GEOMETRICAL ANALYSIS. at t -S7/ < sY8A/.zAg;;. 0 0 000 i. A-...; - aft: t:S X C: -. He... 0 f.0s, He at- V OR THE CONSTRUCTION AND SOLUTION OF VARIOUS GEOMETRICAL PROBLEMS FR0OM ANALYSIS, BY GEOMETRY, ALGEBRA, AND THE DIFFERENTIAL CALCULUS; ALSO, THE GEOMETRICAL CONSTRUCTION OF ALGEBRAIC EQUATIONS, AND A MODE OF CONSTRUCTING CURVES OF THE HIGHER ORDER BY MEANS OF POINTS. BY BENJAMIN HALLO WELL, FORMERLY PROPRIETOR AND PRINCIPAL OF THE ALEXANDRIA, VA., BOARDING-SCHOOL. PHILADELPHIA: J. B. LIPPINCOTT & CO. 1872. Entered, according to Act of Congress, in the year 1871, by J. B. LIPPINCOTT & CO., In the Office of the Librarian of Congress at Washington. TO SWARTHMORE COLLEGE, including the Youthful Laborers of both sexes, its successive inmates, who are devoting themselves to the pursuit of a knowledge of the True, the Beautiful, and the Good in every Department of Science and Nature, which is designed to assist in training and strengthening the Intellectual Faculties and thus securing the needed Discipline in the exercise of the fullest freedom in the whole range of human thought, which is their inherent privilege, and so essential to the progress of Truth, and the complete development of humanity, BY THE AUTHOR, with ardent desires that the Blessing of the Good Providence may rest upon the Institution, and the highest hopes for its usefulness be fully realized. BENJAMIN HALLOWELL. SANDY SPRING, MD., 8mo. 17th, 1871. TABLE OF CONTENTS. PAGE DEDICATION to Swarthmore College....... 5 Introductory Note.......... 9 Explanation of some Symbols and Abbreviations employed.. 17 Remarks to the Student.... 19 Various Geometrical Problems in "Triangles, Quadrilaterals, and Parallels," analyzed by Geometry, 64 in number..... 21 Geometrical Problems involving Properties of "the Circle," analyzed by Geometry, 33 in number.. 96 Theorem and Problem proposed by Francis Miller.. 127 Problem to illustrate the 3Iode of " Discussing a Problem"... 129 Construction of Algebraic Equations of one unknown quantity of the First Power, in which each letter represents a right line.. 134 Construction of Algebraic Equations of the Second Degree.. 136 Problems analyzed by Algebra, and constructed and demonstrated by Geometry...141 Problems analyzed by the Differential Calculus, and constructed and demonstrated by Geometry.. 154 Problems in relation to Areas and the Division of Surfaces 181 Demonstration of some Theorems.. 211 " Theorem of Pappus". 214 Theorem proposed in Ladies' Diary for 1735-6. 215 This Theorem extended and applied 217 The same Theorem further applied. (X.).....219 Three Theorems believed to be original..... 221 A Theorem and Singular Property deducible from these. 225 A Problem in Tree-Planting deduced from Theorem X.... 227 Problem in Tree-Planting, proposed in The Agriczlturist. 229 An Attempt at simple Illustrations of some Principles and Problems in the Differential and Integral Calculus...... 230 Finding the Differential of a Simple Function. 230 Finding the Integral of a Simple Function. 231 The Differential Co-efficient.....232 To find the Areas of Plane Figures by the Calculus... 234 To find the Solidity of the Cylinder, Cone, Paraboloid, and Sphere by the Calculus.236 (7) 8 TABLE OF CONTENTS. PAGE Some Elementary Problems in Analytical Geometry 240 The Equation of a Point.. 240 The Equation of a Straight Line.243 To construct an Algebraic Equation of two unknown quantities of the First Power.. 243 Having given two Equations of two unknown quantities of the First Power, to find the values of the unknown quantities by Construction..250 To construct the Circle from its Equation.. 252 Section on Curves, showing the manner of constructing several of them, and some of their prominent Properties.... 253 The Parabola.-To determine Points in the Parabola geometrically, and deduce its Equation........ 253 To find the Axis, Focus, Directrix, Latus Rectum, etc. of a given Parabola by Construction..... 255 To construct a Parabola by Points from its Equation.. 256 The Ellip)se.-To determine Points in the Ellipse geometrically, and find its Equation......... 257 To find the Axes, Foci, Focal Tangents, Tangent, Normal, Subtangent, etc. of a given Ellipse by Construction. 258 To find the Area of an Ellipse by the Calculus. 260 To construct the Ellipse by Points from its Equation.. 260 The Hype7rbola.-To determine Points in the Hyperbola geometrically, and find its Equation when referred to its Asymptotes 262 Having the Equation of the Hyperbola referred to its Asymptotes, to determine its Equation when referred to its Axes.. 264 To construct the Hyperbola from its Equation by means of Points. 265 To construct the Conchoid of Nicomedes.... 270 To construct the Cissoid of Diocles. 271 To construct the Quadratrix of Dinostratus.... 272 To construct the Logarithmic Curve. 274 To construct the Spiral of Archimedes.. 276 To construct the Lemniscate from its Equation.. 277 Table of Square Roots of Numbers from 1 to 200, to facilitate the Construction of Curves by Points from their Equations.... 279 INTROD)UCTORY NOTE. IN an engagement of nearly forty years' duration in teaching Mathematics, first in the boarding-schools of Fair Hill, Maryland, and Westtown, Pennsylvania, and afterwards in a boarding-school of my own, for boys and young men, at Alexandria, Virginia, I became convinced that the analytic or algebraic method of Descartes, Delambre, and Laplace, while it is a most efficient instrument in the hands of a mathematician, is not so well adapted as the geometric or Greek method, to impart to the student a knowledge of mathematical principles, or to inspire such student with an affection and taste for the science. The mind of the young is less capable than that of an older person of abstract thought, and it needs assistance to the concentration of its ideas, such as is afforded by a mathematical diagram; and with this aid continued for some time, the faculties of perception and conception become cultivated and strengthened, till the mind of the student can readily grasp, without a diagram, a problem of considerable intricacy, and be able to apply to it Descartes' method efficiently. So entirely has the analytic method taken the place of the geometric in our prominent schools and colleges, that no work on pure Geometrical Analysis has, to my knowledge, ever been published in this country. Thomas Simpson, F.R.S., and Professor of Mathematics in the Royal Academy of Woolwich, England, in his "Select Exercises for Young Proficients in the Mathematics," which was published in London in 1752, gave a number of geometrical problems, with the method of constructing them. And, in a treatise on Geometry, by the same valued author, published in London in 1760, the fifth and sixth books are entirely devoted to the Construction and Demonstration of Geometrical Problems; and nearly fifty problems, of great variety, and sonte of them of much elegance and beauty, are constructed and demonstrated at the end of the volume. (9) 10 INTRODUCTORY NOTE. To his Algebra, also, which was published in England about the same time, and reprinted in Philadelphia, by Mathew Carey, in 1809, "froiom the Eighth London Edition," Thomas Simpson added an "Appendix, containing the Construction of Geometrical Problems, with the Manner of resolving them numerically."' This work was of very great value to the mathematical student, and of much service to my revered preceptor, John Gummere, in the preparation of his admirable treatise on Surveying, which was published a few years after Carey's edition of Simpson's Algebra was issued. But in all these works, including the problems in Gummere's Surveying, the Rmode of construction was arbitrarily given, without the least clue to the line of thought that had led to it; and the thoughtful student would naturally inquire, and even wonder, in a problem like the seventy-sixth or seventy-seventh of the Appendix to Simpson's Algebra (Problems six and eight in " Algebraic Analysis" in this book), how the author ever came to think of so complicated a method; and he would feel discouraged, in view of its complexity, from an apprehension that his owln mind would never be endowed with a penetration sufficient to accomplish a result of such intricacy and depth. Whereas, had the problems been analyzed, and the student shown that it was all done by taking one step at a time, and that an easy one, -gradually acquiring a knowledge of what we do not icnow by means of what we do know, which Doctor Johnson says is the only way it can be done,-he would have felt increased confidence in his own powers, and been prepared to ma/ce an effort himselfwhich is a great point —in a similar direction. In 1821, Professor John Leslie, of Edinburgh College, published, in Edinburgh, a very valuable treatise on " Geometrical Analysis, and Geometry of Curve Lines." The " Geometrical Analysis" had been previously published, annexed to his " Elements of Geometry," and it greatly "conspired to advance the study of Geometry, by reviving the fine models bequeathed by the Greeks." As has been said with great truth of this work of Professor Leslie, " the study of such a digest, appears admirably fitted to improve the intellect, by training it to habits of precision, arrangement, and close ilnvestigation. The spirit of Geometrical Analysis may be carried, with the happiest effect, into the domains of Inductive Philosophy, which are to be explored by a similar procedure." But, about the time of the publication of this work of Professor Leslie, the algebraic method of Descartes and Leibnitz came rapidly INTRODUCTORY NOTE. 11 into vogue in this country through the popularity of some eminent French writers upon Mathematics, and Professor Leslie's work was never, as far as I know, published in the United States, or, to much extent, placed within the reach of the American student. The want that I experienced, in my engagement of teaching, of a suitable text-book upon Geometrical Analysis, induced me to make a collection of problems, for the benefit of my students, analyzing, constructing, demonstrating, and giving a method of calculation; — solving in this way all I could meet with or frame myself, and difficult ones forwarded to me for solution, through a number of years, by former students, and many other persons, known and unknown, so that the number of such problems became considerable, from which those in the present volume have been selected, as best adapted to supply the want I had experienced. These problems were given to the more advanced mathematical students as "extras," or "trial questions;" and in their solution, much ingenuity was frequently manifested, and also gratifying evidence of progress, in the skill with which the student would proceed in the different steps of the analysis; and it was from the solicitations of many of my former students who had become teachers, to publish the problems from which they had derived, as they believed, great benefit, that the preparation of the present work was undertaken. The practical teaching of young persons consists of two parts: -instructing them how to do something; and giving the reason for doing it in that way. Now, it accords with reason and sound philosophy to do one thing at a time. Hence these two parts of teaching, as a general rule, should, with the young, be kept as separate as possible. Youth should first learn, well, the practical part,-how to do; then the reason In acquiring a knowledge of the primary rules of Arithmetic, for instance, to undertake to give a child the reason for every step he takes in the processes of subtraction, multiplication, division, etc. would but confuse him. It would be too complicated. He needs that the ideas he is to acquire shall be placed, with well-defined outline, in the simplest possible light, and one at a time, so that his mind can clearly and concentratedly comprehend the truth to be imparted. And when this is effected, the sparkling eye and animated countenance will attest the pleasurable sensations of the soul, from the conscious acquisition of a truth previously unknown to him. The practical part-how to do —first; then the why. We must know a fact, before we care about the reason for it. And this is 12 INTRODUCTORY NOTE. the natural mode of the mind's progress in knowledge. Children learn to use words, before they learn the definitions of them. They form phrases, before they are able to " construe" or " parse" them. And the more nearly a teacher keeps to this natural process, the more successful will he be in developing the minds of his students, and in pleasurably educating them. Accordingly, whenever practicable, in teaching young persons, objects, models, maps, globes, diagrams, apparatus, specimens, and other means of illustration of the truths designed to be taught, should be employed, as an aid to assist in concentrating their thoughts, and ilmparting, a clear idea of the subject. In children, the perceptive faculties are most prominent and active, and must principally be brought into requisition; and thus the reflective powers will gradually develop and strengthen, when there will be less need of this educational machinery. Mathematics is an elevated study. Its constant aim is the discovery of truth. There is no new truth. Every fact in science, every mathematical principle, every property of the triangle or the circle, is eternal. No matter when or by whom it was discovered, it pre-existed, and is the living embodiment of a divine thought. All the mathematician or scientist does in this way is to discover what was previously unknown, although eternally existing. When Dr. Herschel and Prof. Leverrier discovered each a "new planet," as it was termed, in Uranus and Neptune, they only saw for the first time what had existed from the period of Creation. The discovery was new, not the object discovered. So of every truth, principle, or property throughout the whole range of mathematics and all science. It seems remarkable, and, as I get older, the astonishment does not in the least abate, that the five plane figures formed by the cutting of a cone by a plane in different positions - the Triangle, the Circle, Parabola, Ellipse, and Hyperbola-should possess such a great number and diversity of singular and interesting properties, many of which are embodied in the following pages, and which, although eternally existing, have been gradually disclosing themselves to the patient research of mathematicians for over two thousand years, and we are by no means at liberty to suppose that all of them are yet known. For instance, that in any triangle, of whatever size or shape, the three perpendiculars let fall from each angular point-upon the opposite sides (the sides and the perpendiculars being produced if necessary) should all pass through the same point. Also, that the three INTRODUCTORY NOTE. 13 lines bisecting the angles respectively, and the three lines drawn from the angular points to the middle point of the opposite sides respectively, should in each case all pass through the same point. (See Scholiums 1, 2, 3 to " Theorem" 6, in the following pages.) Of the same singular character are the properties of the circle and triangle combined, in " Theorems" 11, 12, and 13 of this work, which were original discoveries with the author, although they may have been made by some one before me, whose writings I have not met with. It is deemed proper, in this Introductory Note, to give some explanation of the arrangement of the work, and of the mode in which it is believed it may be used to the greatest advantage. Each problem is first analyzed, then constructed, demonstrated, and the method of calculation by Plane Trigonometry clearly indicated. In many cases the limits are shown within which the problem is restricted. The greatest number of problems are analyzed by pure Geometry. Where they are analyzed by Algebra, or the Differential Calculus, the constructions and demonstrations are by pure Geometry, and the method of calculation is given by Plane Trigonometry, showing the harmonious results by the different modes of investigation. The references in the text (as III. 8* in the first problem of the circle) are to Davies's edition of Legendre's Geometry, -the III. 8 meaning the 8th Proposition of the Third Book of Legendre, and the * referring to a marginal note or foot-note where the Proposition is given in which the same property is demonstrated in Playfair's Euclid. In this example, the note at the bottom of the page is III. 14, meaning the 14th Proposition of the Third Book of Playfair's Euclid. So of other cases. The design of the work is principally as an aid to professors and teachers, to supply a want which the author had severely experienced. The problems are distributed through the volume without much reference to their intricacy or difficulty of solution, in order that the instructor may have a wide field from which to select extra or trial problems, so as to give different problems to different classes in successive years. The importance of the student's constructing the different problems given him for solution, by means of the Scale and Dividers, with the greatestpossible precision, can scarcely be too much insisted upon, as a very improving engagement in training to accuracy, and one of the best preparations for draughting and other duties in the Coast 14 INTRODUCTORY NOTE. Survey or Civil Engineering; or for the business of a machinist, or any mechanical vocation requiring neatness and precision. If this constructing is done thoughtfully, the eye will soon become practiced in judging of distances and positions, so that drawing by hand can be more rapidly and accurately effected. T'he solutions of the problems in relation to surfaces and solids, in the brief article on the Differential and Integral Calculus, and son(le other portions of that article, if read by a student or a class before entering upon, or while pursuing, that study, it is believed will be of material assistance in mastering the subject satisfactorily. The same may be said of the article on Analytical Geometry. As an exercise or lesson preparatory to using a standard text-book upon the subject, it will be a great assistance, and all the problems there given should be carefully constructed, and the results measured with the Scale and Dividers. Or, the teacher or professor may make the explanations, and give these or similar problems, orally, to his class. This method possesses great advantages. The students generally understand and retain what they hear from their instructor, better than what they read in their text-books. Besides, their respect and regard for their instructor are increased when the students discover that the contents of the text-book are not the limits of his knowledge of the subject they are studying, and see his willingness thus to render the knowledge he possesses advantageous to them. In like manner, before entering upon the study of the Properties of the Parabola, in a treatise on Conic Sections, let the three pages on that Curve, in this volume, be read, or equivalent ideas be imparted by the instructor, the student constructing accurately the three problems given, or their equivalents proposed by the teacher or professor. The same course is recommended in regard to the Ellipse and Hyperbola. The student will thus acquire a practical acquaintance fwith the Axes, Ordinates, Abscissas, Asymptotes, Opposite Hyperbolas, etc., which, as these will then be familiar to him, will prove of the greatest advantage in gaining a knowledge of the properties of the Curve, by leaving his mind more free to concentrate its powers upon the particular idea or truth to be acquired. Also, in regard to the remaining Curves, —the Conchoid, Cissoid, Quadratrix, Spiral, etc.,-the student will find it to his decided advantage to construct each Cutrve by the rules herein given, accurately INTRODUCTORY NOTE. 15 with the Scale and Dividers, previous to entering upon the study of its properties in his text-book. A Table of Square Roots, carried to two and three places of decimals, of numbers froml 1 to 200, is added at the close of the volume, to facilitate the construction of Curves from their Equations, by means of points. The author does not expect or desire any pecuniary return frorm his work. It is a labor of love for the youth of our country, ailld of interest and sympathy for those to whom their education may be intrusted, in their most arduous and responsible engagement. If the work shall prove of somle service to the student, and lighten the labor of the instructor, the highest aim of the author wvill be reached. The mlore good it does, and the more service it performs, the more fully will his object be attained. Should anything in this Introductory Note appear like dictating or prescribing to teachers a particular mode of proceeding, the author trusts it will be excused, and understood as only a suggestion from one who has now completed two more than his " threescore years and ten," and who, although he has not been practically engaged in that vocation for a number of years past, still feels a deep interest in the noble profession of Teachinlg, and those actively concerned in it, and who regards the preparation of this work, in which he has spent this, his seventy-second birthday, as the closing and crowning labor of his life in that direction. BENJAMIN HALLO WELL. SANDY SPRING, MARYLAND, Sm10. 17, 1871. Postscript.-It seems only proper to add, that my son, HENRY C. HALLOWELL, read the whole work carefully in manuscript, corrected several clerical errors, and, in some instances, where there appeared to be abstruseness, suggoested one or two intermediate steps in the process, in order that the idea might be more readily comprehended by the student. Note.-For the gratification of his former students, who are widely scattered over our country, some of whom have not seen him for many years, a likeness of the author is prefixed to the volume, which is accompanied by his kindest remembrances. EXPLANATION OF SOME SYMBOLS AND ABBREVIATIONS EMPLOYED. L, angle. L s, angles. A, triangle. As, triangles. I or perp., perpendicular to, or at right angles to. jl, parallel to..*., therefore. Hyp., hypothenuse. < placed between two quantities, implies that the one which comes before the symbol is less than the one which follows it. Thus, L A < L B is to be understood and read, "the angle A is less than the angle B." > placed between two quantities denotes that the one which precedes is greater than the one which follows it. Thus, L B > L A, is to be read, "the angle B is greater than the angle A." It may assist the memory to observe that the greater quantity is always on the side of the open part of the symbol, and the less quantity on the side of the closed part. A', B', x', y', etc. are read, A prime, B prime, x prime, y prime, etc. Al", B", x"y, yll, etc. are read, A second, B second, x second, y second, etc. A"' is read, A third; Aiv, A fourth; BY, B fifth- etc. 2 (V() CASES IN PLANE TRIGONOMETRY AS REFERRED TO IN THE FOLLOWING CALCULATIONS. Case 1. When the angles and one side are given. Case 2. When two sides and an angle opposite one of them are given.* Case 3. When two sides and the included angle are given.* Case 4. When all the sides are given. * In Cases 2 and 3, when the angle is a right one, the solution may be effected by the rules for right-angled triangles. (18) REMARKS TO THE STUDENT. 1. A triangle is said to be determined in an analysis or a construction when a sufficient number of parts are known to solve it by any one of the preceding cases of plane trigonometry. 2. References are made to Davies's translation of Legendre's Geometry, the Roman numerals being put for the book, and common figures for the proposition. With these is an asterisk (*) referring to a marginal or foot-note, where the proposition is designated in like manner in which the same property is demonstrated in Playfair's edition of Euclid's Elements of Geometry. 3. In performing the trigonometrical calculation of a constructed problem, the rule manifestly must be, to begin the calculation where was begun the construction; that is, in the triangle first formed. For, if there was sufficient given to construct the triangle, there must be sufficient to calculate it. Then, with what is found in this triangle, proceed to the next one that was formed, and so on till what is desired is obtained. 4. When a circle, semicircle, or circular are is used in construction, its radius, as a general rule, must be used in the calculation. An exception to this rule exists when a circular segment is described on a line merely to include an angle of a given magnitude. 5. In the analysis of geometrical problems the ingenuity and inventive powers of the student must be brought into close requisition. No definite rules to meet all cases can be given. We must, however, always draw a figure, supposing the problem constructed as required, and then, when practicable, work on those lines which are given in position or length, or both, and continue on until a triangle is obtained which has a sufficient number of parts given to construct it. Then recall, and observe carefully, the process that has been pursued, and construct the figure accordingly. (19) 20 REMARKS TO THE STUDENT. An attempt will be made in the following problems to render this general advice familiar, and to lead the student on to the analysis of problems of considerable intricacy. 6. It will prove of great benefit to the student, in constructing she problems, to assume definite quantities for the parts given, and then work with the scale and dividers with delicate precision, measuring the results accurately from the scale of equal parts, or, if angles, from the scale of chords, and compare these results with those obtained by trigonometrical calculations. This is an admirable preparation for "field work," architectural drafting, machinists, engineering, etc. 7. At the ends of the problems the limits are frequently specified within which the given quantities must be taken; and certain varied conditions of the problem are occasionally referred to, which cannot generally be comprehended to full advantage by the figure that is given. In such cases the student will derive decided benefit, and progress with much greater rapidity and satisfaction by drawing on paper, with pen or pencil, representations of these different conditions, so as clearly to comprehend the idea designed to be conveyed. In this way he will master all that is before him as he goes along, and acquire greater courage and power to overcome future difficulties as they arise. GEOMETRICAL ANALYSIS. PROBLEM I.-In a plane triangle are given one angle, an adjacent side, and the sum of the other two sides, to determine the triangle. D c Given. the side AB, and the sum of B C and A C. A B Analysis. —Let ABC represent the required triangle. The first thing to do is to get B C and A C into one line, and, if practicable, one whose position is given. Now, the position of A C is not given, but that of BC is, because the angle B is given. Hence, produce BC until the produced part, CD, is equal to CAt; then BD becomes known, being equal to the given sum of BC and AC. Join AD, and then the triangle ABD is determined. (Remark 1.) Also, since CD-= CA, the angle CAD- angle CDA (I. 11*). Whence this Construction.-Make AB —the given side, the angle B -the given angle, and the line BD = the given sum of B C and A C. Join AD, and make the angle DA C- < ADC, then will ABC be the triangle required. Demonstration.-All we have to prove is that BC + CA -BD, the given sum. Since, by construction, < DA C < AD C, we J:TI. 5. (21) 22 GEOMETRICAL ANALYSIS. have (I. 12*) CA= CD. Add each to BC, and we have BC + CA —BC+ CD —BD. Q. E. D. D Calculation. —1. In A ABD, by Case 3, find < ADB, which is equal to < DA C. Then (I. 25, cor. 6t) < B CA DAC + ADC-=2ADB. 2. In A ABC, by Case 1, find the sides AC and BC, which together = BD. Limits.-1. The angle B can be any quantity between 0 and 180~. 2. The sum of BC and AC may be any quantity A __~B greater than AB. PROBLEM II.-In a plane triangle are given one angle, an adjacent side, and the diference between the other two sides, to determine the triangle. c.A < Given the side AB, and the difference between AC and B C. D Analysis.-Let ABC represent the required triangle. Now, we must first obtain a line equal to A C - B C, and whose position is known. Since the angle B is given, the line BC is given in position. Therefore, produce CB until CD = CA; then BD is the given difference. Join AD, and the A ABD is determined. Also, since CD - CA, the angle CAD= CGDA, whence the A ADC, and consequently ABC, is determined. Whence this Construction.-Make AB = the given side, and the angle ABC= the given angle. Produce CB, making BD -the given difference. Join AD, and make the angle DA C -- < AD C, and ABC will be the required triangle. Demonstration. —We have to prove, only, that AC - B C BD, the given difference. Since the angles DAC and ADC are equal by construction, we have (12 I.) AC-= CD. Taking BC from each, we have A C- B _ CD - C- = BD, the given difference. Q. E. D. Calculation. —1. In A ABD, by Case 3, find <-ADB = < DA C; then < A CD = 1800 - 2 AD C. 2. In L ABC, Case 1, find sides AC and BC; then AC — BC =BD. * I6. t I. 32. I.6. TRIANGLES, QUADRILATERALS, AND PARALLELS. 23 Limits. —1. The angle B may be any quantity from 0 to 180~. 2. The difference between AC and BC may be any quantity less than AB. PROBLEM III.-The difference between the diagonal of a square and one of its sides being given, to determine the square. D C Given the difference between Given AC and AB. A B E Analysis.-Suppose ABCD to be to be the required square, having the difference between the diagonal AC and the side AB equal to a given quantity; then produce the side AB until AE = AC, and BE will be equal to that given quantity. Join CE. Then in the isosceles triangle AEC, the angle CAE, being half a right angle, is 450; hence, each of the equal angles AE C and A CE - -— (180-45) - 67 2~; and the triangles EB C and EAC are determined. Whence this Construction. —In any line, make EB = the given difference, the angle BE C = 671~, and let EC meet a perpendicular erected at B, in C. On BC describe the square ABCD, which will be the one required. D)emonstration. —Draw the diagonal AC. We have to prove that AC —AB=BE. In A AEC, since Given - points D, F, and E to be the middles of BC, CA, and AB respectively.:H Analysis. —Suppose ABC to be the triangle required. Produce the side B C both ways, meeting lines drawn parallel to BF and CE in I and G respectively. Then, in similar triangles CAI and CFB, since CA _ 2 CF, we have AI — 2 BF, and CI —- 2 B C, whence BI - B-C. Also in similar triangles BCE and B GA, since BA 2 BE, we have A G -- 2 CE, and B G -- 2 B C, whence G C — B C. Complete the parallelogram AGHRI; then GH —AI2 — SBF, and H= —- A G= — 2 CE; whence this Construction. -With the doubles of the three given distances describe the triangle A GH, and complete the parallelogram AGHI. Draw the diagonal GDI, and trisect it in the points B and C. Join CA and AB, bisect them in F and E, and join CE and BF; then will ABC be the required triangle. Demonstration. —We have to prove only that CE= — AG, and that BF=- GH-H AI. Since, by construction, BC-= CG BI, and BE — E A, and CF= FA, we have, by parallel lines, CE — AG, and BF-== AI - GH. Q. E. D. CalculatioT — By 14 IV. cor.,* AH" 2+ GI' — 2 A G2 + 2 GH2. Hence G= /2 AG2 + 2 GH2 —AH2, and BC = GI. ThenMethod 1. —In triangle AGI, Case 4, find the angles. Then < BCE= BGA, and < CBF — CIA. In triangle BCE, Case 3. find BE, then A B= 2 BE; also in triangle CBE, Case 2, find CF, then AC=2 CE. Method 2. —By 14 IV.,y- A C2 + AB2 - 2 AD2 + 2 CD2; that is, 4 AF2 + 4 BE" 2 AD2 - 2 CD2. (1.) Also AC2 + CB2 - 2 CE2 9 2 BE2; that is, 4AF2 + 4 CD2= 2 CE2 ~ 2 BE2. (2.) By subtracting Equation (2) frcim Equation ( ) we have 4 BE2- 4 CD2 - 2 AD2 2 CD2- 2 CE2-2 BE2. By: B. II. t A. I. TRIANGLES, QUADRILATERALS, AND PARALLELS. 33 transposition, 6 BE2 = 6 CD'2 2 AD2 - 2 CE2. Whence BE/3 CD2 + AD2- CE2 AndAB 2 BE. Also, from Equation(2), AF —\1 2 CE+ + 2 BE2-4 CD2 ( CE2 2+ 2BE2-4 CD2). Then A C2 —2 AF, and we have all the sides. Limits.-Of the three given lines AD, BE, and CE, each must be less than the sum of the other two in order that the triangle A GH may be possible. PROBLEM XII.-In a trapezoid, two of whose angles are right angles, and whose diagonals intersect each other at right angles, are given the base, and the diagonal drawn from the extremity of the base which is adjacent to a right angle, to determine the trapezoid. D C the lines AB and BD, the angles G/ iven A ABC and B CD right angles, and BD perpendicular to A G. A 0 B G Analysis.i-Let ABCD represent the required trapezoid, having B and C right angles. Parallel to BD draw CG, meeting AB, produced, in G. Then ACG is a rigtht angle, being =AFB, and CG = BD. Hence, by similar triangles GBC and GCA, GB: GC:: GC: GA. Wherefore GC2= (BD2) = GA x GB, and we must produce the given line AB, so that GA x GB = G C -- B= D. Whence this Construction. —Draw AB = the given base, and erect the perpendicular BE= the given diagonal BD. Bisect AB in 0, join 0E, and produce OB until OG= OE. On AG describe a semicircle, cutting BE in C. Join AC and CG, and parallel to CG and AB draw BD and CD, respectively, intersecting in D. Then AB CD is the trapezoid required. Demzonstration. —The angle ABC being a right angle by construction, and < AFD —< A CG a right angle, being in a semicircle, B CD is a right angle, and the diagonals A C and BD intersect each other at right angles. It remains only to prove that BD or CGG BE. Now, BE2- OE2 OB=0 G2-0B O2 ( OG +- )OB) x VI. 29. B*~J 3 34 GEOMETRICAL ANALYSIS. E (O(G G-OB)=- GA x GB = (IV. 23*) GC2 —-BD2. Hence BD —BE. Q. E.D. P C Calculation. —. In triangle OBE, OE /= /OB2 + BE2 = OG. Then AG= AO+ OG,andBG-OG-OB-CD. 2. BC=VCG2 BG2. Then BC: XA o I G CG::AB:AC. AlsoAD=/BC2+ (AB- CD)2, and all the parts are known. PROBLEM XIII. —In a triangle are given one angle, and the lengths of the two lines drawn from the middle of each including side to the opposite angle, to determine the triangle. Given CD and AE, drawn to the middles [.} I (D 1B the given perimeter. With D B G L H 2 centre L and given radius, describe the semicircle GIH; then CG- the length of the required hypothenuse CA, and AH= the sum of the sides AB and B C. (See Prob. V., this section.) Erect the perpendicular LI, draw HIK, and to it apply CA - CG, let fall the perpendicular AB, which will be parallel to IL, and hence, since IL = LH, we have AB = BH, and consequently CB + AB — CB + BI=- CH; whence the truth of the construction is manifest. In the triangle ABC (III. Prob. 15*) inscribe the circle DEF. Calculation. —In triangle CAH, Case 2, find angle A CH= ACB. In triangle ACB, Case 1, find the sides CB and BA. PROBLEM XXVI.-In a plane triangle are given one angle, and the lengths of the three perpendiculars let fall from the angular points upon the opposite sides, to determine the triangle. FIG. 1. FIG. 2. A F A [ the angle BAC, i and the lengths of / / Given the perpendiculars LAD, BE, and CF. See either Figure. o B 1) C Analysis.-Let ABC, in either figure, represent the required triangle; then, since the double area of a triangle is equal to the base multiplied by the perpendicular height, we have BC x AD AB x CF-=-AC x BE. Putting each of these equations into a proportion, we have AB: BC::AD: CF; AC: BC::AD:BE; and AB: AC:: BE: CF, which proportions are immediately deduci* IV. 4. TRIANGLES, QUADRILATERALS, AND PARALLELS. 49 ble also from the similar right-angled triangles BAD, BCF; AC GD, BCE; and ABE, ACCF. Hence the sides are inversely proportional to the given perpendiculars falling upon them. In AB and AC take AG -— BE, and A — COF, and draw GH. Then, by the last of the above proportions, AB: AC:: BE (AG): CF(AH). Hence (IV. 16*) GHis parallel to BC, and the triangles AB C and A GH are equiangular and similar. But the triangle A GH is known in all its parts; therefore, since AD is known, the triangle ABC is known in all its parts. Whence this Construction.-At A make an angle equal to the given angle, and on its including sides AB and A C, lay AG - the perpendicular to fall on AC, and A/H= the perpendicular to fall on B C. Join Gi, and on GH produced, if necessary, let fall the perpendicular AI, and produce it, making AD = the perpendicular drawn from the point of the given angle to the opposite side BC. Through D, parallel to GH, draw B C, and on AB and AC, respectively, let fall the perpendiculars CF and BE; then ABC will be the required triangle. IDemonstration.-The triangle ABC, being similar to AGE, and having AD== the given perpendicular from the point A, has, from the analysis, the proper angles and sides, and it remains only to show that BE and CF are of the given lengths. In the similar triangles ABD and B CF, AB: BC:: AD: CF; but AD is made of the given length; hence, since the perpendiculars are inversely as the sides upon which they fall, CF is of the given length. Also, in similar triangles, ACD and BCE, AC: BC:: AD: BE; therefore, since AD is the given length, BE is the given length also. Q. E. D. Calculation.-In the triangles AFC and AEB, Case 1, find AC and AB; then, in similar triangles BCF and ABD, we have AD: CF: AB: B C. NOTE.-See Theorem XV. VI. 2. 4 50 GEOMETRICAL ANALYSIS. PROBLEM XXVI1.-In a plane triangle are given the base, the line that bisects the vertical angle, and the diameter of the circumscribing circle, to determine the triangle. A; D the base AB, the line CD1 &\ I /Given bisecting < ACB, and the diameter of the circle A CB. Analysis. —Let A CB represent the required triangle, and A CBF the circumscribing circle, of which the centre is I. Produce CD to F, and join FB. Then (III. 18*), since angle A CF= B CF, the arc AF —= BF, and angle ABF= BCF. Hence (I. 25, cor. 2t) the triangles FBD and FGCB are similar, and we have FD: FB:: FB: FC, in which FB and CD, part of the line CF, are known. Whence this Construction.-About any point I as a centre, with the given radius, describe a circle, in which apply AB-= the given base, and draw EIF perpendicular to AB; then (III. 6t) AE- = EB, and arc AF- FB. Join FB, and draw BG perpendicular to FB and equal to half the given bisecting line CD. With centre G and radius GB, describe the circle BHL, and draw FIHGL; then HL — 2 B Gthe bisecting line. Apply FD C = FL, join A C and CB, and AB C will be the required triangle. Demonstration.-We have to prove only that CD - HL. Now, the triangles FBD and FCB are similar by the analysis; hence FC: FB:: FB: FD. But (IV. 30~) FL (FC): FB:: FB': FIEH; hence D D IFH, and CD - FC- FD = L-HL I — -- IL= 2 BG —the given length of the bisecting line. Q. E. D. Calculation.-Join AI and AF. We have EI-V/AI2 -AE2, EF= EI+ IF, FB2 =FE2 + EB2, FG —/ zFB2 + BG2; then FL - FG + GB- FC, and PFH FG- GB- -FD. Also ED= V/FD2 -FE2, BD —BE -ED, and AD=AE + ED. Now, in the similar triangles FBD and FCB, we have FB: BD:: FC: B C. Also, in similar triangles FAD and FCA, we have FA: AD:: FC: A C, and all the lines are known. NOTE.-If the vertical angle were given instead of the diameter of the " III. 21. - I. 2II. 3.. I l 6. TRIANGLES, QUADRILATERALS, AND PARALLELS. 51 circumscribing circle, we would (III. Prob. 16*) on AB describe a circle whose segment ACB shall contain the given angle, and then proceed precisely as in the above problem. PROBLEM XXVIII.-In a plane triangle are given the perpendicular height, the line bisecting the base, and the line bisecting the vertical angle, to determine the triangle. A c__ B D G~iEen ( the perpendicular CD, { if oh-e C | Given the line CE bisecting AB, and the line CI bisecting < A GCB. F Analysis.-Suppose ACB to be the required triangle. Because the angle ACI- BGCI, the line CI produced will bisect the arc AFB of a circle circumscribing the triangle A CB. At E erect a perpendicular to AB, and it will pass through the centre of the circle and the middle of the arc AFB. Hence CI produced, and the perpendicular EF, will meet at F. Bisect the chord CF by the perpendicular H G, and G will be the centre of the circumscribing circle. Whence this Construction. —Draw any line, as AB, indefinitely, in it take any point D, and erect the perpendicular DC — the given perpendicular. From C apply CE = the given line bisecting the base, and CI= the given line bisecting the vertical angle. At E erect a perpendicular to meet C1, produced, in F, bisect CF by the perpendicular HG, join CG, and with G as centre, and radius GC or GF, describe the circle A CBF. Join CA and CB, also AG, AF, and BF; then ABC will be the required triangle. Demonstration.-Since arc AF - BF, the angle A CI — B CI, and CI bisects the vertical angle. Q. E. D. Calculation. - We have DE -a/CE2 - CD2, and DI/ CI2- CD2; then EI —DE- DI. 1. in similar triangles CDI and FEI, we have D1: CD:: IE: E'F, and DI: CI:: EI: FI; then FC- =FI+- IC. 2. In, similar triangles FCA and FAI, we have FC: FA:: FA:: I[I. 33. 52 GEOMETRICAL ANALYSIS. FI; hence FA - PFC x 1I, and EA=/FA2 —EF2; then AB -- 2 EA. Also AI —=AE - El, and __ BI-=AE+EI. A\ ____1E_ B 3. Again, in similar triangles FCA and GFAI we have FA:AI:: FC: CA; and in similar triangles FBI and FCB we have FB:BI::PF: UB. Lastly,PFE:PI::FH: _FG or GA=- the radius of the circle; when F all the lines are known. PROBLEM XXIX.-In a plane triangle are given the vertical angle, the line bisecting it, and the difference between the including sides, to determine the triangle. A, 7/ n i(the angle BA C, the line AP / Given bisecting BA C, and the difAnalysis.-Let BAC represent the required triangle. Take AD =AC; then BD - the given difference. Draw PPFparallel to AB, and PE parallel to A C; then, since the angle PAF= PAE, AEPF is a rhombus, having all its sides equal and determined. Itence EDAD - AE — A C- A — CUP. Now, in similar trianlgles CFP and PEAB, we have CP (ED) P:P (PB):: PE: E — (ED + DB). Hence E D x (ED + DB) - PE2. Whence this Construction.-Construct the rhombus AEPF, all of whose parts are known, and on PE erect the perpendicular P G-2- the given difference between AB and AC. With the centre G and radius GP, describe a circle, meeting EG, joined and produced, in I and H. Make EB — EiH, and draw BPC; then will AB —A -C 2 P G —the given difference, and BAC will- be the triangle required. Demonstratio.m-By IV. 30*, we have EH (BEB): EP:: EP: EL Also, by similar triangles PEB and CUPP, we have EB: EP:: PF (EP): CF or ED. Ihence ED - IEl, and AB -A C EB- UP El EB- ED -- Ei - El 1-1- 2 PG -- the given dlilference. Calculation.-In triangle AFP, Case 1, find AF= AE= EP; III. 36. TRIANGLES, QUADRILATERALS, AND PARALLELS. 53 then EG= /EP2 PG2, and EH- EG + GP — EB. Also, EI= EG- GP=ED- CF. Hence we have AB-AE + EB, and A C- AF + CF, and all the parts are known. Renmarks.-1. When the angle BA C is a right angle, AEPF will be a square. 2. This problem is the same as having a point P, equidistant from the lines AB and AC, given in position, to draw through P a line so that AB and AC shall have a given difference. See Problem LXIII. of this division. PROBLEM XXX. —To draw through a given point, between two lines forming any given angle with each other, a line, such that the parts of it intercepted between the point and the given lines shall have a given ratio to each other. the angles BA C and BAP, Given { the distance AP, BE L (and the ratio of EP to PG as mn to n. A G B Analysis. —Let AB and AC be the two lines given in position, P the given point between them, and FPG the required line, having FP: PG::n: n. Draw PE parallel to AB, and PD parallel to AC. Then, in similar triangles FEP and PDG, we have FE: PD (EA):: FP: PG:: m: n. Whence this Construction. —Having drawn PE and PD parallel to AB and AC respectively, take EF a fourth proportional (IV. Prob. 2*) to n, m, and AE, so that we shall have ni: m":: AE: EF. Through F and P draw FPG, and it will be the line required. Demonstration. —FP: PG:: FE: EA or PD:: mn: n in the given ratio. Q. E. D. Calculation. —. In the triangle APD, find, by Case 1, PD in. A E AE, and AD - EP; then n: "n:: A~ EF Also m: n:: n. AD AD: DG-. Whence we have AF AE + EF, and AG = AD +DG.: VI. 12. 