Simpson.
Note VIII.-Page 23.
This proposition is capable of being demonstrated directly.
Let the triangles ABC and DEF have the sides AB and BC equal
to DE and EF, but the base AC greater than DF; the vertical angle
ABC is greater than DEF.
From AC cut off AG equal to
DF, construct (I. 1.) the triangle
AHG having the sides AH and GH
equal to AB and BC or DE and EF,
join HB, and produce HG to meet
BC in I.
Because HI is greater than HG,

B
E
I
A
+
I
H
it is greater than the equal side BC, and therefore much greater
than BI. Consequently the opposite angle IBH of the triangle BIH
is (1. 13.) greater than BHI. But AB being equal to AH, the angle
HBA is (I. 11.) equal to BHA, and therefore the two angles IBH and
IIBA are greater than IHB and BHA, that is, the whole angle CBA
is greater than IHA or GHA. And since the sides of the triangle
AGH are by construction equal to those of EDF, the corresponding
angle AHG is equal to DEF (I. 2.); and hence the angle ABC,
which is greater than AIIG, is likewise greater than DEF.-In like
C C
402
NOTES AND ILLUSTRATIONS.
manner this may be demonstrated, if BH should fall without the
figure.
Note IX. Page 26.
It is not difficult to perceive that the whole structure of geometry
is grounded on the simple comparison of triangles. The conditions
which fix the equality of those elementary figures, are all contained
in the 2d, 3d, 21st and 22d propositions of the first Book. These funda-
mental theorems derive their evidence from the mere superposition of
the triangles themselves, which, in reality, is nothing but an ultimate
appeal, though of the easiest and most familiar kind, to external ob-
servation. The same conclusions however might be deduced more con-
cisely, from the circumstances which must determine the constitution
of an individual triangle. Suppose AB, BC, and AC, any one of which
is shorter than the other two conjoined, to be three inflexible rods
moveable at pleasure. (1.) Place them with their ends meeting
each other, and they will evidently rest in the same position, and
contain a distinct triangle,-which corresponds to Prop. 2. (2.) Ha-
ving joined the rods AB and BC at B, continue
to open them at that point, till they form a
given vertical angle ABC; their position then
becomes fixed, and consequently determines the
rod AC which connects their extremities and
completes the triangle. This inference evi-
dently agrees with Prop. 3. (3.) While the
B

rod AC retains its place, let two rods AB and CB of unlimited
length, and applied at the ends A and C, be opened gradually till the
one forms with AC a given angle CAB, and the other a given angle
ACB; it is evident that AB and BC will then rest crossing each
other in those positions, and containing a determinate triangle, of
which the vertex B is their point of mutual intersection. This pro-
perty corresponds with Prop. 21, (4.) Let the rod AB of a given
length make a given angle with the unlimited rod AC, and applying
at the end B another given rod, turn this gradually round till it
meets AC. If BC exceeds the distance of B from AC, it will evi-
dently, after stretching beyond AC, again come to meet that boun-
dary. With such conditions, therefore, the rods might contain two
NOTES AND ILLUSTRATIONS.
403
determinate triangles, the one acute and the other obtuse, and which
are hence distinguished from each other, by the additional character
of affection. This qualified property, omitted in most elementary works,
is yet of extensive application, and was requisite to complete the con-
ditions of the equality of triangles. It corresponds with Prop. 23.
The four preceding theorems are reducible, however, to a single pro-
perty, which includes all the different requisites to the equality of tri-
angles. The sides of a triangle are obviously independent of each other,
being subject to this condition only, that any one of them shall be less
than the remaining two sides. But since all the angles of a triangle
are together equal to two right angles, the third angle must, in every
case, be the necessary result of the other two angles. A triangle
has, therefore, only five original and variable parts-the three sides
and two of its angles. Any three of these parts being ascertained, the tri-
angle is absolutely determined. Thus-when (1.) all the three sides
are given,-when (2.) two sides and their contained angle are given,-
when (4.) two sides and an opposite angle are given, with the affec-
tion of the triangle, or when (3.) one side and two angles, and thence
the third angle are given, -the triangle is completely marked out.
M. Legendre, in a very elaborate note to his Elemens de Geome-
trie, has sought, with much ingenuity, to deduce à priori the radical
properties of triangles, from the theory of functions. But, like other
similar attempts, his investigation actually involves in it a latent as-
sumption. This subtle logician sets out with the principle which would
seem almost intuitive, that a triangle is determined when the base and
its adjacent angles are given. The vertical angle, therefore, depends
wholly on these data-the base and its adjacent angles. Call the
base c, its adjacent angles A, B, and the vertical angle C. This
third angle being derived from the quantities A, B and c, must be a
determinate function of them, or formed from their combination.
Whence, adopting his notation, C≈4 : (A, B, c). But the line c is
of a nature heterogeneous to the angles A and B, and therefore can-
not be compounded with these quantities. Consequently C≈ :(A, B),
or the vertical angle is simply a function of the angles A and B
at the base, and hence the third angle of a triangle must depend
wholly on the other two.
•
To a speculative mathematician this argument is very seduc-
tive, though it will not bear a rigid examination. Many quantities
c c 2
404
NOTES AND ILLUSTRATIONS.
I
in fact appear to result from the combined relation of other quantities
that are altogether heterogeneous. Thus, the space which a moving body
describes, depends on the joint elements of time and velocity, things
entirely distinct in their nature; and thus, the length of an arc of a cir-
cle is compounded of the radius, and of the angle it subtends at the
centre, which are obviously heterogeneous magnitudes. For aught we
previously knew to the contrary, the base c might, by its combina-
tion with the angles A and B, modify their relation, and thence affect
the value of the vertical angle C. In another parallel case, the force
of this remark is easily perceived. Thus, when the sides a, b and their
contained angle C are given, the triangle is determined, as the simplest
observation shows. Wherefore the base c is derived solely from these
data, or c=9:(a, b, C). But the angle C, being heterogeneous to
the sides a and b, cannot coalesce with them into an equation,
and consequently the base c is simply a function of a and b, or it is
the necessary result merely of the other two sides. Such is the ex-
treme absurdity to which this sort of reasoning would lead! In both of
these instances, indeed, the conclusion is admitted by implication,
only in the one it is consistent with truth, while in the other it is
palpably false. That such an acute philosopher could overlook the
fallacy of his argument, can only be ascribed to the influence which
peculiar trains of thought acquire over the mind, and to the ex-
treme facility with which elementary principles insinuate and blend
themselves with almost every process of reasoning.
Admitting, however, what the slightest inspection readily con-
firms, that the third angle is merely derived from the other two,
M. Legendre demonstrates with great elegance, the property that
the three angles of a triangle are equal to two right angles. Let-
ting fall from the right angle a perpendicular on the hypotenuse, he
divides any right-angled triangle into two subordinate triangles,
which have each of them two angles equal to those of the original
triangle; whence the acute angles of that triangle are alternately e-
qual to the angles which compose the right angle. But every tri-
angle may be divided into two right-angled triangles, by letting fall
a perpendicular from the vertex on the base, and consequently the
acute angles of both these triangles, and which form the angles at
the base, and the vertical angle of the primary triangle,—are toge-
ther equal to two right angles.
}
1
NOTES AND ILLUSTRATIONS.
405
A
لا
B
This theorem may be proved somewhat more directly. In the
triangle ABC, let the angle CBA be greater than ACB, and
draw BD, and then DE, making the angles ABD and BDE
each equal to ACB. The triangles
ABC and ADB having the common
angle BAC and the angle ACB equal
to ABD, their third angles ABC and
ADB must be equal. But the tri-
angles BCD and BDE have also a
common angle CBD, and equal angles DCB and BDE; whence the
third angle BDC is equal to BED, and therefore the supplementary
angle ADB, equal to ABC. is equal to DEC. Again, the triangles
ABC and DEC having two common or equal angles, their third
angles BAC and EDC are equal; wherefore the three angles ABC,
BCA and BAC of the original triangle, are respectively equal to
BDA, BDE and EDC, and hence equal to two right angles.—If the
triangle ABC be equiangular, divide it into two scalene triangles
ABD and CBD, the angles of which, or the angles of the original
triangle, together with the adjoining angles ADB and BDC, must
be equal to four right angles, and consequently the angles of that tri-
angle are equal to two right angles.

B
But the proposition is easily derived from another view of the sub-
ject. If we suppose a ruler turning about the
point A, to change its direction AC into AB,
then opening at B till it gains the direction BC,
and finally wearing about the point C till it ac-
quires the opposite position CA; thus changing
its direction with respect to a remote object, by
three successive openings all to the same side, the ruler, being now
reversed, must have performed half a circuit; that is, the three
angles of a triangle, which constitute those openings, are equal to
two light angles.
The profound geometer already quoted, pursuing his refined argu-
ment, has, from the consideration of homogeneous quantities, like-
wise attempted to deduce the proportionality of the sides of equi-
angular triangles. But in this abstruse research, assumptions are
still disguised and mixed up with the process of induction. Such
indeed must be the case with every kind of reasoning on mathemati-
cal or physical objects, which proceeds à priori, without appealing,
406
NOTES AND ILLUSTRATIONS.
at least in the first instance, to external observation. Of this kind,
are some of those ingenious analytical investigations respecting the
laws of motion and the composition of forces. The principle of
sufficient reason, introduced by Leibnitz, appears to be nothing but
an artificial mode of dressing out an assumption, and which the
celebrated Boscovich has well exposed in his excellent notes to a
didactic poem by Stay, entitled Philosophia Recentior.
Note X.-Page 27.
The subject of parallel lines has exercised the ingenuity of modern
geometers; for Euclid had only endeavoured to evade the difficulty,
by styling the fundamental proposition an axiom. The investigation.
now given seems to be one of the best adapted to the natural progress
of discovery. It is almost ridiculous to scruple about admitting the
idea of motion, which I have employed for the sake of clearness.
But even that futile objection might be obviated, by considering
merely the successive positions of the straight line extending through
the given point.
Note XI.-Page 32.
JA
1
That invaluable instrument, Hadley's quadrant, is founded on the
second corollary, annexed as an obvious consequence of the proposi-
tion. A ray of light SA, from the sun,
impinging against the mirror at A, is re-
flected at an angle equal to its incidence;
and now striking the half-silvered glass
at C, it is again reflected to E, where
the eye likewise receives, through the
transparent part of that glass, a direct
ray from the boundary of the horizon.
Hence the triangle AEC has its exterior
angle ECD and one of its interior angles
CAE, respectively double of the exterior
angle BCD and the interior angle CAB,
of the triangle ABC; wherefore the remaining interior angle AEC,
or SEZ, is double of ABC; that is, the altitude of the sun above
the horizon is double of the inclination of the two mirrors. But the
glass at C remaining fixed, the mirror at A is attached to a move-
able index, which marks their inclination.

Z
D'
E
C/
B
NOTES AND ILLUSTRATIONS.
407
}
The same instrument, in its most improved state, and fitted with
a telescope, forms the sextant, which, being admirably calculated
for measuring angles in general, has rendered the most important
services to geography and navigation.
Note XII-Page 35.
This problem is generally constructed somewhat differently.
In AB take any point C, and on BC
(I. 1. cor.) describe an equilateral tri-
angle CDB, on its side DB, another
DEB; and on DE the side of this, a
third equilateral triangle DFE; join
the last vertex F with the point B; and
BF is the perpendicular required.
·
F

D
E
A
C
B
Because the triangles CDB and DBE are equilateral, the angles.
CBD and DBE are each of them equal to two third parts of a right
angle (I. 32. cor.); and the triangles BDF, BEF, having the sides
BD, DF equal to BE, EF, and the side BF common, are (I. 2.)
equal, and consequently the angles FBD and FBE are equal, and
each of them the half of DBE. The angle FBD, being therefore
one-third part of a right angle, and the angle DBA two-third parts,
the whole angle FBC must be an entire right angle, or the straight
line BF is perpendicular to AB.
Note XIII.-Page 42.
From this proposition the following theorem is easily derived:
Straight lines joining the successive middle points of the sides of a
quadrilateral figure, form a rhomboid.
If the sides of the quadrilateral figure ABCD be bisected, and the
points of section joined in their order; EFGH is a rhomboid.
B
G
C
K
For draw AC, BD. And because FG bisects AB, BC, it is
(II. 4. El.) parallel to AC; and for the
same reason, EH, as it bisects AD and DC,
is parallel to AC. Wherefore FG is paral-
lel to EH (I. 30.). In like manner, it is
proved that EF is parallel to HG; and
consequently the figure EFGH is a rhon-
boid or parallelogram.

A
F
408
NOTES AND ILLUSTRATIONS.
!
It is likewise evident, that the inscribed rhomboid is half of the
quadrilateral figure; for IG is half of the triangle ABC (II. 4. cor.),
and IH is half of the triangle ADC.
Note XIV. Page 43.
This problem is of great use in practical geometry. The plan, for
instance, of any grounds however irregular, is divided into a number
of triangles, which are successively reduced to a simple triangle, and
this again is converted (by II. 7.) into a rectangle. Instead of
computing, therefore, each component triangle, it may be sufficient
to calculate the area of the final triangle or rectangle.
Note XV.-Page 46.
On this proposition is founded the method of offsets, which enters
so largely into the practice of land-surveying. In measuring a field
of a very irregular shape, the principal points only are connected by
straight lines forming sides of the component triangles, and the dis-
tance of each remarkable flexure of the extreme boundary is taken
from these rectilineal traces. The exterior border of the polygon is
therefore considered as a collection of trapezoids, which are measured
by multiplying the mean of each pair of offsets or perpendiculars in-
to their base or intermediate distance.
1
Note XVI.-Page 48.
This famous proposition appears to have been brought from the
East by Pythagoras. The method here given of demonstrating it,
from the transposition of the several parts of the figure, is ascribed
to the Persian astronomer Nassir Eddin, who flourished in the thir-
teenth century of our æra, under the munificent patronage of the
conqueror Zingis Khan.
It may gratify the young student in Geometry to see the mode of
NOTES AND ILLUSTRATIONS.
409
E
प्र. म.
F
M
performing this dissection. Having drawn GO parallel to IK, place
the triangle CKA on CFH, invert the tri-
angle GOA or ADG, place the triangle
GOM on AKN, and transfer the small
triangle GIN to HLM. In this way, the
square AGHC is transformed into the two
squares CKLF and ADIK. By reversing
the process, the squares of the sides of the
right-angled triangle may be compounded
into the single square of the hypotenuse.

G
Note XVII.-Page 49.
I
K
D
A
B
It was a favourite speculation with the Greek geometers, to ex-
press numerically the sides of a right-angled triangle. The rules
which they delivered for that purpose are equally simple and inge-
nious. For the sake of conciseness, it will be convenient, however,
to adopt the language of symbols. Let n denote any odd number;
then,
according to Pythagoras, n,
n² -I
2
and ¹²+¹
or
2
according to Plato, 2n, n²-1 and n²+1, will represent.
the perpendicular, the base, and hypotenuse, of a right-angled tri-
angle. Thus, n being supposed equal to 3, the numbers thence re-
sulting are 3, 4, and 5, or 6, 8, and 10. These expressions are fun-
damentally the same, and are easily derived from Prop. 19. Book II.:
For (n²+1)²—(n²—1)² = ((n²+1) + (2²—1)) ((n² + 1)—(n²—1)) =
2n² × 2=(2n)².
Note XVIII.-Page 51.
An elegant proposition derived from this, deserves a place in an
elementary work :
In any triangle, the square described on the base, is equivalent to the
rectangles contained by the two sides and their segments intercepted
from the base by perpendiculars let fall upon them from its opposite ex-
tremities.
410
NOTES AND ILLUSTRATIONS.
A
Let the perpendiculars AP, CN be let fall from the points A, C
upon the opposite sides BC and AB of the triangle ABC; the square
of AC is equivalent to the rectangles contained by AB, AN and by
BC, CP.
For complete the rhomboids ADHB and CFHB, and let fall the
perpendiculars BR and BS upon DH and FH.
D
H
R
F
N
P
It is manifest, (II. 15. El.) that the rhomboids AH and CH are
equivalent to the square of AC. But the rhomboid AH is equivalent
to the rectangle contained by AB and BR
(II. 1. cor.). Comparing the triangles BHR
and ACN; the angle BRH, being a right
angle, is equal to ANC; and the two acute
angles BHR and RBH, being together equal to
a right angle, are equal to DAN and NAC;
but DAB is equal to DHB (I. 27.), whence
the angle RBH is equal to NAC. These tri-
angles BHR and ACN, having thus two angles.
respectively equal, and the corresponding side.
BH in the one equal to AD or AC in the other,
are therefore equal (I. 21.), and consequently
the side BR is equal to AN. The rectangle
AB and BR, which is equivalent to the rhom-
boid AH, is hence equivalent to the rectangle
contained by AB and AN (II. 1. cor.).

A
H

R
F
D
S
P
N
A
C
In the same manner, it may be demonstrated, by comparing the
triangles BHS and PAC, that the rectangle under BC and BS
which is equivalent to the rhomboid CH, is equivalent to the rect-
angle contained by BC and CP. Wherefore the two rectangles of AB,
AN and BC, CP are together equivalent to the square described on
AC.
If the triangle ABC be right-angled at the vertex B, the perpen-
diculars CN and AP will evidently meet at the vertex, and conse-
quently the rectangles AB, AN and BC, CP will become the squares
of AB and BC. And hence the beautiful Proposition II. 11. is de
rived, being only a remarkable case of a much more general property.
NOTES AND ILLUSTRATIONS.
411
1.
Note XIX. Page 51.
Since rectangles correspond to numerical products, the properties
of the sections of lines are easily derived from symbolical arithmetic :
1. In Prop. 16. let AC be denoted by a, and the segments of AB
by b, c and d; then a(b+c+d)=ab+ac+ad.
2. In Prop. 17. let the two lines be denoted by a and b; then
(a+b)²=a²+b²+2ab.
3. In Prop. 18. let the two lines be denoted by a and b ; then
(a—b)²=—a²+b²—Qab.
4. In Prop. 19. let the two lines be denoted by a and b; then
(a+b) (a—b)=a²—b².
5. In Prop. 20. let the segments of the compound line be denoted
by a, b and c; then (a+b+c)²=a²+b²+c²+2ab+2ac+2bc.
6. In Prop. 21. let the two lines be denoted by a and b ; then
2
a² + b² = 3 (a+b)² + 3 (a—b)² = 2 (a+b)² +2
·2
a
+ 2 (16)
7. In Prop. 22. let the whole line be denominated by a, and its
greater segment by a; then ra(a—x), and x²+ax=a², whence
5a² a
4
x=
:±a(√ §—}). Hence, if unit represent the
whole line, the greater segment is .61803398428, &c. and the small-
er segment .38196601572, &c.
From Cor. 1. an extremely neat approximation is likewise obtained.
Assuming the segments of the divided line as at first equal and
denoted each by 1, these successive numbers will result from their
continued summation :
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, &c.
If the original line, therefore, contained 144 equal parts, its greater
segment would include $9, and its smaller segment 55 of these parts
very nearly.
412
NOTES AND ILLUSTRATIONS.
Note XX. Page 59.
This problem may, however, be constructed somewhat differently,
without employing the collateral properties.
D
G
A
B F
For bisect AB in C (I. 7.), draw (I. 5. cor.) the perpendicular
BD equal to BC, join AD and continue it until DE be equal to DB
or BC, and on AB produced take AF equal
to AE: The line AF is the required exten-
sion of AB. For make DG equal to DB or
BC; and because (II. 19. cor. 2.) the rectangle
EA, AG together with the square of DG or
DB, is equivalent to the square of DA, or to the squares of AB and
DB; the rectangle EA, AG, or FA, FB is equivalent to the square
of AB.

Note XXI.-Page 60.
A neat proposition may be subjoined.
If, from the hypotenuse of a right angled triangle, portions be cut
off equal to the adjacent sides; the square of the middle segment thus
formed, is equivalent to twice the rectangle contained by the extreme
segments.
Let ABC be a triangle which is right-angled at B; from the hy-
potenuse AC, cut off AE equal to AB, and CD equal to CB: Twice
the rectangle under AD and CE is equivalent to the square of DE.
For the straight line AC being divided into three portions, the
A D
B
EC
squares of AE and CD, together with twice the
rectangle AD, CE are equivalent to the squares
of AC and DE (II. 20. cor.). But the squares
of AB and BC, or those of AE and CD, are
equivalent to the square of AC (II. 11.). There consequently re-
mains twice the rectangle AD, CE equivalent to the square of DE.
By an inverse process of reasoning it will appear, that if
twice the rectangle AD, CE be equal to the square of DE, the
straight line AC, so composed, is the hypotenuse of a right-angled
triangle, of which AB and BC are the sides.

NOTES AND ILLUSTRATIONS.
413
,
This proposition will furnish another convenient method of disco-
vering the numbers which represent the sides of a right angled tri-
angle: For since DE2 2AD x CE, it is evident that DE²-
ADX CE; and consequently, expressing DE by a whole number, and
resolving DE² into the factors AD and CE, AD+DE and CE+DE
will represent the two sides, and AD+CE+DE the hypotenuse.
Thus, if 2 be taken, the factors of half its square are 1 and 2,
which produce the numbers 3, 4, and 5. Again, if 4 be assumed,
the factors are 2 and 4, or 1 and 8; whence result these numbers,
6, 8 and 10, or 5, 12 and 13. In this way, a very great variety of
numbers can be found, to express the sides of a right-angled triangle.
Note XXII.-Page 61.
This proposition is of great use in practical geometry, since it en-
ables us to divide a triangle, of which all the sides are given into two
right-angled triangles, by determining the position and consequently
the length of the perpendicular.
Note XXIII.-Page 63.
From this corollary is derived a very simple construction of the
problem,“ to find a square equivalent to a given rectangle."
H H
E
F
C
Let ABCD be the given rectangle, of which the side AD is great-
er than AB. In AB or its production, take AE equal to the half of
AD and place it from E to F; then AF
being joined, is the side of the equivalent
square. For (II. 26. cor. El.) since the
sides AE and EF of the triangle AEF are
equal, the square of AF is equivalent to the
rectangle under twice AE and AB, that is,
from the construction, the rectangle under
AD and AB.

