NEW PLANE AND SOLID GEOMETRY BY WEBSTER WELLS, S.B. PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY BOSTON, U.S.A. D. C. HEATH & CO., PUBLISHERS 1909 COPYRIGHT, 1908, BY WEBSTER WELLS. All rights reserved. PREFACE IN the preparation of this text the author acknowledges joint authorship with Robert L. Short. Geometry is approached from the constructive side, all methods of construction needed for drawing any figure in Books I or II being given in the introduction. In cases where a geometric principle is used in any construction, a note at the end tells where the principle is proved. In all figures in the early portions of Book I, the construction lines and arcs are given; afterwards these are dispensed with. In Props. II and III, Book I, and in other places, colored diagrams are given in addition to the regular figures, in which the equal parts in the given triangles are represented by lines of the same color; this scheme will be found of great assistance to the pupil in the earlier portions of the work. Below each figure is a paragraph in smaller type, giving full directions for the construction of the diagram in accordance with the statement of the theorem. This gives the pupil a familiarity with the figure, and what in it is given, and what to be proved, that is of great value, and lessens the tendency to memorize. Many figures are omitted, but complete directions for their construction are given in each case. In all figures given in connection with theorems and problems, given and required lines are made heavy. Attention is invited to the order of theorems in Book I; in this case, the pupil begins with the easier proofs. From the start the student has practice in representing angles and lines by small letters. Only the outline of the proof is given after Book II, except in the more difficult demonstrations. In this outline work the pupil has explicit directions, but develops the demonstration himself. In all portions of the work proofs are omitted in iii iv PREFACE cases where they should present no difficulty; usually, in such cases, hints are given as to the method of demonstration. The numbering of the steps of the proof makes them more easy of reference. In Book I, and in the first part of Book VI, the authority for each statement of the proof will be found directly below, in smaller type, enclosed in brackets, with the number of the section where it is to be found. In other portions of the work only the section number is given, and in some cases only an interrogation point. In all such cases the pupil should be required to give the authority as fully as if it were actually printed on the page. No principles are given as immediate consequences of theorems except such as actually belong in this category. Separate propositions are made for all truths which are not immediate consequences of preceding theorems. The originals are new, and very largely of a practical nature. They are not too difficult, will make the pupil think, and are more than five hundred in number. The smaller number is compensated for by the fact that the pupil has to do some original work in almost every proof after Book II. Exercises coming under Book I will be found scattered through all the following Books of the Plane Geometry; and a similar remark applies to exercises under Book II, Book III, etc. At the end of Book I will be found a list of principles proved, which will be of great assistance in solving originals. A similar list in regard to similar triangles is given in ~ 265. The authors thought that it would be an improvement to put all work under Polyedrons, including their measurement, in Book VII, devoting Book VIII to the Cylinder and Cone and Book IX to the Sphere and its measurement. The authors wish to thank the many teachers whose advice and criticism have been useful in preparing a treatise that should stand the test of class-room work. They are also indebted to Mr. C. W. Sutton. for a part of the exercises. WEBSTER WELLS, BOSTON, 1908. SUGGESTIONS TO TEACHERS IN studying the opening propositions, in Geometry, beginners have difficulty in fixing clearly in mind just what parts, in the figure, are given. To aid this, in Props. II and III, Book I, and in other places, diagrams are given in addition to the regular cuts, in which the equal given parts are printed in the same color. This color-scheme may be advantageously followed in the class-room, in connection with all figures in the earlier portions of Book I. If colored crayons are not available, the beginner may designate equal lines by the marks, o, /i, etc., and equal angles by single, double, or triple arcs. In solving the non-numerical exercises, while it is not practicable to give very much assistance to the pupil, the following suggestions may be found of service:1. Draw an accurate figure, showing all given and required lines. This sometimes suggests the method of proof. 2. Be sure that the figure is the most general one allowable. Thus, in exercises relating to triangles, do not draw them right, isosceles, or equilateral, unless the exercise calls for a right, isosceles, or equilateral triangle; in exercises relating to quadrilaterals, do not draw them with two sides equal or parallel, unless the exercise calls for such a construction. 3. Write down carefully what is given, and what is to be proved. 4. Look up all previous theorems which may possibly have a bearing on what is to be proved. Thus, if two lines are to be proved equal, refer to all previous theorems regarding equal lines. (Compare ~ 141.) In solving exercises in construction, it is advantageous to regard the problem as solved, and draw the given and required lines. Studying the relations between these will frequently suggest the method of construction to be employed. v CONTENTS PAGE INTRODUCTORY.......... 1 PLANE GEOMETRY BOOK I. RECTILINEAR FIGURES... 12 BOOK II. THE CIRCLE....... 66 BOOK III. THEORY OF PROPORTION.-SIMILAR POLYGONS 101 BOOK IV. AREAS OF POLYGONS...... 134 BOOK V. REGULAR POLYGONS.- MEASUREMENT OF THE CIRCLE. - LOCI...... 154 SOLID GEOMETRY BOOK VI. LINES AND PLANES IN SPACE. - DIEDRAL ANGLES.- POLYEDRAL ANGLES. 175 BOOK VII. POLYEDRONS..... 209 BOOK VIII. THE CYLINDER AND CONE..... 240 BOOK IX. THE SPHERE.... 255 vii NEW PLANE GEOMETRY GEOMETRY INTRODUCTORY A material body A geometrical solid 1. A material body, as, for example, a block of wood, occupies a limited portion of space. The boundary which separates such a body from surrounding space is called the snurtfce of the body. If the material which composes such a body could be conceived as taken away from it, without altering the form of the bounding suoface, we should have a portion of space, with the same bounding surface as the material body. We call this portion of space a geometrical solid, or simply a solid. We. call the surface which bounds it a geometrical surface, or simply a surface; it is also called the surface of the solid. 2. If two surfaces intersect each other, we A call that which is common to both a geometrical C liee, or simply a line. Thus, if surfaces AB and CD cut each other, K D their common part, EF, is a line. B 3. If two lines intersect each other, we call that which is common to both a geometrical point, or simply A a point. Thus, if lines AB and CD cut each other, their common part, 0, is a point. C B 1 2 GEOMETRY 4. A solid has extension in every direction; but this is not the case with surfaces and lines. A point has extension in no direction. We may conceive a surface as existing independently in space, without reference to the solid whose boundary it forms. In like manner, we may conceive of lines and points as existing independently in space. A line is produced by the motion of a moving point, a surface by the motion of a moving line, and a solid by the motion of a moving surface. 5. We define a straight line as a line which has the same direction throughout its length. aB CD E A straight line A curve A broken line A straight line is designated by two letters anywhere upon its length, or by a single letter; thus, the straight line in the figure may be designated either AB or a. We define a curve as a line no portion of which is straight; as CD. We define a broken line as a line which is composed of different successive straight lines; as EFGH. The word "line," without qualification, will be used hereafter as signifying a straight line. 6. We define a plane as a surface such that if a straight line be drawn between any two of its points, it lies entirely in the surface. 'M Thus, if P and Q are any two points in sur- / face MN, and the straight line PQ lies entirely / in the surface, then MN is a plane. N 7. We may conceive a straight line as being of unlimited length; we may also conceive a plane as being of unlimited extent in regard to length and breadth. 8. We define a geometrical figure as any combination of points, lines, surfaces, and solids. INTRODUCTORY 3 We define a plane figure as a figure formed by points and lines all lying in the same plane. A figure is called rectilinear when it is composed of straight lines only. A straight line is sometimes called a right line. 9. Geometry treats of the properties, construction, and measurement of geometrical figures. Plane Geometry treats of plane figures only. Solid Geometry treats of figures which are not plane. 10. We define an angle as the figure formed A by two straight lines drawn from a point. The point is called the vertex of the angle, and the straight lines its sides. /a 11. If there is but one angle at a given vertex, we designate it by the letter at the vertex. But if two or more angles have the same vertex, we avoid ambiguity by naming also a letter on each side, reading the vertex letter between the others. Thus, we should read the angle of ~ 10 "angle B"; but if there were other angles at the same vertex, we should read it either ABC or CBA. Another way of designating an angle is by means of a letter placed between its sides; thus, we may read the angle of ~ 10 "angle a." 12. Two figures are said to be equal when one can be applied to the other so that they shall coincide throughout. To prove two angles equal, we do not consider the lengths of their sides. Thus, if angle ABC can be applied B C E / to angle DEF in such a manner that point B shall fall on point E, side AB on side DE, and side BO on side EF, the angles are equal, even if side AB is not equal in length to side DE, and BC to EF. 4 GEOMETRY 13. We call two angles adjacent when they. have the same vertex, and a common side between them; as AOB and BOO. We call angle AOC the sum of angles AOB and BOO. We also regard angle AOC as greater than 0 C angle BOO, and angle BOC as less than angle AOC. 14. If from a point in a straight line a line be drawn in such a way as to make the adjacent angles equal, each of the adjacent angles is called a right angle, and the lines are said to be perpenlicular to each other. B Thus, if from point A in line CD line AB be drawn in such a way as to make angles BAC and BAD equal, each of these angles is a right angle, and AB and CD are perpen-, D dicular to each other. A 15. Let C be any point in straight line AB; and let straight line CD be drawn in such a way as to make angle BCD less than angle ACD. Let line CD be turned about point C D as a pivot towards the position CA. Then, angle BCD will constantly increase, and angle ACD will constantly diminish; and there must be some position of CD where these angles are 0 C^ B equal, and there can evidently be but one such position. If CE is this position, by the definition of ~ 14, CE is perpendicular to AB at C. Then, at a given point in a straight line, a perpendicular to the line can be drawn, and but one. The following are immediate consequences of ~ 15: 16. Any two right angles are equal. 17. If two adjacent angles have their exterior sides in the same straight line, their sum is equal to two right angles. INTRODUCTORY 5 For in figure of ~ 15, the sum of angles ACD and BCD equals the sum of angles ACE and.BE, or two right angles. 18. The sum of all the angles on the same side of a straight line at a given point is equal to two right angles. C 19. The sum of all the angles about a point in a plane is equal to four right angles. For if a side of any angle, as OA, be ex- - --- tended to E, the sum of the angles on either side of straight line AE is, by ~ 18, equal to / two right angles. 20. If the sum of two adjacent angles is equal to two right angles, their exterior sides lie in the samne straight line. If the sum of angles ACD and BCD is two right angles, side AC if extended through C must coincide with CB; for if it did not the sum of the angles would be A C less or greater than 180~ (~ 17). 21. We define a triangle as a portion of a plane bounded by three straight lines; as ABC. We call the lines AB, BC, and CA the sides of the triangle; and their intersections, A, B, and C, the vertices. We define the angles of the triangle as the angles CAB, ABC, and BCA, between the adjacent sides. B C 22. A triangle is called sealene when no two sides are equal; isosceles when two sides are equal; equilateral when all its sides are equal; equiangular when all its angles are equal. Scalene Isosceles Equilateral 6 GEOMETRY A right triangle is a triangle which has a right angle; as ABC, which has a right angle at C. The side AB opposite the right angle is called the hypotenuse, and the other sides, AC and BC, the legs. 23. We define a circle as a portion of a plane bounded by a curve, called the circumtference, all points of which are equally distant from a point within - called the centre. An arc is any portion of the circumference; as AB. A radius is a straight line drawn from circumference; as OA. the centre to the CONSTRUCTIONS 24. At a given point in a straight line, to draw a perpendicular to the line. A,D C E.B Let it be required to draw a line perpendicular to the straight line AB, at the point C. With C as centre, and any straight line less than AC as radius, describe arcs intersecting AC at D, and BC at E. With points D and E as centres, and any radius, describe arcs intersecting at F. Then, the straight line drawn from C through F will be perpendicular to AB at C. The reason for the above construction will be found in ~ 58. INTRODUCTORY 7 25. Fronm a given point without a strcight line to draw a perpendicular to the line. C / "X-F A D, E Let it be required to draw through any point C without the straight line AB a line perpendicular to AB. With C as a centre, and any radius, describe an arc cutting AB at points D and E. With points D and E as centres, and another radius, describe arcs intersecting at F. Then, the straight line CG drawn from C through F will be perpendicular to AB. The reason for the construction will be found in ~ 58. 26. 7o bisect a given straight line. A-. ----,E -,-B / Let it be required to divide the straight line AB into two equal parts. With points A and B as centres, and with the same radius, describe arcs intersecting at C and D. Draw straight line CD intersecting AB at E. Then, E is the middle point of AB. The reason for the construction will be found in ~ 58. 8 (IEOMAETRY 27. To bisect a given angle. o F/ I Let it be required to bisect angle AOB. With 0 as a centre, and any radius, describe an are intersecting OA at C and OB at D; with C and D as centres, and the same radius, describe arcs intersecting at E. Then, the straight line OE will bisect angle AOB. The reason for the construction will be found in ~ 58. 28. With a given vertex, and a givem side, to construct an angle eqtal to a given angle. /.F B DE-A i D Let it be required to construct with E as the vertex, and ED as a side, an angle equal to angle ABC. With B as a centre, and any radius, describe an arc intersecting AB at G and BC at H. With E as a centre, and BG as a radius, describe an arc intersecting DE at IK With K as a centre, and the distance from G to H as a radius, describe an arc intersecting the former arc at L. Draw straight line ELF. Then, angle DEF will equal angle ABC. The reason for the construction will be found in ~ 53. INTRODUCTORY 9 29. Given two sides and the included angle of a triangle, to construct the triangle. /D m/ n/ E A/ m B Let it be required to construct the triangle having for two of its sides the straight lines m and n, and their included angle equal to angle E. Draw line AB equal to mn, and construct angle BAD equal to angle E (~ 28). On AD take AC equal to n, and draw straight line BC. Then, ABC is the required triangle. 30. Given a side and twzo adjacent angles of a triangle, to construct the triangle. D A/ - \ G A pL ~~ E~~,,G m A B Let it be required to construct the triangle having for a side the straight line m, and its adjacent angles equal to angles F and G. Draw line AB equal to m, and construct angle BAD equal to angle F (~ 28). Draw line BE, making angle ABE equal to angle G, intersecting AD at C. Then, ABC is the required triangle. 10 GEOMETRY 31. Given the three sides of a triangle, to construct the triangle. m ___x_______ AZ L -- -4 / m P Let it be required to construct the triangle having for its sides the straight lines m, n, and p. Take the straight line AB equal to m. With A as a centre, and n as a radius, describe an arc. With B as a centre, and p as a radius, describe an arc intersecting the former arc at C. Then, ABC is the required triangle. 32. We define an Axiom as a truth which is assumed without proof as being self-evident. We define a Theorem as a truth requiring proof. We define a Problem as a question proposed for solution. A Proposition is a general term for a theorem or a problem. A Postulate assumes that a certain problem can be solved. A Corollary is a truth which is an immediate consequence of the proposition which it follows. An Hypothesis is a supposition, made either in the statement or the proof of a proposition. 33. Postulates. We assume that the following problems can be solved: 1. A straight line can be drawn between any two points. 2. A straight line can be extended indefinitely in either direction. 34. Axioms. We assume the following as true: 1. Things which are equal to the same thing, or to equal things, are equal to each other. 2. If equals be added to equals, the sums will be equal. INTRODUCTORY 11 3. If equals be subtracted fiom equals, the remainders will be equal. 4. If equals be multiplied by equals, the products will be equal. 5. If equals be divided by equals, the quotients will be equal. 6. But one straight line can be drawn between two points. 7. A straight line is the shortest line between two points. 8. The whole is equal to the sum of all its parts. 9. The whole is greater than cny of its parts. 35. Since but one straight line can be drawn between two points, a straight line is said to be determined by any two of its points. 36. Symbols and Abbreviations. The following symbols are used in the work: +, plus. -, minus. x, multiplied by. =, equals. A, equivalent, is to. >, is greater than. <, is less than..'., therefore. L, angle. As, angles. A, triangle. A, triangles. _L, perpendicular, is perpendicular to. equivalent i, perpendiculars. II, parallel, is parallel to. Ils, parallels. /0, parallelogram. /u, parallelograms. 0, circle. 0, circles. The following abbreviations are used: Ax., Axiom. Sup., Supplementary. Def., Definition. Alt., Alternate. Hyp., Hypothesis. Int., Interior. Cons., Construction. Ext., Exterior. Rit., Right. Corresp., Corresponding. Str., Straight. Rect., Rectangle, recAdj., Adjacent. tangular. BoOK I RECTILINEAR FIGURES DEFINITIONS 37. We define an acute angle as an angle less than a right angle; as ABC. We define an obtuse angle as an angle greater than a right angle; as DEF. Acute and obtuse angles are called oblique angles; and intersecting lines which are not perpendicular, are said to be oblique to each other. We call two angles vertical when the sides of one are the prolongations of the sides of the other; as AEC and BED. 38. If angles AOB, BOC, COD, DOE, are all equal, we say that angle A OB is contained four times in angle AOE; and similarly for any number of equal parts of angle AOE. 39. We measure an angle by finding how many times it contains another angle taken as the unit of measure. A D -E 0 CC B A The usual unit of measure for angles is the degree, which is the ninetieth part of a right angle. To express fractional parts of the unit, we divide the degree into sixty equal parts, called minutes, and the minute into sixty equal parts, called seconds. We represent degrees, minutes, and seconds by the symbols o, ', ", respectively. 40. If the sum of two angles is a right angle, or 90~, we call one the complement of the other; if their sum is two right angles, or 180~, we call one the supplement of the other. 12 RECTILINEAR FIGURES 13 Thus, the complement of an angle of 34~ is 90 ~- 34~, or 56~; the supplement of an angle of 34~ is 180~ - 34~, or 146~. Two angles which are complements of each other are called coimplementary; and two angles which are supplements of each other are called supplementary. 41. It follows from the above that: 1. The complements of equal angles are equal. 2. The supplements of equal angles are equal. 42. Since i ACD, BUD (Fig., ~ 15), are supplementary (~ 40); the principle of ~ 17 may be stated as follows: If two adjacent angles have their exterior sides in the same straight line, they are supplementary. Such angles are called supplementary-adjacent. Ex. 1. How many degrees are there im the complement of 43~? of 85~? Ex. 2. How many degrees are there in the supplement of 17~? of 162~? Ex. 3. Find the supplement of the complement of 75~; of 22~ 30'; of 45~ 121 18"; of A~. Ex. 4. Find the complement of the supplement of 178~; of 144~; of 125~14'15"; of B~. Ex. 5. Given the sum and difference of two lines, to find the lines. (Bisect the sum, also bisect the difference.) Ex. 6. The sum of the lengths of two given lines is 20 inches, and their difference is 4 inches; find the lines. Make a diagram. 43. Note. The demonstration of a geometric truth consists of three parts: 1. The statement of what is given in the figure. 2. The statement of what is to be proved. 3. The proof, depending upon previous theorems, axioms, or definitions. We shall mark these three divisions of the demonstration by the words Given, To Prove, and Proof, respectively. We shall use the symbols and abbreviations of ~ 36, and also number the successive steps of the proof. 14 PLANE GEOMETRY -BOOK I In every proof in Book I, where a preceding theorem is given as the authority for any statement in the proof, it will be found directly after the statement, in smaller type, enclosed in brackets. PROP. I. THEOREM 44. If tto straight lines intersect, the vertical angles are equal. A D C a Draw intersecting str. lines AB and CD forming vertical A a and c, and also b and d. We then have: Given intersecting str. lines AB and CD, forming vertical / a and c, and also b and d, To Prove Za = Z c and L b = Z d. Proof. 1. Z a + Z b= 180~. [If two adj. A have their ext. sides in the same str. line, their sum is equal to two rt. A.] (~ 17) 2. Also, Z b +Zc= 180~. 3. Then, Za+Zb=Lb+Zc. [Things which are equal to the same thing, or to equal things, are equal to each other.] (Ax. 1) 4. Subtracting Z b from the equals Z a + / b and Z b + Z c, Z.a= - c. [If equals be subtracted from equals, the remainders will be equal.] (Ax. 3) 5. In like manner, / b = Z d. 45. The enunciation of every theorem consists essentially of two parts: the Hypothesis, and the Conclusion. Thus, we may enunciate Prop. I as follows: Hypothesis. If two straight lines intersect, Conclusion. The vertical angles are equal. 1 {LViI RECTILINEAR FIGURES 15 PROP. II. THEOREM 46. Two triangles are equal wthen twvo sides and the included angle of one are equal respectively to two sides and the included angle of the other. C, 'F A B /E Draw any A ABC: draw line DE equal to AB: construct Z EDG equal to Z A, the constructed side being DG: on DG take DF equal to AC: draw line FE. We now have: Given, in A ABC and DEF, AB= DE, AC=DF, and ZA = / D. To Prove A ABC = A DEF. Proof. 1. Superpose A ABC upon A DEF in such a way that Z A shall coincide with its equal Z D; side AB falling on side DE, and side AC on side DF. 2. Since AB = DE, point B will fall on point E. 3. Since AC= DF, point C will fall on point F. 4. Then, side BC will fall on side EF. [But one str. line can be drawn between two points.] (Ax. 6) 5. Therefore, the A coincide throughout, and are equal. 47. Since ABC and DEF coincide throughout, we have ZB=Z E, C= Z DFE, and BC= EF. 48. Note. In equal figures, lines or angles which are similarly placed are called homologous. Thus, in the figure of Prop. II, Z A is homologous to Z D; AB is homologous to DE; etc. It follows from the above that In equal figures, the homologous parts are equal. In equal triangles, the equal sides lie opposite the equal angles. 16 PLANE GEOMETRY -BOOK I Ex. 7. If, in triangles ABD and BCD, sides AB and BD and angole ABD are equal respectively to sides CD, BD, and angle BDC, what other sides and angles of the triangles are equal? BA / A -- PROP. III. THEOREM 49. Tzwo tricngles are equcal when a side angles of one are equal respectively to a side angles of the other. and two adjacent and two acdjcent 0 A -- B? I E\,'G 'A 'z:,w' \ % -- -- \ v.4 An i 1 r Draw any A ABC: draw line DE equal to AB: at D construct Z ED(T equal to Z A: at E construct Z DEH equal to Z B. Let F be point of intersection of DG and ElI. We now have: Given, in A ABC and DEF, AB = DE, /A = D, and B = Z E. To Prove A ABC= A DEE. Proof. 1. Superpose A ABC upon A DEF in such a way that side AB shall coincide with its equal DE; point A falling on point D, and point B on point E. 2. Since ZA=LD, side AC will fall on side DF, and point C will fall somewhere on DF. 3. Since Z B = / E, side BC will fall on side EF, and point C will fall somewhere on EF. 4. Since point C falls at the same time on DF and EF, it must fall at their intersection, F. 5. Then, the A coincide throughout and are equal. II 3tlV id RECTILINEAR FIGURES 17 Ex. 8. What other sides and angles in the triangles of Prop. III are equal Ex. 9. If, in triangles ABD and BCD, side BD and angles ABD anll A lDB are equal respectively to side BD and angles CBD and CDB, what other sides. C and angles of the triangles are equal? I) PROP. IV. THEOREM 50. In an isosceles triangle, the angles opposite the equal sides are equal. b/ \a A D B Draw line AB. With any radius greater than ~ AB and with A as a centre draw an arc; with the same radius and B as a centre draw an arc intersecting the first arc at C; draw lines AC (b) and BC (a). We then have: Given a and b, the equal sides of isosceles A ABC. To Prove Z A = / B. Proof. 