54 GEOMETRICAL ANALYSIS. 2. In triangle GAF, Case 3, find FG; then FA: FG:: FE: FP, and FA: FG:: EA: PG. Scholium.-If FP is to be equal to PG, make EF-= EA; if FP is to be 3 PG, make EF-= 3 EiA, etc. PROBLEM XXXI. —In a right-angled triangle are given the perimeter, and the perpendicular let fall from the right angle on tho hypothenuse, to determine the triangle. B Given pendicular BD, and the sue of AB, B C, and A C. Analysis.-Let ABC represent the required triangle. Produce AC both ways, making AI —AB, and CII= CB; then JIH is known, being equal to the given perimeter. Join IB and B1I; then (I. 25, cor. 6*) the angle AIB — ABI- BA C, and < CHB_ CBHt= 2 A CB. Hfence HIB + IB -1 (BAC A CB) - of 90~ - 45~. Therefore angle 1B]H= 1800-45~ _ 135~, and the angle at the centre of a circle passing through 1, B, and H must = twice (180~ - < IBH) = twice (180~ - 1350) = 90~. Whence this Construction.-Bisect the given perimeter III in F; erect the perpendicular FE- B=FfIor FI. Join EHTand EI, and, with centre E and radius EI or ElI, describe the are IiBIH. Produce El, makinlg FG = the given perpendicular; through G draw GB p)arallel to IH, intersecting the are at B'. Join IB and BHi; mlake the angle IBlA - IJIB, and the angle HB C = IEIB; then ABC will be the required triangle. Dencmostration?.-It is evident (I. 12t) that AB + B C + A C A1 + CH+ AC =IH, and that BD= FG the given perpendicular. It remains to prove that the angle ABC is a right angle. Since, by construction, FE- FI= FH, each of the acute angles in the triangles EFI and E:f1 is 450; hence the anffle IEH= 90~. The angle EIB=EBI (I. 114), and angle AIB= ABI by construction; hence angle EBA =-E IF- 45. Also angle EBH= TRIANGLES, QUADRILATERALS, AND PARALLELS. 55 ~EHB, and angle CBH_ CHB; hence angle EBC E- ll - 45~. Wherefore angle AB C -EBA + EB C -90~. Q. E. D. Calculation. -E12 -EF2 + F12- 2 FI2', and El —FIV,/2 2 EB; EG'-EF+ FG; BG —7/EB2-EG2- DF; then IDIF — )F, and HD J-IF+ DF. Now, in triangle DIB, Case 3, find the angle DIB; then angle DAB- 2 DIB, and anglle A CB 900- -DAB. In triangle ADB, Case i, find AB and AD; and in BD C, Case 1, find B C and CD; then A C = AD + D C. Limit. —The perpendicular may be any quantity not greater than the difference between EB and EF, or El and EF; that is, not greater than the difference between half the perimeter multiplied by the square root of 2, and half the perimeter. If a - perpendicular height and p - semi-perimeter, a must be less than p (V 2 - 1). PROBLEM XXXII.-In a plane triangle are given the two sides, and the length of the line drawn fiom the included angle to the centre of the inscribed circle, to determine the triangle. Af B ( the sides AC and CB, Given i and the distance CD to the centre 0 (of the inscribed circle. Analysis.-Let ABC represent the required triangle, of which D is the centre of the inscribed circle (the circle not drawn in the figure); then (III. Prob. 15*) the line AD bisects the angle A, BD the angle B, and CD the angle C. Let a circle be described about the triangle ADB, meeting CD, produced, in E, and join EB. Then (III. 18 and I. 25, cor. 6t) the angles BED + BDE —-BAD + (B CD +- DB C) - the sum of half the angles A, C, and B respectively, which is equal to 900. Hence the angle DBE is a riglht angle. Also, since < CAD DAB - DEB, and < 1 CD - B CE, the triangles A CD and B CE are similar, and we have CD: CA:: CB: CE; whence this Construction. —Take CD - the given distance, and on it, produced, take CE a fourth proportional to CD and the two given sides CA and CB. On DE describe a circle of which the centre is - IV. 4. t III. 20 and I. 32. 56 GEOMETRICAL ANALYSIS. 0, and to the circle apply CA one of the given sides, and CB c the other. Draw AD, DB, EB, and OB, and AB C will be the required triangle, as is evident,A x\ from the analysis. Calculation. -By construction, CD: CA:: CB: CE — CA x CB o GB:G GAXGB- — ~-Then DE — CE — CD CD, OD —ED- OB, and OCG OD+ CD. E Now, in trliangle OBC, Case 4, find the angle 0 CB; then angle A CB = 2 0 CB. In triangle A CB, Case 3, find angles A and B and the side AB. In triangle ADB, Case 1, find AD and DB, then all the parts are know n. Corollary.-If D be the centre of a circle inscribed in a triangle, CD, produced, will pass through the centre of the circle circumscribing the triangle ADB. PROBLEM XXXIII. (By a civil engineer).-l Having a plane triangle given, it is required to draw a line cutting two of the sides, so that a point in that line shall be a given distance from one of the angles at the base, and the part of the line intercepted between such point and each side shall be equal to the lower segment cut from that side. ( all the angles, the side AB, and the E ( ED) Given distance BP, to draw EPD, so that PD- DB, and PE EA. nalysis.Suppose the problem constructed, as in the figure. Analysis.-Suppose the problem constructed, as in the figfure. About the triangle ACB describe a circle, and let 0 be the mliddle of the arc AB. Join OA, OB, and OP. Now, if OP -OB, the angle OPD will evidently be equal to OBD; also, since OB - OA, the angle OPE will be equal to OAE. Whence this Construction.-Having described a circle about the given triangle ABC, and taken 0, the middle of the arc AB, with the centre O and radius equal to OA or OB, describe an arc, in which apply BP.-the given distance of the point from B. Then draw the line TRIANGLES, QUADRILATERALS, AND PARALLELS. 57 1DPE, making the angle BPD — the angle DBP; then will PDDB, and PE -- EA. Demonstration.-Since angle BPD = DBP by construction, we have PD - DB. Hence it only remains to prove that PE = A. Now, the angles OBP and OPB being equal, also BPD and DBP, we have angle OBD- OPD; also the angles OBC and OAC, or OBD and OAE, together (IJI 18, cor. 4*) are equal to 1800. Hence OBD + OAE- OPD + OPE' (the sum of each pair of angles being equal to 180~) — OBD + OPE. Therefore OAE= OPE. Taking the equal angles OAP and OPA from these, we have the angle EAP - PA, and therefore PE = EA. Q. E. D. Calculation.-The angle AOB —180~ — C; hence we have each of the angles OAB and OBA — the given angle C. In triangle ABO, Case 1, find OB = OP -OA. In triangle OBP, find angle OBP. Now, the angle OB C- OBA + ABC, and the angle PBD OBC — OBP=- BPD. In triangle BPD, Case 1, find BD and PD. Again, in triangle ABP, Case 3, find AP, and angle BAP; then angle EAP —- BA C - BAP. Lastly, in triangle EAP, Case 1, find P= -- EA; then E C — A C- EA, and CD -- CB- BD. PROBLEM XXXIV. —In a rhombus are given the perimeter, and the sum of the diagonals, to find the diagonals. D the sum of AC and BD, 1A E i' A G Given.and the sum of the sides (AB, BC, CD, and DA. B Analysis.-Let AB CD be the required rhombus, whose diagonals intersect each other at E. Since in a rhombus the sides are all equal, each side is one-fourth of the given perimeter. Also the diagonals cross each other at right angles. Now, take EF — EB, then AF- 1 the sum of the diagonals, and the angle EFBB_ 450. Whence this Construction.-In the indefinite line AG, take AF- - the given sum of the diagonals. At F, draw a line making with FA an angle -450, to which line apply AB -- the given perimeter. To AG apply BC —BA, draw CD parallel to AB, and AD parallel to BC. Draw the other diagonal BD; then ABCD will be The required rhomnbus. - III. 22. 0* 58 GEOMETRICAL ANALYSIS. Demonstration. —The sides are evidently all equal, and their sum equal to the given perimeter. Now, in the triangle BEF, since the angle E is 900, and EFB —450 D by construction, angle EBF- 450~, and we have EF —EB. Hence A< F ~ —- G AC - BD - 2 AE + 2 BE — 2 AE + 2 EF- 2 AF- the given sum of the diagonals. Q. E. D. Calculation. —. In the triangle AIFB, Case 2, find angle BAF- BA-E. 2. In the triangle BAE, Case 1, find AE and EB; then AC2 AE, and BD - 2 EB. Limit.-Twice the square of one-fourth of the perimeter must not be greater than the square of half the sum of the diagonals. When twice the square of otne-fourth the given perimeter is just equal to the square of half the sum of the diagonals, the rhombus becomes a square. PROBLEM XXXV.-In a right-angled triangle are given the hypothenuse, and the difference between the two lines drawn from its extremities to the centre of the inscribed circle, to determine the triangle.;?E,jAB, the angle C a right angle, and the / Given difference between AG and GB, G being the centre of the inscribed circle. A F B Analysis.-Let ABC be the right-angled triangle required. In AG take GD -- GB, and join BD; then AD - the given difference between AG and GB. Now, since AG and BG bisect the angles BAC and ABC respectively (III. Prob. 15*), the angles GAB + GBA — (BAC+ ABC)-450. Hence AGB- 1350, and GDB and GBD each equal half the supplement of 1350~- - 450 — 22~. Whence this Construction. —Draw an indefinite line AG, i-n which taike AD the given difference. Make the angle GDB -22?0, and apply AB - the given hypothenuse. Make the angle DBG -22~. Also, make the angle GA C GAB, and the angle GBQCz GBiA; then ABC is the triangle required. Demonstration. —Since GD -GB (I. 12t), we have AG- GB!* IV. 4. t I. 6. TRIANGLES, QUADRILATERALS, AND PARALLELS. 59 -A G - GD- AD - the given difference. Also, since, by construction, the angles BAC and ABC are bisected by the lines AG and BG respectively, G is the centre of the inscribed circle. Again, since the angles GDB and GBD, by construction, each 2210~, the angle AGB- 135~; therefore the angles GAB and GBA together - 45~; and we have CAB + CBA - 2 (GAB + GBA) 90~. Whence A CB is a right angle. Q. E. D. Calculation. —1. In triangle ADB, Case 2, find angle DABGAB. Then angle BA C — 2 GAB. 2. In triangle ABC, Case 1, find AC and BC. In triangle AGB, Case 1, find AG, GB, and GF. Limits. —The given difference may be any quantity less than the hypothenuse. When the difference is nothing, the triangle is isosceles. PROBLEM XXXVI. —In a plane triangle are given the base, an angle at the base, and the sum of the squares of the other two sides, to determine the triangle. p the base AB, Given the angle ABC, and the suam of A C2 + BC2 =l2..EA. ) L' Analysis. —Let ABC represent the required triangle. Bisect the given base AB in D, join CD, erect the perpendicular DH- DC, and join AH. Now (IV. 14*), we have AC2 + BC2 -2 AD2 + 2 D C2 2 AD2 + 2 DH12- 2 AH2. Hence the given line m, whose square is equal to the sum of the squares of AC and BC, must be the hypothenuse of an isosceles right-angled triangle, one of the equal sides of which is equal to AH. Whence this Construction.-Draw AB -the given base, bisect it in D, and make the angle ABI~the given angle. With the centre D, and radius equal to half the given line mn, describe a semicircle, meeting AB, produced both ways, in E and F. Then EF-m. At D erect the perpendicular D G, join GE and GF, and to D G, produced, apply AH EG. Also, to BI apply DC C DH. Join AC, BC, - II. A. 60 GEOMETRICAL ANALYSIS. and CD, and let fall the perpendicular CL; then will ABC be the required triangle. Demonstration. -We have to prove only that AC2 + BC2 -EF2 -M2. By H_ cP IV. 14,* we have AC2 + BC2 - 2AD2 / + 2 DC2-2 AD22 DH+2 2 AHI2 2.E G2, by construction, - EG2 2+ GF2 / lEF2-m2. Q.E.D. Calculation. - D C2 D2 - AH2 - 25 A D L B F 2 2 2 2 2 D L B F AD2 - EG2 AD2 EF2 AD2. Hence D C is known. In triangle DBC, Case 2, find angle BD C and side B C; then A C EF/ E2- B C2. Or, in A AD C, < AD C - 1800~- BD C; by Case 3, find A C. NOTE.-If the perpendicular height CL is given, instead of the angle ABC, lay DO = the given perpendicular, draw OP parallel to AB, to which apply DC-= DH, and join AC and CB. The demonstration is the same as before. In the calculation, find DC2 as before, then DL = 1/ DC2- CL2, AL = AD +DL, and BL = AD - DL. Then AC- 1/ AL2+ LC2, and BC= 11/ BL2 + LC"2. Limits in the last case.-The square of the given perpendicular must not be greater than half the square of the given line, diminished by the square of half the base. That is, DO must not be greater than DH. If DO is just equal to DH, the triangle ABC will be isosceles. PROBLEM XXXVII.-In a plane triangle are given the base, an angle at the base, and the difference between the squares of the other two sides, to determine the triangle. C' /A\Fn ( the base AB, Given the angle ABC, and the difference of A C2 and B C2 equal m2. A L' B D Analysis.-Let ABC represent the required triangle, and let fall the perpendicular CL on the base. Then A C2 -AL2 CL2 =: II. A. TRIANGLES, QUADRILATERALS, AND PARALLELS. 61 B C2-BL2. Whence A C2-B C2 - AL2 - BL2 -m2, and we have this Construction.-On the given base AB describe a semicircle, in which apply AD m, and join BD. With A and B as centres, and radii equal to AB and BD respectively, describe arcs intersecting in E. Make the angle ABI- the given angle, and through E draw LEC perpendicular to AB, meeting BI in C. Join AC and BC, and ABC will be the triangle required. Demonstration. —We have to prove only that A C2 - B C2 A- D2 _i_2. We have AC2 -AL2 - CL2 - BC2-BL2. Hence AC2 -B C2- AL2 BL2. Also, AE2 — AL2 - EL2 BE2 BL2. Hence AE2 - BE2 - AL2- BL2. Wherefore we have A C2 B C2 -AE2- BE2 (by construction) AB2 - BD2 A)2 m2. Q. E. D. Corollary. —The difference of the squares of two lines drawn fiom any point in the perpendicular LC, or LC produced either way, to the points A and B, will always be equal to AD2, or A C2BC2 AL2 -- BL2. Calculation. -- BE BD V-/ AB2 _ AD2. Then, in triangle AEB, Case 4, AB: AE + BE:: AE-BE: 2 OL, 0 being the centre of the semicircle. Hence we have AL - AO + OL, and BL -AO- OL. Now, in triangle BL C, Case 1, find CL; then A C- AL2 ~- CL' and BCz/ BL2 + CL2. NOTE.-If the perpendicular LC were given instead of angle ABC, make the perpendicular LEC= the given perpendicular, and join CA and CB. The demonstration and calculation will remain the same. Limit.-The length of m must always be less than the base; that is, AD must be less than AB. 62 GEOMETRICAL ANALYSIS. PROBLEM XXXVIII. —"' In a city are three steeples visible from a station, at which the angle between the first and second steeples was taken, also the angle between the second and third, the second steeple being nearest the station. The three respective distances between the three steeples are known. It is required to find their respective distances from the station or place of observation."' —From Gummere's Surveying. E the distances AB, BC, and AC Given I the angles ADB and BD C, and consequently AD C. Analysis.-Let A, B, and C represent the positions of the first, second, and third steeples respectively, and D the station. Through the three points A, D, and C pass the circumference of a circle. Join DB, and produce it to meet the circumference in E, and join AE and CE. Then (III. 18, cor. 1*) the angle ACE -the given angle ADB between the first and second steeples, and the angle CAE - the given angle BD C between the second and third steeples. Whence this Construction.-With the three given distances form the triangle ABC, having B nearest to D, make the angle ACE the given angle between the first and second steeples, and CAE that between the second and third steeples. About the triangle AEC (III. Prob. 13t) describe a circle; join EB, and produce it to meet the circumference at D, the station required. Join DA and DC. Demonstration. —By III. 18, cor. 1,t the angle ADB A CE, and BD C CAE, which are the given angles by construction. Q. E. D. Calculation. —-1. In triangle BA C, Case 4, find angle BA C; then angle BAE _ BA C + CAE. In triangle E A C, Case 1, find AE. 2. In triangle BAE, Case 3, find angles ABE and AEB - AED A CD; then angle ABD - 1800 - ABE. 3. In triangle ABD, Case 1, find DA and DB, and in triangle ADC, Case 1, find D C. NOTE.-If B is farther from D than A and C are, conktruct the triangle ABC with B above the line AC. Join BE, and produce it to D, and proceed as before, only, in this case, the angle BAE will equal BAC- CAE. -III. 21. t IV. 5. + III. 21. TRIANGLES, QUADRILATERALS, AND PARALLELS. 63 PROBLEM XXXIX. —" There are given the heights of two columns, the distance of the foot of the lower column fiom the middle of the base of a statue in a straight line between them, and the distances from the head of the statue to the top of each column, to determine the distance between the tops of the columns; also the height of the statue." —Gummere's Surveying. G D the heights of the columns AF and CD, Given the distance AB, and Ethe distances EF and E D. D'I II A B C Analysis. —Suppose AF and CD, in the annexed figure, to represent the columns, and BE the statue. Through D, the top of the higher column, draw DG parallel to AC, to meet AF, produced, in G. Then AG- CD, and we have the line GD given in position, and AB, AF, FE, and ED given in length; whence this Conslruction.-On an indefinite horizontal line lay AB equal the distance of the foot of the lower column from the middle of the base of the statue. At A and B erect perpendiculars, and lay AF — the height of the lower column, and AG -that of the higher. Now, the head of the statue is in the perpendicular erected at B. To this perpendicular apply WFE equal the given distance from the top of the lower column to the head of the statue, and BE will be the statue's height. Through G draw a line parallel to AB, and the top of the higher column must be in this line. To this line apply ED equal to the given distance from the head of the statue to the top of the higher column; let fall the perpendicular DC on the horizontal line, and join DF, which is evidently the distance between the tops of the columns required. Calculation.-Through E draw. HEI parallel to ABC. Then HEY - AB, and FH- V FE2_ AB2, AH- AF- FH- BE the height of the statue IC. And DI- DC - IC. Also EIV/ ED2- DI2. Now, HI- -HE + EI AB + E1 — GD. And GF- AG- AF. Hence we have DF- / GD + GF2- the required distance between the tops of the columns, 64 GEOMETRICAL ANALYSIS. PROBLEM XL.-In a plane triangle are given the base and perpendicular height, to determine it, such that one of the angles at the base shall be double the other. a e the base AB of the triangle AB C, Given the height CD, and an gle AB C to be equal to 2 BA C. A e D B Analysis. —Draw BH bisecting the angle ABC; then each of the angles ABH and HBC will be equal to the angle BAC. Through H draw EHG parallel and equal to the perpendicular CD, and join CG, which will be parallel to AB. Then (I. 11 and 28*) we have AH —BH, AE- BE, and GC -ED. Also, since BHt bisects the angle ABC (IV. 17t), we have AB: BC:: AH: HC:: AE:EDor CG. That is, AB:BC::AE:CG. ButAB- 2AE, therefore BC 2 CG. Draw EF parallel to BC, meeting BG, produced, in F. Then the triangles BEEF and GCB will be similar, and, since the side B C - 2 CG, we have FE — 2 BE -AB. Whence this Construction.-Draw AB -the given base, bisect it in E, erect the perpendicular EG the given height, join BG, and to it, produced, apply EF- AB. Draw B C parallel to EF, G C parallel to AB, join AC and BH, and let fall the perpendicular CD; then ABC will be the required triangle. Demonstration.-In similar triangles FEB and B CG, we have FE (AB): BC:: EB: CG:: AE: ED:: AH: HC; that is, AB: B C:: AH: C. Hence (IV. 17t) BH bisects the angle ABC, and we have angle CBH= ABH- (I. 11~) BAH, and AB C - 2 BA C. Q. E. D. Calculation.-I. In triangle BEG, Case 3, find angle EBGEBF. 2. In triangle EBF, Case 2, find angle EFB GB C; then angle AB C EBG + GB C, and angle BA C AB C. 3. In triangle ABC, Case 1, find AC and B C; then ED - CG -2 B C, and AD ~ AE + ED, and BD- AE —ED. Scholium.-As EF- AB is greater than EB, the arc described with EF can cut BG, or BG produced, in but one point, and the I. 5 and 34p t VI. 3. V 7. 3. TRIANGLES, QUADRILATERALS, AND PARALLELS. 65 angle EIB can have but one value. Also, since ABll 2ED + 2 DB - 2 CG + 2 BD B- C + 2 BD, we have constructed the rightangled triangle CDB', having given the base CD, and the sum of the hyIpothenuse B C, and twice the perpendicular BD (equal to AB). PROBLEM XLI. —In an isosceles triangle are given the angle between the equal sides, and the three distances from a point, either within or without the given triangle, to the three angles, to determine the triangle. FiG. 1. FIG. 2. ~A Given, the angle AB C, B the sides BA and B t 6 / -____ _ \ to be equal, and the D distances PA, PB, and PC, in either Figure. B B Analysis.-Suppose ABC (referring to either figure) to represent the required triangle, with ABC the given angle, and BA- B C. At B make the angle PBD the given angle AB' C, lay BD = BP, and join AD. Then, comparing the triangles ABD and CBP, the angle ABD -PBD - PBA ABC — PBA CBP; also the including sides AB and BD, in one triangle, are equal to CB and BP in the other. Hence (I. 6*) AD PC. Whence this Construction.-Draw BP the given distance from the point to the angle B, make the angle PBD - the given angle, lay BD - BP, and join PD. With P and D as centres, and radii equal to the other two given distances respectively, describe arcs, intersecting in a point A on the opposite side of DP from B, when the point P is to be within the triangle (Fig. 1), but on the same side of DP that B is when the point P is to be without the triangle (Fig. 2). Join AD, AP, and AB. Make the angle AB C - DBP, lay B C AB, and join A C and PC; then ABC will be the required triangle, PC' being equal to AD, as is shown in the analysis. Calculation.-1. In triangle DBP, Case 1, find DP. 2. In triangle APD, Case 4, find angle APD; then (in Fig. 1) angle APB APD + DPB - APD + I the supplement of the griven angle AB]C or PBD. And (in Fig. 2) angle APB -DPB - APDthe supplement of the given angle ABC or PBD, —APD. I. 4. 5 66 GEOMETRICAL ANALYSIS. 3. In triangle APB, Case 3, find AB —BC. 4. In triangle A4 B C, Case 1, find A C. Fri. 1. Fio. 2. Limits.-I1. PA + PC must be greater than PB, the longest given distance. 2. The angle ABC may be made on either side of AB; but if made on the other side, the letters P and D must be transposed. Scholium. —When the triangle ABC is to be equilateral, the analysis, construction, and calculation are precisely the same as in the above problem, using the same letters; only in this case it is simpler to describe equilateral triangles on PB and BA, as PD PB and A C — AB. PROBLEM XLII.-Given the three distances from a point, either within or without a square, to the three nearest corners, to determine the square. FIG. 2. FIG. 1. C E A E Given the three distances PA,PB, and PC, drawn to D I A the corners of the D D square. B c See either Figure. Analysis.-Join A C. Then in the isosceles triangle ABC (either figure), the angle AB C, being a right angle, is given, and the three distances PA, PB, and PC; hence we have to construct the triangle ABC precisely as in the last problem, using the same letters, and then complete the square ABCE, which will be the one required. TRIANGLES, QUADRILATERALS, AND PARALLELS. 67 Property in relation to a square.-If from any point lines be drawn to the four corners of a square, the sum of the squares of the lines drawn to the extremities of one diagonal, will be equal-to the sum of the squares of the lines drawn to the extremities of the other diagonal. FIG. 1. D C FIG. 2. / - ~ B P C P -P PA'2 + PC2 PB2 + PD2. A B P Demonstration.-In Figure 1, give attention to the point P without the square, or in the diagonal, or in the diagonal produced, separately, and the lines drawn from the point under consideration to the corners A, B, C, and D of the square; then we have (IV. 14*) PA2 ~ p C" - 2 P 2 + 2 A 02 2 P O2 2D 02 - pB + PD2.'I l like manner, in Figure 2, giving attention to the point P within the square, or in th.e side, or in the side produced, separately, we ha.ve PA2 + PC2 = 2 P02 + 2 AO2- 2 P02 + 2 DO2 PB2 + PD2. When the point P is in the diagonal, or the diagonal produced, it will be observed that PA - the difference between PO and AO, and PC-their sum. Hence PA2 + P C2 (P0 AO)2 + (P 0 + A 0)2 2 PO2+ 2 AP2. Q.E. D. PROBLEM XLIII. —From a given point in the base of a given right-angled triangle to draw a line to a point in the hypothenuse, such, that if a line be drawn from this last point, parallel to the base, to meet the perpendicular, these two last lines may be equal. D E (the triangle ABC, BF, and AF, A B Given to draw FD, and DE parallel to AB (so that DE shall be equal to PD. Analysis.-Suppose PFD drawn so that DE parallel to AB shall be equal to FD. Draw CFH, meeting a line through A, parallel to'* II. A. 68 GEOMETRICAL ANALYSIS. FD, in G. Then, by similar triangles CDE and CAB, we have CD: DE:: CA: AB; also, in similar triangles CDF and CAG, we have CD: DF: * CA: AG; whence DE: DF::AB': AG; and in order to have DF- DE we must have AG — AB; wherefore this Construction.-To CF, produced, apply AGC AB, draw FD parallel to AG, and DE parallel to AB, and they will be the equal lines required. E Demonstralion.-By similar triangles, CA: A':: B CD:DE; also, CA: AG (AB):: CD:DF. Hence DE - DF. Q. E. D. Calculation.-i. In triangle BFC, Case 3, find angle BCF and CF; then angle ACG-BCA-.B CF. 2. In triangle A CG, Case 2, find CG; then CG: AB x CF AG (AB):: CF: FD - DE Scholium 1. —D is the centre of a circle that would pass through F, and be tangent to the perpendicular BC, at E. Scholium 2.-If it is desired that 1)F 2 DE, apply A G= 2AB; and so of any other ratio; take AG to AB in the same ratio that it is desired to have DF to DE, and then proceed as in the problem. NOTE. —The same analysis, constuction, demonstration, and calculation, also scholium 2, apply when the angle B is acute or obtuse. PROBLEM XLIV. —It is required to find a point in the perpendicular of a given right-angled triangle, such that two lines drawn from said point, one to a given point in the base, the other parallel to the base and meeting the hypothenuse, shall be equal. (the triangle ABC, and DL1 1-P Given - the point G, and PD parallel to ALB, A B to be equal to PG. Analysis.- Suppose PD drawn parallel to AB and equal to PG. Join CG, and produce it to meet BH, drawn parallel to PG, in H. Then, by similar triangles CPD and CBA, we have CP: CB:: TRIANGLES, QUADRILATERALS, AND PARALLELS. 69 PD: BA. Also, by similar triangles CPG and CBH, we have CP: CB::PG: BH. Hence PD:PG::AB:BH, and, in order that PD shall eqcual PCG, we must take BH- AB. Whence this Construction. —Join CG, and to it, produced, apply BH — BA. Draw GP parallel to BH, and PD parallel to AB; then PG PD. Demzonstration.-By similar triangles, CB: CP:: BA: PD; also, CB: CP::BH(BA): PG. Hence PG PD. Q. E. D. Calculation. —1. In triangle BCG, Case 3, find angle BCG and CG. 2. In triangle BCH, Case 2, find CII. Then, by similar triangles BCH and PCG, CH: CG:: BH(AB):PG - PD. Scholium. —If it is desired that PG shall be equal to 2 PD, apply BIH-2 AB; and so of any other ratio, take BH to AB in the same ratio that it is desired to have PG to PD, and proceed as in-the problem. NoTE.-The same analysis, construction, demonstration, and calculation, as also scholium, apply when the angle B is acute or obtuse. PROBLEM XLA. —In a plane triangle are given the base, an angle at the base, and the sum of the side opposite this angle, and the perpendicular height of the triangle, to determine the triangle. H E F in the triangle BAC the base AB, A B Given the angle BA C, and the su?n of B3C and CD. Analysis. —Suppose AB C to represent the required triangle. Produce the perpendicular DC, making CE- CB; then DE is known, being equal to the given sum of BC and CD. Through E, parallel to AB, draw a line, meeting AC, produced, in F; and through A, draw a line parallel to BC to meet FB, joined and produced, in G. Now, by similar triangles FCE and A CD, FC: CA:: EJ C: CD. By composition, FC: FA:: EC: ED. Also, by similar 70 GEOMETRICAL ANALYSIS. triangles FCB and FA G, FC: FA:: CB or E C: AG. Therefore AG — ED; and we have this Construction.-Draw AB =the given base, and Xi X P make the angle BAP equal to the given angle. Erect the perpendicular AH- the given sum of the side and perpendicular, draw lF parallel to AB, X c join FB, and to it, produced, apply A G -AI. Draw BC parallel to AG, then ABC will be the triangle required, having B C + CD AIl. A D\ 1~i~ B IDemonstration.- Through C, parallel to AH, draw D CE, and draw CI parallel to AB; then we must prove that BC- CE. By parallel lines we have AF: FC:: All: HI — CE. Also, AF: FC:: c A. Gor AH: BC; whence BCC-CE. Andc BC + CD-EBC + C — ED — AH. Q. E. D. Calculation. —1. In triangle AITF, Case 1, find AF. 2. In triangle AFB, Case 3, find angle AFB and FB. 3. In triangle AFG, Case 2, find FG; then FGP: WY:: AG: BC. Also DC DE B- C, and D)E: DC:: AF:AC. PROBLEM XLVI. —In a plane triangle are given one angle, an adjacent side, and the angle made with the side opposite the given angle by a line drawn firom the middle point of said side to the given angle, to determine the triangle. the angle ACB, the side BC, B Given and the angle BDC, D being the middle of AB Analysis.-Suppose the figure constructed. Draw DE parallel to A C. Then, since BD- DA, we have BE- EC. Also, since the side BC and the angle BDC are known, the segment BDC is known (III. Prob. 16'). Whence this Construction.-Make B C -the given side, and B CA- the given angle of the triangle. On BC (III. Prob. 16t) describe a circular segment to contain the angle to be made with the side opposite the given angle. Bisect BC in E, and through E, parallel to CA, draw a line to meet the segment in D. Join BD and CD, and produce BD, to complete the required triangle ABC. - III. 33. t III. 33. TRIANGLES, QUADRILATERALS, AND PARALLELS. 71 Demonstration.-Since BC, and the angles ACB and BDC, are the given quantities by construction, we have to prove only that D is the middle point of AB. Now, since ED is parallel to A C, and BE - E C by construction, we have (IV. 15*) BD — DA. Q. E. D. Calculation.-From O, the centre of the circular segment, draw OB, OD, and OE; then, 1. In right-angled triangle BOE, angle BOEz BDC; hence, Case 1, find OE, and OB —- OD; then the angle OED is equal to the difference between the given angle BCA or BED, and 90~. 2. In triangle OED, Case 2, find ED; then CA = 2 ED. 3. In triangle A CB, Case 3, find AB; then BD - AB- AD. PROBLEM XLVII. —In a right-angled triangle are given the difference between the base and perpendicular, and also the direrence between the base and hypothenuse, to determine the triangle. F in triangle ABC, E D \ Given B C - BA, and A AC- BA. Analysis. —Let ABC represent the required triangle. Take BD BA, then DC is known, and the angle BAD -- a right angle. Also take AIl A C, then BH is known. Draw HF parallel to B C, and DE parallel to AB; then, since BD - BA, we have HF AHX AC, and EFP ED- HB. Draw DG parallel to AC, meeting FC in G; then we have, by similar triangles, FA:FD:: AHU: DE, also FA: FD:: A C or AHt: DG; wherefore DG — DE, and we have this Construction. —Draw DE and EF at right angles, and each equal to the given difference between the hypothenuse and base, and D C parallel to EF, and equal to the given difference between the base and perpendicular. Join PD and FC, and to FC apply DG — DE. Through C, parallel to DG, draw a line meeting FD, produced, in A; and through A, parallel to DE, draw a line meeting PFE and CD, produced, in H and B respectively. Then ABC will be the triangle required, having A C- BA - DE, and BC- BA 1DC. VI. 2. 72 GEOMETRICAL ANALYSIS. Demonstration. —Since, by construction,'EF- ED, we have BD -BA, and HF-AH. Hence B C- BA B C- BD D C, F which is equal to the given difference by con- c struction. Also, by similar triangles, FD: FA:: DE: AH; FD: FA:: DG or DE: AC; whence A C - AK. Hence A C — BA - All - BA - BH_- DE the given difference by construction. Q. E. D. Calculation. —FD %/ 2 ED2. 1. In triH B A " angle FDC, Case 3, find angles, and FC. 2. In triangle FGD, Case 2, find FG; then, by similar triangles, FG:FC:: DG or DE: AC AH. Now, AB - AH- BHE A C - DE, and B C -= BD + D C = AB + DC. PROBLEM XLVIII. —In a right-angled triangle are given the sum of the base and perpendicular, and also the sum of the base and hypothenuse, to determine the triangle. F I C A___ in triangle ABC, GGiven B C BC BA, and AC+ BA. Using the term stinm instead of difference, and to FC, produced, applying DG - DE, as in the above figure, the analysis, construction, demonstration, and calculation will be in the precise terms given for the preceding problem, making proper modification in regard to produced lines. TRIANGLES, QUADRILATERALS, AND PARALLELS. 73 PROBLEM XLIX. —In a plane triangle are given the perpendicular height, and the radii of its inscribed and circumscribed circles, to determine the triangle. G [in triangle ABC, the perpenGiven dicular CK, and the radii OS Given \/A M /LS | AB I and E T, or TG, of its inscribed A and circumscribed circles. P Analysis. —Let ABC represent the required triangle, CK the given perpendicular, 0 the centre of the inscribed circle, and T the centre of the circumscribed. Join BO and CO, producing the latter to meet the circumscribed circle in E, and draw the diameter EMTG. Also, draw ROD and C0 parallel to AB, and join EB. Now (I. 25, cor. 6*), the angle EOB-= OCB + OBC -ACO + OBA _ (III. 18, cor. 1t) ABE + OBA- OBE. Hence the triangle EOB is isosceles, and we have EO -EB. Again, by similar triangles EBL and ECB, EL: EB or EO: EB or EO: EC. By division,:EO-EL:EO::EC-EO:EC; that is, OL:EO:: CO:EGC, or EC:EO:: CO: OL:: CR: OS. Hence EC x OSEO x CR EO x CR, and EC OS In similar triangles EDO and ENC, ED: EO::" EN: EC; that isx CEB is, EM+ OS: EO::"EM+ (' O:. Or, by dividing the ~0 second and fourth terms by O, we have EM+ OS: OS:: EMfq. CK: CR. From which EM- OS x (CK- B) OS CR — KR CK-2 O0S That is, EM is a third proportional to the excess of the perpendicular height of the triangle above the diameter of the inscribed circle, and the radius of the same circle. Whence this Construction. —With the given radius ET describe the circle AEBC, and draw the diameter ETG. Take EI- the diameter of D III. 21. D 74 GEOMETRICAL ANALYSIS. the inscribed circle, draw EQ perpendicular to EI and equal the radius of the inscribed circle, and IP- the given height of the G required triangle. Through P and Q describe the semicircle PQM, and draw AMB parallel to E Q. Take MNIP, the given height, draw NC parallel to AB, and join AC and BC; then will 2/ tT ABC he the required triangle. Join E C and EB, and take E 0O EB; then A\ KM~I /LS SUKB will 0 be the centre of the inscribed circle. Let fall the perpendiculars OS Q E and CK, join CG, and draw ROD parallel to AB. w' Demonstration.-PE - IP IE= EQ2 OS2 CK —2 OS. Also, EQ- OS. Hence EM- EQ - 0 EP CK- 2 OS Whence the construction becomes evident from the analysis. OS2 Calculation.-EM- (I 3E= CK- 2 OS;EB(IV.23) EGx EM a s, EOx CR; =EO; CRBCK- OS; EC, by the analysis, EO; EN EC x EJMLVEL2_ EM2 EXM+ CK; EN: EC::EM: EL - EN ML AL - AM 4- ML; BL - BM- ML; AB - AL + BL. Lastly, LB x EB EL LB:: EB: BC-, and EL: LA:: EA or EB: AC EL LA x EB - LAEL B, when all the parts are known. * VI. 8, cor. TRIANGLES, QUADRILATERALS, AND PARALLELS. 75 PROBLEM L.-From the vertical angle of a given triangle to draw a line to a point in the base, such that the square of this line shall be equal to the rectangle of the segments of the base made by that point. A DR M B the triangle AB C, Given to draw CD such that gF B, b CD2 shall equal AD X DB. Analysis.- About the given triangle A CB describe a circle whose centre is 0, and, supposing CD to be the required line, produce it to meet the circumference at F. Then (IV. 28*) we have AD x DB CD x DF. And, by the conditions of the problem, AD x DB- CD2; whence CD x DF — CD2, or DF — CD. Whence this Construction.-Produce CB, making BE - CB. Through E draw a line parallel to AB, meeting a circle through ABC in F and F', and draw CDF and CDF'; then CD or CD' will be the line required. Demonstration.-Since BE — CB, we have, by parallel lines, DF — CD, and D'F' — C. Hence CD2- CD x DF — AD X DB, and CD2 - CD' X D'F = AD' X D'B. Q. E. D. Calculation.-Draw the diameter COG, join GB, OB, and OF, and draw CML and ROS perpendicular to AB; then CM- ML ROS. 1. In right-angled triangle CGB, the angle G the given angle CAB; find GC, Case 1. Then, in similar triangles GCB, A CM, G C: CB:: A C: CM- ML- RS. And BM —/ B C2_ CM2, EL - 2 BM, and RM_ BRB- BM= - AB- BM= SL. 2. OR -/OB2- BR2, OS RS- OR CM- OR, and FS -F'S-V-/ OF2 OSQ2. 3. ES — EL + SL, FE=- ES + FS, FE- ES- FS, BD= 2 FE, BD - EF', AD - AB -BD, AD' = AB BD', CD / AD x DB, and CD' — /ADt X D'B. Scholium 1.-When EF does not meet the circle, the problem is impossible. Scholiunm 2.-If we want AD X DB: CD2:: m: n, make BE: B C:: m: n, draw EF parallel to AB, and draw CDF or C D'F'; then CD or * III. 35. 76 1GEOMETRICAL ANALYSIS. CD' will be the line required. Forn: In:: B C: BE:: CD: DF:: CD': D'F. Hence DE —m CD, and D'F - CD'. Also AD x DBn n CD x DF- -. CD2, and AD' x DtB CD' x D'F - - CD"2 9d n Whence AD x DB: CD2::rn: n, and AD' x D'B: CD2:: 1n: n. PROBLEM LI.-Given, a point between two parallel lines, and a point in one of these parallels, to find another point in this parallel, fromn which, if a line be drawn through the given point between the parallels to meet the other parallel, such line shall be equal to the distance from the point thus found to the given point in that parallel. E D G the parallels EG and LA, ii a rivenand the points B and A, to Given ( L A \ n ifid P such that PBE shall L be equal to PA. Analysis.-Suppose P to be the required point from which, if PBE be driawn, it will equal PA. Join AB, and on it describe the semicircle A CB, of which the centre is 0. Draw Cb'D, which will be perpendicular to each of the two parallels, and hence, as the point B is given, BC and BD are known. Produce EP to F, and join AF. Then, by parallel lines and similar triangles, we have CD: PE:: C: BP:: A:AP; that is, CD): PE:::AF: AP; hence, since PE- AP, AS= CD. Whence this Construction. —On AB, the line joining the given points, describe a semicircle A CB. Through B draw CBD, apply AF- CBD, and draw FPBE, and P will be the required point front which PBE is drawn equal to AP, as is evident from the analysis. Calculation.-1. Having the parallels and the points B and A given, and AF= CBD, which is given, the two triangles A CB and AFB, Case 2, are known in all their parts. Subtract the angle BAC from HAF, and we have the angle CAF- (III. 18, cor. 1*) CBP. 2. In triangle CBP, Case 1, find CP and PB; then PA CA - CP - PBE, and BEZ - PE-PB. -* III. 21. TRIANGLES, QUADRILATERALS, AND PARALLELS. 77 Scholium 1.- If it is desired to have PE: PA:: z: n, take m: n:: CD: AF, and proceed as above. Scholium 2.-If we take m: n:: BD: AF, and draw FPBE, we have EB: PA:: m: n, as is evident by comparing the similar triangles BED and APF. PROBLEM LII.-Given, a right-angled triangle, —it is required to draw a line from the acute angle at the base to meet the perpendicular let fall from the right angle on the hypothenuse, in a point such that, if a line be drawn from this point, parallel to the base, to meet the perpendicular of the given triangle, these two lines thus drawn shall have to each other a given ratio. c { the triangle ABC, to find the point P in the perpendicular BD, such that, PG D iE being drawn parallel to AB, we shall [ have AP: PG:: m:n. F A B Analysis.-In the given right-angled triangle ABC, suppose AP and PG to be the required lines, such that AP: PG:: m: n. Draw DE parallel to PG or AB, and DFparallel to AP. Then, by parallel lines, DF: DE:: AP: PG:: m: n. Whence this Construction.-Draw DE parallel to AB, and take n: m:: DE: (a fourth term) DE, and apply this distance to nieet BA, produced, in F. Draw AP parallel to DF, and PG parallel to AB; then P is the point required. Demonstration.-By parallel lines, AP: PG:: DF: DE:: m: n by construction. Q. E. D. Calculation.-l. By similar triangles CBA and CDB, we have GTB2 CA: CB:: CB: CD- C. Also, CA: CD::AB:DE; then n: m:: DE: DF by construction. 2. In triangle FDB, Case 2, find angle BFD- BAP. 3. In triangle BAP, Case 1, find AP and BP; then im: n:: AP: PG, and PD BD-BP. Scholium.-When AP is to be equal to PG, apply DF- DE. 78 GEOMETRICAL ANALYSIS. PROBLEM LIII.-Given, the perimeter and perpendicular height of a triangle, to construct it, such that one of the angles at the base shall be double the other. Given, the perimeter of AB C, the perpendicular height CD, and angle ABC to be equal to 2 BA C. E A G D B F Analysis. —Let ABC represent the required triangle. Produce AB both ways, making AE - A C and BF- B C, and join E C and FC. Then EF is known, being equal to the given perimeter. Draw ICH parallel to AB, meeting a perpendicular from G, the middle point of EF, in H. Then GOH — CD. Join FO. Then (I. 11, and cor. 6, 25*) angle AFC - ~ AB C- BA C - 2 E C - 2 EFO. Hence FO bisects the angle EFC, and we have (IV. 17t, and parallel lines) F': FC:: EO: OC:: EG: GD or CH; that is, EF: EG or GF:: FC: CH; whence, since ElF — 2 GF, we have FC - 2 CH; and thence this Construction.-Make EF the given perimeter; bisect it in G; erect GiH — the given perpendicular; draw HCI parallel to EF; join FH, and to it, produced, apply GL —EF; draw FC parallel to GL; join E C; make the angle FCB - CFIB, and the angle E CA CEA; let fall the perpendicular CD; then ABC will be the required triangle. Demonstration.-It is evident that the perimeter and perpendicular are equal to the given quantities, and it remains to prove that the angle AB C - 2 BA C. By similar triangles LGF and ECH, LG (EF): GF(EG):: FC: CH(GD); that is, EF: FC::EG: GD:: (by parallel lines)EO': OC. Wherefore (IV. 17t) FO bisects the angle FC, and angle EFC — 2 EFO 2 FEO --- BA C. Also, AB C - 2 EFC - 2 BA C. Q. E. n. Calculation.-1. In triangle FGH, Case 2, find angles, and FH. 2. In triangle FGL, Case 2, fi dld angles; then angle HFC - FLG, and angle DB C - 2 BFC - 2 (DFH + HFC), and DA C - 1DB C by the question. 3. In triangles ADC and i DC, Case 1, find A-C, AD, BC, and BD; then AB -- AD + BD. Corollary.-Since angle CFA 2 CBA - CAB, we have CF-=-. I. 5 and 32. t VI. 3. $ VI. 3. TRIANGLES, QUADRILATERALS, AND PARALLELS. 79 CA AE. Hence, if we have the line EF given, and a line HI parallel thereto, we can find, as above, the points C and A such that UFC, CA, and AE shall all be equal. Scholium.-In the above solution we construct the triangle EFC, having the base EF and perpendicular CD given, so that one of the angles at the base is double the other; that is, angle EFC 2 FEC. PROBLEM LITV.-Given, the vertical angle of a triangle, the angle made with the base by a line drawn from the vertical angle to the base, and the radii of the two circles inscribed in the two triangles into which the whole triangle is divided by this line, to determine the triangle. C the angles ACB and ADC, and the radii Given \EG and FL of the circles GPTHand LIQ. -AG D L B Analysis. —Suppose ABC to be the required triangle, having the angles A CB and AD C given. Join C with the centres E and F; then (III. Prob. 15*) angle ECF — the given angle A CB. Construction.-Draw any line AB, and at any point D, in AB, make the angle ADC-the given angle; and, touching the lines AB and DC with the given radii (see scholium next page), describe the circles E, GH and F,IL, E and F being the respective centres. Join EF, cutting the line CD in R, and on EEF (III. Prob. 16t) describe a segment of a circle, of which 0 is the centre, to contain an angle equal to half the given vertical angle, cutting D C in C. Join CE and CF, also OC, OE, OR, and OF, and (III. Prob. 14t) draw the lines CPA and CQB, touching the circles -GH and IL in P and Q respectively. Then ABC will be the required triangle. Demonstration. —By III. Prob. 15,~ angle ACB- 2 CE, and all the rest is manifest from the analysis and construction. Calculation. —On CRD let fall the perpendicular OS.' 1. In the triangles DGE and DLF, Case 1, find DE and DGIV. 4. tT III.3. IIr. I. ITV.4. 