.^\- '•-^^ i^i I^^J^^ -TV^ r) Digitized by the Internet Arciiive in 2008 with funding from IVIicrosoft Corporation http://www-archive.org/details/bostongeometryOOwentrich TEXT-BOOK OF GEOMETEY REVISED EDITION. BY G. A. WENTWORTH, A.M., AUTHOR OF A SERIES OF TEXT-BOOKS IN MATHEMATICS. BOSTON, U.S.A.: GINN & COMPANY, PUBLTSHEKS. 1893. 0/ n Entered, according to Act of Congress, in the yearr 1888, by G. A. WENTWORTH, In the OflSce of the Librarian of Congress, at Washington. All Rights Rbsebvbd. Typography by J. 8. Cushing & Co., Boston, U.S.A. Prbsswork by Ginn & Co.. Boston, U.S.A. PREFACE. ni T'OST persons do not possess, and do not easily acquire, the power -^^-*- of abstraction requisite for apprehending geometrical concep- tions, and for keeping in mind the successive steps of a continuous argument. Hence, with a very large proportion of beginners in Geom- etry, it depends mainly upon the form in which the subject is pre- sented whether they pursue the study with indifference, not to say aversion, or with increasing interest and pleasure. In compiling the present treatise, the author has kept this fact con- stantly in view. All unnecessary discussions and scholia have been avoided; and such methods have been adopted as experience and attentive observation, combined with repeated trials, have shown to be most readily comprehended. No attempt has been made to render more intelligible the simple notions of position, magnitude, and direc- tion, which every child derives from observation ; but it is believed that these notions have been limited and defined with mathematical precision. A few symbols, which stand for words and not for operations, have been used, but these are of so great utility in giving style and per- spicuity to the demonstrations that no apology seems necessary for their introduction. Great pains have been taken to make the page attractive. The figures are large and distinct, and are placed in the middle of the page, so that they fall directly under the eye in immediate connec- tion with the corresponding text. The given lines of the figures are full lines, the lines employed as aids in the demonstrations are short- dotted, and the resulting lines are long-dotted. IV PREFACE. In each proposition a concise statement of what is given is printed in one kind of type, of what is required in another, and the demon- stration in still another. The reason for each step is indicated in email type between that step and the one following, thus preventing the necessity of interrupting the process of the argument by referring to a previous section. The number of the section, however, on which the reason depends is placed at the side of the page. The constituent parts of the propositions are carefully marked. Moreover, each distinct assertion in the demonstrations and each particular direction in the construction of the figures, begins a new line; and in no case is it necessary to turn the page in reading a demonstration. This arrangement presents obvious advantages. The pupil perceives at once what is given and what is required, readily refers to the figure at every step, becomes perfectly familiar with the language of Geom- etry, acquires facility in simple and accurate expression, rapidly learns to reason, and lays a foundation for completely establishing the science. Original exercises have been given, not so difficult as to discourage ihe beginner, but well adapted to afford an effectual test of the degree in which he is mastering the subjects of his reading. Some of these exercises have been placed in the early part of the work in order that the student may discover, at the outset, that to commit to mem- ory a number of theorems and to reproduce them in an examination is a useless and pernicious labor ; but to learn their uses and appli- cations, and to acquire a readiness in exemplifying their utility is to derive the full benefit of that mathematical training which looks not so much to the attainment of information as to the discipline of the mental faculties. G. A. WENTWORTH. Phillips Exeter Academy, 1878. PREFACE. TO THE TEACHER. When the pupil is reading each Book for the first time, it will be well to let him write his proofs on the blackboard in his own lan- guage ; care being taken that his language be the simplest possible, that the arrangement of work be vertical (without side work), and that the figures be accurately constructed. This method will furnish a valuable exercise as a language lesson, will cultivate the habit of neat and orderly arrangement of work, and will allow a brief interval for deliberating on each step. After a Book has been read in this way, the pupil should review the Book, and should be required to draw the figures free-hand. He should state and prove the propositions orally, using a pointer to indicate on the figure every line and angle named. He should be encouraged, in reviewing each Book, to do the original exercises ; to state the converse of propositions ; to determine from the statement, if possible, whether the converse be true or false, and if the converse be true to demonstrate it ; and also to give well-considered answers to questions which may be asked him on many propositions. The Teacher is strongly advised to illustrate, geometrically and arithmetically, the principles of limits. Thus a rectangle with a con- stant base 6, and a variable altitude a;, will afford an obvious illus- tration of the axiomatic truth that the product of a constant and a variable is also a variable ; and that the limit of the product of a constant and a variable is the product of the constant by the limit of the variable. If x increases and approaches the altitude a as a limit, the area of the rectangle increases and approaches the area of the rectangle ab as a limit; if, however, x decreases and approaches zero as a limit, the area of the rectangle decreases and approaches zero for a limit. An arithmetical illustration of this truth may be given by multiplying a constant into the approximate values of any repetend. If, for example, we take the constant 60 and the repetend 0.3333, etc., the approximate values of the repetend will be ^, -j^. VI PREFACE. ToW^ tVAV' 6^Cm and these values multiplied by 60 give the series 18, 19.8, 19.98, 19.9998, etc., which evidently approaches 20 as a limit; but the product of 60 into J (the limit of the repetend 0.333, etc.) is also 20. Again, if we multiply 60 into the different values of the decreasing series y^, j^, j^jW* Tuhns* ®tc., which approaches zero as a limit, we shall get the decreasing series 2, ^, -^V. sh^ etc.; and this series evi- dently approaches zero as a limit. In this way the pupil may easily be led to a complete compre- hension of the subject of limits. The Teacher is likewise advised to give frequent written examina- tions. These should not be too difficult, and sufficient time should be allowed for accurately constructing the figures, for choosing the best language, and for determining the best arrangement. The time necessary for the reading of examination-books will be diminished by more than one-half, if the use of the symbols employed in this book be allowed. G. A. W. Phillips Exeter Academy, 1879. PREFACE. VU NOTE TO REVISED EDITION. The first edition of this Geometry was issued about nine years ago. The book was received with such general favor that it has been neces- sary to print very large editions every year since, so that the plates are practically worn out. Taking advantage of the necessity for new plates, the author has re-written the whole work ; but has retained all the distinguishing characteristics of the former edition. A few changes in the order of the subject-matter have been made, some of the demonstrations have been given in a more concise and simple form than before, and the treatment of Limits and of Loci has been made as easy of comprehension as possible. More than seven hundred exercises have been introduced into this edition. These exercises consist of theorems, loci, problems of con- struction, and problems of computation, carefully graded and specially adapted to beginners. Nq_geometry can now receive favor unless it provides exercises for independent investigation, which must be of such a kind as to interest the student as soon as he becomes acquainted with the methods and the spirit of geometrical reasoning. The author has observed with the greatest satisfaction the rapid growth of the demand for original exercises, and he invites particular attention to the systematic and progressive series of exercises in this edition. The part on Solid Geometry has been treated with much greater freedom than before, and the formal statement of the reasons for the separate steps has been in general omitted, for the purpose of giving a more elegant form to the demonstrations. A brief treatise on Conic Sections (Book IX) has been prepared, and is issued in pamphlet form, at a very low price. It will also be bound with the Geometry if that arrangement is found to be gen- erally desired. Vlll PREFACE. The author takes this opportunity to express his grateful appre- ciation of the generous reception given to the Geometry heretofore by the great body of teachers throughout the country, and he confidently anticipates the same generous judgment of his efforts to bring the work up to the standard required by the great advance of late in the sci- ence and method of teaching. The author is indebted to many correspondents for valuable sug- gestions ; and a special acknowledgment is due, for criticisms and careful reading of proofs, to Messrs. C. H. Judson, of Greenville, S.G. ; Samuel Hart, of Hartford, Conn. ; J. M. Taylor, of Hamilton, N.Y. ; W. Le Conte Stevens, of Brooklyn, N.Y. ; E. R. Offutt, of St. Louis, Mo. ; J. L. Patterson, of Lawrenceville, 'N.J. ; G. A. Hill, of Cam- bridge, Mass. ; T. M. Blaksiee, Des Moines, la. ; G. W. Sawin, of Cam- bridge, Mass. ; and Ira M. De Long, of Boulder, Col. Corrections or suggestions will be thankfully received. G. A. WENTWORTH Phillips Exetee Academy, 1888. COl^TENTS. GEOMETRY. PACK Definitioits 1 Straight Lines 5 Plane Angles 7 Magnitude of Angles 9 Angular Units 10 Method of Superposition 11 Symmetry 13 Mathematical Terms . 14 Postulates 15 Axioms 16 Symbols 16 PLANE GEOMETRY. BOOK I. The Straight Line. The Straight Line 17 Parallel Lines 22 Perpendicular and Oblique Lines .... 33 Trianqlf^ 40 Quadrilaterals 66 Polygons in General ........ 66 Exercises 72 X CONTENTS. BOOK II. The Circle. • PAOB Definitions . 75 Arcs and Choeds 77 Tangents . . .^ 89 Measueement 92 Theoey of Limits 94 Measure of Angles 98. Problems of Construction 106 Exercises 126 BOOK III. Proportional Lines and Similar Polygons. Theory of Proportion '^ . . 131 Proportional Lines 138 Similar Triangles 145 Similar Polygons 153 Numerical Properties of Figures 156 Problems op Construction 167 Problems of Computation ''. . . . . , .173 Exercises ''. 176 BOOK IV. Areas of Polygons. Areas of Polygons ...'..... 181 Comparison of Polygons . 188 Problems of Construction 19i^ Problems of Computation . , 204 Exercises 205 BOOK V. Regular Polygons and Circles. Regular Polygons and Circles , . ... . 209 Peoblems of Construction . . • . , . . . 222 Maxima and Minima • 230 Exercises 237 Miscellaneous Exercises 240 CONTENTS. ■ XI SOLID GEOMETKY. BOOK VI. Lines and Planes in Space. PAGE Definitions 243 Perpendicular Lines and Planes 246 Parallel Lines and Planes 253 Dihedral Angles 260 Measure of Dihedral Angles . . . . . . 262 Planes Perpendicular to Each Other .... 265 Angle of a Straight Line and a Plane .... 270 A Perpendicular Between Two Straight Links . 271 Polyhedral Angles 272 Symmetrical Polyhedral Angles 273 BOOK VII. Polyhedrons, Cylinders, and Cones. Prisms and Parallelopipeds 279 Pyramids 294 Similar Polyhedrons 308 Regular Polyhedrons 312 General Theorems of Polyhedrons 317 Cylinders 319 Cones 325 Numerical Exercises 334 BOOK VIII. The Sphere. Plane Sections and Tangent Planes . Figures on the Surface of a Sphere Measuremei^t of Spherical Surfaces. Volume of a>ySphere . Numerical Exercises .... MiSCELLANECUr J^ERCISES 339 350 366 375 380 383 GEOMETRY. DEFINITIONS. 1. If a block of wood or stone be cut in the shape repre sented in Fig. 1, it will have six flat faces. Each face of the block is called a surface; and if these faces are made ^ smooth by polishing, so that, when a straight-edge is applied to any one of them, the straight edge in every part will touch the surface, the faces 2i,VQ coWedi plane surfaces^ ov planes. 2. The edge in which any two of these surfaces meet is called a line. 3. The corner at which any three of these lines meet is called a point. 4. For computing its volume, the block is measured in three principal directions : From left to right, A to B. From front to back, A to C. From bottom to top, A to D. These three measurements are called the dimensions of the block, and are named length, breadth (or width), thickness {height or depth). 2 ^ GEOMETRY. A solid, therefore, has three dimensions, length, breadth, and thickness. 6. The surface of a solid is no part of the solid. It is simply the boundary or limit of the solid. A surface, there- fore, has only tivo dimensions, length and breadth. So that, if any number of flat surfaces be put together, they will coin- cide and form one surface. 6. A line is no part of a surface. It is simply a boundary or limit of the surface. A line, therefore, has only one dimen- sion, length. So that, if any number of straight lines be put together, they will coincide and form one line. 7. A point is no part of a line. It is simply the limit of the line. A point, therefore, has no dimension, but denotes position simply. So that, if any number of points be put together, they will coincide and form a single point. 8. A solid, in common language, is a limited portion of STp&ce filled with matter ; but in Geometry we have nothing to do with the matter of which a body is composed ; we study simply its shape and size; that is, we regard a solid as a lim- ited portion of space which may be occupied by a physical body, or marked out in some other way. Hence, A geometrical solid is a limited portion of space. 9. It must be distinctly understood at the outset that the points, lines, surfaces, and solids of Geometry are purely ideal, though they can be represented to the eye in only a material way. Lines, for example, drawn on paper or on the black- board, will have some width and some thickness, and will so far fail of being true lines ; yet, when they are used to help the mind in reasoning, it is assumed that they represent per- fect lines, without breadth and without thickness. DEFINITIONS. 3 10. A point is represented to the eye by a fine dot, and named by a letter, as A (Fig. 2) ; a line is named by two letters, placed one at each end, as £F; a surface is represented and named by the lines which bound it, as BCDF\ a solid is represented by the faces which bound it. Fig. 2. 11. By supposing a solid to diminish gradually until it vanishes we may consider the vanishing point, a jpomi in space, independent of a line, having position but no extent, 12. If a point moves continuously in space, its path is a line. This line may be supposed to }M of unlimited extent, and may be considered independent of the idea of a surface. 13. A surface may be conceived as generated by a line moving in space, and as of unlimited extent, A surface can then be considered independent of the idea of a solid. 14. A solid may be conceived as generated by a surface in motion. Thus, in the diagram, let the up- d right surface ABCD move to the / nght to the position EFOH. The "^X"" points A, B, C, and D will generate [c the lines AE, BF, CO, and DH, ^""" l^Il^ respectively. ' The lines AB, BC, Fig. 3. CD, and AD will generate the sur- faces AF, BO, CH, and AH, respectively. The surface ABCD will generate the solid AO. 15. Geometry is the science which treats of position, form, and magnitude. 16. Points, lines, surfaces, and solids, with tneir relations, constitute the subject-matter of Geometry. 4 GEOMETRY. 17. A straight line, or right line, is a line which has the same direction throughout its * g whole extent, as the line AB. 18. A curved line is a line no part of which is straight, as the line CD. 19. A broken line is a series of different successive straight lines, as the line EF. 20. A mixed line is a line composed of straight and curved lines, as the line GS. A straight line is often called simply a line, and a curved line, a curve. * 21. A plane surface, or a plane, is a surface in which, if any two points be taken, the straight line joining these points will lie wholly in the surface. 22. A curved surface is a surface no part of which is plane. 23. Figure or form depends upon the relative position of points. Thus, the figure or form of a line (straight or curved) depends upon the relative position of the points in that line ; the figure or form of a surface depends upon the relative posi- tion of the points in that surface. 24. With reference to form or shape, lines, surfaces, and solids are colled, figures. With reference to extent, lines, surfaces, and solids are called magnitudes. » 25. A plane figure is a figure all points of which are in the same plane. 26. Plane figures formed by straight lines are called rec- tilinear figures; those formed by curved lines are called curvilinear figures ; and those formed by straight and curved lines are called mixtilinear figures. DEFINITIONS. O 27. Figures which have the same shape are called sirmlar figures. Figures which have the same size are called equiva- lent figures. Figures which have the same shape and size are called equal or congruent figures. 28. Geometry is divided in two parts, Plane Geometry and Solid Geometry. Plane Geometry treats of figures all points of which are in the same plane. Solid Geometry treats of figures all points of which are not in the same plane. Straight Lines. 29. Through a point an indefinite number of straight lines may be drawn. These lines will have difierent drections. 30. If the direction of a straight line and a point in the line are known, the position of the line is known ; in other words, a straight line is determined if its direction and one of its points are known. Hence, All straight lines which pass through the same point in the same direction coincide, and form hut one line. 31. Between two points one, and only one, straight line can be drawn ; in other words, a straight line is determined if tw is (or are) greater than, Iden. . . . identical. < is (or are) less than. Cons, . . . construction. .*. therefore. Sup supplementary. Z angle. Sup.-adj. supplementary-adjacent. A angles. Ext.-int. exterior-interior. X perpendicular. Alt.-int. alternate-interior. Ji perpendiculars. Ex exercise. II parallel. . rt right. lis parallels. st straight. A triangle. q.e.d quod erat demonstrandum, A triangles. which was to he proved O parallelogram. q.e.p. . . . quod erat faciendum, v)hich [EJ parallelograms. was to he done. PLAIN^E GEOMETRY BOOK I. THE STRAIGHT LINE. ^ Proposition I. Theorem. 84. Ml straight angles are equal, F A ^ B E Let /.EC A and Z. FED he any two straight angles. To pT(yve ABCA^A FED. Proof. Apply the ABCA\.q the A FED, so that the vertex C shall fall on the vertex E, and the side CB on the side EF. Then CA will coincide with ED^ (5ecaws6 BCA and FED are straight lines and have two points common). . Therefore the Z BOA is equal to the Z FED. § 59 85. Cor. 1. All nght angles are equal. 86. Cor. 2. The angular units, degree, minute, and second, have constant values. 87. Cor. 3. The complements of equal angles are eqvxxl. 88. Cor. 4. The supplements of equal angles are equal. 89. Cor. 5. At a given point in a given straight line one perpendicular, and only one, can he erected. 18 PLANE GEOMETRY. — BOOK I. Proposition II. Theorem. 90. If two adjacent angles have their exterior sides in a straigM line, these angles are supplements of each other. Let the exterior sides OA and OB of the adjacent AAOD and BOD be in the straight line AB. To prove A AOD and BOD supplementary. Proof. AOB is a straight line. Hyp. .-. the Z^O^ is a St. Z. §46 But the AAOB+ BOB = the st. Z AOB. Ax. 9 .*, the A AOB and BOB are supplementary. § 55 aE. D. 91. Scholium. Adjacent angles that are supplements of each other are called supplementary-adjacent angles. 92. Cor. Since the angular magnitude about a point is neither increased nor diminished by the number of lines which radiate from the point, it follows that, ITie sum of all the angles about a point in a plane is equal to two straight angles, or four right angles. The sum of all the angles about a point on the same side of a straight line passing through the point, is equal to a straight angle, or two right angles. THE STRAIGHT LINE. 19 Peoposition III. Theorem. 93. CoNVEBSELY : If two adjoceiit angles are supple^ merits of en/^h other, their exterior sides lie in the ^ame straight line. AC B F Let the adjacent A OCA + OCB = a straight angle. ^ To prove A C and CB in the same straight line. Proof. Suppose CF to be in the same line with AC. § 81 Then /. OCA + Z. OCF is a straight angle. § 90 But Z. OCA + Z OCB is a straight angle. Hyp. .-. Z OCA + Z OCF=^ Z OCA + Z OCB. Ax. 1 Take away from each of these equals the common Z OCA. Then Z OCF= Z OCB, Ax. 3 .*. CB and 4-P'i> = 2Pi), and PC -\-CP' =2 PC Cons. But PC + CP' OA + OB. Proof. Produce AO to meet the line CB at E. Then AC'\-OE> 0A-\- OE, (a straight line m the shortest distance between two "pointfl^ and BE-\-OE> BO. Add these inequalities, and we have CA + CE-\- BE-\-OE>OA-\- 0E-\- OB, Substitute for CE-\- BE its equal CB, and take away OE from each side of the inequality. We have CA + CB>OA-\- OB. Ax. 5 q. e. d. 36 PLANE GEOMETRY. — BOOK I. Proposition XX. Theorem. 119. Of two oblique lines drawn from the same point in a perpendicular, cutting off unequal dis- tances from the foot of the perpendicular, the more remote is the greater. \ , D Let OC be perpendicular to AB, OG and OE two oblique lines to AB, and GE greater than CG. To prove OE > 00. Proof. Take OF equal to CG, and draw OF. Then OF=OG, §116 {two oblique lines drawn from a point in a ±, cutting off equal distances from the foot of the ±, are equal). Prolong OC to D, making CD = OC. Draw ED and ED. Since AB is J_ to OD at its middle point, EO = ED, and EO = ED, § 116 But OE-\-ED> 0E-{- ED, § 118 {the sum of two oblique lines drawn from a point to the extremities of a straight line is greater than the sum of two other lines similarly drawn, but included by them). .-. 20E> 20E, or OE > OE But 0E= OG. Hence OE > OG. q. e. d. 120. Cor. Only two equal straight lines can be drawn from a point to a straight line ; and of two unequal lines, the greater cuts off the greater distance from the foot of the perpendicular. PERPENDICULAR AND OBLIQUE LINES. 37 Proposition XXL Theorem. 121. Two equal oblique lines, drawn from the same point in a perpendicular, cut off equal distances from the foot of the perpendicular. Let CF be the perpendicular, and CE and CK be two equal oblique lines drawn from the point C to AB. To prove FE= FK. Proof, Fold over CFA on (TFas an axis, until it comes into the plane of CFB. The line FE will take the direction FK, {since Z CFE = Z CFK, each being a rt. Z by hyp.). y Then the point E must fall upon the point K, zxi^FE^FK. Otherwise one of these oblique lines must be more remote from the perpendicular, and therefore greater than the other ; which is contrary to the hypothesis that they are equal. § 119 a E. D. Ex. 11. Show that the bisectors of two supplementary adjacent angles are perpendicular to each other. Ex. 12. Show that the bisectors of two vertical angles form one straight line. Ex. 13. Find the complement of an angle containing 26° 52'' 37''''. Find the supplement of the same angle. 38 PLANE GEOMETRY. — BOOK I. Proposition XXII. Theorem. 122. Every point in the perpendicular, erected at the jniddle of a given straight line, is equidistant from the extremities of the line, and every point not ijv the perpendicular is unequally distant from the extremities of the line. Let PR be a perpendicular erected at the middle oi the straight line AB, any point in PR, and C any point without PR. Draw OA and OB, CA and CB. To prone OA and OB equal, CA and CB unequal. Proof. PA = PB. Hyp. ,'.OA = OB, §116 {two oblique lines drawn from the same point in a ±, cutting off equal dis- tances from the foot of the ±, are equal). Since C is without the perpendicular, one of the lines, CA or CB, will cut the perpendicular. Let CA cut the _L at I), and draw PB. Then PB = PA, {two oblique lines drawn from the same point in a ±, cutting off equal dis- tances from the foot of the ±, are equal). But CBAO, for a straight line is the shortest distance between two points ; and by taking away ^Cfrom both sides, AB>AC-BO,oi AC-BC AE. Proof. Place the A so that AB of the one shall coincide with AB of the other. Suppose ^i^ drawn so as to bisect 4- EBQ. Draw EF, In the A EB F s^ndi QBE EB = BC, Hyp. BF=BF, Iden. Z.EBE=Z.CBF. Cons. .-. the A EBE and CBEine equal, § 150 {having two sides and the included Z of one equal respectively to two sides and the included Z of the other). .'. EE= EC, {being homologous sides of equal A). Now • AE+EE> AE, §137 {the sum of two sides of a A is greater than the third side). .'. AEi-EO AE; or, AC> AE. <^E.tt TRIANGLES. 47 Proposition XXVIII. Theorem. 153. Conversely. If two sides of a triangle are equal respectively to two sides of another, hut the third side of the first triangle is greater than the third side of the second, then the angle opposite the third side of the first triangle is greater than the angle opposite the third side of the second. D B C K F In the triangles ABC and DBF, let AB = DE, AC = DF, but let BG be greater than FF, To prove Z ^4 greater than Z D. Proof. Now Z. A \^ equal to A D, or less than Z Z), or greater than Z. D. But Z ^ is not equal to Z D, for then A ABC would be equal to A DBF, § 150 ijuiving two sides and the included Z. of the owe respectively equal to two aides and the included Z of the other\ and ^C would be equal to EF. And Z ^ is not less than Z D, for then BC would be less than EF. § 152 .-. Z ^ is greater than Z B. 48 PLANE GEOMETRY. — BOOK I. Peoposition XXIX. Theorem. 154. In an isosceles triangle the angles opposite the equal sides are equal. A B T) C Let ABO be an isosceles triangle, having the sides AB and AC equal. To prove Z B = Z C. Proof. Suppose AD drawn so as to bisect the Z BAC. In the A ABB and ABC, AB-^AC. Hyp. AD = AD, Iden. Z BAB = Z CAD. Cons. .\ A ABB = A ABC, §150 {two A are equal if two sides and the included Z of the one are equal respectively to two sides and the included Z of the other). »'» Z B = Z C, Q. E. D. 155. Cor. An equilateral triangle is equiangular, and each angle contains 60°. Ex. 14. The bisector of the vertical angle of an isosceles triangle bisects the base, and is perpendicular to the base. Ex. 15. The perpendicular bisector of the base of an isosceles triangle passes through the vertex and bisects the angle at the vertex. i^RTANOLES. 49 Proposition XXX. Theorem. 156. If two angles of a triangle are equal, the sides opposite the equal angles are equal, and the triangle is isosceles. In the triangle ABC, let the ZB = ZC. To prove AB = AC. Proof. Suppose AD drawn _L to BC, In the rt. A ADB jxnd ADC, AD = AD, Iden. ZB^ZC Hyp. .-. rt. A ADB = rt. A ADC, § 149 {having a side and an acute. Z. of the one equal respectively to a side and an homologous acute Z of the other). .\AB = AC, {being homologous sides of equal A). aE.D. 157. Cor. An equiangular triangle is also equilateral. Ex. 16. The perpendicular from the vertex to the base of an isosdfeles triangle is an axis of symmetry. ^^,„^ OO PLANE GEOMETRY.. — BOOK I. Proposition XXXI. Theorem. 158. If two sides of a triangle are unequal, the an- gles opposite are unequal, and the greater angle is opposite the greater side. C B In the triangle AGB let AB be greater than AC, To prove Z A CB greater than Z B, Proof. Take AE equal to AC. Draw EC. ZAEC=ZACE, §154 {being A opposite equal sides). But Z ^^C is greater than Z^, §146 (an exterior Z of a A is greater than either opposite interior Z). and Z ACB is greater than Z ACE. Ax. 8 Substitute for Z ACE its equal Z A EC, then Z ACB is greater than Z A EC. Much more, then, is the Z ACB greater than Z B. aE.D. Ex. 17. If the angles ABC And ACB, at the base of an isosceles tri- angle, be bisected by the straight lines BD, CD, show that BBCmW be an isosceles triangle. TRIANGLES. 61 Proposition XXXII. Theorem. « 159. Conversely : If two angles of a triangle are unequal, the sides opposite are unequal, and the greater side is opposite the greater angle. In the triangle ACB, let angle ACB be greater than angle B. To prove AB > AC. Proof. Now AB is equal to AC, or less than AC, or greater than AC. But AB is not equal to AC, for then the Z C would be equal to the Z B, § 154 {being A opposite equal sides). And AB is not less than AC, for then the Z C would be less than the Z ^,. §158 {if two sides of a A are unequal, the A opposite are unequal, and the greater Z is opposite the greater $ide). .-. AB is greater than AC. aE. D. Ex. 18. ABC a.B.d ABD are two triangles on the same base AB, and on the same side of it, the vertex of each triangle being without the other, li AC equal AD, show that BC cannot equal BD. Ex. 19. The sum of the lines which join a point within a trianRln to the tliroe vertices is less than the perimeter, but greater than half the perimeter. ^ 62 PLANE GEOMETRY. — BOOK! I. Peoposition XXXIII. Theoeem. 160. Two triangles are equal if the three sides of the one are equal respectively to the three sides of the other, B B' In the triangles ABC and A'BC, letAB=^A'B', AC^A'O, BG=B'C', To prove A ABC= A A'B'Q', Proof. Place A A^B'C in the position AB^C, having its greatest side A'C in coincidence with its equal AC, and its vertex at J3', opposite B ; and draw BB'. Since AB = AB\ Hyp. A ABB' = Z AB'B, § 154 (in an isosceles A the A opposite the equal sides are equal). Since CB=CB\ ' Hyp. /.OBB' = ACB'B, §154 Hence, Z ABQ= Z AB'Q, Ax. 2 /.A ABQ= A AB'0='A A'B'C' § 150 {two /^ are equal if two sides and included Z of one are equal to two sides and included Z of the other). TRIANGLES. 53 Proposition XXXIV. Theorem. 161. Two right triangles are equal if a side and the hypotenuse of the one are equal respectively to a side and the hypotenuse of the other. C B' In the right triangles ABC and A^B'C, let AB= A'B', and AC = A' C. To prove AABC=AA'B*C\ Proof. Apply the A ABC to the A A'B'C, so that AB shall coincide with A'B\ A falling upon A\ B upon B\ and C and C" upon the same side of A'B*, Then BC will take the direction of B'€\ {for /L ABC== Z A'B'C, each being a rt. Z), Since AC=A'0\ the point C will fall upon C\ § 121 {two equal oblique lines from a point in a ± cut off equal distances from the foot of the ±). '. the two A coincide, and are equal. Q.s.a 54 PLANE GEOMETRY. — BOOK I. Proposition XXXV. Theorem. 162. Every point in the bisector of an angle is equi- distant from the sides of the an^le. Let AD be the bisector of the angle BAC, and let be any point in AD. To prove that is equidistant from AB and AC. Proof. Draw Oi^ and OG L to AB and -irrespectively. In the rt. A ^Oi^and AOG AO=AO, Iden. ZBAO = Z^AO, Hyp. .\AAOF=AAOG, § 148 [two rt. ^ are equal if the hypotenuse and an acute Z of the one are equal respectively to the hypotenuse and an acute Z of the other). ... 0F= OG, {homologous sides of equal ^). .'. JA^quidistant from AB and AC. What is the locils of a point : Ex. 20. At a given distance from a fixed point ? § 57. Ex. 21. Equidistant from two fixed points? § 119. Ex. 22. At a given distance from a fixed straight line of indefinite length ? Ex. 23. Equidistant from two given parallel lines ? Ex. 24. Equidistant from tiie extremities of a given line? TRIANGLES. 55 Proposition XXXVI. Theorem. 163. Every -point within an angle, and equidistant from, its sides, is in tlie bisector of the angle. Let be equidistant from the sides of the angle BAC, and let AO join the vertex A and the point 0. To prove that AO is the bisdcior of Z BAC. Proof. Suppose OF and OQ drawn X to AB and AG^ respectively. In the rt. A ^Oi^and AOG 0F= 00, AO = AO. Hyp. Iden. .'.A AOF^AAOG, §161 {two rt. A are equalif the liypotenuse and a side oJ»^ one arc equal to the hypotenuse and a side of the otJ^r)/ .\ZFAO = ZGAO, {homologous A oj equal A). .'. ^0 is the bisector of Z BAC. aE.D. 164. Cor. The locus of a point within an angle, and equi- distant from its sides, is the bisector of the angle. 66 PLANE GEOMETRY. — BOOK I. Quadrilaterals. 165. A quadrilateral is a portion of a plane bounded by four straight lines. The bounding lines are the sides, the angles formed by these sides are the angles, and the vertices of these angles are the ver^jc^rs, of the quadrilateral. 166. A trapezium is a quadrilateral which has no two sides parallel. 167. A trapezoid is a quadrilateral which has two sides, and only two sides, parallel. 168. ^A parallelogram is a quadrilateral which has its oppo- site sides parallel. Trapezium. Trapezoid Parallelogram. 169. A rectangle is a parallelogram which has its angles right angles. 170. A rhomboid is a parallelogram which has its angles oblique angks. 171. A square is a rectangle which has its sides equal. 172. A rhombus is a rhomboid which has its sides equal. Square. Rectangle. Rhombus. Rhomboid. 173. The side upon which a parallelogram stands, and the opposite side, are called its lower and upper bases. QUADRILATERALS. 67 174. The parallel sides of a trapezoid are called its bases^ the other two sides its legs, and the line joining the middle points of the legs is called the median. 175. A trapezoid is called an isosceles trapezoid when its legs are equal. 176. The altitude of a parallelogram or trapezoid is the perpendicular distance between its bases. 177. The diagonal of a quadrilateral is straight line joining two opposite vertices. Proposition XXXVII. • Theorem. 178. Tlie diagonal of a parallelogram divides the figure into two equal triangles. C A E Let ABCE be a parallelogram and AC its diagonal. To prove A ABC= A AEC, In the A ABC and AEC, AC=AC, Iden. ^ACB = ZCAE, . §104 and ZCAB = ^AC£!, {being alt.-int. AofW lines.) .\AABC=AAECi §147 {having a side and two adj. A of the one equal respectively to a side and two adj. A of the other.) a E. D. 68 PLANE GEOMETRY. — BOOK I. Proposition XXXVIII. Theorem. 179. In a parallelogram the opposite sides are equalt and the opposite angles are equal. Let the figure ABCE be a. parallelogram. To prove B0= AU, and AB= EC, also, ZB = ZE,&ndZ BAE -= /. BCE. Proof. Draw AC. AABO=AAEO, §178 {the diagonal of a O divides the figure into two equal A). /. BC=^ AE, and AB = CE, {being homologous sides of equal A). Also, Z B = ZE,s.ndZBAE=ZBCE, §112 {having their sides 11 and extending in opposite directions from their vertices). Q. E. D. 180. Cor. I 'arallel lines comprehended between parallel lines are equal. A B 181. Cor. 2. Two parallel lines are everywhere equally distant. For if AB and DC are parallel, D C Js dropped from any points mAB to DC, measure the distances of these points from DC. But thepe Js are equal, by § 180; hence, all points in AB are equidistant from DC, QUADRILATERALS. 69 Proposition XXXIX. Theorem. 182. If two sides of a quadrilateral are equal and parallel, then the other two sides are equal and par- allel, and the figure is a ■parallelogram. Let the figure ABCE be a quadrilateral, having the side AE equal and parallel to BG, To prove A B equal and II to EC. Proof. Draw AC, In the A ABC Q.ndi AEC BC=^ AE, Hyp. AC^AC, Iden. ZBCA = ZCAE, §104 {being alt.-int A qfW lines). .'.AABC=AACE, §150 '(having two sides and the included Z of the one equal respectively to two sides and the included Z of the other). ,\AB=EC, (being homologous sides of equal A). Also, ZBAC=ZACE, (being homologous A of equal A). .-. AB is II to EC, . § 105 (when two straight lines are cut by a third straight line, if the alt.-int. A are equal, the lines are parallel). .'. the figure ABCE is a O, § 168 (the opposite sides being parallel), a e. d. 60 PLANE GEOMETRY. — BOOK I. Proposition XL. Theorem. 183. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram. A Let the figure ABCE be a quadrilateral having BC=: AE and AB = EC. To prove figure ABCE a O. Proof. Draw AQ, In the A ABCM\d. AEQ BC= AE, Hyp. AB=OE, • Hyp. AC= AG. Iden. .-. A ABC= A AEQ, § 160 {having three sides of the one equal respectively to three sides of the other). .:ZACB = ZOAE, and ZBAO=ZACE, {being homologous A of equal A). .'.BClQ W to AE, and AB is II to EC, § 105 {when two straight lines lying in the same plane are cut by a third straight line, if the alt.-mt. A are equal, the lines are parallel). .-. the figure ABCE is a O, § 168 {having its opposite sides parallel). QUADRILATERALS. 61 Proposition XLI. Theorem. 184. The diagonals of a parallelogram bisect each her. other. Let the figure ADCE be a parallelogram, and let the diagonals AC and BE cut each other at 0. To prove AO=OC, and BO = OK In the A AOE and £00 AE=BC, §179 {being opposite sides of a O). ZOAE=ZOCB, §104 and Z OEA = Z OBC, {being alt.-int. A of W lines). .\ A AOE =: A BOO, §147 {having a side and two adj. A of the one equal respectively to a side and two adj. A of the other). .\AO=00,&naBO=OU, {being homologous sides of equal i^). Ex. 25. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. Ex. 26. The diagonals of a rectangle are equal. Ex. 27. If the diagonals of a parallelogram are equal, the figure is a rectangle. -^ ~ Ex. 28. The diagonals of a rhombus are perpendicular to each other, and bisect the angles of the rhombus. Ex. 29. The diagonals of a square are perpendicular to each other, ind bisect the angles of the square. 62 PLANE GEOMETRY. BOOK I. Proposition XLIL Theorem. 185. Two parallclo!irams, having two sides and tlie included arxgle of the one equal respectively to two sides and the zncluded angle of the other, are equal, B' g' 7 A' D In the parallelograms ABCD and AE'C'D', let AB^ A'B', AD = A'D', and AA=AA\ To 'prove that the UJ are equal. Apply O ABCD to O A'WC'B\ so that AD will fall on and coincide with A^D\ Then AB will fall on A'B\ {for ZA==ZA',by hyp.), and the point B will fall on B\ (for AB = A^B\ by hyp.). Now, BC and B'O^ are both II to AD' and are drawn through point B'. .-. the lines BC and B'C coincide, § 101 and (7 falls on B'C or B'C produced. In like manner, DC and D'C are II to A'B' and are drawn through the point D'. .'. DC and D'C' coincide. § 101 .*. the point C falls on D'C, or D'C produced. .-. (7 falls on both B'C and D'C. .*. Cmust fall on the point common to both, namely, C. .'. the two UJ coincide, and are equal. aE. D 186. Cor. Two rectangles having equal bases and altitudes are equal. QUADRILATERALS. 63 Proposition XLIII. Theorem. 187. // three or more parallels intercept equal parts on any transversal, they intercept equal parts on every transversal. Let the parallels AH, BK, CM, DP intercept equal parts BK, KM, MP on the transversal HP. To prove that they intercept equal parts AB, BCy CD on the transversal AD. Proof. From A, B, and C suppose AE, BF, and CO drawn II to HP, Then AE= HK, BF^ KM, CG = MP, § 180 [parallels comprehended between paralleh are equal). ,\AF=BF=CO. Ax, 1 Also ZBAF=Z CBF= Z BCG, § io6 {being ext.-int. A ofW lines); and . ZAFB--=ZBFC=ZCGB, §112 (having their sides li and directed the same way jrom the verticei). .-. A ABE= A BCF= A CDG, § 147 {each having a side and two ad}. A respectively equal to a side and two ad). A of tht others). ,'. AB = BC^ CD, {.hcmclogous ndes of egital A). Q ^- °- 64 PLANE GEOMETRY. — BOOK I. 188. Cor. 1. The line parallel to the base of a triangle and bisecting one side, bisects the other side cdso. For, let DE Le II to BC and bisect AB. . Draw through A a Hne II to BC. Then this line is II to DE, by § 111. The three parallels by hypothesis intercept equal parts on the transversal AB, and there- -^ fore, by §187, they intercept equal parts on the transversal AC \ that is, the line Z)^ bisects AC. 189. Cor. 2. The line which joins the middle points of two sides of a triangle is parallel to the third side, and is equal to half the third side. For, a line drawn through D, the middle point of AB, II to BC, passes through E, the middle point of ^^> l^y § 1S8. Therefore, the line joining Z) and E coincides with this parallel and is 11 to BC. Also, since EF drawn II to AB bisects AC, it bisects BC, by § 188 ; that is, BF=^ EC = \ BC. But BEEF is a O by construction, and therefore DE==BF=^BC. 190. Cor. 3. The line which is parallel to the bases of a trap- ezoid and bisects one leg of the trap- ezoid bisects the other leg also. For if parallels intercept equal parts on any transversal, they intercept equal parts on every transversal by § 187. i''\ 191. Cor. 4. The median of a trapezoid is parallel to the bases, and is equal to half the sum of the bases. For, draw the diagonal EB. In the A AEB join E, the middle point of AE, to F, the middle point of EB. Then, by § 189, ^i^is II to AB and -- ^AB. In the A EEC join i^to O, the middle point of BC. Then FG is II to EC and =^EC. AB and FG, being II to EC, are II to each other. But only one line can be drawn through F II to AB. There- fore FG is the prolongation of EF. Hence EFG is II to AB and EC, and = 1(^^-1" ^C'). EXERCISES. 65 Exercises. 30, The bisectors of the angles of a triangle meet in a point which is equidistant from the sides of the triangle. lliST. Let the bisectors AD and BE intersect at 0. Then being in AD is equidistant from AC and AB. (Why ?) And being in BE is equidistant from BC and AB. Hence is equidistant from AC and BC, and therefore is in the bisector CF. (Why ?) _ 31. The perpendicular bisectors of the sides of a triangle meet in a point which is equidistant from the vertices of the triangle. Hint. Let the J. bisectors EE^ and DD^ intersect D/ at 0. Then being in EE^ JB equidistant from A ^ and C. (Why ?X And being in DD^ is equidistant from A and B. Hence is equidistant from B and C, and therefore is in the J. bisector FF^. (Why ?) 32. The perpendiculars from the vertices of a triangle to the opposite sides meet in a point. a'- - 4l. »/ Hint. Let the _fe. be AH, BP, and CK: '^"^ "*'" ' Through A, B, C suppose B'C, A'C, A'B' drawn II to BC, AC, AB, respectively. Then AH is ± to B^C. (Why?) Now ABCB^ and ACBC^ are ZI7 (why?), and'AB^^BC, and AC^ = BC. (Why ?) That is, A is the middle point of B^O^. In the same way, B and C are the middle points of A^C^ and A^B^, respectively. There- fore, AH, BP, and CiTare the _L bisectors of the sides of the A A^B^O^. Hence they meet in a point. (Why ?) 33. The medians of a triangle meet in a point which is two-thirds of the distance from each vertex to the middle of the opposite side. Hint. Let the two medians AD and CE meet in 0. Take i^the middle point of OA, and O of OC Join GF, FE, ED, and DG. In A AOC, GF ia II to AC and equal to .} AC. (Why ?) DE is II to AC and equal to ^AC (Why?) Hence DGFE is a O. (Why?) Hence AF=^ FO •-= OD, and CG==GO= OE. (Why ?) ^ Hence, any median cutS off on any other median two-thirds of the dis- tance from the vertex to the middle of the opposite side. Therefore the median from B will cut off AO, two-thirds of AD; that is, will pass through ii\77 H. 66 PLANE GEOMETRY. — BOOK I. Polygons in General. 192. A polygon is a plane figure bounded by straight lines. The bounding lines are the sides of the polygon, and their sum is the perimeter of the polygon. The angles which the adjacent sides make with each other are the angles of the polygon, and their vertices are the ver- tices of the polygon. The number of sides of a polygon is evidently equal to the number of its angles. 193. A diagonal of a polygon is a line joining the vertices of two angles not adjacent ; as ^C, Fig. 1. B Fig. 3. 194. An equilateral polygon is a polygon which has all its sides equal. 195. An equiangular polygon is a polygon which has all its angles equal. 196. A convex polygon is a polygon of which no side, when produced, will enter the surface bounded by the perimeter. 197. Each angle of such*^ polygon is called a salient angle, and is less than a straight angle. 198. A concave polygon is a polygon of which two or more sides, when produced, will enter the surface bounded by the perimeter. Fig. 3. 199. The angle FDE is called a re-entrant angle, and is greater than a straight angle. If the term polygon is used, a convex polygon is meant. POLYGONS. 67 200t Two polygQias are equal when they eaji be divided by- diagonals into the same number of triangles, equal each to each, and similarly placed ; for the polygons can be applied to each other, and th© corresponding triangles will evidently coincide. 201. Two polygons are muiuodl'^ equiangulw^ if the angles of the one are equal to the angles of the other, each to each, when tak«n in the same order. Figs. 1 and 2. 202. The eqaal angles in mutually equiangular polygons are called homologous angles ; and the sides which lie bekueen equal angles are called homologous sides. 203. Two polygons are mutually equilateral, if the sides of the one are equal to the sides of the others each to each, wli.en taken in the same order. Figs. 1 and 2. Fio. 4 FiQ 5. Fio. 6. Fia. 7. Two polygons may be mutnially equiangular without beiiig mutually equilateral ; as, Figs. 4 and 5. And, except in the case of triangiles, two polygons may be mutuary equilateral without being mutually equiangular ; as, Figs. 6 and 7. If two polygons are mutually equilateral and equiangular, they cure equal, for they may be applied the one to the other so as to coincide. 204. A polygon of three sides is called a trigon or triangle; one of four sides, a tetragon or quadrilateral ; one of five sides, 9. pentagon ; one of six sides, a hexagon; one of seven sides, a heptagon ; one of eight sides, an octagon : one oi ten sides, a decagon; one of twelve sides, a dodecagoti. 68 PLANE GEOMETRY. BOOK I. Proposition XLIV. Theorem. 205. The sum of the interior angles of a -polygon is equal to two right angles, taken as many times less two as the -figure has sides. Let the figure ABODE F be a polygon having n sides. Te prove ^ A-\-ZB-{-Z C, etc. = (n- 2) 2 rt. A. Proof. From the vertex A draw the diagonals AC, AD, and AU. The sum of the A of the A =■ the sum of the A of the polygon. Now there ar€k(22_— 2) A, and the sum of the A of each A = 2 rt. A. §138 .-. the sum of the zi of the A, that is, the sum of the A of the polygon = (n - 2) 2 rt. A. ^ ^^ ^ 206. Cor. The sum of the angles of a quadrilateral equals two right angles taken (4 — 2) times, i.e., equals 4 right angles; and if the angles are all equal, each angle is a right angle. In general, each angle of an equiangular polygon of n sides is 2 (n — 2) ■ equal to "^ — right angles. POLYGONS. 69 Proposition XLV. Theorem. 207. The exterior angles of a polygon, made hy -pro- ducing each of its sides in succession, are together equal to four right angles. Let the figure ABODE be a polygon, having its sides produced in succession. To prove the sum of the ext. A = ^ rt. A. Proof. Denote the int. A of the polygon hy A^ B^ (7, Z>, E, and the ext. Ahj a,h, c, d, e. ZA + Aa = 2rt.A, §90 and ZB + Zb = 2 rt. A, {being sup.-adj. A). In like manner each pair of adj. A = 2 rt. A .'. the sum of the interior and exterior A = 2rt. A taken as many times as the figure has sides, or, 2 n rt. A. But the interior ^ =f 2 rt. ^ taken as many times as the figure has sides less two,' = (n — 2) 2 rt. A, or, - 2 n rt. ^ — 4 rt. A. .'. the exterior zi = 4 rt. A. 70 PLANE GEOMETRY. — BOOK I. Proposition XLVI. Theorem. 208. A quadrilateral which has two adjacent sides equal, and the other tivo sides equal, is symmetrical with respect to the diagonal joining the vertices of the angles formed hy the equal sides, and the diago- nals intersect at right angles. Let ABCD be a quadrilateral, having AB = AD, and CB:=CD, and having the diagonals AC and BD. To prove that the diagonal A C is an axis of symmetry, and is S^to the diagonal BD. Proof. In the A ^^(7 and ^DC AB = AI), and BC= BO, Hyp. and AC=Aa Iden. .'