1. Draw str. line CD bisecting Z ACB, meeting AB at D. 2. In A ACD and BCD, CD= CD. 3. By hyp., a = b, and / ACD = Z BCD. 4. Then, A ACD = A BCD. [Two A are equal when the two sides and the included Z of one are equal respectively to two sides and the included Z of the other.] (~ 46) 5. Then, ZA= B. [In equal figures, the homologous parts are equal.] (~ 48) 51. It follows from the preceding that An equilateral tricatgle is also equiangular. 18 PLANE GEOMETRY -BOOK I PROP. V. THEOREM 52. Two tricngles are equal when the three sides of one are equal respectively to the three sides of the other. C F' AB D B l)' raw any A ABC, Ilaking' AB the \c d/ longest side: construct A DEF, having side DE = AB, side DF = AC, and side SF' EF = BC (~ 31). We then have: Given in A ABC and DEF, AB = DE, BC= EF, and AC= DF. To Prove A ABC = A DEF. Proof. 1. Let AB be longest side of A ABC; and place A DEF in position ABFt, side DE coinciding with its equal AB, and vertex F falling at F', on the opposite side of AB from C. 2. Draw line CF', and represent A CF1', BCFt, AF'C, and BF'C by a, b, c, and cd, respectively. 3. Since AC= AF', a = Z c. (1) [In an isosceles A, the A opposite the equal sides are equal.] (~ 51) 4. Since BC = BF', b =. (2) 5. Adding (1) and (2), Za+ b=Z c+ Zd; or ZACB =ZAF'B. 6. Since sides AC and BC andl ACB, of A ABC, are equal, respectively, to sides AF' and BF' and Z AF'B, of A ABF', A ABC = A ABF'. [Two A are equal when two sides and the included Z of one are equal respectively to two sides and the included Z of the other.] (~ 46) 7. That is, A ABC = A DEF. RECTILINEAR FIGURES I9 53. Note. We may now see the reason for the construction given in ~ 28. We know that the triangle whose vertices are the points B, G, and H has its sides respectively equal to those of the triangle whose vertices are the points E, K, and L. Then, the triangles are equal by ~ 52, and the homologous angles B and E are equal. PROP. VI. THEOREM 54. If a perpendicular be erected at the middle point of a straight line, I. Any point in the perpendicular is equally distant friom the extremities of the line. II. Any point without the perpendicular is unequally distant from the extremities of the line. E l-a bdlel At the middle point D of any line AB erect a I DC. From E, any point in DC, draw lines to A and B. We now have: I. Given line CD -1 to line AB at its middle point D, E any point in CD, and lines AE (a) and BE (b). To Prove a = b. Proof. 1. In A ADE and BDE, DE = DE. 2. By hyp., AD = DB. 3. Also, Z ADE = Z BDE. [All rt. A are equal.] (~ 16) 20 PLANE GEOMETRY-13OOK I 4. Then, A ADE = A BDE. [Two A are equal when two sides and the included Z of one are equal respectively to two sides and the included Z of the other.] (~ 46) 5. Then, a = b. [In equal figures, the homologous parts are equal.] (~ 48) F A D B Draw line AB. At D, the middle point of AB, draw DC I AB. From F, any point without CD, draw lines FA and FB. We then have: II. Given line CD I to line AB at its middle point D, F any point without CD, and lines AF and BF. To Prove AF and BF unequal. Proof. 1. Let AF intersect CD at E, and draw line BE. 2. We have BE + EF> BF. [A str. line is the shortest line between two points.] (Ax. 7) 3. But, BE = AE. [If a I be erected at the middle point of a str. line, any point in the 1L is equally distant from the extremities of the line.] (~ 54, I) 4. Substituting for BE its equal AE, AE + EF> BF, or AF > BF. 55. It follows from Prop. VI that every point which is equally distant from the extremities of a straight line, lies in the leppendicular erected at the middle point of the line. RECTILINEAR FIGURES 21 56. A straight line is determined by any two of its points (~ 35); whence, Two points, each equally distant from the extremities of a straight line, determine a epependicular to that line at its middle point. 57. From the equal triangles ADE and BDE, in the figure of ~ 54, I, / AED = Z BED; since homologous parts of equal figures are equal. Then, if lines be drawnc to the extremities of a straight line from any point in the perpendicular erected at its middle point, they make equal angles with the perpendicular. 58. TWe may now see the reasons for the constructions given in ~~ 24, 25, 26, and 27. In ~ 24, by construction, points C and F are each equally distant from points D and E; and CF is perpendicular to AB by ~ 56. In ~ 25, points C and F are each equally distant from points D and E; and in ~ 26, points C and D are each equally distant from points A and B. In ~ 27, points O and E are each equally distant from points C and D; then, OE is perpendicular to straight line CD at its middle point (~ 56), and Z AOE = Z BOE by ~ 57. Ex. 10. If in triangles ABD and BCD, sides B C AB, BD, and AD are equal, respectively, to sides CD, BD, and BC, what angles of the triangles are equal? AD Ex. 11. Two lines of unequal length bisect each other at right angles. Show that any point in either line is equidistant from the extremities of the other. (~ 54.) Ex. 12. If lines be drawn from the extremities of a straight line to any point in the perpendicular erected at its middle point, they make equal angles with the line. (Figure of ~ 54, I. Use ~ 48.) 22 PLANE GEOMETRY - BOOK I PROP. VII. THEOREM 59. From a given point wcithout a straight line, but one perpendicular can be dracwn to the line. (It follows from ~ 25 that, from a given point without a straight line, a perpendicular can be drawn to the line.) C A*~bD B a i a'" From point C without line AB, draw CD I AB (~ 25). We then have: Given point C without line AB, and line CD I AB. To Prove CD the only J_ which can be drawn from C to AB. Proof. 1. If possible, let CE be another _I from C to AB. 2. Extend CD to C', making C'D = CD, and draw line EC'; represent Z CED by a, and Z C'ED by b. 3. Since ED is 1. CC' at its middle point D, Z a = Z b. [If lines be drawn to the extremities of a str. line from any point in the I erected at its middle point, they make equal A with the L.] (~ 57) 4. By hyp., Z a is a rt. Z. 5. Then, Z b is a rt. Z, and Z a + b = two rt. As. 6. Then line CEC' is a str. line. [If the sun of two adj. A is equal to two rt. A, their ext. sides lie in the same str. line.] (~ 20) 7. This is impossible, for, by cons., CDC' is a str. line. [But one str. line can be drawn between two points.] (Ax 6.) ftLATE tlf 8, ie':lt.<e, (JE cant:ot e.t...t.:t_..! atnl C:D is th0te only. that ct anl b it dttraw n. lxI,1. 1 I3 f.A.,A (. an. id t )DBl DC a( re thel equal sidle of two iuneqtlal io. seele.s tr'iatngles A.- C, D' I:t' t having eottMnni sie (...ic the litne johning.* anm.d) bisects IBC.P'vpX., N. Vii. l t:o:x: 60* 'Two ri,/ht triaw ftes are eqttual 'Ihen fthle h/t!lpoten.iuse and an'f. attiacen(t nt.: tl ofne f' are ef ql.ti re. e'stie/t'e.y to to fe hfpotetnuse t andt Dvaw any A A4BC. right angled at,C drawN line I)I 11?: eonlst met /D. Z /t:1; draw linne et a,/ D. q' t owt e. Given, rin. A I 1J Band 1Z)(1 hypotenuse (e) e e and 7. 1 Xl)J 2.........S.....ro A; A w li] pon A 1 is(.e.a way that Ihpotven.ise c shall coi.ide width it's quallI; vertex A,1 falling oil vertex I nui vertex.B (ion e..ex I( 4I, /. '.t. Vr t X./ S 3, Sin /:4 A, sid e i will fall.. o.n snide a. To Prove.A BC':... DEt?. 4-..'lhen, side a till f.all ot i side. ta [Fro t. given l tn t sithou e t a sir. lia', Idt uie f. cant be drawn to tiin line. (~ 59) f,. T'lhere! for. e, tlhe A > eoit'.ide' t:hrotulghoutl, and aret ejtqual, ~ ~ ~ ~ ~ -' I'~ 24 P ILANEI ( ) ) - 1f.I 1RY..001.K It Et ~}^.i\.8\ tt..i..it..'..>.}....t. P1:O.Iz P I X. T:: lou ',: 762ot1. 'io nub! tIiinfs are (quit! 0/l he/e t fU/h uet -teof andI a le Of ione a':re e(iual: WretCft. fp.. toh/e fhyplo','ft'se utt.-.I a( le of tiJee othetr.:. ~^^ s~aas~~~ssss ^ - **** *........... ' **....*'*'-..........1.......,.X - ~ " 2 D ",. /' It raw a nv Ai.: t ( ri ' t; analet!. at ( '' draw l ine.t':. 1 Ct const ruc t. ig F(' a;t D;t dr. t w ftit un' l /. We: now lh.t'ave: Oivenh in t rt.1 f a.nd ) tl 1 /' hlv} tetji ist e A.B V -tai) 1t: us..D.-lt. antd B(:t. '': ' To Prove A.N.AI A t /). Proof. I, Pliae A 1.)1'1 in }i.sit:ion.1.1; vel tex 1 fallhin at; 11, 1' at cs ai )l at /I. on op})l:)it sile of iBC.on.1..2. Sit.io e ' a re rIt.lD.Y..no t,. (./) i. a si. line. [If the sut of two adj. is elqual to twio t.. ', teir ext. sidl es lie lit thet sa le iMr. Hlinl ] (~:o1)) ". "le1, A.I)1 is is 5osceles, and Z... / 1)'. [In an isosceles. A. the /I o ppos tite tl e eiuatsides are. equal ] (~ t0) 4, Tlen, A AB.^ /x (D) [Two rt. are tequal wlhe ti lie typ enus ani at at dj. / of {e. ar (si-ual respectively to t: tle hypotenuse atd anl atij. Zof ttle othetr.l (~ {0)) 5. Tlat is, t.A/Xi: - \ At)/ 1 PROX,. X'. 'i" n.c 62, A'tIf sMide oft a triS tIgfle is qa'reat':er thant t/e d(isfference of (the other ttwo s-des. )raw\ A l.AtC with side t (7 > sidte A We now hate t Given AB t any saie of A /.....,.,..... A al B( > AC, d...... -A I, PLATE V RECTILINEAR FIGURES 25 To Prove AB >BC-AC. Proof. 1. We have AB + AC > BC. [A str. line is the shortest line between two points.] (Ax. 7) 2. Subtracting AC from both members of the inequality, AB > BC-AC. PROP. XI. THEOREM 63. The sumi of any two sides of a triangle is greater than the sum of the lines drawn from Cay point within the triangle to the extremities of the remaining side. C -- B C Draw A ABC. From D any point within the A draw lines DC and DB. Represent AC, AB, DC, and DB by b, c, d, and e, respectively. We then have: Given d and e lines drawn from any point D within A ABC to the extremities of side BC. To Prove b + c > d + e. Proof. 1. Extend BD to meet AC at E. 2. We have c + AE > e + DE. (1) [A str. line is the shortest line between two points.] (Ax. 7) 3. For same reason, CE + DE > d. (2) 4. Adding inequalities (1) and (2), and observing that AE + CE b, we have b + c + DE> d + e + DE. 5. Subtracting DE from both members of the inequality, b + c > d + e. 26 PLANE GEOMETRY-BOOK I Ex. 14. Given one side of a triangle, and the perpendicular drawn to it from the vertex of the opposite angle, to construct the triangle. Can more than one such triangle be drawn? Ex. 15. There are six elements in every triangle, -three sides and three angles. How many of these elements and what elements are needed that only one definite triangle can be constructed? Ex. 16. Triangles are to be formed by choosing any three lines from six given lines whose lengths are 2, 3, 4, 5, 6, 7 inches, respectively. How many triangles can you form and what kind of triangles are they? PROP. XII. THEOREM 64. If oblique lines be drauwn from a point to a straight line, I. Two oblique lines cutting off equal distances fromn the foot of the perpendicular from the point to the line are equal. II. Of two oblique lines cutting off unequal distances from the foot of the perpendicular from the point to the line, the more remote is the greater. A'"E_ I) D.F B I. From any point C without line AB, draw CD I AB (~ 25); take line ED = DF and draw lines CE (a) and CF (b). We then have: I. Given CD _ from point C to line AB; and lines a and b cutting off equal distances from the foot of CD. To Prove a = b. Proof. Since CD is _1 EF at its middle point D, a = b. [If a _ be erected at the middle point of a str. line, any point in the ~ is equally distant from the extremities of the line.] (~ 54) RECTILINEAR FIGURES 27 C C' II. Construct the figure in accordance with the statement. We then have: II. Given CD the I from point C to line AB; and CE (a) and CF (b) oblique lines from C to AB, cutting off unequal distances from the foot of CD; a being the more remote. To Prove a > b. Proof. 1. Produce CD to C', making C'D = CD, and draw lines C'E (c) and C'F (d). 2. Since by cons., AD is I. CC' at its middle point D, a = c, and b = d. [If a I be erected at the middle point of a str. line, any point in the _ is equally distant from the extremities of the line.] (~ 54) 3. But a + c > b + d. [The sum of any two sides of a A is > the sum of the lines drawn from any point within the A to the extremities of the remaining side.] (~ 63) 4. Substituting for c and d their equals a and b, respectively, 2 a > 2 b. 5. Dividing by 2, a > b. Note. The theorem holds equally if CE is on the opposite side of CD from CF. Ex. 17. Two sides of a triangle are 8 inches and 10 inches, respectively. Between what limits must the third side lie? 28 PLANE GEOMETRY -BOOK I PROP. XIII. THEOREM 65. If oblique lines be drazwn from a point to a straight line, I. Two equal oblique lines cut off equal distances from the.foot of the perpendicular from the point to the line. II. Of two unequal oblique lines, the greater cuts off the greater distance from the foot of the perpendicular fromt the point to the line. I. Draw line AB and take C any point without AB. Draw line CD I AB. With C as centre and radius greater than CD describe arc intersecting AD at E, and BD at F; draw lines CE (a) and CF (b). We then have: I. Given CD 1_ from point C to line AB, and a and b equal oblique lines from C to AB. To Prove ED = DF. Proof. 1. In rt. A CDE and CDF, CD = CD. 2. By hyp., a = b. 3. Then, A CDE = A CDF. [Two rt. A are equal when the hypotenuse and a leg of one are equal respectively to the hypotenuse and a leg of the other.] (~ 61) 4. Then, DE = DF. [In equal figures, the homologous parts are equal.] (~ 48) II. Draw line AB and take C any point without AB. Draw line CD I AB; draw lines CE (a) and CF (b) intersecting AD at E and F, respectively, making a> b. We then have: II. Given CD the _L from point C to line AB; and a and b unequal oblique lines from C to AB, a being > b. To Prove DE > DF. Proof. 1. WAe know that DE is either <, =, or > DF. 2. If we suppose DE < DF, a would be < b. [If oblique lines be drawn from a point to a str. line, of two oblique lines cutting off unequal distances from the foot of the L from the point to the line, the more remote is the greater.] (~ 64) RECTILINEAR FIGURES 29 3. But this is contrary to the hypothesis that a is >b; then DE cannot be < DF. 4. If we suppose DE - DF, a would be = b. [If oblique lines be drawn from a point to a str. line, two oblique lines cutting off equal distances from the foot of the I from the point to the line are equal.] (~ 64) 5. This is contrary to the hypothesis that a is > b; then DE cannot be = DF. 6. If DE cannot be < DF, nor = DF, we must have DE > DF. 66. Note I. The method of proof used in Prop. XIII is called the Indirect Method, or the Redulctio ad abslcrdum. We prove a proposition by making every possible supposition in regard to it; and showing that, in every case except the one we wish to prove, the supposition leads to something contrary to the hypothesis. 67. Note II. We may state Prop. XII, I, as follows: Hypothesis. If two oblique lines be drawn from a point to a straight line, cutting off equal distances from the foot of the perpendicular from the point to the line, Conclusion. They are equal. Again, we may state Prop. XIII, I, Hypothesis. If two equal oblique lines be drawn from a point to a straight line, Conclusion. They cut off equal distances from the foot of the perpendicular from the point to the line. We call one proposition the converse of another when the hypothesis and conclusion of the first are, respectively, the conclusion and hypothesis of the second. It follows from the above that Prop. XIII, I, is the converse of Prop. XII, I. Prop. XIII, II, is the converse of Prop. XII, II; also ~ 20 is the converse of ~ 17. 30 PLANE GEOMETRY-BOOK I PARALLEL LINES 68. Def. Two straight lines are said to be parallel (II) when they lie in the same plane, and cannot meet how- A B ever far they may be produced; as AB and CD. C D69. Ax. We assume that but one straight line can be drawn through a given point parallel to a given straight line. PROP. XIV. THEOREM 70. Two perpendiculars to the same straight line are parallel. Draw a line EG; at points A and C, respectively, on EG, draw lines ABand CD EG. We now have: Given lines AB and CD 1 to line AC. To Prove AB 11 CD. Proof. 1. If AB and CD are not II, they will meet in some point if sufficiently produced (~ 68). 2. We should then have two Is from this point to AC, which is impossible. [From a given point without a str. line, but one I can be drawn to the line.] (~ 59) 3. Therefore, AB and CD cannot meet, and are II. PROP. XV. THEOREM 71. Two straight lines parallel to the same straight line are parallel to each other. Draw line EF. Draw lines AB and CD II to EF (~ 70). We then have: Given lines AB and CD 11 to line EF. To Prove AB II CD. Proof. 1. If AB and CD are not II, they will meet in some point if sufficiently produced. (~ 68) 2. We should then have two lines drawn through this point II to EF, which is impossible. [But one str. line can be drawn through a given point 11 to a given str. line.] (~ 69) 3. Therefore, AB and CD cannot meet, and are II. RECTILINEAR FIGURES 31 PROP. XVI. THEOREM 72. A straight lin e perpendicular to one of two parallels is perpendicular to the other. A B C ----- Draw line AB 1I CD (~ 70); draw line AC 1 AB. We then have: Given lines AB and CD II, and line AC L AB. To Prove AC 1 CD. Proof. 1. If CD is not i AC, let line CE be I AC. 2. Then since AB and CE are I AC, CE II AB. [Two I to the same str. line are 11.] (~ 70) 3. But by hyp., CD 11 AB. 4. Then, CE must coincide with CD. [But one str. line can be drawn through a given point II to a given str. line.] (~69) 5. Now by cons., AC i C E. 6. Then since CE coincides with CD, we have AC 1 CD. DEFINITIONS 73. An exterior angle of a triangle is A the angle at any vertex formed by any side of the triangle and the adjacent side produced; as ACD. If any side of a triangle be taken and v C D called the base, we define the corresponding altitude as the perpendicular drawn from the opposite vertex to the base, produced if necessary. In general, any side may be taken as the base; but in an isosceles triangle, unless otherwise specified, the side which is not one of the equal sides is taken as the base. When any side has been taken as the base, we call the 32 PLANE GEOMETRY-BOOK I opposite angle the vertical angle, and its vertex the vertex of the triangle. Thus, in triangle ABC, BC is the base, AD the altitude, and BAC the vertical angle. 74. If two straight lines, AB and CD, are cut by a line EF, called a transversal, the angles are named as follows: c, cl, e, and f are called interiort angles, and a, b, g, and h exterior angles. c and f, or d and e, are called alternateAinterior angles. a and h, or b and g, are called alternateexterior angles. a and e, b and f, c and g, or d and h, are called corresponding angles. A E a/b B e/f D g/h D F PROP. XVII. THEOREM 75. If two parallels are cut by a transversal, the alternateinterior angles are equal. Draw line AB II CD (~ 70). Draw any line EF (not I AB) intersecting AB and CD at G and H, respectively. We then have: Given Ils AB and CD cut by transversal EF at points G and H, respectively, forming alt. int. As AGH (a) and GHD (c), also BGH (b) and CHG (d). To Prove Z a = Z c, and Zb - d. Proof. 1. Through K, the middle point of GtI, draw a line L AB, meeting AB at L, and CD at M. 2. Then, LMIJ CD. [A str. line 1 to one of two Ils is I to the other.] (~ 72) 3. In rt. A GKL and IIKM, by cons., hypotenuse GK = hypotenuse HIK. 4. Also, / GKL = / HKlif. [If two str. lines intersect, the vertical A are equal.] (~ 44) RECTILINEAR FIGURES 33 5. Then A GKIL- A HKIM. [Two rt. A are equal when the hypotenuse and an adj. Z of one are equal respectively to the hypotenuse and an adj. Z of the other.] (~ 60) 6. Then, / a = / c. [In equal figures, the homologous parts are equal.] (~ 48) 7. Again, Z a is the supplement of / b, and Z c of Z d. [If two adj. A have their ext. sides in the same str. line, their sum is equal to two rt. A.] (~ 17) 8. Then, since Z a = Z c, we have / b= / d. [The supplements of equal A are equal.] (~ 41) 76. We have Z a = Z BGE. (Fig. of Prop. XVII.) [If two str. lines intersect, the vertical A are equal.] (~ 44) Then, since / a = / c, we have / BGE = / c. [Things which are equal to the same thing are equal to each other.] (Ax. 1) In like manner, AGE = Z d, Z CIIF= a, and Z DHF = Zb. That is, if two parallels are cut by a transversal, the correspodcling angles are equal. 77. We have / a + Z b = two rt. As. (Fig. of Prop. XVII.) [If two adj. A have their ext. sides in the same str. line, their sum is equal to two rt. A.] (~ 17) Putting for / b its equal Z cl, we have / a + d c two rt. is. In like manner, / b + Z c = two rt. As. That is, if two parallels are cut by a transversal, the sum of the interior a.ngles on the same side of the transversal is equal to two right angles. \ Ex. 18. A line AB intersects line XY at Y 0, and through 0 any line MlV is drawn; if through points C and D on AB equidistant 0 D from 0, parallels to JLNh be drawn, the tri- X \ angles formed are equal. \ 34 PLANE GEOMETRY-BOOK I PROP. XVIII. THEOREM 78. (Converse of Prop. XVII.) If two straight lines are cat by a transversal, and the alternate-interior angles are equcal the two lines are parallel. A Ba E K --- —^-B F / -D F Draw line AB. Draw line EF cutting AB at G. Through any point H, in EF, draw line CD making GIHD (c) equal ZAGH (a). We then have: Given lines AB and CD cut by transversal EF at points G and H, respectively, and a = / c. To Prove AB II CD. Proof. 1. If CD is not II AB, draw line KL through H II AB. 2. Then since 11s AB and KL are cut by transversal EF, Za= / _GHL. [If two Ils are cut by a transversal, the alt. int. A are equal.] (~ 75) 3. But by hyp., Za = Lc. 4. Then, Z GHL = Lc. [Things which are equal to the same thing are equal to each other.] (Ax. 1) 5. But this is impossible unless KL coincides with CD. 6. Then, CD 11 AB. In like manner it may be proved that if AB and CD are cut by EF, and Z B GII = CHG, then AB 11 CD. RECTILINEAR FIGURES 35 79. (Converse of ~ 76.) If two straight lines are cut by a transversal, and the corresponding angles are equal, the two lines are parallel. Suppose Z a = Z CHF. (Fig. of Prop. XVIII.) Now, Zc= O CHF. [If two str. lines intersect, the vertical A are equal.] (~ 44) Then, La = Zc. [Things which are equal to the same thing are equal to each other.] (Ax. 1) Then, by ~ 78, AB 11 CD. In like manner we may prove that if Z BGE = c, or ZAGE = Z CHG, or Z BGH = Z DHF, then, AB 11 CD. 80. (Converse of ~ 77.) If two straight lines are cut by a transversal, and the sum of the interior angles on the same side of the transversal is equal to two right angles, the two lines are parallel. Suppose Z BGII + Z c = two rt. As. (Fig. of Prop. XVIII.) Then, both a and / c are supplements of Z BGH, and Z a - c. [The supplements of equal A are equal.].(~ 41) Then, by ~ 78, AB II CD. In like manner it may be proved that if La + Z CHG = two rt. A, then, AB II CD. Ex. 19. If two parallels are cut by a transversal, the alternate exterior angles are equal. Ex. 20. If two straight lines are cut by a transversal, and the alternate-exterior angles are equal, the two lines are parallel. 36 PLANE GEOMETRY-BOOK I PROP. XIX. THEOREM 81. Two angles whose sides are parallel, each to each, are equal if both pairs of parallel sides extendc in the same direction, or in opposite directions, from their vertices. K bE F /II Draw lines AB and BC II to lines DlI and KF, respectively, intersecting at E. We then have: Given Z DEF (a), with its sides 11 to and extending in the same direction as those of L B, and lZ IEK (b), with its sides II to and extending in the opposite direction to those of Z B. To Prove Z B= -Z a, and / B= Z b. Proof. 1. Let BC and DH intersect at G; denote Z DGC by c. 2. Since Ils AB and DE are cut by BC, ZB =Zc. [If two Ils are cut by a transversal, the corresp. A are equal.] (~ 76) 3. For same reason, since 11s BC and EF are cut by DE, L c = a. 4. Then, B = Za. [Things which are equal to the same thing are equal to each other.] (Ax. 1) 5. Again, / a = Z b. [If two str. lines intersect, the vertical A are equal.] (~ 44) 6. Then, ZB=Zb. [Things which are equal to the same thing are equal to each other.] (Ax. 1) Note. The sides extend in the same direction if they are on the same side of a straight line joining the vertices, and in opposite directions if they are on opposite sides of this line. RECTILINEAR FIGURES 37 82. We have Za the supplement of ZDE7K. (Fig. of Prop. XIX.) [If two adj. A have their ext. sides in the same str. line, their sum is equal to two rt. A.] (~ 17) Then its equal, Z B, is the supplement of Z DEK. That is, two angles whose sides are parallel, each to each, are supplementary if one pair of parallel sides extenc in the same direction, and the other pair in opposite directions, from their vertices. PnOP. XX. THEOREM 83. Two angles whose sides are perpendicular, each to each, are either equal or supplementary. A F /H a.Bi /C b. -------— K Draw line FG; also, line DE meeting FG at E. Draw lines AB and BC I to lines D.E and FG, respectively. We then have: Given As DEF (a), DEG (b); DE 1 AB, and FG 1 BC. To Prove Z B equal to Z a, and supplementary to Z b. Proof. 1. Draw line EH I DE, and line EK I EF; denote Z HEK by c. 2. Since EH and AB are _ DE, EH II AB. [Two Is to the same str. line are II.] (~ 70) 3. Since EK and BC are IL FG, E I1I BC. 4. Then, Z c = Z B. [Two A whose sides are 1I, each to each, are equal if both pairs of II sides extend in the same direction from their vertices.] (~81) 5. Since, by cons., A DEII and FEK are rt. As, each of the s a and c is the complement of Z FEI. 6. Then, a = Zc. [The complements of equal A are equal.] (~ 41) 38 PLANE GEOMETRY- B3OOK I 7. Then, ZB= a. [Things which are equal to the same thing are equal to each other.] (Ax. 1) 8. Again, Z a is the supplement of b. [If two adj. A have their ext. sides in the same str. line, they are supplementary. ] (~ 17) 9. Then, its equal, Z B, is the supplement of Z b. Note. The angles are equal if they are both acute or both obtuse; and supplementary if one is acute and the other obtuse. PROP. XXI. THEOREM 84. The sum of the congles of anyi,/ triacgle is equal to two right angles. _B A C /D Given A ABC. To Prove Z A + Z B +Z C= two rt. A. Proof. 