A
D
H
K
I
The same construction might likewise be deduced from the second
demonstration of the celebrated property of the right-angled triangle.
For, in the figure of page 48, suppose BO were drawn to the hypo-
tenuse AC, making an angle ABO equal to BAO or BAC; since the
two acute angles are together equal to a right angle, the angle BCA
4
414
NOTES AND ILLUSTRATIONS.
1
1
is equal to the remaining portion CBO of the right angle at B, and
consequently the triangles AOB and COB are isosceles, and the
sides OA, OB and OC all equal. Wherefore AB, the side of a
square equivalent to the rectangle ADMN or that under AK and
AN, is determined by making AO equal to the half AK or AC and
inserting it from O to B.-The inspection of the same figure also
points out the mode of dissecting the rectangle, and thence com-
pounding the square; for a perpendicular let fall from K on AB is
evidently equal to GB or AB. Hence, on AF, in the original con-
struction, let fall the perpendicular DG, transpose the triangle FBA
in the situation DHI, and slide the quadrilateral portion into the
place of KAHI; the rectangle ABCD is now transformed into the
square KGDI.-A slight modification will be required when AB is
less than the half of AD.
In this construction of the problem, the application of the circle
which (III. 33. El.) is indispensably required, is only not brought
into view.—When the side AD is double of AB, the point G coincides
with F, and the rectangle is resolved into three triangles, which com-
bine to form a square.
Note XXIV. Page 64.
The following theorem is demonstrated from the same princi-
ples:
If straight lines be drawn from the angulur points of a triangle to
bisect the opposite sides, thrice the squares of these sides are together
equivalent to four times the squares of the bisecting lines.
Let the sides of the triangle ABC be bisected in D, E, and F, and
straight lines drawn from these points to the opposite vertices; thrice
the squares of the sides AB, BC, and AC are together equivalent to
four times the squares of BD, CE and AF.
E
B
T
For, by Prop. II. 25. the squares of AB, BC are equivalent to
twice the square of BD and twice the square
of AD, that is, half the square of AC; the
squares of BC, AC are equivalent to twice the
squares of CE and half the square of AB ; and
the squares of AC, AB are equivalent to twice
the square of AF and half the square of BC.
Whence the squares of the sides of the triangle, repeated twice, are

A
D
C
NOTES AND ILLUSTRATIONS.
415
equivalent to twice the squares of BD, CE, and AF, with half the
squares of the sides of the triangle. Consequently four times the
squares of AB, BC, and AC are equivalent to four times the squares
of BD, CE, and AF, with once the squares of AB, BC, and AC;
wherefore thrice the squares of the sides AB, BC, and AC are toge
ther equivalent to four times the squares of the bisecting lines BD,
CE, and AF.
Note XXV. Page 65.
This general theorem seems to have been first given by the illus-
trious Euler in the Petersburg Memoirs. It evidently comprehends.
the proposition which stands immediately before it; for when the
quadrilateral figure becomes a rhomboid, the diagonals bisect each
other, and the middle points E and F coincide; whence the squares
of all the sides are equivalent simply to the squares of those diagon-
als.-If this rhomboid again becomes a rectangle, it will have equal
diagonals, and consequently, as in the 11th Proposition of the second
book, the squares of the sides of a right angled triangle are equiva-
lent to the square of the hypotenusc.
Note XXVI.-Page $1.
Hence angles are, sometimes measured by a circular instrument,
from a point in the circumference, as well as from the centre.-On
the next proposition depends the construction of amphitheatres; for
the visual magnitude of an object is measured by the angle which it
subtends at the eye, and consequently the whole extent of the stage
will be seen with equal advantage by every spectator seated in the
same arc of a circle.
Note XXVII.-Page S3.
If, on each side of any point in the circumference of a circle, equal
arcs be repeated; the chords which join the opposite points of section
will be together equal to the last chord extended till it meets a straight
line drawn through the middle point and either extremity of the first
chord.
416
NOTES AND ILLUSTRATIONS.
:
Let DAG be the circumference of a circle, in which the arcs AB,
BC, CD on the one side of a point A, and the corresponding arcs
AE, EF, FG on the other side, are all assumed equal; the chords
BE, CF, and DG, are together equal to the line GH, formed by ex-
tending GD' till it meets the production of AB.
I.
For join FD and CE, and produce this to meet GH in the point
Because the arcs BC and
CD are equal to EF and

Р
E
FG, the chords BE, CF, and
F
H
I
D
DG are parallel; but, for the
same reason, since the arcs
BC and CD are equal to AE
and EF, the chords BA, CE and DF are likewise parallel. Hence
the figures HBEI and ICFD are rhomboids, and therefore the ex-
tended chord GH, being composed of the segments HI, ID, and DG,
is equal to the sum of their opposite chords BE, CF and DG. It is
obvious that the same train of reasoning may be pursued to any num-
her of equal arcs.
Note XXVIII.-Page 84.
་
This proposition is of some utility in practice, for an angle may be
hence measured by help of a circular protractor, without the trouble
of applying the centre to its vertex, or the point of concourse of the
sides.-The same principle is likewise applicable to the construction
of some optical instruments, calculated to measure lateral angles by
the intersection of micrometer wires.
Note XXIX. Page 85.
To erect a perpendicular, any point D is taken, as in Prop. 36.
Book I., and from it a circle is described passing through C and B ;
the diameter CDF determines the position of the perpendicular BF.
To let fall a perpendicular, draw to AB any straight line FC, which
bisect in D, and from this point as a centre describe a circle through
C, B and F, FB is the perpendicular required.
NOTES AND ILLUSTRATIONS.
417
1
Note XXX. Page 96.
The rectangle under the segments of a chord is greater or less than
the rectangle under the segments into which a perpendicular from the
point of section divides a diameter, by the square of that perpendicular
-according as it lies without or within the circle.
Let the perpendicular CF be let fall from a point C in the chord
ACB upon a diameter DE; the rectangle BC, CA, is greater or less.
than the rectangle EF, FD, by the square of the perpendicular CF,
according as this lies without or within the circle,
First, let the perpendicular CF lie without the circle, and join CE
and DG.
C
A
G
E
F D
H
The square of the hypotenuse CE is equivalent to the squares of
FE and CF (II. 11.). But the square of CE is composed of the rect-
angles CE, EG, and CE, CG (II. 16.);
and the square of FE is composed of
the rectangles FE, ED, and FE, FD:
Wherefore the rectangles CE, EG and
CE, CG are equivalent to the rect-
angles FE, ED and FE, FD, together
with the square of CF. And since
EGD, standing in a semicircle, is a
right angle (III. 22.), its adjacent angle
CGD is also right, and the angle opposite to this at F is right; con-
sequently (III. 19. cor.) a circle might be described through the four
points C, G, D, F. Whence (III. 32.) the rectangle CE, EG is
equivalent to FE, ED; and taking these from the terms of the for-
mer equality, there remains the rectangle CE, CG, that is, (III. 32.)
AC, CB, equivalent to the rectangle FE, FD, together with the
square of CF.
Next, let the perpendicular CF lie within the circle.
The same construction being made, the rect-
angle CE, EG is still equivalent to the rect-

B

angle FE, ED. But the rectangle CE, EG is
(II. 16.) equivalent to the rectangle CE, CG, D
E
F
and the square of CE, or the squares of FE
and CF; and the rectangle FE, ED is equi-
valent to the rectangle FE, FD and the square
Dd
1
B
418
NOTES AND ILLUSTRATIONS.
of FE. From these equal quantities, therefore, take away the com-
mon square of FE, and there remains the rectangle CE, CG, or AC,
CB, with the square of CF, equivalent to the rectangle FE, FD.
Lastly, if the perpendicular CF lie partly without and partly
within the circle, the Proposition must be slightly modified.
The former construction being retained:
Because the square of CE is equivalent to
the squares of CF and FE, the rectangles.
CE, EG and CE, CG are together equiva-
lent to the square of CF and the difference
between the rectangle FE, ED and FE,
ED; but the rectangle CE, EG is equiva- D
lent to the rectangle FE, ED, and conse-

ANG
E
quently the rectangle CE, CG, or the rect-
angle AC, CB, is equivalent to the diffe-
B
rence between the square of CF and the rectangle FE, FD.
In the first case, if the square of FH be equivalent to the rect-
angle FD, FE, the square of CH will be likewise equivalent to the
rectangle CG, CE; for the rectangle AC, CB, being equivalent to
the rectangle FD, FE, or the square of FH, together with the
square of CF, must (II. 11. El.) be equivalent to the square of CH.
Note XXXI.-Page 118.
The square of the side of a regular octagon inscribed in a circle, is
equivalent to the rectangle contained by the radius and the difference
between the diameter and the side of the inscribed square.
B
E
I
C
Let ABCD be a square inscribed in a circle, and AEBFCGDH
an octagon, which is formed evidently by the bisection of the qua-
drants AB, BC, CD, and DA: The square of AE is equivalent to
the rectangle under AO and the difference between AB and AC.
For draw the diameter EG. It is
manifest, that the triangles AIO and
BIO are right-angled and isosceles;
and because AO is equal to EO, and
AI perpendicular to it,—the square of A
AE (II. 26. cor. El.) is equivalent to
twice the rectangle under EO and EI, or
the rectangle under AO and twice EI.
But El is the difference of EO and 10,

D
NOTES AND ILLUSTRATIONS.
419
and twice EI is, therefore, equal to the difference of twice EO or
AC and twice IO or AB. Whence the square of AE, the side of
the octagon, is equivalent to the rectangle under the radius and the
difference of the diameter and AB the side of the inscribed square.
Note XXXII-Page 119.
Such were the only regular polygons known to the Greeks. The
inscription of all the rest has for ages been supposed absolutely to
transcend the powers of elementary geometry. But a curious and
most unexpected discovery was lately made by Mr Gauss, who has
demonstrated, in a work entitled Disquisitiones Arithmetica, and
published at Brunswick in 1801, that certain very complex polygons
can yet be described merely by help of circles. Thus, a regular
polygon containing 17, 257, 65537, &c. sides, is capable of being in-
scribed, by the application of elementary geometry; and in general,
when the number of sides may be denoted by 2"+1, and is at the
same time a prime number. The investigation of this principle is
rather intricate, being founded on the arithmetic of sines and the
theory of equations; and the constructions to which it would lead
are hence, in every case, unavoidably and most excessively compli
cated. Thus the cosine of the several arcs arising from the division
of the circumference of a circle into seventeen equal parts, are all con-
tained in this very involved expression :
-3/8 +18 √17+ √(34-2№17) -
17+√
† √(17+3 √17−√(S4—2 √ 17)−2 √(34+2 √17))·
As the radicals may be taken either positive or negative, their various
combinations, rightly disposed, will produce eight distinct results. Let ä
2F
32T
denote the semicircumference; then cos
= COS-
9324722294,
17
17
4x
30%
6%
28T
COS
COS- = .7390089172, cos
=cos.4457383558,
17
17
17
17
ST
267
10%
24T
COS
COS = .0922683595, cos =COS
17
17
17
17
-.27366229901, cos = cos
12%
17
22T
14
.6026346364, cos
17
17
20%
16%
COS
.8502171357, and cos
- Cos
17
17
18
17
.9829730997.
D d 2
4.20
NOTES AND ILLUSTRATIONS.
Note XXXIII-Page 120.,
Pythagoras was the first who remarked the simple property, that
only three regular figures,-the square, the equilateral triangle
and the hexagon,-can be constituted about a point. Here the
mystic philosopher might again admire the union of the monad with
the triad. It may not be superfluous perhaps to observe, that on
this property is founded the adaptation of patchwork, and the con-
struction of tessellated pavement.
4799
Note XXXIV.-Page 123.
The words, λoyos in Greek and ratio in Latin, signifying reason or
manner of thought, indicate vaguely a philosophical conception. The
compound term avaλoyia comes nearer to this idea; but its correlative,
proportio, marks very distinctly a radical similarity of composition.
The doctrine of proportion has been a source of much controversy.
In their mode of treating that important subject, authors differ wide-
ly; some rejecting the procedure of Euclid as circuitous and embar-
rassed, while others appear disposed to extol it as one of the hap-
piest and most elaborate monuments of human ingenuity. But, to
view the matter in its true light, we should endeavour previously to
dispel that mist which has so long obscured our vision. The fifth
book of Euclid, in its original form, is not found to answer the pur-
pose of actual instruction; and this fact alone might justify a suspi-
cion of its intrinsic excellence. The great object which the framer
of the Elements had proposed to himself, by adopting such an artifi-
cial definition of proportion, was to obviate the difficulties arising
from the consideration of incommensurable quantities. Under the
shelter of a certain indefinitude of principle, he has contrived rather
to evade those difficulties than fairly to meet them. Euclid seems
not indeed to grasp the subject with a steady and comprehensive
hold. In his seventh book, which treats of the properties of num-
ber, he abandons his former definition of proportion, for another that
is more natural, though imperfectly developed. Through the whole
contexture of the Elements, we may discern the influence of that
mysticism which prevailed in the Platonic school. The language
sometimes used in the fifth book would imply, that ratios are not
mere conceptions of the mind, but have a real and substantial essence.
NOTES AND ILLUSTRATIONS.
421
3
The obscurity that confessedly pervades the fifth book of Euclid
being thus occasioned solely by the attempt to extend the definition
of proportion to the case of incommensurables, the theory of which
is contained in his tenth book-the pertinacity of modern editors of
the Elements in retaining such an intricate definition, appears the
more singular, since, omitting all the books relative to the properties
of numbers, they have not given the slightest intimation respecting
even the existence of incommensurable quantities.
The notion of proportionality involves in it necessarily the ideas
of number. The doctrine of proportion hence constitutes a branch
of universal arithmetic; and had I not on this occasion yielded to
the prevalence of custom, I should have deferred the consideration
of the subject till I came to treat of Algebra, where it is sometimes
given, but in a very contracted form. The properties themselves
are extremely simple, and may be regarded as only the exposition
of the same principle under different aspects. The various transfor-
mations of which analogies are susceptible, exactly resemble the
changes usually effected in the reduction of equations.
According to Euclid, "The first of four magnitudes is said to
have the same ratio to the second which the third has to the fourth,
when any equimultiples whatsoever of the first and third being
taken, and any equimultiples whatsoever of the second and fourth;
if the multiple of the first be less than that of the second, the mul
tiple of the third is also less than that of the fourth; or, if the mul-
tiple of the first be equal to that of the second, the multiple of the
third is also equal to that of the fourth; or, if the multiple of the
first be greater than that of the second, the multiple of the third is
also greater than that of the fourth." This definition, however per-
plexed and verbose, is yet easily derived from that which appears to
furnish the simplest and most natural criterion of proportionality:
For, let A : B :: C: D; it was stated as a fundamental principle,
that, if the mth part of A be contained n times in B, the mth part
of C will likewise be contained n times in D. Whence nA=mB,
and nC=mD; which is the basis of Euclid's definition. But when
the terms are incommensurable, such equality cannot absolutely sub-
sist. In this case, no single trial would be sufficient for ascertaining
proportionality. It is required that, every multiple whatever, mA,
being greater or less than nB, the corresponding multiple, mC, shall
422
NOTES AND ILLUSTRATIONS.
likewise be constantly greater or less than D. Actually to apply
the definition is therefore impossible; nor does it even assist us at all
in directing our search. In the natural mode of proceeding, by as-
suming successively a smaller divisor, we are, at each time, brought
nearer to the incommensurable limit. But Euclid's famous definition
leaves us to grope at random after its object, and to seek our escape,
by having recourse to some auxiliary train of reasoning or induction.
The author of the Elements has likewise given what Dr Barrow
calls a metaphysical definition of ratio: "Ratio is a mutual relation
of two magnitudes of the same kind to one another, in respect of
quantity." This sentence, as it now stands, appears either tautologi-
cal, or altogether void of meaning; and Dr Simson, anxious for the
credit of Euclid, considers it, in his usual manner, as the interpola-
tion of some unskilful editor. I am inclined to think, however, that
the passage will admit of a version which is not only intelligible,
but conveys a most correct idea of the nature of ratio. The origi-
nal runs thus: Λογος εστι δυο μεγεθῶν ὁμογενῶν ἡ καλα Πηλικότητα προς
Now the term λixos, on which the whole evi-
dently hinges, though commonly rendered quantus, may be trans-
lated quotus, as expressing either magnitude or multitude. In its
primitive sense, it probably denoted number, and came afterwards to
signify quantity, as this word itself has, in the French language,
undergone the reverse process. In confirmation of this opinion, it
may be stated, that the relative term λx properly denotes age,
and thence stature or size. According to this interpretation, there-
fore,
"Ratio is a certain mutual habitude of two homogeneous mag-
nitudes with respect to quotity, or numerical composition."
αλληλα ποια σχεσις.
ท
Note XXXV.-Page 134.
This proposition is easily derived from geometry; for, since of
proportional, lines the rectangle under the extremes is equal to that
of the means, the segments AG and AH of the diameter in the
figure of page 93 are (III. 7. El.) the greatest and least terms of an
analogy, of which AB and AD are the intermediate terms, and con-
sequently (III. 6. El.) the diameter GH, or the sum of AG and AH,
is greater than the chord BD, or the sum of AB and AD.
}
NOTES AND ILLUSTRATIONS.
4.23
Note XXXVI.-Page 143.
It is more convenient, however, to derive the numerical ratio, from
the quotients of subdivision in their natural order; and this method
has besides the peculiar advantage of exhibiting a succession of ele-
gant approximations.
The quantities A, B, C, D, &c. are determined, as before, by
these conditions: A=mB+C, B=nC+D, C=pD+E, D=9E+F,
&c. But other expressions will arise from substitution: For,
1. A=mB+C=m(nC+D)+C=(mn+1)C+mD, or, putting
m.n+1=m', A=m'C+mD.
2. A=m'C+mD=m'(pD+E)+mD=(m'p+m) D+m'E, or, put-
ting m'.p+mm", A=m"D+m'E.
3. Am"D+m'E=m"(gE+F)+m'E=(m"q+m') E+m"F, or,
putting m"q+m'=m"", A=m″E+m"F.
Again, the successive values of B are developed in the same man-
ner:
1. B=nC+D=n(pD+E)+D=(np+1) D+nE, or, putting
n.p+1=n', B='D+n.E.
2. B=n'D+nE=n'(qE+F)+nE=(n'q+n) E+n'F, or, putting
n'.q+n=n", B="E+n'F,
These results will be more apparent in a tabular form:
A=mB+C,
=m'C+mD,
=m"D+m'E,
=m""E+n"F,
&c.
The substitutions are thus arranged:
m.n+1=m',
m'.p+m=m",
m".q+m'=m"",
&c.
B=nC+D,
=n'D+nE,
=n"E+n'F,
&c.
n.p+1=n',
n'.q+n=n",
&c.
Whence, the law of the formation of the successive quantities, is
easily perceived.
:
424
NOTES AND ILLUSTRATIONS.
But, to find the ratio of A to B, it is not requisite to know the
values of the remainders C, D, E, &c. Suppose the subdivision to
terminate at B; then AmB, and consequently A: B, as mB : B,
or m: 1. If the subdivision extend to C, then Am'C, and B="C;
whence A: B, as m' : n, In general, therefore, the second term, in
the expressions for A and B, may be rejected, and the letter which
precedes it considered as the ultimate measure, and corresponding to
the arithmetical unit. Hence, resuming the substitutions and com-
bining the whole in one view, it follows, that the ratio of A to B
may thus be successively represented:
1. m: 1.
2. mn+1: n, or m' : n.
3. m'p+m: np+1, or m" : n'.
4. m"q+m': n'q+n, or m"": n".
&c.
&c.
&c.
The formation of these numbers will evidently stop, when the cor-
responding subdivision terminates. But even though the successive
decomposition should never terminate, as in the case of incommen-
surable quantities, yet the expression thus obtained must constantly
approach to the ratio of A: B, since they suppose only the omission
of the remainder of the last division, and which is perpetually dimi-
nishing.
Note XXXVII.-Page 144.
The same conclusion is derived from the division of surds. Thus
'N2
=1+
1
√2-1
1
1
√2-1
2 + 1,
√2+1=2+
√2-1
and then
1
1
1
continually the expansion of the same residue
which there-
√2-1
Hence m being 1 and
fore gives 2 as a repeated integral quotient.
n, p, q, r, &c. all equal to 2, the successive approximations are, by
the last note, 1:1, 2:3, 5:7, 12:17, 29:41, 70:99, &c.
Note XXXVIII.-Page 150.
The consideration of diverging lines furnishes the simplest and rea-
diest means, for transferring the doctrine of proportion to geometrical
figures. The order which Euclid has followed, beginning with pa-
rallelograms, and thence passing from surfaces to lines, appears to be
less natural.
NOTES AND ILLUSTRATIONS.
425
Note XXXIX,-Page 153.
It will be proper here to notice the several methods adopted in
practice, for the minute subdivision of lines. The earliest of these-
the diagonal scale-depending immediately on the proposition in the
text, is of the most extensive use, and constituted the first improve-
ment on astronomical instruments.
8
6
Thus, in the figure annexed, the extreme portion of the horizontal
line is divided into ten equal parts, each of which again is virtually
subdivided into ten secondary parts. This subdivision is effected by
means of diagonal lines, which decline from the perpendicular by in-
tervals equal to the primary divisions, and which are cut transversely
into ten equal segments by equidis-
tant parallels. Suppose, for exam-
ple, it were required to find the
length of 2 and 38-100 parts of
a division; place one foot of the
compasses in the second vertical at the eight interval which is mark-
ed with a dot, and extend the other foot, along the parallel, to the
dot on the third diagonal. The distance between these dots may,
however, express indifferently 2.38, 23.8, or 238, according to the
assumed magnitude of the primary unit.

