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ELE MEN T S
© F.
MATHEMATICS.
COMPREHENDING
GEO MET Ŕ Y.
MENSURATION
CONIC SECTIONS.
SPHER. I C S.
ILLUSTRATED WITH 30 COPPER-PLATES.
FOR THE USE OF SCHOOLS.
BY
JOHN WEST, 116 - 1817
1796
ASSISTANT TEACHER OF MATHEMATICS IN THE
UNIVERSITY OF S. ANDREWS.
EDINBURG H:
PRINTED FOR WILLIAM CREECH;
AND SOLD IN LONDON BY
T. LONG MAN AND T. CADELL.
M,DCC,LXXXIV.
..
:
FA A CE.
PRE F
С
i
So
O ME account of the following work, and
of the motives which induced me to pu-
bliſh it, will perhaps be expected by the reader.
I am perfectly difpofed to gratify this reafon-
able expectation; and the more fo, as an at-
tempt, which propoſes great innovations in the
long eſtabliſhed ſyſtem, requires an apology.
My original intention was not to include the
Firft Elements of Geometry. It then appeared
to me, that the Elements of Euclid might con-
tinue to ferve the purpofe which they had done
for many ages. My defign was only to build
upon the foundation which that illuftrious au-
thor had laid, and, under the feveral heads of
Conic Sections, Menfuration, and Spherics, to
complete a fyftem of Geometry for the inftruc-
'tion
J
!
iv
PREFA SC AE. I
!
tion of youth. The expediency and utility of
fuch a work would, I thought, have occurred to
every teacher. I even began to wonder that it
had not been attempted. Though there are
many excellent performances in our own lan-
guage, on every branch of mathematical ſcience,
et the hi
price which books of feience com-
monly bear, renders the purchaſe of them too
expenfive for general ufe. Or so bad
In the execution of this work, I had many
difficulties to encounter. As I propoſed to eſta-
blish it entirely upon geometrical principles, I
could derive no affiftafice from fome of the beſt
authors on the fubjects treat, who have intro-
duced both Algebra and Fluxions, into their de-
monftrations. Befides, the concife plan which
I meant to follow, rendered the works of o-
thers, where diffufe, of little advantage to me.
I had, therefore, many new demonftrations to
inveſtigate. But the greateft difficulty arofe
from the narrownefs of the bafis on which I
had to build. In the prefent advanced ſtate of
Geometry, there are many ufeful Theorems,
which, though not in Euclid, are confidered by
moft Geometers as elementary propofitions.
Thefe
[
A
}
}
PREFA C qE.q
300
Theſe I could not admit without proof; and,
to demonftrate them, I must either have incum-
bered the demonftrations of almoſt every pro-
pofition, or have interfperfed the work with
Lemmas. Both of theſe methods appeared to
me equally unfyftematical and inelegant. To
begin, therefore, from the very foundation,
feemed the best method to communicate con-
fiftency and order to the fyftem I intended to
form.
the ends to moitunas od ni
The Elements of Euclid have continued for
two thouſand years to hold poffeffion of the
Schools. This fingular fact, and the bad fuc-
cefs of the attempts that have been made to
fupplant them, have been reckoned convincing
proofs of their fuperior excellence. But thefe,
attempts may alfo ferve to fupport what I fug-
geſted above, that, confidering the improved ſtate
of mathematics, Euclid's Geometry is now in-
adequate and defective, as an elementary work,
The bad fuccefs of new elements may be owing
partly to the prejudice which ſtrongly prevails
in favour of Euclid, and which ferves to coun-
terbalance the merit of improvements; but
chiefly to the peculiar objections to which their
imper-
{
{
vi
PREFACE.
imperfections feem to be liable. If, befides, it be
confidered, that the most exceptionable part of
the ancient Elements, is that in which fucceeding
writers have moſt failed, it will not appear fur-
prifing that Euclid's name fhould have triumph-
ed over partial improvements.
7
The doctrine of Proportion is perhaps the
most important in mathematics, and, as treated
by Euclid, difcovers great penetration and fa-
gacity. In order to include incommenfurables,
he affumes, in his definition, a general property
of proportionals, very remote from common
apprehenfion. On this property, he builds a
very ingenious and comprehenfive theory; yet
not with all that clearness and folidity for which
the reft of his Elements is fo remarkable.
deed, his manner of treating this difficult and
abſtract ſubject, is fo obfcure, as to render it
It was a
almoft unintelligible to the learner.
defire to render this intricate and uſeful doctrine
more fimple and intelligible, which determined
me to attempt a new theory of proportion, and
to introduce a new fyftem of the Elements of
Geometry as the first part of my work.
267
૪
In-
What
C
PREFACE.
vii
T
What I offer to the Public, on the fubject of
Proportion, will, I truft, fufficiently evince that
my labours have not been altogether unfuccefs-
ful. And the approbation with which they
have been honoured by men of diftinguished
abilities in mathematical refearches, gives me
ground to hope that they may be ufeful. The
principle I have affumed is the moft fimple and
obvious; and the theory I have founded upon
it, is not lefs general than that of Euclid.
As to the plan I have followed in the reft of
the Elements, I have conftantly kept in view the
celebrated Ancient. Indeed, I have never de-
parted from him, unlefs I could give more eafy
demonftrations, or could fubftitute more uſeful
or more general theorems. I reflect, with plea-
fure, that, notwithſtanding confiderable additions
have been made to the former ftock of elemen-
tary knowledge, yet the number of propofitions
is fo much abridged, as to render the study of
it a taſk of much lefs labour and difficulty.
I am aware of an objection, arifing from the
conciſeneſs of this work. It will be faid, per-
haps, that many propofitions are left undemon-
ftrated, and annexed as corollaries to others,
while
$
viii
PREFAC E.
ง
while their dependence on thefe is not obvious.
to a beginner. But I have to obſerve, that ſuch
propofitions ferve to fharpen the genius, and to
exerciſe the invention, of youth; and that, con-
fidering how much they are accustomed to ex-
erciſes of memory in the firft branches of edu-
cation, it is of the utmost importance to engage
them now to exert the powers of the under-
ftanding. This exertion cannot be effected by
long formal demonftrations. Many ſtudents
eſtimate the difficulty of their task by its length;
they wish to continue to lay the burden on their
memory; and they imagine, that to repeat is
the fame thing as to comprehend. That they
may be induced, therefore, to apply all the
powers of reaſon, and may reap the moſt ad-
vantage from fuch application, I have thought
it requifite, not only to make the ftyle of de-
monſtration ſimple and concife, but alſo to leave
fuch propofitions, as eafily follow from thoſe
that are demonftrated, without further evidence
of their truth: For, if the evidence arifing from
thence be explained, when neceffary, by the
teacher, they will be much fooner apprehended
than in the method of formal demonftration.
To illuftrate this remark, fuppofe the learner
perfectly
1
PREFACE.
ix
1
.
perfectly maſter of the Fourth Propofition of
the Firſt Book of Euclid, I affirm that he will
be more eaſily convinced of the truth of the fol-
lowing propofition, by deducing it from thence,
and imagining the triangles to be ifofceles, than
by the long and intricate demonftration of Eu-
clid. I would not, however, be underſtood to
infinuate that equal advantage will be gained in
every inftande in which I have converted into
corollaries the propofitions which Euclid has
demonſtrated. I only prefume to infer, from
this and other inftances which might be addu-
ced, that the method here purfued, inftead of
being incumbered with difficulties, as Euclid's
admirers will be apt to alledge, will tend much
to facilitate rational, and uſeful progreſs in the
Elements. deren gian son eluent
AR
1
A 100 Mwen bas sigmi noiss
In the Sixth Book, which treats of folids, I
have omitted the propofitionis refpecting the
meaſuring of the fphere; firft, becauſe they are
demonftrated, in the Third Book of Menfura-
tion, in a manner, not confiftent perhaps with
the ancient ftrictnefs of Geometry, but equally
convincing; and with much more concifenefs,
than a regard to that fitritnefs, which I have
b
always
i
PREFACE.
1
always endeavoured to preferve in the Ele-
ments, would have admitted; and, fecondly,
becauſe I would refer the young proficient in
Geometry to Archimedes's Treatiſe on the
Sphere and Cylinder, where, in the fimple and
accurate, though prolix method, of ancient de-
monftration, his mind will trace with peculiar
pleaſure the very ſteps by which that illuftrious
philofopher arrived at theſe important difco-
veries.
1
Throughout the whole work, I have endea-
voured to adopt the arrangement which is moſt
fyftematic, not that which is commonly obfer-
ved in teaching; becaufe the latter may be va-
ried, while the former, remaining the fame,
tends to give clear and diftinét views of the fe-
veral fubjects. I have alſo interiperied a great
variety of Problems and Theorems, which, tho'
intended for exercifes to the learner, are not un-
worthy the notice of thoſe who are farther ad-
vanced, and take pleaſure in mathematical fpe-
culations.
;
ELEMENTS
O F
GEOMETR Y.
N
A
ཡར�་་བར
I.
G
воок
во о к І.
DEFINITIONS.
EOMETRY is the fcience in which the
properties and relations of thofe quantities
which have extenfion, called Magnitudes, are invefti-
gated.
Extenfion is diftinguifhed into length, breadth,
and thickneſs.
2. A line is that which has only length. The
bounds or extremes of a line are points.
3. A furface is that which has only length and
breadth.
The bounds of a furface are lines.
A
4. A
ELEMENT S
BOOK I.
4. A right or ftraight line is that which every
where tends the fame way.
5. A plane furface is that in which any two points
being taken, the ftraight line that joins them lies
wholly in that ſurface.
6. The opening between two ftraight lines which
meet in the fame point, is called an angle.
7. When one ſtraight line, ftanding upon another,
makes the adjacent angles equal, each of them is
called a right angle, and the former line is faid to be
perpendicular to the latter.
8. An acute angle is that which is lefs than a right
angle.
9. An obtufe angle is that which is greater than a
right angle.
10. The diſtance of two points is the ftraight line
joining them. And the diſtance of a point from a
ftraight line is a perpendicular drawn to the line.
from that point.
11. Parallel ftraight lines are fuch as lie in the
fame plane, and, though produced to any length both
ways, do not meet.
12. A figure is a bounded ſpace. When it is a
plane furface, it is called a plane figure.
13. A plane rectilineal figure is that which is
bounded by ſtraight lines.
14. Every plane figure, bounded by three ſtraight
lines, is called a triangle, of which the three ſtraight
lines
*
BOOK I. OF GEOMETRY.
lines are the fides. That fide upon which the triangle:
is conceived to ftand, is named the bafe, and the op-
poſite angular point the vertex.
15. An ifofceles triangle is that which has two fides
equal.
16. An equilateral triangle is that which has all
its fides equal.
17. A right angled triangle is that which has a
right angle. The fide oppofite to the right angle is
called the hypothenufe.
18. An obtuſe angled triangle is that which has an
obtufe angle.
19. Every plane figure, bounded by four ftraight
lines, is called a quadrangle or quadrilateral, of
which the ſtraight line joining two oppofite angles is
called a diagonal.
20. A parallelogram is a quadrilateral, of which
the oppofite fides are parallel.
21. A rectangle is a parallelogram which has all its
angles right.
22. A rhombus is a parallelogram which has all
its fides equal.
23. A fquare is a parallelogram which has all its
fides equal, and all its angles right.
24. Every plane figure, bounded by more than
four ftraight lines, is called a polygon.
25. A circle is a plane figure bounded by one line,
called the circumference, which is fuch, that all
ftraight
•
4
BOOK I.
ELEMENTS
ſtraight lines drawn to it, from a certain point with-
in the figure, called the center, are equal; and theſe
ftraight lines are named the radii of the circle.
POSTULATE S.
1. Let it be granted, that, from any given point, to
any other given point, a ſtraight line may be drawn.
2. That a terminated ftraight line may be produ
ced to any length.
3. That a circle may be deſcribed about any center
at any diſtance from that center.
4. And that any magnitude, however ſmall, may
be multiplied fo as to exceed a given magnitude of
the fame kind.
AXIOM S.
1. Thoſe magnitudes which, if applied to one ano-
ther, would coincide, or exactly fill the ſame ſpace,
are equal.
2. Things that are equal to the fame are equal to
one another.
3. If equals be added to equals, the wholes are
equal; and, if equals be taken from equals, the re-
mainders are equal.
4. IF
.BOOK I.
5
OF GEOMETRY.
+
1
4. If equals be added to unequals, or taken from
them, the fums, or remainders, will ftill be unequal,
and have the fame difference.
5. The doubles, or halves of equal things, are e-
qual.
6. Every whole is greater than its part, and equal
to all its parts taken together.
7. From any one point, to any other point, more
than one ſtraight line cannot be drawn.
8. All right angles are equal to one another.
9. A ftraight line cannot firft come nearer to a
ſtraight line, and then go farther from it, without
cutting it; nor can a ftraight line firſt recede from
another, and then come nearer to it.
PROPOSITION
6
Book I.
ELEMENTS
Fig. 1.
દી
poſt. 3.
b poft. 1.
c def. 25.
d ax. 2.
PROPOSITION I. PROBLEM.
Upon a given straight line to defcribe an equilateral
triangle.
Lit
ET AB be the given ftraight line, upon which
it is required to defcribe an equilateral triangle.
About the centers A, B, with the fame radius AB,
let the circles BCD and ACE be deſcribed a ; and,
from the point C, where they interfect, let the ſtraight
lines CA, CB, be drawn to the points A, B'; then
ABC is an equilateral triangle.
For AC is equal to AB °, becauſe they are radii of
the fame circle; BC is equal to AB for the fame
reafon; and, by confequence, AC is equal to BC d.
Fig. 2.
2 poſt. 1.
PROP. II. PROB.
From a given point to draw a straight line equal to a
given straight line.
Let A be the given point, and BC the given
ftraight line; it is required to draw from A a
ſtraight line equal to BC. Let the ſtraight line AB
be drawn, and the equilateral triangle ADB defcri-
bed
1
Book I.
7
OF GEOMETRY.
:
b
с
d
e
prop. 1.
poft. 3.
poſt. 2.
ed upon it. About the center B, with the radius
BC, let the circle CEF be defcribed, interfecting DB
produced in Eª. And about the center D, with the
radius DE, let the circle EGH be deſcribed, inter-
fecting DA produced in G. Then AG is a ſtraight
line drawn from A equal to BC. For, the ftraight
lines DG, DE are equal, becauſe they are radii of the def. 25.
fame circle, and DA, DB parts of theſe are equal,
becauſe they are fides of an equilateral triangle; there-
fore the remainder AG is equal to the remainder BE. fax. 3
But BC is likewife equal to BE; for BC, BE are
radii of the circle CEF; confequently AG is equal
to BC 8.
8 ax. 2.
•
PROP. III. PROB.
From the greater of two given straight lines to cut off a
part equal to the lefs.
Let AB and C be the two given ftraight lines, of Fig. 3.
which AB is the greater; it is required to cut off
from AB a part equal to C. From the point A let
there be drawn the ftraight line AD equal to C, a prop. 2.
and about the center A, with the radius AD, let the
circle DEF be defcribed, interfecting AB in E:
Then AE is a part cut off from AB, equal to C.
For
8
Book 1,
ELEMENTS
ax. 2.
For AE is equal to AD, becauſe they are radii of
the fame circle; and C is equal to AD by conſtruc-
tion; therefore AE is equal tỏ C³.
Fig. 4.
a
EDF
ax. I
bax. 7.
PROP. IV.
THEOREM.
If two triangles have two fides of the one equal to two
fides of the other, each to each, and have likewife the
angles contained by thefe fides equal, they fhall be equal
in every respect.
Let ABC, DEF be two triangles, having the fide
AB equal to DE, AC equal to DF, and the angle
BAC equal to DEF; then fhall the remaining fides
BC, EF be equal, and the angles fubtended by equal
fides equal, viz. ABC to DEF, and ACB to DFE ;
and the whole triangle ABC fhall be equal to the
whole triangle DEF.
For, fince the angles BAC, EDF are equal, if they
were applied to one another they would coincide;
the fide AB would fall upon DE, and AC upon
DF2. At the fame time, the point B would fall up-
on E, becauſe AB is equal to DE; and the point C
upon F, becauſe AC is equal to DF. Confequently
the ſtraight line BC would coincide with EF, the
angle ABC with DEF, the angle ACB with DFE,
and the whole triangle ABC with the whole triangle
DEF.
•
Book I. OF GEOMETRY.
9
DEF. And therefore the triangles muſt be equal in
every refpect".
COROLLARY. If two fides of a triangle be equal,
the angles oppofite to them are equal.
a
ax,
PROP. V. THEO R.
If two triangles have two angles of the one equal to two
angles of the other, each to each, and have likewife
the fides between thefe angles equal, they fhall be equal
in every respect.
Let ABC, DEF be two triangles, having the angle Fig 4-
ABC equal to DEF, the angle ACB equal to DFE,
and the fide BC equal to EF; then fhall the remain-
ing angle BAC be equal to the remaining angle
EDF, and the fides fubtending equal angles equal;
viz. AB to DE, and AC to DF, and the whole
triangle ABC fhall be equal to the whole triangle
DEF.
For, fince the fides BC, EF are equal, if they
were applied to one another they would coincide. ax. 1.
And, at the fame time, the triangle ABC being ap-
plied to DEF, the fide AB would fall upon DE,
becauſe the angle ABC is equal to DEF, and the ſide
AC upon DF, becauſe the angle ACB is equal to
DFE. Therefore the point A, where AB and AC
B
meet,
10
BOOK L
ELEMENT S
A
• def. 4.
Fig. 5.
6.
7.
}
ax. 1.
meet, would fall upon D. And thus the fide AB
would coincide with DE, the fide AC with DF, the
angle BAC with EDF, and the whole triangle ABC
with the whole triangle DEF. Confequently theſe
triangles are equal in every refpect.
COR. If two angles of a triangle be equal, the fides
oppoſite to them are equal.
44
PROP. VI., THEO R.
If two triangles have the three fides of the one equal to the
three fides of the other, cach to cach, they fhall be c-
qual in every respect.
Let ABC, DEF be two triangles, having the fide
AB equal to DE, AC to DF, and BC to EF; then
the angles fubtended by equal fides fhall be equal;
viz. ABC to DEF, ACB to DFE, and BAC to EDF
and the whole triangle ABC fhall be equal to the
whole triangle DEF.
For, fince the baſes BC, EF are equal, if applied
to one another they would coincide; and then, the
triangles being on different fides of their common
bafe, the ftraight line DA, which joins their verti-
ces, would interfect the common bafe EF, or that
line produced, or would pafs through one of its ex-
tremities.
7
}
If
Book I. OF GEOMETRY.
I I
6.
If DA interfected EF, or EF produced, it would Fig. 5.
form with the two equal fides AB, DE a triangle
DBA, having the angle BAD equal to BDA; and b cor. 4.
with the two et fides AG, DF a triangle DCA,
having the angle CAD
to CDA". Therefore
the whole or remainder, the angle BAC, is equal to
the whole or remainder, the angle EDF ©.
C ax. 3.
If DA paffed through the point F, the fides AC, Fig. 7.
DF would be in one ftraight line, and the angle
BAD equal to BDA ", that is, BAC to EDF, as be-
fore.
Confequently thefe triangles are equal in every re-
fpect.
d
pr. 4.
PRO P. VII. PRO B.
To bifect a given angle.
Let ABC be the given angle; it is required to Fig. 8.
divide it into two equal parts.
In AB let any point D be affumed, and from BC
let BE be cut off equal to BD; let DE be drawn,
the equilateral triangle DFE be defcribed upon DE, ь
and the points B, F joined. The ftraight line BF bi-
fects the angle ABC.
For, let the triangles BDF, BEF be compared; the
fide BD is equal to BE by conſtruction, DF, FE are
equal,
a
pr. 3.
pr. I.
12
Book I.
ELEMENT S
;
pr. 6.
equal, becauſe they are fides of an equilateral
triangle, and BF is common to both; therefore
theſe triangles are equal in every refpect c. Thus,
the angles DBF, FBE fubtended by DF, FE are
equal, and the whole angle ABC is bifected by the
line BF.
Fig. 9.
b
pr. 1.
pr. 7.
pr. 4.
PROP. VIII. PROB.
To bifect a given straight line.
Let AB be the given ftraight line; it is required to
divide it into two equal parts.
Upon AB let the equilateral triangle ABC be de-
fcribed, and let the angle ACB be bifected by the
ſtraight line CD; CD alſo biſects AB.
For the triangles ACD, BCD having AC equal to
CB, CD common to both, and the angle ACD equal
to BCD, are equal in every refpect. Thus, AD is
equal to DB, and the whole line AB biſected in D.
COR. The ftraight line which biſects the vertical
angle of an ifofceles triangle, bifects alfo the bafe at
right angles.
1
PROP.
Book I. OF GEOMETRY.
13
PROP. IX. PROB.
To draw a straight line perpendicular to a given ftraight
line from a given point in the fame.
Let AB be the given ftraight line, and C the given Fig. i.
point in it; it is required to draw from C a ſtraight
line perpendicular to AB.
2
In AC let any point D be affumed, and from CB
let CE be cut off equal to CD. Let the equilateral
triangle DFE be deſcribed upon DE, and the points ↳
F, C be joined. The ftraight line FC is perpendicu-
lar to AB.
For the triangles FDC, FEC having FD equal to
FE, DC equal to CE, and FC common to both, are
equal in every reſpect. Thus, the angle DCF is e-
qual to ECF, that is, the angles at C are right, and
FC is perpendicular to DE or AB ª.
COR. A line drawn from the vertex of an iſofceles
triangle to bifect the bafe, is perpendicular to it, and
bifects the vertical angle.
pr. 3.
pr. 1.
c pr. 6.
a def. 7
PROP.
14
Воок І.
ELEMENTS
+
Fig. 11.
pr. 8.
PROP. X. PROB.
To draw a straight line perpendicular to a given straight
line of an unlimited length, from a given point with-
out it.
Let AB be the given ſtraight line, and C the given
point without it; it is required to draw from Ca
ftraight line perpendicular to AB.
Let any point D be affumed on the other fide of
AB, and about the center C, with the radius CD,
let the circle EDF be defcribed, meeting AB in the
points E, F. Let EF be bifected in G, and the
points G, C be joined; then CG is perpendicular to
AB.
For, the radii EC, CF being drawn, ECF is an
def. 25. ifofceles triangle, and CG a line drawn from the
vertex to bifect the bafe; CG is therefore perpendi-
15.
cor. 9. dular to EF, or AB.
PROP. XI. THEOR.
The angles which one ftraight line makes with another,
upon one fide of it, are together equal to two right
angles.
Let
Book I. OF GEOMETRY.
15
Let the ſtraight line AB make with CD, upon one Fig. 12.
fide of it, the angles ABC, ABD; theſe ſhall be to-
gether equal to two right angles.
If AB be perpendicular to CD, then ABC, ABD
are two right angles. If not, let BE be perpendicu-
lar to it, and BE divides the greater angle ABC pr. 9.
into the two parts EBC, EBA. Now, the one part EBC
с
C 21. 6.
being a right angle, and the other part EBA, added ь def. 7.
to the leſs angle ABD, being equal to another right
angle; it is plain that the whole angle ABC, added
to ABD, is equal to two right angles.
COR. 1. All the fimple angles formed upon one
fide of a ſtraight line, at the farne point in it, are to-
gether equal to two right angles.
COR. 2. All the fimple angles, formed about the
fame point, are together equal to four right angles.
COR. 3. The converfe of the propofition. If two
ftraight lines, drawn from the extremity of another,
make with it two angles, which are together equal to
two right angles, they are in the fame ftraight line.
If ABC, ABD be together equal to two right Fig. 13.
angles, BD is in the fame ftraight line with CB.
For, if not, let BF be the continuation of CB, then
ABC, ABF would be equal to two right angles, and
therefore equal to ABC, ABD; which is impoffible.
}
PROP.
?
16
Book I.
ELEMENTS
Fig. 14.
PROP. XII. THEO R.
If two straight lines cut one another, the oppofite angles
are equal.
Let the two ſtraight lines AB, CD cut each other
in E, the angle AEC fhall be equal to BED, and
CEB to AED. For the angles which AE makes
* pr. 11. with CD are together equal to two right angles'; and
the angles which DE makes with AB are alfo equal
to two right angles"; therefore the two former AEC,
AED are together equal to the two latter AED,
DEB b. And the angle AED, which is common,
being taken away, there remains the angle AEC e-
qual to DEB. In the fame manner, CEB may be
proved equal to AED.
ax. 2.
COR. If from the fame point, in a ftraight line,
two others be drawn on different fides of it, making
the oppofite angles equal, they fhall be in the fame
cor. 3. ftraight line .
II.
*
PROP. XIII.
PROB.
From a given point, in a given straight line, to draw an-
other straight line, that ſhall make with the former an
angle equal to a given angle.
Let
4
Book I.
17
OF GEOMETRY.
1
:
Let AB be the given line, and CDE the given Fig. 15.
angle; it is required to draw a line from the point
A, that ſhall make with AB an angle equal to CDE.
If CDE be a right angle, let AK be drawn per-
pendicular to AB, and it is done. If not, from a pr. 9.
any point G, in CD, let a perpendicular GF be
drawn to DF, produced if neceffary. Then, from Þ
let the perpen-
AB let AH be cut off equal to DF, let the
b
pr. 10.
C pr. 3.
.
dicular HK be drawn
points A, K be joined.
in this manner, will be
equal to FG, and let the
The triangle KAH, formed
equal in every reſpect to the
w
triangle DGF. Therefore the angle KAB bed ax. 8.
equal to CDE, and confequently KAb to Ce
pr. 4.
pr. 11.
PROP. XIV.
THEOR.
If a straight line, falling upon two other ftraight lines,
make the alternate angles equal, thefe two lines arė
every where equally distant from one another.
Let the ftraight line EF, falling upon the two Fig. 16.
17.
•
ftraight lines AB, CD, make the alternate angles
AEF, EFD equal; then AB, CD are every where
equidiftant.
CASE I. When EF is perpendicular to the lines Fig. 16.
AB, CD; the diſtance AC of any point A in one of
them, from the other CD, fhall be equal to EF.
C
For,
18
Book I.
ELEMENTS.
ax. I.
b ax. 7.
C ax. 9.
Fig. 17.
d
pr. 8.
pr. 10.
f pr. 4.
& cor. 12.
def. 11.
In like man-
Therefore the
For, let EB be equal to EA, FD to FC, and B, D
joined; then, if the angle EFC were applied to EFD,
and FEA to FEB, FC would fall upon FD, becauſe
the angles at F are equal, and the point C upon Dª,
becauſe the lines FC, FD are equal.
ner, the point A would fall upon B.
ftraight line AC would coincide with BD, and the
angle ACF with BDF. Confequently BD is perpen-
dicular to CD, and equal to AC. And, fince the
diſtances AC, BD are equal to one another, it is evi-
dent that each of them is equal to EF .
CASE II. When EF is not perpendicular to AB,
CD; let EF be bifected in the point G, from which
let GH be drawn perpendicular to AB, alſo let FK
be equal to HE, and the points G, K joined. The
triangles GEH, GFK, having the angles at E, F e-
qual by hypothefis, and the containing fides equal by
conſtruction, each to each, are equal in every re-
fpect. Hence the angles at G are equal, and con-
fequently HGK a ftraight line; alfo the angles at
H and K equal, and therefore the ftraight line HK
perpendicular to both the lines AB, CD. And thus,
by the Firſt Cafe, the diſtance of any point, in one
of the lines AB, CD, from the other, is equal to
HK.
COR. 1. If a ſtraight line, falling upon two other
ftraight lines, make the alternate angles equal, the
two lines are parallel ". Hence,
1
COR.
i
!
FIG.1.
H
FIG.2.
F
"D
A
B
E
D
GE OME TRY
FIG.3.
FIG. 4.
BA
B
Ꮐ
F
E
A
D
FIG. 5.
E
FIG. 6.
E
F
B
F
C B
C
B
C
B
C
А
FIG. 7.
D
B
FIG.9.
E
P.L. I. pa je.
D
A
FIG.10.
F
A A A A
>
B
FIG.8.
E
F
B
C
D
FIG.II.
F
Α
C
E
FIG.12.
WH
D
A-
D
B
C
E
FIG.13.
F
-D
B
B
D
A
-B
E
Ꮐ
D
FIG.14.
A-
-B
E
K
FIG.15.
FIG.16.
E
B
C
F
D:
*
Б
Be
-E
A
H
D
F
Durs Scalp' F'din?
BOOK I.
19
OF GEOMETRY.
COR. 2. If it make the exterior angle equal to the
interior and oppofite angle on the fame fide, the two
lines are parallel . And,
COR. 3. If it make the two interior angles upon
one fide of it together equal to two right angles, the
two lines are parallel *.
COR. 4. Through a given point a ftraight line may
be drawn parallel to a given ftraight line. For, if
E be the given point, and CD the given line, let EF
be drawn perpendicular to CD', and EA perpendi-
cular to EF then EA is parallel to CD ".
>
i
pr. 12,
cor. 1. 14.
k
I
pr. 11.
cor. I. 14.
101
pr. 10.
pr. 9.
Acor. 1.1
PROP. XV.
THEOR.
If a straight line, falling upon two other straight
lines, make the two interior angles upon one fide of it,
together, less than two right angies, thefe two lines
being indefinitely produced, meet each other on that
fide.
Let the ftraight line AC make, with the two
ſtraight lines AB, CD, the two interior angles BAC,
ACD, upon one fide of it, together, lefs than two
right angles, the lines AB, CD, being indefinitely
produced, meet each other on the fame fide of AC.
For, through the point A, let the ftraight line FL
be drawn to make the angle FAC equal to the alter-
nate
Fig. 18.
20
Book. I.
ELEMENTS
nate Angle ACD; then the angles LAC, ACD are
pr. 11. together equal to two right angles, and therefore
6
pr. 10
d pr. 4.
e pr. 14.
the angle LAC greater than BAC.
Hence AB is on
the fame fide of FL with CD. From any point E in
CD, and from B in AB, let the perpendiculars EF,
BH be drawn to FL. In AB and HB produced,
let BK be taken equal to BA, and BG equal to BH;¸
alfo let G, K be joined, and KL be perpendicular to
AL.
The triangles AHB, GBK have the oppofite
c pr. 12. angles at B equal, and the containing fides equal by
conftruction, each to each; therefore the angles at
H and G are equal, and confequently KL equal to
GH; that is, the diftance of K from AL is the
double of BH. In like manner, a point may be
found in AK produced, whofe diſtance from AL
ſhall be the double of KL; and fo on. Confequently
a point may be found in the fame line AB fuch, that
its diſtance from AL fhall be greater than EF. Let
M be fuch a point, and let the perpendicular MN,
upon AL, meet CD in O. Then, becauſe EF and
ON are equal, MN is alfo greater than ON; there-
fore the point M is on the other fide of CD with re-
fpect to the point A, and the line ABM paffes through
CDO.
I poſt. 4.
:
PROP
3
Book I. OF GEOMETRY.
21
!
PROP. XVI. THEOR.
A ſtraight line falling upon two parallels makes the alter-
nate angles equal, the exterior angle equal to the inte-
rior oppofite angle upon the fame fide, and the two in-
terior angles upon the fame fide together equal to two
right angles.
Let the line EF fall upon the two parallels AB, Fig. 19.
CD, the alternate angles AGH, GHD fhall be equal,
the exterior angle EGB fhall be equal to the interior
oppofite angle GHD, and the two interior angles
upon the ſame fide BGH, GHD fhall be equal to two
right angles.
For, if poffible, let AGH be greater than GHD,
then BGH, GHD are together lefs than AGH,
BGH, that is, lefs than two right angles; and there-
fore the lines AB, CD, being produced towards B, D,
meet one another, which is contrary to the hypothe-
fis. Confequently the angles AGH, GHD muſt
be equal. From which it follows that EGB is equal
to GHD, and BGH, GHD together equal to two s
right angles.
Cor. 1. A ftraight line perpendicular to one of
two parallels, is alſo perpendicular to the other.
COR. 2. A ftraight line which meets one of two
parallels will alſo meet the other.
a
pr. 15.
pr. 12.
ax. 2.
с
pr. II.
ax. 2. 3.
COR.
22
Book I.
ELEMENTS
Cor. 3. Straight lines, parallel to the ſame ſtraight
line, are parallel to one another.
}
Fig. 20.
21.
a cor.2.16.
b
1
PROP. XVII. THE O R.
If two lines which meet, be refpectively parallel to two
other lines, which also meet, and all of them being
confidered as drawn from the points of concourfe, if
both the former be in the fame direction with both the
latte, or both in oppofite directions, the angles they
contain fhall be equal.
Let the two lines AB, AC, which meet in A, be
reſpectively parallel to the two lines DE, DF, which
meet in D, and being confidered as drawn from the
points A, D, let them be all in the ſame direction, or
both AB, AC in oppofite directions to DE, DF; the
angle BAC ſhall be equal to EDF.
a
For, let DF meet AB in G. Then each of the
pr. 16. angles BAC, EDF is equal to the fame angle BGF,
and therefore they are equal to one another.
PROP.
!
BOOK I. OF GEOMETRY.]
23
PROP. XVIII. THEO R.
If one fide of a triangle be produced, the exterior angle is
equal to the two interior oppofite angles; and all the
three angles of a triangle are together equal to two
right angles.
Let ABC be a triangle, having the fide BC pro- Fig. 22.
duced to D; the exterior angle ACD is equal to the
two interior oppofite angles CAB, ABC.
For, through the point C let CE be drawn paral-
lel to AB. Then the line CE divides the angle
ACD into the two parts ACE, ECD, of which the
former is equal to CAB, and the latter to ABC'.
Therefore the whole ACD is equal to CAB and
ABC taken together. And to thefe equals the com-
mon angle ACB being added, the three angles of
the triangle are equal to the two angles ACD, ACB,
and confequently equal to two right angles .
COR. I. The exterior angle of every triangle is.
greater than either of the two interior oppofite angles
-any two angles are together lefs than two right
angles-and two of the angles at leaſt are acute.
COR. 2. If two angles of one triangle be, either
ſeparately or together, equal to two angles of ano-
ther, the remaining angles are equal.
COR.
a
cor. 4.
14.
b
pr. 16.
pr. II.
i
*
+
24
BOOK I.
ELEMENTS
COR. 3. In a right angled triangle, the two acute
angles are together equal to a right angle-and, in
an ifofceles right-angled triangle, each of them is
half a right angle.
COR. 4. Each of the angles of an equilateral
triangle is one third of two right angles, or two
thirds of one right angle.
COR. 5. The four angles of every quadrilateral are
together equal to four right angles.
Fig. 23.
b
cor. 4.
cor. I.
18.
PROP. XIX. THEOR.
In every triangle the greater fide fubtends the greater
angle.
Let ABC be a triangle, having the fide AB greater
than the fide AC; then fhall the angle ACB oppofite
to the former be greater than the angle ABC oppo-
fite to the latter. For, let AD be cut off from AB, e-
qual to AC, and let C, D be joined. Then the angle
ACD is equal to ADC, and ADC greater than
ABC; therefore ACD (which is only a part of
ACB,) is greater than ABC.
whole ACB is greater than ABC.
Confequently the
COR. 1. Converſely, the greater angle is fubtend-
ed by the greater fide.
COR.
BOOK I.
25
OF GEOMETRY.
COR. 2. The hypothenufe is the greateſt fide of
a right angled triangle.
COR. 3. Of all ſtraight lines drawn from the fame Fig. 24i
point, to terminate in the ſame ſtraight line; the per-
pendicular is the leaft, and that which is nearer the
perpendicular is greater than that which is more re-
mote.
PROP. XX. THEOR.
Any two fides of a triangle are together greater than the
third fide.
Two fides AB, AC of the triangle ABC are toge Fig. 25.
ther greater than the third fide BC. For, in AB
produced, let AD be taken equal to AC, and C, D
joined. Then the angle BCD is greater than ACD ²,
and ACD is equal to ADC, therefore BCD is
greater than ADC or BDC. Confequently BD (which
is the fum of AB, AC,) is greater than BC.
2
COR. Any fide of a triangle is greater than the
difference between the other two:
ax. 6,
b
cor: 4,
C
cor. I.
19.
D
PROF.
26
BOOK I.
ELEMENTS
b
Fig. 26.
17.
Fig. 26.
a pr. 20.
PRO P. XXI. THE OR.
If there be two triangles conftituted upon the ſame baſe,
and the one be included in the other, the two fides of
the former fhall be together lefs than the two fides of
the latter, but ſhall contain a greater angle.
Let ABC, DBC, be two triangles conftituted upon
the fame bafe BC, of which DBC is included in
ABC, the two fides BD, DC, fhall be together lefs
than the two fides BA, AC, but the angle BDC
greater than BAC.
CASE I. When D, the vertex of the triangle BDC,
is in AC one of the fides of the triangle ABC; be-
cauſe BA, AD, are together greater than BD, if
DC be added to both, BA, AC, are together greater
than BD, DC; and becauſe BDC is an exterior
angle of the triangle BDA, it is greater than the in-
cor. I. terior oppofite angle BAC".
18.
Fig. 27.
CASE II. When the point D is within the triangle
ABC. Let BD be produced to meet AC in E; then,
by CASE 1. BA, AC, are greater than BE, EC, and
BE, EC greater than BD, DC; confequently BA,
AC, are greater than BD, DC. In like manner, the
angle BDC is greater than BEC, and BEC great-
er than BAC; confequently BDC is greater than
BAC.
COR.
1
Book I. OF GEOMETRY.
27
W
COR. If BA be equal to BD, then CA is greater
than CD.
1
1
PRO P. XXII. THE O R.
If two triangles have two fides of the one equal to two
fides of the other, each to each, but the included angles
unequal, that which has the greater angle fhall like-
wife have the greater baſe.
Let ABC, DEF, be two triangles, having AB e- Fig. 28.
qual to DE, and AC to DF, but the angle BAC
greater than EDF; then fhall the baſe BC be greater
than the baſe EF.
For, let the angle BAG be made equal to EDF
the fide AG equal to DF or AC, and B, G joined;
,
2 pr. 14.
then BG and EF are equal. Now, if the triangle pr. 4.
BGA be included in the triangle BCA, fince AC is
equal to AG, BC muſt be greater than BG or EF. cor. 21.
If not, let G, C be joined, then the angle AGC is
is equal to ACG; therefore the angle BGC, which
is greater than the former, will be ftill greater than
BCG, which is only part of the latter. And hence
BC, as before, is greater than BG ª, or EF.
d
COR. If two triangles have two fides of the one c-
qual to two fides of the other, each to each, but
their bafes unequal, that which has the greater baſe
will alfo have the greater vertical angle.
PROP.
d
cor. I.
19.
1
28
Book I.
ELEMENTS
Fig. 29.
3
b
PRO P. XXIII. THE O R.
If two triangles have two fides of the one equal to two
fides of the other, each to each, and have likewife the
angles fubtended by two equal fides equal, and if the
angles fubtended by the other equal fides be both acute,
or both obtufe, or one of them right; then ſhall the
two triangles be equal in every refpect.
Let ABC, DEF be two triangles, having the fides
AB, AC equal to the fides DE, DF, each to each,
and the angles at B and E oppofite to the equal fides
AC, DF, alfo equal; then, if the angles C, F, op-
pofite to the other equal fides, be both acute, the
triangles fhall be equal in every refpect. For, if the
remaining fides BC, EF be not equal, let BC be the
greater, and the part BG be cut off equal to EF, and
A, G joined. The triangles ABG, DEF having the
angles at B and E equal, and the including fides e-
qual, each to each, are equal in every refpect. Thus,
the fide AG, being equal to DF, will be equal to
AC, and the angle AGB being equal to DFE, will
be an acute angle: Hence AGC will be an obtufe
pr. 11. angle; but ACG is an acute angle; therefore AG
is not equal to AC, which is repugnant. The fame
contradiction will follow from fuppofing BC, EF un-
equal, when the angles C, F are both obtufe, or
pr. 4.
cor. I.
19.
when
BOOK I. OF GEOMETRY.
29
¿
when one of them is a right angle. The fides BC,
EF must therefore be equal, and confequently the
triangles equal in every reſpect.
COR. Two right angled triangles having equal hy-
pothenuſes, and alfo one fide equal to one fide, are
equal in every reſpect.
pr. 4.
PRO P. XXIV. THE O R.
Straight lines which join the extremities of two equal and
parallel lines towards the fame parts, are themſelves
equal and parallel.
Let AB, CD be two equal and parallel ftraight Fig. 30.
lines, joined towards the fame parts by the lines
AC, BD; then fhall AC, BD be alſo equal and pa-
rallel. For, let AD be drawn; the two triangles
ABD, ACD have the fide AB equal to CD, by hy-
pothefis, and AD common to both, and the included
angles BAD, ADC are equal, becauſe they are alter-
nate angles; therefore thefe triangles are equal in
every reſpect. Thus, AC is equal to BD, and the pr. 4.
angle CAD to ABD; whence AC is alfo parallel to ADB
BD:
c
1
pr. 16.
cor. t.
14.
F
APPEN.
30
Book I.
ELEMENTS
APPENDI X.
PROP. 1. If two ftraight lines be drawn to bifect the
angles at the baſe of a triangle, the ftraight line
which joins the point, where they interfect, and the
vertex, will bifect the vertical angle.
PROP. 2. To defcribe a triangle, having its fides.
reſpectively equal to three given ftraight lines, any
two of which are together greater than the third.
PROP. 3.
If the two fides of a triangle be bifected
by perpendiculars, the baſe will alſo be bifected by a
perpendicular falling upon it from the point where
the two former interfect.
PROP. 4. To defcribe a right angled triangle, of
which the hypothenufe and one fide taken together
fhall be equal to a given ftraight line, and the re-
maining fide equal to another given line.
PROP. 5. In any triangle, the angle contained by
the perpendicular falling from the vertex upon the
baſe, and the ſtraight line bifecting the vertical angle,
is equal to half the difference of the angles at the
bafe.
PROP. 6. In a given ſtraight line, of an unlimited
length, to find a point equally diſtant from two given
points, which are fo fituated, that the ftraight line
paffing through them does not cut the given line at
right angles.
PROP.
Book I.
31
OF GEOMETRY.
1
PROP. 7. If, from the vertex of a triangle, two
ftraight lines be drawn to meet the bafe, fo as to
form two ifofceles triangles, having, for their vertical
angles, the angles at the bafe of the triangle, they
ſhall contain an angle equal to half the fum of the
angles at the baſe.
PROP. 8. To conftruct a triangle, having the angles
at the baſe equal to two given angles, (which are to-
gether lefs than two right angles), each to each, and
the fum of all its fides equal to a given ftraight
line.
PROP. 9. If, from the vertex of a triangle, two
ftraight lines be drawn to meet the baſe, ſo as to
form two ifcofceles triangles, having for their vertical
angles one of the angles at the bafe, and the exterior
angle adjacent to the other, they fhall contain an
angle equal to half the vertical angle of the triangle.
PROP. 10. A ftraight line, of an unlimited length,
and two points without it being given, to find the
point or points in the given line, at which ftraight
lines from the given points fhall make equal angles.
with it.
PROP. II. All the interior angles of any recti-
lineal figure are together equal to twice as many
right angles, abating four, as the figure has fides.
PROP. 12. All the exterior angles of any rectili-
neal figure, formed by producing one fide from each
of the angular points, are equal to four right angles.
PROP.
33
BOOK I
ELEMENTS
PROP. 13. If every two adjacent fides of any po-
lygon be produced to interfect the fide or fides oppo
fite to them; the fum of all the faliant angles at the
points of interfection fhall be equal to four right
angles, if the number of fides be even, but to two
right angles only, if the number of fides be odd.
PROP. 14. The ſtraight line which joins the middle
of the baſe, and the vertex of a triangle, is greater,
equal, or leſs than half the bafe, according as the
vertical angle is acute, right, or obtuſe.
PROP. 15.
If the baſe of a right angled triangle
be divided into any number of equal parts, and, from
the points of fection, ftraight lines be drawn to the
oppoſite angular point, theſe will divide the angle
contained by the hypothenufe and perpendicular in-
to unequal parts, of which that next the perpen-
dicular is the greateſt, and the nearer to it is greater
than the more remote.
PROP. 16. If two fides of a triangle contain two
thirds of a right angle, the equilateral triangle de-
ſcribed upon the fum of theſe fides will be equal to
that defcribed upon the bafe, together with triple
the original triangle.
воок
:
ELEMENTS
O F
GEOMETRY.
*�*
ང*་》མ་»“ང་བརམར་བབར''n�ན"*
i.
BY
воок II.
DEFINITION S.
Y the rectangle under two ftraight lines is
meant the rectangle, of which theſe lines (or
others equal to them,) are two adjacent fides.
2. If through any point, in the diagonal of a pa-
rallelogram, two ſtraight lines be drawn parallel to its
fides, theſe divide the whole figure into four other
parallelograms; the two through which the diago-
nal paffes are called the parallelograms about the dia-
gonal, and the remaining two the complements.
3. If a ftraight line be cut at any point (between
or beyond its extremes,) the diſtances of that point
from its extremities are called the fegments of the
line, and its diſtance from the middle of the line the
mean diſtance of the point of ſection.
E
PROP.
¿
34
BOOK II.
ELEMENTS
Fig. 31.
↑ 16. 1.
b 5. 1.
Fig. 32.
PRO" I THEOR.
The oppofite fides and oppofite angles of a parallelogram
are'equal, and the diagonal divides it into two equal
parts.
L
ET ABCD be a parallelogram, of which BD
is the diagonal, the oppofite fides fhall be e-
qual, AB to CD, and BC to AD; the oppoſite
angles equal, DAB to DCB, and ADC to ABC;
and the diagonal BD fhall divide the parallelogram
into two equal triangles DAB, DCB.
For, in theſe triangles, the angles ADB, DBC are
equal, becauſe they are alternate angles; and the
angles ABD, BDC equal for the fame reafon; alſo
the fide BD, between the angles of each, is common
to both; therefore theſe triangles are equal in every
reſpect. Thus, AB is equal to CD, BC to AD,
the angle DAB to DCB, the angle ADC to ABC,
(becauſe their parts are equal), and the whole triangle
DBA to the whole triangle DCB.
COR. In every parallelogram the complements are
equal.
PROP.
+
FIG 17
I
PL.II.pa:34.
GEOMETRY
FIG 18
F
A
H
L
N
B
G
K
-D
E
C
D
M
H
E
A
-B
G
C
F K
E
Fro 19
FIG 20
B
E
B
FIG 21
F
D
G
G
-B
A-
GL
-F
D
A
C
A
C
-D
C
H
FIG 25
FIG 23
F
FIG 24
A
E
TIG 22
D
B
B
F
E
D
B
D
C
A
A
-D
B
C
E
FIG 28
A
FIG 27
FIG 26
B
C
B
Tro 29
D
AA
B
Ꮐ
E
F
!
بولا
B
C
E
Ꮐ
F
FIG 31
D
C
FIG 30
B
A
A
FIG 32
H
D
G
C
A
F
E
K
Book II.
35
OF GEOMETRY.
:
PROP. II. THE O R.
Parallelograms or triangles on the fame, or equal bafes,
and between the fame parallels, are equal.
I. PARALLELOGRAMS.
1. Let there be two pa- Fig. 33.
rallelograms AC, AF, on the fame bafe AB, and
between the fame parallels DF, AB, they ſhall be e-
qual to one another. For the triangles ADE, BCF,
having the fide AD equal to BC, AE to BF, and
likewiſe the included angles DAE, CBF equal, are
equal in every refpect. Therefore, each of them
being taken from the fame figure ABFD, the re-
mainders muſt be equal, viz. the parallelogram AC
to the parallelogram AF.
2
a
1. 2.
b
с
17. 1.
4. 1.
2. Let AC, HF, be two parallalelograms upon e- Fig. 34.
qual bafes AB, HG, and between the fame parallels
AG, DF; AC, HF, fhall be equal. For, let AE,
BF, be drawn; then AB, being equal to HG, will
be equal to EF; and it is alfo parallel to EF; hence 1. 2. ax.
the figure AEFB is a parallelogram; and it is
equal to each of the parallelograms AC, HF; confe-
quently they are equal to one another.
2
2.
b 24. I.
def. 20.
II. TRIANGLES. 1. Let ABC, ABE, be two tri- Fig. 35-
angles on the fame bafe AB, and between the fame
parallels AB, CE; they fhall be equal to one ano-
a
ther. For, let AD be drawn parallel to BC, and cor. 4.
BF
14. 1.
36
BOOK II.
ELEMENTS
Fig. 36.
BF to AE, and let them meet CE produced in D, F;
then the parallalelograms DB, AF are equal; and
conſequently their halves, the triangles ABC, ABE,
are alſo equal.
2. When the triangles are upon equal bafes, and
between the fame parallels, they may be proved equal
in the fame manner.
COR. 1. Of triangles between the fame parallels,
the greater triangle has the greater bafe; and con-
verfely.
COR. 2. If a parallelogram and a triangle be up-
on the fame or equal bafes, and between the fame
parallels, the parallelogram is double the triangle.
COR. 3. Equal triangles, upon the fame baſe, or
upon equal baſes, in the fame ftraight line, and up-
on the fame fide of it, are between the fame paral-
lels.
PROP. III.
PROB.
Fig. 37.
2 9. 1.
C
3.
I.
To defcribe a fquare upon a given ftraight line.
Let AB be the given ftraight line, upon which it
is required to defcribe a fquare. From one extre-
mity A, let the perpendicular AE be raiſed, from
which let AD be cut off equal to AB; and through
cor, 4. B, D, let BC, DC be drawn parallel to AD, AB;
14. I.
then
Book II. OF GEOET MR Y.
37
then the figure ABCD is a parallelogram by con-
ſtruction; hence AD is equal to BC, and AB to d 1. 2.
DC; but AD is equal to AB; therefore the four
fides are equal to one another. Again, becauſe the
two interior angles A, B are together equal to two
right angles, and one of them A is a right angle, © 16. 1.
the other B muft alfo be a right angle. In like man-
1
ner, C and D are proved to be right angles.
fequently the figure ABCD is a ſquare f.
Con-
COR. 1. A rectangle, under two given lines, may
be formed in a fimilar manner.
COR. 2. A parallelogram, having one angle right,
is a rectangle.
COR. 3. A parallelogram, having one angle right,
and two adjacent fides equal, is a fquare.
COR. 4. The rectangle, under two equal lines, is
a fquare.
f def. 23.
I.
PRO P. IV. THE O R.
If two adjacent fides of one rectangle be equal to two ad-
jacent fides of another, each to each, the rectangles
are equal in every refpect.
}
Let ABCD, EFGH be two rectangles, having the Fig. 38.
fides AB, BC equal to EF, FG, each to each, theſe
rectangles are equal in every refpect.
}
For,
38
Book II.
ELE MEN T S
!
ax. I.
:
For, if the rectangle AC were applied to EG, fo
that the point A ſhould fall upon E, and the ſide
AB upon EF, then AB, EF would coincide", be-
cauſe they are equal; AD, BC would fall upon EH,
FG, becauſe the angles are right; the point C
would fall upon G, becauſe BC, F are equal; and
the fide CD would fall upon GH, becaufe the angles
C, G are right, Confequently the point D would
fall upon H, and the rectangles coincide.
COR. 1. The fquares of equal lines are equal; and,
converfely, equal fquares have equal fides.
COR. 2. Parallelograms, or triangles of equal bafes
and altitudes, are equal,
?
نم
Fig. 39.
• 3 •
PROP. V. THE OR,
If a straight line be cut at any point, the mean distance of
the point of fection is equal to half the fum, or half the
difference of the fegments, according as the point is
beyond or between the extremities of the line.
1. Let the ftraight line AB, of which C is the
middle point, be cut at any other point D, between
its extremes; then CD is equal to half the difference
of AD, DB. For, fince AC is equal to CB, AD is
equal to CB, CD, taken together; and, from theſe e-
quals, DB being taken away, the difference of AD,
DB is equal to the double of CD'.
2. Let
}
Book II. OF GEOMETRY.
39
2. Let the ftraight line AB, of which C is the Fig. 40.
middle point, be cut at any point. D, beyond its ex-
tremes; then CD is equal to half the fum of AD,
DB. For, fince AC is equal to CB, AD is equal to
CB, CD, taken together; and to thefe equals DB
being added, the fum of AD, DB is equal to the
double of CD".
COR. I. The greater of two unequal magni-
tudes is equal to half their fum added to half their
difference; and the lefs equal to half their difference
taken from half their fum.
!
COR. 2. If the excefs of the firft of three magni-
tudes above the ſecond be equal to the exceſs of the
ſecond above the third, the ſecond is equal to half
the fum of the firſt and third.
a ax. 3;
i
PRO P. VI. THE OR.
The fum of all the rectangles under one ſtraight line, and
the feveral parts of another, is equal to the rectangle
under the two whole lines.
i
Let there be two ſtraight lines, AB, and C, of Fig. 41.
which AB is divided into any number of parts, AE,
EF, FB; then fhall the fum of the rectangles under
C and AE, C and EF, C and FB, be equal to the
rectangle under C and AB.
For,
t
1
40
Book II.
ELEMENTS
‣
c
cor. 4.
14. I.
For, the perpendicular AH being raiſed from the
9.I point A', equal to C, and the parallelogram HB
completed, let the ftraight lines EL, FK be drawn
from the ſeveral points of divifion in AB parallel to
AH. Then the parallelogram AG, and the parts
AL, EK, FG, into which it is divided by theſe lines,
are rectangles; and, becauſe AH is equal to C, and
EL, FK, each equal to AH, AG is the rectangle
under C and AB, AL under C and AE, EK under
C and EF, FG under C and FB. Now, the whole
AG being equal to all its parts AL, EK, FG, taken
together, the propofition is manifeſt.
cor. 2.
3. 2.
1. 2.
ax. 6.
f cor. 4.
3. 2.
COR. I. If a ſtraight line be divided into any num-
ber of parts, the fum of the rectangles under the whole
line, and each of the parts, is equal to the fquare of
the whole line f.
COR. 2. If a ſtraight line be divided into any two
parts, the rectangle under the whole line, and one of
the parts, is equal to the fquare of that part, together
with the rectangle under the two parts.
COR. 3. The rectangle under one ftraight line,
and the double, triple, &c. of another, is equal to
twice, thrice, &c. the rectangle under the two lines.
PROP.
Book II. OF GEOMETRY.
45
?
{
PROP. VII. THEO R.
Twice the rectangle under the fegments of a straight line
is equal to the difference between the fquare of the line,
and the fum of the fquares of its fegments.
Let AB be the ſtraight line, and AD, DB its ſeg- Fig. 42.
ments; twice the rectangle under AD, DB is equal to
the difference between the fquare of AB and the fum
of the fquares of AD, DB.
For, let the fquare ABEF be defcribed upon
AB¹, from BE or BE produced, (according as the 3. 2.
point D is between or beyond A, B), let BG be cut
off equal to BD, and through G, D, let GK, DL be
drawn parallel to AB, BE, interfecting each other in
H, and meeting the oppofite fides of the fquare in
K, L. Then, becauſe AB, BE are equal, and BD,
BG equal, the wholes or remainders AD, GE are e-
qual; but AD is equal to FL, and GE to FK; ↳ 1. 2.
therefore FK is equal to FL.
Hence the figure
b
I.
KHLF is a ſquare ©, and equal to the ſquare of AD d. 3. c. 3.2.
J
I.C. 4. 2.
And DBGH is the ſquare of DB c. Now the difference
between the ſquare ABEF and the fum of the fquares
KHLF, DBGH is the fum of the rectangles AH,
HE, which is equal to twice the rectangle under AD,
DB e.
F
COR.
e
4. 2.
42
Book I.
ELEMENTS
•
cor. 5.
cor. 2. 5.
2. & cor.2.
COR. I. If a ſtraight line be divided into two
parts, the fquare of the whole line is equal to the
fquares of the two parts, together with twice the rec-
tangle under theſe parts.
COR. 2. The fquare of a ftraight line is equal to
four times the fquare of half the line; and half its
fquare is double the fquare of its half.
COR. 3. Twice the rectangle, under two unequal
feraight lines, is equal to the difference between the
fquare of their fum and the fum of their fquares.
COR. 4. Twice the rectangle, under two unequal
ftraight lines, is equal to the difference between the
fum of their fquares and the fquare of their differ-
ence.
COR. 5. Of two unequal ftraight lines the ex-
cefs of the fquare of their fum above the fum of
their fquares, is equal to the excefs of the fum of their
fquares above the fquare of their difference.
COR. 6. Of two unequal ſtraight lines, the fum of
their fquares is equal to double the fquare of half
their fum added to double the fquare of half their dif
ference".
COR. 7. If a ſtraight line be cut at any point, the
fum of the fquares of its fegments is double the fum
of the fquares of half the line, and the mean diſtance
of the point of ſection.
PROP.
Book II. OF GEOMETRY.
43
PROP. VIII. THEOR.
The rectangle under the fum and difference of two une-
qual ftraight lines, is equal to the difference of their
Squares.
b
b
3. C. 3.
2.
Let AB, CD, be two unequal ſtraight lines, of Fig. 43.
which AB is the greater, the rectangle under the fum
and difference of AB, CD is equal to the difference
of their fquares. For, let the fquare ABEF be
defcribed upon AB; from EF, EB, and AB produ-
ced, let EH, EK, and BG, be cut off, each equal to -
CD, and let KL, parallel to AB, interfect GM, HN,
parallel to BE, in M, N, and AF in L. The figure
EKNH is a fquare, and equal to the ſquare of CD ¹;
therefore the fum of the rectangles HL, LB is the 1. c. 4.
difference of the fquares of AB, CD. Again, be-
caufe EB is equal to AB, and EK to CD, BK or GM,
which is equal to it, is the difference of AB, CD,
and AG is their fum; therefore AM is the rectangle
under their fum and difference. Now, the rectangle
AM is equal to the fum of the rectangles HL, LB,
becaufe LB is common to both, and the remaining
rectangles HL and BM are contained by equal
lines.
COR. The rectangle under the fegments of a
ftraight line is equal to the difference between the
fquares
2.
?
44
BOOK II.
ELEMENT S
fquares of the half line, and the mean diſtance of the
point of fection; and the fum of the rectangle, and
one of the fquares, is equal to the other.
PRO P. IX.
THE OR.
Fig. 44.
3. 2.
6 5. 1.
In a right angled triangle, the fquare of the hypothe-
nufe is equal to the fum of the fquares of the other
two fides.
Let ABC be a right angled triangle, of which AC
is the hypothenufe; the fquare of AC is equal to
the fum of the fquares of AB, BC.
For, let the fquares AE, FB, BI be defcribed upon
the three fides; through B let LM be drawn paral-
lel to AD, meeting FG in M, and DE in L; alfo,
let DA be produced to meet FG in K. The triangles
AFK, ABC are equal in every reſpect; for the
angles at F and B are equal, becauſe they are right
angles; the angles FAK, BAC are equal, becauſe
each of them is lefs than a right angle by the fame
angle KAB; and the fides AF, AB are equal, be-
cauſe they are fides of the fame fquare. Thus, AC
or AD is equal to AK, and therefore the paralle-
logram AL equal to the parallelogram AM. But
AM is equal to FB. Confequently the parallelo-
gram AL is equal to the fquare FB. In the fame
manner,
FIG 33
GEOMETRY
D
C
E
F
D
E
F
PL. III. pa. 44.
A
B
D
C
E
F
D
C
FIG 35
D
Α
A
B
FIG 34
FIG 36
H
G
A
B
E
F
E
D
FIG 37
A
B
H
G
C
H
Ꮐ
FIG 39
A
A
с
D
B
FIG38
L
K
G
H
E
F
FIG 40
A
C
L
E
F
F
FIG 42
E
L
H
K
Ꮐ
B
A
A
D
B
K
H
T
L
N
FIG 43
A
C
D
E
B
M
H
A
-D
B
F
K
FIG 41
M
E
F
B
FIG
44
D
I
H
C
B
I
•
BOOK II. OF GEOMETRY.
45
t
manner, the parallelogram CL may be proved e-
Therefore the whole fquare
qual to the fquare BI.
AE is equal to the fum of the fquares FB, BI.
COR. The fquare of one of the fides of a right
angled triangle is equal to the difference of the
fquares, or the rectangle under the fum and differ-
ence, of the hypothenuſe and the other fide.
PROP. X.
THEOR.
In every triangle, the fquare of one of the fides is greater
or lefs, according as the oppofite angle is obtufe or acute,
than the fum of the fquares of the other fide and the
bafe, by twice the rectangle under the bafe and the
distance of the perpendicular from the angular point.
Let ABC be a triangle, having an obtufe or an acute Fig. 45.
angle at B, and let CD be the perpendicular from
the vertex C upon the bafe AB; the fquare of AC is
greater or lefs than the fum of the fquares of AB,
BC, by twice the rectangle under AB, BD.
For, fince the fquare of AC is equal to the fum of
the fquares of AD, DC, and the fquare of BC equal
to the fum of the fquares of BD, DC, the differ-
ence between the fquare of AC, and the fum of
the fquares of AB, BC, is equal to the difference
9. 2.
between
:
46
Book II.
ELEMENTS
...
5
ax. 4.
c 7. 2.
I. c. 18.
I.
between the fum of the fquares of AD, DC, and
the fum of the fquares of AB, BD, DC; which is the
fame with the difference between the fquare of AD
and the fum of the fquares of AB, BD. But the
fquare of AD is greater or lefs than the fun of the
fquares of AB, BD, by twice the rectangle under
AB, BD, according as the point D is on the other
fide of B, or on the fame fide with A, the 15, ac-
CBA
cording as the angle BCA is obtufe or acute d.