1. Extend AC to D, and draw line CE 11 AB; represent Z ECD by a, and Z BCE by b. 2. We have Za + Z b + Z ACB = two rt. s. (1) [The sum of all the A on the same side of a str. line at a given point is equal to two rt. A. (~ 18) 3. Since lis AB and CE are cut by AD, Za= A. [If two lls are cut by a transversal, the corresp. A are equal.] (~ 76) 4. Since lls AB and CE are cut by BC, L b = Z B. [If two Ils are cut by a transversal, the alt. int. A are equal.] (~ 75) 5. Substituting in (1) Z A for / a and / B for / b, Z A + / B + Z ACCB = two rt. Z. RECTILINEAR FIGURES 39 85. We have, from ~ 84, ZBCD =Zca+1Zb=ZA+ZB. Hence, an exterior angle of a triangle is equal to the snum of the two opposite interior angles. The following are immediate consequences of ~~ 84, 85: 86. An exterior angle of a triangle is greater than either of the opposite interior cagles. 87. If two triangles have two angles of one equal, respectively, to two angles of the other, the third angle of the first is equal to the third angle of the second. 88. A triangle cannot have two right angles, nor two obtuse angles. 89. Two right triangles are equal when a leg and an acute angle of one are equal, respectively, to a leg and the homologous acute angle of the other. For the remaining acute angles are equal by ~ 87, and the triangles are equal by ~ 46. Ex. 21. Can a triangle be formed whose angles are 35~, 65~, and 55~, respectively? 85~, 50~, and 75~? 50~, 75', and 55~0? 45~ 11' 20", 61~ 52' 48", and 72~ 55' 52"? Ex. 22. The sum of one of the base angles of an isosceles triangle and the vertical angle equals 140~; find the degrees in the exterior angle formed by producing one of the equal sides through the vertex. PROP. XXII. THEOREM 90. (Converse of Prop. IV.) If two angles of a triangle are equal, the sides opposite are equal. Draw AABC with ZA = ZB. Represent BC by a and AC by b. Given, in A ABC, Z = Z B. To Prove a = b. The proof is left to the pupil.. Draw line CD1 AB. The A ACD and BCD are equal by ~ 89. 91. From equal A ACD and BCD (Fig. of Prop. XXII), AD = BD, and / ACD = Z BCD. Hence, the perpendicular from the vertex to the base of an, isosceles triangle bisects the base ald also the vertical angle. 40 PLANE GEOMETRY - BOOK I Pipor. XXIII. THEOREM 92. If two sides of a triangle are unejaal, the angles opposite a)re uneqtual, cod the grieater amgle lies oppIosite the greater side. Draw AABC, with BC (a) > AC (b). We then have: Given A ABG with side a > side b. To Prove Z BAC > / B. Proof. 1. On GB take GD - b, and draw line AD. 2. In isosceles A AGD, A GAD = Z GDA. [In an isoseeles A the As opposite the equal sides are equal.] (~ 60) 3. Since CDA is an ext. Z of A ABD, / CDA > ZB. [An ext. Z of a A is > either of the opposite int. A.] (~ 86) 4. Then its equal, Z CAD, is >Z B. 5. Now, Z BAG4C is > / CAD. 6. Then, Z B-A must be > / B. PROP. XXIV. THEOREM 93. (Converse of Prop. XXIII.) If two angles of a triangle cue unequal, the sides o2pposite are umequtal, and the greater side lies oppj~osite the grecater angle. C D 7a A -B Draw A ABC having Z BAC> > ZB. We then have: Given, in A ABG, ZBAGC > Z B. To Prove BC(a) > A1G(b). Proof. 1. Draw line A)D making / BAD = / B, meeting BC at D. RECTILINEAR FIGURES 41 2. In A ABD, AD = BD. [If two A of a A are equal, the sides opposite are equal.] (~ 90) 3. But CD+AD>b. [A str. line is the shortest line between two points.] (Ax. 7) 4. Putting for AD its equal BD, CD+- BD> b, or a> b. The following are immediate consequences of ~ 93: 94. The hypotenuse of a right triangle is its greatest side. 95. The perpendicular is the shortest line which A can be drawn from a point to a straight line. For, if AC is the I, and AB any other line, from A to CD, AB is the hypotenuse of rt. A ABC; and hence is > AC (~ 94). ~D B C' Ex. 23. Two isosceles triangles are equal, if their vertical angles and their bases are respectively equal. PROP. XXV. THEOREM 96. Two parallel lines are everywhere equally distant. Note. By the distance of a point from a line, we mean the length of the perpendicular from the point to the line. Draw lines AB and CD 11 (~ 70). At E and F, any two points in AB, draw lines EG and FHi CD. We then have: Given 11s AB and CD, and also EG (a), and FH (b) I CD. To Prove a = b. Proof. 1. Draw line FG. 2. We have a L AB. [A str. line I to one of two Ils is I to the other.] (~ 72) 3. Then, in rt. A EFG and FGH, FG = FG. 42 PLANE GEOMETRY-BOOK I 4. And since 11s AB and CD are cut by FG, Z EFG = Z FGH. [If two 11s are cut by a transversal, the alt. int. A are equal.] (~ 75) 5. Then A EFG=AFHG. [Two rt. A are equal when the hypotenuse and an adj. Z of one are equal respectively to the hypotenuse and an adj. Z of the other.] (~ 60) 6. Then a = b. [In equal figures, the homologous parts are equal.] (~ 48) PROP. XXVI. THEOREM 97. If straight lines be drawn from a point within a triangle to the extremities of any side, the angle included by them is greater than the angle included by the other two sides. A ci Draw any A ABC; from any point D within the A draw lines DB and DC. We then have: Given D, any point within A ABC; and lines DB and DC forming Z BDC (a). To Prove Z a > Z A. Proof. 1. Extend BD to meet AC at E; call Z DEC b. 2. Since a is an ext. Z of A CDE, Za> b. [An ext. Z of a A is > either of the opposite int. A.] (~ 86) 3. Since b is an ext. Z of A ABE, /b> ZA. 4. Since Z a > Z b, and Z b > L A, we must have a>ZA. RECTILINEAR FIGURES 43 PROP. XXVII. THEOREM 98. Any point in the bisector of an angle is equally distant from the sides of the angle. Construct any Z ABC. Draw BD, the bisector of Z ABC; from any point P in the bisector, draw lines PM (a) and PN (b) I AB and BC, respectively. We then have: Given P, any point in BD, the bisector of Z ABC, and a and b lines from P L AB and BC, respectively. To Prove a = b. Proof. 1. In rt. A BPM and BPN, BP = BP. 2. By hyp., Z PBM= ZPBX. 3. Then, A BPM = A BPN. [Two rt. A are equal when the hypotenuse and an adj. Z of one are equal respectively to the hypotenuse and an adj. Z of the other.] (~ 60) 4. Then P M= PN. [In equal figures, the homologous parts are equal.] (~ 48) PROP. XXVIII. THEOREM 99. (Converse of Prop, XXVII.) Every point which is within an angle and equally distant from its sides lies in the bisector of the angle. D B C Given point P within Z ABC, equally distant from sides AB and BC, and line BP. To Prove Z PBM = Z PBY. (Prove A BPM and BPN equal by ~ 61.) 44 PLANE GEOMETRY-BOOK I PROP. XXIX. THEOREM 100. If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, the third side of the first is greater than the third side of the second. C F b B E G Draw A ABC. Construct Z F< Z ACB. Draw lines FD (e) and FE (d) equal to CA (b) and CB (a), respectively. Draw line DE. We then have: Given, in A ABC and DEF, b = e, a = d, and / ACB > L F. To Prove AB > DE. Proof. 1. Place A DEF in position ACG; side e coinciding with its equal b, and vertex E falling at G. 2. Draw line CHlbisecting / GCB, meeting AB at H; also, line GH. 3. In A CGHI and CBH, CH= CH. 4. By hyp., d = a, and Z GCH= Z BCH. 5. Then, A CGH= A CBH. [Two A are equal when two sides and the included Z of one are equal respectively to two sides and the included Z of the other.] (~ 46) 6. Then, GH= BH. [In equal figures, the homologous parts are equal.] (~ 48) 7. Now, AHi+ GH > AG. [A str. line is the shortest line between two points.] (Ax. 7) 8. Putting for GH its equal BH, AH+- BH> AG, or AB > DE. RECTILINEAR FIGURES 45 PROP. XXX. THEOREM 101. (Converse of Prop. XXIX.) If two triangles have two sides of one equal resjpectively to two sides of the other, but the third side of the first greater than the third side of the second, the included angle of the first is greater than the included ca gle of the second. C F A c B D E Draw A ABC. Draw A DEF making side DF = AC, side EF = BC, and side DE (f) < AB (c). We then have: Given, in A ABC and DEF, AC= DF, BC= EF, and c >f. To Prove Z C > Z F. Proof. 1. Z Cmust be <, =, or > Z F. 2. If we suppose Z C = Z F, A ABC would equal A DEF. [Two A are equal when two sides and the included Z of one are equal respectively to two sides and the included Z of the other.] (~ 46) 3. Then, c would equal f. [In equal figures, the homologous parts are equal.] (~ 48) 4. Again, if we suppose Z C < Z F, c would be < f [If two A have two sides of one equal respectively to two sides of the other, but the included Z of the first > the included Z of the second, the third side of the first is > the third side of the second.] (~ 100) 5. Each of these conclusions is contrary to the hypothesis that c is > f. 6. If / C cannot be = Z F, nor < Z F, we must have Z C > Z F. Ex. 24. ABC is an isosceles triangle. Lines AD and CE, intersecting at 0, bisect the equal angles A and C. Prove A AOE = A DOG. 46 PLANE GEOMETRY-BOOK I Ex. 25. By drawing a line through the vertex of a triangle parallel to the base, prove that the sum of the angles of a triangle is equal to two right angles. Ex. 26. The line which joins the vertex of an isosceles triangle to the intersection of the bisectors of the exterior angles at the base is a perpendicular bisector of the base. Ex. 27. If one of the equal sides of an isosceles triangle be extended through the vertex, a line through the vertex parallel to the base is a bisector of the exterior angle at the vertex. Ex. 28. The bisectors of the base angles of an isosceles triangle meet in a point and form with the base an isosceles triangle. Ex. 29. Through the middle point of one of the equal sides of an isosceles triangle a line parallel to the other equal side is drawn to meet the base and the bisector of the exterior angle at the vertex. Prove the two triangles formed equal. Ex. 30. The bisectors of the base angles of an isosceles triangle form an angle of 110~. Find its vertical angle. Ex. 31. Prove Prop. XXIX when point E falls within triangle ABC. Ex. 32. The line which joins the vertex of an isosceles triangle to the intersection of the bisectors of the base angles, bisects the vertical angle. Ex. 33. The line which bisects the vertical angle of an isosceles triangle bisects the base at right angles. Ex. 34. Is it always possible to find a point in a given line which is equidistant from two given intersecting lines? Is there ever more than one such point? QUADRILATERALS DEFINITIONS 102. A quadrilateral is a portion of a plane bounded by four straight lines; as ABCD. B The bounding lines are called the sides of the quadrilateral, and their points of intersection the vertices. The angles of the quadrilateral are the / angles formed by the adjacent sides, ABC, BCD, etc. D RECTILINEAR FIGURES 47 A diagonal is a straight line joining two opposite vertices; as AC. 103. A Trapezium is a quadrilateral no two of whose sides are parallel. A Trapezoid is a quadrilateral two, and only two, of whose sides are parallel. A Parallelogram is a quadrilateral whose opposite sides are parallel. Trapezium Trapezoid Parallelogram An isosceles trapezoid is a trapezoid whose non-parallel sides are equal. The bases of a trapezoid are its parallel sides; the altitude is the perpendicular distance between them. If either pair of parallel sides of a parallelogram be taken and called the bases, the altitude corresponding to these bases is the perpendicular distance between them. A Rhomboid is a parallelogram whose angles are not right angles, and whose adjacent sides are unequal. A Rhombus is a parellelogram whose angles are not right angles, and whose adjacent sides are equal. A Rectangle is a parallelogram whose angles are right angles. A Squtare is a rectangle whose sides are equal. Rhomboid Rhombus Rectangle Square Ex. 35. Bisect the exterior angles of a given triangle and join the vertices of the triangle thus formed to the opposite vertices of the given triangle. The lines thus drawn are the altitudes of the triangle formed by the bisectors. 48 PLANE GEOMETRY -BOOK I PROP. XXXI. THEOREM 104. In any para7lelogrytm, the opposite sides are equal, and the opposite angles are equal. 'B 7.-, --- — ---- — 7 -Draw the figure in accordance with the statement of the proposition. We then have: Given L7 ABOD. To Prove AB = CD, BC = AD, Z B = D, and Z BAD = / BCD. Proof. 1. Draw diagonal AC; denote As BAC, BCA, ACD, and CAD by a, b, c, and d, respectively. 2. In A ABC and ACD, AC = AC. 3. Since 11s AB and CD are cut by AC, Za=Zc. (1) [If two Ils are cut by a transversal, the alt. int. A are equal.] (~ 75) 4. Since 11s BC and AD are cut by AC, Zb=Zd. (2) 5. Then, AABC A ACD. [Two A are equal when a side and two adj. A of one are equal respectively to a side and two adj. A of the other.] (~ 49) 6. Then, AB = CD, BC = AD, and Z.B = D. [In equal figures, the homologous parts are equal.] (~ 48) (Why are AB and CD, and BC and AD, homologous lines?) 7. Adding (1) and (2), we have Za+ Zb = c + Zd, orZBAD = ZBCD. The following are consequences of ~ 104: RECTILINEAR FIGURES 49 105. 1. Parallel lines included between parallel lines are equal. 2. A diagonal of a parallelogram divides it into two equal triangles. PROP. XXXII. THEOREM 106. If the opposite sides of a quadrilcteral are equal, the figure is a parallelogram. - C d D Draw quadrilateral ABCD having side BC AD, and side AB DC. We then have Given, in quadrilateral ABCD, AB = CD and BC-= AD. To Prove ABCD a D. Proof. 1. Draw diagonal AC; represent As BAC, BCA, ACD, and CAD by a, b, c, and cd, respectively. 2. In A ABC and ACD, U AC= AC. 3. By hyp., AB = CD and (BC = AD. 4. Then, A ABC = A ACD. [Two A are equal when the three sides of one are equal respectively to the three sides of the other.] (~ 53) 5. Then, Zac-= c, and Zb= d. [In equal figures, the homologous parts are equal.] (~ 48) 6. Since Z a = Z c, AB II CD. [If two str. lines are cut by a transversal, and the alt. int. A are equal, the two lines are II.] (~ 78) 7. Since b == Z d, BC II AD. 8. Then, by def., ABCD is a /1. Ex. 36. The perpendiculars from two opposite vertices of a parallelogram to a diagonal are equal. 50 PLANE GEOMETRY - BOOK I PROP. XXXIII. THEOREM 107. If two sides of a quadrilateral are equal and parallel, the figure is a parallelogram. B -- ---— C Draw line BC = and II to AD, and lines AB and DC. We then have: Given, in quadrilateral ABCD, BC equal and II to AD. To Prove ABCD a 17. (Prove A ABC and ACD equal by ~ 46; then, AB = CD. Compare ~ 106.) PROP. XXXIV. THEOIREMr 108. The diagonals of a parallelogram bisect each other. Draw 0 ABCD as in ~ 107; and lines AC and BD. We then have: Given diagonals AC and BD of 7 ABCD intersecting at E. To Prove AE = EC and BE = ED. (Prove A AED = A BEC, by ~ 49.) Note. The point E is called the centre of the parallelogram. PROP. XXXV. THEOREM 109. (Converse of Prop. XXXIII.) If the diagonals of a quadrilateral bisect each other, the figure is a parallelogracn. Given AC and BD, the diagonals of quadrilateral ABCD, bisecting each other at E. (Fig. of Prop. XXXIV.) RECTILINEAR FIGURES 51 To Prove ABCD a l, (Prove A AED = A BEC, by ~ 46; then AD = BC; in like manner, AB = CD; then use ~ 106.) PROP. XXXVI. THEOREM 110. Two parallelograms are equal when two adjacent sides and the included angle of one are equal respectively to two adjacent sides and the included angle of the other. B b cf s/ - " c '/ -g A / E 1h Draw L7 ABCD and EFGH in accordance with the statement of the proposition. Let a, b, c, d, represent sides AB, BC, CD, and DA, respectively; and e, f, g, h, sides EF, FGP, GH, and HlE, respectively. We then have: Given, in E7 ABCD and EFGH, a= e, d=h, and /A= ZE. To Prove ( ABCD = g EFGH. Proof. 1. Superpose D ABCD upon / EFGH in such a way that Z A shall coincide with its equal Z E; side a falling on side e, and side d on side h. 2. Since a = e, point B will fall on point F. 3. Since d = h, point D will fall on point H. 4. Since b I! d, and f II h, side b will fall on side f, and point C will fall somewhere on f. [But one str. line can be drawn through a given point II to a given str. line.] (~ 69) 5. Since c II a, and g II e, side c will fall on side g, and point C will fall somewhere on g. 6. Since C falls on both f and g, it must fall at their intersection, G. 7. Then, the LE coincide throughout, and are equal. 52 PLANE GEOMETRY.-BOOK I The following is an immediate consequence of ~ 110: 111. Two rectangles are equal if the base and altitude of one are equal respectively to the base and altitude of the other. PROP. XXXVII. THEOREM 112. Thie diagonals of a rectangle are equal. Draw rectangle ABCD; draw lines AC and BD. We then have: Given AC and BD the diagonals of rect. ABCD. To Prove AC= BD. (Prove rt. A ABD = rt. A ACD, by ~ 46.) The following is an immediate consequence of ~ 112: 113. The diagonals of a square are equal. PROP. XXXVIII. THEOREM 114. The diagonals of a rhombus bisect each other at right angles. G Given AC and BD the diagonals of rhombus ABCD. To Prove that AC and BD bisect each other at rt. As. (Compare ~ 56.) Ex. 37. The lines which join the middle points of the opposite sides of a parallelogram bisect each other. Ex. 38. Are the diagonals of a parallelogram ever equal? Ex. 39. When do the diagonals of a parallelogram bisect the opposite angles? Ex. 40. If perpendiculars be drawn from the extremities of the upper base of an isosceles trapezoid to the lower base, they are equal and cut off equal segments from the extremities of the lower base. RECTILINEAR FIGURES 53 Ex. 41. The lines which join the middle points of the bases of an isosceles trapezoid to the middle points of the non-parallel sides form, with the sides of the trapezoid, two pairs of equal triangles. Ex. 42. If lines be drawn from the middle point of the base of an isosceles triangle to the middle points of the equal sides, the triangle is divided into two equal triangles and a parallelogram. POLYGONS DEFINITIONS 115. We define a polygon as a portion of a plane bounded by three or more straight lines; as ABCDE. B We call the bounding lines the sides of / - the polygon, and their sum the perimeter. The angles of the polygon are the AX D angles EAB, ABC, etc., formed by the adjacent sides; their vertices are called the E vertices of the polygon. A diagonal of a polygon is a straight line joining any two vertices which are not consecutive; as AC. 116. Polygons are named with reference to the number of their sides, as follows: No. OF No. OF SIDES. DESIGNATION. SIDES. DESIGNATION. 3 Triangle. 8 Octagon. 4 Quadrilateral. 9 Enneagon. 5 Pentagon. 10 Decagon. 6 Hexagon. 11 Hendecagon. 7 Heptagon. 12 Dodecagon. 117. An equilcateral polygon is a polygon all of whose sides are equal. An equiangular polygon is a polygon all of whose angles are equaL 54 PLANE GEOMETRY-BOOK I 118. A polygon is called convex when B no side, if extended, will enter its surface; as ABCDE; in such a case, each angle of the polygon is less than two right angles.. Every polygon considered hereafter will be understood to be convex, unless the - contrary is stated. 119. A polygon is called concave when at G least two of its sides, if extended, will enter \i its surface; as FGHIIK. In such a case, at least one angle of the _ polygon is greater than two right angles. F K Thus, in polygon FGTIlK, the interior angle GIII is greater than two right angles; such an angle is called re-entrant. 120. We call two polygons mutually equilateral when the sides of one are equal re- C' spectively to the sides of the C B' other, when taken in the same B order. Thus, polygons ABCD and A D A' A'B'C'D' are mutually equilateral if AB = A'B', BC = B'C', CD = C'D', and DA = D'A'. 121. We call two polygons mutually equiangular when the angles of one are equal respec- c tively to the angles of the other, when taken in the same order. Thus, polygons EFGH and - D E H E'F'G'H' are mutually equiangular if ZA= ZE, B = =ZF, Z C=Z/ G, and Z D= H. 122. In polygons which are mutually equilateral or mutually equiangular, sides or angles which are similarly placed are called homologous, RECTILINEAR FIGURES 55 123. If two polygons are both mutually equilateral and mutually equtiangular, they are equal. For they can evidently be applied one to the other so as to coincide throughout. 124. Two polygons are equal when they are composed of the same number of triangles, equal each to each, and similarly placed. For they can evidently be applied one to the other so as to coincide throughout. PRop. XXXIX. THEOREM 125. The sum of the angles of any polygon is equal to two right angles taken as many times, less two, as the polygon has sides. Draw any polygon. We then have: Given a polygon of n sides. To Prove the sum of its A equal to n -2 times two rt. Zs. Proof. 1. The polygon may be divided into n - 2 A by drawing diagonals from one of its vertices. 2. The sum of the A of the polygon is equal to the sum of the A of the A. 3. Now the sum of the A of each A is two rt. As. [The sum of the A of any A is equal to two rt. A.] (~ 84) 4. Then, the sum of the As of the polygon is n - 2 times two rt. ZA. 126. We have (n - 2) X 2 rt. As equal to 2 n rt. A - 4 rt. A. Then, the sum of the angles of any polygon equals twice as many right angles as the polygon has sides, less four right angles. 56 PLANE GEOMETRY- BOOK I PROP. XL. THEOREM 127. If the sides of any polygon be produced so as to form an exterior angle at each vertex, the sum of these exterior angles is equal to four right angles. Draw a polygon in accordance with the statement of the proposition. We then have: Given a polygon of n sides with its sides extended so as to form an ext. / at each vertex. To Prove the sum of the ext. As equal to 4 rt. As. Proof. 1. The sum of the ext. and int. A at each vertex is two rt. A. [If two adj. A have their ext. sides in the same str. line, their sum is equal to two rt. A.] (~ 17) 2. Hence, the sum of all the ext. and int. As is 2 n rt. A. 3. But the sum of the int. A alone is 2 n rt. A - 4 rt. A. [The sum of the A of any polygon equals twice as many rt. A as the polygon has sides, less 4 rt. A.] (~ 126) 4. WIhence, the sum of the ext. A is 4 rt. As. Ex. 43. Find the sum of the angles of a quadrilateral; of a pentagon; of a hexagon; of an octagon. Ex. 44. An exterior angle of an equiangular polygon is one-fifth of a right angle. How many sides has the polygon? Ex. 45. How many sides has a polygon each of whose interior angles is eight-fifths of a right angle? Ex. 46. The difference between two consecutive angles of a parallelogram is 75~; find the angles of the parallelogram. Find the angles when this difference is 10~. RECTILINEAR FIGURES 57 Ex. 47. How many sides has the polygon the sum of whose interior angles is 14 right angles? Find the magnitude of each angle if the polygon is equiangular. MISCELLANEOUS THEOREMS PROP. XLI. THEOREM 128. If a series of parallels, cutting two straight lines, intercept equal distances on one of these lines, they also intercept equal distances on the other. C X A ' Draw a series of parallels (~ 70). Draw lines AB and A'B' cutting these 11s at C, D, E, F, and C', D', El, F', respectively. We then have: Given lines AB and Ai'B' cut by Ils CC', DD', EE', and FF' at points C,, D, EF, and C', D', E', F', respectively, so that CD =DE = EF. To Prove C'D' = D'E' = E'F'. Proof. 1. Draw lines CG, Df, and EK I1 A'B', meeting DD', EE, and FF' at points G, H, and K, respectively. 2. In A CDG, DEH, EFK, by hyp., CD = DE= EF. 3. Again, lines CG, DH, and EK are II to each other. [Two str. lines 11 to the same str. line are 11 to each other.] (~ 71) 4. Then, Z DCG = / EDH= Z FEK. [If two 11 lines are cut by a transversal, the corresp. A are equal.] (~ 76) 5. Also, Z CDG = ZDEH= EFKt. [If two II lines are cut by a transversal, the corresp. s are equal.] (~ 76) 58 PLANE GEOMETRY -BOOK I 6. Then, A CDG = A DET = A EFI. A A' [Two A are equal when a side and two adj. A of Cf C' one are equal respectively to a side and two adj. A i D' of the other.] (~ 49) E' 7. Then, CG = DH = EIK. l [In equal figures, the homologous parts are equal.] (~ 48) 8. But, CG = C'D', DH = D'E', and EK = E'F'. [In any 7, the opposite sides are equal.] (~ 104) 9. Then, C'D' = D'E' = E'F'. [Things which are equal to equal things, are equal to each other.] (Ax. 1) 129. The line zwhich bisects one side of a t)riangle, and is parctllel to another side, bisects also the third side. If line DE is II side BC of A ABC, and A bisects side AB, it also bisects AC. This is the particular case in the figure of Prop. XLI where CG coincides with C'D'. B - 130. In the figure of Prop. XLI, CC'E'E is a trapezoid, and DD' bisects its sides CE and C'E'. Hence, the line which is parallel to the bases of a trapezoid and bisects one of the non-parallel sides, bisects the other also. PROP. XLII. THEOREM 131. The line joining the mniddle points of twzo sides of a triangle is parallel to the third side, and equal to one-half of it. Draw any A ABC. Through D and E, middle points of AB and AC, respectively, draw line DE. We then have: Given line DE joining middle points of sides AB and AC, respectively, of A ABC. To Prove DE II BC, and DE = - BC. Proof. 1. A line from D II BC will bisect AC. [The line which bisects one side of a A, and is 11 to another side, bisects also the third side.] (~ 129) RECTILINEAR FIGURES 59 2. Then it must coincide with DE, and DE 1\ BC. [But one str. line call be drawn between two points.] (Ax. 6) 3. Draw line EF II AB, meeting side BC at F. 4. In A ADE and EFC, by hyp., AE = EC. 5. Also, / A = L CEF, and Z AED = Z C. [If two 11s are cut by a transversal, the corresp. A are equal.] (~ 76) 6. Then, A ADE = A EFC. [Two A are equal when they have a side and two adjacent A of one equal respectively to a side and two adjacent A of the other.] (~ 49) 7. Then, DE = FC. [In equal figures, the homologous parts are equal.] (~ 48) 8. But, DE = BF. [In any 0, the opposite sides are equal.] (~ 104) 9. Then, DE = - BC. PROP. XLIII. THEOREM 132. The line joining the middle points of the non-parallel sides of a trapezoid is parallel to the bases, and equal to one-half their sum. B CY E/, E-/ - -\ Draw. trapezoid ABCD, BC and AD being II sides. Let E, F be middle points of AB and CD, respectively. Draw line EF. We then have: Given line EF joining middle points of non-i] sides AB and CD, respectively, of trapezoid ABCD. To Prove EF 1[ to AD and BC, and EF= 1 (AD + BC). Proof. 