80 GEOMETRICAL ANALYSIS. D)I, and FD and LD —DI; then EF —/IDE2+FD2, hEDF being evidently a right angle, IH- DII- DI. 2. By similar triangles EHR and FIR, we have EH: FI: ER:FR, and EH: Fl:: HR: IR. By composition, BEH+ FI: EH:: (ER + FR) / 1\s< \\ or EF: ER, and EH~- lF: / H::(HR +R IR) or IH: H1i; then RF - EF- ER, JR IH- HR, and DR- DH — A H D B HR. In triangle EHR, Case 2, find angle ERH — ERS ERC; hence the angles REC, REO- REC- OCE, and EOF2 BECF, are known. 3. In triangle EOF. Case 1, find OE- OC. In triangle OER, Case 3, find angle EBO, and RO; then angle SRO-EROERS. 4. In triangle SRO, Case 1, find RS and SO; then CS-./ OC2 - S02, RC-RS + CS, and DC-DR +R C. 5. In triangle ERC, Case 3, find EC, and angle ECR - ECA. In triangle FRC, Case 3, find FC, and the angle FCR-FCB; then angle A CD - 2 E CA, and angle D CB - 2 FCB. 6. In triangles ACD and DCB, Case 1, find AC and AD, and B C and BD; then AB - AD + BD. Scholium.-To describe a circle, as required in the preceding problem, which shall have a given radius, and touch two lines given in position, as ADB and DC, it is only necessary to bisect the \/ 0 angles ADC and BDC by the i 0 I / \ straight lines DE and DF respectively; erect the perpendicular DI equal to the given radius, A G D L B and parallel to AB, through I, draw a line to meet the bisecting lines in E and F, either of which will be the centre of a circle that will touch both lines. Demonstration.-Let fall the perpendiculars EG, EH, and FL, FT; then (I. 17*) the triangles DGE and DHIE are equal in all — I. 26. TRIANGLES, QUADRILATERALS, AND PARALLELS. 81 their parts, as also the triangles DLF and D TF. Hence EG —.EH- DI, and FL - FT- DI. The radii EG and FL may be of different lengths, and the construction modified accordingly. Corollary.- The angle EDF'= ED C + CDF — (AD C + CDB)2= half of 1800 - 900. Hence angle EDF is a right angle. PROBLEM LV. —In a plane triangle are given the vertical angle, an adjacent side, and the length of a line drawn fiom the vertical angle to the base, dividing the base in a given ratio, to determine the triangle. Uivn ({the angle A CB, the side AC, and Given the line CD dividing AB so that AD: DB::m:n. A DH B Analysis.-Suppose A CB to be the required triangle. Draw DE parallel to CB; then (IV. 15*) AE: EC:: AD: DB::2 m: n. Whence this Construction. —Make ACB equal to the given angle, and AC equal to the given side. Divide AC (IV. Prob. it) in E so that AE: EC::m': n. Draw ED parallel to CB, and to it apply CD the given line. Join CD and AD, and produce the latter to meet CB in B; then A CB is the triangle required. Demonstration. —By IV. 15,4 AD: DB:: AE: EC:: mn; whence all is manifest. Q. E. D. Calculation.-By construction, m: n:: AE: EC. By composition, ma. AC m + n:m::AE + ECorAC:AE - Also, m +n:n: n. AC AE + EC or AC: ECm + n 1. In triangle CED, the angle CED =1800 - AED 180~A CB (I. 25, cor. 6~); find ED, Case 2; then AE: ED:: AC: CB. 2. In triangle ACB, Case 3, find AB. Then, since m: n:: AD: m. AB D)B, we have, by composition, m + n: m:: AB: AD +'79, - n Also, n + n: n:: AB: DB - AB rn + n * VI. 2. t vI. 10. I. 2.. I. 32. D* 6 82 GEOMETRICAL ANALYSIS. PROBLEM LVI.-In a plane triangle are given the vertical angle, the perpendicular let fall on the base, and the difference of the squares of the perpendiculars let fall from the foot of this perpendicular upon the two sides of the triangle, to determine the triangle. FIG. 1. [ the angle AGB, the perpendicular CD, and the GivenI difference of the squares of DF and DE; that is, 2 X B DF2-DE2- L2, Fig. 2. FrIG. 2. A B Li — i Analysis.-Suppose the triangle A CB constructed, as in the above figure. Bisect the given perpendicular CD in 0, and about the centre 0 describe the circle CFDE, which, as CFD and CED are right angles, must pass through the points F and E. Join OF, OE, and FE, and on FE let fall the perpendicular DG; then the angle FOE - 2 A CB; also (cor. to Prob. 36), FG2 - GE2 2= FD2 _DE L2. Whence this Construction.-With 0 as a centre, and radius equal to half the given perpendicular, describe a circle. At 0 make the angle FOE twice the given vertical angle, and join FE. On FE describe a semicircle, in which apply FI —L, and let fall the perpendicular IH. Bisect _HE in G, draw GD parallel to HI, and draw the diameter DOC, which will be equal to the given perpendicular. Draw ADB perpendicular to D)C, meeting CF and CE, joined and produced, in A and B; then ABC will be the required triangle. Demonstration. —Join DF and DE, which will be (III. 18, cor. 2*) at right angles to AC and CB. The angle A CB - FOE - the given vertical angle. Also, FD2 - DE2 - (cor. to Prob. 36) FG-2GE-2 - (FG + GE) x (FG —GE) FE x FH= (IV. 18, cor.t) FI2 L2. Q. E. D. Calculation. —1. In triangle FOE, Case 1, find FE; then FHFl2 LL2 FE =E and E G - EH (EF- FH). 2. On EF let fall the perpendicular OL, and produce it to meet DP, drawn parallel to FE, in P; then DP-LG —EL-EG. Also OL= OE2 -EL2, OP - %/ OD2 -- DP, DG=PL *- III. 31. t VI. 8, cor. TRIANGLES, QUADRILATERALS, AND PARALLELS. 83 PO-OL, DE-/DG2 + EG2, FD V -/IDG 2 + GF2, CF V CD2 -FD2, and CE / CD2 -ED 2. 3. By similar triangles CFD, CDA, we have CF: FD:: GCD: DA, and CE: CD:: CD: CA. 4. By similar triangles CED, CDB, we have CE: ED:: CD:DB, nd CE: CGD:: CGD: CB, and AB - DA + DB. Lixmit.-L must be less than FE. PROBLEM LVII. —In a plane triangle are given the vertical angle, the perpendicular let fall on the base, and the sum of the perpendiculars let fall from the foot of this perpendicular upon the two sides of the triangle, to determine the triangle. C the angle A CB, the perpendicGiven ular CD, and the sum of the /r (perpendiculars FD and DE. A B Analysis.-Suppose ABC to be the triangle required. On CD describe the circle CFDE, of which 0 is the centre, and join OF, O.E, and FE; then the angle FOE = twice the given angle A CB, and hence the triangle FOE is known in all its parts. Produce FD, making DG = DE, and join EG; then FG = FD + DE is known, and the angle FGE- ~ FDE- the supplement of the given angle A CB is known. Whence this Construction. —With a radius equal to half the given perpendicular, describe a circle of which the centre is 0. At 0 make an angle FOE = twice the given vertical angle, and join FE. On F.E (III. Prob. 16*) describe a segment FGE to contain an angle equal to half the supplement of the given vertical angle, in which apply the line EFDG - the given sum of the perpendiculars, D being the intersection of FG with the circumference of the circle first described. Draw DOC, GE, and DE. Also, through D, perpendicular to CD, draw AB, meeting CF and CE, joined and produced, in A and B respectively; then ABC will be the required triangle. * III. 33. 84 GEOMETIRICAL ANALYSIS. Demonstration.-By III. 18, cor. 2,* DFandDE are perpendicular to A C and BC respectively. The angle A CB FOE the given vertical angle, angle FGE, by construction, equal to 1 FDE. Hence iDEG 1 FDE - FGE, and DE - D G; and we have FD +- DE- FG -the given sum. Q.E. D. F \\ ) Calculation. —1. In triangle FOE, Case 1, find FE. 2. In triangle FGE, BH D al B Case 2, find angles, and EG. In triangle _ED G, Case 1, find ED; then, FD - FG - ED. 3. In triangle FDE, all the angles being known, we have (III. 18, cor. It) angle A CD _ FCD _ FED, and angle BCD - ECD EFD. In triangles ACD and BCD, Case 1, find CA and AD, CB and DB; then AB - AD + DB. PROBLEM LVIII.-In a plane triangle are given the base, one angle at the base, and the segments into which the base is divided by a line bisecting the vertical angle, to determine the triangle. T in the triangle AEC, the base AC, the angle ACE, D~ / | Giveng and the segments AB and I B C, made by EB, bisecting the angle AEC. A B C F Analysis. —Let AEC represent the required triangle. Make the angle EBD = EB C; then BD = B C, and angle BDE - B CE the given angle. Produce A C to F; then angle ADB FCE, they being supplements of the equal angles BDE and BCE, the given angle. Whence this Construction.-Draw an indefinite line AF, in which take AB and B C the given segments of the base, and make the angle ACE —the given angle. On AB (III. Prob. 16t) describe a segment to contain an angle equal to FCE, in which apply BDB C. Join AD, and produce it to meet CE in E, and join EB; then A CE will be the required triangle. Demonstration.-It remains to prove only that EB bisects the angle AEC. Since, by construction, BD —B C, and angle ADB - II. 31. t III. 21. T III. 33. TRIANGLES, QUADRILATERALS, AND PARALLELS. 85 FCE, we have angle EDB = ECB; wherefore ED - EC, and angle DEB or AEB= CEB, and EB bisects the angle AEC. Q. E. D. Calculation. —. In triangle ADB, Case 2, find angles, and AD. 2. In triangle ACE, Case 1, find AE and EC=-ED; then AD — AE-ED. 3. In triangle ABE, Case 1, find BE, if desired. PROBLEM LIX. —In a right-angled triangle are given the base, and the rectangle of the hypothenuse and perpendicular, to determine the triangle. FIG. 1. D i in the triangle AB C, the base AB, Given and the rectangle A C x CB L2, Fig. 2. Fro. 2. A E O B F L Analysis. —Let ABC represent the required triangle. Produce BC, and take BD a third proportional to AB and L (IV. Prob. 2, cor.*); then AB: L::L:BD, and AB x BD - L2 A C CB by the question. Draw CF at right angles to AC, meeting AB, produced, in F. Then, by similar triangles ABC and CBF, we have AB:A C:: CB: CE. Hence AB x CF- A C x CB=- AB x BD; fromr which CF- BD. Whence this Construction.-Bisect the given base AB in E, and, having found BD as in the analysis, take EF- ED. On AF describe the semicircle A CF, and join A C and CF; then AB C will be the triangle required. Demonstration.-By IV. 23,t we have CF2 E AF x FB = (EF + EA) x (EF - EA) - (ED + EB) x (ED- EB) - ED-2 EB2 BD2; hence CF BD. By similar triangles ACB, BCF, we have AB:AC:: CB: CF or BD; hence AC x CB - AB x BDl-L2. Q.E. D. Calculation. —By construction, EF= ED S/ EB2 E + ]BD2; then AF EF ~ AE, and FB EF — AE EF- EB, and (IV. 23t) FC=F/ x A x F, AC / AF AB, and B C -W/AB X FB. * VI. 11. t VI. 8, cor.: VI. 8, cor. 86 GEOMETRICAL ANALYSIS. PROBLEM LX. —Having given the lengths of two lines at right angles to each other at a common extremity, to draw a line from the other extremity of one to meet the other line produced such that the parallel to this line so drawn intercepted between the two given lines, and drawn through the extremity of the line that was produced, shall be equal to the sum of this line and the part produced. A GDive i the lines AB and B C at right angles Given {at B, to draw AE such that the parallel CF shall be equal to BE. Anasis.Suppose the problem constructed as in the Analysis.-Suppose the problem constructed as in the figure. On AB describe the semicircle ADB. Then (IV. 30*) AE: EB:: EB: ED. But, by similar triangles AEB and FCB, we have AE: EB:: CF or EB: GB; whence ED = CB; and we have this Construction.-On the given lines AB and BC describe semicircles, G being the centre of the one on BC. Draw AGI; and to BC, produced, apply AE —AGI, and draw CF parallel to AE; then will CF= BE, and AE will be the line required. Demonstration. — AB2 - A G2 -- GB2 = (A G + GB) x (AG - GB) = (A G + GI) x (A G - GB) == A x (AI- B C) = AE x (A E B C); that is, AE: AB:: AB: AE — B C. But, by similar triangles ABE and ABD, we have AE: AB:: AB: AD; hence AE- B C - AD —AE- ED. Whence ED — BC. Now (IV. 30t) AE: EB:: EB ED or CB. Also, by similar triangles AEB and FCB, we have AE: EB:: CF: CB; hence CF= BE. Q. E. D. Calculation.-AG - V/ AB2 + BG2, AE = AI= AG + GiI, AD=AG-BG, ED=BC, BE ~=/ AE x ED; then BE: EA:: BC: CE, and BE: BA:: BC: BF. -.t III. 36. t III. 36. TRIANGLES, QUADRILATERALS, AND PARALLELS. 87 PROBLEM LXI.-From two given points to draw two right lines to meet in a line of any kind given in position, so that the difference of the squares of these lines shall be equal to a given quantity. FIG. 1. C. GGA rf the points A and B, and the line GH Given of any kind, to draw B C and AC to meet in GHsuch that B C2 A C2L2, Fig. 2. A -,, B FIG. 2. II ~ ~ ~ ~ ~ ~ L L K Analysis. —S uppose AC and B C to be the lines required. Join AB, and on it let fall the perpendicular CD. Take DF —-DA; then L2 BC2- AC2-(cor. to Prob. 36) BD2 -AD2= (BD + AD) x (BD - AD) = AB x BF. Whence this Construction.-On the given distance AB describe a semicircle, in which apply BE = L. Let fall the perpendicular EF on AB; bisect AF in D; erect a perpendicular at D to meet the given line GH in C, and join A C and B C, which will be the lines required. Demonstration.-B C2 - A C2 - BD2 - AD2 = (by analysis) AB x BF= (IV. 23*) BE2 L2. Q. E. D. BE2 Calculation. —BF -BAF= AB - BF, AD —AF, DB DF + BF. Now, combining AD and DB with the given position and equation of the line G CH, A C and B C are determined. Scholiuml. —If the given line GH is the arc of a circle whose centre is O and radius OC, produce CD to meet a line drawn through 0 parallel to AB, in I. Then, because the position of GH is given, the centre O is given, and DI and OI are known, as also the radius OC. Then CI= / 0C2- 012, CD= CI —DI; also, A C -=/ AD2 + CD2, and B C =-/ DB2 + CD2. * VI. 8, cor. 88 GEOMETRICAL ANALYSIS. PROBLEM LXII. —"Having, at an unknown distance from the foot of a steeple, taken its elevation, I advanced sixty yards towards it on level ground, and then observed the angle of elevation, which was just the complement of the former. Advancing twenty yards still nearer, the angle of elevation was now just double of the first. Required, the altitude of the steeple, and the distances of the stations fiom its foot." the distances AB and BC, the angle Given 4EBD to be the complement of EAD, and ECD to be the double of FAD, to find ED. A B C D F Analysis.-Let A, B, C represent the three stations at which the angles of elevation were taken, and DE the steeple. Since the angle at C is, by the problem, double that at A, it is evident that C is the centre and CA the radius of a semicircle AEF passing through the first station A and the top of the steeple E. Describe the semicircle, and join FE; then EBD, by the question, being the complemlent of EAD, is equal to AFE; and hence BE = EF, and BD DF. Whence this Construction.-On an indefinite right line lay AB - 60, and B C - 20. With centre C and radius CA describe the semicircle AEF. Bisect BF in D; erect the perpendicular DE, and join E with the points A, B, C, and F; then JDE will represent the steeple. Demonstration.-The angle ECD (III. 18) = 2 EAD; also (I. 5t), the triangles BDE and FDE are equal. Itecce angle EBD = EFD EFA = complement of EAF or AD.. E. D. Calculation. -BF= B C + CF or CA = 20 + 80 100, BD ~ BF- 50, CD= BD - BC-= 30, CE B CA =80, DE / CE"- CD2 / 802- 302 =V 5500 10, /55. Also, ADAB + BD -- 110. III. 20. t I. 4. TRIANGLES, QUADRILATERALS, AND PARALLELS. 89 PROBLEM LXIII.-In any plane triangle are given the vertical angle, the line bisecting this angle, and the sum of the including sides, to determine the triangle. D M N "A~ ~"~ the angle BA C, the GeF BA C, and the suml L of AB and AC. L Analysis.-Let BAC represent the required triangle. T'hrough P draw PF parallel to AB, and PE parallel to AGC; then AEEPF is a rhombus, having all its sides equal, and known. Produce AB both ways, making AD- AF, and BL- CF; then DL AIB + AC, the given sum of the sides, is known; and EL -DL -2 AE is known. Also, by similar triangles CFP, PEB, we have CF (BL): PF (EP):: EP: EB; wherefore BL X EB - EP2. Whence this Construction.-Having laid AD 2 AF, and DL — the given sum of the sides, on EL describe a semicircle of which the centre is 0; erect the perpendicular E37 —EP; draw MN parallel to AL, and let fall the perpendicular NB; then through B and P draw the line BPC, and BA C will be the triangle required. Demonstration.-By similar triangles, CF: FP (EP):: EP: ElB; hence CF x EB EP2- EM2 - BN2- BL x EB; whence CF BL, and AB + AC - AB + AF+ CF — AB AD - BL DL - given sum of sides. Q. E. D. Calculation. —Join N and 0. In triangle APE, Case 1, find EP - EM AE - AD - NB; then EL - DL-2 AE. In triangle OBN, find OB; whence BE and BL are known; then AB_ A E + EB, and A C — AF + BL. Limits.-I. It will be seen that the line MiN cuts the semicircle, in another point N7. If the point N' had been taken instead of AT, the point B would have come between E and 0, and the sides would have been of the same lengths as above obtained, but AB would have been the shorter. 2. When MN is just tangent to the semicircle, the sides 4AB and AC will be equal, B C will be perpendicular to AP, and the shortest possible, or minimum, line that can be drawn through P to meet the lines AB and A C. 90 GEOMETRICAL ANALYSIS. 3. When MN does not meet the semicircle, the problem is impossible. Remarks. —. When BAC is a right angle, the rhombus AEPF becomes a square. 2. This problem is the same as having a point P given in position equidistant from the lines AB and AC, to draw through P a line such that AB and AC shall have a given sum. (See Prob. XXIX. of this division.) PROBLEM LXIV.-In a trapezium, of the six parts-viz., two opposite sides, and the angle formed where these sides meet when produced, the two diagonals, and the angle formed at their intersection-any five being given,-to construct the trapezium. C CASE 1. D/;; (the two opposite sidcs AB and CD, E Given the angle P, and the two diagonals p (A C and BD. A B F Analysis. —Suppose AB CD to be the trapezium constructed as required. Produce the giver sides CD and BA to meet in P. Draw BE parallel to DC, and CE parallel to DB, and produce AB to F; then angle FBE (I. 20, cor. 3*) the given angle P, BE CD, and CE - BD. Whence this Construction.-In any line, as PBF, take AB its given length, and make the angle FBE _ the given angle P, and BiE - the length of CD. Then, with A and E as centres, and radii equal to the diagonals AC and BD respectively, describe arcs intersecting in C. Join A C, E C, and AE. Draw BD parallel to GE, and CD parallel to BE, producing it to meet BA, produced, in P. Join AD, and ABGCD will be the trapezium required. Demonstration.-Since PC is parallel to BE, the angle P - FBE -the given angle by construction. Also, being opposite sides of a parallelogram, CD - BE, and BD CE, the given lengths of the diagonal and side by construction. Q. E. D. Calculation. —l1. In triangle ABE, Case 3, find angle BAE, and side AE. 2. In triangle A CE, Case 4, find angle CAE; then angle CAB CAE + BAE. 3. In triangle CAB, Case 3, find angle ABC, and side BC. 4. In triangle PBC, Case 1, find PC and PB; whence we have PA and PD. I. 29. TRIANGLES, QUADRILATERALS, AND PARALLELS. 91 5. In triangle PAD, Case 3, find the angles A and D, and the side AD; we then have all the angles and all the sides of the trapezium. NOTE.-Either pair of opposite sides may be given, only it must be observed to produce those sides that are given, and that it is the angle at their intersection that is given.'If the given sides, in any case, are lettered AB and CD, and the angle formed, where they meet on their being produced, be BPC, the same letters used in the preceding analysis, construction, etc. will apply in each case. PROBLEM LXIV., continued. (Case 2.)-In a trapezium are given the two opposite sides, the angle formed where they meet on being produced, one diagonal, and the angle formed at the intersection of the diagonals, to determine the trapezium. (the two opposite sides AB and Given. 1CD, one diagonal, as AC, and the angles P and 0. P A B F Analysis. —As in Case 1, let ABCD represent the required trapezium. Produce the given sides BA and CD to meet in P, and draw BE and CE parallel to DC and DB respectively; then BE = the given side CD, and angle ACE — (I. 20, cor. 3*) the given angle AOB. Whence this Construction.-In any line, as PF, take AB -its given length, make the angle F1B'EE the given angle P, and BE-the given length of CD. Join AE, and on it describe (III. Prob. 16t) a segment to contain an angle equal to the given angle A OB, in which apply AC-the given diagonal. Join BC and EC; draw CP parallel to BE, and BD parallel to EC, and join AD; then ABCD will be the required trapezium. Denmonstration.. —By parallel lines, CD- BE, angle P -- FBE, and angle A OB'-A CE the given angles. Q. E. D. Calculation. —I. In triangle ABE, Case 3, find angles BAE, AEB, and side AE. 2. In triangle ACE, Case 2, find angles CAE, CEA, and side CE — DB. 3. In triangle CAB, angle CAB= CAE- + BAE, find, Case 3, angles ABC, A CB, and side B C. 4. In triangle PBC, Case 1, find PC and PB, thenuce find PA, PD, and AD, and we have all the sides and all the angles of the trapezium. 5' I. 29. t III. 33. 92 GEOMETRICAL ANALYSIS. Scholium 1.-If the diagonal BD were given instead of AC, the other given parts being the same, then, after describing the segment on AE as above, apply EC_ the given diagonal BD, and proceed with the construction and calculation upon the same principle as already directed. Scholium 2.-Either pair of opposite sides may be given, only it mnust be observed to produce the sides that are given, and that it is the angle at their intersection that is given. (See note to Case 1.) PROBLEm LXIV., continued. (Case 3.)-Having given in the trapezium the two opposite sides, the two diagonals, and the angle at their intersection, to determine the trapezium. E (the opposite sides AB and CD, the Given - diagonals AC and BD, and the angle (0 at their intersection. A B Analysis. —Suppose ABCD to be the required trapezium, and, as in the preceding cases, produce the given sides BA and CD to meet in P. Draw BE parallel to AC, and CE parallel to AB; then, by parallel lines, BE A C, E C AB, and the angle DBE - the given angle DOC. Whence this Construction.-DI)raw any line, in which take DB-its given length; make the angle DBE -the given angle DOC, and BE= the given diagonal AC. Join DE, and with E and D as centres, and radii respectively equal to the given sides AB and CD, describe arcs intersecting in C. Join EC, CD, and CB, draw CA parallel to EB, and BAP parallel to CE, meeting CA and CD, produced, in A and P respectively. Join AD; then ABCD will be the trapezium required. Demonstration.-By parallel lines, A.B CE, A C- BE, and angle DOC DBE - the given angle. Q. E. D. Calculation.-I. In triangle DBE, Case 3, find angles BDE and BED, and side DE. 2. In triangle DCE, Case 4, find angle CDE; then angle CDB)CDE + BDE. 3. In triangle CDB, Case 3, find angle D)CB, and side BC. 4. In triangle CAB, Case 4, find angle ABC. 5. In triangle PBC, Case 1, find PC and PB, and thence find Pi), PA, and DA; and we then have all the sides and all the angles of the trapezium. (See note to Case 1.) TRIANGLES, QUADRILATERALS, AND PARALLELS. 93 PROBLEM LXIV., continued. (Case 4.)-When in the trapezium are given one of the opposite sides, both diagonals, the angle at their intersection, and the angle where the given side and the side opposite to it meet on being produced, to determine the trapezium. _c< ( one side of the trapezium, as AB, both Given diagonals AC and BD, the angle 0 at their intersection, and the angle P. P A B Analysis.-Suppose ABCD to be the required trapezium. Draw BE parallel to A C, and CE parallel to AB; then, producing DC to F, the angle FCE, by parallel lines, will be equal to the given angle P, and the angle DBE equal to the given angle DOC. Wherefore BE, being equal to A C, is known in length and position with respect to the given diagonal DB, and angle ECD equal to the supplement of FCE or given angle P. Whence this Construction.-Draw DB its given length; make the angle DBE -the given angle DOC, and BE the other given diagonal AC. Join DE, and on it (III. Prob. 16*) describe a segment to contain an angle equal to the supplement of the given angle P, in which apply E C the given length of AB. Draw CA parallel to EB, and BA parallel to EC; join AC, AD, and CD, and produce CD and BA to meet in P. Demonstration.-By parallel lines, AC -BE, AB EC, and angle P - FCE -the supplement of D CE. Q. E. D. Calculation.-1. In triangle DBE, Case 3, find angles BDE, BED, and side DE. 2. In triangle D CE, Case 2, find angles CDE, CED, and side CD; thence the angles CEB, CAB, and CDB are known. 3. In triangle CAB, Case 3, find angles ACB and ABC, and side B C. 4. In triangle PBC, Case 1, find PB and PC, thence PA, PD, and AD; when we have all the sides and all the angles of the trapezium. Remark.-If the side CD were given instead of AB, then in the segment on DE apply CD- its given length, and proceed with the construction and calculation as above. This concludes all the varieties of the problem where the opposite sides meet. III. 33. 94 GEOMETRICAL ANALYSIS. Scholium.-We now come to consider the Cases, in the same order, where the opposite sides are parallel. CASE 1.-In Case 1 of the preceding problem, where the opposite sides AB and CD are given, and also the diagonals AC and BD, if the opposite sides AB and CD are parallel, then the angle P vanishes, and BE, in that figure, will be the prolongation of AB. Then we have this Construction.-Produce AB (Fig. 1), making the produced part FI. i. BE — the length of CD; and with A and E as centres, D C and radii respectively equal to the given diagonals AC and BD, describe arcs intersecting in C'. Join CA, CB, ~ and CE; draw CD parallel to BE, and BD parallel A B E to EC, and join AD; then AB CD will be the required trapezoid, as is evident by parallel lines. It is manifest that in this case the angle 0 need not be given. Calculatio. —l. In triangle ACE, Case 4, find the angles; then angle AOB —A CE, and angle D CA- CAB. 2. In triangles A CD and CAB, Case 3, find angles, and sides AD and CB. CASE 2.-In Case 2 of the preceding problem, where AB, CD, one diagonal, as A C, and the angles 0 and P are given, if the sides AB and ClD (see Fig. 1) are parallel, then, as in Case 1, the angle P vanishes, and on AB, produced, lay BE = CD, and on AE (III. Prob. 16*) describe a segment to contain an angle equal to the given angle AOB, in which apply the given diagonal AC; or, if BD is given, apply EC -BD, and finish the construction of AB CD, which will evidently be the trapezoid required. Calculation.-1. In triangle A CE, Case 2, find angles CAE and CEA, and the unknown diagonal. 2. The angle ABD — BDC-AEGC. In triangles ABD and BD C, Case 3, find AD and B C. CASE 3. —In Case 3 of the preceding problem, where the two opposite sides AB and CD, the two diagonals, and the angle 0 at their intersection are given, if the opposite sides AB and CD are parallel, the angle P vanishes; then on AB, produced (see Fig. 1), take BE — CGD, and with centres A and E, and?adii respectively equal to AC and BD, describe arcs intersecting in C. Join AC,'III. 33. TRIANGLES, QUADRILATERALS, AND PARALLELS. 95 CB, and CE; draw CD and BD parallel to BE and EC respectively, and join AD and A C; then AB CD is evidently the trapezoid required. Calculation.-The same as in Case 1. NOTE.-It is evident the angle 0 in this case need not be given. CASE 4.-In Case 4 of the preceding problem, where one side, both diagonals, and the angles 0 and P are given, if the sides AB and CD are parallel, the angle P vanishes, and we FIG. 2. then make the angle DBE - (see Fig. 2) the given D c angle D 0 C, DB = its given length, and BE - the given length of AC. Join DE; then, if DC is given, lay its length from D to C; but if AB is given, lay its length from E to C. Join BC, draw A B CA parallel to EB, BA parallel to EC, and join AD; then ABCD is evidently the trapezoid required. Calculation.-1. In triangle DBE, Case 3, find angles, and side DE; then the sides AB and CD are known. 2. In triangles BA C and ACD, Case 3, find AD and BC. PROBLEMS INVOLVING PROPERTIES OF THE CIRCLE. PROBLEM I.-Through a given point within a given circle to draw a chord of a given length. F A tB4 aGive~ tthe diameter AB of the circle, the A p B Given-~ point P, that is, AP and PB, and rG \ \/ (the length of the chord GPF. Analysis. —Through the given point P, and the centre C, draw the diameter APB. Apply the chord BE the given chord GF; and from the centre C let fall the perpendiculars CD and CH; then, since BE- GF, we have (III. 8*) CD = GE. Whence this Construction. — Having drawn the diameter APB, from its extremity B apply the chord BE the given chord, and from the centre C let fall on it the perpendicular CD. On CP describe a semicircle, in which apply CHI- CD, and through P and H draw the chord GPHF, which will be equal to BE, and be the chord required. Demonstration.-The angle PHC, being in a semicircle, is a right angle. Hence CHis perpendicular to the chord GPF, and, it being equal to CD by construction, we have (II-I. 8t) GPF-= BE. Q. E. D. Calculation. —. The perpendiculars CD and CH (III. 6t) bisect the chords BE and GF respectively; hence BD -- BE _- I FG_ FH — GI, and CH2 — CD2 BC2- BD2, PH- - PC2- CH2. 2. Now, GP - GHf- PH, and PF- GH + PH — FH + PH. 9 I ). 14. t III. 14. tIII. 3. (96) THE CIRCLE. 97 Limits.-1. The given chord cannot be greater than the diameter AB. 2. The distance of the given chord from the centre cannot be greater than CP. When it is equal to CP, the required chord will be a perpendicular to the diameter through P. Hence the shortest chord that can be drawn through a given point is the one which is perpendicular to the diameter at that point. PROBLEM II. —Through a given point without a given circle to draw a right line so that the chord intercepted by the circumference shall be of a given length. F the diameter AB of the circle, 0PX S jB Livens the point P; that is, CP and A B Given, AP, and the length of the chord GF. Analysis.-Through the given point P and the centre C draw the diameter PAB. Apply the chord BE -the given chord GF, and from the centre C let fall the perpendiculars CD and CH; then, since BE- GF, we have (III. 8*) CD - CH. Whence this Construction.-Having drawn PA CB, from B, apply the chord BE-the length of the given chord, on which let fall from the centre C the perpendicular CD. On GP describe a semicircle, in which apply GH- CD, and through the points P and H draw the line PGF; then GF will be equal to BE, and PGF be the line required. Demonstration.-The angle PHC, being in a semicircle, is a right angle. Hence CH is perpendicular to the chord GF, and, being equal to CD by construction, GF: BE (III. 8t). Q. E. D. Calculation.-The perpendiculars CD and CH (III. 6t) bisect the chords BE and GF respectively; hence BD- - BE -FG _ F t — GH. Now, CH2 - CD2 BC2- BD2, and PH/COp2-O CH2. Then GP PH — GH, and PF PH + GH — PH + FH. Limit.-The given chord GF must not be greater than thbe diameter AB. II. 14. - III. 14. III. 3. E 7 98 GEOMETRICAL ANALYSIS. Scholium. —To draw a tangent from the given point P without the circle, join P with the point T where the semicircle on PC intersects the semicircle AFB, and P' will be a tangent to the circle AFB, for (III. 18, cor. 2*) angle P TC is a right angle. PROBLEM III.-With a given radius it is required to describe three equal circles which shall touch one another, and then to describe another circle which shall touch them all three. G L the radii of the three equal circles whose Given centres are A, B, < and C. Analysis.-Suppose the problem constructed as in the above figure. Join the centres A, B, and C, and an equilateral triangle ABC will evidently be formed, having each of its sides equal to twice the given radius. Bisect two of the angles A and B by the lines AO and BO, and join OC, which three lines will evidently be equal. Produce these three lines; then they will intersect the equal circles in the points G, H, and I; also, before being produced, in the points GC, HI, and It. Now, since the lines OA, OB, and OC are all equal, we have OH, 01, and OG all equal, as also OI01, OI', and OG' all equal. W\Thence this Construction.-Describe the equilateral triangle ABC, having each of its sides double the given radius. Then, with A, B, and C as centres, and the given radius, describe the three circles, which will evidently touch each other at D, E, and F, the middle points of the sides. Bisect the two angles A and B by the lines A0 and BO, intersecting in 0, and join OC. These three lines, and these lines produced, intersect the given circles in the points G', II', and It and G, H, and I. Then, with centre 0, and radii OH' and OH, describe the circles H', I', G' and H, I, G respectively, and the problem is constructed as required. Q III. 31. THE CIRCLE. 99 Tile demonstration is all evident from the analysis. Calculation. —Iu triangle AOB, Case 1, findd OA; then OH —OA + AH, and OHt' - OA - AH', the radii required. NOTE. —From the above is immediately deduced the following PROBLEM IV. —Having given the radius of a circle, to inscribe therein three equal circles which shall touch one another and also touch the given circle. (See figure on opposite page.) Analysis. —Suppose the problem constructed with OH or 0I the given radius. Join GH, HI, and IG, and HIG will evidently be an equilateral triangle. Join OG, OH, 01, and CI, and draw GL parallel to CI, meeting OI, produced, in L. Then, in similar triangles CBI and GIL, since IB - BC, we have IL- - IG. Whence this Construction.-Having described the circle with the given radius, draw the diameter GOK, and in the circle apply the chords KI and KH, each equal to the given radius KO. Join GH, HI, and IG; then (V. 4*) HIG will be an inscribed equilateral triangle, having the side GI~ / GK2- K12-_ 4 OK2 _ OK2 - / 3 OKA - OKV/3. Also, join OG, OH, and 0I, and produce 0c, making IL! - 1G. Join LG, draw IC parallel to LG, CB parallel to GI, and CA parallel to GH, and join AB. XVith the centres A, B, and C, and the radii AH, BI, and CG respectively, describe the circles HED, IFE, and GDF, which will be the equal circles required, touching each other at the points D, E, and F. Demonstration. —Because of the parallel lines, it is evident that AH, BI, and CG are all equal, that ABC is an equilateral triangle, and that the triangles GIL and CBI are similar. Hence it only remains to be proved that the three circles touch each other. In the similar triangles GIL, CBI, since GI- 2 IL by construction, we have CB- 2 BI- CF + FB - CD + DA- AE + EB. Hence, the distances between the centres of the circles being equal to the sum of the radii (III. 13t), the circles touch each other externally in the points D, E, and F. Q. E. D. Calculation.-By parallel lines, OL: IL:: OG: CG_ Bl- AH, and 2 CG AC AB - BC; also, OG —CG —OC -OAOB. 100 GEOMETRICAL ANALYSIS. PROBLEM V.-To describe a circle which shall pass through two given points and have its centre in a straight line given in position. [ the points A and B, and the D\p A I line FE in position; that is, the allgle P which it forms Given with the line joining A and B, or this line produced, and the - i-E l distance AP or BP. Analysis.-Suppose the figure constructed as above. Join AB, and from C, the centre of the circle, let fall on AB the perpendicular CD; then (III. 6*) AD —Z)B. Whence this Construction. —Join the given points A and B, bisect the line AB in D, and draw DCG perpendicular to AB, and the point in which this perpendicular cuts the given line FE wvill be the centre of the circle required. In the figure this point is C. Join CB, and with it as a radius, and centre C, describe the circle ABIt, which will be the one required. The demonstration is evident fiom the preceding. Calculation.-Let the line joining A and B be produced, if necessary, to meet the given line FE in P; then the angle P, and the distance PA or PB, are known from the given position.of FE. Thence PA, PD, and PB are known, by adding or subtracting, as the point P is in AB produced, or between A and B. And DB or DA -I AB. In triangle PDC, Case 1, find PC and CD; then radius CB — V. CD2 + DB2 - CA. Limits.-1. When FE is parallel to AB, CD will be the perpendicular distance between the two parallel lines, and, as FE is given in position, the distance CD is known, and thence the radius CA or CB / CD2 + DB2, as before. 2. When the given angle at P is a right angle, FE and DG will be parallel, and the problem is impossible; that is, in the language of mathematicians, "the radius would be infinite, and the are of a circle with an infinite radius is a straight line," so that the straight line passing through A and B would fulfil the conditions of the prtlElem mathematically. * III. 3. THE CIRCLE. 101 PROBLEM VI.-To divide a given circle into two segments, such that the angle contained in one shall be double of the angle contained in the other. A Bthe'radius OA of the circle ABDC, to draw Given the chord BC, such that the angle in the segment BA C shall be double that in BD C. D Analysis.-Suppose in the annexed figure the angle BA C to be double BDC; then BAC + BDC -three times BD C. But (III. 18, cor. 4*) the angles BA C + BD C 180~; hence 3 BDC — 180~, or angle BDC 60~, and consequently BAC -1200. Whence this Construction.-Draw the diameter AOD, and lay AB and AC each equal to the radius AO. Join BC, BD, and CD; then angle BA C - 2 BD C. Demonstration.-By III. 18t and V. 4,T it is manifest that the arcs AB and AC each contain 600; hence the arcs BD and CD each contain 120~, the arc BDC contains 240~, or twice the arc BAC; and, as the angle is measured by half the arc on which it stands, the angle BAC in the segment BAC=twice the angle BDC in the segment BD C. Calculation.-In the right-angled triangle ABD, we have AD 2 AB - 2 AO, and BD -/ AD2 - AB2 C / 4 A 02 A O/ 3 AO'- AOV/ 3 -BC CD. Scholium 1.-If the angle in one segment were required to be three times that in the other segment, the first segment would be a quadrant, and the other three quadrants, and the angles 135~ and 450 respectively. Scholium 2.-If the angle in one segment were required to be four times that in the other segment, the circle would be divided by the side of an inscribed pentagon (V. 5~), and the contained angles would be 360 and 1440~. Scholium 3. —If the angle in one segment were —required to be five times that in the other segment, the circle would be divided by the side of an inscribed hexagon (V. 411), and the contained angles would be 300 and 150~. * III. 22. t III. 21. J IV. 15. ~ IV. 11. IV. 15. 102 GEOMETRICAL ANALYSIS. PROBLEM VII.-To describe a circle which shall touch a given circle at a given point, and also touch a straight line given in position. the circle OA, the point A, and Io the position of cA ~ Gie the line GH; AGiven j thatis, the angle P, and OP or the perpendiculIar OB. Analysis. —Suppose ADIto be the required circle, passing through the given point A, and touching the given line GH at the point D. Draw the radius CD, and the line OB, both perpendicular to GH. Join DA, and draw BF parallel thereto; then the triangles ACD and FOB will be similar; and, since CA - CD, we have OFOB. Whence this Construction.-Having let fall on GHthe perpendicular OB, and produced the line joining the centre 0 of the given circle with the given point A, to meet the given line GH in P, produce PO, making OF — OB; join FPB; draw AD parallel to FB, and DC parallel to BO; then, with the centre C and radius CD, describe a circle ADI, and it will pass through the given point A, and touch the line GH in D. Demonstration.-By III. 9,* the circle touches the given line GH at the point D. By similar triangles FOB and A CD, we have FO-_ OB by construction; hence CA- CD, and the circle ADI passes through the given point A, and evidently touches the given circle. Q. E. D. Calculation. —1. In triangle POB, Case 1, find PB and PO; then PA - PO — OA, and PF —-PO + OB. 2. By similar triangles PFB and PAD, we have PF: PA:: PB: PD; also, PB: PD:: OB: CD - CA. Limits.-I. When the line GH is parallel to OA, then OB, the perpendicular distance between the two parallels, will be the radius of the required circle, and, laying this radius on the line OA, produced from A to C, C will be the centre of the required circle. 2. When the given line GH is perpendicular to-OA, or OA produced, the difference between OA and OB will be the diameter of the required circle. * III. 16, cor. THE CIRCLE. 103 PROBLEM VIII.-With a given radius to describe a circle which shall pass through a given point and touch a line given in position. l the radius CP, the line AB, and the Gi point P in position; that is, PD, the K L d Gi perpendicular distance of the point L P from AB. A B D G Analysis.-Suppose PLG to be the circle required, of which the centre is C. Through C draw KHE parallel to AB, cutting the perpendicular DPFin E; then ED - CG - CP. Construction.-Having drawn through the given point P the line DPF perpendicular to AB, on it lay DE — the given radius, and through E, parallel to AB, draw KEH, to which apply PC- the given radius, and let fall on AB the perpendicular CG. Then, with C as a centre, and radius CP, describe the circle PLG, which will be the one required. Demonstration. — CG ED- CP by construction, and CCG is perpendicular to AB, and hence the circle touches the line AB at the point G (III. 9*). Calculation. - We have PE B the difference between the given distance PD and the given radius. Then DG-EC/ Cp2 - PE2. Limits.-1. The given radius may be of any length not less than half DP. When the radius is just half DP, the required circle must be described on DP as a diameter. 2. The radius PC may be applied to KH, on either side of PD, at the point C or I, either of which may be taken as the centre of the circle to fulfil the conditions of the problem. We have EIl EC. If a perpendicular be let fall on AB from I, and IP be joined, such perpendicular will be equal to CG - PC- PI. * III. 16, cor. 104 GEOMETRICAL ANALYSIS. PROBLEM IX.-Through two given points to describe a circle which shall touch a line given in position.;L the points A and X /k~ FB, and the line IK HI in position; A thiat is, the angle J __ BAW Given { p which it forIls with AB, produced, ald the I distance P B or II I h l' ) 1 L PA. Analysis.-Suppose the circle AKBF, whose centre is C, to be drawn to pass through the given points A and B, and to touch the given line HI at F. Join the given points A and B, and through the centre C draw a line LD C perpendicular to AB, nmeeting HI in R; then (III. 16*) AD - DB. Let fall on HI the perpendiculars DE and CF, and join CB and RB, and produce RB to 0. Through D draw DG parallel to CB. Now, by parallel lines and similar triangles, since CB -- CF, we have DG- DE. Whence this Construction.-Join the given points A and B, bisect AB in D, and draw LDCR perpendicular to AB, meeting HI in R. On Il let fall the perpendicular DE; draw RBO, and to it apply DG DE. Draw BC parallel to DG, and CF parallel to DE; thlen, with the centre C and radius OF, describe the circle AKBF. Demonstration. —By III. 9,t the circle AKBF touches the given line HI at F. By parallel lines and si-milar triangles, since DG_ DE by construction, we have CB CF- CA; hence the circle passes through the points A and B. Q. E. D. Calculation.-l. Producing AB to meet HI in P, we have the angle P and the distance PB given, because the line HI is given in position; then PD-PB + 2 AB. In triangles PDR and PDE, find, Case 1, DR and DE - DG. 2. In triangle DRB, Case 3, find RB, and angle DBR; then angle DBG — 180 - DBR. 3. In triangle DBG, Case 2, find BG; then RG-C RB + BG, and we have RGC: RB:: GD: BC CF. Or, after finding RB akd angle DRB, we have sin CRF: sin F:: CF: OR:: CB: COR: sin CRB: sin CBR; then in triangle CBR, Case 1, find BC CF.'- III. 3. t III. 16, cor. THE CIRCLE. 105 Limits. —1. DE may be applied to RBO at G', on the other side of AB. Then, if a line be drawn through B, parallel to GtD, to meet RDL, produced, it will give the centre C' of another circle passing through the given points A and B, and touching the given line HI. 2. When ABP is perpendicular to HI, DR and HI will be parallel, DP will be the radius of the required circle, which apply from B to LR to obtain the centre. PROBLEM X.-To inscribe a square in a given segment of a circle. E i B Given the radius OA or OB, and the ang,~le AOB. E' Analysis.-Draw the chord AB, and perpendicular to it draw the diameter DCOD'; then A C - CB, and the arc AD — DB. Now, taking the smaller segment, let ITFGH represent the required square. Join CF, and produce it to meet BE, perpendicular to AB, in E. Then, by similar triangles CIF and OBE, since IF- IXH 2 IC, we have BE - 2 B C - BA. Whence this Construction.