Therefore alfo the fquare of AC is accordingly
greater or lefs than the fum of the fquares of AB,
BC, by twice the rectangle under AR BD.
In the caſe in which the point D coincides with
A, CAB is a right angle, and confequently CBA an
acute angle; and, fince the fquare of AC is lefs than
cor. 9. z. the fquare of BC, by the fquare of AB, it will be
leſs than the fum of the fquares of AB, EC, by twice
the fquare of AB, which is the fame with twice the
4.c.3.2. rectangle under AB, BD.
COR. 1. If a ftraight line be drawn from the ver-
tex of a triangle to the middle of the baſe, the fum
of the fquares of that line, and half the bafe, is
equal to half the fum of the fquares of the two
* 2. c. 5.2. fides 2.
a
COR. 2. The vertical angle of a triangle is obtufe,
right, or acute, according as the fquare of the bafe is
greater,
}
ין
OF GEOMETRY.
47
greater, equal, or leſs, than the fum of the fquares of
the two fides.
PRO P. XI. THE O R.
The difference of the fquares, or the rectangle under the
fum and difference, of the two fides of a triangle, is e-
qual to twice the rectangle under the baſe, and the di-
Stance of the perpendicular from the middle of the
bafe.
Let ABC be any triangle, of which the bafe BC is Fig. 46.
bifected in E, and let AD be the perpendicular; the
difference of the fquares of BA, AC is equal to twice.
the rectangle under BC and ED.
b
9.2.
For, fince the fquare of BA is equal to the fquares
of BD, DAª, and the fquare of AC equal to the
fquares of DC, DA; the difference of the fquares of
BA, AC is equal to the difference of the fquares of
BD, DC. But the difference of the fquares of BD, ax. 3. 4.
DC, is equal to the rectangle under their fum and
difference, that is, to the rectangle under BC, and
twice ED, or twice the rectangle under BC and
ED c. Therefore the difference of the fquares of
BA, AC, or the rectangle under their fum and dif-
ference, is equal to twice the rectangle under BC and
ED.
e
c.
d
8. 2.
ª 5. 2.
e
3. c. 6. 2.
1
In
48
ELEMENT S
In the cafe in which the point D coincides with B
or C, the propofition is evident.
PROP. XII. PROB.
2
Fig. 47.
3. 2.
9. 2.
To divide a given straight line into two parts, fo that the
rectangle under the whole line, and one of the parts,
may be equal to the fquare of the other part.
Upon the given ftraight line AB let the fquare AC
be deſcribed; let the fide AD be bifected in E, and
the points E, B joined; in EA produced let there be
taken EF equal to EB, and let the fquare AFHG be
defcribed upon AF: The line AB will be divided in
the point G, fo that the rectangle under AB, BG
fhall be equal to the ſquare of AG.
produced to meet DC in K; the rectangle under
DF, FA, together with the
For, let HG be
fquare of AE, is equal to
But the fquare of EB is
AE. Therefore the rec-
c. 8. 2. the ſquare of EF, or EB.
equal to the fquares of AB,
tangle under DF, FA, together with the fquare of
AE, is equal to the fquares of AB, AE; and the
rectangle under DF, FA, equal to the fquare of AB,
that is DH equal to AC; from which the common
part AK being taken away, there remains FG equal
to GC, that is, the fquare of AG, equal to the rec-
tangle under AB, BG.
COR.
OF GEOMETRY.
49
COR. The line that is compounded of the given
line, and its greater ſegment, is alfo divided in the
fame manner.
APPENDIX.
PROP. 1. Every quadrilateral figure, whofe oppo-
fite fides or oppofite angles are equal, is a parallelo-
gram.
PROP. 2. If two fides of a triangle be bifected, the
ſtraight line which joins the points of fection will be
parallel to the remaining fide, and equal to the half
of it.
PROP. 3. To make a triangle that fhall be equal to
a given quadrilateral figure.
PROP. 4. If, from any point, in an equilateral tri-
angle, perpendiculars be let fall upon the fides, thefe
perpendiculars, taken together, fhall be equal to the
perpendicular of the triangle.
PROP. 5. If a ſtraight line be drawn from the ver-
tex, to any point in the baſe of an ifofceles triangle,
the fquare of that line, together with the rectangle
under the fegments of the bafe, will be equal to the
fquare of one of the equal fides of the triangle.
PROP. 6. To find the fide of a fquare that fhall be
equal to two or more given ſquares.
G
PROP.
.....
50
B
ELEMENT
PROP. 7. If, from the vertex of a triangle
ftraight line be drawn to bifect the bafe, it will allo
bifect any other ftraight line parallel to the bafe, and
terminating in the two fides of the triangle.
PROP. 8. The two diagonals of a parallelogram
mutually bifect each other, and the fum of their
fquares is equal to the fum of the fquares of all the
fides.
PROP. 9. To make a rectangle that fhall be equal
to a given rectilineal figure.
PROP. 10. If, from any point, ftraight lines be
drawn to the four angular points of a rectangle, the
fums of the fquares of thofe drawn to the oppofite
angles will be equal.
PROP. 11. In any quadrilateral figure, the fum of
the fquares of the diagonals, and four times the ſquare
of the line joining their middle points, is equal to the
fum of the ſquares of all the fides.
PROP. 12. To find the fide of a fquare that ſhall be
equal to the difference between two given ſquares.
PROP. 13. In any quadrilateral figure, the ſum of
the ſquares of the diagonals is equal to double the
fum of the fquares of the two lines that join the
middle points of its oppofite fides.
PROP. 14. If the vertical angle of a triangle be two
thirds of two right angles, the fquare of the baſe
will be equal to the rectangle under the two fides
added to the fum of their fquares: But, if it be two
thirds
OF GEOMETR Y.
rds of one right angle, the ſquare of the baſe will
be equal to the difference of the rectangle under the
two fides, and the fum of their fquares.
PROP. 15. To bifect a given triangle by a ſtraight
line drawn from a given point in one of its fides.
PROP. 16. If, from the extremities of the baſe of a
triangle, ftraight lines be drawn to bifect the oppo-
fite fides, another ftraight line drawn through their in-
terſection, from the vertex, will bife&t the baſe.
PROP. 17. If a ſtraight line be drawn parallel to the
baſe of a triangle, and the diagonals of the quadrila-
teral figure be drawn, then a ftraight line, drawn
through their interfection, from the vertex, will bi-
fect the bafe.
PROP. 18. To find the fide of a fquare that fhall be
equal to a given rectangle.
PROP. 19. The equilateral triangle, defcribed upon
the hypothenufe of a right angled triangle, is equal to
the fum of the equilateral triangles defcribed upon
the two fides.
PROP. 20. If, through one of the angular points of
a fquare, a ſtraight line be drawn, to cut the two op-
pofite fides, and from one of the points of fection a
perpendicular to that ftraight line be raiſed to meet
the fide oppofite to that from which it is drawn ;
then ſhall the fquare of the produced part of that op-
ofite
52
ELEMENTS
pofite fide be equal to the fquare of the line between
the points of fection added to the original fquare.
PROP. 21. To divide a given ſtraight line into any
propoſed number of equal parts.
PROP. 22. If a ſtraight line be divided into three
parts, fo that the fquares of the lines, compounded
of two adjacent parts, be together equal to the ſquare
of the whole line, then fhall the rectangle under the
two extreme parts be equal to double the rectangle
under the whole line, and the middle part; and con-
verfely.
PROP. 23. If a parallelogram be conſtituted upon
the bafe of a triangle, and on the fame fide with the
triangle, and, upon the other two fides, parallelo-
grams be likewife conftituted, fo that their oppofite
fides paſs through the angular points of the former;
then ſhall the parallelogram on the baſe be equal to
the fum or difference of thofe on the fides, accor-
ding as they fall both without the triangle, or one of
them, upon it.
PROP. 24. Let there be four points, A, B, C, D,
in the fame ftraight line, and let E be the point in
that line that is equally distant from the middle of
the fegments AB, CD; then, F being any other
point whatever, the fum of the fquares of the diftan-
ces of F from the four points A, B, C, D, fhall ex-
ceed the fum of the fquares of the diſtances of E from
the fame points, by four times the fquare of EF.
E LE-
E MENTS
O F
1
GEOMETR
1
i
1.
воок III.
• DEFINITION S.
A
Diameter of a circle is a ſtraight line drawn
through the center, and terminated on both
fides by the circumference. It divides the circle into
two parts, called femicircles.
2. An arch of a circle is any part of its circumfe-
rence.
3. The chord of an arch is the line joining its ex-
treme points.
4. A fegment of a circle is a figure bounded by an
arch and its chord.
5. A fector of a circle is a figure bounded by two
radii, and the arch between them: When the radii are
perpendicular to one another, it is called a quadrant.
6. An angle in a fegment is that which is con-
tained by two ſtraight lines, drawn from any point in
its arch, to the extremities of its chord.
7. An
!
54
ELEMENT
7. An angle is faid to ftand upon the arch is
intercepted between its fides.
8. A tangent to a circle is a ftraight line, which
meets the circumference in one point, and every
where elſe falls without it.
9. Two circles are faid to touch one another when
the circumference of one of them meets the circum-
ference of the other in one point, and every where
elfe falls without it.
10. A ftraight line is faid to be applied in a circle
when both its extremes are in the circumference.
11. When all the angular points of a rectilineal
figure are in the circumference of a circle, the figure
is faid to be infcribed in the circle, or the circle de-
fcribed about the rectilineal figure.
12. When all the fides of a rectilineal figure touch
the circumference of a circle, the figure is faid to be
defcribed about the circle, or the circle infcribed in
the rectilineal figure.
13. When all the angular points of a rectilineal fi
gure are placed in the fides of another, and the one
figure wholly included in the other, then the former
is faid to be infcribed in the latter, or the latter de-
ſcribed about the former.
!
PROP.
}
OF GEOMETRY.
55
PROP. I. PRO B.
To find the center of a given circle.
Let ADB be the given circle, of which it is required Fig. 48.
to find the center.
Having affumed two points, A, B, in the circumfe-
rence, let the ſtraight line AB which joins them be
a
a
bifected at right angles by the line CD meeting the 8. 1.
circumference in D, E, then DE being bifected in
F, the point F will be the center of the circle.
For, if not, let any other point G, without the
line DE, be the center, and let GA, GB, GC, be
drawn. The triangles GAC, GBC have the fides
GA, GB equal, as being radii of the circle, alſo
AC, CB equal by conftruction, and CG common;
therefore they are equal in every refpect. Hence ↳ 6. 1.
the angles GCB, GCA are equal, and each of them
a right angle. But FCB is a right angle: Therefore
GCB muſt be equal to FCB, a part to the whole,
which is abfurd.
COR. 1. A ftraight line, which bifects a chord at
right angles, paffes through the center of the circle.
COR. 2. If two chords be bifected by perpendicu-
lars, their interfection will be the center: And thus
the center of a given fegment may be determined.
PROP.
b
56
ELEMENT
}
Fig. 49.
a 6. 1.
b ax. I.
!
PROP. II. THE OR.
If a straight line, drawn through the center of a circle,
bifect any chord of the fame, not paffing through the
center, it will also bifect the arches which the chord
fubtends.
In the circle AFB, let the ftraight line FCE, paf-
fing through the center C, bifect the chord AB in D,
and it fhall alfo bifect the arches AFB, AEB.
For, let AC, CB, be drawn; then the triangles
ADC, BDC, having AC, CB equal, as being radii of
the circle, AD, DB equal by hypothefis, and CD
common to both, are equal in every reſpect. Hence
the angle ACE is equal to BCE. Now, if the femi-
circle FAE, by revolving upon FE, were applied to
the femicircle FBE, the line AC would fall upon
CB, becauſe the angles ACE, BCE are equal 6, the
point A would fall upon B, becaufe AC, CB are e-
qual, and every point in the arch FA would fall .up-
on FB, becauſe every point in each of them is e-
qually diftant from C, that is, the arches FA, FB
would coincide; confequently they are equal.
COR. I. If a ſtraight line do any two of theſe five,
paſs through the center of a circle, bifect an angle at
the center, bifect the arch upon which it ſtands, bi-
fect the chord of that arch, cut the chord at right
angles;
OF GEOMETRY.
57
; it muſt neceffarily do the other three, provi-
the chord be not a diameter in that cafe which is
tated in the propofition.
COR. 2. If a ſtraight line paſs through the center
of a circle, it bifects the circle, and its circumference;
and converſely.
COR. 3. In the fame circle, or in circles of equal
radii, equal angles at the center ftand upon equal
arches, and the greater angle ſtands upon the greater
arch; and converſely.
COR. 4. A right angle at the center of a circle
ſtands upon a fourth part of the circumference; and
converfely.
COR. 5. Whatever part an angle at the center is of
four right angles, the fame part will the arch upon
which it ftands be of the whole circumference.
COR. 6. When the radii are equal, fectors which
have equal angles are equal; alfo femicircles are e-
qual, and circles equal.
COR. 7. The method of bifecting a given arch, or
a given fegment, is evident.
COR. 8. A ſtraight line, biſecting two parallel
chords, paffes through the center.
G
PROP.
58
ELEMENT S
Fig. 50.
a 6. I.
PROP. III. THEOR.
In the fame femicircle, or in femicircles of equal radii, e-
qual chords fubtend equal arches, the greater chord
fubtends the greater arch; and converfely.
Let AHB, DKE be two femicircles of equal radii
AC, DF, and in theſe let AH, DG be two equal
chords; the arches which they fubtend will be equal.
For the radii CH, FG being drawn, the triangles
ACH, DFG have all their fides equal, each to each;
therefore the angle HCA is
confequently the arches upon
equal to GFD, and
which they ftand are
b cor. 3. equal; alfo converſely, if the arches be equal, their
2. 3.
4.
I.
chords will be equal: For, fince the arches AH, DG
are equal, the angles ACH, DFG are equal. But
the fides AC, CH are equal to DF, FG. Therefore,
in the triangles ACH, DFG, the bafe AH is equal to
the bafe DG.
Next, let the chord DK be greater than the chord
AH, and the arch DK will alſo be greater than the
arch AH. For the triangles DFK, ACH have the
fides DF, FK equal to AC, CH, but the baſe DK
greater than the bafe AH: Therefore the angle DFK
d cor. 23. is greater than the angle ACH, and confequently
1.
the arch DK greater than AH.
The converfe is
↑
proved in like manner.
PROP.
OF GEOMETRY. 59
!
PROP. IV. THE OR.
!
The diameter is the greatest chord in a circle; thofe
chords which are equally distant from the center are
equal; and that which is nearer the center is greater
than that which is more remote; and converfely.
Let AFE be a circle, of which AB is a diameter, Fig. 51.
AB is greater than any other chord, as DE, that does
not paſs through the center. For DC, EC being
drawn to the center, AB is equal to the fum of DC,
CE, and therefore greater than DEª.
C
C
20. I.
1.
cor. I,
2. 3.
Next, let AF, DE be two chords equally diſtant
from the center, that is, let the perpendiculars CG,
CH be equal, then AF is equal to DE. For the
right angled triangles AGC, DHC having their hy-
pothenufes AC, CD equal, and the fides CG, CH
equal, are equal in every refpe&t. Thus, AG is e-cor. 23.
qual to DH, and their doubles AF, DE alfo equal.
Laftly, Let DE be nearer to the center than AK,
that is, let the perpendicular CH be lefs than CL,
then DE is greater than AK. For the fum of the
fquares of CH, HD is equal to the fum of the fquares
of CL, LA, becauſe each of them is equal to the
fquare of the radius. And from theſe the unequal
fquares of CH, CL being taken away, there remains
the
9. 20
60
ELEMENT S
the fquare of HD greater than the fquare of
therefore the line HD is greater than LA, and DE, the
double of HD, greater than AK, the double of LA.
PROP. V.
THEOR.
Fig. 52.
ว 20. I.
Of all the straight lines that can be drawn to meet the cir-
cumference of a circle, from any point that is not the
center, that which paffes through the center is the
greatcft; that which, when produced, paffes through
the center, is the least; those which intercept equal
arches, from the extremity of the greatest, are equal,
and that which intercepts the less arch is the greater ;
and converfely.
Let C be the center of the circle, and A any other
point. Of all the lines that can be drawn from A to
the circumference, AB, which paffes through the cen-
ter, is the greateft. For, if any other, as AE, be
drawn ; then, CE being joined, AB is equal to the
fum of AC, CE, and therefore greater than AE ².
a
Next, AD, which, when produced, paffes through
the center, is the leaft. For, if any other, as AG,
be drawn, and CG joined; then CD, being equal to
CG, will be less than the fum of CA, AGª, and
therefore AD lefs than AG.
Further,
F GEOMETR Y.
6I
her, AE, AF, which intercept equal arches
from the point B, are equal. For, fince EB, BF are
equal, the angles ECB, BCF, and confequently
ACE, ACF ©, are alfo equal. But, in the triangles
ACE, ACF, the fides EC, CF are equal, and AC
common to both; therefore the bafe AE is equal to
the baſe AF d.
J
Laſtly, AE, which intercepts the leſs arch, is great-
er than AG. For, fince the arch EB is leſs than
GB, the angle ECB will be lefs than GCB, and
confequently ACE greater than ACG. But, in the
triangles ACE, ACG, the fides CE, CG are equal,
and AC common to both; therefore the baſe AE is
greater than the baſe AG e.
COR. I. There cannot be drawn more than two e-
equal ſtraight lines, to the circumference of a circle,
from any point that is not the center.
COR. 2. If there be three equal ftraight lines drawn
from a certain point, to the circumference of a circle,
that point is the center.
ร
cor. 3.
2. 3.
C
]
• I.
d 4. I.
e
€ 23.
23. I.
PROP. VI. THEOR.
An angle at the center of a circle is double the angle at
the circumference which ſtands upon the fame arch.
Let
62
ELEMENT S
Ph.
1
Fig. 54.
Fig. 53.
Let ACB be an angle at the center, and
angle at the circumference, upon the fame arch AB,
the angle ACB is the double of ADB.
CASE 1. When the center is in DB, one of the lines
that contain the angle at the circumference: The
triangle ACD is ifofceles; therefore the angles at A
a cor. 4. 1. and D are equal, and the angle ACB, which is equal
to both, will be double one of them.
ว
b 18. I.
Fig. 54.
Fig. 55.
CASE 2. When the center is between the lines AD,
DB, let the diameter DCE be drawn. The angle
ACE is the double of ADE, and the angle ECB the
double of EDB; therefore the whole angle ACB is
double the whole angle ADB.
CASE 3. When the center is without the lines AD,
DB; then, as before, the angle ECB is the double of
EDB, and the part ACE the double of ADE; there-
fore the remaining angle ACB is double the remain-
ing angle ADB.
Fig. 56.
• 10.1.
PROP. VII. THEOR.
All angles in the fame fegment of a circle are equal.
Let ADEB be any fegment of a circle, all the
angles in that fegment are equal. For, from the cen-
ter C, let CF be drawn perpendicular to AB ", meet-
ing the circumference of the other fegment AGB in
G₁
---
PL. I. pa. 69.
IG.45.
GEOMETRY
C
FIG.48.
D
C
C
A
B
D
A
D
B
A
A.
A
FIG.46.
C
K
B
FIG.47.
F
Ꮐ
C
E
B
Ꮐ
Ꮐ
H
B
E
C
D
B
E
D
C
F
D
E
A
F
K
FIG 49.
H
Ꮐ
FIG50.
A
Ꮐ
E
FIG.51.
A
D
A
C
B D
F
E
D
A
FIG.52.
FIG.52:
Ꮐ
E
E
B
B
D
D
A
FIG.53.
G/
E
F
B
FIG.54.
A
B
E
G/
K
C
E
F
FIG.55.
C
A
H
E
B
D
B
i
OF GEOMETR Y.
63
;
Clet AC, CB and DG be joined. Then, be-
caufe CF paffes through the center, and is perpendi-
cular to AB, it bifects the arch AGB 6. Hence the
b cor. I.
2. 3.
C
2. 3.
d 6. 3.
angles ACG, BCG are equal. But the former is cor. 3.
the double of the angle ADG, and the latter the
double of BDG. Therefore the angle ADG is equal
to BDG; and the whole angle ADB equal to ACG,
or BCG. Thus, the angle ACG is equal to any angle
in the fegment ADEB. Confequently all the angles
in that fegment are equal.
COR. 1. All angles at the circumference, which
ftand upon the fame arch, are equal.
COR. 2. An angle at the circumference, ftanding
upon any arch, is equal to an angle at the center, up-
on half that arch.
COR. 3. In the fame circle, or in circles of equal
radii, if two angles at the circumference ſtand upon
equal arches, they are equal to one another; and
converfely.
PRO P. VIII. PRO B.
To defcribe a circle about a given triangle.
Let ABC be the given triangle, about which it is Fig. 57.
required to defcribe a circle. Let the two fides AB,
BC be bifected by the perpendiculars DE, FE 2, inter- 28. 1.
fecting
64
Bo
ELEMENTS
i
4. I.
fecting each other in E, and E is the center pamě
circle that will circumfcribe the triangle. For EA,
EB, EC being drawn, the triangles ADE, BDE are
equal in every reſpect, fince AD, DB are equal,
DE common to both, and the angles at D right:
therefore EA is equal to EB. In the fame manner,
EB is proved equal to EC. If, therefore, a circle be
deſcribed about the center E, at the diftance EA, it
will paſs through the three angular points A, B, C,
and circumfcribe the triangle.
COR. The circumference of a circle may be defcri-
bed through any three points not in the fame ftraight
line.
Fig. 56.
PROP. IX. THEOR.
The angle in a femicircle is a right angle; the angle in
a fegment, greater than a femicircle, is an acute angle;
and the angle in a fegment, less than a femicircle, is an
obtufe angle.
Let ADEB be any fegment of the circle AEG, ha-
ving the line AB for its baſe. From the center C
let CG be drawn perpendicular to AB, meeting the
circumference of the other fegment AGB in G.
Then, according as ADEB is a femicircle, a greater
fegment, or a lefs fegment, the center C will fall up-
on, above, or below the line AB; and therefore the
angle
OOK III. OF GEOMETRY.
65
་
angle ACG will accordingly be right, acute, or ob-
tufe. But the angle ACG is equal to any angle, as
ADB, in the fegment ADEB 2. Confequently the
angle ADB is right, acute, or obtufe, according as
ADEB is a femicircle, a greater fegment, or a leſs
ſegment.
COR. 1. The center of the circumfcribing circle
falls within the triangle, upon one of its fides, or
without, according as it is an acute-angled, right-
angled, or obtufe-angled triangle.
COR. 2. In every right-angled triangle, the diftance
of the middle of the hypothenufe from the right-
angle is equal to half the hypothenufe.
COR. 3. The angle at the vertex of any triangle
infcribed in a circle, differs from a right angle by the
angle which its bafe forms with the radius or diame-
ter at one extremity of the bafe.
2
703.
PROP. X.
THEOR.
The oppofite angles of any quadrilateral figure, infcribed
in a circle, are together equal to two right angles; and
converfely.
Let ABCD be any quadrilateral figure infcribed in Fig. 58.
the circle AED, the two oppoſite angles at B and D are
together equal to two right angles. For, let the diagonal
I
AC
66
Book
ELEMENTS
AC be drawn and bifected by the perpendicular EF,
meeting the circumference in E, F. Then EF paffes
through G, the center of the circle, and bifects the
d1.c.2.3. arches ABC, ADC. Hence the angle AGF is equal
to the angle at B, and AGE to the angle at D. Con-
fequently the fum of the angles at B and D is equal
to the fum of the angles AGF, AGE, which is two
right angles.
b 7. 3.
Fig. 59.
C
c cor. 8.
3.
Alfo converſely, if the oppofite angles at B and D,
of a quadrilateral ABCD, be together equal to two
right angles, a circle may be defcribed about it. For
the circumference of a circle may be defcribed through
the three points A, B, C; and, if that do not paſs
through the point D, let it meet the diagonal DB in
any other point, as H. Then AH, HC being drawn,
the angles at B and H would be together equal to two
right angles, and therefore equal to the angles at B
and D. Confequently the angle at H would be equal
21. 1. to the angle at D, which is impoffible '.
d I.
COR. 1. If one of the fides of a quadrilateral fi-
gure, infcribed in a circle, be produced, the exterior
angle will be equal to the interior oppofite angle.
COR. 2. If, when the diagonals of a quadrilateral
figure are drawn, the angles which ſtand upon one
of the fides be equal; the circumference of a circle.
may be defcribed through the four angular points.
PROP..
OK III. OF GEOMETRY. 67
.
PROP. XI. THEOR.
If a straight line, paffing through any point in the circum-
ference of a circle be perpendicular to the radius at
that point, it will be a tangent to the circle; and con-
verfely.
Let the ftraight line DBE, paffing through any Fig. 60.
point B, in the circumference of the circle AFB, be
perpendicular to the radius CB; DBE is a tangent to
Then
the circle. For, let any point D be affumed in it,
from which let DC be drawn to the center.
CD is greater than CB, and confequently D a point a 3. c. 19,
without the circle.
Alfo converſely, If DE be a tangent at B, it is per-
pendicular to CB. For, fince every point in DE,
except B, is without the circle, CB is the fhorteſt
line that can be drawn to it from the center, and is
therefore perpendicular to DE".
COR. 1. At the fame point, in the circumference of
a circle, there cannot be more than one tangent.
COR. 2. A perpendicular to a tangent, from the
point of contact, paffes through the center.
COR. 3.
ference of a circle in more than one point, without
cutting it; or, between the circle and its tangent, a
A ftraight line cannot meet the circum-
Straight
I.
68
Book
ELEMENTS
ftraight line cannot be drawn from the point of con-
tact.
COR. 4. A ftraight line cutting the circle, and pa-
rallel to a tangent, intercepts equal arches from the
point of contact.
COR. 5. If two tangents paſs through the extremi-
ties of a diameter, they are parallel; and converſely.
COR. 6. Parallel lines, meeting the circumference
of a circle, interfect equal arches; and converfely.
COR. 7. A ftraight line, drawn through the middle.
of an arch, parallel to its chord, is a tangent to the
circle.
Fig. 61.
20. 4.
cor, ib.
PROP. XII. THEOR.
Two circles, whofe circumferences pass through the fame
point, in the line paffing through their centers, touch
one another.
Let the circumferences of two circles, whoſe cen-
ters are A, B, paſs through the fame point C, in the
ftraight line AB; the circles fhall touch one another
in that point. For, if poffible, let them meet in any
other point, as D, and let AD, BD be drawn: Then,
becauſe AD is equal to AC, and BD to BC; the ſum
or difference of AD, DB, two fides of a triangle,
would be equal to the third fide AB, which is abſurd².
COR,
FOR III. OF GEOMETRY.
69.
COR. 1. Two circles, whofe circumferences pafs
through the ſame point, without the line, joining their
centers, cut one another.
COR. 2. The circumferences of two circles cannot
meet in more than one point without cutting.
COR. 3. If two circles touch one another, the
ftraight line which joins their centers will pafs through
the point of contact.
PROP. XIII.
PROB.
To draw a tangent to a given circle, from a given point
without it.
Let EBC be the given circle, and A the given Fig. 62.
point, from which it is required to draw a tangent.
Let AC be drawn to the center, and upon AC, as a
diameter, let a circle be defcribed, interfecting the
given circle in the two points B, E. Then a ſtraight
line drawn from A, to either of theſe points, B, E
will be a tangent to the circle EBC. For, AB, BC
being drawn, becauſe the angle in a femicircle is a
right angle, AB is perpendicular to BC, and be- 29. 3.
cauſe AB is perpendicular to the radius BC, AB is a
tangent to the circle.
COR. 1. From the fame point, without a circle,
there cannot be drawn more than two tangents.
COR
DII. 3.
70
BOOK
ELEMENTS
COR. 2. Two tangents, drawn from the fame point,
without a circle, are equal.
PROP. XIV.
THEO R.
Fig. 63.
1
If a straight line touch a circle, and from the point of
contact a straight line be drawn to cut the circle, the
angles which the fecant makes with the tangent, are
equal to thofe in the alternate fegments; and con
verfely.
Let CD be a tangent to the circle ABE, and BA
any fecant drawn from the point of contact, the angle
ABD fhall be equal to any angle in the fegment AEB,
on the other fide of AB, and
angle, as F, in the fegment AFB.
ABC equal to any
If BA be
If BA be perpen-
dicular to CD, it divides the circle into two femi-
circles, each of which contains a right angle, and the
angles at B are alfo right angles. If not, let AE be
drawn parallel to CD, to meet the circumference in
E, and let EB be joined. The angle ABD is equal
to BAE, becauſe they are alternate angles ", and BAE
bcor. 3.7. is equal to AFB, becauſe they ftand upon equal
arches; therefore the angle ABD is equal to AEB.
a 16. I.
ન
3.
cor. 4.
II. 3.
10. 3.
a
c
Again, becauſe EAFB is a quadrilateral figure, in-
fcribed in the circle, the angles at E and F are toge-
ther equal to two right angles, and therefore equal to
the
BOOK III. OF GEOMETRY. 71
с
the two angles ABD, ABC. From theſe equals let the 11. 1.
equal angles E and ABD be taken away, and there
remains the angle at F equal to ABC.
Converſely, If the fecant AB make, with the
ftraight line CD, meeting the circumference in B,
the angle ABD, equal to any angle in the alternate
fegment, CD will be a tangent to the circle. For,
the fame conſtruction being made, fince the angles
BAE, AEB are each equal to ABD, they are equal
to one another; therefore the arch EB is equal to
BA, and confequently CD a tangent f.
f cor. 7.
II.
3.
!
PROP. XV. PROB.
Upon a given straight line, to defcribe a fegment of a circle
that ſhall contain a given angle.
Let AB be the given ftraight line, and C the gi- Fig. 64,
ven angle, it is required to defcribe upon AB a feg-
ment of a circle that ſhall contain an angle equal to
C. If C be a right angle, let a femicircle be deſcri-
a
bed upon AB. If not, let the angle ABD be made
equal to C, and BF drawn at right angles to BD, to 2 13. t.
meet the perpendicular EF, that bifects AB in the
point F: About the center F, at the diſtance FB, let
a circle be deſcribed. Its circumference will paſs
through
3
72
Book
ELEMENTS
C
4. 1
c II. 3.
d 14 3
through the point A, becauſe FA and FB are equal,
being the bafes of the triangles AEF, BEF, which are
equal in every refpect'; and BD will be a tangent to
that circle, becauſe it is perpendicular to the radius
FB. Hence the angle ABD, and confequently C, is
equal to the angle in the alternate fegment defcribed
upon AB
d
Fig. 65.
b
cor.
6.
II. 3.
cor. 2.
7. 3.
PROP. XVI. THE OR.
Any angle, whofe fides meet the circumference of a circle,
is equal to an angle at the center of that circle, ftand-
ing upon half the fum, or half the difference of the in-
tercepted arches, according as the angular point is
within or without the circle.
Let the fides AB, AC, of the angle BAC, meet the
circumference of the circle BDE, in the points B, E,
and C, D; BAC fhall be equal to an angle at the
center, ftanding upon half the fum, or half the dif-
ference of the arches BC, DE. For, through E,
let EF be drawn parallel to AC.
The arches CF,
DE, intercepted by the parallels AC, EF, are equal'.
Hence the arch BF is the fum or difference of BC,
DE, according as A is within or without the circle.
But the angle BEF, at the circumference, is equal to
an angle at the center upon half the arch BF ь.
There-
:
%
A
F
FIG. 57.
B
D
E
Ꮐ
D
E
FIG. 56.
A
C
D
闷
​FIG.56.
G
FIG. 59.
FIG. 58.
E
B
F
Gr
D
D F
F
C
D
E
B
Ꮐ
PR
B
FIG. 60.
H
F
C
C
H
A
C
D
K
G
E
FIG.61.
E
F
E
C
A
C
B
FA
B
A
E
D
C
FIG. 62.
F
FIG.63.
C
A
FIG.64.
F
E
D
B
-D
C
B
*
!
:
Book III.
OF GEOMETRY.
73
Therefore the angle BAC, which is equal to BEF,
becaufe of parallel lines, is equal to an angle at the
center, upon half the fum, or half the difference, of
the intercepted arches BC, DE.
COR. If two chords interfect one another at right
angles, within a circle, the oppofite arches are toge-
ther equal to half the circumference.
PROP. XVII.
THEOR.
If two chords interfect one another, within or without a
circle, the rectangle under the fegments of the one will
be equal to the rectangle under the fegments of the other.
Let AB, CD be two chords interfecting one ano- Fig. 66.
ther in any point E, within or without the circle
CBA, the rectangle under AE and EB will be equal
to the rectangle under CE and ED.
Let one of them, as AB, pafs through the center
F, from which let FG be drawn perpendicular to the
other CD, and let FC be joined. In the triangle
CFE, the rectangle under the fum and difference of
the fides CF, FE, is equal to the rectangle under the
fum and difference of the fegments of the bafe CG,
GE. But the fum of CF, FE is AE, and their ª 11. 2
difference EB, becauſe AF, FB are each equal to
CF; alfo the fum of CG, GE is CE, and their dif-
ference ED, becauſe CD is bifected by FG . There- cor. I.
fore
K
º.
a
b
2. 3.
74
Book III.
ELEMENTS
Fig. 67.
C II. 3.
fore the rectangle AEB is equal to the rectangle
CED. Confequently the rectangles under the feg-
ments of any two chords which interfect in E are e-
qual.
COR. 1. If, from the fame point, without a circle,
two ſtraight lines be drawn, the one to cut the circle,
and the other to touch it, the rectangle under the
ſegments of the former will be equal to the ſquare of
the latter.
For, if EC be a tangent, CFE is a right angled
triangle, and the difference of the fquares of CF, FE
d cor.9.2. is equal to the fquare of EC.
d
cor. 2.
6. 2.
Fig. 63.
COR. 2. If, from the fame point, without a circle,
two ſtraight lines be drawn, the one to cut the circle,
and the other to meet the circumference, fo that the
rectangle under the fegments of the former be equal
to the fquare of the latter, the latter will be a tan-
gent.
If EC be the line that meets the circle, and EBA
the line that cuts it, fo that the fquare of EC be equal
to the rectangle AEB; then EC is a tangent: For, if
it met the circumference in any other point, as G,
the rectangle GEC would be equal to the rectangle
AEB, and therefore equal to the fquare of EC,
which is impoffible *.
COR. 3. If, from any point in the circumference of
a circle, a perpendicular be let fall upon a diameter,
the
:
Book III. OF GEOMETRY.
75
the rectangle under the fegments of the diameter will
be equal to the fquare of the perpendicular.
COR. 4. Alfo, if from the fame point, a ftraight
line be drawn to one extremity of the diameter, its
fquare will be equal to the rectangle under the whole
diameter, and the adjacent fegment.
PROP. XVIII. THEOR.
If, through one extremity of a diameter of a circle, a
a ftraight line be drawn, to meet the circumference, and
any perpendicular to that diameter, the rectangle under
its fegments, from that point, will be equal to the rec-
tangle under the diameter, and the distance of the fame
point from the perpendicular.
Let ABE be a circle, of which AB is a diameter; Fig. 69,
let CD be any ftraight line perpendicular to AB,
within or without the circle, and AD any ſtraight
line drawn through the point A, to meet the perpen-
dicular CD in D, and the circumference in E; the
rectangle DAE will be equal to the rectangle BAC.
For E, B being joined, the angles DCB, DEB are
equal, becauſe they are right angles ; therefore a
circle may be defcribed through the four points, D,
C, E, B; and confequently the rectangle DAE is 10. 3.
equal to the rectangle BAC .
a
A
a hyp. &
b
9.3
cor. 2. ib.
C 17. 3.
A
COR.
76
BOOK III.
ELEMENTS
COR. If, through any point, in the circumference
of a circle, two ftraight lines be drawn, to meet the
circumference, and any perpendicular to the ſtraight
line pafling through that point and the center, the
rectangles under their fegments, from the fame point,
will be equal.
Fig. 70.
7. I.
b IO. I.
C
c cor. 5.
18. I. &
5. I.
d II. 3.
PROP. XIX. PROB.
To infcribe a circle in a given triangle.
a
Let ABC be the given triangle, in which it is re-
quired to infcribe a circle. Let the two angles at B
and C be bifected by the lines BD, CD meeting
each other in the point D, from which let the per-
pendiculars DE, DF, DG, be drawn to the fides of
the triangle". The triangles BDE, BDG having the
angles at B equal, the angles at E and G right, and
the fide BD oppofite to the right angles common to
both, are equal in every reſpect . Hence DG is e-
qual to DE. In like manner, DF may be proved e-
qual to DE. Thus, the three perpendiculars are e-
qual. Let then a circle be defcribed about the cen-
ter D, at the diftance of one of them, and it will
paſs through the three points E, F, G, and be infcri-
bed in the triangle, becaufe the fides of the triangle
being each perpendicular to a radius of the circle,
touch its circumference .
COR.
Book III. OF GEOMETRY.
77
:
COR. 1. Straight lines, drawn from the angular
points to the center of the infcribed circle, bifect the
angles of the triangle.
!
COR. 2. In the fame manner, may be found the
center of a circle that fhall touch any three ftraight
lines given in pofition, provided they be not all pa-
rallel, nor all tend to the fame point.
PROP. XX.
PROB.
To infcribe a fquare in a given circle.
Let ABCD be the given circle, in which it is re- Fig. 71.
quired to inſcribe a fquare. Let two diameters AC,
BD be drawn to cut each other at right angles, and
let their extremities be joined by the lines AB, BC,
CD, DA. The quadrilateral formed by theſe lines is
2
b
14. I.
a fquare For all its angles are right, becauſe each 9. 3.
of them is an angle in a femicircle; it is a parallelo-
gram, becauſe any two adjacent angles are together 2. cor.
equal to two right angles; and all its fides are equal,
becauſe they are the bafes of the triangles AEB, BEC,
CED, DEA, which are equal in every refpect: And
that fquare having its angular points in the circumfe-
rence, is infcribed in the circle.
C
And © 4. E
PROP.
...
Book III.
78
ELEMENTS
Fig. 72.
b
12. 2.
PROP. XXI. PROB.
To infcribe a regular pentagon in a given circle.
Let EGF be the given circle, in which it is re-
quired to infcribe a regular pentagon. Let the ra-
dius AB be divided in the point C, fo that the rec-
tangle ABC may be equal to the fquare of AC. A-
bout the center B, with a radius equal to AC, let a
circle be deſcribed to interfect EGF in D; and let
AD, DC, DB, be joined, and DC produced to meet
the circumference in E. The arch BE is a fifth part
of the whole circumference. For, let the diameter
EAF be drawn ; then,
1. The triangles ABD, DBC, are mutually equi-
angular. For, fince BD is equal to AC, the rec-
tangle ABC will be equal to the fquare of BD;
therefore BD is a tangent to the circle that paffes
cor. 2. through the three points A, C, D, and the angle
BDC equal to DAC, an angle in the alternate feg-
ment. Thus, the triangles ABD, DBC have the
angles at D and A equal, the angle at B common to
both, and therefore the remaining angle BDA equal
to the remaining angle BCD d.
17. 3.
C 14. 3.
cor. 2.
18. I.
2. In the ifofceles triangle ABD, each of the angles
at the bafe is double the angle at the vertex. For,
fince the angle CBD is equal to BDA, and BDA has
been
Book III. OF GEOMETRY.
79
been proved equal to BCD, the angle CBD is equal
e
to BCD, and therefore the fide CD is equal to BD. cor. 5. 1
But BD is equal to CA. Therefore CD is equal to
CA, and the angle CDA to CAD f.
BDC was alfo proved equal to CAD.
But the angle f cor. 4. 1.
Confequently
the angle BDA is double the angle CAD or BAD.
3. The chord BD is parallel to the diameter EF.
For, fince the angles BDC, CDA are each equal to
CAD, they are equal to one another. But CDA is
equal to AEC. Therefore BDC is equal to AEC,
and they are alternate angles; confequently BD is
parallel to EF.
g
£ 15. I.
h
I.
4. The arches EB, DF are each of them double
the arch BD", becauſe the angles EAB, DAF (being cor. 3.2.
equal to their alternate angles at B and D,) are each
of them double the angle at BAD.
Hence the arch BD is a fifth part of the femicir-
cumference EDF, and confequently BE a fifth part
of the whole circumference.
3.
Now, let the three arches EG, GH, HK, be made
each equal to BE, and the remaining arch BK will be
equal to the fame. Alfo, let the chords of thefe five
arches BE, EG, GH, HK, KB be drawn, and the
polygon which they form will be a regular pentagon
infcribed in the circle. For all its fides are equal,
becauſe they ſubtend equal arches, and all its angles
equal, becauſe they ftand upon equal parts of the cor. 3. 7.
i 3. 3.
Κ
3.
circumference.
COR.
$
80
Book III.
ELEMENTS
COR 1. If the circumference of a circle be divided
into any greater number of equal parts, the chords of
theſe parts form a regular polygon.
COR. 2. If the radius of a circle be divided, fo
that the rectangle under the whole, and one of the
parts, may be equal to the fquare of the other, the
greater fegment is equal to the fide of a regular deca-
gon inſcribed in that circle.
Fig. 73.
a I. I.
b cor. 4.
C
18. I.
cor. I.
12. I.
PRO P. XXII. PR O B.
To infcribe a regular hexagon in a given circle.
Let EDB be the given circle, in which it is requi-
red to inſcribe a regular hexagon. Let A be the
center, and B any point in the circumference. About
the center B, at the diſtance BA, let another circle
be deſcribed, to interfect the former in C, G, and
from the points C, B, G, let the three diameters CF,
BE, GD, of the given circle, be drawn. Then the
chords of the fix arches, into which thefe diameters
divide the circumference, will form a regular hexa-
gon. For, fince ABC and ABG are equilateral
triangles, each of the angles GAB, BAC is a third
part of two right angles. But the three angles
GAB, BAC, CAD make two right angles. There-
fore the angle CAD is alſo a third part of two right
angles,
י,
:
GE OMETRY
A FIG.65.
B
D
D
FIG. 65.
E
E
A
E
B
G
C
E
F
B
C
F
E
B
FIG.68.
F
B
D
D
B
B
A
F
FIG. 66.
FIG. 67.
A
A
D
C
C
A
G
E
FIG.66.
A
FIG.69.
E
B
E
FIG:69.
E
B
C
ете
FIG. 71.
B
A
G
FIG.72.
D
E
FIG70.
G
D
A
B
C
E
FIG.73.
Ꭰ
C
H
0 0 0
B
10
C
D
E
A
F
A
F
G
B
D
B
BOOK III. OF GEOME T R Y.
angles, and theſe three angles equal to one another.
And, becauſe each of theſe is equal to its oppofite
angle, all the fix angles at the center A are equal.
Confequently the arches upon which they ſtand are
equal; and the figure formed by the chords of theſe
arches, a regular hexagon.
f
COR. In every circle, a chord equal to the radius
fubtends a fixth part of the circumference—or, the
radius of the circle is equal to the fide of the infcri-
bed regular hexagon.
d
€
12. I.
cor. 3.
2. 3.
f cor. I.
21. 3.
PROP. XXIII. THEOR.
The Square of the fide of a regular pentagon, infcribed
in a circle, is equal to the fum of the fquares of the
fides of a regular hexagon and decagon inſcribed in the
fame circle.
Let BF be the fide of a regular pentagon, and BD Fig. 74.
the fide of a regular decagon, infcribed in the circle
EDB, of which A is the center. The fquare of BF
ſhall be equal to the fquares of AE, BD.
Let the radius AB be divided in C, fo that the
rectangle ABC may be equal to the fquare of AC, 12. 2.
and let FD, DC, CF, FE be joined. Then each of
the lines DC, DB, DF, being equal to AC, they b 21. 3.
are equal to one another. Hence the angle DBC is
L
equal
·
82
Book III.
ELEMENTS
€
d
equal to DCB. But the two angles EFD, DBC
15. 3. being equal to two right angles, are equal to the
two angles ECD, DCB.
Therefore the angle EFD
d
г. 5. 1. is equal to ECD. From theſe let the equal angles.
DFC, DCF be taken away, and there remains the
angle EFC equal to ECF. Confequently the lines
4. I.
EF, EC are equal º.
Again, becaufe EC is equal to the fum of AB, AC,
£ cor. 12. it is divided fimilarly to AB, that is, the rectangle
f
2.
ECA is equal to the fquare of EA: Therefore twice
the fquare of EA is equal to twice the rectangle
ECA. Let twice the fquare of EA be added to both :
Then four times the fquare of EA is equal to twice
the rectangle ECA, with twice the fquare of EA. But
four times the fquare of EA is equal to the fquare of
EB, or to the fquares of EF, FB". Therefore the
12. & fquares of EF, FB are equal to twice the rectangle
ECA, with twice the fquare of EA. Inftead of twice
the rectangle ECA, with once the fquare of EA, let
as
& cor. 2.
C
2.
9. 3.
2.
75.
i
the fquares of EC, CA be taken. Then the fquares
of EF, FB are equal to the three fquares of EC, CA,
and EA. From thefe equals let the equal fquares of
EF, EC be taken away, and there remains the ſquare
of FB, equal to the fquares of CA and EA.
COR. Hence the fide of a regular pentagon may
be found by the following conftruction. Let the di-
ameter EAB be drawn, the radius AD perpendicular
to it, and the radius EA bifected in F. About the
center
Book III. OF GEOMETRY.
83
center F, at the diſtance FD, let a circle be defcri-
bed to interfect EB in C, and let CD be joined. The
ſtraight line CD is equal to the fide of a regular pen-
tagon inſcribed in the circle.
PRO P. XXIV. PR O B.
About a given circle to defcribe a regular polygon, of the
fame number of fides, with a given one infcribed in the
fame.
Let ABCDEF be the given regular polygon, infcri- Fig. 76.
bed in the given circle ACD: It is required to de-
ſcribe about the circle ACD a regular polygon of the
fame number of fides with ABCDEF. Through the
angular points of the given polygon, let tangents to
the circle be drawn, interfecting one another in the
points G, H, K, L, M, N; and the figure formed by
theſe tangents is a regular polygon, deſcribed about
the given circle.
2
13.3.
For, fince two tangents, drawn from the fame
point, are equal, GA is equal to GF; and thus all car. 2.
the triangles formed upon the fides of the polygon
ABCDEF are ifofceles. They are alfe equal to one
another. For the angle GAF is equal to an angle in
the alternate fegment, that is, to an angle at the 14. 3.
circumference ſtanding upon the arch AF, and the
angle
A
84
BOOK III.
ELEMENTS
1:
3. 3.
7. 3.
e 5. I.
angle HAB equal to an angle at the circumference,
ſtanding upon the arch AB. But the arches AB,
AF, fubtended by equal lines, are equal, and the
angles at the circumference, which ſtand upon them,
d cor. 3. alfo equal". Therefore the angle GAF is equal to
HAB, and confequently the angle GFA equal to
HBA. Whence the triangles AGF, AHB, having
the fides AB, AF equal, are equal in every reſpect .
Thus, the angles at G and H are equal. And, in like
manner, may it be proved that all the angles of the
polygon GHKLM are equal to one another.
the fides GA, AH, are equal. And, in like manner,
may it be proved, that all the fides of the polygon
are bifected in the points of contact; and, confe-
quently, fince their halves are equal, all the fides are
equal to one another.
Alfo,
COR. If a regular polygon be deſcribed about a
circle, the fides of the polygon are bifected in the
points of contact.
Fig. 77.
PRO P. XXV. PRO B.
To defcribe a circle about a given regular polygon, and to
infcribe another in the fame.
Let ABCDEF be a given regular polygon, in and
about which it is required to defcribe circles. Let
the
BOOK III. OF GEOMETRY. 85
1
the two angles at A and B be bifected by the lines
AG, BG meeting in the point G: Then G is the
center of both circles. For, let the perpendiculars
GH, GK be drawn from the point G to the fides
AB, AF, and let F, G be joined. Then, comparing
the triangles AGF, AGB, the fides AF, AB are e-
qual by hypotheſis; AG is common to both, and the
included angles are equal by conftruction; therefore
the bafe FG is equal to the bafe GB, and the angle
AFG equal to ABG 2. But the angle ABG is by 4. 1.
conſtruction half the angle ABC. Therefore the
angle AFG is half the equal angle AFE, that is, the
angle AFE is bifected by the line FG. In like man-
ner, may it be proved, that all the angles of the po-
lygon are bifected by lines drawn from G to the an-
gular points.
2
b
Again, becauſe the angles BAF, AFE are equal,
their halves GAF, FAG are alſo equal; therefore the
In the fame manner, it cor. 5. 1.
fide GF is equal to GA “.
may be demonſtrated, that all the lines drawn from
G to the angular points of the figure are equal to one
another. Therefore the circle defcribed about the
center G, at the diftance GA, will pafs through the
points A, B, C, D, E, F, and circumfcribe the given
polygon.
Next, comparing the triangles AGH, AGK, the
angles at A are equal, becauſe GA bifects the angle
BAF, the angles at H and K are equal, becauſe they
are
*
86
Book III.
ELEMENTS
C
cor. 5.
18. 1. &
I.
5. I.
are right angles, and the fide GA is common to both.
Hence the triangles are equal in every refpect: The
fide GH is equal to GK. In like manner, may it be
proved that all the perpendiculars which fall from G,
on the fides of the figure, are equal to one another.
Therefore the circle defcribed about the center G, at
the diſtance GH, will pafs through their extremities:
And it will be infcribed in the polygon, becauſe the
fides of the polygon, being each of them perpendicu-
lar to a radius of the circle, touch its circumference.
COR. Of any regular polygon, the circumfcribing
and infcribed circles, have the fame center.
APPENDI
DIX.
PROP. 1. Perpendiculars falling from the extremities.
of a diameter of a circle, upon any chord in the fame,
cut off equal fegments.
PROP. 2. The fquare of the fide of an equilateral
triangle, infcribed in a circle, is triple the fquare of
the radius.
PROP. 3. If two chords in a circle interfect one an-
other at right angles, the fum of the fquares of the
four fegments will be equal to the fquare of the dia-
meter.
PROP. 4. The baſe, the perpendicular, and the fum
or difference of the fides, being given, to defcribe the
triangle.
PROP.
BOOK III.
87
OF GEOMETRY.
PROP. 5. If an equilateral triangle be infcribed in
a circle, and from the three angular points, ftraight
lines be drawn to the fame point in the circumfe-
rence, one of them fhall be equal to the fum of the
other two.
PROP. 6. The perpendicular of an equilateral
triangle is triple the radius of the infcribed circle.
PROP. 7. A regular octagon, infcribed in a circle,
is equal to the rectangle under the fides of the infcri-
bed and circumfcribing fquares.
PROP. 8. The vertical angle, the difference of the
fegments of the bafe, and the fum or difference of the
fides are given, to conftruct the triangle.
PROP. 9. If, from the extremities of any chord in
a circle, ftraight lines be drawn, to any point in the
diameter to which it is parallel, the fum of their
fquares will be equal to the fum of the fquares of the
fegments of the diameter.
PROP. 10. If, from two points, in the diameter of
a circle, equally diſtant from the center, two ftraight
lines be drawn to meet each other in the circumfe-
rence, the fum of their fquares will be equal to the
fum of the fquares of the fegments into which the
diameter is divided in one of theſe points.
PROP. 11. If, from a point without a circle, two
tangents be drawn, and from one of the points of
contact, a perpendicular be let fall upon the diameter
paffing through the other, it will be bifected by a
traight
88
Book III.
ELEMENTS
ftraight line joining the point without the circle, and
the fartheft extremity of the diameter.
PROP. 12. The difference of the angles at the baſe,
the difference of the fegments of the bafe, and the
fum or difference of the fides being given, to con-
ftruct the triangle.
PROP. 13. If, from the extremities of the bafe of
any triangle, perpendiculars be let fall upon the op-
poſite fides, a ſtraight line drawn through their inter-
fection from the vertex, will be perpendicular to the
bafe.
PROP. 14. From a point, without a circle, let two
tangents be drawn, and from the points of contact
two chords, to the fame point in the circumference;
then, if a fecant be drawn from the point without the
circle, parallel to the one chord, and meeting the o-
ther, a perpendicular to the parallel chord, from the
point of concourſe, will bifect it.
PROP. 15. If, from any point in the circumference
of a circle, two chords be drawn, and the intercepted
arch be bifected by a perpendicular falling upon ei-
ther chord, that perpendicular will add to the le´s, or
cut off from the greater chord, a part equal to half
their difference.
PROP. 16. The difference of the angles at the bafe,
the difference of the fegments of the bafe, and the
perpendicular being given, to conftruct the triangle.
PROP.
BOOK III. OF GEOMETRY.
89
*
PROP. 17. If, from a point without a circle, two
ftraight lines be drawn, to touch or cut the circum-
ference, and from the fame point a fecant be drawn,
through the line joining the points of contact, or the
interfection of the lines joining the oppofite points of
fection, it will be ſo divided, that the rectangle under
the whole line, and its middle fegment, fhall be equal
to the rectangle under the extreme fegments.
PROP. 18. If the oppofite fides of a quadrilateral,
infcribed in a circle, be produced to meet, the fquare
of the line joining the points of concourfe will be e-
qual to the fum of the fquares of two tangents drawn
from theſe points.
PROP. 19. If, from a point without a femicircle,
two tangents be drawn, and the points of contact be
joined to the oppoſite extremities of the diameter, the
ſtraight line which paffes through their interfection,
and the point without the femicircle, will be perpen-
dicular to the diameter.
}
PROP. 20. The vertical angle, the perpendicular,
and the perimeter, being given, to conftruct the
triangle.
M
ELEMENTS
ELEMENTS
O F
GEOMETRY.
воок
IV.
1 .
DEFINITIONS.
W
HEN one magnitude contains another
a certain number of times, without a re-
mainder, the former is faid to be a multiple of the
latter, and the latter a part of the former.
2. When ſeveral magnitudes are multiples of as ma-
ny others, and each contains its part the fame number
of times, the former are faid to be equimultiples of
the latter, and the latter like parts of the former.
3. Betwixt any two finite magnitudes of the fame
kind, there fubfifts a certain relation, in reſpect to
quantity,
Book IV. OF GEOMETRY.
91
quantity, which is called their Ratio. The two mag-
nitudes compared are called the Terms of the Ra-
tio; the firft, the Antecedent; and the laſt, the
Confequent.
4. Of four magnitudes, the ratio of the firſt to the
fecond is faid to be the fame with (or equal to) the
ratio of the third to the fourth, when the firſt con-
tains any part whatever of the fecond, juſt as oft as
the third contains the like part of the fourth.
5. But the ratio of the firſt to the fecond is faid to be
greater than the ratio of the third to the fourth, when
the firſt contains a part of the fecond oftener than the
third contains a like part of the fourth, and lefs when
not fo often.
6. When two ratios are equal, their terms are ſaid
to be proportional.
7. The equality of ratios, or proportionality of
magnitudes, is expreffed thus, A: B = C: D, (the
ratio of A to B is equal to the ratio of C to D), or
A : B :: C : D, (A is to B as C to D). And the
whole expreffion is named an Analogy or Propor-
tion.
8. The homologous terms of an analogy, are the
Antecedents or the Confequents of the Ratios.
9. Magnitudes are faid to be in continued propor-
tion, when the ratios of the firft to the fecond, of
the ſecond to the third, of the third to the fourth, &c.
are all equal.
10.
92
Book IV.
ELEMENTS
10. Of magnitudes in continued proportion, the
ratio of the firſt to the third is called the duplicate
ratio of the firſt to the ſecond, the ratio of the firſt to
the fourth, the triplicate of the first to the fecond,
&c.; alfo, the ratio of the firft to the fecond is called
the fubduplicate of the first to the third, the fubtripli-
cate of the firſt to the fourth, &c.
11. Of any number of magnitudes of the fame
kind, the ratio of the firft to the laft is faid to be
compounded of the ratios of the firft to the fecond,
the fecond to the third, the third to the fourth, and
fo on to the laſt.
12. Of any analogy, the magnitudes are faid to be
proportional by Inverfion, when the fecond is to the
firſt as the fourth to the third.
13. By Alternation, when the first is to the third as
the fecond to the fourth.
14. By Compofition, when the fum of the firſt and
fecond is to the ſecond, as the fum of the third and
fourth is to the fourth.
15. By Divifion, when the difference of the firſt
and fecond is to the ſecond, as the difference of the
third and fourth is to the fourth.
16. By Conversion, when the firft is to the fum of
the firit and fecond, as the third to the fum of the
third and fourth. Alfo, when the firft is to the dif
ference of the firit and fecond, as the third to the
difference of the third and fourth.
17.
BOOK IV. OF GEOMETR Y.
93
17. By Mixing, when the fum of the firft and fe-
cond is to their difference as the ſum of the third and
fourth to their difference.
18. If there be any number of magnitudes, and as
many others, which, taken two and two, are conti-
nually proportional, either in a direct order, (that is,
if the firſt be to the fecond of the former, as the firſt
to the fecond of the latter, the fecond to the third of
the former, as the fecond to the third of the latter,
&c.) or in a crofs order, (that is, if the firft be to
the fecond of the former, as the laft but one to the
laft of the latter, the fecond to the third of the former,
as the laft but two to the laft but one of the latter,
&c.), they are faid to be proportional by Equality,
when the firſt is to the laſt of the former, as the firft
to the laſt of the latter.
AXIOM S.
1. Equal magnitudes contain the fame or equal
magnitudes, the fame number of times.
2. Of two magnitudes, that which contains a third
magnitude the greater number of times is the greater.
3. Equimultiples, or like parts of the ſame, or e-
qual magnitudes, are equal.
4. A multiple, or part of a greater magnitude, is
greater than the fame multiple, or part of a leſs.
5.
94
BOOK IV
ELEMENTS
5. All ratios of equality are the fame.
6. Ratios that are equal to the fame, or to equal
ratios, are equal to one another.
7. Two equal magnitudes have the fame ratio to
the fame magnitude; and the ſame magnitude has the
fame ratio to each of two equal magnitudes.
8. If one ratio be greater than another, any ratio
equal to the former will be greater than the latter;
alfo the former will be greater than any ratio equal
to the latter.
}
Fig. 78.
LEM M
MMA I.
If there be any number of magnitudes equimultiples of as
many, the fum of the former will be the fame mul-
tiple of the ſum of the latter, that any one of them is of
its part.
Let any number of magnitudes AB, CD be equi-
multiples of as many E, F; the fum of AB, CD will
be the fame multiple of the fum of E, F, that AB is
of E. For, fince AB is a multiple of E, it confifts
* def. 1. 4. of a certain number of parts, each equal to E, let
theſe be Am, mn, nB. In like manner, CD confifts.
of a certain number of parts, each equal to F, let
theſe be Cp, pq, qD. And, becauſe AB, CD, are
equimultiples of E, F, the number of parts in each is
the
Book IV. OF GEOMETRY.
95
the fame. But the ſum of Am, Cp is equal to the def. 2.4
fum of E, F; the fum of mn, pq is equal to the fum
of E, F; and the fum of nB, qD equal to the fum of
E, F. Therefore the fum of AB, CD contains the
fum of E, F the fame number of times, without a re-
mainder, that one of them AB contains its part E,
COR. If AB contains E juft as oft as CD contains
F, the fum of AB, CD will alfo contain the fum of
E, F the fame number of times, although AB, CD
be not equimultiples of E, F.
LEMMA
A II.
If, of two magnitudes, there be any number of equimul-
tiples, the fum of the multiples of the one, and the fum
of the multiples of the other, will also be equimultiples
of the fame magnitudes.
Of the two magnitudes A, B, let there be any Fig. 79.
number of equimultiples, as CD, EF; DG, FH, the
fum of CD, DG, and the fum of EF, FH will alfo
be equimultiples of A, B. For, fince CD, EF are
equimultiples of A, B, there are as many mag-
nitudes equal to A in CD, as there are magnitudes
equal to B in EF. In like manner, there are
as many magnitudes equal to A in DG as there
are
• def. 2. 4.
96
Book IV.
ELEMENT S
are magnitudes equal to B in FH. Confequently the
whole CG contains A the fame number of times that
EH contains B, and there is no remainder; therefore
CG, EH are equimultiples of A, B.
COR. If two magnitudes be equimultiples of other
two, any equimultiples of the former will alſo be e-
quimultiples of the latter. And any like parts of the
And
latter will be like parts of the former.
Fig. So.
a poſt. 4.
I.
LEM MA III.
If, from any magnitude, its half be taken away, and
from the remainder its half, and fo on continually;
there ſhall at length remain a part of that magnitude
less than any magnitude propoſed.
From AB let its half BC be taken away, and from
the remainder AC let its half CD be taken away, and
fo on continually, there fhall at length remain a part
of AB lefs than any magnitude propofed, as FG.
For, let magnitudes GH, HK, KL, each equal to
FG, be continually added to FG, till the whole LF
fhall be a multiple of FG, greater than AB. And,
let the number of divifions BC, CD, DE, EA, in
AB, be the fame with the number of parts in LF.
Then, becauſe AB is leſs than LF, if BC, which is
the half of AB, and LK, which is not greater than the
half
Book IV.
97
OF GEOMETRY.
half of LF, be taken from them, the remainder AC
will be lefs than the remainder FK. In like manner,
AD is lefs than FH, and AE, (which is evidently
part of AB), leſs than FG.
COR. If, from any magnitude, more than its half
be taken away, and from the remainder more than its
half, and fo on continually, there fhall at length be a
remainder leſs than any magnitude propoſed.
PROPOSITION I.