1. A line from E 11 to AD and BC will bisect CD. [The line which is II to the bases of a trapezoid, and bisects one of the non-Il sides, bisects the other also.] (~ 130) 60 PLANE GEOMETRY - BOOK I 2. Then it coincides with EF, and EF is ii to AD and BC. [But one str. line can be drawn between two points.] (Ax. 6) 3. Draw diagonal AC, intersecting EF at G. B ---- X' 4. Since EG bisects side AB of A ABC, E7F and is II BC, it bisects side AC. A [The line which bisects one side of a A, and is 11 to another side, bisects also the third side.] (~ 129) 5. Then, EG = BC (1) [The line joining the middle points of two sides of a A is 11 to the third side, and equal to one-half of it.] (~ 131) 6. In like manner, GF= 1 AD. (2) 7. Adding (1) and (2), EF= ~ BC + AD = - (BC + AD). Ex. 48. A straight line is drawn from the vertex of a triangle to any point in the base. Prove that this line is bisected by the line joining the middle points of the sides of the triangle. Ex. 49. E and F are middle points of sides B E C BC and AD, respectively, of the parallelogram ABCD. Draw lines AE and CF. Prove that these lines trisect the diagonal joining the other A --- D two vertices of the parallelogram. PROP. XLIV. THEOREM 133. The bisectors of the angles of a triangle intersect at a common point. Draw any A ABC. Let a, b, c, be the bisectors of A A, B, and C, respectively. We then have: Given a, b, and c, the bisectors of the angles of A ABC. To Prove that a, b, and c intersect at a common point. Proof. 1. Let a and b intersect at 0. 2. Since 0 is in bisector a, it is equally distant from sides AB and AC. [Any point in the' bisector of an Z is equally distant from the sides of the Z.] (~ 98) RECTILINEAR FIGURES 61 3. Since 0 is in bisector b, it is equally distant from sides AB and BC. 4. Then 0 is equally distant from sides AC and BC, and therefore lies in bisector c. [Every point which is within an Z, and equally distant from its sides, lies in the bisector of the Z.] (~ 99) 5. Then, a, b, and c intersect in the common point 0. 134. It follows from ~ 133 that The point of intersection of the bisectors of the angles of a triangle is equally distant from the sides of the triangle. Ex. 50. The lines joining the middle points of the bases of an isosceles trapezoid to the middle points of the non-parallel sides form a rhombus. Ex. 51. If through the vertex A, of isosceles triangle ABC, a parallel to BC be drawn, and through the middle points of AB and AC parallels, respectively, to AC and AB, a triangle is formed equal to the given triangle, and its vertex will lie in the base of the given triangle. Ex. 52. Two sides of a quadrilateral are parallel. The other sides are equal, but not parallel. Prove the opposite angles supplementary. PROP. XLV. THEOREM 135. The perpendiculars erected at the middle points of the sides of a triangle intersect at a common point. Draw any A ABC; at D, E, and F, middle points of BC, CA, and AB, respectively, of A ABC, draw lines DG, EH, and FKI to BC, CA, and AB, respectively. We now have: Given DG, EH, and FI, the _s erected at the middle points of the sides of A ABC. To Prove that DG, EH, and FK intersect at a common point. (If DG and EHi intersect at 0, 0 is equally distant from B and C, by ~ 54; and also from A and C.) 136. It follows from ~ 135 that The point of intersection of the perpendiculars erected at the middle points of the sides of a triangle, is equally distant from the vertices of the triangle. PLANE GEOMETRY - BOOK I PROP. XLVI. THEOREM; 137. The perpendiculars from the vertices of a triangle to the opposite sides intersect at a conmmon point. A s- ----- A,......- ss F\ B D G Draw any A ABC. From A, B, and C draw lines AD (a), BE (b), and CF (c) 1- to BC, CA, and AB, respectively. We then have: Given a, b, and c, the three Is from the vertices of A ABC to the opposite sides. To Prove that a, b, and c intersect at a common point. Proof. 1. Through A, B, and C draw lines IlKI, KG, and GH II B C, and AB, respectively, forming A GIIK. 2. Since a is I BC, it is also 1 HK. [A str. line I to one of two lis is L to the other.] (~ 72) 3. Since, by cons., ABCH and ACBKi are L7, AH= BC and AK== BC. [In any ~7, the opposite sides are equal.] (~ 104) 4. Then, ' AH= AK. [Things which are equal to the same thing, are equal to each other.] (Ax. 1) 5. Then, a is 1_ IIK at its middle point. 6. In like manner, b and c are I_ to ]KG and GH, respectively, at their middle points. 7. Since a, b, and c are -_ to the sides of A GHK at their middle points, they intersect at a common point. [The Is erected at the middle points of the sides of a A intersect at a common point.] (~ 135) RECTILINEAR FIGURES 138. Def. A median of a triangle is a line drawn from any vertex to the middle point of the opposite side. PROP. XLVII. THEOREM 139. The line drawn from any vertex of a triangle through the point of intersection of medians drawn from the other two vertices, is itszlf a median. Draw A ABC; draw medians BE and AD intersecting at 0; through O draw line CO meeting AB at F. We then have: Given AD and BE, two medians of A ABC intersecting at 0. To Prove CF also a median, or AF = BF. Proof. 1. Draw line AG II BE, meeting CF prolonged at G; also, line BG. 2. Since EO is II to side AG of A ACG, and bisects side CA, 0 is the middle point of CG. [The line which bisects one side of a A, and is II to another side, bisects also the third side.] (~ 129) 3. Since 0 is the middle point of CG, and D of CB, OD is 11 to side BG of A BCG. [The line joining the middle points of two sides of a A is II to the third side.] (~ 131) 4. Since A 11 BE, and AO II BG, AOBG is a E, and AF- =BF. [The diagonals of a 0 bisect each other.] (~ 108) 140. In the figure of Prop. XLVII, OF = FG. (~ 108) Since 0 is the middle point of CG, OG = OC. Then, OF= 1 OG = 10 C; and hence OC= - CF. In like manner, OA = AD, and OB = BE. That is, the point of intersection of the medians of a triangle lies two-thirds the way from each vertex to the opposite side. 141. As an aid in the solution of original exercises, the following list of certain principles proved in Book I will be found useful. 64 PLANE GEOMETRY-BOOK I Two triangles are equal: When two sides and the included angle of one are equal respectively to two sides and the included angle of the other (~ 46). When a side and two adjacent angles of one are equal respectively to a side and two adjacent angles of the other (~ 49). When the three sides of one are equal respectively to the three sides of the other (~ 52). Two right triangles are equal: When the hypotenuse and an adjacent angle of one are equal respectively to the hypotenuse and an adjacent angle of the other (~ 60). When the hypotenuse and a leg of one are equal respectively to the hypotenuse and a leg of the other (~ 61). When a leg and an acute angle of one are equal respectively to a leg and the homologous acute angle of the other (~ 89). Two straight lines are equal: When they are homologous sides of equal triangles (~ 48). When they are opposite equal angles in a triangle (~ 90). When they are opposite sides of a parallelogram (~ 104). When they are diagonals of a rectangle (~ 112). Two angles are equal: When they are the complements or supplements of equal angles (~ 41). When they are vertical (~ 44). When they are homologous angles of equal triangles (~ 48). When they are opposite the equal sides of an isosceles triangle (~ 50). When they are alternate-interior angles (~ 76). When they are corresponding angles (~ 76). When their sides are parallel each to each (~ 81). When their sides are perpendicular each to each (~ 83). When they are opposite angles of a parallelogram (~ 104). Two lines are parallel: When they are perpendicular to the same line (~ 70). When they are parallel to the same line (~ 71). When the alternate-interior angles are equal (~ 78). When the corresponding angles are equal (~ 79). When the sum of the interior angles on the same side of the transversal is equal to.two right angles (~ 80). When they are opposite sides of a parallelogram. When one line is a side of a triangle, and the other joins the middle points of the other two sides (~ 131). RECTILINEAR FIGURES 65 A quadrilateral is a parallelogram: When the opposite sides are equal (~ 106). When two sides are equal and parallel (~ 107). When the diagonals bisect each other (~ 109). One line is greater than another: When it is the hypotenuse of a right triangle of which the other line is a leg (~ 94). When it is opposite a greater angle in a triangle (~ 93). One angle is greater than another: When it is an exterior angle of a triangle in which the other is an opposite interior angle (~ 86). When it is opposite a greater side in a triangle (~ 92). Ex. 53. Is it always possible to find a point equidistant from three given straight lines? Ex. 54. Is it possible to find a point equidistant from four given straight lines? Ex. 55. If two medians of a triangle intersect the sides at right angles, the triangle is equiangular. Ex. 56. If a line joining two opposite vertices of a parallelogram bisects the angles at these vertices, the parallelogram is equilateral. Ex. 57. Prove that lines drawn from two vertices of a triangle and terminating in the opposite sides cannot bisect each other. BOOK II THE CIRCLE DEFINITIONS 142. A diameter of a circle is a straight line drawn through the centre, having its extremities in the circumference. A 143. By the definition of ~ 23 All radii of a circle are equal. Also, all its diameters are equal, since each is the sum of two radii. 144. Two circles are equal when their radii are equal. For they can evidently be applied one to the other so that their circumferences shall coincide throughout. 145. The radii of equal circles are equal. (Converse of ~ 144.) 146. Concentric circles are circles having the same centre. 147. A chord is a straight line joining the extremities of an arc; as AB. The are is said to be subtended by its chord. Every chord subtends two arcs; thus chord A/B\B AB subtends arcs AMB and ACDB. When the arc subtended by a chord is / spoken of, the one which is less than a semicircumference is understood, unless the con- C trary is specified. A segment of a circle is the portion included between an arc and its chord; as AMBN. 66 THE CIRCLE A sector of a circle is the portion included between an arc and the radii drawn to its extremities; as OCD. 148. A straight line cannot meet a circumference at more than two points. For by ~ 64, not more than two equal straight lines can be drawn from a point (in this case, the centre) to a straight line. 149. A central angle is an angle whose vertex is at the centre, and whose sides are radii; as AOC. An inscribed angle is an angle whose vertex is on the circumference, and whose sides are B chords; as ABC. An angle is inscribed in a segment when its vertex is on the arc of the segment, and its sides pass through the extremities of \ the subtending chord; thus, angle B is inscribed in segment ABC. 150. A straight line is said to be tangelt to, or to touch, a circle when it has but one point in common with the circumference; as AB. In such a case, the circle is said to be tangent to the straight line. The common point is called the point D of contact. A secant is a straight line which intersects the circumference in two A B points; as CD. 151. Two circles are said to be tcagent to each other when they are tangent to the same straight line at the same point. They are said to be tangent internally or externally according as one circle lies entirely within or entirely without the other. A common tangent to two circles is a straight line which is tangent to both of them. 68 PLANE GEOMETRY-BOOK II 152. A polygon is said to be inscribed in B a circle when all its vertices lie on the circumference; as ABCD. In this case, the circle is said to be circumtscribed about the polygon. A C A polygon is said to be inscriptible when D it can be inscribed in a circle. A polygon is said to be circumscribed F about a circle when all its sides are tangent to the circle; as EFGH. In this case, the circle is said to be inscribed in the polygon. G PRop. I. THEOREM 153. Every diameter bisects the circle and its circumference. B D Given AC(d) a diameter of 0 ABCD. To Prove that d bisects the 0, and its circumference. Proof. 1. Superpose segment ABC upon segment ADC, by folding it over about d as an axis. 2. Then, arc ABC will coincide with arc ADC; for otherwise there would be points of the circumference unequally distant from the centre. 3. Hence, segments ABC and ADC coincide throughout, and are equal. 4. Therefore, d bisects the 0, and its circumference. THE CIRCLE 69 154. Defs. A semi-circumference is an arc equal to one-half the circumference. A quadrant is an arc equal to one-fourth the circumference. A semicircle is a segment equal to one-half the circle. Ex. 1. If two non-perpendicular diameters be drawn in the same circle, the perpendiculars from the extremities of the one upon the other are equal. PROP. II. THEOREM 155. In equal circles, or in the same circle, equal central angles intercept equal arcs on the circumference. Md h g c s C ad C' Draw I an AMB B', hdaving centres C and ', respectively, and same radius. In (3 AMB, draw Z ACB; in 0 AtM'B', construct Z A'C'B' = ACB. We then have: Given ACB and A'C'B' equal central As of equal ~ AMB and A'M'B', respectively. To Prove arc AB= arc A'B'. Proof. 1. Superpose sector ABC upon sector A'B'C' in such a way that L C shall coincide with its equal Z C'. 2. We have AC= 1A'C' and BC= B'C'. (~ 145) 3. Then, point A will fall at A', and point B at B'. 4. Then, arc AB will coincide with arc A'B'; for all points of either are equally distant from the centre. 5. Whence, arc AB = arc A'B'. Ex. 2. Draw lines dividing a circumference into eight equal parts. 70 PLANE GEOMETRY -BOOK II PROP. III. THEOREM 156. (Converse of Prop. II.) In equal circles, or in the same circle, eqtual arcs are intercepted by equal central angles. M M' Draw ~ AMB and A'M'B', having centres C and C', respectively, and same radius. Take arc AB= arc A'B', and draw lines AC, BC, A' C, and B' '. We then have: Given ACB and A'C'B' central As of equal ( AMB and A'M'B', respectively, and arc AB = arc A'B'. To Prove L C= z C'. Proof. 1. Since the ) are equal, we may superpose 0 AMB upon 0 A'M'B' in such a way that point A. shall fall at A', and centre C at C'. 2. Since arc AB = arc A'B', point B will fall at B'. 3. Then, radii AC and BC will coincide with radii A'C' and B'C', respectively. (Ax. 6) 4. Then, / C will coincide with Z C'; that is, Z C= Z C'. 157. Note. In equal circles, or in the same circle, 1. The greater of two central angles intercepts the greater arc on the circumference. 2. The greater of two arcs is intercepted by the greater central an7gle. PROP. IV. THEOREM 158. In equal circles, or in the same circle, equal chords subtend equal arcs. Draw ~ AMB and A'MI'B', having same radius. In 0 AMB, draw chord AB; il 0 A'/M'B', draw chord A'B' = chord AB. We then have: Given, in equal ( AMB, A'M'B', chord AB chord A'B'. THE CIRCLE 71 To Prove arc AB = arc A'B'. Proof. 1. Draw radii AC, BC, A'C', and B'C'. 2. In A ABC and A'B'C', by hyp., AB = A'B'. 3. Also, AC = A'C', and BC = B'C'. (~ 145) 4. Then, A ABC= A A'B'C'. (~ 52) 5. Then, Z C=Z C'. (?) 6. Then, arc AB = arc A'B'. (~ 155) Ex. 3. If A, B, C, D are points in succession on a circumference, and arcs AB, BC, and CD are equal, prove chord AC = chord BD. PROP. V. THEOREM 159. (Converse of Prop. IV.) In equal circles, or in the same circle, equal arcs are subtended by equal chords. Given, in equal ( AMB and A'M'B', arc AB = arc A'B'; and chords AB and A'B'. (Fig. of Prop. IV.) To Prove chord AB = chord A'B'. (The A ABC and A'B'C' are equal by ~ 46.) PROP. VI. THEOREM 160. In equal circles, or in the same circle, the greater of two arcs is subtended by the greater chord; each arc being less than a semi-circumference. Draw O AMB and A'3liB', having same radius, and centres at C and C', respectively. Take arc AB > arc A'B', each arc < a semi-circumference. Draw lines AB and A'B'. We then have: Given, in equal ~ AMB and A'M'B', arc AB > arc A'B', each arc < a semi-circumference, and chords AB and A'B'. To Prove chord AB > chord A'B'. Proof. 1. Draw radii AC, BC, A'C', and B'C'. 2. In A ABC and A'B'C', AC = A'C', and BC = B'C'. (?) 3. Since, by hyp., arc AB > arc A'B', C> > C'. (~157, 2) 4. Then, chord AB > chord 'B'. (~ 100) 72 PLANE GEOMETRY-BOOK II PROP. VII. THEOREM 161. (Converse of Prop. VI.) In equal circles, or in the same circle, the greater of tzwo chords subtends the greater arc; each arc being less than a semi-circumference. Given, in equal ( AMB and A'M'B', chord AB > chord A'B', arcs AB and A'B' being < a semi-circumference. (Fig. of Prop. VI.) To Prove arc AB > arc A'B'. (We have Z C > Z C', by ~ 101; then use ~ 157, 1.) Ex. 4. A central angle AOB is bisected by a radius OC. Let Al, C', B' be the middle points of OA, OC, and OB, respectively. Prove the length of the broken line A' C'B' equal to the length of the chord AC. PROP. VIII. THEOREM 162. The diameter perpendicular to a chord bisects the chord and its subtended arcs. D Draw ( ABD with centre at 0; draw diameter D C; through point E on OC draw chord AB L D C. We then have: Given, in 0 A BD, diameter CD 1L chord AB. To Prove that CD bisects chord AB, and A arcs ACB and ADB. C Proof. 1. Let 0 be centre of 0; draw radii OA and OB. 2. Since OA = OB, A OAB is isosceles. 3. Then, CD bisects AB, and Z AOB. (~ 91) 4. Since Z AOC = / BOC, arc AC = arc BC. (~ 155) 5. Again, Z AOD = BOD. (~ 41, 2) 6. Then, are AD = are BD. (?) 163. It follows from the above that the perpendicular erected at the middle point of a chord passes through the centre of the circle, and bisects the arcs subtended by the chord. THE CIRCLE 73 PROP. IX. THEOREM 164. In the same circle, or in equal circles, equal chords are equally distant from the centre. Draw a 0 with centre at 0; draw equal chords AB and CD. Draw lines OE and OFI AB and CD, respectively, and meeting these lines at E and F, respectively. We then have: Given AB and CD equal chords of ( ABC, whose centre is 0, and lines OE and OF I AB and CD, respectively. To Prove OE = OF. Proof. 1. Draw radii OA and OC. 2. In rt. A OAE and OCF, OA = OC. (?) 3. E is the middle point of AB, and F of CD. (~ 162) 4. Then, AE = CF; for they are halves of equal chords AB and CD, respectively. 5. Then, A OAE = A OCF. (?) 6. Then, OE = OF. (?) Ex. 5. If two equal chords, not diameters, intersect within a circle, and perpendiculars be drawn from the centre to these chords, the line joining the centre of the circle with the intersection of these chords is a bisector of the angle formed by the perpendiculars. Ex. 6. If two circles are concentric, and a chord of the greater is a secant of the smaller, then the portions of the chord between the two circumferences are equal. Ex. 7. The line which bisects a chord of a circle and its subtended arc will, if extended, pass through the centre of the circle. PROP. X. THEOREM 165. (Converse of Prop. IX.) In the same circle, or in equal circles, chords equally distant from the centre are equal. Given 0 the centre of ( ABC, and AB and CD chords equally distant from 0. (Fig. of Prop. IX.) To Prove chord AB = chord CD. (The rt. A OAE and OCF are equal (?), and AE = CF; E is the middle point of AB, and F of CD.) 74 PLANE GEOMETRY-BOOK II PROP. XI. THEOREM 166. In the same circle, or in equal circles, if two chords are unequally distant fiom the centre, the more remote is the less. D Draw figure in accord- H_ __-1_ —X ance with the statement. E - We then have: Given 0 the centre of 0 ABC, and chord AB more remote from 0 than chord CD. To Prove chord AB < chord CD. Proof. 1. Draw lines OG l AB and OH11 CD; on OG take OK= OH, and through K draw chord EF_1 OK. 2. Then, chord EF = chord CD. (~ 165) 3. Now, chord AB II chord EF. (~ 70) 4. Then, arc AB must be < arc EF. 5. Then, chord AB < chord EF. ( 160) 6. That is, chord AB < chord CD. 167.' By ~ 166, a diameter is greater than any other chord. PROP. XII. THEOREM 168. (Converse of Prop. XI.) In the same circle, or in equal circles, the less of two chords is at the greater distance from the centre. Draw 0 with centre at 0; draw chords a and b, a being < b, their 1 distances from 0 being c and d, respectively. We then have: Given, in 0 whose centre is 0, chords a and b, respectively, at distances c and d from 0, and chord a < chord b. THE CIRCLE 75 To Prove c > d. Proof. 1. If c is < d, chord a will be > chord b. (~ 166) 2. If c= d, chord a will equal chord b. (~ 165) 3. Both of these results are contrary to the hyp. that chord a is < chord b. 4. Then, c > d. PROP. XIII. THEOREM 169. A straight line perpendicular to a radius of a circle at its extremity is tangent to the circle. Draw a 0 with centre at 0, and radius 00. Through point C draw line AB 1 OC. We then have: Given line AB 1 radius OC of 0 O at C. To Prove AB tangent to the 0. Proof. 1. Let D be any point of AB except C. 2. Draw line OD; then, OD > OC. (?) 3. Therefore, point D lies without the 0. 4. Then, every point of AB except C lies without the 0, and AB is tangent to the 0. (~ 150) PROP. XIV. THEOREM 170. (Converse of Prop. XIII.) A tangent to a circle is perpendicular to the radius drazw to the point of contact. A i B Draw 0 with centre at 0. Draw line AB tangent to the 0 at C. Draw line 00. We then have: Given line AB tangent to ( EC at C, and radius OC. To Prove OC 1 AB. (0C is the shortest line from 0 to AB.) 76 PLANE GEOMETRY - BOOK II 171. It follows from ~ 170 that a line perpendicular to a tangent at its point of contact passes through the centre of the circle. PROP. XV. THEOREM 172. Two parallels intercept equal arcs on a circumference. Case I. When one line is a tangent and the other a secant. Draw a 0. Through E, any point on the circumference, draw diameter EF, and tangent AB I EF (~ 169). Draw secant CD II AB, intersecting the circumference at C and D. We then have: Given AB a tangent to 0 CED at E, and CD a secant 11 AB, intersecting the circumference at C and D. To Prove arc CE = arc DE. Proof. 1. Draw diameter EF; then EF - AB. (~ 170) 2. Then, EF CD. (?) 3. Then, arc CE = arc DE. (~ 162) Case II. When both lines are secants. Draw a 0. Draw 1I secants, AB and CD, intersecting the circumference at A and B, and C and D, respectively. We then have: Given, in 0 ABC, AB and CD II secants, intersecting the circumference at A and B, and C and D, respectively. To Prove arc AC= arc BD. Proof. 1. Draw tangent EF II AB, touching the O at G. 2. Then, EF II CD. (?) 3. By Case I, arc AG = arc BG, (1) and arc CG = arc DG. (2) 4. Subtracting (2) from (1), arc AG - arc CG = arc BG - arc DG, or arc AC = arc BD. THE CIRCLE 77 Case II. When both lines are tangents. In (OEGF, draw 11 tangents, AB and CD, touching ( at E and F, respectively. We then have: Given, in ( EGF, AB and CD li tangents, touching the 0 at E and F, respectively. To Prove arc EGF = arc EHF. (Draw secant GH II AB, and apply Case I.) 173. It follows from ~ 172 that the straight line joining the points of contact of two parallel tangents is a diameter. Ex. 8. The diameters of two concentric circles are 11 and 15 inches, respectively. Can a tangent to either circle be drawn through a point 5 inches from the common centre? Why? Ex. 9. The lines joining the extremities of two unequal parallel chords form, with the chords, an isosceles trapezoid. Is the centre within or without the trapezoid? Ex. 10. If two equal chords, AB and CD, of a circle are parallel, the points A, B, D, and C being in succession on the circumference, the chords AC and BD are equal and the figure formed is a rectangle. Can the centre be without the rectangle? PROP. XVI. THEOREM 174. The tangents to a circle from an outside point are equal. JB A- - -- -- -- Draw 0 with centre at 0. Draw radii OB and OC, and tangents I OB and OC, respectively, intersecting at A. We then have: Given AB and AC tangent at points B and C, respectively, to 0 BC whose centre is 0. To Prove AB = AC. (Rt. A OAB and OAC are equal by ~ 61.) PLANE GEOMETRY BOOK II 78 175. From equal A OAB and OAC (Fig. of Prop. XVI), Z OAB = Z OAC and/ AOB = Z AOC. Then, the line joining the centre of a circle to the point of intersection of tvo tancgents makes equal angles with the tccgents, and also with the radii drawn to the points of contact. Ex. 11. The line through the middle points of two parallel chords passes through the centre of the circle. Ex. 12. If two equal circles intersect and parallel lines be drawn through the points of intersection, the portion cut from the first line by one circumference is equal to the portion cut from the second by the other circumference. Ex. 13. The angle formed by two tangents to a circle is the supplement of the angle formed by joining the centre to the points of tangency. Ex. 14. Can a trapezoid be constructed whose sides are 8, 10, 11, 12, respectively? Give method showing the result. Can such a trapezoid be inscribed in a circle? PROP. XVII. THEOREM 176. Through three points, not in the same straight line, a circumference can be drawn, and but one. A Given points A, B, and C, not in the same straight line. To Prove that a circumference can be drawn through A, B, and C, and but one. Proof. 1. Draw lines AB and BC, and lines DF and EG I AB and BC, respectively, at their middle points, meeting at 0. 2. Then 0 is equally distant from A, B, and C. (~ 136) 3. Hence, a circumference described with 0 as a centre and OA as a radius will pass through A, B, and C. THE CIRCLE 79 4. Again, the centre of any circumference drawn through A, B, and C must be in each of the Is DF and EG. (~ 56) 5. Then as DF and EG intersect in but one point, only one circumference can be drawn through A, B, and C. 177. Two circumferences can intersect in but two points; for if they had three common points, they would have the same centre, and coincide throughout. PROP. XVIII. THEOREM 178. If two circumferences intersect, the straight line joining their centres bisects their common chord at right angles. Draw (, with centres at 0 and 0', whose circumferences intersect at A and B. Draw lines AB and 00'. We then have: Given 0 and 0' the centres of two ~, whose circumferences intersect at A and B, and lines 00' and AB. To Prove that 00' bisects AB at rt. As. (Use ~ 56.) PROP. XIX. THEOREM 179. If two circles are tangent to each other, the straight line joining their centres passes through their point of contact. Draw two 0, whose centres are at 0 and 0', tangent to line AB at A. We then have: Given 0 and 0' the centres of two (, which are tangent to line AB at A. To Prove that str. line joining 0 and 0' passes through A. (Draw radii OA and O'A; these lines are J AB, and OAO' is a str. line by ~ 20. Then use Ax. 6.) 