.A ABC =- A ABO, §160 {having three sides of the one equal to three sides of the other). .'.Z BAC=rZBAO, and Z BCA - Z BOA, {homologous A of equal A). Hence, if ABC is turned on ^ C as an axis, AB will fall upon AB, CB on CB, and OB on OB. Hence AC is an axis of symmetry, § 65, and is J_ to ,BD.^ Q. E. O. POLYGONS. 71 Pkoposition XLVII. Theorem. 209. If a figure is symmetrical with respect to two axes perpendicular to each other, it is symmetrical with respect to their intersection as a centre. B • c i^--A .....:::^ A r' (\ -'" -'"^^ D X H B ^ G E Let the figure ABODE FGH be symmetrica,! with respect to the two axes XX', YY\ which intersect at 0. To prove the centre of symmetry of the figure. Proof. Let N be acy point m the perimeter of the figure. Draw NMIA. to YT, end IKL J. to XX\ Join LO,^OJV, ^nd KM. Now KI=KL, {the figure being symmetrical with respect to XX): But KI= OM, (lis comprehended between lis are equal). .-. KL = OM, and KLOM is a O, {having two sides eqvxil and parallel). .'. LO IS equal and parallel to KM. In like manner we may prove OiV equal and parallel to KM. Hence the points L, 0, and iVare in the same straight line drawn through the point H to K3f; and L0= ON, since each is equal to KM. .'. any straight line LON", drawn through 0, is bisected at 0. .'. is the centre of symmetry of the figure. § 64 a E. D. §61 §180 §182 §179 72 PLANE GEOMETRY. — BOOK I. Exercises. - 34. The median from the vertex to the base of an isosceles triangle is perpendicular to the base, and bisects the vertical angle. ] -. 35. . State and prove the converse. ' "'36. The bisector of an exterior angle of an isosceles triangle, formed by producing one of the legs through the vertex, is parallel to the base. "-37. State and prove the converse. \ ^38. The altitudes upon the legs of an isosceles triangle are equal. ^ 39. State and prove the converse. ^ 40. The medians drawn to the legs of an isosceles triangle are equal. ^41. State and prove the converse. (See Ex. 33.) M2. The bisectors of the base angles of an isosceles triangle are equal. 43. State the converse and the opposite theorems. 44. The perpendiculars dropped from the middle point of the base of an isosceles triangle upon the legs are equal. 45. State and prove the converse. '^ '» 46. If one of the legs of an isosceles triangle is produced through the vertex by its own length, the line joining the end of the leg produced to the nearer end of the base is perpendicular to the base. 47. Show that the sum of the interior angles of a hexagon is equal to eight right angles. ' 48. Show that each angle of an equiangular pentagon is f of a right angle. "^ 49. How many sides has an equiangular polygon, four of whose dngles are together equal to seven right angles ? 50. How many sides has a polygon, the sum of whose interior angles is equal to the sum of its exterior angles ? "•51. How many sides has a polygon, the sum of whose interior angles is double that of its exterior angles ? ^52. How many sides has a polygon, the sum^f whose exterior anglee is double that of its interior angles ? EXERCISES. 73 ^. BAC IS a triangle having the angle B double the angle A. If BD bisect the angle B, and meet AC in D, show that BD is equal to AD. ^54. If from any point in the base of an isosceles triangle parallels to the legs are drawn, show that a parallelogram is formed whose perimeter is constant, and equal to the sum of the legs of the triangle. ^ 65. The lines joining the middle point? of the sides of a triangle divide the triangle into four equal triangles. "^56. The lines joining the miadle points of the side of a square, taken in order, enclose a square. ""57. The lines joining the middle points of the sides of a rectangle (not a square), taken in order, enclose a rhombus. "• 58. The lines joining the middle points of the sides of a rhombus, taken in order, enclose a rectangle. *'v59. The lines joining the middle points of the sides of an isosceles trapezoid, taken in order, enclose a rhombus or a square. ^ 60. The lines joining the middle points of the sides of any quadri- lateral, taken in order, enclose a parallelogram. ^- 61. The median of a trapezoid passes through the middle points of the two diagonals. ' « 62. The line joining the middle points of the,diagonals of a trapezoid 18 equal to half the difference oi the bases. "^ 63. In an isosceles trapezoid each base makes > P equal angles with the legs. / \ \ Hint. Draw CEW DB, / \ \ "^64. In an isosceles trapezoid the opposite angles ^ ^ ^ are supplementary. -- 65. If the angles at the base of a trapezoid are equal, the other ' angles are equal, and the trapezoid is isasceles. n66. The diagonals of an isosceles trapezoid are equal. ^ 67. If the diagonals of a trapezoid are equal, the c D trapezoid is isosceles. i^^\/'\ Hint. Draw CE and DF JL to CD. Show that ^ / ''/^J \ ADF and BCE are equal, that ^ COD and AOB are jX ]\\ iposceles, and that i^ ^OC and i?0 7) are equal. a E f b 74 PLANE GEOMETRY. BOOK I. '^di. ABCD is a pai-allel®grara, E and F the middle points of AD and i? (7 respectively : show that BE and Di^will trisect the diagonal AC. " 69. If from th« diagonal BD of a square ABCD, BE is cut off equal to BC, and EF is drawn perpendicular to BD to meet DO at F, show that DE is equal to EF, and also to FC •^ 70. The bisector- »f the vertical angle ^ of a triangle ABC, and the bisectors of the exterior angles at the base formed by producing the sides AB and AC, nieet ki a point which is equidistant fr©m the base and the sides produced. ^71. If the two angles at the base of a triangle are bisect'ed, aod through the point of meeting of the bisectors a line is drawn parallel to the base, the length of this parallel between the sides is equal to the sum of the segments of the sides between the parallel and the base. \ 72. If one of the acute angles of a right triangle is double the otiier, the hypotenuse is double the shortest side. *~73. The sum of the perpendiculars dropped from any point in the base of an isoscelee triangle to the legs is constant, o and equal to tlie altitude upon one of the legs. Hint. Let PD and PE be the two Jfi, BE the altit-ude upon AC. Draw PO JL to BF, and prove / ^; AE, and that AB bisects the circle and the circuviference. Proof, I. From C, the centre of the O, draw CE. CE^CB, • {being radii of the same circle). But AC-\-CE> AE, §137 {the sum of two sides of a A is "^ the third side). Then AC+CB> AE, or AB > AE. Ax. 9 II. Fold over the segment A MB on AB as an axis until it falls upon ABB, § 59. The points A and B will remain fixed; therefore the arc A MB will coincide with the arc ABB ; because all points in each are equally distant from the centre C. § 210 Hence the two figures coincide throughout and are equal. § 59 Q.E. D. 78 PLANE GEOMETRY. — BOOK II. Proposition II. Theorem. 228. A straight line cannot intersect the cirewm- ference of a circle in more than two points. Let HK be any line cutting the circumference AMP, To prove that UK can intersect the circumference in only two points. Proof. If possible, let UK intersect the circumference in three points H, P, and K. From 0, the centre of the O, draw OH, OP, and OK Then OS, OF, and OJTare equal, {being radii of the same circle). Hence, we have three equal straight lines OIT, OF, and OK drawn from the same point to a given straight line. But this is impossible, § 120 {only two equal straight lines can he d^rawn from a point to a straight line). Therefore, KK can intersect the circumference in only two points. Q. E. a ARCS AND CHORDS. 79 Proposition III. Theorem. 229. In the same circle, or equal circles, equal an- gles at the centre intercept equal arcs; conversely, equal arcs subtend equal angles at the centre. p P' In the equal circles ABP and A'B'P' let ^0=^/.O. To prove arc R8 = arc E'S*. Proof. Apply O ABP to O A'B'P, so that Z shall coincide with Z (7. a will fall upon B', and 3 upon S\ § 226 (for 0R= 0^R\ and OS^ 0^8^, being radii of equal ©). Then the arc B8 will coincide with the arc B'S\ since all points in the arcs are equidistant from the centre. §210 .-. arc i?.S'=arc B'8'. Conversely : Let arc RS= arc R'S^. To prove Z.O = Z.a. Proof. Apply O ^^Pto O A'B'F\ so that arc BS shall fall upon arc B'jS', R falling upon B\ 8 upon 8\ and upon 0'. Then RO will coincide with i?'6>', and 80 with aS"0'. .". A O and 0' coincide and are equal. q. e. d. 80 PLANE GEOMETRY. — -BOOK II. Proposition IV. Theorem. 230. In the same circle, or equal circles, if two chords are equal, the arcs which they subtend are equal; conversely, if two arcs are equal, the chords which subtend them are equal. p P' In the equal circles ABP and A'B'P', let chord RS = chord R'S'. To prove arc US— arc B'S'. Proof. Draw the radii OB, OS, O'li', and O'S'. In the A OBS smd O'B'S' IiS= E'S', Hyp. the radii OB and OS^ the radii O'B' and O'S'. § 226 .'.ABOS=AB'0'S', §160 {three sides of the one being equal to three sides of the other). .'.ZO^-=ZO', .'. &VC BS= arc B'S', §229 (in equal ®, eqiuil A at the centre intercept equal arcs). aE.o. Conversely : Let arc RS = arc R^S'. To prove ' chord BS= chord B'S'. Proof." ZO^ZO', § 229 (equal arcs in equal © subtend equal A at the centre), and OB and 0S= O'B' and O'S', respectively. § 226 .■.AOBS=AO'B'S', §150 (having two sides equal each to each and the included A equal). .-. chord BS= chord B'S'. q.e.d. ARCS AND CHORDS. 81 Proposition V. Theorem. 231. In the same circle, or equal circles, if tivo arcs are unequal, and eaxih is less than a semi-circumfer- ence, the greater arc is subtended hy the greater chord; conversely, the greater chord subtends the greater arc. In the circle whose centre is 0, let the arc AMB he greater than the arc AMF. To prove chord AB greater than chord AF. Proof. Draw the radii OA, OF, and OB. Since F is between A and B, OF will fall between OA and OB, and Z AOB be greater than Z AOF. Hence, in the A AOB and AOF, the radii OA and OB = the radii OA and OF, but Z ^0^ is greater than Z AOF. .\AB>AF, §152 {the ^ having two sides equal each to each, bid the included A unequal). Conversely: Let AB be greater than AF. 'I / To prove arc AB greater than arc AF. t ' ,• In the A AOB and AOF, '* ) OA and 0B= OA and Oi^ respectively. ;. ' • But AB is greater than AF. "' Hyp. .-. Z ^0^ is greater than Z. AOF, § 153 (i/i€ A having two sides equal each to each, but the third sides unequal). .-. OB falls without OF. .•. arc AB is greater than arc AF. q e. d. 82 PLAJTE GEOMETRY. — -ROOK II. Proposition VI. Theorem. 232. The radius perpendicular to a cJwrd bisects the chard and the arc subtended by it. E Let AB be the chord, and let the radius OS be per- pendicular to AB at M. To prove AM— BM, and are AS = arc BS. Proof. Draw OA and OB from 0, the centre of the circle. In the rt. A 0AM md OBM the radius OA — the radios OB, and 0M= OM. Iden. .'.A0AM=^/10BM, §161 {having the hypotenu»e and a side of one equal to the hypotermse and a side of the other). .-. AM= BM, Sind Z AOS = Z B08. .'. arc AS— arc BS, {equal A at the centre intercept equal arcs on the eireumfer&nce). aE.D. 233. Cor. 1. Tke perpendicular erected at the middle of a chord passes through the centre of the circle. For the centre is equidistant from the extremities of a chord, and is therefore in the perpendicular erected at the middle of the chord. § 122 234. Cor. 2. The perpendicular erected at tke middle of a chord bisects the arcs of the chord. 235. Cor. 3. The locus of the middle pemts of a system of parallel chords is the dia7ncter perpendicular to thein. ARCS AND CHORDS. . ., 83 Proposition VII. Theorem. 236. In the same circle, or equal circles, equal chords are equally distant from tJie centre ; and CONVERSELY. Let AB and CF be equal chords of the circle ABFC. To prove AB ctnd CF eqiddisUxniJroin the centre 0. Proof. Draw OPl. to AB, OS A. to CF, and join OA and OC. OP and OH bisect AB and CF, § 232 (a radius A. to a chord bisects it). Hence, in the rt. A OF A and OHC AP^CH, Ax. 7 the radius OA = the radius OC. .\AOPA = AOIK\ §161 {having a side and hypotenuse of the one equal to a side enid hypotenuse of the other). .'. OP =011. .*. ^^ and CF 'Ave equidistant from O. Conversely : Let OP = OH. To prove AB = CF. Proof. In the rt. A OPA and OHC the radius OA = the radius OC, and 0P=- OH{hj hyp.). .-. A OPA and OHC sue equal. § 161 ..AP=CH. ,',AB^CF. Ax. 6. % s< o< 84 PLANE GEOMETRY. — BOOK II. Proposition VIII. Theorem. 237. In the same circle, or equal circles, if two chords are unequal, they are unequally distant from the centre, and the greater is at the less distance. In the circle whose centre is 0, let the chords AB and CD be unequal, and AB the greater; and let OE and OF be perpendicular to AB and CD respectively. To prove OE < OF. Proof. Suppose AG drawn equal to CD, and Oil 1. to AG. Then OII=r OF, § 236 {in the same O two equal chords are equidistant from the centre). Join FH. OF and Off bisect AB and AG, respectively, § 232 (a radius L to a chord bisects it). Since, by hypothesis, ^^ is greater than CD or its equal A G, AF, the half of AD, is greater than Aff, the half of AG. .*. the Z AffF is greater than the Z AFIl § 158 {tJie greater of two sides of a A has the greater Z opposite to it). Therefore, the Z OlIF, the complement of the Z AffF, is less than the Z OEff, the complement of the Z AFff. .\OF CD. Proof. Suppose AO drawn equal to CD, and OH 1. to A G. Then 011= OF, § 236 {in the same O two equal chords are equidistant from the centre). Hence, OE < OH, Join EH In the A OEHWiQ Z OHE is less than the Z OEH, § 158 {the greater of two sides of a A has the greater Z opposite to it). Therefore, the Z AHE, the complement of the Z OHE, is greater than the Z A EH, the complement of the Z OEH :.AE>AH, §159 {the greater of two A of a A has the greater side opposite to it). But AE=iAB, and AH=iAG. ,'.An> .1(7; hence AB> CD, the equal of AG. 86 PLANE GEOMETRY. —BOOK TI. Proposition X. Theorem. 239. A straight line perpendicular to a radius at its extremity is a tangent to the circle. II A -^ Let MB be perpendicular to the radius OA at A, To prove MB tangent to the circle. Proof. From draw any other line to MB, as OCR. OR>OA, §114 (a ± is the shortest line from a point to a straight line). .'. the point ^is without the circle. Hence, every point, except A, of the line MB is without the circle, and therefore MB is a tangent to the circle at A. § 213 a E. D. 240. Cor. 1. A tangent to a circle is perpendicular to the radius drawn to the point of contact. For, if MB is tangent to the circle at A, every point of MB, except A, is without the circle. -Hence, OA is the shortest line from to MB, and is therefore perpendicular to MB (§ 114) ; that is, MB is per- pendicular to OA. 241. Cor. 2. A perpendicular to a tangent at the point of contact passes through the centre of the circle. For a radius is perpendicular to a tangent at the point of contact, and there- fore, by § 89, a perpendicular erected at the point of contact coincides with this radius and passes through the centre. 242. Cor. 3. A perpendicular let fall from the centre of a circle upon a tangent to the circle pafises through the point of contact ARCS AND CHORDS. 87 Proposition XI. Theorem. 243. Parallels intercept equal arcs on a circuni- ference. Fig. 2. Let AB and CD be the two parallels. Case I. When AB is a tanf/ent, cmd CD a svcatd. Fig. 1. Suppose AB touches the circle at F. To prove arc CF= arc DF. Proof. Suppose FF^ drawn _L to AB. This J_ to AB at i^ is a diameter of the circle. § 241 It is also ± to CD. § 102 .-. arc CF-= arc DF, § 232 (a radius ± to a chord bisects the chord and its subtended arc). Also, arc FCF' = arc FDF\ § 227 .-. arc (FCF' - FC) ■= arc (FDF' - FD), § 82 arc CF' DF' that is, Case II. When AB and CD are secants. Fig. 2. Suppose ^i'^ drawn 11 to CD and tangent to the circle at J/. Then arc AM — arc BIT and arc C3I = arc DM Case I. .•.by subtraction, arc AC = arc BD Case III. When AB and CD are tangents. Fig. 3. Suppose AB tangent at E, CD at F, and GH 11 to AB. Then arc OF = arc FR Case I. and arc GF = arc JIF .". bv addition, arc EGF= arc EHF ' q. e. d. 88 PLANE GEOMETRY. — BOOK II. Proposition XII. Theorem. 244. Through three points not in a straight line, one cireumference, and only one, ean he drawn. Let A, B, G be three points not in a straight line. To prove that a circumference can be drawn throiujlt A, B, and (7, and only one. Proof. Join v4i5and ^C. At the middle points ^i^AB and .SCstippose Js erected. Since BC\^ not the prelongation of AB, these Js will inter- sect in some point 0. The point 0, being in the J_ to ^^ at its mi^le point, is equidistant from A and B\ and being in the J_ to BC 2X its middle point, is equidistant from B and C, § 12:^ {every point in the perpendicular bisector of a straight line is equidista it from the extremities of the straight line). Therefore is equidistant ivom/ A, B, and C\ and a cir- cumference described from as a centre, and with a radiu^ OA, will pass through the three given points. Only one circumference can be made to pass through these points. For the centre of a circumference passing through the three points must be in both perpendiculars, and hence at their intersection. As two straight lines can intei*- Rect in only one point, is the centre of the only circumfer- ence that can pass through the three given points. ;i. e. d. 245. Cor. Two circumferences can intersect in only two points. For, if ^wo circumferences Imve three points common, they coincide and form one circumference. Tangents. 89 Proposition XT IT. Theorem. 246. The tangervts to a circle drawn fronv an exte- rior point are equal, and make equal angles with, the line joining the point to th£^ centre. <: Let AB and AC be tangents from A to the circle whose centre is 0, and AO the line joining A to 0. To prove AB = AC, and ZBAO = ZCAO. Proof. Draw 0£ and OC. AB is X to OB, and AC ± to OC, § 240 (a tangent to a circle, is A. to the radius drawn to the jwiiit of contact). In the rt. A 0^^ and OAC OB^rOa {radii of the same circle). OA = OA. Iden. .\AOAB = AOAC, U61 {having a side and hypotenuse of the one equal to a side and hypotenme of the other). .'.AB^-AC, and ZBAO^ZCAO. ' q.e.d. 247. Def. The line joining the centres of two circles is called tiio luie of centres. 248. Def. A common tangent to two circles is called a common cxtei'ior imige^it when it does not Cut the line of cen- tres, and a coTnmon interior tangent when it cuts the line of centres. 90 PLANE GEOMETEY. BOOK II. Proposition XIV. Theorem. 249. If two circumferences intersect each others the lijie of centres is perpendicular to their coTnmon chord at its middle point. Let C and C be the centres of two circumferences which intersect at A and B. Let AB be their common chord, and CC Join their centres. To prove CC^ 1. to AB at its middle point. Proof. A J_ drawn through the middle of the chord AB passes through the centres C and C\ § 233 (« ± erected at the middle of a chord passes through the centre of the O). .'. the line CC, having two points in common with this J_, must coincide with it. .'. CC is JL to AB at its middle point. q. e. d. Ex. 78. Describe the relative position of two circles if the line of centres : (i.) is greater than the sum of the radii ; (ii.) is equal to the sum of the radii ; (iii.) is less than the sum but greater than the difference of the radii ; (iv.) is equal to the difference of the radii ; (v.) is less than the diflference of the radii. Illustrate each case by a figure. TANGENTS. 9l Proposition XV. Theorem. 250. If two circumferences are tangent to each otioer, tlie line of centres passes through the point of contact. Let the tvro circumferences, 'whose centres are C and C, touch each other at O, in the straight line AB, and let GC be the straight line joining their centres. To prove is in the straight line CC Proof. A ± to AB, drawn through the point 0, passes through the centres C and C", § 241 {a ± to a tangent at the point of contact passes through the centre of the circle). .-. the line CC\ having two points in common with this J, must coincide with it. .'. is in the straight line CC. ae-o. Ex. 79. The line joining the centre of a circle to the middle of £■ chord is perpendicular to the chord. Ex. 80. The tangents drawn through the extremities of a diameter are parallel. xEx. 81. The perimeter of an inscribed equilateral triangle is equal io half the perimeter of the circumscribed equilateral triangle. y Ex. 82. The sum of two opposite sides of a circumscribed quadri- lateral is equal to the sum of the other two sides. 92 PLANE GEOMETRY. — BOOK II. Measurement. 251. To 7)ieasure a quantity oi' any kind is to find how many times it contains another known quantity of the same kind. Thus, to measure a line is to find how man}^ times it con- tains another known line, called the linear unit. The number which expresses how many times a quantity contains the unit, joined with the name of the unit, is called the numerical measure of that quantity ; as, 5 yards, etc. 252. The magnitude of a quantity is always relative to the magnitude of another quantity of the same Ichid. No quantity is great or small except by comparison. This relative magni- tude is called their ratio, and is expressed by the indicated quotient of their numerical measures when the same unit of measure is applied to both. The ratio of a to 5 is written -, o\ a:h. * 253. Two quantities that can be expressed in integers in terms of a common unit are said to be eommerisurablc. The common unit is called a common 'measure, and each quantity is called a multiple of this common^measure.^ Thus, a common measure of 1\ feet and 3|- feet is -J- of a foot, which is contained 15 times in 2\ feet, and 22 times in 3-| feet. Hence, 2-^ feet and 3f feet are multiples of -J- of a foot, 2^ feet being obtained by taking \ of a foot 15 times, and 3| by taking -J of a foot 22 times. 254. When two quantities are incomrncnsurabl^ th^t is, have no common unit in terms of which both quantities can be expressed in integers, it is impossible to find a fraction that will indicate the exact value of the ratio of the given quanti- ties. It is possible, however, by taking the unit sufliciently small, to find a fraction that shall difiPer from the true value of the ratio by as little as we please. KATIO. U3 Thus, suppose a uiid h to denote two lines, such tliaF a a ,- I. -- , r^^- Now Vil-- 1.414^350...,.., a value greater than 1.414^13^ but less than 1.41^214^ If, then, a viilliorith part of h be taken as the unit, the value of the ratio - lies between IvoHiff ^^"^ fHlftv' ^"^^ there- fore differs from either of these fractions by less than y^T^iTrcir- By carrying the decimal farther, a fraction may be found that will differ from the true value of the ratio by less than a billionth, a trilUonth, or any other as^si'jne.d value whatever. Expressed generally, when a and h are incommensurable, and b is divided into any integral number (n) of equal parts, if one of these parts is contained in a more than m, times, but less than m + 1 times, then «>l^but AC, and take AD = AC. Then the variable AM ma,y assume values between ^Z>and ABj while the variable AI^ must always be less than AB. But this is contrary to the hypothesis that the variables should continue equal. ' .'. ^^ cannot be > AC In the same way it may be proved that ^C cannot be >^ ^. .'. AB and AC are two values neither of which is greater than the other Hence A H -= A C. 98 PLANE GEOMETRY, — BOOK II. Measure of Angles. Proposition XVI. Theorem. 261. In tJve same circle, or equal circles, two angles at the centre have the same ratio as their intercepted arcs. Case I. When the arcs are commensurable. In the circles whose centres are C and D, let ACB and EDF be the angles, AB and EF the intercepted arcs. rp Z ACB arc AB 10 prove -^-z— =— - = — — ,• ^ Z UBF arc EF \ Proof. Let m be a common measure oi AB and FF. Suppose m to be contained in AB seven times, and in FF four times. Then ^^^-l (1) arc ^i^ 4 ^ ^ At the several points of division on AB and FF dY9i,w radii. These radii will divide Z A CB into seven parts, and ^ FDFinto four parts, equal each to each, § 229 (in the same O, or equal (D, equal arcs subtend eqiial A at the centre). .ZACB 7 (2) From (1) and (2), ZFBF 4 ZACB ^&rcAB Z FDF arc FF Ax. 1 MEASURE OF ANGLES. Case II. When the arcs are incommensurable, p P' 99 In the equal circles ABP and A'B'P' let the angles ACB and A^C'B' intercept the incommensurable arcs AB and A!Bf. rj, Z. ACB arc AB To prove ____ = ___, Proof. Divide AB into any number of equal parts, and apply one of these parts as a unit of measure to A^B^ as many times as it will be contained in A'B\ Since AB and A^B^ are incommensurable, a certain number of these parts will extend from -4' to some point, as D, leav- ing a remainder DB' less than one of these parts. Draw C^D. Since AB and A^D are commensurable, Z ACB arc AB /. A'C'D arc A'B Case I. If the unit of measure is indefinitely diminished, these ratios continue equal, and approach indefinitely the limiting ratios Therefore Z ACB . arc AB and — -• arc A'B' arc AB Z A'C'B' ZACB §260 Z A'C'B' arc A'B' {If two vai'iables are constantly equal, and each approaches a limit, theit limits are equal.) Q. E. O. 100 PLANE GEOMETRY. — BOOK II. 262. The circumference, like the angular magnitude about a point, is divided into 360 equal parts, called degrees. The arc-degree is subdivided into 60 equal parts, called minutes ; and the minute into 60 equal parts, called seconds. Since an angle at the centre has the same number of angle- degrees, minutes, and seconds as the intercepted are has of arc- degrees, minutes, and seconds, we say : A71 angle at the centre is measured by its intercepted arc; meaning, An angle at the centre is, such a part of the whole angular magnitude about the centre as its intercepted arc is of the whole circumference. Proposition XVII. Theorem. 263. J^n inscribed angle is measured iy onechalf of the arc intercepted between its sides, B B Case I. When one side of the angle is Xi^diameter. In the circle PAB (Fig. 1), let 'ilie centre C be in - one of the sides of the inscribed angle B. To prove /. B is measured by ^arc PA. Proof. Draw CAl. Radius CA — radius CB. ,'.ZB = ZA, . §154 {being opposite equal sides of the A CAB). But ZFCA=ZB + ZA, §145 {the exterior Zofa A is equal to the sum of the two opposite interior A). .'.ZFCA = 2ZB. But Z PC A is measured by PA, § 262 {the A at the centre is measured by the intercepted arc). .'. Z B'm measured by \ PA. MEASURE OF ANGLES. 101 Case II. When the centre is within the angle. In the circle BAE (Fig. 2), let the centre C fall within the angle EBA. To prove Z EBA is measured hy \ arc EA. Proof. Draw the diameter EC P. Z PEA is measured by \ arc PA, Case I. Z. PEE is measured by -J arc PE, Case 1. .-. Z PEA -r Z PEE is measured by \ (arc PA H- arc PE), or Z EEA is measured by \ arc EA. Case III. When the centre is without the angle. In the circle BFP (Fig. 3), let the centre C fall without the angle ABF. • To prove Z AET 'is measured by \ arc AF. Proof. , . Draw the' diameter ECP. Z PEE is'measm-ed by ^ arc PF, Case I. Z ^EA is measui-ed by ^ arc PA. Case I. .'.ZPEF-ZPEA is-measured by | (arc PF- &rc PA), or Z AEF '\8'm'ea,iiUTed \>y ^ arc AF. aE.D. Fig. 1. Fia. 2. Fig. 3. 264. Cor. 1. An angle inscribed m a semicircle is a right angle. For it is measured by one-half a semi-circumference. 265. Cor. 2. An angle inscribed in a segment greater than a semicircle is an acute angle. For it is measured by an arc less than half a semi-circumference; as. Z CAD. Fig. 2. 266. Cor. 3. An angle inscribed in a segment less than a semicircle is an obtuse angle. For it is measured by an arc greater than half a semi-circumference ; as, Z CEE. Fig. 2. 267. Cor. 4. All angles inscribed in the same segment are equal. For they are measured by half the same arc. Fig. 3. 102 PLANE GEOMETRY. BOOK II. Peoposition XVIIL Theorem. 268. An angle formed hy two chords intersecting ivibhln the circumference, is measured by one-half the sum of the intercepted arcs. Let the angle AOC be kormed by the chords AB and CD. To prove A A 00 is measured hy \ {A C-\- BD). Proof. Draw AD. ZCOA = ZD + AA, §145 {the exterior /.ofa/\is equal to the sum of the two opposite interior A). But Z D k measured by -J arc -4(7, § 263 ami Z A is measured by ^ arc BD, {an inscribed Z is measured by ^ the intercepted arc). .-. Z CO A is measured by i {AC+ BD). Q.E,0. Ex. 83. The opposite angles of an inscribed quadrilateral are sup- plements of each other. s5v Ex. 84. If through a point within a circle two perpendicular chords ''itre drawn, the sum of the opposite arcs which they intercept is equal to a semi-circumference. Ex. 85. The line joining the centre of the square described upon the hypotenuse of a rt. A, to the vertex of the rt. Z, bisects the right angle. Hint. Describe a circle upon the hypotenuse as diameter. MEASURE OF ANGLES. 1U:J Froposition XIX. Theorem. 269. An angle formed hy a tangent arul a chord is me and chord AH. To prove Z. MA His rneasured hy \ arc A EH. Proof. ' Draw the diameter ACF. Z Jf^i^ is art. Z, §210 {the radius drawn to a tangent at the 'point of contact is JL to it). /L MAF being a rt. Z, is naeasured by \ the semi-circuni- ference AEF. But Z //yli^is measured by J arc //i^, §263 {an inscribed Z. is measured by h the intercepted arc). .'. Z MAF-Z HAFia measured by ^{AFF- HF)\ ^ or Z MAH v^ measured by \ A EH. Q. e. O. Ex. 80. If two circles touch each other and two secants are drawn through the point of contact, the chords joining their extremities are parallel. Hint. Draw the common tangent. 104 PLANE GEOMETRY. — BOOK II. Proposition XX. Theorem. 270. An angle formed by two secants, two tangents, or a tangent and a secant, intersecting without the circumference, is measured by one-half the difference of the intercepted arcs. Fig. 3. Cake I. Angle formed by two secafits. Let the angle (Fig. 1) be formed by the two se- cants OA and OB. To prove Z. is measured by ^ (A^ — JEC). Proof. Draw CB. /L ACB^ A 0-Vl.B, § 145 {the exterior A oj a IS is equal to the sum of the two opposite interior A). By taking away Z B from both sides, ZO^ZACB-ZB. ^But and Z A CB is measured by \ AB, 263 Z i? is measured by ^ CE, {an inscribed Z is measured by } the intercepted arc). Z is measured by ^(AB— CE), MEASURE OF ANGLES. 105 §ASE II. Angle formed by two tangents. Let the angle (Fig. 2) be formed by the two tan- gents OA and OB. To prove Z is measured by ^ {AMB — A8B). Proof. ^ Draw AB. /. ABC= ZO + Z OAB, % 145 {the exterior Z of a A is equal to the sum of the two opposite interior A). By taking away Z OAB from both sides, ZO = ZABC-ZOAB. But • Z ABC is measured by ^ AMB, § 269 and Z OAB is measured by -J- ASB, Can Z formed hy a tangent and a chord is measured by I the intercepted arc). .-. Z is measured by \ (AMB— ASB). Case III. Angle formed by a tangent and a secant. Let the angle (Fig. 3) be formed by the tangent OB and the secant OA. To prove Z is measured by -J- (ADS — CES). Proof. Draw C8. ZACS=ZO + ZCSO/ §145 (the exterior Zofa/\is equal to the sum of the two opposite interior A). By taking away Z C80 from both sides, Z0 = ZAC8-ZC80. But ZACSi^ measured by ^ ADS; § 263 (being an inscribed Z), and Z CSO is measured by | CES, § 269 (being an Z formed by a tangent and a chord). .-. Z is measured hj i(ADS— CES). Q.E.O. 106 PLANE GEOMETRY. BOOK II. Problems of Construction. Proposition XXI. Problem. 271. At a given point in a straight line, to erect a perpendicular to that line. :^ * 1 \ ! 1 H n C A- £; " -- - tM Fig. I. Fig. 2. I. Let be the given point in AC. (Fig. 1). To erect a JL to the line AC at the point 0. Construction. From as a centre, with any radius OB, describe an arc intersecting ACin two points .^Tand J5. From JTand J5 as centres, with equal radii greater than OB, describe two arcs intersecting at B. Join OB. Then the line OB is the _L required. Proof. Since and B are two points at equal distances from jETand B, they determine the position of a perpendicular to the line ITB at its middle point 0. § 123 aE.F. II. When the given point is at the end of the line. Let B be the given point. (Fig. 2). To erect a JL to the line AB at B. Construction. Take any point C without AB ; and from C as a centre, with the distance CB as a radius, describe an arc intersecting AB Sit ^. Draw BC, and prolong it to meet the arc again at D. Join BD, and BB is the X required. ' Proof. The Z ^ is inscribed in a semicircle, and is therefore a right angle. § 264 Hence BB is -L to AB. o. e. f. PROBLEMS. 107 Proposition XXII. ' Problem. 272. From a point without a straight line, to let fall a perpendicular upon that line. KO H M Let AB be a given straight line, and C a given point without the line. To lei fall a ± to the line AB from ike •point C. Construction. From C as a centre, with a radius sufiiciently great, describe an arc cutting AB in two points, ^and K. From ^and K a.^ centres, with equal radii greater than \HK, describe two arcs intersecting at 0. Draw CO, and produce it to meet AB at M. CM'm the ± required. Proof. Since Cand are two points equidistant from ^and K, they determine a _L to UK at its middle point. § 123 a E. F. Note. Given lines of the figures are full lines, resulting lines are long-dotted, and auxiliary lines are short-dotted. 108 PLANE GEOMETRY. — BOOK II. Proposition XXIII. Problem. 273. To bisect a given straight line. ^I:i< 4-^ # Let AB be the given straight line. To bisect the line AB. Construction. From A and B as centres, with equal radii greater than J AB, describe arcs intersecting at Cand E. Join CE. Then the line CE bisects^^^. Proof. C and E are two points equidistant from A and B. Hence they determine a X to the middle point of AB. § 123 a E. F. Ex. 87. To find in a given line a point X which shall be equidis- tant from two given points. Ex. 88. To find a point X which shall be equidistant from two given points and at a given distance from a third given point. Ex. 89. To find a point X which shall be at given distances from two given points. Ex. 90. To find a point X which shall be equidistant from three given points. PROBLEMS. 109 Proposition XXIV. Probt.em. 274. To bisect a $lven arc. \ I / Let ACS be the given arc. To bisect the arc ACB. Oonstruction. Draw the chord AB, From A and B as centres, with equal radii greater than \ A B, describe arcs intersecting at I) and U, Draw BR BE bisects the arc ACB. Proof. Since B and B are two points equidistamt from A and B, they determine a JL erected at the middle of chord AB. § 123 And a X erected at the middle of a chord passes through the centre of the O, and bisects the arc of the chord. § 234 aE. F. Ex. 91. To construct a circle having a given radius and passing through two given points. Ex. 92. To construct a circle having its centre in a given line and passing through two given points. 110 PLANE GEOMETHY. — BOOK II. Proposition XXV. Problem. 275. To bisect a given angle. Let AEB he the given angle. To bisect Z. AEB. Construction. From ^as a centre, with any radius, as EA, describe an arc cutting the sides of the A E at A and B. From A and B as centres, with equal radii greater than one-half the distance from A to -6, describe two arcs inter- secting at O. Jom EC, AC, and BC. ^C bisects the Z^. Proof. In the A ^i?C and ^^(7 AE= BE, and AG= BO, Cons. and EC=EC. Iden. .-. A AEC== A BEC, § 160 {having three sides equal each to each). ,:ZAEC=ZBEa Q. E. F. Ex. 93. To divide a right angle into three equal parts. Ex. 94. To construct an equilateral triangle, having given one side. Ex. 95. To find a point Xwhich shall be equidistant from two given points and also equidistant from two given intersecting lines. PROBLEMS. Ill Proposition XXVI. Problem. 276. At a given point in a given straight line, to construct an angle equal to a ^iven angle. Let C he the given point in the given line CM, and A the given angle. To construct an A at Q equal to the Z A. Oonstmctior. From -4 as a centre, with any radius, as AE, describe an arc cutting the sides of the Z ^ at ^ and F. ■ From Cas a centre, with a radius equal to AE, describe an arc cutting CJf at H, From -ff as a centre, with a radius equal to the distance EF, describe an arc intersecting the arc HG at m. Draw Cm, and HCm is the required angle. Proof. The chords ^i^and Hm are equal. Cons. .-. arc EF= arc Hm, § 230 {in equal AB equals one-third the Z BOB. 125. All chords of a circle which touch an interior concentric circle are equal, and are bisected at the points of contact. 126. If two circles intersect, ^nd a secant is drawn through each point of intersection, the chords which join the extremities of the secants are parallel. Hint. By drawing the common chord, two inscribed quadrilaterals are obtained. 127. If an equilateral triangle is inscribed in a circle, the distance of each side from the centre of the circle is equal to half the radius of the circle. 128. Through one of the points of intersection of two circles a diameter of each circle is drawn. Prove that the straight line joining the ends of the diameters passes through the other point of intersection. EXEKCISES. 127 129. A circle touches two sides of an angle BAC a,t B, C; through any point D in the arc BC a tangent is drawn, meeting AB at J57and AC at F. Prove (i.) that the perimeter of the triangle AEF is constant for all positions of D in BC;J^ that the angle EOF is also constant. Loci. 130. Find the locus of a point at three inches from a given point. 131. Find the locus of a point at a given distance from a given circumference. 132. Prove that the locus of the vertex of a right triangle, having a given hypotenuse as base, is the circumference described upon the given hypotenuse as diameter. 133. Prove that the locus of the vertex of a triangle, having a given base and a given angle at the vertex, is the arc which forms with the base a segment capable of containing the given angle. 134. Find the locus of the middle points of all chords of a given length that can be drawn in a given circle. 135. Find the locus of the middle ]K)int.s of all chords that can be • Irawn through a given point -4 in a given circumference. 136. Find the locus of the middle points of all secants that can be lirawu from a given point A to a given circumference^ 137. A straight line moves so that it remains parallel to a given line, and touches at one end a given circumference. Find the locus of the other end. 138. A straight rod moves so that its ends constantly touch two fixed rods which are ± to each other. Find the locus of its middle point. 139. In a given circle let A.OB be a diameter, OC any radius, CD the jierpendicular from Cto AB. Upon OC take OM^CB. Find the locus of the point 1/as OC turns about 0. Construction of Polygons. To construct an eqtftklteral A, having given : 140. The perimeter. 141. The Vadius of the circumscribed circle. 142. The altitude. 143. The radius of the inscribed circle. To construct an isosceles triangle, having given : 144. The angle at the vertex and the ba«e. 128 PLANE GEOMETRY. — BOOK II. 145. The angle at the vertex and the altitude. 146. The base and the radius of the circumscribed circle. 147. The base and the radius of the inscribed circle. 148. The perimeter and the alti- tude. Hints. Let ABC be the A re- quired, and EF the given perimeter. The altitude CD passes through the ^^'< middle of EF, and the ^ AEC, BFC&re isosceles. E A D B F To construct a right triangle, having given : 149. The hypotenuse and one leg. 150. The hypotenuse and the altitude upon the hypotenuse. 151. One leg and the altitude upon the hypotenuse as base. 152. The median and the altitude drawn from the vertex of the rt. Z. 153. The radius of the inscribed circle and one leg. 154. The radius of the inscribed circle and an acute angle. 155. An acute angle and the sum of the legs. - 156. An acute angle and the difference of the legs. To construct a triangle, having given : 157. The base, the altitude, and the Z at the vertex. 158. The base, the corresponding median, and the Z at the vertex. 159. The perimeter and the angles. ^160. One side, an adjacent Z, and the sum of the other sides. 161. One side, an adjacent Z, and the difference of the other sides. 162. The sum of two sides and the angles. 163. One side, an adjacent Z, and radius of circumscribed O. 164. The angles and the radius of the circumscribed O. 165. The angles and the radius of the inscribed O. 166. An angle, the bisector, and the altitude drawn from the vertex 167. Two sides and the median corresponding to the other side. 168. The three medians. To construct a square, having given : 169. The diagonal. < 170. The sum of the diagonal and one side. EXERCISES. 129 To construct a rectangle, having given : 171. One side and the Z formed by the diagonals. 172. The perimeter and the diagonal. 173. The perimeter and the Z of the diagonals. 174. The difference of the two adjacent sides and the Z of the diagonals. To constmct a rhombus, having given : 175. The two diagonals. 176. One side and the radius of the inscribed circle. 177. One angle and the radius of the inscribed circle. 178. One angle and one of the diagonals. To construct a rhomboid, having given: 179. One side and the two diagonals. 180. The diagonals and the Z formed by them. 181. One side, one Z, and one diagonal. 182. The base, the altitude, and one angle. To construct an isosceles trapezoid, having given: 183. The bases and one angle. 184. The bases and the altitude. 185. The bases and the diagonal. 186. The bases and the radius of the circumscribed circle. To construct a trapezoid, having given : 187. The four sides. ' 188. The two bases and the two diagonals. 189. The bases, one diagonal, and the Z formed by the diagonals. Construction of Circles. Find the locus of the centre of a circle : 190. Which has a given radius r and passes through a given point P. 191. Which has a given radius r and touches a given straight line AB 192. Which passes through two given points Pand Q. 193. Which touches a given straight line ^5 at a given point P. 194. Which touches each of two given parallels. 195. Which touches each of two given intersecting lines. 130 PLANE GEOMETEY. — BOOK II. To construct a circle which has the radius r and which also : 196. Touches each of two intersecting lines AB and CD. 197. Touches a given line AB and a given circle K 198. Passes through a given point P and touches a given line AB. 199. Passes through a given point P and touches a given circle K. To construct a circle which shall : 200. Touch two given parallels and pass through a given point P. 201. Touch three given lines two of which are parallel. 202. Touch a given line AB at P and pass through a given point Q. 203. Touch a given circle at Pand pass through a given point Q. 204. Touch two given lines and touch one of thorn at a given point P 205. Touch a given line and touch a given circle at a point P. 206. Touch a given line AB at P and also touch a given circle. 207. To inscribe a circle in a given sector. 208. To construct within a given circle three equal circles, so that each shall touch the other two and also the given circle. 209. To describe circles about the vertices of a given triangle as centres, so that each shall touch the two others. Construction of Straight Lines. 210. To draw a common tangent to two given circles. 211. To bisect the angle formed by two lines, without producing the lines to their point of intersection. 212. To draw a line through a given point, so that it shall form with the sides of a given angle an isosceles triangle. 213. Given a point P between the sides of an angle BAC. To draw through P a line terminated by the sides of the angle and bisected at P. 214. Given two points P, Q, and a line AB ; to draw lines from P and Q which shall meet on AB and make equal angles with AB. Hint. Make use of the point which forms with P a pair of points svmmetrical with respect to AB. 215. To find the shortest path from Pto Q which shall touch a,VmeAB. 216. To draw a tangent to a given circle, so that it shall be parallel to a given straight line. Jl^^ -U P IV 9 ^« BOOK III. PROPORTIONAL. LINES AND SIMILAR POLYGONS. The Theory of Proportion. 292. A proportion is an expression of equality between two equal ratios. A proportion may be expressed in any one of the follow- ing forms : 7= — ; a:6 = c:a; a\h '. '. c : d\ b a and is read, " the ratio of a to 5 equals the ratio of e to d." 293. The terms of a proportion are the four quantities com- pared ; the first and third terms are called the antecedents, the second e^nd fourth terms, the consequents; the first and fourth terms are called the extremes, the second and third terms, the means. 294. In the proportion a:h = c\d, d is a fourth propor- tional to a, h, and c. In the proportion a:h = h:c, c is a third proportional to a and h. In the proportion a:b — h\c, h is a mean proportional between a and c. 132 PLANE GEOMETRY. — BOOK III. Proposition I. 295. In every proportion the product of the extremes is eqii^al to the product of the means. Let a: b = c- d. To prove ad^bc. Now ^ = ^. b d whence, by multiplying both sides by bdy ad = be. Q, E, o^ Proposition II. 296. A mean proportional between two quantities Is equal to the square root of their product. In the proportion (t:b —b :c, h" = ac, § 296 {tilt product of the extremes is equal to the product of the means). Whence, extracting the square root, b = -Vac. Q. E. D. Proposition III. 297. // the product of two quantities is equal to the product of two others, either two may be made the extremes of a pj^oportion in which the other two are made the means. ^_ Let. ad^ he. To prove a : b = c : d. Divide both members of the given equation by bd. Then' ^=|, b d ©r, a:b = c: d. a e. o. THEORY OF PROPORTION. 133 Proposition IV. 298. If four quantities of the same hind are in pro- portion, they will he in proportion by alternation ; that is, the first term will he to the third as the sec- ond to the fourth. * Let a:h = c:d. Now "- " To prove a: c = b :d. a _c h~ d Multiply each member of the equation by -. c Then ? = 1 c d . a:c = b:d. aE.D. Proposition V. 299. If four quantities are in .proportion, they will he in proportion hy inversion ; that is, the second term will he to the first as th\fourth to the third. Let a: b = c: d. To prove b:a = d:c. Now be = ad. § 295 Divide each member of the equation by ac. Then . *=^, a c T or, h : a — d : c. Q.E.O. 134 PLANE GEOMETRY. — BOOK III. Proposition VI. 300. If four quantities are in proportion, they will he in proportion by composition ; that is, the sum of the first two terms will he to the second term as the sum of the last two terms to the fourth term. Let a:h = G:d. To jprove a -\-h \h ^= c -\- d \ d. Now f = 2 b d Add 1 to each member of the equation. Then l+l^l+l; b d a-\-h __e-\-d Q.E. D. that is, —^ , . b d or, a -\- h \h = c -\- d \ d. In like manner, a-^h \a=^c-\- d\c. ^ Proposition VII. 301. If foy^ quantities are in proportion, they will he in proportion by division ; that is, the difference of the first two terms will be to the second term as the difference of the last two terms to the fourth term. Let a: b = c:d. To prove a~b :b = c~ d: d. Now a _c ■ b~d Subtract 1 from each member of the eq Then b d that is, a— b c — d b d ' ' or, a~b :b = c — d : d. In like manner, a — h\a = c~d\c. Q.E.DI THEORY OF PROPORTION. 