Magnitudes have the fame ratio as their equimultiples.
Let CG, DL be equimultiples of A, B ; A: B:: Fig. 81.
CG: DL. For, let Bn be any part of B, lefs than
A; CE, EF, FG, each equal to A; DH, HK, KL,
each equal to B; and Dn, Hn, Kn, each equal to
Bn. Then, as often as A contains Bn, fo often does
8
b
I.
cor. lem.
I.
CE contain Dn; EF, Hn; FG, Kna; and fo often ax. 1. 4.
does CG contain the fum of Dn, Hn, Knº. But the
fum of Dn, Hn, Kn, is the fame part of DL that Dn
is of DH, or Bn of B. Therefore A: B:: CG:
DL d.
COR. Magnitudes have the fame ratio as their like
parts.
C
lem. 1.
d def. 4. 4.
N
PROP.
98
Book IV.
ELEMENTS
Fig. 82.
a
PROPOSITION
II.
If, instead of two homologous terms of an analogy, any
equimultiples, or like parts of them, be taken, the mag-
nitudes will ſtill be proportional.
Let A : B :: C: D. And, ft, inſtead of B, D,
let Bm, Dm, like parts of them, be taken, then
A: B:: C: Dm.
For, let Bn, Dn be any like parts of Bm, Dm;
cor. lem. they will alfo be like parts of B, D; therefore A
2.
def. 4.4. contains BВn as often as C contains Dn. Whence
A: Bm :: C: Dm b.
Fig. 83.
2. Inſtead of A, C, let Am, Cm, like parts of them,
be taken, then Am: B:: Cm : D.
For, if one of theſe ratios, as Am: B, be greater
than the other Cm : D, Am contains fome part of B,
fuppofe Bn, oftener than Cm contains Dn, the like
e def. 5. 4. part of D. Therefore alſo mp = Am contains Bn of-
Fig. 84.
Confe-
tener than mq = Cm contains Dn; and ſo on.
fo
quently the whole of A contains Bn oftener than the
whole of C contains Dn; which is contrary to the
hypothefis.
3. Inſtead of B, D, let BM DM, equimultiples of
them, be taken; then A: BM :: C: DM.
For, let AE, CF, be the fame multiples of A, C,
that BM, DM are of B, D. Then AE: BM (:: A:
B
BOOK IV. OF GEOMETRY.
99.
Bf::C: D) :: CF : DM f.
But A, C are like 1. 4.
f
parts of AE, CF; therefore A: BM:: C: DM.
4. Inſtead of A, C, let AM, CM, equimultiples of
them, be taken, AM : B :: CM : D.
1
For, let BE, DF be the fame multiples of B, D, Fig. 85,
that AM, CM are of A, C.
f
: B' :: C: D) :: CM : DF £.
Then AM: BE (:: A
But B, D are like
parts of BE, DF; therefore AM : B :: CM : D.
PROPOSITION III.
If four magnitudes be proportional, according as the first
is greater, equal, or less, than the fecond, the third
will be greater, equal, or lefs, than the fourth.
Let ABC: D. And, ift, Let A be greater Fig. 86.
than B, then C will be greater than D.
Bn, a part of B, leſs
a
a
For, let AE be equal to B;
than EG, the difference of A, B ; and Dn the like lem. 3.
part of D. Then it is evident that A contains Bm
oftener than (AE, that is,) B contains Bn. But C
contains Dn juſt as often as A contains Bn", and D
contains Dn juſt as often as B contains Bn.
fore C contains Dn oftener than D contains Dn; and
confequently C is greater than Dª.
D.
There-
def. 4. 4.
c def. 2. 4.
dax. 2. 4.
2. Let A be leſs than B, then C will be leſs than Fig. 87.
J
For,
100
BOOK IV.
ELEMENTS
For, let AE be equal to B, Bn, a part of B, lefs
* lem. 3. than either A or EGª, and Dn the like part of D.
© def. 2.4.
b def. 4.4
Then it is evident that (AE,
oftener than A contains Bn.
as often as B contains Bn,
as often as A contains Вn".
that is) B contains Bn
But D contains Dn juft
and C contains Dn juſt
Therefore D contains
Dn oftener than C contains Dn; and confequently D
ax. 2.4. is greater than Cd, or C lefs than D.
3. Let A be equal to B, then C will be equal
to D.
For, from what has been demonftrated, C can
neither be greater nor lefs than D.
1
COR. 1. The greater of two unequal magnitudes
has a greater ratio to the fame magnitude than the
lefs has. And converfely, of two magnitudes, that
which has the greater ratio to any third magnitude is
the greater. If A be greater than B, A C is greater
than B C; for A contains a part of Clefs than
A-B, oftener than B contains the fame. If A : C
be greater than B: C, A is greater than B, becauſe
A contains fome part of C oftener than B does.
•
COR. 2. Thoſe magnitudes which have the fame
ratio to the fame magnitude are equal. If A: C =
B: C, A is equal to B, becauſe otherwife the ratios
would not be equal.
14
PROPO.
Book IV.
OF GEOMETRY. 1or
PROPOSITION IV.
If four magnitudes be proportional, they are also propor-
tional by inverfion.
Let A B
C D, then B: A:: D: C.
:
Fig. 88.
For, let Am, Cm be like parts of A, C. Then
Am : B :: Cm : D', and any multiple of Am is to B ² 2. 4.
as the fame multiple of Cm is to D. Therefore, ac-
cording as B is greater, equal, or lefs, than any mul-
tiple of Am, D is greater, equal, or lefs, than the ſame
multiple of Cm. Confequently B contains Am jufſt ³ 3. 4.
as oft as D contains Cm, and B: A:: D: C.
COR. I. Thoſe magnitudes, to which the fame
magnitude has the fame ratio, are equal. If C: A
:: C: B, A is equal to B. For, by inverfion, A:C
:: B: C; and therefore A=B©.
:
b
с
c 2. C. 3.4.
COR. 2. The fame magnitude has a greater ratio
to the lefs of two unequal magnitudes than it has to
the greater. And converfely, Of two magnitudes,
that to which any third magnitude has the greater
ratio is the lefs. If A be greater than B, C B is
greater than C : A. For theſe ratios cannot be equal,
becauſe from thence it would follow that A, B are
alfo equal, which is contrary to hypothefis. Neither 1. c.4.4.
can C A be greater than C: B; becaufe then C
would contain fome part of A oftener than the like
part
₫
102
ELEMENTS Book IV.
part of B; that is, it would contain a greater magni-
tude oftener than a lefs; which is abfurd. If C: B
be greater than C: A, A is greater than B. For,
fince C contains fome part of B oftener than it con-
tains the like part of A, that part of B muſt be leſs
than the like part of A, and confequently the whole
ax. 4. 4. of B lefs than the whole of A.
Fig. 89.
* cor. lem.
2.
3
Fig. 90.
PROPOSITION
V.
If the first be the fame multiple, or part, of the fecond,
that the third is of the fourth, the four magnitudes
fhall be proportional. And converfely, Of four pro-
portional magnitudes, if the first be a multiple, or part,
of the fecond, the third ſhall be the fame multiple, or
part, of the fourth.
Let A be the fame multiple of B that C is of D,
then A: B:: C: D.
For, let Bn, Dn be like parts of B, D. Then, be-
cauſe A, C, are equimultiples of B, D, and B, D e-
quimultiples of Bn, Dn; A, C will alſo be equimul-
tiples of Bn, Dn; that is, A contains any part what-
ever of B as often as C contains the like part of D.
Therefore A:B:: C: D.
Next, let A be the fame part of B that C is of D
then B is the fame multiple of A that D is of C
;
;
there-
Book IV. OF GEOMETRY.
103
therefore B: A :: D: C; and, by inverfion, A: B
:: C: Db.
A. 4.
Converſely, If A: B:: C: D, and A a multiple or
part of B, C is the fame multiple or part of D. For,
let E be the fame multiple or part of D that A is of
B. Then, A: B:: E:D; but A: B:: C: D;
therefore CD:: E: D. Hence C is equal to E, 2. c.3.4.
and the fame multiple or part of D that A is of B.
COR. If four magnitudes be proportional, and the
firſt a multiple of any part of the fecond, the third
fhall be the fame multiple of the like part of the
fourth. And converſely.
C
d 2. 4.
!
PROPOSITION VI.
If four magnitudes be proportional, according as the first
is greater, equal, or less than the third, the fecond
will be greater, equal, or less than the fourth.
Let A: B:: C: D, then, if A be greater than C, Fig. 91.
B will be greater than D.
For, fince A is greater than C, A: B is greater
than C : B³; but A: B = C: D; therefore C: D is
greater than C: Bb, and confequently B greater than
D. In like manner, if A be equal to C, it may be
proved that B is equal to D; and, if A be leſs than
C, that B is lefs than D.
b
1.c.3.4
ax. 8.
2. c.4.4.
PRO.
104
BOOK IV.
ELEMENTS
Fig. 91.
c. I. 4.
ax. 6.
Þ 1. 4.
• 6. 4°
PROPOSITION VII.
If four magnitudes of the fame kind be proportional, they
are alſo proportional by alternation.
Let A: B::C: D, then A: C : : B : D.
For, let Cn, Dn be like parts of C, D. Then
A: B (:: C: D):: Cn: Dn, and A: B:: any
multiple of Cn the fame multiple of Dn b.
any
There-
fore, according as A is greater, equal, or leſs, than
multiple of Cn; B is greater, equal, or leſs than
the fame multiple of Dn. Confequently A con-
tains Cn juft as often as B contains D- and A: C
:: B : D.
Fig. 92.
PROPOSITION VIII.
If four magnitudes be proportional, they are also propor-
tional by compoſition.
·
Let A B C D, then A + B: B:: C+D: D.
For, let Bn, Dn be like parts of B, D; then, by
hypothefis, A contains Bn juſt as often as C contains
Dn. If, therefore, a certain number of times Bʼn be
added to A, and the fame number of times Dn to C,
the former fum will contain Bn as often as the latter
con
}
F
D
PL. VII. p.104.
OMETRY
Ꭰ
H
A
B
G
E
B
E
B
A
C
F
A
C
K
FIG.74.
E
FIG.77.
FIG.75.
C
D
F
Ꮐ
D
C
A
B
B
K
*
F
n
FIG.76.
L
D
M
G
I
F
In
H
K
E
m
E
FIG.78.
୯
9
D
A
G
E
FIG.79.
G
n
D
Hi
A
H
B
E
B
E
m
A
In
In
n
FrG.80.
D
A
B
A
BI
C DI
FIG.81.
FIG.82.
E
In
771
In
In
B
C
D
A
B
FIG. 83.
F
F
M
M
M
H
E
G
M
F
E
C
FIG.84.
B
C+ D
FIG.85.
n
B
C
FIG. 86.
D
E
G
n
B
172
FIG.87
D
E
المال المال النار عليل المبا
A B
FIG.92.
m
A
B
C D
A B
C D
A B
FIG. 88.
FIG. 89.
C D
FIG.90.
106
BOOK IV.
ELEMENTS
..
Fig. 94.
c 2. 4.
£ 4. 4.
&
fame multiple of Fn. Wherefore A contains Cn as
often as D contains Fn, and A: C:: D: F.
Next, let there be three magnitudes A, B, C, and
other three, D, E, F, which, taken two and two, are
proportional in a crofs order, viz. A: B:: E: F, and
B: C:: D: E, then A: C:: D: F.
For, let Bn, Cn, Fn, be any like parts of B, C, F,
and BO, CP, FQ, any equimultiples of Bn, Cn, Fn,
reſpectively. Then, becauſe A:B::E: F, A : Bn
E: Fn, and A: BO:: E: FQ. Again, be-
caufe B: C: D: E, C: B:: E: D,
cor. I. 4. E: D', and CP: BO:: E:D".
h I. 4.
Cn: Bn::
But, fince A is
1. c. 3.4. greater than CP, A: BO is greater than CP BO;
*ax. 8. 4. therefore alfo E: FQ greater than E: D, and con-
fequently D greater than FQ, as before.
Now, let there be any number of
magnitudes,
and a like number
A, B, C, D, E
F, G, H, K, L
which, taken two and two, are continually propor-
tional in a direct order, A: BF: G, B: C:: G: H,
CD :: HK, &c. ; then, by equality, A: E :: F: L.
For,
taking them three and three, A, B, C, and
F, G, H, ACFH; A, C, D, and F, H, K,
AD: FK, &c.
Alfo, let there be any number of
magnitudes
and a like number
A, B, C, D, E
F, G, H, K, L
which, taken two and two, are continually propor-
!
tional
Book IV. OF GEOMETRY.
107
1
tional in a croſs order, A : B :: K: L, B: C:: H:K,
C D G H, &c. then, by equality, AE :: F: L.
For,
taking them three and three, A, B, C, and
H, K, L; A: C:: H: L; A, C, D, and G, H, L,
A: DG: L, &c.
COR. I. If the confequent terms in one analogy be
antécedents in another, the remaining terms of the
former will be the antecedents, and the remaining
terms of the latter the confequents of a third ana-
logy.
COR. 2. If either the extreme or mean terms of
one analogy be the extremes of another, the remain-
ing terms of the former will be the extremes, and the
remaining terms of the latter the means of a third a-
nalogy.
COR. 3. Two ratios are equal, if each be com-
pounded of the fame number of ratios that are equal,
each to each, either in a direct or croſs order.
PROPOSITION
X.
If four magnitudes be proportional, they are alſo propor-
tional by divifion.
:
Let A B C : D, then A-BBC-D D. Fig. 92.
1. When A is greater than B. Let Вn, Dn, be
like parts of B, D.
Then, fince A contains Вn as
Bʼn
often
108
BOOK IV.
ELEMENT S
* def. 4.4. often as C contains Dn, if any number of times B
be taken from A, and as many times Dn from C, the
former remainder will contain Вn as often as the lat-
ter contains Dn. Thus, A-B contains Bn as often
as CD contains Dn: Therefore A-BB::
C-D: D.
D 4. 4.
©
9. 4.
2. When A is lefs than B. By inverfon B: A::
D: C, and, from what has been demonſtrated,
B—A : A :: D—C ; C
But
ABC: D
Therefore, by equality B-A: B:: DC Dc.
a
4. 4.
b S. 4•
Ĉ
IQ. 4.
PROPOSITION XI.
If four magnitudes be proportional, they are alſo propor-
tional by converfion.
:
:
Let A B C D, then A: A+B C C + D,
and A: A-B:: C: C-D.
For, by inverfion, B: A:: D: C.
;
By compofition, A+ B: A: CD: C, or, by
diviſion, A — B: A::C-D: C. And by inver-
fion A: A + B : : C : C + Dª, and A : A−B:: C
:C-D2.
PROPO-
OOK IV.
OF GEOMETRY. 109
PROPOSITION
XII.
If four magnitudes be proportional, they are alfo propor-
tional by mixing.
Let A B :: C : D, then A + B : A−B :: C÷D
:C-D.
For, by compofition ", A + B : B:: C+D: D.
Alfo, by divifion and inverfion, B: A-B :: D
: C-D.
b
a 8. 4.
b 10. 4.
C 4. 4.
Therefore by equality, A + B : A-B::C+D d 9. 4.
: C-D.
!
PROPOSITIO N. XIII.
If there be any number of equal ratios, whofe terms are
all of the fame kind, as one antecedent is to its confe-
quent, fo is the fum of all the antecedents to the fum of
all the confequents.
Let A B C D E: F, then A: B:: A+ C Fig. 95.
+E:B+D+ F.
For, let Bn, Dn, Fn, be like parts of B, D, F, and
the ſum of Вn, Dn, Fn, will be the fame part of the
fum of B, D, F, that Bn is of B. But, fince C con-
tains Dn, and E, Fn, juft as often as A contains Bn;
it
a
Lem. 1.
110
BOOK IV
ELEMENTS
b
it is evident that the fum of A, C, E, will contain the
cor.lem. fum of Bn, Dn, Fn, the very fame number of times .
Therefore A: B:: A+ C+E: B+D+ F.
I.
* 7. 4.
b II. 4.
PROPOSITION XIV.
If there be two equal ratios, whofe terms are of the fame
kind, as one antecedent is to its confequent, fo is the
difference of the antecedents to the difference of the
confequents.
Let A: B:: C: D, then A:B:: A-C: B-D.
For, by alternation 2, A: C:: B: D, by conver-
b
fion A: A-C:: B: B-D, and by alternation
A: B::A-C:B-D.
a
COR. From this, and the preceding propoſition, it
follows, that a ratio is not changed by either aug-
menting or diminiſhing each of its terms by the ho-
mologous terms of an equal ratio.
PROPOSITION XV.
If the confequent terms in two analogies be the fame, the
fum of the first antecedents will be to their confequent
as the fum of the laſt antecedents to their confequent.
Let
1
!
bok IV. OF GEOMETRY.
III
Let A B C D, and E: B:: F: D, then A+E:
:
B::C+F: D.
For, fince
A:B::CD
And (by inverfion ª)
B:E::D:F
By equality b
A:E::C: F.
By compofition ©
A+E:E::C+F:F
E {
4. 4.
b.9. 4.
€ 8. 4,
But
E:B::F:D
b
Therefore, by equality A+E : B :: C+ F: D.
COR. I. The difference of the firſt antecedents is
to their confequent as the difference of the laft ante-
cedents to their confequent.
COR. 2. If the confequent terms, in any number
of analogies, be the fame, the fum of the firſt antece-
dents is to their confequent as the fum of the laſt an-
tecedents to their confequent.
COR. 3. If the antecedent terms be the fame, the
firft antecedent is to the fum or difference of its con-
fequents as the laft antecedent to the fum or differ-
ence of its conſequents.
PROPOSITION XVI.
If four magnitudes of the fame kind be proportional, the
Sum of the greatest and leaft is greater than the fum of
the other two.
Let
I 12
BOOK I
ELEMENTS
Fig. 96.
b
Let AE: BG::C: D, and AE the greateſt term;
3. 4. & then D is the leaſt 2, and A+D greater than B + C.
For, let AFC and BH = D, then AE: BG::
6.4.
cor. 14. EF: GH. Hence EF is greater than GH'. There-
4.
fore AE is greater than GH added to C, and the ſum
of AE and D greater than the fum of BG and C.
COR. If three magnitudes be proportional, the
middle term is leſs than half the ſum of the extremes,
PROPOSITION XVII.
If there be any number of ratios, and as many others, e-
qual to them, each to each, the ratio that is compound-
ed of ratios that are the fame with the former, each
with each, taken in direct order, is equal to the ratio
that is compounded of ratios, which are the fame with
the latter, each with each, taken in any order whate-
ver.
Let there be any number
of ratios
And as many others
A: B, C: D, E: F, G: H
K:L, M:N, O:P,Q : R
equal to them, each to each, in direct order.
M, N, p, q, r
a, b, c, d, e
Let the ratio of m to r be com-
pounded of the ratios m:n, n: p,p: q
g:r equal to the first ratios, each to each, in direct
order. And let the ratio of a to c be compounded
of
BOOK IV. OF GEOMETRY.
113
of a:b, b:c, c:d, d: e, equal to the fecond ratios,
each to each, in any order whatever. Then fhall the
ratio of m to r be equal to the ratio of a to e.
=
: b,
1. Let there be two ratios, | A: B, C: D; m, n, p
and let ab K: L, b:c=K: L, M: N; a, b, c
M:N. Then, becauſe m : n (= A: B = K: L)
and np (C: D= M: N) = b: c, by equality a
m:p a:c.
Again, let a b=M:N, b:c=K: L. Then, be-
cauſe mn (A: B=K: L) = b:c; and n:p (=
C: D=M: N) a: b, by equalitym: pa: c.
Hence the ratio that is compounded of K: L, M:N*
is the fame with that which is compounded of M : N,
K: L.
=
2. Let there be three
ratios; then, if a: b =
A: B, C : D, E:F, m, n, p, q
K: L, M:N, O: P, a, b, c, d
K: L, b: c=M: N, c:d=O: P; or, if a : b= 0: P,
b:c=M: N, c: d=K: L; it follows, in like man-
ner, by equalityª, that m : q = a: d.
Hence the ratio that is compounded of K: L, M:N,
OP, is the fame with that which is compounded of
O: P, M: N, K : L.
Alſo, the ratio that is compounded of the firſt or
laft of thefe, and of the compound ratio of the other
a 9. 4.
P
two,
* That is, of ratios that are the fame with thefe, according
to the definition of compound ratio.
!
་
114
Book IV.
ELEMENTS
:
}
two, is the fame with that which is compounded of
all the three.
Then,
Again, let them be taken in any other order, as
ab= M: N, b:c O: P, c: d=K: L.
m:n=c:d, n:p = a: b, p : q = b : c; therefore the
ratio of m to q is the fame with that which is com-
pounded of cd, a:b, b: c; or of c: d, a: c; or of
a:c, c:d; that is, m: q = a: d.
3.
Let there be | A: B, C: D,E: F,G:H, m, n, p,q, r
four ratios; and K:L,M:N, O:P, Q: R, a, b, c,d,e
let the fecond ratios be taken in any order whatever,
as ab=M: N, b:c=Q: R, c: d=K: L, d: e=
O: P. Then, becauſe m: n = c:d, n: p = a:b, p : q
=d:e, the ratio of m to q is the fame with that which
is compounded of a: b, c : d, d: e; or of a : b, c : e.
But
9 : y bc. Therefore the ratio of m to r is
the fame with that which is compounded of a: b, c : e,
b:c; or of a:b, b: c,c:e, which is the ratio of a
to e.
The fame reaſoning will hold for any greater num
ber of ratios.
COR. Instead of any one of the ratios, of which a-
ny ratio is compounded, an equal ratio may be fubfti-
tuted; inſtead of two or more, the ratio compounded
of them may be ſubſtituted; alfo, the order of the
ratios may be changed in any manner whatever; and
the compound ratio will ſtill be the fame.
PROPO.
BOOK IV.
115
OF GEOMETRY.
1
PROPOSITION XVIII.
If, from two unequal magnitudes, two magnitudes, having
a given ratio, may be taken away, fo that either of the
remainders fhall be less than any magnitude propofed,
the two wholes fhall have to each other the fame ra-
tio as the magnitudes taken from them.
Let AB, CD be the two unequal magnitudes, of Fig. 97°
which AB is the greater, and M: N the given ratio.
It is ſuppoſed that, from AB, CD, there may be taken
two magnitudes in the ratio of M to N, fo that the
remainder of AB fhall be less
propoſed, and likewiſe that from
than any magnitude
AB, CD there may
be taken two magnitudes in the ratio of M to N, fo
that the remainder of CD fhall be leſs than any mag-
nitude propofed. It is to be proved that AB: CD::
M: N.
1. The ratio of AB to any magnitude CO, lefs than
CD, is greater than the ratio of M to N. For, fince there
may be taken two magnitudes from AB, CD, in the
ratio of M to N, fo that the remainder of CD fhall be
leſs than DO, let theſe be AE, CF; CF will therefore
be greater than CO. Let AR: CO :: AE : CF,
then AR will be lefs than AE, and confequently lefs 4.4.6. 4.
than AB. Hence the ratio AB: CO is greater than
ARC:0, or M: N.
2. The
b I.C. 3.4.
116
BOOK IV.
ELEMENTS
:
2. The ratio of AB to any magnitude CQ, greater
than CD, is lefs than the ratio of M to N.
For, fince there may be taken two magnitudes.
from AB, CD, in the ratio of M to N, fo that the re-
mainder of AB fhall be lefs than either DQ or DL,
(CL being equal to AB) let thefe be AG, CH; GB
will therefore be lefs than either HQ or HL, and con-
fequently AG greater than CH. Let AR: CQ::
cor. 14. AG: CH, then GR: HQ :: AG : CH º. Hence
GR is greater than HQ, and therefore greater than
GB, which is lefs than HQ. Confequently AR is
greater than AB, and the ratio AB: CQ lefs than
61.c.3.4. AR: CQ, or AG: CH or M: N.
c
4.
3. 4.
Confequently the ratio of AB to CD is equal to the
ratio of M to N.
Fig. 98.
OTHER WISE,
If magnitudes may be taken from the first and fecond, in
the ratio of the third to the fourth, ſo that either re-
mainder fhall be less than any magnitude propoſed, the
four magnitudes are proportional.
If AB be not to CD, as M to N, let AR: CD::
M: N. And,
1. Let AR be lefs than AB. Then, fince there
may be taken two magnitudes from AB, CD in the
ratio
Book IV. OF GEOMETRY.
117
ratio of M to N, fo that the remainder of AB fhall be
lefs than RB, let theſe be AE CF. AE, therefore, is
greater than AR. But AE: CFM: NAR: CD.
Wherefore CF is greater than CD, which is con-
trary to hypothefis. Confequently no magnitude lefs
than AB is to CD in the ratio of M to N. And, in
like manner, no magnitude lefs than CD is to AB in
the ratio of M to N.
2. Let AR be greater than AB, and let AB: CO
= AR: CD, then CO will be leſs than CD.
4. 4.
6. 4.
But, b 6.4
fince AB CO AR: CD = M: N, by inverfion
=
CO: AB :: N: M, which is contrary to what has
been demonſtrated. Confequently no magnitude
greater than AB is to CD in the ratio of M to N.
COR. If magnitudes may be taken from the firſt
and third, proportional to the fecond and fourth, fo
that either remainder fhall be lefs than any magni-
tude propofed, the four magnitudes are proportional,
2
ELEMENTS
f
:
?
E LE MEN T S
O F
GEOMETR Y.
#
1.
воок V.
DEFINITION S.
W O rectilineal figures are faid to be fimilar,
when all the angles of the one are equal to
the angles of the other, each to each, in order; and
the fides about the equal angles are proportional in
the fame order.
2. Two magnitudes, and other two, are faid to be
reciprocally proportional, when one of the former is
to one of the latter, as the remaining one of the latter
to the remaining one of the former.
3. A ftraight line is faid to be divided in extreme
and mean proportion, when the whole is to the great-
er fegment as the greater fegment to the lefs.
4. A
Book V. OF GEOMETRY.
119
4. A ftraight line is faid to be harmonically divi-
ded when it is divided into three fegments, fo that
the whole is to one extreme as the other extreme to
the middle fegment.
5. Two ftraight lines are faid to be fimilarly divi-
ded when any two correfponding fegments of them
have the fame ratio as the lines themſelves.
6. The fum of all the fides of any rectilineal figure
is called its perimeter.
AXIO
M.
The circumference of a circle is greater than the
perimeter of its infcribed polygon, but lefs than the
perimeter of its circumfcribing polygon.
PROPOSITION I
Triangles between the fame parallels are to one another as
their bafes.
Let ABC, DEF be two triangles between the fame Fig. 99.
parallels AD, BF, the triangle ABC: DEF::BC
: EF.
For, let EG be any part whatever of the baſe EF
lefs than BC, and GH, HK, KF, the remaining parts
of
>
120
Book V.
ELE MEN T S
به
al
2 2. 2.
b
def. 4. 4.
4.4.
of the fame. Alfo, let the lines BL, LM, MN, NO,
OP, be taken in the bafe BC, Jeach equal to EG, and
as many as the line BC contains. From A and D
let ſtraight lines be drawn to the ſeveral points of di-
viſion in the bafes BC, EF. Then, becauſe the baſes
BL, LM, &c. EG, GH, &c. are all equal, the tri-
DEG, DGH, &c. are alſo
angles ABL, ALM, &c.
equal. Hence DEG is the fame part of the triangle
DEF that EG is of the baſe EF, and the triangle
ABC contains DEG juſt as often as BC contains EF.
Therefore ABC: DEF :: BC: EF.
COR. 1. Triangles having the fame or equal alti-
tudes, are to one another as their bafes; and con-
verfely.
COR. 2. Parallelograms between the fame parallels,
or having equal altitudes, are to one another as their
baſes; and converſely.
COR. 3. Rectangles, and confequently triangles,
which have equal bafes, are to one another as their
altitudes; and converfely.
PROPOSITION II.
If a ſtraight line be parallel to one of the fides of a tri-
angie, it cuts the other fides proportionally; and con-
verfely.
Let
Book V. OF GEOMETRY.
121
Let the ſtraight line DE be parallel to BC, one of Fig. 100
the fides of the triangle ABC, and cut the other fides
AB, AC, or them produced in D, E, AD: DB::
AE: EC.
a
2. 2.
bax. 7. 4.
For BE, DC being drawn, the triangles, DEB,
EDC, upon the ſame baſe, and between the fame pa-
rallels, are equal; therefore the triangle ADE has
the fame ratio to each of them', ADE: DEB:: ADE
: EDC. But the former ratio is the fame with that of
the bafes AD, DB, becauſe the triangles have the 1. 5.
fame altitude, and the latter is the fame with that of
the bafes AE, EC, for the fame reafon.
AD: DB:: AE: ECd.
AD:DB:
C I.
Therefore
d ax. 6. 4t
Converſely, let AD: DB:: AE: EC, then DE is
parallel to BC.
I.C. 4. 4-
For, fince AD:DB:: ADE : DEB ©, and AE:
EC:: ADE: EDC, ADE: DEB :: ADE : EDC 4;
therefore the triangles DEB, EDC are equal˚, and
confequently between the fame parallels f, that is, DE 3. c. 2. 2.
is parallel to BC.
f
COR. 1. Straight lines, which meet three parallel Fig. 101.
I.
ftraight lines, are cut proportionally.
COR. 2. If two ſtraight lines be cut by any number
of parallel lines, the fegments intercepted by two
of the parallels, have the fame ratio as the feg-
ments intercepted by any other two.
COR. 3. If any number of ſtraight lines be parallel
to the baſe of a triangle, the fegments intercepted by
Q
any
122
Book V.
ELEMENT S
any two of the parallels will have the fame ratio as
the two fides of the triangle; and thefe fides will be
fimilarly divided.
COR. 4. Hence the method of dividing a given
ſtraight line fimilarly to a given divided line-of divi-
ding a given ſtraight line in the ratio of two given
lines of dividing a given ftraight line into any pro-
pofed number of equal parts-of finding a fourth pro-
portional to three given ſtraight lines-and of finding
a third proportional to two given ftraight lines.
PROPOSITION
III.
If a straight line bifect the vertical angle of a triangle, or
the angle adjacent to it, and meet the bafe, the feg-
ments of the baſe will be directly proportional to the o-
ther two fides of the triangle; and converſely.
Fig.
102. Let the vertical angle ABC, or the angle aBC ad-
jacent to it, be biſected by the line DB meeting the
baſe AC in D; AD: DC::AB: BC.
2. 5.
b 16. I.
For, through C let CE be drawn parallel to BD, to
meet AB in E. Becauſe BD is parallel to a ſide of
the triangle ABC, it cuts the other fides proportion-
ally. Thus, AD: DC :: AB : BE ª. But BE is
equal to BC, becauſe the angle BEC ABD
= BCE, Therefore AD: DC :: AB: BC.
=
DBC
Con-
BOOK V. OF GEOMETRY.
123
Converſely, If AD: DC: AB: BC, the angle
ABC, or its adjacent angle, will be bifected by the
line BD.
For the fame conftruction remaining, AD: DC::
AB: BE; therefore AB: BC::AB: BE. Whence
BE BC, and the angle BEC BCE. Confe-
quently the angle ABD = BEC BCE = DBC.
b
=
COR. If the two vertical angles of any triangle be
bifected by ſtraight lines meeting the bafe, the greateſt
ſegment of the bafe is harmonically divided by the in-
termediate lines.
2. 5.
c
ax. 6. 4.
dI. c. 4. 4.
e cor. 4.1.
b 16. I.
If
PROPOSITION IV.
two triangles have two angles of the one equal to two
angles of the other, each to each, they are fimilar, and
have the homologous fides oppofite to equal angles.
Let ABC, DEF be two triangles, having the angle Fig. 103.
A =D, B = E, and confequently CF. The fides
about any two equal angles fhall be proportional, ſo
that the homologous terms fhall fubtend equal angles,
BA: AC:: ED: DF.
a 5. I.
For, if the fides AB, DE be equal, the triangles
ABC, DEF are equal in every refpect, and there-
fore fimilar. If not, let AG = DE, and GH paral- bax. 5. 4.
lel to BC. The triangles AGH, DEF, having the
angle
def. 1. 5.
124
BOOK V.
ELEMENTS
c'16. I.
5. IA
a 5.
angle A = D, the angle G ( = B°) = E, and the fide
AG = DE are equal in every refpect. Thus, AH
DF. But, fince GH is parallel to BC, BA : AC
3. c. 2. 5. AG AH, therefore BA: AC:: ED: DF.
BA:AC::ED:DF. In
like manner, AC: CB:: DF: FE, and CB: BA::
FE: ED.
COR 1. A ftraight line, parallel to one of the fides
of a triangle, cuts off another triangle fimilar to it.
COR. 2. The fegments of two parallel ftraight lines,
intercepted by the two fides of a triangle, have to
each other the fame ratio, as the fegments of one of
thefe fides intercepted by the parallels from the ver-
tex.
COR. 3. The fegments of two parallel ftraight lines,
intercepted by the two fides of a triangle, are cut
proportionally by any ftraight line drawn to meet
them from the vertex.
Fig. 103.
PROPOSITION V.
If two triangles have one angle of the one equal to one
angle of the other, and the fides about the equal angles
proportional, the triangles fhall be fimilar.
Let ABC, DEF be two triangles, having the angle
A = D, and BA: AC:: ED: DF, ABC fhall be fi-
milar to DEF.
For,
J
Book V. OF GEOMETRY.
125
= =
For, if the fide BAED, then AC DF, and the
triangles are equal in every refpect".
=DE, and GH parallel to BC.
a 6. 4°
I.
·4.
If not, let AG b
Then AG
(:: BA : AC) ED: DF, but AG = ED,
AH
ED, therefore
AH = DFª. Hence the triangles AGH, DEF are
equal in every reſpect", and confequently the triangles
ABC, DEF fimilar ©.
COR. Two ifofceles triangles, which have equal
vertical angles, are fimilar.
CI. C. 4. 5.
PROPOSITION VI.
If two triangles have the three fides of the one directly
proportional to the three fides of the other, the triangles
fhall be fimilar.
Let the two triangles ABC, DEF have the fides Fig. 103.
BA, AC, CB, of the one, directly proportional to the
fides ED, DF, FE, of the other, BA: AC::ED: DF,
and AC: CB:: DF: FE; ABC, DEF fhall be fi-
milar.
For, if the fide BAED, then ACDF, CB =
{
a 6. 4.
TE 2, and the triangles equal in every reſpect'. If º 6. 1.
not, let AG = DE, and GH parallel to BC. Then
AG:AH (:: BA : AC) :: ED: DF, but AG = ED,
therefore AH = DF. Alfo AH:HG(::AC:CB)
:: DF: FE, but AH = DF, therefore HG = FE².
Hence
126
Book V.
ELEMENT S
Hence the triangles AGH, DEF are equal in every
6. 1. reſpect, and confequently the triangles ABC, DEF
1. c.4.5. fimilar .
C
:
Fig. 103.
• 6. 4.
↳ 23. I.
}
PROPOSITION
VII.
If two triangles have two fides of the one proportional to
trvo fides of the other, the angles fubtended by two ho-
mologous fides equal, and the angles oppofite to the other
two homologous fides, both obtuſe, both acute, or one of
them right, theſe triangles ſhall be ſimilar.
Let the two triangles ABC, DEF have the fides
BA, AC proportional to ED, DF, viz. BA: AC : :
ED: DF, the angles B, E oppofite to the homolo-
gous fides AC, DF equal, and the angles C, F oppo-
fite to the other homologous fides AB, BE both ob-
tufe, both acute, or one of them right; the triangles
ABC, DEF fhall be fimilar.
For, if BA ED, then AC = DF2, and the
triangles equal in every reſpect. If not, let AG =
DE, and GH parallel to BC.
BA: AC)
Then AG AH (::
=
ED: DF, but AG ED, therefore
AH DF. Thus the triangles AGH, DEF have
two fides of the one GA, AH, equal to two fides of
the other, ED, DF, each to each, the angle G (= B
E, and the angle H (=C) and F both obtufe, both
acute,
Book V. OF GEOMETRY.
127
acute, or one of them right; therefore theſe triangles
are equal in every refpect, and confequently the 23. 1.
triangles ABC, DEF fimilar º.
CI.C. 4. 5?
PROPOSITION
VIII.
If two parallelograms, which have one angle of the one
equal to one angle of the other, be equal to one another,
the fides about the equal angles are reciprocally proper-
tional; and converfely.
Let the two parallelograms AC, GE, having the Fig. 104.
angle ABC = EBG, be equal to one another, the
fides AB, BC, and the fides EB, BG, are reciprocally
proportional, AB: BG:: EB: BC.
a
For, let AB, BG, be in the ſame ſtraight line, and
EB, BC will alſo be in one ftraight line". Then the cor.12.1.
parallelogram CG being completed, becaufe the pa-
rallelograms AC, GE are equal, AC: CG::GE:
CG. But AC: CG:: AB: BG, and GE: CG
:: EB: BC. Therefore AB: BG:: EB: BC4.
Converſely, If the parallelograms AC, GE, having
the angle ABC=EBG, have the fides about theſe
angles reciprocally proportional, they are equal to one
another.
For, by hypothefis, AB: BG:: EB: BC. But
AB: BG:: AC: CG, and EB: BC:: GE: CG.
There-
b
ax. 7. 4.
C 2. c. I.5.
ax. 6. 4.
d
6.4°
128
Book V.
ELEMENTS
:
dax. 6. 4. Therefore AC: CG:: GE: CG 4, and confequently
1. c. 4. 4. AC GE".
=
COR. 1. If two triangles, which have one angle of
the one equal to one angle of the other, be equal to
one another, the fides about the equal angles are re-
ciprocally proportional; and converſely.
COR. 2. Equal rectangles have their fides recipro-
cally proportional; and converfely.
COR. 3. If four ftraight lines be proportional, the
rectangle under the extremes is equal to the rectangle
under the means; and converfely.
COR. 4. If a ſtraight line be divided into three
fegments, fo that the rectangle under the whole line,
and middle fegment, be equal to the rectangle under
the extreme fegments, it is divided harmonically; and
converfely.
}
COR. 5. If three ftraight lines be proportional, the
rectangle under the extremes is equal to the fquare of
the mean; and converſely.
COR. 6. If a ftraight line be fo divided, that the
rectangle under the whole, and one of the fegments
be equal to the fquare of the other, it is divided in
extreme and mean proportion; and converfely.
COR. 7. Equal triangles, or equal parallelograms,
have their bafes and altitudes reciprocally propor-
tional; and converſely.
PROPO
BOOK V. OF GEOMETRY.
129
PROPOSITION IX.
Two parallelograms, which have one angle of the one e-
qual to one angle of the other, are to each other in a
ratio compounded of the ratios of the fides of the one to
the fides of the other, cach to each, about the equal
angles.
Let AC, GE be two parallelograms, having the Fig. 104.
angle ABC = EBG, AC is to GE, in a ratio com-
pounded of the ratios of AB to BG, and CB to BE.
a def. II.
ს
4.
For the fame conftruction remaining, as in the laft
propofition, AC: GE is the ratio that is faid to be
compounded of the ratios AC: CG, and CG: GE.
But AC: CG AB: BG", and CG: GE CB: BE. 2. c. 1.5.
Therefore the ratio AC: GE is the fame with that
which is compounded of the ratios AB: BG, and
CB: BE.
COR. I. Two triangles, which have one angle e-
qual to one angle, are to each other in a ratio com-
pounded of the ratios of the fides of the one to the
fides of the other, each to each, about the equal
angles.
COR. 2. Any two rectangles are to each other in
the compound ratio of their fides-and any ratio
compounded of two ratios, (whofe terms are ftraight
R
lines,)
€ 17. 4.
130
BOOK V.
ELEMENTS
:
!
lines,) is the fame with the ratio of the rectangles un-
der their homologous terms.
COR. 3. Two triangles, which have one angle equal
to one angle, are to each other as the rectangles under
the fides about the equal angles.
COR. 4. The rectangles under the correfponding
terms of two analogies are proportional.
COR. 5.
If four ſtraight lines be proportional, their
fquares are alfo proportional; and converfely.
i
Fig. 105.
PROPOSITION
X.
Two fimilar triangles are to each other in the duplicate
ratio of their homologous fides.
Let BAC, EDF be two fimilar triangles, having
the angle ABC = DEF, and AB : BC
DE: EF;
the triangle BAC is to the triangle EDF, in the du-
plicate ratio of BC to EF.
For, let GH be a third proportional to BC, EF, ſo
24.c.2.5. that BC: EF = EF : GH. Then, becauſe the
angles at B, E are equal, the ratio BAC: EDF is the
fame with that which is compounded of BC: EF, and
b1.c.9 5. AB DE b. But, fince AB BC: DE EF, the
C 7. 4.
:
C
:
ratio AB : DE BC: EF = EF: GH. Therefore
=
the ratio BAC : EDF is the fame with that which is
d cor. 17. compounded of BC: EF, and EF: GH, that is, the
4.
ratio
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1
Book V. OF GEOMETRY.
131
ratio BAC: EDF is the fame with the ratio BC: GH,
which is the duplicate ratio of BC to EF º.
COR. 1. Two fquares, or, in general, two fimilar
parallelograms, are to each other in the duplicate ra-
tio of their homologous fides.
COR. 2. Two fimilar triangles, or two fimilar pa-
rallelograms, are to each other as the fquares of their
homologous fides.
COR. 3. If three ſtraight lines be proportional, the
firft is to the third as the fquare of the firft to the
ſquare of the ſecond, (or as the fquare of the ſecond
to the ſquare of the third); and converſely.
e def. 10.
4.
!
PROPOSITION XI.
Upon a given straight line to defcribe a rectilineal figure
Similar to a given one, and fimilarly fituated.
Let ABCDE be the given rectilineal figure defcri- Fig. 106.
bed upon AB, and FG the given ftraight line upon
which it is required to defcribe a figure fimilar to the
given one, and ſimilarly fituated. Let the angle FGH
be made equal to ABC, and let GH be a fourth
proportional to AB, BC, FG . In like manner, let u
the angle GHK = BCD, and HK a fourth propor-
tional to BC, CD,
CDE,
GH; the angle HKL =
and KL a fourth proportional to CD, DE, HK; and
let
13. I.
4. c.2.5.
132
Book V,
ELEMENTS
ព
સ
5. 5.
9. 4.
let L be joined; the figure FGHKL will be fimilar
to ABCDE.
=
For, fince the angles B, G are equal, and AB : BC
:: FG: GH, the triangles ABC, FGH are fimilar ©.
Hence the angle ACB FHG, but BCD = GHK,
therefore ACD FHK. Hence alfo AC : BC :: FH
: GH, but BC: CD :: GH: HK; therefore, by e-
quality, AC CD :: FH: HK. Thus, the triangles
:
ACD, FHK, having the angles at C and H equal, and
AC: CD :: FH: HK, are fimilar. In the fame
manner, ADE, FKL are proved fimilar. Hence the
angles at E and L are equal, and DE: EA :: KL : LF.
But the whole angles at A and F are equal, becauſe
they are compoſed of angles that are equal, each to
each, and, fince the lines AE, AD, AC, AB, from
the fimilarity of the triangles, are directly proportion-
al to FL, FK, FH, FG, by equality AE AB :: FL
: FG. Therefore the rectilineal figures ABCDE and
* def. 1.5. FGHKL are fimilar
e
COR. 1. Similar polygons may be divided into the
fame number of triangles fimilar each to each, and
having the fame ratio each to each that the polygons
have.
i
COR. 2. Thoſe polygons, which confift of the fame
number of triangles, fimilar each to each, and fimi-
larly fituated, are fimilar.
COR. 3. Similar rectilineal figures are to one ano-
ther in the duplicate ratio of their homologous fides.
COR.
Book V.
133
OF GEOMETRY.
COR. 4. Similar rectilineal figures are to one ano-
ther as the fquares of their homologous fides.
COR. 5. If three ſtraight lines be proportional, the
firſt is to the third as a rectilineal figure deſcribed up-
on the first to a fimilar rectilineal figure defcribed
upon the fecond.
COR. 6. If two rectilineal figures of the fame num-
ber of fides have as many angles wanting two, of the
one, equal to the correſponding angles of the other,
each to each, and the fides about the equal angles
proportional in the fame order, the two figures are fi-
milar; and, if the fides be equal, each to each, the
figures are both equal and fimilar.
PROPOSITION XII.
:
The perpendicular from the right angle, upon the hypothe-
nufe of a right angled triangle, divides the triangle in-
to two others, fimilar to the whole and to one another.
Let ABC be the right angled triangle, and BD the Fig. 107.
perpendicular upon the hypothenuſe, the triangles
ABD, DBC are fimilar to the whole, and to one ano-
ther.
For, fince the triangles ABD, ABC have the angle
at A common to both, and the angles at D, B equal,
as
!
134
BOOK V.
ELEMENTS
-
* 4. 5.
ax. 2. 1.
ax, 6.
4.
as being right angles, they are fimilar 2. In like
manner, the triangles DBC, ABC are fimilar. Con-
fequently the triangles ABD, DBC are fimilar to one
another ".
COR. 1. The vertical angles are alternately equal
to the angles at the bafe.
COR. 2. Each fide is a mean proportional between
the hypothenufe and the adjacent fegment; and the
fquare of the former is equal to the rectangle of the
latter.
}
COR. 3. The perpendicular is a mean proportional
between the fegments of the hypothenufe.
COR. 4. A perpendicular to the diameter of a cir-
cle, from any point in the circumference, is a mean
proportional between the fegments of the diameter.
COR. 5. Hence the method of finding a mean
proportional between two given ftraight lines is ob-
vious.
Fig. 108.
PROPOSITION XIII.
The rectangle under the two fides of a triangle is equal to
the rectangle under the perpendicular, and the diameter
of the circumfcribing circle.
Let ABC be a triangle, of which BD is the per-
pendicular on the baſe AC, and let BE be a diameter
of
Book V. OF GEOMETRY.
135
L
of the circumfcribing circle, the rectangle under AB,
BC is equal to the rectangle under BD, BE.
2
b
For AE being joined, the triangles ABE, CBD
have the angles at E, C equal, becauſe they ſtand 1. c. 7.3
upon the fame arch, and the angles at A, D equal,
becauſe they are right angles". Wherefore theſe tri- 9. 3.
angles are fimilar °, and AB: BE::DB: BC. Con-
fequently the rectangle under AB, BC is equal to the
rectangle under BD, BE.
4. 5.
d 3.c.8.5.
PROPOSITION XIV.
a ftraight line bifect the vertical angle of a triangle,
and meet the bafe, the fquare of that line, together with
the rectangle under the fegments of the bafe, is equal to
the rectangle under the two fides of the triangle.
Let the vertical angle ABC, of the triangle BAC, Fig. 109.
be bifected by the line BD meeting the baſe in D,
the fquare of BD, together with the rectangle ADC,
fhall be equal to the rectangle under AB, BC.
For, let a circle be defcribed about the triangle,
and BD being produced, to meet the circumference
in E, let A, E be joined. The triangles ABE, CBD,
having the angles at B equal by hypothefis, and the
angles at E, C equal, as ftanding upon the fame arch,
are fimilar. Therefore EB: BA:: CB: BD, and
the
2
4. 5.
138
Book V.
ELEMENT S
i
1
}
C
3. 5.
d 12. 5.
€
5. c.9.5.
equal to BCE, or to the fum of CAE, CEA, and ta-
king away the equal angles BED, CAE, there re-
mains the angle DEC equal to CEA. Thus, the angle
AED is bifected by EC; therefore AC CD: AE
: ED. But AE: ED:: AB: BE, or BC. There-
fore alfo, AB : BC :: AC: CD, and AB²: BC²: :
AC: CD **, Confequently AB: BD:: AC²:
2
f3.c.10.5. CD * f.
ze
2
Converſely, If AB: BD :: AC: CD, then
AB: BC BC BD. For, the fame construction re-
maining, AB BD :: AB*: BE*f; therefore AC ::
CD:: AB': BE, and AC: CD:: AB: BE::
AE: ED. Hence the angle AED is bifested by
EC. But the angle BED is equal to CAE. There-
fore the angle BEC is equal to the fum of the angles
CEA, CAE, that is, to the angle BCE. Confequent-
2.c.12.5. ly BC is equal to BE, and AB : BC :: BC : BD².
&5.c.8.5.
COR. If three ftraight lines be proportional, the
first is to the third as the fquare of the fum of the
firſt and fecond to the fquare of the fum of the ſecond
and third.
1
PROPOSITION XVII.
To find a point in a given straight line, or in that line
produced, ſo that the rectangle under the ſegments may
be equal to the rectangle under two given lines.
Let
Book V.
139
OF GEOMETR Y.
t
Let AB be the given ſtraight line, in which it is re- Fig. 112.
quired to find a point, fuch, that the re tangle under
the ſegments ſhall be equal to a given retangle. Let
AC, BD, be perpendicular to AB, and equal to the
fides of the given retangle, each to each, Let a ſe-
micircle be deſcribed upon AB, and let the ſtraight
line that joins the points C, D, meet its circumference
in E. Then the perpendicular to CD from E will
meet AB in the point F required.
For, let AE, EB be joined. Then, becauſe DEF,
DBF are right angles, the angle EFA is equal to the
angle EDB, and becauſe AEB, FED are right
angles, the angle AEF is equal to the angle DEB.
Therefore the triangles EAF, EBD are fimilar. In ь
like manner, the triangles EBF, EAC are fimilar.
с
10. 3.
2. c. ib.
I. c. ib.
I.C. 7.3.
9. 3.
4. 5.
From the former
From the latter
AF: AE :: BD : BE.
AE AC: BE: FB.
d
Therefore, by equality AF: AC :: BD: FB.
Confequently the reflangle AFB is equal to the
d 9. 4.
rectangle under AC, BD ‘.
€ 3. c. 8. 5.
PROPOSITION
XVIII.
To find a point in a given straight line, or in that line
produced, ſo that the ſquare of one of the fegments may
be equal to the rectangle under the other fegment and
a given line.
Let
140
BOOK V.
ELEMENT S
Fig. 113.
Б
9. 2.
Let AB be the given ſtraight line, in which it is re-
quired to find a point, fuch, that the fquare of one of
the ſegments ſhall be equal to the re Rangle under the
other fegment, and the given line M. Let AB be
produced to C, fo that BC == M, and upon AC let
the femicircle ADC be deſcribed. Let BD be per-
pendicular to AC, BC bifested in E, and E, D joined.
Then a circle deſcribed about the center E, with the
radius ED, will interfect AB in the point F required.
For, fince 'BC is bifested in E, the rectangle CFB,
together with the fquare of BE, is equal to the fquare
a cor. 8.2 of EF, or to the fum of the fquares of BD, BE "; and
taking away the common fquare of BE, there remains
the rectangle CFB equal to the fquare of BD. But
c.3.17.3. the fquare of BD is equal to the reangle ABC.
Therefore the rectangle CFB is equal to the retangle
ABC. Hence, by adding or taking away the com-
mon rectangle CBF, we fhall have the fquare of FB
equal to the rectangle under AF and BC.
e
PROPOSITION
XIX.
In the fame circle, or in equal circles, angles at the cen-
ter are to one another as the arches upon which they
Stand.
鎏
​Let
A
',
B
A
B
1
C
FIG. 105.
A
E
A
FIG. 109.
E
D
B
C
F
D
FIG. 106.
D
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Ꮐ
F
a
FIG. 1.
GEOMETRY
H
B
1
FIG. 107
-H
F
G
E
K
C
D
A
FIG.110.
E
A
В
D
C
A
A
E
A
F
PL. IX. pa. 140.
B
D
E
B
C
D
FIG. 108.
FIG. 12.
I
FIG.112.
D
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F
B
Book V. OF GEOMETRY.
;
141
Let ABC, DEF, be any two angles, at the centers
of the equal circles AGC, DHF; the angle ABC :
DEF:: AC : DF.
For, let Dn be any part whatever of the arch DF,
leſs than AC, and let Am, mo, oq, qr, be the arches
which AC contains, each equal to Dn. The angles
ABm, mBo, oBq, qBr, which ftand upon theſe arches,
are each equal to DEn. Wherefore the angle ABC ² 3. c. 2. 3.
contains DEn juft as often as the arch AC contains
Dn. But DEn, it is evident, is the fame part of the
angle DEF, that Dn is of the arch DF. Confequently
the angle ABC: DEF:: AC: DF.
COR. 1. In the fame, or equal circles, angles at the
circumference, are to one another as the arches upon
which they ſtand.
COR. 2. An angle at the center of a circle has the
fame ratio to four right angles, that the arch upon
which it ftands has to the whole circumference.
COR. 3: Sectors of the fame,
to one another as their arches.
or equal circles, are
For this may
For this may be pro-
1
ved in the very fame manner.
PROPOSITION XX.
Circles are to one another in the duplicate ratio of their
diameters.
Let
142
Book V.
ELEMENTS
:
Let ABC, EFG be two circles, of which AC, EG
are diameters, and let AC EG :: EG: RQ, the
circle ABC EFG :: AC: RQ.
:
For, let the fquares ABCD, EFGH be infcribed in
the circles. Let the arches which the fides of thefe
fquares fubtend be biſected, and the regular polygons
KL, MN be formed by joining each point of ſection
to the adjacent angular points of the fquare. Let the
arches which the fides of theſe polygons fubtend be
bifected, and other polygons formed, by joining each
point of fection to the adjacent angular points of the
former; and ſo on.
Any one of theſe polygons, in the circle ABC, as
KL, is to the polygon MN, of the fame number of
fides, in the circle EFG, as AC to RQ. For, fince
each of the polygons KL, MN confifts of the fame
number of ifofceles triangles, and magnitudes have
the fame ratio as their like parts, KL: MN :: AOL
: EPN. But the ifofceles triangles AOL, EPN, ha-
ving equal vertical angles, (each of them being the
a 5.c.
5. c. 2. 3. fame part of four right angles, are fimilar ". There-
ь
cor. 5. 5.
C c 10. 5.
d 1. 4.
fore AOL : EPN (:: AO : RQ ©) :: AC : RQ.
Confequently KL MN:: AC: RQ 4.
:
Again, becauſe the difference between one of the
circles and its infcribed fquare ABCD is the fum of
the four fegments AB, BC, CD, DA, and each of the
triangles inſcribed in theſe fegments is greater than
half its fegment, (being equal to half the circumfcri-
bing
BOOK V. OF GEOMETRY.
143
bing parallelogram), by taking the polygon KL from
the circle ABC, there will be taken away more than
half that difference. In like manner, by taking the
next polygon from the fame circle, there will be ta-
ken away more than the half of what laſt remained;
and ſo on. Therefore, from the two circles ABC,
EFG, there may be taken away two polygons in the
ratio of AC to RQ, fo that either of them fhall leave
a remainder leſs than any magnitude propoſed.
Confequently the circles themſelves are in the fame
ratio f.
}
COR. 1. Similar fetors (that is, fectors of equal
angles,) are to one another in the duplicate ratio of
their radii.
COR. 2. Circles, and fimilar fetors of theſe circles,
are to one another as the fquares of their diameters or
radii.
• lem. 3.
4.
[ 18. 4.
PROPOSITION XXI.
Every circle is equal to a rectangle under the radius, and
a ftraight line, equal to half the circumference.
Let ABC be a circle, of which O is the centre, and Fig. 116.
let the ſtraight line BR be equal to half its circumfe-
rence; the circle ABC fhall be equal to the rectangle
OBR.
For.
144
Book V.
ELEMENT S
1
b
2. c. 13.
3. 6. 1.
1. c. 2.3.
CII. 3.
d
2. c. 19.
For, let the fquare ABCD, and the polygon KL be
infcribed in the circle, as in the preceding propofition.
Alfo, let the fquare EFGH, and polygon MQ, be
defcribed about the circle, fo that their fides touch
the circumference in the angular points of the former.
Then, becaufe OH bifects the angle DOC, it paffes
through the point K, and is perpendicular to MN,
a fide of the circumfcribing polygon. Hence MH is
greater than MK d
than MK or MD; therefore the triangle
MKH is greater than MKD, and confequently the
triangle MNH greater than half the triangular fpace
DHC, without the circle. Thus, the fides of the po-
lygon MQ cut off from the fquare EG more than
half its excefs above the circle. In like manner, the
next circumfcribing polygon cuts off more than half
the remainder; and ſo on. Therefore a polygon may
be deſcribed about the circle, fo that their difference
£ lem. 3.4. fhall be lefs than any magnitude propofed f.
e
I.
1. c. 2. 2.
Now, if the reЯangle OR be greater than the cir-
cle ABC, a polygon may be defcribed about the circle
that ſhall differ lefs from it than OR, that is, that
ſhall be leſs than OR. Let MQ, then, be a circum-
fcribing polygon lefs than OR. Becaufe MQ may
be divided into iſofceles triangles, each of which is e-
qual to the rectangle under the radius of the infcribed
2 c. 2. 23 circle, and half its bafe, the whole polygon is equal
to the rectangle under the radius of the infcribed
circle and half its perimeter ¹. But half its perimeter
b 6 2.
is
BOOK V.
145
OF GEOMETRY.
is greater than BR, half the circumference. There- i ax. 5.
fore alſo that retangle, or the polygon MQ, is
greater than the rectangle OR, which is contradic-
tory.
Again, if the rectangle OR be lefs than the circle
ABC, a polygon may be infcribed in the circle that
ſhall differ leſs from it than OR, that is, that ſhall
be greater than OR. Let KL then be an infcribed
polygon, greater than OR. Then, as before, KL is
equal to the rectangle under the radius of its inſcribed
circle, and half its perimeter. But the radius of its
inſcribed circle is leſs than OB, and half its perimeter
is leſs than BR . Therefore alſo that rectangle, or
the polygon KL, is lefs than the rectangle OR, which
is contradictory.
Conſequently the rectangle OR is equal to the
circle ABC.
COR. Hence any fector is equal to a rectangle un-
der the radius, and a ftraight line equal to half its
circumference.
PROPOSITION XXII:
The circumferences of circles are to one another as their
radii.
T
Let
146
Book V.
ELEMENT S
Fig. 117..
21. 5.
b 2. c.20.
5.
Let ABC, DEF be any two circles, of which OB,
HE are radii, the circumference ABC DEF:: OB
: HE.
For, let the ftraight lines BR, EK, touching the
circles in the points B, E, be each equal to half the
circumference of its circle, and let the rectangles OR,
HK, and fquares OG, HI be completed. Becauſe the
circles are equal to the rectangles OR, HK, and
have the fame ratio as the fquares OG, HI; OR:
HK:: OG: HI, and, by alternation, OR OG ::
HK HI. But rectangles of the fame altitude are to
one another as their bafes; therefore BR BG::
EK: EI, and alternately BR: EK :: BG : EI.
Confequently 2BR 2EK:: OB: HE.
:
:
:
:
COR. I. The arches of fimilar fectors are to one
another as their radii.
COR. 2. The arches of any two ſectors are to one
another, in the ratio compounded of the ratios of
their radii and angles.
Fig. 118.
PROPOSITION XXIII.
If fimilar rectilineal figures be defcribed upon the three
fides of a right angled triangle, the figure upon the
hypothenufe is equal to the other two taken together.
Let ABC be a right angled triangle, of which AC
is the hypothenufe, and let AE, FB, BG, be fimilar
rectilineal
Book V. OF GEOMETRY.
147
rectilineal figures defcribed upon the fides; the figure
AE upon the hypothenufe is equal to the fum of the
figures FB, BG, upon the other two fides.
For, let BD be perpendicular to AC; then, be-
cauſe AC: AB: AB: AD', AC: AD:: AE: FB.
In like manner, AC: CD :: AE : BG.
Therefore
AC: AD + CD:: AE: FB+BG c.
But AC = AD + CD, confequently AE÷FB+BG ª.
COR. If circles be defcribed upon the three fides of
a right angled triangle, as diameters, or with theſe
fides as radii, the circle of the hypothenufe is equal to
the ſum of the other two.
a
b
C
d
2. c. 12.
5. c. 11.
5.
5.
3. C. 15.
4.
3. 4.
APPENDI X.
THEOREM 1. If, from two angular points of any
triangle, ftraight lines be drawn, to bifect the oppo-
fite fides, they will divide each other into fegments,
having the ratio of two to one.
THEOR. 2. If two triangles have two angles equal,
and other two angles together equal to two right
angles, the fides about the remaining angles will be
proportional.
THEOR. 3. If, from any point in the circumference.
of a circle, ftraight lines be drawn through the extre-
mities of a chord, to meet the diameter that is per-
pendicular
148
BOOK V.
ELEMENTS
pendicular to it, the radius will be a mean propor-
tional between the fegments of that diameter intercep-
ted from the center.
THEOR. 4. If, from the extremities of the baſe of a
triangle, two ſtraight lines be drawn, each parallel to
one of the fides, and equal to the other; then ſtraight
lines, joining their other extremes with the other ex-
tremes of the bafe, will cut off equal fegments from
the fides, and each of theſe will be a mean propor-
tional between the other two fegments.
THEOR. 5. If a regular polygon be deſcribed about
a circle, and another, of the fame number of fides,
infcribed in it; then a third regular polygon, of
double the number of fides, infcribed in the fame.
circle, will be a mean proportional between the two
former.
THEOR. 6. If, from the extremities of the baſe of
a triangle, perpendiculars be let fall upon the oppo-
fite fides, and likewife ftraight lines drawn to bifect
the fame, the interfection of the perpendiculars, the
interfection of the bifecting lines, and the center of
the circumfcribing circle, will be in the fame ftraight
line.