4
2
8642
Nunez, or Nonius, proposed one more complicated. He placed a
number of parallel scales, differently divided, and forming a regular
descending gradation. An index laid any where across these scales
would, therefore, cut at least one of them at some division, and hence
the intercepted space would be expressed by a corresponding frac-
tion.
But the method of subdivision afterwards introduced by Vernier,
is much simpler and far more ingenious. It is founded on the dif-
ference of two approximating scales, one of which is moveable.
Thus, if a space equal to ❀ 1 parts on the limb of the instrument
be divided into a parts, these evidently will each of them be smaller
than the former, by the nth part of a division. Wherefore, on shift-
ing forward this parasite scale, the quantity of aberration will dimi-
nish at each successive division, till a new coincidence obtaius, and
426
NOTES AND ILLUSTRATIONS.
then the number of those divisions on that scale will mark the
fractional value of the displacenient.
Thus in the annexed figure, nine divisions of the primary scale
forming ten equal parts in the attached or sliding scale, the move-
able zero stands

beyond the first
interval between
9 8 7 6 5 4 3 2 1 0
F
the third and fourth division. To find this minute difference, ob-
serve where the opposite sections of the scales come to coincide,
which occurs under the fourth division of the sliding scale, and there-
fore indicates the quantity 1.34.
Note XL-Page 158,
This corollary is easily deduced by a direct process; for CD: DE::
DE: OD, and CD : OD :: CD² : DE² or AD× db.
Note XLI.-Page 160.
If two straight lines be inflected from the extremities of the base of
a triangle to cut the opposite sides proportionally, another straight line,
drawn from the vertex through their point of concourse, will bisect the
base.
In the triangle ABC, let AE and CD, drawn from the extremi-
ties of the base to cut the opposite sides proportionally, intersect
each other in F, join BF, which produce if necessary to meet the
base in the point G; AG will be equal to GC.
For join DE. And because the sides AB and BC are cut propor-
tionally, DE is parallel to AC(VI. 1. cor.),
whence BD BA ; : BH : BG (VI. 1.);
but BD : BA :: DE: AC (VI. 2.), and
therefore BH: BG :: DE : AC. Again,
the parallels DE and AC being cut by the
diverging lines AE and CD, DE: AC : :
DF: FC (VI. 2.), and DF: FC : : FH :
FG (VI, 1.); wherefore BH: BG :: FH :
A.
G
B
-I
E
FG, or BF is cut internally and externally in the same ratio. But
1
f

NOTES AND ILLUSTRATIONS.
427
DH being parallel to AG, BH: BG:: DH: AG; and since DII
is also parallel to GC, HF: FG :: DH: GC; whence DH: AG : :
DH: GC, and consequently AG is equal to GC.
Note XLII.-Page 165.
If a semicircle be described on the side of a rectangle, and through
its extremities two straight lines be drawn from any point in the cir-
cumference to meet the opposite side produced both ways; the altitude
of the rectangle will be a mean proportional between the segments thus
intercepted.
Let ABED be a rectangle, which has a semicircle ACB described
on the side AB, and the straight lines CA and CB drawn from a
point C in the circumference to meet the extension of the opposite.
side DE; the altitude AD of the rectangle will be a mean propor-
tional between the exterior segments FD and EG.
For, the angle ADF, being evidently a right angle, is equal to
the angle ACB,
which stands
in a semicircle
(III. 22.), and

the angle DFA
is equal to the
exterior angle
BAC (I. 23.);
3
H I
wherefore (VI.
D
E
G
12.) the triangle FAD is similar to ABC. In the same manner, it
is proved that the triangle BGE is similar to ABC; whence the tri-
angles DFA and BGE are similar to each other, and consequently
(VI. 12.) FD : AD :: BE or AD : EG.
If the straight lines CD and CE be drawn, they will (VI. 2.)
divide the diameter AB into segments AH, HI, and IB, which
are respectively proportional to the segments FD, DE, and EG
of the extended side DE. Consequently when ABED is a square,
and therefore DE a mean proportional between FD and EG, it must
follow that HI is likewise a mean proportional between AH and IB.
If the rectangle ABED have its altitude AD equal to the side.
423
NOTES AND ILLUSTRATIONS.
of a square inscribed within the circle, the square of the diame-
ter AB is equivalent to the squares of the two segments AI and BH.
For FD : AD :: AD: EG, whence (V. 6.) FD.EG=AD², or
2FD.EG=2AD²; but (IV. 16. cor.) 2AD²=AB² or DE², and con-
sequently 2FD.EG-DE²; wherefore (VI. 2.) 2AH.IB=HI², and,
hence, by Note XXI. the segments AI, BH are the sides of a right-
angled triangle, of which AB is the hypotenuse, or AB²=AI²+BH².
Note XLIII.-Page 166.
A chord of a circle is divided in continued proportion, by straight
lines inflected to any point in the opposite circumference from the ex-
tremities of a parallel tangent, which is limited by another tangent ap-
plied at the origin of the chord.
Let AB, AC be two tangents applied to a circle, CD a chord
drawn parallel to AB, and AE, BE straight lines inflected to a point
E in the opposite circumference; then will the chord CD be cut in
continued proportion at the points F and G, or CF : CG:: CG: CD.
For join BD, BC, and CE. Because the tangent AB is equal
to AC (III. 32. cor. 2.), the angle ABC is equal to ACB (I. 11.);
but ABC is equal to the angle BCD (I. 23.), and to the angle BDC
(III. 25.); whence (VI. 12.)
the triangles BAC and BDC
are similar, and AB: BC: :
BC: CD, and consequently (V.
6.)BC2=AB.CD. Again, the
triangles CBG and CBE are si-
milar, for they have a common
angle CBE, and the angle BCG,

B
D
G
F
E
or BCD, is equal to BDC, or BEC (III. 18.): Wherefore BG : BC
:: BC: BE, and BCBG.BE. Hence AB.CD=BG.BE, and
AB: BE:: BG : CD; but FG being parallel to AB, AB: BE ::
FG: GE (VI. 2.), and consequently FG GE:: BG: CD; there-
fore (V. 6.) FG.CD=BG.GE; and since (III. 32.) BG.GE=CG.GD,
it follows that CG.GD=FG.CD, and FG: CG:: GD: CD, and
hence (V. 10.) CF : CG :: CG : CD.
:
·
NOTES AND ILLUSTRATIONS.
429
Note XLIV.-Page 166.
The chord DG in the second construction is hence equal to the
tangent in the third. But the tangent being at right angles to the
radius GO, is less than DO; and therefore the geometrical, is less
than the arithmetical, mean.-It may be observed, that, in geometri-
cal constructions, the transition from the sine to the tangent frequently
takes place, each of these lines being perpendicular to a limiting ra-
dius.
Note XLV.-Page 170.
This well-known proposition is now rendered more general, by its
extension to the case of the exterior angle of the triangle. The two
cases combined afford an easy demonstration of the corollary to
Prop. 7. Book VI.; for the straight lines bisecting the vertical and
its adjacent angle form a right-angled triangle, of which the hypote-
nuse is the distance on the base between the points of internal and
external section.
A
Note XLVI.-Page 170.
If, from the vertex of a triangle, two straight lines be drawn, ma-
king equal angles with the sides and cutting the base; the squares of
the sides are proportional to the rectangles under the adjacent segments
of the base.
In the triangle ABC, let the straight
lines BD and BE make the angle ABD
equal to CBE; then AB² : BC² :
DA.AE: EC.CD.
For (III. 10. cor.) through the points
B, D, and E describe a circle, meeting
the sides AB and BC of the triangle in F
and G, and join FG.
B

F
A
Τ
E
G
430
NOTES AND ILLUSTRATIONS.
J
Because the angles DBF and
EBG are equal, they stand
(III. 18. cor.) on equal arcs DF
and EG, and consequently
(III. 20. cor.) FG is parallel to
DE. Whence (VI. 1.) AB :
BC: AF: CG, and therefore
(V. 13.) AB²: BC² :: AB.AF :
BC.CG; but (III. 32.) AB.AF
= DA.AE, and BC.CG =
B

D
A
E
F
G
EC.CD. Wherefore AB² : BC² :: DA.AE: EC.CD.
:
B
!
If the triangle ABC be right-angled at C, and the vertical
lines BD and BE cut the base internally; then BC²+AC.CE :
BC² :: AE: CD.
AE CD. For make AH equal to EC. Because AB2:
BC² :: DA.AE : EC.CD, and (II. 11.) AB²—
AC²+BC2, therefore AC+BC* : BC² ::
DA.AE EC.CD, and, by division, AC² :
BC² :: DA.AE-EC.CD : EC.CD. But, by
successive decomposition, DA.AE—EC.CD=
DA.AC — DA.EC EC.CD DA.AC
EC.AC-AC.HD; whence AC: BC²:: ĄC.HD: EC.CD, and
(V. 13. and cor.) AC.EC : BC² : : EC.HD: EC.CD, or (V. 3.)
HD: CD; consequently (V. 9.) BC+AC.EC: BC: HC: CD;
but, AH being equal to EC, HC is equal to AE; wherefore BC2+
AC.EC: BC :: AE: CD,

A
HDE C
:
If the vertical lines BD, BE cut the base AC of a right-
angled triangle ACB externally; then will BC-AC.EC : BC² : :
AE: CD. For make AH-EC. It is demonstrated as before, that
AC²: BC² :: DA.AE-EC.CD: EC.CD; but DA.AE-EC.CD=
DA.AC + DA.EC ·
EC.CD = DA.AC
EC.AC = AC.HD;
wherefore AC2: BC² ::
AC.HD EC.CD, and
AC.EC: BC :: EC.HD
D
A
B

C E
: EC.CD :: HD: CD, and consequently BC-AC.EC: BC2² ::
HC or AE: CD.
:
NOTES AND ILLUSTRATIONS.
431
:
Note XLVII.-Page 175.
The latter part of the scholium was added, with a view to explain
the principle of the construction of the pantagraph, a very useful in-
strument contrived for copying, reducing, or even enlarging plans.
It consists of a jointed rhombus DBFE, framed of wood or brass,
and having the two sides BD and BF extended to double their
length; the side DE and the branch DA are marked from D
with successive divisions, DO being made to BO always in the ratio
of DP to BC; small sliding boxes for holding a pencil or tracing point.
are brought to the corresponding graduations, and secured in their po-
sition by screws; the point O is made the centre of motion, and rests
on a-fulcrum or support of lead; and the tracer is generally fixed at
C, while the crayon or drawing point is lodged at P. From the pro-
perty of diverging lines intersect-
ing parallels, the three points O,
P and C must evidently range in
the same straight line, and which
is divided at P in the determinate
ratio. While the point C, there-
fore, is carried along the bounda-
ries of any figure, the intermediate
point P will, by the scholium,
trace out a similar figure, reduced
in the proportion of OC to OP
D
B

D
or of OB to OD, and which, in the present instance, is that of
three to one.
But the point P may be placed on the fulcrum, the tracer insert-
ed at O, and the crayon held at C; in which case, C would deli-
neate a figure which is enlarged in the ratio of OP to PC or of OD
to DB. If the points O and P were now brought to coincide with
A and E, the distances AE and EC being equal, the original figure
would be transferred into a copy exactly of the same dimensions.
In reducing small figures, however, artists commonly prefer ano-
ther method, which is partly mechanical. The original is divided
into a number of small squares, by means of equidistant and inter-
432
NOTES AND ILLUSTRATIONS.
secting parallels. Other reduced squares are drawn for the copy,
which is then filled up, by observing the same relative position and
form of the boundaries.-One material advantage results from this
practice; for if oblongs be used in the copy instead of squares,
the original figure will be more reduced in one dimension than ano-
ther, which is often very convenient where height and distance are
represented on different scales.
Note XLVIII.-Page 181.
The curious properties of the crescents, or lunula, contained in
the first corollary, were discovered by Hippocrates of Chios, in
his attempts to square the circle. But a beautiful extension
of the same principle was briefly suggested by Mr Lawson, and
afterwards explained and demonstrated in Dr Hutton's Mathemati-
cal Tracts. It is a mode of dividing a given circle into equal
portions and contained within equal circular boundaries. For ex-
ample, let it be required to cut the circle APBQ into five equal
spaces: Divide the diameter AB into five equal parts at the points
C, D, E, and F; on AC, AD, AE, and AF describe the semicircles
AGC, AID, ALE, and ANF, and on BC, BD, BE, and BF, to
wards the opposite side, describe the semicircles BHC, BKD, BME,
and BOF; the circle APBQ will be
divided into five equal portions, by
the equal compound semicircumfe-
rences AGCHB, AIDKB, ALEMB,
and ANFOB.
For the diameter AB is to the
diameter AD, as the circumference
of AB to the circumference of AD,
or (V. 3.), as the semicircumference
APB to the semicircumference AID;
P