135 Proposition VIII. 302. In any propoiiiion the terms are in proportion by composition and division ; that is, the sum of the first two terms is to their difference as the sum of the last two term^s to their difference. Let a:b = c:d. a-\-b _ c-\-d ' a c a — b c — d Then, by § 300, And, by §301, By division, a c ar-\-b _ c-\-d a ~b c — d or, a-\-b:a — b—c-\-d\c — d. aE. D. Proposition IX. ^ 303. In a series of equal ratios, the sum of the an- tecedents is to the sum of the consequents as any antecedent is to its consequent. Let a:h = c:(l=e :f= g : ft. To prove a-^c-\-e-{-g :b-\-d-{-f-\- h = a:b. Denote each ratio by r. Whence, a = br, c = dr, e =fr, g = hr. Add these equations. Then a-^c-\-e^ g = (b-\-d-\-f-\-h)r. Divide by {b + d-\-f-\- h). Then pLp:A^^r = ^, b+d+f+h b or, a-\-c-\-e-}-g:b-\-d-{-f-\-h = a:b. Q. E. D 136 PLANE GEOMETRY. — BOOK III. Proposition X. 304. The products of the corresponding terms of two or more proportions are in proportion. Let a:b = c:d, e:f=g:h, k:l = m:n. To prove aeh : bfl = cgm : dhn. Now ? = ^, 1 = 2, h=^. d f hi n Whence, by multiplication, aeh _ cgm bfl dhn or, ■ aeh : bfl = cgm : dhn. Q.E. D. Proposition XI. 305. Like powers, or like roots, of the tenrbs of a proportion are in proportion. Let aih — cd. To prove a'\b''^c^: d'\ - L 11 and a» : bri = c^ : c?«. Now ? = 1. b d By raising to the nth power, ^ = ~; or a^:b^ = 'c^:di 6** d^ By extracting the nth root, -y = — -] or, a*^ : 0^ = c^ : an. m a E. D. 306i Equimultiples of two quantities are the products ob- tained by multiplying each of them by the same number. Thus, ma and mb are equimultiples of a and b. THEORY OF PROPORTION. 137 Proposition XII. 307. Equimultiples of two quantities are in the same ratio as the quantities themselves. Let a and b be any two quantities. To prove ma : 7nb =a:b. m Now ? = ?. Multiply both terms of first fraction by m. Then V^^=% mb b or, ma : mb ^= a -.b. Q. E. D. 308. Scholium. In the treatment of proportion it is as- sumed that fractions may be found which will represent the ratios. It is evident that the ratio of two quantities may be represented by a fraction when the two quantities compared can be expressed in integers in terms of a common unit. But when there is no unit in terms of which both quantities can be expressed in integers, it is possible to find a fraction that will represent the ratio to any required degree of accuracy. (See §§ 251-256.) Hence, in speaking of the product of two quantities, as for instance, the product of two lines, we mean simply the product of the numbers which represent them when referred to a com- mon unit. An interpretation of this kind must be given to the product of any two quantities throughout the Geometry. 138 PLANE GEOMETRY. — BOOK III. Proportional Lines. Proposition I. Theorem. 309. If a line is drawn through two sides of a tri- angle parallel to the third side, it divides those sides proportio nally. E/ \V E/ \F B Fig. 1. C Fig. 2. In the triangle ABC let EF be drawn parallel to BG. rj. EB FC To prove aE= AF Case I. When AE and EB (Fig. 1) are commensurable. Find a common measure of AE and EB, as BM. Suppose BM to be contained in BE three times, and in AE four times. E=i (^) At the several points of division on BE and AE draw straight lines II to BC. These lines will divide ^Cinto seven equal parts, of which i^Cwill contain three, and ^i^will contain four, § 187 {if parallels intercept equal parts on any transversal, they intercept equal parts on every transversal). ' . !£ = ?. (2) " AF 4: ^ ^ Compare (1) and (2), EB^IV^ j^^^ AE AF PROPORTIONAL LINES. 139 Case II. When AE and EB (Fig. 2) are incommensurable. Divide AE into any number of equal parts, and apply one of these parts as a unit of measure to EB as many times as it will be contained in EB. Since ^^ and EB are incommensurable, a certain number of these parts will extend from ^ to a point K^ leaving a remainder KB less than t he unit of measure. Draw KH II to BQ. Then ff=Tf- ^^«« I' AE AF Suppose the unit of measure indefinitely diminished, the ratios -^-— and — - continue equal ; and approach indefi- EB FG nitely the limiting ratios -— — and — --, respectively. AE AF Therefore ^ = ^' §260 ae. D. 310. Cor. 1. One side of a tiiangle is to either part cut off by a straight line parallel to the base as the other side is to the corresponding part. For EB:AE=FC: AF, by the theorem. .-. EB:\- AE\ AE= FC+ AF: AF, § 300 or AB:AE=AC:AF 311. Cor. 2. If two lines are cut by any number of parallels, the corresponding intercepts are proportional. Let the lines be AB and CD. Draw AN II to CD, cutting the lis at L, M, and N. Then AL=CO, LM=^GK, MN=KD. §180 By the theorem, B n d AS'. AM= AF: AL = FH: LM=^ HB \ MN. That is, AF:CG= FH: QK^ HB : KD. If the two lines AB and CD were parallel, the correspond- ing intercepts would be equal, and the above proportion be true. A c FJ\L V H \m V 140 PLANE GEOMETRY. — BOOK III. Proposition II. Theorem. 312. If a straight line divide two sides of a tri- angle proportionally, it is parallel to the third side. In the triangle ABC let EF be drawn so that AB^AC AE AF To prove EFWtoBC. Proof. From E draw EH II to BC. Then AB:AE=^AC:AII, ' §310 \ {one side of a A is to either part cut off by a line II to the base, as the other side is to the corresponding part). But AB : AE=AC: AF. Hyp. •The last two proportions have the first three terms equal, each to each ; therefore the fourth terms are equal ; that is, AF= AH. .'. EFsiTid -£'.5" coincide. But . ^^is II to BC. Cons. .'. EF, which coincides with EH, is (j to BC. a ED PROPORTIONAL LINES. 141 Proposition III. Theorem. 313. The bisector of an angle of a triangle divides the opposite side into segments proportional to the other two sides. -.0 B Let CM bisect the angle C of the triangle GAB. To prove MA : MB = CA : CB. Proof. Draw AE II to CMio meet -B(? produced at E. Since CM is II to AE of the A BAE, we have §309 MA.MB^CE'.CB. (1) Since CM\^ II to AE, ZACM-=ZCAE, §104 (being alt. int. A of II lines) ; id ZBCM=ZCEA, §106 {being ext.-int. A o/ll lines). But theZAC3f=ZBCM. Hyp. .-.the 2:: CAE = A CEA. Ax. 1 .: CE= CA, §156 [if two A of a A are equal, the opposite sides are equal). Putting CA for CE in (1), we have MA:MB=CA:C^. aE.o 142 PLANE GEOMETRY. — BOOK III. Proposition IV. Theorem. 314. The bisector of an exterior angle of a triangle meets the opposite side produced at a point the dis- tances of which from the extremities of this side are proportional to the other two sides. Let CM' bisect the exterior angle ACE of the tri- angle CAB, and meet BA produced at M'. To prove M'A : M^B ^ CA : CB. Proof. Draw AF II to CM" to meet BC at R Since AF\^ II to CM' of the A BCM\ we have § 309 M'A'.M'B=CF-CB. (1) Since ^i^ is II to CM\ t\iQAM'CE = A.AFC, §106 {being ext.-mt. AqfW lines) ; and the Z M'CA = Z CAF, § 104 {being alt.-mt AofW lines). Since OM' bisects the Z FCA, ZM'CF=ZM'CA. .'. the ZAFC = Z CAF. Ax. 1 .-. CA = CF, § 156 {if two A of a A are equal, the opposite sides are equal). Putting CA for Ci^in (1), we have M'A:M'B=CA:CB. Q. E. D, PROPORTIONAL LINES. 143 315. Scholium. If a given line AB is divided at M, a point between the extremities A and B, it is said to be divided internally into the segments MA and MB ; and if it is divided at M\ a point in the prolongation of AB, it is said to be divided externally into the segments M^ A and M^B. JI//C , B A M In either case the segments are the distances from the point of division to the extremities of the line. If the line is divided internally, the sum of the segments is equal to the line ; and if the line is divided externally, the difference of the segments is equal to the line. Suppose it is required to divide the given line AB inter- nally and externally in the same ratio ; as, for example, the ratio of the two numbers 3 and 5. \^ 1 M' m' ' ' ' h "We divide AB into 5 + 3, or 8, equal parts, and take 3 parts from A ; we then have the point M, such that MA:MB = S:6. (1) Secondly, we divide AB into two equal parts, and lay off on the prolongation of AB, to the left of A, three of these equal parts ; we then have the point M\ such that M'A:M'B = S:b. (2) Pomparing (1) and (2), MA:MB=^ MA : M'B. 316. If a given straight line is divided internally and externally into segments having the same ratio, the line is said to be divided harmonically. 144 PLANE GEOMETRY. — BOOK III. 317. Cor. 1. The bisectors of an interior angle and an exte- rior angle at one vertex of a triangle p divide the opposite side harmoni- cally. For, by §§ 313 and 314, each bisector divides the opposite side into segments proportional to the other two sides of the triangle. 318. Cor. 2. If the points M and M' divide the line AB harmonically, the points A and B divide the line MM^ har- monically. For, if MA: MB = M'A : M'B, by alternation, MA : MA = MB : M^B. §' 298 That is, the ratio of the distances of A from M and M^ is equal to the ratio of the distances of B from M and M\ The four points A, B, M, and M are called harmonic points, and the two pairs, A, B, and M, M, are called con- jugate harmonic points. Similar Polygons. 319. Similar polygons are polygons that have their homol- ogous angles equal, and their homologous sides proportional. E 1) -£" Thus, if the polygons ABODE and A*B^C^D^E^ are similar the A A, B, C, etc., are equal to A A', B\ C\ etc. , AB BQ CD , 320. In two similar polygons, the ratio of any two iiomol- ogous sides is called the ratio of similitude of the polygons. SIMILAR TRIANGLES. 145 Similar Triangles-. Proposition V. Theorem. 321. Two mutually equiangular triangles are sim- ilar. A A ^' In the triangles ABC and A'B'C let angles A, B, C be equal to angles A', B', C respectively. To prove A. ABO and A}B^C^ similar. Proof. Apply the A A'B'C to the A ABC, so that Z A' shall coincide with Z A. Then the A A'B'C^ will take the position of A AEIT. Now Z AEir(s&me as Z.B') = Z B. .-. ^^is II to BC, §108 {when two straight lines, lying in the same plane, are cut by a third straight line, if the ext.-int. A are equal the lines are parallel). :. AB : AE= AC: AH, § 310 oi- AB'.A'B'^AC'.A'C. In like manner, by applying A A'B'C* to A ABC, so that Z J5' shall coincide with Z B, we may prove that AB-.A'B'^BC.B'C. Therefore the two A are similar. § 319 Q.E.D. 322. Cor. 1. Two triangles are similar if two angles of the one are equal respectively to two angles of the other. 323. Cor. 2. Two right triangles are similar if an acute angle of the 07ie is equal to an acute angle of the other. 146 PLANE GEOMETRY. BOOK III. Proposition VI. Theorem. 324. If two triangles have their sides respectively proportional, they are similar. In the triangles ABC and A'B'G' let AB ^ AG ^ EG A'B' A'Gf B'G'' To prove A ABO and A^B^C^ similar. Proof. Take AE^ A'B', and AII= A'C. Draw EH. Then from the given proportion, AB^AC_ i ae' ah .-. EHis II to BC, § 312 {if a line divide two sides of a A proportionally, it is II to the third side). Hence in the A ABC &nd AEH ' "^ Z ABO= Z. AEH, § 106 and ZACB = ZAHE, {being ext.-int. AofW lines). .-. A ABO siud AEH are similar, § 322 {two A are similar if two A of one are equal respectively to two A of the other). .'. AB: AE ^ BO: EH; that is, AB : A'B' = BO: EH. SIMILAR TRIANGLES. 147 But by hypothesis, The last two proportions have the first three terms equal each to each ; therefore the fourth terms are equal ; that is, ub:= b'C. Hence in the A ^^.fi'and A'B'C, EE:=B'C\ AE=A'B\ and^^=^'C'. .-. A AEB:= a A'B'C\ § 160 {having three sides of the one equal respectively to three sides of the other). But A AEH is similar to A ABC. .-. A A'B^C is similar to A ABC. ^ e. d. 325. Scholium. The primary idea of similarity is likeness u/form ; and the two conditions necessary to similarity are : I. For every angle in one of the figures there must be an equal angle in the other, and II. The homologous sides must be in proportion. In the case of triangles, either condition involves the other, but in the case of other polygons, it does not follow that it one condition exist the other does also. Q' Thus in the quadrilaterals Q and Q\ the homologous sides are proportional, but the homologous angles are not equal. In the quadrilaterals R and B! the homologous angles are equal, but the sides are not proportional. 148 PLANE GEOMETRY. BOOK III. Pkoposition VII. Theorem. 326. If two triangles have an angle of the one equal to an angle of the other, and the including sides pro- portional, they are similar. In the triangles ABC and A'B'C\ let ZA = ZA', and AB ^ AC A'B' A'C' To prove A ABC and A'B'C similar. Proof. Apply the A A'B'O' to the A ABC, so that Z .4' diall coincide with Z A. Then the A A'B'C will take the position of A A EH. AB _ AC A'B' A'C ^T Jil^ Ji.U XT That is, ^= A£- — AE AH Therefore the line UH divides the sides AB and ^C pro- portionally ; .-. ^^is II to BC § 312 {if a line divide two sides of a A proportionally, it is II to the third side). Hence the A ABC and AEH are mutually equiangular and similar. .-. A A'B'C is similar to A A BC. Q. E. D. SIMILAR TRIANGLES. 149 Proposition VIII. Theorem. 327. If two triangles have their sides respectively parallel, or respectively perpendicul/ir, they are siin- ilar, A In the triangles A'B'C and ABC let A'B', A'C, B'C be respectively parallel, or respectively perpendicular, to AD, AC, BC. To prove A A'B'C and ABC similar. Proof. The corresponding A are either equal or supplements ..f each other, §§112,113 (if two A have their sides II, or JL, they are equal or supplementary). Hence we may make three suppositions : 1st. A + A' = 2vtA, B + B' = 2rtA, C+Cf==2rtA. 2d. A = A', B-}-B' = 2rt.A, C+C"=2rt.Zs. 3d. A=^A\ B = B\ .'. C= C\ § 140 Since the sum of the A of the two A cannot exceed four right angles, the third supposition only is admissible. § 138 /. the two A ^5(7 and A'B'C* are similar. § 321 {tiuo mutually emdangular ^ are similar). a E. o 150 PLANE GEOMETRY. BOOK III. Proposition IX. Theorem. 328. The homologous altitudes of two similar tri- angles have the saine ratio as any two homologous sides. A o uV In the two similar triangles ABC and A'B'C, let the altitudes be CO and CO'. To prove CO AC AB CO' A'C A'£' Proof. In the rt. A COA and CO' A', A A^Z. A\ % 319 {being homologous A of the similar A ABC and A'B'C). .-.A COA and CO' A' are similar, s 323 {two rt. A having an acute Z of the one equal to an acute Z of the other are similar), * * ca A'c' In the similar A ^5(7 and A^B'C\ AC _ AB A'C' A'B' §319 Therefore, CO _ AC ^ AB CO' A'C A'B' Q. E. D. SIMILAR TRIANGLES. 151 Proposition X. Theorem. 329. Straight lines d^awn through flie sajne point intercept proportional segments upon two parallels. Let the two parallels AE and A'E' cut the straight lines OA, OB, OC, OD, and OE. AB BC CD DE Topi A'B' BC CD' D^I? Ax. 1 Proof. Since A'E' is II to AE, the pairs of A OAB and OA'B\ OBC and OB^C, etc., are mutually equiangular and similar, . AB OB ^. BC OB goio {homologous sides of similar A are proportional). • • A'B' B'C' In a similar way it may be shown that BC _ CD . CD _ DE B'C CD' ^^ CD' D'E'' Q. E. D Remark, A condensed form of writing the above is A^=.(OB^\^ BC f 00\ _ CD ^ fO^\- DE A'Bf \OB'J B'C [OCJ CI/ \0b') D'E'^ where a parenthesis about a ratio signifies that this ratio is used to prove the equality of the ratios immediately preceding and following it- 152 PLANE CtEOMETRY BOOK in. Proposition XI. Theorem. 330. Conversely : // three ' or more non-parallel straight lines infercej)t proportional segments upon two parallels, they pass through a coTninon point. BL J) IF Let AB, CD, EF, cut the parallels AE and BF so that AC : BD=GE '. DF. To prove that AB, CD, EF prolonged meet in a point. Proof. Prolong AB and CD until they meet in 0. Join OE. If we designate by F the point where OE cuts BF, we shall have by § 329, AC:BD=CE'.DF\ But by hypothesis AC:BD=CE:DF. These proportions have the first three terms equal, each to each ; therefore the fourth terms are equal ; that is, DF^DF, .'. F' coincides with F. .'. ^i*^ prolonged passes through O. .'. AB, CD, and ^i^ prolonged meet in the point 0. "-^ Q. E. Q SIMILAR POLYGONS. 153 Similar Polygons. Proposition XII. Theorem. 331. If two polygons are composed of the sniy\j>, nurn' ber of triangles, similar each to eaxih, arbd similarly placed, the polygons are similar. E B C B' In the two polygons ABODE and A'^'CU'E', let the triangles AEB, BEC, CED be similar respectively to the triangles A'E'B', B'E'C, C'E'D'. To prove ABODE similar to A'B'C'D'E. Proof. Z.A--=Z.A\ §319 {being homologotis zi of similar A). /.ABE-=/.A'B'E' §319 Z EBC = Z EB'C Also, and By adding, Z ABC = Z A'B'C. In like manner we may prove Z BCD — Z B'C'D', etc. Hence the two polygons are mutually equiangular. Now AE __ AB _ f EB\ _ BC^f EC\ CD B'C \ A'E' ED E'D' A'B' \E'B'J B'C \E'C'J CD' {the homologous sides of similar A are proportional). Hence the homologous sides of the polygons are proportional. Therefore the polygons are similar, §.319 (having their homologous A equal, and their homologous sides proportional). 164 PLANE GEOMETRY. — BOOK III. Proposition XIII. Theorem. 332. If two -polygons are siinilar, they are composed of the same rvwinher of triangles, similar each to each, and, similarly plax^ed. B C B' C Let the polygons ABODE and A'B'C'D'E' be similar. From two homologous vertices, as U and U', draw diagonals jEB, EC, and E'B\ E^CK To prove A EAB, EBC, ECD similar respectively to A E'A'B', E'B'C, E'C'I)'. Proof. In the A EAB and E'A'B',. ZA = ZA', {being homologous A of similar polygons) ; AE __ AB A'E' and §319 § 319 A'B' {being homologous sides of similar polygons). .'. A EAB and E'A'B' are similar, § 326 (having an Z of the one equal to an Z of the other, and the including sides proportional). Also, ZABC=ZA'B'C', (1) {being homologous A of similar polygons). And ZABE--=ZA'B'E', (2) {being homologous A of similar A), Subtract (2) from (1), ZEBC=Z E'B'C. Ax. 3 Now And SIMILAR POLYGONS. EB _ AB E'B' A'B'^ (being homologous sides of similar ^). BO _ AB B'C A'B'' {being homologous sides of similar polygons). . EB BO 155 Ax. 1 §326 ' • E'B' B'O* .-. A EBO ?ind E'B'C* are similar, {having an /. oj the one equal to an Z of the other, and the including sides proportional). In like manner we may prove A EOD and E^O^D^ similar. a E. D. Proposition XIV. Theorem. 333. The perimeters of two similar polygons have the same ratio as any two homologous sides. E B C B' C Let the two similar polygons be ABODE and A'B'C D'E', and let P and F* represent their perimeters. To prove F:F' = AlB:A'B'. AB : A'B' = BO: B'O' - OB : O'D', etc., § 319 {the homologous sides of similar polygons are proportional). .-. AB + BO, etc. : A'B' + B'O', etc. = AB : A'B', § 303 {in a series of equal ratios the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent). That is, F:P = AB:A'Br Q.E,a 156 PLANE GEOMETRY. — BOOK III. Numerical Properties of Figures. Proposition XV. Theorem. 334. If in a right triangle a perpendicular is drawn from the vertex of the right angle to the hypotenuse : I. The perpendicular is a mean proportional be- tween the segments of the hypotenuse. II. Each leg of the right triangle is a mean pro- portional between the hypotenuse and Us adjacent segment. In the right triangle ABC, let BF be drawn from the vertex of the right angle B, perpendicular to AC. I. To prove AF: BF= BF: FC. Proof. In the rt. A ^^i^and BAC the acute Z. A is common. Hence the A are similar. § 323 in the rt. A ^CT^and BCA the acute A C\s common. Hence the A are similar. § 323 Now as the rt. A ^-Si^and CBFsltb both similar to ABC, they are similar to each other. In the similar A ^^i^and CBF, AF, the shortest side of the one, BF, the shortest side of the other, BF, the medium side of the one, FC, the medium side of the other. • II. and To prove ACiAB AC: BO AB:AF, BC : FC. NUMERICAL PROPERTIES OF FIGURES. 157 In the similar A ABC and ABF, AC, the longest side of the one, AB, the longest side of the other, AB, the shortest side of the one, AF, the shortest side of the other. Also in the similar A ^J5Cand FBC, AC, the longest side of the one, : BC^ the longest side of the other, : : BC, the medium side of the one, : FC, the medium side of the other. q. e. o. 335. Cor. 1. The squares of the two legs of a rif/ht triangle are proportional to the adjacent segments of the hypotenuse. The proportions in II. give, by § 295, AB'^ACxAF, and BC' = ACx CF. By dividing one by the other, we have A^_ ACxAF _AF -BC^ ACxCF CF' 336. CoR. 2. The squares of the hypotenuse and either leg are propo7'iional to the hypotenuse and adjacent segment. ^ AC" _ ACxAC _AC °'' AB" ACxAF AF 337. Cor. 3. An angle inscribed in a semicircle is a right angle (§ 264). Therefore, I. The perpendicular from aiiy point in the circumference to the diameter of a circle is a mean proportional between the segments of the diameter. "^ ^^ II. The chord drawn from the point to eithe)' extremity of the diamete)' is a mean proportional between the diameter and the adjacent segment. Remark. The pairs of correspoftding sides in similar triangles may be called longest, shortest, medium, to enable the beginner to see quickly these pairs ; but he must not forget that two sides are homologous, not because they appear to be the longest or the shortest sides, but because they lie opposite corresponding equal angles. 158 PLANE GEOMETRY, BOOK III. 4 Proposition XVI. Theorem. 338. The sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. Let ABC be a right triangle with its right angle at B. To prove AB' + BC' = AC\ Proof. Draw BF 1. to AC. Then AF = AC x AF § 334 and BC" = ACx CF By adding, AB" + BC" ^- AC(AF-j- CF) = AC\ q. e. d. 339. Cor. The square of either leg of a right triangle is equal to the difference of the squares of the hypotenuse and the other leg. 340. Scholium. The ratio of the diagonal of a square to the side is the incommensurable num- ber V2. For if ^ C is the diagonal of the square ABCD, then AB' + BC\ or IC" = 2 AB\ AC Divide by AB , we have AC' 'Z, or AC V2. AB' ^B Since the square root of 2 is incommensurable, the diagonal and side of a square are two incommensurable lines. 341, The projection of a line CD upon a straight line ^^ is that part of the line AB comprised q between the perpendiculars CP and DR let fall from the extremities of CD. Thus, PR is the projection of CD upon AB. '' F R A- 'B NUMERICAL PROPERTIES OF FIGURES. 159 Proposition XVII. Theorem. 342. In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides diminished hy twice the product of one of those sides and the prqjection of tJie other upon that side. Let C be an acute angle of the triangle ABC, and DC the projection of AC upon BC. To prove AB' - BC''-{- AC"-2BCx Da Proof. If D fall upon the base (Fig. 1), I)B = BC-I)C; If B fall upon the base produced (Fig. 2), nB=^BC-Ba In either case, BB' ^ BC'-{- BC' -2B0x BC. Add AB^ to both sides of this equality, and we have AB" + BB". ^ BC'+ 10 ■\- BC'- 2BCx BC} But ^^c{-BB' = AB'^,^ §338 and A^+^'^AC', (the Sinn of the squares ofihe two legs of a rt. A is equal to the square of the hypotenuse). Put AB^ and AC' for their equals in the above equality, AB' - BC'' + AC"- 2BCx BC. 160 PLANE GEOMETRY, BOOK III. Proposition XVIII. Theorem. 343. In any obtuse triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides increased by twice the product of one of those sides and the projection of the other upon that side. Let C be the obtuse angle of the triangle ABC, and CD be the projection of AG upon BC produced. To prove AB' = BC'' -\- AO^ -^2BCx DC. Proof. DB=^BC-^ DC. Squaring, DB\^ BC'' + DO" -\-2BCx DC . Add AD to both sides, and we have Aff + DB' = BC'-\-AD'+DC"-^2BC x DC But AD' + DF^AB", §338 and AD'+D0' = AG\ (the sum of. the squares of the two legs of a rt. A is equal to the square of the hypotenuse).: Put AB and AC for their equals in the above equality, AB'=:^BC' + AG' + 2BCxDC a E. o. Note. The last three theorems enable us to compute the lengths of the altitudes if the lengths of the three sides of a triangle are known. NUMERICAL PROPERTIES OF FIGURES. 161 Proposition XIX. Theorem. 344. I. The sum of the squares of two sides of a tri- angle is equal to twice the square of half the third side increased by twice the square of the median upon that side. II. The difference of the squares of two sides of a triangle is equal to twice the -product of the third side hy the projection of the median upon that side. A In the triangle ABC let A }f be the median, and MD the projection of AM upon the side. BC. Also let AB be greater than AC. Toprove I. AF + AC^ = 2BM' -{- 2AM^. II. AB" ~ AC" = 2£Cx MD. Proof. Since AB>AC, tlie Z. AMB will be obtuse, and the Z AMC will be acute. . § 153 Then AB" = Blf\ + AM' + 2 BM X MB, § 343 {in any obtuse A the square of the side opposite the obtuse Z is equal to the sum of the squares of the other two sides increased by twice the product of one of those sides and the projection of the other on that side) ; and AC" = MC' + -Sf ' - 2 MCx MD, § 342 {in any A the square of the side opposite an acute Z is equal to the sum of the squares of the other two sides diminished by twice the product of one of those sides and the projection of the other upon that side). Add these two equalities, and observe that BM= 3fC. Then AB" + AO" = 2BM\-h2AM\ Subtract the second equality from the first. Then AB" - AC' = 2BCx MD. q.e.d. Note. This theorem enables us to compute the lengths of the mediam< if the lengths of the three sides of the triangle are known. 162 PLANE GEOMETRY. — BOOK III. Proposition XX. Theorem. 345. If any chord is drawn through a fixed point within a circle, the product of its segments is con^ stant in whatever direction the chord is drawn. iLet any two chords AB and CD intersect at 0. To prove OA X OB = OD X 00. Proof. Draw ^C and ^i>. In the ^ AOO and BOD, AQ^AB, §263 (eacA heing measured by ^ arc AD). ZA = ZIl, §263 (each being measured by ^ arc BO). .'. the A are similar, § 322 (two A are similar when two A of the one are equal to two A of the other) Whence OA, the longest side of the one, : OB, the longest side of the other, : . 00, the shortest- side of the one, : OB, the shortest side of the other. .-. OAxOB^OBx 00. § 295 aE. D 346. Scholium. This proportion may be written OA^qO OA^J_, OB OB' ^^ OB OB' \ 00 that is, the ratio of two corresponding segments is equal to the reciprocal of the ratio of the other two corresponding segments. In this cast the segments are said to be reciprocally proportional. NUMERICAL PROPERTIES OF FIGURES. 163 Proposition XXI. Theorem. 347. If from a fijced point without a circle a secant is drawn, ths product of the secant and its external segment is constant in whatever direction the secant is drawn, O Let OA and OB be two secants drawn from point 0. To prove OAxOO=OBx 01). Proof. Draw J5C smd AD. In the A OAD and OBC Z is common, ZA = Z.B, §263 {each being measured by J arc CD). .'. the two A are similar, § 322 {two i^ are similar when two A of the one are equal to two A of the other). Whence OA, the longest side of the one, OB, the longest side of the other, OD, the shortest side of the one, OC, the shortest side of the other. .-. OA X 00= OB X OD. § 295 Q. E. D. Remahk. The above proportion continues true if the secant OB turns about until B and D approach each other indefinitely. Therefore, by the theory of limits, it is true when B and D coincide at H. Whence, OA X 0C=^ 0H\ This truth is demonstrated directly in the next theorem. 164 PLANE aEOMETRY. — BOOK III. Proposition XXII. Theorem. 348. If from a point without a circle a secant and a tangent are drawn, the tangent is a meaTi propor- tional between the whole secant and the external segment. Let OB be a tangent and OC a secant drawn from the point to the circle MBC. To prove 00 : OB = OB: OM. Proof. Draw ^ Jf and ^a In the A Oj^Jf and OBC ^ is common. Z OBM is measured by ^ arc 3fB, § 269 (being an Z formed hy a tangent and a chord). Z C is measured by ^ arc BM, § 263 (being an inscribed £). ,\/.OBM=Z.C. .-. A 0^(7 and OBM ^xq similar, § 322 (having two A of the one equal to two A of the other). Whence 00, the longest side of the one, OB, the longest side of the other, OB, the shortest side of the one, OM, the shortest side of the other. Q. E. D. V NUMERICAL PROI-EJEITIES OF FIGURES. 165 Proposition XXIII. Theorem. 349. The square of the bisector of an angle of a triangle is equal to the product of the sides of this angle diminished by the product of the segments determined by the bisector upon the third side of the triangle. 'K- m E Let AD bisect the angle BAC of the triangle ABC. To prove Iff = ABx AC- DB X DC. Proof. Circumscribe the^O ^jBC about the A ABC. Produce AD \o meet the circumference in Ey and draw EG. Then in the A ABD and AEC, ZBAD = ZCAE, Hyp. ZB = Z E, § 263 {each being measured Sy \ the arc AC). .-. A ABD and AEC sltb similar, § 322 [tivo ^ are similar if two A of the one are equal respectively to two A oj the other). Whence AB, the longest side of the one, AE, the longest side of the other, AD, the shortest side of the one, AC, the shortest side of the other. .\ AB X AC= AD X AE^ § 295 = AD(AD+DE) = AD" + AD X DE. But AD X DE= DB x DC, § 345 (the prodjict of the segments of a chord drawn through a fixed point in a Q is constant). .'. ABx AC= AD' + DBxDC. Whence AD" = ABxAC— DB x DC. q. e. d. Note. This theorem enables us to compute the lengths of the bisectors of the angles of a triangle if the lengths of the sides are known. 166 PLANE GEOMETRY. — BOOK III. Proposition XXIV. Theorem. 350. In any triangle the product of two sides is equal to the product of the diameter of the circurn- scrihed circle by the altitude upon the third side. Let ABC he a triangle, AD the altitude, and ABC the circle circumscribed about the triangle ABC. Draw the diameter AE, and draw EC. To prove ABxAC=AExAD. ,,' Proof. In the A ABD and AEC, /. BDA is a rt. Z, Cons. Z. EC A is a rt. Z, §264 (being inscribed in a semicircle), 2.udZB=^ZE. §263 .-.A ABB and ^^Care similar, §323 wo rt. A hav inq an acute Z of the one equal to an acute Z of the other w are similar). AB, the longest side of the one, AE, the- longest side of the other, AB, the shortest side of the one, A Cf, the shortest side of the other. ".-. ABxAC=AExAB. §295 Q. E. D. Note. This theorem enables us to compute the length of the radius of a circle circumscribed about a triangle, if the lengths of the three sides of the triangle are known. i'KuBLEMS OF CONSTKUCTION. 167 Problems of Construction. Proposition XXV. Problem. 351. To divide a given straight line into parts pro- portional to any number of given lines. A, IL A' B n ■■■•.. \ m- p. . ^ Let AB, m, n, and p, be given straight lines. To divide AB into parts proportixynal to m, w, and p. Construction. Draw AX, making an acute Z with AB. On ^Xtake AC=m, CE=n, EX=p. Draw BX. From ^and Cdraw ^^and CH II to BX. uff'and JTare the division points required. Proof, (AE\^AB^IJK_^KB_^ ^ .^^ \AE) AC CE EX ^ (a line drawn through two sides of a A II to the third side divides those sides proportionally). .-. AJI : UK : KB = AC : CU : EX. Substitute m, n, and p for theii- equals AC, CE^ and EX. Then AH : UK : KB = m : n :p. Q.E.F 168 PLANE GEOMETRY. — BOOK III. 5 Proposition XXVI. Problem. < 352. To find a fourth -proportional to three given straight lines, t-^^ m B n C « f: Let the three given lines be m, n, and p. To find a fourth 'proportional to m, n, and p. Draw Ax and Ay containing any acute angle. Oonstruction. On Ax take AB equal to m, BG= n. On Ay take AD = p. Draw BD. From Cdraw OF \\ to BD, to meet Ay at F. DF is the fourth proportional required. Proof. AB : BC= AD: DF, § 309 (a line drawn through two sides of a A W to the third side divides those sides proportionally). Substitute m, ?i, snidp for their equals AB, BO, and AD. Then m '. n=^ p : DF. PLOBLEMS OF CONSTRUCTION. 169 Proposition XXVII. Problem. 353. To -find a third proportional to two given straight lines. A T>^ - - -----E Let m and n be the two given straight lines. To find a third proportional to m and n. Oonstmction. Construct any acute angle A, and take AB=^m., AC=n. ;^oduce AB to D, making BB =■- AC. Join BC. Through D draw DE II to BO to meet reproduced at E. CE is the third proportional to AB and AC. Proof. * AB: BB = AC: CE. § 309 (a line drawn through two sides of a A W to the third side divides those sides proportionally). Substitute, in the above proportion, ^Cfor its equal BI). Then AB : AC=AC: CE. That is, m:n — n : CE. aE.F. Ex. 217. Construct x, if (1) x = — , (2) or = -• c c Special Cases : (1) a = 2, 6 = 3, c = 4 ; (2) a = 3, & = 7, c = 11 ; (3) a = 2, c -= 3 ; (4) a = 3, c = 5 : (5) a - 2c. 170 PLANE GEOMETRY. — BOOK III. Proposition XXVIII. Problem. 354. To -find, a -mean proportioiial between two given straight lines. II m ^ m c ^ Ji -E Let the two given lines be m and n. To find a mean proportional between m and n. Construction. On the straight line AE take AC^^ m, and CB = n. On AB as a diameter describe a semi-circumference. At C erect the _L CH to meet the circumference at H. CITis a mean proportional between m and 71. Proof. .-. AC -CH =CH : CB, § 337 [the, ± let fall from a point in a circumference to the diaineter of a circle is a mean proportional between the segments of the diameter). Substitute for ^(7 and CB their equals m and 71. Then m : CII ^ CB \ n. Q. E. F. 355. A straight lin« is divided in extre7ne and 7nean ratio, when one of the segments is a mean proportional between the whole line and the other segment. Ex.218. Construct a? if a; ="N/a6. Special Cases : (1) a = 2, 6 - 3 ; (2) a = 1, 6 = 5 ; (3) a = 3, 6 = 7. PROBLEMS OF CONSTRUCTION. 171 Proposition XXIX. Problem. 356. To divide a given line in extreme and mean ratio. \Q E.-" O- A C B Let AB be the given line. To divide AB in extreme and mean ratio. Oonstniction. At B erect a ± BE equal to one-half of AB. From _E'as a centre, with a radius equal to EB, describe a O. Draw AE, meeting the circumference in i^and G. On ^^take^C=^i^. On BA produced take AC ^ AG. Then AB is divided internally at C and externally at C" in extreme and mean ratio. Proof. AG : AB = AB '. AF, § 348 {if froin a point without a O a secant and a tangent are drawn, the tan- gent is a mean proportional between the whole secant and the external segment). Then by § 301 and § 300, AG^ AB : AB = AB - AF : AF, (J) AG + AB:AG = AB + AF : AB. (2J By construction FG = 2EB=^ AB. .-. AG~AB = AG-FG = AF=Aa Hence (1) becomes AC :AB = BC :AC^. or, by inversion, AB : AC= AC : BCJ ^ Again, since C'A = AG = AB + Al\ ^ (2) becomes C'B : C'A = C'A : AB. I ^ Q. E. F. 172 PLANE GEOMETKY. — BOOK III. Propositioit XXX. Problem. 357. Upon a given line Jiomologous to a given side of a given polygon, to construct a polygon similar to the given polygon. E Let A'E' be the given line homologous to AE of the given polygon ABODE. To construct on A'U' a jpolygon similar to the given polygon. Oonstmction. From E draw the diagonals EB and EC. From E' draw E'B\ E'C\ and E' D\ m^VmgAA'E'B\ B'E'C\ and C'E'D' equal respectively to A AEB, BEC, and CED. From A' draw A'B\ making Z. E'A'B'= Z EAB, and meeting E^B^ at B\ From B' draw B'C\ making Z E^BW' = Z EBC, and meeting E'C at C. From C draw O^B', making Z E'C'B' = Z ECB, and meeting E'B' at I)'. Then A'B'C'B'E' is the required polygon. Proof. The corresponding A ABE and A'B'E\ EBC and E'B'C, ECD and E'C'H are similar, § 322 {two A ar& similar if they have two A of the one equal respectively to two A of the other). Then the two polygons are similar, § 331 {two polygons composed of the same number of A similar to each other and similarly placed, are similar). mOBLEMS OF COMPUTATION. 173 Problems of Computation. 219. To comi>ute the altitudes of a triangle in terms of its sides. C C e F D Fig. 2. At least one of the angles ^ or .S is acute. Suppose it is the angle B. In the A CDB, In the A ABC, Whence, Hence /t2 = a2 - BI)\ b^ = a' + c^-2cxBD. 338 342 BD 4c2 "" 4c2 ;^ + b^) __ (2ac + a^ + c^- b^){2ac - a' 4c2 ^ {(a + c)'-6'}{6'-(a-c)^} 4c2 _ (g + 6 + c) (g + c - 6) (6 + g - c) (6 - g + c) 4c2 g + 6 + c = 2s. a + c - b = 2{s - b), b + a — c = 2{s — c), b — a + c = 2{s — a). ^2_ 2s X 2(s - g) X 2(s - 6) X 2(s - c) 4c2 By simplifying, and extracting the square root, h = - Vs'(s-g)(8-6)(s-c). Let Then Hence 220. To compute the medians of a triangle in terms of its sides. By I 344, g2 + 62 = 2 m2 + 2 /'^ y. (Fig. 2) Whence 4m2 = 2(g2 + 6»)-c2. .-. m = J V2(g» + 62)_c2. Uwyi::*^^-':)" '^ 174 PLANE GEOMETRY. BOOK III. 221. To coiupuLe the bisectors of a triangle in terms of the sides. By g 349, By 'i 313. ^2 = b .-. AD = ah-ADx BD. BD__AD + BD a he a + 6 Whence t' =-ah~ a + b and BD abc'^ c_ a + b ac a + b 4; ^^ \T^ c D! A ■' '} J N^ / ^-.,^_ .ji'-''' Whence (a + by \ (« + if) ab{{a + bf~c^} (a + bf — <^^ (c^ + ^ -f- c) (<^ + ^ — c) (a + bf abx2sx2(s- c) {a + hf Vabs {s^ — c). o a+b 222. To compute the radius of the circle circumscribed about a tri- angle in terms of the sides of the triangle. ^ ^^ t By §350, ABxAO=-AExAD, or bc = 2Bx AD. But AD Whence Ji^ - Vs (s — a) {s — 6) (s — c). a abc a) (s — b){s — c) 223. If the sides of a triangle are 3, 4, and 5, is the angle opposite 5 right, acute, or obtuse ? ^-^ 224. If the sides of a triangle are 7, 9, and 12, is the angle opposite 12 right, acute, or obtuse ? ' ' 225. If the sides of a triangle are 7, 9, and 11, is the angle opposite 11 right, acute, or obtuse ? ■ 226. The legs of a right triangle are 8 inches and 12 inches; find the lengths of the projections of these legs upon the hypotenuse, and the dis- tance of the vertex of the right angle from the hypotenuse. 227. If the sides of a triangle are 6 inches, 9 inches, and 12 inches, find the lengths (1) of the altitudes ; (2) of the medians ; (3) oi the bisec- tors ; (4) of the radius of the circumscribed circle. EXERCISES. 175 Theorems. ^ 228. Any two altitudes of a triangle are inversely proportional to the corresponding bases. V 229. Two circles touch at P. Through P three lines are drawn, meet- ing one circle in A, B, C, and the other in A^, B\ C, respectively. Prove that the triangles ABC, A'B'C are similar. Si^SO. Two chords AB, CD intersect at M, and A is the middle point of the arc CD. Prove that the product AB x -4i/ remains the same if the chord AB is made to turn about the fixed point A. Hint. Draw the diameter AE, join BE, and compare the triangles thus fvyrmed. s 231. The sum of the squares of the segments of two perpendicular chords is equal to the square of the diameter of the circle. If AB, CD are the chords, draw the diameter BE, join AC, ED, BD, and prove that AC=^ ED. Apply g 338. ^232. In a parallelogram ABCD, a line DE is drawn, meeting the diagonal AC \n F, the side BC in G, and the side AB produced in E. Prove that Df = FOx FE. ^233. The tangents to two intersecting circles drawn from any point s/ in their common chord produced, are equal. (§ 348.) "'^ ^234. The common chord of two intersecting circles, if produced, will bisect their common tangents. (§ 348.) ■- 235, If two circles touch each other, their common tangent is a mean proportional between their diameters. Hint. Let AB be the common tangent. Draw the diameters AC, BD. Join the point of contact P to A, B, C, and D. Show that APD and BPC are straight lines JL to each other, and compare A ABC, ABD. 236. If three circles intersect one another, the common chords all pass through the same point. Hint. Let two of the chords AB and CD meet at 0. Join the point of intersection E to G, and suppose that EO produced meets the same two circles at two different points P and Q. Then prove that 0P= OQ; hence, \ that the points P and Q coincide. 176 PLANE aEOMETRY. — BOOK III. ""237. If two circles are tangent internally, all chords of the greater circle drawn from the point of contact are divided proportionally by the circumference of the smaller circle. Hint. Draw any two of the chords, join the points where they meet the circumferences, and prove that the A thus formed are similar. ^238. In an inscribed quadrilateral, the product of the diagonals is equal to the sum of the products of the opposite sides. nmT. Dr&vf DJS, making ZCDU=ZADB. The ^^^ A ABD and CDE are similar. Also the A BCD and ADE are similar. 239. The sum of the squares of the four sides of any quadrilateral is equal to the sum of the squares of the diagonals, increased by four times the square of the line joining the middle points of the diagonals. Hint. Join the middle points F, E, of the diag- onals. Draw EB and ED. Apply ^ 344 to the A ABC and ADC, add the results, and eliminate BE^ + DE'' by applying g 343 to the A BDE. '" 240. The square of the bisector of an exterior angle of a triangle is equal to the product of the external segments deter- mined by the bisector upon one of the sides, dimin- ished by the product of the other two sides. Hint. Let CD bisect the exterior Z BCH of the A ABC. Circumscribe a O about the A, pro- duce DC to meet the circumference in F, and draw BF. BCF similar. Apply § 347. " 241. If a point is joined to the vertices of a triangle ABC, and through any point A^ in OA a line parallel to AB is drawn, meeting OB at B^, and then through B^ a line parallel to BC, meeting OC at C^, and (7 is joined to A^, the triangle A^B'C will be similar to the tri- angle ABC 242. If the line of centres of two circles meets the circumferences at the points A, B, C, D, and meets the common exterior tangent at P, then FAxFD = PBx FC "^ 243. The line of centres of two circles meets the common exterior tangent at P, and a secant is drawn from P, cutting the circles at the consecutive points E, F, G, H. Prove that FExPH^- PFx PG. Trove ^ACD, ■7i ft EXEBCISES. . 177 L Numerical Exercises. ^ - i'- "^244". A line is drawn parallel to a side AB of a triangle ABC, and cutting *AC in D, BC in E. U AD. DC =2: 3, and AB = 20 inches, find DE. l^j ^ ^ J - ^245. The sides of a triangle are 9, 12, 15. Find the segments made by' '^'' bisecting the angles. (^ 313.) " . ^^ ^^ 246. A tree casts a shadow 90 feet long, when a vertical rod 6 fe©^ *^'^*'. high casts a shadow 4 feet long. How high is the tree ?/'!-' ^ 247. The bases of a trapezoid are represented by a, b, and the altitude by h. Find the altitudes of the two triangles formed by producing the legs till they meet. "^ 248. The sides of a triangle are 6, 7, 8. In a similar triangle the side homologous to 8 is equal to 40. Find the other two sides. **!.'''' _ - ' 249. The perimeters of two similar polygons are 200 feet and 300 feet. If a side of the first polygon is 24 feet, find the homologous side of the second polygon, 3 w 250. How long must a ladder be to reach a window 24 feet high, if the lower end of the ladder is 10 feet from the side of the house ? '^ w *-^ \^251. If the side of an equilateral triangle = a, find the altitude, jf'7' ^ 2o2. If the altitude of an equilateral triangle = h, find the side. - " ^ 253. Find the lengths of the longest and the shortest chord that can be drawn through a point 6 inches from the centre of a circle whose radius is equal to 10 inches. T / C> ^ 254. The distance from the centre of a circle to a chord 10 inches long r- is 12 inches. Find the distance from the centre to a chord 24 inches long. ^ " 255. The radius of a circle is 5 inches. Through a point 3 inches from the centre a diameter is drawn, and also a chord perpendicular to the diameter. Find the length of this clil^^, and the distance from one end of the chord to the ends of the diametter. 'V — *■ '^/; -^ '^56. The radius of a circle is 6 inches. Through a point 10 inches from the centre tangents are drawn. Find the lengths of the tangents, Vi^6 and also of the chord joining the points of contact. XO • ^ - 257. If a chord 8 inches long is 3 inches distant from the centre of the circle, find the radiuinand the distances from the end of the chord to the ends of the diameter which bisects the chord. ■Sf' ^ •' //, '-/-^ 178 PLANE GEOMETRY. — BOOK III. 258. The radius of a circle is 13 inches. Through a point 5 inches from the centre any chord is drawn. What is the product of the two seg- mer^^^ii%^e chord ? What is the length of th)j sljoftest chord that can be drawn through the point ? ' / 259. From the end of a tangent 20 inches long a secant is drawn through the centre of the circle. If the exterior segment of this secant is 8 inches, find the radius of the circle. "^J^ / — ^260. The radius of a circle is 9 inches ; the length of a tangent is 12 inches. Find the length of a secant drawn from the extremity of the tangent to the centre of the circle. 261. The radii of two circles are 8 inches and 3 inches, and the dis- tance between their centres is 15 inches. Find the lengths of their com- mon tangents. / - ' •^' "^ 262. Find the segments of a line 10 inches long divided in extreme and mean ratio. 263. The sides of a triangle are 4, 5, 6. Is the largest angle acute, right, or obtuse ? t^- Problems. >-- 264. To divide one side of a given triangle into segments proportional to the adjacent sides. (§ 313.) ^ 265. To produce a line AB to a point Cso that AB -. AC=3:5. '" 266. To find in one side of a given triangle a point whose distances from the other sides shall be to each other in a given ratio. 267. Given an obtuse triangle ; to draw a line from the vertex of the obtuse angle to the opposite side which shall be a mean proportional between the segments of that side. ■^ 268. Through a given point P within a given circle to draw a chord AB so that AP: BP= 2:3. 269. To draw through a given point P in the arc subtended by a chord AB 2u chord which shall be bisected by AB. 270. To draw through a point P, exterior to a given circle, a secant PAB so that PA : ^P = 4 : 3. 271. To draw through a point P, exterior to a given circle, a secant PAB so that A^ ^PAx PB. 272. To find a point Pin the arc subtended by a given chord AB so that PA:PB = 3: 1. EXERCISES. 179 ■273. To draw through one of the points of intersection of two circles a secant so that the two chords that are formed shall be to each other in the ratio of 3 : 5. ^274, To divide a line into three parts proportional to 2, |, ^. v275. Having given the greater segment of a line divided in extreme and mean ratio, to construct the line. ~276. To construct a circle which shall pass through two given points and touch a given straight line. 277. To construct a circle which shall pass through a given point and touch two given straight lines. 278. To inscribe a square in a semicircle. 279. To inscribe a square in a given triangle. Hint. Suppose the problem solved, and DEFO the inscribed square. Draw Clf II to -4-5, and let j4i^ produced meet • r^ -^^ CM'm M. Draw Off and MN Jl to AB, and produce A Bio meet MN at N. The A ACM, AOF are similar; also the A AMN, AFE Ql are similar. By these triangles show that /l the figure CMNH is a square. By construct- LA ing this-square, the point F can be found. A L 280. To inscribe in a given triangle a rectangle similar to a given rectangle. 281. To inscribe in a circle a triangle similar to a given triangle. 282. To inscribe in a given semicircle a rectangle similar to a given rectangle. 283. To circumscribe about a circle a triangle similar to a given triangle. 284. To construct the expression, x - ^^-^ ; that is, — x c de d e 285. To construct two straight lines, having given their sum and their ratio. 286. To construct two straight lines, having given their difference and their ratio. 287. Having given two circles, with centres and 0\ and a poiiit A in their plane, to draw through the point A a straight line, meeting'the <:ircumferences at B and C, so that AB -. AO^ 1:2. Hint. Suppose the problem solved, join OA and produce it to D. making OA. AD^} -.2. Join DC; ^ OAB. A DC are similar. BOOK IV. AREAS OF POLYGONS. 358, The area of a surface is the numerical measure of the surface referred to the unit bf surface. The unit of surface is a square whose side is a unit of length ; as the square inch, the square foot, etc. 359. Equivalent figures are figures having equal areas. Proposition I. Theorem. 360. The areas of two rectangles having equal alti- tudes are to each other as their bases. Di : ! ! ' 1 ^ ^c n B " - o /^ Let the two rectangles be AC and AF, having the same altitude AD. rrj red. AC AB 10 prove -— 1 = — -— • ^ rect.^i^ AE Proof. Case I. When AB and AE are commensurable. Suppose AB and AE have a common measure, as AO, which is contained in AB seven times and in AE four times. Then if = 4- ' W Apply this measure to AB and A ll\ and at the several points of division erect Js. The rect. ^C'will be divided into seven rectangles, and the rect. ^jPwill be divided into four rectangles. AREAS OF POLYGONS. These rectangles are all equal. rect. AC 7 Hence From (1) and (2) rect.