-Erect the perpendicular BE BA; join CE, cutting the arc BD in F; draw FG parallel to AB, and FI and GH each parallel to BE; then IFGH will be the square required. Demonstration.-IFGH (I. 20, cor.*) is rectangular. Also, since, by construction, BE- BA 2 B GC, we have, by similar triangles CIF and BCE, FJi- 2 IC IH FG - GH. Q. E. D. Calculation. —1. Join OF. In triangle OCGB, Case 1, find OC and CB. 2. In triangle BCE, Case 3, find CE, and angle EGB; then angle O CF 900 + E CB. 3. In triangle OCF, Case 2, find CF; then, by similar triangles CIF and CBE, CE:EB:: CF: F = IH. Scholium.-When the square is to be inscribed in the larger segment AD'B, the perpendicular BE' must be made equal to BA; - I. 29. E* 106 GEOMETRICAL ANALYSIS. and, using the marked letters instead of the corresponding ones unmarked, the construction, demonstration, and calculation are just the same as that given for the smaller segment, only the angle O CF' in this case- 90- El CB. If the point El falls on the circumference, BA is the side of the square required, and the arc ADB is a quadrant. When the point E' falls within the circumference; a square cannot be inscribed in the larger segment, having one side on the chord ziAB. When each segment is a semicircle, BE and BE' are each equal to the diameter, and the two squares are equal. PROBLEM XI.-To inscribe a circle in a given quadrant. F D/ (-iven in the quadrant ABC, the radius AB 7 Given or A C, to inscribe a circle in the is (quadrant. G E B Analysis.-Suppose DGH to be the required inscribed circle, then D is evidently the middle point of the arc BC, and the radii OD, OG, and OH are all equal. Let fall the perpendicular 1DE; join GD, and parallel to it draw EF to meet AD, produced, in F; then the triangles OGD and DEF will be similar, and, since OD OG, we have JDF —DE. Whence this Construction.-Bisect the are B C in D; let fall the perpendicular DE; join AD, and produce it, making DF= DE. Join EF; draw DG parallel to EF, GO parallel to AC, and OH parallel to AB; then, with the centre 0 and radius OD or OG, describe the circle D GH, and it will be inscribed in the quadrant AB C. Demonstration.-By parallel lines, the triangles ODG and DFE are similar, and, since DF- DE by construction, we have OD — OG_ OH, and AB and AC, being perpendicular to the radii OG and OH, are tangents to the circle at the points G and H (III. 9*). Q. E. D. Calculation.-A E- ED. Hence AE2 -- AD2, and AE - AID2 DRE= DF. Also, A F- AD + DF; then, by similar triangles AFE and ADG, we have AF: AD:: AE: AG OHOD. * III. 16, cor. THE CIRCLE. 107 Remnark. —If DI be drawn parallel to AB, AEDI will be a square inscribed in the given quadrant ABC, whose side AE or ED -I/' AD! 2 - -1 AB2 2 AB 2. PROBLEM XII.- To describe a circle which shall touch three lines given in position. D F (the side BC, and the angles Given. BCD and CBA, to describe a circle touching the three lines. A G B Analysis. —Suppose the circle FGH, of which 0 is the centre, to touch the three given lines Al?, BC, and CD in the points G, H, and F respectively. Draw the radii 0G, OH, and OF, and they will be perpendicular to AB, BC, and CD respectively (III. 9*). Also, join OB and OC; then, in the right-angled triangles OGB and OHB, since OG- OH, we have the angle OBG- OBH. For a like reason, in the triangles OCF and O CII, we have the angle OCF- OCH. Hence the lines BO and CO bisect the angles ABC and BCD respectively. Whence this Construction.-Bisect the angles ABC and BCD by the lines BE and CL, intersecting in 0. From 0, on the given lines, let fall the perpendiculars OG, OH1, and OF. Then, with the centre 0 and any one of these lines a.s radius, describe a circle, and it will pass through the extremities of the others, and touch the given lines as required. Demonstration. —Since BO bisects the angle GBH by construe tion, the triangles OBG and OBH are equal, and OG - OH. For a like reason, OH= OF; hence OG, Of, and OF are all equal. Also, since AB, BC, and CD are respectively perpendicular to the radii OG, OH, and OF at their extremities, they touch the circle at the points G, H, and F by reference in analysis. Q. E. D. Calculation.-I. In triangle BOC, Case 1, find BO and OC. 2. In triangle BOH, Case 1, find OH- OG, and BH- BG; then CF- CHZ- CB' Bf-. Scholium 1.-When the three lines on being produced do not meet, each one must be parallel to both the others, and the problem is impossible. Scholium 2.-When the sum of the given angles ABC and BCD III. 16, cor. 108 GEOMETRICAL ANALYSIS. is equal to 1800, the lines CD and BA are parallel. When the sum of the given angles is less than 180~, the lines CD and BA will meet, if produced, and form a triangle, when the problem becomes the same as III. Prob. 15,* to inscribe a circle in a triangle. In both cases the construction, etc. are the same as that given above. PROBLEM XIII.-To draw a common tangent to two circles given in magnitude and position. [ the radii OA and CB, and the 0 }\ e Givenp idistance 0 C between the centres, to draw a common Bn L ~tangent AB. B Analysis. —Let AB be the common tangent required; then (III. 9t) the radii OA and CB are perpendicular to AB, and consequently parallel to each other. Through C, the centre of the less circle, draw CD parallel to AB; then AD - B C, and OD is the difference between the given radii, and the triangle ODC is known. Whence this Construction.-On the given distance OC describe a circle, in which apply OD — the difference of the given radii. Produce OD to meet the circumference at A; draw the radius CB parallel to OA, and join AB, which will be the common tangent required. Demonstration. —Since, by construction, OD is the difference between the radii, and OA the greater radius, we have DA equal and parallel to CB, and hence AB is equal and parallel to CD; and consequently, since ODC is a right angle, being in a semicircle, AB is perpendicular to the radii OA and CB at their extremities, and is hence a tangent to both circles (III. 9t_). Calculation.-In triangle OD C, we have CD / CO2 - OD2 AB. Scholium. —Another common tangent may be drawn to the same circles. Suppose it done, as A'B', and on it let fall the perpendiculars CB' and OA', and produce OA' to meet CD' drawn parallel to A'B', in D'; then AID'- CB', the angle D' is a right angle, and consequently the point Dt is in the semicircle described on the given line OC, and the line OD' is equal to the sum of the given; IV. 4. t III. IS. c III. 16, cor. THE CIRCLE. 109 radii. In the semicircle described on OCt apply OD' _ the sum of the given radii; join CD'; draw CB' parallel to OD', and join A'Bt, and it will be a common tangent, as is evident from the preceding demonstration. The length of A'B' — D' / 2OC OD'2. Remark.- AB is called the external common tangent, and A'B' the internal common tangent. PROBLEM XIV.-On a given line to describe two semicircles touching each other such that their contimmon tangent may be of a given length. /0 the line AB, and the length of the common A Given tangent IL, to.B describe the se in icircles AIE and ~~~D [~ ELB. Analysis. —Suppose AIE and ELB to be the required semicircles described on the given line AB, G and F, their respective centres, and IL their common tangent, I and L being the points of contact. Since GE — GA, and EF FB, we have GF- AB. On GF describe a semicircle, draw the radii GI and FL, which will be perpendicular to IL, and therefore parallel, and join F and H, the point where GI intersects the semicircle on GF. Now, since the angle GHIF is a right angle, being in a semicircle, HF is parallel and hence equal to JL, and HI- FL, and GH is the difference between the two radii of the required semicircles. Whence this Construction.-Bisect the given line AB in C, on AC describe a semicircle, in which apply AD — the given length of the comnllll tangent, join CD, and make CE- CD. On AE and EB describe semicircles of which the centres are G and F, and on GF describe a semicircle of which the centre is 0. Draw FH p'arallel to AD, draw GHl, and draw FL parallel to it, and join IL, and it is done. Demonstration.-By last problem, IL is a common tangent to the two semicircles, and equal to HF. Because A C — GF,-each being half of AB, and FH is parallel to AD by construction, the triangles ADC and FHG are equal, and we have AD1 FlH= IL; also, GH — CD - CE. Now, since GE- I and CE GH, we hlar'e 110 GEOMETRICAL ANALYSIS. G C HIZ- EF- FL, and GE - EF GI — FL. Hence FL HI, and GH is the difference between the radii GI and FiL, and is -- the difference of the diameters AE and EB. Calculation.-CE - GH- CD -/ A C2- AD2 -/VA C2-IL2; then AE - A C + CE, and EB A C- CE. Scholium 1.-If the P given common tangent is just half' the'H given line AB, the semicircles will be G e o E F B equal. This is its superior limit. Scholiurrnt 2. The ZD common tangent IL touches the semicircle GHF1 at P, the middle point of IL. If OP be joined, it will be perpendicular to IL, and equal to half GI+ FL = OG or OFT AB; also, GE - CF, GC -- EF, and O C — OE. PROBLEM XV.-Of two circles which touch each other externally are given the radius of one, and the length of their common tangent, to determine the radius of the other. F < z[A rthe radius AB, and the B E a / | J Gi length of the common Given tangent BC, to find the A G B Lradius EC. D Analysis. —Suppose the figure constructed as.above, AB the given radius, and E C the required radius. Produce the radius EC, making ED -EA, and join AD, intersecting BC in G, and join E G. Then, since ED — EA and E C - EF, we have CD - AFAB. By similar triangles DCG and ABG, since6 CD -AB, we have CG -GB, and DG- GA, and EG perpendicular to DA. Whence this Construction.-Draw B C equal the given common tangent, and THE CIRCLE. 111 draw BA and D)C perpendicular to B C, on opposite sides, and each equal to the given radius. Join DA, intersecting BC in G; then, evidently, by similar triangles DCG and ABG, CG- GB and DG GA. Draw G.E perpendicular to DA, meeting DC, produced, in E; draw EFA, and with the centres E and A and radii EC and AB, respectively, describe circles, and they will touch each other at 6he point F, and also touch the line B C at the points B and C. Demonstration.-B C is the common tangent by construction, and it only remains to prove that the circles are tangent to each other; that is, that AE -AB + CE. Since GE is perpendicular to AD at its middle point G, we have EA — ED-)- CD + CE AB + CE -the sum of the radii. Hence (III. 13*) the circles touch each other externally. Q. E. D. Calculation. —DG -/ CD)2 + CG2; then, by similar triangles DGC and D)EG, we have DC:DG:: DG:1)E -EA, and DEDC - E C - EF. Limit.-The common tangent can be taken any length. Scholium. —When two circles touch each other externally, half the common tangent is a mean proportional between the radii; that is, in the figure, EC: CG or GB:: CG:AB. PROBLEM XVI.-Given, the radii of two concentric circles, to find the radius of a circle which shall touch one of these circles externally and have a common tangent of a given length with the other. X _ _ the radii AB and AI, and /Given the length of the common / tangent B C, to find the a radius E C. D P Analysis.-Suppose the figure constructed as above. Produce EC, making CD AI, join AD and AE, and let fall the perpen dicular EG; then ED=-EA, DG= GA, and BC, Iecrpendicular to the radii E C and AB at their extremities, is the given common tangent. Whence this Construction. —With the given radii AB and AI having described X III. 12. 112 GEOMETRICAL ANALYSIS. the concentric circles, draw BC perpendicular to ARBI, and equal to the given length of the common tangent. Draw CD perpendicular to CB and equal to AI. Join AD; bisect it in G; and draw GE perpendicular to DA, meeting DC, produeed,inE; and drawEFA. With centre E and radius EC describe 1X J\ \ / J / a circle, and it will touch one of < / Bthe given circles at F, and have the common tangent BC with the other circle of the given length. Demonstration.- BC is eviD P dently the common tangent to the circles AB and EC, and it was made the given length by construction. Since GE is perpendicular to DA at its middle point G, we have EA - ED Z EC + CD - D EC + AI= the sum of the radii. Hence (III. 13*) the circles touch each other at F. Q. E. D. Calculation.-Through D, parallel to CB, draw a line to meet Al, produced, in P; then DP- CB, and AP -AB + CD - AB + AI; also, AD-_ AP2 DP-2, and D G=-AD. By similar triangles APD, DGE, we have AP: PD:: DG: DE; that is, AI + AB: B C:: AD: DE; then DE- CD - DE - AI E C-_EF. Scholium. —If A B AI, it reduces to the preceding problem. If CA and D1 be joined, they will be parallel. PROBLEM XVII.-Of two circles which intersect each other there are given the radius of one, their common chord, and their common tangent, to determine the radius of the other. H %G~~ the radius DB or Gi chord EF, and the a liven~ common tangent L/GH, to find the L radius CG or CE. Analysis. —Suppose the problem constructed, D being the centre of the given circle, and C the centre of the required circle. From - III. 12. THE CIRCLE. 113 the extremities of the common tangent GH, draw GC, GE, and GF; HD, ]HE, and HF; and produce the common chord EF to meet the common tangent in I; then (IV. 30*) IH21- IE x IFIG2; hence IH- IG, and the common chord, produced, bisects the common tangent. Whence this Construction.-In the given circle, whose centre is D, apply the given chord EF, through the middle point P of which draw the diameter BDL. Draw FO perpendicular to FE, or parallel to LB, and equal to half the length of the given commonn tangent. Join PO, and produce EF until PI-PO, and apply IHX FO, half the given common tangent. Produce HI, making IG - IH; draw G C parallel to DH, meeting BDL, produced, in C; and draw CE, GC, GE, and GF; HED, HE, and HF; and with centre C and radius CG, describe the circle AGFE, and it will be the one required. Demnonstration.-IE x IF= (IP + PF) X (IP - PF) - IP2PF2- OP2 - PF2 F0O2 - 1-12- IG2. Hence (IV. 30t) IG is a tangent to the circle GFE passing through F and E, and IT is a tangent to the circle IFE passing through the same points; consequently, GHis a common tangent to both. circles, and it is of the given length. Q. E. D. Calculation. —Draw IT parallel to HD or GC; then, since GI= IH- by construction, we have C T- TD and IT- - (DH + CH). Now, PI: PO — / F02 + FP2, and PD V DE2 _- EP2. Join ID; then in the rigiht-angled triangle IHD, Case 3, find angle IDH; and in triangle IDP, Case 3, find angle IDP. Their sum — angle PDH- P TI. Then, in triangle PTI, Case 1, find IT — EHD + Q CG. Whence we have CG O 2 IT- HD. * III. 36. - III. 36. 8 114 GEOMETRICAL ANALYSIS. PROBLEM XVIII.lFrom one extremity of the diameter of a given semicircle, to draw a right line such that the part intercepted between the circumference and a tangent at the other extremity shall be of a given length. te dia of te sei(the diameter AB of the semiG ix en circle ADB, and the intercepted distance DC of the line AD C. A B Analysis.-Let ADB represent the given semicircle, BHla tangent, and DC the given intercepted distance. Join BD; then the triangles ABC and ADB are similar, and we have (IV. 18*) AC:AB:: AB: AD. Whence this Construction.-On the tangent BHlay BI- the given intercepted line, and on BI describe a semicircle of which E is the centre. Draw the line AFEG, and to BH apply A C — A G, cutting the semicircle in D; then AD will equal AF, and DC - FG _ BI. Demonstration.-By similar triangles and IV. 30,t we have AC: AB::AB:AD, and AG(AC):AB::AB: AF; hence AD -AF; and we have C AC —AD -- AG — AF FG - BI. Q. E. D. Calculation.-AE-/ AB2 + BE2; then AG - AE + EB AC, and AF — AE- FA AD. Also, BC AC2 —AB2. Limit.-D C may be of any length. V- VI. 4. t III. 36. THE CIRCLE. 115 PROBLEM XIX. —Having a circle given, it is required to inscribe in it an equilateral triangle, and also to circumscribe an equilateral triangle about the circle. c F_ X Given {ithe diameter DH — 2 R, or the radius Given (ID - R. A D B Analysis. —Let EDFH represent the given circle, of which the centre is I, EDF the inscribed triangle, and ABC the circumscribed triangle. Join CD, which will be perpendicular to AB; and, since AB is a tangent to the circle at D, CD will pass through the centre I, and through G and H, the middle points of the chord EF and of the arc EHE, respectively. Also, since A C and B C are tangents to the circle at the points E and F, respectively, a circle described about H as a centre, with the radius HI, will pass through the points E, C, and F. Whence this Construction.-Having the circle, perpendicular to its diameter DIH draw an indefinite line ADB. With the centre H and radius HI describe the circle FIE, meeting DII, produced, in C. Join CE and CF, and produce them to meet the perpendicular through D, in A and B. Also join EF, ED, and DF; then EDF will be the inscribed triangle, and ABC the circumscribed one, required, as is evident from the analysis and V. 4.* Calculation. —CD is evidently equal to 3 R, and DG —C CD - and DG2 —. Now, in triangle DGF, DGI2- DF — FG2 2 4 4 DF2_ (Ie DF)2 - DIT2 DF - 3 DF2. Hence I DF-2 4 Whence DF~2 3 R2, or DF- RE /3 - EB. Also, ABBC- 2BF- 2 DF- 2 3. Corollary.-The side AB being equal to twice EF, the area, of the circumscribed triangle AB C will be four times the area of the inscribed triangle DFE. The four partial equilateral triangles conposing ABC are all equal, and each one-fourth the aria of ABC; and each of the twelve sides of these four triangles is equal to half of each side of the triangle ABC. - IV. 15. 116 GEOMETRICAL ANALYSIS. PROBLEM XX.-Given, the lengths of two chords cutting each other at right angles in a circle, and the distance of their point of intersection from the centre of the circle, to determine the radius of the circle. D [ the lengths of the chords AFB, F Give EFD, the angle at F a right angle, _Gi ve I and the distance FC, C being the L centre of the circle. Analysis.-Suppose ADBE to be the circle constructed as required. Then (IV. 28, cor.*) AF x FB - EF x FD. Bisect the given chords AB and DE by the perpendiculars CM and CG, which will be parallel to DE and AB respectively. Then AF x PB - (AM + MF) x (AM - MF) AM2 - IF2. Also, EF x PFD- (EG + GF) x (EG - GF) EG2- GF2. Hence we have AM2 — MF2 EG2 - GF2, or AM22 _ EG2 MF2- GF2 = MF2 - CM2. Whence this Construction.-Draw CF- the given distance from the centre to the intersection of the chords, and on it describe a semicircle. From C and F as centres, with radii equal to half the given chords DE and AB respectively, describe arcs intersecting in H. Join CH and HF, and draw HMI perpendicular to CF. Join CM and PM, and produce FM both ways, making MB and MA each equal to PFE, half the given chord AB; then AB -the given length. With the centre C and radius equal to CA or CB describe a circle, and through F, parallel to CM, draw a line to meet the circumlference in D and E; then DE will equal 2 CH — the given length. Demnonstration.-Draw CG parallel to AB. The angle CMF, being in a semicircle, is a right angle, and hence all the angles in the parallelogram CM'G are right angles, and C G is perpendicular to DE, and bisects it. Now, by the analysis, we have EG2 - GF2 A M2- MFP2, or A - 2 - E M G2 - F2- GF2 - MF2 - C1M2(Prob. 36; cor., in "Triangles," etc.) F2 — IC2 FPH2- CH'2= A M2- CH2; hence EUG2 - CH2, and G- CH, and DE- 2 E G -2 CH- the given length. Calculation. —We have MFP2 + CM2 CFP2, and MF2 -- CM2* III. 35. THE CIRCLE. 117 (MF2 - FG2) - AM2- EG2 by analysis. Their sum gives 2 MF' CF2 + AM2 - EG2, or MF_4 CF2 Al2 —EG2 Their difference gives 2 CM2- CF2 —AM2 + EG2, or CM== CF2 -AM2 + EG2 -.FG; whence EF, FD, and AFl, PB are 2 known. Also, radius CE- V CG2 + EG2. PROBLEM XXT.-Given, the three distances from a point either within or without a circle to the three nearest extremities of two diameters which intersect each other at right angles, to find the diameter of the circle. (See Problems XL. and XLI., "Triangles," etc.) FiG. 1. FIG. 2. Given, the (listances PA, r. ~ PB, and PC, in either fig- D D ure. nrc. Analysis. —Suppose AC and BE (referring to either figure) to be the two diameters crossing each other at right angles at the centre 0, and that we have the distances PA, PB, and PC given. Join the points AB, B C, CE, and EA, and AB CE will be a square, and ABC an isosceles triangle, having the angle ABC included between the equal sides given, and also the three distances from the point P, PA, PB, and P C, to the three angular points, to construct the triangle and find the sides, which is done in Problem XL. of "Triangles, Quadrilaterals, and Parallels." Then A C will be the diameter required. Complete the square ABCE, and describe the circle with the radius OA or OC. The same letters used in Problem XL. apply to these figures in the construction, demonstration, etc. Property of the circle.-If from any point within or without a circle lines be drawn to the extremities of diameters of the circle, the sum of the squares of the two lines drawn to the extremities of one diameter, will be equal to the sum of the squares of the two lines drawn to the extremities of any other diameter. 118 GEOMETRICAL ANALYSIS. Demonstration.-Join PE and PO (Fig. 1); then (in either figure) (IV. 14') PA2 - PC2 ~ 2 PO2 + 2 AO2 2 P0O2 2 OB2FIG. 1. FIG. 2. D PB2 + PE2. And if two lines be drawn from P to the extremities of any other diameter, they together with such diameter regarded as the base will form a triangle of which PO will always be a line drawn from the vertex to the middle of the base, and hence the sum of the squares of these two lines (IV. 14t) will be equal to twice P02 + twice the square of the radius -2 P02 + 2 OA2 2 PA2 + AC2. Q. E. D. PROBLEM XXII. —Having given the positions of two lines, and of a point between them, it is required to describe a circle which shall pass through the given point and touch each of the lines given in position. Q Fthe angles BAGC and Q G \ Given BAI, and the distance G /, I iAAP of the point P from A. H D H B Analysis.-Suppose GC to be the centre of the required circle passing through the given point P, and touching the lines AB and A C given in position in the points Hi and Q respectively. Through the centre G draw the line AGL indefinitely; then (III. Prob. 15~) AL bisects the given angle BAG. Join GP, let.fall the perpen-II. A. t II. A. 2 The student may give attention to either circle, but he should confine it to one at a time. The analysis, construction, etc. apply to both circles alike, taken separately. # IV. 4. THE CIRCLE. 119 dicular GH, and through the given point P draw ED parallel to GH, and draw EF parallel to GP. Then, by parallel lines, we have AG: GH:: AE: ED, and AG: GP (GH):: AE: EF; wherefore F —Z ED; and we have this Construction.-Through the given point P draw ED perpendicular to AB, the line AEL bisecting the given angle BAG. Apply EF E —D; draw PG parallel to EF; let fall the perpendicular GH; and with the centre G and radius GH describe the circle HQ, which will pass through the given point P and touch the lines AB and AC. Demonstration.-By the principles and reasoning in the analysis, GP- GH, and hence the circle passes through P, and by III. Prob. 15* it touches AB and A C. Q. E. D. Calculation.-In triangle APD, Case 1, find AD. Then, in triangle ADE, Case 1, find AE and ED- EF. Now, angle EAF - BAC- BAI. In triangle AEF, Case 2, find AF, which has two values. Taking the less value of AF, we have AF: FE (ED):: AP: GP, the radius of the greater circle. And taking the greater value of AF, we have AF: FE (ED):: AP: GP, the radius of the less circle. Limzit.-If P coincide with D, G will coincide with E, ED will be the required radius, and there will be but one circle. a IV. 4. 120 GEOMETRICAL ANALYSIS. PROBLEM XXIII.-Having given the positions of two right lines, and also a circle whose centre is between them is given in position and magnitude, it is required to describe another circle which shall touch each of the said lines and also the given circle. Q ]\7'1\T~~~~~~~~~~1 ID IH B K O K T Given ~the angles BA C and BAP, the perpendicular DP, and i radius PM of the circle MLN. Analysis. —Suppose G* to be the centre of the required circle, touching the lines AB and A C in the points H and Q, and the given circle, whose centre is P, in L. Through the centre G draw the line AGR indefinitely, and (III. Prob. 15t) the line AGR will bisect the given angle BA C. Draw GLP; let fall on AB the perpendicular GH; through P draw EPD parallel to GH, and draw EF parallel to GP. Also, produce ED, making DO — the radius PM, and through 0, parallel to AB, draw a line, meeting RA, produced, in W, and GH, produced, in K; then GK — GH+ HK- GL + PM or PL- GP. Also, by parallel lines, WG: GK:: TFE: EO, and WG: GP (GK):: WE: EF; hence EF- EO; and we have this Construction.-Draw A B bisecting the given angle BA C; through the centre P of the given circle draw EPD perpendicular to AB, and produce it, making DO PM. Through 0, parallel to AB, draw a line WI, meeting RA, produced, in W. Through P draw WPI, to which apply EFz EO. Draw PG parallel to EF; let fall the perpendicular GHKIC; and with the centre G and radius GH describe the circle HQL, which will be the one required. Demonstration. —By parallel lines, since EF — EO by construction, we have GP - GKz GH + HK — GH + PL - GL + PL. Hence the circles touch externally at the point L. And by III. * The student may give attention to either of the required circles, but he should confine it to one at a time. The analysis, demonstration, etc. apply to both circles (QLH) alike, taken separately. t IV. 4. THE CIRCLE. 121 Prob. 15* HIQL touches the lines AB and A C, the angle BA C being bisected by the line AR. Q. E. D. Calculation.-In the triangle PAD, Case 1, find AD. Then, in the triangle AED, Case 1, find AE and ED; then EO —~ED + PM; also, PO =PD + P1. By similar triangles EDA, EOW, we have ED:'EO::AD: WO, and ED:'EO::AE: WE. In triangle P WO, Case 3, find angle P WO and side WP; then angle EWE WF- EWO - P WO - (I. 20,'cor. 3t) EAB - P WO. In triangle E WF, Case 2, find WE, which will have two values. Taking the less value of WF, we have, by parallel lines, WF: WP:: EF (EO): PG, from which take PL or PM, and we have GL =GH, the radius of the greater circle HLQ, the point Q being on the line AC. Also, taking the greater value of WF, we have WF: WP:: EF(EO): PG; then GL -GP — PL-= GP - PM the radius of the less required circle HLQ. PROBLEM XXIV.-To divide the arc of a given semicircle into two segments such that the radius shall be a mean proportional between the chords of the segments. the radius CA or CB of the semicircle ~D~E Given ADB, to find the point D so that we shall have AD: AC:: CB: BD. A C FB Analysis. —Suppose the figure constructed as above. Draw DE parallel to AB to meet the perpendicular BE at E; then the angles DAB and DBE, being each the complement of DBA, are equal, and tl-e triangles DAB and DBE are consequently similar, and we have AB: AD:: BD: BE. Whence this Construction.-Having the semicircle described on the diameter A CB, since, in the above proportion, the first term AB is twice the radius, we must erect the perpendicular BE 7 half the radius, draw ED parallel to AB, and join AD and DB, and they will be the chords of the required segments. Demonstration.- In similar triangles DAB and DRBE, we have AB (2BC):AD::BD:BE(IBC). Hence ADX DB 2 BC x C C-=BC2 A C x BC. Wherefore we have AD:AC::BC: BD. Q. E. D. Calcilation. -Join CD, and on AB let fall the perpendicular DF; then CGF- CD2 _ F2 _ D/C 2 (-(I2 GD)2 / 4 dT7 " IV. 4. t I. 29. F 122 GEOMETRICAL ANALYSIS. 2 CD! 3. Also AF AC CF, and BF AC- CF; then we have AD =- / AF2 + DF2, and BD _ / BF2 + DF2. PROBLEM XXV. —Through two given points to describe a circle which shall touch another circle given in position and magnitude. Given, the points A and Q, the v, / angle QAB, the distance AB, and the radius;i/rP /:TL -— ~ BL or BT, to describe the circle AQT. Analysis.-Suppose A Q T to be the required circle, of which C is the centre, touching the given circle BL in T. Through the centre C draw ROS perpendicular to A Q, and draw CA and CTB. Then on the base AB we have to construct a triangle ABC, whose vertex C shall be in the line ROS, and the difference of whose sides BC and CA shall be equal to the radius BL, which is done by Problem VIII. in "Analysis by Algebra." Calculation.-See calculation in Problem VIII. "Analysis by Algebra." By construction, QAB: BL:: BL:DH; then AHAD - DH; also HI — BL by construction. Draw _ENM parallel to HI; then ENz- HI. Also draw AK perpendicular to AB; then angle MEK- OAK- the complement of the given angle QAB. Now, 1. In triangle ENF, Case 1, find EF; and in triangle EMK, having EM -AH, find KM, Case 1. 2. In triangle AOK, Case 1, find AK; then EEH- AM -AK — Kill. 3. In triangle AHE, Case 3, find angle HAE- AEJi; then angle AEF= AEM + MEKW GEF. 4. In triangle EGF, Case 2, having FG — AB, find angle FGE CAE; then angle CAP CAE + HAE, and angle OA C OAB- CAP. In triangle OAC, Case 1, find AC=_ CT -the required radius. THE CIRCLE. 123 PROBLEM XXVI. —From two given points without a circle which is given in position and magnitude, it is required to draw two lines to meet in the circumference of the given circle such that the angle formed by these lines at the circumference shall be the greatest possible. (the same things as in the last problem (see figure to Given Problem XXV.), to draw the lines A Tand Q T such that the angle A TQ shall be the greatest possible. Analysis.-Join AT and Q T, T being the point in which the circle A QT touches the circle LT, in the figure on the opposite page; then A TQ is the angle required. Demonstration. —In the circumference of the given circle L T, take any other point, as T', and draw A T' and Q Tt. One of these lines must cut the circumference ATQ in a point, as V. Draw a line from V to the other extremity of the base, as VA; then (III. 18, cor. 1*) the angle A TQ - A VQ. But (I. 25, cor. 61) the angle A VQ is greater than A TIQ. Hence the angle A TQ is greater than A TQ, and, as T' is any point in the circumference of the given circle, the angle A TQ is the greatest possible angle that can be formed by two lines drawn from the points A and Q to meet in the circumference of the given circle L TT'. Wherefore, as in the last problem, we must describe a circle to pass through the points A and Q, and touch the given circle in the point T, and draw the lines A T and Q T, which will be the lines required. PROBLEM XXVII. —Having given the chord and tangent of a circle, to determine the radius. the chord AD, and the tangent Given AB, to find the radius AC of the A F arc A GD. A c E Analysis. —Let AD be the chord and AB the tangent of the arc AGD, and CDB the secant. Complete the semicircle ADE; join DE; and on BC let fall the perpendicular AF; then- (IV. 23T) angle BAF — BCA- DCA — 2DEA - (III. 21~) 2 DAI;; hence M III. 21. t r. 16. + IV. 8. g III. 32. 124 GEOMETRICAL ANALYSIS. AD bisects the angle BAF. Wherefore, in the right-angled triangle B BFA, we have given the hypothenuse AB, and the line AD bisecting the acute angle BAF, to determine the triangle D BFA by Problem XII. in "Analysis by Algebra." Then draw AC perpendicular to AB to meet BF, produced, in C, and AC will A C E be the required radius. Calculation.-After finding BF and BA, as in the problem referred to, we have BF: FA:: BA: A C -the radius required. PROBLEM XXVIII. —Having given the angle formed by two lines, the length of a line bisecting this angle, and the radius of a circle whose centre is the remote extremity of this bisecting line, to draw a tangent to the circle such that the part intercepted between the two lines, whose inclination to each other is given, may be of a given length. FIG. 1. 1. FIG. 2. 1 J Q A' Al G' 0' \ the angle BA C, the line AO= c Given ~A'O', the radius OF, and the B Hie length of the tangent DE, in\ tercepted between AB and A C. Analysis.-Suppose the problem constructed as in figure 1. About the triangle ADE describe the circle AIEGD. Through 0 draw B C parallel to DE, and through G, where the- circumference cuts AO, draw IIGPI perpendicular to BC or DE; then GI is the diameter of AIEGD; DP -PE, arc DG — GE, and OG x GA =HG x GL Whence this Construction. - On the given tangent DE (III. Prob. 16*) describe a circle such that the segment DAE shall contain the given * III. 33. THE CIRCLE. 125 angle. Bisect DE by the diameter IPG, which produce until PH — the given radius. Draw BHC parallel to DE. On A'O' (Fig. 2), equal to the given line AO, describe a semicircle of which 11 is tho centre; draw the radius MJ perpendicular to A'O'; lay MN a mean proportional between HG and GI. Draw NK parallel to A'O', and KG' parallel to MJ; then A'G' x G'O' - G'K2 - MN2 - HG x GI. Apply G'O' from G to O in BC, and produce OG to meet the circumference in A. Draw ADB, AEC; let fall the perpendicular OFon DE, and with centre O and radius OF describe the circle O,FL, to which the line DFE will be tangent. Demonstration.-It remains to show only that AG - A'G'. Join AI; then, in similar triangles GAI and GIHO, GH: GO:: GA: GI; hence GO x GA (or G'O' x GA) - GI x GH- (by construction) G'O' x G'A'. Hence GA - G'A' and OA - O'A'. Calculation. —OH- OG2 — GH2 PF. By similar triangles GHO and OFR, GH:GO::OF: OR, and GH:HO::OF:FR. Whence RP, RE, DR, and AR become known; and by similar triangles, B O, O C, and BC also. Then in triangles ABC and ACO we have, Case 2, AB and AC; and thence, by similar triangles, we have AD and AE. Scholiumn.-If the point 0 comes between DE and the arc DGE, we must produce A'O' to G"' (IV. Prob. 4*) so that the rectangle A'G" x G"O' GH x HIz MN2, and proceed in every other respect as already directed. (See Problem XXXII. of "The Circle. ") IIST Prob. 11. 126 GEOMETRICAL ANALYSIS. PROBLEM XXIX.-From a given point within a given semicircle, it is required to draw a line to a point in the diameter of the semicircle such that that line may be equal to the ordinate of the semicircle at the point at which the line meets it. H r the semicircle AHB, of which R - 1B is the centre, and the point C, to draw CD such that CD shall be -F F HE equal to the ordinate DH. G Analysis.-Since AB and the point C are given, the triangle A CB is given. About A CGB describe a circle of which the centre is 0, and suppose CD drawn so that CD -DH; then CD2 = D12 (by the property of the semicircle) AD X DB; so that this problem -resolves into Problem L. of "Triangles, Quadrilaterals, and Parallels," and the construction, demonstration, and calculation are the same as in that problem. Then at D and D1 erect the perpendiculars, or draw the ordinates DH and D'H', which will be equal to D C and D' C respectively, as is evident from the demonstration of that problem. Scholium.-If it were desired that CD should have to DH any given ratio, as m to n, then produce CB, so that CB: BE:: m2: n2 (IV. Prob. 11*); draw EF'F parallel to AB; then draw CDF and CD'F', and CD or CD' will be the line required. For, since m2:' n2:: CB: BE:: CD: DF, we have DF= — 2. CD.?~2 Now, DH2 = AD x DB = CD X DF — 2 CD2. Hence DH== n~~~~~~~~~~'n -. CD; that is, m:n:: CD:DH. In the same way m:n:: CD': DIH'. Q. E. D. -:- See Problem XXXIII. of " The Circle."' THE CIRCLE. 127 THEOREM AND PROBLEM XXX. (Proposed by Francis Miller.) -If from any point, either within or without a circle, lines be drawn to the extremities of diameters to the circle, the sum of the squares of the two lines drawn to the extremities of one diameter will be equal to the sum of the squares of the two lines drawn to the extremities of any other diameter. And having given three of these lines (and consequently the fourth) drawn from the given point to the extremities of two diameters making a given angle with each other, it is required to find the diameter of the circle. FIG. 1. B the diameters AB and CD, and P any point, to prove Be I the di meter ABjPB, and PC given, to find L the diameter AB. Demonstration of the Theorenm.-Join PO; then, in each of the three figures, we have PA2 + PB2 =(IV. 14*) 2 PO2 ~ 2 0B22 P02 + 2 OD2 = PC2 + PD2. Q. E. D. Corollary.-PD PA2 + PB _ — P C2. Analysis of the Problem. (Case 1.) —When the diameters cross at right angles (Figure 1), draw CE perpendicular and equal to CP, and join EB; then, in triangles ECB and PCA, we have EC== PC, CB = CA, and angle ECB ECP + PCB A CB + PCB -P CA; hence EBB- PA; and we have this Construction.-Draw EC and CP at right angles, and each equal to the shortest of the given lines. With P and E as centres, and radii respectively equal to the given lines PB and PA, describe arcs intersecting in B. Join EB, PB, and BC. On BC describe the square BCAD; draw the diagonals AOB, GOD; and with centre 0 and radius OB or OC, describe the circle BCAD, which will be the one required. Demonstration.-In the triangles P CA, E CP (I.. t), PA = EB; and PB and PC were made of the given lengths. Q. E. D. Calculation.-Join EP; then EP = / PC2 + CE2 = 2 PC2. In triangle EPB, Case 4, find angle EPB; then angle CPB= EPB - 450. In triangle CPB, Case 3, find BC; then AB or CD — /2 B C2. *.- II. AI.4. 128 GEOMETRICAL ANALYSIS. Scholiun 1. —When the given point is within the circle, let the two arcs described about P and E meet on the other side of PE, as at B'; join B'C, and on it describe the square, the diagonal of which will be the diameter required. In the calculation, the angle CPB will sometimes equal 450 - EPB. Scholium 2.-This problem solves the one for constructing a square when the three distances from a given point, either within or without the square, are given to the three nearest corners of the square. The fourth distance is then also given. (See Problem XLI., "Triangles," etc.) CASE 2.-When the diameters cross at any angle (which of course includes Case 1). Fro4. 3. FIG. 2. P E [the distances PA, PB, and PC, and Given the triangle AOC A B any quantity, to aX,x B find the diameter LAB or CD. Analysis.-Join OP; and in OP, or OP produced, take any point p. Parallel to PA, PB, PC, and PD, respectively, draw the lines pa, pb, [OA [Oa pc, and pd; then, since, by similar triangles, PO: pO:: O C )b OD L Od we have Oa, Ob, Oc, and Od all equal, and the points a, b, c, d lie in the circumference of a circle of which 0 is the centre. Also, by similar triangles, we have PA:pa::PB:pb; PC:pc:: PD:pd; PA: pa:: PC:pc, etc. Whence this Construction.-With centre 0 and any radius, describe the circle adbc, and draw the diameters aOb and cOd, making the angle aOcthe given angle. Then, as in Problem XXIV. of "Triangles, Quadrilaterals, and Parallels," describe an arc such that two lines drawn from a and b to any point in this are shall have the ratio of the two given lines PA and PB. In the same manner describe another arc such that two lines drawn from the points c and d to any point in this arc shall have the ratio of the two given lines PC THE CIRCLE. 129 and PD, and let these arcs intersect in a point p. Join pa, pb, pc, pd, and pO. On ap produced, if necessary, lay aE equal to the given length of AP, and parallel to ab draw EP, meeting Op produced, if necessary. Then draw PA, PB, PC, and PD, respectively, parallel to pa, pb, pc, and pd, meeting the diameters ab and cd, produced, if necessary, in the points A, B, C, and D respectively. With centre 0 and radius OA describe a circle, and it will pass through the points C, B, and D, and be the circle required, as is evident from the analysis. Calculation. —As in Problem XXIV. of "Triangles, Quadrilaterals," etc., find the lines ap and cp; then ap: AP:: aO: A O, the double of which is AB or CD, the diameter of the circle required. Scholium.-Tl'he two circles or arcs which intersect at p would intersect also in another point on the opposite side of ab, showing that there are two points that will fulfil the conditions of the problem. The construction, demonstration, etc. are the same in both instances. PROBLEM XXXI. (To illustrate the mode of "discussing a problem."-(Two circles being given in magnitude and position, it is required to apply a line which shall have an extremity in the circumference of each circle and be parallel to a given line. F IG. 1. FIG. 2. C Given, AB, the C angle AB C, the radii AG and BH, andtheline BD, to apply A.EFparallel and equal to BD. _' F Analysis.-Suppose the line EF (either figure) applied parallel and equal to BD. Join BF and DE; then (I. 30*) BDE'F will be a parallelogram, and we will have DE-= BF — BH. Whence this Construction. —From D apply DE, DE' to the circumference of AG, equal to the radius BH, then apply EF or EI'F equal to the given line BD, either of which will be the applied line required, equal and parallel to BD. Demonstration. —Join BF, BF', DE and DE', and irA; then, since DE - BF and EF- BD by construction, EF is parallel to'I. 33. F* 9 130 GEOMETRICAL ANALYSIS. BD (I. 30*). Also, since DE' - BF' and EtF'= BD, E'F' is parallel to BD. Q. E. D. Calculation.-What is.required is the angle BDE or DEF, and the angle B)DE' or DE'F'. Join AE and AE'; then in triangle FIG. 1. FIG. 2. C C D E' ft F ABD, Case 3, find angle ADB, and side AD. In triangle ADE, Case 4, find angle ADE — ADE'; then angle BDE ADB — ADE, and angle DEF 1800 - BDE. Also, angle BDE'ADB + ADE', and angle DE'F' - 1800 - BDE'. Discussion of the problem. 1.-If the radius BH, applied from D, does not reach the circumference of A G, the problem is impossible, as is evident. Discussion 2.-Regarding Figure 1, it will be seen that, as BD must always be of the length of the given applied line, and DE equal to the radius BH, as the applied line is taken shorter and shorter, D approaches B. But as D approaches B it gets more remote from A, and E and El approach each other, till the angles ADE and ADE' vanish, the points E and E' meet at E"l, and DEI" D-E' IDE — BH, in which case the applied line will be a minimum, or the shortest line possible, to have an end in the circumference of each circle and be parallel to the line B C. FIG. 3. To obtain this minimum line by conC struction, with centre A (Fig. 3) and radius AD equal to the sum of the radii 0\oX AG and BH, describe an arc cutting EB'AR ED- BC in a point D nearer B than the foot of a perpendicular AO let fall from A on B C. Draw AED; also draw F' EF parallel to B C, and join BF; then will EF evidently be equal and parallel L to BD, and be the minimum line required. To find its length, in triangle ABD, Case 2, find BDEF. *I. 33. THE CIRCLE. 131 Discussion 3.-In like manner, regarding Figure 2, as BD must always be of the length of the given applied line, and DE equal to the radius BH, it will be seen that, as the applied line is taken longer and longer, D recedes from B, AD increases, E and E' approach each other, till the angles ADE and ADE' vanish, E and E' meet at EB", when DE" — DE - DE' - BEH, and then the applied line will be the longest possible, or a maximum. To obtain this maximum line by construction, with centre A (Fig. 3) and radius equal to the sum of the two radii AG and Bi, describe an arc, cutting BC in D', further from B than the foot of the perpendicular AO let fall from A on B C. Draw AE'D', and draw E'Ft parallel to B C. Join BFE; then will B D'E'F' be a parallelogram, and EIF' will evidently be equal and parallel to BD', and be the required maximum line. To determine its length by calculation, in triangle ABD', Case 2, find BD-' EBE'. NoTE. —Angle ADB -- 180~ - ADB. Discussion 4.-Through A (Fig. 3), parallel to BC, draw a line meeting BF and BF', produced, in I and L respectively; then, since AD -AD', and AE AE', EE' is parallel to B C. But EF is parallel to BC. Hence E'EF is a straight line. Again, since B L -- AD' and BF' - D'E', we have F'L AE'. In like manner BIJ AD and BF- ED. Therefore FI AE A E- A - F'L. Hence FF' is parallel to IL or BC. But EF is parallel to BC; wherefore the three lines E'E, EF, and FF' are in one and the same straight line, parallel to BC. Therefore, the shortest line possible (EF), and the longest line possible (E'F'), that can be applied between the circles, parallel to a line given in length and position, are in the same straight line, the minimum (EF) being part of the maximum (E'Fl). Discussion 5.