THEOR. 7. If, from a point without a circle, two
tangents be drawn, and from the fame point, and one
of the points of contact, two ftraight lines be drawn,
to meet in the circumference, then fhall the chord,
which is drawn from the other point of contact, pa-
rallel
F
FIG. 113.
EOME
Y
D
A
F
B
E
C
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FIG. 114.
D
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D
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FIG. 115
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PL.X.a. 148.
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C
L
B
R
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E
D
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FIG. 117
FIG. 119.
A
E
D
A
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FIG. 116.
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FIG. 118.
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Book V. OF GEOMETRY.
149
rallel to one of the tangents, be divided by theſe
ftraight lines, ſo that the fegment intercepted by the
latter from the point of contact, will be a mean pro-
portional between the whole chord and the fegment
intercepted by the former.
THEOR. 8. Let there be two concentric circles, and
let a tangent to the lefs cut the circumference of the
greater; from the point of contact let two equal
chords be applied in the former, and from the point
of fection let others parallel to them be applied in the
latter; the fum of the chords in the lefs circle will
be equal to the fum or difference of thofe in the
greater, according as theſe laſt are on the fame, or on
different fides of the tangent.
THEOR. 9. Let there be a ſtraight line harmoni-
cally divided, and from any point without it let
ftraight lines be drawn through its extremities, and
the points of divifion; then any ftraight line drawn
from the one extremity, to terminate in the line that
paffes through the other, will alſo be harmonically
divided by the intermediate lines.
THEOR. 10. If, from the extremities of the baſe of
a triangle, ftraight lines be drawn through the fame
point in the perpendicular, to interſect the oppofite
fides, ftraight lines, joining the points of fection with
the foot of the perpendicular, will make equal angles
with the bafe.
THEOR.
150
Boo V.
ELEMENTS
1
THEOR. II. If, from the extremities of the diame-
ter of a circle, two perpendiculars be raiſed, each e-
qual to the fide of the infcribed fquare, and from their
extremities ſtraight lines be drawn through any point
in the circumference, to meet that diameter, thefe
ftraight lines will each intercept a fegment from the
fartheft extremity of the diameter, fuch that the fum
of their fquares fhall be equal to the fquare of the
diameter.
THEOR. 12. If, upon the baſe of a triangle, a right
angled triangle be conſtituted, having its hypothenufe
equal to the fum of the two fides of the triangle, and
through the vertex a ftraight line be drawn parallel
to the bafe; the fegment of the hypothenufe between
the parallels, half the ſum of the two fides, and half
the difference of the fegments of the bafe, will form a
right angled triangle.
i
PROBLEM I. The perpendicular, the difference of
the fegments of the bafe, and either the fum or dif-
ference of the two fides of a triangle, are given to
conſtruct it.
PROB. 2. The perpendicular, the difference of the
angles at the bafe, and either the fum or difference of
the two fides of a triangle, are given to conſtruct it.
PROB. 3.
In a quadrilateral figure, infcribed in a
circle, two adjacent fides, one of which is equal to the
diagonal drawn from their interfection, and the ratio
of
i
Book V. OF GEOMETRY.
151
of the fegments of the other diagonal are given to
conftruct the figure.
PROB. 4. To defcribe a quadrilateral figure, ha-
ving its fides of a given magnitude, and its oppofite
angles together equal to two right angles.
PROB. 5. To divide a given triangle into three o-
thers, proportional to three given lines, by ftraight
lines drawn from a point within the triangle.
PROB. 6. From a given triangle to cut off another,
having a given ratio to the whole, by a straight line
drawn through a given point.
PROB. 7. A circle, and a point within or without
it being given, to find another point, fuch, that any
ſtraight line being drawn through it, to interfect the
circumference, the diſtances of the points of ſection
from the given point fhall be to one another directly
as the fegments of the ftraight line intercepted by
the circumference from the point required.
PROB. 8. A circle, and a point without it, being
given, to find another point, fuch, that any ftraight
line being drawn through it, to interfect the circum-
ference, the diſtances of the points of fection from the
given point fhall be to one another in the duplicate
ratio of their diſtances from the point required.
ELEMENTS
•
ELEMENT S
O F
GEOMETRY.
H
H
1.
воок
VÍ.
DEFINITIONS.
A Straight line is faid to be perpendicular to a
plane, when it is perpendicular to all the
ftraight lines that can be drawn in that plane from the
point on which it infiſts.
2. The inclination of a ftraight line to a plane is
the acute angle which it forms with the line joining
its interfection with the plane and the extremity of
a perpendicular to the plane, drawn from any point
in the ſtraight line.
3. Two planes are perpendicular to one another,
when any ſtraight line drawn in either of the planes,
at right angles to their line of common fection, is per-
pendicular to the other.
4. The
OF GEOMETRY. 153
1
4. The inclination of two planes is the angle form-
ed by ſtraight lines drawn in theſe planes, at right
angles to their line of common fection, from the fame
point in it.
5. Two planes, or a ſtraight line and a plane, are
parallel, when being produced indefinitely they do
not meet.
6. A folid is that which has length, breadth, and
thickneſs.
7. A folid angle is that which is formed by more
than two plane angles at the fame point, but not in
the fame plane.
8. Two folids, bounded by planes, are fimilar,
when their folid angles are equal, and their plane fi-
gures fimilar, each to each.
9. A parallelopipedon is a folid bounded by fix
planes, of which the oppofite ones are parallel. If the
adjacent planes be perpendicular to one another, it is
a rectangular parallelopipedon.
10. A cube is a rectangular parallelopipedon, of
which the fix fides are ſquares.
11. A priſm is a folid, of which the fides are pa-
rallelograms, and the ends are plane rectilineal fi
gures.
12. A pyramid is a folid, of which the fides are
triangles, having a common vertex, and the bafe any
plane rectilineal figure. If the baſe be a triangle, it
U
is
154
ELEMENTS
}
is a triangular pyramid. If a fquare, it is a ſquare
pyramid.
13. A cylinder is a folid defcribed by the revolu-
tion of a rectangle about one of its fides remaining
fixed; which fide is named the axis; and either of
the circles defcribed by its adjacent fides the baſe of
the cylinder.
14. A cone is a folid defcribed by the revolution of
a right angled triangle about one of its fides remain-
ing fixed; which fide is named the axis, and the
circle defcribed by the other fide, the baſe of the
cone.
15. Cones or cylinders are fimilar, when they are
defcribed by fimilar figures.
PROPOSITION I
Fig. 119.
If two straight lines interfect each other, they are both in
one plane.
The two ftraight lines AB, CD, which interfect
each other in the point E, are both in one plane.
For, let any plane meet the ftraight line CD, in the
points E, D; and becaufe CD paffes through theſe
* def. 5. 1. points, it is wholly in that plane. Let the fame plane
always touching the line CD revolve about it, until it
meet the point A; then the ſtraight line AB, paffing
through
OF GEOMETRY.
155
through the points A, E, is likewife wholly in that
•
plane. Therefore the ftraight lines AB, CD are in def. 5. 1.
one and the fame plane.
COR. I. If two planes interfect each other, their
common fection is a ſtraight line. For, if two planes
pafs through the points E, D, the ftraight line CD is
wholly in each of them, and is therefore their line of
common fection.
COR. 2. Three ftraight lines, which interfect one
another in three points, are all in one plane. Or,
a plane may be extended through any three points
whatever.
PROPOSITION II.
If a straight line be perpendicular to two straight lines in-
terfecting each other, it is perpendicular to their plane.
Let the ſtraight line EF be perpendicular to each of Fig. 120.
the ſtraight lines AB, CD, which interfect each other
in the point E; EF is perpendicular to their plane.
For, let any ſtraight line GH be drawn through the
point E, in the plane of AB, CD. Let the lines EA,
EB, EC, ED, be affumed equal.
From any point F
in the perpendicular let FA, FB, FC, FD, be drawn;
and let AC, BD be joined. The triangles AEC,
BED, are equal in every reſpect, for they have two
fides of the one equal to two fides of the other, each
to
4. I
156
ELEMENTS
3
5. I.
4. I.
c 6. I.
d def. 1.6.
Fig. 121.
to each, and the included angles equal; therefore
AC =
BD, and the angle CD. The triangles.
CGE, EHD, having the angle C
E equal, and the fide CE = ED,
fidė CG = DH, and GE = EH ".
triangles FEA, FEB, FEC, FED,
D, the angles at
will alſo have the
The right angled.
have the fide FE
common to all, and the other fides about the right
angles equal, therefore the hypothenuſes FA, FB, FC,
FD, are equal. Thus, the triangles FCA, FBD, have
the three fides of the one equal to the three fides of
the other, each to each, therefore the angle FCG =
FDH. But the triangles FCG, FDH, have alfo the
fides about theſe angles equal, each to each; there-
fore the bafe FG FH 2. Hence the triangles
FEG, FEH, are mutually equilateral ; therefore
FEH are equal, and each of
the angles FEG,
them a right angle.
Confequently the ftraight line
EF is perpendicular to any ftraight line drawn in the
plane of AB, CD, from the point E, that is, it is per-
pendicular to the plane of theſe lines º.
COR. I. There cannot be more than one ſtraight
line perpendicular to a plane at the fame point in it.
For, if the lines FE, EK be both perpendicular to the
plane AB at the point E, let their plane meet AB in
the line of common fection GEH, and the lines FE,
EK are both perpendicular to GH, which is impof-
fible.
4
COR
OF GEOMETRY.
157
}
COR. 2. There cannot be more than one ſtraight Fig. 122.
line perpendicular to a plane from the fame point
without it. For, if the lines FE, FG be both perpen-
dicular to the plane AB from the point T, let their
plane meet AB in the line of common fection EG,
and the angles FEG, FGE, are both right angles;
which is impoffible.
COR. 3. There cannot be more than one plane per- Fig. 123.
pendicular to a ſtraight line at the fame point. For,
if each of the planes CD, EF be perpendicular to the
line AB, at the point B, let a plane touching AB in-
terfect them both in the lines of common fection BC,
BE, then ABC, ABE are in one plane, and both
right angles; which is impoffible.
COR. 4. If a ſtraight line be perpendicular to three
or more ſtraight lines at the fame point, they are all
in one and the fame plane.
PROPOSITION III.
Every ftraight line parallel to a ſtraight line that is per-
pendicular to a plane, is perpendicular to the fame ;
and converſely.
Let the ftraight line AB be perpendicular to the Fig. 124.
plane FG, then any ſtraight line CD, parallel to AB,
will alſo be perpendicular to FG.
For.
-1
›
158
ELEMENTS
* def. 1. 6.
4. I.
c 6. I.
في
2. 6.
For, let the plane of the parallels meet FG in the
line BD; let DE be perpendicular to BD, in the
plane FG, and equal to AB, and let AD, AE, BE be
joined. Since AB is perpendicular to FG, ABD,
ABE are right angles, and becauſe CD is parallel to
AB, CDB is a right angle; it is to be proved that
CDE is alſo a right angle.
The triangles ABD, BDE have the fide AB = DE,
BD common, and the angles at B, D right; there-
fore the baſe AD BE. Hence the triangles ADE,
=
ABE are mutually equilateral; therefore the angle
ADE = ABE, and confequently a right angle. Thus,
ED being perpendicular to both the lines AD, DB, is
perpendicular to their plane BC; therefore CDE is a
right angle, and confequently CD being perpendi-
cular to both the lines BD, DE, is perpendicular to
their plane FG ‘.
d
Converſely, If CD be perpendicular to FG, it is
parallel to AB. For it may be proved in the fame
manner that ADE is a right angle. But BDE is a
right angle by conftruction, and CDE a right angle
by hypothefis. Therefore, fince ED is perpendicular
to the three lines BD, AD, CD, at the fame point,
4. c. 2. 6. thefe lines are all in one plane. Confequently AB,
CD are in the fame plane; and fince they are alſo
perpendicular to the fame line BD, they are parallel.
PROPO.
OF GEOMETRY. 159
PROPOSITION IV.
Straight lines parallel to the fame are parallel to one ano-
ther, though they be not all in the fame plane.
Let the ftraight lines AB, CD, be each of them pa- Fig. 125.
rallel to the ſtraight line EF, but not both in the fame
plane with EF, AB is parallel to CD.
For, from any point H, in the line EF, let GH be
drawn perpendicular to it in the plane AF, and HK
perpendicular to it in the plane FC. Then EF is
perpendicular to the plane of the lines GH, HK'; ² 2. 6.
therefore AB, CD, which are parallel to EF, are per-
pendicular to the fame plane GHK, and confe. 3. 6.
quently they are parallel to one another ".
b
PROPOSITION V.
To draw a straight line perpendicular to a given plane
from a given point without it.
Let AB be the given plane, and C the given point Fig. 126.
without it, from which it is required to draw a
ſtraight line perpendicular to AB. Let any ſtraight
line DE be drawn in the plane AB. In the plane
CDE let CF be drawn perpendicular to DE, in the
plane
1
160
ELEMENTS
E ME
་
2. 6.
3. 6.
plane AB, let FG be alfo perpendicular to DE, and
in the plane CFG, CG perpendicular to FG: Then
CG fhall be perpendicular to AB.
For, let GH be parallel to DE. Then, becauſe
DE is perpendicular to both the lines CF, FG, it is
perpendicular to their plane; therefore GH, which
is parallel to DE, is perpendicular to the fame plane
CFG. Thus, CGH is a right angle; but CGF is a
right angle by conftruction; confequently CG is
perpendicular to the plane of the lines FG, GH',
that is, to AB.
COR. Hence a ſtraight line may be drawn perpen-
dicular to a given plane, from a given point in the
fame. For it may be drawn parallel to a perpendicu-
lar drawn from any point affumed without the plane.
Fig. 127.
PROPOSITION VI.
If two planes interfect each other, and from points in the
line of common fection ftraight lines be drawn perpendi-
cular to it in each plane, the angles formed by thefe
perpendiculars fhall be equal.
Let BE be the line of common ſection of the two
planes AE, BF. Let AB, DE be perpendicular to
BE in the former, and BC, EF perpendicular to it in
the latter; the angle ABC fhall be equal to DEF.
For,
BOOK VI.
161
OF GEOMETRY.
a
For, let AB, BC, DE, EF, be all equal, and AC,
CF, FD, DA, be joined. Becauſe AB, DE are both
perpendicular to BE, and in the fame plane, they are
parallel. But AB, DE are alfo equal. Therefore
AD is equal and parallel to BEª. In like manner,
CF is equal and parallel to BE. Confequently AD is
equal and parallel to CF, and AC equal and paral. Þ
lel to DF2. Thus, the triangles ABC, DEF are mu-
tually equilateral; therefore the angle ABC is equal
to the angle DEF ©.
b
23.
4. 6.
c 6. I
PROPOSITION VIL.
If a straight line be perpendicular to a plane, any plane
touching it will be perpendicular to the fame.
Let the ftraight line AB be perpendicular to the Fig. 128.
plane CD, then any plane CE, touching AB, will be
perpendicular to CD.
For, let CBG be their line of common fection, and
from any point G, in CG, let GE be perpendicular
to it in the plane CE. Alfo, let BF, GH be perpen-
dicular to CG, in the plane CD. Then, becauſe AB
is perpendicular to CD, ABF is a right angle; there-
fore the angle EGH, contained by the perpendiculars
at G, is alſo a right angle. But EGC is a right = 6. 6.
angle by conſtruction. Wherefore EG is perpendi-
X
cular
162
Book
ELEMENTS
b 2.6.
cular to the plane CD ³. In like manner, HG is per-
pendicular to the plane CE. Confequently the planes
* def. 3. 6. CE, CD are perpendicular to each other.
COR. A ftraight line, drawn at right angles to one
of two perpendicular planes, through any point in the
other, meets the former in the line of common fec-
tion, and lies wholly in the latter.
Fig. 129.
a c. 7. 6.
PROPOSITION VIII.
If two planes cutting each other be perpendicular to the
fame plane, their line of common fection is alfo perpendi-
cular to it.
Let the two planes AB, CD, cutting each other
in the line EF, be both perpendicular to the plane
GH, then EF is perpendicular to the fame plane.
a
For the perpendicular to GH, at the point F, where
EF meets it, is wholly in each of the planes AB, CD ²,
and confequently is the fame with their line of com-
mon fection.
PROPOSITION
IX.
If two planes, or a straight line and a plane, be each of
them perpendicular to the fame ftraight line, they are
parallel to one another.
Let
BOOK VI. OF GEOMETRY.
163
Let the two planes AB, CD be each of them per- Fig. 130.
pendicular to the ſame ſtraight line EF, at the points
E, F; AB, CD are parallel.
a
For, if not, let them meet in the line GH, and let
EG, GF be joined. Then, becauſe EF is perpendi-
cular to AB, CD, it is perpendicular to the lines EG,
FG, which are in their planes. Thus, EGF is a def. 1. 6.
triangle, having two of its angles right, which is im-
poffible. Therefore the plane CD does not meet AB.
But any ſtraight line, as FL, perpendicular to EF at
the point F, is in the plane CD. Confequently the
ftraight line FL is alfo parallel to AB.
¹6.
6. c. 2.6.
PROPOSITION
X.
If two parallel planes be cut by a third plane, the lines of
common fection are parallel.
Let AB, CD, be two parallel planes, cut by a third Fig. 131.
plane EG, the lines of common fedion EF, GH are
parallel.
For, if not, let them be produced, and meet in the
point K. Then, becauſe the ſtraight line FEK every
where touches the plane AB, and the ftraight line
GHK the plane CD, the point K is in each of the
planes. Thus, the planes AB, CD meet at the def. 5. 1.
point K, which is contrary to hypotheſis.
COR.
164
BOOK VI.
ELEMENTS
COR. 1. If two planes, cutting each other, be pa-
rallel to other two planes alfo cutting each other, the
lines of common fection are parallel.
COR. 2. If a plane be parallel to a ſtraight line, its
line of common fection, with any plane touching that
line, will alſo be parallel to the fame.
COR. 3. If two ftraight lines be parallel, each of
them is parallel to any plane touching the other, ex-
cept that which is common to both.
COR. 4. A ftraight line, perpendicular to one of
two parallel planes, is perpendicular to the other, and
the planes are everywhere equidiftant: Alfo any plane
perpendicular to the one is perpendicular to the o-
ther.
COR. 5. If equal perpendiculars be erected from
points in the fame plane, their other extremes are alſo
in one plane, parallel to the former.
COR. 6. Planes parallel to the fame plane are paral-
lel to one another.
COR. 7. If a ſtraight line be parallel to a plane,
any ſtraight line which meets them both, and is per-
pendicular to the latter, will alfo be perpendicular to
the former.
i
PROPO.
BOOK VI. OF GEOMETRY.
165
PROPOSITION XI.
If the fides of two angles which are not in the fame plane
be parallel, each to each, the angles are equal, and
their planes parallel.
Let ABC, DEF be two angles not in the fame Fig. 132.
plane, which have their fides parallel, each to each,
AB to DE, and BC to EF, the angles ABC, DEF
fhall be equal, and their planes parallel.
For, from the point B, let BG be drawn perpendi-
cular to the plane DEF, and from G, GH parallel 5. 6.
to DE, and GK to EF. Then, becauſe BG is per-
pendicular to the plane DEF, BGH is a right angle.
But, fince AB, GH are both parallel to DE, they are
parallel to one another. Therefore GBA is alſo a
right angle. Thus, BG is perpendicular to the lines
BA, GH. In like manner, it is perpendicular to the
lines BC, GK. Therefore BG is perpendicular to
the planes ABC, DEF, and conſequently theſe planes
are parallel.
Again, becauſe AB, GH are parallel, they are in
the fame plane, and, for the ſame reaſon, BC, GK
are both in one plane. Thus, ABGH, and CBGK,
are two planes interfecting each other in the line BG,
to which lines AB, GH are perpendicular in the one
plane, and BC, GK in the other; therefore the angle
ABC
b
2. 6.
166
Book VI.
ELEMENTS
e 6. 6.
d 17. 1.
I.
DEF.
DEF.
ABC is equal to HGK. But HGK is equal to
Confequently the angle ABC is equal to
Fig. 133.
3. 6.
PROPOSITION XII.
If perpendiculars be drawn to a plane, from two points in
any ſtraight line inclined to it, they ſhall have the fame
ratio as the fegments of the line which the plane inter-
cepts from theſe points.
From the points A, B, in the ftraight line AB, in-
clined to the plane CD, and meeting it in the point.
E, let the perpendiculars AG, BF be drawn to CD,
AG: BF: AE: EB.
Becauſe AG, BF are both perpendicular to CD,
they are parallel. Hence the ſtraight lines AG,
AB, BF, are all in one plane, and F, E, B are
points in its line of common fection with CD. But
1.c.1.6. the common fection of two planes is a ſtraight line ".
Therefore F, E, B, are in the fame ftraight line;
and the right angled triangles AEG, BEF, being
equiangular, AG: BF:: AE: EB ©.
€ 4. 5.
COR. 1. The ſtraight line which joins the extremi-
ties of the perpendiculars, paffes through the point
where the inclined line interfe is the plane; and is
there divided into fegments, having the fame ratio as
the perpendiculars.
A
COR.
F
F
B
A
GEOMETRY
PL.XI. pa
F
FIG. 123.
166
LA
B
F
K
FIG121.
B
G.
-H
E
G
Ꮐ
A-
E
H
-B
E
E
G
13
C
A
FIG.122.
FIG.128.
FIG. 120.
H
E-
-F
A
D
FIG. 125.
A
C
A
B
FIG. 124.
B
-D
K
A
D
F
FIG.126.
F
E
·
C
B
Ꮐ
G
F
IT
D
FIG.129.
E
H
E
Ꮐ
B
E
A
FIG.127.
E
FIG. 132.
Ꮐ
B
A·
C
D
H
E
H
K
T
E
G
FIG.130.
D
B
C
L
F
T
FIG. 131.
K
B
A
C
E
H
F
В
t
BOOK VI. OF GEOMETRY.
167
COR. 2. Straight lines, which meet three parallel
planes, are cut proportionally.
PROPOSITION XIII.
!
If a folid angle be contained by three plane angles, any
two of them are together greater than the third.
Let there be a folid angle at the point A, contained Fig. 134-
by the three plane angles BAC, CAD, DAB; any
two of theſe are together greater than the third.
AE,
If all the angles be equal, or if the two greater be
equal, the propofition is evident. In any other cafe, let
BAC be the greateſt angle, and let BAE be cut off
from it, equal to DAB. Through any point E, in
AE, let the ſtraight line BEC be drawn, in the plane
of the angle BAC, to meet its fides in B, C. Let
AD be equal to AE, and BD, DC joined. Becauſe
the triangles BAD, BAE have the fide AD
BA common, and the included angles equal, the bafe
BD is equal to BE. But the ſum of BD, DC is
greater than the fum of BE, EC. Therefore DC is 20. 1.
greater than EC; and, fince the fide DA is equal to
AE, and AC common to the two triangles ACD,
ACE, the angle CAD is greater than EAC. Con- c 22. 1.
ſequently, the ſum of the angles BAD, CAD is greater
than the angle BAC.
b
4 I.
PROPO.
เ
168
Book VI.
ELEMENTS
Fig. 135.
a 13. 6.
PROPOSITION XIV.
All the plane angles which form any folid angle, are toge-
ther less than four right angles.
Let there be a folid angle at the point A, contained
by the plane angles BAC, CAD, DAE, EAB, theſe
angles, taken together, are lefs than four right
angles.
For, through any point B, in AB, let a plane be
extended to meet the fides of the folid angle in the
lines of common fection BC, CD, DE, EB, thereby
forming the pyramid BCDE-A: And from any
point O, within the rectilineal figure BCDE, which is
the bafe of the pyramid, let the ftraight lines OB,
OC, OD, OE, be drawn to the angular points of the
figure, which will thereby be divided into as many
triangles as the pyramid has fides. Then, becauſe
each of the folid angles at the bafe of the pyramid is
contained by three plane angles, any two of them are
together greater than the third ". Thus, ABE, ABC
are together greater than EBC. Therefore the angles
at the baſes of the triangles, which have their vertex
at A, are together greater than the angles at the
bafes of the triangles which have their vertex at
O. But all the angles of the former triangles are
together equal to all the angles of the latter. There-
a
fore
Book VI. OF GEOMETRY.
169
fore the remaining angles at A are together lefs than
the remaining angles at O, that is, lefs than four
right angles".
b 2. C. IJ
I.
PROPOSITION XV.
If two folid angles be each of them contained by three
plane angles, and have theſe angles equal, each to each,
and alike fituated, the two folid angles are equal.
Let there be a folid angle at the point A, contained Fig. 136.
by the three plane angles BAC, CAD, DAB, and a
folid angle at the point E, contained by the three
plane angles FEG, GEH, HEF, equal to the former,
each to each, and alike fituated; theſe ſolid angles
are equal.
For, let the ftraight lines AB, AC, AD, EF, EG,
EH, be all equal, and let their extremes be joined
by the lines BC, CD, DB, FG, GH, HF: And thus
there are formed two ifofceles pyramids, BCD-A and
FGH-E. Upon the bafes BCD, FGH, let the perpen-
diculars AK, EL fall from the vertices A, E. Then,
becauſe the right angled triangles AKB, AKC, AKD
have equal hypothenuſes, and the fide AK common,
their other fides KB, KC, KD, are alfo equal: 2. c.23.
Therefore K is the center of the circle that circum-
fcribes the triangle BDC. In like manner, L is the
Y
center
I.
170
BOOK VI.
ELEMENTS
4
6. I.
4. I.
center of the circle that circumfcribes the triangle
FHG. But theſe triangles are equal in every reſpect º.
For the fides BC, FG are equal, becaufe they are
the bafes of equal and fimilar triangles BAC, FEG;
and, for the fame reafon, CD is equal to GH, and
DB to HF. If, therefore, the pyramid BCD-A be
applied to the pyramid FGH-E, their baſes BCD and
FGH will coincide. Alfo, the point K will fall upon
L, becauſe, in the plane of the circle FHG, no other
point than the center is equally diftant from three
1. c. 5.3. points in its circumferenced; the perpendicular KA
will be in the fame ftraight line with LE, becaufe there
a C. 23. I.
=
cannot be more than one perpendicular to a plane at
1. c. 2.6. the fame point; and the point A will coincide with
E, becauſe the right angled triangles AKB, ELF ha-
ving equal hypothenufes, and EK FL, their other
fides AK, EL are alfo equal. But the points B,
D, C coincide with F, H, G, each with each; there-
fore the ftraight lines AB, AC, AD coincide with
EF, EG, EH, and the plane angles BAC, CAD,
DAB, with FEG, GEH, HEF, each with each. Con-
fequently, the folid angles themfelves coincide, and
are equal.
f 11. 6.
I.c.10.6.
COR. 1. If two folid angles, each contained by three
plane angles, have their linear fides, or the planes that
bound them, parallel each to cach, the folid angles
are equal.
COR.
;
Book VI. OF GEOMETRY.
171
COR. 2. If two folid angles, each contained by the
fame number of plane angles, have their linear or
plane fides parallel, each to each, the folid angles are
equal.
For each of the folid angles may be divided into
folid angles, each contained by three plane angles,
and the parts being equal and alike fituated, the
wholes are equal.
PROPOSITION
XVI.
If two triangular prifms have the plane angles and lincar
fides, about two of their folid angles equal, cach to cach,
and alike fituated, the prifms are equal and fimilar.
Let ABC-DEF and GHK-LMN, be two triangular Fig. 137.
prifms, having the plane angles BAC, CAD, DAB,
about the folid angle at A, equal to the plane angies
HGK, KGL, LGH, about the folid angle at G, each
to each, and the linear fides AB, AC, AD, equal to
the linear fides GH, GK, GL, each to each; thefe
priſms are equal and fimilar.
2
For, fince the folid angles A, G are each contained
by three plane angles, which are equal each to each,
thefe folid angles, being applied to each other, coin-
cide. Therefore the ftraight lines AB, AC, AD, 15. 6.
coincide with the ftraight lines GH, GK, GL, each
with each. But AB coinciding with GH, and the
point D with L, the ftraight line DE muft fall upon
LM, ·
172
Book VI.
ELEMENT S
b 16. I.
LM, becauſe there cannot be more than one parallel
to the fame line drawn from the fame point; fo
likewiſe the ſtraight line BE falls upon HM: There-
fore the point E coincides with M. In like manner,
the point F coincides with N. Confequently, the rec-
tilineal figures which bound the one priſm coincide
with, and are equal and fimilar to the rectilineal fi-
gures which bound the other, each to each; the folid
angles of the one coincide with, and are equal to the
folid angles of the other, each to each; and the folids
themſelves are equal and fimilar.
COR. 1. If two triangular pyramids have the plane
angles, and linear fides, about two of their folid
angles, equal each to each, and alike fituated, the py-
ramids are equal and fimilar.
COR. 2. If two triangular prifms have the linear
fides about two of their folid angles both equal and
parallel, each to each, the prifms are equal and fimi-
lar.
COR. 3. If two pyramids have the linear fides a-
bout their vertical angles both equal and parallel,
each to each, the pyramids are equal and fimilar.
PROPOSITION XVII.
The oppofite fides of a parallelopipedon are fimilar and e-
qual parallelograms, and the diagonal plane divides it
into two equal and fimilar prifms.
Let
BOOK VI. OF GEOMETRY.
173
Let ABCD-EFGH be a parallelopipedon, or folid, Fig. 138.
bounded by fix planes, of which the oppofite ones are
parallel, any two of its oppofite fides, as ABCD and
EFGH, are fimilar and equal parallelograms.
a
10. 6.
For, fince the oppofite planes AF, DG are paral-
lel, and cut by a third plane BD, the lines of com-
mon ſection AB, DC are parallel. Alfo, becauſe the
oppofite planes BG, AH are parallel, and cut by the
plane BD, the lines of common fection, AD, BC, are
parallel. Thus, the figure ABCD is a parallelogram.
In like manner, all the plane figures which bound the
folid are parallelograms. Hence AD is equal to EH,
and DC to HG, and the angle ADC, by reaſon of ↳ 1. 2.
parallel lines, is equal to EHG; therefore the tri- c 11. 6.
angles ADC, EHG, and conſequently the parallelo- a
grams ABCD, EFGH are equal and fimilar.
C
4. I.
e 23. 1.
Again, becauſe AE, CG are each equal and paral-
lel to DH, they are equal and parallel to each other ;
therefore the figure ACGE is a parallelogram, and
divides the whole parallelopipedon AG into two tri-
angular priſms ABC-EFG, and ACD-EGH. And
theſe priſms are equal and ſimilar f, becauſe the plane f 16. 6.
angles and linear fides about the folid angles at F and
D are equal, each to each: For the plane angles
EFG, ADC are equal to one another, as being each
equal to EHG, and the linear fides FG, DA equal
to one another, as being each equal to EH, and fo of
the others.
COR.
174
Book VI.
ELEMENTS
COR. I. The oppofite folid angles of a parallelopi-
pedon are equal.
COR. 2. Every rectangular parallelopipedon is
bounded by rectangles.
COR. 3. The ends of a prifm are fimilar and equal
figures, and their planes parallel.
COR. 4. Every parallelopipedon is a quadrangular
prifm, of which the ends are parallelograms; and
converſely.
COR. 5. If two parallelopipedons have the plane
angles, and linear fides, about two of their folid angles
equal, each to each, and alike fituated, or the linear
fides, about two of their folid angles, both equal and
parallel, each to each, the folids are equal and fimi-
lar.
COR. 6. If, from the angular points of
any rectili-
neal plane figure, there be drawn ftraight lines above
its plane, all equal and parallel, and their extremes be
joined, the figure formed thereby is a priſm. If the
rectilineal figure be a parallelogram, the prifm is a
parallelopipedon. If the rectilineal figure be a rec-
tangle, and the ftraight lines be perpendicular to its
plane, the prifm is a rectangular parallelopipedon.
COR. 7. Hence the method of forming a paralle-
lopipedon, of which one of the folid angles, and its
linear fides, are given; alfo of forming a parallelo-
pipedon, from a given triangular prifm, as one of its
halves.
PROPO.
BOOK VI. OF GEOMETRY. 175
PROPOSITION XVIII.
Parallelopipedons upon the fame bafe, and between the
fame parallel planes, are equal.
a
Let ABCD-EFGH, and ABCD-KLMN, be two pa- Fig. 139.
rallelopipedons upon the fame bafe ABCD, and be-
tween the fame parallel planes AC, EM; the paralle-
lopipedons AG, AM are equal. Becaufe the ftraight
lines EF, GH, KL, MN, are parallel to AB, CD,
they are parallel to one another, and, being all in the
fame plane, NK, ML, will therefore meet both EF
and GH; let them meet the former in O, P, and the z.c.16.1.
latter in R, Q, and let ÀO, BP, CQ, DR, be joined.
The figure ABCD-OPQR is a parallelopipedon: For,
by hypothefis, the plane EQ is parallel to AC, the
plane of the parallels DC-HQ is parallel to the plane
of the parallels AB-EP, and the plane AD-NO paral-
lel to BC-MP. Hence AEO-DHR and BFP-CGQ
are two triangular prifms, which have the linear fides
AE, AD, AO, about the folid angle at A, both equal
and parallel to the linear fides BF, BC, BP, about the
folid angle at B, each to each. Confequently, thefe
prifms are equal, and each of them being taken a-
way from the whole folid ABCD-EPQH, the remain-
ders, the parallelopipedons AG, AQ are equal. In the
fame manner, the parallelopipedon AM may be proved
equal
↳ 17. 6.
c 16. 6.
}
}
:
i
176
ELEMENTS
BOOK VI.
equal to AQ. Therefore the parallelopipedons AG
and AM are equal to one another.
Fig. 140.
17. 6.
II.
PROPOSITION XIX.
Parallelopipedons, upon equal bafes, and between the fame
parallel planes, are equal.
CASE I. When the bafes have one fide common,
and lie between the fame parallel lines. Let ABCD-
EFGH and ABKL-OPQR be two parallelopipedons,
between the parallel planes AK, EN, and upon the
bafes ABCD, ABKL, which have the fide AB com-
mon, and lie between the parallel lines AB, DK ;
theſe parallelopipedons AG, AQ are equal.
For, let the planes EAL, FBK meet the fide HC of
the former in the lines of common fection LM, KN,
and the fide HF in the lines EM, FN. Then, fince
EA, AL are parallel to FB, BK, each to each, their
11. 6. planes are parallel, and the figure AN is a paralle-
lopipedon. Hence ADL-EHM and BCK-FGN are
two triangular prifms, which have the linear fides AD,
AE, AL about the folid angle at A, both equal and
parallel to the linear fides BC, BF, BK, about the fo-
lid angle at B, each to each. Confequently, theſe
priſms are equal, and each of them being taken a-
way from the whole folid ABKD-EFNH, the remain-
e 16. 6.
ders,
BOOK VI. OF GEOMETR Y.
177
ders, the parallelopipedons AG, AN are equal. But
AN is equal to AQ, becauſe they are upon the fame
bafe, and between the fame parallel planes, There-
fore the parallelopipedons AG, AQ are equal to one
another.
e
d 18. 6,
c. 12. 1.
I.
CASE II. When the bafes are equiangular. Let Fig. 141.
ABCD, CEFG be equal bafes of two parallelopipe-
dons, between the fame parallel planes, and let them
have the angles BCD, GCE equal. Let DC, CE be
in the fame ſtraight line, then BC, CG are alfo in one
ftraight line Alfo, let their fides be produced to
meet in the points H, K. Since the parallelograms f 2. c. 16.
AC, CF are equal, they are complements of the pa-
rallelogram AF, and the ftraight line HCK is its dia-
gonal. Upon the bafe AF, let a parallelopipedon be c. 1. 2.
erected of the fame altitude with thofe upon AC,
CF; and let it be cut by planes parallel to its fides, c. 6. 17.
and touching the lines DCE, GCB: Thefe planes
divide the whole parallelopipedon into four other pa-
rallelopipedons upon the bafes DG, AC, BE, CF.
But the diagonal plane touching HK divides each of
the parallelopipedons upon AF, DG, BE into two
equal prifins. Therefore the prifms upon RGC, i 17. 6.
CEK, are together equal to the prifms upon HDC,
CBK; and thefe being taken away from the equal
priſms upon HFK, HAK, the remainders, the paral-
lelopipedons ereted upon AC, CF are equal. Con-
Z
fequently
b
6.
178
ELEMENTS Book VI.
k 18. 6.
Fig. 142.
ſequently any two parallelopipedons upon theſe bafes,
and between the fame parallel planes, are equal *.
CASE III. When the bafes are neither equiangular,
nor have one fide common. Let ABCD, EFGH be
the equal bafes of two parallellopipedons, between the
fame parallel planes. At the point E, let the angle
FEL be made equal to BAD, let GH meet EL in L,
and let the parallelogram FL be completed. Becauſe
the parallelograms AC, FL are each equal to EG,
they are equal to one another; and they are alſo e-
quiangular Therefore the parallelopipedon upon AC
is equal to any parallelopipedon upon FL, between
the fame parallel planes. But the parallelopipedon
upon EG is equal to the fame. Therefore the paral-
lelopipedons upon AC, EG, are equal to one ano-
ther.
COR. 1. Parallelopipedons of equal bafes and equal
altitudes are equal.
COR. 2. Triangular priíms, upon equal bafes, which
are either both triangles, or both parallelograms, and
of equal altitudes, are equal.
COR. 3. Two triangular prifms, upon a parallelo-
gram and a triangle, as their bafes, and of equal al-
titudes, are equal to one another, if the parallelogram
be double the triangle.
COR. 4. A triangular prifm, and confequently any
other priſm, upon one of its ends, as a baſe, is equal
to a parallelopipedon of equal bafe and altitude.
PROPO.
I
4
1
C
FIG.133.
GE OMETRY
A
D
FIG. 134.
F
Ꮐ
E
B
D
B
C
E
E
A
FIG.136.
H
L
B
C
F
Ꮐ
$
D
B
F
A
E
FIG. 138.
C
D
Ꮐ
R
P
H
G
M
H
B
A
ތ
E
H
FIG. 137.
H
Ꮐ
K
F
E
F
B
E
FIG. 135.
Ꮐ
PL. XI.pa. 178,
A
N
M
K
FIG.139.
D
FIG.141.
J
Ꮐ
I'
T
H
E
F
FIG.140.
D
B
E
A
K
P
E
/R
P
M
12
H
K
>
FIG. 142.
B
Book VI.
OF GEOMETRY. 179
PROPOSITION XX.
Parallelopipedons of equal altitudes are to one another as
their bafes.
!
Let ABCD, EFGH be the bafes of two parallelopi- Fig. 143.
pedons AK, EL, of equal altitudes, AK: EL :: AC
: EG.
a
19. G.
For, let the baſe AB, of the parallelogram AC, be
produced to M, fo that BM be a fourth proportional
to the altitudes of AC, EG, and the bafe EF, and let
the parallelogram CM, and the parallelopipedon MK
be completed. Let Bn be any part whatever of the
line BM, let no be parallel to BC, and let the paralle-
lopipedon MK be cut by the plane noqr parallel to
BK. Then, becauſe parallelopipedons of the fame
altitude, upon equal bafes, are equal, the parallelo-
pipedon MK will contain Bq as often as the parallelo-
gram CM contains Bo, or the line BM contains Bn.
Thus, Bq may be any part whatever of KM. But,
for the fame reafon, the parallelopipedon AK con-
tains Bq just as often as the parallelogram AC con-
tains Bo. Conſequently AK: KM :: AC : CM.
But, from the conſtruction, it is evident that the pa-
rallelogram EG is equal to CM; therefore the pa- 67. c. 8.5.
rallelopipedon EL is equal to KM, and AK: EL
e
67.c.8.5.
1.c.19.6.
:: AC: EG.
COR.
180
Book VI.
ELEMENT S
COR. Priſms ſtanding upon their ends, and of e-
qual altitudes, are to one another as their bafes.
Fig. 144.
PROPOSITION XXI.
Two parallelopipedons, which have a folid angle of the one
equal to a folid angle of the other, are to one another in
the ratio compounded of the ratios of the linear fides of
the one to the linear fides of the other, each to each, a-
bout theſe ſolid angles.
Let DE, KL be two parallelopipedons, of which
the folid angle at B, in the former, is contained by
plane angles ABC, ABE, EBC, reſpectively, equal to
the plane angles FGH, FGL, LGH, containing the
folid angle at G in the latter, and alike fituated; then
the ratio DE: KL is the fame with that which is
compounded of the ratios AB: FG, CB: HG, and
EB: LG.
For, let AB, CB, EB, be produced to M, N, O, fo
that BM FG, BN =
=
GH, and BO = GL. With
the lines EB, BM, and BN, let the parallelopipedon
* 7. c. 17. EP be completed, and with the lines OB, BM, and
BN, the parallelopipedon OP. Then, becauſe the
ratio DE OP is compounded of the two ratios DE
D
C
6.
20. 6.
2. c. I.5.
: EP, and EP : OP, of which the former is equal to
AC: MN, and the latter to EM: MO", or EB: BO;
the
Book VI. OF GEOMETRY.
181
the ratio DE: OP is the fame with that which is
compounded of AC: MN, and EB: BO, or of dc. 17.4.
AB: BM, CB : BN, and EB: BO. But OP is e- 9. 5.
f
qual to KL, becauſe theſe parallelopipedons have the 5.c.17.6.
plane angles, and linear fides about their folid angles
at B, G reſpectively equal and alike fituated. Con-
fequently the ratio DE: KL is the fame with that
which is compounded of AB: BM, CB: BN, and
EB: BO; or of AB: FG, CB: HG, and EB : LG.
COR. I. Two rectangular parallelopipedons are to
one another in the ratio compounded of the ratios of
the linear fides of the one to the linear fides of the
other, each to each: And any ratio compounded of
three ratios (whofe terms are ftraight lines,) is the
fame with the ratio of the rectangular parallelopipe-
dons, under their homologous terms.
COR. 2. Two cubes, or, in general, two fimilar pa-
rallelopipedons, are to one another in the triplicate
ratio of their homologous linear fides.
COR. 3.
Similar parallelopipedons are to one ano-
ther as the cubes of their homologous linear fides.
COR. 4. If four ſtraight lines be in continued pro-
portion, the firſt is to the fourth as the cube of the
firſt to the cube of the ſecond.
COR. 5. The rectangular parallelopipedons, under
the correſponding terms of three analogies, are pro-
portional.
COR.
ว
182
ELEMENTS BOOK
COR. 6. If four ſtraight lines be proportional, their
cubes are alſo proportional; and converſely.
COR. 7. Rectangular parallelopipedons, and confe-
quently any other parallelopipedons, are to one ano-
ther in the ratio compounded of the ratios of their
bafes and altitudes.
COR. 8. Parallelopipedons of equal bafes are to one
another as their altitudes.
COR. 9. Parallelopipedons, whoſe baſes and alti-
tudes are reciprocally proportional, are equal; and
converſely.
COR. 10. Prifms are to one another in the ratio
compounded of the ratios of their baſes and altitudes.
Therefore the 8th and 9th Cor. may be applied to
prifm's.
Fig. 145.
PROPOSITION XXII.
Every triangular pyramid may be divided into two equal
prifms, which are together greater than half the whole
pyramid, and two equal pyramids, which are fimilar to
the whole, and to one another.
Let BCD-A be any triangular pyramid. Let its
linear fides AB, AC, AD, be bifected in E, F, G,
and the points of fection joined ; and let the fides of
of the baſe CB, BD, DC, be bifected in H, K, L, and
the
!
Book VI. OF GEOMETRY.
183
b
the points of fection joined: Alfo, let EH, EK, and
LG, be drawn. Becaufe the two fides AB, AC of
of the triangle ABC, are biſected in E, F, the ftraight
line EF is equal and parallel to CH or HB, half the
remaining fide BC; and fo of the other ftraight 22. 5.
lines that join the points of fection. Hence EC, CG,
and confequently EL, are parallelograms, and the
planes EFG, BCD parallel. Therefore the folid
EFG-HCL, upon the triangular baſe HCL, is a prifm.
And the folid EHK-GLD is a triangular priſm of the
fame altitude, upon the parallelogram HD, which is
double the triangle HCL. Confequently thefe priſms
are equal. The two remaining folids EFG-A, and
and BHK-E, are triangular pyramids. They are e-
qual and fimilar to one another, and alſo fimilar to
the whole, becauſe the triangles which bound the one
are equal and fimilar to the triangles which bound the
other, and alfo fimilar to thofe which bound the
whole, each to each, and alike fituated. But either
of theſe pyramids is, for the fame reaſon, equal to
the pyramid KLD-G, and therefore leſs than either of
the two prifms. Confequently, the two prifms EFG-
HCL, and EHK-GLD, are together greater than the
two pyramids EFG-A, and BHK-E, and greater than
half the whole pyramid BCD-A.
COR. By taking from the whole pyramid the two
equal prifms, and from each of the remaining pyra-
mids two equal prifms, formed in like manner, there
will
3. c. 19
6.
184
BOOK VI.
ELEMENTS
will at length remain a magnitude lefs than any pro-
pofed.
Fig. 146.
12. 6.
b c. 20.
6.
PROPOSITION XXIII.
Pyramids of equal altitudes are to one another as their
bafes.
CASE I. Let BCD-A, and NOP-M, be two trian-
gular pyramids of equal altitudes, BCD-A: NOP-M
:: BCD: NOP.
There-
For, fince the priſms EFG-HCL and QRS-TOX
have eqnal altitudes, (each half the altitude of its py-
ramid), they are to one another as their baſes ↳,
EFG-HCL : QRS-TOX :: HCL: TOX. But HCL
: TOX :: 4HCL : 4TOX :: BCD: NOP.
fore the ratio of the two priſms, in the pyramid
BCD-A, to the two prifms in the pyramid NOP-M, is
the fame with that of the baſes BCD, NOP. In like
manner, the ratio of the two prifms in the pyramid
EFG-A, to the two priſms in the pyramid QRS-M, is
the fame with that of (EFG, QRS, or HCL, TOX,
or) BCD, NOP. Confequently, the fum of all the
prifms taken from the pyramid BCD-A, after the
fame manner, by any number of divifions, is to the
fum of all the prifms taken from the pyramid NOP-M,
the fame number of divifions, as the baſe BCD is to
by
the
Book VI. OF GEOMET R Y. 185
the baſe NOP. And, fince the remainder of either
may
be leſs than any magnitude propoſed, the pyra-
mids themſelves are in the fame ratio °.
CASE II. Let ABCDE-F and GHKL-M be two py-
ramids of equal altitudes, having any rectilineal fi-
gures for their bafes, ABCDE-F GHKL-M ::
ABCDE: GHKL.
For, let the bafe ABCDE be divided into triangles
by the diagonals AC, AD, then the whole pyramid
will be divided into triangular pyramids, by planes
touching thefe lines AC, AD, and paffing through
the vertex F: Alfo, let NOP-Q be any triangular py-
ramid of equal altitude.
Then, fince the pyr. ABC-F : NOP-Q :: ABC : NOP
ACD-F: NOP-Q :: AED: NOP
And
And
ADE-F: NOP-Q :: ADE: NOP
c 18. 4°
Fig. 147.
The whole pyr. ABCDE-F: NOP-Q :: ABCDE:NOPad 15. 4.
In like manner GHKL-M: NOP-Q:: GHKL: NOP.
Hence, the terms of the laſt analogy being inverted,
it follows, by equality, that ABCDE-F: GHKL-M ;:
ABCDE: GHKL.
PROPOSITION XXIV.
Every triangular prifm may be divided into three cqual
triangular pyramids.
A a
Let
186
Book VI.
ELEMENTS
Fig. 148.
23.6.
Let ABC-DEF be a triangular prifm, and let the
diagonals BD, BF, of two of its fides, be drawn from
the fame point B ; the plane DBF cuts off a triangular
pyramid DEF-B, and there remains a quadrangular
pyramid DACF-B, which may be divided into two
triangular pyramids DAC-B, and DFC-B, by a plane
paffing through its vertex B, and touching DC, the
diagonal of its bafe. Thus, by the two planes DBF,
DBC, the whole prifm is divided into three triangu
lar pyramids, having a common vertex at B, DEF-B,
DAC-B, and DFC-B. Of theſe the ſecond and third
are equal ª, becauſe they are upon equal bafes, and
have the fame altitude. But the firft and third are
alfo equal, for they are the fame with BEF-D and
BCF.D, which are upon equal baſes, and have the
fame altitude. Therefore the prifm is divided into
three equal triangular pyramids.
a
COR. 1. Every pyramid is the third part of a priſm
of the fame (or equal) baſe and altitude.
COR. 2. Pyramids are to one another in the ratio
compounded of the ratios of their bafes and alti-
tudes.
COR. 3. Similar pyramids or prifms are to one an-
other in the triplicate ratio of their altitudes, or their
homologous linear fides.
COR. 4. Similar pyramids, or prifins, are to one
another as the cubes of their homologous fides.
COR.
BOOK VI. O GEOMETR Y.
187
COR. 5. The ruftum of a triangular pyramid may
be divided, in the fame manner, into three triangular
pyramids, which are in continued proportion. For
DAC-B : DFC-B = (DAC: DFC AC: DF = BC
=
: EF = BCF: BEF =) BCF-D: BEF-D.
PROPOSITION XXV.
A cone, and a pyramid, of equal altitudes, are to one ano-
ther as their bafes.
Let BD-H be a cone, and EF-G a pyramid of e- Fig. 149.
qual altitudes, BD-H : EF-G :: BD : EF.
For, in the bafe of the cone, let the fquare ABCD
be infcribed; and let the regular polygon KL be
formed by bifecting the arches which the fides of the
fquare fubtend. The difference between the fquare
pyramid ABCD-H and the cone BD-H is the ſum of
the conic ſegments AHB, BHC, CHD, DHA, cut off
by the fides of the pyramid: But the triangular pyra-
mid AKB-H is greater than half the conic fegment
AHB, in which it is infcribed, becauſe it is equal to
half the triangular pyramid BHM, that circumfcribes
AHB. If, therefore, the pyramid KL-H be taken 23. 6.
from the cone BD-H, there will be taken away more
than half that difference. In like manner, by taking
from the fame cone the infcribed pyramid of double
the
189
Book VI.
ELEMENTS
the number of fides, there will be taken away more
than half the remainder; and fo on.
Therefore a
pyramid may be taken from the cone that fhall leave
a remainder lefs than any magnitude propofed. And,
fince pyramids of equal altitudes are to one another
as their bafes, of the four magnitudes BD-H, EF-G,
BD, EF, there may be taken from the first and third,
magnitudes proportional to the fecond and fourth, fo
that the remainder of either fhall be lefs than any
propofed magnitude. Confequently thefe four mag-
bc. 18. 4. nitudes are proportional".
I.
·
COR. 1. It may be demonftrated, in like manner,
that a cylinder, and a prifm, of equal altitudes, are to
one another as their baſes.
COR. 2. A cone is equal to a pyramid, and a cy-
linder equal to a prifm, of equal bafe and altitude.
COR. 3. A cone is the third part of a cylinder of
the fame (or equal) baſe and altitude.
COR. 4. Cones or cylinders are to one another, in
the ratio compounded of the ratios of their bafes and
altitudes.
COR. 5. Similar cones or cylinders are to one ano-
ther in the triplicate ratio of their axes, or of the di-
ameters of their baſes, or in the fimple ratio of the
cubes of theſe dimenfions.
1
CONIC
1
M
K
D
D
K
OMETRY
FIG.143.
H
G
A
PL. XII.a, 189
FIG.145
A
B
n
E
M
E
Ꮐ
F
FIG.144.
E
A
M
K
B
H
L
B
N
P
F
G
N
T
H
K
M
F
M
H
G
D
FIG. 149.
B
C
E
B
D
FIG.147.
R
FIG. 146.
S
E
L
P
K
A
FIG. 148.
H
P
F
'
CONIC SECTIONS.
BOOK I.
PARABOL A.
DEFINITION S.
I.
"L
ET there be a ſtraight line, and a point with-
out it, given in pofition, and let a point move
always equally diftant from both, it will defcribe a
curve, which is called a Parabola.
2. The given line is named the Directrix, and the
given point the focus.
3. The vertex of the parabola is the middle of the
perpendicular, which falls upon the directrix from the
focus: And the axis, or principal diameter, is that
part of the perpendicular produced indefinitely which
falls within the curve.
4. Any
190
BOOK I
CONIC SECTIONS.
4. Any ftraight line, drawn from a point in the
curve, parallel to the axis, and in the fame direction,
is called a diameter, and the point in the curve its
vertex.
5. An ordinate to a diameter is a ſtraight line,
terminated on both fides by the curve, and biſected
by that diameter: The part of the diameter, which it
cuts off, is called an abfcifs.
6. The parameter of a diameter is four times the
diſtance of its vertex from the directrix.
7. A tangent is a ſtraight line, which meets the
curve in one point only, and every where elfe falls
without it.
8. A ſubtangent is that part of a diameter which is
intercepted by a tangent, and an ordinate, from the
point of contact.
Fig. 1.
PROPOSITION I
The diftance of a point from the focus is greater than its
diſtance from the directrix, if the point be without the
parabola, but lefs if it be within.
Let GVN be a parabola, whofe directrix is AB,
vertex V, and focus F; and let P be a point without
the curve, that is, on the fame fide of the curve with
the directrix: Then, if PF be joined, and PQ be
drawn
!
1
1
BOOK I.
CONIC SECTIONS.
191
drawn perpendicular to AB, PF will be greater than
PQ.
For, as PF neceffarily cuts the curve, let G be the
point of fection, GD perpendicular to AB, and P, D
joined. Then, becauſe GF = GD ², PF = PG + GD.
Hence PF is greater than PD, and confequently ſtill
greater than PQ ʻ.
Again, let O be a point within the curve. The
perpendicular OD, upon AB, neceffarily cuts the
curve; let G be the point of ſection, and let GF,
FO be joined: Then OD is equal to the fum of OG,
GF 2, and therefore greater than OF ы.
a
COR. 1. A point is without, in, or within the
curve, according as its diſtance from the focus is
greater, equal, or lefs, than its diftance from the di-
rectrix.
COR. 2. A perpendicular to the axis, at its vertex,
is a tangent to the parabola.
a def. I.
20. 1. G
3.c 19.1.
PROPOSITION II.
Every Straight line perpendicular to the directrix meets
the parabola, and every diameter falls wholly within
it.
Let DG be perpendicular to the directrix, at any Fig. 1.
point of it, then ſhall DG meet the curve.
For,
192
Book I.
CONIC SECTIONS.
+
For, DF being joined, let the angle DFG be made
22.c.16.1. equal to GDF, and let FG meet DG, which is pa-
rallel to FC, in G. The triangle DGF, having the
angles at D and F equal, will alſo have the fides GD,
GF equal; and therefore G is a point in the curve".
Again, the diameter GE falls wholly within the
curve b.
b 1. c. I.
For, if any point O be affumed in it, it is evident
that OD is greater than OF.
COR. Hence, the two legs of the curve continually
diverge from the axis.
1
Fig. 2.
PROPOSITION
III.
A ſtraight line bisecting the angle formed by two lines
drawn from the fame point in the curve, the one to the
focus, and the other perpendicular to the directrix, is
a tangent to the parabola in that point.
The ſtraight line GE, bifecting the angle DGF, is
a tangent to the parabola in G.
For, let H be any other point in GE, from which
let there be drawn HF and HD, alſo HA perpendicu-
lar to AB. Then, becauſe GE bifects the vertical
angle of the ifofceles triangle GDF, it will alfo bi-
fect the baſe DF at right angles. Hence the triangles
HED, HEF are equal in every refpect. Thus, HF
3.c.19.1. is equal to HD, and therefore greater than HA,
Confequently, the point H is without the curve.
$ 4. I.
CI, C. I.
COR.
}
A
Book I. CONIC SECTIONS..
193
COR. I. Hence the method of drawing a tangent
from any point in the curve.
COR. 2. If a ſtraight line be drawn from the focus,
to any point in the directrix, the perpendicular which
bifects it will touch the parabola; alſo, every perpen-
dicular to it, which cuts the curve, will be nearer to
the focus than to the point in the directrix.
COR. 3. The parabola is concave towards the
axis.
PROFOSITION IV.
Every ſtraight line, drawn through the focus of a para-
bola, except the axis, meets the curve in two points.
Let FQ be a line paffing through the focus, FD Fig. 3.
perpendicular to it, DA, DE each equal to DF; AG,
•
EH parallel to CF, and interfected by FQ, in G, H³; 2.c.16.1.
and let the points AF be joined. Then, becauſe DA
is equal to DF, the angles DAF, DFA are equal;
and thefe being taken from the right angles DAG,
DFG, the remainders, the angles GAF, GFA are e-
qual: Whence the fides GA, GF are alfo equal, and
therefore G a point in the curve. In the fame man- I. c. I.
ner, it may be fhewn that H is a point in the curve.
COR. A ftraight line, making an indefinitely finall
angle with the axis, being produced, meets the curve;
hence the rate of divergency muft be very fmall.
B b
PROPO..
5
{
194
Book I.
CONIC SECTIONS.
Fig. 4.
!
PROPOSITION V.
If, from any point in a parabola, a straight line be drawn,
not parallel to a diameter, nor bifecting the angle form-
ed by two lines drawn from that point, the one to the
focus, and the other perpendicular to the directrix, it
will meet the curve in one other point, and not in more
than one.
Let H be a point in the curve, and HG a line not
parallel to CF, nor bifecting the angle EHF; then
HG fhall meet the curve in another point.
For, let FL be perpendicular to GH, and it will
meet AB, fince GH is not parallel to CF; alſo, the
point D, in which it meets AB, will be different from
E; fince, if D, E were the fame, it would follow that
HG bifects the angle EHF, contrary to the hypo-
thefis.
Let DA DE, and AG parallel to EH, meet HG
in G, G will be a point in the curve. About the
center H, with the radius HE or HF, let a circle be
defcribed interfecing FD in K, and let another circle
be defcribed through the three points A, K, F. Then,
becauſe AB touches the circle EKF in E, the rec-
1.c.17.3 tangle FDK = DE *° = DA²; hence DA is a tan-
gent to the circle AKF, and therefore AG paffes
z, c. ib.
C.
I
1113 through its center; but HL, which bifects the chord
FK,
BOOK I. CONIC SECTIONS.
195
"
I.
FK, at right angles. alfo paffes through its center; 1. c. 2.3,
confequently G is the center of the circle AKF.
Whence GA = GF, and G a point in the parabola §. £ 1. c.
If GH were to meet the curve in another point, that
point would be the center of a circle paffing through
F, K, and touching the line AB in a point different
from A, E, which is evidently impoffible.
COR. 1. If a straight line be drawn through any
point within the parabola, not parallel to a diameter,
it will meet the curve in two points.
COR. 2. At the fame point, in the curve, there
cannot be more than one tangent.
COR. 3. A tangent bifects the angle formed by a
ſtraight line drawn to the focus, and another perpen-
dicular to the directrix from the point of contact; it
alfo bifects, and is perpendicular to the line that ſub-
tends that angle.
COR. 4. If a ſtraight line from the focus be perpen
dicular to a chord, it will biſect that part of the di-
rectrix which is intercepted by perpendiculars falling
upon it from the extremities of the chord; and con-
verſely. If FL be perpendicular to GH, it will bi-
fect AE. For the circle EKF being defcribed about
the center H, FL = LK, hence GK GF, but GF =
ཟ
GA; therefore G the center of the circle AKF, and
DA² = FDK = DE ²,
PROPO
"
196 CONIC SECTIONS.
Book I.
Fig. 5.
23. c. 5.
b 4. c. 5.
1.c.2.5.
PROPOSITION VI.
Aftraight line terminated by the parabola, and parallel
to a tangent, is an ordinate to the diameter that paffes
through the point of contact.
The chord GH, parallel to the tangent MQ, will
be biſected by the diameter MN.
For, let GA, HE be perpendicular to AB, and let
FD meet MQ, GH in Q, L. Then, becauſe FD is
perpendicular to MQ, it is alfo perpendicular to
GH, and therefore bifects AE in D. Whence, fince
AG, DN, EH, are parallel, GH is bifected in N.
COR. 1. An ordinate to any diameter is parallel to
the tangent at the vertex of that diameter. For, if
GN NH, then AD = DE; therefore DF
therefore DF perpen-
=
dicular to GH, and GH parallel to MQ.
COR. 2. The ftraight line which bifects two paral-
lel chords is a diameter.
PROPOSITION
VII.
The fquare of a femi-ordinate, to any diameter is equal to
the rectangle under the parameter of that diameter, and
the correfponding abfcifs.
CASE
Book I.
197
CONIC SECTIONS.
CASE I.
then GN
Let GNH be an ordinate to the axis; Fig. 6.
=4CV ×
4CV x VN.
= ª
For, GN=GF-FN2a: But GF GA CN; c. 9. 2.
therefore GN-CN-FN-CN + FN x CN-FNb b 8.2.
= 2VN × CF, and becaufe 2 VN: VN:: 4CV : CF,
2VN x CF = 4CV x VN. Hence GN = 4CV × VN.
CASE II. Let GNH be an ordinate to any other Fig. 5.
diameter MK Then GN = 4DM × MN.
For, let the tangent MQ, the perpendiculars GA,
GK, HE, and the lines DH, HF, FD, be drawn.
Then, becauſe the right angled triangles FCD, DLN
are equiangular, and MQ, LN parallel, DF: FC::
DN: DL:: MN: LQ, therefore DF x LQ= FC ×
MN, and 2DF × LQ (= 2FC × MN) 4CV x
MN. But 2DF × LQ = DH-HF² d = DE ² =
DA²= GK. Confequently, GK² = 4CV × MN.