^i^ 4 rect. ^6^ _^^ rect.^i^ AU Case II. When AB and AE are incommensurable. D> ^ f^- 181 §186 (2) Ax. 1 B D K E' Divide AB into any number of equal parts, and apply one of them to AE as often as it will be contained in AE. Since AB and AE are incommensurable, a certain number of these parts will extend from ^4 to a point K, leaving a remainder KE less than one of the parts. Draw KH II to EF. Since AB and ^iTare commensurable, xQGi.AH ^AK ab' Case I. rect. A C These ratios continue equal, as the unit of measure is indefi- nitely diminished, and approach indefinitely the limiting ratios rect.^i^ . AE ,. , — T and -^ respectively. rect.^C AB ^ ^ . rect. ^i^ AE ab' §250 rect. AC {if two variables are constantly equal, and each approaches a limit, the limits are equal). q ^ q 361. Cor. The areas of two rectangles having equal bases are to each other as their altitudes. For AB and AE may be con- sidered as the altitudes, AD smd AB b.s the bases. Note. In propositions relating to areas, the words "rectangle," "triangle," etc., are often used for "area of rectangle," "area of tri- angle," etc. 182 PLANE GEOMETRY. BOOK IV. Proposition II. Theorem. 362. The areas of two rectangles are to each other as the products of their bases hy their altitudes. R h h' b Let 11 and B' be two rectangles, having for their bases h and h', and for their altitudes a and a'-. R _ aX b ■ El a' X h^' Proof. Construct the rectangle 8, with its base the same as that of R, and its altitude the same as that of R\ R a To prove Then 8 §361 and {^rectangles having equal bases are to each other as their altitudes) 8_^h R' V {rectangles having equal altitudes are to each oilier as their bases) By multiplying these two equalities, R __ aXb R' a'xb'' Q.i § 360 Ex. 288. Find the ratio of a rectangular lawn 72 yards by 49 yards to a grass turf 18 inches by 14 inches. Ex. 289. Find the ratio of a rectangular courtyard 18^ yards by ]5h yards to a flagstone 31 inches by 18 inches. Ex. 290. A square and a rectangle have the same perimeter, 100 yards. The length of the rectangle is 4 times its breadth. Compare their areas. Ex. 291. On a certain map the linear scale is 1 inch to 5 miles. How many acres are represented on this map by a square the perimeter of whirli is 1 inch ? AREAS OF ruLYGoNS. 183 Proposition III. Theorem. 363. The area of a rectangle is equal to the product of its base and altitude. s Let R be the rectangle, b the base, and d the alti- tude; and let U be a square whose side is equal to the linear unit. To prove the area of R --- axh. • R ^ axh 'U 1x1 {ixioo rectan(/les are to each other as the product of their bases and altitudes). = axb, §362 But — = the area of R. §358 .*. the area of R = a X b. q. e. d. 364. Scholium. When the base and altitude each contain che linear unit an integral number of times, this proposition is rendered evident by dividing the figure into squares, each equal to the unit of measure. Thus, if the base contain seven linear units, and the altitude four, the figure may be divided into twenty-eight squares, each equal to the unit of measure ; and the area of the figure equals 7x4 units of surface. 184 PLANE GEOMETRY. — BOOK IV. Proposition IV. Theorem. 366. The area of a parallelogram is equal to the product of its base and altitude. BE C F A h D A h D Let AEFD be a parallelogram, AD its base, and CD its altitude. To prove the area of the O AEFB =-- AD X CB. Proof. From A draw AB II to BCio meet i^.^ produced. Then the figure ABCB Avill be a rectangle, with the same base and altitude as the O AEFB. In the rr. A ABE and BCF AB-^CB and AE^ BE, § 179 {heing opposite sides of a CJ). .'.AABE^ABCF, §161 {two rt. ^ are equal when the hypotenuse and a side of the one are equal respectively to the hypotenuse and a side of the other). Take away the A BCF, and we have left the rect. ABCB. Take away the A ABE, and w^ have left the O AEFB. .-. rect. ABCB o O AEFB. Ax. 3 But the area of the rect. ABCB = axb, § 363 .-. the area of the O AEFB = axb. Ax. 1 Q. E. O. 366. Cor. 1. Parallelogravis having equal bases and equal altitudes are equivalent. 367. Cor. 2. Parallelograms having equal bases are to each other as their altitudes ; parallelograms having equal altitudes are to each other as their bases ; any two parallelograms are to each other as the products of their bases by their altitudes. AREAS OF POLYGONS. 185 Proposition V. Theorem. 368. The area of a triangle is equal to one-half of the product of its base by its altitude. §168 §178 B D Let ABC be a triangle, AB its base, and DC its altitude. To prove the area of the A ABC— ^ AB X DC. Proof. , From C draw CH II to AB. From A draw J ^ II to BC. The figure ABCH'i^ a parallelogram, {having its opposite sides parallel), and AC ia its diagonal. .■.AABC=AAJa:C, (the diayonal of a CD divides it into two ec^ual ^). The area of the O ABCJET is equal to the product of its base by its altitude. § 365 Therefore the area of one-half the CJ, that is, the area of the A ABC, is equal to one-half the product of its base by its altitude. Hence, the area of the A ABG=\AB X DC. Q. E. D. 369. Cor. 1. Triangles having equal bases and equal alti- tudes are equivalent. 370. Cor. 2. Tnangles having equal bases are to each other as their altitiides ; triangles having equal altitudes are to each other as their bases ; any two triangles are to each other as the product of their hoses by their altitudes. 186 PLANE GEOMKTEY. BOOK IV. Proposition VI. Theorem. 371. The area of a trapezoid is equal to one-half the sum of the parallel sides multiplied hy the alti- tude. H E h' C qL 1 "\ \r a \ A. FOB Let ABCH be a trapezoid, and EF the altitude. To prove area of ABCH = i {HC-\- AB) EF. Proof. Draw the diagonal AG. Then the area of the A ABC= \ (AB X UF), § 368 and the area of the A AirO= I {HCx EF). By adding, area of ABCH= J {AB + HC) EF. q. e. d. 372. Cor. The area of a trapezoid is equal to the product of the tnedian hy the altitude. For, by § 191, OF is equal to \{HC-\-AB)\ and hence the area of ABCH= OP X EF. 373. Scholium. The area of an irregular polygon may be found by dividing the poly- gon into triangles, and by finding the area of each of these triangles separately. But the method generally employed in practice is to draw the. longest diagonal, and to let fall perpendiculars upon this diagonal from the other angular points of the polygon. The polygon is thus divided into right triangles and trape- zoids ; the sum of the areas of these figures will be the area of the polygon. L AREAS OF POLYGONS. 18; Proposition VII. Theorem. 374. The areas of two tidangles which have an angle of the one equal to an. angle of the other are to each other as the products of the sides including tlxe equal angles. Let the triangles ABC and ADE have the common angle A. n. A ABC ABxAC 10 prove Proof. Now and A ADE AD X AE Draw BE. A ABC AC A ABE AE A ABE_ _AB A ADE AD {^ having the same altitude are to each other as their bases). By multiplying these equalities, AABC _ ABxAC A ADE AD X AE §370 Q. E. D. Ex. 292. The areas of two triangles which have an angle of the one supplementary to an angle of the other are to each other as the products 01 the sides including the supplementary angles. 188 PLANE GEOMETEY, — BOOK IV. Comparison of Polygons. Proposition VIII. Theorem. 375. The areas of two similar triangles are to each other as the squares of any two homologous sides. A o D A' O' Ji' Let the two triangles be ACB and A'C B'. AAOB AB" To prove A A^C^B^ JTb'' Draw the perpendiculars CO and CO'. A ACB AB X CO AB Then CO A A'C'B' A'B' X Ca A'B' ^ C'O'^ §370 {two A are to each other as the products of their bases by their altitudes). But ^=^: §328 (the homologous altitudes of similar A have the same ratio as their homolo- gous bases). CO • AB Substitute, in the above equality, for ^ its equal then CO' AB AB" A ACB AB A A'C'B' A'B' A'B' jJB' A'B'' Q. E. D COMPARISON OF POLYGONS. 189 Pboposition IX. Theorem. 376. The areas of two similar polygons are to each other as tJie squares of any two homologous sides. ir c' Let S and S^ denote the areas of the two similar polygons ABC, etc., and A'B'C, etc. To prove 8:8' = IF : IB^. Proof. By drawing all the diagonals from the homologous \rertices E and E\ the two similar polygons are divided into triangles similar and similarly placed. § 332 AB' A ABE A'B' AA'B'E' Kc'E'V rBE!_\ KWe'V A CBE A BCE A B'C'E' AC'B'E' ^ ' {similar A are to each other as the squares of any two homologous sides). ■ , . A ABE _ A BCE _^ A CBE inatis, '^j^,^,^, '^B'C'E AC'jD'E'' . A ABE-\- BCE+ CBE __ A ABE ^ AB^ . oqq " AA'B'E' -{-B'C'E'-^C'B'E' AA'B'E' A'W {in a series of equal ratios the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent). 8:8' = AB":A'B' Q. E. O. 377. Cor. 1. The areas of two similar polygons are to each of he)' as the squares of any two homologous lines. 378. Cor. 2. The homologous sides of two similar polygons have the same ratio as the square roots of their areas. 190 PLANE GEOMETRY. BOOK IV. Proposition X. ^Theorem. 379. The square described on the hypotenuse of a right triangle is equivalent to the sum of the squares on the other two sides., ' Let BE, GH, AF, be squares on the three sides of the right triangle ABC. To prove BC' =0= AB' + AC\ Proof. Through A draw ^X II to CE, and draw AB and FC. Since A BAG, BAG, and CAII are rt. A, CAG and BAIT are straight lines. Since BB ~ BC, being sides of the same square, and BA — BT] for the same reason, and since Z ABB = ZFBC, each being the sum of a rt. Z and the Z ABC, the A ABB = A FBC. Now the rectangle BB is double the A ABB, [having the same base BD, and the same altitude, the distance between the lis AL and BD), and the square^i^is double the A FBC, {having the same base FB, and the same altitude, the distance between the lis FB and OC). Hence the rectangle BB is equivalent to the square AF. In like manner, by joining ^^and BK, it may be proved that the rectangle CB is equivalent to the square CH. Therefore the square BE, which is the sum of the rectangles BL and CB, is equivalent to the sum of the squares CH and ^F. . Q. E. D. 380. Cor. The square on either leg of a right triangle is equivalent to the difference of the squares on the hypotenuse and the other leg, X §15U COMPARISON OF POLYGONS. 191 Ex. 293, The square constructed upon the sum of two straight lines is equivalent to the sum of the squares constructed upon these two line£, increased by twice the rectangle of these lines. Let AB and BC be the two straight lines, and AC their sum. Con- struct the squares ACOK and ABED upon ^Cand AB respectively. Prolong BE and DE until they meet KG and CO respectively. Then we have the square EFGH, with sides each equal to BC. Hence, the square ACOK is the sum of the squares ABED J) and EFOH, and the rectangles DEHK and BCFE, the dimensions of which are equal to AB and BC. A K BC E H O Ex. 294. The square constructed upon the difference of two straight lines is equivalent to the sum of the squares constructed upon these two lines, diminished by twice the rectangle of these lines. Let AB a,nd AC he the two straight lines, and BC their difference. L C D B E Construct the square ABFG upon AB, the square jj^ j^ ACKH Vi^on AC, and the square BEDC upon BCios shown in the figure). Prolong ED until it meets AG in L. The dimensions of the rectangles LEFG and HKDL are AB and AC, and the square BCDE is evidently the difference between the whole figure and the sum of these rectangles ; that is, the square constructed (^ '^ upon BC is equivalent to the sum of the squares constructed upon AB and AC diminished by twice the rectangle of AB and AC. Ex. 295. The difference between the squares constructed upon two straight lines is equivalent to the rectangle of the sum and difference of these lines. Let ABDE and BCGF be the squares constructed upon the two straight lines AB and BC. The difference between these squares is the polygon ACOFDE, which poly- gon, by prolonging CO to H, is seen to be composed of the rectangles ACHE and OFDH. Prolong AE and CHto J and A" respectively, making j^JJand I£K each equal to BC, and draw IK. The rectangles OFDH and EHKI are equal. The difference between the squares ABDE and BCOF is then equivalent to the rectangle ACKI, which has for dimensions AI= AB + BC, and EH = AB-BC K E H a D A C B 192 PLANE GEOMETRY, BOOK IV. Problems of Construction. , Proposition XI. Problem. 381. To construct a square equivalent to the sum of two given squares. Bl A^ Let R and B' be two given squares. To construct a square equivalent to R' + Ti. Construction. Construct the rt. Z A. Take A G equal to a side of H', AB equal to a side of R\ and draw BC. Construct the square S, having each of its sides equal to BC. S is the square required. Proof. BO' - AC' + AB\ § 379 {the square on the hypotenuse of a rt. A is equivalent to the sum of the squares on the two sides). .'.S-B' + R. Q. E. F. Ex. 296. If the perimeter of a rectangle is 72 feet, and the length is equal to twice the width, find the area. Ex. 297. How many tiles 9 inches long and 4 inches wide will be required to pave a path 8 feet wide surrounding a rectangular court 120 feet long and 36 feet wide ? Ex. 298. The bases of a trapezoid are 16 feet and 10 feet ; each leg is equal to 5 feet. Find the area of the trapezoid. PROBLEMS OF CONSTRUCTION. 193 Proposition XII. Problem. 382. To construct a square equivalent to the differ- ence of two given squares. I \ / \ A R Let B be the smaller square and B' the larger. To construct a square equivalent to PJ — R. Oonstmction. Construct the rt. Z A. Take AB equal to a side of R. From ^ as a centre, with a radius equal to a side of R\ describe an arc cutting the line ^Xat C. Construct the square 8, having each of its sides equal to A 0. 8 is the square required. Proof. AC^ =0= J^ _ aW; § 380 {the square on either leg of a rt. A is equivalent to the difference of the squares on the hypotenuse and the other leg). .-. 8^ R' - R. Q. E. F. Ex. 299. Construct a square equivalent to the sum of two squares whose sides are 3 inches and 4 inches. Ex. 300. Construct a square equivalent to the difference of two squares whose sides are 2^ inches and 2 inches. Ex. 30L Find the side of a square equivalent to the sum of two squares whose sides are 24 feet and 32 feet. Ex. 302. Find the side of a square equivalent to the difference of two squares whose sides are 24 feet and 40 feet. Ex. 303. A rhombus contains 100 square feet, and the length of on( diagonal is 10 feet. Find the length of the other diagonal. 194 PLANE GEOMETRY. BOOK IV. Proposition XIII. Problem. 383. To construct a square, equivalent to the sum of any number of given squares. II y \ \ / \ / \ n / -V...^ \ \ o • / ■■■--.. \ \ r ::i^; Let m, n, o, p, r be sides of the given squares. To construct a square o= m'^ + n^ + o^ -i-p^ + r\ Oonstruction. Take AB = m. Draw AC =n and ± to AB at A, and draw BC. Draw CU = a and _L to BC at C, and draw BU. Draw TJF =p and ± to BE at E, and draw BF. Draw FII= r and ± to BE at E, and draw BH. The square constructed on BITis, the square required. Proof. BE'- - EH -^ BE\ -fh'-^ef' + eb'} - EH" + EF'' -+(:^' + C^} <^ EH" -{- EG!-\- EF -\^CAl + aS\ § 379 (^"' 376 {similar polygons are to each other as the squares of their homologous sides). P^B _ AP^ + A^P^ P'' By addition, 1. A"P"' Q. E. F 196 PLANE GEOMETRY. — BOOK IV. Probositlon XV. Problem. 385. To construct a polygon similar to two given similar polygons and equivalent to their difference. Al B' A B A" B" P O Let R and R' be two similar polygons, and AB and A'B' two homologous sides. To construct a similar polygon equivalent to H' — H. Construction. Construct the rt. Z F, and take FO = AB. From as a centre, with a radius equal to A'B', describe an arc cutting P-Z at JI, and join OIL Take A".B" = FH, and on A"F", homologous to AF, construct F" similar to F. Then F^' is the polygon required. Proof. FB" = 0If'-6F', .'. A^^W = A^ - IS. F' AJB''' Now B" A'^W' and ^^-A£=. §376 {similar polygons are to each other as the squares of their homologous sides). By subtraction, R'-F ^ AJF^-AF '__. " F" A^r^'' PEOBLEMS OF CONSTRUCTION. 197 Proposition XVI. Problem. 386. To construct a triangle equivalent to a given polygon. I A ■-STC Let ABCDHE be the given polygon. To construct a triangle equivalent to the given polygon. Construction, From D draw DE, and from ^draw IIFW to DE. Produce AE to meet IFF at E, and draw DE. Again, draw CE, and draw EE II to CE to meet AE pro- duced at K, and draw CK. In like manner continue to reduce the number of sides of the polygon until we obtain the A CIE. Proof. The polygon ABCBE has one side less than the polygon ABCDHE, but the two are equivalent. For the part ^-SCi)^ is common, and the A DEE^ A DEH, § 3G9 (Jor the base DE is common, and their vertices F and JI are in the line FH II to the base). The polygon ABCK has one side less than the polygon ABODE, but the two are equivalent. For the part ABCEx?, common, and the A CEK^ A CED, § 369 {^Jor the base CF is common, and their vertices K and D are in the line KD II to the base). In like manner the A CIK^ ABCK. Q. E. F. 198 PLANE GEOMETEY. — BOOK IV. Proposition XVII. Problem. 387. To construct a square which shall have a given ratio to a given square. A"^. m- 11- /' ^<^ "-^ y' X / \ \ / a h \-( IB Hi ~-^^ / / E-^«-. / n •-,y Let R be the given square, and ~ the given ratio. m To construct a square which shall be to JR as n is to 7)i. Construction. Take AB equal to a side of H, and draw Ai/, making any acute angle with AB. On At/ take AU=m, EF= n, and join EB. Draw FC II to EB to meet AB produced at C. On J. (7 as a diameter describe a semicircle. At B erect the JL BD, meeting the semicircumference at D. Then BD is a side of the square required. Proof. Denote AB by a, BC by h, and BD by x. Now a:x-='x'.h\ that is, x^ = ah. "§ 337 ^^ence, a^ will have the same ratio to x^ and to ao^ Therefore a^ \ a^ — a^ : ah =^ a : h. But a '. h ■= m : n, §309 (a straight line drawn through two sides of a A, parallel to the third side, divides those sides proportionally). Therefore a^ : x^ = m \ n. By inversion, x^ : a^ n : m. Hence the square on BD will have the same ratio to B as n has to m. q. e. f. PROBLEMS OF CONSTRUCTION. 199 Proposition XVIII. Problem. 388. To construct a polygon similar to a given poly- gon and having a given ratio to it. / \ \ \ \ V / A' D' Let R be the given polygon and -- the given ratio. m To construct a polygon shnilar to JR, which shall be to It as nis to m. Construction. Find a line A'B\ such that the square con- structed upon it shall be to the square constructed upon AB as 71 is to w. § 387 Upon A^Ii' as a side homologous to AB, construct the poly- gon S similar to B. Then S is the polygon required. Proof. S:B = A^'':AB', §376 {similar polygons arc to each other as the squares of tlieir homologous sides). But A^''':AB'' = n:m. Cons. Therefore S: B = n:'m. aE.F. Ex. 301. Find the area of a right triangle if tlie length of the hypote- nuse is 17 feet, and the length of one leg is 8 feet. Ex. 305. Compare the altitudes of two equivalent triangles, if the base of one is three times that of the other. Ex. 306. The bases of a trapezoid are 8 feet and 10 feet, and the alti- tude is 6 feet. Find the base of an equivalent rectangle having an equal altitude. ,J^'' "^ 200 PLANE GEOMETRY. BOOK IV. Proposition XIX. Problem. 389. To construct a square equivalent to a given parallelogram. B R m. N O ^ Let ABCD be a parallelogram, b its base, and a its altitude. To construct a square equivalent to the O ABCD. Oonstmction. Upon the line MX take MJ^= a, and NO = b. Upon MO as a diameter, describe a semicircle. At iV erect NF ± to MO, to meet the circumference at P. Then the square R, constructed upon a line equal to NP, is equivalent to the O ABCD. Proof. MN :NP=NP: NO, § 337 [a JL let fall from any point of a circumference to the diameter is a mean proportional between the segments of the diameter). That is, .'.NP'=MNxJSrO = P^ a ABCD. axb. Q. E. F. 390. Cor. 1. A square may he constructed equivalent to a given triangle, hy talcing for its side a mean 'pro'portional be- tween the base and one-half the altitude of the triangle. 391. Cor. 2. A square may be constructed equivalent to a given 'polygon, hy first reducing the -polygon to an equivalent triangle^ and then constructing a square equivalent to the triangle. a PROBLEMS OF CONSTRUCTION. 201 X Proposition XX. Problem. 392. To construct a parallelogram equivalent to a ^iven square, and having the sum of its base and altitude equal to a given line. R W- Let R be the given square, and let the sum of the base and altitude of the required parallelogram be equal to the given line MN. To construct a O equivalent to H, with the sum of its base and altitude equal to MIT. ,| Oonstraction. Upon MN as a diameter, describe a semicircle. At M erect a J. MP, equal to a side of the given square It. Draw PQ II to MN, cutting the circumference at 8. "Dt^vj 8C 1. to MN Any O having CM for its altitude and CNiov its base is equivalent to M. Proof. SC=PM. §§100,180 .\'8C' = PM"=B. But MC: 80= 80: ON, § 337 (a ± let fall from any point in the circumference to the diameter is a mean proportional between the segments of the diameter). Then So'^MOxON a e. f. Note. This problem may be stated : To construct two straight lines the sum and product of which are known. 202 PLANE GEOMETRY. — BOOK IV. Proposition XXI. Problem. 393. To construct a parallelograin equivalent to a given square, and having tlie difference of its base and altitude equal to a given line, s /^ / / / / / / n' / / / / Let R he the g:iven square, and let the difference oi the base and altitude of the required parallelogram be equal to the given line MN. To construct a O equivalent to H, with the difference of the base and altitude equal to MN. Oonstmction. Upon the given line MN2.^ a diameter, describe a circle. From M draw MB, tangent to the O, and equal to a side of the given square II . Through the centre of the O draw BB intersecting the cir- cumference at (7 and B. Then r.ny O, as B) , having BB for its base and BC for its altitude, is equivalent to R. Proof. BB-BM^BM: BO, § 348 (if from a point without aOa secant and a tangent are drawn, the tangent is a mean proportional between the whole secant and the part without the O). Then BM'-BBxBC, and the difference between BB and BC is the diameter of the O, that is, J[fiV. Q.E.F. Note. This problem may be stated: To construct two straight lines the difference and product of which are known. PROBLEMS OF CONSTRUCTION. 203 Proposition XXII. Problem. 394. To construct a polygon similar to a given poly- gon P, and equivalent to a given polygon Q, \ \ \ \ \ 1 / / A! B' 7j w,. A Let P and Q be two polygons, and AB a side of P. To construct a polygon similar to P and equivalent to Q. Oonstrnction. Find squares equivalent to P and Q, § 391 and let m and n respectively denote their sides. Find A'B*, a fourth proportional to m, n, and AB. § 351 Upon A'£*, homologous to AB, construct P* similar to P. Then P* is the polygon required. Proof. m:n — AB : A^B*, Cons .-. nv'-.n'^ AB" : JJW, But Pom^, and Qon^. Cons, ,'. P: Q^m'' '.n^ = A^ : A^'\ But P:P==A^xAJB^, §376 {mnilar "polygons are to each other as the eqv-ares of their homologous si-des). /.P'.Q^P.r. Ax. 1 .'. P' is equivalent to Q, and is similar to P by construction. acp. ^ 1 HA :^ 204 PLANE GEOMETRY. — BOOK IV. / Problems of Computation. Ex. 307. To find the area of an equilateral triangle in terms of its side. Denote the side by a, the altitude by A, and the area by S. 4 4 * Then Ji' = a' 2 But axh 2 2 aVS a^VS Ex. 308. To find the area of a triangle in terms of its sides By Ex. 219 A = 7 Vs (s — a) (s — 6) (s — c). Hence, S=^X^ Vs{s-a){s-b){s-c) = Vs (s — a){s — b) (s — c). Ex. 309. To find the area of a triangle in terms of the radius of the circumscribing circle. If B denote the radius of the circumscribing circle, and h the altitude of the triangle, we have, by Ex. 222, bxc = 2Bxh. Multiply by a, and we have axbxc = 2Exaxh. But axh = 2S. .•.axbxc = 4:IiX S abc .■.s= 4:B Note. The radius of the circumscribing circle is equal to 4/Sf EXERCISES. / i 205 Theorems. ^310. In a right triangle the product of the legs is equal to the product of the hypotenuse and the perpendicular drawn to the hypotenuse from the vertex of the right angle. \311. If ABC is a right triangle, C the vertex of the right angle, BD a line cutting AG in D, then BD" + IC' = AB' + DO". "^12. Upon the sides of a right triangle as homologous sides three similar polygons are constructed. Prove that the polygon upon the hypotenuse is equivalent to the sum of the polygons upon the legs. 313. Two isosceles triangles are equivalent if their legs are equal each to each, and the altitude of one is equal to half the base of the other. 314. The area of a circumscribed polygon is equal to half the product of its perimeter by the radius of the inscribed circle. 315. Two parallelograms are equal if two adjacent sides of the one are equal respectively to two adjacent sides of the other, and the included angles are supplementary. 316. Every straight line drawn through the centre of a parallelogram divides it into two equal parts. 317. If the middle points of two adjacent sides of a parallelogram are joined, a triangle is formed which is equivalent to one-eighth of the entire parallelogram. 318. If any point within a parallelogram is joined to the four vertices, the sum of either pair of triangles having parallel bases is equivalent to one-half the parallelogram. 319. The line which joins the middle points of the bases of a trape- zoid divides the trapezoid into two equivalent parts. 320. The area of a trapezoid is equal to the product of one of the legs and the distance from this leg to the middle point of the other leg. 321. The lines joining the middle point of the diagonal of a quadri- lateral to the opposite vertices divide the quadrilateral into two equiva- lent parts. 322. The figure whose vertices are the middle points of the sides of any quadrilateral is equivalent to one-half of the quadrilateral. 323. ABC\b a triangle, M the middle point of AB, P any point in AB between A and M. If MD is drawn parallel to PC, and meeting BC at D, the triangle BPD is equivalent to one-half the triangle ABC 206 PLANE GEOMETRY. — BOOK IV. Numerical Exercises. 324. Find the area of a rhombus, if the sum of its diagonals is 12 feet, and their ratio is 3 : 5. 325. Find the area of an isosceles right triangle if the hypotenuse is 20 feet. 326. In a right triangle, the hypotenuse is 13 feet, one leg is 5 feet. Find the area. 327. Find the area of an isosceles triangle if the base = b, and leg = c. 328. Find the area of an equilateral triangle if one side = 8. 329. Find the area of an equilateral triangle if the altitude = h. 330. A house is 40 feet long, 30 feet wide, 25 feet high to the roof, and 35 feet high to the ridge-pole. Find the number of square feet in its entire exterior surface. 331. The sides of a right triangle are as 3 : 4 ; 5. The altitude upon the hypotenuse is 12 feet. Find the area. 332. Find the area of a right triangle if one leg = a, and the altitude upon the hypotenuse = h. 333. Find the area of a triangle if the lengths of the sides are 104 feet, 111 feet, and 175 feet. 334. The area of a trapezoid is 700 square feet. The bases are 30 feet and 40 feet respectively. Find the distance between the bases. 335. ABCD is a trapezium; AB^ 87 het, BO = 119 ieet, CD = 41 feet, DA = 169 feet, A0= 200 feet. Find the area. 336. What is the area of a quadrilateral pircumscribed about a circle whose radius is 25 feet, if the perimeter of the quadrilateral is 400 feet ? What is the area of a hexagon having an equal perimeter and circum- scribed about the same circle ? 337. The base of a triangle is 15 feet, and its altitude is 8 feet. Find the perimeter of an equivalent rhombus if the altitude is 6 feet. 338. Upon the diagonal of a rectangle 24 feet by 10 feet a triangle equivalent to the rectangle is constructed. What is its altitude ? 339. Find the side of a square equivalent to a trapezoid whose bases are 56 feet and 44 feet, and each leg is 10 feet. 340. Through a point Pin the side AB of a triangle ABC, a line is drawn parallel to BC, and so as to divide the triangle into two equiva- lent parts. Find the value of AF in terms of AB. EXERCISES. 207 341. What part of a parallelogram is the triangle cut off by a line drawn from one vertex to the middle point of one of the opposite sides ? 342. In two similar polygons, two homologous sides are 15 feet and 25 feet. The area of the first polygon is 450 square feet. Find the area of the other polygon. 343. The base of a triangle is 32 feet, its altitude 20 feet. What is the area of the triangle cut off by drawing a line parallel to the base and at a distance of 15 feet from the base ? 344. The sides of two equilateral triangles are 3 feet and 4 feet. Find the side of an equilateral triangle equivalent to their sum. 345. If the side of one equilateral triangle is equal to the altitude ot another, what is the ratio of their areas ? 346. The sides of a triangle are 10 feet, 17 feet, and 21 feet. Find the areas of the parts into which the triangle is divided by bisecting the angle formed by the first two sides. 347. In a trapezoid, one base is 10 feet, the altitude is 4 feet, the area is 32 square feet Find the length of a line drawn between the legs parallel to the base and distant 1 foot from it. 348. If the altitude A of a triangle is increased by a length m, how much must be taken from the base a in order that the area may remain the same ? 349. Find the area of a right triangle, having given the segments p, q, into which the hypotenuse is divided by a perpendicular drawn to the hypotenuse from the vertex of the right angle. Problems. 350. To construct a triangle equivalent to a given triangle, and having one side equal to a given length I. 351. To transform a triangle into an equivalent right triangle. 352. To transform a triangle into an equivalent isosceles triangle. 353. To transform a triangle ABO into an equivalent triangle, hav- ing one side equal to a given length I, and one angWequal to angle BAC. Hints. Upon AB (produced if necessary), take AD = I, draw BE II to CD, and meeting J. C (produced if necessary) at -E"; A BED=o=ABEC. 354. To transform a given triangle into an equivalent right triangle, having one leg equal to a given length. 208 PLANE GEOMETRY. BOOK IV. 355. To transform a given triangle into an equivalent right triangle, having the hypotenuse equal to a given length. 356. To transform a given triangle into an equivalent isosceles tri- angle, having the base equal to a given length. To construct a triangle equivalent to : 357.' The sum of two given triangles. 358. The difference of two given triangles. 359. To transform a given triangle into an equivalent equilateral triangle. To transform a parallelogram into : 360. A parallelogram having one side equal to a given length. 361. A parallelogram having one angle equal to a given angle. 362. A rectangle having a given altitude. To transform a square into : 363. An equilateral triangle. 364. A right triangle having one leg equal to a given length. 365. A rectangle having one side equal to a given length. To construct a square equivalent to : 366. Five-eighths of a given square. 367. Three-fifths of a given pentagon. 368. To draw a line through the vertex of a given triangle so as to divide the triangle into two triangles which shall be to each other as 2 : 3. 369. To divide a given triangle into two equivalent parts by drawing a line through a given point P in one of the sides. 370. To find a point within a triangle, such that the lines joining this point to the vertices shall divide the triangle into three equivalent parts. 371. To divide a given triangle into two equivalent parts by drawing a line parallel to one of the sides. 372. To divide a given triangle into two equivalent parts by drawing a line perpendicular to one of the sides. 373. To divide a given parallelogram into two equivalent parts by drawing a line through a given point in one of the sides. 374. To divide a given trapezoid into two equivalent parts by draw- ing a line parallel to the bases. 375. To divide a given trapezoid into two equivalent parts by draw- ing a line through a given point' in one of the bases. Je the polygon ABC, etc., regular. Proof. The arcs AB, BC, CD, etc., are equal, § 230 (in the same O, equal chords subtend equal arcs). Hence arcs ABC, BCD, etc., are equal. Ax. 6 and the A A, B, C, etc., are equal, {being inscribed in equal segments). Therefore the polygon ABC, etc., is a regular polygon, being equilateral and equiangular. q.e. d. 210 PLANE GEOMETRY. — BOOK V. Proposition II. Theorem. 397. A circle may he circurnscrihed about, and a circle may he inscrihed in, any regular polygon. Let ABODE be a regular polygon. I. To prove that a circle may be circumscribed about ABODE. Proof. Let be tlie centre of the circle passing through A,B,C, Join OA, OB, 00, and OD. Since the polygon is equiangular, and tlie A OBOis isosceles, Z.ABO=^Z.BOD and Z.OBO.^4- OOB By subtraction, Z. OB A = Z OCD Hence in the A OB A and OCD the Z. OB A = /. OCD, the radius OB = the radius OC, and AB--=CD. §395 .•.AOAB = AOCD, ' §150 {having two sides and the included /. of the one equal to two sides and the included Z of the other). .-. OA = OD. Therefore the circle passing through A, B, and C, also passes through D. REGULAR POLYGONS AND CIRCLES. 211 In like manner it may be proved that the circle passing through B, C, and D, also passes through E\ and so on through all the vertices in succession. Therefore a circle described from as a centre, and with a radius OA, will be circumscribed about the polygon. II. To prove that a circle may he inscribed in ABODE. Proof, Since the sides of the regular polygon are equal chords of the circumscribed circle, they are equally distant from the centre. § 236 Therefore a circle described from as a centre, and with the distance from to a side of the polygon as a radius, will be inscribed in the polygon. aE.o. 398. The radius of the circumscribed circle, OA, is called the radius of the polygon. 399. The radius of the inscribed circle, OF, is called the apothem of the polygon. 400. The common centre of the circumscribed and in- scribed circles is called the centre of the polygon. 401. The angle between radii drawn to the extremities of any side, as angle AOB, is called the angle at the centre of the polygon. By joining the centre to the vertices of a regular polygon, the polygon can be decomposed into as many equal isosceles triangles as it has sides. Therefore, 402. Cor. 1. The angle at the centre of a regular polygon is equal to four right angles divided by the number of sides of the polygon. 403. Cor. 2. The radius drawn to any vei-tex of a regular polygon bisects the angle at the vertex. 404. Cor. 3. jThe interior angle of a regular polygon is the supplement of the angle at the centre^ For the Z AB0=2ZAB0={Z ABO-\-ZBAO. Hence the A ABO is, the supplement of tne Z AOB. 212 PLANE GEOMETRY. BOOK V. Proposition III. Theorem. / 405. If the circumference of a circle is divided into any number of equal parts, the chords joining the successive points of division form a regular inscribed polygon, and the tangents drawn at the points of division form a regular circumscribed polygon. T D H F Let the circumference be divided into equal arcs, AB, BC, CD, etc., be chords, FBG, GGH, etc., be tangents. I. To prove that ABCDE is a regular polygon. Proof. The sides AB, BC, CD, etc., are equal, § 230 {in the same O equal arcs are subtended by equal chords). Therefore the polygon is regular, § 396 {an equilateral polygon inscribed in a O is regular). II. To prove that the polygon FGHIK is a regular polygon. Proof. In the A AFB, BOC, CHD, etc. AB = BC= CD, etc. § 395 Also, Z BAF= Z. ABF= Z CBG = Z BCG, etc., § 269 {being measured by halves of equal arcs). Therefore the triangles are all equal isosceles triangles. Hence ZF=ZQ = AH, etc. Aiso, FB^BG^GC = CH, etc. Therefore i^6^ = 6«.Zr, etc. /. FGHIK IB a regular polygon. § 395 aE. D. 406. Cor. 1. Tangents to a circumference at the vertices of a regular inserihed polygon form a regular circumscribed poly- gon of the same number of sides. REGULAR POLYGONS AND CIRCLES. 213 407. Cor. 2. If a regular polygon is inscribed in a circle, the tangents drawn at the middle points of the arcs subtended by the sides of the polygon form a circumscribed regular polygon, whose sides are parallel to the sides of the inscribed polygon and whose vertices lie on the radii {prolonged) of the inscribed polygon. For any two cor- A M n' responding sides, as AB and A'B\ perpendicular to OM, are parallel, and the tangents MB^ and NB\ intersecting at a point equidistant from OJf and OiV(§ 246), intersect upon the bisector of the Z MOJSr(§ 163) ; that is, upon the radius OB. 408. Cor. 3. ^ the vertices of a regular inscribed polygon are joined to the middle points of the arcs sub- tended by the sides of the polygon, the- joining lines form a regular inscribed polygon of double the number of sides. 409. Cor. 4. If tangents are drawn at the middle points of the arcs between adjacent points of contact of the sides of a regular cir- cumscribed polygon, a regular circumscribed polygon of double the number of formed. 410. Scholium. The perimeter of an inscribed polygon is less than the perimeter of the inscribed polygon of double the number of sides; for each pair of sides of the second polygon is greater than the side of the first polygon which they replace (§137). The perimeter of a circumscribed polygon is greater than the perimeter of the circumscribed polygon of double the num- ber of sides ; for every alternate side FO, SI, etc., of the poly- gon FGIII, etc., replaces portions of two sides of the circum- scribed polygon ABCD, and forms with them a triangle, and one side of a triangle is less than the sum of the other two sides. 214 PLANE GEOMETRY. BOOK V. Proposition IV. Theorem. li 411. Two regular polygons of the same nwmber of sides are similar. A! n Let Q and Q' be two regular polygons, each having n sides. To prove Q and Q' similar polygons. Proof. The sum of the interior A of each polygon is equal to (n~2)2 rt. A, § 205 {the sum of the interior A of a polygon is equal to 2 rt. A taken as many times less 2 as the polygon has ddes). Each angle of either polygon = ^ ^ — '- — > § 206 {for the A of a regular polygon are all equal, and hence each Z is equal to the sum of the A divided by their number). Hence the two polygons Q and Q' are mutually equiangular. Since AB = BC, etc., and A'B' = B'G\ etc., § 395 AB\A'B'=^BC\B'Q\^tQ, Hence the two polygons have their homologous sides proportional. Therefore the two polygons are similar. § 319 a E.D. 412. Cob. The areas of two regular polygons of the same number of sides are to each other as the squares of any two homologous sides. § 376 REGULAR POLYGONS AND CIRCLES. 215 Proposition V. Theorem. - -413. The perimeters of two regular polygons of the same number of sides are to each otJier as the radii of tlwir eircumscrihed circles, and also as the radii of their inscribed circles. A M n ^i' M' B' Let P and P' denote the perimeters, and O the centres, of the two regular polygons. From 0, 0' draw OA, O'A', OB, O'B', and Js OM, O'M'. To prove F : P = OA : O'A' = OM: O'M'. Proof. Since the polygons are similar, P:P = AB:A'B'. In the isosceles A OAB and O'A'B' the Z = the Z 0', and OA -'OB = O'A' : O'B'. .-. the A OAB and O'A'B' are similar. .-. AB : A'B' = OA : O'A', Also AB : A'B' = OM: O'M', {the homologous altitudes of similar ^ have the same ratio as their bases). .-. F:F'=OA: O'A'^OM: 0M\ aE, D. 414. Cor. The areas of two regular polygons of the same number of sides are to each other as the squares of the radii of their drcumseribed circles, and also as the squares of the radii of their inscribed circles. § 376 §411 §333 §402 §326 §319 §328 216 PLANE GEOMETKY. BOOK V. Proposition VI. Theorem. 415. The difference hetween the lengths of the perim- eters of a regular inscribed polygon and of a similar circumscrihed polygon is indefinitely diminished as the number of the sides of the polygons is indefinitely increased. Let P and P' denote the lengths of the perimeters, AB and A'B' two corresponding' sides, OA and OA' the radii, of the polygons. To prove that as the number of the sides of the polygons is indefinitely increased, JP ~ F is indefinitely diminished. Proof. Since tlie polygons are similar, F:P=OA':OA. By division, P~ P-. P= OA^ - OA: OA. OA' - OA §413 Whence F-P=PX OA Draw the radius OC to the point of contact of A'B\ In the A OA'C, OA' ~OC A EEC. Proof. From E let fall the JL ED. The A ABC ?iwdi EBC, having the same base BC, are to each other as their altitudes AB and ED. § 370 Now EB > ED. § 114 By hypothesis, EB = AB. .'. AB > ED. :.AABC>AEBQ. q.e.d. [AXIMA AND MINIMA. 231 Proposition XIX. Theorem. 446. Of all triangles having the same base and equal perimeters, the isosceles triangle is the maximum. Let the ^ ACB and ADB have equal perimeters, and let the A ACB be isosceles. To prove AACB>AA DB. Proof. Produce AC io II, making CII= AC, and join HB. ABII'm a right angle, for it will be inscribed in the semi- circle whose centre is C, and radius CA. Produce HB, and take I)F= DB. Draw C^and DM II to AB, and join AP. Now AII= AC-\- CB - AJ)^ DB = AD + DP. '^wt AD -^ DP>AP, hence AII> AP. Therefore HB > BP. § 120 But KB = ^ HB and 3fB =iBP. § 121 Hence KB > MB. By ^ 180, KB-^CE and MB = DF, the altitudes of the A ACB &,udi ADB. Therefore A ABC> A ADB. § 370 Q. E. D 232 PLANE GEOMETRY. — BOOK V. Proposition XX. Theorem. 447. Of all polygons with sides all given hut one, the majcimuryb can he inscribed in a semicircle which has the undetermined side for its diameter. c Let ABODE be the maximum of polygons with sides AD, BC\ CD, DE, and the extremities A and E on the straight line MN. To prove ABCDE can he inscribed in a semicircle. Proof. From any vertex, as (7, draw QA and CE. The A ACE must be the maximum of all A having the given sides CA and CE', otherwise, by increasing or diminish- ing the /. ACE, keeping the sides CA and CE unchanged, but sliding the extremities A and E along the line MN, we can increase the A AGE, while the rest of the polygon will remain unchanged, and therefore increase the polygon. But this is contrary to the hypothesis that the polygon is the maximum polygon. Hence the A ACE ^ffii\l the given sides CA and CE is the maximum. Therefore the Z. ACE i^ 2^ right angle, § 445 (the maximum of A having two given sides is the A with the two given sides including a rt, Z). Therefore Clies on the semi-circumference. § 264 Hence every vertex lies on the circumference ; that is, the maximum polygon can be inscribed in a semicircle having the undetermined side for a diameter. a e o. MAXIMA AND MINIMA. 233 Proposition XXI. Theorem. 448. Of all -polygons with given sides, that which van be inscribed in a circle is the maximum. Let ABODE be a polygon inscribed in a circle, and A'B'C'D'E' be a polygon, equilateral with respect to ABCDEf which cannot be inscribed in a circle. To prove ABODE greater than A'B'C'D'E^ Proof. Draw the diameter AIT. Join (7^ and BH. Upon CD' {= CB) construct the A C'IBB' - A CHB, and draw A' IP. Now ABCH> A'B'C'H\ § 447 and AEBH> A'E'B'W, {of all polygons with sides all given hut one, the maximum can he inscribed in a semicircle having me undetermined side for its diameter). Add these two inequalities, then ABCHBE > A'B'C'BT'B'E'. Take away from the two figures the equal A CB^B and C^H^BK Then ABCBE > A'B'C'B'E'. cle.d. 234 PLANE GEOMETRY. — BOOK V. Proposition XXII. Theorem. 449. Of isoperimetric polygons of the same number of sides, the maximum is equilateral. K Let ABCD, etc., be the maximum, of isoperimetric polygons of any given number of sides. To prove AJB, BC, CD, etc., equal. Proof. Draw AC. The A ABC must be the maximiim of all the A which are forraed upon J. C with a perimeter equal to that of A ABC. Otherwise, a greater A AKC could be substituted for A ABC, without changing the perimeter of the polygon. But this is inconsistent with the hypothesis that the poly- gon ABCD, etc., is the maximum polygon. .-. the A ABCh isosceles, § 446 {of all A having the same base and equal perimeters, the isosceles A is the maximum). In like manner it may be proved that BC= CD, etc. q.e. d. 450. Cor. The maxiTnum of isoperimetric polygons of the same number of sides is a regular polygon. ■ For, it is equilateral, § 449 {the maximum of isoperimetric polygons of the same number of sides is equilateral). Also it can be inscribed in a circle, § 448 {the maximum of all polygons formed of given sides can be inscribed in a O). That is, it is equilateral and equiangular, and therefore regular. § 395 Q. E. D. MAXIMA AND MINIMA. 235 Proposition XXIII. Theorem. 451. Of isoperimetric regular polygons, that which has the greatest number of sides is the Tnaximunv, O A D B Let Q be a regular polygon of three sides, and Q' a regular polygon of four sides, and let the two poly- gons have equal perimeters. To prove Q' greate)' than Q. Proof. Draw CD from Cto any point in AB. Invert the A CDA and place it in the position DCE, let- ting D fall at (7, Cat Z>, and A at E. The polygon DBCE is an irregular polygon of four sides, which by construction has the same perimeter as Q , and the same area as Q. Then the irregular polygon DBCE of four sides is less than the regular isoperimetric polygon Q of four sides. § 450 In like manner it may be shown that Q is less than a regular isoperimetric polygon of five sides, and so on. q. e, d. 452. Cor. The area of a circle is greater than the area of any polygon of equal perimeter. \382. Of all equivalent parallelograms having equal bases, the rec- tangle has the least perimeter. '^SSS. Of all rectangles of a given area, the square has the least perimeter. ^384. Of all triangles upon the same base, and having the same alti- tude, the isosceles has the least perimeter. ^^85. To divide a straight line into two parts such that their product shall be a maximum. ^. 236 PLANE GEOMETRY. — BOOK V. Proposition XXIV. Theorem. 453. Of regular -poly ions having a given area, that which has the greatest number of sides has the least perimeter. Q' Let Q and Q' be regular polygons having the same area, and let Q' have the greater number of sides. To prove the perimeter of Q greater than the perimeter of Q\ Proof. Let Q" be a regular polygon having the same perim- eter as Q\ and the same number of sides as Q. Then Q > Q\ § 461 {of isoperimetric regular polygons, that which has the greoAest number of sides is the maximum). But Q=Q'. .-. Q > Q". .'. the perimeter of Q> the perimeter of Q". But the perimeter of Q' = the perimeter of Q". Cons. .•. the perimeter of § > that of Q'. 454. Cor. The circumference of a circle is less than the perimeter of any polygon of equal area. 386. To inscribe in a semicircle a rectangle having a given area ; a rectangle having the maximum area. ^S»387. To find a point in a semi-circumference such that the sum of its ' distances from the extremities of the diameter shall be a maximum. EXERCISES. 237 Theorems. 388. The side of a circumscribed equilateral triangle is equal to twice the side of the similar inscribed triangle. Find the ratio of their areas. 389. The apothem of an inscribed equilateral triangle is equal to half the radius of the circle. 390. The apothem of an inscribed regular hexagon is equal to half the side of the inscribed equilateral triangle. 391. The area of an inscribed regular hexagon is equal to three- fourths of that of the circumscribed regular hexagon. 392. The area of an inscribed regular hexagon is a mean proportional bulween the areas of the inscribed and the circumscribed equilateral triangles. 393. The area of an inscribed regular octagon is equal to that of a rectangle whose sides are equal to the sides of the inscribed and the cir- cumscribed squares. 