-In case of the minimum and maximum lines EF and EF'' (Fig. 3), the angle EAG - HBF, and arc EG is similar to arc HF. Also, the angle E'AG HBF', and arc E!G is similar to arc HFI. Also, ADBI, AEFI, AD'BL, AE'F'L are parallelograms. Discussion 6. —The maximum angle which the line BC, given in position, can make with the line AB, is that made by the internal common tangent to the two given circles, as EPF, E'PF' (Fig. 4). For it is evident that a line drawn from any point in the circle BII, making with AB an angle greater than BPF or BPF', will not meet the circle AG; and that a line drawn from any point in the 132 GEOMETRICAL ANALYSIS. circle AG, making with AB an angle greater than APE or APE', will not meet the circle BH. These two lines EF and E'F' are FrG. 4. a equal; they intersect each other in Mv T,the same point P on AB; and are the only lines which can be applied between the circles, making with B AB an angle as great as APE or APE'. To find these lines by construc1o'~ ~, tion.-On AB as a diameter describe a circle, in which apply A 0 and A O0 equal to the sum of the radii AG and BE. Draw BOC, and join BO', parallel to which lines, respectively, draw EF and E'F', and join BF and BF'; then (Problem XIII. of "The Circle") EF and E'F' will be tangents to both circles. To find the lengths of these common tangents by calculation.-We have EF- EIF' I BO -- AB2 — A02. Corollary.-If BF and BF' be produced to meet the circle described on AB in I' and I, then Fl' AE -AE -F'I, and EF is parallel to AI' as well as to BO, also E'F' is parallel to Al and to B O'. Discussion 7.-The minimum angle which the line BC, given in position, can make with AB is 00. In this case BC (Fig. 5) will coincide with BA, and, making BD -- the length of the A C B, B given line, and applying DE, DE' each B -\/ equal to BH, and drawing EF and E'F' B'_ —~ zvX..i each parallel to BD, we have the construction just as in Figure 1. Calculation. —In triangle AED, Case 4, find angle ADE - DEP -1 DE'' - ADE'; then EDB - 180~ - ADE - E'DB. FIG. 6. DLiscussion 8.-In Figure 3, when the c perpendicular AO is just equal to the sum of the radii AG and BH, then D and D' coincide at O, OR (Fig. 3) becomes equal IT r B to DE or BH, and the maximum and minimum lines E'F' afnd EF coincide in the common tangent EEF, as in Figure 6, which makes with AB the maximum angle A PE- A-B C', as shown in Discussion 6. THE CIRCLE. 133 PROBLEM XXXII.-To produce a line of a given length so that the rectangle of the whole line thus produced, and the part produced, shall be equal to the square of a given line. (See IV. Prob. 4.*) Given the length of ED, to produce it so that EA x AD AB2. A B Analysis.-It is evident that AB isi tangent to the circle whose radius CB is half the given line ED, to be produced so that EA x AD - AB2. Whence this Construction.-With a radius equal to half the length of the given line to be produced, describe a circle, as BDE. Draw any radius, as CB, perpendicular to which draw the tangent BA- the given line to whose square the rectangle is to be equal. Through A and C draw ADCE, the line required. Demonstration. -By IV. 30,t EA x AD-AB2. Q. E. D. Calculation.-A C V- AB2 + B C2; then AE = A C + CB, and AD- AC CB. PROBLEM XXXIII.-To find a square which shall be to a given square in a given ratio, as m to n. (See IV. Prob. 11.) I the two segments FE and / Given FG, to find two lines whose GB vn I Esquares shall have the same Fp X ratio as EF: FG as m: n. K 1 Analysis. —Having laid EF and FG, or m and n, contiguous on the same line.EG, on EG describe a semicircle; erect the perpendicular FH to meet the semicircle in H. Join HG and HE, and produce them indefinitely. Then, drawing KII, parallel to GE, anywhere above or below GE, we always have (IV. 11, cor.t) EF: FG:: HE2: HG2: HI2: HK2:: H2: HK2 in the given ratio. Construction and demonstration are manifest. Calculation.-If HK is the side of the given square, then GF: FE or n,: m:: HK2: H1'2 - HK2. n If HI is given, then EF: FG or m: n:: HI2: HK2 _ HI2. m -ar11. Il trR II~3B rat41 ANALYSIS BY ALGEBRA. PART I. CONSTRUCTION OF ALGEBRAIC EQUATIONS OF ONE UNKNOWN QUANTITY OF THE FIRST DEGREE, IN WHICH EACH LETTER REPRESENTS A RIGHT LINE. NOTE.-In a fraction, when the sum of the indices in the numerator exceeds the sum of the indices in the denominator by one, the fraction will represent a line; if by two, it will represent a square or surface; if by three, a cube or solid. PROBLEM I. —Given, the equation x -a + b + c + d, to find x by construction. a b c d A,, I i F Draw an indefinite line AF, and on it lay the lines a, b, c, and d from A to E; then x - AE. Limit.-The given lines a, b, c, d, etc. may be of any number. PROBLEM II.-Given, the equation x a + b-c + d + e -f-g, to find x by construction. a b d e Al I I,. I F E g f cB Draw an indefinite line AF, and on it lay all the affirmative quantities a, b, d, e successively from A to B, their extremities being marked by short lines above the line AF. Then, beginning at B, lay all the negative quantities back from B to E, their extremities being marked by short lines below the line AF. Whence we have x a - b + d - e —(c f + g)- ABBE AE. Limit.-The lines a, b, c, d, e, etc. may be continued to any number. When the sum of the negative quantities exceeds the sum of the affirmative, E will be on the line to the left of A, and AE will be negative, or a minus quantity. PROBLEM III.-Given, the equation x - a to find x by conde (134) CONSTRUCTION OF ALGEBRAIC EQUATIONS. 135 struction. Here, the sum of the indices in the numerator being one more than the sum in the denominator, the result will be a line. Lay the lines represented by the letters, as in the annexed figure, AD - d, DF a, AE - b. Draw d EO parallel to AB, and lay EH e, HI= c. Join DE, and draw FG parallel to DE. Join HG, and draw IL parallel to HG. Then GL will equal dabe - c x. For we have, by parallel lines, d: a:: b:x', and e: c:: x': x or GL. By multiplying the corresponding terms, we have de: ac:: b: abe x - -GL. de NOTE.-The figure looks more symmetrical when EO is drawn parallel to AB, as above, but it may be drawn making any angle with EC. PROBLEM IV. —Given, the equation a2bcd2efg x —- h2i l, to find x by construction. Lay the lines represented by the given letters, as in the annexed figure, and draw the corresponding parallel lines, observing to take the first term, or ante- /,. cedent, in the successive proportions, from the denominator, and the consequents from the numerator; and, also, / that the second consequent or last term in each proportion becomes the second V7 antecedent or third term in the next proportion. / a2b Then we have the proportions h': a:: a: x', hence x' a h: b:: x: x", hence x" —- etc.; t- ~q-, etc.; c: d:: xI": xiv c d: xiv xv 1: e:: xv: xvi n f:f xvi: Zvii in: g:: Xvii: x; then, by multiplying the corresponding terms of these proportions, and observing that all 136 GEOMETRICAL ANALYSIS. the third terms but the first, and all the fourth terms but the last, a2bcd2efg cancel each other, we have h2i21lmn: abcd2efg:: a: x - h2ik2lnn GE. Limits.-The terms may be continued to any number, always observing that if the sum of the indices in the numerator is one greater than the sum in the denominator, the result will be a line; if two greater, a square; if equal, it will be unity; if one less, the result will represent unity divided by a line; if two less, unity divided by a square, imaginary quantities. PART II. CONSTRUCTION OF ALGEBRAIC EQUATIONS OF THE SECOND DEGREE. XA PROBLEM V.-Given, the equation albcd~efgp /- i bdefgp, to find x by construction. As in last problem, construct the line a2bed2efg h2i2 abdefg and thus obtain GE. Proh2i/kIlmn' duce GE to L, and on GL lay EC_ V'p. On GC describe a semicircle, and erect the perpendicular EF. Then (IV. Prob. 3*) we have EF2- GE x,* Ea-s a2bcd2efgp EBC- GI X. p- h22, xor h ik lmn GE —x. * VI. 13. CONSTRUCTION OF ALGEBRAIC EQUATIONS. 137 PROBLEM VI.-Given, the equation x2 -a2 + b2 + c2 + d2 + e2 + f 2 + g2, to find x by construction. Draw two lines A B and B Cat right angles; D P make AB -a; B C — b; CD perpendicular to / F A C - c; D)E perpendicular to AD -d; EF perpendicular to AE -e; FG perpen- a dicular to AF-f; GHI perpendicular to B A G - g, and join AH; then will AH be the line required _ x. For A C2 _ AB2 + B C2 I a2 2 b2; AD2 2 A C2 + CD'2- a2 + b2 + c2; AE'2 - AD ~ DE'2 a2 - b2 - c2 + d2; AFP2 - AE2 + EF2 -a2 + b2 + c2 + d' + e2; A G2 - AF2 + FG2 a2 + b ~- c'2 + d2 + e2 +f2; and AH 2- A G3 + GH2 a2 + b2 + c2 + ds + es2 +f2 + g2 x2; whence x - AH. Scholium.-These squares to whose sum x2 is to be equal may be of any number. Also, we have A C2 a2 -+ b2, and hence AC a2 + b2; AD2 = a ~+ b2 + c2, and hence AD -+/ a 2+ b2 + c2; AE2z- a2 + b2 + c2 ~d2, and hence AE -,/ a2 + b2 + c2 + d', etc., etc., to any number, greater or less than the number given in the problem. PROBLEM VII. —Given, the equation x2 - a2 + b2 - c2 + d - e2 + f 2 -g2 2+ h', to find x by construction. Proceed in the same manner as in E Problem VI., and when we come to a C F quantity that is negative (as C2) on the line (as A C) whose square represents the value of all the preceding quantities, B A describe a semicircle, in which apply the line whose square is negative (as c); join A with this point (as AD), and then proceed as in the last problem. It will be observed that producing the line applied in the semicircle will be erecting a perpendicular on tlec li,. drawn from A to the remote end of the applied line. Thus, CD, produced, gives DE perpendicular to AD. Now, we have, as in last problem, A C' AB'2 + B C' a2 + b2; AD' 2 AC2- CD 2 a2 F b2 - C2; AE2 = AD2 + DE2 a' + b2 c- + d'; AF2- AE2 - EFS2 a2 + b2- c2 + d- e2', etc., etc.; and, finally, we have AI2 a2 + b- - c2, d2 _ e2 _. f 2 g2 + h2 2 x2; whence x AI. Scholium.-The squares to whose collected value x2 is to be equal may be of any number. 138 GEOMETRICAL ANALYSIS. PROBLEM VIII. —Given, the equations x2 + ax- b2, and x2- ax b2, to find the value of x in each. Construction. -Draw two lines AB and BD at right angles, and make AB- a and BD - b. On AB describe E a circle whose centre is C, and draw b the secant line D)ECF; then EF — AB - a; and, in the equation X2 + ax = b2, DE = x; while in the equaAL / a B tion x2- ax- b2, DF- x. Demnonstration.-DE x DF- DB.2 (IV. 30*). Now, if DE- -x, DFx- + a, and DE x DE-x x (x - a) X / 2+ax DB2 — b2. Also, if DF x, DE -x-a, and DE x DFx x (x -a)- -ax DB — b2. Limit. —The given lines a and b may be of any length whatever. Examples. —1. Construct the equation x2 + ]22x 82, and find x 4, measured from the scale. 2. Construct the equation X2 - 12x- 82, and find x 16, measured from the scale. 3. Construct the equation x2 + 21x 102, and find x _ 4, measured from the scale. 4. Construct the equation x2 - 21x 102, and find x - 25, measured from the scale. PROBLEM IX.-Given, the equation x2 - ax - b2, or ax xb2, to find the two values of x. Construction.- Draw two lines AB and BD at right angles, making AB - a, and BD - b; and on Gta b AB describe a semicircle of which C is the centre. C -7B rParallel to AB draw DG, meeting the circumference in E, and let fall the perpendicular EF on the diameter AB; then EF- BD-b, and either BF or AF-x. Demonstration. —By IV. 23, cor.,t AF x FB - FEY - BD'2 b. Now, if either BF or AF is x, the other will be a - x, and we always have AF x FB=x x (a-x) = ax-.: b. In an equation of the form ax - x2 = b2, there are always two positive values of X. * III. 36. t VI. 8, cor. CONSTRUCTION OF ALGEBRAIC EQUATIONS. 139 Limit.-BD must always be less than the radius BC; that is, b must be less than 1 a. Examples.-I. Construct the equation x2- 20x - 82, or 20x xx2- 82, and find x - 4 or 16. 2. Construct the equation x2 -299x - 102, or 29x - x2 102, and find x — 4 or 25. PROBLEM X.-Given, the equations x2 + ax = d, and x2 - ax - d, to find the value of x. CASE 1.-When d can be divided into D two factors b and c, whose difference is less than a, then the equations become F X2 2+ ax = be, or X2 - ax - be. Construction. —With a radius equal to B I a, describe a circle, the centre of which is C. In this circle apply the chord AB the difference between b and c. Produce AB, making the produced part BD the less factor, then AD the greater factor. Through C draw DE CF; then DE will be the value of x in the equation x2 + ax d - be, and DF will be the value of x in the equation x2 - ax d be. Demonstration.-For (IV. 29*) DE X DF — DB x DA -bc. Now, if DE x, we have DF= x-+ a. Whence DE X DFx (x + a) - x2 + ax be d. Also, if DF — x, we have DE - x - a, and DE x DF — (x - a) x- x2 - ax - be - d. Examples. —1. Construct the equations x2 + 8x - 48 - 8 x 6, and X2- 8x — 48 - 8 x 6, and find x-=4 and 12. 2. Construct the equations x2 + lOx -24- 6 X 4, and x2 -- 10x 24 6 x 4, and find x. CASE 2.-When d cannot be divided into two factors whose difference is less than a, then take the square root of a to three figures, and, putting this v/d-b, we have d-b2, and the given equations become x2 + ax - b2, and X2 - ax - b2, whence x is found as in Problem VIII. Examples.-1. Construct the equations X2 + 6x - 112, and x2 6x=-112. Now, 112 = 56 x 2 -=28 x 4 16 x 7-=14-x 8, and the difference of no two factors is less than 6 or a. Hence take * III. 36, cor. 140 GEOMETRICAL ANALYSIS. V/112 = 10.9 b; then d= 10.92 —- b2- 112, and we have x2 + 6x 10.92, and x2 6x- 10.92. 2. Construct the equations x2 + 10x - 24, and x2 —10x - 24. Here V/24-= 4.90, and the equations become x2 - 10x- 4.902, to find x as in Problem VII. PROBLEM XI.-Given, the equation x2 - ax = - d, or ax - x2 - d, to find x. CASE 1.- When d can be divided into two factors B EB b and c, whose sum is less than a, then we have x2 -ax — be, or ax- x - bc. Construction.-With a radius equal to - a, and centre C, describe a circle, in which apply the chord A DE the sum of the given lines b and c. On DE lay DF= to one of these lines, then EF will be equal to the other. Through C and F draw the diameter A CFB; then x will be either AF or FB. Demonstration.-If either AF or FB is taken as x, the other will be a - x. Now (IV. 28, cor.*), we always have AF x FB = DF x FE; that is, x (a - x) - bc, or ax - X2- bc d. E xamples. —l. Construct the equation 1 9x - x2 = 60 - 10 x 6, and find x - ] 5 or 4. 2. Construct the equation 19x - x2- 48= 6 X 8, and find x 16 or 3. CASE 2.-When d cannot be divided into two factors whose sum is less than a. Construction.-Take the square root of d to three figures, and put the root =- b; then /d = b, or d- b2. Then x2- ax- - b, or ax -- x2 b2, which construct as in Problem IX. Examples.-i1. Construct the equation x2 - 22x- - 112, or 22x:x2-112 — 10.9, and find x — 14 or 8. 2. Construct the equation 22x - x2 - 85 9.22, and find x = 17 or 5. NoTE. —The results of all these numerical equations should be verified by solving the equations algebraically. Remark.-The preceding problems under the head of " Construction of Algebraic Equations" contain all the forms of equations of one unknown quantity of the first and second degrees, which are all that can be constructed by Plane Geometry. The results may be combined in various ways in a problem. - III. 35. PROBLEMS ILLUSTRATING THE USE OF ALGEBRA IN GEOMETRICAL ANALYSIS. PROBLEM I.-In a plane triangle are given the base, the sum of the other two sides, and the line drawn from the vertex to the middle of the base, to determine the triangle. L ithe base AB, the sum of AC and CB, and the line Cj H being the middle _ of AB. A HD B B G Analysis by algebra.-Let ABC represent the required triangle, having AH — HB. Put b - AH or HB, d CH, s -=the sum of A C and CB, and x - - the difference of A C and CB; then A C -s + x, and CB= s - x. Now (IV. 14*), AC2 + CB2-2 AH2 + 2 CH2; that is, (s + x)2 + (s - x)2 — 2b2 + 2d2; or, by expanding the two quantities in the first member of the equation, 2s2 + 2x2 2b2 + 2d2, whence x2 -- b2 + d2 s2; and we have this Construction.-In any straight line take AG = the given sum of the sides, and AB = the given base, and bisect them in the points D and H respectively; then AD - s and AH= b. At H erect the perpendicular HL - d, the given line to bisect the base; join AL, and to a perpendicular at D apply AI — AL; then D12 -= AI - AD2 = AL2 - AD2 -= AH2 + H2 - AD2 - b2 + d2 _ S2 = hence x -- DI. Make DE — DI; then AE — s + x, and EG = s- x, the two required sides. From Has a centre and a radius HL describe an arc L C, to which apply AC = AE; join CB and CH, and A CB will be the triangle required. Demonstration geometrically. —We have to prove only that B C 2 2 EG. Now, AE2 + EG2= AD DE + AD —DE 2 AD2 + 2 DE2'2 AD2 + 2 DI2 = 2 A1'2 - 2 AL2 - 2 AH2 + 2 HL2 = 2 AH2 + 2 CH2 = (IV. 14t) AC2 + B C2 - AE2 + B C2. Hence BC2- EG2, or B C =EG. Q. E. D. IL A. t II. A. (141) 142 GEOMETRICAL ANALYSIS. Calculation.-By construction, DE = D — = AH 2 + HL2- AD2. Then AC -- AE - AD + DE, and BC = EG -- DG DE AD-DE. Limit.-The sum of the squares of AH and CH can never be less A 11 D E B G than the square of AD, half the sum of the sides. If AH2 + CH-2 AD2, then DI — 0 and DE - 0, and the sides AC and BC will be equal, HC will coincide with HL, and AL will be half the sum of the sides. PROBLEM II.-In a plane triangle are given the ratio of the two sides, and the segments of the base made by a perpendicular let fall from the opposite angle on the base, to determine the triangle. C _/ffin the ratio of BC to AC as m // Given to n, and the segments AD and DB made by the perpendicular CD on the base AB. A D H B F Analysis by algebra.-Let A CB represent the required triangle, having CD perpendicular to AB. Put BD =-a, AD = b, VxBC, and nx-AC; then BC:AC::mx:nx::m:n; and, since B C2 -BD2 = CD 2- A C2 - AD2, we have B C2 - A C2 = BD2 AD2; that is, m2x2 - n2x2 -a2 b2, whence x2- 2 2x X.2 n 2_n2/ a2 _ b2 n V/ a2 _ b2 n / a2 _a b2 _ 2 -n2 x- 2 2 B C, and nx -2 AC. WThence this Construction.-Take DH= n, and apply HG -- m; then D G /2m —n2 2. Apply AE BD =a; then DE= aa"b2. Dl'raw EF parallel to GH. Now, by similar triangles DGH, DEE, DG (a/l2 - 2): GH (m).DE (V/ a _ -b2)~ EF-= = the value of B C as found in the analysis; also DG (m2 —n2): DH (n):: nD ( a/a b D DE (Va/a —b2). DF —- 2 =the value of AC, as found in -V2__ n2 ANALYSIS BY ALGEBRA. 143 the analysis. Apply B C FE, and join AC, then ABC will be the required triangle. Demonstration by geometry. —By construction and parallel lines, we have EF or B C: FD:: GH: HD:: m: n; hence we have only to prove that FD = A C. By the analysis, B C2 - A C2 BD2 - AD2 AE2 AD2 =DE2 — EF2- FD2 — B C2- FD2. Hence AC2 F-D2andAC- FD. Q.E.D. Calculation.-We have DG - = n2m- n2, and DE — =a —b2; then, by similar triangles DGH and DEF, we have DG:DE:: GH x DE GH: EF - - B C, and D G: DDE:: DH: DF or A C DG DII x DE D DG Example.-If BD = 17, AD - 8, m - GHI= 5, n DH —= 4, then DG — = 3, DE -- 15, B C - 25, and A C - 20. NOTE.-The same problem can be analyzed geometrically. (See Problem XXIII. in'' Triangles, Quadrilaterals, and Parallels.") PROBLEM III.-It is required to draw a line from one angle of a given square so that the part intercepted between the side and the side produced, that meet at the opposite angle, may be of a given length. I the square BEFG, and the length of the part G 7' i AC of the line BAC Given intercepted between the A sides EF and GF, proL duced, that meet at F. B E o H L Analysis by algebra.-Let BAC represent the required line of which the length of AC is given. Draw CL perpendicular to AC to meet BE, produced, in L, and on BL let fall the perpendicular CH; then CH= EF — BE, and (IV. 21*) the triangles CHL and BAE are similar and equal, having CL = AB and HL = AE. Now, put AC b,EF —a, EL=-x, and CL =y AB. In similar triangles EBA and CBL, we have AB (y): BE (a):: BL (a 4- x): B C (b + y); wherefore by + y2 = a' + ax, or 2by + 2y2 2a2 + 2ax. Also, BL2 B C2 + CL2; that is, (a + x)2 -(b + y)2 + y2, or a + 2ax + x2= b2 + 2by + 2y2- (from above) b2 + 2ax + 2a2. From which x2- =a2 + b2. Whence this - VI. 8. 144 GEOMETRICAL ANALYSIS. Construction.-On BG, the side of the given square (produced, if necessary), lay BI= the given intercepte'd line b. Join EI, and I nmake EL= — EI — E - BI2 — = I/a _Y -x. On BL describe a semicircle of which the centre G /\ c is 0, meeting GF, produced, in C. Draw BA C; then will AC= B1 / the given length. / Demonstration by geometry.B) b; o H L, Join CL, AL, and OC, and let fall on BL the perpendicular CE; then AC 2+ CL2 - AL2 — AE2 + EL2 _ AE2 + E12= AE2 E2 + EB + BI2 -= AB2 + B12= CL2 - B12. Taking CL2 from the first and last of these equals, we have A C2-=B12, or A C BI. Q.E. D. Calculation.-We have EL- = E= /EB2+BI = /EB2+A C2, BL - BE + EL, OC = OL - BL, OH-4/ O C2- CH2, OL -OH —HL - AE, EO - OB-BE- OL-BE, FC-E H =EO + OH, and G C= GF+ FC, CL =- AB = =VBE2 + AE2. Limit.-The line A C muay be taken any length. PROBLEM IV. —In a right-angled triangle are given the hypothenuse, and the side of the inscribed square, to determine the triangle.,? G c' c I the hypothenuse A C of the triangle ABC, Given and the side BG of E /F f the inscribed square L B GFE. Analysis by algebra.-Let ABC represent the required triangle of which the hypothenuse AC and the inscribed square BGFE are given. Draw CL perpendicular to AC, meeting El, produced, in L, and let fall the perpendicular CH_, which will be equal to GF or EF; then (IV. 21*) the triangles CEHL and EAF are similar and equal, having CL - AF and HL = AE. Now, put A C- b, EF a, EL = x, and CL - y = AF. In similar triangles AFE and CFL, we have AF (y): FE (a):: FL (x - a): FC (b - y). WhereVI. 8. ANALYSIS BY ALGEBRA. 145 fore by - y2 = ax - a2, or -2by + 2y2 = -2ax + 2a2. Also, FL2 -FC2 + CL2; that is, (x - a)= (b - y)2 -t y2, or x2 — 2ax + a2 b2 - 2by + 2y2 = (from above) b2 - 2ax + 2a2. From which we have X2 a2 -+ b2, and this Construction. —On BG, the side of the given square, produced, lay BI- the given hypothenuse. Join El, and make EL -El= EB2 2+ B12 =Va2 ~+ b =x. On FL describe a semicircle of which the centre is 0, cuttiig, B~ in C; draw CFA; then will AC - BI, the given length of the hypothenuse. Demonstration by geometry.-Join CL, AL, and OC, and on EL let fall the perpendicular CH; then A C2 + CL2 = AL2 = AE2 + EL2 AE2 + El2 AE2 + EB2 + B12= AE2 -+ EF2 + Bl2= AFE2 + B =- CL2 + B12. Whence AC2=BI2, or AC=BL Q. E. D. Calculation. —EL = EI= V/EB2 + BI2-= /EB2 + AC2, FL EL - -EF, OF or 0 C -=- FL, OH= /O C2 -_ CH 2; then FI1 - FO + OH- G C, and HL - FO - OH- AE. Also, B CBG + GC, and AB BE E+ AE. Limit. —When the semicircle described on FL does not meet BI, the problem is impossible. If the semicircle just touches GI, that is, when the radius O C-= FG, the triangle AB C will be isosceles, having BA = B C, and A C will be the shortest possible line which can be drawn through Fto meet BE and BG produced. Scholium.-As the semicircle on FL cuts GI in the point C', as well as C, if through C' and F a line had been drawn to meet BE, produced, in a point A', the triangle A'B C' thus formed would have been equal in all its parts to AB C, having A C'- A C, A'B = BC, and B C' AB. PROBLEM VI.-In a plane triangle are given the base, the perpendicular height, and the sum of the other two sides, to determine the triangle. the base AB, the perpendicular L] "iF W-XL4JIB Given CP, and the sum of AC and Analysis by alebra.Let A CB represent the required triangle. Analysis by algebra. —Let A CB represent the required triangle. Bisect the given base AB in D. Produce BA indefinitely, and lay G 10 146 GEOMETRICAL ANALYSIS. BL — the given sum of AC and CB. Let fall the perpendicular CP. Now, put L-=a, AB=b, CP-p, AC x, and AP= y; then we have BC —=a —x, and BP=b —y. And, since (cor. to Prob. XXXVII., "Triangles," etc.) B C2 - AC2 = BP2 - A2, we have (a - x)2- x2= (b - y)2 y2, or a2- 2ax- =b2- 2by; whence axb (2 2 Now, construct this known quantity 2b]qC F Take a third proportional (IV. Prob. 2, cor. *) to 2b and a, and lay it from D to H; then 2AB(2b): BL (a):: BL (a): a2 a2 b H — L B A D BDH- And 2b-2-DH AD /~ = —=AH/. Then, AH being known, erect the perpendicular HE to meet a line through C, parallel to AB, and we have E- = CP. Now, the above equation y b ( ) becomes ax ax y b~-AH, or — y + AH AP AH- P- H= C; that ax is, we have - E(C, or b: a:: x:EC; or, by inversion, a (BL): b(AB):: EC: x(CA). Hence, if EF is made equal to BL (a), and FG is drawn parallel to A C, we shall have FG = AB. Whence this Construction.-Take DH a third proportional to 2 AB and B L. Erect the perpendicular HE the given perpendicular. Draw EF parallel and equal to BL. To EA, joined and produced, apply FG AB. Draw AC parallel to FG, join CB, and let fall the perpendicular CP; then A CB is the required triangle. Demonstration by geometry.-We have to prove only that A C + BC —BL. By parallel lines we have FG (AB): EF(BL):: AC: EC(PH). HenceAB x PH- BL x A C, or 2 BL x A C- 2 AB x PH. By construction, 2AB: BL:: BL: D; hence BL-= 2 AB x DH. Taking the first of these equations from the second, we have BL2 _ 2 BL x A C- 2 AB x DH- 2 AB x PI — 2 AB (DH — PH)= 2 AB x DiP - AB x 2 DP — AB (BP - AP) (BP + AP) x (BP- AP) = BP2 _ 4p2 B C2- A C2. To the first and last of these equals add A C2, and we haved BL2-2 BL x A( -+ A C2 B C2; that is, (BL - A C) BC, or B' L-, = BC. WhenceAC+BCC BL. Q. E.D. * VI. 11. ANALYSIS BY ALGEBRA. 147 Calculation.-By construction, 2 AB: BL:: BL: DH; then All =-DH- AD. 1. In trianglle HAE, Case 3, find < EAH= < GEF. 2. In triangle EFG, Case 2, find , pointA, to construct the triantle B A" / au~- I —--- ~F i ACB, such that the base A l, a of L shall be a minimum. Analysis.-Let F'CE' be the given circle, whose centre is O, a This question was proposed as a "Prize Problem" in the fourth number of the Mathematical Diary, edited by Robert Adrain, in 1825, and the above solution was given at the time by the author of this work, and published the next month in No. 5, page 132, and it shared the prize with the late Dr. Charles Farquhar, then of Alexandria, D.C., and J. H. Swale, of Liverpool, England. Dr. Bowditch and Prof. Strong were correspondents of the same mathematical periodical. ANALYSIS BY THE DIFFERENTIAL CALCULUS. 179.EF* the line given in position, A the given point in EF to be one extremity of the base, and FAD equal to the given vertical angle. Draw AH and OL at right angles to AD and EF respectively. Then it is evident (III. Prob. 16t) that the centre of the circle circumscribing the required triangle will lie in AH. In HA, produced, lay Al(in a direction from the perpendicular OL) equal to the radius of the given circle. Join I0, and draw AC parallel thereto. Also, draw 0 C, and produce it till it meets AH in G; then (IV. 15t), since AI — OC, the radius of the given circle, we have GA - GC. Hence, with the centre G and radius _-GA or GC, describe the segment A CB, and join CB; then will A CB be the triangle required, having its base AB a mninimum. Demonstration.-By III. Prob. 16,~ the angle A CB BAD the given vertical angle. Now, since the radius of any other circular segment whose centre is in AH, to pass through the point A (as must be to make the contained angle equal to FAD) and meet the given circle, must necessarily be greater than GA, it is evident that AB is the least possible base. Q. E. D. Calculation.-Since the position of the line EF, and of the point A, are given, if we join OA, the triangle OAL is given in all its parts. Hence, angle OA G -O0AL -t LAG. In triangle OAG, find angles, and GA- GC. In triangle AGB, Case 1, find AB; then (III. 1811) angle ABC=~AGO. In triangle ABC, Case 1, find AC and JBC. Corollary.-If the point A coincides with L, AI may be laid either way, and there will then be two similar and equal triangles formed with the shortest possible bases. PROBLEM XXII.-Having given the two parallel sides of a trapezoid, and one diagonal, to construct it such that its area may be a maximum. (See figures to scholium under Problem LXIV., "Triangles, Quadrilaterals, and Parallels.") BD rC the parallel sides AB and CD of the trape-! zoid ABCD, and the diagonal AC, to Given { construct it such that its area shall be the EB A B greatest possible. Analysis. —The trapezoid is made up of the two triangles A CB * If the corresponding marked or "prime" letters are used, the samne construction, demonstration, and calculation apply when the given line and base are taken woithin the circle. t III. 33.: vI. 2. II. 33. III. 20. 180 GEOMETRICAL ANALYSIS. and ACD, the bases AB and CD of which are given, and hence constant quantities. Therefore, the greater the perpendicular heights of these triangles, the greater will be their areas, and, hence, the area of the trapezoid. But, evidently, their altitudes will be greatest when the diagonal A C is perpendicular to AB, and, consequently, to DC. Whence this D c Construction.-In any straight line lay BA D c one of the given sides, and AE the length of the other side. Erect the perpendicular AC - the given diagonal. Draw CD parallel and E A B equal to AE, and join DE, DA, BD, and BC; then ABCD will evidently be the maximum trapezoid required. Calculation.-The area of AB CD -- A C x (AB + CD). The side CB- AB2 + A C2, the side AD -/CD2 AC2, BEA BA + AE, and BD = BEV 2 - ED2 = /BE2 +- A C2; whence the angles may be found at pleasure. Corollary.-The triangle DEB _ triangle CEB (AB +- AE) x A C = (AB + CD) x A C - the area of the trapezoid AB CD. NOTE.-See note to Problem VI. of this division, as alike applicable to these two problems. PRtOBLEMS IN RELATION TO AREAS AND THE DIVISION OF SURFACES. PROBLEM I.-To find the area of a triangle when two sides and the included angle are given. angle BA C. A D B Calculation.-Let fall the perpendicular CD. In triangle A CD, Case 1, find CD; then the area of ABC - (AB x CD). In practice, the following method is convenient: Sin D or rad.: sin A:: AC: CD:: (multiplying the last couplet by AB) AB x AC: AB x CD - 2 area; that is, 2 ABCsinA x AB x AC, or rad.:sin included angle::product of the rad. sides including that angle: double the area of the triangle. PROBLEM II.-In a triangle the area, one angle, and a side adjacent to that angle are given, to find the other adjacent side; that is, in the above figure, are given the area of ABC, the angle BA C, and the side AB, to find A C. Calculation.-By the last problem we have rad.: sin A:: AB x A C: 2 area. Hence rad. x 2 area = sin A x AB X AC, where all the terms are known but AC, which is determined by converting this equation into a proportion so as to have A C for the fourth term, thus:-sin A x AB:rad.:: 2 area: AC; that is, as sin given angle multiplied by the given side: rad.:: 2 area: the other side adjacent to the given angle. NOTE 1. —The double area of a triangle being equal to the product of the (181) 182 GEOMETRICAL ANALYSIS. base by the altitude or perpendicular height, the altitude will be found by reducing the double area to the same denomination as the base and dividing the result by the base. Hence, in the triangle ABC the perpendicular CD — 2 area ABC divided by AB, and the figure for this problem is readily constructed. NOTE 2.-Since the area of a parallelogram is double the area of a triangle of the same base and altitude, we have sin given angle Xgiven side: ad.:: area of parallelogram: the other side of the parallelogram aodjacent to the given angle. NOTE 3. —The perpendicular height of a parallelogram is equal to its area divided by the base, the area and base being reduced to their respective units of the same name. PROBLEM III.-Having given the angles and one side of a triangle, to find the area. (: D Given Jin the triangle ABC, all the angles and the side AB, to find the area. A B Calculation.-Let fall the perpendicular BD from one extremity of the given side on one of the other sides, as AC; then we have sin C: sin B:: AB: AC. Also, sin D (rad.): sin A:: AB: BD. By multiplying the corresponding terms of these proportions, we have rad. x sin C: sin A x sin B:: AB2: A C x BD - the double area of the triangle ABC. That is, the product of rad. x sin angle opposite to any side: the product of the sines of the other two angles:: the square of that side: double the area of the triangle. We have the three proportions, rad. x sin C: sin A x sin B:: AB: 2 area ABC. (1.) rad. x sin B: sin A x sin C:: A C2: 2 area ABC. (2.) rad. x sin A: sin B x sin C:: B C2: 2 area AB C. (3.) Corollary.-The fourth term in each of these proportions is evidently the area of a parallelogram constructed with the sides AB and AC and the included angle A. PROBLEM IV. —Having given the angles and area of a triangle, to find a side. (See above figure.) Calculation.-Take the proportions in the preceding problem by PROBLEMS IN RELATION TO AREAS. 183 inversion. To find any side, take the sines of the angles adjacent to that side for the first term. Thus: To find AB, we have sin A x sin B: rad. x sin C::2 area ABLC: AB2. To find AC, we have sinA x sin C:rad. x sinB::2areaABC:AC2. To find BC, we have sin B x sin C:rad. X sinA: 2 area4 B C: BC2. NOTE.-The area will always be in the same denomination as the side, and the side as the area. That is, if the area is in square chains, perches, yards, or feet, the side will be in linear chains, perches, yards, or feet; and, if the side is in linear chains, perches, yards, or feet, the area will be in square chains, perches, yards, or feet. PROBLEM V.-HIaving given the angles and area of a plane triangle, to construct it. K H' the angles A, B, and C, and the Given. area of the triangle ABC, to conD4, (struct it. Analysis and construction.-Draw any line ASM, and make the angle MAKr-the given angle A. Let the given area, reduced if necessary, be in chains, perches, or some denomination of which there is a linear unit, and make AD - the square root of the given area. On AD describe the square ADIIH, which will then contain the given area. Let the side HI, produced if necessary, cut AK in L, and join LD. Then the triangle ALD, being equal to half the square (IV. 2, cor. 1*), contains half the given area. Lay LG AL, and join GD; then, since the bases AL and LG are equal, the triangles ADL and LDG are equal, and ADG - 2 ADL -the square ADIH-the given area. At G make the angle AGF- the given angle C; then the triangle A GF will be similar to the required triangle. Now, we have to construct a triangle similar to one' VI. 1. 184 GEOMETRICAL ANALYSIS. triangle, and equal in area to another, both having the same altitude; that is, to construct ABC similar to AFG, and equivalent to ADG (IV. Prob. 15*). To do which, find a mean proportional between:K their bases AF and AD by describing a c/ semicircle upon the greater base AF, at D, the end of the less base AD, erecting C / \ a perpendicular to meet the semicircle in L E, and joining AE, which will be the / \ l~ /\ \ mean proportional required (IV. 23T). A Lay AB AE, and draw BC parallel to D B aF GF; then ABC will be similar to AFG, and equivalent to ADG. Demonstration.-By IV. 25,t we have BE AFG: BC:: AFl2: AB2:: AlF2: AE2:: AF'2: AF x F:: AD:: AFG: ADG (IV. 6, cor.~). That is, AFG: AB C:: AFG: ADG. Whence AB C -ADG -- 2 ADLthe square ADIHt- the given area by construction. Calculation.-By Prob. IV. A., that is, the last problem, find the sides. Scholium.-Since (IV. 23, cor.,l1 and the above demonstration) we always have AF2: AE2:: AF: AD, AF2: FE2:: AF: FD, and AE2: El2::AD: DF, and similar figures being to each other as the squares of their homologous sides, we have AF: AD:: fig. on AlF: sim. fig. on AE or its equal AB, and AF: FD:: fig. on AF: sim. fig. on FE or its equal, and AD: DF:: fig. on AE: sim. fig. on El' or their equals. Remarle. -The preceding five problems are fundamental problems in areas, and the principles involved in them are in so frequent requisition that the student will find it to his advantage to be familiar with them. In solving the following problems they will be referred to, when used, as Prob. I. A., Prob. II. A., Prob. III. A., etc.; that is, Problem I. in Division on "Areas," Problem II. in Division on "Areas," etc. - VI. 25. t VI. 8, cor. $ VI. 19. VI VIVI. 8, cor. PROBLEMS IN RELATION TO AREAS. 185 PROBLEM VI. —To bisect a given triangle by a line drawn from a given point in one of the sides. c..nF i(the triangle AB C, and the distance BP, em/7 \ 1 \ Given to draw the line PF so that PFB shall //~ \i \ (be equivalent to A CEP. A 1' B Analysis. —Suppose P' to be the required line dividing the triangle ABC into two equal parts. Join CP; bisect AB in E, and join CE; then, since BE- 2 AB, the triangle BCE B CA _ BPE. Take triangle BEF from the first and last of these equals, and we have the triangles EPF and ECF equal; hence (IV. 2, cor. 2*) EF is parallel to CP. Whence this Construction. —Bisect AB in E; join CP; draw EF parallel to CP, and join PF and CE; then PF will be the division line required. Demonstration. —By IV. 2, cor. 2,t the triangles EPF and E CF are equal. To each add BEE, and we have BPF-= BE C BA C. Q. E. D. Calculation. —By parallel lines, BP: BE:: BC: BF; then CFB C — BF, and BA A C:: BP: PF. Limits.-If BPE- BE -- -AB, the line CP will coincide with CE and be the bisecting line. If BP were less than BE, the point would fall on the line A C, but the method of construction and calculation would be the same. PROBLEM VII.-Having a square given, it is required to construct a rectangle which shall have the same perimeter as the square, and contain half the area. A B G H the square ABCD, to construct the rectangle AE1FG to be equal (Given half AB CD, and have the sides AG + GF= AB + BC. D c Analysis.-Let ABCD be the given square, of which the diagonals are AC and BD, and AEFG the required rectangle. In AG, produced, take GIl= GF; then AHI half of the perimeter * I. 39. t I. 37. 186 GEOMETRICAL ANALYSIS. of the rectangle- (by the prol)lem) half the perimeter of the square AB + B C; hence BH= B C. Again, the area of AEYFG - AG x G-= AG x G H=(AB + BG) x (AB- BG) AB2BG2 (by the problem) - AB2 B' 02 or A 02. Wherefore B G2BO2, or BG —BO; and we have this Construction.-Havin g drawn the diagA 1 Ga H onals AC and BD to the given square, intersecting in 0, produce the side AB, making BH=AB; then At- 2 the perimeter of the required rectangle. On ~ \_BH lay B G BO, erect the perpendicular GF= GH, and draw FE parallel to AH; then BEFG will be the rectangle required. Demonstration.-We have A G + GF- A G - G+ -- A11 AIJ + BC. Hence the perimeters are equal. Also, area ABEFG - AG x GF- AG x GH- (AB + BG) x (AB - BG)AB2 -BGC2 _-AB2- B02=-A02= —AB'2 Q. E. D. PROBLEM VIII.-Having given the position of three consecutive zigzag lines, and the lengths of the first two, to draw a line from the starting-point to the third line so as to cut off equal triangles in the two given angles. G I,, C ~ the angles ABC and BO G, and the lengo-ths of the Given sides AB and B C, to draw the line AED to make the triangle ABE - E CD. B A Analysis.-Suppose the line AED so drawn as to make the triangles ABE and ECD equal. Join AC. Add the triangle AEC to each of these equals, and we have the triangles ABC and ADC equal. Whence (IV. 2, cor. 2*) BD is parallel to AC; and we have this Construction.-Having drawn AB, BC, and CG as given, and joined A C, draw BD parallel to A C, meeting the third side CG in D. Join AD, which will be the required line. Demonstration.-By IV. 2, cor. 2,* the triangles ABC and ADC are equal. Take AEC from each, and we have tim triangles ABE and E CD equal. Q. E. D. * I. 37. PROBLEMS IN RELATION TO AREAS. 187 Calculation. —1. In triangle ABC, Case 3, find AC, and angle A CB. 2. In triangle CBD, Case 1, find CD and BD. Angle ABD ABC+ CBD. 3. In triangle ABD, Case 3, find AD, and angle ADB. 4. In triangle BE1D, Case 1, find DE and EB. Then AE -AD - DE, CE - B C - BE; and we know every line and every angle, and can readily find the areas of the equal triangles AEB and CED by Prob. I. A. or II. A. NOTE.-When AB and CG are parallel, CD must equal AB, and AD= CGB, and the triangles AEB and CED will be equal in all their parts. PROBLEM IX.-To divide a given triangle into any number of equal parts by lines parallel to one of the sides. / B' u" Given the triangle ABC, to divide it into any. 2 X ~ 3 number of equal parts, say four. Construction.-In accordance with Problem V. A., scholium, divide AB into the same number of equal parts that the triangle is to be divided into, in the points 1, 2, 3, etc., and at these points erect perpendiculars to meet a semicircle described on AB, in the points D', D", D"', etc. Join AD', AD", AD"', etc., and make AB' - AD', AB" - AD)", etc., and parallel to BC draw B'C', B"I CII, etc., and they will divide ABC as required. Demonstration. —By scholium to Prob. V. A., ABC: AB'C', ABII"C", etc.:: AB: Al, A2, etc.; hence the parts AB'C', B' C' C"B", B"I C"'B C t', etc. are all equal. Q. E. D. Calculation.- AB' - AD' _-V B AB/ x Al, AB" - ADo 4/AB x A2, etc. Also, AB: AB'::BC: BC', anid AB:AB":: B C: B" C", etc. NOTE.-If it is desired to have the parts of the triangle in any given ratio, divide AB in that ratio, and proceed as above. 188 GEOMETRICAL ANALYSIS. PROBLEM X.-To divide a given circle into any number of given concentric rings. -the circle whose radius is CA, to, AfS a C I Given divide it into any number of concentric rings whose radii are CG, CH, etc. Construction.-Divide CA into the same number of equal parts at the points B, D, etc. that the given circle is to be divided into, and at B, D, etc. erect perpendiculars to meet the semicircle on the radius CA, in E,F, etc. Join CE, CF, etc., and with these lines as radii, and centre C, describe circles, and they will divide the given circle as required, as is evident from scholium to Prob. V. A., and the preceding problem. NOTE. —If it is desired to have the rings in a given ratio, divide the radius in that ratio, and proceed as above. PROBLEM XI.-In a quadrilateral figure there are given the four angles, and one pair of the opposite sides, to find the area. Fio. 1. B Given, the angles A, -c/ \ B, C, and D, and the c,? sides AB and CD, to find the area. i), /) A / (; 1 2A Analysis.-Let AB CD (Fig. 1) represent the quadrilateral figure, of which AB and CD are the knownr sides. Produce the unkcnown sides BC and AD till they meet, as in E. Draw E1 parallel to D C, and CF parallel to DE; then EF= DC, and the angle AEF.the given angle ADC. Whence this Construction. —Draw AB- the longer given side, and make the angles EAB and ABE —the given angles A and B respectively. At the point E where these lines meet, make the angle AEF1-the given angle D, and lay EF equal to the shorter given side. Draw FC parallel to AE, and CD parallel to EF, and ABCD will be the quadrilateral required. The demonstration is evident from the analysis. PROBLEMS IN RELATION TO AREAS. 189 Calculation.-By Prob. III. A., in triaOngle AE B, we have rad. x sin E: sin A x sin B:: AB2: 2 area ABE, and in triangle DEC, rad. x sin E:sin D x sin C::DC: 2 area DEC. Half the difference of these two results gives the area of AB CD, in the same denomination as the sides AB and CD. If the lengths of the sides AD and BC are desired, in the triangle AEB, Case 1, find AE and BE; and in triangle DEC, Case 1, find D)E and CE; then AD - AE — DE, and B C - BE — CE. Limits. —If the two given sides AB and CD are parallel, as in Fig. 2, the problem is unlimited; for, if we draw any line, as CID', parallel to CD, the quadrilaterals AB C'D' and AB CD will be equiangular, and have the sides AB and C'D' respectively equal to AB and CD, and hence either of them will fulfil the conditions of the problem. PROBLEMr XII.-In a quadrilateral are given two angles, and the three including sides, to find the area. (See above figure, where the angles A and B and the sides DA, AB, and BC are given.) Construction. —Draw AB - to the second given side, and make the angles A and B -the given angles. Lay AD - the first given side, and B C -the third, and join CD, and AB CD will evidently be the figure required. Calculation.