–
2
Again, from the ſimilar triangles DFC, DQM, we
have CF FD: DQ: DM, therefore CV: DQ::
DQ: DM, and CV: DM:: DQ * : DM ² f.
2
3.c. 5.5°
d II. 2.
e
1. 4.
But, f3. c.10.5.
from the fimilar triangles GKN, DQM, GK: GN *
2
2
:: DQ: DM² 5. Whence GK: GN :: CV: DM- 85. c. 9.5ª
:: 4CV × MN : 4DM × MN. But GK = 4CV ×
2
MN, confequently GN 4DM × MN.
=
2
COR. I. The fquare of a perpendicular, upon any
diameter, from a point in the curve, is equal to the
rectangle under the parameter of the axis, and the
abfcifs correſponding to the ordinate from the fame
point.
COR.
ļ
198
CONIC SECTIONS. Book I.
COR. 2. If there be two diameters, and from the
vertex of each a femi-ordinate be applied to the other,
the abfciffes will be equal.
COR. 3. The fquare of that part of a tangent, be-
tween the point of contact and any diameter, is equal
to the rectangle under the external fegment of that
diameter, and the parameter of the diameter which
paffes through the point of contact.
COR. 4. The fquares of ordinates, or femi-ordi-
nates, to any diameter, are to one another as their
correſponding abfciffes; and the ſquares of perpendi-
culars, from the fame points, are in the fame ratio.
COR. 5. If, from points in the curve, two tangents
be drawn to meet, their ſquares will be to each other
as the parameters of the diameters paffing through
the points of contact.
COR. 6. If the fquares of parallel lines, drawn from
certain points, to meet a line given in pofition, be to
one another as the parts they cut off towards one ex-
tremity, theſe points will be in the curve of a para-
bola, which has the given line for a diameter.
}
:
PROPO
Book I. CONIC SECTION S.
199
PROPOSITION VIII.
Afubtangent, upon any diameter, is bifected in the vertex
of that diameter.
Let the tangent GM meet any diameter VN in M, Fig. 6,
and let GN be an ordinate to it, from the point of
contact, the ſubtangent MN is bifected in V.
For, let the diameter GL, and its femi-ordinate.
b
VL be drawn, then is the abfcifs GL = VN; but, a 2. c. 7.
fince LM is a parallelogram. GL = MV, therefore 1. c. 6.
MV = VN.
COR. 1. That part of the axis, between the focus
and any tangent, is of the parameter of the diameter
paffing through the point of contact.
COR. 2. If MV = VN, and GN a femi-ordinate,
then GM is a tangent, or, if GM be a tangent, GN is
a femi-ordinate.
!
PROPOSITION IX.
That ordinate of a diameter which paffes through the
focus is equal to its parameter.
Let GE be any diameter, and RE the femi-ordinate Fig. 6.
to it, which paffes through the focus; then RE
4GD.
For,
200
Book I.
CONIC SECTIONS.
a
27. c.
5 I. c. 8.
For RE 2
2
4GD × GEª; but GE = FM = GD ':
Therefore RE = 4GD, RE = 2GD, and 2RE =
4GD.
COR. If an ordinate to any diameter paſs through
the focus, the abfcifs will be equal to the diſtance of
the vertex from the focus.
Fig. 7.
b
7.
v 8. 2.
PROPOSITION
X.
If, from any point in the parabola, a parallel to a diameter
be drawn to meet an ordinate to the fame, the rectangle
under the parameter of the diameter and the parallel
will be equal to the rectangle under the fegments of the
ordinate.
From any point H, in the curve, let HE be drawn
parallel to the diameter VN, to meet its ordinate GQ
in E; then P x HE = GE x EQ.
For, let the femi-ordinate HK be drawn ; then
2
2
GN * = P x VN, and HK P × VK ¹,
therefore
GN-HK' = P x KN, or GN +HK × GN-HK b
= PX HE, that is PX HE=GE × EQ.
COR. 1. The parameter of any diameter is to the
fum of two femi-ordinates as their difference to the
difference of their abfciffes.
COR. 2. Straight lines, drawn parallel to a diameter,
from points in the curve, to meet any chord, are to
one
P
G
D
V
Fic.1.
E
2
J
D
E
D
H
L
I
CONIC SECTIONS
B
A D
C
PL.I. po.200.
B
୯
下​:
B
FIo.3.
M
FIG. 5.
K
172
H
G
FIG. 2.
A
D
E
C
B
K
L
V
H
Fig. 4.
M
A
D
B
R
L
E
N
FIG. 6.
H
B
1
Book I.
201
CONIC SECTIONS.
one another as the rectangles under the fegments of
the chord.
COR. 3. If two parallel chords meet any diameter,
the rectangles under their fegments will be to each
other directly as the parts of the diameter which they
intercept from the vertex.
PROPOSITION XI.
If a diameter be cut by any straight line paſſing through
two points in the parabola, the part intercepted from
the vertex will be a mean proportional between the ab-
fciffes correfponding to the two ordinates drawn from
the fame points in the curve.
?
Let G, H be two points in the curve, and the line Fig. 7. 8.
GH cut the diameter VN in O; alfo, let the femi-
ordinates GN, HK be drawn; then VN: VO:: VO
: VK.
For VN VK (:: GN: HK:: GO: HO*)::
ON: OK; therefore VN: VO :: VO : VK °.
COR. If two tangents, and the line joining their
points of contact, meet the fame diameter, the feg-
ment of the diameter, intercepted by this line from
the vertex, will be a mean proportional between thoſe
intercepted by the tangents.
CC
PROPO.
ª
4. c.7.
d 16. 5.
cor. ib.
202
Book I.
CONIC SECTIONS.
Fig. 9.
9.
b 10.
PROPOSITION XII.
Any two chords which interfect each other in the focus of
a parabola, are to one another directly as the rectangles
under their fegments.
Let the chords GH and PQ interfect one another
in the focus F, then GH: PQ:: GFH: PFQ.
For, fince GH and PQ are equal to the parameters
of the diameters to which they are ordinates, GH
× VF = GFH ®, and PQ × VF = PFQ; conſequently,
GH: PQ:: GFH : PFQ.
COR. If two chords interfect in any point, the rec-
tangles under their fegments are to one another as
the parameters of thofe diameters to which they are
ordinates.
APPENDI X.
PROP. 1. Any ſtraight line, drawn from the focus
of a parabola to a point in the directrix, is a mean
proportional between half the parameters of the dia-
meters which paſs through its extremities.
PROP. 2. If, from any point in the parabola, a
tangent, femi-ordinate, and perpendicular, be drawn
to
Book I.
203
CONIC SECTIONS.
to meet the ſame diameter, their ſquares will be to one
another as the parameters of three diameters, that
which paffes through the point of contact, that which
they meet, and the axis.
PROP. 3. If, through the focus of a parabola, a fe-
mi-ordinate be applied to the axis, and from its extre-
mity a tangent be drawn to meet another femi-ordi-
nate produced; then fhall the produced femi-ordinate
be equal to the line joining its extremity in the curve
and the focus.
PROP. 4. If a tangent be drawn from any point in
the curve, to meet the axis produced, and from the
point of contact a perpendicular to the tangent, and a
ſemi-ordinate to the axis, be drawn ; then the ſeg-
ment of the axis between the perpendicular and femi-
ordinate will be equal to half the parameter of the
axis, and the ſegment between the perpendicular and
tangent equal to half the parameter of the diameter
which paffes through the point of contact.
PROP. 5. To find the directrix and focus of a pai
rabola given in pofition.
PROP. 6. If, from the vertex of any diameter, a
ftraight line be drawn to the extremity of a femi-
ordinate meeting another femi-ordinate, the latter
will be a mean proportional between its fegments
next the diameter and the former. ·
i
PROP. 7.
A diameter of a parabola, the tangent
at the vertex of that diameter, and a point in the
curve,
204
CONIC SECTIONS. Book I.
curve, being given, to find the directrix and fo-
cus.
PROP. 8. If, from any point in a tangent, a parallel
to the diameter paffing through the point of contac
be drawn to meet an ordinate of the fame, the rec-
tangle under the parallel, and its external fegment,
will be equal to the fquare of that part of it that is
between the curve, and the line joining the vertex
of the diameter, and either extremity of its ordi-
nate.
PROP. 9. The focus and directrix of a parabola
being given, to draw a tangent to the curve parallel to
a line given in pofition that is not perpendicular to
the directrix.
PROP. 10. If there be two tangents to a parabola,
fuch, that the ſtraight line joining their points of con-
tact paſs through the focus, they will cut each other
at right angles, their interfection will be in the di-
rectrix, and the line joining it with the focus will be
perpendicular to the line joining the points of con-
tact.
PROP. 11. From a given point, in a given para-
bola, to draw a tangent without finding the focus.
PROF. 12. Two parabolas of equal parameters, ha-
ving their axes in the fame line, and in the fame di-
rection, but their vertices at different points, being
produced indefinitely, continually approach, but ne-
ver meet.
PROP.
BOOK I.
205
CONIC SECTIONS.
PROP. 13. In a given parabola to find a diameter
that makes a given angle with its ordinates.
PROP. 14. If, from any point in a tangent, a
ſtraight line be drawn, to meet the curve and the di-
ameter paffing through the point of contact, the ſquare
of its fegment, between the tangent and the diame-
ter, will be equal to the rectangle under its fegments
between the tangent and the points in the curve.
PROP. 15. The focus and directrix of a parabola
being given, to draw a tangent to the curve from a
given point without it.
PROP. 16. If, from a point in a parabola, a ſemi-
ordinate be applied to a diameter, and from the ſame
point any other ſtraight line be drawn to meet it, the
fquare of this line will be equal to the rectangle un-
der the abſciſs of that diameter, and the parameter of
the diameter to which the ftraight line, when produ-
ced, is an ordinate.
CONIC
CONIC SECTIONS.
2
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.
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воок
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DEFINITIONS.
ET there be two given points, and, in the
fame plane, let a point move around them
in fuch a manner, that the fum of its diftances from
them may be always the fame, it will defcribe a curve
called an ellipfe.
2. The given points are named the foci, and the
middle of the line that joins them, the center of the
ellipfe.
3. The diſtance of the center from one of the foci
is called the excentricity.
4. A diameter is a ftraight line drawn through the
center, and terminated on both fides by the curve.
5. The
BOOK II.
207
CONIC SECTIONS.
5. The diameter which paffes through the foci is
named the tranfverfe axis, and that which is perpen-
dicular to it, the conjugate axis.
6. An ordinate to a diameter is a ſtraight line
not paffing through the center, but terminated by the
curve, and bifected by the diameter.
7. Two diameters are faid to be conjugate to one
another when each is parallel to the ordinates of the
other.
8. The parameter of a diameter is a third pro-
portional to that diameter and its conjugate.
PROPOSITION I.
'
Ïf, from any point in an eilipſe, ſtraight lines be drawn to
the foci, their fum is equal to the tranfverfe axis.
Let EFGH be an ellipfe, of which EF is the tranf- Fig. 10.
verſe, and GH the conjugate axis, A, B the foci, and
D any point in the curve; then AD + DB = EF.
a
For EA+EB = FA+FB'; hence AE = BF, and def. 1.
AD + DB = EA + EB² = EF.
COR. 1. If two ftraight lines be drawn to the foci,
from a point without the ellipfe, their fum is greater
than the tranfverfe; but, if from a point within it,
lefs.
COR.
'
208
BOOK II.
CONIC SECTIONS.
COR. 2. A point is without, in, or within the curve,
according as the fum of the lines drawn to the foci is
greater, equal, or lefs, than the tranſverſe axis.
COR. 3. The distance of either extremity of the
conjugate axis, from one of the foci, is equal to half
the tranfverſe.
COR. 4. The tranfverfe and conjugate axis are bi-
fected in the center.
COR. 5. A perpendicular to the tranfverfe at one
of its extremities is a tangent to the ellipfe.
COR. 6. The fquare of half the conjugate axis is
equal to the rectangle under the fegments into which
the tranſverſe is divided in one of the foci.
COR. 7. The diſtance of the foci is a mean propor-
tional between the fum and difference of the tranſverſe
and conjugate axis.
1
PROPOSITION
II.
Fig. 11.
The Straight line which bifects the angle adjacent to that
which is contained by two straight lines, drawn from
any point in the ellipfè to the foci, is a tangent to the
curve in that point.
Let D be any point in the curve, from which AD,
DB are drawn to the foci, and let the angle BDI ad-
jacent to ADB, be bifected by the line DT; then is
DT a tangent to the ellipfe in the point D.
For,
Book II. CONIC 209
CONIC
SECTIONS.
For, let any other point R be affumed in DT, and
DI being made equal to DB, let BI, RA, RB, and
RI be drawn. The line DNT, which bifects the ver-
tical angle of the ifofceles triangle BDI, alfo bifects
the baſe BI at right angles. Hence RB RI, and ª 4. 1.
AR + RB = ARRI, and therefore greater than
AI or EF. Confequently the point R is without
the ellipfe d.
ΑΙ
b
COR. I. A perpendicular to the conjugatė axis, at
one of its extremities, is a tangent to the ellipſe.
COR. 2. The method of drawing a tangent from a
given point in the curve, alfo, of drawing a tangent
parallel to a line given in pofition, is evident.
COR. 3. There cannot be more than one tangent
to the ellipfe at the fame point. For D is the point
in the line DT, the fum of whofe diitances from A,
B is the leaft poffible: If any other line were drawn
through D, a point (different from D) may be found
in it, fuch that the fum of its diftances from A, B
fhould be the leaſt poffible, and confequently leſs than
AD +DB; whence that point would be within the
ellipfe, and the line paffing through it would cut, and
not touch the curve.
COR. 4. Every tangent bifes the angle adjacent
to that contained by ftraight lines drawn to the foci
from the point of contact, or, which is the fame, theſe
lines make equal angles with the tangent.
D d
COR.
a
b
20. I.
C I.
₫ 2. c. I
210
BOOK II.
CONIC SECTIONS.
2
{
1
COR. 5. A ftraight line drawn from the center to
meet a tangent, and parallel to the line joining the
point of contact, and one of the foci, is equal to half
the tranfverfe axis.
For, fince BC CA, and BN NI, the line that
joins CN will be parallel to AD, and equal to the
half of AI.
COR. 6. A perpendicular to a tangent, from one
of the foci, and a parallel to the line joining the other
focus, and the point of contact from the center, meet
the tangent in the fame point.
COR. 7. If a chord pafs through one of the foci,
and the tangents at its extremities be produced to
meet, the ſtraight line that joins the point of con-
courfe and the focus, will be perpendicular to the
chord.
Let the tangents at the extremes of the chord
DBO, paffing through the focus, meet in the point
T, and let TI be perpendicular to AD; then, becauſe
DT, TO bifest the exterior angles of the triangle
1
2.c.19.3. ADO, Alis equal to half its perimeter AD + DB.
& 2. c. 13.
3.
Hence DB = DI, and TDI, TDB equal in every re-
fpect.
PROPO
Book II. CONIC SECTION S. 211
PROPOSITION III.
From any point in an ellipfe, a perpendicular being let fall
upon the tranfverfe, and straight lines drawn to the
foci, as half the tranfverfe is to the excentricity, fo is
the distance of the center from the perpendicular to
half the difference of the straight lines drawn to the
foci.
Let D be the point in the curve, from which DA, Fig. 12.
DB are drawn to the foci, and DK perpendicular to
EF, then CF : CB :: CK:
AD — DB.
2
For, let EL AD, then LF = DB, and 2CL = AD
— DBª. But AD + DB × AD — DB = AB × 2CK³,
or EF × 2CL = ABX 2CK, therefore EF: AB::2CK
: 2CL, and CF: CB:: CK: CL.
5. 2.
b II. 2.
PROPOSITION
IV.
Any two chords which paſs through one of the foci of an
ellipfe, are to one another directly as the rectangles
under their ſegments.
Let DO, MI be the two chords paffing through the Fig. 13.
focus B, DO: MI : : DB × BO : MB × BI.
For,
;
1.
212
Book II.
CONIC SECTIONS.
@5.c.2.
6 3.
C
7. C. 2.
For, let the tangents at D, O meet each other in
R, through Clet NQ be drawn parallel to DO,
meeting the tangents in N, Q, and let RB meet NQ
in P. It has been proved that CN or CQ CF;
therefore NO = EF; that CF: CB :: CK: CL',
therefore CFX CL CB x CK, and that DBR, or
=
DEP is a right angle, therefore the triangles DBK,
BCP, are fimilar, and DB x CP = BC x BK.
BCX BK.
Hence
d 1. c. 6. 2. CF2 = CFX CL + CF x LF = BC x CK + DBxCQ
c. 8. z.
2 d
= BC × CK + BC × BK + BD × PQ = BC ² + DB
£ 3 c. 4.5. x PQ, and DB
f
لام
9. 4.
a def. 8.
b b. c. I.
²
2
× PQ = CF 2 — BC ² - EB × BF.
But DO: NQ:: BO: PQ : : DB × BO: DB × PQ:
Therefore
DO: EF:: DB× BO: EBX BF
In like manner EF: MI :: EB X BF : MI × BI
Confequently
DO: MI :: DBX BO: MB × BI g
COR. The rectangle under any chord paffing thro'
the focus and the parameter of the tranfverfe, is equal
to four times the rectangle under its fegments.
For, fince EF: DO:: EBx BF: DB x BO, if P
be the parameter of the tranfverfe, EF × P: DO × P
:: 4EB × BF: 4DB x BL. But EF x P = * GH * *
= 4EB × BF°; therefore DL × P = 4DB × BO ‘.
b
€ 6. 4.
PROPO.
* That EF, GH are conjugate diameters, will be proved
independent of this Cor.
i
E
G
R
H
K
1
FIG. 7.
E
N
т
P
H
FIG. 9.
FIG.11.
A
୯
B
I
CONIC
SECTIONS
▼
G
H
K
IN
FIG. 8.
FIG. 10.
G
P
D
୧
E
A
IC
B
Ο
T
H
FIG. 12.
Ꮐ
D
PL.II.pa. 22.
F
E
A
C
I
K
B..
H
BOOK II. CONIC SECTIONS.
213
PROPOSITION V.
If a tangent to the ellipfe meet either axis produced, and
a perpendicular to the axis be drawn from the point of
contact, then ſhall the femi-axis be a mean proportional
between its fegments intercepted from the center by the
tangent and perpendicular.
CASE I. Let the tangent DT, and the perpendicu- Fig. 14.
lar DK, meet the tranfverfe in T, K; then CT : CF
:: CF: CK.
For, fince DT bifects the angle BDI, AT: TB::
AT+TB AT — TB
AD: DBª, and
:
,
2
2
AD — DB ¹.
2
That is,
CT : CB :: CF : CL
But
CB: CF :: CL: CK©
Therefore
:
CT CF CF: CK d
AD + DB
a
3. 5.
2
b
CASE II. Let the tangent DT, and the perpendicu-
lar DN, meet the conjugate in M, N; then MC: CG
:: CG: CN.
For, let DO, perpendicular to DT, meet the tranf
verſe in O, then DO evidently biſects the angle ADB,
therefore AO:OB:: AD:DB¹, hence AT; TB::
AO OB, and CT: CB:: CB: CO, therefore
CT
12. 4.
& I. 4.
c 3.
d
9. 4.
214
Book II.
CONIC SECTIONS.
* I. c. 12.
5.& 4.5.
CT × CO = CB ', but CT × CK = CF ², conſequently
CT x
KO CF CB = CG².
- ?
Again, from the fimilar triangles CTM, KDO,
CT × KO = CM × DK = CM x CN, therefore CM
× CN = CG², and CM: CG :: CG: CN.
COR. 1. The rectangles under the fegments of the
axis, and the fegments of the femiaxis produced, both
being divided by the perpendicular, are equal.
For EKF CF-CK=TCK - CK² = CKT.
-
COR. 2. The fegments into which the axis is di-
vided by the tangent, are directly proportional to the
fegments into which it is divided by the perpendicu-
lar.
For, fince CT: CF::CF: CK, CT + CF : CT-
CF :: CF + CK: CF-CK, that is, ET: TF: ÉK
: KF.
COR. 3. The fegments CT, CB, CO, are propor-
tional; and, in like manner, it may be proved that
CK, CL, CO, are proportional.
PROPOSITION
VI.
If, from any point in an ellipfe, a perpendicular fall upon
either axis, as the fquare of that axis is to the fquare
of the other, fo is the rectangle under the fegments of
the former, to the fquare of the perpendicular.
Let
Book II.
215
CONIC SECTIONS.
:
Let the perpendicular DK fall from any point in Fig. 16.
the curve upon EF, then EF*: GH²:: EKF : DK³.
a
For, fince CT CF CF: CK, CF: CK²:: ª 5.
CT: CK, and, by converſion, CF²: EKF :: CT : TK. b3.c.10.5.
But CT TK:: CG: CN, or DK2, therefore
CF*: EKF :: CG: DK2, and alternately CF: CG *
Z
:
:: EKF: DK. Confequently EF GH:: EKF
I.
: DK. If the perpendicular DN fall upon GH, it 1. 4.
may be proved in the fame manner that GH: EF ²
::GNH:DN
COR. 1. The fquares of perpendiculars, falling from
points in the curve, upon either axis, are to one ano-
ther as the rectangles under the fegments into which
they divide the axis.
COR. 2. If a circle be defcribed upon the tranfverfe, Fig. 15.
and another upon the conjugate axis, the former will
contain, and the latter will be contained in the ellipſe,
fo that the curve lines fhall touch one another only
in the extremities of their common diameter.
COR. 3. The two axes are the greateſt and leaſt
diameters of the ellipfe.
COR. 4. If a circle be defcribed upon either axis of
an ellipfe, perpendiculars to the common diameter
are cut proportionally by the curves.
COR. 5. Every ftraight line terminating in the
curve, and parallel to one axis, is an ordinate to the
other; and converfely.
COR.
!
216
Book II.
CONIC SECTIONS.
:
Fig. 23.
i
COR. 6. The two axes are conjugate diameters.
COR. 7. The ordinate through the focus is the pa-
rameter of the tranfverfe.
COR. 8. Equal ordinates to either axis are equally
diſtant from the center; the greater ordinate is nearer
the center; and converfely.
COR. 9. If, from any point in the curve, a ſtraight
line be drawn to the conjugate axis, equal to half the
tranfverfe, its fegment intercepted from the fame
point by the tranfverfe, will be equal to half the con-
jugate; and converfely.
PROPOSITION VII.
A ftraight line, not paſſing through the center, but termi-
nated by the ellipfe, and parallel to a tangent, is an or-
dinate to the diameter that paffes through the point of
contact.
Fig. 17.
The chord MN, parallel to the tangent at D, is an
ordinate to the diameter CD.
For, let MN meet CT in L, and from the points
M, N, let MI, NO be drawn perpendicular to EF,
MR, NQ parallel to DC. Then, by fimilar tri-
angles, MI: DK :: RI: CK.
And MI: DK :: IL: KT.
Hence
!
Book II. CONIC SECTIONS.
217
Hence a
But
Therefore
And
By compofition
By alternation and
converfion
In like manner
Hence d
And
By divifion
EIF: EKFb:: RIL: CKT
EKF
CKT ©
C
EIF RIL
=
EI: IL:: RI: IF
EL: IL: RF : IF
EL: ER+ FL:: IL: FL
EL: EQ+FL:: OL: FL
a
4. C. 9.5.
b I. c. 6.
C
I. C. 5.
IL: OL:: EQ + FL: ER+FL d2. c. 9.4.
RL : QL :: EQ +FL : ER + FL
RQ:QL:: RQ:ER + FL.
Confequently QL = ER + FL, ER = FQ, CR = CQ₂
and MP PN.
COR. 1. Every ordinate to a diameter is parallel to
the tangent at its vertex.
For, if not, let a tangent be drawn parallel to it,
then the diameter through the point of contact would
bifect it; and thus the fame line would be bifected in
two different points, which is abfurd.
COR. 2. All the ordinates to the fame diameter are
parallel to one another.
COR. 3. A ftraight line, drawn through the vertex
of a diameter, parallel to its ordinate, is a tangent to
the curve in that point.
COR. 4. The diameter which bifects one of two
parallel chords, will alfo bifect the other.
COR. 5. The ſtraight line which bifects two parallel
chords, and terminates in the curve, is a diameter.
E e
COR.
i
218
CONIC SECTION S. Book II.
n
COR. 6. If two tangents be at the vertices of the
fame diameter, they are parallel; and converſely.
COR. 7. Every diameter divides the ellipfe into
two equal parts.
PROPOSITION VIII.
Fig. 18.
ㄖ
​.0
b
I. c. 6.
1. C. 5.
* I. C. 7.
Two diameters, one of which is parallel to the tangent,
at the vertex of the other, are conjugate to one another.
Let the diameter PQ be parallel to the tangent DT,
at the vertex of the diameter DI; then fhall PQ, DI
be conjugate diameters.
For, let DK, QN be perpendicular to EF, and let
the tangents at D, Q_meet EF in TR. The triangles
TDK, CQN, by reafon of parallel lines, are equiangu-
lar, therefore TK; CN :: DK : QN; hence TK²:
CN¹ :: DK: ON:: EKF: ENF:: CKT: RNC,
and TK: CN :: CK: RN, or DK: QN:: CK: RN.
Conſequently the triangles DKC, RNQ, are equian-
gular, and DC parallel to RQ. Whence PQ, DI are
each parallel to the ordinates of the other, that is,
they are conjugate diameters.
ON
COR. 1. If, from the extremities of two conjugate
femi-diameters, perpendiculars be let fall upon either
axis, the rectangle under the fegments, into which
the axis is divided by one of the perpendiculars, is
equal
:
E
A
D
CLK
B
FIG. 13.
P
FIG. 15.
M
R
G
N
I
P
CONIC SECTIONS
R
M
M
Ꮐ
N
E
C
F
K
IL
:
H
I
FIG. 1.
M
P
PL.W.p.218,
F
T
A
C
O L
K B
F
FIG. 14.
H
M
Ꮐ
N
FIG. 16.
E
T
C
K
F
FIG. 18.
I
H
D
R
N
C
K
N
E
T
R
C I
K Q
OF
L
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P
•T
Book II.
219
CONIC SECTIONS.
equal to the fquare of the fegment, which the other
intercepts from the center.
For, fince RQ is parallel to CD, CT:TK::CR
: CN, and CT × CK : CK X KT:: CR x CN: CN².
But CT × CK CF² = CR × CN, therefore CK × KT,
or EK x KF CN".
X
=
COR. 2. Alfo, the fum of the fquare of the feg-
ments, intercepted from the center, is equal to the
fquare of half the axis upon which the perpendiculars
fall; and the fum of the fquares of the perpendiculars
is equal to the ſquare of half the other axis.
COR. 3. The fum of the fquares of any two con-
jugate diameters is equal to the fum of the fquares of
the tranſverſe and conjugate axis.
PROPOSITION IX.
Any femi-diameter which meets an ordinate and tangent
drawn from the fame point in the ellipfe, is a mean
proportional between the fegments intercepted from the
center.
CASE I. Let the femi-diameter CN meet a tangent Fig. 19.
and ordinate drawn from F, one extremity of either
axis in M, P: Then CM: CN :: CN: CP.
For, let the tangent and ordinate from N meet the
aixs EF in T, K. It has already been demonſtrated,
with
220
Book II.
CONIC SECTIONS.
5.
b
with reſpect to either axis, that CT: CF :: CF: CKª;
hence CM: CN:: CF: CK :: CT: CF :: CN : CP.
CASE II. Let the femidiameter CN meet a tangent
and ordinate from any other point D in M, P,
CM: CN: CN: CP.
For, let DP, the tangent at N, DM, and NR paral-
lel to DM, meet the axis EF in L, Q, R, T. Alſo,
let DK, NO, be the femi-ordinates to EF from D, N.
The lines DK, DT, DL, are reſpectively parallel to
NO, NR, NO.
2. c. 6. Hence CO×OQ:CK×KT(::'NO': DK')::OQ :: KL*
By alternation CO: OQ:: CK × KT : KL'
& I. c. 5.
© 5.
But
Therefore
OQ: OR: KL: KT
CO: OR: CK: KL
By conversion CO: CR:: CK: CL
e
Again
X
CT × CK = CF ** = CQ × CO°
Therefore
CT : CO :: CQ : CK
Hence
CT: CR :: CQ: CL
And
Fig.22.23.
CM: CN:: CN: CP
COR. I. Tangents, at the extremities of an ordi-
nate, meet its diameter produced in the fame point.
COR. 2. A ftraight line, drawn through the cen-
ter, from the point of interſection of two tangents, bi-
fects the line that joins the points of contact.
COR. 3. If a circle be defcribed upon any diameter
of an ellipfe, and through any point of it an ordinate
of
Book II. CONIC SECTIONS.
221
of each curve be drawn, tangents, at their extremes,
will meet one another in the common diameter pro-
duced.
COR. 4. The fegments into which any diameter is
divided by a tangent, are directly proportional to the
fegments into which it is divided by an ordinate from
the point of contact.
COR. 5. If a femi-diameter be produced to meet a
tangent, the rectangle under the fegments into which
it is divided, by an ordinate from the point of con-
tact, is equal to the rectangle under the fegments,
into which the whole diameter is divided by the fame
ordinate.
PROPOSITION X.
21.
If a femi-ordinate be applied to any diameter of an ellipfe,
the fquare of that diameter is to the fquare of its con-
jugate, as the rectangle under its ſegments to the ſquare
of the femi-ordinate.
Let PQ, DI be two conjugate diameters, and MN
any femi-ordinate to DI; DI: PQ :: DNI : MN *.
For, let the tangent at M meet DI, PQ in R, O,
and let ML be a femi-ordinate to PQ. Then, becauſe
Fig.
CR
.
222
CONIC SECTIONS. Book II.
!
. 9.
b II. 4.
CR: CD :: CD: CN³, CD² : CN² :: CR: CN, and
CD: DNI:: CR: RN :: OC: CL:: CQ: CL2.
¹ ²
Hence CD: CQ:: DNI ; MN, and DI: PQ'
:: DNI: MN 2.
COR. 1. The fquares of femi-ordinates are to one
another as the rectangles under the fegments of the
diameter.
COR. 2. Any diameter of an ellipfe is to its para-
meter as the rectangle under its fegments to the
fquare of the femi-ordinate that divides them.
Fig. 24.
5.
3.
PROPOSITION XI.
:
If, through any point in the ellipfe, a parallel to the con-
jugate axis meet the tranfverfe, and the focal tangent,
its ſegment between theſe will be equal to the distance
of that point from the focus.
At the focus B, let BD be a femi-ordinate to the
tranfverfe; then DT, touching the ellipfe in D, is na-
med the focal tangent. Through any point R in the
curve, let RQ, parallel to GC, meet DT, EF in P, Q;
PQ is equal to RB.
For, let FLRB.
Then, becauſe CT: CF :: CF: CB² :: CQ : CL
Alternately CT: CQ:: CF CL
By converſion CT : TQ : : CF : FL
b
Hence
Book II. CONIC SECTIONS. 223
Hence MC: PQ : : CF : RB.
But MC = CF, confequently PQ = RB.
ÇOR. 1. A perpendicular to the tranfverfe, at either
extremity, limited by the focal tangent, is equal to
the diſtance of that extremity from the focus.
COR. 2. The distance of any point in the curve,
from the perpendicular to the tranſverſe, at the point
where it interſects the focal tangent, is to its diſtance
from the focus as half the tranfverfe is to the excen-
tricity.
c 5. c. 2.
PROPOSITION XII.
If four ſtraight lines touch an ellipfe at the vertices of two
conjugate diameters, the parallelogram which they con-
tain will be equal to the rectangle under the tranfverſe
and conjugate axis.
a
Let DI, PQ be two conjugate diameters, and LM, Fig. 25.
MN, NO, OL tangents at their vertices. Then LMNO
is a parallelogram, divided into four equal parallelo- 6. c. 7.
grams by PQ, DI; and the whole MO is equal to the
rectangle under EF and GH, or the part CN= CF
× CG.
For, let DK, PR be perpendicular to EF, and let
EF meet the tangent DN in T, and TP be joined.
Then,
224
CONIC SECTIONS. Book II.
b 6.
C
c 1.c.8.5.
d
.5. c. 9.
5.
9. 4.
b
Then, fince CF2: CG:: (ERF) CK: PR *
But
Therefore
CF: CG:: CK: PR d
C
CT : CF :: CF : CK ©
CT: CG:: CF: PR f
Hence CF × CG = CT × PR = double the
triangle CTP = the parallelogram CN.
COR. 1. If perpendiculars fall from the extremities.
of two conjugate femi-diameters upon either axis, the
axis upon which they fall is to the other as the feg-
ment intercepted by one of the perpendiculars from
the center is to the other perpendicular.
COR. 2. All parallelograms whofe fides touch an
ellipfe in the vertices of two conjugate diameters, are
of the fame magnitude.
COR. 3. Every parallelogram whofe angular points
are the vertices of two conjugate diameters, is equal to
half the rectangle under the axis.
APPEND I X.
PROP. 1. The rectangle under perpendiculars to the
tranfverſe paffing through its extremities, the rectangle
under perpendiculars from the foci to any tangent,
the rectangle under perpendiculars to the tranfverfe,
from the center and the point of contact, and the rec-
tangle under perpendiculars to the tangent, from the
point
!
E
F
E
FIG. 19.
FIG. 20.
FIG. 22.
P
M
CONIC SECTIONS
L
T
C
K
F
P
M
KRF T
I
C
E
I
C
P
M
N
FIG. 24.
M
P
G
R
I
E
CL P
B
T
F
I
I
G
N
FIG. 23.
Ꮐ
FIG. 21.
M
M
PL.N.pa.224.
N
P
B
C
K
F
T
R
FIG. 25.
H
M
K
R
N
T
Book II. CONIC SECTIONS.
225
point of contact, and the center, (all the perpendicu-
lars being limited by the tranfverfe and tangent),
are each equal to the fquare of the femiconjugate
axis.
PROP. 2. If, from any point in an ellipfe, two
ftraight lines be drawn through the extremities of a
diameter, any other ftraight line parallel to the tan-
gent at that point, and terminating in thefe lines, will
be bifected by an ordinate to that diameter, from the
point of contact.
PROP. 3. If the diagonals of a quadrilateral figure,
infcribed in an ellipfe, interfect one another in one of
the foci, and two oppofite fides be produced to meet,
the ſtraight line that joins the point of concourfe and
the foci will bifect the angle formed by thefe diago-
nals.
PROP. 4. If two tangents, at the vertices of a dia-
meter, meet any third tangent, the rectangle under
the two former, and the rectangle under the fegments
of the latter, from the point of contact, are refpeively
equal to the fquares of the femi-diameters to which
the tangents are parallel.
PROP. 5. If, through the extremities of the tranf-
verfe axis, two tangents be drawn to meet a third,
the circle defcribed upon the intercepted tangent will
pafs through the foci.
Ff
PROP.
226
Book II.
CONIC SECTIONS.
PROP. 6. If, from the vertices of two conjugate
femi-diameters, two femi-ordinates be applied to any
diameter, the fquare of its fegment, between the
center and either femi-ordinate, will be equal to the
rectangle under the fegments into which it is divided
by the other femi-ordinate.
PROP. 7. If, from the vertex of any diameter,
ftraight lines be drawn to the foci, their rectangle is
equal to the fquare of its femi-conjugate.
PROP. 8. If a tangent meet two conjugate diame-
ters, the rectangle under its fegments, from the
point of contact, will be equal to the fquare of the
femi-diameter to which it is parallel.
PROP. 9. Every diameter is a mean proportional
between the ordinate to its conjugate, which paſſes
through the focus and the tranfverſe axis.
PROP. 10. If, from any point, in one tangent, a
ftraight line be drawn parallel to another, to meet
the curve in two points, the rectangle under its feg-
ments, between the tangent and the curve, is equal
to the fquare of its fegment between the tangent and
the line joining the points of contact.
PROP. 11. The rectangles under the fegments of
two chords which interfect, are to one another di-
rectly as the fquares of the diameters to which they
are parallel.
PROP.
Book II. CONIC SECTIONS.
227
PROP. 12. If, from any point, in a given ellipfe,
ftraight lines, parallel to lines given in pofition, be
drawn to meet the fides of a given quadrilateral in-
ſcribed in it, the rectangles under thoſe drawn to
oppofite fides, will have to each other a given ra-
tio.
CONIC
די
Fig. 26.
*
CONIC SECTION S.
་་་བར་་་་
воок
III.
HYPERBOL A.
DEFINITIONS.
I.
"L
ET there be two given points, and let any
point in the line which joins them, except
the middle, move in the fame plane with theſe, in
fuch a manner, that the difference of its diftances.
from the given points may be always the fame, it
will defcribe a curve, called an hyperbola. And e-
very two points that are equidiftant from the middle
defcribe two equal and fimilar curves, called oppo-
fite hyperbolas.
2. The given points are named the foci; that part
of the line that joins them, intercepted by oppofite
hyperbolas,
t
BOOK III.
229
CONIC SECTIONS.
hyperbolas, the tranfverfe axis; and the middle of
the fame line, the center.
3. The conjugate axis is a ftraight line, paffing thro'
the center, perpendicular to the tranfverfe, and limited
by a circle defcribed from one extremity of the tranf-
verſe with the diſtance of either focus from the cen-
ter as a radius.
4. If two other hyperbolas be defcribed, having
the conjugate for their tranfverfe axis, and their foci
at the fame diftance from the center as the foci of the
former, thefe are alfo called oppofite hyperbolas, and
have the tranfverfe of the former for their conjugate
axis.
5. A diameter is a ftraight line drawn through the
center, and terminated by oppoſite hyperbolas.
6. An ordinate to a diameter is a ftraight line, ter-
minated by the hyperbola, and bifested by that dia-
meter produced.
7. An external ordinate to a diameter is a ftraight
line, terminated by oppofite hyperbolas, and bifected
by that diameter, or the fame produced.
8. Two diameters are faid to be conjugate to one
another when they are mutually parallel to each o-
ther's ordinates.
9. A third proportional, to any diameter, and its
conjugate, is called its parameter.
10. An affymptote of the hyperbola is a ſtraight
line, which, being produced indefinitely, does not
meet,
230
Book II.
CONIC SECTIONS.
Fig. 27.
a def 1.
b
5. 2.
meet, but continually approaches the curve, fo as to
come within lefs than any given diftance from it.
PROPOSITION I.
If, from a point in any hyperbola, ſtraight lines be drawn
to the foci, their difference is equal to the tranfverfe
axis.
Let DF, EP be oppofite hyperbolas, of which A, B
are the foci, C the center, EF the tranfverfe axis, and
D any point in one of the curves; then AD - DB
= EF.
DB AF - FB
a
For AD
2CF b EF.
COR. 1. If two ftraight lines be drawn to the foci,
from a point without the oppofite hyperbolas, their
difference is leſs than the tranſverſe axis, but, if from
any point within either of them, it is greater.
Let M be a point within the hyperbola DF, and N
a point without, between the curve and its conjugate
axis. The lines AM, NB neceffarily meet the curve;
let them meet it in the points D, D; then AM-MB is
greater than AM MD DB, or EF, but AN --
+
NB lefs than AD DN
-NB, or EF.
COR. 2. A point is within, in, or without the
curve, according as the difference of its diftances from
the foci, is greater, equal, or lefs, than the tranſverſe
axis.
COR
BOOK III.
CONIC SECTIONS. 231
ter.
COR. 3. The conjugate axis is bifected in the cen-
COR. 4. The rectangle under the fegments, into
which the tranfverfe is divided in one of the foci, is
equal to the fquare of the femi-conjugate axis.
COR. 5. The fquare of the diſtance of the foci is
equal to the fum of the fquares of the tranfverfe and
conjugate axis. This, and the two preceding Corol-
laries follow from the third definition.
COR. 6. A perpendicular to the tranfverfe, at its
extremity, is a tangent to the hyperbola.
PROPOSITION II.
The Straight line which bifects the angle, contained by
two ſtraight lines, drawn from any point in the hyper-
bola to the foci, is a tangent to the curve in that point.
Let D be any point in the hyperbola DFO, from Fig. 28,
which AD, DB are drawn to the foci, and let the
angle BDI be biſected by the line DT, then is DT a
tangent to the curve in the point D.
For, let any other point R be affumed in DT, and
DI being made equal to DB, let BI, RA, RB, RI be
drawn. The line DNT, which bifects the vertical
angle of the ifofceles triangle BDI, alfo bifects the
bafe BI at right angles. Hence RB RI, and
AR
1
24
232
CONIC SECTIONS. OOK TH
AR — RB = AR RI, and therefore lets than
AI or EF. Confequently, the point R is without the
* 2. c. 1. hyperbola
I.
COR. 1. The method of drawing a tangent from a
given point in the curve, alfo, of drawing a tangent
parallel to a line given in pofition not parallel to the
tranfverfe, is evident.
COR. 2. There cannot be more than one tangent
to the hyperbola at the ſame point.
COR. 3. Every tangent bifects the angle contained
by ftraight lines drawn to the foci from the point of
contact; or, which is the fame, theſe lines make e-
qual angles with the tangent.
COR. 4. Every tangent to the fame hyperbola meets
the tranfverfe between its vertex and the center.
COR. 5. A ſtraight line drawn from the center to
meet a tangent, and parallel to the line joining the
point of contact, and one of the foci, is equal to half
the tranſverſe axis.
COR. 6. A perpendicular to a tangent, from one
of the foci, and a parallel to the line joining the other
focus, and the point of contact from the center, meet
the tangent in the fame point.
COR. 7. If a chord pafs through one of the foci,
and the tangents at its extremities be produced to
meet, the ſtraight line that joins the point of con-
courfe and the focus, will be perpendicular to the
chord.
Let
;
III CONIC SECTIONS.
233
–
Let the tangents at the extremes of the chord DBL,
paffing through the focus, meet in the point O, and
let Ol be perpendicular to AD: Then, becauſe
AD DB AL — LB, AD + LB = AL + DB =
half the perimeter of the triangle ADL = AL + DI,
becauſe O is the center of the infcribed circle. Hence
DB = DI, and the triangles DBO, DIO equal in e-
very reſpect.
PROPOSITION III.
From
any point in an hyperbola, a perpendicular being let
fall upon the tranfverfe, and straight lines drawn to
the foci; as half the tranfverfe is to the distance of
the center from the focus, fo is the distance of the cen-
ter from the perpendicular to half the fum of the
Straight lines drawn to the foci.
Let D be the point in the curve, from which DA, Fig. 27.
DB are drawn to the foci, and DK perpendicular to
EF; then CF: CB:: CK:
AD+DB.
2
For, let EFL = AD, then LF = DB, and 2CL =
AD + DB. But AD + DB × AD-DB= ABX
2CK', or EF × 2CL = AB × 2CK; therefore EF:
AB:: 2CK; 2CL, and CF : CB :: CK : CL.
G g
PROPO.
5. 2.
11. 2.
234
CONIC SECTION
PROPOSITION
IV
Fig. 29.
Any two chords which interfect each other in the focus of
an hyperbola, are to one another directly as the rec-
tangles under their fegments.
Let the chords DO, MI interfect one another in the
focus B of the hyperbola DFO, DO: MI: : DB ×
BO: MB × BI.
For, let the tangents at D, O meet each other in
R, through C let NQ be drawn parallel to DO,
meeting the tangents in N, Q, and let BR meet NQ
5. c. 2. in P. It has been proved that CN or CQ = CF¹,
b
3.
1
therefore NQ = EF; that CF: CB:: CK: CL,
therefore CF x CL CB x CK, and that DBR is
7. c. 2. a right angle, therefore the triangles DBK, BCP
are fimilar, and DB x CP =
2
CF CF Xx CLCF x FL
=
BC x BK. Hence
BC x CK-DB x
²
CQ = BC × CK - BCX BK-DB × PQ = BC --
2
X
× ².
DB × PQ, and DB PQ - BC - CF = EB × BF.
But DO: NQ :: BO: PQ :: DB× BO: DB × PQ.
Therefore DO: EF: DB x BO: EB × BF.
In like manner, EF: MI:: EB × BF : MB × BI.
Confequently
DB
DO: MI:: B x BO: MB x BI.
COR. The rectangle under any chord paffing thro'
the focus and the parameter of the tranfverfe, is equal
to four times the rectangle under its fegments.
PROPO
1
་
}
F
A
E
P
a+
FIG. 26.
G
A E
C
F B
FIG.
Ꮐ
H
b
N
M
27.
D
A
E
C
F
B
K
L
I
R
P
FIG. 28.
R
C
T
F
B
I
N
D
PL.V. pa. 234.
FIG. 29.
}
IF
L
I
BOOK III
235
CONIC SECTIONS.
PROPOSITION V.
If a tangent meet either axis of the hyperbola, and a per-
pendicular to that axis be drawn from the point of
contact, then ſhall half the axis be a mean propor-
1
tional between its fegments intercepted from the center
by the tangent and perpendicular.
CASE I. Let the tangent DT, and perpendicular Fig. 30
DK, meet the tranfverfe in T, K, CT: CF :: CF ;
CK.
For, fince DT bifects the angle ADB, AT: TB::
AT+TB
AT-TB AD + DB
AD DB, and
2
2
2
AD
DB.
2
That is,
But
CF: CT: CK: CF
CB CT :: CL: CF.
CF : CB :: CK : CL ª
Therefore CF: CT
And
CT : CF :: CF : CK.
CASE II. Let the tangent DT, and the perpendi-
cular DN meet the conjugate in M, N, MC: CG ::
CG: CN.
For, let DO, perpendicular to DT, meet the tranf-
verſe in O, DO evidently bifects the exterior vertical
angle of the triangle ADB, therefore AO: OB ::
AD : DB :: AT: TB. Hence CB: CO :: CT : CB,
and
2.
3.
3
235
CONIC SECTIONS. BooK TH
and CT × CO = CB.
But CT x CK
CF ², con-
*
fequently CT > KO=CB - CF² = CG ².
Again, from the fimilar triangles CTM, DKO,
CT × KO = MC × DK = MC x CN; therefore
MC x CN=CG, and MC: CG:: CG: CN.
COR. 1. The rectangle under the fegments of the
tranfverfe, into which it is divided by the perpendi-
cular, is equal to the rectangle under the diſtances of
the perpendicular from the center, and the point.
where the tangent meets the tranſverſe.
COR. 2. The fegments into which either axis is
divided by the tangent are directly proportional to
the fegments into which it is divided by the perpen-
dicular,
PROPOSITION
VI.
Fig. 30.
a
a 5.
ம்
If, from any point in an hyperbola, a perpendicular fall
upon the tranfverfe axis, as the fquare of that axis is
to the fquare of the other, fo is the rectangle under the
Segments of the former, to the ſquare of the perpendicu
lar.
Let the perpendicular DK fall from any point D in
the curve, upon EF, EF: GH*:: EKF : DK ².
For, fince CT: CF:: CF: CK 2, CF *: CK 2 :: CT
: CK, and, by converfion, CF: EKF:: CT: TK. But
CK,and,
a
CT
BOOKHIL ICONIC SECTIONS.
237
1
CT: TK ::"CG: CN2, or DK, therefore CF2:
EKF :: CG': DK, and alternately CF: CG *::
EKF : DK². Confequently EF 2: GH*: EKF: DK².
COR. 1. The fquare of the conjugate axis is to the
fquare of the tranfverfe, as the fum of the fquares of
the femi-conjugate, and the diſtance of the center
from a perpendicular falling upon the conjugate, from
any point in the curve, to the fquare of that perpen-
dicular.
COR. 2. The fquares of perpendiculars, falling
from points in the curve upon its tranfverfe axis, are
to one another as the rectangles under the fegments
of the axis; and the fquares of perpendiculars upon
the conjugate, are to one another as the fums of the
fquares of the femi-conjugate, and the diſtance of
each from the center.
COR. 3. Every ſtraight line, terminating in the
hyperbola, or in oppofite hyperbolas, and parallel to
one axis, is an ordinate to the other; and converfely.
COR. 4. The two axes are conjugate diameters.
COR. 5. The ordinate through the focus is the pa-
rameter of the tranfverfe.
COR. 6. Equal ordinates to either axis are equally
diſtant from the center, the lefs ordinate is nearer
the center; and converfely.
PROPO-
¿
238
CONIC SECTION_B
}
I
Fig. 31.
A
5.
3.
PROPOSITION VII.
If, through any point in the hyperbola, a parallel to the
conjugate axis meet the tranfverfe and the focal tan-
gent, its fegment between thefe will be equal to the di-
Stance of that point from the focus.
Through any point R in the curve, let RQ, paral-
lel to GC, meet the focal tangent DE, and tranſverſe
axis EF in P, Q; then PQ = RB.
}
For, let FL =RB. Then, becaufe CT: CF ::
૧
b
CF : CB³ :: CQ : CL ³.
Alternately CT: CQ:: CF: CL
By converſion CT: TQ :: CF : FL
Hence
But
MC: PQ :: CF : RB
MC CF, confequently PQ = RB.
COR. 1. A perpendicular to the tranfverfe, at one
extremity, limited by the focal tangent, is equal tợ
the diſtance of that extremity from the focus.
COR. 2. The diſtance of any point in the curve,
from the perpendicular to the tranſverſe, at the point
where that axis meets the focal tangent, is to its di
ftance from the focus, as half the tranfverfe is to the
diſtance of the center from the focus.
COR. 3. Every perpendicular to the tranfverfe be-
yond its vertices meets the curve on each fide of the
axis.
PROPO.
NIC SECTION S. 239
PROPOSITION
VIII.
Straight lines drawn through the center, parallel to the
lines which join the extremities of the axes, are affymp-
totes to the hyperbolas.
The lines PQ, RS, drawn through C, parallel to Fig. 32.
FH, HE, do not meet, but continually approach the
curve, fo as to come within leſs than any given di-
ftance from it.
For, let FQ be perpendicular to EF, and through
the points K, q, affumed in the tranſverſe produced,
let IS, mo be drawn parallel to GH, meeting PQ in
I, m, the curve in D, n, and RS in S, 0. Then,
fince CF: CG :: EKF: DK**, and FQ= CH, or 6.
2
24
CG, CF²: FQ * :: EKF : DK ². Hence CK: KI
:: EKF: DK; but CK is always greater than
EKF, therefore KI is greater than DK. Thus, every
point of PQ is without the curve.
Again, by alternation and converfion,
2
CK: CF :: KI: IDS
2
2
But CK: CF :: KI * : FQ ².
Therefore IDS = FQ. In like manner, mno =
FQ.
Hence ID : mn :: no : DS, and, fince no is greater
than DS, ID is greater than mn, whence the point n
is nearer to PO than the point D.
Laftly,
240
BOOK
CONIC SECTIONS
Laftly, to ſhow that the diſtance of the curve from
PQ, when both are produced, becomes leſs than any
given line, however ſmall, let Dy be drawn parallel to
PQ, at the diſtance of the given line from it, on the
fame fide with the hyperbola, and meeting FQ in y;
then, to Qy and FQ, let DS, parallel to FQ, and ter-
minating in Dy, CS, be a third proportional, and let
it be produced to meet PQ in I; ID is equal to
therefore ID: FQ :: FQ: DS, and IDS =
FQ 2.
Whence, it is plain, from what has been demon-
ftrated, that D is a point in the curve. Now, D
is at the diſtance of the given line from PQ; confe-
quently the curve, at any point beyond it, is at a leſs.
diſtance from PQ.
COR. 1. There cannot be more than two afſymp-
totes to the hyperbola; and the fame lines are alſo
affymptotes to the other three hyperbolas.
COR. 2. The four hyperbolas approach indefinitely
near, but do not meet one another.
COR. 3. Every ſtraight line, paffing through the
center, except the affymptotes, meets two oppofite
hyperbolas, each in one point only.
COR. 4. If a ſtraight line be drawn through any
point in the hyperbola, parallel to either axis, and
meeting the affymptotes, the rectangle under its ſeg-
ments from that point is equal to the fquare of half
the other axis.
For,
N
FIG.30.
CONT
D
0
K
C
F
B
E
M
H+
}
FIG. 32.
R
G
P
E
i
C
M
T F
13
K
ss
9
PL.VI. na 240,
FIG.31.
D
B
I
Book III. CONIC SECTIONS.
241
*
For IDS = FQ
CG; and it may be proved
in the fame manner that RDO = CF².
COR. 5. According as the conjugate axis is great-
er, equal, or lefs, than the tranfverfe, the angle of the
affymptotes is greater, equal, or lefs, than a right
angle. When it is a right angle, the hyperbola is-
named an equilateral or rectangular hyperbola.
*
PROPOSITION IX.
If, from any point in an hyperbola, a straight line be
drawn parallel to any diameter, to meet the affymptotes,
the rectangle under its fegments, from that point, is e-
qual to the fquare of half the diameter.
Let D be any point in the hyperbola, and DO Fig. 33.
drawn parallel to the diameter LCM, meet the af-
ſymptotes in Q, V.
Then VDQ = & LM 2.
For, through D, and either extremity of LM, let
IDS, PLR, be drawn parallel to GH, to meet the
affymptotes. Then, becauſe of fimilar triangles,
ID: DQ:: PL: LC
DS: DV:: LR: LC
And
:
Therefore IDS DQ × DV :: PLR LC. But
IDS = PLR'; therefore DQ x DV = LC. In the 2
fame manner, DV × DQ = CM². Hence LC = CM,
and DQ x DV = the fquare of half the diameter
LM.
Hh
COR
4. c. 8.
242
Book III.
CONIC SECTIONS.
.
COR. 1. Every diameter is bifected in the center.
COR. 2. If a ſtraight line be drawn through two
points, in the fame, or in oppofite hyperbolas, its feg-
ments from theſe points, intercepted by the affymp-
totes, will be equal.
For, if DV meet the curve in O, OQ × OV = LC
- DV x DQ, therefore OQ: DQ:: DV: VO, and,
OD: DQ:: OD: VO. Whence DQ VO, and
=
DV = QO.
COR. 3. Every tangent limited by the affymptotes
is bifected in the point of contact, and is equal to that
diameter to which it is parallel. Alfo, converfely, a
ſtraight line, terminated by the affymptotes, and ha-
ving its middle point in the curve, is a tangent to the
hyperbola.
COR. 4. If, from two points in the fame, or in op-
pofite hyperbolas, parallel ftraight lines be drawn, to
meet the affymptotes, the rectangles under their feg-
ments will be equal.
•
PROPOSITION
X.
Fig. 34.
Aftraight line, terminated by the hyperbola, and parallel
to a tangent, is an ordinate to the diameter that paffes
through the point of contact.
The chord DO, parallel to the tangent XLY, is an
ordinate to the diameter LM.
For,
:
.
BOOK III.
243
CONIC SECTION.S.
For, let the affymptotes meet the tangent in X, Y,
and the chord in Q, V ; alſo, let LM meet DO in R.
Then, becauſe XL = LY, RQ = RV. But,
DQ = OV³. Confequently RD = RO.
COR. 1. A ſtraight line, terminated by oppofite
hyperbolas, and parallel to a tangent, is an external
ordinate to the diameter that paffes through the point
of contact.
COR. 2. Every ordinate, and every external ordi-
nate, to a diameter, is parallel to the tangent at its
vertex.
COR. 3. All ordinates to the fame diameters are
parallel to one another.
COR. 4. A ftraight line, drawn through the ver-
tex of a diameter, parallel to its ordinates, is a tan-
gent to the hyperbola.
COR. 5.
The diameter which bifects one of two
parallel chords, will alſo bifect the other.
COR. 6. The ſtraight line, which bifects two pa.
rallel chords, paffes through the center of the hyper-
bola.
COR. 7. If two tangents be at the vertices of
the fame diameters, they are parallel; and con-
verfely.
a
3 c. 9.
b z. c. g:
¡
!
!
مر
PROPO
244
CONIC SECTION S. Book III. no
Fig. 34.
PROPOSITION
XI.
Two diameters, one of which is parallel to the tangent at
the vertex of the other, are conjugate to one another.
Let LCM and ICN be two diameters, of which
IN is parallel to the tangent XY, at the vertex of
LM; IN, LM are conjugate diameters.
For, let XI be drawn and produced to meet the
other affymptote in Z. Then, becaufe XL is equal
and parallel to IC, XZ is parallel to LM. Now, the
triangles ICZ, LCY, by reafon of parallel lines,
3. c. 7. are equiangular, and they have the fide IC LY;
therefore IZ = LC = IX. Whence XZ is a tan-
gent at I; and LM has been proved parallel to it.
Confequently, IN, LM are mutually parallel to each
other's ordinates.
COR. 1. Tangents to the four hyperbolas, at the
vertices of two conjugate diameters, form a parallelo-
gram, whofe diagonals are coincident with the affymp-
totes.
COR. 2. Any two conjugate diameters of an equi-
lateral hyperbola are equal to one another.
For, the angle of its affymptotes being a right
* 2. c. 9. 3. angle, CL = LX = CI.
9.3.
PROPO
Book III.
245
CONIC SECTIONS.
PROPOSITION
XII.
If an ordinate be applied to any diameter, the fquare of
that diameter is to the fquare of its conjugate as the
rectangle under its fegments to the fquare of the femi-
ordinate.
Let LCM be any diameter, IN its conjugate, and Fig. 34-
DRO an ordinate to LM, in the hyperbola DLO;
LM² : IN² :: MRL: DR².
F
For, let DO meet the affymptotes in Q, V, and the
tangent at L meet them in X, Y.
24
2
Then, from fimi-
lar triangles, CL : CR² :: LX² : RQ², by conver-
fion and alternation, CL²: LX :: MRL: RQ² -
LX². =
But LX CI = QDV', and RQ — LX²
= DR²; therefore CL: CI' :: MRL: DR², and
LM² : IN² :: MRL: DR².
COR. 1. If an external ordinate be applied to any
diameter, the ſquare of that diameter is to the ſquare
of its conjugate, as the fum of the fquares of the fe-
mi-diameter and fegment intercepted by the external
ordinate, from the center, to the fquare of the ex-
ternal femi-ordinate.
COR. 2. The fquares of femi-ordinates are to one
another as the rectangles under the fegments of the
diameter.
COR.
246 CONIC SECTIONS.
BOOK III.
COR. 3. Any diameter of an hyperbola is to its
parameter, as the rectangle under its fegments to the
fquare of the femi-ordinate that divides them.
COR. 4. In an equilateral hyperbola, the rectangle
under the fegments of any diameter is equal to the
fquare of the femi-ordinate that divides them; and
the fquare of an external femi-ordinate is equal to the
fum of the fquares of the femi-diameter and fegment
intercepted from the center.
`
Fig. 35.
PROPOSITION
XIII.
If from any point in an hyperbola, two straight lines be
drawn, to meet the affymptotes, and from any other
point in the fame, or in the oppofite hyperbola, Straight
lines, parallel to thefe, be alfo drawn to the aſſymp-
totes; the rectangle under the former fhall be equal to
the rectangle under the latter.
From the point D, in the hyperbola, let there be
drawn any two ſtraight lines DQ, DR to terminate
in the affymptotes CQ, CS, and from any other point
N let NP be drawn parallel to DQ, and NS parallel
to DR, to terminate in the fame lines: Then RDQ
= PNS.
For, let LDK, and MNO be drawn parallel to the
conjugate axis, to meet the affymptotes. Then, from
the
Book III.
247
CONIC SECTIONS.
ť
the fimilar triangles RDK, SNO, RD:DK:: NS
NO, and from the fimilar triangles, LDQ, MNP,
DQ: LD :: PN: MN: Therefore RDQ: LDK ::
PNS MNO. But LDK = MNO.
:
RDQ = PNS.
Confequently 4. c. 7.
COR. 1. If, from two points, in the fame, or in op- Fig. 36.
pofite hyperbolas, ftraight lines be drawn, each pa-
rallel to one affymptote, and meeting the other, they
are to one another inverfely as the parts they inter-
cept from the center.
COR. 2. If, from any point in a given hyperbola,
two ftraight lines be drawn parallel to the affymp-
totes, the parallelogram formed thereby is of a given
magnitude.
COR. 3. Every ſector of an hyperbola is equal to
the quadrilateral figure contained by the curve, by
one affymptote, and by parallels to the other, through
the extremities of the bafe of the ſector.
For, fince the parallelograms RQ, PS are equal,
the two triangles COD, CSN, are together equal to
PS, and theſe equals being taken from the figure
CQDNS, there remains the fetor CDN equal to the
quadrilateral figure PQDN.
PROPO.
248 CONIC SECTIONS.
BOOK III.
:
PROPOSITION
XIV.
Fig. 37.
3. c. 9.
b 2. c. 9.
10.
If, in one of the affymptotes of an hyperbola, any number
of points be affumed, fuch, that their distances from the
center be in continued proportion, and straight lines be
drawn from theſe points to the curve, parallel to the
other affymptote, the quadrilateral figures formed there-
by will be equal.
Let the points A, B, D, E, be affumed in the af-
fymptote CS, ſo that CA: CB :: CB: CD :: CD :
CE, and let AF, BG, DH, EK, be drawn to meet the
curve parallel to the other affymptote CQ; the qua-
drilateral figures AFGB, BGHD, DHKE fhall be e-
qual.
=
3
For, let the tangent at G, and the line that joins
H, F meet CQ in N, L, and CS in O, M. Then,
becauſe OG GN', OB=BC, and becauſe MH=FL
MD AC; therefore MD: OB:: CB: CD :: DH
: BG. Hence, the triangles MDH, OBG, which
have the angles at D and B equal, are equiangular,
and LM is parallel to NO. The diameter CG, there-
fore, bifects the chord HF, and every chord parallel to
HF. Confequently CG bifes the fegment FGH
of the hyperbola. But it alfo bifects the triangle
FCH;
1.