394. The area of an inscribed regular dodecagon is equal to three times the square of the radius. 395. Every equilateral polygon circumscribed about a circle is regu- lar if it has an odd number of sides. 396. Every equiangular polygon inscribed in a circle is regular if it has an odd number of sides. 397. Every equiangular polygon circumscribed about a circle is regular. 398. Upon the six sides of a regular hexagon squares are constructed outwardly. J rove that the exterior vertices of these squares are the ver- tices of a regular dodecagon. 399. The alternate vertices of a regular hexagon are joined by straight lines. Vrove that another regular hexagon is thereby formed." Find the ratio of the areas of the two hexagons. 400. The radius of an inscribed regular polygon is the mean propor- tional between its apothem and the radius of the similar circumscribed regular polygon. 401. The area of a circular ring is equal to that of a circle whose diameter is a chord of the outer circle and a tangent to the inner circle. 402. The square of the side of an inscribed regular pentagon is equal to the sum of the squares of the radius of the circle and the side of the inscribed regular decagon. 238 PLANE GEOMETRY. — BOOK V. If R denotes the radius of a circle, and a one side of a regular inscribed polygon, show that : R 403. In a regular pentagon, a = -y iq — 2"\/5. 404. In a regular octagon, cb = R-^ 2— \/2. 405. In a regular dodecagon, a = R Vz-^Vl. 406. If on the legs of a right triangle, as diameters, semicircles are described external to the triangle, and from the whole figure a semicircle on the hypotenuse is subtracted, the remainder is equivalent to the given triangle. Numerical Exercises. 407. The radius of a circle == r. Find one side of the circumscribed equilateral triangle. 408. The radius of a circle = r. Find one side of the circumscribed regular hexagon. 409. If the radius of a circle is r, and the side of an inscribed regular polygon is a, show that the side of the similar circumscribed regular )olygon is equal to 2ar V'4r^-a2 "N410. The radius of a circle = r. Prove that the area of the inscribed regular octagon is equal to 2r'^V2. ■^411. The sides of three regular octagons are 3 feet, 4 feet, and 5 feet, respectively. Find the side of a regular octagon equal in area to the sum of the areas of the three given octagons.— y^-^ M12. What is the width of the ring between two concentric circum- ferences whose lengths are 440 feet and 330 feet ? 0» t) -f ^413. Find the angle subtended at the centre by an arc 6 feet 5 inches long, if the radius of the circle is 8 feet 2 inches. ^ ^"^ -f- >414. Find the' angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle. 6'"'^« ^^ ' •415. What is the length of the arc subtended by one side of a regular/ «|L dodecagon inscribed in a circle whose radius is 14 feet? -v-.'V^ "i^i" §^ v416. Find the side of a square equivalent to a circle whose radius is 86 feet fi EXERCISES. 239 ^417. Find the area of a circle msc^Pibed in a square containing 196 square feet. / 3' ^ ,9 ^ ^ V ^Jt f^ * M:18. The diameter of a circular grass plot is 28 feet. , Eind the diam- eter of a circular plot just twice as large. _j ^« ^ "* f-^ *419. Find the side of the largest square that can be cut out of a cir- cular piece of wood whose radius is 1 foot 8 inches. ^ / "^ / • M20. The radius of a circle is 3 feet. What is thejadius of a circle 25 times as lp,rge ? | as large ? -jJjy as large ? -, O'T* ^ •-421. The radius of a circle is 9 feet. What are the radii of the con- centric circumferences that will divide the circle into three equivalent parts? ^.1- c^y^A^ "^i^" ^422. The chord of half an arc is 12 feet, and the radius of the circle is 18 feet. Find the height of the arc. 423. The chord of an arc is 24 inches, and the height of the arc is.-9 inches. Find the diameter of the circle. 424. Find the area of a sector, if the radius of the circle is 28 feet, and the angle at the centre 22J°. 425. The radius of a circle = r. Find the area of the segment sub- tended by one side of the inscribed regular hexagon. 426. Three equal circles are described, each touching the other two. If the common radius is r, find the area contained between the circles. Problems. To circumscribe about a given circle : 427. An equilateral triangle. 429. A regular hexagon. 428. A square. 430. A regular octagon. 431. Totdraw through a given point a line so that it shall divide a given circumference into two parts having the ratio 3 : 7. 432. To construct a circumference equal to the sum of two given circumferences. 433. To construct a circle equivalent to the sum of two given circles. 434. To construct a circle equivalent to three times a given circle. 435. To construct a circle equivalent to three-fourths of a given circle. To divide a given circle by a concentric circumference : 436. Into two equivalent parts. 437. Into five equivalent parts. 240 plane geometry. — book v. Miscellaneous Exercises. Theorems. 438. The line joining the feet of the perpendiculars dropped from the extrertiities of the base of an isosceles triangle to the opposite sides is parallel to the base. 439. If AD bisect the angle J. of a triangle ABC, and BD bisect the exterior angle CBF, then angle ADB equals one-half angle ACB. 440. The sum of the acute angles at the vertices of a pentagram (five- pointed star) is equal to two right angles. 441. The bisectors of the angles of a parallelogram form a rectangle. 442. The altitudes AD, BE, CF of the triangle ABC bisect the angles of the triangle DEF. Hint. Circles with AB, BC, ^C as diameters will pass through E and D, E and F, D and F, respectively. 443. The portions of any straight line intercepted between the cir- cumferences of two concentric circles are equal. 444. Two circles are tangent internally at P, and a chord AB of the larger circle touches the smaller circle at C. Prove that PC bisects the angle AFB. Hint. Draw a common tangent at P, and apply §g 263, 269, 145, 445. The diagonals of a trapezoid divide each other into segments which are proportional. 446. The perpendiculars from two vertices of a triangle upon the opposite sides divide each other into segments reciprocally proportional. 447. If through a point P in the circumference of a circle two chords are drawn, the chords and the segments between P and a chord parallel to the tangent at P are reciprocally proportional. 448. The perpendicular from any point of a circumference upon a chord is a mean proportional between the perpendiculars from the same point upon the tangents drawn at the extremities of the chord. 449. In an isosceles right triangle either leg is a mean proportional between the hypotenuse and the perpendicular upon it from the vertex of the right angle. 450. The area of a triangle is equal to half the product of its perim- eter by the radius of the inscribed circle. MISCELLANEOUS EXERCISES. 241 451. The perimeter of a triangle is to one side as the perpendicular from the opposite vertex is to the radius of the inscribed circle. 452. The sum of the perpendiculars from any point within a convex equilateral polygon upon the sides is constant. 453. A diameter of a circle is divided into any two parts, and upon these parts as diameters semi-circumferences are described on opposite sides of the given diameter. Prove that the sum of their lengths is equal to the semi-circumference of the given circle, and that they divide the circle into two parts whose areas have the same ratio as the two parts into which the diameter is divided. 454. Lines drawn from one vertex of a parallelogram to the middle points of the opposite sides trisect one of the diagonals. 455. If two circles intersect in the points A and B, and through A any secant CAD is drawn limited by the circumferences at C and D, the straight lines BC, BD, are to each other as the diameters of the circles. 456. If three straight lines AA^, BB\ CC^, drawn from the vertices of a triangle ABC to the opposite sides, pass through a common point within the triangle, then OA' OB' oa ^. AA' BB' CC ' 457. Two diagonals of a regular pentagon, not drawn from a common vertex, divide each other in extreme and mean ratio. Loci. 458. Find the locus of a point P whose distances from two given points A and B are in a given ratio (m : n). 459. OP is any straight line drawn from a fixed point to the cir- cumference of a fixed circle ; in OP a point Q is taken such that OQ: OP is constant. Find the locus of Q. 460. From a fixed point A a straight line AB is drawn to any point in a given straight line CD, and then divided at P in a given ratio (m : n). Find the locus of the point P. 461. Find the locus of a point whose distances from two given straight lines are in a given ratio. (The locus consists of two straight lines.) 462. Find the locus of a point the sum of whose distances from two given straight lines is equal to a given length k. (See Ex. 73.) 242 PLANE GEOMETRY. BOOK V. Problems. 463. Given the perimeters of a regular inscribed and a similar circum- scribed polygon, to compute the perimeters of the regular inscribed and circumscribed polygons of double the number of sides. 464. To draw a tangent to a given circle such that the segment inter- cepted between the point of contact and a given straight line shall have a given length. 465. To draw a straight line equidistant from three given points. 466. To inscribe a straight line of given length between two given circumferences and parallel to a given straight line. (See Ex. 137.) 467. To draw through a given point a straight line so that its dis- tances from two other given points shall be in a given ratio (m : n). Hint. Divide the line joining the two other points in the given ratio. 468. Construct a square equivalent to the sum of a given triangle and a given parallelogram. 469. Construct a rectangle having the difference of its base and altitude equal to a given line, and its area equivalent to the sum of a given triangle and a given pentagon. 470. Construct a pentagon similar to a given pentagon and equiva- lent to a given trapezoid. 471. To find a point whose distances from three given straight lines shall be as the numbers m, n, and p. (See Ex. 461.) 472. Given two circles intersecting at the point A. To draw through A a secant B AC such that AB shall be to ^Cin a given ratio (m : n). Hint. Divide the line of centres in the given ratio. 473. To construct a triangle, given its angles and its area. 474. To construct an equilateral triangle having a given area. 475. To divide q, given triangle into two equal parts by a line drawn parallel to one of the sides. 476. Given three points A, ^, C. To find a fourth point P such that the areas of the triangles AFB, APC, BFC, shall be equal. . 477. To construct a triangle, given its base, the ratio of the other sides, and the angle included by them. 478. To divide a given circle into any number of equivalent parts by concentric circumferences. 479. In a given equilateral triangle, to inscribe three equal circles tangent to each other and to the sides of the triangle. SOLID GEOMETRY. BOOK VI. LINES AND PLANES IN SPACE. Definitions. 455. A plane has already been defined as a surface such that a straight line joining any two points in it lies wholly in the surface. A plane is considered to be indefinite in extent, so that however far the straight line is produced, all its points lie in the plane ; but a plane is usually represented by a quadrilat- eral supposed to lie in the plane. 456. A plane is said to be determined by lines or points, if no other plane can contain these lines or points without being coincident with that plane. 457. A plane can be n\ade to turn about any straight line in it as an axis, and be made to assume as many different posi- tions as we choose. Hence it is evident that a plane is not deter- -^^^~~^ J mined by a straight line. y ' / In making a complete revolu- tion about a straight line as an axis the plane passes successively through all points of space. 458. A plane is determined by a straight line and a point without thai line. I N 244 SOLID GEOMETEY. — BOOK VI. If a plane containing the straight line AB revolve about this line as an axis until it con- t^j tains the point C, the plane is determined. For if the plane y^ A^ revolve either way about the ''^ line AB as an axis, it will cease to contain the point C 459. Three points not in a straight line determine a plane. For, by joining any two of the points we have a straight line and a point without it, and these determine a plane. § 458 460. Two intersecting straight lines determine a plane. For, a plane containing one of these straight lines and any point of the other line in addition to the point of intersection is determined. § 458 461. Two parallel straight lines- determine a plane. For, two parallel straight lines lie in iiiSr same ;£lane, and a plane containing either of these parallels and any point in the other is determined. § 458 462. A straight line is perpendicular to a plane if it is per- pendicular to every straight line of the plane drawn through' its foot ; that is, through the point where it meets the plane. In this case the plane is perpendicular to the line. 463. A line is oblique to a plane if it is not perpendicular to all straight lines drawn in the plane through its foot. 464. The distance from a point to a plane is the perpen- dicular distance from the point to the plane. 465. A line is parallel to a plane if it cannot meet the plane however far both are produced. In this case the plane is parallel to the line. 466. Two plpLnes are parallel if they cannot meet however far they are produced. LINES AND PLANES IN SPACE. 245 467. The projection of a point on a plane is the foot of the perpendicular from the point to the a. B plane. 468. The projection of a line on a plane is the locus of the projections of all its points. 469. The angle which a line makes with a plane is the angle which it makes with its projection on the plane. 470. The intersection of two planes is the locus of all the points common to the two planes. Proposition I. Theorem. 471. If two planes cut each other, their intersection is a straight line. Let MN and PQ be two planes which cut one another. To prove their intersection a straight line. Proof. Let A and B be two points common to the two planes. Draw a straight line through the points A and B. Since the points A and B are common to the two planes, this straight line lies in both planes. § 455 No point out of this line can be in both planes ; for only one plane can contain a straight line and a point without the line. Therefore the straight line through A and B is the locus of all the points common to the two planes, and is consequently the intersection of the planes (§ 470). q. e.d. 246 SOLID GEOMETEY. BOOK VI. Perpendicular Lines and Planes. Proposition II. Theorem. 472. If a straight line is perpendicular to each of two other straight lines at their point of intersection, it is perpendicular to the plane of the two lines. \l' Let AB be perpendicular to BG and BD at B. To prove A B perpendicular to the plane MN of these lines. Proof. Through B draw in MJV any other straight line BE, and draw CD cutting £C, BE, BD, at O, E, and D. Prolong ^^ to i^, making BF=j= AB, and join J. and i^to each of the points C, E, and D. Since BO and BD are each J_ to ^i^at its middle point, • AC=='FC^rid.AD = FD. §122 .-. AACD = A FCD (§ 160), and hence AACD^Z FCD. Now in the A ACE and FCE AC= EC, CE= CE, and Z. ACE== A FCE. .-. A ACE^- A FCE {% 150), and hence AE= FE. .'. BE is ± to AF Sit B. §123 Hence ^i? is _L to BE, any, that is, every, straight line drawn in MN through B, and therefore is ± to MN. ^ 462 Q E. D. PERPENDICULAR LINES AND PLANES. 247 Proposition III. Theorem. 473. Every perpendicular that can he drawn to a straight line at a given point lies in a plane perpen- dicular to the line at tlve given point, A Let the plane MN be perpendicular to AB at B. To prove that BE,, any perpendicidar to AB at B, lies in the plane MN.- Proof. Let the plane containing ABi^ and BE intersect MN in the line BE' ; then AB is ± to J^\ . § 462 Since in the-p^ttrne ABE only one JL can be drawn to AB at- B (§ 89), BE and BE^ coincide, and BE lies in MN Hence every ± to AB at B lies in the plane MN. •^ Q. E. D. 474. Cor. 1. At a given point' m a straight line one plane perpendicular to the line can he draivn, and only one. 475. Cor, 2. Through a given point without a straight line, one plane can he drawn perpendicular to the line, and only one. Let AC hQ the line, and the point without it. In the plane OCA draw OC J_ to AC, .and in another plane contain- ing ^ a draw CD A. to AC 2X C. Then CO and CD determine a plane l.io AC x Every plane JL to AC B,nd. passing through cuts the plane OCA in a line ± to ^C and containing 0. This ± coincides, then, with OC, and every such plane is ± to ^C at 0. But only one plane can be ± to ^(7at C(§ 474). Hence only one plane can be drawn from X to ^Cat 0. 248 SOLID GEOMETET. — BOOK VI. Proposition IV. Theorem. 476. Through a given point one perpendicular can he drawn to a given plane, and only one, Pr K H \ ' 1 "/ ^ / $ / 'B f / -^ M Fig. 1. Fig. 2. Q Case I. When the given point is in the given plane. Let A be the given point in the plane MN (Fig. 1). To prove that one perpendicular can be erected to the plane MN at A, and only one. Proof. Draw in JfiVany line -SC'througli A, and pass through A a plane DEHK JL to BC, and cutting MN'm DE. At A erect in the plane DEHK a line AF X to DE. The line BQ, being J_ to the plane DEHKhj construction, is J_ to ^i^ which passes through its foot in the plane. § 462 That is, AF is JL to BC) and as it is JL to DE by con- struction, it is X to the plane MN. § 472 Moreover, every other line AG drawn from A, is oblique to MN. For ^i^and AO intersecting in A determine a plane DEHK, which cuts MN'm the straight line DE; and as AF is X to MN, it is X to DE (§ 462) ; hence ^G^ is oblique to DE (§ 89), and therefore to Mn[^ 463). Therefore ^jPis the only X to ifiV at the point A. PERPENDICULAR LINES AND PLANES. 249 Case II. When the given point is without the given 'plane. Let A be the given, point, and MN the plane. To prove that one perpendicular can be drawn from A to MN, and only one. Proof. Draw in 3£N any line HK, and pass through A a plane FQ L to HK, cutting MN in FO, and HK in C. Let fall from A, in the plane FQ, ^ A. AB upon FO. Draw in the plane MN B.ny other line BF from B. Prolong AB to F, making BF= AB, and join A and F to each of the points Cand F. Since i)C is _L to FQ by construction, and CA and CF lie in FQ, the A FCA and DC^ are right angles. § 462 In the rt. A FCA and FCF, FCis common, and CA = CF. § 122 .-. A FCA = A FCE (^ 151), and hence FA = FF. .-. BF is ± to AF&tB. § 123 That is, AB is JL to BF, any straight line drawn in MN through its foot, and therefore _L to MN Moreover, every other straight line AI, drawn from A, is oblique to MN For the lines AB and ^/determine a plane FG which cuts the plane MN in the line FG. The line AB being ± to the plane MN, is _L to FG (§ 462). Therefore AI oblique to FG, and consequently to MN (^ 463). Therefore AB is the only ± from A to MN. Q. E. O. 477. Cor. The perpendicular is the sho7'testUne from a point to a plane, for it is the shortest line from the point to any straight line of the plane passing through its foot (§ 114). y [ 250 SOLID GEOMETEY. BOOK VI. Proposition V. Theorem. 478. Oblique liiies draivn from a point to a plane, and meeting the plane at equal distances froin . the foot of the perpendicular, are equal; and of two ob- lique lines meeting the plane at unequal distances from the foot of the perpendicular the more remote is the greater. Let AC and AD cut off the equal distances BC and BD from the foot of the perpendicular AB, and let AD and AE cut off the unequal distances BD and BE, and BE be greater than BD. To prove AC= AD, and AE > AD. Proof. The right A ABCd^ndi ADD have AB common, and £0= BD by hypothesis. Therefore they are equal, and AC =^ AD. The right A ^^^, ^^C'have AB common, and BE> BC. Therefore AE> ACi% 119), and hence AE> AD. Q. E. D, 479. Cor. 1. Equal oblique lines from a point to a plane vieet the plane at equal distances from the foot of the perpen- dicular ; and of two unequal oblique lines the greater meets the plane at the greater distance from the foot of the perpendicular. 480. Cor. 2. The locus of a point in space equidistant from all points in the circumference of a circle is a straight line pass- ing through the centre and perpendicular to the plane of the circle. PERPENDICULAR LINES AND PLANES. 251 Proposition VI. Theorem. 481. If from the foot of a perpendicular to a plane a straight line is drawn at right angles to any line in the plane, the line drawn from its intersection with the line in the plane to any point of t,%e per- pendicular is perpendicular to the line of the plane. M ^N — _ '» \,--'' ■"^' v B -•-... '• r Let AB-be a perpendicular to the plane MN, BE a perpendicular from B to any line CD in MN, and EA ^ line from E to any point A in AB. To prove AE perpendiculo.r to CD. Proof. Take ^(7 and ED equal ; draw BC\ BD, AC, AD. Now BC= BD (§ 116), and AC= AD (§ 478). :.AE\^l.\oCD, §123 482. Cor. The locus of a point in space equidistant from the extremities of a straight line is the plane perpendicular to this line at its middle point. For, if the plane MN is ± to ^^ at its middle point 0, and^any point C in this ^^fc^ plane is joined to A, 0, and B, CO is ± to AB ; therefore CA and CB are equal. § 116 ^ Also, since all the Js to the line AB at the point lie in the plane 3/iV^(§473), any point D without the plane MN cannot lie in a ± to AB at 0, and therefore is unequally distant from A and B. § 122 252 SOLID GEOMETKY. BOOK VI. Proposition VII. Theorem. 483. Two straight lines perpendicular to the same plane are parallel. / A C §462 §481 lie in §473 i^ Let AB and CD bs perpendicular to MN. To prove AB and CD parallel. Proof. Let A be any point in AB-, join AD and BD, and tlirough D draw ^i^in the plane UN 1. to BD. Then CD is ± to ER Also, AD is ± to EF. Therefore CD, AD, and BD, being _L to EF at D the same plane. Therefore AB and CD lie in the same plane ; and since, by hypothesis, they are ± to MN, they are J. to BD. § 462 Therefore AB and CLD are parallel. q. e. d. 484. Cor. 1. If one of two parallel lines is perpendicular to a plane, the other is also perpendicular to the plane. For, if AB and CD are II, and AB is ± to the plane MN, and if through any point of CD a line is drawn ± to MN, it will be II to AB (§ 483). Since through the point only one line can be drawn II to AB (§ 101), CD will coincide with this J_ and be J_ to MN. 485. Cor. 2. If two straight lines AB and EF are parallel to a third line CD, they are parallel to each other. For, a plane MN _L to CD, is J_ to AB and ^i^(§484). Hence AB and EF, being ± to MN, are parallel (§ 488). A G M n ^=^^ ^ /B Df N parallel lines and planes. Parallel Lines and Planes Proposition VIII. Theorem. 486. // two straight lines are parallel, every plane containing one of the lines, and only one, is parallel to the other line. 4 ™™B Let AB and CD be two parallel lines, and MN any plane containing: CD and not AB. To prove AB and MN parallel. Proof. The lines AB and CJD are in the same plane A BCD, which intersects the plane MNin the' line CD. Since AB is in the plane AD, it must meet the plane 3IN, if at all, in a point common to the two planes ; that is, in a point of their intersection CD. But since AB is II to CD, it cannot meet CD. Therefore AB cannot meet the plane MN, and hence is II to MN. q. e. d. 487. Cor. 1. Through a given straight line a plane can be passed parallel to any other given straight line in space. For, if a plane is passed through one of the lines AB and any point G of the other line CD, and a line CE is drawn in this plane 11 to AB, the plane MN determined by CD and CE is II to .45. §486 254 SOLID GEOMETEY. — BOOK VI. 488. Cor. 2. Through a given point a plane can he passed parallel to any two given straight lines in space. For, if is the given point, and AB and CD the given lines, by drawing through a line A'B' II to AB in the plane determined by AB and 0, and also a line CD' II to CD in the plane determined by CD and 0, we shall have two lines A'B^ and C^D^ which determine a plane passing through and liaes AB and CD. I trj^Q—p' I to each of the §486 Proposition IX. Theorem. 489. If a given straight line is parallel to a given plane, the intersection of the given plane with any plane passed through the given line is parallel to that line. A B \1| Let the line AB be parallel to the plane MN, and let CD be the intersection of MN with any plane passed through AB. To prove AB and CD parallel. Proof. The lines AB and CD are in the same plane ABCD, and therefore if the line AB meets the line CD, it must meet the plane MN. But AB is by hypothesis II to MN, and therefore cannot meet it ; that is, it cannot meet CD, however far they may be produced. Hence AB and CD are parallel. Q.E. D. PARALLEL LINES AND PLANES. 2d5 490. Cor. If a given straight line and a plane are 'parallel, a parallel to the given line drawn through any point of the plane lies in the plane. For the plane determined by the given line AB and any point (7 of the plane cuts MN in a line CD II to AB (§ 489) ; but through C only one parallel to AB can be drawn (§ 101) ; therefore a line drawn through C II to AB coincides with CD, and hence lies in the plane MN. Proposition X. Theorem. 491. Two planes perpendicular to the same straight line are parallel. Let MN and PQ be two planes perpendicular to the straight line AB. To prove MN and PQ parallel. Proof. ^i*v''and PQ cannot meet. For if they could meet, we should have two planes from a point of their intersection J^ to the same straight line. But this is impossible, § 475 {through a given point vnthout a straight line, only one plane can he passed ± to the given line). Therefore JfiVand PQ are parallel. Q. E. D. Ex. 480. Find the locus of a point in space equidistant from two given parallel planes. Ex. 481. Find the locus of a point in apace equidistant from two given points and also equidistant from two given parallel planes. 256 SOLID GEOMETEY. — BOOK VI. Proposition XI. Theorem. 492. The intersections of two parallel planes hy a third plane are parallel lines. R \y Let the parallel planes MN and PQ be cut by B8. To prove the intersections AB and CD parallel. Proof. AB and CD are in the same plane R8. They are also in the parallel planes MN and FQ, which cannot meet however far they extend. Therefore AB and CD cannot meet, and are parallel. a E. D. 493. Cor. 1. Parallel lines included between parallel planes are equal. For, if the lines ^(7 and BD are II, the plane of these lines will intersect JfiV^ and PQ in the II lines AB and CD (§ 492). Hence ABDCis a parallelogram, and AC said BD are equal. 494. Cor. 2. Tivo parallel planes are everywhere equally distant. For'Js dropped from any points in MN to PQ measure the distances of these points from PQ. But these Js are parallel (§ 483), and hence equal (§ 493). Therefore all points in MN are equidistant from PQ. PAKALLEL LINES AND PLANES. 257 Proposition XII. Theorem. 495. A straight line perpendicular to one of two parallel planes is perpendicular to the other. M ^^'""^^ ]D..[.r-~-~ ' f-«::;3 / p \ N /^ ^ ..—-'-"""2 i 1 Let AB be perpendicular and PQ parallel to MN. To prove that AB is perpendicular to FQ. Proof. Pass through the line AB any two planes intersect- ing JO^in the lines ^Cand AD, and PQ in BE and BF. Then .4(7 and AD are II to BE and .5i^ respectively. § 492 But AB is ± to 4Cand 4Z> (§ 462), and is therefore ± to their parallels BE and BF. § 102 Therefore, AB is ± to PQ. § 472 Q.E.D. 496. Cor. 1. Through a given point A one plane, and only 07ie, can be drawn parallel to a given plane FQ. For, if a line is drawn from 4 X to FQ, a plane passing through A Xto this line is II to FQ (§ 491) ; and since only one plane can be drawn through a point X to a given line (§ 474), only one plane can be drawn through A II to FQ. 497. Cor. 2. Jf two intersecting lines AC and AD are each parallel to a plane FQ, the plane of these lines MN is parallel to FQ. For draw AB Xio FQ, and through the point B draw BE and BF W to AC and AD. Then BE and BF lie in the plane FQ (§ 490). Hence AB is ± to BE and BF. Therefore AB is ± to 4C and AD (? 102), and hence to the plane MN {^ 472). Hence JfiVand FQ are parallel. § 491 258 SOLID GEOMETRY. BOOK VI. Proposition XIII. Theorem. 498. If two angles not in the same plane have their sides respectively parallel and lying in the same di- rection, they are equal, and their planes are parallel. Let the angles A and A' be respectively in the planes MN and PQ and have AD parallel to AD' and AC par- allel to A'C and lying in the same direction. To prove /. A = Z.A\ and MN II to PQ. Proof. Take AB and A^ D^ equal, also J.Cand ^'(7' equal. Join AA\ DD\ CC\ CD, C'D\ Since AD is equal and II to A^D\ the figure ADD^A^ is a parallelogram, and AA^ = DD\ § 182 In like manner AA^ = CC\ Also, since (7(7' and DD^ are each II to AA\ and equal to AA\ they are II and equal. Therefore CDD'O' is a parallelogram, and CD = CD'. .-. A ADC=--- A A'D'C, and Z A--=ZA\ § 160 Also, since FQ is II to each of the lines AC and AD (§ 486). FQ is II to the plane of these lines MN{^ 497). q. e. d. PARALLEL LINES AND PLANES. 259 Proposition XIV. Theorem. 499. If two straight lines are intersected by three parallel planes, their corresponding segments are proportional, M C ^' --- ^^/^ Let AB and CD be intersected by the parallel planes MN, PQy RS, in the points A, E, B, and C, F, D. To prove AE : EB = CF: FD. Proof. Draw AD cutting the plane PQ in G. Join EO and EG. Then EG is II to BD, and GE'\^ II to AC. § 492 .\AE'.EB = AG: GD, §309 and CF:FD=AG.GD. :.AE:EB=CF:FI). aE. D. Ex. 482. The line AB meets three parallel planes in the points A, E, B ; and the line CD meets the same planes in the points C, F, D. If AE= 6 inches, BE=S inches, (7i> = 12 inches, compute CFand FD. Ex. 483. To draw a perpendicular to a given plane from a given point without it. Ex. 484. To erect a perpendicular to a given plane at a given point in it. 260 SOLID GEOMETRY. — BOOK VI. Dihedral Angles. 500. The opening between two intersecting planes is called a dihedral angle. The line of intersection AB of the planes is the edge, and the planes MA and NB are the /aces of the dihedral angle. 501. A dihedral angle is designated by its edge, or by its two faces and its edge. Thus, the dihedral angle in the margin may be designated by AB, or by M-AB-N. 502. In order to have a clear notion of the magnitude of the di- hedral angle AB, suppose a plane at first in coincidence with MA to turn about the edge AB, in the direction indicated by the arrow, until it coincides with the face NB. The amount of rotation of this plane is the dihedral angle AB. 503. Two dihedral angles are equal when they can be made to coincide. I 504. Two dihedral angles M-AB-JSf and JSf-AB-P are adja- cent if they have a common edge ^ ^ AB, and a common face NAB, be- tween them. 505. When a plane meets another plane and makes the adjacent dihe- dral angles equal, each of these an- gles is called a right dihedral angle. 506. A plane is perpendicular to another plane if it forms with this second plane a right dihedral angle. DIHEDRAL ANGLES. 507. Two vei^tical dihedral angles are angles that have the same edge and the faces of the one are the prolongations of the faces of the other. 508. Dihedral angles are acute, obtuse^ complementary, supplementary, under the same conditions as plane angles are acute, obtuse, complementary, supplementary, respec- tively. 509. The demonstrations of many properties of dihedral angles are identically the same as the demonstrations of anal- ogous theorems of plane angles. The following are examples : 1. If a plane meets another plane, it forms, with it two adjacent dihedral angles whose sum is equal to two right dihedral angles. 2. If the sum of two adjacent dihedral angles is equal to two right dihedral angles, their exterior faces are in the same plane. 3. If two planes intersect each other, their vertical dihedral angles are equal. 4. If a plane intersects two parallel planes, the exterior- interior dihedral angles are equal ; the alternate-interior dihe- dral angles are equal ; the two interior dihedral angles on the same side of the secant plane are supplements of each other. 5. When two planes are cut by a third plane, if the exterior- interior dihedral angles are equal, or the alternate-interior dihedral angles are equal, and the edges of the dihedral angles thus formed are parallel, the two planes are parallel. 6. Two dihedral angles whose faces are parallel each to each are either equal or supplementary. 7. Two dihedral angles whose faces are perpendicular each to each are either equal or supplementary. 262 SOLID GEOMETRY. — BOOK VI. Measure of Dihedral Angles. 510. The plane angle of a dihedral angle is the plane angle formed by two straight lines, one in each plane, perpendicular to the edge at the same point. 511. The plane angle of a dihedral angle has the same magnitude from whatever point in the edge the perpendiculars are drawn. For any two such angles, as CAB, GIIT, have their sides respec- tively parallel (§ 100), and hence are equal (§498). Proposition XV. Theorem. 512. Two dihedral angles are equal if their plane angles are equal. \ c H F G E Let the two plane angles ABD and A'B'D' of the two dihedral angles CB and C'B' be equal. To prove the dihedral angles CB and C'B^ equal. Proof. Apply B'C^ to BC, making the plane angle A'B'B' coincide with its equal ABD. The line B^C being ± to the plane A'B'B' will likewise be X to the plane ABB at B, and take the direction BC, since at B only one _L can be erected to this plane. § 476 The two planes A'B'C and ABC, having in common two intersecting lines AB and BC, coincide. § 460 In like manner the planes B'B'C and Z>^C coincide. Therefore the two dihedral angles coincide and are equal. Q. E. D. DIHEDRAL ANGLES. 263 Proposition XVI. Theorem. * 513. Two dihedral angles have tJie same ratio as their plane angles. B' A " " Case I. When the plane angles are commensurahle. Let A-BC-D and A'-B'a-D'be two dihedral angles, and let their plane angles ABD and A'B'D' be commensur- able. To prove A-BC-D : A'-B'C'-D^ = /. ABD : Z A'B'D'. Proof. Suppose the A ABD and A'B'D' have a common measure, which is contained three times in Z. ABD and five times in Z A'B'D'. Then " A ABD'.AA'B'D'^2>:b. Apply this measure to Z ABD and Z A'B'D', and through the lines of division and the edges ^Cand B'C pass planes. These planes divide A-BC-D into three parts, and A'-B'C'D' into five parts, all equal because they have equal plane angles. Therefore A-BC-D : A'-B'C'-D' = 3:5. Therefore A-BC-D : A'-B'C'-D' - Z ABD : Z A'B'D'. 264 SOLID GEOMETRY. — BOOK VI. Case II. When the plane angles are incommevisurahle. Let A-BC-D, A'-B'C-D' be dihedral angles, and let their plane angles ABD, A'B'D' be incommensurable. To prove A-BC-D : A'-B^C'-D' ^ZaBD.Z A'B'D\ Proof. Divide the Z ABD into any number of equal parts, and apply one of these parts to the Z. A'B'D^ as a measure. Since ABD and A'B'D' are incommensurable, a certain number of these parts will form the Z A'B'U, leaving a remainder JSB'D', less than one of the parts. Pass a plane through ^'^and B'C Since the plane angles of the dihedral Singles. A-BC-D and A'-B'C'-BJ Sive commensurable, we have by Case I., A-BC-D : A'-B'C'-U=Z ABD : Z A'B'K If the unit of measure is indefinitely diminished, these ratios continue equal, and approach indefinitely the limiting ratios, A-BC-D : A'-B'C'-D, and Z ABD : Z A'B'D'. .-. A-BC-D : A'-B'C'-D' - Z ABD : Z A'B'D'. § 260 Q. E. D. 514. Scholium. The plane angle is taken as the measure of the dihedral angle. (Compare § 262.) DIHEDRAL ANGLES. 265 Planes Perpendiculab, to Each Otheb. Proposition XVII. Theorem. 515. If two 'planes are perpendicular to each other, a straight line drawn in one of them perpendicular to their intersection is perpendicular to the other. A N Let the plane PAB be perpendicular ta MN, and let CD be drawn in PAB perpendicular to their intersec- tion AB. To prove CD perpendicular to MN. Proof. In the plane MN Am\N DEI. to AB at D. Then CDE is the plane angle of the right dihedral angle P-AB-N, and is therefore a right angle. By construction CD A is a right angle. Therefore CD is X to DA and DE at their point of inter- section, and consequently -L to their plane MN. § 472 Q. E. D. 516. Cor. 1. If two planes are perpendicular to each other ^ a perpendicular to one of them at any point of their intersection will lie in the other. For, a line CD drawn in the plane PAB _L to AB at the point D will be X to MN {% 515). But at the point D only one J_ can be drawn to MN (§ 476). Therefore a X to MN erected at D will coincide with CD and lie in the plane PAB, 266 SOLID GEOMETRY BOOK VI. 617. Cor. 2. If two planes are perpendicular to each other, a perpendicular to one of them from any point of the other will lie in the other. For, a line CD drawn in the plane PAB from the point Q L to AB will be ± to MN{^ 515). But from the point C only one ± can be drawn to MN (§ 476). Therefore a ± to MN drawn from Cwill coincide with CD and lie in PAB. Proposition XVIII. Theorem. 518. If a straight line is perpendicular to a plane, every plane passed through the line is perpendicu- lar to the first plane. cA li aX r-S I Let CD be perpendicular to MN, and PAB be any plane passed through CD intersecting MN in AB. To prove the plane PAB perpendicular to the plane MN. Proof. Draw DE in the plane MN, and J. to AB. Since CD is ± to MN, it is JL to AB. Therefore Z CDE is the plane angle of P-AB-N But Z CDEh a right angle, and therefore PAB is ± to MN § 514 Q. E. D. . 519. Cor. A plane perpendicular to the edge of a dihedral angle is perpendicular to each of its faces. DIHEDRAL ANGLES. 267 Proposition XIX. Theorem. 520. If two intersecting -planes are each perpendicu- lar to a third plane, their intersection is also perpen- dicular to that plane. Let the planes BD and BC intersecting in the line AB be perpendicular to the plane PQ. To prove AB perpendicular to the plane PQ. Proof. A X erected to FQ at B, a point common to the three planes, will lie in the two planes ^Cand £D. § 516 And since this JL lies in both the planes BC and BI), it must coincide with their intersection AB. .-. AB is ± to the plane PQ. aE. D. 521. Cor. 1. If a plane PQ is perpendicular to each of two intersecting! planes ABC and ABI), it is perpendicular to their intersection AB. 522. Cor. 2. If a plane PQ is perpendicular to two planes ABC and ABD, which include a right dihedral angle, the intersection of any two of these planes is perpendicular to the third plane, and each of the three intersections is popendicular to the other two. 268 SOLID GEOMETRY. — BOOK VI. Proposition XX. Theorem. 523. Through a given straight line not perpendicu- lar to a plane, one plane, and only one, can be passed perpendicular to the given plane. J? Let AB be the given line not perpendicular to the plane MN. To prove that one plane can he parsed through AB perpen- dicular to MN, and only one. Proof. From any point A of AB draw AC l^io MN, and through AB and J^Cpass a plane AD. The plane AD is _L to MN, since it passes through AC, a. line ± to MN § 518 Moreover, if two planes could be passed through AB J_ t(y the plane MN, their intersection AB would be ± to MN. § 520 But this is impossible, since AB is by hypothesis oblique to the plane MN. Hence through AB only one plane can be passed X to 3fN. OLE D. 624. Cor. If a straight line is oblique to a plane, its projec- tion IS a st7^aight line. For, the plane passed through ^^ _L to MN contains all the Js let fall from different points of AB upon MN (§ 516). Therefore the intersection CD of these planes is the locus of the projections of the points in AB. But the intersection CD\s a straight line ; that is, the projection of AB is a, straight line. DIHEDRAL ANGLES. 269 Proposition XXI. Theorem. 625. Every point in a plane which bisects a dihe- dral angle is equidistant from the faces of the angle. A Let plane AM bisect the dihedral angle formed by the planes AD and AC ; and let PE and FF be perpen- diculars drawn from any point P in the plane AM to the planes AO and AD. To prove FE=PF. Proof. Through FE and FF pass a plane intersecting the planes ^(7 and AD in the lines O^and OF, and join FO. The plane FEE is ± to ^ (7 and to AD. § 518 Hence the plane FEE is JL to their intersection AO. ^ 521 .:/.FOE=AFOE, [being measures respectively of the equal dihedral A M-OA- C and M- OA-D). .'. rt. A FOE= rt. A FOE. § 148 .'.FE=FF. a E. D. vEx. 485. Find the locus of a point in space equidistant from three given points not in a straight line. Ex. 486. Given two points A and B on the same side of a given plane MN; find a point in this plane such that the sum of its distances from A and B shall be a minimum. 270 SOLID GEOMETRY. — BOOK VI. Angle of a Straight Line and a Plane. Proposition XXIL Theorem. 526, The acute angle which a straight line makes with its own projection upon a plane is the least an- gle which it makes with any line of the plane. Let BA meet the plane MN at A, and let AC "be its projection upon the plane MN, and AD any other line drawn through A in the plane. To prove A BA C less than Z BAB. Proof. Take AD = AC, and join BB. In the A B AC SiTid BAB BA = BA, AC= AB, but BC< BB. § 477 .-. Z BACis less than Z BAB, § 153 Q. E. a 527. Scholium. If the straight line AC turns about the point A, the angle ^^C increases ; it is a right angle when ^C is perpendicular to its initial position; then it becomes obtuse, and reaches its maximum value when AC falls upon AGHhe prolongation of CA. Afterwards the angle passes through the same values in reverse order. DIHEDRAL ANGLES. 271 * A Perpendicular between Two Straight Lines. Proposition XXIII. Theorem. ^^^^ Between tivo straight lines not in tl%e same plane, one common perpendicular can be drawn, and only I one, C E D M_Al_,/ ! ...; ^ Let AB and CD be the given lines. To prove that one common perpendicular can be drawn be- tweert them^ and only one. Proof. Through any point B of AB draw BG II to CD, and let JfiVbe the plane determined by AB and BG. Through CD pass the plane CD' X to MN, and intersecting AB at C". At C" erect a ± C'Cto the plane MN. C'C will lie in the plane CD (§ 516), and be X to AB and CD' (§ 462). Since CCis ± to CD', it is JL to CD (§ 102). Hence CC is a common perpendicular to CD and AB. Moreover, CC is the only common perpendicular. For, if any other line UB could be'± to CD and AB, it would be ± to BG and AB (§ 102), and hence J. to MK But .EB: in the plane CD' and J. to CD', is JL to MN (§ 515), and we should have two Js from U to MN. But this is impossible. § 476 Hence CC is the only common ± to CD and AB. an. D. 272 SOLID GEOMETRY, — BOOK VI. Polyhedral Angles. 529. A polyhedral angle is the opening of three or more planes which meet at a common point. 530. The common point 8 is the vertex of the angle, and the intersections of the planes 8A, SB, etc., are « its edges; the portions of the planes included A between the edges are its faces, and the angles /7 \ formed by the edges are it^face angles. a /-. \ '/- --------Ac 531. The magnitude of a polyhedral angle B \ depends upon the relative position of its faces, and not upon their extent. 532. In a polyhedral angle, every two adjacent edges form a face angle, and every two adjacent faces form a dihedral angle. These face angles and dihedral angles are the parts of the polyhedral angle. 533. Two polyhedral angles ? ^ can be made to coincide and yy \ /7 \ are equal if their corresponding A// \ X// \ parts are equal and arranged / l/'^ ''"-■vAg A "■"---Ap' in the same order. ^B ^ B' A 534. A polyhedral angle is convex if any section made by a plane cutting all its faces is a convex polygon. 535. A polyhedral angle is called trihedral, tetrahedral, etc., according as it has three isices, four faces, etc. 536. A trihedral angle is called rectangular, bi-rectangular, tri-rectangular, according as it has one, two, or three right dihedral angles. Two adjacent walls and the floor of a rectangular room form a tri-rectangular trihedral angle. 537. A trihedral angle is called isosceles if it has two of its face angles equal. SYMMETRICAL POLYHEDRAL ANGLES. 273 Symmetrical Polyhedral Angles. 538. If the edges of a given polyhedral angle 8-ABCD are produced through the vertex S, another polyhedral angle S-A'B'C'D' is formed, symmetrical with respect to S-AB CD. The face angles A8B, B80, etc., are equal respectively to the face angles A'8B\ B'8C', etc., since they are ver- tical angles. Also the dihedral angles whose edges are 8A, 8B, etc., are equal ^ respectively to the dihedral angles whose edges are 8A', 8B\ etc., since they are vertical dihedral angles. (The second figure shows a pair of vertical dihedral angles.) The edges of 8-ABCD are arranged from left to right in the order 8B, 8C, 8D, but the edges of 8-A'B'C'D' are arranged from right to left in the order 8B\ 8C', 8D' ; that is, in an order the reverse of the order of the edges in 8-ABCD. Two symmetrical polyhedral angles, therefore, have all their parts equal, each to each, but arranged in reverse order. In general, two symmetrical polyhedral angles are not super- posable. Thus, if the trihedral angle 8-A'B'C' is made to turn 180° about the bisector xy of the angle A'8C, the side 8A' will coincide with 8C, 8G' with 8A, and the face A'8C' with A8C; but the dihedral angle 8 A, and hence the dihedral angle 8A', not being equal to 8C, the plane A'8B' will not coincide with B8C; and, for a similar reason, the plane C'8B^ will not coincide with A8B. Hence the edge 8B' takes some posi- tion 8B" not coincident with 8B ; that is, the trihedral angles are not superposable. J 074. SOLID GEOMETRY. — BOOK VI. Proposition XXIV. Theorem. X 539. The sum of any two face angles of a trihedral angle is greater than the third face angle. S In the trihedral angle S-ABG let the angle ASC be greater than A8B or BSG. To prove Z. A8B -f- Z BBC greater than Z ASC. Proof. In A8C draw SD, making Z ASD = Z A8B. Through any point D of BB draw ABCixi the plane ABC Take SB = BD. Pass a plane through the line ^Cand the point B. In the A ABB and ABB, AB= AB, BB = BB, and Z ABB = Z ABB. .-. A ABB = A ABB (§ 150), and AB = AB. InihQAABC; AB-\-BC> AC §137 But AB =AB By subtraction, BC> BC In the A BBC and BBC, BC= BC, and BB = BB, but BC> BC. Therefore Z BBC is greater than Z BBC. § 153 :•. A ABB + BBC are greater than A ABB + BBC That is, A ABB + BBC are greater than Z ^/^C. . f Q. E. D. POLYHEDKAL ANGLES. 275 Proposition XXV. Theorem. 540. The sum of the face angles of any convex poly- hedral angle is less than four right angles. Let S be a convex polyhedral angle, and let its faces be cut by a plane, making the section ABCBE a convex polygon. To prove Z A8B -\- Z B8C, etc., less than four rt. A. Proof. From any point within tbe polygon draw OA, OB^ OC, OD, OE. The number of the A having their common vertex at will be tl|3 same as the number having their common vertex at 8. Therefore the sum of all the A of the A having the common vertex at 8 is equal to the sum of all the A of the A having the common vertex at 0. But in the trihedral A formed at ^, B, C, etc. Z 8AE+ Z 8AB is greater than Z EAB, and Z 8BA + Z 8BC is greater than Z ABC, etc. § 539 Hence the sum of the A at the bases of the A whose com- mon vertex is 8 is greater than the sum of the A at the bases of the A whose common vertex is 0. Therefore the sum of the A at the vertex 8 is less than the sum of the A at the vertex O. But the sum of the ^ at = 4 rt. A. § 92 Therefore the sum of the zi at xS' is less than 4 rt. A. Q.E. D. 276 SOLID GEOMETRY. — BOOK VI. Proposition XXVI. Theorem. 