80 PLANE GEOMETRY-BOOK II Ex. 15. If two circumferences intersect, the distance between their centres is greater than the difference of their radii. Ex. 16. If a tangent has its extremities in two parallel tangents to the same circle, the angle which is formed by the lines joining the centre of the circle to the extremities of the first tangent is a right angle. [Draw radii to the points of tangency.] MEASUREMENT 180. The ratio of one magnitude to another of the same kind is the quotient of the first divided by the second. Thus, if a and b are quantities of the same kind, the ratio of a a to b is - We measure a magnitude by finding its ratio to another magnitude of the same kind, called the unit of measure. If the quotient can be obtained exactly as an integer or fraction, we call it the numerical measure of the magnitude. 181. Two magnitudes of the same kind are said to be commensurable when a unit of measure (called a common, measure) is contained an integral number of times in each. Thus, two lines whose lengths are 23 and 3- inches are commensurable; for the common measure I inch is contained an integral number of times in each; i.e. 55 times in the first line, and 76 times in the second. Two magnitudes of the same kind are said to be incommensurable when no magnitude of the same kind can be found which is contained an integral number of times in each. For example, if AB and CD are two lines such that AB CD = /2; since V2 can only be obtained approximately as a decimal, no line, however small, can be found which is contained an integral number of times in each line, and AB and CD are incommensurable. 182. A magnitude which is incommensurable with respect to the unit has no exact numerical measure (~ 180). THE CIRCLE 81 AB Still if CD is the unit of measure, and -B _ 2, we shall call V/2 the numerical measure of AB. 183. It is evident from the above that the ratio of two magnitudes of the same kind, whether commensurable or incommensurable, is equal to the ratio of their numerical measures when referred to a common unit. THE METHOD OF LIMITS 184. We define a variable as a quantity which, under the conditions imposed upon it, may have an indefinitely great number of different values. We define a constant as a quantity which remains unchanged throughout the same discussion. 185. A limit of a variable is a constant quantity, the difference between which and the variable may be made less than any assigned quantity, however small. In other words, a limit of a variable is a fixed quantity to which the variable approaches indefinitely near. 186. Suppose, for example, that a point moves from A towards B under the condition that it A C D E B shall move, during successive equal inter- I vals of time, first from A to C, halfway between A and B; then to D, halfway between C and B; then to E, halfway between D and B; and so on indefinitely. In this case, the distance between the moving point and B can be made less than any assigned distance, however small. Then, the distance from A to the moving point is a variable which approaches the constant distance AB as a limit. Again, the distance from the moving point to B is a variable which approaches the limit 0. As another illustration, consider the series t t i is oe-f te p. where each term after the first is one-half the preceding. 82 PLANE GEOMETRY- BOOK II In this case, by taking terms enough, the last term may be made less than any assigned number, however small. Then, the last term of the series is a variable which approaches the limit 0 when the number of terms is indefinitely increased. Again, the sum of the first two terms is 1~; the sum of the first three terms is 1-; the sum of the first four terms is 17; etc. In this case, by taking terms enough, the sum of the terms may be made to differ from 2 by less than any assigned number, however small. Then, the sum of the terms of the series is a variable which approaches the limit 2 when the number of terms is indefinitely increased. 187. The Theorem of Limits. If two variables are always equal, and each approaches a limit, the limits are equal. A M C B L I I i A' l3' B L I I Given AM and A'M' two variables, which are always equal,. and approach the limits AB and A'B', respectively. To Prove AB = A'B'. Proof. 1. If possible, suppose AB > A'B'. 2. Take, on AB, AC equal to A'B'. 3. Then, variable AMnmay assume values > AC, while variable A'M' is restricted to values < AC. 4. This is contrary to the hypothesis that the variables are always equal; then, AB cannot be > A'B'. 5. In like manner, we may prove that AB cannot be < A'B'. 6. Since AB cannot be >, nor < A'B', we have AB = A'B'. Note. In the above demonstration, we have supposed AM and A'M' to be increasing variables; the theorem may be proved in a similar manner if AM and A'M' are decreasing variables. THE CIRCLE 83 MEASUREMENT OF ANGLES PROP. XX. THEOREM 188. In the same circle, or in equal circles, two central angles are in the same ratio as their intercepted arcs. Case I. WTien the arcs are commensurable (~ 181). B E.4 sI Draw figure in accordance with the statement. We then have: Given, in 0 ABC, AOB and BOC central As intercepting commensurable arcs AB and BC, respectively. P AOB are AB To Prove / BOC are BC Proof. 1. By hyp., arcs AB and BC are commensurable; let arc AD be a common measure of arcs AB and BC, and suppose it to be contained 4 times in arc AB, and 3 times in arc BC. arc AB 4 2. Then, (1) arc BC 3 3. Drawing radii to the several points of division of arc AC, / AOB will be divided into 4 s, and Z BOC into 3 As, all of which As are equal. (~ 155) /AOB 4 4. Then, Z AOB 4 (2).Z A.OB are AB 5. From (1) and (2), B a(?) ZBOC arc BC Note. The theorem may be proved in a similar manner, whatever the number of subdivisions of arcs AB and BC. 84 PLANE GEOMETRY -BOOK II Case II. When the arcs are incommensurable (~ 181). Draw figure in accordance with the statement. We then have: Given, in 0 ABC, AOB and BOC central /s intercepting incommensurable arcs AB and BC, respectively. / AOB arc AB Torove Z BOC are BC Proof. 1. Let arc AB be divided into any number of equal arcs, and let one of these arcs be applied to arc BC as a unit of measure. 2. Since arcs AB and BC are incommensurable, a certain number of equal arcs will extend from B to C', leaving a remainder C'C less than one of the equal arcs. 3. Draw radius OC'; since arcs AB and BC' are commensurable, Z AOB arc AB BOC' are BC' (Case I) 4. Now let the number of subdivisions of arc AB be indefinitely increased. 5. Then the unit of measure will be indefinitely diminished; and the remainder C'C, being always less than the unit, will approach the limit 0. 6. Then L BOC' will approach the limit Z BOC, and arc BC' will approach the limit arc BC. 7. Hence, AOB will approach the limit AOB, Z BOC Z aBOC' and arc will approach the limit arc B are BC' ' are BC THE CIRCLE 85 LZAOB an are AB 8. Now, Zv AOB and arc AB are two variables which are ZBOC' are BC' / AOB always equal; and they approach the limits ZABO and Z BOC arc AB arc BC" respectively. 9. By the Theorem of Limits, these limits are equal. (~ 187) 10 Then Z AOB arc AB 10. Then, BOO B Z BOC are BC' 189. The usual unit of measure for arcs is the degree, which is the ninetieth part of a quadrant (~ 154). The degree of arc is divided into sixty equal parts, called minutes, and the minute into sixty equal parts, called seconds. If the sum of two arcs is a quadrant, or 90~, one is called the complement of the other; if their sum is a semi-circumference, or 180~, one is called the supplement of the other. 190. By ~ 155, equal central As, in the same 0, intercept equal arcs on the circumference. Hence, if the angular magnitude about the centre of a 0 be divided into four equal A, each Z will intercept an arc equal to one-fourth of the circumference. That is, a right central angle intercepts a quadrant on the circumference. (~ 19) 191. By ~ 188, a central Z of n degrees bears the same ratio to a rt. central Z that its intercepted arc bears to a quadrant. But a central Z of n degrees is n of a rt. central Z. Hence, its intercepted arc is 9- of a quadrant, or an arc of n degrees. The above principle is usually expressed as follows: A central angle is measured by its intercepted arc. The above statement signifies simply that the number of angular degrees in a central angle is equal to the number of degrees of arc in its intercepted arc. 86 PLANE GEOMETRY-BOOK II PROP. XXI. THEOREM 192. An inscribed angle is measured by one-half its intercepted arc. Case I. When one side of the angle is a diameter. Draw 0 ABC with centre at 0. Draw diameter AC and chord AB. We then have: Given AC a diameter, and AB a chord, of 0 ABC. To Prove that Z A is measured by I arc BC. Proof. 1. Draw radius OB; denote Z BOC by a. 2. We have, OA OB. (?) 3. Then, B Z A. (~ 50) 4. Since a is an ext. L of A OAB, Za = Z A+Z B. (~ 85) 5. Then, Za=2 A,or ZA= —Za. 6. But, Z a is measured by arc BC. (~ 191) 7. Whence, Z A is measured by ~ are BC. Case II. When the centre is' within the angle. Draw 0 ABC; draw an inscribed Z BAG with the centre within the Z. We then have: Given AB and AC chords of 0 ABC, and the centre of the 0 within / BAC. To Prove that / BAC is measured by - arc BC. Proof. 1. Draw diameter AD; denote / BAD by a and Z BAC by b. 2. By Case I, Z a is measured by I arc BD, and Z b is measured by 1 are CD. 3. Then, Z a + / b is measured by - arc BD. + - arc CD. 4. Then, Z BAC is measured by - arc BC. Case III. When the centre is without the angle. Draw ( ABC. Inscribe Z BAC, the centre lying without the Z. Draw diameter AD. (The proof is left to the pupil.) THE CIRCLE 87 193. Angles inscribed in the same segment B are equal. / For, if A, B, and C are s inscribed in seg- A ment AED of 0 ABC, ZA=ZL =Z7C; for each L is measured by I arc DE (~ 192). 194. An angle inscribed in a semicircle is a right angle. B --- For if BC be a diameter, and AB and AC chords, of 0 ABD, LA is measured by I an arc of 180~, or 90~ (~ 192). D B 195. The opposite angles of an inscribed quadrilateral are supplementary. For their sum is measured by i of 360~, or 180~. (?) A o D Ex. 17. A quadrilateral ABCD is inscribed in a circle. Arc AB is 84~, arc CD is 96~, and Z C is 80~. Find remaining arcs and angles. Ex. 18. A quadrilateral ABCD is inscribed in a circle. Z B is 104~, Z C is 92~, and arc AB 70-. Find remaining angles of the quadrilateral and the arcs subtended by its sides. Ex. 19. A regular polygon has seven sides. Find the number of degrees in each angle. Ex. 20. The median drawn to the hypotenuse of a right triangle divides it into two isosceles triangles. Ex. 21. If angles be inscribed in two segments whose sum is equal to the whole circle, they are supplements and each angle is a supplement of one-half the central angle which its arc subtends. Ex. 22. A triangle is inscribed in a circle; one of its angles is 74~, and one side of this angle subtends an arc of 42~; find the remaining angles of the triangle and the subtended arcs. Ex. 23. One of the equal sides of an isosceles triangle ABC is 30 inches, the'vertical angle A is 120~. A circle described on one of the equal sides as a diameter intersects BC at D. Find BD in terms of BC. 88 PLANE GEOMETRY-BOOK II PROP. XXII. THEOREM 196. The angle formed by a tangent and a chord is measured by one-half its intercepted arc. A B C D ----------— ^gr Draw a 0. Draw tangent AC, touching the 0 at B, and chord BD making an acute Z with AB. We then have: Given AC tangent to 0 BDE at B, and BD a chord. To Prove that L ABD is measured by I arc BD. Proof. 1. Draw chord DE II AC; then, / ABD = Z D. (~ 75) 2. Now / D is measured by - arc BE. (~ 192) 3. Also, arc BE = arc BD. (~ 172) 4. Then, Z ABD is measured by -1 arc BD. Since Z CBD is sup. to Z ABD, it is measured by ~ (a circumference - arc BD), or ~ arc BED. (~ 192) PROP. XXIII. THEOREM 197. The angle bfrmed by two chords, intersecting within the circumference, is measured by one-half the sunm of its intercepted arc, and the arc intercepted by its vertical angle. Draw 0 ABC. Draw chords AB and CD intersecting within the circumference at E; denote Z AEC by a. We then have: Given, in 0 ABC, chords AB and CD intersecting within the circumference at E. To Prove that Z a is measured by - (arc AC + arc BD). Proof. 1. Draw chord BC; since a is an ext. Z of A BCE, /a=-B + Z C. (?) THE CIRCLE 89 2. But, LB is measured by I arc AC, and / C is measured by I arc BD. (?) 3. Substituting in result of (1), / a is measured by 1 (arc AC + arc BD). Ex. 24. In an inscribed quadrilateral ABCD, Z A = 75, Z B = 40~, arc BC=95~; find the angle formed by the diagonals of the quadrilateral. PROP. XXIV. THEOREM 198. The angle formed by two secants, intersecting without the circumference, is measured by one-half the difference of the intercepted arcs. E D Draw 0 ABC. Draw secants AE and CE intersecting without the circumference at E, and intersecting the circumference at A and B, and C and D, respectively. We then have: Given, in 0 ABC, secants AE and CE intersecting without the circumference at E, and intersecting the circumference at A and B, and Cand D, respectively. To Prove that Z E is measured by 1 (arc AC - arc BD). Proof. 1. Draw chord BC; denote Z ABC by a. 2. Since a is an ext. L of A BCE, Z a = Z E + Z C. (?) 3. Then, ZE=Za-ZLC 4. Now Z a is measured by I arc AC, and Z C is measured by 1 arc BD. (?) 5. Then, LE is measured by 1 (arc AC- arc BD). 90 PLANE GEOMETRY-BOOK II 199. If the opposite angles of a quadrilateral are supplementary, the quadrilateral can be inscribed in a circle. (Converse of ~ 195.) D Suppose, in quadrilateral ABCD, Z A sup. to Z C, and Z B to Z D; and a circuniference drawn through A, B, and C. (~ 176) If ZD is sup. to ZB, it is measured by A C 2 arc ABC. (~ 192) B Then, D must lie on the circumference; for if it were within the 0, Z D would be measured by 2 an arc > ABC; and if it were without the 0, Z D would be measured by - an arc < ABC. (~~ 197, 198) Ex. 25. Two equal chords bisect in a circle. What are the angles formed? Ex. 26. Two chords intersect within a circle, and form an angle of 75~; one of the intercepted arcs is 60~: find the other. Ex. 27. Two chords intersect within a circle; one of the angles formed is A~, and one of the intercepted arcs is B~: find the other. PROP. XXV. THEOREM 200. The angle formed by a secant and a tangent, or two tangents; is measured by one-half the difference of the intercepted arcs. E E D /D A C 6"\ /~~\ x ^ < >c FIG. 1. FIG. 2. FIG. 1. FIG. 2. Draw figures in accordance with the statement, AE tangent to ( BDF at B, and CE a chord in Fig. 1, a secant in Fig. 2. We then have: THE CIRCLE 91 1. Given AE a tangent to 0 BDC at B, and EC a secant intersecting the circumference at C and D. (Fig. 1.) To Prove that / E is measured by I (arc BGC - arc BD). (We have / E= Z ABC - C.) 2. (In Fig. 2, Z E = / ABD - / BDE; then use ~ 197.) Ex. 28. In Ex. 27 find the point of intersection of these chords when A and B are equal. Ex. 29. Two chords intersecting in a circle intercept opposite arcs on the circumference 58~ and 14~, respectively. Find the angle formed by the chords. Ex. 30. Two unequal chords intersect at right angles. Find the sum of the opposite intercepted arcs. Will this sum remain constant if the lengths of the chords vary? Ex. 31. A chord subtends an arc of 144~. Through one extremity of the chord a tangent is drawn; find the angles which it forms with the chord, Ex. 32. The angle formed by a tangent and a secant is 94~, and one of the intercepted arcs is 210~; find the other. Ex. 33. Find the angle formed by two secants intersecting without a circle, if the intercepted arcs are 36~ and 144~, respectively; also if the angle is 75~ and one of the intercepted arcs is 44~, find the other. Ex. 34. An inscribed angle is 32~. How many degrees in the arc in which it is inscribed? Ex. 35. In how large a circle can a triangle whose sides are 12, 13, and 5 be inscribed? What kind of a triangle is it? Ex. 36. The angle formed by two tangents is 76~; find the intercepted arcs; also for 92~. Ex. 37. An inscribed quadrilateral subtends arcs of 70~, 80~, 100~, and 110~, respectively. Find the angles of the quadrilateral, and also the angles of the quadrilateral formed by the intersections of the tangents through the vertices of the quadrilateral. Ex. 38. The sum of the degrees in the angle formed by two tangents and the smaller of the intercepted arcs is 180~. Ex. 39. Choose any two of the following statements for the hypothesis and any one of the remaining three for the conclusion, and prove each of the ten theorems thus formed, not already proved in ~~ 162, 163: 92 PLANE GEOMETRY-BOOK II A straight line that 1. Passes through the centre of a circle; 2. Bisects the chord; 3. Is perpendicular to the chord; 4. Bisects the smaller arc; 5. Bisects the greater arc. Ex. 40. Give a general rule for the measurement of angles formed by two lines intersecting (a) within a circumference, (b) on a circumference, (c) without a circumference. Ex. 41. What figure is formed by the bisectors of the angles of a parallelogram? Give proof. Ex. 42. Two equal chords intersect either within a circumference, on the circumference, or without the circumference. Prove that in either case the bisector of one of the angles formed by the chords passes through the centre of the circle. Ex. 43. Given a chord CD and A and B, two points on the circumference, such that line AC = AD and line BC = BD, prove that line AB is a diameter. CONSTRUCTIONS PROP. XXVI. PROBLEM 201. At a given point in a straight line to erect a perpendicular to that line. In ~ 24, we gave a method for drawing a perpendicular to a straight line at a given point; the following method is advantageous when the point is near the end of the line. \,.A B iven any point in line AB. Given C any point in line AB. THE CIRCLE 93 Required to draw a line _ AB at C. Construction. 1. With any point 0 without line AB as a centre, and distance OC as a radius, describe a circumference intersecting AB at C and D. 2. Draw diameter DE, and line CE. 3. Then, CE is L AB at C. Proof. 4. Z DCE, being inscribed in a semicircle, is a rt.. (~ 194) PROP. XXVII. PROBLEM 202. To bisect a given arc. E I — ---- AlA Given arc AB. Required to bisect arc AB. Construction. 1. With A and B as centres, and with equal radii, describe arcs intersecting at C and D. 2. Draw line CD intersecting arc AB at E. 3. Then E is the middle point of arc AB. Proof. 4. Draw chord AB. 5. Then, CD is _L chord AB at its middle point. (?) 6. Whence, CD bisects arc AB. (~ 163) Ex. 44. The hypotenuse of an isosceles right triangle is 18. Construct the triangle. Ex. 45. Given two angles of a triangle, to construct the third. (Compare ~ 84.) 94 PLANE GEOMETRY-BOOK II PROP. XXVIII. PROBLEM 203. Throgh a given point without a given straight line, to dr aw a parallel to the line. /F /, /E A 1 — / Given C any point without line AB. Required to draw through C a line 1I AB. Construction. Through C draw any line EF, meeting AB at E, and construct Z FCD = Z CEB (~ 28); then, CD II AB. (The proof is left to the pupil.) PROP. XXIX. PROBLEM 204. Given two sides of a triangle, and the angle opposite to one of them, to construct the triangle. Given m and n sides of a A, and A' the L opposite to n. Required to construct the A. Construction. 1. Construct / DAE = Z A' (~ 28); on AE take AB = m. 2. With B as a centre and n as a radius, describe an arc. Case I. When A' is acute, and in > n. /E,' n a/ y A/e -th \caseD There may be three cases: THE CIRCLE 95 I. The arc may intersect AD in two points. 3. Let C1 and C0 be the points in which the arc intersects AD, and draw lines BCI and BC2. 4. Then, either ABC, or ABC2 is the required A. This is called the ambiguous case. II. The arc may be tangent to AD. In this case there is but one A. 5. Since a tangent to a ( is 1 the radius drawn to the point of contact (~ 170), the A is a right A. III. The arc may not intersect AD at all. In this case the problem is impossible. Case II. When A' is acute, and m = n. 6. In this case, the arc intersects AD in two points, of which A is one; there is but one A, an isosceles A. Case III. When A' is acute, and m < n. /.E m,, n,,, /n o/ — Z- A-Cy\- /i +2D A '-C A2 7. In this case, the arc intersects AD in two points. 8. Let C1 and C2 be the points in which the arc intersects AD, and draw lines BC1 and BC2. 9. Now A ABC1 does not satisfy the conditions of the problem, since it does not contain Z A'. 10. Then there is but one A; A ABC2. Case IV. When A' is right or obtuse, and mn < n. 11. In each of these cases, the arc intersects AD in two points on opposite sides of A, and there is but one A. The pupil should construct the triangle corresponding to each case of ~ 204. 96 PLANE GEOMETRY-BOOK II PROP. XXX. PROBLEM 205. To divide a given straight line into any number of equal parts.,GC x-\. = \...-.~\ A H K L B Given line AB. Required to divide AB into four equal parts. Construction. 1. On the indefinite line AC, take any convenient length AD. 2. On DC take DE = AD; on EC take EF= AD; on FC take FG = AD; and draw line BG. 3. Draw lines DH, EK, and FL 11 BG, meeting AB at H, K, and L, respectively. 4. Then, AH= HK= KL = LB. (~ 128) PROP. XXXI. PROBLEM 206. To circumscribe a circle about a given triangle. S U U Given A ABC. Required to circumscribe a 0 about A ABC. Construction. 1. Draw lines DF and EG - AB and AC, resptively, at their middle points, intersectivelyng at. (~ 26) THE CIRCLE 97 2. With 0 as a centre, and OA as a radius, describe a 0. 3. The circumference will pass through A, B, and C. (The proof is left to the pupil; see ~ 136.) Note. The above construction serves to describe a circumference through three given points not in the same straight line, or to find the centre of a given circumference or arc. Ex. 46. Through any given point within a circumference to draw a chord of given length. Ex. 47. To inscribe a circle in a given triangle. (Compare ~ 134.) PROP. XXXII. PROBLE-i 207. To draw a tangent to a circle through a given point on the circumference. B A C Given A any point on the circumference of OAD. Required to draw through A a tangent to 0 AD. Construction, 1. Draw radius OA. 2. Through A draw line BCLI OA. (~ 24) 3. Then, BC will be tangent to 0 AD. (?) Ex. 48. Given two sides and the included angle of a parallelogram, to construct the parallelogram. Ex. 49. Circumscribe a parallelogram about a circle and find relations between the lengths of the adjacent sides. Ex. 50. To draw a tangent to a given circle which shall be perpendicular to a given line. Ex. 51. To draw the shortest chord through a given point within a given circumference. 98 PLANE GEOMETRY-BOOK II PROP. XXXIII. PROBLEM 208. To draw a tangent to a circle through a given point with. out the circle. A o Given A any point without BC. Required to draw through A a tangent to 0 BC. Construction. 1. Let 0 be the centre of 0 BC. 2. Draw line OA, and with OA as a diameter, describe a circumference, cutting the given circumference at B and C. 3. Draw lines AB and AC. 4. Then, AB and AC are tangents to ( BC. Proof. 5. Draw line OB. 6. Since / ABO is inscribed in a semicircle, it is a rt. Z. (?) 7. Then, AB is tangent to 0 BC. (?) In like manner, we may prove AC tangent to ( BC. PROP. XXXIV. PROBLEM 209. Upon a given straight line, to describe a segment which shall contain a givenb angle. G G \M D "\\ 1V_, Given line AB, and / A'. THE CIRCLE 99 Required to describe upon AB a segment such that every Z inscribed in the segment shall equal / A'. Construction. 1. Construct Z BA = Z A'. (~ 28) 2. Draw line DE I AB at its middle point. (~ 26) 3. Draw line AF I AC, intersecting DE at 0. 4. With 0 as centre and OA as radius, describe 0 AMBI. 5. Then, AMB will be the required segment. Proof. 6. If AGB be any Z inscribed in segment AMB, it is measured by ^ are ANB. (?) 7. But, by cons., AC 1_ OA; and AC is tangent to 0 AMB. (?) 8. Therefore, Z BAC is measured by I arc ANB. (~ 196) 9. Then, AGB = / BAC = ZA'. (?) 10. Hence, every / inscribed in segment AMB equals / A'. (~ 193) Ex. 52. From a given point without a circle to draw two equal secants terminating in the circumference. Can more than two be drawn? Ex. 53. Trisect a right angle. (In general, an angle cannot be trisected by methods of Euclidean geometry.) Ex. 54. In a circle whose radius is 12, draw a chord whose length is 8, parallel to a given line. Ex. 55. One of the equal sides of an isosceles triangle is 8, and one of the equal angles 30~. Construct the triangle. Ex. 56. The distance between two parallel lines is a, and a line is drawn intersecting these two parallels. Construct a circle which shall be tangent to the three lines. How many such circles are possible? Ex. 57. The sides of a triangle are a, b, and c, respectively. The angle opposite b is twice, and the angle opposite c three times, the angle opposite a. Construct the triangle. Ex. 58. The base of an isosceles triangle is 10, and the perimeter 18. Construct the triangle. Ex. 59. In a rhombus ABCD, ZA is twice XB. The shorter diagonal is 9. Construct the rhombus. 100 PLANE GEOMETRY-BOOK II Ex. 60. Two angles of an inscribed quadrilateral are 90~ and 60~, respectively. Find the remaining angles and inscribe the quadrilateral in a given circle. Can more than one such quadrilateral be inscribed? Ex. 61. One side of a triangle is three-fourths another, and threefifths the third side. The perimeter of the triangle is 48. Construct the triangle. Ex. 62. A given point lies within, without, or on the circumference of, a circle. With the given point as centre, describe a circumference passing through the extremities of a given diameter of the circle. Ex. 63. We define the angle between two intersecting curves as the angle formed by tangents to the curves at their point of intersection. With a point outside a given circle as a centre, describe an arc which shall intersect the given circumference at right angles. BooK III PROPORTION. - SIMILAR POLYGONS DEFINITIONS 210. A Proportion is a statement that two ratios are equal. 211. The statement that the ratio of a to b is equal to the ratio of c to d is written a c b d 212. In the proportion -a = d we call a the first term, b the b d second term, c the third term, and d the fourth term. 213. We call the first and fourth terms the extremes, and the second and third terms the means. We call the first and third terms the antecedents, and the second and fourth terms the consequents. Thus, in the above proportion a and d are the extremes, b and c the means, a and c the antecedents, and b and d the consequents. 