5
c
R
P
M
L
N
B
FIG. 33.
P
FIG.36.
S
G
N
FIG. 37.
K
A
B
D
OME
T
Բ
R
CONI SECTIONS
M
E
I
S
**
Z
R
FIG. 34.
G
I
H
P
is
S
区
​PL.VII.pa. 248.
I
M
D
FIG. 35.
O
N
i
R
d
III. CONIC SECTIONS.
249
FCH; therefore the fector CFG = CGH, and the
Quadrilateral AFGB = BGHD º.
N. B. The 9th and 12th propofition of the ellipfe
may be applied to the hyperbola, and demonſtrated
in the fame manner.
d 3. c. 13.
APPENDIX.
PROF. 1. The fquare of any femi-diameter of an
hyperbola is equal to the rectangle under the diftan-
ces of its vertex from the foci, added to the differ-
ence of the fquares of the femi-tranfverfe, and femi-
conjugate axis.
PROP. 2. Every tangent of an hyperbola is harmo-
nically divided by the tranſverſe axis and perpendicu-
lars falling upon it from the foci.
PROP. 3. The difference of the fquares of any two
conjugate diameters of an hyperbola is equal to the
difference of the fquares of the two axes,
PROP. 4. If, from any point in an hyperbola,
ftraight lines be drawn through the vertices of a dia-
meter, to limit the tangents at theſe points, the rec-
tangle under the tangents will be equal to the fquare
of the ſemi-conjugate diameter,
I i
PROP
250
Boo
CONIC SECTIONS.
PROP. 5. A femi-ordinate to any diameter is a
mean proportional between its fegments, intercepted
from the diameter by two ftraight lines interfecting.
each other in any point of the curve, and paffing
through the vertices of the diameter.
PROP. 6. If a quadrilateral figure be formed by
tangents to the four hyperbolas, a ſtraight line thro'
the center, parallel to that which joins two oppofite
points of contact, will divide the two oppofite fides.
of the figure, fo that the fegments of the one ſhall
be inverſely proportional to the fegments of the o-
ther.
PROP. 7. Alſo, the ftraight line which joins the
middle points of its diagonals will pafs through the
center of the hyperbolas.
PROP. 8. If, through a fixed point, any ſtraight
line be drawn, to meet the hyperbola, or oppofite hy-
perbolas, in two points, the rectangle under its feg-
ments, from the fixed point, will be to the rectangle
under its fegments, intercepted by the affymptotes,
from either point in the curve, in a conftant ratio.
PROP. 9. If, from a point in one of the affymptotes
of an hyperbola, any ftraight line be drawn to in-
terfect the curve (or oppofite curves,) in two points,
and from the points of fetion, lines parallel to the
fame affymptote be drawn to meet the other, the fum
(or
III. CONIC SECTIONS.
251
difference) of the parallels will always be of the
e magnitude.
ROP. 10. A ftraight line, drawn from any point
in the curve, parallel to either affymptote, to termi-
nate in the directrix, is equal to a ſtraight line drawn
from the fame point to the focus.
N. B. The propofitions in the Appendix to the pre-
ceding Book may be applied to the hyperbola.
CONIC
!
CONIC SECTION
...........རར་རར་མ་ར་ར་ར་ར་རན་རན་རརར་……
BOOK
IV.
DEFINITION S.
Fig. 39.
1
"L'
ET there be a circle, and a fixed point, with-
out its plane, and let a ftraight line, always
paffing through that point, and indefinitely extended
both ways, revolve about the circle in its circumfe-
rence; the two furfaces thus defcribed are each of
them named a conic furface, the fixed point its ver-
tex, the circle its bafe, and the ſtraight line paffing
through the vertex and the center its axis.
2. A cone is a folid bounded by a circle and a co-
nic furface. The fixed point is called the vertex, and
the circle the baſe of the cone. Alſo, the ſtraight line
paffing through the fixed point, and the center of the
circle, is named the axis of the cone.
3. Let
CONIC SECTIONS.
SECTIONS. 253
Get there be a circle, and any ſtraight line in- Fig. 40.
ts plane at the center, and let another
ftrae always parallel to the former, revolve
about the circle in its circumference; the furface thus
deſcribed is named a cylindric furface, of which the
circle is the baſe, and the ftraight line through its
center the axis.
4. A cylinder is a folid bounded by two equal
and parallel circles, and a cylindric furface. Either
of the circles is named the baſe, and the ſtraight line
joining their centers the axis of the cylinder.
5. A cone or cylinder is termed right or oblique,
according as the axis is perpendicular or oblique to
the bafe.
6. The ſection of a cone or cylinder is termed pa-
rallel or oblique, according as its plane is parallel or
inclined to the baſe.
7. When a cone or cylinder is cut by a plane
touching the axis, and perpendicular to the bafe, and
by another plane perpendicular to the former, in fuch
a manner, that the common fection of the two planes
make angles with the common fections of the firft
plane and the furface, alternately equal to thoſe which
the common ſection of the firſt plane and the baſe
makes with the fame lines on the ſame fide, the fec-
tion of the ſecond plane is called a fubcontrary fec-
tion.
8. When
254
CONIC SECTIONS.
I
8. When the circumference of a circle, de
through any point in the curve of a parabola,
bola, or ellipfe, fo nearly coincides wideve,
that the circumference of no other circle can be de-
ſcribed between them, the conic ſection is ſaid to be
of the fame curvature with the circle at that point,
and the radius of the circle is termed the radius of
curvature.
!
COROLLARIES FROM THE DEFINITIONS.
COR. 1. A ſtraight line joining the vertex, and any
point in a conic furface, lies wholly in that furface
its continuation one way is in the fame, and its conti-
nuation the other way in the oppoſite ſurface; alſo,
every fuch line meets the circumference of the baſe.
COR. 2. Any plane touching the axis of a conic
furface cuts that furface in two ſtraight lines.
COR. 3. If a cone be cut by a plane touching the
axis, or by a plane through the vertex, and any two
points in the circumference of the baſe, the ſection is
a triangle.
COR. 4. If a plane, touching a tangent to the baſe,
paſs through the vertex, it touches the conic furface
in the ſtraight line joining the vertex and the point of
con-
CONIC SECTIONS.
255
every point in the plane, except in that
ftrationes being without the furface.
COR. 5. A ftraight line, drawn from any point in
a cylindric furface, parallel to the axis, lies wholly in
that furface.
COR. 6. Any plane touching the axis of a cylindric
furface cuts that furface in two parallel ftraight lines.
PROPOSITION I.
If a conic or cylindric ſurface be cut by a plane parallel to
the baſe, their line of common fection is the circumfe-
rence of a circle, having its center in the axis.
1. Let ABDCA be a conic furface, of which BCD Fig. 39.
is the bafe, and AE the axis, and let it be cut by the
plane GOL, parallel to BCD, the line of common
fection GKHL is the circumference of a circle ha-
ving its center in AE.
For, let any two planes, ABC, AFD, touching the
axis AE, cut the furfaces in the ſtraight lines AB, AC;
AF, AD, the baſe BCD, in the diameters BEC,
DEF, and the parallel feЯion GOL, in the ftraight
lines GOH, KOL. Then BC is parallel to GH, and
FD to KL. Hence, by fimilar triangles, ED : OK
(:: AE: AO) :: EC: OH. But ED = EC, therefore
OK OH. Confequently all ftraight lines drawn
=
from
a co
cor. 2.
}
256
CONIC SECTIONS.
Fig. 40.
or. 6.
Tec-
from the point O, where the axis meets themtraile
plane GOL, to terminate in the line of com
tion GKHL, are equal to one another, and GKHL is
the circumference of a circle, of which O is the cen-
ter.
2. Let MBDCN be a cylindric ſurface, of which BCD
is the baſe, and AE the axis, and let it be cut by the
plane GOL, parallel to BCD, the line of common
fection GKHL is the circumference of a circle, ha-
ving its center in AE.
For, let any two planes, ABC, AFD, touching the
axis AE, cut the furface in the ftraight lines MB,
NC; LF, KD, the bafe BCD, in the diameters BEC,
DEF, and the parallel fection GOL, in the ſtraight
lines GOH, KOL. Then BC is parallel to GH, and
FD to KL. But NC, KD are each parallel to AE.
Therefore OD, OC are parallelograms, and OK =
ED = EC = OH. Whence GKHL is the circumfe-
rence of a circle, of which O is the center.
COR. I. If a conic furface be cut by a plane, paral-
lel to the bafe, the folid betwixt that plane and the
vertex is a cone.
COR. 2. If a cylindric ſurface be cut by two planes
parallel to the baſe, the folid betwixt them, and the
folids between each of them and the baſe, are cylin-
ders.
COR.
3.
rallel to the bafe, the fetion is a circle.
If a cone or cylinder be cut by a plane pa-
COR
IV. CONIC SECTIONS.
257
COR. 4. Any plane touching the axis of a cone or
cylinder, cuts every parallel fection in its diameter.
PROPOSITION II.
Every fub-contrary fection of an oblique cone or cylinder is
a circle.
Let the cone or cylinder MBCN be cut by the Fig.41.42
plane MC, perpendicular to the bafe, touching the
axis, and meeting the furface in the ftraight lines
MB, NC, and the bafe in the diameter BC. Let it
be cut by another plane DFG, perpendicular to the
former, fo that their line of common fection DF may
form with one of the lines MB, NC, the angle NFD,
equal to the angle MBC, which BC forms with the
other on the fame fide. The latter fection DGF,
which is called a fub contrary ſection, is a circle, and
DF its diameter.
For, let the parallel fection HGI país through any
point K in DF. Becaufe HGI is parallel to the baſe,
a
it is perpendicular to MC, and therefore GK, its line 4.c.10.6,
of common fection, with DGF, is perpendicular to
the fame. Thus, GK is at right angles to DF, and 8. 6.
HI, and fince HI is a diameter of the parallel fection,
the rectangle HKI = GK *.
GK. But the triangles HDK,
JKF, being fimilar (for the angles at K are equal, and
K k
the
b
4. C. I,
258
Boo IV
CONIC SECTIONS.
2
the angle NFD= MBC = DHK) DK: HK:: KI: KF,
and the rectangle DKF = HKI.
Confequently the
rectangle DKF = GK, and the fection DGF a circle,
having DF for a diameter.
COR. Every fub-contrary fection of an oblique cy-
linder is equal to its baſe.
For the triangles HDK, IKF are ifofceles.
PROPOSITION III.
Fig.43.44.
Every oblique circular fection of a cone or cylinder is a
Sub-contrary fection.
Let EHFG be an oblique circular fection of the
cone or cylinder MNP, the circle EHFG is a fub-
contrary fection.
For, let any plane ODRL, parallel to the baſe,
meet it in the line DKL, and let OKR be a diameter
of the parallel ſection perpendicular to DKL. Alſo,
let the plane MNP, of the axis, and the diameter
OKR, meet the furface in the lines MN, PQ, the bafe
in NP, and the circular fection in EKF. And through
any point C, in EF, let AGBH be another parallel
fection, which the plane MNP interfects in the diame-
ter ACB, and EHFG in the chord GCH. Then,
fince the lines of common fection of two parallel
10. 6. planes, with any third plane, are parallel, AB is
4. C. I.
parallel
BOOK IV. CONIC SECTIONS.
259
c
parallel to OR, and GH to DL. But OR is perpen-
dicular to DL, therefore AB is perpendicular to GH. 11. 6.
And, fince the diameter of a circle bifects the chord to
which it is perpendicular, the chords GH, DL are
bifected in the points C, K. Confequently the line
EF, which bifects two parallel chords of the circular
fection EHFG, is a diameter of that circle, and per-
pendicular to thefe chords. Thus HCE is a right angle,
and, fince HCA is alſo a right angle, the line HC is
perpendicular to the plane MNP 4. And confequently
the plane MNP is perpendicular to each of the planes
AGBH, and EHFG, touching the line HC, and alfo e 7. 6.
perpendicular to the bafe which is parallel to the for-
iner f.
Again, becauſe HC is perpendicular to the diame-
ters EF and AB, the rectangle ECF = CH² =
CH² = ACB,
and EC:CA::BC: CF: Therefore the triangles
AEC, CBF are fimilar, and the angle CFB = EAC
= ENP. Whence, fince EHFG is perpendicular to
the plane MNP, touching the axis, and MNP perpen-
dicular to the bafe, EHFG is a fub-contrary fection.
COR. No other than a parallel, and a fubcontrary
fection of a cone, or cylinder, is a circle.
PROPOSITION IV.
Every fection of a cone or cylinder, by a plane meeting the
conic or cylindric furface on every fide, that is neither
a parallel nor a fub-contrary fection, is an ellipfe.
Let
d
f
2. 6.
£4. c. 10.6.
260 CONIC SECTIONS.
BOOK IV.
Fig.43.44.
Let EHFG be a fection of a cone or cylinder
MNP, by a plane meeting the furface on every fide,
but neither a parallel nor a fub-contrary fection,
EHFG is an ellipfe.
For, the fame conftruation remaining, as in the
preceding propofition, only let C be the middle of
EF, it is fhewn, in the fame manner, that HC is per-
pendicular to AB, and that EF bifects GH and DL.
By fimilar triangles EC: EK:: AC: OK,
And
Therefore
CF: KF:: CB: KR.
EC²: EKF :: CG²: DK².
Confequently EHFG is an ellipfe, of which EF and
GH are two conjugate diameters.
Fig. 45.
a cor. 4.
PROPOSITION
V.
If a cone be cut by a plane parallel to another, touching
the conic furface, the fection is a Parabola.
Through any point B, in the circumference of the
bafe BGC, let the tangent DE be drawn, the plane
ADE, touching that line, and the vertex A, touches
the conic furface in the ftraight line AB. Let the
cone be cut by the plane VGN, parallel to ADE, and
meeting the bafe in GN; the ſection GVN is a para-
bola.
For,
1
}
FIG. 38.
F
G
L
B
E
1
H
M
A
D
FIG. 41.
B
E
G
H
B
N
YF
F
CONIC SECTIONS
E
M
I
FIG. 39.
D
A
N
R
D
M
G
FIG./40.
B
E
N
E
FIG. 42.
A
H
M
I
Ꮐ
B
E
M/?
H
K
I
FIG. 43.
G
B
P L.VIIIpa260.
K
N
Book IV. CONIC SECTIONS.
261
For, let ABC, the plane of AB and the axis, in-
terfect GVN, in the line VK, and the baſe in the di-
ameter BKC. Alfo, let MHL be a parallel fection,
through any point I in VK, interfecting ABC, in the
diameter ML, and VKG in the ftraight line HI.
Then, by reaſon of parallel planes, the ftraight line
AB is parallel to VK, BD to GK, GK to HI, and
BC to ML. Hence MK is a parallelogram; and, 10. 6.
fince DBC is a right angle, GK is perpendicular to
BC, and HI to ML.
VK: VI : KC: IL;
VK: VI:: BKC
Now, by fimilar triangles, 11. 6.
but BK = MI. Therefore
MIL :: GK: HI'.
Whence
GVN is a parabola, of which VK is a diameter, and
GK its femi-ordinate.
PROPOSITION VI.
Every other fection of a cone is a hyperbola.
Let DFO be a ſection of the cone ABDC, different Fig. 46.
from any that has been mentioned, meeting the baſe
in the line DO. Let DO be cut at right angles by
the diameter BC, and let ABC, the plane of the axis,
and BC, interſect the furface, and the plane FDO, in
the ſtraight lines AB, AFC, and FK. Then FK is
not parallel to AB; otherwiſe, as DK is parallel to
the tangent BG, the plane FDO would be parallel to
the tangent plane AGB, contrary to the hypothefis;
let
11. 6.
262 CONIC SECTIONS.
Book IV.
4. C. I.
€ 10. 6.
а
II. 6.
let them meet, therefore, in the point E, which will
be in the oppofite ſurface, becauſe, by hypotheſis, the
plane FDO does not meet the other on every fide.
Alfo, let LMI be a parallel fection, through any point
N in FK, interfealing ABC in the diameter LI, and
DKF, in the ftraight line MN. Then LI is parallel
to BC, and MN to DK, and therefore MN perpen-
dicular to LI. Now, by fimilar triangles,
And
EKEN:: BK: LN
KF: NF:: KC: NI
b
Therefore EKF: ENF:: BKC:LNI:: DK2: MN³,
Whence DFO is a hyperbola, of which EF is a dia-
meter, and DK, MN femiordinates.
PROPOSITION VII.
Fig. 47.
To find the radius of curvature at a given point, in a
given conic fection.
CASE 1. Let D be the given point, in the given pa-
rabola DVL, at which it is required to find the radius
of curvature.
From D, let the tangent DM, the perpendicular
DR, and the diameter DF, be drawn. Alſo, through
any point Q, in the curve near to the point D, let the
circle DOO be defcribed, to touch DM in D, and
meet DF in P. Let PQ, QD be joined, and QN
drawn
Book IV. CONIC SECTIONS.
263
1
*
f
drawn parallel to MD, to meet DF in N. Then, be-
cauſe the angle DPQ = MDQ = DQN, the triangles
DNQ, PQD, having a common angle at D, are equi-
angular.
Hence PD: DQ:: DQ : DN, and PD × DN = DQ₂•
Alfo PD² : PQ : : DQ : QN"
Therefore PD. PQ:: PD × DN: P× DN
Or
PD²: PQ :: PD: P, (P being the para-
meter of DF). Now, it is manifeft that the nearer
the point Q is to the point D, the nearer will the cir-
cumference of the circle be to a coincidence with the
curve at that point; and, therefore, as no portion of
thefe curves, however fmall, can be the fame, the
circumference of the circle will have approached the
neareſt poffible to a coincidence with the curve at the
point D, when the point Q falls upon it; in which
cafe the laſt analogy, which holds univerfally, becomes
PD: PD' :: PD: P, where it is plain that PD is e-
qual to P, the parameter of DF. Hence, if DP be
taken equal to four times the diſtance of the given
point from the directrix, and bifected by the perpen-
dicular IR meeting DR, perpendicular to MD in R,
then RD is the radius of curvature.
CASE II. In the ellipfe, or hyperbola, let D be the Fig. 48.
given point, DM a tangent, DF, EG conjugate dia-
meters, and DHO perpendicular to the two parallels
DM, EG. Through any point Q, in the curve, near
to the point D, let the circle DOO be defcribed to
touch
{
264
BOOK IV.
CONIC SECTIONS.
:
touch the line DM, and meet DF in P. Let PQ,
QD be joined, and QN drawn parallel to DM, to
meet DF in N. The triangles DNQ, PQD, are fi-
milar, DN : DQ :: DQ: DP, hence DN: DP :: DN²
: DQ',
Or
But
DN: DP :: QN : PQ².
QN²
DF: P:: FN × DN: QN²
Therefore DF × DN: P × DP :: FN × DN : PQ².
DF: FN: P x DP : PQ* ;
And
×
which analogy, when Q coincides with D, becomes
DF: DF :: P× DP: DP, where DP is equal to the
parameter of DF, as before.
COR. 1. If a circle touch a tangent to a conic fec-
tion, at the point of contact, on the fame fide, and cut
off from the diameter which paffes through that point,
a fegment, equal to its parameter, the conic ſection is
of the fame curvature with the circle at the point of
contact.
COR. 2. If, from any point in the curve of an el-
lipfe or hyperbola, a diameter be drawn, and a per-
pendicular be let fall upon its conjugate, the radius
of curvature, at that point, will be a third propor-
tional to the perpendicular and the femi-conjugate
diameter.
For, fince DH:DC:: DP: DO,
And
DC: EC:: EG: DP.
DH: EC::EC: RD.
PROPO-
BOOK IV.
i
;
CONIC SECTIONS.
PROPOSITION
VIII.
265
If, from any point, in the curve of a conic fection, an or-
dinate be applied to the axis, and likewife an ordinate
to the diameter which paffes through the other extre-
mity of the first ordinate, the circle defcribed through
the extremities of the fecond, to touch the tangent to the
curve, at the affumed point, "will be of the fame curva-
ture with the conic fection at that point.
From the point D, in the curve, let the ordinate Fig.49.50.
DR be applied to the axis AB, and the ordinate DHL
to the diameter RG; then the circle DPL deſcribed
through D, L, to touch MD, the tangent to the curve
at D, will be of the fame curvature at that point.
For, let its circumference meet DF again in P, let
PL be joined, and MD, RG interfect each other in E.
CASE I. In the parabola DAL, becauſe the diame- Fig. 49.
ters are parallel, the angle DHE = PDL, and the
angle MDP = DEH. But, fince MD is a tangent to
the circle, the angie MDP PLD. Therefore the
angle DEH = PLD, and the triangles EDH, DPL,
LD: DP. But
equiangular. Hence EH HD :: LD: DP.
the fubtangent EH is bifected in R,
and the ordinate
LD in H. Therefore RH: HD:: HD: DP. Con-
fequently DP is the parameter of RG, or DF and
LI
DPL
266
BOOK IV.
CONIC SECTIONS.
i
Fig. 50.
DPL the circle of curvature correfponding to the
point D.
CASE II. In the ellipſe and hyperbola, becauſe DR
is an ordinate to the axis AB, the diameters DF and
RG are equal, and the rectangle ECHCR = CD².
Hence the triangles ECD and DCH are fimilar.
Therefore the angle PDL DEH, and the angle PLD
(= MDP) = DHE; and confequently the triangles.
DPL and EDH are fimilar: Whence DP: DL:: ED
:EH::EDXCH:EH× HC. But ED × CH =
CD × DH, and EH × HC = GH × HR.
X
X
Therefore DP: DL:: CD x DH: GH: x HR.
And
By alterna-
tion
And by in-
DPXDH:DLXDH::CDXDH: GHXHR.
DP: CD:: 2DH: GH × HR.
verfion DF: DP :: GHX HR: DH.
Confequently DP is the parameter of DF, and DPL
the circle of curvature, correfponding to the point D.
i
APPENDIX.
PROP. I. Let there be a ftraight line, and a point
without it, and let another point without it move in
fuch a manner, that the ratio of its diſtances from
theſe ſhall always be the fame; that point fhall de-
fcribe
H
B
FIG. 44.
PL.IX. ja 266.
CONIC
SECTIONS
B
E
A
A
M
I
M
E
F
G
B
N
K
D
G
FIG. 45.
L
N
E
P
K
D
FIG./47.
I
R
H
M
G
M
C
C
B
D
FIG. 46.
E
I
R
P
M
F
G
F
P
F
D
R
}
FIG. 48.
FIG. 49.
P
H
I
M
1
H
G
A
Book IV. CONIC SECTIONS.
267
ſcribe an ellipfe, pa.abola, or hyperbola, according as
its diſtance from the other point is greater, equal, or
lefs, than its diſtance from the ftraight line.
PROP. 2. Let a ftraight line touch the circumfe-
rence of a circle; the point which moves equally di-
ſtant from both deſcribes a parabola. Let two circles
touch one another, internally or externally; the
point which moves equally diftant from both their
circumferences deſcribes accordingly an ellipfe or hy-
perbola.
PROP. 3. If two tangents to a conic fection inter-
fet one another, and, from any point in the one, a
line be drawn parallel to the other, to interfest the
curve, and the line joining the points of contact, its
fegments from that point will be continually pro-
portional.
PROP. 4. If two parallel chords of a conic feion
be interfected by any third chord, the rectangles un-
der the fegments of the former will be to each other
direaly as the rectangles under the fegments of the
latter.
PROP. 5. If two chords of a conic feaion interfect,
and be always parallel to two lines given in pofition,
the rectangles under their fegments will have to each
other an invariable ratio.
PROP. 6. If, from the vertex of two tangents, a
ftraight line be drawn through the focus, it will bi-
fet
+
268 CONIC SECTIONS.
BOOK IV.
fect the angle formed by lines drawn through the fo-
cus, and the points of contact, or will be perpendicu
lar to theſe lines, if they be in the fame ftraight line.
PROP. 7. Any straight line, drawn from the vertex
of two tangents, through the line joining their points
of contact, to interfe&t the conic fection in two points,
is harmonically divided.
PROP. 8. A ftraight line drawn from the vertex of
two fecants, through the interfection of the lines
which join the oppofite points of fection, to meet the
conic fection in two points, is harmonically divided.
PROP. 9. Any tangent interfesting two other tan-
gents, and the line paffing through their points of
contact, is harmonically divided.
PROP. 10. If a ſtraight line be drawn through the
vertex of two tangents, parallel to the line joining
their points of contact; any ftraight line drawn thro'
the middle of the laft line to meet the firft, and in-
terfect the conic fection in two points, is harmoni-
cally divided.
PROP. 11. If two ftraight lines, not parallel, be in-
ſcribed in a conic fection, and their extremes joined
by four lines; the interſections of theſe and the ver-
tices of the tangent triangles, erected upon the two
chords, will be all four in one ſtraight line.
PROP. 12. If a perpendicular to a tangent, from the
point of contact, meet the axis of a conic fection, and
from
Book IV CONIC SECTIONS.
1
269
from the point of concourfe a perpendicular be let fall
upon the line joining the focus and the point of con-
tact, the fegment of this line between the perpendicu-
lar and the tangent, is equal to half the parameter of
the axis.
PROP. 13. Let there be any two chords in a conic
fection, and, from two extremes, which are not of the
fame chord, let two lines be drawn through any point
in the curve, and from the other two extremes let two
lines be drawn through any other point in the curve;
the interfections of thofe lines which are not drawn
from the fame chord, nor meet each other in the
curve, will be in a ſtraight line, parallel to the chords,
if the chords be parallel, or, if not, in the fame
ſtraight line with their interfection.
MENSURATION.
1
MENSURATION.
་་་་་
¡
+
ར་ ་ འ ་ ར ་་་་
воок
BOOK I.
OF LINES AND ANGLES.
I.
E
VERY magnitude is meaſured by another
of the fame kind, called the meaſuring unit.
Thus, a line is meaſured by a line, an angle by an
angle, a furface by a ſurface, and a folid by a folid.
2. Certain magnitudes being given, that is, their
meaſures being determined by an actual application
of the meaſuring unit, it is the bufinefs of menfura-
tion to fhew how the meaſures of others which depend
on theſe may be obtained.
3. The first part of menfuration, which treats of
lines and angles, being chiefly concerned about mea-
furing the fides and angles of a triangle, is commonly
named trigonometry.
4. Any
1
{
OOK 1 MENSURATION.
271
4. Any ftraight line of a determinate length may
be affumed as the meafuring unit of lines; but the
90th part of a right angle is the meaſuring unit of
angles, and it is called an angle of one degree; and
fo every right angle is an angle of 90 degrees.
5. An angle at the center of a circle has the fame
ratio to four right angles, that the arch intercepted
between its fides has to the whole circumference.
Therefore the angle will be determined when it is
known what part the arch is of the whole circumfe-
And hence the arch has been called the mea-
rence.
fure of the angle.
6. The circumference of every circle is fuppofed to
be divided into 360 equal parts, called degrees, each
degree into 60 minutes, each minute into 60 feconds,
&c. marked thus, 25° 14′ 12" 8" (25 deg. 14 min.
12 fec. 8 thirds). Thus, a quadrant, or 4th part of
the circumference, contains 90°, and an angle is of fo
many deg. min. &c. as are contained in its arch.
7. The complement of an arch, or angle, is its dif. Fig. 1.
ference from a quadrant, or a right angle. Thus, the
arches AB, BD, alfo the angles ACB, BCD, are the
complements of each other.
8. The fupplement of an arch or angle is its diffe-
rence from a femi-circle or two right angles.
Thus,
the arches AB, BL, alfo the angles ACB, BCL, are
the fupplements of each other.
9. The
:
272
MENSURATIO N.
9. The fine of an arch is a perpendicular falling
from one extremity of the arch upon a diameter paſ-
fing through the other. BE is the fine of AB.
10. The verfed fine of an arch is that part of the
diameter intercepted between the fine and the arch.
AE is the verfed fine of AB.
11. The tangent of an arch is a ſtraight line touch-
ing it in one extremity, and limited by a line drawn
from the center through the other extremity. AG is
the tangent of AB.
12. The fecant of an arch is the line which limits
the tangent. CG is the fecant of AB.
13. Theſe are alſo faid to be the fine, tangent, &c.
of the angle ACB to the radius CA.
14. By the cofine, cotangent, &c. of an arch or
angle, is meant the fine, tangent, &c. of its comple-
ment.
15. Hence, The chord of 60°, the fine of 90°, and
the tang. of 45°, are each equal to the radius-An
arch, and its fupplement, have the fame fine-The
fine of an arch is half the chord of double the arch-
The fine of an angle at the circumference is half the
chord of the arch upon which it ftands to the radius
of that circle-The cofine of an arch is equal to that
part of the diameter which is intercepted between the
center and the fine.
16. The meaſures of three lines being given, the
meaſure of the fourth proportional to them may be
found
1
+
MENSURATION.
273
found by multiplying the 2d and 3d numbers together,
and dividing their product by the firft-The mea-
fures of two lines being given, that of a mean pro-
portional between them, by multiplying the given
numbers, and extracting the fquare root of their pro-
duct-The meaſures of the two fides of a right angled
triangle being given, that of the hypothenufe may be
found, by adding the fquares of the given numbers,
and extracting the fquare root of the fum.
Let a, b, c, repreſent the meaſures of the hypothe- Fig. 2.
nufe, and the two fides of a right angled triangle, and
m, n, thoſe of the fegments of the hypothenufe, then
a:b::b: m³, and a:c::c:n, hence am = b², an = c²,
and am + an
= b² + c², or a² = b² + c².
quently a √ b² + c², alfo b
a²
c².
Confe-
17. The folution of the three following problems
is the object of the first part of menfuration.
PROBLEM I. The meaſure of the diameter of a
circle being given, to find that of its circumference.
PROB. 2. The meafure of the radius being given,
to find from thence the meaſures of the fine, tangent,
fecant, and verfed fine, of every arch of a quadrant.
PROB. 3. The meaſures of three parts of a triangle
being given, to find from thefe the meaſures of the
remaining parts.
One method of meafuring the circumference of a
circle, and of conftructing the trigonometrical ca-
non, which is a table exhibiting the meafures of the
M m
finę,
2.C 12.6.
274
MENSURATION.
fine, tangent, fecant, and verſed fine ef every arch of
a quadrant, containing an exact number of minutes,
is contained in the following propoſitions.
PROPOSITION I.
Fig. 3.
The fupplemental chord of any arch is a mean proportion-
al between the radius, and the fum of the diameter and
Supplemental chord of double the arch.
The fupplemental chord BE, of the arch AE, is a
mean proportional between the radius BC, and the
fum of the diameter AB, and fupplemental chord
BD, of the arch AD, the double of AE.
For, let BA be produced to F, fo that AF BD,
and let ED, EC, EA, EF, be drawn. The triangles
BDE, EAF, having the fide BD = AF, DE EA,
1.c.10.3. and likewife the angle BDE FAE, are equal in e-
6 4. 5.
–
=
very reſpect. Thus, the angle at F = DBE = EBA,
-
and the triangle FEB ifofceles. Hence the triangles
ECB, FEB are equiangular: Therefore BC: BE::
BE: BF, or BC: BE:: BE: AB+ BD.
COR. I. The chord of any arch is a mean propor-
tional between the radius and the difference of the
diam. and fupplemental chord of double the arch.
For, if BG be made equal to BD, and E, G joined,
the triangles BGE, BDE are equal in every refpect,
and
MENSURATION. 275
and CAE, EAG fimilar. Hence CA: AE :: AE:
AG AB
AB — BD.
COR. 2. If the radius BC= 1, and the fupplemen-
tal chord BD be given in decimal parts of the radius,
then,
2 + BD = BE the fupplemental chord of half
the arch AD.
2 - BD AE the chord of half the arch AD.
=
PROPOSITION II.
4
The diameter of any circle is to its circumference as I to
3.1416 nearly.
Let ABC be a circle, in which AD is the fide of a Fig. 4.
hexagon, and let the radius AC (AD) = 1; then
BD = /AB — AD = √3. Thus, the fupplemen-
tal chord of
I 2
24
I
48
I
38
768
The chord of 7536
of circumf. is√3 = 1.7320508075=a
2+ a=1.9318516525=b
√2+6= 1.9828897227-c² 2. c. 1.
2 + c = 1.9957178464=d
2+d=1.9989291749=m
√2+m=1.9997322758=n
√2+n=1.9999330678=p
√2+p=1.9999832669=q
2
9.004090612.
The
1
276
Boo
MENSURATION.
囊
​མར་བ,
I.C.21.3.
22.5.
i
The fide of a regular polygon ", therefore, of 1536
fides, inſcribed in the circle, is .004090612. Now,
let EF reprefent one of its fides, and GH one of the
fides of the fimilar circumfcribed polygon; then the
✓CE EF
perpendicular CK =
3
√2+9
2
•999997908; but CK: CL:: EF: GH, therefore
•999997908 : 1 ::.004090612 : GH = .004090618,
hence the perimeter of the infcr. polygon= .004090612
× 15366.283180032, the perimeter of the circumf.
polygon .004090618 x 1536 = 6.283189248.
confequently the circumference of the circle will be
nearly 6.283185, and the ratio of the diameter of
any circle to its circumference 2: 6.283185, or
1: 3.1416 nearly.
I
1
COR. 1. Let D, C reprefent the diameter and cir-
cumference of a circle, then C = 3.1416 D, and
D = .3183 C.
COR. 2. To the radius 1,.01745 is the length of
1 degree, hence .01745 RN is the length of an arch
of N degrees to the radius R.
COR. 3. An arch of a circle of the fame length
with the radius contains 57.295 degrees.
PROPO.
MENSURATION.
277
PROPOSITION
III.
To find the fine and cofine of 1 minute to the radius 1.
It is evident that a very ſmall arch, fuch as 1 mi-
nute, and its fine, will be very nearly equal; therefore
6.283185
360 x 60
= .0002908882 is the fine of 1 minute, and
I
confequently 1.0002908882.9999999577 is
'its cofine, or the fine of 89° 59'.
PROPOSITION IV.
Twice the rectangle under the fine of half the fum, and
the cofine of half the difference of two arches, is equal
to the rectangle under the radius and the fum of their
fines.
Let AB,,AD be the two arches, C the center, Fig. 5]
AF a diameter, DL parallel to it, BG perpen-
dicular to DL, CE perpendicular to the line join-
ing B, D, and AH parallel to BD. Then the tri-
angles BGD, CHA, which have the angles at G, H
right, and the angles at D, A equal, by rea-
fon of parallel lines, are equiangular; therefore
BD
278
MENSURATION.
!
BD: BG: CA: CH, and 2BI × CH CA x BG.
Now, BI is the fine of BE, half the fum of the arches,
CH is the cofine of AE, half their difference, and BG
the fum of their fines. Whence the propofition is
manifeft.
COR. I. If the fine of the mean of three equi differ-
ent arches, the cofine of the common difference, and
likewife the fine of one extreme be given in numbers,
(the radius = 1) the fine of the other extreme may be
found, by fubtracting the latter of the given numbers
from the product of the two former.
COR. 2. The fum of the fines of two arches, which
together make 60°, is equal to the fine of an arch that
is greater than 60°, by either of the two arches.
For, if AB+ AD = 60°, then BDCA, and con-
fequently BG=CH S,AN, or S,AMS,AD+60°,
or S,AB + 60°.
PROPOSITION
V.
To find the fines of all arches from 1 minute to 90 de-
grees.
Since the fine and cofine of 1 minute are given, let
e denote the latter. Then, by Cor. 1. Prop. 4.
20
MENSURATION.
279
2c x fine '
fine o'
=
fine 2'.
2c x fine 2'
—
fine 1'
fine 3'.
X
2c × fine 3'
fine 2'
fine 4'.
2c × fine 4'
fine 3'
=
fine 5'.
Thus, the fines of all arches of an exact number
of minutes may be fucceffively derived from one ano-
ther. But the fines of thofe above 60° may be found
by addition only, by CoR. 2. in this manner.
Sine 'fine 59°59′ = îine 60°1′.
Sine 2 + fine 59°58' = fine 60°2'.
Sine 3' + fine 59°57′ = fine 60°3'•
Sine 4+ fine 59°56′ = fine 60°4
With refpe to thofe arches which do not conſiſt
of an exact number of minutes, for the odd feconds in
each, a proportional part of the difference between
the fines of the next greater and lefs arches may be
taken and added to the fine of the lefs, and that will
give the fine of ſuch an arch nearly.
4
M
PROPOSITION VI.
The fines of every arch of the quadrant being given, to
find the tangents, fecants, and verfed fines.
Let
280
MENSURATION.
Fig. 6.
Let AB be any arch of the quadrant AD, BE its
fine, CE its cofine, AE its verfed fine, AG its tan-
gent, and CG its fecant. Then AE CA CE,
and, by fimilar triangles, CE: EB:: CA: AG, alfo
CE: CB:: CA: CG. Thus, the verfed fine is e-
qual to the difference between the radius and the co-
fine; the tangent is a fourth proportional to the co-
fine, the fine, and the radius; and the fecant a third
proportional to the cofine and the radius.
But the fecants may be more eaſily found from the
tangents. For the fecant of any arch is equal to its
tangent added to the tangent of half its complement.
Let GF be made equal to GC, and CF joined: Then
the angle ACF =
BCD, therefore the arch
AH=RD, and AF the tangent of half the com-
plement of AB.
G
2
On thefe principles may the trigonometrical canon
be conftructed; and, by means of it, the four cafes of
the third problem, both with refpect to right angled
and oblique angled triangles, may be refolved, the
proportions for finding the unknown parts being firſt
deduced from the following Theorems.
THEOREM
MENSURATION.
281
1
THE ORE M I.
In any right angled triangle, the hypothenufe is to one of
the fides as the radius (in the Tables) to the fine of
the angle oppofite to that fide; also, one of the fides is
to the other as the radius to the tangent of the angle
oppofite to the latter.
Let ABC be a right angled triangle, DE the radius Fig. 7-
of the tables, the angle EDF - BAC, FG its fine,
and EH its tangent. Then the triangle ABC is fi-
milar to each of the triangles DGF, DEH, becauſe
the angles at A, D are equal, and the angles at B, G,
E right. Hence AC: CB :: DF: FG, and AB: BC
:: DE: EH, that is, AC: CB:: R: S, A, and
AB: BC: R: T, A.
COR. 1. To the hypothenufe as a radius, each fide
is the fine of its oppofite angle; and to one of the
fides, as a radius, the other fide is the tangent of its
oppofite angle, and the hypothenufe is the fecant of
the fame angle.
COR. 2. The fine, tangent, &c. of an arch or angle
have the fame proportion to their radius, and to one
another, that the fine, tangent, &c. of a fimilar arch,
or equal angle, have to their radius, and to one ano-
ther.
N n
Cafes
282
Book
MENSURATION
!
:
Cafes of Right-angled Triangles.
CASE 1. When the hypothenufe and one of the ob-
lique angles are given.
CASE 2. When a fide and one of the oblique angles
are given.
CASE 3. When the hypothenufe and one of the
fides are given.
CASE 4. When the two fides are given, to find the
remaining parts of the triangle.
Proportions for folving thefe Cafes.
CASE 1. Given the hypothenufe AC, and the angle
A, to find the other parts of the triangle. As the ob-
lique angles are the complements of one another,
when the one is given or found, the other will alſo be
known. Now, to find AB, BC,
R: S,A: AC: CB
R: S,C: AC : AB
Or, S',A : T,A :: AC: CB
S',A: R :: AC: AB
Or, St,C: R :: AC : CB
S,C: T,C:: AC: AB
The
1
BOOK I.
283
MENSURATION.
The proportions for the other cafes are to be ftated
in like manner. Thoſe proportions which have ra-
dius for one of their terms are generally uſed in prac-
tice.
THEOREM
II.
In any plane triangle, as one of the fides is to another, fo is
the fine of the angle oppoſite to the former to the fine of
the angle opposite to the latter.
Let ABC be a triangle, then AB: BC:: S,C: S,A. Fig. 8.
About the center A, with the radius AB, let the fe-
micircle DBE be defcribed upon AC produced. Let
the radius AF be drawn parallel to BC, and FG, BH
perpendicular to the diameter DE. Then FG, BH
are the fines of the angles (FAD) BCA, BAC, and
the triangles FGA, BHC are fimilar: Hence AF :
BC:: FG: BH, and therefore AB: BC::S,C: S,A'. 2. c.1.th.
THEOREM III.
}
In any plane triangle, as the fum of two unequal fides is
to their difference, fo is the tangent of half the fum of
the oppofite angles to the tangent of half their differ-
ence.
Let
}
284
Book I
MENSURATION.
:
:
Fig. 9.
* 18. I.
3 6.3.
1. c. 5. 2.
Let ABC be a triangle, of which the fide AC is
greater than AB; then AC AB AC AB :: T,
B+C
: T,
2
B-C
2
:
About the center A, with the
radius AB, let the femicircle DBE be defcribed upon
AC produced. Let DB, BE, and EF parallel to BC,
be drawn. Then DC is the fum of AC, AB, and EC
is their difference. Alfo DAB is the fum 2, and DEB
half the fum' of the oppofite angles ABC, ACB;
and BEF is half their difference. Now, to the fame
radius EB, DB is the tangent of DEB, and FB the
tangent of BEF. And, becaufe EF is parallel to BC,
DC:CE::DB: BF; hence AC+AB: AC— AB::T,
B-C
B+C
: T,
2
2
THE ORE M IV.
Fig. 10.
II. 2.
b2.c.8.5.
As half the baſe of any plane triangle is to half the fun
of the two fides, fo is half their difference to the di-
ftance of the middle of the bafe from the perpendicular.
a
Let ABC be a triangle, of which the bafe AC is bi-
fected in E, and BD is the perpendicular upon AC.
Then, becauſe AC × 2ED = AB+BĊ × AB—BCª,
AC: AB + BC :: AB- BC: 2ED, and AC:
AB + BC
يم
AB
::
BC
: ED.
2
Ι
2
COR,
BOOK I.
285
MENSURATION.
COR. This fourth proportional ED added to half
the baſe gives the greater fegment; alfo AC × ED =
AB BC AB
2
X
BC
2
N4
THE ORE M V.
As the rectangle under any two fides of a plane triangle
is to the rectangle under half the fum and half the dif-
ference of the bafe, and the difference of the two fides,
fo is the fquare of the radius to the fquare of the fine of
half the included angle.
Let ABC be a triangle, having the fide AB greater Fig. 11.
than BC. From AB let BD be cut off equal to BC,
and from BC produced, BE equal to BA.
Let AE,
CD, and CG perpendicular to AE, be drawn. Alfo,
let BHF be drawn to bifect the angle ABC, and it
will bifect the bafes AE, CD, at right angles. Now,
from the right angled triangles ABF, DBH we have
thefe proportions:
AB: AF: R: S, ABC.
BD:DH::R: S, ABC.
I
b
Confequently AB BD: AF × DH :: R²: S³, ABC. 4. c. 9.5.
AB × BC: AF × FG :: R²: S,²
Or
½ ABC,
There-
286 MENSURATION.
Book I.
1
Therefore (Cor. Theo. 4.) AB × BC :
AC+ CE
2
AC
CE
X
:: R²: S,² ABC.
2
2
I
COR. Hence ACCE² + 4AB × BC × S,² ABC.
R²
Cafes of Oblique angled Triangles.
CASE 1. When two angles and a fide are given.
CASE 2. When two fides, and an angle oppofite to
one of them, are given.
CASE 3. When two fides and the included angle
are given.
CASE 4. When the three fides are given, to find
the remaining parts of the triangle.
An angle of a triangle, and the fum of the other
two, are the fupplements of each other; therefore,
when the one is given or found, the other will alſo be
known. The proportions for the 1ft and 2d cafes are
deduced from the 2d Theorem; but the data in the
2d cafe are not fufficient, when it is the leſs ſide that
fubtends the given angle.
In the 3d Cafe, half the difference of the required
angles is found by the 3d Theorem, and that half dif-
ference added to the half fum, gives the greater angle;
then the unknown fide is found by the 1ft Theorem.
{
The
.
}
>
A
FIG.1.
G
MENSURATION
万
​FIG. 2.
c
FIG. 3.
E
D
F
M
m'
B
H
D
α
n
F
JA
Ꮐ
с
E
K
N
FIG. 4.
D
FIG. 6.
B
C
E
H
B
M
B
FIG. 5.
K
F
E
G
H
L
F
C
F
FIG: 9.
A
B
I
Ꮐ
N
D
PL. I. pa. 286.
A
Ꮐ
с
H
A
B
F
FIG. 8.
HE
I
A
F
FIG./7
D
G
B
A
D
G
A
H
E
B
FIG. 12.
C
E
B
B
FIG.II.
FIG. 10.
D
B
H
F
Ꮐ
A
C
C
D
B
D
E
•
Book I.
287
MENSURATION.
The laſt Caſe will be refolved into the 3d Cafe of
right angled triangles, if the fegments of the bafe be
found by the 4th Theorem, or it may be folved di-
"ectly by the 5th Theorem.
Practical Solution of the Cafes.
To avoid the tedious operations of Multiplication
and diviſion, in finding a fourth proportional, loga-
rithms are generally uſed. Theſe are a fet of num-
bers, fo adapted to the natural numbers, that the ad-
dition and fubtraction of the former correfpond to
multiplication and divifion in the latter. However,
in the proportions of the cafes of right angled tri-
angles, where radius is always one of the terms, the
4th term may be found with as much expedition by
the natural as by the more artificial method of work-
ing.
Right angled Triangles.
CASE 1. In the right angled triangle ABC, given Fig. 7.
the hypothenuſe AC 324, and the angle BAC 48°17′,
to find the other parts of the triangle.
The angle ACB=90° 48° 17′ = 41° 43'. Then,
to find AB, BC.
R:
!
288
Book I.
MENSURATION.
}
R: S,A: AC: CB
R: S, 48° 17;: 324: CB
R: S,C:: AC : AB
R S, 41° 43' :: 324: AB.
Solution by Natural Sines.
1.7464446
324 BC= 241.8480504.
1: .6654475: 324 AB =
:: 324: AB = 215.60499.
Solution by Logarithmic Sines.
L. S, 48° 17′
9.8729976
L. 324
2.5105450
Sum 12.3835426
L. R.
10.0000000
L. BC
2.3835426
L. S, 41° 43′
9.8231138
L. 324
2.5105450
Sum 12.3336588
L. R.
10.0000000
L. AB
2.3336588
Whence BC 241.84805, and AB= 215.605.
i
CASE
Book I. MENSURATION.
289
%
CASE 2. Given the fide AB 125, and the angle
A 51° 19'
The angle C = 90° · 51° 19′ = 38° 41′
R: St. A :: AB : AC
R: St. 51° 19' 125: AC
RT, A
::
:
AB BC
R : T, 51° 19′ :: 125: BC
Solution by Natural Secants and Tangents.
I: 1.5964824:: 125: AC = 199.5603.
I: 1.2489484: 125: BC 156.11855.
=
Solution by Logarithmic Secants and Tangents.
L. St 51° 19′
10.2031641
L. 125
2.0969100
L. AC = 199.5603
2.3000741
L. T, 51° 19
10.0965445
2.0969100
2.1934545
L. 125
L. BC = 156.1186
CASE 3. Given the hypothenufe AC 415, and the
fide AB 249.
AB: AC :: R : St. A
249: 415 :: R: St. A
R: T, A ::
AB : BC
R: T, A :: 249: BC
Oo
249
290
BOOK I.
MENSURATION.
=
249: 415::1: 1.666 St. A 53° 7′48″.3.
L. 415 + L. R.
12.6180481
L. 249
2.3961993
L. St. A 53° 7′ 48″.3
10.2218488
I: 1.333
:
249 BC= 332.
L. T, 53° 7′ 48″
10.1249384
L. 249
2.3961993
L. BC= 332
2.5211377
2
BC may be found without finding an angle,
AC2
AB² = BC².
172225-
620001 = 110224.
"
AC + AB × AC AB BC2
664 × 166
Therefore BC= √110224 = 332.
-
= 110224:
CASE 4. Given the fide AB 53 and BC 67.
AB : BC :: R:T, A
53
67: R: T, A
R: St, A :: AB : AC
R: St, A :: 53: AC
A:
53: 67 :: 1: 1.26415094 = T,A 51° 39′ 15.″9.
L. 67 + L. R
11.8260748
L. 53
1.7242759
L. St. A 51° 39′ 15″. 9
10.1017989
:
I : I.
虐
​:
Book Ì. MENSURATION.
291
1: 1.16118553 :: 53: AC = 85.4283309.
L. St. 51° 39′ 15″.9
L. 53
L. AC = 85.42834
10.2073260
1.7242759
1.9316019
Or, fince AC² = AB² + BC, AC/7298
$5.428332, &c.
Oblique angled Triangles.
CASE I. Given the angle A 49° 25', the angle C Fig. 12.
63° 48′, and the fide AB 275.
The angle B 180° -
=
49° 25′ +03° 48′ = 66° 47′.
S, C: S, A :: AB : BC
S, 63° 48′ : S, 49° 25′
:: 275: BC
L. S. 49° 25' + L. 275
12.3198379
L. S, 63° 48′
ة
9.9529175
2.3669204
L. BC=.232.7665
S, C: S, B :: AB : AC
S, 63° 48′ : S,66° 47′ :: 275 : AC
L. S, 66° 47' + L. 275
L. S,63° 48'
L. AC 281. 67
|
12.4026580
9.9529175
2.4497405
CASE 2.
292*
Book I..
MENSURATION.
↑
CASE 2. Given the fide AB 532, BC 358, and the
angle C 107° 40'.
AB : BC :: S,C: S,A
532
358 S, 107° 40′ S, A
:
L. S, 107° 40', or 72° 20' + L. 358 12.5329022
L. 532
L. S, A 39° 52′ 51″:7
2.7259116
9.8069906
S, C: S, B
:: AB : AC
S. 107° 40′ :
S, B ::
B:
532 : AC
L. S. 32° 27′ 8″.3 + L. 532
12.4555601
L. S. 72° 20′
L. AC 299.6
=
9.9790192
2.4765409
CASE 3. Given the fide AB 176, BC 133, and the
included angle ABC 73°.
A+ C
C-A
AB + BC : AB
BC:: T,
I,
2.
2
}
309 43 T, 53° 30′ : T,
C—A
2
L. T. 53° 30′ + L. 43
L. 3༠9.
11.7642596
2.4899585
C-A
L. T.
10° 39′ 2″.7
2
77
1
9.2743011
S, C
Book I.
293
MENSURATION.
S, C: S, B :: AB : AC.
S, CS, 73°
L. S, 73° + L. 176
L. S, 64° 9′ 2″.7
176
AC.
12.2261090
9.9542156
2.2718934
L. AC = 187.022
CASE 4. Given the three fides AB 150, AC 140,
and BC 130.
½
AC: AB + BC :: AB — BC: ED.
1 1
70: 140
::
10: ED = 20.
AB AD: R: Cos. A
150: 90
:: R: Cos. A 53° 7′ 48″.4
In like manner, we find the angle C 67° 22' 48'.4,
whence B = 59° 29′ 23″.2.
Otherwife,
BC+D BC D
AB × AC:
X
:: R²: S², A,
2
AC.
: : R² : S² ± A.
where D AB
150 × 140: 60 × 70
L. 60 + L. 70 + L. R²
L. 150+ L. 140
L. S², A
L. S. A 26° 33′ 54″ 18.
N4
23.6232493
4.3222193
19.3010300
9.6505150
Hence the whole angle A = 53° 7′ 48″.36. In like
manner, we find C = 67° 22′ 48″.5, and therefore
B = 59° 29′ 23″.14.
Thefe
'
:
294
MENSURATION. Book I.
:
Theſe Cafes or Problems may be constructed geo-
metrically, and the unknown parts meaſured by
means of a ſcale of equal parts, and a line of chords.
A fimple ſcale is a ſtraight line divided into a certain
number of equal parts, and having the firſt of theſe
fubdivided into 10 equal parts: So that, if the fmaller
diviſions be taken for units, the larger will be tens;
or, if the former be tens, the latter will be hundreds.
Therefore, any number expreffed by two digits may
be taken from fuch a fcale.
A diagonal fcale is that which is defcribed upon a
fimple one, in fuch a manner, as to exhibit the 10th
parts of the ſmaller divifions, fo that any number of
three figures may be taken from it. The method of
conſtruction is evident from the figure.
A line of chords is a chord of 90° taken from any
circle, and fo divided, as to exhibit the chord of every
arch of the quadrant, confifting of an exact number
of degrees. The chord of 60° gives the radius of the
circle from which it was taken.
EXAM PL E.
AMPLE.
Fig. 13.
To conftruct the third cafe in oblique angled Triangles.
part
With the chord of 60°, as a radius, defcribe
of the circumference of a circle. Apply the chord of
73°
1
Book I.
295
MENSURATION.
73° from m to n. From the center B draw the lines.
Bm, Bn. Upon the one ſet off 176 equal parts from
B to A, and 133 upon the other, from B to C. Join
A, C, and the triangle is conftructed. Then AC,
meaſured upon the fame diagonal fcale is 187, the
angle A, meaſured by the fame line of chords, is 43°,
and the angle C 64°,
The meafuring of Heights and Distances.
The inftruments commonly uſed for meaſuring
heights and diſtances are, a chain, a quadrant, a
ſquare, and a theodolite.
A chain is uſed for meaſuring thoſe diſtances or
lines which are to be given fides of triangles. The
Engliſh chain is in length 4 poles, or 66 feet. It con-
fifts of 100 equal links, made of iron; each link,
therefore, fhould be 7.92 inches long. Every ten
links, from one end to the middle of the chain, is di-
ftinguiſhed with a piece of brafs.
A quadrant is uſed for determining vertical angles.
It is made of brafs or wood, the radius being of any
convenient length, the circumference is divided into
90 equal parts, and thefe again fubdivided, as far as
the dimenfions of the quadrant will admit. Alſo, a
plummet is fufpended by a thread from the center.
A
296
Book I.
MENSURATION.
2
Fig. 14.
A fquare is uſed for finding the ratio of the fides
of a right angled triangle. It is made of the fame
materials. Two of its fides are divided each into
100 equal parts. And a plummet hangs from the op-
pofite angle.
A theodolite is ufed for meaſuring horizontal as
well as vertical angles. It is à circle of braſs divided
into degrees, &c. having an index moveable about its
center, and is furniſhed with ſights.
1
PROBLEM I.
To find the height of an acceffible object ſtanding upon
level ground.
I. By the QUADRANT.
Let any convenient diftance BA be meaſured by
the chain, in a direct line from the foot of the per-
pendicular BC, that falls from the top of the object.
Then, ftanding at the point A, let the quadrant be
held as reprefented in the figure, fo that the eye at D
may fee the top of the object C, along the fide of the
quadrant DF. Now, if the plummet hang freely, the
line FP will be perpendicular to the horizon, and
therefore parallel to BC; hence the angles DFP,
DCE are equal, and their complements GFP, CDE
alfo
Book I.
297
MENSURATION.
alfo equal. Thus GN, the arch of the quadrant that
is remote from the eye, will give the number of de-
grees in the angle of elevation CDE. Whence, in
the right angled triangle CED, the fide DE (= AB,)
and the angle CDE being given, we may find CE, to
which EB, (= DA), the height of the eye above the
ground being added, we get the whole height of the
object.
If the angle of elevation be 45°, then DE EC,
that is, the diſtance meaſured is equal to the height
of the object above the eye.
II. By the SQUARE.
Having meaſured AB, as before, hold the fquare to Fig. 15.
the eye at D, as in the figure. Then, the plummet
hanging freely, the line FP cuts off from the fquare a
fmall triangle fimilar to CDE. Therefore we ſhall
have the ratio of DE to EC, and the former being
given, the latter may be found by the rule of propor
tion. Let n reprefent the number of equal parts,
which the plummet cuts off from the fide DH or HG,
towards the end D or G. Then,
1. When the plummet cuts the fide GH, remote
from the eye, it is as 100: n :: DE: EC. Hence,
if in this cafe, DE = 100, then EC = n.
2. When it paffes through the oppofite angle H,
we have a ratio of equality, DE = EC.
Pp
3. When
298
BOOK I.
MENSURATION.
}
3. When it
cuts the fide DH contiguous to the
eye, it is as n
:
100 :: DE: EC.
PROBLEM
II.
Fig. 16.
To find the Height of an Inacceffible Object.
I.
1. By the QUADRANT.
From any convenient ftation B, meaſure the diſtance
BA in a direct line with the foot of the object, and at
both ftations A, B, take the angles of elevation DAC,
DBC. The difference of thefe angles will give the
angle ADB.
Then, in the triangle ABD, S,ADB:S,A:: AB : BD
And in the triangle BCD, R: S,DBC:: BD: DC
Whence Rx S,ADB: S,A × S,DBC :: AB : EC,
that is, as the rectangle under the radius, and the fine
of the difference of the angles of elevation, is to the
rectangle under their fines, fo is the diftance mea-
fured to the height of the object above the eye.
ANOTHER SOLUTION.
In the right angled triangles DCB, DCA, to the
fame radius DC, AC, CB are the cotangents of the
angles
Book I.
299
MENSURATION.
.
angles of elevation, and their difference is AB; hence
this proportion. As the difference of the natural co-
tangents of the angles of elevation is to the radius, fo
is the diſtance meaſured to the height of the object
above the eye.
II. By the SQUARE.
At the ſtation A find by the fquare the ratio of AC
to CD, and, at the ſtation B, the ratio of BC to CD ;
hence the ratio of AC to BC, and confequently that
of AB to BC will be given, from which BC, and
confequently CD may be found.
Let AB = d, AD: CD:: mn, and BC: CD ::
1,
P:q; then CD =
nqd
mq
пр
9
If, at both ſtations,
the plummet cut the fide of the fquare remote from
the eye, CD =
nq
d
X
If the fide conti-
오
​72
100
Food
If the plummet
การ
p
guous to the eye, CD =
cut the oppofite angles of the fquare at the firft ftation
nd
B, CD =
100
n
PROBLEM
*
300°
MENSURATION. BOOK I.
Fig. 17.
PROBLEM
III.
To find the distance of a given place from an inacceffibie
object.
Let A be the inacceffible object, it is required to
find its diſtance from the "given ftation B. Meaſure
any convenient diftance BC, as the bafe of a triangle,
whofe vertex is at A. Then, the theodolite being
placed at B, let the diameter be directed towards the
ſtation C, and the moveable index towards the object
A, and the intercepted arch fhews the number of de-
grees in the angle ABC. In like manner, let the
angle BCA be meaſured, and the angle at A will be
known. Then, S, A: S, C ::.CB: BA.
Fig. 18
PROBLE
E M
FRIOS
IV.
To find the distance between two inacceffible objects.
Let a proper diftance CD be meaſured as the baſe
of two triangles, whofe vertices are at the objects
A, B. Then the angles at C and D being meaſured
by the theodolite, we find, as in the laft problem,
the fides AD, DB: And, as the included angle ADB
is
Book I.
301
MENSURATION.
is given, the other angles of the triangle DBA may be
found by the 3d theorem, and then the fide AB by
the 2d theorem.
APPENDI X.
Properties of Signs, Tangents, &c.
PROPOSITION I. As the fum of the tangents of two
arches is to their difference, fo is the fine of the fum
of thoſe arches to the fine of their difference.
PROP. 2. Twice the rectangle, under the fine and
cofine of any arch, is equal to the rectangle under the
radius, and the fine of double the arch.
PROP. 3. As the fum of the fines of two arches is
to their difference, fo is the tangent of half their fum
to the tangent of half their difference.
PROP. 4. As the fum of the cofines of two arches.
is to their difference, fo is the cotangent of half their
fum to the tangent of half their difference.
PROP. 5. Half the rectangle under the radius, and
the ſum of the fines of two arches, is equal to the rec-
tangle under the fine of half their fum, and the cofine
of half their difference.
PROP. 6. Half the rectangle under the radius, and
the difference of the fines of two arches, is equal to
the
302
Book I.
MENSURATION.
K
the rectangle under the cofine of half their fum, and
the fine of half their difference.
PROP. 7. Half the rectangle under the radius, and
the verfed fine of any arch, is equal to the fquare of
the fine of half that arch.
PROP. 8. Half the rectangle under the radius, and
the difference of the verfed fines of two arches, is e-
qual to the rectangle under the fines of half their fum
and half their difference.
PROP. 9. As the fine of half the difference of two
бо
arches, which together make 20
difference of their fines, fo is 1 to
degrees is to the
2
PROP. 10. As the fine of half the fum of two
arches is to the fine of half their difference, fo is the
fine of their fum to the difference of their fines.
PROP. 11. The difference between the tangent and
cotangent of any arch is double the tangent of the
difference between the arch and its complement.
PROP. 12. The fum of the tangent and cotangent
of any arch is double the fecant of the difference be-
tween the arch and its complement.
Trigonometrical
Боок I.
303
MENSURATION.
Trigonometrical Theorems.
THEOREM I. In any right angled triangle, the fum
of the hypothenufe and one fide, is to the other
fide as the radius to the tangent of half the angle op-
pofite to the latter.
THEOR. 2. The difference of the hypothenufe and
one fide is to the other fide as the tangent of half the
angle oppofite to the latter is to the radius.
THEOR. 3. In any plane triangle, as the greater of
the two fides is to the lefs, fo is radius to the tangent
of an angle; and radius is to the tangent of the ex-
ceſs of 45°, above this angle, as the tangent of half
the fum of the angles at the bafe is to the tangent of
half their difference.