541. Two trihedral angles are equal or symmetrical when the three face angles of the one are respectively equal to the three face angles of the other. B' E' ABE A A E' B In the trihedral angles S and S^ let the angles ASB A8C, BSC, be equal to the angles A'S'B', A'S'C, B'S'C, respectively. To prove S-ABQ and S^-A^B'C^ equal or si/mmetrical. Proof. On the edges of these angles take the six equal dis- tances SA, SB, SO, S'A\ S'B\ S'C\ Draw AB, BO, AC, AB\ BW\ AOK The isosceles A BAB, 8 AC, SBC, are equal respectively to S'A'B',S'A'C', S'B'C^. § 150 AB, AC, BCsiYo equal respectively to A'B', AC, B'C. .\AABC=AA'B'C'. § 160 At any point D in SA draw BU and i)i^ in the faces ASB and ^/S'C respectively, and JL to SA. These lines meet AB and J^C respectively, {since the A SAB and 8 AC are acute, each being one of the equal A of an isosceles A). Join UF. OuS'A' tBke A'B' = AB. poly;eedeal angles. 277 Draw Z)'^' and D'i^' in the faces of A'S'B' and A'jS'C respectively, ± to S'A', and join JE'F'. In the rt. A ADU and A'D'U', AD = A'B', and Z ZlAU= Z B'A'U'. .-. rt. A ^i>^= rt. A A'D'E'. § 149 .-. AE= A'E' and BE= D'E'. In like manner we may prove AF=A'F' and DF=D'F'. Hence in the A ^^i^and A'F'F' AE= A'F', AF= A'F', and Z EAF= Z E'A'F'. :. A AEF= A A'E'F\ and EF= E'F\ § 150 Hence, in the A EDF a,nd E'D'F' we have ED = E'D', DF= D'F', and EF= E'F'. .'. A EDF= A ^'X>'i^' and Z EDF= Z E'D'F'. § 160 Therefore the dihedral angle B-AS-C equ&h dihedral angle D'-A'S'-C, {since A EDF and E'D'F\ the measures of these dihedral A, are equal). In like manner it may be proved that the dihedral angles A- BS-C and A-C8-B are equal respectively to the dihedral angles A'-D'S'-C and A'~C'8'-B'. o. e. d. This demonstration applies to either of the two figures de- noted by S'-A'B'C, which are symmetrical with respect to each other. If the first of these figures is taken, /S and S' are equal. If the second is taken, S and S' are symmetrical. 542. Scholium. If two trihedral angles have three face angles of the one equal to three face angles of the other, then the dihedral angles of the one are respectively equal to the dihe- dral angles of the other. '^x. 487. An isosceles trihedral angle and its symmetrical trihedral angle are superposable. ^ Ex. 488. Find the locus of a point equidistant firom the three edges of a trihedral angle. *Ex. 489. Find the locus of a p®int equidistant from the three faces of a trihedral angle. BOOK VII. POLYHEDRONS. CYLINDERS. AND CONES. Polyhedrons. 543. A polyhedron is a solid bounded by planes. The bounding planes, limited by each other, are the faces, their intersections are the edges, and the intersections of the edges are the vertices, of the polyhedron. 544. A diagonal of a polyhedron is a straight line joining any two vertices not in the same face. 545. A polyhedron of four faces is called a tetrahedron ; one of six faces, a hexahedrbn; one of eight faces, an octahedron; one of twelve faces, a dodecahedron; one of twenty faces, an icosahedron. 546. A polyhedron is convex if the section made by any plane cutting it is a convex polygon. Only convex polyhedrons are considered in this work. 547. The volume of a solid is its numerical measure, referred to another solid taken as the unit of volume. 548. A polyhedron of six faces, each face a square, is called a cube; and the cube whose edge is the linear unit is generally taken as the unit of volume. 549. Two solids are equivalent if their volumes are equal. 550. Two polygons are parallel if their sides are respectively parallel. POLYHEDRONS. 279 Prisms and Parallelopipeds. 551. A prism is a polyhedron of which two opposite faces, called bases, are parallel polygons, and the other faces, called lateral faces, intersect in parallel lines, called lateral edges. The lateral edges are equal (§ 493), the lateral faces are parallelograms (168), and the bases are equal (§§ 179, 498). 552. The sum of the areas of the lateral faces of a prism is called its lateral area. 553. The altitude of a prism is the length of the perpen- dicular between the planes of its bases. 554. Prisms are called triangular, quadrangular, etc., ac- cording as their bases are triangles, quadrilaterals, etc. 555. A right section of a prism is a section made by a plane perpendicular to its lateral edges. Truncated Prism. Right Prism. Rectangular Parallelopiped. Parallelopiped. 556. A truncated prism is the part of a prism included between the base and a section made by a plane inclined to the base and cutting all the lateral edges. 557. An oblique prism is a prism whose lateral edges are not perpendicular to its bases ; a right prism is a prism whose lat- eral edges are perpendicular to its bases ; a regular prism is a right prism whose bases are regular polygons. 558. A prism whose bases are parallelograms is called a parallelopiped. If its lateral edges are perpendicular to the bases, it is called a right parallelopiped. If its six faces are all rectangles, it is called a rectangular parallelopiped. 280 SOLID GEOMETRY. — BOOK VII. Proposition I. Theorem. 559. The sections of a prism made hy parallel planes are equal polygons, ^"^ J) Let the prism AD be intersected by the parallel planes GK, G'K'. To prove GHIKL = G'TT'I'X'L'. Proof. Since the intersections of two parallel planes by a third plane are parallel (§ 492), the sides GIT, HI, IK, etc., are parallel respectively to the sides G^H^, II'I\ I'K', etc. The sides GS, SI, IK, etc., are equal respectively to G^H\ Wr, VK, etc., since parallel lines comprehended between parallel lines are equal. . § 180 The A GUI, HIK, etc., are equal respectively to A &WI\ WVK, etc., since two A not in the same plane, having their bides respectively parallel and lying in the same direction, are equal. § 498 Therefore GHIKL = G'HI'K'H, § 203 because they are mutually equiangular and equilateral. Q. E. D. 560. Cor. Aut/ section of a prism parallel to the base is equal to the base; and all right sections of a prism are equal. PRISMS. 281 Proposition II. Theorem. 561. The lateral area of a prism is equal to the ■product of a lateral edge by the perimeter of. the right section. A E' Ji Let GHIKL be a right section of the prism AU. To prove lateral area of AD' = AA\GH-\- III+ etc.). Proof. Consider the lateral edges AA\ BB\ etc., to be the bases of the HI AB\ BC\ etc., which form the lateral surface of the prism. Then the bases of these UJ are all equal. § 551 Since the sides of the right section, GH, HI, etc., are X to AA\ BB', etc. (§ 462), they are the altitudes of these ZI7, and the sum of the altitudes OS", HI^ IK, etc., is the perimeter of the right section. The area of each O is the product of its base and alti- tude. § 365 Hence, the sum of the areas of the /17 is the product of a lateral edge AA^ by the perimeter of the right section. But the sum of the areas of the JJ] is the lateral area of the prism. Therefore the lateral area of the prism is equal to the product of a lateral edge by the perimeter of a right section. q. e. d. 562. Cor. The lateral area of a right prism is equal lo the altitude multiplied by the perimeter of the base. 282 SOLID GEOMETRY. BOOK VII. Proposition III. Theorem. 563. Two prisms are equal if three faces including a trihedral angle of the one are respectively equal to three faces including a trihedral angle of the other, and are similarly placed. J f In the prisms AI and AT, let AD, AG, AJ, be respec- tively equal to A'D', A'G', A' J', and similarly placed. To prove prism AI=^ prism. A^I'. Proof. By hypothesis the face A BAE, BAF, EAF, are equal to the face A B'A'E', B'A'F', F'A'F', respectively. Therefore the trihedral angle A=A'. § 641 Apply A to its equal A' ; then the faces AD, AG, A J, will coincide with the equal faces A'B', A'G', A'J^, respec- tively, the points Cand D falling at C and D'. As the lateral edges of the prisms are parallel, CIST will take the direction of O^E^, and DIoi D'F. Since the points F, G, and J coincide with F', G', and J', each to each, the planes of the upper bases will coincide. Hence ^will coincide with EC', and 7" with J'. Therefore the prisms coincide and are equal. q.e.d. 564. Cor. 1. Too truncated prisms are equal if three faces including a trihedral of the one are respectively equal to three faces including a trihedral of the other, and are similarly placed, 565. Cor. 2. Two right prisms having equal bases and alti- tudes are equal. If the faces are not similarly, placed, one of the prisms can be inverted and applied to the other. A PEISMS. 283 Proposition IV. Theorem. 566. An oblique -prism is equivalent to a right prism whose base is equal to a right section of the oblique prisma, and whose altitude is equal to a lateral edge of the oblique prism. Let FI be a right section of the oblique prism AD'. Produce AA' to F', making FF'= AA', and at F' pass the plane F'F ± to FF', cutting all the faces of AZ>' produced, and forming the right section F'l' equal and parallel to FI. To prove AD'^FF. Proof. In the solids' AI and A'F, AD= A'D'. § 551 Also AG = A'G'] for, AF= A'F', and BG = B'G', since AA' = FF' and BB'= GG'; and AB and FG are equal and parallel to A'B' and F'G' respectively, since AB' and FG' are parallelograms (§ 551). Therefore J. 6^ and A'G' are mutually- equilateral and equiangular, and hence equal. § 203 In like manner we may prove ^-^and B'^' equal. Hence the truncated prisms ^/and A' I' are equal. § 564 Taking each in turn from the whole solid, we have AD'^FF. ^^.D. y 284 SOLID GEOMETRY. BOOK VII. Proposition V. Theorem. 567. Any two o-pposite faces of a parallelopiped are equal and parallel, E H 1 Let AG he a, parsbUelopiped. To prove faces AF and DO equal and 'parallel. Proof. ^^is II to Z)(7and ^^is II to D^ §§658,168 Hence Z EAB - Z HDC. § 498 Also AB = .DC^TidiAE=DS. ^ §179 Therefore the face AF== face DG. § 185 Moreover, the face ^i^is II to DG, § 498 {if two A not in the same plane have their sides II and lying in the same direction, their planes are parallel). In like manner any two opposite faces may be proved equal and parallel. q. e. d. 568. Scholium. Any two opposite faces of a parallelepiped may be taken for baaes, since they are equal arfd parallel parallelograms. Ex. 490. Show that any lateral edge of a right prism is equal to the altitude. Ex. 491. Show that the lateral faces of right prisms are rectangles. Ex. 492. Prove that every section of a prism made by a plane par- allel to the lateral edges is a parallelogram. PRISMS. 285 Proposition VI. Theorem. 569. The plane passed through two diagonally opposite edges of a parallelopiped divides the paral- lelopiped into two equivalent triangular prisms. G Let the plane AEG C pass through the opposite edges AE and CG of the parallelopiped AG. To prove that the parallelopiped AG is divided into two equivalent triangular prisms ABC-F and ACD-S. Proof. Let IJKL be a right section of the parallelopiped made by a plane JL to the edge AE. Since the opposite faces are parallel, § 567 IJ'-is. II to LK, and IL to JK. § 492 Therefore IJKL is a parallelogram. § 168 The intersection IK of the right section with the plane AEQG\% the diagonal of the O IJKL. :.AIKJ=AIKL. §178 But the prism ABC-F i^ equivalent to a right prism whose base is /t/lfiTand whose altitude is AE^ and the prism ACD-H is equivalent to a right prism whose base is ILK, and whose altitude is AE. § 566 But these two right prisms are equal. § 565 .-. ABC-F- ACD-K. o. e. a 286 SOLID GEOMETEY. BOOK: VII. Proposition VII. Theorem. 570. Two rectangular parallelopipeds having equal bases are to each other as their altitudes. B / p / / / B- / P' / / / / / / / / / / / — — /^ / /^ / 7 Let AB and A'B' be the altitudes of the two rectan- gular parallelopipeds P and P\ having equal bases. To prove P \ P^ = AB : A^BK Case I. When AB and AJB^ are commensurable. Proof. Find a common measure of AB and A^B\ Suppose this common measure to be contained in AB m times, and in A^B^ n times ; then we have AB : A^B' =-m:n. (1) At the several points of division on AB and A'B^ pass planes JL to these lines. The parallelepiped P will he divided into m, and P^ into n, parallelopipeds, equal each to each. § 565 Therefore P\P^ = m\n. (2) From (1) and (2), P : P^ ^ AB : A'B'. PRISMS. 287 Case II. When AB and A'B' are incommensurable. ^ s p' s \ s \J Let AB he divided into any number of equal parts, and let one of these parts be applied to A'B' as a unit of measure as many times as A'B' will contain it. Since AB and A'B' are incommensurable, a certain number of these parts will extend from A' to a point D, leaving a re- mainder JDB' less than one of the parts. Through I) pass a plane J. to A'B', and denote the parallel- epiped whose base is the same as that of F', and whose altitude is A'B, by Q. Now, since AB and A'B are commensurable, Q:F=A'I):AB. Case I. If the unit of measure is indefinitely diminished, these ratios continue equal, and they approach indefinitely the lim- iting ratios F' : F and A'B' : AB respectively. Therefore F:F= A'B' : AB, § 260 {if two variables are constantly equal, and each approaches a limit, their limits are equal). a E. D. 571. Scholium. The three edges of a rectangular parallelo- piped which meet at a common vertex are its ditnensions. Hence two rectangular parallelopipeds which have two dimen- sions in common are to each other as their third dimensions. 288 SOLID aEOMETRY. — BOOK VII. Proposition VIII. Theorem. 572. Two rectangular parallelopipeds having equal altitudes are to each other as their bases» Let a, 6, and c, and a\ h', c, be the three dimensions respectively of the two rectangular parallelopipeds P and P'. rp P axh 10 prove ~:;r.=—, r/ Let Q be a third rectangular parallelepiped whose dimen- sions are a\ b, and c. Now Q has the two dimensions b and c in common with P, and the two dimensions a' and c in common with P'. Q a'' {two rectangular parallelopipeds which have two dimensions in common are to each other as their third dimensions). The product of these two equalities is Then and P P' ax Xb' aE. o. 573. Scholium. This proposition may be stated as follows : Two rectangular parallelopipeds which have one dimension in common are to each other as the products of the other two dimensions. PRISMS. 289 Proposition IX. Theorem. 574. Two rectangular parallelopipeds are to each other as the products of their three dimensions. \ \ \ "' ^ c s \ c k ' \ \ a' Let a, 6, c, and a', 6', c', be the three dimensions re- spectively of the two rectangular parallelopipeds P and P'. rn F axbxc Proof. Let Q be a third rectangular parallelepiped whose dimensions are a, b, and c\ P^c_ Q c'' Q ^ axb P' a' X b'' {two rectangular parallelopipeds which have one dimension in common are to each other as the products of the other two dimensions). The product of these equalities is _, ^ r-^ ^ aXbXc ^^ Then and §571 §573 /^e-.T, t' >< b' X c' 1/ a. ly^ Ex. 493. j^ind the ratio of two rectangular parallelepipeds if their altitudes are each 6 inches, and their bases 5 inches by 4 inches, and 10 inches by 8 inches, respectively. Ex. 494. Find the ratio of two rectangular parallelopipeds, if their dimensions are 3, 4, 5, and 9. 8, 10, respectively. 290 SOLID GEOMETEY. — BOOK VII. Peoposition X. Theorem. 575. The volume of a rectangular parallelopiped is equal to the product of its three dimensions, the unit of volume being a cube whose edge is the linear uniL TZT / / / f / / / / 1 Let a, b, and c be the three dimensions of the rec- tangular parallelopiped P, and let the cube U be the unit of volume. To prove that the volume of P=aXh X c. P aXbx c Proof. U 1x1x1 aXbXc. F . Since Uis the unit of volume, — is the volume of P. §574 §547 Therefore the volume oi F=aXb X c. a e. d. 576. Cor. 1. The volume of a cube is the cube of its edge. bllt Cor. 2. The product axb represents the area of base when c is the altitude ; hence : The volume of a rectangular parallelopiped is equal to the product of its base by its altitude. 578. Scholium. When the three dimensions of the rectan- gular parallelopiped are each exactly divisible by the linear unit, this proposition is rendered evident by dividing the solid into cubes, each equal to the unit of volume. Thus, if the three edges which meet at a common vertex contain the linear unit 3, 4, and 5 times respectively, planes passed through the several points of division of the edges, and perpendicular to them, will divide the solid into cubes, each equal to the unit of volume ; and there will evidently be 3 X 4 X 5 of these cubes. PEISMS. 291 Proposition XI. Tseorem. 679. The volume of any parallelopiped is equal to the product of its base by its altitude. Let n denote the altitude of the parallelopiped AG. To prove that the volume AG — A BCD X jff". Proof. Consider ADHE the base of AG, and prolong the lateral edges AB, DC, EF, HG. The right parallelopiped A^G\ determined by two right sec- tions A'D6), and A'B'CD' = A^KZB'(n^^)- But ^'0 is a rectangular parallelopiped, for the right sec- tions A*]^, KO, are rectangles, since the opposite faces A}P, B^O, are ± to A'B'LK Hence the volume A'O = A'B'LKx IT. § 577 Therefore the volume AG = ABCD X ^ Q. E. 0. 292 SOLID GEOMETRY. BOOK VII. Proposition XIL Theorem. 580. The volume of a triangular prism is equal to the product of its base by its altitude. Let V denote the volume, B the base, and H the altitude of the triangular prism AEC-E', To prove V= B X H. Proof. Upon the edges AU, EC, EE\ construct the paral- lelepiped .^^OZ)-^'. Then, since a plane passed through two diagonally opposite edges of a parallelepiped divides it into two equivalent trian- gular prisms, § 669 AEC-E' -\AECB-EK Since the volume of any parallelepiped is equal to the prod- uct of its base by its altitude, AECD-E* = AECD x H. § 579 But AECD = 2B. §178 .-. V=i(2Bxir) = Bxir. Ex. 495. Find the volume of a right triangular prism, if its height is 14 inches, and the sides of the base are 6. 6, and 5 inches. PRISMS. 293 Proposition XIII. Theorem. 581. The volume of any ■prisin is equal to the prod- uct of its base by its altitude. Let V denote the volume, B the base, and H the altitude of the prism DA'. To prove F= B X IT. Proof. Planes passed through the lateral edge AA', and the diagonals AC, AD of the base, will divide the given prism into triangular prisms. The volume of each triangular prism is equal to the product of its base by its altitude (§ 580) ; and hence the sum of the volumes of the triangular prisms is equal to the sum of their bases multiplied by their common altitude. But the sum of the triangular prisms is equal to the given prism, and the sum of their bases is equal to the base of the given prism. Therefore the volume of the given prism is equal to the product of its base by its altitude. That is, V=Bxir. aE.D. 582. Cor. The volumes of two prisms are to each other as the products of their bases and altitudes ; prisms having equivalent bases are to each other as their altitudes ; prisms having equal altitudes are to each other as their bases ; prisms having equiv- alent bases and equal altitudes are equivalent. 294 SOLID GEOMETRY. — BOOK VII. Pyramids. 683. A pyramid is a polyhedron of wWch one face, called the base, is a polygon, and the other faces, called lateral faces, are triangles having a common vertex, called the vertex of the pyramid. 584. The intersections of the lateral faces are called the lateral edges of the pyramid. ^ ^- 585. The sum of the areas of the lateral faces is called the lateral area of the pyramid. 586i The altitude of a pyramid is the length of the perpen- dicular let fall from the vertex to the plane of the base. 587. A pyramid is called triangular, quadrangular, etc., according as its base is a triangle, quadrilateral, etc. 588. A triangular pyramid, having four faces, is called a tetrahedron, and any one of its faces can be taken for its base. 589. A pyramid is regular if its base is a regular polygon whose centre coincides with the foot of the perpendicular let fall from the vertex to the base. 590. The lateral edges of a regular pyramid are equal, since they, cut off equal distances from the foot of the perpendicular let fall from the Regular Pyramid, vertex to the base (§ 478). Therefore the lateral faces are equal isosceles triangles. 591. The slant height of a regular pyramid is the length of the perpendicular from the vertex to the base of any one of its lateral faces. It is the common altitude of all the lateral faces, and bisects the base of the lateral face in which it is drawn. 592. A frustum of a pyramid is the portion of a pyramid PYRAMIDS, 295 included between its base and a plane parallel to the base and cutting all the lateral edges. 593. The altitude of a frustum is the length of the perpendicular between the planes of its bases, 694. The lateral faces of a frustum of a regu-^ lar pyramid are equal trapezoids. 595. The slant height of the frustum of a regular pyramid is the altitude of one of these trapezoids. Proposition XIV. Theorem. 696. The lateral area of a regular pyramid is equal to one-half the product of the slant height by the perimeter of its base. Let S denote the lateral area at the regular pyra- mid V-ABCDE, and VII its slant height To prove that 8-=^ VII{AB -\'B0-\- etc.). Proof. The A VAB, VBO, etc., are equal isosceles A. §590 Theareaof the sum of these A=|F^(^^+^C+etc.) § 368 But the sum of their areas equals the lateral area of the pyramid. ...^=|KS-(^5+£C+etc.). aE.a 597. Cor. 1. The latcn^al area of the frustum of a regular pyramid, is equal to one-half the sum of the perimeters of the bases multiplied by the slant height of the frustum, § 371 296 SOLID GEOMETRY. — BOOK VII. Proposition XV. Theorem. 598. If a pyramid is cut hy a plane parallel to its I. Th^ edges and altitude are divided propor- tionally ; II. The section is a polygon similar to the hase. V Let V-ABCDE be cut "by a plane parallel to its base, intersecting the lateral edges in a, b, c, d, e, and the altitude in o. rp J Va Vb Vo - To prove I. _:=^....^ — ; IL The section abode similar to the base ABODE. I. Proof. Suppose a plane passed througli the vertex V 11 to the base. Since the edges and the altitude are intersected by three parallel planes, §499 Vo^ vd Va ^ Vb VA vb" II. Since the sides ab, be, etc., are parallel respectively to AB, BC, etc., § 492 the A abc, bed, etc., are equal respectively to the A ABC, BCD, etc. § 498 Therefore the two polygons are mutually equiangular. PYEAMIDS. 297 Also, since the sides of the section are parallel to the corre- sponding sides of the base, A Vab, Vbc, etc., are similar respectively to A VAB, VBC, etc. • A. = fZ^\ = Il=fJ^\ = ^ etc " AB \VBj BO \VC) CD' " Hence the polygons have their homologous sides propor- tional ; Hence section abode is similar to the base ABODE. § 319 599. Cor. 1. Any section of a pyramid parallel to its base is to the base as the square of its distance from the vertex is to the square of the altitude of the pyramid. Since i^ = ^i^V— . .•.-^=-^. §305 VO \VBj AB VO' AB' But _abcde_^;^^ g 3^g ABODE aS abode Vo ' ' ABODE vo' 600. Cor. 2. If two pyramids having equal altitudes are cut by planes parallel to their bases, and at equal distances from their vertices^ the sections will have the same ratio as their bases. -p abode _ Vo °'' ABODE~ vo' and j±_c_ ^ _K_o_ « ^gg A'BO' v^i' ^ But Vo = V'o\ and VO^Va. . . abode : ABODE = a'b'o' : A'B'O^. Whence abode : a'b'o' = ABODE : A'B'O^. § 298 601. Cor. 3. If two pyramids have equal altitudes and equivalent bases, sections made by planes parallel to their bases^ and at equal distances from their vertices, are equivalent. 298 SOLID GEOMETRY. — BOOK VII. Proposition XVI. Theorem. 602. Two triangular pyramids having equivalent hases and equal altitudes are equivalent. Let S-ABC and S'-A'B'C have equivalent bases situ- ated in the same plane, and a common altitude. To prove 8-ABC<> jS'-A'B'O'. Proof. If the pyramids are not equivalent, suppose S-ABC the greater. Divide the common altitude into n equal parts. Through the points of division pass planes II to the plane of their bases. The corresponding sections of the pyramids are equivalent. § 601 On the base of S-ABC, and on each section, as lower base, construct a prism with lateral edges equal and parallel to AB. Similarly, construct a prism on each section of S^-A'B^C^^ as upper base. The sum of the first series of prisms is greater than S-ABC, and the sum of the second series is less than S-A'B'C; there- fore the difference between S-ABC si,nd S^-A^B'C is less than the difference betvf een the sums of these two series of prisms. Each prism in S'-A'B^C is equivalent to the prism next above it in S-ABC (§ 582). Hence the difference between the two series of prisms is the lowest prism of the first series. But by increasing n indefinitely this can be made less than any assigned volume, however small. Therefore the two pyramids cannot difier by any volume how- ever small ; therefore the pyramids are equivalent. q. e. d. PYRAMIDS. • 299 Proposition XVII. Theorem. 603. The volume of a triangular pyramid is equal to one-third of the product of its base and altitude. ^r--v.-- Let V denote the volume, and H the altitude, of the triangular pyramid S-ABC. To prove V= J ABCx H. Proof. On the base ABC construct a prism ABC-8ED, having its lateral edges equal and parallel to SB. The prism will be composed of the triangular pyramid 8-ABC2JCidi the quadrangular pyramid S-ACDE. Through 8 A and 8D pass a plane 8 AD. This plane divides the quadrangular pyramid into the two triangular pyramids 8-ACD and 8-AED, which have the same altitude and equal bases. § 178 .-. 8-ACD =o 8-AED. § 602 Now the pyramid 8-AED may be regarded as having E8D for its base and J. for its vertex. .'.8-AED 0^8- ABO. Hence the three pyramids into which the prism ABC-8ED is divided are equivalent ; the pyramid 8- ABC is equivalent to one-third of the prism. But the volume of the prism is equal to the product of its base and altitude. § 580 .•.V=\ABCxn. Q.E.Q 300 SOLID GEOMETRY. — BOOK VII. Proposition XVIII. Theorem. 6p4'. The volume of any pyramid is equal to one^ third the product of its base and altitude, S Let V denote the volume of the pyramid S-ABCDE, To prove F= ^ ABODE x SO. Proof. Through the edge SI>, and the diagonals of the hase' DA, DB, pass planes. These divide the pyramid into triangular pyramids, whose bases are the triangles which compose the base of the pyramid, and whose common altitude is the altitude 80 of the pyramid. The volume of the given pyramid is equal to the sum of the volumes of the triangular pyramids. But the sum of the volumes of the triangular pyramids is equal to one-third the sum of their bases multiplied by their common altitude. § 603 That is, V=- 1 ABODE x SO. a e. d. 605. Cor. The volumes of two pyramids are to each other as the products of their bases and altitude^; pyramids having equivalent bases are to each other as their altitudes; pyramids having equal altitudes are to each other as their bases; pyramids having equivalent bases and equal altitudes are equivalent 606i Scholium. The volume of any polyhedron may be found by dividing it into pyramids, computing their volumes separately, and finding the sum of their volumes. PYRAMIDS. 301 Proposition XIX. Theorem. 607, The volumes of two teiTrahedrons, having a tri- hedral angle of the one equal to a trihedral angle of the other, are to each other as the products of the three edges of these trihedral angles. & Let V and V^ denote the volumes of the two tetra- hedrons S-ABC and S-A'B'C, having the common tri- hedral angle S. To prove — = Proof. Draw CD and CD' ± to the plane SA'B', and let their plane intersect SA'B' in SDD'. The faces SAB and SA'B' may be taken as the bases, and CD, CD' as the altitudes, of the triangular pyramids SAB-C and SA'B'-C. SAB X CD SAB .. CD §605 SA'B'x CD' SA'B' CD' {any two pyramids are to each other as the products of their bases and altitudes). But .^^=^4x^. §374 and SA'B' SA' X SB' CD _^SC CD' SC' ■ {being homologous sides of the similar ^ SBC and 8D^C^). . V _ SAxSBxSC ' ' V SA' X SB' X SO*' §319 aE. D. 302 SOLID GEOMETRY. BOOK VII. Proposition XX. Theorem. 608. The frustum of a triangular pyramid is equiv- alent to the sum of three pyramids whose common altitude is the altitude of the frustum and whose bases are the lower base, the upper base, and a mean proportional between the two bases of the frustvMm. Let B and b denote the lower and upper bases of the frustum ABC-DEF, and H its altitude. Through the vertices A, U, C and U, I), C pass planes dividing the frustum into three pyramids. Now the pyramid E-ABC has for its altitude H, the alti- tude of the frustum, and for its base B, the lower base of the frustum. And the pyramid C-EDF has for its altitude H, the alti- tude of the frustum, and for its base h, the upper base of the frustum. Hence it only remains To prove E-ADC equivalent to. a pyramid, having for its altitude H, and for its base -\/ B X h. Proof. E-ABQ 2.T\di E-ADO, regarded as having the com- mon vertex (7, and their bases in the same plane BD, have a common altitude. PYRAMIDS. 303 .-. C-ABE : C-ADE = A AEB : A AEE, § 605 {pyramids having equal altitudes are to each other as their bases). Now since the A AEB and AED have a common altitude, {that is, the altitude of the trapezoid ABED), we have A AEB : A AEB = AB : BE, § 370 .-. C-ABE : C-ABE = AB : BE. That is, E-ABC : E-ABC= AB : BE. In like manner E-ABC and E-BFC, regarded as having the common vertex E, and their bases in the same plane DC, have a common altitude. .-. E-ABC : E-BFC ^ A ABC : A BFC § 605 But since the A ABC and BFC have a common altitude, {that is, the altitude of the trapezoid ACFD), we have A ABC : A BFC = AC : BE. §370 .-. E-ABC : E-BFC= AC : BF. But -A BEE is similar to A ABC, § 598 {the section of a pyramid made by a plane II to the base is a polygon similar to the base). .\AB:BE=AC:BF. §319 .-. E-ABC : E-ABC = E-ABC : E-BFC Now E-ABC ^ilTxB, §603 and E-BFC = C-EBF=\Hx b. .'.E-ABC=-ViE:xBxiirxb = iJI-VB> J Pkoposition XXI. Theorem. j^ 610. T/z/g voluTne of the frustum of any pyramid is equal to the sum of the volumes of three pyramids whose common altitude is the altitude of the frus- tum, and whose bases are the lower hase, the upper hase, and a mean proportional between the bases of the frustum. S T Let B and b denote the lower and upper bases, B the altitude, and V the volume of ABCD-EFGL To prove V=\II{B + 5 + VB~xh). Proof. Let T-KLM be a triangular pyramid having the same altitude as 8-ABCD and its base KLM=^ ABCD, and lying in the same plane. Then T-KLM ^ 8-ABCD. § 605 Let the plane UFG I cut T-KLM in NOP. Then NOP =o= EFGL § 601 Hence T-NOP=o= 8-EFGL Taking away the upper pyramids leaves the frustums equivalent. But the volume of the frustum of the triangular pyramid is equal to ^ ^(^ + ^ + VB^b). § 609 .-. v= \b:{b + h + -VB^b). Q. E. D. NUMERICAL EXERCISES. -' ""*? 305^ v' ' ^ ^ ' • Numerical Exercises. 71-% ' 496. Find the length of an edge of a cubical vessel which will hold 2-ton8 of water. 497. How many square feet of lead will be required to line a cistern, open at the top, which is 4 feet 6 inches long, 2 feet 8 inches wide, and contains 42 cubic feet ? i ^'T ^498. An open cistern is made of iron 2 inches thick. The inner o dimensions are: length, 4 feet 6 inches; breadth, 3 feet ; depth, 2 feet '^ 6 inches. What will the cistern weigh (i.) when empty ? (ii.) when full ^ of water? Specific gravity of iron =- 7.2. IfYxAAjf^ ^U/.«t " iT/ ^^ ' 499. An open cistern 6 feet long and 4J feet wide holds 108 cubic feet of water. How many cubic feet of lead will it take to line the sides and bottom, if the lead is \ inch thick? 500. The three dimensions of a rectangular parallelopiped are a, ?», c ; find the surface, the volume, and the length of a diagonal. 501. The base of a right prism is a rhombus, one side of which is 10 inches, and the shorter diagonal is 12 inches. The height of the prism is 15 inches. Find the entire surface and the volume. 502. Find the volume of a regular hexagonal prism whose height is 10 feet, each side of the hexagon being 10 inches. 503. A pyramid 15 feet high has a base containing 169 square feet. At what distance from the vertex must a plane be passed parallel to the base so that the section may contain 100 square feet? /? ' ^ " 504. The base of a pyramid contains 144 square feet. A plane par- allel to the base and 4 feet from the vertex cuts a section containing 64 square feet ; find the height of the pyramid. ' 505. A pyramid 12 feet high has a square base measuring 8 feet on a side. What will be the area of a section made by a plane parallel to the base and 4 feet from the vertex ? / 1 T 506. Two pyramids standing on the same plane are 14 feet high. The f\ = first has for base a square measuring 9 feet on a side ; the second a reg- ular hexagon measuring 7 feet on a side. Find the areas of the sections , ^, made by a plane parallel to their bases and 6 feet from their vertices. 'T**'' 507. The base of a regular pyramid is a hexagon of which the side measures 3 feet. Find the height of the pyramid if the lateral area is equal to ten times the area of the base. jj/ 306 SOLID GEOMETRY. — BOOK VII. Proposition XXII. Theorem. ' 611. A truncated triangular prism is equivalent to the sum of three pyramids whose common base is the base of the prism, and whose vertices are the three vertices of the inclined section. Let ABC-DEF be a truncated triangular prism whose base is ABC, and inclined section DEF. Pass the planes AEC and DEC, dividing the truncated prism into the three pyramids E-ABC, E-ACD, and E-CBF. To prove ABC-DEF equivalent to the sum of the three pyra- mids, E-ABC, D-ABC, and F-ABC Proof. E-ABC ho.^ the base ^^Cand the vertex E. The pyramid E-A CD =o B-A CD, § 602 (for they have the same base A CD and the same altitude, since their vertices E and B are in the line EB II to the base ACD). But the pyramid B-ACD may be regarded as having the base ABC and the vertex D ; that is, as D-ABC. The pyramid E-CDF<^ B-ACF, for their bases CDF and ACF, in the same plane, are equiva- lent, § 369 {since the A CDF and A CF have the common base CF and equal altitudes, their vertices lying in the line AD II to CF), PYRAMIDS. 307 and the pyramids liave the same altitude, {since their vertices E and B are in the line EB II to the plane of their bases ACDF). But the pyramid B-ACF may be regarded as having the base ^^Cand the vertex F\ that is, as F-ABC. Therefore the truncated triangular prism ABC-DEF is equivalent to the sum of the three pyramids F-ABC, D-ABC, and F-ABC. a e. d 612. Cor. 1. The volume of a truncated right triangular prism is equal to the product of its base by one-third the sum of its late)'al edges. For the lateral edges DA, EB, FC, being perpendicular to the base, are the altitudes of the three pyra- mids whose sum is equivalent to the truncated prism. And, since the volume of a pyramid is one-third the product of its base by its altitude, the sum of the volumes of these pyramids = ABGx\{DA-\-EB + FC). 613. Cor. 2. The volume of any truncated triangular prism is equal to the p7'oduct of its right section by one-third ^e sum of its lateral edges. For let ABC-A'B'C be any truncated triangular prism. Then the right section DEF divides it into two truncated right prisms whose volumes are DEFx J {AD^ BE-^ CF) and DEFx \ {A*I)-{-B'E-\- C'F). Whence their sum is I)EFxi(AA'+ BB'-\-CC^). 308 SOLID GEOMETRY. BOOK VII. Similar Polyhedrons. 614. Similar polyhedrons are polyhedrons that have the same number of faces, respectively similar and similarly placed, ar^ their corresponding polyhedral angles equal. Homologous faces, lines, and angles of similar polyhedrons are faces, lines, and angles similarly placed. 615. Cor. 1. The homologous edges of similar polyhedrons are proportional. § 319 616. CoR. 2. Two homologous faces of similar polyhedrons are proportional to the squares of two homologous edges. § 377 617. CoR. 3. The entire surfaces of two similar polyhedrons are proportional to the squares of two homologous edges. § 303 618. Cor. 4. The homologous dihedral angles of similar polyhedrons are equal. Proposition XXIII. Theorem. 619. Two similar polyhedrons may he decomposed into the same number of tetrahedrons similar, each to eojch, and similarly placed. Let P and P' be two similar polyhedrons. SIMILAR POLYHEDRONS. 309 To prove that the similar polyhedrons P and P' can he decomposed into the same number of tetrahedrons, similar each to each, and similarly placed. Proof. Through the vertices A, G, C, and the homologous vertices A\ G', 0', pass planes. The tetrahedrons G-ABC and G'-A'B'C have the faces ABC, GAB, GBC, similar respectively to A'B'C\ G'A'B', G'B'C. § 332 Hence in the faces G^^Cand G'A'C AG ^( AB\_ AC _f BC\^ GO § oiq A'G' \A'B'J A'C \B'cO G'C' ^ Therefore the face G AC is similar to G'A'C. § 324 Hence the faces of these tetrahedrons are similar, each to each. Also, any two corresponding trihedral A of these tetrahe- drons are equal. § 541 Therefore the tetrahedron G-ABC is similar to G'-A'B'C. §614 If G-ABC and G'-A'B'C be removed, the polyhedrons remaining will continue similar ; for the new faces GAC and G'A'C have just been proved similar, and the modified faces ^6^i^ and A'G'F*, CGJI &ud C'G'IT', will be similar (§ 332) ; also the modified polyhedral A G and G', A and A\ (7 and C, will remain equal each to each, since the corresponding parts taken from them are equal. The process of removing similar tetrahedrons can be carried on until the polyhedrons are reduced to tetrahedrons ; that is, until the two similar polyhedrons are decomposed into the same number of tetrahedrons similar each to each, and simi- larly situated. a E. D. 620. Cor. Any two homologous lines in two similar polyhe- drons have the same ratio as any two homologous edges. 310 SOLID GEOMETRY. — BOOK VII. Proposition XXIV. Theorem. 621. The volumes of two similar tetrahedrons are to each other as the cubes of their homologous edges. Let V and V denote the volumes of the two similar tetrahedrons S-ABC and S'-A'B'C, To prove V_^ SB" Proof. Since the homologous trihedral angles 8 and >S" are equal, we have V SBxSCxSA V jS'B' X S'C X jS'A' ^ SB SO SA S'B' S'C S'A'' But SB SO V v S'B' SB S'O' xMr.X SA S'A'' SB SR S'B' S'B' S'B' -^T^'' §607 §615 Q. E. D. Ex. 508. The homologous edges of two similar tetrahedrons are as 6 : 7. Find the rat^ tf the'i/ s&faces and of tij/ei^ volumes. Ex. 509. If the edge of a tetrahedron is a, find the homologous edge of a similar tetrahedron twice as large. ^^ ■ ^ ^ /7^- "ira} : o, : s'<^b). Proof. Let V* denote the volume, £' and b^ the lower and upper bases, and -ff"the altitude, of an inscribed frustum of a pyramid. Then r = i B-(B' + b'-\- V^' X b'). §610 Let the number of lateral faces of the inscribed frustum be indefinitely increased, the new edges continually dividing the arcs in the bases of the frustum of the cone. Then, however great the number of lateral faces of the frustum of the pyramid. yf^;^ ^^^r _^ j, _^ VBxI'y .'.V =-iB:(B '\-b-\-VBxb). § 260 a E. D. 678. Cor. If the frustum is that of a cone of revolution, and R and r are the radii of its bases, we have B = tt^', b = ttt^, and ^B xb — irRr. ... F= 4 irHiE' + r' + Rr). 334 "^ SOLID GEOMETRY. — BOOK VII. Numerical Exercises. The Pyramid. Find the volume in cubie feet of a regular pyramid : 513. When its base is a square, each side measuring 3 feet 4 inches, and its height is 9 feet. 514. When its base is an equilateral triangle, each side measuring 4 feet, and its height is 15 feet. 515. When its base is a regular hexagon, each side measuring 6 feet, and its height is 30 feet. Find the total surface in square feet of a regular pyramid : 516. When each side of its square base is 8 feet, and the slant height is 20 feet. 517. When each side of its triangular base is 6 feet, and the slant height is 18 feet. 518. When each side of its square base is 26 feet, and the perpendic- ular height is 84 feet. Find the height in feet of a pyramid when : 519. The volume is 26 cubic feet 936 cubic inches, and each side of its square base is 3 feet 6 inches. 520. The volume is 20 cubic feet, and the sides of its trls-ngular base are 5 feet, 4 feet, and 3 feet. 521. The base edge of a regular pyramid with a square base meas- ures 40 feet, the lateral edge 101 feet ; find its volume in cubic feet. 522. Find the volume of a regular pyramid whose slant height is 12 feet, and whose base is an equilateral triangle inscribed in a circle having a radius of 10 feet. 523. Having given the base edge a, and the total surface T, of a regular pyramid with a square base, find the volume V. 524. The base edge of a regular pyramid whose base is a square is a, the total surface T\ find the height of the pyramid. 525. The eight edges of a regular pyramid with a square base are equal in length, and the total surface is T; find the length of one edge. 526. Find the base edge a of a regular pyramid with a square base, having given the height h and the total surface T. NUMERICAL EXERCISES. 335 Cylinders and Cones. ^ 527. If the total surface of a right circular cylinder closed at both ends is o, and the radius of the base is r, what is the height of the cylinder ? ^ 528. If the lateral surface of a right circular cylinder is a, and the volume is b, find the radius of the base and the height. { 529. How many cubic yards of earth must be removed in construct- ing a tunnel 100 yards long, whose section is a semicircle with a radius of 10 feet ? 530. If the diameter of a well is 7 feet, and the water is 10 feet deep, how many gallons of water are there, reckoning 7J gallons to the cubic foot? *" 531. When a body is placed under water in a right circular cylinder 60 centimeters in diameter, the level of the water rises 30 centimeters ; find the volume of the body. ^532. If the circumference of the base of a right circular cylinder is c, and the height h, find the volume V. 533. Having given the total surface T of a right circular cylinder, in which the height is equal to the diameter of the base, find the volume V. ^ 534. If the circumference of the base of a right circular cylinder is c, and thedotal surface is T, find the volume V. ^ 535. The slant height of a right circular cone is 2 feet. At what distance from the vertex must the slant height be cut by a plane parallel to the base, in order that the lateral surface may be divided into two equivalent parts ? y^ 536. The height of a right circular cone is equal to the diameter of its base ; find the ratio of the area of the base to the lateral surface. >»-^ 587. "What length of canvas f of a yard wide is required to make a conical tent 12 feet in diameter and 8 feet high ? ^ 538. The circumference of the base of a circular cone is 12^ feet, and its height 8\ feet ; find its volume. /^ 539. Given the total surface T of a right circular cone, and the radius r of the base ; find the volume V. ^>^540. Given the total surface T of a. right circular cone, and the lateral surface 8; find the volume V. 336 SOLID GEOMETRY. — BOOK VII. Frustums of Pyramids and Cones. ^^,---^541. How many square feet of tin will be required to make a funnel if the diameters of the top and bottom are to be 28 inches and 14 inches respectively, and the height 24 inches ? 542. Find the expense of polishing the curved surface of a marble column in the shape of the frustum of a right cone whose slant height is 12 feet, and the radii of the circular ends are 3 feet 6 inches and 2 feet 4 inches respectively, at 60 cents a square foot. "^43. The slant height of the frustum of a regular square pyramid is 20 feet, the length of each side of its base 40 feet, of each side of its top 16 feet ; find its volume. ^ 544. If the bases of the frustum of a pyramid are two regular hexa- gons whose sides are 1 foot and 2 feet respectively, and the volume of the frustum is 12 cubic feet : find its height. 545. The frustum of a right circular cone is 14 feet high, and has a volume of 924 cubic feet. Find the radii of its bases if their sum is 9 feet. - 546. From a right circular cone whose slant height is 30 feet, and circumference of whose base is 10 feet, there is cut off by a plane parallel to the base a cone whose slant height is 6 feet. Find the convex surface and the volume of the frustum. \ 547. Find the difference between the volume of the frustum of a pyramid whose bases are squares, measuring 8 feet and 6 feet respec- tively on a side, and the volume of a prism of the same altitude whose base is a section of the frustum parallel to its bases and equidistant from them. • 548. A Dutch windmill in the shape of the frustum of a right cone is 12 meters high. The outer diameters at the bottom and the top are 16 meters and 12 meters, the inner diameters 12 meters and 10 meters, respectively. How many cubic meters of stone were required to build it? 549. The chimney of a factory has the shape of a frustum of a regu- lar pyramid. Its height is 180 feet, and its upper and lower bases are squares whose sides are 10 feet and 16 feet respectively. The flue is throughout a square whose side is 7 feet. How many cubic feet of material does the chimney contain ? P"^ 550. Find the volume V of the frustum of a cone of revolution, ^ having given the slant height a, the height \ and the convex surface S. NUMEEICAL EXERCISES. 337 Equivalent Solids. 551. A cube whose edge is 12 inches long is transformed into a right prism whose base is a rectangle 16 inches long and 12 inches wide. Find the height of the prism, and the difference between its total surface and the surface of the cube. ^ 552. The dimensions of a rectangular parallelopiped are a, 6, c Find (i.) the height of an equivalent right circular cylinder having a for the radius of its base ; (ii.) the height of an equivalent right circular cone having a for the radius of its base. '^ 553. A regular pyramid 12 feet high is transformed into a regular prism with an equivalent base ; what is the height of the prism ? ^ 554. The diameter of a cylinder is 14 feet, and its height is 8 feet; find the height of an equivalent right prism, the base of which is a square with a side 4 feet long. ^ 555. If one edge of a cube is a, what is the height h of an equivalent right circular cylinder whose diameter is 5 ? "" 556. The heights of two equivalent right circular cylinders are as 4: 9. The diameter of the first is 6 feet; what is the diameter of the other? ^ 557. A right circular cylinder 6 feet in diameter is equivalent to a right circular cone 7 feet in diameter. If the height of the cone is 8 feet, what is the height of the cylinder ? '^558. The frustum of a regular four-sided pyramid is 6 feet high, and the sides of its bases are 5 feet and 8 feet respectively. What is the height of an equivalent regular pyramid whose base is a square with a side 12 feet long ? ^ 559. The frustum of a cone of revolution is 5 feet high, and the diameters of its bases are 2 feet and 3 feet respectively ; find the height of an equivalent right circular cylinder whose base is equal in area to the section of the frustum made by a plane parallel to its bases, and equidistant from the bases. 560. Find the edge of a cube equivalent to a regular tetrahedron whose edge measures 3 inches, - 561. Find the edge of a cube equivalent to a regular octahedron whose edge measures 3 inches. 338 SOLID GEOMETRY. BOOK VII. Similar Solids. "^ 562. The dimensions of a trunk are 4 feet, 3 feet, 2 feet. What are the dimensions of a trunk similar in shape that will hold four times as much ? - 563. By what number must the dimensions of a cylinder be multiplied in order to obtain a similar cylinder (i.) whose surface shall be n times that of the first; (ii.) whose volume shall be n times that of the first? 564. A pyramid is cut by a plane which passes midway between the vertex and the plane of the base. Compare the volumes of the entire pyramid and the pyramid cut off. ^ 565. The height of a regular hexagonal pyramid is 36 feet, and one side of the base is 6 feet. What are the dimensions of a similar pyramid whose volume is -^^ that of the first ? 