214. If the means of a proportion are equal, either mean is called a mean proportional between the first and last terms, and the last term a third proportional to the first and second terms. Thus, in the proportion = b b is a mean proportional beb c tween a and c, and c a third proportional to a and b. 215. In the proportion a -= d is called a fourth proporb d tional to a, b, and c. 101 102 PLANE GEOMETRY-BOOK III PROP. I. THEOREM 216. In any proportion, the product of the extremes is equal to the product of the means. Given the proportion a (1) b d To Prove ad = bc. Proof. Multiplying both members of equation (1) by bd, ad = be. 217. Applying the above theorem to the proportion a b a b- we have b2 = ac, or b = Vac. b c That is, the mean proportional between two numbers is equal to the square root of their product. PROP. II. THEOREM 218. (Converse of Prop. I.) If the product of two numbers is equal to the product of two others, one pair may be made the extremes, and the other pair the means, of a proportion. Given ad = be. (1) To Prove c b d Proof. Dividing both members of (1) by bd, ad be a c bd -bd' b d In like manner, we may prove a b = d; etc. c d a c PROP. III. THEOREM 219. In any proportion, the terms are in proportion by ALTERNATION; that is, the first term is to the third as the second term is to the fourth. Given the proportion = (1) b d PROPORTION 103 To Prove a c d Proof. From (1), ad = be. (~ 216) Then, ga _ b. Then, a= (~ 218) c d PROP. IV. THEOREM 220. In any proportion, the terms are in proportion by INVERSION; that is, the second term is to the first as the bfurth term is to the third. a c Given the proportion. (1) b d To Prove b d a c Proof. From (1), ad = be. (?) Then, b (? a c PROP. V. THEOREM 221. In any proportion, the terms are in proporton by COMPOSITION; that is, the sum of the first two terms is to the first term as the sum of the last two terms is to the third term. a _ c Given the proportion -c (1) b ' To Prove a + b c + d To Prove a c Proof. From (1), ad = be. (?) Adding both members of this equation to ac, ac + ad = ac + be, or a(c + d) = c(a + b). Then, a + c- d (~ 218). (2) a c In like manner, we may prove = c +d. b d 104 PLANE GEOMETRY-BOOK III PROP. VI. THEOREM 222. In any proportion, the terms are in proportion by DIVISION; that is, the difference of the first two terms is to the first term as the difference of the last two terms is to the third term. Given the proportion a= (1) a -b c -d To Prove a-b a c Proof. From (1), ad = bc. (?) Subtracting both members from ac, ac — ad= ac- be, or a(c- d) = c(a- b). a- b o- d( Then, ( 218). (2) a c a-b c-d In like manner, we prove b = -d b d PROP. VII. THEOREM 223. In any proportion, 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 terms is to their difference. Given the proportion a b d To Prove a~b c~d a-b c —d Proof. Dividing equation (2), ~ 221, by equation (2), ~ 222, a + b c+d a-b c-d Ex. 1. What is the ratio of 50 cents to $1? of I hour to 1 day? of 150 rods to 1 mile? Ex. 2. If x is a fourth proportional to a, b, c, find x. Ex. 3. Apply composition and division to the following: x + x/-a _ 2a + b x- /x2a2 2a-b PROPORTION 105 PROP. VIII. THEOREM 224. In a series of equal ratios, the sum of all the antecedents is to the sum of all the consequents as any antecedent is to its consequent. a c e Given a (1) b df To Prove a + c + e _ a To Prove a +~ea b+d+f b Proof. We have ba = ab. And from (1), be = ad, and be = af. (?) Adding, ba + be + be = ab + ad + af. Or, b(a + c + e) =a(b + d +f). Then, a + c + e a Then, b'~ (?) b'+ d +f b 225. The ratio of two magnitudes of the same kind is equal to the ratio of their numerical measures when referred to a common unit (~ 183). Then, in every proportion involving the ratio of magnitudes of the same kind, we shall regard the ratio of the magnitudes as replaced by the ratio of their numerical measures when referred to a common unit. Thus, if AB and CD are lines, and AB m CD n ' where m and n are numbers, this means simply that the ratio of the numerical measures of AB and CD, when referred to a common unit, is equal to the ratio of m to n. Ex. 4. Find a third proportional to -4 and -1. Ex. 5. Find a mean proportional between 5 and 24. Ex. 6. If the ratio of the hypotenuse of a right triangle to one of the legs is 2, the acute angles are 30~ and 60~. Ex. 7. Each of two angles of a triangle is 45~. Find the ratio of the sides opposite these angles. 106 PLANE GEOMETRY-BOOK III Ex. 8. Two angles of a triangle are 30~ and 60~, respectively. Find the ratio of the sides opposite these angles. Is the ratio commensurable? PROPORTIONAL LINES 226. Def. Two straight lines are said to be divided proportionally when their corresponding segments are in the same ratio as the lines themselves. A E B Thus, lines AB and CD are divided pro- C D portionally if l_ I I AE BE AB CF DF CD PROP. IX. THEOREM 227. A parallel to one side of a triangle divides the other two sides proportionally. Case I. When the segments of each side are commensurable. A 1/ - E Draw A ABC. Take D, a point in AB, such that AD and DB are commensurable; draw DE II BC, meeting AC at E. We then have: Given, in A ABC, segments AD and BD of side AB commensurable, and line DE II BC, meeting AC at E. AD AE To Prove BD= CE Proof. 1. Let AFbe contained 4 times inAD, 3 times in BD. 2. Find ratio of AD to BD. 3. Through points of division of AB draw Ils to BC. 4. Find ratio of AE to CE. (~ 128) 5. From steps (2) and (4), AD (? 'BD CE PROPORTIONAL LINES 107 Case II. When the segments of each side are incommensurable. A B', Draw figure in accordance with the statement of the proposition. We then have: Given, in A ABC, segments AD and BD of side AB incommensurable, and line DE li BC, meeting AC at E. AD AE To Prove m=. BD CE Proof. 1. Let AD be divided into any number of equal parts, and let one of these parts be applied to BD as a unit of measure. 2. Since AD and BD are incommensurable, a certain number of the equal parts will extend from D to B', leaving a remainder BB' < one of the equal parts. 3. Draw line B'C' II BC, meeting AC at C'. 4. Then, AD and B'D are commensurable, and AD AE BA'D C_-AE (Case I) B'D C'E 5. Now, if the number of subdivisions of AD be indefinitely increased, the unit of measure will be indefinitely diminished, and the remainder BB' will approach the limit 0. 6. Then, AD- will approach the limitAD B'D BD' and AE will approach the limit AE C'E CE 7. By the Theorem of Limits, these limits are equal. (?) AD_ AE. 8. Then, A AE BD CE PLANE GEOMETRY -BOOK III The above result simply means that the ratio of the numerical measures of lines AD and BD, when referred to a common unit, is equal to ratio of the numerical measures of lines AE and CE, when referred to a common unit. A similar meaning is attached to every proportion involving the ratio of two geometrical magnitudes of the same kind. (Compare ~ 225.) 228. Applying the theorem of ~ 221 to the proportion of ~ 227, we have AD + BD AE + CE AB A AD AE ' AD AE ) AB AC In like manner, AB (2) BD CE Again, applying the theorem of ~ 220 to the proportions (1) and (2), we obtain the proportions AB AD AB BD AC-AE' -AC CE Then by Ax. 1, AB AD BD. (3) AC AE CE The proportions (1), (2), and (3) are all included in the statement of Prop. IX: A parallel to one side of a triangle divides the other two sides proportionally. PROP. X. THEOREM 229.' (Converse of Prop. IX.) A line which divides two sides of a triangle proportionally is parallel to the third side. A B C Draw A ABC. Draw line DE dividing AB and AC proportionally at D and E, respectively. We then have: PROPORTIONAL LINES 109 Given, in A ABC, line DE meeting AB and AC at D and E, respectively, so that ABAC AD AE To Prove DE II BC. Proof. 1. If DE is not I BC draw line DF 1i BC', meeting AC at F. 2. Then, ABAC228 AD A ABAC 3. But by hyp., AB AC AD AE 4. Then AC AC or AE = AP. (?) AE AF 5. Then,' line DF coincides with DE, and line DE is II BC. (Ax. 3) Pnop. XI. THEOREM 230. In any trianyle, the bisector of an angle divides the oppos-ite side into segments proportional to the adjacent sides. Draw A ABC. Draw line AD, bisecting /A, meeting BC at D. We then have: Given line AD bisecting Z A of A ABC, meeting BC at D. DB AB To Prove DCAC DC AC' Proof. 1. Draw line BE II AD, meeting CA extended at E; represent zs BAD, CAD, and ABE by a, b, c, respectively. 2. We have Zc = Za. (?) 3. Also, IE=ZLb. (?) 4. Byhyp., Za=Z b. 5. Then, Zc=zLE. (?) 6. Then, AB = AE. (?) DB AE 7. Now DOA(~ 228) DJC AC 110 PLANE GEOMETRY-BOOK III o mDB AB 8. Then, A(?) DC-A C' 231. Def. The segments of a line by a point are the distances from the point to the extremities of the line, whether the point is in the line itself, or in the line extended. Ex. 9. Three parallels cut segments 5 and 3, respectively, from a line drawn at random across them. What is the ratio of the segments cut from another line drawn across these three parallels? Ex. 10. A line 15 inches long is parallel to the base of a triangle. It divides the sides in the ratio of 3 to 2. Find the base. PROP. XII. THEOREM 232. In any triangle the bisector of an exterior angle divides the opposite side externally into segments proportional to the adjacent sides. Draw A ABC. Extend CA to E and draw line AD bisecting Z BAE, meeting CB extended at D. We then have: Given line AD bisecting ext. Z BAE of A ABC, meeting CB extended at D. DB AB To Prove DB AB DC AC' (Draw BF II AD; then Z ABF = Z AFB, and AF = AB; BF is II to side AD of A ACD.) The above theorem is not true for the exterior angle at the vertex of an isosceles triangle. Ex. 11. The base of a triangle is 25, and the other sides are 15 and 16, respectively. Draw a line which shall be parallel to the base of the triangle, terminate in the sides, and be 21 in length. Ex. 12. One of the parallel sides of a trapezoid is double the other. In what ratio do the diagonals intersect? Ex. 13. In a triangle ABC, AB = 12, BC= 15, CA = 13. Find the segments into which the bisector of / B divides the opposite side. Ex. 14. The sides of a triangle are 8, 10, and 12; find the segments of the side 12 made by the bisector of the opposite angle. SIMILAR POLYGONS Ill Ex. 15. The sides of a triangle are AB -- 6, BC = 7, CA = 8. Find the length of the segments into which the bisector of the exterior angle at B divides AC. SIMILAR POLYGONS 233. Def. Two polygons are said to be similar if they are mutually equiangular (~ 121), and have their homologous sides proportional. B B' A - A' C' E D E' D' Thus, polygons ABCDE and A'B'C'D'E' are similar if Z A = A', Z B = Z B', etc., aAB BC CD and, etc. A'B' B' C' C'D' 234. The following are given for reference: 1. In similar polygons, the homologous angles are equal. 2. In similar polygons, the homologous sides are proportional. PROP. XIII. THEOREM 235. Two triangles are similar when they are mutually equiangular. A - E B C B' C' Draw A ABC; also, AA'B'C' having ZA' = ZA, ZB =ZB, and Z C' = C. We then have: Given, in A ABC and A'B'C', Z A = /A', B = Z B', and / C= ZC'. PLANE GEOMETRY-BOOK III 112 To Prove A ABC and A'B'C' similar. Proof. 1. Place A A'B'C' in position ADE; Z A' coinciding with Z A, vertices B' and C' falling at D and E, respectively, and side B'C' at DE. AB AC 2. We have AB=. (~ 228) 3. Then, AB' (?) A'B' -.4' CA 4. In like manner, by placing B' at B, AB BC A'B' B' C' AB AC BC 5. Then, AB AC B(? A'1Bt A' C- B' C' 6. Then A ABC and A'B'C' have their homologous sides proportional, and are similar. (~ 233) The following are consequences of ~ 235: 236. Two triangles are similar when two angles of one are equal respectively to two angles of the other. For their remaining As are equal each to each. (~ 87) 237. Two right triangles are similar when an acute angle of one is equal to an acute angle of the other. 238. If a line be drawn between two sides of a triangle parallel to the third side, the triangle formed is similar to the given triangle. For the triangles are mutually equiangu- lar. (~ 76) D 239. Note. In similar triangles, the ho- B mologous sides lie opposite the equal angles. Ex. 16. Find the sides of a triangle similar to that in Ex. 14, one side being 9. Does more than one triangle satisfy this condition? Ex. 17. Each of two isosceles triangles has a vertical angle of 32~. Do their homologous sides form a proportion? Why? .................. -- -- -- -- -........... — -- - -- -- -- -- - - - -- -- -- - .V~~~~<' >'.~~~~~~'. ~ c ~ ~ '0~~ /! 3~~~~~ 3r ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~~~~/ SIMILAR POLYGONS 113 PROP. XIV. THEOREM 240. Two triangles are similar when their homologous sides are proportional. A B C B' C' Draw A ABC and A'B' C' having their homologous sides proportional. We then have: Given, in A ABC and A'B'C', AB AC BC A'B'I A'C' B'C' To Prove A ABC and A'B'C' similar. Proof. 1. On AB and AC, take AD = A'B' and AE= AC'; and draw line DE. 2. Since A = A C DE IIBC. (~ 229) AD AE' 3. Then A ADE and ABC are similar. (~ 238) AB BC AB BC 4. Then A = - E' or AB' = DE (~ 234, 2) AD DE' A'B' DE' 5. Then DE = B'C' by comparing this with (1). 6. Then A ADE = A A'B'C'. (?) 7. Then, AA'B'C' is similar to A ABC. Ex. 18. The sides of a triangle are 7, 12, and 14. If the side of a similar triangle homologous to 12 is 18, find the remaining sides of the second triangle. Ex. 19. The diagonals of any trapezoid divide each other proportionally. Ex. 20. A post a feet high casts a shadow b feet long. At the same time, a tree casts a shadow c feet long. Find the height of the tree. What theorem of proportion is involved? 114 PLANE GEOMETRY- BOOK III 241. NOTE. To prove two polygons in general similar, it must be shown that they are mutually equiangular, and have their homologous sides proportional (~ 233); but in the case of two triangles, each of these conditions involves the other (~~ 235, 240), so that it is only necessary to show that one of the tests of similarity is satisfied. PROP. XV. THEOREM 242. Two triangles are similar when they have an angle of one equal to an angle of the other, and the sides including these angles proportional. Draw A ABC and A'B' C' having Z A -= A', and the sides including these sides proportional. We then have: AB AC Given, in A ABC and A'B'C', / A = / A', and = A A'B' A C' To Prove A ABC and A'B'C' similar. (Place A A'B'C' upon A ABC so that Z A' coincides with Z A and vertices B' and C' fall at D and E, respectively; by ~ 229, DE II BC; the theorem follows by ~ 238.) PROP. XVI. THEOREM 243. Two triangles are similar when their sides are parallel each to each, or perpendicular each to each. A G~ A/ DaB 1. C BI Ce A't FIG. 1. FIG. 2. FIG. 3. Draw A ABC and A'B'C' in accordance with the statement of the proposition. We then have: Given sides AB, AC, and BC, of A ABC, II respectively to sides A'B', A'C', and B'C' of A A'B'C' in Fig. 2, and I respectively to sides AB', ADC', and B'C' of A A'B'C' in Fig. 3. SIMILAR POLYGONS 115 To Prove A ABC and A'B'C' similar. Proof. 1. A A and A' are either equal or supplementary, as also are B and B', and C and C'. (~~ 81, 82, 83) 2. We may then make the following hypotheses with regard to the A of the A: I. A +A'=2 rt. s, B +B'=2 rt. As, + C=-2 rt. Zs. II. A+A'=2 rt. As, B+B'=2 rt. s, C= C'. III. A + A' = 2 rt. s, B = B', C+ C 2 rt. s. IV. A=A', B + B' 2 rt. A, C +C' = 2 rt. As. V. A=A', B=B', whence C=C'. (~87) 3. The first four hypotheses are impossible; for, in either case, the sum of the As of the two A would be > 4 rt. As. (~ 84) 4. We then have only A = A', B = B', and C = C'. 5. Therefore, A ABC and A'B'C' are similar. (~ 235) 244. Note. 1. In similar triangles whose sides are parallel each to each, the parallel sides are homologous. 2. In similar triangles whose sides are perpendicular each to each, the perpendicular sides are homologous. Ex. 21. The bisectors of the equal angles of an isosceles triangle extended to meet the equal sides are divided proportionally. Ex. 22. The altitudes to the equal sides of an isosceles triangle intersect in a point such that the product of segments of one equals the product of the segments of the other. Ex. 23. The diagonals of inscribed quadrilateral ABCD intersect at 0. If OR and OS are the altitudes of triangles OBC and OAD, respectively, prove OR - OS AD PROP. XVII. THEOREM 245. The homologous altitudes of two similar triangles are in the same ratio as any two homologous sides. Draw A ABC, and line AD I BC. Draw A AB' C', making ZB'A'C' = Z BAC and Z B' = Z B; draw line A'D' B' C'. We now have: Given AD and A'D' homologous altitudes of similar A ABC and A'B'C'. PLANE GEOMETRY -BOOK III AD AB AC BC To Prove A)D' A'B' A'C' B'C' (Rt. A ABD and A'B'D' are similar by ~ 237.) PROP. XVIII. THEOREM 246. Twvo polygons are similar when they are composed of the same number of triangles, similar each to each, and similarly placed. A 4A' TIs? CD DEoB T~c n' Draw polygon ABCDE; from E draw diagonals EC and EB. Con. struct A E'D'C' similar to A EDC, sides DC and D'C' being homologous; on E' C', homologous to EC, construct A El C'B' similar to A ECB; on E'B', homologous to EB, construct A E'B'A' similar to A EBA. We then have: Given, in polygons AC and A'C', A ABE similar to A A'B'E', A BCE to A B'C'E', and A CDE to A C'D'E'. To Prove polygons AC and A'C' are similar. Proof. 1. A ABE and A'B'E' are similar; find equal parts of these A. 2. A BCE and B'C'E' are similar; find equal parts of these A. 3. Now, L ABE+ EBC= L A'B'E' + Z E'B'C' or ABC==Z A'B'C'. 4. In like manner, Z BCD = Z B'C'D', etc.; and AC and A'C' are mutually equiangular.. AB BE and BE BC w AB BC 5. Then -- and -; whence A'B' BE' B'E' B'C" A'B'B'C' AB BC CD 6. In like manner, A- = C =, etc. A ' a Be s l' C(~ 7. Then AC and A'C' are similar. (~ 233) SIMILAR POLYGONS 117 PROP. XIX. THEOREM 247. (Converse of Prop. XVIII.) Two similar polygons may be decomposed into the same number of triangles, similar each to each, and similarly placed. A A' B E B' Draw similar polygons ABCDE, A'B'C'D'E', vertices E, E' being homologous; and diagonals EB, EC, E'B', E' C'. We then have: Given E and Ef homologous vertices of similar polygons AC and A'C', and lines EB, EC, E'B', and E'C'. To Prove A ABE similar to A A'B'E', A BCE to A B'C'E', and A CDE to A C'D'E'. Proof. 1. In similar polygons, the homologous As are equal, and the homologous sides proportional. 2. Then A ABE and A'B'E' are similar. (~ 242) 3. Using other As and sides of AC and A'C', and of A ABE and A'B'E', prove A BCE and B'C'E' similar. 4. In like manner, prove A CDE and C'D'E' similar. PROP. XX. THEOREM 248. The perimeters of two similar polygons are in the same ratio as any two homologous sides. Draw similar polygons ABCDE, A'B'Ct'DE', vertices A, A' being homologous. We then have: Given AB and AB', BC and B'C', CD and C'D', etc., homologous sides of similar polygons AC and A'C'. To Prove AB + BC+ CD + etc. AB BC CD A'B' + B'C' + C'D' + etc. A'B' - B'C' C''D" e (Apply ~ 224 to the equal ratios of ~ 233.) PLANE GEOMETRY-BOOK III Ex. 24. From any point in the base of an isosceles triangle perpendiculars to the equal sides are drawn. Prove that these perpendiculars form equal angles with the base and that they are proportional to the segments they cut off from the equal sides; the segments being measured from the extremities of the base towards the vertex. Ex. 25. A straight line bisecting the exterior angle at the vertex of a triangle is parallel to the base; find the ratio of the sides of the triangle. PROP. XXI. THEOREM 249. If a perpendicular be drawn from the vertex of the right angle to the hypotenuse of a right triangle, I. The triangles formed are similar to the whole triangle, and to each other. II. The perpendicular is a mean proportional between the segments of the hypotenuse. III. Either leg is a mean proportional between the whole hypotenuse and the adjacent segment. Draw A ABC, with rt. Z at C; also, line CD I AB. We then have: Given line CD J1 hypotenuse AB of rt. A ABC. To Prove A ACD and BCD similar to A ABC, and to each other; also, AD CD AB AC d AB BC CD BD C D' AC D' BC BD Proof. 1. Rt. A ACD and ABC are similar. (~ 237) 2. In like manner, A BCD and ABC are similar. 3. Then, A ACD and BCD are similar. 4. In similar A, homologous sides lie opposite equal As (~ 239); find sides of A BCD homologous to sides AD and CD of A ACD. AD CD 5. Then, D (~234,2) 6. In like manner, from A ABC and ACD, and A ABC and BCD, AB AC a AB BC AC AD' BC BD SIMILAR POLYGONS 119 250. Since an angle inscribed in a semicircle is a right angle (~ 194), it follows that: If a perpen dicular be drawn from any point in the circumference of a circle to a diameter, 1. The perpendicular is a mean propor- A D B tional between the segments of the diameter. 2. The chord joining the point to either extremity of the diameter is a mean proportional between the whole diameter and the adjacent segment. 251. We have from the three proportions of ~ 249, C2 = AD x BD, AC2 = AB x AD, and B2 = AB x BD. (?) Hence, if a perpendicular be drawn from the vertex of the right angle to the hypotenuse of a right triangle, 1. The square of the perpendicular is equal to the product of the segments of the hypotenuse. 2. The square of either leg is equal to the product of the whole hypotenuse and the adjacent segment. These equations mean that the square of the numerical measure of CD equals the product of the numerical measures of AD and BD; etc. (Compare ~ 225.) Ex. 26. The non-parallel sides AD and BC of trapezoid ABCD intersect at 0. If AB= 15, CD =24, and the altitude of the trapezoid is 8, what is the altitude of triangle OAB? Ex. 27. Given two similar triangles, and the altitude of the first double the homologous altitude of the second. Find the ratio between the radii of the circles circumscribing A these triangles. Ex. 28. If two circles are tangent internally, C and a chord of the greater is a tangent to the B smaller, and if through the points of contact a chord of the greater circle be drawn, then the chords joining the extremities of the tangent p and the chord form two pairs of similar triangles. 120 PLANE GEOMETRY-BOOK III Ex. 29. The perimeters of two similar polygons are A and B feet, respectively. If a side of the first is a feet, find the homologous side of the second. Ex. 30. The perimeters of two similar polygons are in the ratio of 4 to 5. If the perimeter of the second is 200, what is the perimeter of the first? If the ratio is A to B and a side of the second is b, what is the homologous side of the first? Ex. 31. Give a method for stretching strings which mark out a rectangular building for the excavators. Ex. 32. The legs of a right triangle are 12 and 16. Find the length of the perpendicular from the vertex of the right angle to the hypotenuse. Ex. 33. In any right triangle if a perpendicular be drawn from the vertex of the right angle to the hypotenuse, the hypotenuse is to either leg as the other leg is to the perpendicular to the hypotenuse. PROP. XXII. THEOREM 252. In any right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs. Draw A ABC with rt. Z at C. We then have: Given A ABC, with rt. Z at C. To Prove AB2 = AC2 + fBC". Proof. 1. Draw line CD 1 AB, 2. Then, AC2 = AB x AD, and BC2= AB x BD. (~ 251, 2) 3. Adding, AC2 + BC = AB x (AD + BD) = AB x AB. 4. Then, AB2 = AC2 + BC2. 253. It follows from ~ 252 that AC = AB2 -BC2, and BC' = AB2 - AC. That is, in any right triangle, the square of either leg is equal to the square of the hypotenuse, minus the square of the other leg. Ex. 34. If D is the middle point of side BC of right triangle ABC, and DE be drawn perpendicular to the hypotenuse AB, prove AB2 - CB2 = AE2 _ EB2. SIMILAR POLYGONS 121 Ex. 35. One leg of a right triangle is double the other. Find the ratio of the segments of the hypotenuse formed by a perpendicular drawn from the vertex of the right angle to the hypotenuse. Ex. 36. In an isosceles triangle ABC, the altitude CE is extended to D, so that Z DAE = 30~. It is then found that AE is a mean proportional between CE and ED. Find AB in terms of AC. Ex. 37. If from the sum of the squares of the diagonals of a square we subtract the sum of the squares of the four sides of the square, the result is zero. 254. Defs. The projection of a point upon a straight line of indefinite length, is the foot of the perpendicular from the point to the line. ' Thus, if line AA' be perpendicular to line CD, the projection of point A on line CD is __ point A'. CA B'D The projection of a finite straight line upon a straight line of indefinite length, is that portion of the second line included between the projections of the extremities of the first. Thus, if lines AA' and BB' be perpendicular to line CD, the projection of line AB upon line CD is line A'B'. PROP. XXIII. THEOREM 255. 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, minus twice the product of one of these sides and the projection of the other side upon it. A A C D B C B D FIG. 1. FIG. 2. Draw acute-angled A ABC; draw also A ABC having an obtuse angle at B. Let CD be projection of CA upon CB. We then have: Given C an acute Z of A ABC, and CD the projection of side AC upon side CB, extended if necessary. (~ 254) 122 PLANE GEOMETRY-BOOK III To Prove AB2 =BC2 +A -, BC x CD. Proof. 1. Draw line AD; then, AD A CD. (~ 254) 2. There will be two cases according as D falls ^ on CB (Fig. 1), or on CB extended (Fig. 2); in each / figure express BD in terms C D z~n G-. >C D B 0 B D of BC and CD. 3. Square both members of each of these equations. 4. Add AD2 to both members of the result. 5. Find values of AD+ BD2 and AD)+ CD2 from the figures. (~ 252) 6. Then, AB2- B B2 + AC2-2BCx CD. PROP. XXIV. THEOREM 256. In any triangle having an obtuse angle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides, plus twice the product of one of these sides and the projection of the other side upon it. Draw A ABC having an obtuse angle at C; draw AD I BC, meeting BC extended at D. We then have: Given C an obtuse Z of A ABC, and CD the projection of side AC upon side BC extended. To Prove AB2 = B 2 + A 2 + 2 BC x CD. (We have BD = BC + CD; square both members, using the algebraic rule for the square of the sum of two numbers, and then add AD2 to both members.) Ex. 38. In any triangle, if a perpendicular be drawn from the vertex to the base, the sum of the other sides of the triangle is to the sum of the segments of the base as the difference of the segments of the base is to the difference of the sides. Ex. 39. In right triangle ABC, BC = 3 AC2. If line CD be drawn from the vertex of the right angle to the middle point of AB, prove Z A CD equal to 60~. Ex. 40. If in triangle ABC, Z C = 120~, prove AB2 = BC2+ AC2 +AC x BC. SIMILAR POLYGONS 123 Ex. 41. CD is a chord of the greater of two concentric circles, intersecting the smaller cir- C A-D cumference at A and B. CD is 40, Z A OB is 60~, and AB equals BD; find distance CO and the distance from 0 to CD. PROP. XXV. THEOREM 257. In any triangle, if a median be drawvn from the vertex to the base, I. The sum of the squares of the other two sides is equal to twice the square of half the base, pius twice the square of the median. II. The difference of the sqluares of the other two sides is equal to twice the product of the base and the projection of the median upon the base. C 6/T\\ a A c D E B Draw A ABC having Z B > / A, and CD a median meeting AB at D. Draw line CE L AB. We then have-: Given DE the projection of median CD upon base AB (c) of A ABC; and AC (b) > BC (a). To Prove b2 + a' 2 = ~ 2 A CD, (1) and b2- a2 = 2 c x DE. (2) Proof. 