-In triangle ABE, by Prob. III. A. and Case 1, find double area, and sides AE and BE; then ED- ALE - DA, and EC - BE- B C. Also, in triangle ECD, Prob. I. A., find double area E CD; then A B CD — (2 ABE - 2 E CD). Limit.-When BC and AD are parallel, as in Figure 2, let fall the perpendicular BI, and in triangle ABI, Case 1, find BI; then area ABCD I BI x (AD + B C). 190 GEOMETRICAL ANALYSIS. PROBLEM XIII. —Having given the position of three consecutive zigzag lines, and the length of the second one, to draw a line making a given angle with the first side and cutting off equal triangles in the two given angles.2 [the angles FB C and B CI, and the side Givend BC, to draw'AD, making the angle BAD equal to a given angle, and so as to - G A }B cut off equal triangles ABE and E CD. Analysis. —Suppose AD to be the required line, making BAD the given angle and cutting off the equal triangles ABE and ECD. Produce the first and third sides, BA and DC, to meet in F. To the equal triangles ABE and ECD, add the figure AECF, and we have the triangles FCB and FDA equal. Draw CG parallel to DA; then the triangle FCG is similar to FDA. Now, we must construct the triangle FDA similar to FCG, and equal to FCB, they having the same altitude. Construction.-(See Prob. V. A.) Having drawn the three sides FiVB, BC, and CI, making BC its given length, and FB1C and BC1 the given angles, and produced the sides BF and IC to meet in F, draw CG (Prob. XVIII. "Triangles," etc.), making the angle BGC-the given angle BAD; find a mean proportional FlY between the bases FB and FG; make FA -FH, and draw AD parallel to CG, which will be the line required. Demonstration.-The angle BAD-BGC the given angle. By Prob. V. A., the triangles FDA and FBC are equal. Take the quadrilateral AE CF from each, and we have the triangles ABE and ECD equal. Q. E. D. Calculation.-l. In triangle FBC, Case 1, find FB and FC. In triangle FGC, Case 1, find FG and GC; then FA —FH /FXB X FCG, and AB = FB- FA. 2. In similar triangles FGC and FAD, we have-FG: IFA:: FC: FD; then CD —FD -FC. Also, FG: FA:: GC: AD. Now, in triangle ABE, Case 1, find AE and EB, and by Prob. III. A., the area of ABE -area CED; then CE —BC —EB, and DE PROBLEMS IN RELATION TO AREAS. 191 AD - AE, and we know all the sides and angles, and the area of each triangle. NoTE. —When CI is parallel to FB, bisect BCin E, and through Edraw AED (Prob. XVIII. "Triangles," etc.), making the angle BAD = the given angle; then BA will be equal to CD, and the triangles ABE and ECD equal in every respect. PROBLEM XIV.-To divide a given triangle into two parts having a given ratio to each other, by a line drawn parallel to one of the sides. c /\ D i" n(the sides and angles of the triangle ABC, (BDGC:: m: n. LI Analysis.-Divide AB in F so that AF may be to FB in the ratio of m:n (IV. Prob. 1*). Join CF; then ACF and B CF are to each other as m to n, and hence A GD —A CF, and BDG C - B CF. Wherefore we have to construct a triangle A GD similar to ACB and equal to A CF, they having the same altitude. Construction.-(See Prob. V. A.) Find AE a mean proportional between the bases AB and AF, make AD-AE, and draw DG parallel to BC, and it will be the division line required. Denzonstration.-By IV. 25,t ABC: ADG:: AB2: AD2:: A B2: AB x AF:: AB: A:: AB C: A CE. Hence ADG - A CF. Take each from ACB, and we have BDGC B CF. Therefore we have ADG: BDGC: ACE: BCF:: AF: FB::m: n by construction. Q. E. D. Calculation.-By construction, m: n:: AF: FB. By composition, m+ n: m:: (AF + FB) AB:AF x AB. Also, m+-n: m - n n:: AB BFE - x AB. AndAD-AE AB x A ABA b. Then BD AB - AD, and AB: A C:: D:A G, A B: BC::AD:DG, and CG —AC-AG. Now, areaADJlG In + To x area AB C, and area BDG C x area AB C. rn + n Scholium.-If on AB, BD were laid equal to BE, and DG drawn _,:: VI.- 10n t VI.r 19. 192 GEOMETRICAL ANALYSIS. parallel to A C, the triangle would be divided in the given ratio by a line parallel to A C. If the dividing line is to be parallel to AB, the semicircle must be described on one of the other sides, AC or BC. When the triangle is to be divided into equal parts, AB must be bisected in F. PROBLEM XV.-To divide a given triangle into two parts which shall have a given ratio to each other by a line which shall make a given angle with the base. c the triangle AB C, and the angle ADE, G Given to draw DE so that the ratio of ADE lF (to EDBC shall be as m to n. Analysis.-Suppose DE to make the given angle with AB, and to divide the triangle ABC so that ADE: EDB C:: mv: n. Divide the base AB in F so that AF: FB:: me: n, and join CF; then (IV. 6, cor.*) ACF:FCB::AF: FB::n:n. Hence ACF and FCB are equivalent to the required areas ADE and EDBC. Parallel to ED draw CG; then angle AGC z angle ADE, and we have to construct a triangle similar to A CG and equal to A CF, both having the same altitude, by Prob. V. A., thus: Construction.-Through C (Prob. XVIII., "Triangles," etc.) draw CG, making with AB, produced if necessary, the angle AGC.the given angle ADE. Make AD —AH- a mean proportional between the bases AG and AF, and draw DE parallel to CG, and it will be the division line required. Demonstration.-By I. 20, cor. 3,t the angle ADE angle AGC the given angle by construction. By Prob. V. A., A CG: AED:: AG2': AD2:: AG2': AG x AF:: AG: AF:: A CG:A CF. Hence AED - A CF. Taking each from A CB, we have EDB C - FCB. Wherefore AED: EDB C:: A CF: FCB:: AF: FB:: m: n by construction. Q. E. D. Calculation. —We have AF —m X AB, and FB x m+ n m+n: VI. 1. t I. 29. PROBLEMS IN RELATION TO AREAS. 193 AB. Also, in triangle ACG, Case 1, find AG; then AD -AH=*C/AG x AF, and DB -AB-AD. In triangle ADE, Case 1, find AE and ED; then EC AC- AE; area AED - x fl + n area A CB, and area ED CB x area A CB. m +n Limit.-AH1 must be less than AB. PROBLEM XVI.-One side and the two adjacent angles of a quadrilateral figure being given, to lay off a given area by a line which shall make given angles with the two unknown sides. FIG. 1. FIG. 2. E A cII Given, the line p AB, the angles G I E BAFand ABE, A. < G and the angles FD C and E CD. Analysis.-Produce the lines EB and FA, either way, to meet inl P. Divide twice the given area, reduced to the same denominationki as AB, by AB, and on AB erect the perpendicular B~, in Fig. 1, and Al, in Fig. 2, equal to the quotient; draw IH parallel to AlI, and join HA; then (IV. 2*) the triangle ABH contains the given area, and is equal to the quadrilateral ABCD. Consequently the triangles PAH and PCCD are equal. In Fig. 1 draw HG, and in Fig. 2 AG, parallel to CD, meeting the lines EB and FA, produced(l if necessary. Then we have to construct a triangle PCD equal to PAH, and similar to PHG in Fig. 1 and PAG in Fig. 2, both havihi, the same altitude, by Prob. V. A., thus: Construction.-Having made the triangle ABH to contain tlt, given area, as in the analysis, and drawn HG (Fig. 1), making IliI, angle EHG — the given angle ECD, take PD - PL, a mean priportional between the bases PG and PA, and draw DC parallel to, HG. Then ABCD will contain the given area-ABH. The construction of Fig. 2 is precisely similar. Demonstration.-(Prob. V. A.) The triangles PCD and PA] X VI. 1. I 13 194 GEOMETRICAL ANALYSIS. are equal. Take the difference between each and PA B, and we have AB CD — ABH- the given area, and CD is parallel to fHG. FIG. 1. FIG. 2. E A Calculation.-In triangle PBA, Case 1, and Prob. III. A., find PB and PA, and area PBA; whence the area of the triangle PCD is known by adding or subtracting the given area to or from PBA. In triangle PDC, Prob. IV. A., find side DC, and thence, Case 1, the sides PD and PC, from which AD and BC are obtained by subtraction. Scholium I. —When CD is to be parallel to AB, HIG, in Fig. 1 must be drawn parallel to AB, and AG, Fig. 2, need not be drawn, G coinciding with B. Scholium. 2.-When AF and BE are parallel, bisect AH in 0, and through 0 draw a line CD, making the angle E CD - the given angle, and it will be the division line required. For the triangles AOD and HOC will be equal (I. 5*), and consequently ABCD_ ABH — the given area. PROBLEM XVII.-To divide any given quadrilateral figure into two parts which shall have a given ratio to each other, by a line which shall make given angles with the two sides it intersects. C Given the quadrilateral figure ABCD, Given and the angles AFE and XBEF, to IPa<- A 9 ]?1)t1 draw FE so as to divide AIB CD LB L [ l in the ratio of n: n. L Analysis.-Produce the sides CB and DA which the division line * I. 4. PROBLEMS IN RELATION TO AREAS. 195 FE is to cut, to meet in P. Then reduce the quadrilateral ABCD to an equivalent triangle by joining CA, drawwing BG parallel to CA, and joining CG. Then (IV. 2, cor. 2*) triangle AGC -ABC. Add A CD to each, and we have GC) — AB CD. Divide the base GD in-I so that GI may be to ID in the given ratio of m to n, and join CI; tlhen triangle GCI: ICD:: GI: ID:: m: n. Tence ABEF must equal GCI, and FE CD must equal ICD. Draw CH parallel to the division line EPF; then we have to make a triangle PEF equal to PCI and similar to PCH, both having the same altitude, by Prob. V. A., thus: Constru'tion. —Make the angle PCH —the given angle PEF, and let CH meet AD, produced if necessary, in H. Take PF — PL - a mean proportional between the bases PH and PI, and draw FE parallel to CI, and it will be the division line required. Demonstration. We have (Prob. V. A.) PCH: PEEI:: PI1T: PE~2:: PH2: PI x PI:: PH: P1:: P CI: PC1. Hence PEF — PC1. Take each from PCD, and we have ECD- ICD. But, by the analysis, AB CD- G CD. HIence, taking equals fi'rom equals, we have ABE — GCI. But GCI: ICD:: nm:z. Hence their equals ABEl and FEL CD are to each other as m to n, in the given ratio. Q. E. D. Calettlation. —In triangl(e PBA, Case 1, and Prob. III. A., find PB and PA, and area PA-1. In triangle PCD, find area. Th'enI area ABI(CD the difference of these areas; and since m: n:: ABLE: FEYCD, we have, by composition, m +:' m~::ABCD: AEI;, and In + it: - -:: A1 CD: FECD. Whlence we have --- _J + a x area A B CD - area of ABEF, and x area /AB CD -- area I-2 + it FE~CD. Also, area PEF-PAB + ABEF. Now, in triangle PEF, by Prob. V. A., find EE, and thence, Case 1, find PE and PF; then, by subtraction, find AF, FD, BE, and BEC. Scholium 1.-If the given quadrilateral is a parallelogram, divide either of the sides which the division line is to cut, in the ratio of mn to n, and through the point thus obtained draw a line parallel to one of the other sides, and this line will divide the given parallelogram into two parallelograms which have to each other the givenj ratio of mn to n. Now through the middle point of this dividing line draw * VI. 1. 196 GEOMETRICAL ANALYSIS. FE (Prob. XVIII., "Areas," etc.), making the angle AFEB=the given angle; then FE will be the division line required. Scholium 2.-If only the lines BC and AD which the division line is to cut are parallel, divide each of these lines in the ratio of m to n, the parts corresponding to m in each being adjacent to AB. Join the points of division, and this line will evidently divide the given trapezoid into two partial trapezoids s / A/ \,1 1 \ < which shall have to each other the ratio of m to n. Then, as before, through the middle point of this dividing line draw FE, vmlaking the angle AFE the given angle, and FE will be the division line required. Scholium 3. —If AB and CD are parallel, and BC and AD, which the division line is to cut, are not parallel, it is the same as this problem. Scholium 4.-When EFis to be parallel to CD, CH will coincide with CD, and the semicircle must be described on PD, and GD ~must be divided in I so that GI: ID: n: mn. PROBLEM XVIII.-To divide a given trapezium into four equal parts by two lines, one of which shall be parallel to the third side, and the other cut the second and fourth sides. K A the sides and angles of the f jA c: trapezium ABCD, the sides being numbered in H the order of the letter s, Given ( AB being the first, to divide \ it into four equal parts by G J A L the lines EF and IH, of which El is to be parallel to CD. N Analysis.-The same as in the preceding problems. Construction and demonstration. -Produce AB and D C to meet in K, and DA and CB to meet in G. By last problem, scholiumn 4, PROBLEMS IN RELATION TO AREAS. 197 divide ABCD into two equal parts by a line EF parallel to CD, thus: —Join CA, draw BJ parallel to it, and join CJ; then JCD - AB CD. Bisect JD in L, and join C L; then JCL —LCDZ AB CD. Now, make a triangle GEF equal to G CL, and similar to GCD (Prob. V. A.), by describing a semicircle on the greater base GD, erecting a perpendicular LN fr'om the end of the less base to meet the semicircle, joining G.~, making GE- GN, and drawing EF parallel to CD; then EF is one of the required division lines. For (IV. 25*) GOD: GFE:: GD2: GE2:: GD2: G2:: GD2: GD x GL:: GD: GL:: GCD: GCL. Hence GFE -GCL. Take each fromn GCD, and we have EFCD — L CD -- - AB CD. Talkingr these equals from the equals JCD and ABCD, we have ABFE JCL — ABCD - EFCD. In a similar manner, precisely, divide ABCD into two equal parts by the line PO parallel to AD, and divide ABFEE into two equal parts by the line h'Q parallel to AE. Join OQ, and produce it to meet AB in I, draw P1 parallel to IQO, and join III, which will be the other division line required. For triangle 1HO - IPO. Add AIOD to each, and we have AIHD- APOD - AB CD. Now, putting Z at the point where the division lines El7 and 1H intersect each other, we have (IV. 25t) IIO: IZQ:: I2: IQ2:: IPO: IRQ. But IHO-IPO. Hence IZQ- IQ. Add AIQE to each, and we have AIZE-ARQE - -AB1FE Y- ABCD. Hence IBFQ, FCHZ, and EZHD are each equal to I AB CD also, and the trapezium AB CD is divided into four equal parts, as required, by the lines EF and 1H. Calculation. —Through Q, parallel to AB, draw Q IY, meeting OP in Y; then PY-R Q. 1. In triangle GBA, Prob. III. A. and Case 1, ffind area GBA, and GB and GA; then G - GB + BC, and GD- GA + AD. 2. In triangle GCD, Prob. III. A., find area GCD; then area ABCD — GCD- GBA, ABE -EABCD, and GFE1- GBA ABFE. 3. In triangle GFE, Prob. IV. A., find EE, and thence, Case 1, GE and GF; then AE - GE - GA, ED AD - AE, BE- GF - GB, and FC- BC- BF. 4. In triangle KB C, Prob. III. A. and Case 1, find area KB C, and sides KB and KCC; then KA- KB + BA, and KD -- - C + CD. 5. In triangle K'PO (-KBC + I A]BCD) find (Prob. IV. A.) P O, and thence, Case 1, KP and KO; then AP —KA - -KP. VI VI. t VI. 19. 198 GEOMETRICAL ANALYSIS. 6. In triangle MBF, Prob. III. A. and Case 1, find area MBF, and sides MB and MEF; then MA M JIB + BA, and ME -MF + FE. 7. In triangle IR Q(- IIBF+ K ABCD) find RQ, and thence, Case 1, BIR and MQ; then AR A )lA \\MA -MR. Now, OY- OPFvrA/'a d RTQ, and QY —PR - AP - AR; A/T /v\7,/ then (IV. 18*) OY: YQ:: OP: PI. And AI —AP -P. Also, o IB - AB — AI, and KI — KB + G VI-, I BL P.By similar triangles KJO, J A L - D KPH, we have KI: KP:: KO KH. Then DH — zKD- KH, and CH\\ CD- DH. 8. In triangle KIll, Case 3, find angle KllH, and side IJt; then, by parallel lines, we have KI: MI:: IH: IZ, and KI: MI:: KH: MZ; and thence, by subtraction, we have ZH, FZ, and ZE, and then all the sides, angles, and area of each of the four equal parts are known. PROBLEM XIX.-Through a given point within a given trapezium to draw a line which will cut off a given area adjacent to the first side. cthe trapezium ABCD, of which A B is the first Pf GiN H e side, the angle ABP, and Given n G/iven \the distance BP, to draw G A Q D BR~PQ, cutting off the QF /L biQ D area ABR Q equal to L2. K L Analysis, construction, and demonstration.-Produce the sides DA and CB to meet in F, and through the given point P, parallel to FD, draw an indefinite line EPH. Now, to the line FE apply a parallelogram FENM, equivalent to the triangle& FBA, thus:Bisect FA in G, and join BG; then FBG - ABG -= BA. Join EG, draw BIM parallel to EG, and join EM; then EMG = EBG *VI. 4. PROBLEMS IN RELATION TO AREAS. 199 Add FE G to each, and we have triangle FEM- FBG = G FBA. Draw MN parallel to FE; then parallelogram FEIYN (I. 28, cor.*) 2 FEM- 2 FB G - triangle FBA. Next, to the line MN apply the parallelogram MN1THI to contain the given area. of ABRQ -L2, thus: —Perpendicular to MN draw MS - L, and also on MN lay M7lL L. Join JNS, draw L T parallel to NS, and TIt parallel to MN. Then, by similar triangles MNS and ML T, we have MN: MS(L):: MS(L):MT. Hence MNx MT(which is the area of the parallelogram MNEI(IV. 1, cor,-))L2 the given area. Lastly, we must draw QPR so as to make the triangle FRQ the parallelogram FEELt; then, since FENM -EBA, MNHUI will be equal to A BRQ. To do this, observe that of the three similar triangles 10Q, EPR, and POH, the first two, IOQ and EPR, are portions of the triangle FRQ, but not of the parallelogram FE-TI, while the third, POH, is part of the parallelogram, but not of the triangle; hence the first two, IOQ and EPR, must be equivalent to the third, POH, and (IV. 27, cor.t) their homologous sides EP, P1H, and IQ will form a right-antgled triangle, of which PHEis the hypothenuse. Wherefore, perpendicular to FD draw IK- EP, from K apply KQ -PH, and draw QOPR, which will be the line required. For, since the homologous sides IQ, EP, and PH form a right-angled triangle IKQ, by construction, we have (IV. 27, cor.~) IOQ + EPR -POH. To each add FEPOI, and we have FRQ — FEMI. Take the equals FBA -FENM from these, and we have ABR Q- -INEHI- L2 the given area to be cut off. Q. E. D. Calculation.-1. In the triangle FBA, Prob. III. A. and Case 1, find area FBA, and the sides FB and FA; then FG - l FA. Angle EBP _- EBA + ABP. 2. In triangle EBP, Case 1, find EB and EP; then FE=- FB -EB - =MN. 3. In similar triangles FEG, FBM, we have FE: FB::FG: FiM 4. In parallelogram MNHI, we have MN, the area, and angle M1 -F, to find MI (Prob. II. A., Note 2), thus:-sin lNI x MN: rad.::area MNlI: MI; then FIr FM + MI — EE, and PH= EH- EP. 5. In triangle IKQ, IQ =V/Q2 IIK- /PH- - EP2; then FQ=- F1 + IQ, and A Q FQ - FA, also QD - AD Y - A Q. 6. Area FQR-FBA + ABBQ, the given area. In triangle I. 34. t VI. 1. I Vr. 31. VI. 31. 200 GEOMETRICAL ANALYSIS. FQR, Prob. II. A., find FR, and thence, Case 3, find BQ; then BR B- FIR -F B, and / C B- B C- BR. 7. In triangle PBR, Case 3, find PB/; then P Q- RQ — PR. PROBLEM XX.-Having given the area of any triangle, and the sides and included angle of its inscribed parallelogram, to determine the triangle. o C [ Cthe area of the triangle ABC, and the D Given sides AF and AD and the angle A of Given (ts its inscribed parallelogram ADE lF, to A p L construct the triangle. Analysis. —Produce AF and AD indefinitely to P and 0. Divide the given area by AD, draw DK perpendicular to AD and equal to the quotient, and through K, parallel to AD or Elh, draw GLII; then the parallelogram ADCGH-the triangle A BC. Hence the triangle ELG, which is the part of the parallelogram withou.t the triangle, must be equivalent to the sunm of the two triangles HLB and D CE, which are the parts of the triangle without the parallelogram. Now, these three triangles are manifestly similar; hence (IV. 27, cor.*) the homologous sides HB, DE, and E G will form a right-angled triangle, of which EG is the hypothenuse. Whence this Construction.m-At II erect the perpendicular HI= — AF- DE; from I apply IB - CG or FEl, and through B and E draw the line BLE C, and ABC will be the triangle required - AD GII. Demozstration.-Since the homologous sides DE, EG, and HB of the similar triangles CDE, E GL, and LHIL are sucli as to form a right-angled triangle HIR, of which EG - the hypothenuse IB, we have (IV. 27, cor.t) LH1f + CDE- EGL. To each add the irregular figure AJrILED, and we have the triangle AB C'-the parallelogram AD G-I -the given area. Q.E. D. Calculation.-By Prob. II. A., Note 2, we have AD x sin A: given area ADGH:: rad.: AlH; then HII= AF, IB — FIT= AHAF, and HB R /Ib'2 — HlI. Whence AB zA-4 t+ MHI, FBIFIi + HB, and FB: A B:: FE: A C. Limit. —i can never be less than AF. If FHl is equal to _AF, * VI. 31. t VI. 31. PROBLEMS IN RELATION TO A'REAS. 201 AH will be the base of the required triangle, of which one side AB would equal twice AFi, and the other A C equal twice AD, and the area of the triangle ABC would be equal to twice ADEF. PROBLEM XXI. —Having given the difference of the radii of two concentric circles, to determine them such that the area of one shall be double that of the other. B/ ~ ~ \ln i,(lBD, the difference of the radii CB c G Given and CD, and the area of JBG to be \equal twice ADH. Analysis. —Let ADf and IB G represent the two circles, of which the area of IBG - 2 ADH, and BD - the given difference of their radii. Now, since the area of IBG- 2ADH, we have CB2 - 2 CD2. Make the angles BCA and CBA each equal to half a right angle; then CA - AB, and CB2'- CA2 + AB` 2 CYA2- 2 CD2. Hence CD= CA- AB. Lay BE -BA; then CE-_ BD. Erect EF perpendicular to CB; join BF; then, since BA BE, AFz FE - EC - BD, the given difference. Whence this Construction.-Draw an indefinite line, in which take CE - the given difference of the radii. Erect the perpendicular EF- EC, join CF, and produce it, making FA -_FE. Erect the perpendicular AB to meet the line CE, produced, in B; then AB — AC. Join BF, and with the centre C and radii CA and CB, respectively, describe the circles ADH and IBG, and they will be the circles required. Demonstration.-We have AB - A C; hence CB2 - AB2 + A C2 2 A C'. Wherefore the circle IBG - 2 ADH. Again, since AF FE, the triangles BAF and BEF are equal; hence BE_ BA _ CA - CD, and BD - CB - CD = CB — BE- CE — the given difference of the radii by construction. Calculation. - CF V/ CE2 + EF2 2 C/2 CEV/2, and CA CF +FA CF+ CE - CD. And CB - CD+ DB= CF+ 2 FA- CF + 2 BD. Scholiumr. —BA is a tangent to the circle ADH at the point A. r* 202 GEOMETRICAL ANALYSIS. PROBLENM XXII. —Having given the difference of the sides of two concentric squares, to determine them such that the area of one shall be double that of the other. C P D Ii E ( the difference between the sides AB and Given | GF, and the area of the square AB CD to I p< (be twice FGHE. B L I A Analysis. —Suppose ABCD and FGHE to be the concentric squares required. Draw the diagonals FH and GE, and produce them, and they will form the diagonals AC and BD, both pairs of diagonals intersecting in the common centre 0. Now, since ABCD —2FGHE, AB2 must equal 2 FG2. But AB' - A O" + B O2 - 2 B 02; hence FG - B O. Make B L — BO or FG, erect the perpendicular LM, and join BJl; then, in triangles BOJIF and BLIVI, since BO —BL, we have O1_- JIL- -AL- the given difference of the sides. Whence this Construction.-Draw AB and AD at right angles, and A C bisecting the angle BAD. Lay AL-=the given difference of the sides, erect the perpendicular LM, lay MIO - ML, and erect the perpendicular O.B; then AB will be the side of the larger sqoare, and BO or A 0 equal to the side of the smaller square. On AB describe the square ABCD, and produce AO and BO, forming thie diagonals AC and BD, and ABCD will be the larger square required. Now, take BL BO, bisect AL in I, through I, parallel to AD, draw IFEP, draw FG, GH, and HE parallel to AB, BC, and CD respectively; then FGHE will be the smaller required square. Demonstration.-It is evident that AB CD, FGTIE are concentric squares. We have to show that ABCD - 2 FGHE, and that AB - FG- AL - 2 AI 2 IF. In triangles BOM and BLM, since MO - ML by construction, we have BO= —BL. Also AB - BL+AL -BL +2AI —'BL + 2 IF. But IP, which is equal to ABl EP + IF + EP - El? + 2IF. Hence BL-EF, and AL —AB —BL- AB —EF; thatis, AB —FG -AL- 2A- 2 IF. Also, AB- B02 ~ A 02 2 B 2- 2 BL2'- 2 FG2- 2 EFi2; hence area AB CD - 2 FGI1E. Q. E. D. Calculation. —AM-I V'AL2 + LP[2 - AL,/2. Also, AO = PROBLEMS IN RELATION TO AREAS. 203 AtJflMO-AM+AL- EF- GF. And AB13BL+AL_ AO + AL - GF+ AL. PROBLEM XXIII.-Having given two paralielograms with a common vertical angle, it is required to draw a line fiom the remote extremity of a side forming the common angle, cutting the other side of that parallelogram, and both the sides produced of the other parallelogram, so as to form two equal trapezoids. D M UC fthe parallelograms AB CD and EBGF, to draw the line CIL so A HI B-.Given. as to make the trapezoids ADCI L G F I and IEFL equal. Analysis.-Suppose CIL to be the required line, making AD CI -IEFL. Now, if from the larger parallelogram ABCD we cut off ADMIH — B GF, we shall have IB GL -IHM3TC. To each add IBC, and we have the triangle CGL-BIHJIC. Whence this Construction.-Take Dill such that AD: BE:: BG': DMi, and dlraw MI- parallel to AD; then (IV. 24 and 2*) the parallelograms AD1JIH and BEFG are equal. Now, we have to make the triangle CGL- BHtJIGC, to do which, take CG: 2 BII":: CB: GL, and draw CIL; then (IV. 24 and 2t) CGL BHIIC, and CIL is the line required. Demonstration. —By construction, ADM -H -BEFG, and CGL BHtIMGC. From these last equals take the common part CBI, and we have IBGL- IHMC. To these, respectively, add the equals BEFG and ADMIT, and we have the trapezoids IEFL and AD CI equal. Q. E. D. Calculation.-By construction, AD: BE:: BG: DM; then BIT GCM — CD — DM. Also, CG: 2 BH: B C: GL; then FLZ FG + GL. And, by similar triangles CGL, CBI, we have CG: GL:: CB: BI; then EI- EB + BI, and A[I AB BI. In triangle CGL, Case 3, find CL; then CG: CB:: CL: Cl, and IL CL- CI. -VI. 14. t VI. 14. 204 GEOMETRICAL ANALYSIS. PROBLEM XXIV.-Given, the base of a plane triangle, and one angle at the base, to construct the triangle such that its area shall be double the area of its inscribed square. o F E' E (the base AB, the angle ABO, and the Given area of the triangle ABC to be double that of the inscribed square GIlIL. A L D I B Analysis.-Suppose the triangle constructed as in the adjacent figure. Let fall the perpendicular CPD; then, as the triangle ABC is equal to twice the square GHL, the four small triangles ALG, GPC, BIIH, and HPC must together be equal to the square. The first two of these triangles are similar, and the last two are similar; and when CP PD, the first two are equal, and each half of GPDL, and the last two are equal, and each half of PHID. Also, AL - GP - LD, and BI- HP - ID, and hence AB - 2 GH2 LI — 2 PD- CP. Whence this Construction. —Make AB-the given base, and angle ABO= the given angle. Erect the perpendicular BE — AB, draw ECF parallel to AB, join A C, let fall the perpendicular CD, and bisect it in P. Through P, parallel to AB, draw GPH, and through G and H, parallel to CD, draw GL and HI; then ABC will be the triangle required, and GHIL its inscribed square. Demonstratian.-Since CD= AB and CP 2- CD, we have GH AB 2 CD = GL; hence GHIL is a square. Also, area ABC - AB x 2 CD - 2 LI x LI- 2 LI2 -twice the area GHIL. Calculation.-1. In triangle BDC, Case 1, find BC. 2. In triangle ABC, Case 3, find AC. LI — IH- AB. Area ABC- AB x BE AB'; area GIL -- AB CZ- AB2. Scholium.-In any triangle in which a square is inscribed, we have CD: AB:: CP: GH. By composition, AB + CD: AB:: GH AB x CD AB' + CP (or CD): GH j- -$B (when CD AB)ABAB_ CD. 2 2 PROBLEMS IN RELATION TO AREAS. 205 PROBLEM XXV.-In a plane triangle are given its area, the vertical angle, and the length of the line drawn from the vertical angle to the middle of the base, to construct the triangle. C the area of ABC, the vertical angle ACB, Given and the line CD drawn from C to D, the B middle point of AB. Analysis. —Let A CE represent the required triangle. Complete the parallelogram ACBE. Perpendicular to CD draw GDF, meeting AG and BF, drawn parallel to the diagonal CGE, in G and F respectively. Now, since AD DB, the triangles ACD and DCB are equal, and each is equal to half the given area of A GB. Hence each of the perpendiculars DG and DF is equal to the given area of AB C divided by CD. Also, the diagonal CEG- 2 GD, the given line, and the angle CBE the supplement of the given angle A CB. Whence this Construction. —Draw GE -twice the given line, and through D, its middle point, draw the perpendicular GDF, making DG and DF each equal to the quotient of the given area of A CB divided by the given line CD. On CE (III. Prob. 16*) describe the circular segment CBE to contain the supplement of the given angle ACB. Draw FB and AG parallel to CE, join AC, CGB, BE, and EA; then will ACB be the required triangle. The demonstration is evident from the analysis. Calculation.-On CE let fall the perpendicular BI, and join E and B with 0, the centre of the arc CBE. Then, supposing the circle completed, it is evident (III. 18, and cor. 4t) the angle EOD the given angle A GCB. 1. In triangle ODE, Case 1, find OD and OEB- OB; then OF_ OD + DF. 2. In triangle OFB, Case 2, find FB -DI; then CI — CD + DI, and E -- ED — DI. 3. In triangle CBI, CB:-Sv/I- IB 2, and in triangle EIB, BE —=z/EI2 + IB2 A. C. 4. Also, in triangle DIB, DB — V/D12 + IB2, and AB - 2 DB. Scholium. —The triangle AEB is similar and equal to ACB in every respect, as is evident from the properties of a parallelogram. ~ III. 33. t III. 20 and 22. 206 GEOMETRICAL ANALYSIS. PROBLEM XXVI.-Given, three points in a right line, it is required to find a fourth point in that line, such that the rectangle of the distances of the fourth point from the first and second points shall be equal to the square of its distance from the third point. the three points A, B, and C in the right line A C, to P B A Given find a fourth point P such that PB x PA shall equal ~~~~D L~~P C2. F Analysis. —Suppose the thing done, and that, in the above figure, PB x PA- PC2. On AB describe a semicircle, to which (Prob. II. of "The Circle") draw the tangent PE; then (IV. 30*) PE2PB x PA —PC2 by the problem. Hence PE PC; and we have this Construction. —Erect the perpendicular CF- OB, join OF, and on it describe a semicircle FCEO, cutting the semicircle on AB in E. Draw OE and FPE; then will P be the point required. Demonstration.-In similar triangles PCF and PEO, since FC -EO, we have PC-PE, and PC2-PE2 -(IV. 30t) PA x PB. Q. E. D. Calculation.-The triangles OEF and OCF are equal in every respect; hence FE- OC, and the angle COF- OFF. 1. In triangle OCF, Case 2, find angle COFz_ POF; then angle OPE - OFP + POF- 2 COF. 2. In triangle OPE, Case 1, find OP and PE- PC; then PB -BC- CP, and PA- PB + BA. Scholium. —If we wish CP2 to be to PB x PA in a given ratio, say as m2 to n2, take n: m:: OE: CF. - x OE, then proceed as n above. For, z: n:: CF: OE:: (by similar triangles) CP: PEz n n2 - x CP. Hence - x CP2 PE2 -PB x PA, and we have m2: m mn2 n2: CP2:PB X PA. - III. 36. t III. 36. PROBLEMS IN RELATION TO AREAS. 207 PROBLEM XXVII.-To divide a given line into two parts such that the square of one part shall be equal to the rectangle contained by the other part, and a given line. D the lines CB and BA, to find the Given point P in CB such that PB2 CP x BA. C P Q 0 A Analysis.-Let the given lines CB and BA be placed contiguous, forming the straight line CA, and suppose P to be the required point. Bisect BA in 0. Then, since PB2- CP x BA, we have CP: PB:: PB: BA. By composition, CP + PB: PB:: PB + BA: BA; that is, CB: PB:: PA: BA. Hence CB x BA - PA x PB -(P OB) x (PO-OB) -XOB)_ P02- OB2. Whence this Construction.-On CA describe a semicircle, erect the perpendicular BD, bisect BA in 0, join OD, and make OP - OD; then P will be the point required. Demonstration.-By analysis and construction, CB x BA - BD2 D 02 _ OB2 - Po2 _ OB2 -(PO + OB) x (PO - OB) - PA x PB. Hence CB: PB:: PA: BA. By division, CB - PB: PB:: PA -- BA: BA; that is, CP: PB:: PB: BA. Hence PB2- CP x BA. Q. E. D. Calculation.-Join OD; then OP - OD V- /BD2 + B02 - */CBX BA+ ( AB)2. And PB -OP-BO, and CP GCB-. PB. PROBLEM XXVIII.-In a plane triangle are given the area, an angle at the base, and a line from the vertex to the middle of the base, to determine the triangle E g =F rthe area of ABC, the angle A' / \}t BAG, and the line CD from Given the vertex to the middle of the X = H G B 0 base, to construct the triangle ~~A B L ABC. Analysis.-Suppose ABC to be the required triangle. On CD erect the perpendicular CE to meet AE drawn parallel to CD; then CE is known, being equal to the double area of ADC divided by 208 GEOMETRICAL ANALYSIS. GD, or equal to the given area of ABC divided by CD. The angle DAC is also known. Whence this Construction.-Draw CD equal to the given bisecting line, on it (III. Prob. 16*) describe a circular segment, of which the centre is 0, to contain the given angle; on CD erect A' F the perpendicular CE-= the given area divided by CD; draw EA'A parallel G \ B to GD; join CA and AD; produce the A )D latter until DB -AD, and join CB; then A CB will be the required triangle. Denmonstration.-Since AD - DB, area A CB 2 A CD G CD x CE — the given area by construction. Q. E. D. Calculation.-Join OA, OC, and OD; on CD let fall the perpendiculars OF and AG, and draw OH parallel to GD; then AG= E C. 1. In triangle OCF, CF- CD, and angle COF-the given angle BAG; find, Case 1, OC, and OF - HG; then AH- A G 1G. Also, OH-= /AO -2 AH-Z FG, and CG CF ~ FG, and DG - DF- FG. 2. CA=V — G2 +AG2, and AD=V/DG +AG2, and AB —z 2 AD. 3. In'triangle CAB, Case 3, find CB. Scholium.-If A' had been taken instead of A as the end of the base of the triangle at which the angle is given, and CA' and A'D had been joined, and the latter produced until DB' should equal A'D, and CB' been joined, we should have had another triangle A'CB', fulfilling all the conditions of the problem, in which CA' wo uld - AD, AID - A C, and A'B' = 2 A C. * III. 33. PROBLEMS IN RELATION TO AREAS. 209 PROBLEM XXIX.-Given, four consecutive sides of a survey in length and position, it is required to draw a line from the remote extremity of the fourth side to cut the second and first sides such that the triangle formed by the first part of this line, the fourth side, and the adjacent segment of the third side, shall be equal to the quadrilateral formed by the remaining part of said line, the second side, and the adjacent segments of the first and third sides. O L F A H,:-\ [in length and position, the four lines \H>\/ /->H LA, AB, B C, and CD, to draw the \\G B a/ Givent line DGF such that the triangle DCG shall be equal to the quadrilateral AB GF. Analysis.-Suppose the figure constructed so that DCG ABGF. Producing DC and AL to meet in 0, and adding OFGC to each of these equal quantities, we have the triangle ODF- the quadrilateral OAB C, a given quantity. Whence this Construction. —First reduce the quadrilateral OAB C to a triangle, thus:-Join OB, draw AH parallel to OB, and join OH; then OHB - OAB. Add OCB to each, and we have OHC- -OABC. Now, it remains to make the triangle ODF= the triangle OHC. To do this, join HD, draw CS parallel to HD, and SF parallel to OD, and draw DGFJ the line required. Demonstration.-Join DS; then (IV. 2, cor. 2*) ODF- ODS OCS + SCD - OCS + S'CHz- OHC - OAB C, as shown in the construction. From the first and last of these equals take the common part OFGC, and we have DCG-G-ABGF. Q.E. D. Calculation. —1. In the quadrilateral OABC, we have all the angles, and the sides AB and BC, to find the area, and the sides CO andAO; then OD -- DC+ CO. 2. In the triangle ODF, we have the area — OABC, the angle O, and the side OD, to find OF (Proh. II. A.), and thence the side DF, and the angle ODF; then AF, FL, and LO are known. 3. In the triangle CDG, Case 1, find CG, DG, and the area of C DG AB GE; then B G and GF are known by subtraction. - VI. 1. 14 210 GEOMETRICAL ANALYSIS. PROBLEM XXX.-To construct a lune that shall be equivalent in area to a given isosceles right-angled triangle. B (the isosceles right-angled triangle // E \ Given FACB, to construct a lune AEBF equivalent to the triangle A CB. D C A Analysis.-On the hypothenuse AB of the given isosceles rightangled triangle A CB describe the semicircle AFPB, and with centre C describe the quadrantal are AEB. Now, since the area of the lune AEBF is to be equal to the area of the triangle A CB, by adding the segment AEBA to each, we have the area of the semicircle ABF- the area of the quadrant CAEB. Whence this Construction.-With centre C, and radius CA or CB, describe the quadrantal arc AEB, and on AB describe the semicircle AFB; then AEBF will be the lune required. Demnonstration.-Complete the semicircle ABD, and join BD; then AB - BD. Now, since AB2 - AD2, the semicircle AFB = I the semicircle ABD - the quadrant A CB. Take the' segment AEBA from each, and we have the lune AEBF — the triangle ACB. Q.E. D. Scholium.-A lune can be constructed equal in area to any rectilinear figure by reducing the figure to an equivalent triangle, then forming an isosceles right-angled triangle equivalent to this triangle, and, lastly, constructing the lune as in the problem, equivalent to this right-angled triangle. VU1I LIb Y Z5l DEMONSTRATION OF SOME THEOREMS. THEOREM I.-The lines drawn from each of the three angles of a plane triangle to the middle of the opposite side, intersect in the same point; also, the distance from the middle of either side to this point is one-third of the distance to the opposite angle. A [ the triangle ABC, and D, E, F - Given the middle points of the sides; then AD, BF, and CE intersect at a common point 0. (; C( D B I Demonstration.-Let ABC be a triangle, D, E, and F the middle points of the sides BC, AB, and A C respectively; then, if AD, BE, and CF be joined, they will intersect in a common point 0. Produce BC both ways, making CG and BI each equal to BC, and join AG, AI; then DG-DI, and (IV. 16*) AG is parallel to CE, and AI to BF. Also, DB 1 DI, and DC — DG. Now, in the triangle ADI, whose sides are cut by the parallel BF, since DB 3 D], we have the distance from D to the point in which BF cuts AD equal to IDA. Also, in the triangle ADG, whose sides are cut by the parallel CE, since D C D G, we have the distance from D to the point in which CE cuts AD equal to - DA. Hence BF and CE cut AD in the same point, which call O; then all the three bisecting lines intersect in the common point 0, and DO DA. Also, join DF, FE, and DE; then 1F- AB, FE B C, and DE A C. Now, in similar triangles COB, EOF, since FE - ABC, we have FO= —OB —FB, and EO-1OC —1EC. Q. E. D. NOTE.-See Scholium 3, Theorem VII. Corollary 1. —Since AD bisects BC, it bisects all lines parallel to BC; hence the centre of gravity of the triangle AB C is in the line VI. 2. (211) 212 DEMONSTRATION AD. Also, since CE bisects AB, it bisects all lines parallel to AB; wherefore the centre of gravity of the triangle AB C is in the line CE. Consequently, as the centre of gravity is ill both the lines AD and CE, it must be at their intersection 0, and AO - AD, CO - CE, and BO - 3 BF. Corollary 2.-If the points F, E, and D) be joined, the lines will be respectively \ be half the length of the side in the triangle ABC to which it is parallel, and these lines will divide the given triangle into four equal triangles, all similar to each other and to the whole triangle. Moreover, each side of the triangle DEF is the base of two equal rhombuses, which have their upper bases halves of the side to which it is parallel in the triangle ABC, and each rhombus is half of ABC. THEOREM II.-If a straight line be bisected and produced to any point, the rectangle of the whole line thus produced and the part produced, with the square of half the line bisected, are together equal to the square of the line which is made up of the half and the part produced.* (Let AB be bisected in C, and produced to D; -A / B D then will AD x DB + CB2- CD2, or CB' ( CD2 AD x DB. Demonstration.-With centre C, and radius CB or CA, describe a circle, to which draw the tangent DE by describing on DC a semicircle DEC, and join CE; the.n (IV. 30t) AD x DB -DE'. To each add CB2 - CE2, and we have AD X DB + CB - DE' + CE' = CD2, or CB' - CD2 —AD x DB. Q. E. D. - This theorem is the Sixth Proposition of the Second -Book of Euclid; but, as it is not in Legendre's Geometry, and the property is a very useful one, it is inserted here with a different demonstration from that given in Euclid. t III. 36. OF SOME THEOREMS. 213 THEOREM III: —If upon the radius of a quadrant a semicircle be described, any radius drawn in the quadrant will intercept equal arcs on the semicircle and on the quadrant. - (Draw the radii APG and Apg; then the | 9~ g arcs AP and EG are equal; also the arcs Pp and Gg, and pD and gD. A C D Demonstration.-Since the quadrantal are EGD and the semicircle APD are equal in length (V. 11*), we have, 90~: angle EAG:: arc EGD: arc EG. Also, 180~: angle A CP:: are APD: are AP; that is, 900: A CP (= ADP- EAG):: are APD (- are EGD) arc AP. Hence arc EG - are AP, and, consequently, arc DG arc DP, and, in like manner, the are A Pp = are EGg, and are Dp are Dg; then Pp — Gg. Q. E. D. THEOREM IV.-If in an isosceles triangle a circle be described, and a tangent be drawn to the circle parallel to the base, then the diameter of the circle will be a mean proportional between the base and the part of the tangent which is intercepted by the sides of the triangle.,F c -E!r \G {Let the tangent EFG be parallel to AB; then FD2- AB X EG. A D B Demonstration.-Let 0 be the centre of the circle; join OB and OG, and let fall the perpendiculars CFD and OH on the sides AB and BC respectively; then H will be the point of contact. Now, it is evident that the angles FGH and DOH are equal,; because they are respectively the supplements of the equal angles EGC and DBH. Hence their halves OGF and DOB are equal, and the rightangled triangles 0 GF and D OB are similar, and we have OF: I G:: BD: DO; whence 2 OF: 2 FG::2 BD: 2 D0; that is, FD: EG:: AB: FD. Hence FD2- AB x EG7. Q.E. D.. I. Supp. B. 214 DEMONSTRATION THEOREM V. ("Theorem of Pappus.")-In any triangle, any parallelograms described upon the two sides are together equivalent to a parallelogram described on the base, and limited by the opposite sides of the parallelograms described on the sides of the triangle, and by parallels to the line which joins the vertex of the triangle and their point of concourse. Prove this, and show how I. 41 of Euclid (or IV. 11, Legendre) can be deduced therefrom. FIG. 2. Fira. 1. B P I 0 L D/ I A - I C F A c Demonstration.-Let ABC (Fig. 1) be any triangle, ABED be any parallelogram described at pleasure on AB, and BCFG be any parallelogram described at pleasure on BC, and let DE and FG be produced to meet in P, which will be their point of concourse. Draw PBH, and draw AI and CL parallel thereto, meeting DE and FG in the points I and L, and join IL. Then, since the opposite sides of a parallelogram are equal, we have AI- BP, and CLz BP; hence AIJ- CL, and IL is parallel to A C, and AILC is a parallelogram described on the side AC, and we have to prove that AILC -ABED + BCFG. Now (IV. 1*), ABED - ABP — AHOI. Also, B CFG - B CLP - CL OH. Therefore, AILC - AHOI + CL OH-ABED + BCFG. Q. E. D. Next, let the triangle ABC (Fig. 2) be right-angled at B. On AB and BC describe the squares ABED and BCPG, and let the sides DE and FG be produced and meet in P, which will be the point of concourse. Draw PBH. Then the triangles CBH and PBE are evidently equiangular, and angle BHC BEP - a right angle, and hence PBH is perpendicular to A C. Draw AI and CL parallel to PBEH, meeting DE and FG in the points 1 and L, and join IL; then AI- BP z CL. Whence IL is parallel to A C, and AIL C is a rectangle. But the triangles ADI and AB C are evidently similar, and since AD-AB, AIl AC, and AILC is a square described on the hypothenuse A C, and we have to prove that AIL C * VI. 1. OF SOME THEOREMS. 215 ABED + B CFG. Now (IV. 1*), ABED - ABPI- AEOI. Also, BCFG - BCLP - CLOH. Hence AILC - AHOI + CLOH- ABED + B CFG. Q. E. D. THEOREM VI. (From Ladies' Diary.t) —If through any point within a triangle lines be drawn from the three angles to cut the opposite sides, the product of the three alternate segments, beginning at any angle, and going round in one direction, will be equal to the product of the three alternate segments, beginning at the same angle, and going round in the opposite direction. T C S F Through the given point E, in the triangleABC, draw the three lines AEB', BEGC', and CEA'. Prove that AA' x BB' x CC' — A C' x A A' B L CB' x BA'. Denlonstration.