THEOR. 4. As the bafe of any plane triangle is to
the difference of the two Ades, fo is the fine of half
the fum of the angles at the bafe to the fine of half
their difference.
THEOR. 5. As the bafe is to the fum of the two
fides, fo is the cofine of half the fum of the angles at
the baſe to the cofine of half their difference.
THEOR. 6. In any plane triangle, twice the rec-
tangle under the two fides is to the excets of the fum
of their fquares, above the fquare of the bafe, as ra-
dius to the cofine of the vertical angle.
THEOR.
1
304
Book I.
MENSURATION.
.
THEOR. 7. As the rectangle under the two fides is
to the rectangle under half the perimeter, and its ax-
is above the bafe, fo is the fquare of the radius to the
fquare of the cofine of half the vertical angle.
THEOR. 8. As the rectangle under half the peri-
meter, and its excefs above the bafe, is to the rec-
tangle under its exceffes above the two fides, fo is the
fquare of the radius to the fquare of the tangent of
half the vertical angle.
THEOR. 9. As half the perimeter is to its excefs a-
bove the baſe, ſo is the cotangent of half either angle
at the baſe to the tangent of half the other angle.
THEOR. 10. As the excefs of half the perimeter a-
bove the leſs fide is to its exceſs above the greater, fo
is the tangent of half the greater angle at the baſe to
the tangent of half the leſs.
THEOR. 11. As the bafe is to the difference of its
II.
fegments, fo is the fine of the vertical angle to the
fine of the difference of the angles at the bafe.
THEOR. 12. As radius to the fum of the cotangents
of the angles at the baſe, fo is the perpendicular to the
perimeter.
Examples to the Cafes.
EXAMPLE 1. In a right angled triangle, given the
hypotherufe 185, and one of the acute angles 32° 40',
to find the other parts of the triangle.
Ex.
BOOK I.
305
MENSURATION.
Ex. 2. Given one of the fides 64, and the oppofite
acute angle 23° 15'.
3
{
Ex. 3. Given the hypothenufe 324, and one of the
fides 265.
Ex. 4. Given one of the fides 78, and the other
fide 59.
Ex.
5.
In an oblique angled triangle, given one of
the angles 69° 14, another 46° 27', and the fide be-
tween them 248, to find the remaining parts of the
triangle.
Ex. 6. Given one of the angles 28°, another 114°,
and the fide between them 57400 at
Ex. 7. Given one of the fides 215, another 169,
and the angle oppoſite to the former 74°.
Ex. 8. Given one of the fides 329, another 248,
and the angle oppofite to the latter 26°.
Ex. 9. Given one of the fides 79, another 67, and
the included angle 85° 16'.
Ex. 10. Given one of the fides 241, another 173:
and the included angle 103°.
Ex. 11. Given one of the fides 126, another 148,
and the third 96.
Ex. 12. Given one of the fides 119, another 196,
and the third 175.
10
Q q
Examples
P
306
i
MENSURATION.
BOOK I
Examples in meafuring Heights and Diſtances.
Ex. 1. Required the height of a tower, whoſe
angle of elevation, at the diſtance of 160 feet from
the bottom, is 51° 30.
Ex. 2. What is the height of a hill, whoſe angle of
elevation at the bottom is 34°, and, 120 yards farther
off, 29° 15′.
Ex. 3.
To find the diſtance between two inaccef-
fible objects, there are given the angles which it fub-
tends, at two affumed ftations, 60° and 66°, the di-
ftance between thefe ftations 320 yards, and the o-
ther angles, at the fame points, 43° and 44° 30′ re-
fpectively.
Ex. 4. To find the height of an object on the top
of a hill, there are given the elevation of the hill 33°,
and the elevation of the object at the fame ſtation
50°, alſo the elevation of the object, at another ſta-
tion, 80 yards, in a direct line from the former, 30%
Ex. 5. From the top of a tower, whofe height wast
360 feet, the angles of depreffion of the top and bot-
tom of an object, upon the fame horizontal plane,
were obferved to be 41° and 54° refpectively. The
height of that object is required.
Ex. 6. From the fummit of a hill, the angles of
depreffion of the top and foot of a fpire, upon the
plane below, whoſe height was known to be 120 feet,
were
i
BOOK I.
307
MENSURATION.
were obſerved to be
height of the hill?
30° and 33°.
Required the
Ex. 7. Two ftations,
at the diſtance of 60 yards,
were taken on the fide of a hill, from whence the
angles of depreffion of the foot of a tower, on an op-
pofite hill, were obſerved to be 70° and 80° 30′; alſo,
at the upper ſtation, the depreffion of the tower was
found to be 18°, and the elevation of the tower 5°,
and all the angles were taken in the fame vertical
plane. Required the height of the tower?
Ex. 8. On the top of a monument, whofe height is
60 feet, ftands a ftatue 12 feet high. At what diſtance
from the monument may the ftatue be viewed, under
an angle of 3° ? and what is the greateft angle under
which it can be viewed?
Ex. 9. The elevation of a tower, at one ſtation, is
20° 45', at another ſtation, 60 yards diftant from the
firft, (but not in the fame ftraight line with it, and
the foot of the tower), 19° 40, and, at a third ftation,
30 yards diftant from each of the former 21° 48'. Re-
quired the height of the tower?
Ex. 10. If a ftatue, 12 feet high, on the top of a
monument, appear to an obferver under an angle
of 4° 46', and, at the fame ftation, a part of the mo-
nument, of 20 feet from the top, fubtend an angle of
8° 30'; What is the height of the tower, and its di-
ftance from the place of obfervation ?
Ex.
܃
308
BOOK I..
MENSURATION.
1
}
Ex. 11. At three ftations, in the fame ftraight line
with the foot of a tower, the angles of elevation are
fuch, that the firft is the double, and the fecond the
complement of the third; alſo the diſtance of the firſt
and fecond ſtations is 27,
cond and third 100 yards.
the tower?
i
and the diſtance of the fe-
Required the height of
Ex. 12. The elevations of a tower, at three feveral
ſtations, (no two of which are in the ſame ſtraight
line with the foot of the tower), are 50° 45, 58° 15',
and 46° 45'; alfo, the diſtance of the first and fecond
ftations is 24, the diſtance of the fecond and third
38, and the diſtance of the firſt and third 50 yards:
What is the height of the tower?
MENSURATION.
?
!
**
MENSURATION.
*****
воо к
II.
OF PLANE SURFACES.
A Straight line being affumed, as the meaſuring
unit of lines, its fquare is affumed, as the mea-
furing unit of furfaces. The number of meaſuring
units which a furface contains is called its area. And
the general problem, in this part of menfuration, is to
find the area of any given plane figure.
The area of a rectangle may be found, by multi- Fig. 19.
plying the numbers which expreſs its length and
breadth. Thus, if AB contain 5 times the meaſuring
unit x,
and BC 3 times the fame; then the rectangle
ABCD will contain 15 times the meaſuring unit y,
which is the fquare of x.
For,
**
310
BOOK II.
MENSURATION
:
Fig. 20.
**
For, let AE be equal to x, and EF parallel to AB;
then the ratio AC: AF = AD : AE = 3 : 1;
3: there-
fore the rectangle AC is equal to 3 times AF; but,
fince AF: y = AB: x = 5: 1, AF is equal to 5 times.
7. Confequently AC is equal to 15 times y.
Hence the areas of all other rectilineal figures, alſo
the areas of circles, ellipfes, and parabolas, may be
obtained by means of the following Theorems.
THE ORE M I..
Every parallelogram is equal to the rectangle under its
bafe and altitude.
THE ORE M II.
1
Every triangle is equal to half the rectangle under its
bafe and perpendicular.
THEOREM
III.
As radius is to the fine of one of the angles of a triangle,
fo is half the rectangle under the containing fides to the
triangle.
THEOREM
Book II. MENSURATION. 311
THE ORE M IV.
1.
Every triangle is a mean proportional between the rec-
tangle under half the perimeter and its excefs, above
one of the fides, and the rectangle under its cxceffes,
bove the two remaining fides.
Let the angles A and B, of the triangle ABC, be Fig. 21.
bifected by the lines AG, BG meeting in G. Alſo,
let the exterior angle CBH be biſected by the line
BK meeting AG in K. From the points G, K, let
perpendiculars be drawn to each of the fides of the
triangle, and let GC, CK be joined.
Steps of the Demonftration.
ift, The perpendiculars GD, GE, GF, are equal;
and the fegments AD, AE; BD, BF; CE, CF e-
qual.
2d, The perpendiculars KH, KM, KL, are equal:
And the fegments AH, AL; BH, BM; CM, CL
equal.
3d, AH or AL is half the perimeter. Alfo BF
= CM, and BM = CF,
4th,
312
BOOK II.
MENSURATION.
Fig. 22.
Fig. 23.
:
4th, BH is the exceſs of half the perimeter above
AB, DB its exceſs above AC, and AD its exceſs a-
bove BC.
1
5th, The rectangle under AH and DG is equal to
the triangle ABC.
6th, From the fimilar triangles ADG, AHK ;
AD : DG :: AH: HK; therefore AH × AD: AH
X DG AH x DG: HK x DG.
7th, The triangles GDB, BHK are fimilar; there-
fore HK × DG = DB × BH.
8th, Hence AH × AD : ABC:: ABC: DB × BH.
COR. Hence, if a, b, c repreſent the numbers which
meaſure the three fides of a triangle, and s half their
fum, the area of that triangle will be,
S X S
a X s
b x s
C
THEOREM V.
Every quadrilateral figure is equal to half the rectangle,
under one of its diagonals, and the fum of the perpendi-
culars falling upon it from the oppofite angles.
THE ORE M VI.
Every quadrilateral figure, having two of its fides paral-
lel, is equal to half the rectangle under the fum of the
parallel fides, and the distance between them.
I
THEOREM
Book II.
313
MENSURATION.
7
#
THEOREM
VII.
In any quadrilateral figure, radius is to the fine of the Fig. 244
angles formed by its diagonals, as half the rectangle
under thefe diagonals to the quadrilateral.
THE OREM VIII.
In
any regular polygon, radius is to the tangent of half
the angle of the polygon, as half its fide to the radius
of the infcribed circle; and the polygon itſelf is equal to
half the rectangle under its perimeter, and the radius
of the infcribed circle.
SCHOLIU M.
By this theorem, the areas of the following regular
polygons (the fide of each being 1) have been calcu-
lated, and put down in the Table oppofite to their
names. Hence it will be eafy to find the area, when
the fide is expreffed by any other number.
For, fince fimilar polygons are to one another as
the fquares of their homologous fides, the area of any
of thefe polygons may be found, by multiplying the
fquare
Rr
314
Book II.
MENSURATION.
!
fquare of its fide by the number ſtanding oppofite to
its name in the Table.
No. of Sides.
Names.
Areas.
3
Trigon.
0.433013
4
Tetragon.
1.000000
A
5
Pentagon.
1.720477
{
6
Hexagon.
2.598076
7
Heptagon.
3.633912
:
8
Octagon.
4.828427
9
Nonagon.
6.181824
10
Decagon.
7.694209
II
Undecagon.
9.3656 41
1
1 2
Duodecagon.
11.196152
i
THE ORE M
IX.
Every circle is to the fquare of its diameter as 7854 to 1
nearly.
For, the diameter of a circle being 1, its circumfe-
rence is 3.1416 nearly, and every circle being equal
to
A
FIG. 13.
B
FIG. 14.
C
F
1
D
THE G
E
B
E
A
}
B
C
B
D
FIG. 18.
E
FIG. 20.
MENSURATION
FIG. 15.
PL. II.a.314.
T
DAG
F
Ꮐ
D
FMP
A
E
G
TH
E
T
DA
BA
PH
E
B
A
B
FIG. 16.
B
C
F
بتا
D
E
C
FIG. 17.
A
B
FIG. 19.
B
C
F
M
X
y
B
H
C
B
FIG. 21.
F
E
D
FIG. 23.
E
E
FIG. 22.
FIG. 24.
K
;
B
}
:
Book II.
MENSURATION.
2
315
to half the rectangle under the radius, and a ſtraight
line equal to the circumference, its area is X = X
3.1416.7854. But circles are to one another as
the ſquares of their diameters. Therefore .7854:1
is the ratio of any circle to the fquare of its diameter.
COR. I. Let D, C, A reprefent the diameter, cir-
cumference, and area of a circle. Then A .7854D,
A =.07958 C³‚D = √1.2732 A, and C=/12.5664A.
COR. 2. Let N be the number of degrees in the arch
of a ſector. Then .01745 RN × R, or .008726 R-N
24
is the area of the fector to the radius R.
1
THE ORE M X.
Every ellipfe is to the rectangle, under its tranfverfe and
conjugate axis, as .7854 to 1 nearly.
Let ACBD be an ellipfe, of which AB is the tranſ- Fig. 25,
verfe axis.
From F, any point in the curve, let FG be perpen-
dicular to AB, and meet the circle defcribed upon it
in E. Let FT, ET be the tangents at F, E, and thro'
the point n, indefinitely near F, let mnp be drawn pa-
rallel to EG. Then,
The triangle EGT:FGT::EG: FG
And
mpT: npT: EG: FG
Therefore Ep: Fp EG: FG::AB: CD.
:
Hence
316
Book II.
MENSURATION.
Hence every portion of the circle, between two pa-
rallel ordinates, is to the correfponding portion of the
ellipfe, in the conſtant ratio of AB to CD. Confe-
quently the whole circle AHBK: ellipſe ACBD::
AB: CD:: AB² : AB × CD. But AHBK : AB' ::
7854 1. Therefore ACBD: AB x CD::.7854:1.
COR. 1. Ellipfes are to one another as the rec-
tangles under their axes.
COR. 2. Every ellipfe is equal to a circle whofe di-
ameter is a mean proportional between the tranfverfe
and conjugate axis.
COR. 3. If a circle be defcribed upon either axis.
of an ellipfe, as that axis is to the other, fo is the
circle to the ellipfe.
!
COR. 4. In the fame ratio, are the fegments, (and
half fegments), cut off by an ordinate to the common
diameter. The parts intercepted between two ordi-
natės (or femi-ordinates) to the common diameter.
The fegments formed by joining the extremities of
thefe femi-ordinates. And the ſectors formed by
ftraight lines, drawn from any point in the common
diameter, to the points where two perpendiculors to
it meet the curves.
Thus, AB: CD = EBG : FBG = HOGE : COGF
-HE: CF = OLB : OMB OLK: OMI. '
=
THEOREM
Book II. MENSURATION.
317
THE ORE M
XI.
!
Every fegment of a parabola is two thirds of its circum-
fcribing parallelogram.
Let ABC be a fegment of a parabola, cut off by Fig. 26.
the ftraight line AC. Let AC be bifected in F, the
tangent DBE drawn parallel to it, and B, F joined
Then is BF a diameter, and AC its ordinate.
Alfo,
let AD, CE be drawn parallel to BF. The parabola
ABC ſhall be two thirds of the parallelogram AE.
For, let AG, the tangent at A, meet BF, produced
in G, and through the points o, t, in the curve, or
tangent, equidiftant from the point A, and indefinitely
near it, let mn, xy, pq, rs, be drawn parallel to AF,
AD. Then, becaufe the parallelograms An, Ap, have
AF: AD + Am:
a common angle at A; An : Ap
==
Aq. But Am oq, and oq: Aq=FG: AF. There-
fore An Ap AF: AD + FG AF FG: AD.
And, fince the fubtangent FG is always bifected in
B, the vertex of the diameter, FG 2BF = 2AD.
Whence An: Ap 2: 1. Thus, the parallelogram
An is the double of Ap, and the parallelogram mnxy
the double of pqrs. But otxn = mnN", and otsp = pqrs.
Confequently otxn the portion of the parabola, be-
tween two indefinitely near femi-ordinates, is always
the
318
MENSURATION. Book II.
Fig. 27.
i
:
the double of otsp, the correfponding portion of the
external ſpace BDA. Therefore alfo the whole ABF
is double the whole BDA. Whence ABF is of
FD, and ABC of AE.
COR. 1. In like manner, may the quadrature of a-
ny curve, whofe fubtangent and abfcifs are in a con-
ftant ratio, be obtained.
COR. 2. Let AB, CD, EF, be three equidiftant di-
ameters, cut at right angles by the ſtraight line BDF,
the parabolic fpace ACEFB is equal to a reflangle
under AB + 4CD+ EF, and BD.
I
For, let the diameters AB, EF meet the tangent at
Then the quadrilateral
C in m, n.
AF AB EF × BD.
And 2mF4CD 'x BD.
Confequently (AF 2mF =) 3ACEFB
AB+4CD + EF x BD.
+
COR. 3. All parabolas, on the fame, or equal ba-
fes, and having equal altitudes, are equal.
Cor. 4. If two parabolas, on the fame, or equal
bafes, be between the fame parallels, any two parts,
intercepted by lines parallel to the bafe, are equal.
If C, c be two parallel chords, in any pa-
rabola, and D the diftance between them; then is
COR, 5.
C³
C
xD,
C²
с
C+
C²
D, or C + C + c
xD, the expref-
fion for the area of that part of the parabola which
they intercept.
I
THEOREM
Book II. MENSURATION.
319
THE ORE M XII.
Every fector of a parabola, formed by two straight lines,
drain from the focus to the curve, is equal to half the
external ſpace between perpendiculars falling from the
extremitics of the bafe upon the directrix.
D
From the extremities A, B, of the baſe of the fec- Fig. 28.
tor AFB, let the perpendiculars AD, BE fall upon
I
the directrix CD; the AFB = ADEB.
For, let two points, p, q be affumed in the curve,
or in the tangent AG, equidiftant from the point A,
and indefinitely near it. Let Fp, Fq be joined,
mDn drawn parallel to AG, and the parallelogram
mnpq completed. Then, becaufe the parallelogram
mp, and the triangle Fpq ftand upon the fame baſe
pq, and have equal altitudes, (for the tangent AG bi-
fects FD at right angles), the former is double the
But pqrs =*mp; therefore pqrs = 2Fqp, and
conſequently the whole external ſpace ADEB double
the whole fector AFB.
latter. But pqrs =
!
}
PROBLE M.
A
320
MENSURATION. Book II.
1
PROBLEM.
Fig. 29.
To find an approximation to the area of any curve dif
ferent from a Parabola.
Let AGN be any curve, different from a parabola,
and BO any ſtraight line, upon which, from the curve,
there fall two perpendiculars, AB, NO, it is required
to find an approximation to the area of the quadrila-
teral ABON. Let BO be divided into an even num-
ber of equal parts, by the perpendiculars CD, EF,
GH, IK, LM, falling upon it from the curve, and let
the number of thefe perpendiculars be fuch, that any
two adjacent parts of the curve may nearly coincide
with the curve of a parabola paffing through their ex-
tremities, and having its axis parallel to the perpen-
diculars. Then the ſpace
AEFB AB + 4CD + EF x
EIKF = L F
INOK
BD nearly.
EF + 4GH + IK × BD.
I
= IK + 4LM + NO × 3 BD.
Therefore ABON AB + 4CD + 2EF + 4CH + 2IK
÷ 4LM + NO × BD nearly.
उ
Hence, if F and L. denote the first and laft terins,
in the ſeries of numbers that meafure the perpendi-
culars, E the fum of all the even terms, R the fum of
all the reft, and D their common diftance, we have
this
1
•
Book II. MENSURATION.
321
this general expreffion F+L+4E+2RxD for an
approximation to the area of the quadrilateral ABON,
whatever be the curve AN.
EXAMPLE I.
Let it be required to find the area of the quadrant ABC,
whereof the radius AC = 1.
Let AC be bifected by the perpendicular DE, and Fig. 30.
CD divided into 4 equal parts by the perpendiculars
mn, pq, rs.
I
Then BCDE = BC + DE + 4mn + 4rs
+2pq x CM nearly. Now, BC = 1, DE
77272
8
√55, rs=√√63, pq =
2
√√√39
÷.
√60, and Cm =
Thus, BC+DE = 1 + ÷ √√√3 = 1.8660
I
4mn + 4rs = ½ √55+÷√√/63=7.6767
2pq =
2
15
*
1.9365
I
11.4792 X 24= •4783 =
fubtracting the
BCDE, from which,
triangle CDE
There remains the fector BCE
.2165
= .2618
The triple of which is the quadrant ABC .7854
Or the area of a circle, whofe diameter is 1I.
S s
EXAMPLE
322
MENSURATION. Book II.
EXAMPLE II.
Fig. 31.
To find the area of the hyperbola FDM, of which the
abfcifs FM = 10, femiordinate MD= 12, and femi-
tranfverfe CF = 15.
Let FM be divided into 5 equal parts, by the femi-
ordinates H, mn, pq, rs.
Then HI, n= $ √3+, pq = 5/6, and rs
= √19.
I 2
HI + MD = 16.8
!
4mn + 4rs
= 68.8339
209
17.6363
= 3
X
The figure HIMD 103.2702 x 68.8468
to which adding HIF
6.4, confidered as a portion
of a parabola, we have 75.2468 for the area of the
hyperbola.
EXAMPLE
III.
Fig. 32.
Let AF, AE be the affymptotes, and C the vertex
of an equilateral hyperbola; let CB be perpendicular
to AE, and AB or BC= 1, required the area of the
ſpace BCDE, when BE 1, and DE perpendicular.
to
3
Book II.
323
MENSURATION.
6
C
to AE. Let BE be divided into 6 equal parts, and
the feries of perpendiculars at the points of fection
will be,,, TO IT, F + L = 1.5, 4E8.27705,
2R = 2.7, and the figure BCDE 12.47705 × 75
= .69316.
=
Ι
X J S
OF SURVEY IN G.
The moft ufeful inftruments for furveying, are the
chain and plane table. A ftatute acre of land being
160 ſquare poles, the chain is made 4 poles in length,
that 10 fquare chains (or 100,000 ſquare links) may
be equal to an acre. The plane table is ufed for
drawing a plan of the field, and taking fuch angles
as are necellary to calculate its area. It is of a rec-
tangular form, and is furrounded with a moveable
frame, by means of which a fheet of paper may be
fixed to its furface. It is furniſhed with an index,
by which a line may be drawn on the paper, in the
direction of any object in the field, and with ſcales of
equal parts, by which fuch lines may be made pro-
portional to the distances of the objects from the plane
table, when meafured by the chain; and its frame is
divided into degrees, for obferving angles.
PROBLEM
*
f
•
324 MENSURATION.
BOOK I
Fig. 33.
PROBLEM
To measure a field with the chain.
I.
Let AmBCDq reprefent the field to be meaſured.
Let it be refolved into the triangles AmB, ABD,
BCD, AqD. Let all the fides of the large triangles
ABD, BCD, and the perpendiculars of the fmall ones,
AmB, AqD, from their vertices m, q, be meaſured by
the chain, and the areas calculated: Their amount is
the area of the whole.
But if, on account of the curvature of its fides,
the field cannot be wholly refolved into triangles, then
either a ſtraight line may be drawn over the curve
fide, fo that the parts cut off from the field, and thofe
added to it, may be nearly equal, or, without going
beyond the bounds of the field, the curvilineal ſpaces
may be taken fo fmall, that they may be confidered
as fegments of a parabola, and meaſured according-
ly.
Fig. 34.
PRO BLE M
II.
To measure a field with the plane table.
Let the plane table be fixed at F, about the middle
of the field ABCDE, and its diftances FA, FB,
FC,
Book II.
'325
MENSURATION.
FC, &c. from the feveral corners of the field meaſu-
red by the chain. Let the index be directed from
any point affumed on the paper to the points A, B,
C, &c. fucceffively, and the lines Fa, Fb, Fc, &c.
drawn in thefe directions. Let the angles which
thefe lines contain be obferved, and the lines them-
felves made proportional to the diftances meaſured.
Then, their extremities being joined, there will be
formed a figure abcde, fimilar to that of the field 1;
and the area of the field may be found by calculating
the areas of the feveral triangles of which it confifts.
PROBLEM
III.
هر
To plan a field from a given bafe line.
Let two ſtations, A, B, be taken within the field, Fig. 35.
but not in the fame ftraight line, with any of its cor-
ners, and let their distance be meafured. Then, the
plane table being fixed at A, and the point a affumed
on its 'furface directly above A, let its index be di-
rected to B, and the ſtraight line ab drawn along the
fide of it, to reprefent AB; alfo, let the index be di-
rected from a, to an object at the corner c, and an in-
definite ftraight line drawn in that direction; and fo
of
}.
326
Book I
MENSURATION.
I
of every other corner fucceffively. Next, let the
plane table be fet at B, fo that b may be directly over
B, and ba in the fame direction with BA, and let a
ftraight line be drawn from b, in the direction BC.
The interfection of that line with the former, it is evi-
dent, will determine the poſition of the point C, and
the triangle abc, on the paper, will be fimilar to ABC
in the field. In this manner are all the other angular
points to be determined; and theſe being joined,
there will be formed a reprefentation of the field.
If the angles at both ſtations were obſerved, as the
diſtance between them is given, the area of the field
might be calculated from theſe data; but the opera-
tion is too tedious for practice; it is uſual, therefore,
to meaſure fuch lines in the figure that has been con-
ſtructed, as will render the calculation eaſy.
APPENDIX.
THE ORE M S.
i
THEOREM 1. Every right angled triangle is equal,
J
to the rectangle under half its perimeter, and the ex-
cefs of half the perimeter above the hypothenufe.
THEOR..
:
In
H
FIG. 25.
D
FIG. 27.
F
72
m
K
E
MENSURATION
T
IN
B
M
FIG. 27
A
{72
C
B
D
F
B
D
F
FIG. 29.
I
I
G
E
pq
B
π q s
E
T
D 72
B
عا
F
FIG. 26.
n
F
m
E
FIG. 31.
FIG. 30.
H
B
D
F H
K
M
Ċ m p r D
A
F I
m
р
7
FIG. 32.
C
F
D
B
FIG.34.
D
77
E
FIG. 33.
72
A
B
E
A
A
FIG. 28.
M
E
E
BOOK II. MENSURATION.
327
THEOR. 2. As radius is to the fine of double one
of the acute angles of a right angled triangle, fo is
the fquare of half the hypothenufe to the triangle.
THEOR. 3. As radius to the tangent of half the
vertical angle of a triangle, fo is the rectangle under
half the perimeter, and its exceſs above the bafe to
the triangle.
THEOR. 4. Every triangle is a mean proportional
between the excefs of the fquare of half the fum of
the two fides above the fquare of half the baſe, and
the excefs of the fquare of half the bafe above the
fquare of half the difference of the two fides.
THEOR. 5. Every quadrilateral figure infcribed in
a circle is a mean proportional between the excefs of
the fquare of half the fum of two fides above the
fquare of half the difference of the other two, and the
exceſs of the fquare of half the fum of the latter a-
bove the fquare of half the difference of the former.
THEOR. 6. Every quadrilateral figure infcribed in
a circle, is a mean proportional between the rec-
tangle under the exceffes of half the perimeter, above
two of the fides, and the rectangle under its exceffes,
above the other two fides.
:
NUMERICAL
?
:
328
BOOK II.
MENSURATION.
:
;
{
1
NUMERICAL PROBLEMS.
PROBLEM I. Given two fides of a triangle 732,
and 456, and the included angle 112°; to find its
area.
PROB. 2. To find the area of a triangle, whofe fides
are 273, 294, and 315.
PROB. 3. In the quadrangular field ABCD, given
the fide AB 83, BC 31, CD 9, and DA 10 chains,
alſo ABC a right angle: Required the area?
PROB. 4. From the ſtation F, within the five-fided
field ABCDE, are meaſured FA 10.27, FB 9.35,
FC 7.48, FD 7.85, and FE 7.15 chains; alfo the
angle AFB 73°, BFC 75°, CFD 52°, and DFE 96:
Required the area?
PROB. 5. At two ftations, A, B, within the field
CDEFGH, diftant 1250 links, were obferved the fol-
lowing angles, ABC 137°, BAC 21° 30′, ABD 93° 15′,
BAD 52°, ABE 25°, BAE 136°, EAF 70°, ABF
15° 20′, BAG 81°, ABG 70° 30', BAH 26° 40′, and
ABH 136° 30': Required the area ?
PROB. 6. The fides of a triangle are 252 and 240,
and the baſe 156, alfo the fides of another triangle
cut off from it next the vertex, are 60 and 47. Re-
quired the area of the remaining part of the tri-
angle?
PROB.
Book II. MENSURATION.
329
PROB. 7. Given the fides of a triangle 169, 195,
Required the areas of the infcribed and
and 182
circumfcribing circles?
PROB. 8. The fides of a quadrilateral infcribed in
a circle, are 68, 43, 55, and 40: Required the a-
rea?
PROB. 9. The baſe of a triangle, and the perpendi-
culars falling from its extremities, upon the fides, are
123, 120, and 112.2, refpectively: Required the
fides and area of the triangle?
PROB. 10. The vertical angle of a triangle, whofe
baſe is cut in extreme and mean proportion by the
perpendicular, is 86°, and the fum of the including
fides 112 Required the fides and area?
PROB. II. The area of a triangle is 200, and the
baſe, which is a mean proportional between the two
fides, is 40: Required theſe fides ?
PROB 12. The two fides of a triangle are 640 and
360, and the bafe 800: Required the line bifecting
the vertical angle, and the areas of the triangles into
which it divides the whole?
PROB. 13. Required the fides and area of a tri-
angle, of which the perpendicular is a mean propor-
tional between the whole bafe and one of its feg-
ments, the vertical angle being 70°, and the fum of
the including fides 100?
PROB. 14. A triangle whofe fides are 123, 187,
and 200, is divided into three triangles, which are to
Fareed
Tt
one
339
Book II.
MENSURATION.
!
one another as 1, 2, 3, by ftraight lines drawn to the
extremities of theſe fides, from the fame point within
the triangle, the areas and unknown fides of theſe
triangles are required?
PROB. 15. The fides of a triangle are 300, 376,
and 484, alſo the angles fubtended by theſe fides, at
a certain point within the triangle, are 104°, 112°,
and 144°, refpectively Required the diftances of
that point from the angular points of the triangle.
PROB. 16. The fides of a quadrilateral infcribed in
a circle are 195, 93, 149, 85: Required its area and
diagonals, and likewife the area of the circle?
PROB. 17. In a right angled triangle, there is gi-
ven the diſtance of one of the extremities of the hy-
pothenufe from the center of the infcribed circle 4,
and the produced part of the fame, when continued,
to meet the perpendicular 2; to find the hypothe-
nufe and area?
PROB. 18. The fides of one triangle are 105, 252,
273, and the fides of another 396, 671, 715: Re-
quired the fides of a 3d triangle, fimilar to the 1ſt,
and equal to the 2d?
PROB. 19. The angles at the bafe of a triangle
are 70° and 80°, and the fides and baſe of another
are 460, 510, and 830: Required the fides and area
of a 3d, fimilar to the 1ft, and deſcribed about the
ad, fo that their bafes are parallel?
PROB.
1
BOOK II.
331
MENSURATION.
PROB. 20. Required the area of an ellipſe, whereof
the focal diſtance is 12, and the diſtance of the focus
from the directrix 27: Alfo, the areas of the feg-
ments of a parabola and hyperbola, cut off by an or-
dinate to the axis, the abfcifs being equal to the
tranſverſe axis of the ellipfe, and the focal diſtance
and parameter the fame in each curve?
1
MENSURATION:
*
!
{
MENSURATION.
воок
III.
OF SOLIDS AND CURVE SURFACES.
Fig. 36.
A
Cube, whofe linear fide is the meaſuring unit
of lines, is affumed as the meaſuring unit of
folids; and the number of meafuring units which a
folid contains is called its content or folidity.
The content of a rectangular parallelopipedon may
be found by multiplying the three numbers which
exprefs its length, breadth, and thickneſs, continually
together.
Thus, let ABCD-EFGH be a rectangular paralle-
lopipedon, of which the linear fide AB contains the
meaſuring unit x 5 times, the fide AD 3 times, and
the
T
*
BOOK III.
333
MENSURATION.
the fide AE 4 times; then the rectangle AC con-
tains
y, or the fquare of x, 15 times, and the folid
AG fhall contain z, or the cube of x, 60 times.
For, let the folid AG be cut by the plane KLM,
parallel to the baſe AC, and diftant from it by AK,
equal to x. Then, becauſe AC: y = AM : ≈, and
AC = 15y, AM = 15Z. But AE: AK (= AH: AL)
= AG : AM, and AE = 4AK, conſequently AG
4AM 60z.
Hence the folidities of priſms, pyramids, cylinders,
cones, and ſpheres, alfo of parabolic, elliptic, and hy-
perbolic conoids, may be obtained by means of the
following Theorems.
.
Τ Η Ε Ο REM I.
Every prifm or cylinder is equal to a rectangular paral-
lelopipedon of equal baſe and altitude.
THE ORE M II.
Every pyramid or cone is one third of the prifm or cylin
der of the fame baſe and altitude.
THEOREM
III.
The frustum of any pyramid or cone is equal to three
whole pyramids or cones of the fame altitude with the
fruftum,
334
MENSURATIO N. Book III.
fruftum, of which the greatest and least have their ba-
fes equal to the ends of the fruftum, and the other is a
mean proportional between them.
SCHOLIU M.
.
The demonftrations of theſe theorems are given in
the fixth book of Geometry; and they may alſo be
eafily deduced from the following axiom: Of two
folids, between two parallel planes, if the fections,
• by a plane parallel to theſe, at any altitude, be in
the fame ratio with the bafes, the folids themſelves
<
<
are to one another as their bafes,' upon which a-
xiom the demonftrations of the remaining theorems
depend.
Fig. 37.
THEOREM
IV.
Every sphere is two thirds of its circumfcribing cylinder.
Let the quadrant ACB, revolving around the ra-
dius AC, defcribe a hemifphere; the circumfcribing
fquare ADBC will at the fame time defcribe a cylin-
der, and the triangle CAD a cone. Of thefe folids
the
Book III MENSURATION.
335
the cylinder fhall be equal to the fum of the other
two.
-
For, through any point E, in AC, let EF be drawn
parallel to BC, meeting the diagonal CD, the cir-
cumference AB, and the oppofite fide BD, in the
points G, H, and F; and let C, H be joined. Be-
cauſe CEH is a right angled triangle, the circle de-
fcribed with the radius CH is equal to the fum of the
circles deſcribed with the radii CE, EH. But CH
EF, and, fince CA = AD, CE =
AD, CE = EG. There-
fore the circle defcribed with the radius EF is equal
to the fum of the circles defcribed with the radii EG,
EH; that is, of the three folids, the cylinder, cone,
and hemiſphere, which are between the fame parallel
planes, the fection of the cylinder, at any altitude, is
equal to the correfponding fections of the cone and
hemiſphere taken together. Confequently the cylin-
der is equal to the fum of the cone and hemifphere.
But the cone is one third of the cylinder; therefore
the hemifphere is two thirds of the fame, and the
whole fphere two thirds of the whole cylinder.
COR. 1. Hence .5236 D³ expreffes nearly the content
of a ſphere, whofe diameter is D.
COR. 2. Every fphere is equal to a cone, of which
the diameter of the bafe is equal to the diameter of
the fphere, and the altitude double the fame.
i
THEOREM
336
MENSURATION. BOOK III.
:
!
Fig. 37.
THEOREM
V.
The fruftum of a hemifphere is equal to the difference be-
tween a cylinder and a cone of the fame altitude with
the fruftum, the former having the radius of the
Sphere, and the latter the altitude of the fruftum for
the radius of its baſe.
COR. If D be the diameter of the hemifphere, and
A the altitude of the fruftum, the folidity of the
fruftum is .2618 A x D² — A².
THE ORE M
VI.
Every Sector of a sphere, confifting of a fegment and a
cone, upon the fame baſe, having its vertex in the
center, is two thirds of a cylinder, having the fame
altitude with the fegment, and the radius of the fphere.
for the radius of its bafe.
1
The fpherical fector, defcribed by AHC, is two
thirds of the cylinder defcribed by AF.
For, fince the circle defcribed by EF, is equal to
the fum of the circles defcribed by EH and EG, the
cone
J
Book III.
MENSURATION. 337
cone CEF is equal to the fum of the cones CEH and
CEG. Let the double of the cone CEF be added to
both, and the cone CEG be taken away, then the dif
ference between the cylinder EB, and the cone CEG,
that is, the fruftum EHBC, is equal to double the
cone CEF added to the cone CEH. Let the ſpheri-
cal fegment AHE be added to both, then, the he-
miſphere ABC is equal to double the cone CEF, ad-
ded to the ſpherical fector ACH, that is, two thirds.
of the cylinder AB equal to two thirds of the cylin-
der EB, added to the ſpherical fector AHC. Confe-
quently the ſpherical fector AHC is two thirds of the
cylinder AF.
COR. Every fector of a ſphere is equal to a cone,
of which the diameter of the bafe is equal to the dia-
meter of the ſphere, and the altitude double that of
its fegment.
:
THE ORE M VII.
Every parabolic conoid is the half of its circumfcribing
cylinder.
Let there be a cylinder, and a parabolic conoid, Fig. 38.
deſcribed by the revolution of the rectangle ADCB,
and the femi-parabola ACB, around the axis AB;
the cylinder is double the conoid.
U u
For,
338
MENSURATION. BOOK III.
For, let BHD be the fimilar and equal curve,
which paffes through B, D, having the fame axis
AB, and the equal femi-ordinate AD. Through any
point E, in AB, let EF be drawn parallel to BC,
meeting the two curves, and the oppofite fide of the
rectangle in G, H, F; and let P be the parameter of
the axis.
Then, fince EG P x AE
And
EH² = P × EB
EG² + EH² = P x AB = BC² = EF²
Therefore, of the folids deſcribed by ADCB, ACB,
ADB, between the fame parallel planes, the fection
of the cylinder, at any altitude, is equal to the cor-
refponding fections of the conoids taken together,
Confequently, the cylinder is equal to both the co-
noids, and the double of either of them.
OTHER WIS E.
Fig. 39.
Let GH be the femi-ordinate that bifects AB, and
let EGF be parallel to AB. Then twice the fquare
of GH is equal to the fquare of CB; and the ſum of
the fquares of any two femi-ordinates, equally diftant
from GH, is equal to the fquare of CB. Hence the
conoid AGH and fruftum CGHB taken together,
that is, the whole conoid ACB, is equal to the cylin
der AF, which is half the cylinder AC.
THEOREM
BOOK III. MENSURATION. 339
THE ORE M
VIII.
The fruftum of a parabolic conoid is equal to a cylinder of
the fame altitude, having its bafe equal to half the
fum of the circular ends of the fruftum.
THEOREM
IX.
Every Segment of a sphere is equal to the difference of a
cone and cylinder, each of which has the altitude of
the fegment for the radius of its bafe, the former ha-
ving alfo the altitude of the fegment, but the latter the
radius of the fphere for its altitude.
Let AHBC be a quadrant, C its center, ADBC the Fig. 40,
circumfcribing fquare, AB, CD, its diagonals, and
EF a ftraight line, drawn from any point E in AC,
parallel to BC, and meeting AB, CD, AHB, BD, in
the points M, G, H, F; and let the whole figure re-
volve around AC. The fpherical fegment, defcribed-
by AHE, is equal to the difference between the cy-
linder and conic fruftum, defcribed by AF and AG. th. 4.
But that difference is equal to the difference between
the cylinder and cone defcribed by AF and AME
revolving upon EF.
For
a
340
MENSURATION. Book III.
For, let the parabola AKB be defcribed through
A, B, having AE for its axis, and meeting EF in K.
The radius AC, it is evident, is equal to the parame-
==
ter of the axis; therefore AC x AE EK². But
2AC × AE + EC² = AC + AE, EM = AE,
EG EC, and EF AC. Therefore 2EK² + EG²
-
= EF² + EM.
-
Hence the double of the conoid
AKE, added to the conic fruitum ADGE, is e-
qual to the cylinder AF, added to the cone AME,
fince all thefe folids, defcribed by their refpective fi-
gures revolving around the common fide AE, lie be-
tween the fame parallel planes. But the double of
the conoid AKE is equal to its circumfcribing cylin-
der AK, and the cylinder AK is equal to the cylinder
AF, having FE for its axis, becauſe their bafes and
altitudes are reciprocally proportional. Therefore the
cylinder AF, of which FE is the axis, added to the
conic fruftum ADGE is equal to the cylinder AF, of
which AE is the axis, added to the cone AME, and
the difference between the firſt and laſt of theſe folids,
equal to the difference between the other two.
Confequently the ſpherical fegment defcribed by
AHE, revolving upon AE, is equal to the difference
of the cylinder and cone defcribed by AF and AME,
revolving upon EF.
THEOREM
BOOK III.
MENSURATIO N. 341
THE OREM X.
Every elliptic conoid is two thirds of its circumfcribing
cylinder.
Let the elliptic quadrant ABC, and its circumfcri- Fig. 41.
bing rectangle AKBC, revolving around the femi-
axis AC, deſcribe a conoid and cylinder; the for-
mer folid fhall be two thirds of the latter.
For, let the quadrant AGD be deſcribed about the
center C, with the radius AC, let AFDC be its cir-
cumfcribing fquare, and let EG, parallel to CD, from
any point E, in AC, meet the circle and ellipfe in the
points G, H. Then, becauſe EG: EH :: CD: CB,
the circles defcribed by thefe lines are alfo propor-
Hence the hemifphere AGDC conoid
cylinder AFDC: cylinder AKBC. Con-
fequently the conoid AHBC is two thirds of the cy-
linder AKBC.
tional.
AHBC
་་
:
COR. The fegment of the conoid defcribed by
AHE is equal to the difference of the cylinder and
cone defcribed by AN and AME revolving around.
EN.
:
THEOREM
342
MENSURATION. Book III.
THE ORE M
XI.
Every Spheroid, (defcribed by the revolution of a femi-
ellipfe about either axis), is two thirds of its circum-
fcribing cylinder.
THE ORE M XII.
Every fegment or fruftum of the elliptic conoid is to the
correfponding fegment or fruftum of the hemifphere as
the fquare of the fixed axis of the ellipfe is to the ſquare
of the other.
Fig. 42.
THE ORE M XIII.
¿
If an hyperbolic conoid, a parabolic conoid, and a cone,
have the fame bafe and altitude, the difference between
the first and fecond of thefe folids, is to the difference
between the fecond and third, as their common altitude
is to the fum of that altitude, and the tranſverſe axis
of the generating hyperbola.
Let the three folids be defcribed by the hyperbola
AGCB, the parabola AFCB, and the triangle AHCB
revolving
BOOK III.
343
MENSURATION.
revolving upon AB; let DA be the tranfverfe axis
of the hyperbola, and let the ſtraight line EHGF be
drawn parallel to the femi-ordinate BC, from any
point E in AB.
i
BC: EG²:: DB x BA: DE × EA
Then
And
EF²: BC²::
EA:
BA
Therefore
EF: EG:: DB
: DE
Hence
EF EG²: EF:: BE: DB
Again, becauſe
EF: BC:: AE: AB
And
Therefore
Hence
:
EF²: EH²::AB: AE
Confequently EF-EGLF-EH²:: AB: DB,
from which the propofition is manifeſt.
BC: EH :: AB AE
EF EF EH²::AB: BE
THE ORE M XIV.
The fruftum of a hyperbolic conoid is equal to the excess
of the corresponding frujcum of the afſymptotic cone, a-
bove a cylinder of the jame altitude, having the dia-
meter of its bafe equal to the conjugate axis of the
nerating hyperbola.
ge
Let AHD be a hyperbola, of which A is the ver- Fig. 43.
tex. C the center, and HK, DB femi-ordinates to the
axis, meeting the affymptote CE in G, F; let AE be
perpendicular to AC, and EL parallel to AB; and
let
344
MENSURATION. BOOK III.
1
:
let the whole figure revolve about the axis CA. The
hyperbolic fruftum HKBD is equal to the excefs of
the conic fruftum GKBF, above the cylinder KL,
becauſe the excefs of the fquare of GK, above the
ſquare of AE, is every where equal to the ſquare of
HK.
:
COR. The whole hyperbolic conoid is equal to the
exceſs of the circumfcribing fruftum of the affympto-
tic cone, above a cylinder of the fame altitude, ha-
ving the diameter of its baſe equal to the conjugate
axis of the hyperbola.
THE ORE M
XV.
The curve furface of a right cone is equal to the fector of
a circle, of which the radius is the hypothenufe of the
generating triangle, and the arch equal to the circum-
ference of the baſe of the cone.
THE ORE M XVI.
The curve furface of the fruftum of a right cone is equal
to the rectangle under half the fide of the fruftum and
a ftraight line equal to the fum of the circumferences of
its ends.
THEOREM
1
:
H
FIG.35.
D
MENSURATION
:
FIG.36.
H
E
G
F
PI.IN.la:344.
E
M
C
K
A
B
A
B
G
F
?
J
1
N
FIG.37.
D
H
F
B
F
FIG.38.
G
F
E
E
FIG.39.
D E
FIG. 40.
F
K M
白梅​路​齿
​K
FIG.41.
FIG.42.
D
F
FIG.43.
E
EZ Z
L
D B
N H
F
H
HE
G H
F D
I
C
J
BOOK III. MENSURATION.
345
THE ORE M
XVII.
The curve furface of a right cylinder is equal to the rec-
tangle under its altitude, and a straight line equal to
the circumference of its bafe.
THE ORE M XVIII.
The furface of a sphere is equal to the rectangle under
the diameter and circumference of the generating
circle.
!
For, as the fphere may be conceived to be made
up of an infinite number of cones, having a common
vertex in the center, and their bafes in the ſurface of
the ſphere, the whole fphere will be equal to one cone,
whoſe baſe is the furface, and altitude the radius of
the ſphere. But the fphere is alfo equal to a cone,
of which the diameter of the baſe is equal to the dia-
meter of the ſphere, and the altitude double the fame.
Wherefore thefe cones are equal to each other, and
their bafes and altitudes reciprocally proportional.
That is, R 2D::RC: S
Hence R x C: DX C:: RXC:S
Confequently S = D × C.
X X
COR.
1
346
MENSURATION. Book III.
COR. The furface of a fphere is equal to a circle.
whoſe radius is the diameter of the ſphere.
THE ORE M XIX.
i
The curve furface of the fegment of a sphere is equal to
the rectangle under its altitude, and the circumference
of the generating circle.
This theorem may be demonftrated after the fame
manner.
COR. The curve furface of the fegment of a ſphere
is equal to a circle whofe radius is the chord of the
generating arch.
THE ORE M XX.
The curve furface of the fruftum of a sphere is equal
to the rectangle under its altitude, and the circumfe-
rence of the generating circle.
APPENDIX.
J
BOOK III. MENSURATION.
347
APPENDIX.
NUMERICAL PROBLEMS.
PROBLEM I. To find the furface and folidity of
an upright triangular prifm, of which the fides of
the baſe are 36, 40, 68, and the altitude 80.
PROB. 2. To find the furface and folidity of an e-
quilateral pyramid, of which the linear fide is 60.
PROB. 3. Required the furface and folidity of a
cylinder, infcribed in an upright triangular prifin,
whereof the fides of the bafe are 17, 25, 28, and al-
titude 30.
PROB. 4. Of a triangular pyramid, infcribed in a
right cone, the fides of the bafe are 312, 90, 360,
and the altitude 468: Required the furface and fo-
lidity of the conè?
PROB. 5. To find the furface and folidity of the
fruftum of a right cone, whofe diameters are 18 and
8, and altitude 12.
PROB. 6. To find the furface and folidity of a
ſphere, whofe diameter is 60, and of a ſegment there-
of, whofe altitude is 10.
PROB. 7. To find the folidity of the three conoids,
the axis and diameter of the baſe in each being 45
and
348
MENSURATION. Book III.
}
and 30 refpectively, 45 being alfo the tranfverfe axis
of the generating hyperbola.
PROB. S. To find the furface and folidity of a
fphere, whereof the circumference of the generating
circle is 76.
PROE. 9. To find the furface and folidity of a
cone, whereof the fide is 30, and the circumference
of the baſe 49.
PROB. 10. Required the altitudes of the three e-
qual parts, into which the fruftum of a cone, whofe
diameters are 36 and 8, and altitude 42, may be di-
vided by fections parallel to the baſe.
PROB. 11. Required the weight of lead in a pipe
600 yards long, the diameter of the bore being 11
inches, the thickneſs of the metal, and the weight
of a cubic inch .40917 lb. Averdupois.
PROB. 12. A ftone in the form of a rectangular
parallelopipedon, meaſures 12 feet long, 7 feet
broad, and 8 inches thick. A pound weight, of the
fame kind of ftone, immerfed in a cylindrical veffel of
water 3.8 inches diameter, raiſes the fluid therein .85
of an inch. Required the weight of the ftone?
!
SPHERICS.
1
}
SPHERIC S.
རར་ས་ ་རན་་ར་རནར་་་་་་ར ར་སར་་ར
в о о к
I.
SPHERICAL GEOMETRY.
DEFINITION S.
I.
:
A
diameter.
Sphere is a folid conceived to be generated
by the revolution of a femicircle about its
2. The center of the femi-circle is equally diftant
from every point on the furface of the ſphere, and is
therefore called the center of the ſphere.
3. Circles of the fphere, whoſe planes paſs through
the center, are called great circles, and all others
fmall circles.
4. A ftraight line, drawn through the center of a-
ny circle of the fphere, perpendicular to its plane,
and
350
BOOK I.
SPHERIC S.
1
and limited on both fides, by the furface of the fphere,
is called the axis of that circle.
5. The poles of a circle of the fphere are the ex-
tremities of its axis.
6. By the diſtance of two points on the furface of
the ſphere is meant an arch of a great circle inter-
cepted between them.
7. By a ſpherical angle, or triangle, is always
meant that which is formed on the furface of the
fphere, by arches of great circles.
S. A quadrantal triangle is that of which one of
the fides is a quadrant.
9. A lunary furface is a part of the furface of the
fphere, contained by the halves of two great circles.
10. A fegment of a fphere is a part cut off by a
plane.
Fig. 1.
PROPOSITION
Every fection of a fphere is a circle.
I.
Let a plane cut the fphere AGH in any direc-
tion. If it pafs through the center, the fection is
evidently a circle. But, if it do not pass through
the center, let ABCD be the fection, and from E,
the center of the fphere, let EF be drawn perpendi-
cular to its plane; alſo, let FA, FB, FC, be drawn
in the plane, to meet the furface of the fphere. Then
EAS
ย
BOOK I. SPHERIC S.
351
EA, EB, EC, being joined, the right angled triangles
EAF, EBF, ECF, have equal hypothenuſes, EA,
EB, EC, becaufe they are radii of the fphere, and
one fide EF common to all; therefore their other
fides FA, FB, FC, are alſo equal. Confequently
ABCD is a circle, whofe center is F.
COR. 1. Any two great circles cut one another in
a diameter of the fphere, and therefore mutually bi-
fect each other.
COR. 2. Only one great circle can paſs through
the fame two points, in the furface of the ſphere, that
are not diametrically oppofite.
COR. 3. Any fide of a fpherical triangle is lefs than
a femicircle, and any two fides being produced, in-
terfect again at the diftance of a femicircle.
COR. 4. The two poles of any circle, its center,
and the center of the ſphere, are always in the fame
ftraight line, and that ftraight line is perpendicular
to the plane of the circle.
COR. 5. And therefore, if a line or plane be per-
pendicular to a circle of the ſphere, and paſs through
one of theſe points, it will pass through the other
three: Or, if it paſs through two of them, it will be
perpendicular to the circle, and alfo pafs through the
remaining two.
COR. 6. Hence, two great circles, whofe planes
are perpendicular, país through each other's poles;
and converfely.
COR.
3
352
Book I.
SPHERIC S.
COR. 7. And, if one great circle paſs through a
pole of another, the latter will paſs through the poles
of the former.
COR. 8. All parallel circles have the fame axis and
the fame poles; and converfely.
PROPOSITION
II.
Fig. 2.
Each pole of any circle of the sphere is equally distant on
the furface from every point in its circumference.
Let ABCD be any circle of the fphere, whofe cen-
ter is F, and axis GFH: Its poles G, H are each of
them equally iftant from its circumference.
For, let the bere be cut by planes paffing through
G, H, and let the fections, which will be great circles,
becauſe the line GH paffes through the center of the
ſphere, meet ABCD in the lines of common fection
FA, FB, FC. Then GA, GB, GC, being joined,
the right angled triangles GFA, GFB, GFC, have
the fides FA, FB, FC, equal, becauſe they are radii
of the fame circle, and one fide GF common to all;
therefore the hypothenufes GA, GB, GC, and con-
fequently the arches which they fubtend, are likewife
equal.
COR. 1. The pole of a great circle is at the diſtance
of a quadrant from its circumference.
COR.
BOOK I.
353
SPHERIC S.
COR. 2. Hence any plane paffing through the cen-
ter of the fphere, divides it into two equal parts,
which are therefore called hemifpheres.
COR. 3.
If a point in the furface of the ſphere be
at the diftance of a quadrant from other two points
not diametrically oppofite, it will be the pole of the
great circle paffing through them.
COR. 4. The radius of a ſmall circle is the fine of
its diftance from either pole to the radius of the
fphere, or the cofine of its diſtance from the parallel
great circle.
COR. 5. Hence, thofe fmall circles, whofe planes
are equally diſtant from the center, are equal; and
converſely; and of two circles unequally diftant, that
which is nearer the center is the greater; and con-
verfely.
COR. 6. Parallel circles intercept equal arches on
thofe great circles which pafs through their poles.
PROPOSITION
III.
The intercepted arch of a great circle, whofe pole is the
angular point, is the meaſure of a spherical angle.
Let ABC be a ſpherical angle, of which the angu. Fig. 3.
lar point B is the pole of the great circle ACD. Then
is the intercepted arch AC the meaſure of ABC.
Y Y
For,
354
Book I.
SPHERIC S.
RICS.
I. C. 2.
67. c. I.
For, let the tangents MB, NB, and the radii of the
fphere EA, EB, EC, be drawn. The angle MBN is
the fame with the fpherical angle ABC; but MBN is
equal to AEC; becauſe, fince AB, BC are qua-
drants, and BEA, BEC right angles, MB, BN are
parallel to AE, EC. Wherefore the ſpherical angle
ABC is equal to AEC, the meaſure of which is the
arch AC to the radius of the fphere.
COR. I. The circumferences of two great circles
cut each other at right angles, when their planes are
perpendicular; and converfely.
COR. 2. At the point of interfection of two great
circles, the oppofite angles are equal, two adjacent
angles are together equal to two right angles, and
each angle is equal to its oppofite one, at the other
point of interfection.
COR 3. The distance of the adjacent poles of two
great circles is the meafure of their inclination, or of
the fpherical angle.
For, fince AB, BC pafs through the poles of ACD,
ACD paffes through the poles of AB, BC; let P
be the pole of AB and Q, the adjacent pole of BC.
Then AP, CQ are quadrants, and AC = PQ.
COR. 4. The intercepted arch of any circle, whofe
pole is the angular point, is the meaſure of the fphe-
rical angle to the radius of that circle.
COR. 5. Two great circles, which pafs through the
poles of parallel circles, intercept fimilar arches.
PROPO.
:
Book I. SPHERIC S.
355
*
PROPOSITION
IV.
:
If a plane be perpendicular to the diameter of a fphere, at
one of its extremities, it touches the sphere.
Let the plane CD be perpendicular to AB, the dia- Fig. 4.
meter of the ſphere ABG, at its extremity B; then
CD touches the fphere in that point.
For, let F be any other point in CD, and EF, FB
be drawn. The angle EBF is, by hypothefis, a right
angle. Hence EF is greater than EB, and confe-
quently Fa point without the fphere. Thus the plane
CD meets the ſphere only in the point B, and there-
fore touches it.
1
COR. I. A ſphere and a plane can touch one ano-
ther only in one point.
COR. 2. If a plane touch a ſphere, the radius at the
point of contact is perpendicular to it.
COR. 3. If a plane touch a ſphere, a perpendicular
to it, at the point of contact, paffes through the cen-.
ter.
COR. 4. If a plane touch a ſphere, its line of com-
mon fection, with the plane of any circle of the
ſphere paffing through the point of contact, is a tan-
gent to that circle.
COR.
·
t
356
Book I.
SPHERIC S.
COR. 5. A tangent, to any circle of the ſphere, is
the common tangent of all the circles in whofe plane
it is.
Fig. 5.
PROPOSITION V.
If the angular points of any ſpherical triangle be made
the poles of three great circles, another triangle will be
formed by their interfections, fuch, that the fides of the
one triangle will be respectively the fupplements of the
measures of the angles oppofite to them in the other.
Let the angular points of the triangle ABC be the
poles of three great circles, which, by their interfections,
form the three lunary furfaces, DQ, FR, and EO;
A being the pole of EF, B the pole of DF, and C the
pole of ED. Then the triangle DEF, which is com-
mon to thefe three lunary furfaces, will be, in every
reſpect, fupplemental to the triangle ABC.
For, let each fide of ABC be produced, to meet
the fides that contain the angle oppofite to it, in the
triangle DEF. Then, becaufe BC paffes through the
poles of ED, DF; ED, DF muſt alſo paſs through
7. c. 1. the poles of BC: Therefore the points D, Q are the
b 1. c. I.
poles of BC. In like manner, R, F are the poles of
AB, and E, O, the poles of AC.
Hence EL, FK are
quadrants, and therefore EF the ſupplement of KL.
But,
1
Book I.
4
SPHERIC S.
357
But, fince A is the pole of EF, KL is the meaſure of
the angle at A. Thus, EF is the fupplement of the
meaſure of the angle at A. In like manner, FD is
the ſupplement of the meafure of the angle at B,
and DE the fupplement of the meaſure of the angle
at C.
Further, it will appear in the fame manner, that BC
is the fupplement of HM, the meaſure of the angle at
D, that AB is the fupplement of NK, the meaſure of
the angle at F, and that AC is the fupplement of GL,
the meaſure of the angle at E.
COR. I. In each of the three triangles, which, with
the ſupplemental triangle, make up the three lunary
furfaces, two of the fides are the meaſures of two
angles in the original triangle; and converfely. Alſo,
the third fide, and its oppofite angle, are the ſupple-
ments of the third angle, and its oppofite fide. They
are named femi-fupplemental triangles.
с
3.
COR. 2. Let a great circle paſs through D, A, the
vertices of the triangles ABC, DEF, it will cut the
baſes BC, EF at right angles ", becauſe it paffes thro' d5.c. 1.
their poles, and the fegments BP, PC, of the baſe, in
the one triangle, will be the complements of the mea-
fures of the vertical angles IDF, IDE, in the other,
taken alternately.
COR. 3. If the original triangle be right angled,
the fupplemental and femi-fupplemental, are quadran-
tal
358
BOOK I.
SPHERIC S.
і
tal triangles; or, if the former be quadrantal, the
latter are right angled.
!
:
PROPOSITION VI.
}
Of all the arches of great circles that can be drawn, from
point in the furface of a fegment of the fphere that is
not the pole of the bafe, to meet its circumference, that
which paffes through the pole is the greatest, the other,
in the fame plane with it, is the leaft, and that which
makes a leſs angle with the former is the greater of two
others.
Fig. 6.
5.c. I.
Let AGC be a fegment of the ſphere, cut off by the
plane of the circle ABCD; alſo, let G be the pole
of the bafe, and E any other point in the furface of
the fegment. Of all the arches of great circles which
fall from E, upon the circumference of the bafe,
EGA, which paffes through the pole, is the greateſt,
EC, in the fame plane with EGA, is the leaſt, and
EK, nearer to EGA, is greater than EB, more re-
mote.
Becauſe the plane AGEC paffes through the cen-
ter of the ſphere, and one of the poles of the circle
ABC, it is perpendicular to the plane of that circle,
and paſſes through its center F. Thus, AFC, their
line
BOOK I.
359
SPHERIC S.
line of common ſection, is a diameter of ABC. Let
EH be perpendicular to AC, and EA, EK, EB, EC,
HK, HB be drawn. Then, of all the ftraight lines
drawn from H to the circumference ABCD, HFA is
the greateſt, HC the leaft, and HK greater than HB.
Hence, in the right angled triangles EHA, EHK,
EHB, EHC, which have the fide EH common, EA
is the greatest hypothenufe, EC the leaft, and EK
greater than EB. Confequently the arch EA is the
greateft, EC the leaft, and EK greater than EB.
COR. 1. Of all the arches of great circles that fall
on the circumference of another great circle, the per-
pendiculars are the greateſt and leaſt, and only two
equal arches can fall from the fame point, one on
each fide of the plane of the perpendiculars, and they
meet it at equal diſtances from the perpendicular.
COR. 2. In every ifofceles triangle, of which the e-
qual fides are not quadrants, the perpendicular from
the vertex bifects the bafe, and, converfely, the arch
from the vertex which bifects the bafe is perpendicu
lar to it.
PROPOSITION VII.
Any two fides of a ſpherical triangle are together greater
than the third, and all the three fides are together less
than a circle.
In
!
360
Book I.
SPHERIC S.
*
Fig. 7.
a 6.
b
I.
In the ſpherical triangle ABC, any two fides are
together greater than the third. If the three fides be
equal, or if two of them be equal, and each of them
greater than the third, the propofition is evident. In
every other cafe, let AC be the greateſt, and let A be
the pole of the circle DCE, meeting AB produced in
D, E. Then BC is greater than BE, and confe-
quently the fum of AB, BC, greater than AE or AC.
Further, all the three fides AB, BC, CĂ, are to-
gether lefs than a great circle.