N
L
I
A
B
CI
D
ESFO
M
K
H
Q
and AB is to BD, as the semicircumference APB to the semicir-
cumference BKD. Wherefore (V. 20.) AB is to AD and BD to-
gether as the semicircumference APB to the compound boundary
AIDKB; and consequently these interior boundaries AGCHB,
NOTES AND ILLUSTRATIONS.
435
AIDKB, ALEMB, and ANFOB, are all equal to the semicircum-
ference of the original circle.
Again, the circle on AB is to the circles on AE and AF,
as the square of AB to the squares of AE and AF; and con-
sequently (V. 20.) the circle on AB is to the difference between the
circles on AE and AF, as the square of AB to the difference be-
tween the squares of AE and AF, that is (II. 19.), the rectangle
under the sum and difference of AE and AF, or twice the rectangle
under EF and AS, the distance of A from the middle point of EF.
Whence the circle APBQ is to the difference of the semicircles ALE
and ANF, or the space ALEFN, as the square of AB to the rect-
angle under AS and EF; and, for the same reason, the circle APBQ
is to the space FOBME, as the square of AB is to the rectangle
under BS and EF; consequently (V. 20.) the circle ABPQ is to
the compound space ALEMBOFN, as the square of AB to the
rectangles under AS and EF and BS and EF, or the rectangle under
AB and EF; but the square of AB is to the rectangle under
AB and EF, (V. 25. cor. 2.) as AB to EF, which is the fifth part of
AB; wherefore (V. 5.) any of the intermediate spaces, such as
ALEMBOFN, is the fifth part of the whole circle.
Note XLIX. Page 183.
This elegant theorem admits of an algebraical investigation.
Put AC-a, AB≈b, BC=c, and let s denote the semiperimeter,
and T the area of the triangle; then, by Prop. 26. Book II.,
2AC. CD a² + c²-b², consequently
B
=
a² + c²—b
CD=
and BD-BC²—CD²=
2a
+c²_b² 3
-), and, therefore, by
D
2a
Prop. 6. Book II. T-
AC2.BD2
4
4a²c² — (a²+c²—62)²
16
But this expression, consisting of the difference of two squares,
may be decomposed, by Prop. 19. Book II.; whence T²=
Qac + a² + c²—b² Qac—a²—c² + b² _ (a + c)²—b² b²—(@—c)³
4
4 ¿
E e
434
.
NOTES AND ILLUSTRATIONSLATIONS.
and, decomposing these factors again,
T²_a+b+c a―b+c a+b―c a+b+c
2
2
2
2
a+b+c
a-b+c
Now
= 8,
S -b,
2
a+b—c .
2
S
- c,
and
a+b+c
2
S
a; wherefore we obtain, by substitution,
T= √(s(sa) (s—b) (s—¿`) ),
1
This most useful proposition was known to the Arabians, but
seems to have been re-invented in Europe about the latter part of
the fifteenth century.
Note L.-Page 186.
This ingenious and concise approximation to the quadrature of the
circle was first published at Padua, in the year 1668, by my illustri-
ous predecessor James Gregory. It is the more deserving of atten-
tion, as it seems to have led that original author to the invention of
the method of series.
The Appendix to the books of Geometry cannot fail, by its no-
velty and singular beauty, to prove highly interesting. The first part
is taken from a scarce tract of Schooten, who was professor of Ma-
thematics at Leyden, early in the seventeenth century. But the se-
cond and most important part is chiefly selected from a most inge-
nious work of Mascheroni, a celebrated Italian mathematician, which
in 1798 was translated into French, under the title of Geometrie du
Compas. It will be perceived, however, that I have adapted the ar-
rangement to my own views, and have demonstrated the proposi-
tions more strictly in the spirit of the ancient geometry.
Note LI.-Page 211.
These three books are designed to exhibit a distinct and compre-
hensive view of the mode by which the Greek geometers conducted
their Analysis. For that purpose, I have chosen a series of propo-
NOTES AND ILLUSTRATIONS.
435
sitions, at first extremely simple, but gradually rising in difficulty as
the train of investigation proceeds. The first book, being rather of
a miscellaneous nature, is drawn from a variety of sources. The
25th and 26th Propositions contain the different analyses of the two
problems so famous in the Platonic school-the trisection of an angle
--and the duplication of the cube-which led immediately to the cul-
tivation of the higher geometry. The concluding theorem is the on-
ly one supplied by the Data of Euclid.
In the second and third books, I have endeavoured to comprise
all that relates to the ancient analysis in its most improved state, as
extended by the labours of Apollonius and his illustrious contempo-
raries. Without omitting any material proposition, I have yet avoid-
ed the prolixity of pursuing in detail their numerous subdivisions.
Our system of modern education, embracing such a wide range,
would scarcely indeed afford leisure for indulging in those easy tasks.
The method of analysis, so deservedly valued in the ancient schools,
was regularly studied after the Elements of Geometry. According
to Pappus, it consisted of eight distinct treatises:
1. The Data— gì v dedoμv-in a single book of considerable
length, but containing propositions only of the very simplest kind.
2. The Section of Ratio—ñigì roys 'axoloµñs—in two books, which
Dr Halley, with much sagacity and incredible labour, restored, from
a MS. in the Bodleian library. The object of the tract was the so-
lution of this problem, branched out into a multitude of cases, and
marked with various limitations: "Through a given point to draw
a straight line intercepting segments on two straight lines which are
given in position, from given points and in a given ratio." It forms
the first four propositions of the second book.
3. The Section of Space—rigi xwgis 'andloµãs-in two books. Of
these no vestige remained; but Dr Halley, guided by a few hints
from Pappus, very successfully exerted his ingenuity in divining the
original structure. It was proposed-" Through a given point, to
draw a straight line cutting off segments from given points on two
straight lines given in position, and which shall contain a rectangle
equal to a given space." This occupies the propositions from the
5th to the 10th inclusively of the second book.
4. The Determinate Section—zigi diwgioµévns Toµns—in two books
E e 2
436
NOTES AND ILLUSTRATIONS.
These were also lost; but Dr Simson, assisted by the attempts of
Schooten, has restored them in the most luminous manner. Their
object was-" To find a point, the rectangles or squares of whose
distances from given points in the same straight line should have a
determined ratio. They form Prop. 10-19. Book II.
5. Inclinations—ę vévoɛw-in two books. It was proposed-
"To insert a straight line, of a given magnitude, and tending to a
given point, between two lines which are given in position." This
problem was restored by Marinus Ghetaldus, a patrician of Ragusa ;
and other investigations were given by Hugo de Omerique, in his
ingenious treatise on Geometrical Analysis, printed at Cadiz in 1698.
Two solutions of the case of the rhombus, remarkable for their ele-
gance, appeared in the posthumous works of Huygens, who was im-
bued with the finest taste for the ancient geometry. I have con-
densed the whole in Prop. 19-26. Book II.
6. Tangencies—rıgi inaqur—in two books. Of this tract only
some lemmas were preserved, which enabled the celebrated Vieta in
a great measure to restore it. Some of the cases which had escaped
him were solved by Marinus Ghetaldus; and farther improvements
were made in 1612, by Alexander Anderson of Aberdeen, an an-
cestor of the Gregorys. The general problem occupies the remain-
der of the second book.
7. Plane Loci regi Tómwv exinídwr—in two books. The object
was" To find the conditions under which a point, varying in its
position, is yet confined to trace a straight line or a circle given in
position." This beautiful train of investigation was partly restored
by Schooten in 1650, though after a sort of algebraical form. The
ingenious Fermat succeeded in bestowing greater simplicity on the
subject. But all these attempts have been eclipsed by Dr Simson,
whose treatise De Locis Planis, published at Glasgow in 1749, is a
model of geometrical strictness and elegance. The first 16 propo-
sitions of the third Book include all the principal theorems, which I
have selected with additions.
The six preceding branches of analysis were all the creation of
Apollonius of Perga, the most assiduous and inventive of the Greek
geometers.
8. Porisms-rigì tãv Togicμalavin three books, composed by
NOTES AND ILLUSTRATIONS.
437
Euclid. No trace of these now remains, except some obscure hints
of Pappus, rendered still more perplexed by the corrupt and mutila-
ted state of his text. The subject had long proved an ænigma
which it baffled the efforts of the ablest and most learned mathema-
ticians to unravel. Fermat advanced some steps; but the honour
of completing the discovery was reserved for our countryman Dr
Simson, whose restoration of the Porisms was given to the scientific
world in 1776, in a posthumous volume, printed at the expence of
the late Earl Stanhope. From that work I have extracted what
seemed the best suited to my purpose; and I have likewise availed
myself of the judicious remarks and illustrations of my distinguished
colleague, Professor Playfair. These porisms, with some additions,
are contained in Prop. 18-25. Book III,
The remaining propositions of the third book relate to the subject
of Isoperimeters; which I have treated with the conciseness of the
moderns, without departing, I hope, from the spirit of the ancient
geometry.
Note LII.-Page 241.
This proposition is only a very limited case of the general problem
of inclinations, which occupies inclusively from the 19th to the 25th
Propositions of the Second Book of Analysis. The construction gi-
ven in the text is immediately deduced from the second solution of
Prop. 25. Book II.
Note LIII.—Page 247.
This and the next problem, which has been ascribed to a response
of the Delian oracle, called forth the powers of the ancient analysis,
and transcending the limits of elementary construction, led to
the discovery of some of the higher curves, and essentially contribu
ted to the extension of geometrical science. For the trisection of an
angle, Nicomedes proposed the conchoid, a curve of such a nature,
that every straight line drawn from a given point called its pole has
the same portion intercepted between the curve and a straight line
given in position and termed the directrix. An elegant solution of
the problem is given in Newton's Universal Arithmetic, by means of
an hyperbola whose asymptotes form an angle of 120°.
438
NOTES AND ILLUSTRATIONS.
Note LIV.-Page 250.
In this proposition, I have condensed and endeavoured to simplify
the fine speculations of the Greeks, respecting the duplication of the
cube. The first analysis is that given by Hero, in his Mechanical
Institutions; and the variation of it was proposed by Philo of By-
zantium. The second analysis of the problem was given by Nico-
medes, and the third by Pappus of Alexandria. In the first and
second modes, the solution may be performed by the conchoid; in
the third, it is effected by the cissoid of Diocles, which is so consti-
tuted, that any straight line, drawn from its cusp, has an equal por-
tion intercepted by the curve, and by the generating circle and the
directrix. Menechmus solved the problem in two ways-either by
combining two parabolas-or by combining a parabola with a rect-
angular hyperbola.
Note LV.-Page 252.
Since the angle BDF is half of the angle ABC, and DF: BF ::
R: tanBDF, it follows that, 4R: tanABC:: (AB+BC)²-AC²:
area of the triangle, or, by decomposition, R tan}ABC : :
AB+ BC + AC (AB+BC-AC): area of the triangle. It hence
2
2
follows that, assuming the former notation, T=s(s—AC)tan}ABC.
The same property might also be deduced by comparing Prop. 31.
Book VI. of the Elements with Prop. 12. of the Trigonometry.
In Prop. 20. Book I. of Geometrical Analysis, it may be ob-
served, that the limit occurs when the points F and F' coincide;
in which case HF=FK, HF²=GE²=AG.AH, and consequently
AE+AF, at its greatest contraction, is equal to AG + AH+
2'AG.AH.
1
Note LVI.-Page 288.
This and the six preceding propositions include those cases of the
problem of inclinations which admit of an elementary construction.
The first solution is borrowed from the geometrical analysis of Hugo
de Omerique, and the second from the posthumous works of the cele-
brated Huygens. To solve the general problem would require the
application of the conchoid.
NOTES AND ILLUSTRATIONS.
439
Note LVII.-Page 297.
The first solution of this problem is taken from Dr Simson's post-
humous works. But the second investigation, which is obviously
shorter and more elegant, was communicated to me by my respected
pupil Mr Wildig of Liverpool, to whose ingenuity and accurate taste
in geometrical science I am glad to have this opportunity of bearing
testimony, and to whose judicious remarks this edition is indebted
for various improvements, as it owes much of its typographical cor-
rectness to his obliging and very patient revision of the sheets.
Note LVIH.-Page 311.
This proposition, extended to points in different planes, furnishes a
legitimate demonstration of the remarkable property of projected
masses, which forms, in Newton's Principia, the fourth corollary to the
laws of motion; namely, that of any system of bodies impressed
with uniform and rectilineal motions, the centre of gravity either re-
mains at rest or travels uniformly in a straight line.
Note LIX.-Page 323.
It is easily perceived from the mode of successive construction,
that the centre of the circle which terminates this process, must
likewise be the centre of gravity of the several points. This cu-
rious property is noticed in Huygen's elegant tract, entitled Horolo-
gium Oscillatorium, and furnishes another example of the application
of the principle of the conservatio virium vivarum, which has such
extensive influence in the mutual action of bodies.
Note LX. Page 325.
The porismatic point D is the centre of gravity of particles of
matter situate at A, B and C; for MN being any straight line
drawn through D, the distance CG is equal to the combined dis-
tances AH and BI, and consequently the opposite efforts of those
particles, to turn their plane, must, about the centre D, maintain
every way an exact equipoise.
440
NOTES AND ILLUSTRATIONS.
7
The proposition might easily be extended to any number of points
in the same plane; but it is true universally, if the points only have
a determined position. The writers on Statics, however, have com-
monly assumed, what they were not entitled to do, the existence of
an individual centre of gravity. This fundamental property of mat-
ter is simply and elegantly demonstrated by the ingenious Boscovich,
in his Theoria Philosophic Naturalis, a work of very great and ori-
ginal merit.
Note LXI.-Page 329.
The composition of this problem is readily derived from Note.
XXX.; for CE.CFCG²+GH.GI=CG²+GD²=CD².
Note LXII.-Page 337.
This problem was first proposed by Sir Isaac Newton, for deter-
mining the path of a comet, from four observations made at given
short intervals of time. But unfortunately it was afterwards found
in practice to give uncertain or even erroneous results. This unex-
pected failure led Boscovich to examine closely the circumstances
which might affect the solution, and he discovered that the problem
becomes indeterminate or porismatic, in the very case where its aid
is wanted to guide astronomical observation.
Note LXIII.-Page 343.
All the comparisons in geometry being originally founded on the
properties of the triangle, and thence transferred to other rectilineal
figures, it is evident that we can never reason directly concerning
the circle, which can only be viewed in that respect as a polygon
having innumerable sides. The consideration of limits, more or less
disguised, must therefore unavoidably enter into every investigation
which has for its object the mensuration of the circle.
Note LXIV. Page 350.
The French philosophers have, at the instance of Borda, lately
proposed and adopted the centesimal division of the quadrant, as
easier, more consistent, and better adapted to our scale of arithnie-
NOTES AND ILLUSTRATIONS.
441
tic. On that basis, they have also constructed their ingenious sys-
tem of measures. The distance of the Pole from the Equator was
determined with the most scrupulous accuracy, by a chain of trian-
gles extending from Calais to Barcelona, and since prolonged to the
Balearic Isles. Of this quadrantal arc, the ten millionth part, or
the tenth part of a second, and equal to 39.371 English inches, con-
stitutes the metre, or unit of linear extension. From the metre again,
are derived the several measures of surface and of capacity; and wa-
ter, at its greatest degree of contraction, furnishes the standard of
weights.
It would be most desirable, if this elegant and universal system
were adopted, at least in books of science, Whether, with all its
advantages, it be ever destined to obtain a general currency in the
ordinary affairs of life, seems extremely questionable. At all events,
its reception must necessarily be very slow and gradual; and, in the
meantime, this innovation is productive of much inconvenience, since
it not only deranges our habits, but tends to displace our delicate
instruments and elaborate tables. The fate of the centesimal divi-
sion may finally depend on the continued merit of the works framed
after that model.
Note LXV.-Page 351.
The remarks contained in the preliminary scholium, will obviate
an objection which may be made against the succeeding demonstra-
tions, that they are not strictly applicable, except when the arcs.
themselves are each less than a quadrant. But this in fact is the
only case absolutely wanted, all the derivative arcs being at once
comprehended under the definition of the sine or tangent. To fol-
low out the various combinations, would require a fatiguing multipli-
city of diagrams; and such labour would still be quite superfluous,
because the mode of extending or accommodating the results from
the general principle is so easily perceived.
Note LXVI.-Page 356.
The general properties of the sines of compound arcs may be
derived with great facility from Prop. 22. of Book VI. of thè
1
:
4.4.2
NOTES AND ILLUSTRATIONS.
་
Elements. For, since AB.CD+BC.AD=AC.BD, it is evident
that AB.CD+1BC.}AD={AC.BD; but (cor. 1. def. Trig.)
the semichord of an arc is the same as the sine of half the arc, and
consequently, by substitution, sinABsin CD + sin BC sinABCD=
sin ABC sin BCD. Let AB=L, BC=M, and CD-N; where-
fore ABCDL+M+N, ABC=L+M, and BCD=M+N,
and hence the general result; sinL sinN + sin Msin(L+M+N)=
sin(L+M)sin(M+N), in which L, M and N are any arcs what-
ever. This expression, variously transformed, will exhibit all the
theorems respecting sines. For the sake of conciseness, let the radius
be denoted as usual by 1, and the semicircumference by 7.
1. Put A=M, BN, and let L be the complement of A. Then,
cosAsinB+sinAsin(A+B+2—A)=sin(~—A +A) sin(A+B);
that is, since the sine of an arc increased by a quadrant is the same
as its cosine, sinA cosB+cos A sinB=sin(A+B).
2. Let the arc B be taken on the opposite side, or substitute-B
for it in the last expression, and sinAcosB-cosAsinB=sin(A-B). ·
3. In art 1, for A substitute its complement; then sin(A+B)=
sin(",
—A+ B)=sin(~+A—B)= cos(A-B), and hence cos▲ cosB
+sinA sinB cos(A-B).
4. In art 2, likewise substitute for A its complement, and the re-
sult will become cosA cosB-sinA sinB≈cos(A+B.)
5. In art. 1, let AB, and 2sinAcosA=sin2A.
6. In art. 4, let A=B, and cosA-sin A²=cos2A.
7. In art. 3, let A=B, and cos A+ sin A²=1.
8. Add the formula in art. 1 and 2, and 2sinAcosB=sin(A+B)+
sin(A—B).
9. Subtract the formule of art. 2. from that of art. 1, and 2cos Asin B
=sin(A+B) sin(A-B).
10. Conjoin the formula of art. 3 and 4, and 2cosA cosB=
cos(A+B)+cos(A—B).
11. Take the formula of art. 4. from that of art. 3, and 2.sinAsinB=
cos (A-B)-cos (A+B).
12. In art. 8, let B be the complement of A, and 2sinA²—
sin(A +2,—A)+sin(A—+A)=1—cos2A=vers 2A.
NOTES AND ILLUSTRATIONS.
443
"
2
13. In art. 9, let B be the complement of A, and 2cos A²=
sin(A+,—A)—(sinA— —~+A)=1+cos2A=suvers2A.
14. In art. 5, instead of A substitute its half, and 2sin}A× cos}A=
sinA.
15. In art. 6, likewise substitute the half of A for A, and
(cos&A)²—(sin} A )² — cosA.
16. In art. 12, for A substitute its half, and 2(sinA)²—1—cosA,
or sin§A= √ (½( 1—cosA))=√½versA.
17. Make the same substitution in art. 13, and 2(cos}A)²=
1+cosA, or cos§A=√(}(1+cosA))=√‡suversA.
18. In art. 8, transform A and B'into A+B and A-B, and con-
sequently, for A+B and A-B, substitute 2A and 2B; then
2sin(A+B)cos(A—B)=sin2A+sin2B, or sin (A+B)cos(A—B)=
}(sin2A+sin2B).
19. Make the same transformation in art. 9, and 2cos(A+B)
sin (A — B) = sin2A sin2B, or cos(A + B)sin(A — B)=
(sin2A-sin2B).
20. Repeat this transformation in art. 10, and 2ros (A+B)
cos (A — B) = cos2A + cos2B, or cos(A + B)cos(A — B) =
(cos2A+cos2B ).
21. The same transformation being still made in art. 11,
2sin(A+B)sin (A-B) cos2B-cos2A, or sin (A+B)sin(A—B)=
(cos2B-cos2A).
22. Suppose L=N=B, and M-A-B; then the general expres-
sion becomes sin B²+sin(A—B) sin(A+B)=sinA², or sin(A+B)
sin (A—B)=sinA²—sinB².
23. Instead of A in the last article, take its complement, and
2
2
sin(—— A+B)sin(—— A — B) = cos A² — sinB², or cos(A — B)
cos(A+B)=cos A²—sin B²,
24. Compare art. 21. with 22, and (cos2B-cos2A)=sinA²—
sin B2,
25. Comparing likewise art. 20. with 23, and (cos2A+cos2B)=
cos A²-sin B².
26. Resolve the difference of the squares in art. 22. into its fac-
tors, and sin(A+B)sin (A—B)=(sinA+sinB) (sinA—siu B).
:.
}
444
NOTES AND ILLUSTRATIONS.
27. Make a similar decomposition in art. 23, and cos(A+B)
ćos (A—B)=(cos A+sinB) (cosA—sinB).
24. In art. 18, instead of A and B take their halves, and
sinA+sinB=2sin}(A+B)cos}(A—B).
25. Make the same change in art. 19, and sinA
2sin}(A-B)cos}(A+B).
sin B
26. Change likewise art. 20, and cosB+cos A = 2cus}(A+B)
cost (A-B).
27. Do the same thing in art. 21, and cosB—cos A=2sin}(A—B}
sin (A+B).
From Note XXVII: a very simple expression may be derived for
the sum of the sines of progressive arcs. Suppose the diameter AO
were drawn; then BE+CF+ DG=HG=HO+DO, or 2sinAB+
2sinAC+2sinAD HO+sinAD, and sinAB + sinAC+sinAD=
¿HO+{sinAD=}AO.tanBAO+‡sinAD. Wherefore, in general,
sin a+sin2a+sin3a ..... sin na=vers na.cota + sin na. Hence
the sum of the sines in the whole semicircle is cota. Thus, if
the sines for each degree up to 180°, the radius being unit, were
added together, the amount would be 114,58866.
Note LXVII-Page 358.
On examining the formation of the successive terms of the first
and second tables, it will appear that the coefficients are certain mul-
tiples of the powers of 2, whose exponents likewise at every step
decrease by two. It is farther manifest that if 1, A, B, C, &c.
1, A', B', C', &c. and 1, A', B', C', &c. denote the multiples corre-
sponding to the arcs n.a, n↓ 1.a, and n-1.a; then A + 1 = A',
B+A=B', C+B'=C', &c. Whence the values of A, B, C, &c.
are determined, either by the method of finite differences,
adopting the appropriate notation, or from the theory of func-
tions. Thus, in the first table,
the first table, ▲A=1, and A
A=n N — 2;
n-3 n-4
2
AB=A'=n—3, and B=-
C=
n—4.n-5.n—6.
2.3
n-4 n-5
; AC=B'=
"
and
Wherefore in general
NOTES AND ILLUSTRATIONS.
445
(1.) Sin na=2″-I. CM-IS
's — n—2,2″--3--35 +
2—3.1— 4
2
Q7--5 M--55--
n—4.n—5.n—6
2.3
2~7 (78 + &c.
(2.) Cos na—2—¹‚c”—µ.2″--³.c²−2+
n.n-3
2
97--5CM--4-
n.n—4.n— 5
2.3
2n~~7.C~+ &c.
The third and fourth tables are evidently formed by multiplying
constantly by 2cos 2a or 2—4s², and subtracting the term preceding;
or the multiplication by 4s² produces the second differences of the suc
cessive quantities. Hence in the former, ▲▲A≈4n", ^^B=4A", &c.;
N.N—I.N+I
wherefore ▲ An+1.n+1, and A=
AB=2(-
and B =
2.3
2. n+2. n+1. n+3) _ n + 1. n + 1. n—1. n +3
3
N.N—I.NI. n-3. n + 3
2.3.4.5
3.4
But in the fourth table,
^^A=4, ▲▲В=4A", AAC'=4B"; and consequently AA=2n+2,
n. n+I. n+2
and A= ; AB=Σ(2.n+2.n+2)=
and B =
3
N.
n². n—2.n+2
2.3.4
Wherefore in general,
(3.) Sin na=n.S—N.
822 I
2.3
n²-I n²-9 n²-25
၇
n²-1 n²-985.
2.3
4.5
-s² + &c.
2.3 4.5
6.7
na
n² n².
n² 4
n² n²-4 n² —16
2
+
2 3.4
5.6
-s°+,&c.
(4.) Cos na=1—
2 3.4
In the fifth and sixth tables, the coefficients are evidently the
same as those of the power of a binomial, only proceeding from
both extremes to the middle terms. Hence, according as n is odd
or even,
(5.) 2″-1, sin a"=±sin naĦn. sin(n—2)a±n.
NI
sin(n—1)α——
N—I N-2
9.
sin(n-6)a &c.; and
2
ર
3
446
NOTES AND ILLUSTRATIONS.
2″-1 sin aª= cos na■, cos(n—2)a ±n.
n-I
·cos (n=4)a
કર
N—I N▪
——2
N.
cos (n—6)α, &c.
B
2
3
Again,
NI
(6.) 2″-1 cos a”—cos na +n. cos(n—2)u +n.
cos (n—4a) +
2
і
N—I N-2
n.
.cos (n-6)a, &c.
2
3
In these three expressions, half the last term, which corresponds
to the middle in the expansion of the binomial, is to be taken, when
n is an even number.
It will be satisfactory likewise to subjoin an investigation of the
sine of the multiple arc, as derived from the Theory of Functions.
It appears from inspecting the successive formation of the sines
of the multiple arcs, 1. that the odd powers only of s occur; 2. that
the coefficient of the first term is only n, and the other coefficients
are its functions of third, fifth, &c. orders; and 3. that since, in the
case when "=1, the rest of the coefficients evidently vanish, those
coefficients in general, as affected by opposite signs, must in each
term produce a mutual balance.
///
Let therefore sin nu=n.s + n.s³ + n.ss &c.; where s denotes the
1
/////
sine of the arc a, and n, n, n, &c. the successive odd orders of the
functions of n. It is evident, from (Prop. 3. cor. 2. Trig.) that, by
substitution
3
((n+1)+(n−í))s+ ((n+"')+(n—“"))) s³+ ((n+1)+(n−1)) s³+ &c.
/////
= 2√ (1—5²) sin na = (2— s² — 454, &c.) (ns + ns³ + ns³, &c.)
///
11/11
=2ns + (2n−n)s³ + (21——4n)ss, &c. Now, equating corre-
sponding terms, and rejecting the powers of s, we obtain these ge-
neral results:
1
ፀሀ
“በሀ
m
2n'=2n'; (n+1)+(n+1)=2n−n; (n+i)+(n−1)=2n−n—‡n.
NOTES AND ILLUSTRATIONS.
447
It remains hence to discover the several orders of the functions
of n.
1. The equation 2n'-2n' contains a mere identical proposition;
but other considerations indicate that n must always denote the first
term, or that the first function of n is n itself.
!!!
ii)
2. The equation (n+1) + (2—1) = Qu
n fixes the conditions
of the third function of n, which, from the nature of the
relation, is obviously imperfect, and wants the second term.
Put, therefore, n" = an³ + Br; and, by substitution, 2n3 +
Dan+2ẞn=2an³+2ßn-n. Equating now the corresponding terms,
and 6α-1, or e-; but
; but &+B=0, and therefore ß+.
10
Whence n±n³+&n=
+-
n²
2.3
4511
ባቡ ///// //
3. Again, in the third equation, (n+i)+(n−1)—2n—n—‡n',
Wit
substitute nan³ +ßn³+yn, and the conditions of the fifth order of
the function of a will be determined by this compound expression:
Qans + (20x + 28)n³ + (10 +63 +2y)n = 2an³ + (2B + 1 ) n³
+ (22
Equate the corresponding terms, and
ž) n.
20α +26= 2ß + ž, or α=
120
1
2.3.4.5
In like manner,
I
10α +68+2y= 2y-, and. 8=-=
-10
; but a + ß + y = 0, whence y
2.3.4.5
9
2.3.4.5
ns — 10n³ +9n
/////
N
[ n²
n²-9
=R.
N.-
Collectively, therefore, n= 2.3.4.5
Whence, resuming all the terms, sin na = ns — N.
2.3
4.5
N
n².
+
2.3
1.
n² ~ 1 n² —9
2.3
-&c. as before.
4.5
From the expression for the sine of a multiple arc, may be de-
duced the series for the sine of any arc, in terins of the arc itself,
A
and conversely. Let na A, and therefore a ; if n be sup-
12
posed indefinitely great, then a must be indefinitely small, and
consequently in a ratio of equality to s. Whence, substituting
448
NOTES AND ILLUSTRATIONS.
A
A for na, and for s in the general expression, there results,
n
sin A=A—
n²—I A³
n².
+
n² —I n² —9 As
n²—9 A³
2.3 n
&c.
2.3 4.5 N4
But n being indefinitely great, the composite fractions.
n
-9
na
&c. are each in effect equal to unit, which forms their ex-
treme limit. Consequently, assuming that modification,
sinA A
A3
As
+
&c.
2.3 2.3.4.5
Again, putting a=A and s=S, suppose n to be indefinitely small,
and sin na=na=nA; whence, by substitution,
nAnS-n.
n².
2.3
S³+n.-
A=S
S³ + n.
2.3
n³—t n²—9
2.3 4.5
n² — 1 n²-9
2.3 4.5
S5--, &c. and
SS-&c.
But, if n vanish from all the terms, the series will pass into a
simpler form.
1
A =S+ s+
2.3
1.9
2.3.4.5
ss +
1.9.25
2.3.4.5.6.7
S7+, &c.
By a similar investigation, the series for the cosine of an arc is
likewise found.
A$
+, &c.
A² A+
1,2 2.3.4 2.3.4.5.6
CosA=1 +
These series' are very commodious for the calculation of sines,
since they converge with sufficient rapidity when the arc is not a
large portion of the quadrant. Though the method explained in the
text is on the whole much simpler, yet as the errors of computation
are thereby unavoidably accumulated, it would be proper at intervals
to calculate certain of the sines by an independent process.
The series' now given furnish also various modes for the rectifi-
cation of the circle. Thus, assuming an arc equal to the radius,
its sine is, I
1
}
is, 1
2
G
1
1
2.3 2.3.4.5
+
&c. .841471, and its cosine
i+ -&c. .440302 But that arc evidently approaches
2.3.4
}
NOTES AND ILLUSTRATIONS.
449
to 60°, of which the sine is .866025, and the cosine .500000.
Wherefore (Pr. 1. Trig.) the sine of the difference of these two arcs
is.866025X.540302-.841471xX.500000=.0471S, and consequent-
ly, by the series, that interval itself is .0472. Hence the length of
the arc of 60° is 1.0472, and the circumference of a circle which
has unit for its diameter is 3X 1.0472-3.1416; an approximation
extremely commodious.
Note LXVIII. Page 360.
This may be otherwise demonstrated from the corollaries to the pro-
position contained in Note XLVI.
Let AB and BC, or BC', be two arcs, of which AB is the
greater; make AD, or AD', equal to BC,
and apply the respective tangents. Be-
cause OAE is a right-angled triangle, and
OG', OF, are drawn, making equal angles
with OA and OE, it follows, that OA²-
AE.AG' : OA²:: EG': AF, and conse-
quently R2-tanAB.tanBC:R2: :tanAB+
tanBC: tan(AB + BC.). Again, since
OG and OF' make equal angles with
OA and OE, it is evident, that ÓA+
AE.AG: OA':: EG: AF', and hence
R2+tanAB tanBC: R:: tanAB-tan BC:
tan(AB-BC).
Note LXIX. Page 360.

E
B
D
1
G
The radius being expressed by unit, the sum of the tangents of
the angles of any triangle, is equal to the number arising from their
continued product. For, let A, B, and C, denote the several angles
of the triangle; and since two of these, such as A and B, are supple-
mentary to the remaining one C, the tangent of A+B is the same
(schol. def. Trig.) as that of the third angle in an opposite direction.
tan A-tanB
Whence
1–tan A.tanB
tan C, and therefore tan A + tan B =
-tanC+tanA tanB tanC, or tanA+tanB+tanC≈tanAtan B tanC:
F f
450
NOTES AND ILLUSTRATIONS.
1
Note LXX. Page 360.
The properties of the tangents are easily derived from those of the
sines.
1, TanA+tanB=
sinB sinAcosB+cosAsinB
sin A
+
cosA
cos B
cos AcosB
sin(A+B)
(art. 1. Note LXVI.)
cos AcosB
2. Change the sign of B in the last article, and tanA-tanB-
sin(A-B)
CosAcosB
3. Instead of A and B in art. 1. substitute their complements and
cot A+ cotB
sin(A+B)
sinAsinB
4. Make the same substitution in art. 2, and cot B-cotA=
sin(A+B)
sin(A-B)
sin Asin B
5. Tan(A+B)= cos(A+B)=(art. 1. and 4. Note LXVI.)
sinAcosB+cusAsin B
cosAcosB-sinAsinВ'
which, being divided by cos A cos B or
tanA+tanB
cot B+cot A
sinA sin B, gives tan(A+B)=tanA tanB-cot B cot A-1'
6. Change the sign of B in the last article, and tan(A-B)—
tan Atan B
1+tunA tan B
cot B-cot A
cot B cotA l
7. Divide the expression in the first article by that in the second,
and
sin(A+B)___tan AtanB
sin(A—B) tanA—tanB
cotB+cot A
cot B-cot A
8. In the last article, change the sign of B, and instead of A take its
complement, and
cos(A+B) cut B-tanA__cot A-tanB
cos(A-B)cot B+tanA cot A+tan B
9. Divide the expression of art. 12. in Note LXVI. by that of
1-cos2A QsinA² sin A
sin2A 2sinAcosAcosA
art. 5., and
:tanA.
10. Divide the expression of art. 5. in the same Note, by that of
art. 13., and
sin2A
1+cos2A
2sin Acos A sin A
2cosa²
=tanA.
COSA
NOTES AND ILLUSTRATIONS.
451
11. Multiply the expressions of the two preceding articles, and
1-cos2A
1+cos2A
tanA², or tanA=
tanA=√
1—cos2A
1+cos2A
1
cos2 A
12. Decompose the expression in art. 9., and tanA= sin2A sin2A
=cosec2A-cot2A.
13. In the last article, change A into its complement, and cotA=
cosec2A+cot2A.
14. Subtract the last expression from the one preceding it, and
tan A-cot A=-2cot2A, or tanA=cot A-2cot2A.
15. In art. 9. 10. and 11. for 2A and A, take A and A, and
I-COSA
tan }A = sinA
sin A
1+cosA
I-COSA
I+COSA
16. Multiply the expressions of art. 1. and 2., and (tanA+tanB)
(tan A-tan B)=tan A²-tan B²
2
sin(A+B) sin(A—B)
cos A² cosB²
17. Multiply the expressions of art. 3. and 4., and (cotB+cot A)
(cot B—cot A) — cotB²—cotA³— sin(A—B) sin(A+B)
sin A² sin B²
sin A+ sin B
sinA sinB
18. Divide art. 28. of Note LXVI. by art. 29,, and
2sin(A+B) cos}(A-B) _tan (A+B)
2cos}(A+B)sin‡(A—B)—tan}(A—B)'
19. Divide art. 30. of the same Note by art. 31.,
=
2cos(A+B) cos(A-B)_cot (A+B)
2sin(A+B) sin(A—B)
tan(A—B)*
cos B+cos A
and
cosВ cosA