1. Since b > a, E lies between B and D; and Z ADC is obtuse, and Z BDC acute. 2. In A ADC, b2 D2 -C2 2 AD X DE (~ 256) = AD C + CD- + c x DE. (3) 3. In A BDC, a= BD CD -2BD x DE (~ 255) = AAD2+ CD -c x DE. (4) 4. Adding and subtracting (3) and (4) gives (1) and (2). 124 PLANE GEOMETRY-BOOK III Ex. 42. One side of an inscribed quadrilateral is 14. The diameter of the circle is 24. Find the distance of this side from the centre of the circle correct to three decimal places. Ex. 43. The radii of two concentric circles are a and b, respectively. Find the length of a chord of the greater which is tangent to the smaller. Ex. 44. Two unequal circles are tangent externally at A. BC and DE are secants drawn through A, and terminating in one circumference at B and D, in the other at C and E. Chords BD and CE are drawn, forming triangles ABD, ACE. Prove AB x AE AC x AD. Ex. 45. If a median of a triangle is perpendicular to the side, the triangle is isosceles. Ex. 46. One of the equal sides of an isosceles right triangle is 12 inches in length; find the length of the median drawn to the hypotenuse. Ex. 47. Two parallel chords are 6 inches apart. The length of the smaller chord is 14 inches, and that of the greater is 32 inches. Find the distance of the smaller chord from the centre. Ex. 48. If D is the middle point of side BC of triangle ABC, rightangled at C, prove AB2 - AD2 = 3 CD2. PROP. XXVI. THEOREM 258. If any two chords be drawn through a fixed point within a circle, the product of the segments of one chord is equal to the product of the segments of the other. Draw a (; let P be any point within it. Through P draw two chords AB and A'Bt. We then have: Given AB and A'B' any two chords passing through fixed point P within 0 AA'B. To Prove AP x BP= A'P x B'P. Proof. 1. Draw lines AA' and BB'. 2. A A and B' are measured by one-half the same are, as also are A A' and B. (?) 3. A/ AA'P and BB'P are similar. (~ 236) 4. Find sides of A BB'P homologous to sides AP and A'P of AAAAP.p B (~ 239) 5. Then, A- = Bj; and AP x BP= A'P x B'P. (?) A'P BP () SIMILAR POLYGONS 125 Note. If, in the figure of Prop. XXVI, chord AB be supposed to revolve about point P as a pivot, the variable segments of the chord will have a constant product; and if one segment increases, the other decreases in the same ratio. If, for example, AP were doubled, BP would be halved. If two variable magnitudes are so related that, if one increases, the other decreases in the same ratio, they are said to be reciprocally proportional. Thus, the segments of a chord by a fixed point are reciprocally proportional; and the theorem may be written: If any two chords be drawn through a fixed point within a circle, their segments are reciprocally proportional. Ex. 49. In a triangle ABC, CD is a median and 4 CD2=Aj-2 CB2. Prove that AB2 equals AC2 plus four times the square of the projection of CD upon CB. Ex. 50. The sides of a triangle are 21, 20, and 18, respectively. Find the length of the median drawn to the side 18. Ex. 51. The sides of a triangle are 21, 20, and 18. A median is drawn to the side 18. Find the projection of the median upon this side. Ex. 52. Using values found in Ex. 50, find the altitude of the above triangle drawn to side 18. Ex. 53. The median drawn to the base of a triangle is 3-. The other sides of the triangle are 12 and 5. Find the base and altitude. Ex. 54. Given a circle and a chord CD. Find a point P in the cirPC 5 cumference such that PC- 5 PD 3 Ex. 55. If, in triangle ABC, the altitudes AD, BE, and CF intersect OA OE OB OF at 0, prove A = E and O OF OB OD OC OE Ex. 56. Two sides of a parallelogram are a and b, respectively. Prove that the bisectors of the angles of the parallelogram fqrm a rectangle, whose diagonal is b - a. Ex. 57. Two chords intersect in a circle. The segments of the first are 8 and 9. One segment of the second is 41; find the other. Also, if the segments of the first are a and b, and a segment of the second is c. 126 PLANE GEOMETRY-BOOK III PROP. XXVII. THEOREM 259. If through a fixed point without a circle a secant and a tangent be drawn, the product of the whole secant and its external segment is equal to the square of the tangent. Draw a 0. Through P, any point without the 0, draw line PC tangent to the 0 at C; through P draw a secant intersecting the circumference at B, and terminating in the circumference at A. We then have: Given AP a secant, and CP a tangent, passing through fixed point P without 0 ABC. To Prove APx BP= P2. (Draw lines AC, BC. Z A= BCP, for each is measured by 1 arc BC (?); then A ACP and BOP are similar, and their homologous sides are proportional.) 260. It follows from ~ 259 that if through a fixed point without a circle a secant and a tangent be drawn, the tangent is a mean proportional between the whole secant and its external segment. 261. If any two secants be drawn through a fixed point without a circle, the product of one and its external segment is equal to the product of the other and its external C segment. - For if P be any point without O ABC, AP and A'P secants intersecting the A B circumference at A and B, and A' and B', respectively, and CP tangent to the 0 at C, then AP x BP= A'P x B'P; for both AP x BP and A'P x B'P equal CP. (~ 259) Note. If any two secants be drawn through a fixed point without a circle, the entire secants and their external segments are reciprocally proportional. (Compare Note, ~ 258.) Ex. 58. Two chords, AB and CD, intersect at K. The centre of the circle is 0. CK = 8, KD = 5, OIK= 3. Find the radius of the circle. SIMILAR POLYGONS Ex. 59. A tangent to a circle at a fixed point is 12 feet, and the diameter of the circle is 12 feet. Locate the point from which the tangent is drawn. Ex. 60. One side of an equilateral triangle is 16; find the altitude, correct to three decimal places. Ex. 61. One side of a square is 1; find the diagonal of the square, correct to three decimal places. Ex. 62. A tangent 12 inches long and a secant 16 inches long are drawn from an external point to a circle. If the secant passes through the centre, find the radius of the circle. Ex. 63. If a secant be divided by the circumference in the ratio of 1 to 2, what is the length of the tangent drawn to its extremity? PROP. XXVIII. THEOREM 262. In any triangle, the product of any two sides is equal to the diamieter of the circumscribed circle, multiplied by the perpendicular drawn to the third side from the vertex of the opposite angle. B 'I Draw AABC. Circumscribe a 0 about it (~ 206); draw diameter AD, and line AE _ BC. We then have: Given AD a diameter of the circumscribed 0 ACD of A ABC, and line AE 1 BC. To Prove AB x AC = AD x AE. (In rt. AABD and ACE, ZD-Z C; then, the A are similar, and their homologous sides are proportional.) 263. It follows from ~ 262 that in any triangle, the diameter of the circumscribed circle is equal to the product of any two sides divided by the perpendicular drawn to the third side from the vertex of the opposite angle. 128 PLANE GEOMETRY-BOOK III PROP. XXIX. THEOREM 264. In any triangle, the product of any two sides is equal to the product of the segments of the third side formed by the bisector of the opposite angle, plus the square of the bisector. Draw A ABC, and line AD bisecting Z A, and meeting BC at D. We then have: Given, in A ABC, line AD bisecting Z A, meeting side BC at D. To Prove AB x A C= BD x DC+ AD2. Proof. 1. Circumscribe a 0 about A ABC; extend AD to meet the circumference at E, and draw line CE. 2. In A ABD and ACE, Z BAD = Z CAE, by hyp. 3. Also B= /E. (?) 4. Then, A ABD and ACE are similar. (?) 5. Find sides of A ACE homologous to sides AB and AD of A ABD. (~ 239) 6. Then, A A or AB x AC= AD x AE. (?) AD - AC 7. Then, AB x AC AD x (DE AD) =AD x DE + AD2. 8. Express AD X DE in terms of BD and DC. (~ 258) 265. The following cases in which two triangles are similar will be found useful in solving original exercises. Two triangles are similar: When they have two angles of one equal to two angles of the other (~ 236). When their homologous sides are proportional (~ 240). When they have an angle of one equal to an angle of the other, and the sides including these angles proportional (~ 242). When their sides are parallel each to each, or perpendicular each to each (~ 243). Two right triangles are similar: When an acute angle of one is equal to an acute angle of the other (~ 237). Ex. 64. CD is a common chord of two equal circles. At C tangents are drawn to each circle and meet the circumferences in E and F. Prove A EDC equal to A DFC. SIMILAR POLYGONS 129 Ex. 65. In a circle whose centre is 0, CD is a chord perpendicular to a diameter at any point F not the centre. At C a tangent is drawn, CD DE and DE perpendicular to the tangent. Prove = D.-* OC CF Ex. 66. AB is a diameter of a circle with centre at 0. CD is a chord perpendicular to AB at E. DF is a line through 0, and CF is perpendicular to DF. OC is drawn. Prove A COE similar to A DFC. Ex. 67. ABC is a right triangle, 0 any point in the hypotenuse AB. Through 0 two perpendiculars are drawn, one to the leg AC, the other to AB. These perpendiculars are extended to meet any line DE parallel to AB, A o _. at x and y, respectively. Prove xy- Ox D- - E BO AB CONSTRUCTIONS PROP. XXX. PROBLEM 266. To construct a fourth proportional (~ 215) to three given straight lines. m n p Given lines m, n, and p.,B E,-" - -, -D, \ \w _w X, % ~ ~ ~ ~~~~~~~~~~~~~~~~~~~~ I d P F G Required to construct a fourth proportional to m, n, and p. Construction. 1. Draw lines AB and AC, making any L. 2. On AB take AD = m, and DE = n; on AC take AF = p. 3. Draw line DF, and line EG II DF, meeting AC at G. 4. Then, m_ p n FG (?) 267. If AF = n, the above proportion becomes = m n n FG In this case, FG is a third proportional (~ 214) to m and n. 180 PLANE GEOMETRY -BOOK III PROP. XXXI. PROBLEM 268. To construct a mean proportional (~ 214) between two given straight lines. D m ^ —/ \ Given lines m and n. A m B n C ~ Required to construct a mean proportional between m and n. Construction. 1. On line AE, take AB = m, and BC= n. 2. With AC as a diameter, describe a semi-circumference, and draw line BD I AC, meeting the arc at D. 3. Then, BD=BD (~ 250, 1) BD n 269. By aid of ~ 268, a line may be constructed equal to ca, where a is any number whatever. Thus, to construct a line equal to -/3, we take AB equal to 3 units, and BC equal to 1 unit. Then, BD = VAB X BC (~ 217) = /3 X 1 = V3. PROP. XXXII. PROBLEM 270. To divide a given straight line into parts proportional to any number of given lines. F.C __ m P.,-' / n-/ 6 E H Given line AB, and lines m, n, and p. Required to divide AB into parts proportional to m, n, and p. Construction. 1. On line AC take AD = m, DE = n, and EF= p. 2. Draw line BF; and lines DG and EH II BF, meeting AB at G and H, respectively. SIMILAR POLYGONS 131 3. Then, AG GOH B (1) m n p Proof. 4. In A AEIH A AG=GII (?) AE m n 5. In AABF, AH B; whence, equation (1). (?) AE p Ex. 68. Construct a fourth proportional to three lines whose lengths are 8, 9, and 12, respectively. Ex. 69. Construct a fourth proportional to 12, 8, and 6. Ex. 70. Construct a third proportional to lines whose lengths are 8 and 6, respectively. Ex. 71. Construct a mean proportional to lines whose lengths are 9 and 4. Ex. 72. Construct a line whose length is /5. PROP. XXXIII. PROBLEM 271. Upon a given side, homologous to a given side of a given polygon, to construct a polygon similar to the given polygon. D' A B A' B' Given polygon ABCDE, and line A'B'. Required to construct upon side A'B', homologous to AB, a polygon similar to ABCDE. Construction. 1. Divide polygon ABCDE into A by drawing diagonals EB and EC. 2. Construct A A'B'E', ABE similar, with / A' =- Z A and Z A'B'E' = / ABE. (?) 3. In like manner, construct A B'C'E' similar to A BCE, and A C'D'E' similar to A CDE. PLANE GEOMETRY-BOOK III 272. Def. A straight line is said to be divided by a given point in extreme and mean ratio when one of the segments (~ 231) is a mean proportional between the whole line and the other segment. D A C B Thus, line AB is divided internally in extreme and mean ratio at C if AB AC AC BO' and externally in extreme and mean ratio at D if AB AD AD BD PROP, XXXIV. PROBLEM 273. To divide a given straight line in extreme and mean ratio (~ 272). // / 2.'~ 1 D -------— ' B Given line AB. Required to divide it in extreme and mean ratio. Construction. 1. With radius BE = 1 AB, draw 0BFG tangent to AB at B; and line AE cutting circumference at F and G. 2. On AB take AC = AF; on BA extended take AD = AG; then, AB is divided at C internally, and at D externally, in extreme and mean ratio. Proof. 3. By ~ 2619 AG AB AG AB AB A oAB -AC (1) SIMILAR POLYGONS -133 AG - AB_ AB -AC 4. Then, (?). (2) AB AC 5. By cons.. AR =2 BE =FG. 6. Then, AG - AB = AG - FG = AF= AC. AC BC ARB AC 7. Substitute in (2), - or A =A (~ 220). (3) AR AC' AC BC S. Again from (1), AGA-AB+AC (? (4) AG AR 9. But AG~+ AB =AD +AB = BD; ad AB + AC = FG + AF = AD. (?) 10. Substitute in (4), BDAD or ADAR. AD -AR BRD AD 274. Putting AB = m, AC = x, BC =mz - xin (3), x MXz (M- (~ 216) Then, x2 = mB - mx; or x2 + MX = -)L2 Multiplying by 4, and adding mIL2 to both members, 4 x2 + 4 mnx ~ 9),2 = 4 m2 + m22 = 5M2 Extracting the square root of both members, 2 x + m = ~ m V 5. Since x cannot be negative, we take the positive sign before the radical sign; then, 2x=mV= - m and x (or AC) n m(V5-1) BOOK IV AREAS OF POLYGONS PROP. I. THEOREM 275. Two rectangles having equal altitudes are to each other as their bases. The words "rectangle," 'parallelogram," "triangle," etc., in the propositions of Book IV, mean the amount of sulface in the rectangle, parallelogram, triangle, etc. Case I. When the bases are commensurable. B r, C F G AA E DT Draw rectangles ABCD and EFGH having equal altitudes and the bases AD and Eli commensurable. We then have: Given rectangles iABCD and EFGH, with equal altitudes AB and EF, and commensurable bases AD and Eli. ABOD AD To Prove EF E (1) EFGH EH 0 Proof. 1. Let AK be contained 5 times in AD, and 3 times in EH. 2. Find ratio of AD to EIT. 3. Through points of division of AD and EH draw Js to these lines. 4. Find ratio of ABCD to EFGH. (~ 111) 5. Prove equation (1). (?) 134 AREAS OF POLYGONS 135 Case II. When the bases are incommensurable. B' L B E- 1 G D E Draw rectangles in accordance with the statement. We then have: Given rectangles ABCD and EFGH, with equal altitudes AB and EF, and incommensurable bases AD and EH. ABCD AD To Prove AB(2) EFGH EH Proof. 1. Divide AD into any number of equal parts. 2. Let one of these parts be contained exactly in EK, with a remainder KH < one of the parts. 3. Draw line KL 1- EIl, meeting FG at L. ABCD AD 4. By Case I, AFEK E.EFLI -- EK' 5. Let number of subdivisions of AD be indefinitely increased. 6. Find limits ABOD and AD EFLK EK 7. Prove equation (2) by Theorem of Limits. 276. Since either side of a rectangle may be taken as the base, it follows that Two rectangles having equal bases are to each other as their altitudes. PROP. II. THEOREM 277. Any two rectangles are to each other as the products of their bases by their altitudes. a a a R V I --- —-------------- b b' 136 PLANE GEOMETRY-BOOK IV Draw any two rectangles M and N. We then have: Given M and iV rectangles, with altitudes a and a', and bases b and b' respectively. X a'xb' (1 a a) at| N a ct To Prove a 2. M and R have = altitudes; and bases b and b', respectively; find ratio M to R by ~ 275. 3. R and have = bases; and altitudes a and a, respectively; find ratio R to V by ~ 276. 4. Multiplying results of steps (2) and (3), gives (1). DEFINITIONS 278. The area of a surface is its ratio to another surface, called the unit of surface, adopted arbitrarily as the unit of measure (~ 180). The usual unit of surface is a square whose side is some linear unit; for example, a square inch or a square foot. 279. We call two surfaces equivalent (a) when their areas are equal. The dimensions of a rectangle are its base and altitude. PROP. III. THEOREM 280. The area of a rectangle is equal to the product of its base and altitude. In all propositions relating to areas, the unit of surface (~ 278) is understood to be a square whose side is the linear unit. a M1 1 N b 1 Draw any rectangle M, and a square N whose side is 1. We then have: Given a the altitude and b the base, of rect. M; and N the unit of surface, a square whose side is the linear unit. AREAS OF POLYGONS 137 To Prove that, if N is the unit of surface, area M=a X b. (1) Proof. 1. We have V a x b ( 277) 2. being unit of surface, V is area of M; whence equation (1). Note. The statement of Prop. III is an abbreviation of the following: If the unit of surface is a square whose side is the linear unit, the number which expresses the area of a rectangle is equal to the product of the numbers which express the lengths of its sides. An interpretation of this form is always understood in every proposition relating to areas. 281. By ~ 280, the area of a square equals the square of its side. Ex. 1. Find the ratio of the area of a square to the product of its diagonals. Ex. 2. The area of a rectangle whose base is 24 is 482. Find the diagonal of the rectangle. Ex. 3. How many shingles will it take to cover a roof 36 feet by 18 feet (one side) with shingles that average 4 inches wide, 16 inches long, and are laid 6 inches to the weather? PROP. IV. THEOREM 282. The area of a parallelogram is equal to the product of its base and altitude. Draw / ABCD, AD (b) being the base; draw DF (a) the altitude meeting BC at F. We then have: Given b the base, and a the altitude, of EZ ABCD. To Prove area ABCD = a X b. (1) Proof. 1. Draw line AE 11 DF, meeting CB prolonged at E. 2. Then, rt. A ABE and DCF are equal. (~~ 61, 104) 3. If from figure ADCE we take A ABE, there remains 7 AC; if we take A DCF, there remains rect. AF. 4. Then, area ABCD = area AEFD; whence equation (1). 138 PLANE GEOMETRY -BOOK IV 283. It follows from ~ 282 that: 1. Two parallelograms having equal bases and equal altitudes are equivalent (~ 279). 2. Two parallelograms having equal altitudes are to each other as their bases. 3. Two parallelograms having equal bases are to each other as their altitudes. 4. Any two parallelograms are to each other as the products of their bases by their altitudes. Ex. 4. The area of a parallelogram is 288, the base is twice the altitude. Find the dimensions. Construct the parallelogram. Can more than one such parallelogram be drawn? How many and what parts are necessary for a definite figure? Why? Ex. 5. Find the ratio of the area of a rhombus to the product of its diagonals. PROP. V. THEOREM 284. The area of a triangle is equal to one-half the product of its base anid altitude. Draw A ABC having base BC (b); draw AE I BC, meeting BC, or BC extended, at E. We then have: Given b the base, and a the altitude of A ABC. To Prove area ABC = I a x b. (Draw line AD II BC, and line CD II AB. By ~ 105 AC divides 0 ABCD into two equal A.) 285. It follows from ~ 284 that: 1. Two triangles having equal bases and equal altitudes are equivalent. 2. Two triangles having equal altitudes are to each other as their bases. 3. Two triangles having equal bases are to each other as their altitudes. 4. Any two triangles are to each other Us the products of their bases by their altitudes. AREAS OF POLYGONS 139 5. The area of any triangle is one-half that of a parallelogram having the same base and altitude. Ex. 6. Draw a line dividing a given right triangle into two equivalent isosceles triangles. Ex. 7. Prove that each of its medians divides a triangle into two equivalent parts. Ex. 8. The base of a triangle is 37 feet, the altitude is 32 feet. How many square yards in its area? Ex. 9. The area of a triangle is 216, the altitude is 12; find the base. Ex. 10. If E and F are the middle points of sides A E B AB and CD, respectively, of parallelogram ABCD, the lines AF, EF, and CE divide the parallelogram into four equal triangles. Ex. 11. If the middle point of any side of a D F C parallelogram be joined to the opposite vertices, the triangle included by these lines and the opposite side is equivalent to one-half the parallelogram. Ex. 12. If the middle point of a diagonal of any quadrilateral be joined to the opposite vertices, the quadrilateral is divided into two pairs of equivalent triangles, and into two equivalent parts. PROP. VI. THEOREM 286. The area of a trapezoid is equal to one-half the sum of its bases multiplied by its altitude. Draw trapezoid ABCD, AB (b) and DC (b') being II sides. From D draw altitude DE (a) meeting AB at E. We then have: Given AB (b) and DC (b') the bases, and DE (a) the altitude, of trapezoid ABCD. To Prove area ABCD = a x - (b + br). (Draw diagonal BD. The trapezoid is composed of two A whose altitude is a, and bases b and b', respectively.) 287. Since the line joining the middle points of the nonparallel sides of a trapezoid is equal to one-half the sum of the bases (~ 132), it follows that The area of a trapezoid is equal to the product of its altitude by the line joining the middle points of its non-parallel sides. 140 PLANE GEOMETRY-BOOK IV Note. The area of any polygon may be obtained by finding the sum of the areas of the triangles into which the polygon may be divided by drawing diagonals from any one of its vertices. But in practice it is better to draw the longest diagonal, and draw perpendiculars to it from the T —L ---remaining vertices of the polygon. The polygon will then be divided into right triangles and trapezoids; and by measuring the lengths of the perpendiculars, and of the portions of the diagonal which they intercept, the areas of the figures may be found by ~~ 284 and 286. PROP. VII. THEOREM 288. Two similar triangles are to each other as the squares of their homologous sides. Draw A ABC; and construct A A'B' C' similar to A ABC, AB (c) and A'B' (cl) being homologous sides. We then have: Given c and c' homologous sides of A ABC and A'B'C', respectively. ABC c2 To Prove A' () Proof. 1. Draw altitudes CD(h) and C'D'(h') 1 AB and A'B', respectively. 2. Find ratio ABC to A'B'C' in terms of h and c, and h' and c'. (~ 285, 4) 3. Express ratio h to h' in terms of c and c'. (~ 245) 4. Substituting in result of step 2, gives (1). 289. Note. Two similar triangles are to each other as the squares of any two homologous lines. Ex. 13. A field has two sides parallel, the lengths of these sides being 160 yards and 302 yards, respectively. The distance between the parallel sides is 48 rods. How many acres has the field? Draw a diagram. Ex. 14. A field has two parallel sides whose lengths are 322 and 168 yards, respectively. The area of the field is 12 acres. Find the distance between the parallel sides and make a diagram of the field. Ex. 15. A square field contains 3 acres 12 square rods. Find its perimeter in rods, correct to three decimal places. AREAS OF POLYGONS 141 Ex. 16. A square field contains one-half a square mile. Find its diagonal in rods, correct to three decimal places. Ex. 17. The base of a rectangle is 462 feet. One diagonal is 504 feet; find the area of the rectangle. Ex. 18. One-quarter of a government tree claim, containing 160 acres, is to be planted with trees 4 feet apart each way. How many trees will be required? Will the same number of trees be required whether the part selected for planting be rectangular or square? ~Ex. 19. The area of a trapezoid is 2774, the altitude is 38, the upper base is 29; find the lower base. Ex. 20. One of the equal sides of an isosceles trapezoid is 10 feet, the lower base is 12 feet longer than the upper base; find the altitude. Ex. 21. One side of a rhombus is 20, the longer diagonal is 32; find the other diagonal, the area, and the altitude. Ex. 22. If from any point within an equilateral triangle perpendiculars to the sides be drawn, the area of the triangle is equal to one-half the sum of the perpendiculars multiplied by one side of the triangle. Ex. 23. Find the area of a trapezoid whose parallel sides are a and b, and whose altitude is 2, divided by the sum of the parallel sides. If a and b vary, will the area vary? Will the form of the trapezoid change? Draw the trapezoid. PROP. VIII. THEOREM 290. Two triangles having cn angle of one equal to an angle of the other, are to each other as the products of the sides including the equal angles. A B B OC ' Draw A AB'C' and line BC meeting AB' at B, and AC' at C. We then have: Given Z A common to A ABC and AB'C'. 142 PLANE GEOMETRY-BOOK. IV ABC AB x AC A To Prove ' B x C' () R'cY — =AnB' x AUC" Proof. 1. DrawlineB'C; AABCand B AB'C have same vertex C, and bases AB and AB' in same str. line. 2. Express ratio ABC to AB'C in terms of AB and AB'. (~ 285, 2) 3. A AB'C and AB'C' have same vertex B', and bases AC and AC' in same str. line. 4. Express ratio of AB'C to AB'C' in terms of AC and AC'. 5. Multiplying results of steps (2) and (4) gives equation (1). Ex. 24. If the equal angles of an equilateral triangle be bisected, the bisectors and the base form a triangle whose area is one-third that of the given triangle. Ex. 25. Any two altitudes of a triangle are inversely as the sides to which they are drawn. Ex. 26. The altitude of a triangle is 8, its area is 48, and the perpendicular to the base divides the base in the ratio of 2 to 1. Find the sides of the triangle. Construct the triangle. Ex. 27. Two triangles have equal areas and equal vertical angles. Prove that the products of the sides including the equal angles are equal. PROP. IX. THEOREM 291. Trwo similar polygons are to each other as the squares of their homologous sides. D Draw polygon ABCDE; construct polygon A'B'C'D'E' similar to ABCDE, AB (a) and A'B' (a') being homologous sides. We then have: AREAS OF POLYGONS 143 Given a and a' homologous sides of similar polygons AC and A'C', whose areas are K and K', respectively. K a2 To Prove ' (1) K' a Proof. 