566. The length of one of the lateral edges of a pyramid is 4 meters. How far from the vertex will this edge be cut by a plane parallel to the base, which divides the pyramid into two equivalent parts ? ^ 567. The length of a lateral edge of a pyramid is a. At what dis- tances from the vertex will this edge be cut by two planes parallel to the base, which divide the pyramid into three equivalent parts ? -- 568. The length of a lateral edge of a pyramid is a. At what dis- tance from the vertex will this edge be cut by a plane parallel to the base, and dividing the pyramid into two parts which are to each other as 3: 4? 569. The volumes of two similar cones are 54 cubic feet and 432 cubic feet. The height of the first is 6 feet ; what is the height of the other ? ' 570. In each of two right circular cylinders the diameter is equal to the height. The volume of one is f that of the other. What is the ratio of their heights ? 571. Find the dimensions of a right circular cylinder ^f as large as a similar cylinder whose height is 20 feet, and diameter 10 feet. '*^572. The height of a cone of revolution is h, and the radius of its base is r. What are the dimensions of a similar cone three times as large ? ^573. The height of the frustum of a right cone is | the height of the entire cone. Compare the volumes of the frustum and the entire cone. '^ 574. The frustum of a pyramid is 8 feet high, and two homologous edges of its bases are 4 feet and 3 feet respectively. Compare the vol- ume of the frustum and that of the entire pyramid. / BOOK VIII. THE SPHERE. Plane Sections and Tangent Planes. 679. A sphere is a solid bounded by a surface every point of which is equally distant from a point called the centre. 680. A sphere may be generated by the revolution of a semicircle ACB about its diameter AB as an axis. 681. A radius of a sphere is a straight line drawn from its centre to its surface. < 682. A diameter of a sphere is a straight line passing through the centre and limited by the surface. Since all the radii of a sphere are equal, and a diameter is equal to two radii, all the diameters of a sphere are equal. 683. A line or plane is tangent to a sphere when it has one, and only one, point in common with the surface of the sphere. 684. Two spheres are tangent to each other when their surfaces have one, and only one, point in common. 340 / SOLID GEOMETRY. — BOOK VIII. Proposition I. Theorem. 685. Every section of a sphere made hy a plane is a circle. Let be the centre of a sphere, and ABD any sec- tion made by a plane. To prove that the section ABD is a circle. Proof. Draw the radii OA, OB, to any two points A, B/\n the boundary of the section, £lnd draw OQ 1. to the section. In the rt. A OAC, OBC, OC is common. Also OA = OB, (being radii of the sphere), .'.AOAO=AOBO, §161 .-. CA = CB. In like manner any two points in the boundary of the sec- tion may be proved to be equally distant from C. Hence the section ABB is a circle whose centre is C. .q. e. d. 686i Cor. 1. The line joining the centre of a sphere to the centre of a circle of the sphere is_ perpendicular to the plane of the circle. 687. Cor. 2. Circles of a sphere made hy planes equally dis- tant from the centre are equal. For AC = AO ~ OC ; and AO and 0(7 are the same for all equally distant circles; there- fore ^C is the same. THE SPHERE. 341 688. Cob. 3. Of two circles made hy planes unequally distant from the centre, the Tiearer is the larger. For, in the expression AG = AO —00 , as (9 C decreases, ^Cincreases. i9. A great circle of a sphere is a section made by a plane which passes through the centre of the sphere. 690. A small circle of a sphere is a section made by a plane which does not pass through the centre of the sphere. — ^_691r The axis of a circle of a sphere is the diameter of the sphere which is perpendicular to the plane of the circle. The ends of the axis are balled the poles. , 692. Parallel circles have the same axis and the same poles. 693. All great circles of a sphere are equal. 694. Every greai circle bisects the sphere. For the two parts into which the sphere is divided can be so placed that they will coincide ; otherwise there would be points on the surface unequally distant from the centre. 695. Two great circles bisect each other. For the intersec- tion of their planes passes through the centre, and is a diameter of each circle. 696. Two great circles whose planes are perpendicular pass through each other s poles; and conversely. 697. Through two given points on the surface of a sphere an arc of a great circle may always be drawn. For the two given points together with the centre of the sphere determine the plane of a great circle whose circumference passes through the two given points. If the two given points are the ends of a diameter, the posi- tion of the circle is not determined ; for through a diameter an indefinite number of planes may be passed. 698i Through three given points on the surface of a sphere one circle m,ay be drawn, and only one. For the three points determine one, and only one, plane. y SOLID GEOMETRY. — BOOK VIII. Proposition II. Theorem. ?99P T he shortest distance on the surface of a sphere between any two points on that surface is the arc ^ not greater than a semi-circumference, of the great circle which joins them. Let AB be the arc of a great circle which Joins any two points A and B on the surface of a sphere; and let ACPQB be anj other line on the surface between A and B. To prove ACFQB > AB. Proof. Let P be any point in ACFQB. Let arcs of great circles pass through A, P, and P, B. § 697 Join A, P, and B with the centre of the sphere 0. The A AOB, AOP, and FOB are the face A of the trihe- dral angle whose vertex is at 0. ' The arcs AB, AF, and FB are measures of these A. § 262 Now Z AOF+Z FOB is greater than Z AOB, § 539 .-. arc AF-\- arc FB > arc AB. In like manner, joining any point in ^CP with A and P, and any point in FQB with P and P, by arcs of great (D, the sum of these arcs will be greater than arc AF-\- arc FB ; and therefore greater than arc AB. If this process be indefinitely repeated, the sum of the arcs of the great (D will increase and always be greater than AB. Therefore ACFQB, which is the limit of the sum of these arcs, is greater than AB. q.e.d. THE SPHERE. 343 700. By the distance between two points on the surface of a sphere is meant the arc of a great circle joining them. Proposition III. Theorem. 701. The distances of all points in the circumfer- ence of a circle of a sphere from its poles are equal. Let P, P' be the poles of the circle ABC, and A, B, C, any points on its circumference. To prove that the great circle arcs PA, PB, PC are equal. Proof. The straight lines PA, PB, PC are equal, § 478 Therefore the arcs PA, PB, PC sere equal. § 230 In like manner, the great circle arcs PA, PB, PC may be proved equal. q^ e. d. 702. The distance from the nearer pole of a circle to any point in the circumference of the circle is called the polar dis- tance of the circle. 703. Cor. 1. The polar distance of a great circle is a quad- rant-arc. For it is the measure of a right angle whose vertex is at the centre of the sphere. 704. Scholium. The distances of all points in the -circum- ference of a circle of a sphere from any point in its axis are equal. 344 SOLID GEOMETRY. — BOOK VIII. Proposition IV. Theorem. 705. A -point on the surface of a sphere, which is at the distance of a quadrant from each of two other points, not the extremities of a diameter, is a pole of the great circle passing through these points* Let the distances PA and PB be quadrants. To prove P a pole of the great circle which passes through A and B. Proof. The A POA and POP are rt. A, {because each is measured by an arc equal to a quadrant). .-. PO is X to the plane of the O ABO, § 472 Hence P is a pole of the O ABO. § 691 a E. D. 706. Cor. The above theorem enables us to describe with the compasses an arc of a great circle through two given points A and B of the surface of a sphere. For, if with A and B as centres, and an opening of the compasses equal to the chord of a quadrant of a great circle, we describe arcs, these arcs will cut at a point P, which will be the pole of the great circle passing through A and B. Then with P as centre, the arc passing through A and B may be described. In order to make the opening of the compasses equal to the chord of a quadrant of a great circle, the radius or the diam- eter of the sphere must be given. THE SPHERE. 345 Proposition V. Problem. Given a material sphere to find its radius. :->5 C> V Let PBP'G represent a material sphere. It is required to find its diameter, Oonstniction. From any point P of the given surface, with any opening of the compasses, describe the circumference ABC on the surface. Then the straight line PB is known. Take any three points A, B, and C in this circumference, and with the compasses measure the chord distancea^^, BC^ and CA. Construct the A A'B'C\ with sides equal respectively to AB, BC, and CA, and circumscribe a O about the A A'B'C. The radius B'B' of this is equal to the radius of O ^^C. Construct the rt. A bdp, having the hypotenuse hp = BP, and one side bd= B^U. Draw hfp^ X to h'p, and meeting pd produced in p'. Then jop' is equal to the diameter of the given sphere. Proof. Suppose the diameter PP and the straight line PB drawn. r^^^ ^ ^^p ^^^ ^^^ ^^^ ^^^^1^ g ^g^ Hence the A PBP and pip' are equal. § 149 Therefore />/ = PP'. And \pp^ is equal to the radius. q. e.f. 346 SOLID GEOMETRY. — BOOK VIII. Proposition VI. Theorem. 708. A plane perpendicular to a radius at its ex- tremity is tangent to the sphere. Let be the centre of a sphere, and MN a plane perpendicular to the radius OP, at its extremity P To prove MN tangent to the sphere. Proof. From draw any other straight line OA to the plane MN. OF equally distant from the faces ^jSCand ADC. Therefore 0, the common intersection of these three planes, is equally distant from the four faces of the tetrahedron. ^=**' Hence a sphere described with as a centre, and with the radius equal to the distance from to any face, will be tangent to each face, and will be inscribed in the tetrahedron. § 713 a&o. 716. Cor. The six planes which bisect the six dihedral angles of a tetrahedron intersect in the same point. 348 SOLID GEOMETRY. — BOOK VIII. Proposition VIII. Theorem. 717. A sphere may he circumscribed about any given tetrahedron. Let ABCD be the given tetrahedron. To prove that a sphere may he circumscrihed about ABCD. Proof. Let M^ N, respectively be the centres of the circles circumscribed about the faces ABC, ACD. Let also MR be ± to face ABC, NS 1. to face ACD. MR is the locus of points equidistant from A, B, C, and N8 is the locus of points equidistant from A, C, D.^ 480 Also MR and JVS lie in the same plane. For, if a plane -L to AC he passed through its 'middle point, this plane will contain all points equidistant from A and O. § 482 .*. MR and I^S must lie in this plane. Also MR and iV/S', being X to planes which are not II, can- not be II, and must therefore meet at some point 0. .'. O is equidistant from A, B, C, and D, and a spherical surface whose centre is 0, and radius OA. will pass through the points A, B, C, and D. a e. o 718. Cor. 1. The four perpendiculars erected at the centres of the faces of a tetrahedron meet at the same point. 719. Cor. 2. The six planes perpendicular to the edges of a tetrahedron at their middle points intersect at the same point. THE SPHERE. 349 Proposition IX. Theorem. 720. The intersection of two spherical surfaces is the circumference of a circle whose plane is perpen- dicular to the line joining the centres of the surfaces and whose centre is in that line. Let 0, 0' be the centres at the spherical surfaces, and let a plane passing through 0, 0' cut the sphere in great circles whose circumferences intersect each other in the points A and B. To prove that the spherical surfaces intersect in the circ7im- ference of a circle whose plane is perpendicular to 00', and whose centrt is the point where AB meets OO. Proof. The common chord AB is X to 00' and bisected at (7, § 249 {when two circumferences intersect each other, the line joining their centres is J- to the common chord at its middle point). If the plane of the two great circles revolve about 00', their circumferences will generate the two spherical surfaces, and the point A will describe the line of intersection of the surfaces. But during the revolution ^C will remain constant in length and ± to 00'. Therefore the line of intersection described by the point A will be the circumference of a circle whose centre is C and whose plane is ± to 00', § 473 Q.E.D. 350 SOLID GEOMETRY. BOOK VIII. Figures on the Surface of a Sphere. 721. The angle of two curves passipg through the same point is the angle formed by the two straight lines tangent to the curves at that point. If the two curves are arcs of great cir- cles, th^ angle is called a s'pherical angle. Proposition X. Theorem. (723/ A spherical angle is measured hy the arc of a great circle described from its verteoc as a pole and included between its sides (produced if necessary). Let ABy AC be arcs of great circles intersecting at A; AB' and AC, the tangents to these arcs at A; EC an arc of a great circle described from A as a pole and included between AB and AC. To prove that the spherical A BACis measured hy arc BC. Proof. Draw the radii OA, OB, 00. In the plane AOB, AB' is ± to AO, and §240 §100 OB is ± to AO. .: AB' is II to OB. Ada II to 00. .'.Z B'AC'=A BOO. Z BOO is measured by arc BO. .'. Z B'AO' is measured by arc BO. .'. Z BAO is measured by arc ^(7. q. e. o. 723. Cor. A spherical angle has the same measure as the dihedral angle formed hy the planes of the two circles. Similarly, But §498 §262 THE SPHERE. 351 Proposition XL Problem. 724. To describe an arc of a great circle through a given point perpendicular to a given arc of a great circle. Let A be a point on the surface of a sphere, CHD an arc of a great circle, P its pole. To describe an arc of a great circle through A perpendicular to CHD. Oonstniction. From ^ as a pole describe an arc of a great circle cutting CHD at E. From ^ as a pole describe the arc AB through A. Then ^^ is the arc required. Proof. The arc ^^ is the arc of a great circle, and E is its pole by construction. § 703 The point E is at the distance of a quadrant from P. § 703 Therefore the arc AB produced will pass through P. And since the spherical Z PBE is measured by an arc of a great circle extending from P to E, § 722 the Z ABD is a right angle. Therefore the arc AB is J- to the arc CHD. q e d Ex. 575. Every point in a great circle which bisects a given arc of a great circle at right angles, is equidistant from the extremities of the given arc. 352 SOLID GEOMETRY. — BOOK VIII. "725. A spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles. The bounding arcs are the ddes of the polygon ; the angles which they form are the angles of the polygon ; their points^ of intersection are the vertices of the polygon. The values of the sides of a spherical polygon are usually expressed in degrees, minutes, and seconds. 726. The planes of the sides of a spherical polygon form a polyhedral angle whose vertex is the centre of the sphere, whose face angles are measured by the sides of the polygon, and whose dihedral angles have the same numerical measure as the angles of the polygon. Thus, the planes of the sides of the polygon ABCD form the polyhedral angle 0-ABCD. The face angles AOB, BOC, etc., are measured by the sides AB, BC, etc., of the polygon. The dihedral angle whose edge is OA has the same measure as the spherical angle BAD, etc. Hence, /rom any property of polyhedral angles we may infer an analogous property of spherical poly- gons; and conversely. 727. A spherical polygon is convex if the corresponding polyhedral angle is convex (§ 534). Every spherical polygon is to be assumed convex unless otherwise stated. 728. A diagonal of a spherical polygon is an arc of a great circle connecting any two vertices which are not adjacent. 729. A spherical triangle is a spherical polygon of three sides ; like a plane triangle, it may be right or oblique, equi- lateral, isosceles, or scalene. 730. Two spherical polygons are equal if they can be applied, the one to the other, so as to coincide. THE SPHERE. 353 Proposition XII. Theorem. 731 Each side of a spherical triangle is less than the sum of the other two sides. LetABG be a spherical triangle, AB the largest side. To prove AJB < A0+ BO, Proof. In the corresponding trihedral angle 0-ABC, ^ AOB is less than Z AOC+ Z. BOO. § 539 .•.AB-= 90°. Adding, we have B'E+ C*B = 180°. That is, B'l) +DE-\^C'n = 180°. Or Z)^+5'e' = 180°. 'EutBW^ = a!. Also DE measures /.A, § 722 .-. A-\-a' = 180°. In a similar way all the other relations are proved. q, e. d. 736. Scholium. Two polar triangles are sometimes called supplemental triangles. 356 SOLID GEOMETRY. — BOOK VIII. Proposition XVI. Theorem. 7§7^ The sum of the angles of a spherical triangle is greater than 18(P and less than 540^. a' Let ABC be a spherical triangle, and let A, B, C denote the values of its angles, and a', b', c', respectively, the values of the opposite sides in the polar triangle A'B'CK To prove A-\-B-\- C> 180^ and < 540°. Proof. Since the A ABC, A^B^C\ are polar A, A^a^= 180°, B-i-b'= 180°, C+ c' = 180°. § 735 By addition, A-i- B-\-0+a'+b^ + c' = 540°. .-. ^ + ^+(7-540°-(a'+^»' + c'). Now a' + 5' + c' is less than 360°, § 732 .\A-\-B-{-0= 540° - some number less than 360°. .\A + B+C> 180°. And since a' + Z>' -j- AB. Proof. Draw the arc £D of a great circle, making Z OJBD equal Z ACB. Then I>C= BB, § 755 Now AB-{-DB>AB, § 731 ,\AI) + I)C> AB, or AC > AB. n. Let AC be greater than AB. To prove A ABQ greater than Z. A CB. Proof. The Z ABC must be equal to, less than, or greater than the Z ACB. If /. ABC= Z C, then AC= AB, § 755 and if Z ABCk less than Z C, then A0< AB. Case I. But both of these conclusions are contrary to the hypothesis. /. Z ABOb greater than Z C. q,e.b. 366 SOLID GEOMETRY. BOOK VIII. Measurement of Spherical Surfaces. 757. A zone is a portion of the surface of a sphere included between two parallel planes. The circumferences of the sections made by the planes are called the bases of the zone, and the distance between the planes is its altitude. 758. A zone of one base is a zone one of whose bounding planes is tangent to the sphere. If a circle (Fig. 1) be revolved about a diameter PQ, the arc AD will generate a zone, the points A and D will gen- erate its bases, and (91P is its altitude. The arc I* A will generate a zone of one base. A 759. A lune is a portion of the surface of a sphere bounded by two semi-circumferences of great circles. 760. The angle of a lune is the angle between the semi- circumferences which form its boundaries. Thus (Fig. 2), ABEQA is a lune, BA is its angle. 761. As in Plane Geometry it is convenient to divide a quadrant of a circle into 90 equal parts, called degrees, so in Solid Geometry it is convenient to divide each of the eight equal tri-rectangular triangles of which the surface of a sphere is composed (§ 744) into 90 equal parts, and to call these parts spherical degrees. The surface of every sphere therefore contains 720 spherical degrees. THE SPHERE. 367 Proposition XXVI. Theorem. 762. The area of the surface generated hy a straight line revolving about an axis in its plane is equal to the product of the projection of the line on the axis hy the circumference whose radius is a perpen- dicular erected at the middle point of the line and terminated hy the a^xis. J. 4 M B B j M^^^\ 1 1 \ Y^^^ — ^^ ! i i i,\ I X c 7 DC R D R 1) * Let XY be the axis, AB the revolving line, CD its projection on XY, M its middle point, MO perpendicu- lar to XY, and MB perpendicular to AB. To prove that area AB = CD X ^ttMM. Proof. (1) If AB is 11 to XY, then CD = AB, MR coincidea with MO, area AB is the surface of a right cylinder, and the truth of the theorem follows at once from § 646. (2) If AB is not )l to XY, area AB will be the surface of the frustum of a cone of revolution. .-. area AB = ABx 2irM0, § 676 Draw AE II to XY. The A ABE, MOR are similar. § 327 .-. MO: AE^MR:AB. .'. ABxMO= AExMR. Or, since AE= CD, § 180 ABxMO=CDxMR. Substituting this value of AB X MO in the first equation, we obtain area AB = CD X 2TrMR. (3) If ^ lies in the axis XY, the above reasoning still holds good ; only AE and CD coincide, and the truth follows from §670. aE.a ^ 368 SOLID GEOMETRY. — BOOK VIIL Proposition XXVII. Theorem. ^7§^ The area of the surface of a sphere is equal to the product of its diameter by the circumference of a great circle. Let the sphere be generated by the semicircle ABODE revolving: about the diameter AE, and let be the centre, and B the radius. To prove that the area of the surface = AU X 27ri?. Proof. Inscribe in the semicircle half of a regular polygon having an even number of sides, as ABODE. From the centre draw Js to the chords AB, BC, etc. These Js bisect the chords (§ 232) and are equal (§ 236). Let a denote the length of each of these ^, From B and D drop the Js ^i^and DO to AE. When the semicircle revolves about AE, the sides of the polygon generate surfaces whose areas are as follows : area AB = AFx 2'jra. § 762 area BC = FO X 27ra. area CD = OG X 27ra. area DE = GE X^ira. Adding, area ABCDE= AEx 2ira. Now suppose the number of sides of the semi-polygon to be indefinitely increased ; then the limit of the area ABODE is the area of the surface of the sphere, and the limit of a is B. Hence the area of the surface of the sphere = AEx ^ttR. ? 260 aE.D. THE SPHERE. 369 764. Cor. 1. If /S' denotes the area of the surface of a sphere, then by § 763, But ttH^ is the area of a great circle ; therefore, The surface of a sphere is equivaleni to four greed circles. ^765. Cor. 2. Let R and R^ denote the radii, D and ly the diameters, and xS'and 8' the areas of the surfaces of two spheres; then, by § 764, 8=^'jrR', 8' = 4:7rR'\ . S _ 4:7rR' ^R' _^ (il)y ^D' "8' 47ri2" R" (iJDy D"' Therefore, the areas of the surfaces of two spheres are as the squares of their radii, or as the squares of their diameters. 766. Cor. 3. If we apply the reasoning of § 763 to the zone generated by the revolution of the arc BCD, we obtain for the result, area of zone BCD = FG X 27rR. Now FO is the altitude of the zone ; therefore, ■ The area of a zone is equal to the product of its altitude hy the drcumf&rence of a great circle. 767. Cor. 4. Zones on the same sphere, or equal spheres, are to each oiher as their altitudes. 768. Cor. 5. The arc AB generates a zone of one base; and zone AB = AFx2TTR=^irAFxAE. Now since AFx AF= AB" (§ 337), the zone AB = 7r AB\ That is, a zone of one hose is equivalent to a circle whose radius is the chord of the generating arc. Ex. 583. Find the area of the surface of a sphere whose radius is 6 inches. -■*' .' 1 y Ex. 584. Find the area of a zone if its altitude is 3 inches, and the radius of the sphere is 6 inches. . 370 V/ SOLID GEOMETRY. — BOOK VIII. Proposition XXVIII. Theorem. 769, The area of a lune is to the area of the sur- face of the sphere as the number of degrees in its angle is to S60, Let ABEC be a Inne, BCDF the great circle whose pole is A; also let A denote the number of degrees in the angle of the lune, L the area of the lune, and S the area of the surface of the sphere. To prove that Z: S= A: 360. Proof. The arc ^(7 measures the Z A of the lune. § 722 Hence, arc £0: circumference £CDF= A : 360. (1) If BO and BCDF are commensurable, let their com- mon measure be contained tyi times in ^(7, and n times in BCDF. Then arc BQ\ circumference BCDF— m : n. .'. A:S60 =m:n. § 262 Pass arcs of great (D through the diameter AF and all the points of the division of BCDF. These arc^ "will divide the entire surface into n equal lunes, of which th^. lune ABEC will contain tti. .*. L : 8='m : n. ,\L:8=A:Zm. (2) If BC and BCDF are incommensurable^ the theorem can be proved by the method of limits as in § 261. a e. a THE SPHERE. 371 770. Cor. 1. If L and S are expressed as spherical degrees (§ 761), then since S contains 720 spherical degrees, Z:720 = J[:360^ Whence L = 2A. That is, The numerical value of a lune expressed in spherical degrees is twice the numerical value of its angle expressed in angle-degrees. 771. Cor. 2. If L and 8 are expressed in ordinary units of area (as square inches, etc.), then, since 8~ ^irB^^ L-.^-nR^^A :360°. Whence ^ = ''^- 90° 772. Cor. 3. If we compare two lunes on the same sphere, or equal spheres, R is constant ; hence, if Z, V denote the lunes, A, A' their angles, L-n-'^EA- E^^'-A -A' ^''^~ 90° • 90^~^-^- That is, Txoo lunes on the same sphere, or equal sphet^es, have the same ratio as their angles. TIZ. Cor. 4. If we compare two lunes X, L\ which have the same Z A, but are situated on unequal spheres whose radii are R and R\ then L : L' --^llEA : I^IA = E' : R'\ 90° 90° Two lunes on unequal spheres which have the same angle may be called similar lunes. Therefore, * Si'inilar lunes have the same ratio as the squares of the radii of the spheres on which they are situated. v Ex. 585. Given the radius of a sphere 10 inches ; find the area of a lune whose angle is 30°. . .- ... \ _, Ex. 586. Given the diameter of a sphere 16 inches ; find the area of a lune whose angle is 75°. kL SOLID GEOMETEY. — BOOK VIII. \ Proposition XXIX. Theorem. 774. The area of a spherical triangle, expressed Mn spherical degrees, is numerically equal to the spherical excess of th& triangle. .0 c Let A, B, C denote the values of the angles of the spherical triangle ABC, and E the spherical excess. To prove that the number of spherical degrees in A ABC— E. Proof. Produce the sides oi /\ ABCio complete circles. These circles divide the surface of the sphere into eight spherical triangles, of which any four having a common ver- tex, as A, form the surface of a hemisphere, and therefore con- tain 360 spherical degrees. Now A /li?e+ A y1'/?(7=olune ABA'C. And the A A^BC, AB'C are symmetrical. .-.AA'BC-AAB'C. ^ §748 .-. A ABC+ A AB'C =0= lune ABA'C. Also A ABC-\- A AB'O =0= lune BAB'C. - ,And A^^(7+A^i?e' =o=lune e^C''^. ' ,^ Add and observe that in spherical degrees ^ A ABO-}- AB'C + AB'C+ ABC'=Sm, and lunes ABA' C-{- BAB'C-}- CAC'B are numerically equal to 2 (^ + ^ + C), and we have § 770 2 A A B0-\- Se>0 ^ 2(A-}- B -I- C). Whence A A BC^-^ A -\- B + C- 180 - E. q. e. o THE SPHERE. 373 776. Cor. 1. Since in Bpherical degrees A ABC= E, and the entire surface of the sphere = 720, therefore, A ABC : entire surface =- E : 720. That is, The area of a spherical triangle is to the area of the surface of the sphere as the number which expresses its spherical excess is to 720. 776. Cor. 2. Hence we may easily express the value of A ABC in ordinary units of area (as square inches, etc.). For, let S denote the area of the surface of the sphere. Then A ABC: 8^ E : 720. .'.AABC=4^^ 720 Bnt 8= i'jrB' (^ 764). AttB'E it re '.AABC=^ 720 180 Ex. 587. What part of the surface of a sphere is a triangle whose angles are 120°, 100°, and 95° ? "What is its area in square inches, if the radius of the sphere is 6 inches ? Ex. 588. Find the area of a spherical triangle whose angles are 100°, 120°, 140°, if the diameter of the sphere is 16 inches. Ex. 589. If the radii of two spheres are 6 inches and 4 inches respec- tively, and the distance between their centres is 5 inches, what is the area of the circle of intersection of these spheres ? Ex. 590. Find the radius of the circle determined in a sphere of 6 inches diameter by a plane 1 inch from the centre. Ex, 591. If the radii of two concentric spheres are R and R\ and if a plane is drawn tangent to the interior sphere, what is the area of the section made in the other sphere ? Ex. 592. W^koints A and B are 8 inches apart. Find the locus in space of a point^l'nnches from A and 7 inches from B. Ex. 593. The radii of two parallel sections of the same sphere are a and h respectively, and the distance between these sections is d; find the radius of the sphere. 374 SOLID GEOMETRY. — BOOK VIII. Proposition XXX. Theorem. 777. If T denotes the sum of the angles of a spherieal polygon of n sides, the area of the poly- gon expressed in spherical degrees is numerically equal to T-(n-2)18(P, Let ABODE be a polygon of n sides. To prove that the area of ABODE is numerically equal to 2'_(^_2)180°. Proof, Divide the polygon into spherical triangles by draw- ing diagonals from any vertex, as A. These diagonals will divide the polygon into n — 2 spherical triangles, and the area of each triangle in spherical degrees is numerically equal to the sum of its angles minus 180°. § 774 Hence the sum of the areas of all the w — 2 triangles is nu- merically equal to the sum of all their angles minus (n — 2) 180°. Now the sum of the areas of the triangles is the area of the polygon, and the sum of their angles is the sum of the angles of the polygon, that is, T, Therefore the area of the polygon is numerically equal to r- (71-2) 180°. Ex. 594. Find the area of a spherical quadrangle whose angles are 170°, 139°, 126°, and 141°, if the radius of the sphere is 10 inches. Ex. 595. Find the area of a spherical pentagon whose angles are 122° 128°, 131°, 160°, 161°, if the surface of the sphere is 150 square feet. Ex. 596. Find the area of a spherical hexagon whose angles are 9G°, 110°, 128°, 136°, 140°, 150°, if the circumference of a great cir'^le of tho sphere is 10 inches. THE SPHERE. 375 The Volume of a Sphere. 778. A spherical pyramid is the portion of a sphere bounded by a spherical polygon and the planes of its sides. -^^f^^^^^^^Q The centre of the sphere is the vertex of the pyramid. The spherical polygon is its base. Thus, 0-ABCD is a spherical pyramid. 779. A spherical sector is the portion of a sphere generated by the revolution of a circular sector about any diameter of the circle of which the sector is a part. The base of a spherical sector is the zone generated by the arc of the circular sector. Thus, the circular sector AOB revolving about the line JfiV generates a spherical sec- tor whose base is the zone generated by the arc AB\ the other bounding surfaces are the conical surfaces gen- erated by the radii OA and OB. The sector generated by AOMi?, bounded by a conical surface and a zone of one base If 00 is perpendicular to OM, the sector generated by AOC is bounded by a conical surface, a plane surface, and a zone. 780. A spherical segment is a portion of a sphere contained between two parallel planes. 781. The bases of a spherical segment are the sections made by the parallel planes, and the altitude of a spherical segment is the distance between its bases, 782. If one of the parallel planes is tangent to the sphere, the segment is called a segment of one bo^e. 783. A spherical wedge is a portion of a sphere bounded by a lune and two great semicircles. 376 SOLID GEOMETEY. BOOK VIII. Proposition XXXI. Theorem. 784. The volume of a sphere is equal to the product of the area of its surface hy one-third of its radius. Let R be the radius of a sphere whose centre is 0, >S its surface, and V its volume. To prove V=SxiB. Proof. Conceive a cube to be circumscribed about the sphere. Its volume will be greater than that of the sphere, because it contains the sphere. From O, the centre of the sphere, conceive lines to be drawn to the vertices of the cube. These lines are the edges of six quadrangular pyramids, whose bases are the faces of the cub^e, and whose common alti- tude is the radius of the sphere. The volume of each pyramid is equal to the product of its base by J its altitude. Hence the volume of the six pyramids, that is, the volume of the circumscribed cube, is equal to the area of the surface of the cube multiplied by J i?. Now conceive planes drawn tangent to the sphere, at the points where the edges of the pyramids cut its surface. We shall then have a circumscribed solid whose volume will be nearer that of the sphere than is the volume of the circum- scribed cube, because each tangent plane cuts away a portion of the cube. THE SPHERE. 377 From cjiiceive lines to be drawn to each of the polyhe- dral angles of the solid thus formed, a, h, c, etc. These lines will form the edges of a series of pyramids, whose bases are the surface of the solid, and whose common altitude is the radius of the sphere ; and the volume of each pyramid thus formed is equal to the product of its base by \ its altitude. Hence the sum of the volumes of these pyramids, that is, the volume of this new solid, is again equal to the area of its surface multiplied by -^ i?. Now this process of drawing tangent planes may be consid- ered as continued indefinitely, and, however far this process is carried, the volume of the solid will always be equal to the area of its surface multiplied by \ R. But the volume of the circumscribed solid will approach nearer and nearer to that of the sphere ; and as the volumes approach coincidence, the surfaces also approach coincidence. Hence, Y and 8 are the limits of the volume and the sur- face respectively, of the circumscribed solid. .'. V=By.\R. §260 Q. E. D. 785. Cor. 1. Since B-^^-kR (§ 764), and R-=\I), we obtain by substitution the formulas V=\itR\ and V=\itD\ 786. Cor. 2. The volumes of iivo spheres are to each other as the cubes of their radii. For, if i?, R' denote the radii, T^and V the volumes, V= i-n-R', and V = i7rR'\ .', V : V = iirR' : ^ttR" = R' : R'\ 787. Cor. 3. The volume of a sphetncal pyramid is equal to the product of its base by one-third of the radius of the sphere. For, it is obvious that the reasoning employed iu § 784 applies equally well to a spherical pyramid. 378 SOLID GEOMETRY. — BOOK VIII. 788. Cor. 4. The voluine of a sphericojl sector is equal to the product of the zone which forms its base by one-third of the radius of the sphere. 789. Cor. 5. If R denotes the radius of a sphere, C the cir- cumference of a great circle, -S'the altitude of the zone, Zthe surface of the zone, and Y the volume of the corresponding sector ; then, since C= 27ri?, and Z= 2'irIiII, we have Proposition XXXII. Problem. 790. To find the volume of a spherical segment. D B M Let AC and BD be two semi-chords perpendicular to the diameter MN of the semicircle NGDM. Let OM=R, AM=a, BM^b, AD=a-b = h, AC=-r, BD^r'. Case I. To find the volume of the segment of one base gen- erated by the circular semi-segment ACM, as the semicircle revolves about NM as an axis. The sector generated by OCM= ^^rRa. § 789 The cone generated by OCA = \Trr''{R - a). § 672 Hence segment ACM= ^irE^a — ^Trr^ (B — a) = '^(2I^a-Br' + ar'). Now 7^ = a(2Ii — a) (§ 337) ; therefore by substitution, the segment ACM=^ ira' (R-f)- (1) THE SPHERE. 379 If from the relation r' = a (2i? — a) we find the value of R, and substitute it in (1), we obtain the volume in terms of the altitude and the radius of the base. The segment ACM= iirr'a + ^ 7^a^ (2) Case II. To find the volume of the segment of two bases gen- erated hy the circular semi-segTuent AjBDC, as the semicircle revolves about NM as an axis. Since the volume is obviously the difference of the vol- umes of the segments of one base generated by the circular semi-segments ^C'Jf and BDM, therefore by formula (1), segment ABDC=^ ira^ (r - -") - irh'' (r - -^ = irRh{a-^b)-'^{a^^ab-\-b') , o = irh [{Ra -\-Rb)~:^ (a' -{-ab-j- b')]. Since a-b = h, a'' -2ab + b^ = h^ ; therefore a'' + ab -{- b^ = h"" + S ab -, also since (2R-a)a = r\ and (2R~b)b = r", Ra-i-Rb=^±tj:l + ^±^. 2 2 Hence th^Begment ABBC^ ^h {"^^^ + "' + '>' ~^-ab] L 2 2 3 J J r' + r" , A' , , h' ,'\ 380 SOLID GEOMETRY. — BOOH: VIII. Numerical Exercises. ^ 597. Find the surface of a sphere if the diameter is (i.) 10 inches ; (ii.) 1 foot 9 inchefc (iii.) 2 feet 4 inches ; (iv.) 7 feet ; (v.) 4.2 feet ; (vi.) 10.5 feet. ^ 698. Find the diameter oi u. sphere if the surface is (i.) 616 squari. inches ; (ii.) 38^ square feet ; (iii.) 9856 square feet. V 599. The circumference of a dome in the shape of a hemisphere is 6Q feet ; how maijy square feet of lead are required to cover it ? ^ 600. If the ball on the top of St. Paul's Cathedral in London is 6 feet in diameter, what would it cost to gild it at 7 cents per square inch ? "^ 601. What is the numerical value of the radius of a sphere if its sur- face has the same numerical value as the circumference of a great circle ? ^602. Find the surface of a lune if its angle is 30°, and the total sur- face of the sphere is 4 square feet. "^ 603. What fractional part of the whole surface of a sphere is a spher ical triangle whose angles are 43° 27^ 81° 57^ and 114° 36^? 604. The angles of a spherical triangle are 60°, 70°, and 80°. The radius of the sphere is 14 feet. Find the area of the triangle in square feet 605. The sides of a spherical triangle are 38°, 74°. and 128°. The radius of the sphere is 14 feet. Find the area of the polar triangle in square feet. 606. Find the area of a spherical polygon on a sphere whose radius is 10^ feet, if its angles are 100°, 120°, 140°, and 160°. 607. The planes of the faces of a quadrangular spherical pyramid make with each other angles of 80°, 100°, 120°, and 150° ; and the length of a lateral edge of the pyramid is 42 feet. Find the area of its base in square feet. 608. The planes of the faces of a triangular spherical pyramid make with each other angles of 40°, 60°, and 100°, and the area of the base of the pyramid is 4ir square feet. Find the radius of the sphere. 609. The diameter of a sphere i;; 21 feet. Find the curved surface of a segment whose height is 5 feet. -610. What is the area of a zone of one base whose height is A, and the radius of the base r \ What would be the area if the height were twice as great ? NUMEHicAL p:xeiicises. 381 611. In a sphere whose radius is r, find the height of a zone whose area is equal to that of a great circle. "" 612. The altitude of the torrid zone is about 3200 miles. Find its area in square miles, assuming the earth to be a sphere with a radius of 4000 miles. "^613. A plane divides the surface of a sphere of radius r into two zones, such that the surface of the greater is a mean proportional between the entire surface and the surface of the smaller. Find the distance of the plane from the centre of the sphere. . 614. If a sphere of radius r is cut by two planes equally distant from the centre, po that the area of the zone comprised between the planes is equal to the sum of the areas of its bases, find the distance of either plane from the centre. 615. Find the area of the zone generated by an arc of 30°, of which the radius is r, and which turns around a diameter passing through one of its extremities. "^ 616. Find the area of the zone of a sphere of radius r, illuminated by a lamp placed at the distance h from the sphere. *^ 617. How much of the earth's surface would a man see if he were raised to the height of the radius above it? ^^618. To what height must a man be raised above the earth in order that he may see one-sixth of its surface ? ^19. Two cities are 200 miles apart. To what height must a man ascend from one city in order that he may see the other, supposing the circumference of the earth to be 25,000 miles ? ^^ 620. Find the volume of a sphere if the diameter is (i.) 13 inches ; (ii.) 3 feet 6 inches ; (iii.) 10 feet 6 inches ; (iv.) 17 feet 6 inches ; (v.) 14,7 feet ; (vi.) 42 feet. V 621. Find the diameter of a sphere if the volume is (i.) 75 cubic feet 1377 cubic inches; (ii.) 179 cubic feet 1152 cubic inches ; (iii.) 1047.816 cubic feet ; (iv.) 38.808 cubic yards. "" 622. Find the volume of a sphere whose circumference is 45 feet. "^ 623. Find the volume F of a sphere in terms of the circumference of a great circle. '^ 624. Find the radius r of a sphere, having given the volume F. ^625. Find the radius r of a sphere, if its circumference and its volume have the same numerical value. 382 SOLID GEOMETRY. — BOOK VIII. ^^26. If an iron ball 4 inches in diameter weighs 9 pounds, what is the weight of a hollow iron shell 2 inches thick, whose external diameter is 20 inches ? ^ 627. The radius of a sphere is 7 feet ; what is the volume of a wedge whose angle is 36° ? ^ 628. What is the angle of a spherical wedge, if its volume is one cubic foot, and the volume of the entire sphere is 6 cubic feet ? ^ 629. What is the volume of a spherical sector, if the area of the zone which forms its base is 3 square feet, and the radius of the sphere is 1 foot? ^ 630. The radius of the base of the segment of a sphere is 16 inches, and the radius of the sphere is 20 inches ; find its volume. ^631. The inside of a wash-basin is in the shape of the segment of a sphere ; the distance across the top is 16 inches, and its greatest depth is 6 inches ; find how many pints of water it will hold, reckoning 7j gal- lons to the cubic foot. 632. What is the height of a zone, if its area is S, and the volume of the sphere to which it belongs is F? ^ 633. The radii of the bases of a spherical segment are 6 feet and 8 feet, and its height is 3 feet ; find its volume. 634. Find the volume of a triangular spherical pyramid if the angles of the spherical triangle which forms its base are each 100°, and the radius of the sphere is 7 feet. 635. The circumference of a sphere is 28 tt feet: find the volume of that part of the sphere included by the faces of a trihedral angle at the centre, the dihedral angles of which are 80°, 105°, and 140°. 636. The planes of the faces of a quadrangular spherical pyramid make with each other angles of 80°, 100°, 120°, and 150°, and a lateral edge of the pyramid is 3 J feet ; find the volume of the pyramid. '^ 637. The radius of the base of the segment of a sphere is 40 feet, and its height is 20 feet ; find its volume, 638. Having given the volume V, and the height h, of a spherical segment of one base, find th% radius r of the sphere. ^ 639. Find the weight of a sphere of radius r, which floats in a liquid of specific gravity s, with one-fourth of its surface above the surface of the liquid. The weight of a floating body is equal to the weight of the liquid displaced. , MISCELLANEOUS EXERCISES. 383 Miscellaneous Exercises. '^ 640. Determine a point in a given plane such that the difference of its distances from two given points on opposite sides of the plane shall be a maximum. 641. In any warped quadrilateral, that is, one whose sides do not all lie in the same plane, the middle points of the sides are the vertices of a parallelogram. " 642. In any trihedral angle, the three planes bisecting the three dihe- dral angles intersect in the same straight line. ^ 643. To draw a line through the vertex of any trihedral angle, mak- ing equal angles with its edges. ^ 644. In any trihedral angle, the three planes passed through the edges and the respective bisectors of the opposite face angles intersect in the same straight line. 645. In any trihedral angle, the three planes passed through the bisectors of the face angles, and perpendicular to these faces respec- tively, intersect in the same straight line. " 646. In any trihedral angle, the three planes passed through the edges, perpendicular to the opposite faces respectively, intersect in the same straight line. 647. In a tetrahedron, the planes passed through the three lateral edges and the middle points of the sides of the base intersect in a straight line. "^48. The lines joining each vertex of a tetrahedron with the point of intersection of the medial lines of the opposite face all meet in a point called the centre of gravity, which divides each line so that the shorter segment is to the whole line in the ratio 1 : 4. 649. The straight lines joining the middle points of the opposite edges of a tetrahedron all pass through the centre of gravity of the tetrahedron, and are bisected by the centre of gravity. "" 650. The plane which bisects a dihedral angle of a tetrahedron divides the opposite edges into segments proportional to the areas of the faces including the dihedral angle. ^51. The altitude of a regular tetrahedron is equal to the sum of the four perpendiculars let fall from any point within it upon the four faces. WD 884 SOLID GEOMETRY. — BOOB! VIII. 652. Within a given tetrahedron, to find a point such that the planes passed through this point and the edges of the tetrahedron shall divide the tetrahedron into four equivalent tetrahedrons. 653. To cut a cube by a plane so that the section shall be a regular .aexagon. " 654. To cut a tetrahedral angle so that the section shall be a parallelogram. 655. The portion of a tetrahedron cut off by a plane parallel to any fac« is a tetrahedron similar to the given tetrahedron. 656. Two tetrahedrons, having a dihedral angle of one equal to a dihedral angle of the other, and the faces including these angles respec- tively similar, and similarly placed, are similar. 657. Two polyhedrons composed of the same number of tetrahedrons, similar each to each and similarly placed, are similar. 658. If the homologous faces of two similar pyramids are respec- tively parallel, the straight lines which join the homologous vertices of the pyramids meet in a point. -^659. Two symmetrical tetrahedrons are equivalent. 660. Two symmetrical polyhedrons may be decomposed into the same number of tetrahedrons symmetrical each to each. 661. Two symmetrical polyhedrons are equivalent. 662. If a solid has two planes of symmetry perpendicular to each other, the intersection of these planes is an axis of symmetry of the solid. 663. If a solid has three planes of symmetry perpendicular to each other, the three intersections of these planes are three axes of symmetry of the solid ; and the common intersection of these axes is the centre of symmetry of the solid. 664. The volume of a right circular cylinder is equal to the product of the lateral area by half the radius. 665. The volume of a right circular cylinder is equal to the product of the area of the rectangle which generates it, by the length of the cir- cumference generated by the point of intersection of the diagonals of the rectangle. 666. If the altitude of a right circular cylinder is equal to the diam- eter of the base, the volume is eqnal to the total area multiplied by a third of the radius. MISCELLANEOUS EXERCISES. 386 Construct a spherical surface with given radius : 667. Passing through three given points. 668. Passing through two given points and tangent to a given plane. 669. Passing through two given points and tangent to a given sphere. 670. Passing through a given point and ^ngent to two given planes. > 671. Passing through a given point and tangent to two given spheres. 672. Passing through a given point and tangent to a given plane and a given sphere. 673. Tangent to three given planes. 674. Tangent to three given spheres. 675. Tangent to two given planes and a given sphere. 676. Tangent to two given spheres and a given plane. ' 677. Find the area of a solid generated by an equilateral triangle turning about one of its sides, if the length of the side is a. 678. Find the centre of a sphere whose surface shall pass through three given points, and shall touch a given plane. 679. Find the centre of a sphere whose surface shall touch two given planes, and also pass through two given points which lie between the planes. 680. Through a given point to pa.ss a plane tangent to a given cir- cular cylinder. 681. Through a given point to pass a plane tangent to a given cir- cular cone. 682. Through a given straight line without a given sphere, to pass a plane tangent to the sphere. 683. The volume of a sphere is two-thirds of the volume of a circum- scribing cylinder, and its surface is two-thirds of the total surface of the cylinder. 684. Given a sphere, a cylinder circumscribed about the sphere, and a cone of two nappes inscribed in the cylinder ; if any two planes are drawn perpendicular to the axis of the three figures, the spherical seg- ment between the planes is equivalent to the difference between the corresponding cylindrical and conic segments. 