Since by art. 14. cot A-2cot2A➡tanA, if the arc A and its com.
pound expression be continually bisected, there will arise:
cotA cot Atan}A
cot A-cot Atan A
cot} A-cotA=}tan} A
&c. &c. &c.
Wherefore, collecting these successive terms, and observing the ef-
fects of the opposite signs, the general result will come out,
1
A
-cot
2* 2*
A
-tan-
2". 9
cotA=}tan}A+}tan}A+}tan}A ... +
1
r f 2
452
NOTES AND ILLUSTRATIONS.
If n be supposed to become indefinitely large, then
1
A 1
·cot
=
Q"
2" 2"
A
is ultimately
1 1 2" 1
2 A 2 A
or
; and con-
A
tan
1
2"
Qr
sequently ==cotA+}tan‡A+{tan}A+}tan}A+ŕtan&zA+ &c.
This neat and very simple investigation is given in the third edi-
tion of Cagnoli's Trigonometry, printed at Paris in 1808, and forming
the completest treatise which has yet appeared on the subject. It
was also, and nearly about the same time, communicated by my
friend Mr Wallace of Great Marlow, a geometer of the first order,
to the Royal Society of Edinburgh; another instance of that acci-
dental coincidence which has occurred so frequently in the history of
mathematical discovery.
Note LXXI.-Page 361.
It is obvious that the terms of the series for the tangent of the
multiple arc are formed out of the coefficients of the powers of a bi-
Wherefore,
nomial.
nt-n.
NI N
2
-2
פא
·t³ + &c.
3
(7.) Tan na=
I-N.
27---- I
2
+1.
N—I n—2 n—3
2 3 4
Hence also,
(8.) Sin na=c"(nt—n.
-14 &c.
N—I N~2
13+n.
3
N-I N 2 N-3⋅n—4
2 3
—
15 ~ &c.)
4
5
and
ท
I
N—I n—2 n—3
(9.) Cos na = c”(I—n.
2
2
3 4
n.
t°— &c.)
N—I n▬▬2 n—3 n—Ą N—5
2 3 4 5 6
10
The series for the tangent in terms of the arc, is easily derived,
by the theory of functions, from the expression of the tangent of
2t
the double arc.
Since tan2a=
= 2t + 2ts + 215 + &c.
Putta
Aa³ + Ba³ + &c. and, by substitution, tan2a=
2a + 8Aa³ + 32Ba³ + &c. = 2a + (2A + 2) a³ +
NOTES AND ILLUSTRATIONS.
453
2
(2B6A + 2)as+, &c: Equating, therefore, the corre-
sponding terms, we obtain, 8A 2A + 2, or A, and
32B = 2B + 6A + 2, or 30B = 4, and B&. Whence, in
general, tan aa+a³+a³, &c. Again, revert this series and
a=t———ť³ +} 15—4t7+&c.
C.
The last series affords the most expeditious mode for the recti-
fication of the circle. Assume an arc a, whose tangent t is one-fifth
part of the radius, and tan4a=
(Prop. 5. Trig.) tan(4a—45°)—
1
239
4t-413 120
1—6t²+t^—119
; consequently
:.004,184,100,418. Wherefore,
computing the terms of the series, a=.197,395,559,850, and
4a=.789,582,239,400. In like manner, we find 4a 45°=
.004,184,076,000, and hence the difference between these values,
or .785,398,1634 exhibits the length of the octant; which number,
multiplied by 4, gives 3,1415926536 for the circumference of a cir-
cle whose diameter is 1.
The next proposition, with its corollaries, would furnish a simple
quadrature of the circle. The sine of a semiarc being equal to half
the chord, it follows that the ratio of an arc to its chord is com-
pounded of the successive ratios of the radius to the cosines of the
continued bisections of half that arc. Assuming therefore the arc
of 60°, whose chord is equal to the radius, the logarithm of the ra-
tio of the circumference of a circle to its diameter will be thus com-
puted:
Arith. comp. log. Cos 15°
Cos 7° 30'
= .0150562219
= .0037314339
Cos 3° 45'
Cos 1° 52′ 30″
Cos 0° 56' 15"
Cos 0° 28' 7"
One-third of the last term.
Logarithm of 3,
.0009308547
= .0002325891
= .0000581395
= .0000145344
= .0000048448
.4771212547
.4971493730, which
454
NOTES AND ILLUSTRATIONS.
exceeds only by 3 in the last place the logarithm of 3,141592654.
As the successive terms come to form very nearly a progression that
descends by quotients of 4, the third of the last one is, for the rea-
son stated in page 363, considered as equal to the result of the con-
tinued addition.
Note LXXII.-Page 367.
An elegant mode of forming the approximate sines corresponding
to any division of the quadrant, may be derived from the same prin-
ciples: For the successive differences of the sines of the arcs A—B,
A, and A+B, are sin A-sin (A-B,) and sin (A + B)-sinA; and
consequently the difference between these again, or the second dif-
ference of the sines, is sin (A+B) +sin (A—B) —2sinA = (Prop. 3.
cor. 3. Trig.)—2vers B sinA. The second differences of the progres-
sive sines are hence subtractive, and always proportional to the sines
themselves. Wherefore the sines may be deduced from their second
differences, by reversing the usual process, and recompounding their
separate elements. Thus, the sines of A-B and A being already
known, their second and descending difference, as it is thus derived
from the sine of A, will combine to form the succeeding sine of
A+B, which is—2versBsinA + ( sinA—sin(A—B) ) + sinA. It
only remains then, to determine, in any trigonometrical system, the
constant multiplier of the sine, or twice the versed sine of the com-
ponent arc. Suppose the quadrant to be divided into 24 equal parts,
each containing 3° 45', or 225'. The length of this arc is nearly
22 1 11
7 48 168
and consequently twice its versed sine =
11
(168) 2
(1) in approximate terms. If the successive sines, corresponding
233
to the division of the quadrant into 24 equal parts, be therefore con-
tinually multiplied by the fraction , or divided by the number
1
233
233, the quotients thence arising will represent their second diffe-
rences. But, since 233 is nearly equal to 225, or the length in mi-
nutes of the primary or component arc, and which differs not sensi-
bly from its sine,-this last may be assumed as the divisor, the small
aberration so produced being corrected by deferring the integral quo-
tients. In this way the following Table is constructed :
NOTES AND ILLUSTRATIONS.
455

Parts of the
quadrant.
Arcs.
Sines. 1st Diff. 2d Diff.
Arcs.
183
225'
225
224
2
450'
449
222
675'
671
219
183
3° 45'
2
7° 30'
11° 15'
4
900/
890
215
4
15°
5
1125/
1105
210
5
18° 45'
6
1350/ 1315
205
5
22° 30′
7
1575/
1520
199
6
26°
15/
8
1800/ 1719
191
*
9
2025/ 1910
183
78
30° O'
33° 45′
10
2250' 2093
174
9
37° 30′
11
2475' 2267
164
10
41° 15′
12
2700' 2431
154
• 10
45°
0
13
2925′ 2585
143
11
48° 45'
14
3150′ 2728
131
12
52° 30′
15
3375' 2859
119
* 12
56° 15'
16
3600' 2978
106
13
60°
17
3825' 3084
93
13
63° 45'
18
4050' 3177
79
14
67° 30′
19
4275' 3256
65
14
71° 15′
20
4500' 3321
51
14 75° 0'
21
4725' 3372
37
# 14
78° 45′
22
4950′ 2409
22
15 82° 30′
23
5175' 3431
7
15
Stº 15'
24
5400/ 3438
0
15
90°
The number 225, which expresses the length of the component
arc, and consequently represents very nearly its sine, is here employ-
ed as the constant divisor. Thus, 225, divided by 225, gives a quo-
tient 1, and this, subtracted from 225, leaves 224, which, being join-
ed to 225, forms 449, the sine of the second arc. Again, 449 divi-
ded by 225, gives 2 for its integral quotient, which taken from 224,
leaves 222; and this, added to 449, makes 671, the sine of the third
In this way, the sines are successively formed, till the qua-
drant is completed. The integral quotients, however, are deferred;
that is, the nearest whole number in advance is not always taken:
arc.
38
Thus the quotient of 1315 by 225, is 5 which approaches near-
45'
er to 6, and yet 5 is still retained. These efforts to redress the er-
rors of computation are marked with asterisks.
It should be observed, that each of the three composite columns
really forms a recurring series, In the second quadrant, the first dif-
·
456
NOTES AND ILLUSTRATIONS.
f
ferences become subtractive, and the same numbers for the sines are
repeated in an inverted order. By continuing the process, these sines
are reproduced in the third and fourth quadrants, only on the oppo-
site side.
Such is the detailed explication of that very ingenious mode, which,
in certain cases, the Hindu astronomers employ, for constructing the
table of approximate sines. But, ignorant totally of the principles
of the operation, those humble calculators are content to follow blind-
ly a slavish routine. The Brahmins must, therefore, have derived
such information from people farther advanced than themselves in
science, and of a bolder and more inventive genius. Whatever may
be the pretensions of that passive race, their knowledge of trigono-
metrical computation has no solid claim to any high antiquity. It
was probably, before the revival of letters in Europe, carried to the
East, by the tide of victory. The natives of Hindustan might receive
instruction from the Persian astronomers, who were themselves taught
by the Greeks of Constantinople, and stimulated to those scientific
pursuits by the skill and liberality of their Arabian conquerors.
:
The same principles lead to an elegant construction of the approxi-
mate sines, entirely adapted to the decimal scale of numeration, and
the nautical division of the circle. Suppose a quadrant to contain
16 equal parts, or half points; the length of each arc is nearly
22 1 11
7 32 112
>
and consequently twice its versed sine is (), or,
in round numbers,
1
103
112
It will be sufficiently accurate, therefore, to
employ 100 for the constant-divisor. The sine of the first being like-
wise expressed by 100, let the nearer integral quotients be always re-
tained, and the sine of the whole quadrant, or the radius itself, will
come out exactly 1000. The first term being divided by 100 gives
I for the second difference, which, subtracted from 100, leaves 99
for the first difference, and this joined to 100, forms the second term.
Again, dividing 199 by 100, the quotient 2 is the second difference,
which, taken from 99, leaves 97 for the first difference, and this add-
ed to 199, gives the third term. In like manner, the rest of the
terms are found,
NOTES AND ILLUSTRATIONS.
457
Half
points.
Correct
Arcs.
Sines.
1st Diff. 2d Diff. Excess.
Sines.
123
5° 37'
100
99
1
3
97
11º 15'
199
97
2
4
195
16° 52
296
94
3
5
291
4 22° 30'
390
90
4
6
384
5678
28°
7
480
85
5
7
473
33° 45'
565
79
6
8
557
39° 22
641
73
6
9
635
8 45° 00'
717
66
7
10
707
9 50° 37
783
58
8
9
774
10
56° 15'
841
50
S
8
833
11 61° 52′
891
41
9
7
884
12
67° 30′
932
32
9
6
926
13
73°
73
964
22
10
5
959
14
78° 45'
986
12
15 84° 22
998
20 20
10
4
982
3
995
16
90° 00'
1000
The errors occasioned by neglecting the fractions accumulate at
first, but afterwards gradually diminish, from the effect of compen-
sation. The greatest deviation takes place, as might be expected,
at the middle arc, whose sine is 707 instead of 717. Reckoning the
error in excess as limited by 10, and declining uniformly on each
side, the correct sines are finally deduced. The numbers thus ob-
tained seldom differ, by the thousandth part, from the truth, and are
hence far more accurate than the practice of navigation ever requires.
This simple and expeditious node of forming the sines is not merely
an object of curiosity, but may be deemed of very considerable im-
portance, as it will enable the mariner, altogether independent of the
aid of books, to the loss of which he is often exposed by the hazards
of the sea, to construct a table of departure and difference of latitude,
sufficiently accurate for every real purpose.
Note LXXIII-Page 367.
In trigonometrical investigations, it is often requisite to determine.
the proportion which the difference of an arc bears to that of its re-
lated lines. With this view, let A denote the increment or finite
difference of the quantity to which it is prefixed.
1. In art. 29. of Note LXVI. change A into A+AA, and B into A ;
then will
458
NOTES AND ILLUSTRATIONS.
AsinA 2sin▲Acos(A+▲A).
1
2. Make the same change in art. 31. of that Note, and
AcosA-2sin▲Asin(A+▲A).
3. In art. 2. of Note LXX. let a similar change be made, and
Atan A
sin▲A
cos Acos(A+AA)
4. Do the same thing in art. 4. and
AcotA=
sin ΔΑ
sin Asin(A+▲A)
5. In art. 22. of Note LXVI. make a like substitution, and
▲sinA²=sin▲Asin(2A+▲A).
6. Let the same change be made in art. 23., and
▲cos A²——sin▲Asin(2A+▲A).
7. Do the same thing in art. 16. of Note LXX. and
Atan A²=
sin▲A(sin2A+SA)
cus A*cos(A+AA)2
8. Lastly, let a similar change be made in art. 17. of that Note, and
sin▲A(sin2A+AA)
Acot A²=
sın A²sin(A+AA)²
If the differences be conceived to diminish indefinitely and pass in-
to differentials, these expressions, in coming to denote only limiting
ratios, will drop their excrescences and acquire a much simpler
form. Thus, adopting the characteristic d, since the ratio of an arc
to its sine is ultimately that of equality, and the sine of A÷dA
may be considered as the same with the sine of A; it follows, that
1. dsinA=+cosAdA.
2. dcos Asin AdA.
dA
3. dtanAt
cosA²
dA
4. dcot A =
sin A²
5. dsinA
+2sinA cos AdA.
6. dcosA'2sin A cos AdA.
2tan AdA
7. dtanA²=+
cos A2
2cot AdA
8. dcot A2:
sin A
NOTES AND ILLUSTRATIONS.
459
Note LXXIV.-Page 379.
Since, by Note LXXIII, d sin AcosAdA, or the variation of the
sine of an arc is proportional to its cosine; it follows that, near the
termination of the quadrant, the slightest alteration in the value of a
sine would occasion a material change in the arc itself. Again,
or the variation of the tangent
from the same Note, dtanA=
dA
COSA²'
is inversely as the square of the cosine, and must therefore increase
with extreme rapidity as the arc approaches to a quadrant.
Note LXXV.-Page 379.
It is convenient to reduce the solution of triangles to algebraic
formula. Let a, b and c denote the sides of any plane triangle, and
A, B, and C their opposite angles. The various relations which
connet these quantities may all be derived from the application of
Prop. 11.
1. Cos A=
¿² fc² —a²
2bc
2. But, since (art. 16. Note LXVI.) sinдA-(1-cosA), it follows,
Qbc—b*—c² taz a²—(b—c)²
by substitution, that sin A*≈
(a+b—c) (a−b+c)
4bc
4bc
and therefore, s denoting the semiperimeter,
4bc
(s——b) (s—c)
Sin A²=
which corresponds to Prop 14.
bc
3. Again, because (art. 17. Note LXVI.) cos A²=}(1+cosA), by
substitution, cos3A2-
2bc+b²+c²—a² (b+c)² —a²
4bc
46c
((b+c)+a ) ( (b+c)—a), and consequently
46c
Cos}A²=5(s—^); which agrees with Prop. 13.
bc
4. The second expression being now divided by the third, gives
Tan A=
(s—b) (s—c)
s(s—a)
, corresponding to Prop. 12.
460
NOTES AND ILLUSTRATIONS.
}
These are the formula wanted for the solution of the first case of
oblique angled triangles. To obtain the rest, another transformation
is required.
5. It is manifest that sinA²-1-cos A²=
4b³c² — (b²+c²—a²)²
462c2
and consequently, by Note XLIX., sinA¹=
4T2
2T
62629
or sin A=
bc
For the same reason, sinB
2T
— , and hence
sin A
a
which cor-
ac
sin B
b
responds to Prop. 9.
6. Again, by composition,
sin A-sinB
sinA+sinB a+b²'
a
=
b
and therefore, by
art. 18. Note LXX.,
a
==
b__tan (A—B), which agrees with Prop. 10.
a+btan(A+B)'
7. Lastly, transforming the first expression, there results,
a = √(b²+c²—2bc cos A)=((bc)²+2bc versA)
=√((b+c)² -2bc(1+cosA)).
The preceding formula will solve all the cases in plane trigonome-
try; but, by certain modifications, they may be sometimes better a-
dapted for logarithmic calculation.
b
1
S. Divide the terms of art. 6. by a, and
tan} (A—B)
tan (A+B)
a
1+
a
h
let
= tan x, and tan(A—B)
a
tun (A+B)
1-tan x
1+tan x
(art. 6. Note LXX.)
b
tan (45°-x).
Wherefore
tan x, and tan(45°—x) =
a
tan C tan (A-B)=tan C cot(C+B)=
tan C(col(C+A)).
9. Again, from art. 7. a=((b-c)²+2be vers A)=
2bc
(b—6) √(1+ versA); consequently find tan x=
(b−c)²
✔2bc
✔ vers A =2-
VDC
✔ bc
-sin}A, and a = (b—c) sec x=
b-c
COS X
10. But the expression in art. 1., by a different decomposition,
NOTES AND ILLUSTRATIONS.
461
gives a =((b+c)²— 2bc suversA) )=(b+c)√(1-
wherefore find sin x =
✔ 2bc
b+c
a = (b+c) cos x.
2bc
(b+c)²
(b+cja suversA);
√suversA=2 cos A, and
Nbc
b+c
11. Other expressions are likewise occasionally used. Thus, by art. 1,
2bc, cosA = b² of c²—a², or c² —2bc, cosA = u²
or c² —2bc, cosA — u² — b², and, solving
this quadratic, we obtain cb cosA±√ (a² · b² + b²cosA²) =
b cos A±√ (a²-b²sin A²), or c=b cos A±√((a+b sinA) (a−b sinA)).
When two sides and an angle opposite to one of them are given, the
third side is thus found by a direct process.
12. From art. 5, c≈a but C being a supplementary angle,
sinC
sin A
its sine is the same as that of A+ B, and consequently
sinAcos B+cos Asin B
c=a
sin A
sinC
ca
). By a similar transformation,
sinc
a
sin(B+C')=ª (sinBcosC+cosBsinC
13. Lastly, from art. 3. of Note LXX, cotA+cotC=
=
cusB+sinBcotC
sin( A + C)
sin AsinC
sin B
sin AsinC a sinC'
b
b
and therefore cotA=
-cotC=
a sinC
b-a cosc
a sin C
a sin C
or tanA =
ba cos C
If the angle A be assumed equal to 90°, the preceding formula
will become restricted to the solution of right-angled triangles.
14. From art. 1, cosA=
b²+c²—a²
26c
; whence, a²b² + c²,
which expresses the radical property of the right angled triangle.
sin B
b
15. From art. 5,
>
and consequently sinB=
sin A
a
b
a
which
corresponds with Prop. 7.
16. Again, from the same article,
b
fore tanB=
=cotC.
C
ს
sin B
C
sinc
sin B
cos B'
and there-
162
NOTES AND ILLUSTRATIONS.
For the convenience of computing with logarithms, other expres
sions may be produced.
17. Thus, from art. 14., ba²-c², and hence
b=√((a+c)(a−c)).
c²
b
18. Since a²=b²(1+1), put ==tan x, and a=b(sec x) = 105
COS I
19. Lastly, because ¿²=a”(1— €²)
), put ——=sin x, and b=a.cos x.
a
Besides the regular cases in the solution of triangles, other com-
binations of a more intricate kind sometimes occur in practice. It
will suffice here to notice the most remarkable of these varieties.
20. Thus, suppose a side, with its opposite angle and the sum or
difference of the containing sides, were given, to determine the tri-
b sin A
sin B
angle. By art. 5., a=
Ɑ=
c sin A
sinC
and therefore
b sinA+c sinA __(b+c)sin(B+C)— (art. 5. and 18. Note LXVI.)
sinB+sinC sin B+sinC
(b+c)2sin}(B+C)cos}(B+C)
(b+c)2sin(B+C) cos (B+C) (b+c)cos(B+C)
2sin(B+C)cos(B-C)

cos(B-C)
But cos}(B+C)=sin}A, and hence cos(B~~C)—(b+c)sin}A
a
and the difference of the supplementary angles B and C being kuown,
these angles themselves are hence found.
In like manner, it will be found that sin}(B—C)
(bc)cos&A
a
21. Let a side with its adjacent angle and the sum of the other
sides be given, to determine the triangle. By art. 4. tunA²=
Sa.s
S- ·a.s—c
and tan B²=
sub.sc
5.81
5.5-6
2
; whence tan A² tan } B²=
s—a.s—b.(s—c)², and consequently tan Atan}B=
Sα.s-b.s²
or cot+B=tan} A (a+b)+c
(4+0)_c
s—c (a+b)—c
S
(a+b)+c'
Again by art. 1, 2bc cos Ab²+c²-a², or a²-b2c22bc.cosA,
and adding 2ab+262 to both sides, a²+2ab+b²—c² 2ab+2b³——--
2bc.cos▲, or (a+b)² = 2b (a + b—c.cos A); whence ((a+b)+c)
1
NOTES AND ILLUSTRATIONS.
463
1