-Through C, parallel to AB, draw a line meeting AB' and BC', produced in S and T. Then, by similar triangles TCE and BA'E, we have TC: CE:: BA': A'E. By similar triangles SCE and AA'E, we have CE: CS:: A'E: AA'. By multiplying the corresponding terms, we have TC: CS:: BA': AA'. (A). Again, by similar triangles ABC' and CTC', we have TC: AB:: CC':AC'. Also, by similar triangles ABB' and SCB', we have AB: CS:: BB': CB'. By multiplying the corresponding terms, we have TC: CS:: BB' x CC': A C' X CB'. (B). Hence, the first couplet being the same in proportions A and B, we have BA':AA'::BB' X CC': AC' x CB'. Wherefore, by equating the products of the extremes and means, we have AA' x BB' x CC' — AC' X CB' x BA/. Q. E. D. THEOREM VII.-Conversely, when the product of the three alternate segments of the sides of a triangle, taken by going round in one direction, is equal to the product of the three alternate segments, beginning at the same place, and going round in the ofiposite direc-' VI. 1. t See Ladies' Mathematical Diary for 1735-6, Leybourn's Collection, vol. i. p. 246. 216 DEMONSTRATION tion, then the three lines drawn from the opposite angles and forming these segments pass through the same point. That is, if AA' x BB' x CC' - A C' x CB' x BA', then the three lines AB', BC', and CA' pass through the same point. Demonstration.-Let AB' and CA' be two of those lines intersecting at E; then, if the third line T 6C X does not pass through E, when the l' products of the alternate segments taken in opposite directions are equal, let it have another direction,, A' B as BF. Now, in this case we have, by the hypothesis, AA' x BB' x AF AA' x BB' CF- AF X CB' X BA', and hence -CB X Join E, and produce it to cut A C in C'; then, by Theorem VI., AA' x BB' A C' AA' x BB' X CC' AC' x CB' x BA'. Hence C B WhereCC't CB' X BA' AF AGC fore AcF- AC, and we have AF: AC':: CF: CC'. But AF is _CF C' C/ less than A C', being a part of it; hence CF must be less than its part CC', which is absurd. Whence the given relation can exist only when BF coincides with BE C', and the three lines pass through the same point. Q. E. D. Scholium 1. —When the three lines AB', BC', and CA' bisect the three angles of a triangle respectively, they pass through the same point. Demonstration.-Let the three lines AB', BC', and CA' bisect the angles A, B, and C respectively; then we have (IV. 17*) AC': CC':: AB: BC. CB': BB':: AC: AB. BA': AA':: B C: A C. Whence, by multiplying the corresponding terms of these proportions, we have A C' x CB' > BA': CC' x B B' x AA':: AB x AC x BC:AB x AC x BC:: I:. Hence AC' CB' x BA'- CC' x BB' x AA', and, by the theorem, the three lines pass through the same point. Scholium 2.-When the three lines AB', B C',' and CA' are perpendicular to the opposite sides respectively, they pass through the same point. VI. 3. OF SOME THEOREMS. 217 Demonstration.- In similar right-angled triangles A C'B and AA' C, we have A C': AA':: B Ct: CA'. In similar right-angled triangles BA' C and BBtA, we have BAt: BB':: CA': AB'. In similar right-angled triangles CB'A and CC'B, we have CB': CC':: AB: B C'. Multiplying these proportions, we have A C' x CB' x BA': AA' x BB' x CC':: AB' x CA' B C': AB' x CA' x BC':: 1: 1. Hence AA' x BB' x CC' -AC' x CB' x BA', and the three lines pass through the same point. (See Prob. XXVI., "Triangles," etc.) Scholium 3. -When the three lines AB', BC', and CA' bisect the opposite sides respectively, they pass through the same point. (See Prob. XI., "Triangles," etc.) Demonstration.-We here have AA' - BA', BB' - CB', and CC' =AC'; hence, by multiplying these three equations together, we have AA' x BB' x CC' - AC' x CB' x BA'; wherefore, by the theorem, the three lines pass through the same point. (See Theorem I.) THEOREM VIII., being an extension of the property of Theorem VI.-If a straight line be drawn to cut any two sides of a triangle, and the third side, one or all the sides being produced if necessary, thus dividing them into six segments (a prolonged side being taken as one segment, and its prolongation as anothor), then will the product of any three of these segments, which are not adjacent, be equal to the product of the other three. FIG. 1. FIG. 2. c s C s Fin. 3. j in either of the three adjacent figures, the triangle ABC, and the points A', A Given B', and C' taken in the A A sides, or the sides produced, in the same straight line, B' to prove that AA' x BB' X CC' - A C' X BA' x CB'. Demonstration.-Through C, parallel to AB, draw CS, meeting K 218 DEMONSTRATION the straight line in which the points A', B', and C' are situated, in S. Then, by similar triangles C'AA' and C' CS, we have A Ct: AA':: CC': CS. Also, by similar triangles A'tBB and CB'S, we have BA': BBt:: CS: CB'. By multiplying the corresponding terms, we have A C' x BA': AA' x BB':: CC': CB'; whence AA' x BB' x CC' -AC' x BA' x CB'. Q. E. D. THEOREM IX.-Conversely, if in the three sides of a triangle, or these sides produced, three points be taken such that the product of any three of the six segments not adjacent, into which the sides with their prolongations are divided, shall be equal to the product of the other three segments, then will the three points so taken be in the same straight line. FIGo. 1. FIG. 2. C, S C S B FIG. 3. C' S C B' Devmonstration.-We have to prove that if AAt X BB' X CCt-' AC' X BA' X CB', then the three points A', B', and C' will be in the same straight line. Parallel to AB draw CS to meet the line in which the two points Bt and Ct are situated, in S. Then, if this line does not cut AB in At (Fig. 1), suppose it to cut AB in some other point, as F. Then, by Theorem VIII., AF X BB' X Ct - BF X CB' X AC', AF CB' X A C' AA' whenceBF-B ct But from the given hypothesis BA B BB'> ( CC'n BA' CB' X AC' AF AA' T. Wherefore - t', and AF: AA.:: BF: BA. BB' x CC' B BA' OF SOME THEOREMS. 219 But AF is greater than AA', wherefore BF is greater than BA', a part greater than the whole, which is absurd. Wherefore the proportion AF: AA': BF: BA' can be true odnly when F coincides with A'. Hence the three points A', B', and C' are in the same straight line. The same may be proved in like manner by either of the other figures. NOTE.-It will be observed in each of the figures under Theorems VI. and VIII. that each side has two segments; and bearing in mind that when a side is produced the whole line thus formed is one segment, and the produced part is the other segment, then taking the sides in order, and beginning at any angle (as A), take the segment (AA/) terminating there of the first side; then the segment (BB') of the second side, terminating at B; then the segment (CC/) of the third side, terminating at C, and place their product equal to the product of the remaining three segments (BA' X CB' X AC'), and we have AA' X BB' X CC' -= BA' X CB' X AC'. By a little attention in this manner, the order in which the segments are to be taken to form the equation in any case, may be readily fixed on the mind of the student. THEOREM X.*-If a quadrilateral figure be in any manner divided into two quadrilateral figures, and diagonals be drawn to the whole quadrilaterals, and also to the two partial ones, then the points of intersection of these three pairs of diagonals will be in the same straight line. Required, the demlonstration. Fia. 1. Fie. 2. D E F F A, C' Demonstration.-We have to prove that the points G, H, and 1, which are the intersections of the three pairs of diagonals of the quadrilaterals ABED, ACED, and BC~EE, respectively, are in the same straight line. CASE 1.-When the lines DF and A C are parallel (Fig. 1). In similar triangles BIC and FIE, we have BI: F:: C: E~F; hence BI X Ei'- FI X BC. In similar triangles FHD and CHA, we have FH: AH:- DF: A C; hence TEL X A C -A- X DE: * See Silliman's Jolurnal of Science, vol. xxiii. p. 224, Second Series. See, also, Gillespie's Laed Surveying, p. 387. 220 DEMONSTRATION In similar triangles AGB and EGD, we have AG:EG:: BG: DG. By composition, we have AG: AB:: BG: BD; hence AG X BD _AE X BG. Since triangle ABG is cut by the line CLD, we have, Theorem VIII., AL X BG X DG —ACX BD X GL. Since triangle DEG is cut by the line AKF, we have, Theorem VIII., GKX DFX AE DKX EFX AG. Since triangle AGiK is cut by the line ILD, we have, Theorem VIII., AH X GL X DK AL X DG X HIK. Fi_. 1. 1. Fi. 2. D E P F A B c( -a By multiplying the terms on each side of these six equations together, and observing that the twelve terms EF, A C, AG, BD, AL, B C, DG, DF, ALE, AH, GL, and DKon one side cancel the same quantities on the other side, we have BIX FHX GK=IFX BG X ILK; whence, by Theorem IX., the points G, H, and I are in the same straight line, the triangle whose sides are cut being KBF. CASE 2. —When the sides DF and AC are not parallel (Fig. 2), let these sides be produced, and they will meet in some point, as P. Then, since triangle BFP is cut by the line CIE, we have, fheorem VIII., BI X EF X CP - B C X IF X EP. Since triangle AFP is cut by the line CID, we have, Theorem VIII., A C X FIX DP-AHX DF X CP. Since triangle ABG is cut by the line CLD, we have, Theorem VIII., AL XBC X DG ACX BDX GL. Since triangle DEG is cut by the line BAP, we have, Theorem VIII., BD X EP X AG -DP X AE X BG. Since triangle DE G is cut by the line AKF, we have, Theorem VIII., DFX AE X GK -DK X EF X AG. Since triangle A GK is cut by the line DLHE, we have, Theorem VIII., AH X GL X DK- AL X DG X HK. By multiplying the terms on each side of these six equations respectively together, and observing that the fifteen ternms EF, CP, OF SOME T-EOREMS. 221 AC, DP, AL, BC, DG, BD, EP, AC, DF, AE, AH, GL, DK, on one side, cancel the same quantities on the other side, we have BI X FR X GK- I' X BG X HK; whence, by Theorem IX., the three points G, H, and I are in the same straight line, the triangle whose sides are cut being KBF. NoTE.-The interesting property involved in this theorem was contained in a problem proposed for demonstration about the year 1830, in the last number of the Mathematical Diary, a monthly periodical, commenced by Prof. Robert Adrain in 1825, and afterwards conducted by James Ryan, of New York; but, the work being discontinued, no demonstration was published. THEOREM XI.* —If in each of the three sides of a plane triangle a point be taken at pleasure, and circles be described through each angular point of the triangle, and the points taken in the two sides forming that angle, these three circles will intersect one another at a common point. Required, the demonstration. e Let the points D, E, F be taken at pleasure in the three sides of the given triangle A ABC; then if the circles passing through / / the points A, D, Faand F, C, E intersect (/P each other in the point P, the circle passA ing through the points E, B, D will pass [ through P also. Demonstration.-Join DP, EP, and FP. Then, since ADPF is a quadrilateral inscribed in a circle, its opposite angles A and P are equal to two right angles (III. 18, cor. 4t). For the same reason, the opposite angles C and P of the qutadrilateral FCEP are equal to two right angles. Whence these four angles A, C, DPF, and FPE are equal to four right angles. But the three angles DPF, FPE, and DPE are equal to four right angles. Whence the angles A + C + DPF + FPE - DPF + FPE + DPE. Wherefore DPE A +- C. To each add the angle B, and we have DPE + B A + C + B -two right angles; consequently the points D, P, E, and B, which designate the quadrilateral DPEB, are in the * In seeking a mode of demonstration for Theorem XIV., following, which was sent to me some years ago by my valued cousin and mathematical correspondent, Benjamin Shoemaker, now of Germantown District, Philadelphia, who was my student in 1826, I discovered the interesting properties in Theorems XI., XII., and XIII. The same properties may, however, have been previously discovered by some one with whose writings I have not had the pleasure to meet. t III. 22. 222 DEMONSTRATION circumference of a circle, and hence the circumference of a circle which passes through the three points D, B, and E must pass through the point P, the intersection of the circles through the points A, D, F, and F, C, E. QE. D. Corollary.-The angle DPF- the sum of the angles B and C, the angle FPE-the sum of tile angles A and', and the angle DPE -the sum of the angles A and C. THEOREM XII.-If in each of the three sides of a plane triangle any point be taken at pleasure, and a circle be described through each angular point of the triangle and the two points taken in the sides that form that angle, the triangle formed by joining the centres of these three circles will be equiangular and similar to the first triangle. C f Prove that the triangle 1G11 formed \L/II\ I | by joining the centres of the circles ADPF, CFPE, and BDPE is H ~~ ir< E similar to the triangle ABC. By Theorem XI., the three circles pass A B l through the same point P. Demnonstration.-The are LP (III. 1]1*) is half the are FLP, and the arc PS is half the arc PSD; hence the whole arc LPS, which measures the angle H, is half the whole arc FPD. But (III. 18t) the angle A is measured by half the same arc FPD; hence the angle H of the triangle HIG is equal to the angle A of the triangle AB C. In like manner the are lRP is half the arc DRP, and the arc PO is half the arc POE; hence the arc RPO, which measures the angle G, is half the arc DPE. But the angle B is measured by half the arc DPE; therefore the angle G of the triangle HIG is equal to the angle B of the triangle ABC. So, also, are KP -1 FKP, and arc PQ- PQE; hence arc KPQ, which measures the angle I, is equal to half FPE. But 2 FPE measures the angle C. Therefore the angle I of the triangle HIG is equal to the angle C of the triangle ABC. Consequently, the two triangles TIG and ABC are equiangular and similar. Q. E. D. e III. 3. t III. 20. OF SOME THEOREMS. 223 THEOREM XIII.-If in each of the three sides of any plane triangle any point be taken at pleasure, all triangles whose sides pass through these points and terminate in the circular arcs passing through the angular points of the triangle and the points taken in the sides which form that angle, will be equiangular and similar. [If the points D, E, and F are taken _F/R+XX at pleasure in the sides of the triangle ABC, and the arcs FAD, / N mu/ // \ 2 X any other side, as A C, drawn throu,1h F. Ilence the area of the triangle N 1\/ E )B circle, which is the generating circle, \ y", supposed to move from and parallel to COD to B. The axis about I' which the revolution is made is called the fixed axe. The equation of the ellipse (see Prob. I. of "Ellipse," p. 253) is a2y2 + b2x2- a2b2, where a = OB, b- OC, x- =OEGE, y=- EF OG. Put p -- 3.1416, the area of a circle whose radius is 1. Now, the area of the generating circle FE11- EF2 X p y2 x p -(a2 - x2)p yf. Hence the solidity y'dx rb2 Cbpdx rb2px2dx~ 31bCpx3 b- (a2- _X2)pdx - b2pd b d b2pX bp (when x a 2 a2 3a2 a) ab2p - 3 abpp 2 the solidity of the senli-spheroid CODB. The whole spheroid ADBC-4ab2p. Now, putting the major axis A B AB- A, and the minor axis CD =B, we have a —, and b — substituting these values for a and b in the equation -4ab2p, it becomes A B2 3.1416 4 x 2 x - x 3.1416-A x B2 x 6- A x B2 x.5236- the solidity of the whole prolate spheroid ADBC, where the ellipse is revolved about AB. Case 2. —When the revolution is about the conjugate axis CD, then the ordinate GF describes the circle IGF, which is the generating circle, moving from and parallel to the circle AOB to C. Hence, as x is the moving line and y the space moved through, the solidity in this case will be fx'dy. The circle IF= GF2 x p- x2 p- (b2- y2)p =- x', and f x'dy a y (a2 _ y2py - faPb2 dY a2py- 3b2 -(wben y p-O b) a2bbp 2atbp 0 C - b) a2bp _ the solidity of the semi-spheroid CALCULUS. 239 A2 B ACBO. The whole spheroid ADBC 4a 2bp 4 x - x 3.1416 3.1416=A2 X B X A2 x B X.5236. In either case the 6 solidity of the whole spheroid equals the square of the revolving axe multiplied by the fixed axe, and the product multiplied by.5236, which is one-sixth of the area of a circle whose radius is 1. SOME ELEMENTARY PROBLEMS IN ANALYTI CAL GE OME TRY. INTRODUCTORY REMARKS. —In many mathematical investigations the operations are much simplified by designating a point, line, or surface by an equation, referring its position to two lines called axes, crossing each other at right angles. This method of representing points, lines, and surfaces by algebraic equations was first introduced by Des Cartes in his Geometry, published in the year 1637; but the principle has been extended and amplified by Lagrange, Delambre, Laplace, and many others, and it has now become one of the most powerful instruments in mathematical research. It is proposed to give here an idea of the principle, accompanied by a few practical problems as illustrations.:.- Let XX', YY', in the adjacent 2B A, t figure, represent the lines crossing each other at right angles in the 2 1 point 0, and let these lines be taken 1-'F 3 4 X as axes, to which it is proposed to refer points and lines on the same D plane. XX' is called the horizontal axis, YY' the vertical axis, and their point of intersection, 0, the origin of the axes, or simply the origin. The perpendicular distance, either way, from the vertical axis, measured on the horizontal axis, or on a line -parallel thereto, is called an abscissa; and such distance is generally represented by the letter x. The perpendicular distance, either way, from the horizontal axis, (240) OF THE POINT. 241 measured on the vertical axis, or on a line parallel thereto, is called an ordinate, and such distance is generally represented by the letter y. Thus, designating the four angles at the origin by 1, 2, 3, 4, as in the figure, and drawing AED, BFC parallel to the axis YY', then taking the point A, in angle 1, OE is the abscissa (x), and EA the ordinate (y). In like manner, of the point B, in angle 2, OF is the abscissa (x), and FB the ordinate (y). Of the point C, in angle 3, OF is the abscissa (x), and FC the ordinate (y). Of the point D, in angle 4, OE is the abscissa (x), and ED the ordinate (y). The abscissa and ordinate of any point are called the co-ordinates of that point. Thus, OE and EA are the co-ordinates of the point A, OF and FB are the co-ordinates of the point B, OF and FC are the co-ordinates of the point C, and OE and ED are the co-ordinates of the point D. The line XX' is called the axis of abscissas, or the axis of x; and the line YYt the axis of ordinates, or the axis of y. The axes are sometimes called diameters. Of the two co-ordinates to a point, the abscissa x always represents the perpendicular distance of the point from the axis YY', or the axis of y; and the ordinate y represents the perpendicular distance of the point from XX', or the axis of x. But these abscissas may measure either way, right or left, from the vertical axis YY'; and the ordinates may measure either way, up or down, fromt the horizontal axis XX'. Hence, in order to designate which way they are to be laid off or measured, mathematicians have adopted the notation of marking the abscissas which are to the right of the vertical axis YY' with the plus sign, +, and those to the left with the.minus sign, -. Also, those ordinates which are above the horizontal axis XX' are marked plus, and those below, minus. Hence, for any point in angle 1, x and y will both be plus. For a point in angle 2, x will be minus and y plus. For a point in angle 3, both x and y will be minus. In angle 4, x will be plus and y minus. For any point on OX or OX', y is 0. For any point on 0OYor OY', x is 0. At.- the origin O, x 0 and y 0. Also, if y 0, the point must be on the axis of x, and if x - 0, the point must be on the axis of y. Hence, if x and y are both given with their signs, the position of the point is determined, and such point is called the point x and y, or the point xy. Wherefore x _ a, and y - b, a and b representing known lines or distances, is the equation of a point. L 16 242 ANALYTICAL GEOMETRY. Example 1. —Determine the point whose equation is x = 9, y -12. The required point must be in angle 1, because both co-ordinates are plus. On OX lay OE- 9; through E, parallel to YY', draw a B G 4A line, and on it lay EA, above the | _____. 211 _______ axis XX', equal to 12, and A will be.I 2_ 1 _ the point required, whose equation.17 F. 3 t is x- OE - 9, y- -EA -- 12. c f D -Example 2. —Determine the point whose equation is x — 9, y — 12. The required point must be in angle 2, because x is minus, and y plus. On OX' lay OF-9; through F, parallel to YY', draw a line, and on it lay FB, above XX', -12; then B will be the point whose equation is x= OFP -9, and y -FB - +12. Example 3.-Determine the point whose equation is x — 9, y -— 12. This point must be in angle 3, because both co-ordinates are minus. On OX' lay OF= 9; through F, parallel to YY', draw a line, on which lay FC, below XX', - 12; then C will be the point whose equation is x OF — 9, y FC — 12. Example 4. —Determine the point whose equation is x 9, y 12. This point must be in angle 4, because x is plus, and y minus. On OX lay OE- 9; through E, parallel to YY', draw a line, and on it, below XX', lay ED - 12; then D will be the required point whose equation is x = OE _ 9, y- ED --- 12. Example 5.-Determine the point whose equation is x - ~9, y 0. This point must be on the axis of x, since y -- 0. When x - +9, y 0, it is E; when x -9, y -- 0, it is F. Example 6.-Determine the point whose equation is x 0, y -12. Since x- 0, this point must be on the axis of y. Lay OG, OH each-=-12: then the equation of the point G is x- 0, y- +12; and of the point H is x -0, y-= -12. OF THE STRAIGHT LINE. 243 Example 7.-Determine the point whose equation is x -- 0, y - 0. It is the origin O. Scholium.-The distance of any point (xy) from the origin is always _'x2 + y2, and, as the square of either a plus or a minus quantity is plus, this distance will always be the same, whatever tile sign of the co-ordinates, if they are of the same value. If D represent the distance, then D _'/x + y2 /'/OE2 + EA2 VOF2 + OB2 VOF2 + OC2 VOE2 + O01)D2 OA OB OC- OD. Hence A, B, C, and D are in the circumference of a circle whose radius is OA, OB, etc. OF THE STRAIGHT LINE. Every algebraic equation with two unknown quantities of the first degree represents a straight line on a plane, of which the unknown quantities are co-ordinates. Hence y -ax, and y- ax + b, are equations of a straight line, of which x and y are the coordinates; and if a and b are known, either integral or fractional, the line can be determined by construction.* Example 1. —Construct the equation y- 3x. Analysis.-Here, when x - 0, y - 0, and vice versa; consequently the line must pass through 0, the origin of the axes XX', YY'. Also, when x is plus, y is plus; and when x is minus, y is minus. Construction. —ln OX take any distance OE. Through E, parallel to YY', draw a line, and on it, above XX', lay EG 3 OE. Through O and G draw the line AOB, and it will be the line required, having its equation y-3x; that is, such that the ordinate of any p -r point in that line will be equal to three times the abscissa of that point. For, take any point in the line AB, as P, A r and parallel to YY' draw the ordinate PF. Then, in similar triangles OEG, OFF, we have OE: EG:: OF: FP. But EG - 3 0E; hence FP- 3 OF; that is, y 3x, 5" Definition.-The equation of a lidne on a plale. whether the line be straight or curved, is an algebraic expression involving two unknown quantities, and such that if one of the unltnown quantities is assumned of any specific magnitude, the corresponding value of the other unknown quantity is determnined front the algebraic expressiotn. 244 ANALYTICAL GEOMETRY. or the ordinate of the point P three times its abscissa. If x -2, the equation gives y - 6; if x 5, y? — 15, etc. In like nianner, take any other point, as p, and draw the ordinate pf parallel to the axis YY'. Then, in similar triangles OEG and B Ofp, we have OE: EG:: Of:fp. But )EG P 3 times OE; hence pf - 3 Of; I- /that is, y - 3x, or the ordinate of the CG ( point p -3 times its abscissa. If x — ], y —-3; if x -— 4, y -_-12, etc.,-. - -- x Scholium 1.-It is manifest that for all points in the part OB of the line h'B, x will be plus and y plus; and for all A' points in the part OA, x will be minus and y minus. Scholium 2.-By trigonometry, in triangle OEG we have OE: E G 3 EG:::rad.: tang EOG_ (when radius is 1) E 1-3. Take the log. of three, and increase its index by 10 for the radius, and we have the logarithmic tangent -10.477121, which gives the angle EOG or XOB -71~ 34', nearly. In the equation y - ax, a is the tangent of XOB to radius 1. Example 2.-Construct the equation 2y - -3x, or y - — x. Analysis.-In this equation, when x — 0, y- 0, and vice versa; hence the required line must pass through the point 0, the origin of the axes XX', YY'. We see, also, that when x is positive, y will be negative, and that when x is negative, y will be positive; and that x: y:: 2:-3. B Construclion. —In Ox take the abscissa OE = 2, and parallel to the axis \P YY' draw the ordinate EG below the axis XX', because y is negatinve whene x is positive, equal to 3. Through O X/. i d, fB i X and G draw the line AOB, which will be the line required, having y — 3x, or 2y- — 3x. For, take any point in that line, as, Pw P, of which the co-orcdinates are OP A and PF, we have, in similar triangles OPEG, OFP, OF(x):FP P(y):: OE(2):EG ( —3). Hence 2y— 3x, or y --— x, so that if x is known y is known. Also, if we OF THE STRAIGHT LINE. 245 take the point p, whose co-ordinates are Of and fp, we have, in similar triangles OQEG, Ofp, Of(x):fp (y):: OBE (2): EG (-3). 2x Hence 2y -3x, or y -—, as before. The same is true for each and every point in the line AOB. Scholium 1. —It is manifest that for all points in the part OA of the line AB, x will be positive and y negative, and for all points in the part OB, x will be negative and y positive. Scholium 2.-By trigonometry, in triangle OEG we have OE (2): EG (-3):rad. (1): tang.-angle EOG or AOX —. To obtain the value of this angle, from the logarithm of 3 subtract the logarithm of 2, and add 10, the logarithm of the tabular radius, to the index of the remainder, and we obtain 10.176091, for the tabular tangent of EOG or AOX, which gives, from the table, 56~ 19, nearly. The tangent being negative shows that the line OA falls below the axis of x. NOTE 1.-The co-efficient of x in the value of y ( —) is properly the tangent of the angle which the required line makes with the axis of x, measuring from OX upwards. When this tangent is negative, it gives the angle XOB, showing that OB falls to the left of OY. NOTE 2.-The student will bear in mind that the point in which a line cuts the axis of x has y — 0, and where it cuts the axis of y, it has x= O. At the origin x_ 0 and y= 0. Example 3.-Construct the equation 4x + 3y 24, or y-8 4x Analysis.-The line represented by the equation 4x + 3y 21 cannot pass through the origin 0, because, when x -0, y- 8, andl when y - 0, x - 6, so that, if either y or x becomes 0, the other has a numerical value. Now, at the point where the line cuts B Y the axis of x, y must 0; hence we P have, from the equation, x 6 for the abscissa of that point - OE, the distance from the origin at which the required line:' -— _,.____ must cut the axis of x. Hence E is a point in the required line. Also, at the point where the line cuts the axis of y, x 0, and then, from the equation, y A 8 OG, the ordinate of that point, or the distance from the origin 246 ANALYTICAL GEOMETRY. at which the required line must cut the axis of y. Hence G is another point in the required line. Construction.-Lay O.E = 6, and O G —8, and through the B y points E and G draw the line AEGB, XP kand it will be the one required. For, take any point in this line, as P, of which H4-G the co-ordinates are OFand FP. Tl'hrough f, SE S G draw GIl parallel to XX'; then GI: F.r - } — OF, and they are both negative, because they are to the left of the axis of y. Now, in similar triangles PHG and GOE, we have -GH or -OF: PH::: OE (6): OG (8). Whence 6PH= — 8 OF, or 8 OF 4 OF 4x PH- And PF — y-PH+ O — 6 3 3 4x -- + 8. Wherefore 3y — 4x + 24, or 3y + 4x = 24, the equation of the line AB. In like manner, if we take any other point, as p, of which the co-ordinates are Of and fp, we have Ef: -fp:: EO (6): OG (8); whence Ef - 6 - 4 Now, x - Of= OE + EJ- 6 4 8 4 4 Hence 4x = 24 3y, or 4xw 3y - 24, as before. Scholium 1. —It is manifest that, for any point in the part EA of the line AB, x will be positive and y negative, and for any point in the part GB of the line AB, x will be negative and y positive, while for any point in the part EG, both will be positive. Scholium 2.-The angle OEG or AE X is found as in Scholium 2 to last problem, its tangent being equal to the co-efficient of x in the value of y in the given equation, -= —; from the logarithm of 4 subtract the logarithm of 3, add 10 to the index of the remainder, and we obtain 10.124939 for the tabular tangent of OEG or AEX, which gives 530 8', or 1260 52', for the angle, and the tangent beings negative shows that the line EA falls below the axis of x, or it is properly the angle XEB. Example 4.-Construct the equation 3y + 5x' 27, or y =- 9 5x 3 In this equation, when x-0, y = 9 = OG, and when y = 0, x OF THE STRAIGHT LINE. 247 OE -— = =7 5.4. Hence make OE - 5.4, and OG - 9, and through BE and G draw the line AB, and it will be B y the line required, whose equation is 3y + 5x — =7. For, take any point, as P, whose co-ordinates are OF and FP, and draw GTH II a parallel to XX'; then GH'- OF- x. And by similar triangles GHP and E OG, we A F X have -Gil or -OF ( —): PH:: OEy -_): OG (9). Hence - x PH - --— 9x, A 45x 5w 5w or P1I — 27 -5 And y= PF-=PH~- OG -3 - 9; that is, 3y = -— 5x + 27, or 3y + 5x- 27, the given equation of the required line. In like manner, if we take the point p, whose co-ordinates are Of and fp, we have, in similar triangles Efp and EOG, Elf: -fp (~\..OEf27n 9Ff= —— y, or Ff= — y. ( ) (): OG (9). Hence y or E 5 5 5 27 3 Now, x - OE + lf-= -- -; whence 5x =27- 3y, or 3y + 5x 5 5 _ 27, the same as before. Exanmple 5. —Construct the equation 3y + 5wx =-27, or y- =-9 5x Here, when x — 0, y -— 9 —0 G, and B 5 P when y-=-0, x- - - 5 OE. kp Hence lay OE to the left of 0 5.4, and OG below 0- 9, and through G and E A'.... draw the line AB, and its equation will be 3y + 5x= —27. For, take any point, as C \I P, in AB, of which the co-ordinates are OF,., \ and FP. Then, in similar triangles GHP A and EOG, we have GH or OF(x): —HP:: EO - 2 ): OGC (-9). Hence + —7 x HP -9x, or HP = 5 45x 5x 5w -217 --. Now, y - P FP - FH + HP — 9 —; or 3y-27- w5x; that is, 3y + 5x- -27, the given equation of the required line. 248 ANALYTICAL GEOMETRY. In like manner, take any other point, as p, whose co-ordinates are Of and fp. In similar triangles Efp, EOG, we have — Ef:fp (y):: OEt ( — ):OG(-9). Hence+9Ef- - yorf Andy or Ef. And 2+ 3y x=Of=OE + 2Ef -- 7-, or 5x- -27 - 3y, or 3y + 5x 5 5 - -27, as before. Scholium 1. —In these two examples the B $ triangles.EOG are equal. Hence, if the two lines were constructed on the same axes, the angle OGE in onme would be equal A, f and alternate to OGE in the other, and the two lines would be parallel. Wherefore ax + by-= c, and ax + by= -c, are equations of two parallel lines constructed r a, on the same axes. a 5x Scholiumn 2. —In Example 4, y 9 - and in Example 5, y — 9. Hence, by Scholium 2 in Examples 2 and 3, the tangent of the angle which the required line makes with the axis of x, in each case, is —. To find its amount, fiom the logarithm of 5 deduct the logarithm of 3, add 10 to the index of the remainder, and we have 10.221849 for the tabular tangent, which gives 590 2' for the angle OEG or AEX below EX, or, more properly, 120~ 58' for the angle XEB. 5x Example 6.-Construct the equation 3y - 5x = 27, or y = w + 9. B 27 y' Here, when y 0, x 5 -- -5.4 OE, and when x =0, y 3 =9= OG.,P, f E/ _ _. Hence lay OE — 5.4 to the left of O, and OG /9 above 0, and through E and G draw the line AB, whose equation will be 3y- 5x= 27. /P For, taking the points P and p, and proceeding A y' as in Example 4, we have PH= - OF-3x, 5x andy-=FP=OG + PH= 9+-. Hence 3y = 27 +5x, or3y-5x 3 OF THE STRAIGHT LINE. 249 - 27, the given equation of the required line. Also, Ef-= -P 3y 27 3y and x= OE + Ef — 27 +. Hence 5x - 27 + 3y, or ~~-5 -5 5 3y- 5x = 27, as before. Example 7.-Construct the equation 3y — 5x= —27, or y 5x -_ - 9. In this equation, when y = 0, x - - 5.4 = OE, and when x 0, y -— 9= OG. Hence lay OE —5.4, and OG=9, below the origin 0, and through G and E draw the Y B line AB, and its equation will be 3y- 5x -27. For, taking the points P and p, and proceeding as in Example 5, we have EF - 3 FP 3y 27 f ) -= -P3, and x- OF= OE + EF= -, I G + ~Y, whence 5x-=-27 + 3y, or 3y - 5x 5 -27, the given equation of the required line. A Y — 5 x —Of 5x Also, we have Hp = - - And y -fp - OG + p 3 3 5x — 9 + 5x. Hence 3y = -27 + 5x, or 3y- 5x- -27, the same as before. Scholiumt 1.-In these two figures, also, the triangles EOG are equal. Hence, if the two lines AB had been constructed on the same axes, the angle OGE in one would have been equal and alternate to OGE in the other, and the two lines would have been parallel. If, therefore, the co-efficient of x and y in two equations are the sayme, or in the same ratio, the two lines which they represent will be parallel. Scholium 2. —In the general equation, ay - bx = -c; if x - 0, y = = the distance from the origin at which the required line a cuts the axis of y; and if y-0, w-=t-, the distance from the origin at which the required line cuts the axis of x, these distances being represented in the preceding figures by the lines OG and OE respectively. Hence, in the construction of a line fromn its equation of this form, we will always have OE: O::: -:: a: b; that L* b a 250 ANALYTICAL GEOMETRY. is, OE is to OG as the co-efficient of y is to the co-efficient of x, whatever c may be, the value of c only influencing the distances from the origin at which the required line cuts the axes, and not the angle it makes with them. Example 8.-Construct the equations 3y — 5x -- 27, or y 5x 5x + - -9; and -3y + 5x -42, or y - + 14, on the same 3 (3 axes. The first equation is constructed in the last figure, and gives the line AEB in the figure hereto annexed, where OE- 5.4 and OG 9. Y B' In the second equation -3y + 5x= B 42 -42, when y=O, x = — - 8.4= /l 42 __ ____ OE'; and when x =0, y = — = 14 X' E 3 OG'. Now, lay OE-5.4, and OG -9, and draw the line AGEB, which repreG sents the equation 3y - 5x - 27. A / Also, lay O.E' 8.4 to the left of 0, /:y' and OG'I 14, and through E' and G' A draw the line A'B', and it will represent the equation -3y + 5x - 42, proved as in Example 6. Now, the triangles OGE and OG'E' are equiangular; for, by'27 construction of AB, OE: OG:: 9::27: 45:: 3: 5; that is, OE: OG:: co-efficient of y: co-efficient of x in the first given equation. 42 Also, by construction of A'B', we have OE: OG':: -:14:: 42: 70:: 3:5; that is, OE': OG'::3:5::the co-efficient of y:to the co-efficient of x in the second given equation. Wherefore the angles OGE and OG'E' are equiangular, having the angle OGE- OGrE', and these are alternate angles, and hence A'B' is parallel to AB. Scholium 1.-Although these equations may be constructed on the same axes, they are not coincident equations; that is, the values of x and y of the unknown quantities cannot be the same in both equations. In the equation 3y - 5; - 27, if y = 6, x = 9. In the equation -3y + 5x -42, if y = 19, x- 3i and if x- 9, y= 29, instead of 6 as in the preceding equation, showing that the unknown quantities in each equation have independent integral values. Scholium 2.-If we have two independent equations involving OF THE STRAIGHT LINE. 251 two unknown quantities, and the two unknown quantities have each a common value in the two equations, the lines can be constructed, and the common values of the unknown quantities determined, as in the following examples: _Example 9.-Given, the equations 5x - 3y-= 27, and -2x + 5y =31, to find the values of x and y by construction. Analysis.-Since the co-efficients of r B x and y in these two equations have not the same ratio, the lines which p these equations represent are not parallel, and therefore they must meet and A_ cross each other at some point, and of c/E' E F this common point of the two lines there will be common co-ordinates which will be the values of x and y A common to both equations. Now, in the first equation, if x-0, y=-9; if y=O0, x- =5.4. In 31 31 the second equation, if x=0, y — 6.2; if y =, x- — 15.5. Therefore, lay OE = 5.4, and OG, below 0, — 9, and through G and E draw the line AB, and it will represent the equation 5x- 3y 27. Also, lay OE', - 15.5, to the left of 0, and OG', - 6.2, above O, and through E' and G' draw the line CD, intersecting the former in the point P. Then CD will be the line represented by -2x + 5y 31. Now, the co-ordinates OF and FP of the common point P will be the values of x and y respectively, common to the two equations. Measured from the same scale by which OE, OG, etc. were measured, we find OF- 12 —x, and FP= 11=y, which are the values found by resolving the equations algebraically. Thus we find algebra and geometry supporting each other, and leading alike to truth. Example 10.-Given, 10x _- 4y - 68, and -8x + 5y = -40, to find the values of x and y by construction. Ans. x=-10, y= 8. Example 11.-Given, 10x - 4y - 68, and -5x + 2y = -26, to find the values of x and y by construction. 252 ANALYTICAL GEOMETRY. Here the co-efficients of x and y in the two equations have the 4 2 same ratio; that is, —. Hence the two lines are parallel, 10 5 (Example 7, Scholium 1), and there can be no common values of x and y, as the lines, being parallel, can have no common point. Example 12. —Given, the equation x2 + y2 = 162 = r2, to construct the line represented by thte equation. Analysis. —Let XXt, YY' be the axes, Y and O the origin, as in the preceding ex/ 9 amples. Now, when y=O, x-=V162= +16 or -.16, which shows that the line XAt F'!< E BR x- which the equation represents cuts the axis of x at two points, distant 16 from the origin, F\. v~ on opposite sides of it. On OX and OX', therefore, lay 16 to B and A, and they will be two points in the required line. Also, when x= 0, y — =-= +16 or -16, showing that the line represented by the given equation cuts the axis of y in two points, one above and the other below the origin, and distant 16 from it. Therefore, on O Y, OY/ lay OC and OD each =16, and C and D will be two other points in the required line. Again, since the distance of any point in a line from the origin (scholium to Example 7 "Of the Point") is the square root of the sum of the squares of the co-ordinates of that point; that is, to /X2 + y2, and as this distance in the present example is constant, bein) for every point, as G, F, G' or Fl- f /x2+y2 V/2.56 or Vr2' it follows that every point in the required line is the samne distance from the origin, and that distance is V/256 = OB, OA, OC, OD 16. Hence, if a circle be described with OB or OC as radius, it will be the line represented by the equation x2 + y2 = 162- r2. Scholium 1. —The line represented by the equation x2 + y2 = r2 is a circle whose radius is r. Scholium 2. —Since y = -/256 - - 2, it is evident that for every value of x, as OE or OEt, there will be two equal values of y, as EG and EF, EG' and EF'. Also, since x — i:/256 - y2, for every value of y there will be two equal values of x. Hence each axis bisects all the double ordinates drawn parallel to the other axis. SECTION ON CURVES: SHOWING THE MANNER OF CONSTRUCTING SEVERAL OF THEM, AND SOME OF THEIR PROMINENT PROPERTIES. TIHE PARABOLA. PROBLEM I.-To construct a parabola and find its equation. Draw DR and GL at right angles. Take any point F in GL. Bisect FG in A. Parallel to DGR, through F, and any number of points in AL, as 1, 2, 3, 4, 5, etc., anyhow taken, draw lines, to which apply the distance G1, D G from F to a and a, on the parallel through the point 1; apply a A GC, from F to b and b, on the b rb parallel through F; apply G2, from F to e and c, on the par. allel through 2; apply G3, from F to d and d, on the parallel ga _ through 3, and so on for each 7 h point to H; and apply GH, B from F to B and C, on the L parallel through H; then the curve passing through the several points A, a, b, c, d, etc., on each side of A, will be-a parabola. It must be observed that the distance from G to any point on GL must be applied from F to the parallel through that point. The curve may be continued in like manner below H indefinitely. The line DGR is called the directrix, AL the axis,F the focus of the parabola, A the vertex; the line bFb, parallel to the directrix, through the focus, is called the latus rectum to the axis, or para( 253 ) 254 ON CURVES. mieter, and it is equal, as seen by the construction, to 2 GF- 4 AF 4 AG. A line from any point of the curve, perpendicular to the axis, is an ordinate to the curve at that point, and the distance from the vertex to the foot of the ordinate is the corresponding abscissa. D G R Thus, for the point b, bF is the ordinate, and AF the abA scissa. For the point 2, c2 is b t_ \b the ordinate, A2 the abscissa. c)/' Z - ~t X %d For the point B, BH is the er? / ~ \Xe ordinate, and AHthe abscissa. Vf// a2f Also BHC, h7h, are double _ _ — _ _ {g ordinates of the points B and h. hi/ 7 DLh Demonstration. -Draw a Br FKNC 0line from the focus F to the L extremity of any ordinate, as HB. Then, by the construction, FB GK-1 AH + AF, and FH=- AH - AF. Now, BH2 FB2- FI2 - (AH+ AF)'(AH-AF) - 4 AF x AH — 2FG x AH-bFb x AH-L x AH (L representing the latus rectum), which is the property of the parabola. L - bFb -_ 2 FG - 4 AF. Putting x any abscissa, as AH, y - its ordinate BH, a.-= latus rectum - 4 AF, we have y2 ax, which is the equation of the parabola, being true for any and every point of the curve. Corollary.-The square of any ordinate being equal to 4 AF multiplied by its abscissa, and 4 AF being constant, the abscissas are to each other as the squares of the ordinates; that is, AH: A4:: B.H2': e42; AH: AF: BH': bF2; A6: A3:: g62': d3', etc. THE PARABOLA. 255 PROBLEM II.-Given, a parabola, to find its axis, focus, directrix, latus rectum, etc., by construction. Let BA C be a parabola. Draw any two parallel chords YL and ab, and bisect them D --. R in V and c; through the points c and V draw the diameter P. I c VO. Perpendicular to EO draw the double ordinate El, which bisect in S. Through S, parallel to the diameter \ E VO, draw the axis GASL. Through E, the extremity of s t the diameter EO, parallel to L the chord YL, draw the tan- L gent E T, meeting the axis, produced, in T. Then TS is the subtangent to the point E. Draw ER perpendicular to E T; then ER is the normal and SR the subnormal to the point E. At E make the angle TEF- angle E TL; then F is the focus of the parabola. Make A G - AF, and perpendicular to GL draw DGR', and it will be the directrix. Since angle TEF was made equal to ETF, and TER is a right angle, a semicircle described with centre F and radius FT will pass through the points E and R. Hence FT, FE, and FR are all equal. Produce OE to any point, as Z; then angle ZE T- E TF (being alternate angles)- TEF; hence the tangent E T bisects the angle ZEF. By the construction of the parabola, GS-FE FT. Take GA from the first, and its equal AF from the second, and we have AS- A T. Hence the subtangent to a point (ST)- twice the abscissa (AS). Also, since.FG- 2AF, the tangent at P or Q, the extremity of the latus rectum, will meet the axis at G, its intersection with the directrix. To draw a tangent to any point, as E, make the subtangent ST_ twice the abscissa AS, and join TE, the tangent. Also, since GS ( FE)- FR, take FS from each, and we have SR- FG- FP FQ; hence the subnormal (SR) - half the latus rectum (PFQ). NoTE.-The area of the parabola, and the solidity of the paraboloid, were found in the article on " The Calculus." 