386 SOLID GEOMETRY. — BOOK VIII. 685. Compare the volumes of the solids generated by a rectangle turning successively about two adjacent sides, the lengths of these sides being a and h. 686. An equilateral triangle revolves about one of its altitudes. Compare the convex surface of the cone generated by the triangle and the surface of the sphere generated by the circle inscribed in the triangle. 687. An equilateral triangle revolves about one of its altitudes. Compare the volumes of the solids generated by the triangle, the inscribed circle, and the circumscribed circle. 688. The perpendicular let fall from the point of intersection of the medial lines of a given triangle upon any plane not cutting the triangle is equal to one-third the sum of the perpendiculars from the vertices of the triangle upon the same plane. 689. The perpendicular from the centre of gravity of a tetrahedron upon any plane not cutting the tetrahedron is equal to one-fourth the sum of the perpendiculars from the vertices of the tetrahedron upon the same plane. 690. The volume of any polyhedron having for its bases any two polygons whose planes are parallel, and for lateral faces trapezoids, is the product of one-sixth the distance between the bases into the sum of the two bases plus four times a section midway between the bases ; that is, if H denotes the distance between the bases B and 6, and B^ a section midway between the bases, V=\H{B-\-h + ^B^\ Note. From any point in the section midway between the bases, draw lines to the vertices of the solid angles of the polyhedron, thus divid- ing the solid into pyramids. The pyramids having B and h as bases, evi- dently equal \H{B-\- h). It remains to be proved that the volume of each pyramid having a lateral face as its base equals \ H into four times that portion of the section midway between the bases intercepted by thie pyramid. This theorem is much used in earth-work. BOOK IX. CONIC SECTIONS. The Parabola. (\J9^ The curve traced by a point which moves so that its distance from a fixed point is always equal to its distance from a fixed line is called a parabola. The curve lies in the plane of the fixed point and line. 792. The fixed point is called the foctis; and the fixed line, the directnx. 793. A parabola may be described by the continuous motion of a point, as lollows : E Place a ruler so that one of its edges shall coincide with the directrix DE. Then place a right triangle with its base edge in contact with the edge of the ruler. Fasten one end of a string, whose length is equal to the other edge BO, to the point B, and the other end to a pin fixed at the focus F, Then slide the triangle BCE along the directrix, keeping the string tightly pressed against the ruler by the point of a pen- cil P. The point P will describe a parabola ; for during the motion we always have PF= PC. 388 GEOMETRY. — BOOK IX. Proposition I. Problem. ,£9p To construct a -parabola by points, having given its focus and its directri.v. Let F be the focus, and CDE the directrix. To construct the parabola by points. Construction. Draw FD ± to CE, and meeting CE at D. Bisect FD at A. Then ^ is a point of the curve. § 791 Through any point M in the line DF, to the right of A, draw a line II to CE. With i^as centre and EM SiS radius, draw arcs cutting this line at the points P and Q. Then P and Q are points of the curve. Proof. Draw FC, QE± to CE. Then FC= EM, and QE= EM, and EM=FF=QF. .-. PC^PF; and QE = QF. Therefore P and Q are in the curve. In this way any number of points may be found ; and a continuous curve drawn through the points thus determined will be the parabola whose focus is Pand directrix CDE. Q. E. F. Cons. §791 THE PARABOLA. 389 795. The point A is called the vertex of the curve. The line DF produced indefinitely in both directions is called the axis. 796. The line FF, joining the focus to any point F on the curve, is called the focal radius of F. 797. The distance AM ia called the abscissa, and the dis- tance FM the ordinate, of the point F. 798. The double ordinate LF, through the focus, is called the latus rectum or parameter. 799. Cor. 1. Since FF= FQ (Cons.), MF= MQ (§ 121) ; hence, the parabola is symmetrical with respect to its axis (§63). 800. Cor. 2. The curve lies entirely on one side of the per- pendicular to the aods ei'ected at the vertex ; namely, on the same side as the focus. For, any point on the other side of this perpendicular is obviously nearer to the directrix than to the focus. 801. Cor. 3. The parabola is not a closed curve. For any point on the axis of the curve to the right of F is evidently nearer to the focus than to the directrix. Hence the parabola QAF cannot cross the axis to the right of F. 802. Cor. 4. The latus rectum is equal to 4:AF. For, draw LG ± to OFF. Then, LF= LG, and ZG = FF. .'.LF=FF=2AF. Similarly, FF= FF=2AF Therefore LF-=4:AF 803. Remark. In the following propositions, the focus will be denoted by F, the vertex by A, and the point where the axis meets the directrix by D. 390 GEOMETEY. — BOOK IX. Proposition II. Theorem. 804. The ordinate of any point of a parabola is a mean proportional between the latus rectum and the abscissa. Let P be any point, AM its abscissa, PM its ordinate. -2 To prove Proof. PM =4:AFxAM. §791 Hence Q. E. D. iW = FP -FM" = Dlt - FM" = {DM- FM) {DM+ FM) = DF{DF+ FM-\- FM) = 2AF{2AF-\-2FM) ~2AF{2AM). FM' = 4:AFxAM. (1) 805. Cor. 1. The (greater the abscissa of a point, the greater the ordinate. For PJIf increases with AM in equation (1), 806. Cor. 2. If P and Q are any two points of the curve, FM' _ 4:AFx AM _AM Q^' 4:AFxan an' Hence, the squares of any two ordinatcs are as the abscissas. THE PARABOLA. 391 Proposition III. Theorem. 807. Every point within the parabola is nearer to the focus than to the directrix; and every point with- out the parabola is farther from the focus than from the directrix. 1. Let Q be a point within the parabola. Draw QC perpendicular to the directrix, cutting the curve at P. Draw QF, PF, To prove QF < QC. Proof. In the A QPF, QF Q'C Proof. In the A Q'FP, Q'F > PF- PQ\ § 137 or QF> PC~PQ. That is, Q'F>Q'a ae-D. 808. Cor. A point is within or without a parabola according as its distance from the focus is less than, or greater than, its distance froon the directrix. 809. A straight line which touches, but does not cut, a parabola, is called a tangent to the parabola. The point where it touches the parabola is called the point of contact. 392 GEOMETRY. BOOK IX. Proposition IV. Theorem. 810. If a line PT is drawn from any point P of the curve, bisecting the angle between FF and the perpendicular from F to the directrix, every point of the line FT, except F, is without the curve, j^ H Let PG be the perpendicular from P to the directrix, the angle FPT equal the angle OPT, and let X he any- other point in PT except P. To prove that X is without the curve. Proof. Draw XE ± to the directrix, and join CX, FX, OF, and let OF meet P^at R. In the isos. A FCF, OR = RF. Ex. 14 Hence CX= FX. § 122 But EX< ex. § 114 Therefore EX < FX. That is, JTis without the curve. § 808 Q. E. D. 811. Cor. 1. FT is the tangent at the point F (§ 809). 812. Cor. 2. FThisects EC, and is perpendicular to EG. 813. Cor. 3. Since the angles EFT and FTP are equal, FT equals EF(I 156). THE PARABOLA. 393 814. Cor. 4. The tangent at A is pe^'pcndicular to the axis. For it bisects the straight angle FAD. 315. Cor. 5. The tangent at A is the locus of the foot of the perpendicular dropped from the focus to any tangent. Since FR = EC, and FA = AD, B, is in AR{^ 811). 816. The line FN drawn through F perpendicular to the tangent PTis called the normal at F. 817. If the ordinate of F meet the axis in M, and the tan- gent and normal at Pmeet the axis in T and JV respectively, then MT'\^ the suhtangent and MN ih.Q subnormal. ^ 818. Cor. 6. The suhtangeni is bisected by the vertex. For, FT= FF, § 813 and FF^DM. §791 Hence FT= DM', also AF= AD. Therefore FT- AF= DM- AD, or TA = AM. V 819. Cor. 7. The subnormal is equal to half the IcUus rectum. For CF = FN, and CF = DM. § ISO Hence PiV- DM, or FM-\- ¥JSr= DF-\- FM. Therefore MN= DF. 820. Cor. 8. The normal bisects the angle between FF and CF produced; that is, bisects the angle FFO. For Z NFT= Z NFK, and Z FFT= Z TFC=- Z OFK. Hence Z NFF= Z NFG. 821. Cor. 9. The circle with F as centre and FF as radius parses through T and N. / 394 GEOMETRY. — BOOK IX. Proposition V. Problem. 822. To draw a tangent to a parabola from an exterior point. Let B be any point exterior to the parabola, QAP. To draw a tang e7it from M to QAP. Oonstmction. With R as centre and RF as radius, draw arcs cutting the directrix at the points B, C. Through ^ and Cdraw lines parallel to the axis, and meeting the parabola in P, Q, respectively. Join RP, RQ. Then RP and RQ are tangents to the curve. Proof. RB = RF, Cons. PB = PF. § 791 Hence Z RPB = Z. RPF. § 160 Therefore RP is the tangent at P. § 811 For like reason, RQ is the tangent at Q. atF. 823. Cor. Since R is without the curve, it is nearer to the directrix than to the focus (§ 807); therefore, the circle with R as centre and RF as radius, must always cut the directrix in two points ; therefore, two tangents can always he drawn to a parabola from an exterior point. 824. The line joining the points of contact P and Q is called the chord of contact for the tangents drawn from R. THE PARABOLA. 395 Proposition VI. Theorem. 825. The line joining the foeus to the intersection of two tangents rnakes equal angles with the focal radii drawn to the points of contact. Let the tangents drawn at P and Q meet in R. To prove Z RFP = Z BFQ. Proof, Draw the Js FjB, QC to the directrix, and join FB, FC, FF. Since FB = FF, and FF=FF, §§ 812, 112 AFFF^AFBF, §160 and ZFFF^ZFBF. Similarly, Z FFQ - Z FCQ. Now, ZFBF=90''+ZFBC, and Z FCQ = 90° i-Z FOB; and since FB = FF,mdFC=FF, therefore FB = FC. Hence ZFBC^ZFCB. §154 Therefore ZFBF=ZFCQ, and ZFFF=ZFFQ. aE.D. 826. Cor. If the chord of contact FQ passes through F, then FFQ is a straight line. Hence FFF + FFQ = 180°, and FFF=FFQ = 90\ Therefore FBF = FCQ = 90°. Therefore, the tangents drawn through the ends of a focal chord meet in the directrix. 396 GEOMETRY. BOOK IX. Proposition VII. Theorem. 827. If a pair of tangents are drawn from a point R to a parabola, the line drawn through R parallel io the axis will hiseet the chord of contact. Let the tangents drawn from li meet the curve in P, Q, and let the line through R parallel to the a:xis meet the directrix in H, the curve in S, and the chord of contact in M. To prove FM= QM. Proof. Drop the Js FR, QC to the directrix, and join RR, RC. RJIis± to RO, §102 RR^RC. . . §823 Hence RJ3'= CH. § 121 Since RR, QC, and i^Jf are II, § 100 therefore RM= QM. § 187 Q. E. D THE PARABOLA. 397 Proposition VIII. Theorem. 828. // a pair of tangents EP, RQ are drawn from a point B to a parabola, and through E a lUie par- allel to the axis is drawn, meeting the curve in 8, the tangent at 8 will he parallel to the chord of con- tact. Let the tangent at S meet the tangents PR, QR in T, V, respectively. To prove TV W to FQ. Proof. Draw TN II to 8M, and let it meet 8F iii N. Then Pi\r= ]^8 § 827 Hence FT= TE. § 188 Similarly, QV= VE. Therefore TV is II to FQ. § 189 aE. D. ^ 829. Cor. 1. If we suppose E to move along E3f towards the curve, then since the point 8 and the direction of the tan- gent TF" remain fixed, the chord FQ will remain parallel to TV, while its middle point M will move along i?Jf towards 8 ; finally, E, M, P, and Q will all coincide at xS'. Hence, the line EM is the bctis of the middle points of all cho7'ds drawn parallel to the tangent at 8. 830. The locus of the middle points of a system of parallel chords in a parabola is called a diameter. The parallel chords are called the ordinates of the diameter. 831. Cor. 2. The diameters of a parabola are parallel to its axis ; and conversely, efoefry straight line parallel to the axis is a diameter ; that is, bisects a system of parallel chords. 832. CoR. 3. Tangents drawn through the ends of an ordi- nate intersect in the diameter corresponding to that ordinate. 833. CoR.4. The point /S' is the middle point of Pif(§ 188); therefore, the portion of a diameter contained between any ordi- nate and the intersection of the tangents drawn through the ends of the ordinate is bisected by the curve. 398 GEOMETRY. BOOK IX. 834. Cor. 5. The point S is also the middle point of the tangent TV; therefore, the part of a tangent parallel to a chord contained between the two tangents drawn through the ends of the chord is bisected by the diameter of the chord at the point of contact. Proposition IX. Theorem. 835. The area of a parabolic segment made hy a chord is two-thirds the area of the triangle formed hy the chord and the tangents drawn through the ends of the chord. X Ti m Q Let PQ be any chord, and let the tangents at P and Q meet in R. To prove segment P8Q = | A PRQ. THE PARABOLA. 399 Proof. Draw the diameter RM, meeting the curve at 8, and at 8 draw a tangent meeting PR in T and QR in V. Join 8P, 8Q. Since FT= TR, and qy=. VR, "^^ § 828 FT is II to PQ, and PQ = 2x VT. § 189 .-. A PQ8= 2 A TFi?. § 370 If now we draw through T, V, the diameters T8', V8", and then draw through 8', 8", the tangents T'8'V', T"8"V", we can prove in the same way that AP88' = 2AT'V'T, and AQ88"=2AT"V"V. If we continue to form new triangles by drawing diameters through the points T', V\ T'\ F", and tangents at the points where these diameters meet the curve, we can prove that each interior triangle formed by joining a point of contact to the extremities of a chord is twice as large as the exterior triangle formed by the tangents through these points. ' And this is true however long the process is continued. Therefore the sum of all the interior triangles is equal to twice the sum of the corresponding exterior triangles. Now if we suppose the process to be continued indefinitely, then the limit of the sum of the interior triangles will be the area contained between the chord PQ and the curve, and the limit of the sum of the exterior triangles will be the area con- tained between the curve and the tangents PR, QR. Hence segment PQ8= twice the area contained by PR, QR, and the curve,* = | A PQR. § 260 aE.D. 836. Cor. If throtcgh P and Q lines are drawn parallel to 8M, meeting the tangent TV produced in the points X and Y, then-the segment PQ8= i O PQ YX. 400 GEOMETRY. — BOOK IX. Proposition X. Theorem. 837. The section of a right circular cone made hy a -plane -parallel to one, and only one, element of the surface is a parabola. Let SB be any element of the cone whose axis is 8Z, and let QAP be the section of the cone made by a plane perpendicular to the plane BSZ and parallel to SB. To prove that the curve PAQ is a parabola. Proof. Let SO he the second element in whicli the plane BSZ cuts the cone, and let BAD be the intersection of the planes ^xS'^ and PAQ. Draw the O tangent to the lines /^, SO, BD, and let G, H, F be the points of contact respectively. Kevolve BSQ and the O OOH about the axis SZ, the plane PAQ remaining fixed. The O will generate a sphere which will touch the cone in the O GKII, and the plane PA Q at the point F. THE PARABOLA. 401 Since 80 is ± to GJE, SO is ± to the plane GKR. § 462 Hence the plane JBSC is ± to the plane GXIT. § 518 Let the plane of the O GKH intersect the plane of the curve PAQ, in the straight line MR ; then will MR be J. to the plane BBQ {% 520), and therefore ± to DR. Take any point P in the curve, and draw SP meeting the O G^//in K\ join FP, and draw PM 1. to RM. Pass a plane through P -L to the axis of the cone. Let it cut the cone in the O EPLQ, and the plane of the curve PAQ in the line PNQ. PJSTk ± to the plane PSO (§ 520), and therefore ± to DR. Since PP and PIT are tangents to the sphere O, they are tangents to the circle of the sphere made by a plane passing through the points P, P, K, and are therefore equal. § 246 That is, PF= PK. But PJr= LG. % 666 .-. PP= LG. (1) Now X^ and PJtf are each II to NR ; hence LG is II to PM. § 485 The planes GKH ^x\^\ LPE are parallel. § 491 .-. LG = PM. § 493 From (1) and the last equation, we have PF= PM. That is, any point P on the curve PAQ is equidistant from a fixed point Pand a fixed line RM'\xv its plane. Therefore the curve PAQ is a parabola. Q. E. D 402 GEOMETRY. — BOOK IX. Exercises. 691. Prove that if the abscissa of a point is equal to its ordinate, each is equal to the latus rectum. 692. To draw a tangent and a normal at a given point of a parabola. 693. To draw a tangent to a parabola parallel to a given line. 694. Show that the tangents at the ends of the latus rectum meet ati). 695. Prove that the latus rectum is the shortest focal chord. 696. The tangent at any point meets the directrix and the latus rec- tum produced at points equally distant from the focus. 697. The circle whose diameter is FP touches the tangent at A. 698. The directrix touches the circle having any focal chord for diameter. 699. Given two points and the directrix, to find the focus. 700. The J. i^ (7 bisects TP. (See figure, page 392.) 701. Given the focus and the axis, to describe a parabola which shall touch a given straight line. 702. If PN is any normal, and A PNF is equilateral, then PF is equal to the latus rectum. 703. Given a parabola, to find the directrix, axis, and focus. 704. To find the locus of the centre of a circle which passes through a given point and touches a given straight line. 705. Given the axis, a tangent, and the point of contact, to find the focus and directrix. 706. Given two points and the focus, to find the directrix. THE ELLIPSE. 403 The Ellipse. 838. The locus of a point which moves so that the sum of its distances from two fixed points is constant is called an ellipse. The fixed points are called the foci, and the straight lines which join a point of the curve to the foci are called focal radii. The constant sum is denoted by 2a, and the distance between the foci by 2 c. c The ratio c : a is called the eccentricity, and is denoted by e. Therefore c = ae. 839. Cor. 2 a must he greater than 2 c (§ 137) ; hence e must he less than 1. 840. The curve may be described by the continuous motion of a point, as follows : Fasten the ends of a string, whose length is 2a, at the foci -Fand F'. Trace a curve with the point Pof a pencil pressed against the string so as to keep it stretched. The curve thus traced will be an ellipse whose foci are F and F\ and in which the constant sum of the focal radii is FF-{- FF'. The curve is a closed curve extending around both foci ; if it cuts i^i^' produced in A and A', it is easy to see that AA^ equals the length of the string. c 404 GEOMETRY. — BOOK IX. Proposition XI. Problem. 841. To construct an ellipse hy points, having given the foci and the constant sum 2a, B Let F and F' be the foci, and 2CD = 2a. To construct the ellipse. Oonstruction. Through the foci F, F' draw a straight line ; bisect FF* at 0. Lay off OA' =0A = CD. Then A, A' are two points of the curve. Proof. From the construction, AA'=2a, and AF= A'F'. Therefore AF+ AF' = A'F-j- A'F'= AA'=2a, and A'F-\- A'F'= A'F+ AF = AA'= 2a. To locate other points, mark any point JT between F and F'. Describe an arc with F as centre and ^X as radius ; also another arc with F' as centre and A'JC SiS radius; let these arcs cut in P, Q. Then P, Q are two points of the curve. This follows at once from the construction and § 838. By describing the same arcs with the foci interchanged, two more points P, S may be found. By assuming other points between F and F', and proceed- ing in the same way, any number of points may be found. THE ELLIPSE. 405 The curve passing through all the points is an ellipse hav- ing F^ F^ for foci, and 2 a for the constant sum of focal radii. aE.F. 842. Cor. 1. By describing arcs from the foci with the same radius OA, we obtain two points B, B' of the curve such that they are equidistant from the foci. Therefore the line BB^ is perpendicular to A A* and passes through (§ 123). 843. The point is called the centre. The line A A* is called the major axis; its ends A, A^ are called the vertices of the curve. The line BB' is called the minor axis. The length of the minor axis is denoted by 2h. 844. Cor. 2. The major axis is bisected at 0, and is equal to the constant sum 2a. 845. Cor. 3. The minor axis is also bisected at (§ 123). Therefore OB = OB' = b. 846. Cor. 4. The values of a, b, c are so related that a' = b' + c\ For, in the rt. A BOF, bf' = ob'+oT. 847. Cor. 5. The axis A A' bisects FQ at right angles (§ 123). Hence the ellipse is symmetrical with respect to its major axis. 848. The distance of a point of the curve from the minor axis is called the abscissa of the point, and its distance from the major axis is called the ordinate of the point. The double ordinate through the focus is called the kUus rectwn or parameter. 849. Remark. In the following propositions F and F^ denote foci of the ellipse, its centre, AA' the major axis, and BB^ the minor axis. 406 GEOMETRY. — BOOK IX. Proposition XIL Theorem. 850. An ellipse is symmetrical with respect to its minor axis. A' F' Let P be a point of the curve, PDQ be perpendicular to OB, meeting OB in D, and let DQ equal DP. To prove that Q is also a point of the curve. Proof. Join P and Q to the foci F, F'. Eevolve ODQF ahout OB ; Pwill fall on P' and Q on P. Therefore Q,F= PF\ and Z PQF= Z QPF'. Therefore A PQF =■ A QPF\ and QF' = PF. Hence QF-\- QF' = PF-\- PF. But PF-\- PF = 2a. Therefore QF-{- QF' = 2a. Therefore Q is a point of the curve. a E. D. 851. Every chord passing through the centre of an ellipse is called a diameter. 862. Cor. 1. From §§ 847, 850, it follows that an ellipse consists of four equal quadrantal arcs symmetrically placed about its centre (§§ 209, 64). 853. Cor. 2. Every diameter is bisected at the centre. §106 Hyp. THE ELLIPSE. 407 Proposition XIIL Theorem. 854. Ifd denotes the abscissa of a point of an ellipse, r and r' its focal radii, then r' = a + ed, r=a — ed. Let P be any point of an ellipse, PM perpendicular to AA\ d equal OM, r equal PF, r' equal PF*. To prove r' = a -\- ed, r = a — ed. Proof. From the rt. A i^PJf and F'PM Therefore r''^ - r' = PW - FM\ Or (r' + r) (r' -r) = (F'M-^ F3I) (F'3I- FM). Now Also, Hence From / + r = 2a, and F^M-\- FM= 2c. F'M- FM= OF'-i-OM- FM=20M= 2d a (r' — r) = 2 cd. t 2cd ci ■, r — r = = 2ed. r' -f r = 2 a, and 2ed, 2r'=2(a + ec?), and 2r = 2{a~ed). Therefore r' = a -f ec?, and r = a — ed. 408 GEOMETRY. — BOOK IX, 855. The circle described upon the major axis of an ellipse as a diameter is called the auxiliary circle. The points where a line perpendicular to the major axis meets the ellipse and its auxiliary circle are called corresponding points. Proposition XIV. Theorem. 856. The ordinates of two corresponding points in an ellipse and its auxiliary circle are in the ratio h:a. Let P be a point of the ellipse, Q the corresponding point of the auxiliary circle, and QP meet AA' at M, To prove Proof. Let then PM: QM^h'.a. OM^d) ' d\ QM - a^ PM' = PF'- FM" = {a- edy-{c - dj § 854 = a' — '2.aed-\-e^d'' — c^ -\- 2cd — d\ Or, since c = ae and c^ — & = J', § 846 PM'^h^-{\-e')d^^^(a^-d-'\ a^ Therefore PW : QW = b' : a\ Or PM .QM ^b :a, aE. D. THE ELLIPSE. 409 Proposition XV. Problem. 857. To construct an ellipse hy points, having given its two axes. A' O M A Let OA, OB be the given semi- axes, the centre. Oonstniction. With as centre, and OA, OB, respectively, as radii, describe circles. From draw any straight line meeting the larger circle at Q and the smaller circle at R, Through Q draw a line II to OB, and through B draw a line II to OA. Let these lines meet at P. Then will P be a point of the required ellipse. Proof. If QP meet A A* at M, PM: QM= OB : OQ. § 309 But OB = b and OQ = a. Therefore PM: QM= h : a. Therefore P is a point of the ellipse. (§ 856) By drawing other lines through 0, any number of points on the ellipse may be found ; a smooth curve drawn through all the points will be the ellipse required. o. e. f. 410 GEOMETRY. BOOK IX. Proposition XVI. Theorem. 858. The square of the ordinate of a point in an ellipse is to the product of the segments of the major axis made hy the ordinate as IP- 1 c^. Let P, Q be corresponding^ points in the ellipse and auxiliary circle, respectively; let PQ meet AA! in M, To prove ¥W : ^Jf X A'M^ b" : a\ Proof. FM'' : QM" = h' : a\ § 856 But 03" = AMY. A'M. § 337 Therefore PM" : AMv. A' 31= h' : a\ q. e. d. 859. Cor. The latus rectum is a third proportional to the major axis and the minor axis. For Now and Therefore Hence and Therefore LF-.AFxA'F=b':a\ A'F= a-c, AF=a + c. AFxA^F=a'~c' = b\ LT :h'' = h': a\ LF:h ^h :a. 2a'.2h=r2h:2LF. §858 §846 THE ELLIPSE. 411 Proposition XVII. Theorem. 860. The sum of the distanees of any point from the foei of an ellipse is greater or less than 2 a, oa^- cording as the point is without or within the curve. Q A' F' 1. Let Q be a point without the curve. To prove QF+ QF' > 2a. Proof. Let P be any point on the arc of the curve between QF and QF'. Draw PPand FF'. Then QF+ QF' > FF+ FF'. § 118 But FF-{-FF=2a. §838 Therefore QF-\- QF' > 2 a. 2. Let Q' be a point within the curve. To prove Q'F-{- Q'F' < 2 a. Proof. Let F be any point of the curve between FQ^ and F^Q' produced. Then Q'F-h Q'F' < FF+ FF'. § 118 But PP+PP'=2a. Therefore Q'F+ Q'F'< 2 a. o. e, d. 861. Cor. Converse^, a point is without or within an ellipse according as the sum of its distances from the foci is greater or less than 2 a. 862. A straight line which touches but does not cut an ellipse is called a tangent to the ellipse. The point where it touches the ellipse is called the point of contact. 412 GEOMETRY. — • BOOK IX. Proposition XVIII. Theorem. If through a -point P of an ellipse a line he drawn bisecting- the angle between one of the focal radii and the other produced, every point in this line except P is without the curve^ Let FT bisect the angle PPG between FT and FP produced, and let Q be any point in PT except P, To prove that Q is without the curve. Proof. Upon FP produced take PG = PF'. Join OF', QF, QF', QG. Then QG+QF>GF'. Now AGPQ = AF'PQ. Therefore QG = QF'. Also GF=2a. Therefore QF' +QF>2a. Therefore Q is without the curve. §137 §150 §638 §861 aE. D. 864. Cor. 1. PTis the tangent at P. § 862 865. Cor. 2. The tangent to an ellipse at any point bisects the angle between one focal radius and the other produced. THE ELLIPSE. , 413 866. Cor. 3. If QF' cuts PT at X, then GX-= F'X, and FT is perpendicular to QFK § 123 867. Cor. 4. The locus of the foot of a perpendicular dropped from the focus of an ellipse to a tangent is the auxiliary circle. For, join OX. Since F'X= GX, and F'0=OF, therefore 0X= \FG= i(2a) = a. § 189 Therefore the point X lies in the auxiliary circle. ' Proposition XIX. Problem. 868. To draw a tangent to an ellipse from an exte^ rior point. -^.^ To draw tangents to the ellipse ORQfroTn the exterior point P. Oonstruction. Describe arcs with P as centre and PF as radius, and with F^ as centre and 2a as radius; let these arcs intersect in G and 8. Join OF^ and 8F', cutting the curve in Q and R respec- tively. Join QP and RP, and they will be the tangents required. Proof. PG = PF, and QG = QF. Cons., § 838 .•.APQG = APQF. §160 .■.ZPQG = APQF. Therefore PQ is the tangent at Q. For like reason Pi2 is the tangent at i?. aE.F. 869. Cor. The (D GFS and GS will always intersect (Ex. 78). Hence, two tangents may always be drawn to an ellipse from an exterior point. 414 GEOMETRY. BOOK IX. Proposition XX. Theorem. 870. The tangents drawn at two corresponding points of an ellipse and its auxiliary circle cut the 7}vajor axis produced at the same point. Af F' N D M F A Let the tangent to the auxiliary circle at Q cut the major axis produced at T, and let the ordinate QM meet the ellipse at P. Draw PT. To prove that FT is the tangent to the ellipse at P. Proof. Through R, any point in PT except P, draw RD A. to AA\ cutting the tangent QT, the auxiliary circle, and the ellipse, in L, K, and xS', respectively. Then RD : FM= DT: MT^ or RB'.LD^PM But FM:QM-=h .-. RD. LD = b Again, 8D:KD = b .'. RD: LD = SD: But DD > KD. :. RD > SD. .'. R is without the ellipse. Hence FT is the tangent at F. § 862 aE. D = LD : QM, §321 QM. §208 a. §856 a. a. §856 KD. THE ELLIPSE. 871. Cor. 1. OTxOM=a\ 415 §334 872. The straight line PN drawn through the point of con- tact of a tangent, perpendicular to the tangent, is called the normal. MT ia called the subtangeni, MJ^ the subnormal. 873. Cor. 2. The normal bisects the angle between the focal radii of the point of contact. For Z TPN= Z GPN=. 90°. § 872 Subtract Z TPF--= A GPF\ % 865 And Z FPN= Z F'PN. Hence a ray of light issuing from F will be reflected to F\ 874. Cor. 2f. If d denote the abscissa of the point of contact, the distances measured on the major axis from the centre to the tangent and the normal are — and id, respectively. §871 §334 (1) Since OM=d, and OTx OM=d therefore d (2) Since OMx MT= QM\ and MNx MT= PM\ therefore OM Qir a' MN PM' b^ Therefore OM~MN a'-h" e ^ OM a' a" ' That is, OM ' Hence ON^e'xOM=--e'd. §301 416 GEOMETRY. — BOOK IX. Proposition XXI. Problem. 875. The tangents drawn at the ends of any diam- eter are parallel to eaeh other. Let POQ be any diameter, PT and QT the tangents at P, Q respectively, meeting the major axis at T, V, To 'prove priito qv. Proof. Draw the ordinates PM, QN. . Then A OPM= A OQJSr. §148 Therefore 0M= ON. But ^^=oV^"^^^'=oV §874 Hence OT=OT\ Therefore AOPT=AOQT\ §150 and AOPT^Aoqr, Hence PT\% 11 to qr. aE.D. 876. One diameter is conjugate to another, if the first is parallel to the tangents at the extremities of the second. Thus if ROB\^ I! to PT, PS is conjugate to Pq. THE ELLIPSE. 417 Proposition XXII. Theorem. 877. If one diameter is conjugate to a second, the second is conjugate to tJie first. Let the diameter POP' be parallel to the tangent RT. To prove that MOJR' is parallel to the tangent PT. Proof. Draw the ordinates PM and RN, and produce tliem to meet the auxiliary circle in Q and 8. Join OP, OQ, OB, OS; and draw the tangents QT, ST'. Now, since OP is II to i?X'. the A 03fP and T'JSTP are similar. § 321 .-. T'JST: OM=NE: MP. But. NR : NS = MP : MQ, §870 or NR : MP= NS : MQ. .-. T'J^: OM^NS :MQ. Hence A T'JSTS and OMQ are similar. § 326 . .'.Z]^T'S=ZMOQ. .'. T'Sk II to OQ. §105 Hence Z QOS = Z OST' = 90°. § 240 .-. S0\^ W io QT § 105 418 GEOMETRY. — BOOK IX. /.A SNO and QMT&re similar. ,\0N: TM=NS :MQ, ^JSrH'.MR §856 .*. A ONR and IMP are similar. .-. OR is a to FT. .'. RR^ is conjugate to PP*. aE.D. 878. Cor. 1. Angle QOS is a right angle. 879. Cor. 2. MP : 0N= b : a. For 08=0Q, and OS is ± to OQ, §878 Also OM is ± to NS. .\ZNSO = ZMOQ. § 113, Rem. Hence A JSrSO = A MOQ, .'. 0N= MQ. .'. MP : 0]Sr= MP : MQ. §148 But MP: MQ = b:a. §856 Hence MP : ON^h: a. THE ELLIPSE. 419 Proposition XXIII. Theorem. The area of an ellipse is equal to Trob, A' M N Let A'PRA he any semi- ellipse. To prove that the area of twice A'PRA is equal to irab. Proof. Let FM, i?iVbe two ordinates of the ellipse, and let Q, S be the corresponding points on the auxiliary circle. Draw PV, QUW to the major axis, meeting iViS'in F, U. Then O PiV= FMx MN, and nQN=-QMxMN. O PN PMxMN PM h a Therefore §856 CJ Q]^ QMxMN QM The same relation will be true for all the rectangles that can be similarly inscribed in the ellipse and auxiliary circle. Hence sum of g? ip elHpse^ft, g 303 sum of £17 in circle a And this is true whatever be the number of the rectangles. But the limit of the sum of the (H in the ellipse is the area of the ellipse, and the limit of those in the is the area of the O. Therefore area of ellipse ^ h g 260 area of circle a Therefore the area of the ellipse — - X tto^ a irab. § 425 aE. o. 420 GEOMETRY. — BOOK IX. Proposition XXIV. Theorem. ^l) The section of a right circular cone made hy a plane that cuts all the elements of the surface of the cone is an ellipse. Let APA* be the curve traced on the surface of the cone SBC by a plane that cuts all the elements of the surface of the cone. To prove that the curve A PA* is an ellipse. Proof. The plane passed through the axis of the cone and X to the secant plane APA* cuts the surface of the cone in the elements 8B, SO, and the secant plane in the line AA'. Describe the © and O* tangent to SB, SO, AA\ Let the points of contact be D, II, F, and B, C, F\ respectively. Turn BSC and the © 0, 0' about the axis of the cone. The lines SB, SO will generate the surface of a right circular cone cut by the secant plane in the curve APA' ; and the © 0, O will generate spheres which touch the cone in the © DNH, BWQ, and the secant plane in the points F, F'^ THE ELLIPSE. 421 Let P be any point on the curve APA\ Draw PF, PF'; and draw jSP, which touches the (D PIT, PC at the points iV^, iV', respectively. Since PF and PiV^ are tangent to the sphere 0, they are tangent to the circle of the sphere made by a plane passing through P, F, and iV. Therefore PF= PK § 246 Likewise PF' = PJV'. Hence PP+ PF' = PN^- PJST' = NN', a constant quantity. Therefore APA' is an ellipse with the points F and F' for foci, and A A' as 2 a. q, e, q, 882, Cor. If the secant plane is parallel to the base, the sec- tion is a circle, which is a particular case of the ellipse. Exercises. 707. Prove that the major axis is the longest chord that can be drawn in an ellipse. 708. If the angle FBF^ is a right angle, prove that a? = 262. 709. To draw a tangent and a normal at a given point of an ellipse. 710. To draw a tangent to an ellipse parallel to a given straight line. 711. Given the foci ; it is required to describe an ellipse touching a given straight line. 712. Prove that OF"^ =OTxON. (See figure, page 414.) 713. Prove that OM : 0N= a^ : c^. (See figure, page 414.) 714. The minor axis is the shortest diameter of an ellipse. 715. At what points of an ellipse will the normal at the point pass through the centre of the ellipse ? 716. Prove that if FR, F^S are the perpendiculars dropped from the foci to any tangent, then FR X F^S-= b'^. 422 GEOMETRY. — BOOK IX. 717. To draw a diameter conjugate to a given diameter in a given 718. Given 2a, 2b, one focus, and one point of the curve, to construct the curve, 719. If from a point Pa pair of tangents FQ and FE be drawn to an ellipse, then FQ and FF will subtend equal angles at either focus. 720. To find the foci of an ellipse, having given the major axis and one point on the curve. 721. To find the foci of an ellipse, having given the major axis and a straight line which touches the curve. 722. A straight line moves so that its extremities are always in con- tact with two fixed straight lines perpendicular to each other. Prove that any point of the moving line describes an ellipse. 723. To construct an ellipse, having given one of the foci and three tangents. 724. To construct an ellipse, having given one focus, two tangents, and one of the points of contact. 725. To construct an ellipse, having given one focus, one vertex, and one tangent. 726. The area of the parallelogram formed by drawing tangents to an ellipse at the extremities of any pair of conjugate diameters is equal to the rectangle contained by the axes of the ellipse. THE HYPERBOLA. 423 The HyrERBOLA. 883. The locus of a point which moves so that the difference of its distances from two fixed points is constant is called an hyperbola. The fixed points are called the foci, and the straight lines which join a point of the locus to the foci are called focal radii. The constant difference is denoted by 2 a, and the distance between the foci by 2c. The ratio c : a is called the eccentricity, and is denoted by e. Therefore c = ae. 884. Cor. 2a micst be less than 2c (§ 137) ; hence c 7)iust be greater than 1. 885. An hyperbola may be described by the continuous motion of a point, as follows : To one of the foci F' fasten one end of a rigid bar F'B so that it is capable of turning freely about F* as a centre in the plane of the paper. 424 GEOMETEY. — BOOK IX. Take a string whose length is less than that of the bar by the constant difference 2a, and fasten one end of it at the other focus F^ and the other end at the extremity B of the bar. If now the rod is made to revolve about F^ while the string is kept constantly stretched by the point of a pencil at P, in contact with the bar, the point P will trace an hyperbola. For, as the bar revolves, F^P and FP are each increas- ing by the same amount ; namely, the length of that portion of the string which is removed from the bar between any two positions of P\ hence the diiference between F^P and FP will remain constantly the same. The curve obtained by turning the bar about F is the right- hand branch of the .hyperbola. Another similar branch on the left may be described in the same manner by making the bar revolve about Pas a centre. If the two branches of the hyperbola cut the line FF' at A and A\ it is easy to see, from the symmetry of the construction, that ^^'= 2 a. The hyperbola, therefore, is not a closed curve, like the ellipse, but consists of two similar branches which are sepa- rated at their nearest points by the distance 2a, and which recede indefinitely from the line PP' and from one another. THE HYPERBOLA. 425 Proposition XXV. Problem. 886. To construct an hyperbola hy -points, having given the foci and the constant difference 2a. ^\Xi/' \^ / \ / \ .1 M > \ 1 rrX \ \ 1 1' A j^ — -4^^ -"K" /s\ /qn \ Let F, F' be the foci, and a =■ CD. To construct the hyperbola. Construction. Lay off OA = 0A'= CD. Then A and A^ are two points of the curve. For from the construction AA^ = 2a and AF— A'F'. Therefore AF' - AF= AF* - A'F' = AA' = 2a. And A^F - A'F' = A'F- AF= AA' = 2a. To locate other points, mark any point JTin i^'i^ produced. Describe arcs with F' and i^as centres, and A'X and ^Xas radii, intersecting in P, Q. Then P, Q are points of the curve. By describing the same arcs with the foci interchanged, two more points P, 8 may be found. By assuming other points in P'P produced, any number of points may be found. The curve passing through all the points thus determined is an hyperbola having FF' for foci and 2 a for the constant difference of the focal radii. o. e. f, 426 GEOMETRY. — BOOK IX. 887» Cor. 1. No point of the curve can be situated on the perpendicular to FF^ erected at 0. For every point of this perpendicular is equidistant from the foci. 888. The point is called the centre ; A A' is called the transverse axis; A and A^ are called the vertices. In the perpendicular to FF^ erected at 0, let B^ B^ be the two points whose distance from A (or J.') is equal to c ; then BB* is called the conjugate axis, and the length BB' is denoted by 2b. If the transverse and conjugate axes are equal, the hyper- bola is said to be equilateral or rectangular. a 889. Cor. 2. Both the axes are bisected at 0. 890. Cor. 3. It is evident thai c^ — a^-\- IP. 891. Cor. 4. The curve is symmetrical ivith respect to the transverse axis.. 892. The distances of a point of the curve from the trans- verse and conjugate axes are called respectively the ordinate and abscissa of the point. The double ordinate through the focus is called the latus rectum or parameter. 893. Remark. The letters A, A\ B, 5', F, F\ and 0, will be used to designate the same points as in the above figure. THE HYPERBOLA. 427 Proposition XXVI. Theorem. 894. An hyperbola is synvinetrical with respect to its conjugate axis. F' A Let P be a point of the curve, PDQ be perpendicu- lar to OB, meeting OB at D, and let DQ equal DP. To prove thai Q is also a point of the curve. Proof. Join P and Q to the foci F, F'. Turn ODQF' about OB ; F' will fall on F, and Q on P. Therefore QF' = FF, and ZFQF' = Z.QFF. Therefore A FQF' = A QFF, and QF= FF'. Hence QF- QF' = FF' - FF. But FF'-FF=2a. Therefore QF- QF' = 2a. Therefore Q is a point of the curve. 895. Every chord passing through the centre is called a diameter. 896. Cor. 1. An hyperbola consists of four equal quadrantal arcs symmetrically placed about its centre 0. § 209 897. Cor. 2. Every diameter is bisected at 0. §150 Hyp. Q. E. O. 428 GEOMETRY. — BOOK IX. Proposition XXVII. Theorem. 898. If d denote the abscissa of a -point of an hy- perbola, r and r' its focal radii, then r^ed — a, and r'= ed+ a. A! A F M Let P be any point at the curve, FM perpendicular to AA\ d equal OM, r equal PF, r' equal PF', To prove r = ed—a, r' = ed-\-a. Proof. From the rt. A FFM, F'FM, F]IF + FM\ FM' + F'IF 2 Therefore r'^ -r' = F'M - FM Or (r' + r) (r' -r) = (F'M-{- FM) {F'M- FM). Now r'-r = 2a, and F'M~FM^2c. Also F'M+ FM= 2 0F-\- 2 FM= 2 0M= 2d. By substituting these values, a (r' + r) = 2 cd. Or r' + r = — = 2cc^. a From r'+ r = 2ec?, and / — 7'=2a, by addition, 2 r' = 2 {ed-\- a) ; by subtraction, 2r =2(ed—a). Therefore r = ed—a, and r* = ed-i-a. o. e. o. THE HYPERBOLA„ 429 899, The circle described upon AA' as a diameter is called the auxiliary circle. Proposition XXVIII. Theorem. 900. Any ordinate of an hyperbola is to the tangent from its foot to the auxiliary circle as h is to a. v/ Let P be any point of the hyperbola, PM the drdi' nate, MQ the tangent drawn from M to the auxiliary circle. FM: QM^h'.a. OM^d. QW = d'- a\ To prove Proof. Let Then Also Or since Therefore Or = (ed-ay-{d-cy §898 = e^d'-2aed-\-d'-d'-{-2cd-c\ c = ae, and a^-c'=-&^ § 890 FM'=(e''-l)d'-b'=^(d'-a:'). FM' : QM' = b' : a\ FM .QM=^h '.a. aE.t>. 430 GEOMETRY. BOOK IX. Proposition XXIX. Theorem. 901. The square of the ordinate of a point in an hyperbola is to the product of the distances from the foot of the ordinate to the vertices as h^ is to a^- P/ Let P be any point of the curve, PM the ordinate, MQ the tangent drawn from M to the auxiliary circle. To prove PM' : AMx A'M= b' : d\ Proof. Now Fif : QM' = b' : a\ But QW = AMx A'M. Therefore FM' : AMx A'M= b' : a\ §348 902. Cor. The laius rectum is a third proportional to the transverse and conjugate axes. For LF^ : AF X A'F= b' : a\ But AF= c — a, and AF'= c + a. Therefore AFx A'F= & - c Hence LT \W = l^\ a\ And LF\b =^h '.a. Therefore 2a : 26 = 25 : 2 LF. b\ §901 §890 THE HYPERBOLA. 431 Proposition XXX. Theorem. 903. The difference of the distances of any point froT)% the foci of an hyperbola is greater or less than 2a, according as the point is on tJie concave or con- vex side of the curve. F' A o ^L F 1. Let Qbe a point on the concave side of the curve. To prove QF' —QF>2a. Proof. Let QF' meet the curve at P. Since F'Q=^F'F+FQ, and FQ F'F - FF. But F'F-FF=-2a. Therefore F'Q-FQ>2 a. 2. Let Q' be a point on the convex side of the curve To j^rove Q'F' — Q'F< 2 a. Proof. Let Q'F cut the curve at F. Since F'Q' < F'Fi-FQ', and FQ' - PP+ FQ', therefore F'Q' - FQ' < F'F - FF. But F'F-FF=2a. Therefore F'Q'~FQ'<2a. o. e. d. 904. Cor, Convei^sely , a point is on the concave or the con- vex side of the hypei^hola according as the differejice of its dis- tances from the foci is greater or less than 2 a. 432 GEOMETRY. BOOK IX. 905. A straight line which touches but does not cut the Jiyperbola is called a tangent, and the point where it touches the hyperbola is called the point of contact. Proposition XXXL Theorem. 906. If through a point P of an hyperbola a line he drawn bisecting the angle between the focal radii^ every point in this line except P is on the convex side of the curve. F' A Let PT bisect the an^le FPF', and let Q be any point in PT except P. To prove that Q is on the convex side of the curve. Proof. On PF' take PG = PF; draw FG, QF, QF\ QG. Then QF' ~QG< GF\ Also APGQ = APFQ. Therefore QG=QF. Also GF'=PF'-PF=2a Therefore QF'-QF<2a. §137 §150 §904 ae. D. §905 908. Cor. 2. The tangent to an hyperbola at any point bisects the angle between the focal radii. 909. Cor. 3. The tangent at A is perpendicular to A A*. 910. Cor. 4. If FG cuts PT at X, then GX^ FX, and PT IS perpendicular to FG^ Therefore Q is on the convex side of the curve. 907. Cor. 1. PTis the tangent at P. THE HYPERBOLA. 433 911. Cor. 5. The locus of the foot of the perpendicular from the focus of an hyperbola to a tangent is the auxiliary circle. For, since FX= GX, and FO = OF', therefore 0X= iF'G = i (FF' - FF) - a. § 189 Therefore the point X lies in the auxiliary circle. Proposition XXXII. Problem. 912. To draw a tangent to an hyperbola from a §iven exterior point. Let P he the given point. Oonstrnction. Describe arcs with F as centre and FF as radius, and with F' as centre and 2 a as radius ; let these arcs intersect in O and F. Draw F'O and F'F, and produce them to meet the curve in Q and D, respectively. Join FQ and FD ; FQ and FD are the tangents required. Proof. FG = FF, QF= QF' -2a=^- QG. .\AFQG--=AFQF. .-. Z FQG - Z FQF. .'. FQ is the tangent at Q. For like reason FD is the tangent at D. o. e. f 913. Cor. Two tangents may always he drawn to an hyper- bola from an extei'ior point. 434 GEOMETRY, BOOK IX. Proposition XXXIII. Theorem. 914. The tangents to an hyperbola drawn from the centre r)xeet the curve at an infinite distance from the centre. \ Let OR be the tangent from 0. To prove that OR meets the curve at an infinite distance. Proof. Let G be the intersection of arcs described from ftnd -F' as centres with Oi^and 2 a as radii. The point of contact is the intersection of F^G and OR. § 912 Join FG, cutting OR at Q. Now OF' - 0F\ ff also QG=QF. §910 Therefore F'G is D to OR. § 189 Therefore the point of contact is at an infinite distance. Q. E. D THE HYPERBOLA. 435 In the same way another tangent OS may be drawn, meet- ing the other branch of the curve at an infinite distance. 915. The lines OB, OS. indefinitely produced in both direc- tions, are called the asymptotes of the hyperbola. 916. Cor. 1. ITie line FG is tangent to the auxiliary circle at Q. For FG is ±\o OB, § 910 Therefore Q lies on the auxiliary circle. § 91 1 Hence FG touches the auxiliary circle at Q. § 239 917. Cor. 2. FQ is equal to the semi-conjugate axis b. For F^ = OF'-0^, §339 and 6« = c»-a*. §890 But 0F= c, and OQ = a. Therefore FQ = b, 918. Cor. 3. If the tangent to the curve at A meets the asymptote OB at B, then AB = b. For A OAB = A OQF § 149 Therefore AB = FQ = b. 919. Cob. 4. The asymptotes of an hyperbola are the diago- nals of the rectangle BSUV constructed with for its centre, and the transverse and conjugate axes for its two sides. 920. A perpendicular to a tangent erected at the point oi contact is called a normal. The terms subtangent and subnormal are used in the hyper- bola in the same sense as in the ellipse. § 872 436 GEOMETRY. — BOOK IX. Proposition XXXIV. Theorem. 921. The section of a right circular cone made hy a plane that cuts both nappes of the cone is an hyper- bola. Q N'l T). /t\ x -- JK ... '\ P Q m Let a plane cut the lower nappe of the cone in the curve FAQ, and the upper nappe in the curve P'A'Q'. To prove that FAQ and F'A'Q' are the two branches of an hyperbola. Proof. The plane passed through the axis of the cone per- pendicular to the secant plane cuts the surface of the cone in the elements BS, CS (prolonged through S), and the secant plane in the line UN', THE HYPERBOLA. 437 Describe the © 0, 0', tangent to BS, OS, JSTJV'. Let the points of contact be D, IT, F, and D', H', F\ respectively. Turn B8C and the (D and 0' about the axis of the cone. B8 and C8 will generate the surfaces of the two nappes of a right circular cone ; and the © 0, 0' will generate spheres which touch the cone in the © DKH, D'K'II\ and the secant plane in the points F, FK Let P be any point on the curve. Draw PF and PF^ ; and draw P8 which touches the © DKH, B^K^W at the points K. K'. Now PjFand P^^are tangents to the sphere from the point P. Therefore PF= PK. Also PF' = PK'. Hence PF' - PF^ PK' - PK ■= KK', a constant qunntity. Therefore the curve is an hyperbola with the poinrs F and F' for foci. * £AUFOQ!£> THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO 50 CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE, r MAR 14 1939 ^eP — 1 1 939 scp n ^» ^ msTT^'^^ ij n ^':AH Jl , w ]90cf59Df! r;ov lo 1: / :\i -: UNIVERSITY OF CAUFOfbriA UBRARY X \ v\ kvvf ^<^y^m Wv 4.