((a+b)—c)=2b(a+b—c.cosA), and b=¿ ( (a+b)+c)((a+b)—c)¸
(a+b)—c.cosA
If the sign of b be changed, and the supplement of its adjacent
angle therefore assumed, we shall obtain
cot} B = tan§Ã©+(ª—b) , and b=3( (c−(a−b) ) (c+(a−b)),

c—(a—b)
c.cosA—(a—b)
The relation of the sides and angles of a triangle might also be
in some cases conveniently expressed by a converging series. Thus
sin B
sinA¯¯sin(B+C)~sinBcosC+cos BsinC'
b
sin B
sin B
and consequently
a
b sinB cosC+b cosB sinCa sinB, or
b sinC
sin B
tanB.
a-b cusc
cos B
62
Wherefore, by actual division, tanB=—-sinC+
sinc cosC+
a
a²
a³
64
at
sin C cos C² + sin cos C³+ &c.; and, substituting the powers of
this expression for those of the tangent in the series of Note LXXI.,
b3
b
b2
we obtain B= sinc+ sinC cosC + 323 (4cosC³—1)sinC+
a
a²
b
{2cosC²-1)sinC cosC+ &c.; or
_sinC +
a4
a
b2
2a*
sin2C+
b3
3a²
64
404
sin3C+ sin4C &c.
In certain extreme cases, approximations can likewise be employ-
ed with advantage. Thus, suppose the angles A and B to be ex-
ceedingly small; then, by the last paragraph of page 364, their
versed sines are very nearly equal to half the squares of the sines.
Wherefore, sin C, or sin (A + B) = (art. 1. Note LXVI.),
sinA (1-sinB) + sin B (1-sin A2) nearly, and consequently,
by art. 5., c=(a+b) (1—žsinA sinB); or, the arcs being nearly
equal to their sines, substitute c for a + b in the second or
differential term, and c=a+b-cAB. Again, put С=-6, or
♦=A+B, and (a+b)({sinAsinB)=sinAsinB
(a+b)²
ab or
= }
a+b a+b
nearly, or ca+6—1
ab oz
a+b'
464
ILLUSTRATIONS.
NOTES AND ILJ
ī
Note LXXVI.-Page 388.
This problem, which is employed with great advantage in mari-
time surveying, admits likewise of a convenient analytical solution.
Let the given distances AB, BC and AC be denoted by a, b and
c, and the observed angles ADB and CDB by m and n; then
(art. 5. Note LXXV.) BD
sin BAD
sin BCD
LXX.)
and
b sin m-
a sin n
b sin m+a sin n
a sinBAD_b sinBCD
sin m
sin n
sinBAD-sinBCD
sinBAD+sin BCD
b sin m
or
a sin n
sinBAD+sinBCD= (art. 18. Note
tan}(BAD—BCD). But the angles ABC and ADC of
tant (BAD+BCD)
the quadrilateral figure DABC being evidently given, the sum of the
remaining angles BAD and BCD is given, and each of them is con-
sequently found. Hence the triangles ABD and CBD are imme.
diately determined,
This most useful problem was first proposed by Mr Townley, and
solved, in its various cases, by Mr John Collins, in the Philosophi
cal Transactions for the year 1671. The second solution given in
the text is borrowed from Legendre.
Note LXXVII.-Page 390.
This useful problem is commonly solved by the help of spheri-
cal trigonometry. It admits, however, of a simple and elegant
general solution, derived from the arithmetic of sines. Let a and
b denote the two vertical angles, or the acclivities of the diverging
lines, A the oblique angle which these contain, and A' the reduced
or horizontal angle. Since the magnitude of an angle depends
not on the length of its sides, assume each of them equal to the
radius or unit, and it is evident that the base of the isosceles tri-
angle thus limited will be the chord of the oblique angle A, the
perpendiculars from its extremities to the horizontal plane, the
NOTES AND ILLUSTRATIONS.
465
}
sines, and the horizontal traces or projections, the cosines, of the
vertical angles a and b. The base of the isosceles triangle forms
the hypotenuse of a right-angled vertical triangle, of which the
perpendicular is the difference between the vertical lines. Conse-
quently the square of the reduced base is equal to the excess of
the square of the chord of A above the square of the difference
of the sines of a and b, or
(cor. 6. def. Trig.) 2-2cos A-(sin a—sin b)²=
(II. 18. El.) 2—2cosA-sin a²—sin b²+2sin a sin b➡
(2. cor. def. Trig.) cos a² + cos b² + 2sin a sin b—2cosA.
Wherefore (Prop. 11. Trig.) in the triangle now traced on the
horizontal plane, 2 cos a cos b cosA' 2cosA — 2 sin a sin b; and
multiplying by sec a sec b, there results (cor. 4. def. Trig.),
1. CosA' sec a sec b cosA-tan a tan b.
This expression appears concise and commodious, but it may be
still variously transformed.
sec a sec b cos A
For vers A' 1-cos A' = 1 + tan a tan b
= sec a sec b (cos a cos b↓ sin a sin b cos A) =
(Prop. 2. Trig.) sec a sec b (cos (a — b) - cos A); whence
2. VersA' sec a sec b (versA-vers (a—b).)
Again, because (2. cor. 1. and 3. cor. 5. Trig.) versA'=2 sin‡A'², and
A+ (a—b)
A-la
versA-vers (a—b)=2sin
sin
2
20
by substitution,
A
6)
we obtain,
t
3. Sin}A?=sec a sec b(sinª + (a−b). sin A—(? +b)).
2
Of these formulæ, the first, I presume, is new, and appears dis-
tinguished by its simplicity and elegance. The last one however is,
on the whole, the best adapted for logarithmic calculation.
When the vertical angles are small, the problem will admit of a
very convenient approximation. For, assuming the arcs a, b as
equal to their tangents, it follows, by substitution, that cosA'=
cosA √ (1 + a²) √ (1 + b²)—ab = cosA ((1 + ža²) (1 + §6²))—ab
=cosA(1 + §a² + 362)-ab, nearly. Whence, by Note LXXIII.,
the decrement of the cosine of that oblique angle is
G g
#.
466
NOTES AND ILLUSTRATIONS.
1
ab--cos A (a²+b²); but
(II. 19. El.) ab=(ª+º)²—(ª—")², and
.2
a
(II. 21. El.) }a²+3b²= (ª+b)² + (² = 6)
wherefore the decrement of cosA'=
2
(a + b) — (~—b)² = cosA ((a+b)² + (~—- ¹³) :) =
a
a
(ª +¹³)² (1 — cos A)—('~—')² (i+cosA).
2
2
Consequently the increment of the oblique angle itself is, by Note
LXXIII,
a+
cos A
sin A
2
sin A
—b)(+0)=(art. 15. Note LXX.)
(a+b)² tan}A—~—b
cot A.
2
Such is the theorem which the celebrated Legendre has given,
for reducing an oblique angle to its projection on the horizontal
plane. It is very neat, and extremely useful in practice. But to
connect it with our division of the quadrant, requires some adap-
tation. Let a and b express the vertical angles in minutes; then
² tan} A—(ª—³)² cot} A);
will ((a+b)² tan} A—
cot) denote, likewise in mi-
1
3438
nutes, the quantity of reduction to be applied to the oblique angle.
!
In computing very extensive surveys, it becomes necessary to allow
for the minute derangements occasioned by the convexity of the sur-
face of our globe. The sides of the triangles which connect the suc-
cessive stations, though reduced to the same horizontal plane, may be
considered as formed by arcs of great circles, and their solution hence
belongs to Spherical Trigonometry. But, avoiding such laborious
calculations, for which indeed our Tables are not fitted, it seems far
better to estimate merely the deviation of those incurved triangles
from triangles with rectilineal sides. For effecting that correction
two ingenious methods have lately been proposed on the Continent.
NOTES AND ILLUSTRATIONS.
467
The first is that employed by M. Delambre, who substitutes the chords
for their arcs, and thus converts the small spherical, into a plane,
triangle. This conversion requires two distinct steps. 1. Each
spherical angle, or that formed by tangents at the surface of the
globe, is changed into its corresponding plane angle contained by the
chords. Let a and express the sides or arcs in miles; and the
angles of elevation, or those made by the tangents and the respective
chords, will be (III. 29. El.) denoted by
21600
24856
21600
A and
16
24856
in minutes, or
1350'
3107
aand
1350'
3107
β.
Insert these values, there-
fore, in place of a and b in the formula of the preceding note, and
the quantity of reduction of the angle A, contained by the small arcs
≈ and §, will be( (∞-8)² tan §A——B)² cot A) in seconds.
1214
2. Each arc is converted into its chord: But, by the Scholium to
Proposition VI. of the Trigonometry, an arc is to its chord,
a²
6D²;
as 1 to 1- ;; wherefore the diminution of that are in passing
into its chord, amounts to the
375,600,000
part of the whole.
These reductions bestow great accuracy, and are sufficiently com-
modious in practice. But the second method of correcting the
effects of the earth's convexity, and which was given by M. Legendre,
is distinguished by its conciseness and peculiar elegance. That pro-
found geometer viewed the spherical triangle as having its curved
sides stretched out on a plane, and sought to determine the variation
which its angles would thence undergo. Analysis led him, through a
complicated process, to the discovery of a theorem of singular beauty.
But the following investigation, grounded on other principles, appears
to be much simpler.
Let A and B denote any two angles in the small spherical tri-
angle, and a and express in miles the opposite sides, or those of its
extension upon a plane. Since (Prop. 9. Trig.) ≈ : ß :: sınA : sınB,
there must exist some minute arc, such that sina: sin ß :: sin(A+0):
sin(B.). But (art. 1. Note LXVI.) sin (A+6)=sin A+ cos A, and
G
← g 2
a
1
468
NOTES AND ILLUSTRATIONS.
03
(Schol. Prop. VI. Trig.) sin aa-
; whence a
B-
6
6
6
sinA+ cosA: sin B+ cosB.
fore, (V. 9. El.) 1—
sinA sinB+ sinA cosB.
៩
Now a sinB: sinA, and there-
ß:
B2
:1 :: sin A sin B + cos A sin B:
6
But the first and second terms being
very nearly equal, and likewise the third and fourth,—it is obvious
that the analogy will not be disturbed, if each of those pairs be
increased equally. Hence 1: 1 +
-B²
6
: : sinA sinB : sinA sinB+
e (sin A cos B-cos A sin B); and since (Prop. I. Trig.) sin A cos B
—cos A sin B = sin(A—B), therefore (V. 10. El.) 1:
sinA sinB: 0 sin(A—B).
02 B2
6
Consequently, since a and ẞ are pro-
portional to sinA and sinB, ◊ (sin A—B) = sin A sin B(ª²
uß
αβ
6
B(¤²—ß²)
-)=
-( sinA² — sinB² ) = (Proposition III. cor. 5. Trigonometry,)
αβ
6 (sin(A+B) sin(A—B)), or e=sin(A+B). But the sine of
the sum of A and B is the same as that of their supplement C, or
of the angle contained by the sides a and ß; and consequently is
αβ
2
the third part of sinC, the area of the triangle, or the third part
of the excess of the angles of the spherical, above those of the plane,
triangle. Wherefore the sines of the sides, which, in the spherical
triangle, are as the sines of their opposite angles, are likewise pro-
portioned, in the plane triangle, to the sines of those angles, increa-
sing each by the common excess. It is hence evident, that the angles
of the plane triangle are obtained from those of the spherical, by
deducting from each the third part of the excess above two right
angles, as indicated by the area of the triangle.
The whole surface of the globe being proportioned to 720°,
720°
that of a square mile will correspond to
24856×7912’
or the
1 part of a second. Hence each angle of the small spherical
75.88
i
1
triangle requires to be diminished by as sin C
in seconds.
455.28
NOTES AND ILLUSTRATIONS.
469
Another problem of great use in the practice of delicate surveying
is to reduce angles to the centre of the station. Instead of planting
moveable signals at each point of observation, it will often be found
more convenient to select the more notable spires, towers, or other
prominent objects which occur interspersed over the face of the
country. In such cases, it is evidently impossible for the theodolite
or circular instrument, although brought within the cover of the build-
ing, to be placed immediately under the vane. The observer ap-
proaches the centre of the station as near, therefore, as he can with
advantage, and calculates the quantity of error which the minute
displacement may occasion. Thus, suppose it were required to de-
termine the angle AOB which the remote objects A and B sub-
tend at O, the centre of a permanent station: The instrument is
placed in the immediate vicinity at the point C, and the distance
CO, with the angle of deviation OCA, are noted, while the principal
angle ACB is observed. The central angle AOB may hence he
computed from the rules of trigonometry; but the calculation is ef-
fected by simpler and more expe-
ditious methods. Since (I.32. El.)
the exterior angle ADB is equal
both to AOB with OAC, and to
ACB with OBC; it is evident that
AOB ≈ ACB + OBC — OAC.
=

But the angles OBC and OAC,
being extremely small, may be
considered as equal to their
sines, and (art. 5. Note LXXV.)
CO
sin OBC = sin BCO, and
OB
CO
OA
D
E
C
B
sin OAC = sin ACO; wherefore the angle AOB at the centre,
is nearly equal to ACB + CO
sin BCO
OB
sinACO
OA
= ACB +
CO (
sin(ACB+ACO) sinACO
OB
UA
).
Call the distances AC and
BC of the point of observation, a and b, the distances AO and
i
·470
NOTES AND ILLUSTRATIONS.
I
BO of the centre a' and '; the displacement CO, and the
angle ACO of deviation m and 9, while the subtended angles
ACB and AOB are denoted by C and C', and the opposite angles,
ABO and OAB by A and B; then C'=C+m (sin(+4) __ sind,
C’=C+m(
b'
a'
) 3438'.
If the centre O lies on AC, the correction of the observed angle,
m
expressed in minutes, will be merely (sinC) 3438.
Let a
But the problem admits of a simpler approximation.
circle circumscribe the points A, O, and B, and cut AC in E.
The angle AOB = (III. 18. El.) AEB ACB + CBE ; but
CE
EB
sin CBE = sin ACB, and sin OEC = sin AEO or ABO is
co
CE
CO sin ACB sin (ABO — ACO)
sin COE or AEO — ACO, and hence by combination
equal to
sin CBE =
EB
sin ABO
Since, therefore, EB
is nearly equal to OB, and the small angle CBE may be regarded
correction to be added to the observed an-
as equal to its sine, the
gle is denoted in minutes by
m_sinC sin(A-Q)
b'
sin A
3438. This quantity,
it is evident, will entirely vanish when & becomes equal to A, or the
angle ABO equals ACO; in which case, the point of observation C
coincides with E, or lies in the circumference of a circle that passes
through the two remote points A and B and centre of the station.
To place the instrument at E therefore, would only require to move
it along CA, till the angle AEO be equal to ABO.
Both these methods for the reduction of an angle to the centre are
given by Delambre; but, in his calculations, he generally preferred
the last one, as being simpler and sufficiently accurate for practice.
The investigation however will be found to be now considerably
shortened.
The accuracy of trigonometrical operations must depend on the
proper selection of the connecting triangles. It is very important,
therefore, in practice, to estimate the variations which are produced
among
the several parts of a triangle, by any change of their mutual
NOTES AND ILLUSTRATIONS,
471
relations. Suppose two of the three determining parts of a triangle
to remain constant, while the rest undergo some partial change; and
let, as before, the small letters a, b and c denote the sides of the tri-
angle, and the capitals A, B and C their opposite angles.
Case I.-When two sides a and b are constant.
Since the angles A and B, after passing into A+▲A and B+AB,
must have their sines still proportional to the opposite sides, it is
evident that
sin A
sin B
sin(A+AA) sin(B+AB)'
sin(A+▲A)—sinA __ sin(B+▲B)— sinB
sin(A+AA)+sinAsin(B+AB)+sinB
tion and art. 7. Note LXX.,
and consequently
; wherefore, by alterna- .
tan▲A tan(A+▲A)
1.
=
tan AB tan(B+^В)
Next, in the incremental triangle formed by the sides c, c+Ac, and
the contained angle AA, (art. 1. Note LXXV.)-
플​AC
C+BAC
2.
tan (B+AB)
and hence reciprocally,
cot▲A
LAC
tan ΔΑ
c++ Ac
cot(B+AB)
In like manner, from the incremental triangle contained by the
sides c, c+Ac and the angle AB, it follows that
3.
Ac
tan AB
c+Ac
col(A+AA)
Again, the base of the incremental isosceles triangle contain-
ed by the equal sides b, b, and the vertical angle AC, is (art. 15.
Note LXXV.) 2b sin AC; wherefore, in the incremental triangle
formed with the same base and the sides c and c+ac, by art. 20.
(c+Ac)sin AB
b sint Ac
Note LXXV., cos(A+4▲A) =
; whence
sin AB
4.
b cos(A+▲A)
==
sin AC
C+AC
After the same manner, it will be found that
sin▲A
5.
sin AC
a cos(B+1B).
c+140
•
472
NOTES AND ILLUSTRATIONS.
6.
7.
Multiply the expressions of art. 4. into those of art 3. and
LAC b sin(A+▲A)
sin AC
COSAB
Multiply likewise the expressions of art. 2. and 5., and
LAC a sin(B+4B),
SAC
cos ΔΑ
If, in all the preceding formula, the increments annexed to the
varying quantities be omitted, there will arise much simpler expres-
sions for the differentials.
dA
tan A
# 1.
dB
tan B
dc
C
2.
dA
cot B
dc
C
3.
dB
cot A
dB
b
4.
cosA.
ac
C
dA
• 5.
dC
ee
cos B.
C
dc
* 6.
=bsinA.
dc
dc
#7.
=a sinB.
dC
}
Case II.-When one side a, and its opposite angle A, are constant.
a
Since (art. 5. Note LXXV.)
b
it is evident that
sin A sin B'
a sinBb sinA, and taking the differences by art. 1. of Note LXXIII.
sin LAB
AbsinA≈2a sing▲В cos (B+▲B), whence A6
and consequently, by art. 5. of Note LXXV.
sin A
a cos(B+AB)'
sin AB
8.
146
sin AC
sin B
Ab b cos(B+4B)*
In like manner, it will be found that
sin AB
9.
플​AC
sin AC
LAC
sinC
c cos(C+AC)
Combine the two last expressions, and
Ab
10.
cos(B+AB)
Ac
cos (C+AC)
NOTES AND ILLUSTRATIONS.
478
The differentials are discovered, by rejecting the modifications of
the variable quantities.
dB
sin B
tan B
#8.
db
b cos B
b
dB
sin C
tanC
*9.
dc
c cos C
C
db
cos B
* 10.
dc
COSC
Case III.-When one side a, and its adjacent angle B, are constant.
In the incremental triangle contained by the sides ь, ь+Aь, and
Ac, it is evident, (art. 5. Note LXXV), that
11.
Ac
SIRAC
12.
Ас
h
b+Ab
sin▲A Sin (A+▲A) sin A
Again, in the same incremental triangle, (art. 6. Note LXXV.)
ΙΔΙ
tan AC
ΔΕ
b+46
tan▲A tan(A+¼^A)`
Or, transforming the preceding expression,
дь
6+Ab
Ab
b
-tan AA(
tan▲A(-
wherefore,
ΙΔΙ
13.
sin AC
tanţ▲A
tan(A+▲A)'
and consequently
tan AA
sin(A+▲A)
tan(A+AA)+tan▲A
cos(A+}▲ Acos}▲A)) = — sin‡▲A(~,
(art. 1. Note LXX.)
)=—
cos(AAA)
sin(A+AA)
146
sin▲A
= b (sin(A+AA)
cos(A+▲A)
·).
Again, in the same incremental triangle, by art. 20. Note LXXV.
Ab
cos(A+AA)= (-cos AC)= cos▲A; whence
Ас
cos(A+AA)
Ab
14.
Ac
cos ΔΑ
Ab
Δε
The differentials are found as before, by the emission of the mi-
nute excrescences.
dc
* 11.
dC
db
* 12.
aC
91313
dc
b
dA
sinA
b
dA tan A
474
NOTES AND ILLUSTRATIONS,
db
db
13.
=b
dc
dA
sinA
(COSA) = b cota,
db
14..
=cosA.
dc
To compute the values of the finite differences, when these differ-
ences themselves are involved in their compound expression, the
easiest method is to proceed by repeated approximations. Thus,
from art. 3. Ac =
tan AB
cot(A+AA)
(2c+Ac); assume, therefore,
first, Ac=
(2c -
tan AB
cot (A+AA)
2c; and then, Ac=
tan AB
cot(A+AA)
tan AB
cot (A+4A)
2c). But it will seldom be requisite to advance
beyond two steps; though the process, if continued, would evidently
form an infinite converging series.
When only one part of a triangle remains constant, the expressions
for the finite differences will often become extremely complicated.
It may be sufficient in general to discover the relations of the dif-
ferentials merely. To do this, let each indeterminate part be sup-
posed to vary separately, and find, by the preceding formula, the
effect produced; these distinct elements of variation being collected
together, will exhibit the entire differential.
The materials of this intricate Note appear in Cagnoli, but the
subject was first started by our countryman Mr Cotes, a mathema-
tician of profound and original genius, in a brief tract, entitled
Estimatio errorum in mixtâ Mathesi. It is unfortunate that I have
not room for explaining the application of those formula to the se-
lection and proper combination of triangles in nice surveys.
HAVING in some of the preceding notes briefly pointed out the se-
veral corrections employed in the more delicate geodesiacal opera-
tions, I shall subjoin a few general remarks on the application of
trigonometry to practice. The art of surveying consists in deter-
mining the boundaries of an extended surface. When performed in
the completest manner, it ascertains the positions of all the promi-
nent objects within the scope of observation, measures their mutual
distances and relative heights, and consequently defines the various
NOTES AND ILLUSTRATIONS.
475
contours which mark the surface. But the land-surveyor seldom
aims at such minute and scrupulous accuracy; his main ob-
ject is to trace expeditiously the chief boundaries, and to compute
the superficial contents of each field. In hilly grounds, however, it
is not the absolute surface that is measured, but the diminished
quantity which would result, had the whole been reduced to a hori-
zontal plane. This distinction is founded on the obvious principle,
that, since plants shoot up vertically, the vegetable produce of a
swelling eminence can never exceed what would have grown from
its levelled base. All the sloping or hypotenusal distances are,
therefore, reduced invariably to their horizontal lengths, before the
calculation is begun.
Land is surveyed either by means of the chain simply, or by com-
bining it with a theodolite or some other angular instrument. The
several fields are divided into large triangles, of which the sides are
measured by the chain; and if the exterior boundary happens to be
irregular, the perpendicular distance or offset is taken at each bend-
ing. The surface of the component triangles is then computed
from Prop. 31. Book VI. of the Elements of Geometry, and that
of the accrescent space by Note XV. to Prop. 10. Book II.
In
this method the triangles should be chosen as nearly equilateral
as possible; for if they be very oblique, the smallest error in the
length of their sides will occasion a wide difference in the estimate
of the surface. The calculation is much simpler from the applica-
tion of Prop. 6. Book II. of the Elements, the base and altitude of
each triangle only being measured; but that slovenly practice ap-
pears liable to great inaccuracy. The perpendicular may indeed be
traced by help of the surveying cross, or more correctly by the box
sextant, or the optical square, which is only the same instrument in
a reduced and limited form; yet such repeated and unavoidable in-
terruption to the progress of the work will probably more than coun
terbalance any advantage that might thence be gained.
The usual mode of surveying a large estate, is to measure round it
with the chain, and observe the angles at each turn by means of
the theodolite. But these observations would require to be made
with great care. If the boundaries of the estate be tolerably regu-
lar, it may be considered as a polygon, of which the angles, being
476
NOTES AND ILLUSTRATIONS.
necessarily very oblique, are therefore apt to affect the accuracy of
the results. It would serve to rectify the conclusions, were such
angles at each station conveniently divided, and the more distant
signals observed. The best method of surveying, if not always the
most expeditious, undoubtedly is to cover the ground with a series of
connected triangles, planting the theodolite at each angular point, and
computing from some base of considerable extent, which has been
selected and measured with nice attention. The labour of transport-
ing the instrument might also in many cases be abridged, by obser-
ving at any station the bearings at once of several signals. Angles
can be measured more accurately than lines, and it might therefore
be desirable that surveyors would generally employ theodolites of a
better construction, and trust less to the aid of the chain.
The quantity of surface marked out in this way, is easily computed.
from trigonometry: Adopting the general notation, the area of a
triangle which has two sides, and their included angle known, it is
ab
evident, will be denoted by.sinC, and the area of a triangle of
which there are given all the angles and a side, is
a² sinB sinC
2 sin A
The English chain is 22 yards, or 66 feet in length, and equivalent
to four poles; it is hence the tenth part of a furlong, or the eightieth
part of a mile. The chain is divided into a hundred links, each oc-
cupying 7.92 inches. An acre contains ten square chains or 100,000
links. A square mile, therefore, includes 640 acres; and this large
measure is deemed sufficient, in certain rude and savage countries, as
the Back Settlements of America, where vast tracts of new land
are allotted merely by running lines north and south, and intersecting
these by perpendiculars, at each interval of a mile.
The Scotch chain consists of 24 ells, each containing 37.069 inches,
and ought therefore to have 74.138 feet for its correct length. The
English acre is hence to the Scotch, in round numbers, as 11 to 14,
or very nearly as the circle to its circumscribing square. But this
provincial measure is gradually wearing into disuse, and already the
statute acre seems to be generally adopted in the counties south of
the Forth.
NOTES AND ILLUSTRATIONS.
477
LEVELLING is a delicate and important branch of general survey-
ing. It may be performed very expeditiously by help of a large
theodolite, capable of measuring with precision the vertical angle
subtended by a remote object, the distance being calculated, and al-
lowance made for the effect of the earth's convexity and the influence
of refraction. But the more usual and preferable method is to em-
ploy an instrument designed for the purpose, and termed a spirit-level,
which is accompanied by a pair of square staves, each composed of
two parts that slide out into a rod of ten feet in length, every foot
being divided centesimally. Levelling is distinguished into two
kinds, the simple and the compound; the former, which rarely ad-
mits of application, assigns the difference of altitude by a single ob-
servation; but the latter discovers it from a combined series of ob-
servations carried along an irregular surface, the aggregate of the se-
veral descents being deducted from that of the ascents. The staves
are therefore placed successively along the line of survey, at suitable
intervals according to the nature of the ground and not exceeding
400 yards, the levelling instrument being always planted nearly in
the middle between them, and directed backwards to the first staff,
and then forwards to the second. The difference between the
heights intercepted by the back and the fore observation, must evi-
dently give at each station the quantity of ascent or descent, and
the error occasioned by the curvature of the globe may be safely
overlooked, as on such short distances it will not amount at each
station to the hundredth part of a foot. To discover the final result
of a series of operations, or the difference of altitude between the ex-
treme stations, the measures of the back and fore observations are
all collected severally, and the excess of the latter above the former
indicates the entire quantity of descent.
As an example of levelling, I shall take the concluding part of a
survey which my friend Mr Jardine, civil engineer, has recently
made for the Town-Council of Edinburgh, with a degree of accuracy
seldom attempted, in tracing the descent from the Black and Craw-
ley springs, near the summits of the Pentland chain, to the Reservoir
on the Castlehill, with a view to the conducting of a fresh supply of
water from those heights, To avoid unnecessary complication,
478
NOTES AND ILLUSTRATIONS.
however, I shall only notice the principal stations. The figure an-
nexed represents a profile or vertical section of the ground, LV is
the level of the Black spring, and the several perpendiculars from it
denote the varying depth of the surface, referred to the base assumed
700 feet below. The stations marked are as follow:
L Lowest point in the Meadow.
M Cleansing cocks on the north side of the Meadow.
N Sunk fence in Lord Wemyss's garden.
Air cock in Archibald's nursery.
P South side of Lauriston road
Q Bottom of Heriot's Green Reservoir.
R Head of Hamilton's close.
S. Strand on south side of Grassmarket.
T Cleansing cock on north side of Grassmarket.
U Gaelic Chapel.
V Upper side of the belt of Castlehill Reservoir.
L
M
N 0 P
R
S
TU
200 300
2000
2000
Back Ob- Fore Ob-
Distance.
servation.
servation.
Ascent.
Stations.
Feet:
Feet,
Feet.
Feet.
-NZOPCRSTUD
L
M
370
4.59
2.04
2.55
640
8.68
3.05
8:18
905
9.12
2.22
15.08
1236
29.43
2.11
42.40
1493
16.24
1.40
57.24
1925
2.54
26.98
32.80
2260
4.69
53.28
-15.79
2352
4.22
4,42
-15.99
2540 32.40
1.25
15.15
2705 94.77
9.97
99.95
1