256 ON CURVES. PROBLEM III. —To construct the parabola from its equation 2 ax. * Analysis.-Let 0 be the origin of the axes XX', YY'. It is evident from the d a~?f-f equation y2 -ax that if y= —, x -0;'a'/b'1 and if x= -0, y - 0. Hence the line represented by the equation passes through Li/ 1 1 2 3 4 5 the origin 0. It is evident, moreover, X that x cannot be negative, or minus, for, if it were, ax would be minus, and we sa \ I would have y -V/ —ax, which is imr/ dpossible. Hence the curve must lie wholly on the right of the axis YY'. B Also, since y-: z:ax, every value of x will give two values of y, numerically equal, but of different signs: hence the axis of x bisects all lines drawn parallel to the axis of y and terminated at both ends by the curve; and it bisects, also, the area of any portion of the parabola cut off by a line parallel to the axis of y. Example 1.-In the equation y2 ax, take x - 5, and it becomes y- 5x, or y = =t — 5x. By the table of square roots, annexed to this volume, if x -1, y — /5 - 2.24; if x - 2, y -_ V10 _ ~:3.16; if x -3, y 15 -: =3.87; if x -4, y =4.47; if x 5, y /25 — = 5; if a-6, y — 5.48, etc. Now, lay 01 1, 02 - 2, 03-3, 04 -4, etc., and through the points 1, 2, 3, 4, etc. draw lines parallel to the axis YY'; and, since all the values of y have both plus and minus signs, they must be laid both ways from the axis of x; lay la = 2.24, 2b — 3.16, 3c- 3.87, 4d - 4.47, etc., laying each up and down, and the curve passing through 0, a, b, c, d, etc., both ways from 0, will be the parabola required. Example 2.-In the equation of the parabola, take x - -3, and it becomes y2 -— 3x, or y -V/'-3x. Here the values of x must be minus in order to have 3x plus and y2 plus. Hence the curve lies wholly to the -eft of the axis YE'. * See remark to Example 2, Analytical Geometry. THE ELLIPSE. 257 If x- - y - /3 l.73. B, Y ".x -2, y /6- — 2.45. ".x -3, y 003. o 0. "x -4, y 3.46. "x -5, y -- 3. 87.2 " Xz —6, y zt4.24. "xz -, y i-+4.58. "xz - 8, y - -+4.90. "x = —9, y -+5.20. d Construction. - Let 0 be the c f origin of the axes XX', YY'; then proceed exactly as in Example 1, laying the values of x, 1, 2, 3, etc. to the left of YY', because all its values are necessarily negative, and hence the curve lies wholly to the left of YY'. NoTE.-By taking x -.5, 1.5, 2.5, 3.5, etc. in these examples, or other intervening quantities, intermediate points at pleasure may be obtained. THE ELLIPSE. PROBLEM I. —To construct the ellipse, having the axes given, and determine its equation. Construction. -Draw the axes AB and CD, bisecting each other at right d - d angles at the point 0, which will be b/ the centre of the ellipse. Draw BF a a parallel and equal to OC or OD, and A 2 3 B join OF. With the centre O and 11l radius O(A or OB, half the greater 1 g2 axis AB, describe the semicircle AEB, 3 E being the prolongation of CD.. 4 Divide the semicircle AEB into six equal parts by laying the radius from A to 4, and from B to 4, and fiom E, both ways, to 2 and 2; then bisect each of these equal parts, and each quadrant will be divided into six equal parts in the points 1, 2, 3, etc., and the whole semicircle into twelve. Through each of these points draw lines parallel to CE, cutting the axis AB in the points 1, 2, 3, etc., and the line OF in the points S, R, c, Q, and P, the point Q coinciding with 2 on the semicircle. Now, take the distance IS in the dividers, and lay it from each 5 which is on the axis, vp and down, to a and a, giving four points in the ellipse. 17 258 ON CURVES. Take the distance 2R, in like manner, and lay it from each 4 on the axis, utp and down, to b and b, giving four more points in the curve. In like manner, take the distance 3c (noting that c on OF is a point in the curve), and lay it from each 3 on the axis, up and down, to c and c. Lay the distance 4 Q from each 2 on the axis, up and down, to d and d; and lay 5P from each 1 on the axis, up and down, to e and e; 7 b lthen the curve passing through these a twenty-four points A, B, C, D, and 5 4 3 2 1 12 3 4"5 each of the points a, b, c, dc, e, taken.,4~j4 ~ S2 ]1- R- \in regular succession from A through, B, and D, will be an ellipse. It 43-. 3 ]~ a -- will be noticed that IS is just as far, a4,,5 a L measured on the semicircle, from OE as 5a is fiom B or A; that is, E5 B1- A1; also, 2R as far from OE as 4b is from B or A, so that the arcs E5-_B1 or Al, E4 - B2 or A2, E2 B4 or A4, etc. Also, since B4 and E2 are equal, each being composed of four equal parts of the semicircle, we have the line 04 equal to the line 2d4. Demonstration.-Taking any point in the curve, as d, in the quadrant BD, 2d, by construction, = 4Q. In similar triangles OBF and 04Q, we have 04 (or 2d4): 4Q (or 2Rd):: OB: BF. Hence (2d4): (2Bd)2:: OB2: BF' or OD2. But (IV. 23, cor.*) (2d4)2 — A2 x 2B; hence A2 x 2B: (2Rd)2:: OB2: OD2; that is, as the rectangle of the two abscissas of the point d (A2 x 2/B) is to the square of the ordinate (2Rd), so is the square of the semi-axis major (OB) to the square of the semi-axis minor (OD), which is the property of the ellipse. If OB - a, OD - b, 02 -x, and 2Rd - y - the ordinate, then A2 a + x, and 2B a —x, the two abscissas, and the above proportion becomes (a + x) x (a-x): y:: a2: b2, or a2x2: y2:: b2 b a2: b2; whence y2a 2(a2 -x2), or y b /a — x; or a2y2 + b2x2 a a a2b2, either of which is the equation of the ellipse. PROBLEM II. —Having an ellipse given, it is required to find its axes, foci, focal tangents, and to draw a tangent to any point in the ellipse. Construction. —In the given ellipse, draw any two'parallel chords, as EG and HI; bisect these in J and K, and through these points * VI. 13. THE ELLIPSE. 259 draw the line LJKJrI, which will be a diameter to the ellipse. Bisect LMin 0, which will be the centre of the ellipse. With the centre 0 and radius O L describe an arc, cutting the ellipse in L L and N; bisect theF arc LN in P, and A _ R A/ through P and 0O T draw APOB, which will be the major axis or principal diameter of the ellipse. Through 0 draw COD perpendicular to AB, and it will be the mi',nor axis. With C as a centre and radius OA or OB describe arcs, cutting the axis AB in the points F and F', which will be the twofoci of the ellipse. Through the centre 0, parallel to either of the chords EG and LH, draw WOX, and it will be the conjugate diameter to LOM, and LX and WX are called a pair of conjugate diameters, of which each one bisects all chords in the ellipse drawn parallel to the other, such chords being called double ordinates to the diameter that bisects them, and the two parts into which the diameter is divided are the corresponding abscissas. The lines through the extremities of either diameter, parallel to its conjugate or double ordinate, are tangents to the ellipse at those points. Thus, J1JQ, drawn through M, parallel to the diameter WOX or the chord HI, is a tangent to the ellipse at the point M. On AB, with O as the centre, describe a semicircle. From the focus F, perpendicular to AB, draw the line FT' V, draw VY a tangent to the semicircle, and join T' Y; then T' Y will be a tangent to the ellipse at I', and it is called the focal tangent. To draw a tangent to the ellipse at any given point, as Z, parallel to COD draw the double ordinate URZ, and produce it to meet the semicircle at S. At S draw a tangent ST to the semicircle, meeting the axis AB, produced, in T, and join LZ, which will be the tangent to the ellipse at Z. The tangents to the semicircle and ellipse, from points in the same ordinate, meet at the same point in the axis. Since, as shown in last problem, A R x RB: Z2:: A 02: OD2, and AlR x RBz1 B 2S2, we have RS2: BZ2:: A 01 OD')2, or RS: RZ::A'O: OD; that is, the ordinate of the circle at any point of the axis is to the ordinate of the ellipse at the same point, as AO to OD, or as the major axis to the minor axis, in a constant ratio. Hence these 260 ON CURVES. ordinates to the circle and ellipse will always be proportionate to each other, and we have SR: ZR:: VF: T'F, etc. PROBLEM IIL. —To find the area of an ellipse by the calculus. (See problem in the article'" Calculus.") Put AO a. O C - b, OR -x, R U- y. Now, to find the area of the quadrant BOCUB of the ellipse, RU is the mnoving line, supposed to commence at OC, and moving parallel to itself throu-hl the distance OB, to generate the surface of the quadrant BOOC). Now, by Problem I., the equation of the ellipse is y2 b 2(a' x'), olr y =-Va2 X2. Hence the area of the quadrant BOCUB = a fydx X- Va2- 2 X dx — as/a — X x dx. But, by Example 4 of the problem in the article "Calculus," f'/a2_-'2 X dx is tile area of a quadrant of a circle whose radius is a. Hence - a- 2_X b x dx- x the area of the quadrant of a circle whose radius is a. a That is, we have a: b:: area of quadrant of circle: area quadrant of ellipse:: area of circle: area of ellipse, the diameter of the circle being the major axis of the ellipse,- 2a. Then the area of the circle - 4a2 x.78539, and a:b::4a2 x.78539:atrea of ellipse 4ab x.78539 -2a x 2b x.78539 -AB x CD x.78539. That is, the a}rea of ant ellipse is equal to the product of the axes of the ellipse multiplied by the area of a circle whose diameter is'unity, or it is equal to the area of a circle whose diameter is a mealn proportional between the two axes, - /4ab. PROBLEM IV.-To construct the ellipse froml its equation y2 b2 - /aR -- x2, or a'2y2 + b2x2 - a262.* Analysis.-It is evident from the equation a 2y2 + b2x2- a2b2 that if y -O, x +a or -a, which shows that the curve represented by the equation cuts the axis of x in two points, distant a from the origin, one to the right, the other to the left of O. Also, if x- 0, y - +b or -b, showing that the line cuts the axis of y in two points, distant b froni the origin, one above and the other below. -- See remark to Example 11, Analytical Geometry. THE ELLIPSE. 261 Construction —Let XX', YY' be the axes, intersecting at the origin 0; in OX', lay OA - a; and in OX, lay OB -a; also in OY and 0 Y', lay 0 C and OD each equal to b, and A, C, B, and D will be four points in the curve, and AB and CD the two axes. Since x andy z, 2 124 enter into the equation of the A B curve in the second power only, each of them may be b D b _ either plus or minus, because the square in either case would b b be plus. Also, since y = -b/a2 - + - - /a 2, or a a -—?a2 — x2, it is evident that for every value of x (which may be a taken any quantity, plus or minus, numerically less than a) there will be two values of y, numerically equal, but of contrary signs. 2 2 o a every And since x - t-bs/b2 _ y 2 _+b N/b2 -2 Y, or- y, every b b b value of y (which may be taken any quantity, plus or minus, numerically less than b) will give two values of x, numerically equal, but of contrary signs. Hence each axis bisects every line drawn parallel to the other axis and having its extremities in the curve whose equation is a2y2 + b2"x2- a2b2. Now, taking any numerical values for a and b, as a -- 5, b -3, with these values we have y -- — t/25 -_x2. As x occurs in the equation only in the form of x2, and as the square of +x and of -x is the same, +x2, x may have either sign. Taking x, therefore, successively -1-, ~t2, -t3, etc., we have (using the table of square roots), If x - 1, y 3V — /24 - 2.94. x -2, y - =-/ 21 — 2.75. " x- -3, y- ~-'/16 — f2.40. "x = 4, y-~ -/ 9 = 1.80. " x — _5, y - 0. Now, on the axis AB, from 0, lay the values taken for x, 1, 2, 3, and 4, both ways, to the points 1, 2, 3, and 4, and through these several points, parallel to CD, draw lines. Then lay the value of y, when x = —1, from I on each side of O, up and down, to a and a, which will give four points in the ellipse. In like manner lay the 262 ON CURVES. values of y, when x -+-2, -3, ~:4, from the points 2, 3, and 4 respectively, on each side of 0, up and down, to b and b, c and c, d and d; then, through these several points A, C, B, D, and a, b, c, d, beginning at A, and taking the points in regular succession, draw the curve, which will be the ellipse required. NorE. —By taking smaller values for x, as -,, ax, 1 1-, 14-, 2, 21, etc., the number of points may be increased at pleasure, and the points in the curve being then nearer together, they can be connected with greater correspondence with the curve. This remark applies to all curves constructed bypoints from their equations. THE HYPERBOLA. PROBLEM I. —To construct a hyperbola, having the axes given, and determine its equation when referred to its asymptotes. Given, the axes AB and CD. Construction. -Draw AB equal to the first axis, which P bisect in 0, and through B, perpendicular to AB, draw a line, and on it lay BC and BD, YEszE each equal to half the second F I F' > axis. Draw OC and OD, and GH Cr ~Hproduce them indefinitely, as to Ki Kz BR and Q. Through B draw Ba J J' parallel to OQ, and Ba' par-.r allel to OR; then Oa, aC, aB, Oa', a'D, and a'B will all be equal. Lay either of these equal distances on the line OCR, from C to b, c, d, etc., and on ODQ, from D to b', c', d', etc. Through the points C, b, c, d, etc., parallel to OQ, draw lines, and through the points D, b', c', d', etc., parallel to OR, draw lines. Then through the points A and a draw a line, to meet the parallel through C, in E. Through A and C draw a line to meet the parallel, through b, in F; and, in like manner, from A, through each point b, c, d, etc., and a', D, b', c', d', etc. draw a line to meet the parallel at the next point, in G, H, K, etc. and in E', F', G, H', KI', etc.; and the curve passing through these points, B, E, F, G, El, etc., Br, E', F', G', HI, etc., will be a hyperbola, of which AB and CD are the axes, 0 the centre, and OR and OQ the asymptotes. THE HYPERBOLA. 263 Demonstration.-Through A, parallel to OD, draw a line to meet CO, produced, in P. Then OP Oa z- OC, and AP - OP - aB Oa'. Also, Oa2- 1 OC2 -- (OB2 + BC23. Now, taking any point, as F, in the curve, and comparing the similar triangles FCb and ACP, we have bF: AP:: bC: CP. But, since OP - Oa bC, awe have bO — CP. Hence bF: Oa':: Oa: bO; wherefore bF x bO Oa' x Oa — Oa2 (OB2 + BC2), which is the equation of the hyperbola referred to its asyzmptotes. The same can, in like manner, be proved of any other point, as G, H, etc., G', H', etc. We, hence, always have Oa2 - Oa x aB- OC x CE = Ob x bF — Oc x cG Od x dI, etc. -OD x DE'-Ob' X b'F', etc. Q. E. D. Corollary.-Since Oa x aB- OC x CE — Ob x bF= Oc x cG, and Ob is greater than OC, bF must be less than CE. For a like reason, since the factors on OG, as Oc, Od, etc., become continually greater, the other factors, cG, dH, eK, etc., must continually become less and less, so that the curve approaches nearer and nearer to the asymptote. Yet we see, by the mode of construction, the curve can never meet the asymptote, because the line drawn from A to a point in the asymptote must cross the asymptote, and be contiinued to meet the parallel through the next point, before it can form a point in the curve.* Scholium 1. —If the asymptotes OR and OQ be regarded as the axes of co-ordinates, and their intersection 0 the origin, and x represent the abscissa of any point in the curve, and y its corresponding ordinate, then of the point H, xz Od, y- dH; of the point G, x — Oc, y -cG; of the point E, x- OC, y z CE, etc. Now, putting OB-a, and BC- b, we have OC a2 + b2, Oa2 or aB2' 4 0C2 — 1(a2 + b2), and xy -Od x dH - Oa x aB Oa2 - (a2 + b2). Hence xy- -(a2 + b2) is the equation of the hyperbola referred to its asymptotes as axes of co-ordinates. ScholiumG 2.-In triangle OBC, Case 2, find angle BOC. Twice angle BOC- COD angle ROQ angle made by the axes of co-ordinates OR and OQ. When the axes of the hyperbola are equal, we have OB - BC, the angles COB and BOD are each half a right angle, and ROQ a right angle. In this case the hyperbola. It frequently perplexes the young student when he is told that one line can approach nearer and nearer to another, forever, without the possibility of ever arriving at it; but the principle may be illustrated by the familiar example of reducing ~ to a decimal==.33333, etc., every additional figure bringing the valve of the decimal nearer to ~; but if the line of figures were continued around the earth, it could never become equal to ~, although it is getting nearer to that value by every fiqure that is added. 264 ON CURVES. is called equilateral, and the axes and all the co-ordinates are rectangular. Scholium 3. — 0C — OB2 + B C2, Oa=- 0C - aB, O' _ 2 Oa, Ob = 3 Oa, Oc -4 Oa, Od -5 Oa, Oe — 6 Oa, etc., all of which are hence known. Now, since Oa2 — OC x CE, we have Oa2 Oa2 C Ey - -C 0 -0 Q Oa. Also, Ob x bFe — Oa2; hence bF Oa2 30a -- I Oa. In like manner, cG =4 Oa, df- 10 Oa, eK- 6 Oa. At the millionth point, the ordinate would be one-millionth of Oa; at the billionth point, one-billionth of Oa; and so on, the ordinate always having a value represented by figures, so that the curve can never meet the asymptote, mathematically regarded. PROBLEM 2. —Having given the equation of the hyperbola referred to its asymptotes, to pass to or determine its equation when referred to its axes or principal diameters. Given, Od X dH = aB2 = Oa2 - 0 OC2) -z(OB2 + BC2), to \4 q~vu4 prove that A Y x YB: YH2:: OB2: B C2:: AB2: CD2. Let the hyperbola be constructed as in Problem A, p I.; then AB is the first A\ ~axis, CD the second, and 0 the centre of the hyperbola. Make OS and OS' each equal to OC; then S -aG, H and S' will be the foci. ebb. r' b Through any points in the':'f hyperbola, as H and F, draw TH YH't and LFIIF' 1.y Q parallel to CD, and consequently perpendicular to ihe axis AX, t and 1 beingf the intersections of OQ, with TY and LI produced. Then HYH', PFif'T are double ordinates to the points Hand F respectively, HY and Fl the ordinates, AY and BY the abscissas of the point H, and Al and BI the abscissas of the point F. By equal triangles TdE, td'H', and A TY, AtY, we have THztH', and TYz- tY; hence HY- H' Y. By similar triangles Tild, CBa, and ZHt, CBa, we have TH: dH:: CB: Ba, and Ht: ZI (Od):: CB: Ca (Ba). By multiplying the corresponding terms of these proportions, we obtain TH x Ht: Od x dl":: CB2: Ba2 or Oa2. THE HYPERBOLA. 265 But Od x dH — Oa2; hence CB2 _ T x lt (t TY- YH) x (TY + YH) - TY2 — Y2. Also, Theorem II., OB2 - OY2 - A Y x YB. Again, by similar triangles OBC, O YT, we have 0 Y2: Ty2:: [OB: C2::0 Y 2 AYx YB: TY2 - YH2. By division, Y2: A Y YB:: TY2: YI12; or by inversion, AYx Yb: YH2::[0Y2: TY2]:: OB2: B C2:: AB: CD'. Q. E. D. Scholium 1.-Since Oa x aB - OC x CE - Ob x bF_- Oc x cG — Od x dH, etc., the parallelograms OdHZ, and all those under Oc, cG; Ob, bF; OC, CE, etc., are equal to one another, and each equal to the parallelogram OaBa', which is a property of the asymptotes of the hyperbola. Scholiumn 2.-It was shown that Tt x Ht- B C2. In the same way it may be shown that LF x Fl BC2, and so of any other point in the hyperbola. Hence these rectangles are all equal, and we have THI: LF:: Fl: lit.' But Hit is greater than Fl; hence TI is less than L', and consequently the fact that the curve continually approaches the asymptote is shown by another method. Scholium 3.-Putting OB - a, BC - b, and taking any point in the hyperbola, as H, whose ordin.ate is YH, and abscissas A Y and YB, and putting the ordinate YHzy, and O Y x, then the abscissa A Y — x + a, and YB- -xa, and A Y x YB- (x + a) x (x — a) x2 — a2; then the equation of the hyperbola, AYx YB: Y2P:: OB2: BC2, becomes x2 _ a2: y2:: a2: b, whence y2 (x2 a2), or ay - bx2X — a2b2, which is the analytical or algebraic equation of the hyperbola referred to its axes. PROBLEM III.-To construct the hyperbola from its equation referred to its axes, which, by last problem, is a2y2 -b2x2 -a2b2, or yS b2 or y2 (X- (x2 a2).* Analysis.-It is evident from the equation that if y = O, x= +a or — a, showing that the line represented by the equation intersects the axis of x at two points, distant a fiom.the origin, one to the right, the other to the left. Also, since when x -O, y V/ —b2, an imaginary quantity, the line cannot intersect the axis of y. Since x and y enter into the equation in the second power only, each of them may be either plus or minus, because the square in - See remark to Example 11, Analytical Geometry. M 266 ON CURVES. either case would be plus. It is evident from the equation y2= a-(x2- a2) that x may be taken any quantity not numerically less than a, plus or minus; and that for every value of x there will be two values of y, numerically equal, but of contrary signs. Also, for every value of y there will be two values of x, numerically equal, but of contrary signs. Hence each axis bisects every line drawn parallel to the other axis, having its extremities in the curve of the hyperbola, or the opposite hyperbolas. In the expression y = I2 (x2 a2), x cannot be less than a, otherwise we would have the square root of a minus quantity, which is impossible. Examjple 1.- Construction. - Taking a - -=5, b -3, the equation becomes 25y2 - 9 -2 = -225, or y -::/x7 - 25. IIF H Draw the axes XXl, YY', intersecting in 0, and take OA, OB, each — 5, on opposite sides of 0; then A and B will be the points in which the curve represented by the equation 25y2-_ 9y2- — 225 cuts the axis of x. Lav also Oc- +3, and OD, — 3. Now, as x cannot be less than 5, take x- -=6, x 7, x- t8, x =-9, etc., and lay these values of x both ways, from 0, on the axis XX', to 6, 7, 8, 9, etc., and through these points, respectively, draw lines parallel to YY'. If x = 6, y =-VE3/11 *.-~1.99, which lay on the parallels through 6, up and down, to a, a and a', a', and these will be points in the curve. THE HYPERBOLA. 267 If x = 7, y — i-324-::2.94, which lay on the parallels through 7, up and down, to b, b and b', b', and these will be other points in the curve. If x -8, y- 5/V39 —-+-3.71; if x-9, y=~ —4.49; if x —= 10, y ~ i-5.20, etc., which lay on the corresponding parallels through 8, 9, etc. to c, c, d, d, etc., and to c', c', d', d', etc., and then each of the curves passing through the points A, a, b, c, d, etc. both ways from A, and through B, a', b', c', d', etc. both ways from B, will be a hyperbola, and the two together are called opposite hyperbolas. O is the centre, AB the first axis, GD the second axis, and A and B are called the vertices of the hyperbolas. As equal ordinates, or values of y, correspond to equal abscissas, or values of x, the opposite hyperbolas are similar figures. Join BC, and lay the distance BC, from 0, on the axis of x, both ways, to F and F'; then these points are the foci of the hyperbola. Parallel to CD draw any double ordinate HGE, H'G'E'; then, of the point H, GH is the ordinate, and BG and AG the two corresponding abscissas; and of the point H/, G'H' is the ordinate, and AG', BG' the two corresponding abscissas. Scholium. —Join AD, and it will be parallel to BC, because the alternate angles OB C and OAD are equal. 268 ON CURVES. Example 2.-Determine the axis, and construct the hyperbola, whose equation is y2 - 3x2 - -5. -5 5 Multiply the given equation by -1 3,* and it becomes 1 x -3 3 5 5 25 5 5y2 5X2 x -5_ - Then a2- b —5, and a2 x b2 3 3 3 3 25 ---, and the equation is of the form a2y2- b2x2 — a2b2, which is the equation of a hyperbola. We have for the semi-axes, a 5.s —4~/15 — — 1.29, > \i/.c I b —/5 — 5-2.24. We have y2- 32 _ -5;' a / hence y-=- 5. 0C / Now, the least value that 3x2 can have is 5. 5 15 Then x2 - and 3 9' GI F'2 I B A a _, Gin+'13 -X 5- 34 / -+-1.29, which lay on the axis of x from O / D b \ I I to A and B, and these / ab' a a\l I will be the vertices of /) \ ~I I the opposite hyperboE' C' a,, Ce \EB las. On the axis of y lay also O C and OD each - V/5 _ — 2.24. If x-1-l, y-==~2.05; if x-=2, y =it2.65; if x-=2-, y ~3.71. It is evident that x cannot be numerically less than = -- ~V'15 = 1.29. Lay 01, 02, 03, etc., -the values of x, on XX', to the right of 0, and also to the left of O, to 1, 2, 3, etc., and through the points 1, 2, 3, parallel to CD, draw lines, on which parallels lay the corresponding values of y to a, b, c, etc., and to a', b', c', etc., and these give points in the required curve, any number of which may be obtained by assuming different values for x. Each of the curves — 5 e It will be seen that the multiplier -1 X is the absolute quantity -5 divided by the product of the co-efficients of y2 and x2 with their respective signs. THE HYPERBOLA. 269 passing through the points A, a, b, c, etc., both ways from A, and through B, a', b', c', etc., both ways from B, will be a hyperbola, of which the equation is y2 - 3x2 -5. The two together are called opposite hyperbolas. Join AD and BC, and they will be parallel. Lay BC on the axis of x, from 0, both ways, to F and F', which points will be the foci, AB is the first axis, and CD the second. If the double ordinates HGE and H'G'E' be drawn, HG is the ordinate and AG and BG the abscissas of the point H; and H'G' the ordinate and AG' and BG' the abscissas of the point H'. Example 3.-Determine the axes, and construct the hyperbola, whose equation is 2y2 - 4x"- 4. Multiply the given equation by 2 +4 =-~,* and we get y2 -y 2x2= -2, where a2 = — b22, a2b2 _-2, and the given equation is reduced to the form a2y2 b2x2 - -a2b2, which is the equation of a hyperbola. The first axis- Y 2/ —i, which shows that the curve _ cannot nleet the axis of x. If y- dd 0, x- -v; if x-O0, y0:V2 2 ~i1.41. Lay OA, OB each equal a 11 a to i; then AB - the first axis. Lay OC and CD each equal to V/2\ 1.41; then CD — the second axis. B A Now, y2 2x2 + 2; hence y ~iV-2x2 + 2. The least value of y a is when x -0; then y ~ —i/2' \ 1.41 OC or OD; x2- or do 4 jd 2 El GA q-~" f —~22 _/22 x - _1,V" - 4. Y/ If y 1-, x= —+-.35; if y, 2, x::-:1.00; if y - 2, x- =E1.46; if' y — 3, x =-+-1.87. Lay the values of y from 0, up and down, to 1, 2, 3, 4, etc., and through these points draw lines parallel to the axis XX', and on these parallels, from the points 1, 2, 3, 4, etc., lay the corresponding values of x to a, a; b, b; c, c; d, d, etc., and to a', a'; b', b'; c', c'; d', d', etc.; then each of the curves passing through the points C, a, b, c. See foot-note to last exan plc. 270 ON CURVES. d, etc., both ways from C, and through D, a', b', c', d', etc., both ways from D, will be a hyperbola, whose equation is 2y2 - 4x2 - 4. Join AC and DB, and they will be parallel. Lay AC on the axis of y, both ways from O, to F and I', which points will be the foci. C and D are the vertices of the opposite hyperbolas. If parallel to XX' the double ordinates HGE and H'G'E' be drawn, HiG will be the ordinate and CG and DG the abscissas of the point H; and H/'G the ordinate and CG' and D G' the abscissas of the point H'. THE CONCHOID OF NICOMEDES.* To construct the conchoid by points. b V 6 b B - b P Construction.-Draw PA V, and EF at right angles to it at A. In PA V take any point P below EAF, and V above EAF. Draw any number of lines, as Pb, Pb, etc., on each side of P V, and make extensions beyond EF from P, as ab, ab, etc., each equal to A V, and the curve passing through the point V, and the several points b on each side of V, will be the superior conchoid. Now, on the same lines drawn from P, lay the same distance equal to AV below EAF, as AV', ab', ab', etc., on each side of AP, and the curve passing through the point V', and the several points b' on each side of V', will be the inferior conchoid. The point P is called the pole; the line EAF, the directrix; V, the rertex of the superior conchoid; and V', of the inferior. Property 1.-The directrix EAF is an asymptote to both curves. For, let fall the perpendiculars bG, b'G' on the directrix, and bH, b'I_' on PI V. Then the triangles baG, PaA, taking Pab the line from the remote extremity of which the pe-rpendiculars bG and bil are let fall, are similar, and we have Pa: PA:: ab (Al V): bG. Hence * For a treatise on the interesting subject of' curves, see Prof. Leslie's Geomnetry of ('rre Lines, or, the "Application of Algebra. to the Doctrine of Curves," at the close of the second volunme o Botnnycastle's Algybro. THE CISSOID OF DIOCLES. 271 Pa x bG -PA x A V, a constant quantity. Wherefore Pa x bG must also be constant; and, consequently, since the factor Pa increases as the point of the curve gets further from PA V, on either side, so the other factor bG must decrease, and the curve approach nearer and nearer to EF without the possibility of ever meeting it. In like manner, using the similar triangles braG', and the same triangle PaA, it may be shown that EAF is an asymptote to the inferior conchoid. Property 2. —The equation of the superior conchoid B VB' is x' + 2bx3 + (b2 - a2 + y2) x2 -2a2bx = a2b2. The equation of the inferior conchoid CV'C' is x - 2bx3 + (b2 - a2 + y2) x2 + 2a2bx - a2b2, where x represents AH or bG, and y represents bH or AG in the superior conchoid, or their correspondents in the inferior. Scholium. —When A V is greater than AP, V' will fall below P, and the curve passing through C, P, V', C', and the several points b', as thus formed, is called the nodated conchoid, and the part P, VI, b', b', etc., is called a node. The student would find it interesting to construct the figure under this condition. THE CISSOID OF DIOCLES.* To construct the cissoid. A h g.f e d c B C D E G' tF I Construction.-Draw XY perpendicular to AB, the diameter of the given circle AVPBp, and from A to the line XY draw any number of lines AC, AD, AE, AG', etc., Ac, Ad, Ae, etc., on each side of AB, cutting the circumference of the circle in the points R, P, Q, etc., y, z, p, etc. Then lay the chord AR on CA from C to L; the chord AP on DA from D to 0; the chord AT on G'A from G' to K, and so proceed to lay the intercepted chord of every line on that line firom its intersection with XY towardls A, and the curve passing through A; and these several points L, 0, Q, K, etc. 9i See Bonnycastle's AlqlTehr,,vol. ii. p. 401, London edition, 1820. 272 ON CURVES. thus given successively from A, both ways, will be the cissoid of Diocles. A h g f e d c B C D E Gt H I The circle A VBp is called the generating circle; AB the axis of the tboo branches, which meet in a cuspj at A, and pass through the middle points Q and p of the two semicircles, drawing continually nearer and nearer to the directrix XY as it extends fiarther froml AB; and the directrix XY is hence their commzon asymptote. Draw P11 and O G parallel to XY; put AB -a, A G = x, GOy; then the equation of the curve is x3 -- (a —x)y2. THE QUADRATRIX OF DINOSTRATUS. Construction.-Let A VB be a semicircle of which the centre is C. At C, perpendicular to AB, draw a line, on which lay CAEl the diameter AB, and draw NAIL parallel to AB. Divide the N NML quadrant; B Y, and,] the -' --- radius C V, into any o but the sane numbi)er \. --— 9 _"~ __of eq.ual parts. sa.v 9, A\ \ in the poinlts 1, 2, 3, x \o etc., numnbering from \-7/ \ZX B on the quadllant KxQN ~'\\\ ~5/,j 7aj ld C on the radlius, I /\\,' V and continlue thle equal divisions, at.A5 c7 & pleasure, be,,ond V "? on the semicicele.. and 2' F\e above V, on CM. Then' from C to ai? number on the radius CV will be the same p)art of the radius C Y. that from B to the same number on the quadrant B V is of th, quadrant BBV; that is, C5: B5::1ad. CV: (qtuad. I V:: dia:lter THE QUADRATRIX OF DINOSTRATUS. 273 CM: semicircle B VA. Also, CT: B7:: rad. CV: quad. BV, and C14: B14:: CMi: BVA. Join C with each of the points 1, 2, 3, etc. on the semicircle, and parallel to AB, through the point I on the radius, draw a line to meet the line C1 in E. Through the point 2 on the radius draw a line parallel to AB to meet the line C2 in F. So, through each point on C1, successively, parallel to AB, draw a line to meet the line drawn from C to the point of the same number on the semicircle, in the points G, H, I, J, V, B, S, 0, etc., and the curve passing through tthese several points E, F, G, H, etc., taken in order, is called the quadratrix. The circle A VB completed is called the generating circle; the line NML is called the directrix, which is also, as will be readily seen from the mode of construction, the asymptote to the quadratrix. For, suppose there be taken on MC a distance Mp a thousandth or a ten-thousandth part of the distance C1, and on the semicircle Ap a like part of B1, join Cp, and produce it to meet a line parallel to MLN, through p, on MC, which will give a point in the curve, and this point will become more and more remote from Mi as the parts Mp and Ap are made smaller, and the limit is the extension of the two parallels CA and IMN, which will never meet. Therefore LiMN, produced, is the asymptote of the curve. Putting the radius CB - a, CD, the base of the quadratrix, = b, the are B5 -z, and C T y, the equation of the quadratrix is ay - bz. NOTE 1.-This curve obtained much notoriety from the fact that if it were possible to form it accurately by a simple geometrical operation, it would enable mathematicians to determine the rectification and quadrature of the circle. It was from this property that the curve was called the quadratrix. It would also afford a means of dividing a given angle, or given arc, into any number of equal parts or in any given ratio.* NOTE 2.-One principal difficulty in constructing the quadratrix is in finding the point D in CB so as to determine the base CD of the quadratrix. The most practical method I have been able to devise is to continue the semicircle below B, and also the line VCbelow AB. Then take B~ a half, a fourth, or as small a part as may be employed of B1, and lay it from B to -1 below B. On VC, produced to X, lay CQ-, a like part of C1, from C t'o A'. Join C and the points 4- and 1' on the arc, and through the p)ints 2 and 4' on VX, parallel to CB, draw the lines from 2 to d and fronm' to d', meeting the lines from C to 4- and 1' in d and d', which will be points in the curve, and in the curve continued below CB, and hence the curve must pass thrbough th-epoint D, and thus give CD, the base of the quadratrix, as accurately as can be obtained by mechanical means.: See Bonnyeastle's Algcbra, vol. ii. p. 403, London edition, 1820. 18 274 ON CURVES. THE LOGARITHMIC CURVE. Construction. M/ - -- Draw the line RQ, on which erect a perpendicutar AlH of any length. Iln RQ lay off from'fo I I 0 r A, both ways, S A' P h g f e d c b a A B C D E F G Q continued at pleasure, the abscissas AB, AC, AD, etc., Aa, Ab, Ac, etc., increasing in an arithmetical progression, having the intervening spaces AB, BC, CD, Aa, ab, be, etc. all equal; and then, parallel to AH, through the points B, C, D, etc., a, b, c, d, etc., draw lines BI, CJ, DK, etc., ai, bj, ck, etc., whose lengths shall be in geometrical progression, increasing from AH towards Q by a common multiplier or ratio, and decreasing from A towards R by a common divisor of the same value. In the present figure AH was taken - 12, the common difference = 5, and the ratio 14. Then the curve passing through the upper extremities of all these lines N, Ml, L, K, etc. will be the logarithmic curve. BI — 12 xI 1X 15, ai A 1H1 —-=12 14 —9.6, CJ = 15 x 1 - 18.75, bj 9.6 1.- 7.68, DK-= 18.75 x 11 23.4375, ck 6.144, EL 29.2969, dl - 4.9152, FM -- 36.621, em - 3.93216, GN= 47.776, fn = 3.14.57. By laying these respective distances on the lines indicated, thle points I, J, K, L, etc. in the curve are obtained, and the curve can readily be drawn through the upper extremities of all these lines, beginning at N, and taking M, L, K, J, I, H, i, j, k, etc. in succession. Scholium 1.- It is evident that the curve approaches nearer and nearer to RP without the possibility of ever arriving at it; for, however mnany timnes we may (livide the value of AHhby-l, or, which is the same thing, mlltiply it I)v -, itt must s/ill j /i'e somne value to lay on aperpen~dicular uboce RQ; htcnce the curve, though approaching THE LOGARITHMIC CURVE. 275 nearer and nearer, can never arrive at RQ, and RQ is the asymptote to the curve. Scholium 2.-Since the ordinates AH, BI, CJ, etc. are in geometrical progression, the square of any one is equal to the product of the two adjacent ones, or of any two lines equally distant from it. Thus, AH -- BI X ai =CJ x bj -DK x ck- EL x dl, etc. Scholium 3.-Putting r - ratio (in this case I ), and AH- a, we have BIz a. r; CJ= a. r2; DK- a. r3; EL a. r4, etc.; and a a a a ait -; bj 2-; c —a; dl --, etc. If now we commence the r r r r curve at H', where A'H' will be just equal to r,* and then lay off the arithmetical ratios, or common differences, A'P, Ph, hg, gf, fe, etc., and erect perpendiculars at these successive points, we have At'H - r; PpI' At I x r -- r2; hq -Pp' x r -r3; go = r4; fn - r5; em - r6, etc. Putting x to represent any number of these arithmetical differences from A' towards Q, calling A' one, P two, h three, etc., and putting y to represent the corresponding ordinate, we have y = rx, which is the equation of the logarithmic curve; and from the form of its equation, the logarithmic curve is sometimes called the exponential curve. A'IH' r Now, if we lay A'iS= —'P, then the ordinate Ss. r r r~ - 1; and beginning at,, the ordinates will be 1, r, r2, r3, r4, r5, etc. — 1, 1, ( 1)2, (l)'3, (1 ), (), ()5, etc.; that is, Ss- 1, A'H'4, Pp'- (1)2 1)3, go (1- )4, fn - (1)5, etc.: To find the distance AA' at which the ordinate A'H' of the curve will be equal to the ratio r, having AH and the arithmetical and geometrical ratios given, let x the number of arithmetical differences in AA', then, by the hypothesis, we have A'H' a (X+l1) - -=}. Hence ar ). Whence, taking the logarithm of each side of the rx log. r equation, we have log. a=-(zx+1) X log. r, or z-f-l log-. Taking, as in the log. 12 1.079181 example, a = 12, and r= I, we have x + 1- log. 1= 11.136 Hence log. 1.25 0.096910 x - 11.136 -1 - 10.136 = the number of arithmetical spaces between A and A'; and as each space in the figure is 5, we have AA' = 10. 136 X 5 = 50.68. 276 ON CURVES. THE SPIRAL OF ARCHIMEDES, OR EQUABLE SPIRAL. To construct the equable spiral. Construction. - Let the ~j33 — 98distance that the spiral gets from the centre C in one complete revolution be de. i g > A noted by r; then with radius CA - any multiple of r by a whole number, describe the 30. 6c4 s6Acircle AFBD, and draw the diameters A CB and D CFat 1\ }<<~ g'gig right angles. Divide each &1\ 5 \ a /@ ~quadrant into three equal parts, at the points X, G, LH, -, If, L, P, Q. Through these points draw diameters, and produce them both ways from the centre. On CE, from C, lay 1. of r to 1; on CG lay 2 of r to 2; on CFlay -, of r to 3, etc., in order, around to 12 on CA, which will be - r and the curve passing through the several points 1, 2, 3, etc. will be one revolution of the spiral of Archimedes. -Example.-Let r 12, and CA 2r -24. From C, on CE, lay 1; on CG, 2; on CF, 3, etc., when 12 will come on CA. Then, as before remarked, the curve passing through the points 1, 2, 3, etc. will be one revolution of the spiral. From the points 1, 2, 3, 4, 5, etc. lay on each of the radii a distance equal to r, around to A; then CA - 2r = 24. Through these points successively, beginning at 12, pfiss a curve through 13, 14, 15, etc., and it will give the second revolution of the spiral. In like manner, from these several points, beginning at 13, lay on the several radii, produced if necessary, distances each equal to r,,to 25, 26, 27, etc., and through these several points pass a curve, hreginning at A, and it will give the third revolution of the equable spiral. In the same way these successive revolutions may be continued at pleasure; the whole, whatever the number of revolutions, being called the spiral of Archimedes. With the centre C, and radius C12-r, describe any arc 12Y, which put -z. Put = —the whole circumference of that circle, and the corresponding ordcinate C4 -y; then it is evident that C4: CY THE LEMNISCATE. 277 or r:: arc 12Y: the whole circumference. That is, y r: z: 7r, rz whence y -- the equation of the spiral THE LEMNISCATE. To construct the lemniscate from the equation a2y2 — a2x2 - x4. Analysis. —When x - O0, y - 0. Hence the curve passes through the origin O. If y-= 0, we have \ 2. 2 x2- a2 or x- =-a, which shows that the curve cuts the axis of x in two points at the distance of a X 9 5d from the origin, one to the right, the other to the left of O. We have y- a —/. ] Z Taking a —10, If x-i 1 y=-JV-0/99 ~0.995. " x=- 2, y=- /96=zl. 96. " x = 3, y =- 9gl-= ~2.86. "x-4, y - -3.666. "x=- 5, y- ~4.33. "z x- i6, y = t4.8., " x -t-7, y - i 4.999. " x —~8, y=- i4.8. "x — ~-9, y- — 3.92. ", x-~ —10,y= 0. Construction.-Draw the axes XX', YY' at right angles, intersecting each other at 0, the origin. Lay OE, OF, each = a=-10, and E and F will be the points in which the curve cuts the axis of x besides the origin. On OX, OX' lay the successive values of x fromr 0, both ways, to the points 1, 2, 3, 4, 5, etc., through which points, parallel to YY', draw lines, and on these parallel lines from where they intersect the axis XX', lay the corresponding values of y, respectively, both ways, up and down, and the curve passing through F. 0, and E, and the remote extremities of these parallel lines in the order of their succession, will be the lemniscate, as in the figure. Commencing at 0, the curve 278 ON CURVES. extends through 0, G, E, H, 0,1, F, L, and back again to 0. AOB and COD are tangents to the curve. NOTE.-A straight line, parallel to the axis of x, will evidently cut the curve in four points, x being in the fourth power in the equation. y being in only the second power, a line parallel to the axis of y cuts the curve in only two points. In the ellipse and hyperbola, where both the co-ordinates x and y are in the second power in the equation, a straight line parallel to either axis will cut the curve in two points, regarding the complete hyperbola as consisting of the opposite hyperbolas. In the parabola, whose equation is y2 =ax, a straight line parallel to the axis of y will cut the curve in two points, but parallel to the axis of x in but one, x being in the first power only. See these three curves in the first part of this section. A TABLE OF SQUARE ROOTS OF NUMBERS FROM 1 TO 200, TO FACILITATE THE CONSTRUCTION OF CURVES BY POINTS FROM THE EQUATIONS. NOS. ROOTS. NOS. ROOTS. NOS. ROOTS. NOS. ROOTS. NOS. ROOTS. NOS. ROOTS. NOS. ROOTS. NOS. ROOTS. 1 1. 26 5.099 51 7.141 76 8.718 101 10.05 126 11.22 151 12.29 176 13.27 2 1.414 27 5.196 52 7.211 77 8.775 102 10.10 127 11.27 152 12.33 177 13.30 3 1.732 28 5.292 53 7.280 78 8.832 103 10.15 128 11 31 153 12.37 178 13.34 4 2. 29 5.385 54 7.348 79 8.888 104 10.20 129 11.36 154 12.41 179 13.38 5 2.236 30 5.477 55 7.416 80 8.944 105 10.25 130 11.40 155 12.45 180 13.42 6 2.449 31 5.568 56 7.483 81 9. 106 10.30 131 11.45 156 12.49 181 13.45 7 2.646 32 5.657 57 7.550 82 9.055 107 10.34 132 11.49 157 12.53 182 13.49 8 2.828 33 5.745 58 7.616 83 9.110 108 10.39 133 11.53 158 12.57 183 13.53 9 3. 34 5.831 59 7.681 84 9.165 109 10.44 134 11.58 159 12.61 184 13.56 10 3.162 35 5.916 60 7.746 85 9.220 110 10.49 135 11.62 160 1.2.65 185 13.60 11 3 317 36 6. 61 7.810 86 9.274 111 10.54 136 11.66 161 12.69 186 13.64 12 3.464 87 6.083 62 7.874 87 9.327 112 10.58 137 11.70 162 12.73 187 13.67 13 3.606 38 6.164 63 7.937 88 9.381 113 10.63 138 11.75 163 12.77 188 13.71 14 3.742 39 6.245 64 8 89 9.434 114 10.68 139 11.79 164 12.81 189 13.75 15 3.873 40 6.325 65 8.062 90 9.487 115 10.72 140 11.83 165 12.85 190 13.78 16 4. 41 6.403 66 8.124 91 9.539 116 10.77 141 11.87 166 12.88 191 13.82 17 4.123 42 6.481 67 8.185 92 9.592 117 10.82 142 11.92 167 12.92 192 13.86 18 4.243 43 6.557 68 8.246 93 9.644 118 10.86 143 11.96 168 12.96 193 13.89 19 4.359 44 6.633 69 8.307 94 9.695 119 10.91 144 12. 169 13. 194 13.93 20 4.472 45 6.708 70 8.367 95 9.747 120 10.95 145 12.04 170 13.04 195 13.96 21 4.583 46 6.782 71 8.426 96 9.798 121 11. 146 12.08 171 13.08 196 14. 22 4.690 47 6.856 72 8.485 97 9.849 122 11.05 147 12.12 172 13.11 197 14.04 23 4.796 48 6.928 73 8.544 98 9.899 123 11.09 148 12.17 173 18.15 198 14.07 24 4.899 49 7. 74 8.602 99 9.950 124 11.14 149 12.21 174 13.19 199 14.11 25 5. 50 7.071 75 8.660 100 10. 125 11.18 150 12.25 175 13.231 200 14.14