For, if AB, BC be produced to meet in F; BAF,
3. c. 1. BCF are femi-circles; and CA being lefs than the
fum of AF, FC, the fum of AB, BC, CA will alfo
be lefs than the fum of BAF, BCF, that is, leſs than
a whole circle.
PROPOSITION
VIII.
Fig. 8.
If two fides of a Spherical triangle be equal, the angles op-
pofite to them are alfo equal and converfely.
In the triangle ABC, if the fides AB, AC be
equal, the angles ABC, ACB will alfo be equal. If
AB, AC be quadrants, ABC, ACB are right angles.
If not, let the tangent to the fide AB at B, meet EA,
the line of common ſection of the planes AB, AC, in
F;
ཡཾ མྨཱམྨཱ ཙ ཙ སཏྟཱ ཏི ཝིཏྭཱཏཱ
BOOK I.
361
SPHERIC S.
F; and let the tangents to the bafe BC, at its extre-
mities, meet each other in G; alfo, let FC, FG, EC,
and EB, be joined.
have FE common, EB
AEC; therefore FB
FBE, a right angle.
Then the triangles FEB, FEC
=
EC, and the angle AEB =
FC, and the angle FCE =
Hence FC is a tangent, and the
triangles FGB, GCF mutually equilateral. There-
fore the angle FBG
= ACB.
FCG, and confequently ABC
Again, if the angles ABC, ACB be equal, the fides
AB, AC are equal.
For, in the femi-fupplemental triangle EQF, the Fig. 5
fides FQ, QE being the meaſures of the angles B, C,
will be equal; therefore the angles at F, E, and con-
fequently their meaſures AB, AC, are alfo equal.
COR. 1. Every equilateral fpherical triangle is alfo
equiangular; and converfely.
COR. 2. In any triangle, the greater angle is fub-
tended by the greater fide; and converfely. If the Fig. 9.
angle ACB be greater than ABC, let BCD ABC,
then BD = DC, and AB AD + DC, which is
greater than AC.
COR. 3. In every ifofceles triangle, of which the e-
qual fides are not quadrants, the perpendicular from
the vertex bifects the vertical angle, alfo, the arch
which bifects the vertical angle bifects the baſe at
right angles.
For, fince the angles B, C are equal, the fupple-
mental triangle DEF will be ifofceles; and therefore
Z z
the
362
Book I.
SPHERIC S.
2
b
ཤ
2. C. I.
2. c. 5.
↳
the fegments of the bafe EI, IF, made by the per-
pendicular, are equal, and their complements IL,
IK equal, which are the meaſures of the vertical
angles PAC, PAB.
The converfe appears in the
fame manner.
:
Fig. 5.
PROPOSITION IX.
All the angles of any fpherical triangle are together great-
er than two, and less than fix right angles.
In the triangle ABC, the three angles are together
leſs than fix right angles, becauſe, when added to the
three exterior angles, they only make fix. And they
are greater than two right angles, becauſe their mea-
fures GH, KL, MN, added to DE, EF, FD, the fides
of the ſupplemental triangle, are equal to three femi-
circles; and DE, EF, FD, being leſs than two ſemi-
circles, GH, KL, MN, muft be greater than one.
COR. I. The fum of all the angles of a fpherical
triangle, and double the fupplement of one of them,
is lefs than fix right angles.
For GH, KL, MN, added to DE, EF, FD, being
equal to three femicircles, or fix quadrants, and EF,
FD, greater than DE; GH, KL, MN, added to
twice DE, will therefore be lefs than fix quadrants.
COR.
Book I.
363
SPHERIC S.
COR. 2. Hence, if one angle of a ſpherical triangle
be right, the fum of the other two is greater than one,
but lefs than three right angles.
COR. 3. If one angle of a ſpherical triangle be
acute, all the three angles are together lefs than four
right angles.
COR. 4. If one angle of a ſpherical triangle be half
a right angle, or leſs, all the three angles are together
leſs than three right angles.
COR. 5. The exterior angle of any ſpherical triangle
is lefs than the fum of the two interior oppofite angles.
PROPOSITION X.
Any two angles of a spherical triangle are together great-
er, equal, or lefs, than two right angles, according as
the sum of the oppofite fides is greater, equal, or leſs,
than a femicircle; and converfely.
Let the fides AB, AC, of the ſpherical triangle
ABC, be produced to meet in D. Then it is mani-
feft that, according as the fum of AB, BC, is greater,
equal, or lefs than the femicircle ABD ; the ſide BC
will be greater, equal, or lefs, than BD; the angle D
(or A) will be greater, equal, or lefs than BCD; and
the fum of the angles BAC, BCA greater, equal,
or lefs, than the fum of BCA, BCD, which is two
right angles.
Cor.
364
Book I.
SPHERIC S.
COR. 1. According as half the fum of any two fides
of a ſpherical triangle is greater, equal, or lefs, than
a quadrant, half the fum of the two oppofite angles
will be greater, equal, or lefs, than a right angle;
and converſely.
COR. 2. According as one of the equal fides of an
ifofceles triangle is greater, equal, or lefs than a
quadrant, its oppofite angle will be greater, equal, of
lefs than a right angle,
Fig. 11.
#
PROPOSITION
XI.
In a right angled spherical triangle, according as either of
the fides about the right angle is greater, equal, or leſs
than a quadrant, its oppofite angle is greater, equal, or
less than a right angle; and converfely.
Let ABC be a triangle right angled at B, and let
the fides AB, BC be produced to meet in D. Then,
becauſe they pafs through each other's poles, E, the
middle of BAD, will be the pole BCD. Let a great
circle pafs through the points CE. The arch EC is
a quadrant, and the angle ECB a right angle. Now,
it is plain that, according as AB is greater, equal, or
lefs, than the quadrant EB, the oppofite angle ACB
will be greater, equal, or lefs, than the right angle
ECB; and converfely.
COR.
Book I.
365
SPHERIC S.
COR. 1. If the two fides be both greater, or both
lefs than quadrants, the hypothenufe will be lefs than
a quadrant; but, if the one be greater, and the o-
ther lefs, the hypothenufe will be greater than a qua,
drant; and converſely.
For, of the triangles ABC, ADC, right angled at
B, D, in which the fides AB, BC, are lefs, and con-
fequently AD, DC greater than quadrants, the hy-
pothenuſe AC is lefs than a quadrant, becauſe it is
nearer to BC than the quadrant CE. But, in the
triangle aBC, of which the fide aB is greater, and
BC lefs than a quadrant, the hypothenufe aC is great-
er than a quadrant, becauſe it is further from CB,
than CE is.
COR. 2. In every fpherical triangle, of which the
two fides are not both quadrants, if the perpendicular
from the vertex fall within, the angles at the baſe
will be both acute or both obtufe; but, if it fall
without, the one will be obtufe, and the other acute;
and converfely.
PROPOSITION
XII.
Any two circles of the fphere, paſſing through the poles
of two great circles, intercept equal arches upon them.
Let
366
SPHERIC S. Book I.
Fig. 12.
|
1
Let AFB, CFD be the two great circles interſect-
ing one another in F, and let F be the pole of the
great circle ACBD, cutting them in the diameters
AEB, CED. The circle ACBD paffes through their
poles; let the diameter MN be perpendicular to
AB, and PQ to CD; then M, N are the poles of
AFB, and P, Q, the poles of CFD. Let the ſmall
circle PGN paſs through the poles P, N, and cut the
circle ACBD in the line of common fection PKLN;
the arches BH, DG, of the circles AFB, CFD, in-
tercepted by the circle paffing through P, N, are e-
qual.
For, let EH, HK, EG, GL, PG, NH, be drawn.
The triangles PEL, NEK, are equal in every re-
fpect; hence PL = NK, and EL = EK.
=
The triangles PEG, NEH, are equal; therefore
PG NH; hence PH NG, and PNH = NPG.
=
The triangles NKH,
HK = LG.
PLG are equal; therefore
=
The triangles EKH, ELG are equal; therefore
the angle HEK = GEL, and HB GD. Hence the
arches intercepted between any two circles paffing
through P, N are equal,
APPENDIX,
A
*
.
!
}
i
C
D
D
E
32
FIG. 1.
H
A
FIG. 4.
FIG.7.
A
C
F
D
E
B
C
Ꮐ
Q
F
SPHERICS
FIG. 3.
PL.I.pa. 366.
B
M
FIG. 2.
B
F
H
R
FIG.5.
D
N
M
H
B
C
P
E
K
L
B
FIG.8.
FIG.1.
F
E
a
A
E
B
Q
FIG. 9.
B
F
G
CA
E
B
D
A
D
C
A
C
D
E
Q
C
P
D
F
B
Ꮐ
E
F
H
FIG. 6.
FIG.10.
B
FIG.12.
M
F
H
B
E
G
Q
N
D
"
BOOK I.
367
SPHERIC S.
:
APPENDIX.
PROP. I. Two ſpherical triangles, which have two
fides of the one equal to two fides of the other, each
to each, and the included angles alfo equal, are equal
in every reſpect.
PROP. 2. Two fpherical triangles, which have two
angles of the one equal to two angles of the other,
each to each, and the included fides alfo equal, are e-
qual in every refpect.
PROP. 3. Two fpherical triangles, which have the
three fides of the one equal to the three fides of the
other, each to each, are equal in every reſpect.
PROP. 4. Two ſpherical triangles, which have the
three angles of the one equal to the three angles of
the other, each to each, are equal in every reſpect.
PROP. 5. Two fpherical triangles, which have two
fides of the one equal to two fides of the other, each
to each, and an angle equal to an angle oppofite to
equal fides, are equal in every refpect, when the o-
ther two angles oppoſite to equal fides, are together
greater or leſs than two right angles.
PROP. 6. Two ſpherical triangles, which have two
angles of the one equal to two angles of the other,
each to each, and a fide equal to a fide oppofite
to equal angles, are equal in every refpect, when
the
368
BOOK I.
SPHERIC S.
the other two fides, oppofite to equal angles, are to-
gether greater or leſs than a femicircle.
PROP. 7. In every ſpherical triangle, if one of the
angles be acute, one of the fides is lefs than a qua-
drant; or, if one of the fides be greater than a qua-
drant, one of the angles is obtufe.
PROP. 8. In every ſpherical triangle, if the three
angles be acute, each fide is lefs than a quadrant ;
or, if each fide be greater than a quadrant, the
three angles are obtufe.
SPHERIC S.
I
SPHERIC S.
་་་ས“མ.
B O O K
:
II.
ངས་་་་རང་
SPHERICAL TRIGONOMETRY.
IN
PROPOSITION I.
N any right angled spherical triangle, as radius is to
the fine of the hypothenufe, fo is the fine of one of the
oblique angles to the fine of its oppofite fide.
Let ABC be a ſpherical triangle, having a right Fig. 13.
angle at B, and let AD, BD, CD, be drawn to the
center of the ſphere. From C, in the plane DCA, let
CE be drawn perpendicular to DA, and from E, in
the plane DBA, EF perpendicular to the fame line,
and let CF be joined. Then, becauſe DA is
perpen-
A a a
dicular
370
Book II.
SPHERIC S.
Fig. 14.
dicular to the two lines CE, EF, it is perpendicular
to the plane CEF, and confequently the plane CEF
perpendicular to the plane DBA. But the plane
DCB is alfo perpendicular to DBA; therefore their
line of common fection CF is perpendicular to the
fame. Hence CFD, CFE, are right angles. Now,
in the right angled plane triangle CFE, R: CE ::
S, E CF. But the angle CEF, being the inclina-
tion of the planes DCA, DBA, is the fame with the
fpherical angle CAB, CE is the fine of AC, and CF
the fine of BC; therefore R: S, AC:: S, A: S, BC.
COR. 1. As radius is to the cofine of one of the
fides, fo is the cofine of the other to the cofine of the
hypothenufe.
For, let the great circle, of which A is the pole,
meet the three fides in D, E, F. Then F is the pole
of AD; and, applying this propofition to the com-
plemental triangle FCE, R: S, FC :: S, F: S, CE,
that is, R: Cos. BC :: Cos. AB: Cos. AC.
COR. 2. As radius is to the cofine of one of the
fides, fo is the fine of its adjacent angle to the cofine
of the other angle.
:
PROPO-
Book II. SPHERIC S.
371
;
1
PROPOSITION
II.
In
any right angled fpherical triangle, as radius is to the
fine of one of the fides, fo is the tangent of its adjacent
angle to the tangent of the other fide.
From B, let BE be drawn perpendicular to DA, Fig. 15.
and from E, EF alfo perpendicular to DA, in the
plane DCA, to meet DC in F, and let FB be joined.
It may be fhewn, as in the preceding propofition, that
FB is perpendicular to the plane DBA. Fence FB
is the tangent of BC, and FBE a right angled tri-
angle; therefore R: EB:: T, E: FB, that is, R:
S, AB :: T, A: T, BC.
COR. 1. As radius is to the cofine of the hypothe-
nufe, fo is the tangent of one of the angles to the co-
tangent of the other.
For, in the complemental triangle FCE, R: S, CE
: : T, C : T, FE; that is, R: Cos. AC:: T, C: Cot.
A, or, R: Cos. AC:: T, A: Cot. C.
1
COR. 2. As radius is to the cofine of one of the
angles, fo is the tangent of the hypothenufe to the
tangent of the fide adjacent to that angle.
For RS, FE :: T, F: T, CE; that is, R: Cos.
A:: Cot. AB: Cot. AC, or R: Cos. A:: T, AC
* T,AB.
SCHO-
372
BOOK II.
SPHERIC S.
Fig. 16.
SCHOLI U M.
Let the hypothenufe, the two angles, and the com-
plements of the two fides of any right angled fpheri-
cal triangle, be named the five circular parts of the
triangle. Any one of thefe being confidered as the
middle part, let the two which are next to it be called
the adjacent parts, and the remaining two the oppofite
parts. Then the two preceding theorems, with their
corollaries, may be all expreffed in one propofition,
adapted to practice, as follows.
PROPOSITION
III.
In any right angled fpherical triangle, the rectangle un-
der radius and cofine of the middle part, the rectangle
under the cotangents of the adjacent parts, and the
rectangle under the fines of the oppofite parts are all
equal.
R Cos. M = Cot. A x Cot. a.
X
Rx Cos. M = S, O × S,∞.
CASE I. Let the hypothenufe AC be the middle
part.
....
Then
Book II.
373
SPHERIC S.
Then
a R: Cos. AC:: T, C: Cot. A.
Therefore (R:T, C::) Cot. C: R:: Cos. AC: Cot. A.
And
a I. c. 2.
R: Cos.AB:: Cos. BC: Cos. AC. b 1. c. I.
CASE II. Let the angle A be the middle part.
Then
CR: Cos. A::T,AC: T,AB.
Therefore (R:T,AC::) Cot. AC:R:: Cos. A: T,AB.
dR: Cos. BC:: S,C: Cos. A.
And
CASE III. Let the complement of the fide AB be
the middle part.
Then
c
C 2. c. 2.
d 2. c. I.
2.
e
© R: S,AB::T,A : T,BC,
£R:S,AC::S,C:S,AB.
f Is
Therefore (R: T,A::) Cot. A:R::S,AB:T,BC,
And
SCHOLI U M.
With reſpect to right angled triangles, the general
problem in ſpherical trigonometry is, When any two
parts beſides the right angle are given, to find the
three remaining parts. Of which there are the fol-
lowing cafes.
CASE 1. When the hypothenufe and one of the
angles are given.
CASE 2. When one of the fides and one of its ad-
jacent angles are given.
CASE
}
374
Book II,
SPHERIC S.
CASE 3. When one of the fides, and its oppofite
angle, are given.
CASE 4. When the hypothenufe and one of the
fides are given.
CASE 5. When the two fides are given.
CASE 6. When the two angles are given.
In applying the above propofition to the folution of
thefe cafes, let each of the given parts be made the
middle part, by which two of the required parts will
be found; and, to find the remaining one, let the
fame be affumed as the middle part.
By Prop. XI. and COR. 1. B. I. each of the un-
known parts is in every cafe, except the 3d, limited
to one value.
Fig. 17.
PROPOSITION IV.
In any ſpherical triangle, the fines of the fides are di-
rectly proportional to the fines of the oppofite angles.
This propofition has been demonftrated in the cafe
of right angled triangles.
Let ABC be any oblique angled triangle, divided
into two right-angled triangles ABD, CBD, by the
perpendicular BD, falling from the vertex upon the
baſe AC.
In
BOOK II.
375
SPHERIC S.
a
In the former, the complement of BD being the
In the
middle part, Rx S, BD = S, AB × S, A'. In the = 3.
latter, the complement of BD being the middle part,
Rx S, BDS, BC × S, C.
= S, BC × S, C, and S,AB:
Hence S, AB × S, A
S,BC :: S, C : S, A.
COR. 1. The cofines of the two fides are to one
another directly as the cofines of the ſegments of the
bafe.
For this may be proved in like manner, by making
AB, BC the middle parts.
COR. 2. The tangents of the two fides are to one
another inverſely as the cofines of the vertical angles.
This will follow from making the angles ABD, CBD
the middle parts.
LE M M A I.
The fum of the tangents of two arches is to their differ-
ence, as the rectangle under the fine and cofine of half
their fum to the rectangle under the fine and cofine of
half their difference.
Let AB and AC (= AE) be the two arches, AG, Fig. 18.
AF their tangents, D the center, DK, DI perpendi-
cular to BC, BE. Then the triangle GDF: BDC
(:: GD × DF: BD x DC:: GD × DH: BD × DE ::)
GDH: BDE.
And
-
376
Book II.
SPHERIC S.
:
•
And alternately GDF: GDH:: BDC: BDE.
FG: GH:: BK × KD : BI × ID.
Therefore
That is, T, AT,a: T, AT, a :: S,
A+ a
2
× Cos.
X
A+ a
2
: S,
A
a
A
a
× Cos.
2
2
Fig. 19.
LE M M A
II.
The fum of the fines of two arches is to their difference
as the rectangle under the fine of half the fum, and
cofine of half the difference of these arches, is to the
rectangle under the fine of half the difference, and co-
fine of half the fum.
Let AB and AD (= AK) be the two arches, C the
center; KO, DL, parallel to CA; CR, CI, BG,
perpendicular to BK, BD, DL; and AH, AM paral-
lel to BD, BK. The right angled triangles BGD,
CHA, alfo BOK, CMA, by reafon of parallel lines,
are equiangular.
Hence BG BD :: CH: CA, and BG x CA =
×
BD × CH.
Alfo, BO : BK :: CM: CA, and BO × CA →
BK X CM.
Confequently BG: BO:: BD × CH: BK × CM
:: BI × CH : BR × CM.
That
Book II.
377
SPHERIC S.
That is, S, A+ S, a : S, A—S,a :: S,
A—as, A—ª × Cos.
2
А
S,
2
A+ a
2
A+ a
× Cos.
2
LE M M A
IIL
The fum of the fines of two arches is to their difference
the tangent of half the fum of theſe arches is to the
tangent of half their difference.
Let AB, AD, be the two arches, C the center, CF, Fig. 20
DO, BQ, perpendicular to CA, alfo DM, BN per-
pendicular to CF, BE = ED, and FEK the tangent
at E, which will be parallel to BD, becauſe they are
both perpendicular to the radius CE. Now, by fimi-
lar triangles,
DO: BQ :: DL: BL:: GK: HK.
Therefore DO + BQ : DÓ — BQ : : 2EK; 2EH
:: EK: EH.
::
That is, S, A + S, a : S, A
a: AS, a :: T,
A+ a
2
:T,
2
Bbb
LEMMA
1378
Book II.
SPHERIC S.
L E
E M M A
IV.
The fum of the cofines of two arches is to their difference
as the co-tangent of half the fum of thefe arches to the
tangent of half their difference.
For, in the fame figure, BN : DM :: BI: DI ::
HF: FG.
Therefore BN + DM : BN
:: EF: EG.
DM:: 2EF : 2EG
That is, Cos. A + Cos. a: Cos. A Cos. a::
A+ a
A
a
Cot.
: T,
2
2
!
Fig. 17.
I. C. 4.
PROPOSITION
V.
In any Spherical triangle, the tangent of half the fum of
the fegments of the base is to the tangent of half the
fum of the two fides, as the tangent of half their dif-
ference to the tangent of half the difference of the feg-
ments of the bafe.
સી
-
For, Cos. AB : Cos. BC :: Cos. AD: Cos. DC.
Therefore Cos. AB Cos. BC: Cos. AB - Cos.
BC: Cos. AD + Cos. DC: Cos. AD-Cos. DC.
And,
Book II.
379
SPHERIC S.
2.
+
b
ADDC Lem. 4
AB+ BC
AB-BC
And Cot.
: T,
:: Cot.
:T,
AD — DC
2
2
2
2
Or, Cot.
AB+BC
: Cot.
AD +
DC
AB-
BC
:: T,
2
2
2
:T,
AD
2
DC
Hence, T,
:T,
AD + DC
AB + BC
AB-BC
: T.
T,
::T, ,
2
2
2
AD - DC
2
COR. I. The cotangent of half the fum of the ver-
tical angles, and the tangent of half their difference,
or the cotangent of half their difference, and the tan-
gent of half their fum, according as the perpendicu-
lar falls within or without, are reciprocally propor-
tional to the tangents of half the fum, and half the
difference of the angles at the baſe.
For, let the vertical angles ABD, CBD, or their
meaſures, be denoted by x and y; then, applying this
propofition to the femi-fupplemental triangle EQF,
EI + IF
we have T,
2
EI IF
: T,
2
CREA
-QF Fig. 21.
EQ
: T, EQ + OF
2
::T,
2
But, when the perpendicular falls
within, EI ——— IF = x — y, and EI + IF
2
is the com-
plement of. And, when it falls without, EI
2
EI + IF
IF = x+y, and
is the complement
of
380.
BOOK II.
SPHERIC S.
}
of
X
2
y
, or, EI + IF = x+y, and
the complement of * = y.
Therefore,
+
2
EI
IF
is
2
I. Cot. *T, A+C::T, AC: T, =
2
2
2
2. Cot. :T, AC :: TAGT,
*
Y
2
COR. 2. Hence,
1. Cot.
2
x
2
+1:T, A+C:: Cot.
2
:T, A+C
2
24
2
2
2
x + 1 x
2
י
×
Ty
2. Cot.² * — !: T,²
A+C
:: Cot.
2
X
2
x +
2
A-
Y
T, AC
2
Y
× T,
2
x
A — C: T¸ A+ C × T, * + y
2
COR. 3
I. Cot.
2
*
:
:T,
Alfo,
2
}
2
Z
2
:T, AC :: Cot. * + 1xT, A+C
T,²
2
X
T, AC × T, *—
2
2
لا
2
X
y
2
=:: Cot. = x T,
2
2. Cot.
x
1: T,
A-C
2
A - CT, AC × T,
X
= 1.
2
2
2
PROPO.
BOOK II.
SPHERIC S. 38£
1
PROPOSITION · VI.
In any ſpherical triangle, the fine of half the fum of the
two fides, is to the fine of half their difference, as the
co-tangent of half the vertical angle to the tangent of
half the difference of the angles at the baſe.
For, fince T, AB: T, BC:: Cos. y: Cos. x.
Therefore T, AB + T, BC: T, AB-T, BC ::
Cos. x + Cos.´y: Cos. x-
Cos. y.
}
Fig. 17,
a
2. C. 4.
Hence ↳ S,
AB+ BC
AB+ BC
b
× Cos.
: S,
2
2
AB- BC
AB — BC
× Cos.
:: Cot.
x+y: T₁ =
T,
2
2
2
2
or :: Cot.
Y: T,
x + y
2
2
Again, fince S, AB : S, BC :: S, C ; S, A.
Therefore S, ABS, BC: S, ABS, BC ::
S, A+S, CS, AS, C.
C
b Lem. 1
4.
AB + BC
AB-BC
Henced S,
× Cos.
: Cos.
a Lem. 3.
2
2
AB + BC
AB BC
A+ C
A-C
× S,
:: T,
2
2
2
1
Confequently S,
B: T, A—Ce
BT,
2
AB + BC
: S,
: T,
AB- BC
2
:: Cot.
2
2
• 3. c. 5.
PROPO
382 SPHERIC S.
Book II.
Fig. 17.
A
;
PROPOSITION VII.
In any Spherical triangle, the cofine of half the fum of the
two fides, is to the cofine of half their difference, as
the co-tangent of half the vertical angle to the tangent
of half the fum of the angles at the bafe.
For, in the demonſtration of the preceding propo-
fition, we have theſe analogies.
AB + BC
AB + BC
AB
BC
S,
× Cos.
: S,
2
2
2
× Cos.
2
AB-BC
+
:: Cot. * y
2
Cot.
x+y.
2
1
AB
BC
S,
× Cos.
AB + BC
AB + BC
: S,
2
2
× Cos.
AB — BC
2
Whence, Cos.
AB+ BC
2
: Cos.
A-C
:: T,
2
I,
A+ C
AB
2
2
2
BC
:: Cot.
A+ C
B: T,
2
PROPO
20
Book II. SPHERIC S.
383
TAM
PROPOSITION
VIII.
In any ſpherical triangle, the fine of half the fum of the
angles at the bafe is to the fine of half their difference,
as the tangent of half the baſe to the tangent of half
the difference of the two fides.
For, in the triangle EQF, whofe fides FQ, QE are Fig. 21.
the meaſures of the angles A, C, and whofe angles
F, E are meaſured by the fides AB, BC; alfo, whoſe
bafe EF is the fupplement of the meaſure of the angle
at B, and whoſe angle at Q is meaſured by the fup-
plement of the baſe AC.
E-F
S,
EQ → QF : S,
EQ - OF
:: Cot. Q: T,
2
2
2
A+ C
A-C
That is, S,
: S,
I
:: T, AC : T,
2
2
AB
BC
2
PROPOSITION IX.
In any ſpherical triangle, the cofine of half the fum of the
angles at the bafe, is to the cofine of half their differ-
ence, as the tangent of half the bafe to the tangent of
half the fum of the two fides.
For,
4
384
Book II.
SPHERI C S.
i
נ.
For, in the femi-ſupplemental triangle EQF,
Cos. EQ + QF
2
:T, E + F
2
: Cos.
EQ -
QF :: Cot. Q
2
A+C
That is, Cos.
A—C
: Cos.
:: T, AC: T,
2
2
AB+ BC
2
SCHOLIU M.
Let one of the fix parts of any ſpherical triangle be
neglected; let the one oppofite to it, or its fupple-
ment, if an angle, be called the middle part, the two
next to it the adjacent parts, and the remaining two
the oppofite parts. Then the four preceding propofi-
tions may be expreffed in more general terms, as fol-
lows :
PROPOSITION. X.
In any Spherical triangle, the fine or cofine of half the
fum of the adjacent parts, is to the fine or cofine of half
their difference, as the tangent of half the middle part
to the tangent of half the difference, or half the fum of
the oppofite parts.
i
S,
:
+
FIG. 13.
SPHERICS
F
E
FIG. 14.
D
C
B
6
F
H
FIG. 16.
C
Α.
A
B
B
B
F
B
FIG. 17.
A
FIG.20.
K
Ꮐ
TIG.19.
M
M
D
HE
C
E
B
C
D
FIG. 15.
E
A
B
PLJ.pa. 385.
K
D
FIG.18.
N
H
B
L
G
D
K
B
C
О
QA
L
FIG. 22
E
F
B
Ꮐ
+
IF
B
N
E
FIG. 21.
E
B
K
Ꮐ
H
E
$ 0 0 €
D
E
F
L
A
Q
A
L
Book II.
385
SPHERIC S.
S,
At a
A
S,
2
2
a
:: T, MT,
2
2
A + a
Cos.
: Cos.
A = 4
a
: : T, ; M : T,
2
2
PROPOSITION
XI.
0 + 0
2
If, from the extremities of the bafe of any spherical tri-
angle, arches of great circles be defcribed to meet the
fides, and to cut off a part on each from the vertex, e-
qual to the other fide; the rectangle under the fines of
half thefe arches fhall be equal to the rectangle under
the fines of half the fum and half the difference of the
baſe, and the difference of the two fides.
Let ABC be a ſpherical triangle, having two une-
Fig. 22.
qual fides AB, BC. From BC, the greater, let BD
qual´fides
be cut off equal to BA, and from BA produced BE
BC; and let great circles pafs thro' A, D, and E, C.
Then, S, AD x S, EC S,
¿
AC + AE
2
X
AC
AE
S,
2
For, in BA produced, let AF
AC, and FG = AE;
alfo, let the ftraight lines AG, GD, DA, EF, FC,
CE, be drawn. The ftraight lines AD and EC are
parallel ;
C c c..
386
Book II.
SPHERIC S.
parallel; for, if a great circle bifect the angle at B,
it alſo bifects the arches AD, EC, and is perpendicu-
lar to their planes; therefore the chords AD, EC are
perpendicular to the lines of common fection, and
confequently both are perpendicular to the plane of
that great circle. But AG and EF are alfo parallel.
Therefore the plane of the triangle DAG is parallel
to the plane of the triangle CEF. Let the plane
AFC, which cuts the latter in FC, cut the former in
HK. Then HK is parallel to FC. Therefore, fince
the chords AF, AC are equal, HK is a tangent to the
circle that paffes through A, F, C, and confequently
it is alfo a tangent to the circle that paffes through
*5.c.4.1. A, D, G. Hence the angle ADG GAH EFC.
CEF. Therefore the triangles
But the angle DAG
=
AGD, EFC are equiangular, AD: AG :: EF: EC
1
AD: AG :: ¦ EF: ¦ EC, and AD × 1/1/ EC
2
2
= AG ×
x
EF.
PROPOSITION
XII.
In any spherical triangle, the rectangle under the fines of
the two fides, is to the fquare of the radius, as the rec-
tangle under the fines of half the fum, and half the
difference of the bafe, and the difference of the two
fides to the fquare of the fine of half the vertical angle.
Let
Book II. SPHERIC
S. 387
X
E
}
Let ABC be the triangle, BD = BA, BE = BC, Fig. 23.
and BFG bifecting the vertical angle B; then BFG
alfo bifects AD, EC, and cuts them at right angles.
Therefore, in the right angled triangles ABF, EBG,
I
S, AB: R:: S, AF : S, 1
S, BER :: S, EG: S,
B.
B.
Hence S, AB × S, BC: R² :: S, AF × S, EG :
2
S,' B.
2
AC+AE
And S, AB x S, BC: R²; : S,
× S,
2
AC-AE
S, B.
2
S CH O LIU M.
Of the general problem in fpherical Trigonometry
there are fix cafes.
CASE 1. When two fides, and an angle oppoſite to
one of them, are given.
CASE 2. When two angles, and a fide oppofite to
one of them, are given.
CASE 3. When two fides, and the included angle
are given.
CASE 4. When two angles, and the included fide
are given.
CASE 5. When the three fides are given.
CASE 6. When the three angles are given.
:
SV
388
Book II.
SPHERIC S
As every oblique angled triangle may be refolved
into two right angled triangles, all thefe cafes may be
folved, by means of the 3d and 5th propofitions only.
But a much better folution may be obtained, from
the 4th, 10th, and 12th propofitions.
be all reduced to three; for, by uſing the ſup-
plemental or femi-fupplemental triangle, in the 2d,
4th, and 6th cafes, theſe may be converted into the
1ſt, 3d, and 5th, reſpectively.
may
And the cafes
By Prop. X. XI. and Cor. each of the unknown
parts is limited to one value in all the cafes, except-
ing fome of the fub-cafes of the ft and 2d.
In the third cafe, the unknown fide may be found,
without finding the angles, by means of the follow-
ing Theorem, which is demonftrated in the Appen-
dix to Book IV.
THEOREM. In any ſpherical triangle, as the fquare
of the radius is to the rectangle under the fines of the
two fides, fo is the verfed fine of the included angle
to the excefs of the verfed fine of the bafe above the
verfed fine of the difference of the two fides.
SPHERIC S.
ધ
**
SPHERIC S.
BOO K
III.
STEREOGRAPHIC PROJECTION
OF THE SPHERE.
"To
DEFINITIONS.
O project an object, is to repreſent every point
of it upon the fame plane, as it appears to the
eye in a certain pofition.
2. That plane upon which the object is projected,
is called the plane of projection; and the point where
the eye is fituated the projecting point.
3. The ftereographic projection of the fphere is that
in which a great circle is affumed as the plane of pro-
jection, and one of its poles as the projecting point.
4. The great circle, upon whofe plane the projec-
tion is made, is called the primitive.
5. By
390
BOOK III.
SPHERIC S.
5. By the femi-tangent of an arch is meant the
tangent of half that arch.
6. By the line of meaſures of any circle of the
fphere is meant that diameter of the primitive, pro-
duced indefinitely, which is perpendicular to the line
of common fection of the circle and the primitive.
A X I O M.
The projection or reprefentation of any point is
where the ſtraight line, drawn from it to the pro-
jecting point, interfects the plane of projection.
PROPOSITION
لحم
Fig. 24.
Every great circle which paffes through the projecting
point is projected into a straight line, paffing through
the center of the primitive; and every arch of it, rec-
koned from the other pole of the primitive, is projected
into its femi-tangent.
1
Let ABCD be a great circle, paffing through A, C,
the poles of the primitive, and interfecting it in the
line of common fection BED, E being the center of
the ſphere. From A, the projecting point, let there
be
BOOK III.
SPHERIC S. 391
be drawn ſtraight lines AP, AM, AN, AQ, to any
number of points, P, M, N, Q, in the circle ABCD.
Theſe lines will interfect BED, which is in the ſame
plane with them; let them meet it in the points p,
m, n, q; then p, m, n, q are the projections of P, M,
N, Q. And thus the whole circle ABCD is project-
ed into the ftraight line BED, paffing through the
center of the primitive.
Again, becauſe the points C and M are projected
into E and m, the whole arch MC will be projected
into the ſtraight line mE, which, to the radius AE, is
the T, m AE = T, MC. Thus, the arch MC is
I
2
projected into its femi-tangent mE; PC into its femi-
tangent pE, &c. All arches, therefore, on the circle
ABCD, reckoned from the pole C, are projected in-
to their femi-tangents.
COR. 1. Each of the quadrants contiguous to the
projecting point is projected into an indefinite ſtraight
line, and each of thofe that are remote, into a radius
of the primitive.
COR. 2. Every finall circle which paffes through
the projecting point is projected into that ſtraight line
which is its common fection with the primitive.
COR. 3. Every ſtraight Ene, in the plane of the
primitive, and produced indefinitely, is the projection
of fome circle on the fphere paffing through the pro-
jecting point.
COR.
392
BOOK III
SPHERIC S.
COR. 4. The stereographic projection of any point in
the furface of the fphere is diftant from the center of the
primitive, by the femi-tangent of that point's distance
· from the pole opposite to the profecting point.
PROPOSITION
II.
Fig. 25.
Every circle on the sphere which does not pass through
the projecting point is projected into a circle.
If the circle be parallel to the primitive, the propo-
fition is evident.
For, a ſtraight line, drawn from the projecting
point to any point in the circumference, and made to
revolve about the circle, defcribes the furface of a
cone, which is cut by a plane, (viz. the primitive),
parallel to the bafe; and therefore the ſection (the fi-
gure into which the circle is projected) is a circle.
If the circle MN be not parallel to the primitive
; let the great circle ABCD, paffing through
the projecting point, cut it at right angles, in the dia-
meter MN, and the primitive in the diameter BD.
Through M, in the plane of that great circle, let MF
be drawn parallel to BD; let AM, AN be joined, and
meet BD in m, n. Then, becaufe AB, AD are qua-
drants, and BD, MF parallel, the arch AM =
BD
AF,
and
Book III.
SPHERIC S. 393
and the angle AMF or Amn = ANM. Thus, the
conic ſurface, deſcribed by the revolution of AM, a-
bout the circle MN, is cut by the primitive in a fub-
contrary poſition; therefore the fection is, in this
cafe, likewiſe a circle.
COR. I. The centers, and poles of all circles, pa-
rallel to the primitive, have their projections in its
center.
COR. 2. The center, and poles of every circle, in-
clined to the primitive, have their projections in the
line of meaſures.
COR. 3.
All projected great circles cut the primi-
tive in two points diametrically oppofite; and every
circle in the plane of projection, which paffes through
the extremities of a diameter of the primitive, or
through the projections of two points that are diame-
trically oppofite on the fphere, is the projection of
fome great circle.
COR. 4. A tangent to any circle of the ſphere,
which does not paſs through the projecting point, is
projected into a tangent to that circle's projection;
alfo, the circular projections of tangent circles touch
one another.
COR. 5. The extremities of the diameter, on the
line of meaſures of any projected circle, are diftant
from the center of the primitive, by the femi-tangents
of the circle on the fphere's leaſt and greateſt die
ſtances from the pole oppofite to the projecting point.
D d d
COR.
394
Book III.
SPHERIC S.
COR. 6. The extremities of the diameter, on the
line of meaſures of any projected great circle, are di-
ſtant from the center of the primitive, by the tan-
gent and cotangent of half the great circle's inclina-
tion to the primitive.
COR. 7. The radius of any projected circle is e-
qual to half the fum, or half the difference, of the ſe-
mitangents of the circle's leaft and greateft diftan-
ces from the pole oppofite to the projecting point,
according as the circle does or does not encompafs
the axis of the primitive.
Fig. 26.
PROPOSITION III.
An angle, formed by two tangents, at the fame point, in
the furface of the Sphere, is equal to the angle formed
by their projections.
Let FGI, and GH, be the two tangents, and A the
projecting 'point; let the plane AGF cut the ſphere
in the circle AGL, and the primitive in the line
BML. Alfo, let MN be the line of common fection,
of the plane AGH, with the primitive. Then the
angle FGH LMN. If the plane FGH be parallel
to the primitive BLD, the propofition is manifeft. If
not, through any point K, in AG produced, let the
plane
1
Book III.
SPHERIC S. 395
plane FKH, parallel to the primitive, be extended to
meet FGH in the line FH. Then, becauſe the plane
AGF meets two parallel planes, BLD, FKH, the lines
of common ſection LM, FK are parallel; therefore
the angle AML = AKF. But, fince A is the pole
of BLD, the chords, and confequently the arches
AB, AL are equal; and the arch ABG is the fum of
the arches AL, BG. Hence the angle AML is equal
to an angle at the circumference, ftanding upon AG,
and therefore equal to AGI, or FGK. Confequently
the angle FGK = FKG, and the fide FG FK. In
like manner, HG = HK. Hence the triangles GHF,
KHF, are equal in every refpect, and the angle FGH
= FKH = LMN.
=
COR. 1. An angle, contained by any two circles of
the ſphere, is equal to the angle formed by their pro-
jections.
For, the tangents to thefe circles on the fphere are
projected into ſtraight lines, which either coincide
with, or are tangents to, their projections on the pri-
mitive.
COR. 2. An angle, contained by any two circles of
the ſphere, is equal to the angle formed by the radii
of their projections, at the point of concourſe.
PROPO.
}
·396: SPHERIC S.
Book III.
Fig. 27.
PROPOSITION
IV.
The center of a great circle's projection is diſtant from
the center of the primitive by the tangent of the
great
circle's inclination to the primitive, and its radius is
the fecant of the fame.
Let BMDN be the primitive, EA its femi-axis, and
A the projecting point; alfo, let MEN be the dia-
meter of any great circle, oblique to the primitive,
and FMGN its projection, the femicircle below the
plane of the primitive being projected into the arch
MFN, and that above into MGN. Let BED be
drawn perpendicular to MN, interfecting the projected
circle in F, G; then H. the middle of FG, is its cen-
ter. Let NH, NF, NG, be drawn, and let NF, NG
meet the primitive in L, Q. Then LE, EQ being
joined, becauſe GNF is a femicircle, FNQ is a right
angle, LBNQ is likewiſe a femicircle, and LEQ a
traight line. Alfo, let K be the point in the propo-
fed circle, of which G is the projection, and let KE
be joined. Then ME, being perpendicular to both
AE and EG, is perpendicular to the plane AEG, and
confequently to EK. Thus, KE and DE are both at
right angles to MN, the line of common fection of
the great circle and the primitive, and therefore KED
is
{
Book III, SP HER I C S. 397
is its angle of inclination to the primitive. Now, the
triangles AEG, NEG, having EG common, AE =
EN, and GEA, GEN right angles, are equal in every
reſpect. Thus, the angle EKA = EAK = ENQ =
EQN; therefore AEK = NEQ; whence DEQ
DEK = angle of inclination. Again, the angles
ENF, EGN are equal; therefore their doubles MEL,
EHN are equal; and confequently the complements
of theſe LEB, ENH alfo equal. Thus, ENH – LEB
DEQ = angle of inclination, and to the radius
EN, EH is the tangent, and HN the fecant of the
angle ENH.
•
"
ANOTHER DEMONSTRATION.
Let A be the projecting point, ABCD a great Fig. 28.
circle paffing through it, perpendicular to the propo-
fed great circle, KEL their line of common ſection,
and BED the line of common fection of ABCD, and
the primitive. Then, becauſe ABCD is perpendicular
both to the propofed great circle, and to the primi-
tive, it is perpendicular to their line of common fec-
tion, and confequently BE, EK are likewife perpen-
dicular to the fame. Hence BEK is the angle of in-
clination of the propofed circle to the primitive. Let
AK, AL be drawn, and meet BD in F, G, the
ftraight line FG is the diameter of the projection.
Let
1
398
Book III.
SPHERIC S.
1
Let it be bifected in H, and let A, H be joined. Be-
caufe FAG is a right angle, HA=HF, and the angle
HAF = HFA = FEK + FKE; from thefe equals,
taking the equal angles EAF, FKE, there remains
HAE = FEK, the angle of inclination. And, in the
right angled triangle AEH, to the radius AE, EH is
the tangent, and AH the fecant of HAE.
отн
OTHERWISE,
Fig. 27.
Let MNG be the projection of a great circle, meet-
ing the primitive in the extremities of the diameter
MN, and let the diameter BD, perpendicular to MN,
meet the projection in F, G. Let FG be bifected in
H, and N, H be joined. Then, becauſe any angle, con-
tained by two circles of the fphere, is equal to the
angle formed by the radii of their projections at the
point of concourfe, the angle contained by the pro-
pofed great circle and the primitive is equal to the
angle ENH, of which EH is the tangent, and NH the
fecant to the radius of the primitive.
COR. 1. All circles which pafs through the points
M, N are the projections of great circles, and have
their centers in the line BG. All circles which paſs
through the points F, G, are the projections of great
circles,
}
Book III.
399
SPHERIC S.
circles, and have their centers in the line HI, perpen-
dicular to BG.
COR. 2. If NF, NH be continued to meet the pri-
mitive in L, T, then BL is the meaſure of the great
circle's inclination to the primitive, and MT = 2BL.
PROPOSITION.
V.
The center of projection of a ſmall circle, perpendicular
to the primitive, is diftant from the center of the pri-
mitive, the fecant of the circle's difiance from its near-
er pole, and the radius of projection, is the tangent of
the fame.
Let ABCD be a great circle, paffing through the Fig. 29.
projecting point, and perpendicular to the propofed
fmall circle, MON their line of common fection, and
BOD the line of common fection of ABCD, and the
primitive. Then BD is the axis, and O the center
of the ſmall circle. Let AM and AN be drawn to
meet BD in G, F; FG is the diameter on the line of
meaſures of the fmall circle's projection. Let it be
bifected in H, and EM, MH joined. Then, becauſe
AE: EF:: NO: OF, CE; EF :: MO; OF; there-
fore the points C, F, M, are in a ſtraight line, and
that ftraight line is perpendicular to AG. Hence
HM = HF, and the angle HMF (= HFM = FEM
+ FME)
490
BOOK III.
AS PHE RICS.
SP
Ą
Fig. 30.
+ FME) = FEM + FCE; taking from thefe the e-
qual angles OMF, FCE, there remains the angle
HMO = FEM. Confequently the angle HOM =
HME, and HME a right angle. Thus HM, which is
equal to the radius of the projection, is the tangent
of MD, the diſtance of the ſmall circle from its near-
er pole, and HE is the fecant of the fame.
COR. 1. Let FPGQ be the projection of a ſmall
circle, perpendicular to the primitive, and interfecting
it in P, Q, and let LEK be the diameter of the pri-
mitive, that is, perpendicular to BG, the line of
meaſures: Then L, P, G, or L, F, Q, are in the
fame ftraight line.
For, fince AE: EF:: NO: OF, LE: EF:: QO
: OF, &c.
COR. 2. The angle EPH is a right angle, and EP,
PH tangents to the two circles.
For, the triangles EMH, EPA are equal in every
refpect
COR 3. Any radius HI is a tangent to the circle
which paſſes through the points B, I, D.
PROPO.
Book III.
SPHERIC S. 401
t
PROPOSITION IV.
The projections of the poles of any circle, inclined to the
primitive, are in the line of meafures diftant from the
center of the primitive, the tangent and cotangent of
half its inclination.
Becauſe ABCD is perpendicular to the plane of the Fig. 28.
great circle KL, it paffes through its poles, (which are
alfo the poles of all its parallel ſmall circles), let theſe
be p, q, and let Ap, Aq, meet BD in P, Q, their pro-
jections. Then, the quadrants pK, CB being equal,
and CK common to both, pC will be equal to BK,
which meaſures the inclination of the great circle, (or
its parallel ſmall circles), to the primitive. Now, EP,
to the radius AE, is the tangent of pC, and EQ the
tangent of qC, or cotangent of pC.
I
2
COR. 1. The projection of that pole which is adja-
cent to the projecting point, is without the primitive,
and the projection of the other within.
COR. 2. The diſtances of either projected pole from
the centers of the primitive and projected great circle,
are directly proportional to the radii of theſe circles.
For AP bifects the angle EAH, and confequently
AQ biſects the external angle EAR. Hence EP: PH
:: EA: AH, and EQ: QH :: EA : AH.
Eee
PROPO.
402 SPHERIC S.
Book III.
}
PROPOSITION
VII.
Fig. 31.
}
If, from either pole of a projected great circle, two straight
lines be drawn to meet the primitive and the projec-
tion, they will intercept correfponding arches of theſe
circles.
From the pole P, of the projected great circle
FMG, let there be drawn any two ftraight lines PL,
PQ, meeting the primitive in R, S, and the projection
in L, Q; then ſhall the arch LQ, be the projection
of an arch equal to RS.
For SS, RR are the projections of two circles, each¨
of which paffes through A, p, a pole of the primitive,
and a pole of the great circle, and which, therefore,
intercept equal arches upon them. Now, RS is one
of the intercepted arches, and the other is projected
into LQ. Hence LQ, RS are correſponding arch-
es.
COR. Hence, if, from the point where the projec-
tions of two great circles interſect one another, twą
ftraight lines be drawn through their adjacent poles,
theſe will intercept on the primitive an arch, which is
the meaſure of their inclination.
41
PROPO
1
!
2
A
E
B
m
M
A
F
B
G
FIG. 23.
D
C
FIG. 25.
بہتر
B
SPHERICS
P
A
72
B
E
E
K
FIG.27.
L
M
C
n
D
A
FIG 24.
N
M
C
B
M
I
PL.II.pa.
FIG.26.
K
F
F
H
B
F
IE
P
H
FIG.28.
Ꮐ
K
B
F
E
H
D
C
M
FIG.29.
FIG.30.
D
G
B
E
F
0
D
H
Ꮐ
B
E
F
0
ID
H
C
R
Ο
Book III. SPHERIC s.
403
ว
PROPOSITION
VIII.
Through two given points, in the plane of the primitive,
to defcribe the projection of a great circle.
PROPOSITION IX.
Through a given point, in the circumference of the primi-
tive, to deſcribe the projection of a great circle that
fhall form with the primitive a given angle.
PROPOSITION X.
Through a given point in the circumference of a project-
ed great circle, to deſcribe the projection of another
great circle, that ſhall make with the former a given
angle.
PROPOSITION XI.
To defcribe the projection of a ſmall circle, parallel to the
primitive, its diftance from the projecting point being
given.
PROPO.
-
404
BOOK III.
SPHERIC S.
1
PROPOSITION
XII.
To defcribe the projection of a ſmall circle, perpendicular
to the primitive, its pole, and distance from its pole,
being given.
PROPOSITION
XIII
To find the poles of a given projected great circle, and the
meaſure of its inclination to the primitive.
1
PROPOSITION
XIV.
}
To measure any part of a projected great circle:
:
:
PROPOSITION
XV.
To measure the projection of a spherical angle:
PROPOSITION
XVI.
To project the feveral cafes in fpherical trigonometry, and
meafure the unknown parts of the triangle.
APPENDIX.
!
1
f
BOOK III. SPHERIC S.
$
APPENDI X.
THEOREM 1. If, from one of the points, in which
a perpendicular ſmall circle meets the primitive, a
ftraight line be drawn to any point in the circumfe-
rence of its projection, and continued to meet the di
ameter of the primitive that is perpendicular to the
line of meaſures, the point of concourfe will be the
pole of the projected great circle, which paffes thro'
the poles of the fmall circle, and that point in the
circumference of its projection.
THEO. 2. If two equal circles, one of which is pa
rallel, and the other inclined to the primitive, be pro-
jected, the diſtances of the pole of the inclined pro-
jected circle, from the centers of the projections, will
be directly proportional to the radii of thefe projected
circles.
THEOR. 3. If two equal circles, one of which is
parallel, and the other inclined to the primitive, be
projected, ſtraight lines, drawn through the pole of
the inclined projected circle, will intercept corre-
ſponding arches on the projection.
PROBLEM I. Given the diſtance of a circle from
its pole, and the projection of that pole, to defcribe
the projection of the circle.
#
PROB.
SPHERIC S.
BOOK III.
PROB. 2. About a given projected pole, to defcribe
the projection of a circle that ſhall pass through a gi-
ven point, in the plane of the primitive.
PROB. 3. To meaſure
To meaſure any part of a given projec-
tion.
PROB. 4. Through a given point, in the plane of
the primitive, to defcribe the projection of a great
circle that ſhall form a given angle with a given pro-
jected great circle.
PROB. 5.
To defcribe the projection of a great circle
that fhall form given angles with two given projected
great circles.
PROB. 6. To defcribe the projection of a great
circle that fhall have a given arch intercepted between
two given projected great circles, and form a given
angle with one of them.
LEMMA 1. The fine of the fum of two arches is to
the fine of their difference as the fum of the tangents
of theſe arches is to their difference.
LEM. 2. The cofine of the fum of two arches is to the
cofine of their difference, as the difference between
the cotangent of the one, and the tangent of the o-
ther, is to their fum.
DEMONSTRATION of PROP. VI. Book II.
Let DEF be the projection of the ſpherical triangle
ABC, the point D being the projection of B, and the
center
BOOK III.
SPHERIC S. 407
center of the primitive. Let EF be joined, and the
tangents EG, FG, drawn. Then
ED + DF: EDDF:: Cot. EDF: T,
DEF DFE DEG DFG
2
or
2
That is, T, AB + T, ½ BC: T, ¦ AB-T, 1 BC
A—C
:: Cot. B: T,
2
AB + BC
-
AB BC
Hence
S
: S,
:: Cot. B:
2
2
P
T, AC
2
1
Prop. VII. may be demonftrated in like manner,
by means of the ſecond lemma.
SPHERICS.
1
SPHERIC S.
в о O K
IV.
ORTHOGRAPHIC PROJECTION OF
THE SPHERE.
IN the orthographic projection of the fphere, the
projecting point is ftill fuppofed to be in the axis,
of a great circle, affumed as the primitive or plane of
projection, but at fo great a diſtance, that a ſtraight
line drawn from it to any point of the fphere, may
be confidered as perpendicular to the plane of the
primitive.
The orthographic projection of any point, there-
fore, is where a perpendicular from that point meets
the primitive.
PROPO
Book IV. SPHERIC S. 409
PROPOSITION I.
Every great circle, perpendicular to the primitive, is pro-
jected into a diameter of the primitive; and every
arch of it, reckoned from the pole of the primitive, is
projected into its fine.
Let BFD be the primitive, and ABCD a great Fig. 33.
circle perpendicular to it, paffing through its poles
A, C; then the diameter BED, which is their line
of common fection, will be the projection of the circle
ABCD.
For, if from any point as G, in the circle ABC, a
perpendicular GH fall upon BD, it will alſo be per-
pendicular to the plane of the primitive. Therefore
H is the projection of G. Hence the whole circle is
projected into BD, and any arch AG into EH =
GI, its fine.
COR. 1. Every arch of the great circle, reckoned
from its interfection with the primitive, is projected
into its verfed fine.
COR. 2. The orthographic projection of any point on
the furface of the ſphere, is, within the primitive, di-
Atant from its center, by the fine of that point's distance
from either pole of the primitive.
Fff
COR.
410
Book IV.
SPHERIC S.
i
COR. 3. Every fmall circle, perpendicular to the
primitive, is projected into its line of common fec-
tion with the primitive, which is alfo its own diame-
ter; and every arch of the femicircle above the pri-
mitive, reckoned from the middle point, is projected
into its fine.
COR. 4. Every diameter of the primitive is the pro-
jection of a great circle, and every other chord the
projection of a ſmall circle.
COR. 5. A ftraight line, perpendicular to the pri-
mitive, is projected into a point, a parallel to the pri
mitive, into an equal line, and one inclined to the
primitive, into a lefs line, fuch that radius is to the
cofine of the inclination, as the inclined line to its
projection.
COR. 6. A ſpherical angle, at the pole of the pri
mitive, alfo any rectilineal angle, whofe plane is pa-
rallel to the primitive, is projected into an equal
angle.
PROPOSITION.
II.
A circle parallel to the primitive is projected into a circle
equal to itſelf, and concentric with the primitive.
Let
Book IV. S PHERIC s. 411
Let the ſmall circle FIG be parallel to the plane of Fig. 34.
the primitive BND. The ftraight line HE, which
joins their centers, is perpendicular to the primitive;
therefore E is the projection of H. Let any radius
HI and IN perpendicular to the primitive, be drawn.
Then IN, HE, being parallel, are in the fame plane;
therefore IH, NE, the lines of common fection of the
plane IE, with two parallel planes, are parallel, and
the figure IHEN is a parallelogram. Hence NE =
IH, and confequently FIG is projected into an equal
circle KNL, whofe center is E.
COR. The radius of the projection is the cofine of
the parallel's diſtance from the primitive, or the fine
of its diftance from the pole of the primitive.
PROPOSITION
III.
An inclined circle is projected into an ellipfe, whofe tranf
verfe axis is the diameter of the circle.
CASE 1. Let ELF be a great circle, inclined to the Fig. 35.
primitive EBF, and EF their line of common fection.
From the center C, and any other point K, in EF,
let the perpendiculars CB, KI, be raiſed in the plane
of the primitive, and CL, KN, in the plane of the
great circle, meeting the circumference in L, N. Let
LG, ND, be perpendicular to CB, KI; then G, D
are
412 S P HERIC S. BOOK IV.
Fig. 36.
Fig. 37.
!
are the projections of L, N. And, becauſe the tri-
angles LCG, NKD are equiangular, CL: CG::
NK²: DK², or EC CG:: EKF: DK; there-
fore the points G, D, are in the curve of an ellipfe, of
which EF is the tranfverfe, and CG the femi-conju-
gate axis.
COR. 1. In a projected great circle, the femi-conju-
gate axis is the cofine of the great circle's inclination
to the primitive.
COR. 2. Perpendiculars to the tranfverfe axis inter-
cept correfponding arches of the projection and the
primitive.
COR. 3. The excentricity of the projection is the
fine of the great circle's inclination to the primitive.
CASE 2. Let AQB be a ſmall circle, inclined to
the primitive, and let the great circle LBM, perpen-
dicular to both, interfect them in the lines AB, LM.
From the center O, and any other point N in the
diameter AB, let the perpendiculars TOP, NQ be
drawn in the plane of the fmall circle, to meet its
circumference in T, P, Q. Alfo, from the points
A, N, O, B, let AG, NI, OC, BH, be drawn per-
pendicular to LM, and from P, Q, T; PE, QD, TF,
perpendicular to the primitive; then G, I, C, H, E,
D, F, are the projections of theſe points. Becauſe
OP is perpendicular to LBM, and OC, PE, being
perpendicular to the primitive, are in the fame plane,
the
?
Book IV.
SPHERIC S. 413
the plane COPE is perpendicular to LBM. But the
primitive is perpendicular to LBM. Therefore the
line of common fection EC is perpendicular to LBM,
and to LM. Hence CP is a parallelogram, and EC
= OP. In like manner, FC, DI, are proved per-
pendicular to LM, and equal to OT, NO. Thus,
ECF is a ſtraight line, and equal to the diameter PT.
Let QR, DK, be parallei to AB, LM; then RO =
NQ = DI = KC, and PR × RT = EK × KF. But
AO:CG::NO: CI; therefore AO² : CG² : : QR*
DK³, and EC² : CG² :: EKF : DK². 10:
COR. 1. The tranfverfe axis is to the conjugate, as
radius to the cofine of the circle's inclination to the
primitive.
COR. 2. Half the tranfverfe axis is the cofine of
half the fum of the greatest and leaft diſtances of the
fmall circle from the primitive.
COR. 3. The extremities of the conjugate axis are
in the line of meaſures diftant from the center of the
primitive, by the cofines of the greateft and leaſt di-
ftances of the ſmall circle from the primitive.
COR. 4. If, from the extremities of the conjugate
axis, of any elliptical projection, perpendiculars be
raifed, (in the fame direction, if the circle do not in-
terfect the primitive, but, if otherwife, in oppofite di-
rections,) they will intercept an arch of the primi-
tive, whofe chord is equal to the circle's diameter.
PROPO.
}
414
Book IV.
SPHERIC S.
PROPOSITION IV.
Fig. 38.
'
The projected poles of an inclined circle are, in its line
of measures, diftant from the center of the primitive,
the fine of the circle's inclination to the primitive.
Let ABCD be a great circle, perpendicular, both to
the primitive and the inclined circle, and interfecting
them in the diameters AC, MN. Then ABCD paf-
fes through the poles of the inclined circle; let theſe
be P, Q, and let Pp, Qq, be perpendicular to AC,
p, q are the projected poles, and it is evident that
¿O = S,BP, or MA, the inclination.
COR. I. The center of the primitive, the center of
projection, the projected poles, and the extremes of
the conjugate axis, are all in one and the fame
straight line.
COR. 2. As radius is to the fine of a ſmall circle's
inclination to the primitive, fo is the cofine of its
diſtance from its own pole to the diſtance of the cen-
ter of its projection from the center of the primitive.
PROP Qu
BOOK IV. SPHERIC S. 415
PROPOSITION V.
To defcribe the projection of a ſmall circle parallel to the
primitive, its diftance from the pole of the primitive
being given.
PROPOSITION
VI.
To project an inclined circle, whofe diſtance from its pole
and inclination to the primitive, are given.
PROPOSITION
VII.
To find the poles of a given projection.
PROPOSITION
VIII.
To measure any part of a given projection.
APPENDI X.
THEOREM I. In any inclined circle, two diame-
ters, which cut each other at right angles, are pro-
jected into conjugate diameters of the ellipfe.
THEOR.
416
SPHERI C S. Book I
:
THEOR. 2. In every elliptical projection, half the
tranſverſe axis is to the excentricity, as radius to the
fine of the circle's inclination to the primitive; and
half the conjugate axis is to the excentricity, as ra-
dius to the tangent of the fame.
PROBLEM I. Through two given points, in the
plane of the primitive, to deſcribe the projection of a
great circle.
PROB. 2. Given the diſtance of a circle from its
pole, and the projection of that pole, to defcribe the
projection of the circle.
PROB. 3. Through a given point, in the plane of
the primitive, to defcribe the projection of a great
circle, having a given inclination to the primitive.
DEMONSRTATION of the laft THEOR.
Book II.
:
Let the ſpherical triangle ABC be projected ortho-
graphically upon ABD, the plane of one of its fides;
and let the projections of the other two fides be con-
tinued to meet the primitive in F, D. Let AEF,
BED, the tranfverfe axes of the ellipfes ACF, BCD,
be drawn, alfo AK, LCM, GEI, perpendicular to
BD,
!
Book IV. S PHERIC s.
417
BD, and LP, CO perpendicular to AF. Then AK,
LM are the fines of the two fides, GH the verſed fine
of the included angle, and PO the excefs of the
verfed fine of the baſe, above the verfed fine of AL,
the difference of the two fides
But
And
;
GE: LM:: GH: LC,
AE: AK :: LC: PO.
Therefore AE × GE : AK × LM :: GH : PO.
That is R²: S,AB × S,BC :: V, B: V, AC — V;
AB BC.
FINI S,
S.
JUL 1 2 1920
:
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1.
FIG.31.
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B
F
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SPHERICS
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A
F
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FIG.3.5.
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C
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FIG.38.
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A
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G
n
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FIG.37.
D
N
E
R
P
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FIG.32.
D
FIG34.
B
M
E
N
FIG.33.
A
I
B
E
FIG36.
B
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C
1
PL.M.
N.pano
H
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L
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C
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FIG.39.
I
ERR AT A.
Page 2. In def. 12. for a furface, read a plane furface.
7. line 7. for equal read equal ".
8. dele *,
|
8.
|
9.
29.
40.
41.
46.
59.
61.
62.
10. for ax. 4. read ax. 3.
11. for DEF read EDF.
19. for ax. 2. read ax. 1.
2. for ax. 2. read ax. 1.
ult. for ABD read ADB.
2. for 10. 1.
read
9. I.
4. for fquare read fquares.
9. for BCA read CBA.
16. for cor. 24. read cor. 23.
4. for 12. 1. read 11. 1.
23. for 11. 1. read 10. I.
UNIVERSITY OF MICHIGAN
!
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Elements of mathe-
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