NOTES AND ILLUSTRATIONS.
479
Black spring, being 620.05 feet above the level of the Meadow,
is therefore 520.1 feet higher than the belt of the reservoir. The
numbers exhibited in the last column, are obtained by taking the
differences of the aggregates of the two preceding columns. Where
the ground either sinks or rises suddenly, some intermediate obser-
vations are here grouped together into a single amount. Thus, three
observations were made between O and P, two between P and Q, three
between Q and R, five between R and S, three between T and U,
and no fewer than nine between U and V. The slight sketch be-
tween the perpendiculars from Q and R, shows the mode of planting
and directing the instrument.
The mode of levelling on a grand scale, or determining the heights
of distant mountains, will receive illustration from the third volume
of the Trigonometrical Survey, which Colonel Mudge has been kind-
ly pleased to communicate to me before its publication. I shall select
the largest triangle in the series, being one that connects the North
of England with the Borders of Scotland. The distance of the sta-
tion on Cross Fell to that on Wisp Hill, is computed at 235018.6
feet, or 44.511 miles, which, reckoning 60944 feet for the length of
a minute near that parallel, corresponds, on the surface of the globe,
to an arc of 38′ 33″.7. Wisp Hill was seen depressed 30′ 48″ from
Cross Fell, which again had a depression of 2′ 31" when viewed from
Wisp Hill. The sum of these depressions is 33′ 19″, which, taken
from 38′ 33″.7, the measure of the intercepted arc, or the angle at
the centre, leaves 5′ 14″.7, for the joint effect of refraction at both
stations. The deflection of the visual ray produced by that cause,
and which the French philosophers estimate in general at .079, had
therefore amounted only to .06805, or a very little more than the
fifteenth part of the intercepted arc. Hence, the true depression of
Wisp Hill was 30′ 48"-16′ 39″.5 14' 8".5; and consequently,
estimating from the given distance, it is 967 feet lower than Cross
Fell.
From Wisp Hill, the top of Cheviot appeared exactly on the same
level, at the distance of 185023.9 feet, or 35.0424 miles. Where-
fore, two-thirds of the square of this last number, or 819, would, from
the scholium at page 391. express in feet the approximate height of
在
​480
NOTES AND ILLUSTRATIONS.
Cheviot above Wisp Hill. But refraction gave the mountain a more
towering elevation than it really had; and the measure being re-
duced in the former ratio of 38' 33".7 to 33′ 19″, is hence brought
down to 708 feet.
Again, the distance 292012.7 feet, or 55.3054 miles, of Cross
Fell from Cheviot, corresponds to an arc of 47′ 54″.8, which, re-
duced by the effect of refraction, would leave 41′ 23″.8 for the sum
of the depressions at both stations. Consequently, Cheviot had,
from Cross Fell, a true depression of only 23′ 44″-20′ 41″.9 or
3' 2.1, and is therefore lower than that mountain by 258 feet.
These results agree very nearly with each other. The height of
Cross Fell above the level of the sea being 2901, that of Wisp Hill
is 1934, and that of Cheviot 2642 or 2643. In the Trigonometrical
Survey, the latter heights are stated at 1940 and 2658; a difference
of small moment, owing to a balance of errors, or perhaps to the
adoption of some other data with respect to horizontal refraction,
and which do not appear on record.
From the same valuable work, I am tempted to borrow another
example, which has more local interest. From Lumsdane Hill, the
north top of Largo Law, at the distance of 189240.1 feet, or 35.84
miles, appeared sunk 9' 32" below the horizon. Here the intercept-
ed arc is 31' 3" and the effect of the earth's curvature, modified
by refraction, is 13′ 24″.8; whence the true elevation of Largo Law
was 13′ 24″.8-9′ 32″, or 3′ 52″.8, which makes it 213 feet higher
than Lumsdane Hill, or 938 feet above the level of the sea. In the
Trigonometrical Survey, this height is stated at 952; but I am in-
clined to prefer the former number, having once found it by a ba-
rometrical measurement, in weather not indeed the most favourable,
to be only 935 feet.
MARITIME SURVEYING is of a mixed nature: It not only deter-
mines the positions of the remarkable headlands, and other conspi-
cuous objects that present themselves along the vicinity of a coast,
but likewise ascertains the situation of the various inlets, rocks, shal-
lows and soundings which occur in approaching the shore. To sur-
NOTES AND ILLUSTRATIONS.
481
vey a new or inaccessible coast, two boats are moored at a proper
interval, which is carefully measured on the surface of the water;
and from each boat the bearings of all the prominent points of land
are taken by means of an azimuth compass, or the angles subtended
by these points and the other boat are measured by a Hadley's sex-
tant. Having now on paper drawn the base to any scale, straight
lines radiating from each end at the observed angles, as in Prop. 21.
of the Trigonometry, will by their intersections give the positions of
the several points from which the coast may be sketched.-But a
chart is more accurately constructed, by combining a survey made
on land, with observations taken on the water. A smooth level piece of
ground is chosen, on which a base of considerable length is measured
out, and station staves are fixed at its extremities. If no such place
can be found, the mutual distance and position of two points con-
veniently situate for planting the staves, though divided by a broken
surface, are determined from one or more triangles, which connect
with a shorter and temporary base assumed near the beach. A boat
then explores the offing, and at every rock, shallow, or remarkable
sounding, the bearings of the station staves are noticed. These ob-
servations furnish so many triangles, from which the situation of
the several points are easily ascertained.-When a correct map of
the coast can be procured, the labour of executing a maritime survey
is materially shortened. From each notable point of the surface of
the water, the bearings of two known objects on the land are taken,
or the intermediate angles subtended by three such objects are ob-
served. In the first case, those various points have their situations
ascertained by Prop. 21. and the second case by Prop. 25. of the
Trigonometry. To facilitate the last construction, an instrument
called the Station-Pointer has been invented, consisting of three brass
rulers, which open and set at the given angles.
The nice art of observing has in its progress kept pace with the
improved skill displayed in the construction of instruments. Sur-
veys on a vast scale have lately been performed in Europe, with that
refined accuracy which seems to mark the perfection of science. Af-
ter the conclusion of the American war, a memoir of Count Cas-
sini de Thury was transmitted by the French government to our
Hh
482
NOTES AND ILLUSTRATIONS.
Court, stating the important advantages which would accrue to
astronomy and navigation, if the difference between the meri-
dians of the observations of Greenwich and Paris were ascertain-
ed by actual measurement. A spirit of accommodation and con-
cert fortunately then prevailed. Orders were speedily given for
carrying the plan into execution; and General Roy, who was
charged with the conduct of the business on this side of the Chan-
nel, proceeded with activity and zeal. In the summer of 1784,
a fundamental base, rather more than five miles in length, was traced
on Hounslow Heath, about 54 feet above the level of the sea, aud
measured with every precaution, by means of deal rods, glass tubes,
and a steel chain, allowance being made for the effects of the vari-
able heat of the atmosphere in expanding those materials. The
same line was, seven years afterwards, remeasured with an im-
proved chain, which yet gave a difference on the whole of only
three inches. The mean result, or 27404.2 feet, at the tempera-
ture of 62° by Fahrenheit's scale, is therefore assumed as the true
length of the base. Connected with this line, and commencing from
Windsor Castle, a series of thirty-two primary triangles was, in 1787
and 1788, extended to Dover and Hastings, on the coast of Kent and
Sussex. Two triangles more stretched across the Channel. The
horizontal and vertical angles at each station were taken with singu-
lar accuracy by a theodolite, which the celebrated artist Ramsden
had, after much delay, constructed, of the largest dimensions and the
most exquisite workmanship. At the same period, a new base of veri-
fication was measured on Romney Marsh, 15 feet above the sea, and
found, after various reductions, to be 28535.6773 feet in length. This
base, computed from the nearest chain of triangles dependent on that
of Hounslow Heath, ought to have been 28533.3; differing scarcely
more than two feet on a distance of eighty miles. The mean, or
28534.5, is adopted for calculating the adjacent and subsequent
triangles. These triangles near the coast were unavoidably confined and
oblique; but their sides are generally deduced from larger and more
regular triangles, expanding over the interior of the country. The
annexed figure exhibits the most interesting portion of this memor-
able survey, and represents the various combination of triangles. At-
tached to it is a scale of English miles.
NOTES AND ILLUSTRATIONS.
483
A Frant Church
B Goodburst Church
C Hollingborn Hill
D Tenterden Church
E Fairlight Down
F Allington Knoll
G Lydd Church
H Ruckinge
I High Nook
K Folkstone Turnpike
L Padlesworth
M Swingfield Church
N Dover Castle
O Church at Calais
P Blancnez Signal
R Fiennes Signal
S Montlambert Signal
KL The base of verification.

}
M
L
B
F
H
S
20
20
30
10
50
60
70
Calculation of the sides of the Triangles.
ACE
BDE
A 70° 23' 2"
C 52 11 3
E 48 25 55
141744.4 B 49° 39' 35.77"
113926 D 94 59 25.81
107895.7 E 35 20
71637.2
93629.2
58.42
ABC
CDF
A 27 4
B 136 27
36.13
71298.5 C 40 0
58.96
61777.5
35.87
D 91 34 22.04
96039.8
C 16 27
48
44391.2 F 48 24 39
ABE
A 43 18 25.87
B 105 39 28.86
DFG
93629.2 D 43
45
23.18
47850.9
F 73 0
27
66169.2
E 31 2
5.27
G 63 14
9.82
BCD
DEG
B 68 13 19.5
71887.5 D 62 32
52.51
71692.2
C 44 38 44.04
D 67
7 56.46
*
54376.5 E 54
59
17.31
G 62 27
50.18
71637.2
иh 2
484
NOTES AND ILLUSTRATIONS.
EFG
KLM
E 21° 18′ 37″
47850.9 K 60° 27′
39.5″
17056.6
F 32
G 125 42
59 23
L 70
54
5.5
18525.S
0
106926.2 M 48
38
15
FGI
KMN
F 33 8
46.1
31363.7 K 19
43
53.5
30560.4
G 26 57
29.9
23185.7 M 75
36
40
31555.7
I 121
121 53
44
N 34
39
26.5
FHI
KLN
F 91 27
19.5
28534.5 K 130
11
33
42562.7
II 54
19
18.5
L 34
29
42.5
I 34 13
22
16053 N 15
18
44.5
FGK
ELN
F 109
50 39.35
G 38
2
23.76
84662.8 E 6 6 39.43
55463.6 L 152
*
15
25.15 186119
K 32
6
56.89
IN 21
37 55.42
EGL
ENP
E 13 38
2.95
G 154
5
54.4
79536.1 E 25
14739.2 N 110
33 55.02
L 12
16
2.65
P 43
55 29.83
30 35.15
116660
252505.6
FIK
ENS
F
76
1
53.25
54708 E
43 19 58.52
168827
I 79 41 0.5
K 24 17 6.25
N 87 30
IS 49
29.58
245786
9
31.9
I
14 48
K 57 2 0
IKL
25.5 *
NPS
14714.3 N 23 25
0.25
77237.2
L 108 9 34.5
48305.2 P 119 41 41.64
S 36 53 18.11
In this register, each angle in the successive triangles is, for the
sake of conciseness, marked by the single letter affixed to it, and the
computed length of its opposite side in feet ranges in the same line.
The addition of an asterisk denotes that an angle was not actually obser-
ved, but only deduced from calculation. The oblique triangles ABC
and ABE have their sides BC and BE derived from other larger tri-
angles, which were nearly equiangular. The triangles ELN and
ENP had their angles discovered from conjoined observations. In ge-
neral the several angles, as affected by the spherical excess, were cor-
rected for computation by a sort of tentative process. It results from
a train of calculations, that Dover Castle lies south 67° 44' 34" east,
NOTES AND ILLUSTRATIONS.
485
and at the distance of 328231 feet or 62.165 miles, from Greenwich
Observatory. On their part, the French astronomers, under the di-
rection of Cassini, carried forward the trigonometrical operations
from Dunkirk to Paris; employing Borda's repeating circle, an in-
strument much smaller and less perfect than Ramsden's theodolite,
but formed on a principle which always procures the observer a near
compensation of errors. From a comparison of the whole, it follows
that the meridian of the Observatory of Paris lies 2° 19′ 51″ east
from that of Greenwich, differing only nine seconds in defect from
what the late Dr Maskelyne had previously determined from combin-
ed astronomical observations.
The success with which that great survey was attended, gave oc-
casion both in France and England to still more extensive projects.
The National Assembly, amidst other essential improvements which
it meditated, having resolved to adopt a general and consistent sys-
tem of measures, the length of a degree of the meridian at the mid-
dle point between the pole and the equator was proposed as a per-
manent basis. But to secure greater accuracy in determining the
standard, it had been decided to prolong the observations on both
sides of the mean latitude, and trace a chain of triangles over the
whole extent from Dunkirk to Barcelona. This bold plan was ex-
ecuted in the course of the years 1792, 1793, 1794 and 1795, with
equal sagacity and resolution, by MM. Delambre and Mechain, who,
during all the horrors of revolutionary commotion, yet pressed for-
ward their operations in spite of obstacles and dangers of the most
sickening kind. After the various triangles, amounting in total to
115, had been observed, they were connected, in the neighbourhood
of Paris, with a base of more than seven miles in length, and
measuring, at the temperature of 161° on the centigrade scale, or
614° by Fahrenheit, 6075.9 toises from Melun to Lieursaint. A base
of verification was likewise traced near the southern extremity of the
line of survey, extending 6006.25 toises along the road from Perpignan
to Narbonne. This base appeared not to differ one foot from the
calculation founded on the other, though separated by a distance of 400
miles,—a convincing proof of the accuracy with which the observa-
tions had been made. A specimen of the French triangulation is
given in the figure below, where the vertical line represents the me-
ridian of Dunkirk, with the distances expressed by intervals of
10,000 toises.
•
486
NOTES AND ILLUSTRATIONS.

A St Martin du Têrtre.
·50
A
11
B
B
Dammartin.
C
Pantheon at Paris.
40
-19
E
Brie.
D
-13
-30
F
E
T
-14
-20
K
H
I
-15
10
L
D Belle Assise.
Montlheri.
G Lieursaint.
H Melun.
I
K
..
Malvoisine.
Torfou.
L Forêt.
Chapelle.
N Pithiviers.
O Bois Commun.
Chatillon.
Q Château-neuf.
R Orleans.
M
M
16
0
N
18
117
P
P
R
.18
i
GH The primary base.
Calculation of the sides of the Triangles.
ABC
FIG
A 76° 2′ 30″.66
B 57 20
17.82
17310.3013 F 49° 34′ 22″.32
15017.3211 76 47
$369.1673
42.98
10703.5616
C 46 37
11.52
G 53 37
54.70
BCD
IGH
B 59 52
2.20
15756.8013 40 36
56.68
6075.8998
C 48 17
34.50
13601.3539 G 75
39
29.67
9042.5510
D 71 50
23.30
H 63
43
33.65
CDE
FIK
C 37 1
40.59
9516.5896 F 55 10
1.03
7357.8627
D 57 21
E 85 37
1.87
13305.8528 43 52
3.25
6212.1595
17.54
K 80 57
55.72
CEF
IKL
C 61 13
47.94
13101.0845 53 22
24.93
8349.1059
E 55 51
48.75
12370.8194K 81 36
49.90
10292.0814
F 62 54
23.31
L 45 0
45.17
EFI
ILM
E 40 32 37.60
8852.8293 70 51
37.77
13438.2345
F 45 18 40.41
12374.2130L 62 47
29.54
12650.5655
I 74 8 41.99
M 46 20
52.69
NOTES AND ILLUSTRATIONS.
487
LMN
OPQ
5
L 68° 35′ 59″.16
M 51
N 60 18
2.0625/0
13.26
47.58
14402.06250 62° 31′ 30″.34
12036.0949 P 93 0
Q 24 28
10446.5520
17.27
11758.3955
12.39
MNO
PQR
M 21 58
52.87
9190.1355 P 50 28
6.42
12053.9075
N 91 55
5.70
17341.8323Q 87 35
8.93
15614.7105
O 56 6
1.43
R 41 56 44.65
NOP
N 31 53
2.40
4877.2386
O 52 33 5.48
P 95 33 52.12
7330.6166
Through the whole process of their survey, the French astrono-
mers have certainly displayed superior science. In deducing the
correct results, they seem to exhaust all the refinements of calcula-
tion. The angles measured by the repeating circle, it was necessa-
ry to reduce, not only to the horizontal plane, but generally besides
to the centre of observation. This would have required much nice
and tedious computation; the labour of achieving such reductions
was however greatly simplified and abridged, by help of concise for-
mule, and the application of auxiliary tables. There is even room
to suspect that those ingenious philosophers have carried the fond-
ness for numerical operations to an excess, and often pushed the de-
cimal places to a much greater length in their estimates than the
nature of the observations themselves could safely warrant.
In the spring of 1799, the registers of all these operations were
referred to a commission, consisting of the ablest members of the
Institute, and some other learned men deputed from the countries
then at peace with France. The various calculations were carefully
examined and repeated; and a comparison of the celestial arc with
that which had been measured in Peru having given 1
334
for the ob-
lateness of the earth, the length of the quadrant of the meridian, or
the distance of the pole from the equator, was finally determined at
5130740 toises, the ten millionth part of which, or the space of
443.295936 lines forms the metre. This standard was afterwards
definitively decreed by the Legislative Body.
488
NOTES AND ILLUSTRATIONS.
Mechain, however, still anxious to realize his early project of cx-
tending the meridian as far as the Balearic Isles, again repaired
to Spain, and conducted with incredible exertions a chain of tri-
angles over the savage heights from Barcelona to Tortosa, and was
about to observe the altitude of the stars, and measure the base of
Oropesa, when, worn out by continued fatigue, he caught an epide-
mic fever, which fatally closed his meritorious labours, at Castellon
de la Plana, in the kingdom of Valentia, about the latter part of Sep-
tember 1805. The prosecution of the plan was subsequently com-
mitted to Biot, who has brought it to a fortunate conclusion. This
ardent philosopher, during his stay on the rocky island of Formen-
tera, had likewise an opportunity of making observations and ex-
periments interesting to physical science. In the winter of 1806 and
the spring of 1807, he continued the series of triangles from Barcelona
to the kingdom of Valentia, and joined that coast with the Balearic
Isles, by an immense triangle, of which one of the sides exceeded an
hundred miles in length. At such prodigious distances, the stations,
however clevated, and notwithstanding the fineness of the climate,
could not be seen during the day; but they were rendered visible at
night, by combining Argand lamps with powerful reflectors. These
observations give a result which agrees almost exactly with what
had been already found by Delambre and Mechain.
If the mean
were adopted, it would yet scarcely affect the length of the metre by
the diminution of a four millionth part. The meridional arc extending
from Dunkirk to Formentera, measures 12° 22′ 13″.395; and from
this ample basis, the circumference of the earth is computed to be
24855.42 English miles.
In England, the prosecution of the trigonometrical survey, without
aiming at such splendid views, has, suitably to the genius of the
people, been directed to objects of more domestic interest, and
perhaps real utility and importance. The perplexing inaccuracy of
our best maps and charts had long been the subject of most serious
complaint. It was in consequence resolved to extend the series of con-
nected triangles over the whole surface of the island. But the death
of General Roy, happening so early as 1790, threatened to prove fatal
to the completion of his favourite scheme, and for which the talents
and experience he possessed had so eminently fitted him. After some
interruption, however, an opportunity was embraced of resuming that
NOTES AND ILLUSTRATIONS.
489
*
noble plan; and it was, under the direction of the Board of Ord-
nance, committed to the care of Colonel Mudge, who, with equal
ability and undiminished ardour, has, during the space of now almost
twenty years, been engaged in carrying on the most extensive and
varied system of operations ever attempted, and in a style of execu-
tion which reflects on him the highest credit. In 1793 and 1794,
the chain of primary triangles was continued from Shooter's Hill to
Dunnose in the Isle of Wight, including a great part of Surrey, Sus-
sex, Hants, Wiltshire and Dorsetshire, and connecting with a new
base of verification measured on Salisbury Plain. This base had, after
correction, a length of 36574.4 feet, or 6.92697 miles, having lost al-
most a whole foot in being reduced from an elevation of 588 feet to
the level of the sea. It differed scarcely an inch from the computation
founded on the base of Hounslow Heath. In 1795, the triangles were
carried into Devonshire; and they were continued in 1796 through
Cornwall to the Scilly Islands. The West of England became the scene
of repeated operations. In 1798, a third base was measured on King's
Sedgemoor near Somerton, and found, after various corrections, to
be 27680 feet, or 5.242425 miles, differing only about a foot from
the result of the caiculation dependent on that of Salisbury Plain.
The survey now advanced to the centre of England, and was extended
in 1803 to Clifton in Yorkshire; another base of verification, 26342.7
feet in length, having been measured at Misterton Carr, on the north
of Lincolnshire. The triangles were next carried towards Wales,
and made to rest on a base of 24514.26 feet, stretching from the
western borders of Flintshire to Llandulas in Denbighshire. From
this last base, numerous triangles have been extended in different di-
rections; one series bending, through Anglesea and by Cardigan Bay,
to the Bristol Channel; another penetrating into the central parts of
England; while a third series stretches northwards, through Lanca-
shire, Cumberland and Westmoreland, into Scotland, and uniting with
the collateral chain of Misterton Carr from Yorkshire and Northum-
berland, is prolonged to the heights immediately beyond the Firth of
Forth. We look forward with anxiety to the conclusion of this ar-
duous undertaking. The mountains and islands near the western
coast of Scotland will furnish triangles of vast extent. Colonel
Mudge will not emit, we are confident, the opportunities that such
stations may afford to determine the quantity of horizontal refraction.
490
NOTES AND ILLUSTRATIONS.
noting at the same time the variable state of the atmosphere. We have
perfect reliance in the accuracy of his observations; yet it would be
desirable in all cases, as in the French operations, that the third
angle of each triangle were actually measured.
Besides the principal triangles thus determined, a multitude of
subordinate ones were ascertained in the progress of the survey, and
which serve to connect all the remarkable objects that occurred over
the face of the country. The capital points were hence established
for constructing the most accurate charts and provincial maps. A
number of royal military surveyors, of approved skill, have since
been constantly employed in filling up the secondary triangles, and
embodying the skeleton plans. The various materials are collected
at the drawing-room of the Tower, and there adjusted, reduced and
combined. Under the same able direction, an extensive establish-
ment has been formed in those spacious apartments, where a volumi-
nous series of maps, and on the largest scale, are not only delineated
but engraved. This truly national work advances with great activi-
ty, and has already proved highly advantageous to the public service.
The Ordnance Maps, in elaborate accuracy, and even beauty of exe-
cution, surpass every thing hitherto designed.
To determine geometrically the altitude of a mountain requires, it
hence appears, a nice operation performed with some large instrument.
The barometrical mensuration of heights, is therefore, in most cases,
preferred, as much easier and often more exact.
This curious ap-
plication was early suggested, by the objections themselves which
ignorance opposed to Torricelli's immortal discovery of the weight of
our atmosphere. But more than a century elapsed before the im-
provements in mechanics had completely adapted the machine to
that purpose, and experiment combined with observation had ascer-
tained the proper corrections. Barometers of various constructions
are now made quite portable, and which indicate with the utmost
precision the height of the mercurial column supported by the pres-
sure of the atmosphere.
The air which invests our globe, being a fluid extremely compres-
sible, must have its lower portions always rendered denser by the
weight of the incumbent mass. To discover the law that connects
NOTES AND ILLUSTRATIONS.
491
the densities with the heights in the atmosphere, it is only requisite,
therefore, to apply the fact which experiment has established,—that
the elasticity counterbalancing the pressure is exactly proportioned
to the density. The elasticity of the air at any point of elevation, is
hence measured by a column possessing the same uniform density,
with a
certain constant altitude. Let AB denote the height
of this equiponderant column, and the perpendicular BI its densi-
ty; and suppose the mass of air below to be distinguished into
numerous strata, having each the same thickness BC. It is evident
that the weight of the minute stratum at B will be expressed by
BC; whence AB is to AC, or BI to CK, as the pressure at B to
the augmented pressure at C, and therefore the density at C is de-
noted by CK.
Again, having
joined IC and
drawn KD pa-
A
rallel, BI: CK::
JKLMNO

BCDEFGH
BC: CD; and consequently CD will, on the same scale of den-
sity, express the weight of the stratum at C. Hence, AC is to
CD, as CK to DL, or as the density at C is to that at D. It thus
appears, that, repeating this process, the densities BI, CK, DL, &c.
of the successive strata form a continued geometrical progression.
But the same relation will evidently obtain at equal though sensible in-
tervals. Thus, the density of the atmosphere is reduced nearly to
one half, for every 3 miles of perpendicular ascent. At 7 miles in
height, the corresponding density is one-fourth; at 10 miles, one-
eighth; and at 14 miles, one-sixteenth,
The difference of altitude between two points in the atmosphere,
is hence proportional to the difference of the logarithms of the cor-
responding densities or vertical pressures. But the heights of
mountains may be computed from barometrical measurement
to any degree of exactness, by a simple numerical approximation.
Since AB, AC, AD, &c. are continued proportionals, it fol-
lows that AB : BC::AB+AC+AD, &c.; BC + CD + DE, &c.
or BH, Let n denote the number of sections or strata contain-
ed in the mass of air, and (AB + AH) will nearly express the
N
2
sum of the progression AB, AC, AD, &c. ; wherefore, AB+ AH
192
NOTES AND ILLUSTRATIONS.
J
BH:: 2AB: nBC, or the absolute difference of altitude. The height
AB of the equiponderant column, reduced to the temperature of
freezing water, is nearly 26,000 feet; and hence this general rule,—·
As the sum of the mercurial columns is to their difference, so is the
constant number 52,000 to the approximate height. This number is
the more easily remembered, from the division of the year into
weeks.
Two corrections depending on the variation of temperature are be-
sides required. Mercury expands about the 5,000th part of its bulk,
for each degree of the centigrade scale; and hence the small addi-
tion to the upper column will be found, by removing the decimal
point four places to the left, and multiplying by twice the difference
of the attached thermometers. But the correction afterwards ap-
plied to the principal computation is of more consequence. Air has
its volume increased by one 250th part, for each degree of heat on
the same scale. If therefore the approximate height, having its
decimal point shifted back three places, be multiplied by twice the
sum of the degrees on the detached thermometers, the product will
give the addition to be made.
An example will elucidate the whole process. In August 1775,
General Roy observed the barometer on Caernarvon Quay at 30.091
inches, the attached thermometer being 15°.7, and the detached
15º.6 centigrade, while on the Peak of Snowdon the barometer stood
at 26.409, the attached thermometer marking 10°.0, and the detach-
ed 8°.8. Here, twice the difference of the attached thermometers is
11°.4, which multiplied into .00264 gives .030, for the correction of
the upper barometer. Next, 30.091 + 26.439: 30.091 — 26.439,
or 56.530: 3.652 :: 52000: 3359. Again, twice the sum of the
degrees marked on the detached thermometers is 48.8, which mul-
tiplied into 3.359 gives 164; wherefore, the true height of Snowdon
above the Quay of Caernarvon is 3359+164, or 3533 feet.
This mode of approximation may be deemed sufficiently near, for
any heights which occur in this island; but greater accuracy is at-
tained by assuming intermediate measures. To illustrate this, I
shall select another example. At the very period when General
Roy was making his barometrical observations at home, Sir George
Shuckburgh Evelyn found the barometer to stand at 24.167 on the
summit of the Mole, an insulated mountain near Geneva, the attach-
NOTES AND ILLUSTRATIONS.
493
•.
=
ed and detached thermometers indicating 140.4 and 130.4, while
they marked 16°.3 and 17°.4 at a cabin below and only 672 feet
above the lake, the altitude of the barometer at this station being
28.132. Now, 3.8X.0024.009, and 24.167 +009 24,176;
the arithmetical mean between which and 28.132 is 26.154;
and hence, separately, 50.330: 1.978: : 52000: 2044, and
54.286: : 1.978 :: 52000: 1895. Wherefore, joining these two parts,
2044 + 1895, or 3939 expresses the approximate height. The final
correction is 61.6×3.939 — 243, and consequently the Mole has its
summit elevated 4354 feet above the lake of Geneva, and 6082
above the level of the sea.
In general, let A and Anb denote the correct lengths of the
columns of mercury at the upper and the lower stations; the ap-
proximate height of the mountain will be expressed by
b
(2A +6+
b
b
+
2A +36 2A+56
b
J
+
2A+2n-1.b
)52000.
If n were assumed a large number, the result would approach to
the accuracy of a logarithmic computation, though such an extreme
degree of precision will be scarcely ever wanted.
To expedite the calculation of heights from barometrical observa-
tions, I have now caused Mr Cary, optician in London, to make for
sale a sliding-rule, of an easy and commodious construction. That
small instrument, which should be accompanied with a barometer
of the lightest and most portable kind, will be found very useful to
mineralogical travellers who have occasion to explore mountainous
tracts. Nothing could tend more to correct our ideas of physical
geography, than to have the principal heights in all countries mea-
sured, at least with some tolerable degree of precision. But the
elevation of any place above the sea may be ascertained very nearly,
from the comparison of even very distant barometrical observations,
especially during the steadiness of the fine season in the happier
climates. In this way, is traced a profile or vertical section, which
exhibits at one glance the great features of a country. As a speci-
men, I have combined and reduced the sections which the celebrated
philosophic traveller Humboldt has given of the continent of Ameri-
ca, running in a twisted direction from Acapulco to Vera Cruz, and
connecting the Pacific with the Atlantic Ocean.
494
NOTES AND ILLUSTRATIONS.
A ACAPULCO.
a Peregrino.
B CHILPANSINGO.
¿ Mescala.
c Tepecuacuilco.
d Puente de Istla.
C CUERNAVACA.
e La cruz del Marques.
D MEXICO.
Z02000
f Venta de Chalco.
g St Martin.
E LA PUEBLA DE LOS ANGELES.
h El Pinal.
i Perote.
k Cruz blanca.
F XALAPA.
10'000
G VERA Cruz.

5000
5000
A
a B b
B b c d Ce
e
D f
D f g Eh
ik F
B
100
200
כני
300
K
E
400

A
G
The divided scale expresses the horizontal distance in miles, while
the parallels, on a much larger scale, mark the elevation in feet. This
profile is really composed of four successive sections, which are dis-
tinguished by opposite shadings. The survey proceeded first along
the road from Acapulco to Mexico, thence to Puebla de los Angeles,
next to Cruz Blanca, and finally to Vera Cruz. These several direc-
tions and distances are expressed in the ground plan.
An attempt is likewise made in this profile, to convey some idea of
the geological structure of the external crust:
Limestone, is represented by straight lines slightly inclined from the
horizontal position.
Basalt, by straight lines slightly reclined from the perpendicular.
Porphyry, by waved lines somewhat reclined.
Granite, by confused hatches.
Amygdaloid, by confused points.
NOTES AND ILLUSTRATIONS.
495
But the easiest way of estimating within moderate limits the
elevation of a country, is founded on the difference between the
standard and the actual mean temperature as indicated by deep
wells or copious and shaded springs. Professor Mayer of Göttingen,
from a comparison of distant observations on the surface of the
globe, proposed a formula, which, with a slight modification, appears
to exhibit correctly the temperature of any place at the level
of the sea. Let denote the latitude; and 29 cosp², or 14½ suvers 24,
will express, in degrees of the centigrade scale, the medium heat on
the coast. But the gradations of climate are more easily conceiv
ed by help of a geome-
trical diagram. From
the centre C, draw
straight lines to the se-
veral degrees of the
quadrant, and cutting
the interior semicircle;
then, the radius CA de-
noting 29°, the perpen-
diculars from the points
of section will intercept
segments proportional to
the mean temperature
expressed on DE.
B
180

70
03
50
10
30
20
10
W
610 50 40 30
20 FO
J
D
5
10
15
20
25 I
The higher regions are invariably colder than the plains; and I
have been able, after a delicate and patient research, to fix the
law which connects the decrease of temperature with the altitude.
If B and b denote the barometric pressure at the lower and upper
B
stations; then will
b
-음​) 25
-) 25 express, on the centigrade scale,
the diminution of heat in ascent. Hence, for any given latitude,
that precise point of elevation may be found, at which eternal frost
b
prevails. Put x = and the standard temperature; then
B
(-—-— — x ) 25 = t, or x²+.04tr=1, which quadratic equation be-
ing resolved, gives the relative elasticity of the air at the limit of con-
gelation, whence the corresponding height is determined. From
these data the following table has been calculated.
496
NOTES AND ILLUSTRATIONS.


Mean temperature at the
Lati-
tude.
level of the Sea.
Centigrade.
Fahrenheit,
Height of Curve
of Congelation
Feet
Lati-
tude.
Mean temperature at the
level of the Sea.
Height of Curve
of Congelation.

Centigrade.
Fahrenheit.
Feet.
0° 29°.00
84°.2
15207
46°
13°.99
57°.
7402
183
28.99
84.2
15203 47 13.49
56.3
7133
2
28.96 84.1
15189 48
12.98
55.4
6865
28.92
84.0
15167 4.9
12.48
54.5
6599
4
28.86
83.9
15135
50
11.98
53.6
6334
567∞
28.78 83.8
15095
28.68 83.6 15047
51
11.49
52.7
6070
52
10.99
51.8
5808
28.57 83.4
14989
53
10.50
50.9
5548
8 28.44 83.2
14923
54
10.02
50.0
5290
9
28.29
82.9
14848
55
9.54
49.2
5034
10
28.13 82.6
14764 56
9.07 48.3
4782
11
27.94 82.3
14672 57
8.60
47.5
4534
12
27.75 82.0
14571 58 8.14
46.6
4291
13
27.53
81.6
14463 59 7.69
45.8
4052
14
27.30
81.1
1 1345
60
7.25
45.0
3818
15
27.06
S0.7
14220
61
6.82
44.3
3589
16 26.80 $0.2
14087
62
6.39
43.5
3365
17 26.52 79.7
13947
63
5.98
42.8
3145
18
26.23 79.2
13798
64
5.57
42.0
2930
19 25.93 78.7
13642
65
5.18
41.3
2722
20
25.61
78.1
13478 66
4.80
40.6
2520
21
25.28
77.5
13308 67
4.43
40.0
2325
22
24.93 76.9
23 24.57
24 24.20
13131
68
4.07
39.3
2136
76.2
12946 69
3.72
38.7
1953
75.6
12755
70
3.39
38. 1
1778
25 23.82 74.9
12557
71
3.07
37.5
1611
26 23.43 74.2
12354
72
2.77
37.0
1451
27 23.02 73.6
12145
73
2.48
36.5
1298
28 22.61 72.7
11930
74 2.20
36.0
1153
29
22.18 71.9
30 21.75 71.1
31 21.31
20.86
32
11710
75 1.94
35.5
1016
70.3
11253 77
69.5
11018 78
33 20.40
68.7
10778 79
2722
11484 76 1.70 35.1
887
1.47
34.6
767
1.25
34.2
656
1.06 33.9
552
34 19.93
67.9
10534
SO
.87
32.6
457
35 19.46
67.0 10287
36
18.93
66.2 10036
81
.71
33.3
371
82
.56
33.1
294
37 18.50
38 18.01 64.4 9523
65.3
9781
$3
.43
32.8
226
$4
.32
32.6
167
39
17.51 63.5
9263
85
.22 32.4
117
40
17.02
62.6
9001
$6
.14
32.3
76
41
16.52 61.7
8738
ST
.08
32.2
44
42
16.02 60.8
8473 SS
.04 32.1
20
43
15.51
59.9
8206
89
.01
32.0
5
44
15.01
59.0
7939
90
.00
32.0
45 14.50
58.1
7671
NOTES AND ILLUSTRATIONS.
497
This table will facilitate the approximation to the altitude of any
place, which is inferred either from its mean temperature or its
depth below the boundary of perpetual congelation. The decrements
of heat at equal ascents are not altogether uniform, but advance
more rapidly in the higher regions of the atmosphere. At moderate
elevations, however, it will be sufficiently near the truth, to assume
the law of equable progression, allowing in this climate one degree
of cold by Fahrenheit's scale for every ninety yards of ascent.
Thus, the temperatures of the Crawley and Black springs on the
ridge of the Pentland hills, were observed by Mr Jardine, where
they first issue from the ground, to be 46°.2 and 45°; which, com-
pared with the standard temperature at the same parallel of latitude,
would give 567 and 891 feet of elevation above the sea. The real
heights found by levelling were respectively 564 and 882; a coin-
cidence most surprising and satisfactory.-This ready mode of esti-
mation claims especially the attention of agriculturists.
The rule stated above for computing the measurements by the ba-
rometer, seems to give results somewhat less, on the whole, than
those which are obtained from geometrical observations. It would
ensure greater accuracy, perhaps, to view the approximate height as
answering to a temperature one degree under the point of congela-
tion; and consequently, in applying the last correction, to add unit to
the indications of the detached thermometers. But the whole subject
demands a more thorough investigation. The elasticity of air is affect-
ed by moisture as well as heat, although the want of an exact in-
strument for measuring the former has hitherto prevented its in-
fluence from being distinctly noticed.
When the hygrometer which I have invented shall become better
known to the public, it may not seem presumptuous to expect, in due
time, more correct data concerning the modifications of the atmos.
phere. Yet, after all, in ascertaining the volume of a fluid subject
to incessant fluctuation, it would be preposterous to look for that
consummate harmony which belongs exclusively to astronomical sci-
ence; nor can I help regarding the introduction of some late re-
finements into the formula for measuring heights by the barometer,
and which would embrace the minutest anomalies of atmospheric
pressure, arising from the influence of centrifugal force, combined
with the diminution of gravity in receding from the earth's centre,-
as an utter waste of the powers of calculation.
I i
*..
3
1
!
1
i
498
*
NOTES AND ILLUSTRATIONS.
I shall now subjoin a concise table of the most remarkable heights
in different parts of the world, expressed in English feet. The al-
titudes measured by the barometer are marked B, while those de-
rived from geometrical operations, and taken chiefly from the last
work of Colonel Mudge, are distinguished by the letter G.
Snæ Fiall Jokul, on the north-west point of Iceland,
4558 G
Hekla, volcanic mountain in Iceland,
Pap of Caithness,
Ben Nevis, Inverness-shire,
Cairngorm, Inverness-shire,
Ben Lawers, Perthshire,
Ben More, Perthshire,.
Schihallien, Perthshire,
Ben Ledi, Perthshire,
Ben Lomond, Stirlingshire,
Lomond Hills, east and west, Fifeshire,
Soutra Hill, on the ridge of Lammer muir,
Carnethy, highest point of the Pentland ridge,
Tintoc, Lanarkshire,
3950 G
1929
4380 B
4080 B
4015 B
3870 B
3281 G
3009 B
3240 B
1466 and 1721 G
1716 G
1700
1720 B
Leadhills, the house of the Director of the mines,
1564
Queensbery Hill, Dumfries-shire,
2259 G
Dunrigs, Roxburghshire,
2408 G
Elden Hills, near Melrose, Roxburghshire,
1364 G
Crif Fell, near New Abbey in the Stewartry of Kirkcudbright, 1831 G
Goat Fell, in the Isle of Arran,
2950 B
Paps of Jura, south and north, in Argyllshire,
2359 and 2470
Snea Fell, in the Isle of Man,
2004 G
Macgillicuddy's Reeks, county of Kerry,
3404
Mourne Mountains, county of Down,
2500
Helvellyn, Cumberland,
3055 G
Skiddaw, Cumberland,
3022 G
Saddleback, Cumberland,
2787 G
Whernside, Yorkshire,
2384 G
Ingleborough, Yorkshire,
2361 G
Shunnor Fell, Yorkshire,
2329 G
Snowdon, Caernarvonshire,
3571 G
Cader Idris, Caernarvonshire,
2914 G
NOTES AND ILLUSTRATIONS.
499
靠
​1
Beacons of Brecknock,
2862 G
Plynlimmon, Cardiganshire,
Penmaen Mawr, Caernarvonshire,
Malvern Hills, Worcestershire,
2463 G
1540 G
*1444 G
Rippin Tor, Devonshire,
Cawsand Beacon, Devonshire,
Brocken, in the Hartz-forest, Hanover,
1792 G
1549 G
3690
Schneekopf, in Silesia,
4950
Priel, in Austria,
6565
Peak of Lomnitz, in the Carpathian ridge,
8640
Mont Blanc, Switzerland,
15646 G
Village of Chamouni, below Mont Blanc,
3367 G
Jungfrauhorn, Switzerland,
13730
St Gothard, Switzerland,
9075
Hospice of the Great St Bernard, on the passage to Italy,
8040 B
Village of St Pierre, on the road to Great St Bernard,
Passage of Mont Cenis,
Ortler Spitze, in the Tyrol,
5338 B
6778 B
15430
Rigiberg, above the lake of Lucerne,
Dole, the highest point of the chain of Jura,
540S
5412 B
Mont Perdu, in the Pyrenées,
11283
Loneira, in the department of the high Alps,
1445!
Peak of Arbizon, in the department of the high Pyrenées,
8344
Puy de Dome, in Auvergne,
5197
Summit of Vaucluse, near Avignon,
2150
Soracte, near Rome,
9271 G
Monte Velino, in the kingdom of Naples,
8397 G
Mount Vesuvius, volcanic mountain beside Naples,
3978
Etna, volcanic mountain in Sicily,
10963 B
St Angelo, in the Lipari Islands,
5260
Top of the Rock of Gibraltar,
1439 B
Mount Athos, in Rumelia,
3353
Diana's Peak, in the Island of St Helena,
2692
Peak of Teneriffe, one of the Canary Islands,
Ruivo Peak, the highest point in the Island of Madeira,
Table Mountain, near the Cape of Good Hope,
12358 B
5162
3520
Chain of Mount Ida, beyond the plain of Troy,
Chain of Mount Olympus, in Anatolia,
4960
6500
500
NOTES AND ILLUSTRATIONS.
Italitzkoi, in the Altaic chain,
Awatsha, volcanic mountain in Kamtchatka,
Taganai, in the Uralian chain,
The Volcano, in the Isle of Bourbon,
Ophir, in the centre of the Island of Sumatra,
St Elias, on the western coast of North America,
Chimborazo, highest summit of the Andes,
10735
9600
4912
7680
13842
12672
21440 B
Antisana, volcanic mountain in the kingdom of Quito,
19150 B
Cotopaxi, volcanic mountain in the kingdom of Quito,
18890 B
Tonguragua, volcanic mountain, near Riobomba, in Quito,
Rucu de Pichincha, in the kingdom of Quito,
Heights of Assuay, the ancient Peruvian road,
16579 B
15940 B
15540 B
Peak of Orizaba, volcanic mountain east from Mexico,
17390 G
Lake of Toluca, in the kingdom of Mexico,
12195 B
City of Quito,
9560 B
City of Mexico,
7476 B
Silla de Caraccas, part of the chain of Venezuela,
8640 B
Blue Mountains, in the Island of Jamaica,
7431
Pelée, in the Island of Martinique,
5100
Morne Garou, in the Island of St Vincent's,
5050
Mount Misery, in the Island of St Christopher's,
3711
The
I shall conclude with briefly stating the French measures.
Parisian foot was to the English, or the toise to the fathom, as
1.065777 to 1, or nearly as 16 to 15. The metre, or base of the
new system, and equal to 39.371 English inches, ascends decimally,
forming the decametre or perch, the hectometre, the kilometre or mile,
and the myriametre or league, equivalent to 6.213856 of our miles ;
and descending by the same scale, it forms successively the decimetre
or palm, the centimetre or digit, and the millimetre or stroke. The
square of the decametre constitutes the are, and that of the hectume-
tre, the hectare or acre, and equal to 2.47117 English acres. The
cube of a metre, or 35.3171 feet, forms the unit of solid measure
or the stere, that of a decimetre, or 61.028 inches forming the litre
or pint; and the weight of this bulk of water at its greatest contrac-
tion makes the kilogramme or pound, equivalent to 2.1133 pounds
Troy, the gramme answering to 15.444 grains.
FINIS.


Early in the ensuing Season will be Published,
BY THE SAME AUTHOR,
In One Volume Octavo,
1
A
VIEW
OF THE
FACTS ASCERTAINED
CONCERNING
HEAT,
AND ITS RELATIONS
WITH
AIR AND MOISTURE.
* The object of this short treatise is to convey, in a po-
pular form, a distinct though general conception of the mo-
dern doctrine of Heat, disengaged as much as possible from
all hypothetical reasonings. It will trace the progress of ex-
periment from the earlier times, to the recent discovery of the
artificial production and conservation of extreme cold.


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UNIVERSITY OF MICHIGAN
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