1. Draw diagonals from E and E'. 2. A ABE is similar to A A'B'E'. (~ 247) 3. Express ratio ABE to A'B'E' in terms of a and a'. (~ 288) 4. Also, BCE BC2 a2 DE CD 2 a2 soB'C'E-' -C2 at; C'D'E' CL2 a2 (~ 234, 2) A. BE BCE CDE 5. Then, ABE' - B' CE(?' ~ A'B'EW B'C'IE1 CI'Dv / 6. Apply ~ 225 to above result. 7. Substituting values of members in above result gives (1). Ex. 28. The area of a polygon having a side 6 feet is 112 square feet; find the area of a similar polygon whose homologous side is 9 feet. Ex. 29. The ratio of the homologous sides of two similar polygons Is 2 to 3. What is the ratio of their areas? Find a side of the first. If the area of the second is A, what is the area of the first? Ex. 30. The area of a polygon is 15 the area of a similar polygon; find the ratio of their homologous sides. PROP. X. PROBLEM 292. To find the area of a triangle in terms of its sides. Draw A ABC, a, b, c, being the sides opposite A A, B, and C, respectively, and Z C acute. We then have: Given sides a, b, and c of A ABC. Required to find area ABC in terms of a, b, and c. Solution. 1. Draw altitude AD. 2. Then, c2= a2+ b2 2a x CD. (~255) 3. Transposing, 2 a x CD = a2 + b2 - c2, or a" + b2 ~ c2 ~or ~CD =a+b2a 2a 144 1PLANE GEOMETRYY-BOOK IV 4. Then, AD2 = AC CD (~ 253) (AC+ CD)(AC - CD) = (b+ C2 + b - c a - (2ab + a2 + b2 - c2)(2 ab - a2 - b2 + C2) 4a 2 -[(a + b)2 - c2][c2 _ (a - b)2] 4 a2 - (a b c)(aA-b-c)(c a-b)(c-a+b) 4 a 2 5. Now leta+b+c=2s. -2 - 2 s(2 s - 2 c)(2 s - 2 b)(2 s - 2 a) 6. Then, ADh 4 2 - 16 s(s - a)(s - b)(s - c) 4 a2 7. Then, AD = 2 Vs(s - a)(s - b)(s - c). a 8. Then, area ABC = I a x AD (?) Vs(s-a)(s- b)(s-c). As an example of ~ 292, let it be required to find the area of a triangle whose sides are 13, 14, and 15. Let a = 13, b 14, and c z 15; then s = - (13 + 14~+ 15) = 21. Whence, s - a _-8, s - b = 7, and s - c = 6. Then, the area of the triangle is V21 x 8 x 7 x 6 =V3 x 7 x 22 x 7 x 2 x 3 =V2 x 32 x 72 22 x 3 x 7 84. Ex. 31. If a line be drawn from the middle point of each side of a quadrilateral to the middle point of the next side in order, the figure formed is a parallelogram whose area is one-half that of the quadrilateral. Ex. 32. Prove that the lines connecting the middle points of the sides of a triangle divide it into four equal triangles. AREAS OF POLYGONS 145 293. Note. Since the area of a square is equal to the square of its side (~ 281), we may state Prop. XXII, Book III, as follows: In any right triangle, the square described upon the hypotenuse is equivalent to the sum of the squares described upon the legs. The theorem in the above form may be proved as follows: F M E Draw A ABC, right-angled at C; on AB, AC, and BC describe squares ABEF, ACGH, and BCKL, respectively. We then have: Given squares AE, AG, and BIT described upon the 'hypotenuse and legs of A ABC. To Prove ABEF == ACGH + BCKL. (1) Proof. 1. Draw line CD L AB, and prolong it to meet EF at lM; also, lines BH and CF. 2. We haveA A ABHI A ACF; since AB = AF, AH = AC, and BAH = COAF, each being a rt. L + Z BAG. (?) 3. Then ABH = -A ACGH, having same base and altitude. (~ 285, 5) 4. Again, ACF- ADMF. 5. Then 1 ACGH C ~ ADMF, or ACGH ADMF. 6. Similarly, drawing lines from A to L and from C to E, BCKL = BDME. 7. Adding results of (5) and (6), gives equation (1). 146 PLANE GEOMETRY-BOOK IV Note. The theorem of ~ 293 is supposed to have been first given by Pythagoras, and is called after him the Pythagorean Theorem. Several other propositions of Book III may be put in the form of statements in regard to areas; for example, Props. XXIII and XXIV. Ex. 33. In a triangle whose sides are a, b, and c, the angle opposite side c is 90~. Find the area in terms of the sides. Ex. 34. The sides of a triangle are a + b, b + c, c + a; find the altitude to side a + b. Ex. 35. The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of the four sides of the parallelogram. [Project one line upon another.] Ex. 36. Given a trapezoid with bases bl and b2, altitude h and area S, find the altitude a of an equivalent equilateral triangle whose area is &2 [1-= 82.] Ex. 37. If lines be drawn through the extremities of the diagonals of a rhombus parallel to the diagonals, they form with the diagonals four rectangles, each equivalent to one-half the rhombus. Ex. 38. The sides of a rectangle are 24 and 36. The altitude of a triangle of the same area is 33; find the base to which this altitude is drawn. Can more than one triangle be drawn which satisfies the conditions of the problem? Ex. 39. AB, CD, EF, tII are parallel lines. EF is equidistant from AB and GI. A B State and prove all theorems suggested by the figure. C a.\) t) CC~ D Ex. 40. Given two chords AB and AC / E\\ -F in a circle of 5-inch radius, such that central c/ angle AOB =- 60', and OB is perpendicular to AC, to find the distance of each chord from / e the centre 0. Ex. 41. If D is the intersection of the perpendiculars from the vertices of triangle ABC to the opposite sides, prove AB2 -AC-2 = B2 -CD-2 Ex. 42. Draw figure and give a geometrical proof of the algebraic theorem: The square of the sum of two numbers equals the square of the first, plus twice the product of the first by the second, plus the square of the second. AREAS OF POLYGONS 147 Ex. 43. Find the ratio of the areas of two triangles which have two sides of one equal to two sides of the other, and the angles formed by the equal sides supplementary. Ex. 44. The square of the altitude of an equilateral triangle is threefourths the square of one side. Ex. 45. A person in a train observes that the direction in which the rain appears to fall makes an angle of 60~ with the direction of the train's motion, which is at the rate of 45 miles an hour. The rain is really falling vertically. What is its velocity? CONSTRUCTIONS PROP. XI. PROBLEM 294. To construct a square equivalent to the sum of two given squares. P A B Given squares M and N. Required to construct a square M -+ N. Construction. Construct a rt. A with legs equal to the sides of M and N, and square P with its side equal to the hypotenuse; then, P M + N. (Prove by ~ 293.) 295. By an extension of ~ 293, a square may be constructed equivalent to the sum of any number of given squares. Thus, suppose we have given three squares whose D sides are m, n, and p, respectively. p, Take line AB = m; draw line AC_1AB, and equal to n, and line BC; draw line CD -L BC, and \ equal top, and line BD. n ' Then, the square having its side equal to BD --- will be the sum of the given square will be ~ the sum of the given squares. 148 PLANE GEOMETRY- BOOK IV PROP. XII. PROBLEM 296. To construct a square equivalent to the difference of two given squares. IV Al m Given squares M and N, 1 > N. Required to construct a square = 1= 1- N. Construction. Draw rt. Z BAD, with AB equal to a side of N; find point C in AD, its distance from B equal to a side of 1V; then, a square with side equal to AC will be - M - V. (Prove by ~~ 253 and 281.) PROP. XIII. PROBLEM 297. To construct a square equivalent to a given parallelogram. H DC A Y~ B F Gc Given 0 ABCD. Required to construct a square = ABCD. Construction. A square having for its side FG a mean proportional between AB and DE (~ 268), the altitude of the 0, will be ABCD. (By ~ 217, FG2 - AB X DE.) 298. A square may be constructed equivalent to a given triangle by taking for its side a mean proportional between the base and one-half the altitude of the triangle. Ex. 46. To find a point within a parallelogram through which if a:.y line be drawn, the parallelogram is divided into two equivalent parts. Are these parts parallelograms? Trapezoids? Triangles? AREAS OF POLYGONS 149 PROP. XIV. PROBLEM 299. To construct a rectangle equivalent to a given square, having the sum of its base and altitude equal to a given line. C zDj.... —.. [ ___...__\ EIYIJ [23 _,, ' A E B Given square M, and line AB. Required to construct a rectangle = M, having the sum of its base and altitude equal to AB. Construction. 1. Construct semi-circumference with diameter AB; and find point D in its arc by drawing II to AB at a distance from AB equal to a side of M. 2. Draw line DE I AB; then rectangle with dimensions AE and BE will be L M. (By ~ 250, 1, D E -; and DE = AE x BE.) DE BE PROP. XV. PROBLEM 300. To construct a rectcangle equlivalent to a given square, having the difference of its base and altitude equal to a given line. \ C,,1 ',,? Given square M, ancd line AB. Required to construct a rectangle - M, having the difference of its base and altitude equal to AB. Construction. 1. Describe 0 with diameter AB. 2. Draw tangent AC equal to a side of M. 150 PLANE GEOMETRY-BOOK IV 3. Through C and centre of 0 draw a secant, and construct a rectangle with dimensions equal to the secant and its exterior segment. (By ~ 259, CD x CE =- A2; also, AB= CE - CD.) PROP. XVI. PROBLEM 301. To construct a square having a given ratio to a given square. L _ _ _ 0___ __ _ L _ 1 m D B'" Required to construct a square having to M the ratio -. Construction. 1. Take AD = m, BD = n; with AB as diameter describe a semi-circulufelrence; find point C where 1_ at D meets arc. 2. On CA take CE equal to a side of M; at i draw 11 to AB meeting CB at F; square N with side equal to CF will be required square. E AC 7 2 ~12 Proof. 3. We have AE=; then E2 =G2 (1) GE AG' GCE AG BGC AB x BAD BAD nm 4. In rt. A ABG, Ac2 AB x AD AD n (~ 0 5. Substitute this in (1); and for CF2 and CE2 areas N and 3M. Ex. 47. To inscribe a parallelogram in an equilateral triangle, one angle of the parallelogram being 1200. Can more than one such parallelogram be drawn? Ex. 48. To inscribe a rhombus in a given triangle, the rhombus and triangle to have one angle in common. AREAS OF POLYGONS 151 PROP. XVII. PROBLEM 302. To construct a triangle equivalent to a given polygon. A G D C Given polygon ABCDE. Required to construct a A ABCDE. Construction. 1. Take three consecutive vertices, A, B, C; draw diagonal AC, and line BF II AC, meeting DC prolonged at F, also line AF. 2. A ACF - A ABC; then, polygon AFDE - ABCDE (having common part ACDE), and has a number of sides less by 1. 3. Draw diagonal AD, and line EG II AD, meeting CD prolonged at G, also line AG; then, A AED =E A AGD, whence A AFG = polygon AFDE, or to ABCDE. Proof. 4. A ACF and ABC have same base and altitude. (~ 96) 5. Then area ACF = area ABC. (?) 6. Similarly, area AED = area AGD. Note. By aid of ~~ 302 and 298, a square may be constructed equivalent to a given polygon. Ex. 49. The bases of a trapezoid are 8 and 10, respectively, the altitude 6. Construct an equivalent equilateral triangle. [An equilateral triangle can be constructed when its altitude is known.] Ex. 50. The bases of a trapezoid are 6 and 8, respectively; the area is 35. Construct the trapezoid and an equivalent equilateral triangle. 152 PLANE GEOMETRY -BOOK IV Ex. 51. Given A ABC. Draw a line from A to BC which shall divide the triangle into two triangles the ratio of whose areas is 1 to 2. PROP. XVIII. PROBLEM 303. To costruct Ca polygon similar to a given polygon, and having a given ratio to it. m - IIII Given polygon AC, and lines m and n. Required to construct a polygon similar to it the ratio -. m to AC, and having Construction. 1. Let A'B' be side of square having to the square described on AB the required ratio. (~ 301) 2. Polygon A'C', similar to AC, will have required ratio to it. ( y a291,area A'C' A'B' 2 \ By ~ 291, - 2, which equals -. area AC AB mv PROP. XIX. PROBLEM 304. To construct a polygon similar to one gons, and equivalent to the other. A B n I A... of two given polyA' B' Given polygons M and N. Required to construct a polygon similar to M, and = — N. Construction. 1. Find m and n sides of squares 3 1f and N. (Note, ~ 302) AREAS OF POLYGONS 153 2. Polygon P, similar to M, with A'B' a fourth proportional to m, n, and AB, will be == N. area if AB2 I2 Proof. 3. By ~ 291, area by cons. area P 1A'B'2 n area M area 1M 4. Then area area and area P = area 1 area P' area N Ex. 52. To inscribe in a given triangle a parallelogram whose area is one-half the area of the triangle. Can a parallelogram be inscribed whose area is still greater? Ex. 53. Through a given point P, either within or without a given angle, to draw a line which shall form with the sides of the angle a triangle of given area. Ex. 54. To construct a triangle whose angles shall be equal, respectively, to the angles of a given triangle, and whose area shall be four times the area of the given triangle. 1300K V REGULAR POLYGONS.-MEASUREMENT OF THE CIRCLE. - LOCI 305. Def. A regular polygon is a polygon which is both equilateral and equiangular. Puop. I. THEOREM 306. A circle can be cirmcunmcribed about any regular polygon. D a BB Given regular polygon ABCDE. To Prove that a 0 can be circumscribed about it. Proof. 1. Draw circumference through A, B, C; also, radii QA, OB, OC, and line OD. 2. In A QAB, OCD, O = OC, AB GD. (?) 3. Z BAnl = ABC- Z OBC, ZOCDZ =Z BCD - LZOCB. 4. Since Z ABC L BCD (?), and Z OBC Z OCB (?), Z OBA = Z OCD. o. Then A OQAB A OCD, and QA4 = OD. (?) 6. Then circumference through A, B, C passes through D; and similarly through E. 154 REGULAR POLYGONS 155 307. Since AB, BC, CD, etc., are equal chords of the circumscribed 0, they are equally distant from 0. (~ 164) Then, a 0 drawn with 0 as a centre, and a line OFl to any side AB as a radius, will be inscribed in ABODE. Hence, a circle can be inscribed in any regutlar polygon. 308. Defs. The centre of a regular polygon is the common centre of the circumscribed and inscribed circles. The angle at the centre is the angle between the radii drawn to the extremities of any side; as AOB. The radius is the radius of the circumscribed circle, Oil. The apothem is the radius of the inscribed circle, OF. 309. From equal A OAB, OBC, etc., we have ZAOB =ZBOC = COD, etc. (?) But the sum of these As is four rt. A. (~ 19) Whence, the angle at the centre of a regular polygon is equal to four right angles divided by the number of sides. PROP. II. THEOREM 310. If a circumference be divided into equal arcs, their chords form a regular inscribed polygon. L A F K H Given circumference ACD divided into five equal arcs, AB, BC, CD, etc., and chords AB, BC, etc. To Prove polygon ABCDE regular. Proof. 1. Sides of polygon are equal by ~ 159. 2. Each Z is measured by one-half the sum of three of the equal arcs. (~ 192) 156 PLANE GEOMETRY-BOOK V 311. Let lines LF, FG, etc., be tangent to 0 ACD at A, B, etc., respectively, forming polygon FGHIIL. In A ABF, BCG, CDH, etc., AB = BC = CD, etc. (~ 159) Also, since arc AB = arc BC = arc CD, etc., we have Z BAF = ABF= Z CBG = BCG, etc. (~ 196) 1 A F K a Whence, ABF, BCG, etc., are equal isosceles A. (~~ 49, 90) Then, Z F= Z G = Z I, etc., and BF = BG = CG = CH, etc. (~ 48) Then, FG= GH = HI, etc., and polygon FGIIKL is regular. Then, if a circumiference be divided into equal arcs, tangents at the points of division obrm a regular circunmscribed polygon. 312. It follows from ~~ 310 and 311 that 1. If from the middle.point of each arc subtended by a side of a regular inscribed polygon lines be drawn to its extremities, a regular inscribed polygon of double the number of sides is formed. 2. If at the middle point of each arc included between two consecutive points of contact of a regular circumscribed polygon tangents be drawn, a regular circumscribed polygon of double the number of sides is formed. 3. An equilateral polygon inscribed in a circle is regular. For its sides subtend equal arcs. (?) REGULAR POLYGONS 157 PROP. III. THEOREM 313. Tangents to a circle at the middle points of the arcs subtended by the sides of a regular inscribed polygon, form a regular circumscribed polygon. A' E Ef D Given ABCDE a regular polygon inscribed in 0 AC, and polygon A'B'C'D'E' with sides A'B', B'C', etc., tangent to ( AC at middle points F, G, etc., of arcs AB, BC, etc., respectively. To Prove A'B'C'D'E' a regular polygon. (Arc AF = are BF = arc BG = arc CG, etc.; use ~ 311.) PROP. IV. THEOREM 314. Regular polygons of the same number of sides are similar. D E A B A' B' Given AC and AC' regular polygons of five sides. To Prove AC and A'C' similar. (Use ~ 233.) Find the angle, and the angle at the centre, Dx. 1. Of a regular pentagon. Ex. 2. Of a regular dodecagon. Ex. 3. Of a regular polygon of 32 sides. Ex. 4. Of a regular polygon of 25 sides. 158 158 ~PLANE GEOMETRY - BOOK V PRpOP. V. THEOREM 315. The perimeters of two simtilar regular polygon~s are to each other as their radii, or as their apotheins. D 0 p A F B Al'F'B' Given 0, 0' the centres, QA (R), O'A' (B') the radii, OF (r-), O'F (r-') the apothems, and P, P' the perimeters, respectively, of similar regular polygons AC and A'C'. P B rTo Prove(1 Proof. 1. Draw lines OR, O'R'; then L AORB Z A'O'R'. (~ 309) 2. Since R - OR As QAB, O'AR' are similar. (~ 242) 3. Then, AB -R (~ 245) A'B' R' r AB P 4. Now A '(~248) 5. Substituting in resnlt of (3) gives equation (1). 316. If K denote the area of polygon AC, and K' of A'C, IK AR'Z2 (~291) But, AR -R = whence, K 11R2 r2 A' R'B'r"'R 2 T hat is, the areas of two similar regutlar polygons are to each other as the squtares of their radii, or as the squares of their apothems. REGULAR POLYGONS 159 PROP. VI. THEOREM 317. The area of a reglar polygon is equal to one-half the prooduct of its perimeter and apothem. D Given P the perimeter, and r the apothem, of regular polygon AC. E — ) To Prove area AC= P x r. (A OAB, OBC, etc., have common altitude r.) A F B Ex. 5. One side of a regular hexagon is 2/v-3. Find the radius of the inscribed circle. Ex. 6. The apothem of a regular polygon is 9, and the area 182. What is the perimeter? Ex. 7. In an equilateral triangle find the area K in terms of the apothem r. Ex. 8. In a regular hexagon find the apothem in terms of B; find the area in terms of R; find the area in terms of the apothem. Ex. 9. Find the ratio of the altitude and apothem of an equilateral triangle. Ex. 10. A triangle whose area is 120 is inscribed in a semicircle whose radius is 13. Construct the triangle. PROP. VII. PROBLEM 318. To inscribe a square in a given circle. Given 0 AC. B Required to inscribe a square in (0 AC. Construction. I)raw 1L diameters AC and BD, and lines AB, BC, CD, and A ----- -- DA; ABCD is a square. (The proof is left to the pupil; see ~ 310.) D 319. Denoting radius OA by R, AB2 = OlA + ~OB = 2 R' (~ 252); whence, AB = R V2. That is, the side of an inscribed square is equal to the radius of the circle multiplied by -/2. 160 PLANE GEOMETRY- BOOK V PROP. VIII. PROBLEM 320. To inscribe a regular hexagon in a given circle. BID A- o D Given 0 AC. Required to inscribe a regular hexagon in 0 AC. Construction. 1. Find point B in circumference at a distance from A equal to radius OA. 2. Line AB is a side of a regular inscribed hexagon; and to inscribe a regular hexagon in a 0 apply radius 6 times as a chord. Proof. 3. A AOB is equiangular. (~ 51) 4. Then Z AOB is i of 2 rt. A (?), or - of 4 rt. s. 321. It follows from ~ 320 that the side of a regular inscribed hexagon equals the radius of the circle. 322. Note. If chords be drawn joining the alternate vertices of a regular inscribed hexagon, there is formed an inscribed equilateral triangle. 323. Let AB be a side of an equilateral B A inscribed in 0 AD, whose radius is R. Drawing diameter AC, and chord BC, BC A ------------- is a side of a regular inscribed hexagon, and therefore equal to R. (~~ 321, 322) Now ABC is a rt. A. (~ 194) D Then by ~ 253, AB2 = AC2- BC2= (2 R)2 _ R2 = 3 R2. Whence, AB = R V3. REGULAR POLYGONS 161 Then, the side of an inscribed equilateral triangle equals the radius of the circle multiplied by -V3. PROP. IX. PROBLEM 324. To inscribe a regular decagon in a given circle. D A Given ( AC. Required to inscribe a regular decagon in 0 AC. Construction. 1. Divide radius OA internally in extreme and mean ratio (~ 273), making OA OM O01 AJM 2. OMis a side of a regular inscribed decagon. Proof. 3. Take chord AB = OM, and draw lines OB, B7M; in A OAB, ABM, A = ZA. ~ OA AB 4. Since OM1= AB, (1) becomes OA=AB~ AB AM 5. A OAB, ABM are similar (~ 242); then Z ABM= / O. 6. A ABJM is isosceles, being similar to A OAB. 7. By Ax. 1, AB = BM= OM; then / OBM= / O. (?) 8. Add results of (5) and (7), OBA = 2 ZO. (2) 9. Find sum of A of A OAB; 2 / OBA + / 0 = 180~. (?) 10. Substitute in this value of / OBA in (2), then 5 0 =180~; and /Ois -- of 4rt. A. 325. Note. If chords be drawn joining the alternate vertices of a regular inscribed decagon, there is formed a regular inscribed pentagon. 162 PLANE GEOMETRY- BOOK V 326. Denoting the radius of the ( by R, we have AB = OM= J1(V5-1). (~ 274) 2 This is an expression for the side of a regular inscribe d decagon in terms of the radius of the circle. Ex. 11. The diameter of a circle is 20. Inscribe a regular hexagon and find its area. Ex. 12. The side of a regular hexagon is 6; find its area. If this hexagon be an inscribed one, what is the area of the regular hexagon circumscribing the same circle? Ex. 13. A regular polygon is inscribed in a circle. Give a general method for finding its area. Ex. 14. An equilateral triangle is inscribed in a circle. If its side is 10, what is its area, and how does it compare with the area of the equilateral triangle circumscribing the same circle? Ex. 15. If the altitude of the triangle in exercise 14 were 10, what would be its area and the area of the equilateral triangle circumscribing the circle? Ex. 16. The centre of a regular hexagon bisects every line drawn through it which terminates in the sides of the hexagon. Ex. 17. The apothem of an inscribed equilateral triangle is one-half the radius of the circle. PROP. X. PROBLEM 327. To construct the side of a regular pentedecagon inscribed in a given circle. C M/t X Given arc MN. Required to construct the side of a regular inscribed polygon of fifteen sides. Construction. If AB is a side of a regular inscribed hexagon (~ 321), and AC a side of a regular inscribed decagon (~ 326), arc BC is 1- -~, or J- of the circumference. REGULAR POLYGONS 163 Ex. 18. An equilateral triangle is inscribed in a circle. The distance from the intersection of the medians to the middle point of the base is i. Find the radius of the circle. Ex. 19. If a circle be inscribed in an equilateral triangle, the altitude of the triangle passes through the centre of the circle, and is three times the radius of the circle. Ex. 20. The area of the square which circumscribes a circle is twice that of the inscribed square. Ex. 21. The area of the equilateral triangle circumscribing a circle is four times the area of the inscribed one. MEASUREMENT OF THE CIRCLE PROP. XI. THEOREM 328. The circumference of a circle is shorter than the perimeter of any circumscribed polygon. C A Given polygon ABCD circumscribed about a (D. To Prove circumference of 0 shorter than perimeter ABCD. Proof. 1. Of the perimeters of the 0, and of all possible circumscribed polygons, there must be some perimeter such that all the others are of the same or greater length. 2. But no circumscribed polygon can have this perimeter; for suppose polygon ABCD to have this perimeter, and draw line EF tangent to the 0, meeting AB at E and AD at F. 3. We know that EF is < (AE + AF). (Ax. 6) 4. Then, the perimeter of polygon BEFDC is < the perimeter of polygon ABCD. 5. Hence, the circumference of the ( is < the perimeter of any circumscribed polygon. 164 PLANE GEOMETRY-BOOK V PROP. XII. THEOREM 329. If a regular polygon be inscribed in, or circumscribed about, a circle, and the number of its sides be indefinitely increased, I. Its perimeter approaches the circumference as a limit. II. Its area approaches the area of the circle as a limit. A' F B' \, / \ // 0 Given p, P perimeters, k, K areas, of two regular polygons of same number of sides, respectively inscribed in, and circumscribed about, a 0, whose circumference is C, and area S. To Prove that, if the number of sides of the polygons be indefinitely increased, P and p approach the limit C, and K and k the limit S. Proof. 1. Let A'B' be side of polygon whose perimeter is P; draw radius OF to point of contact. 2. If OA' and OB' cut circumference at A and B, AB is side of polygon whose perimeter is p. (~ 309) 3. Polygons are similar; then -—. (~~ 314, 315) p 'OF 4. Then, p OF OF P-i)= Oor (?); or P- = x (oA,-0 5. p is < circumference; OA' - OF is < A'F. (Ax. 7, ~ 62) 6. Then, P -p < x A'F. (1) 7. If number of sides of each polygon be indefinitely increased, the polygons continuing to have same number of sides, the length of each side will be indefinitely diminished, and A'F will approach the limit 0. REGULAR POLYGONS 165 8. O being constant, P- p will approach limit 0. OF 9. Now circumference is < P, and > p. (~ 328, Ax. 7) 10. Then, P - Cand C - p will approach limit 0, and P and p will approach limit C. 1K OA-A' K —k OA'2 - OF2 11. Again, by ~ 316,; or -- -2 or k 2 (?) 12. Then, K- = k x X A'F2. (~ 253) OF2 13. If the number of sides of each polygon be indefinitely increased, the polygons continuing to have the same number of sides, A'F will approach limit 0. 14. k X A'F2 being < x AtF2, will approach limit 0. OF OF 15. Then, K- kl will approach limit 0; and S being < K, and > k, K and k will approach limit S. 330. If a regular polygon be inscribed in a circle, and the number of its sides be indefinitely increased, its apothem approaches the radius of the circle as a limit. For, it was shown in ~ 329 that OA'- OF approaches the limit 0, whence OF approaches the limit OA'. OF is the apothem of a regular polygon inscribed in a circle whose radius is OA'. PROP. XIII. THEOREM 331. The circumferences of two circles are to each other as their radii. Given C and C' the circumferences of two (D whose radii are R and R', respectively. 166 I'LAENE GEOMETRY-BOOK V C R? To Prove (1) Proof. 1. Inscribe similar regular \ E, polygons whose perimeters are P and P', in ~ whose radii are R and R'. 2. Then p = R- (~ 315); whence P x R'= P' X R. PR' 3. If number of sides of polygons be indefinitely increased, P x R' approaches limit C X R't and P' x R the limit C' x II. (~ 329, I) 4. Equate these limits (~ 187), and use ~ 218. 332. Multiplying the terms of the ratio - by 2, we have C 2R D C' 2 1R' D" if D and D' denote the diameters of the 0 whose radii are R and R', respectively. That is, the circumferences of two circles are to each other as their diameters. C D 333. The proportion = (~ 332) may be written C C' _I =el_.~ (~ 220) D D' That is, the ratio of the circumference of a circle to its diameter has the same value for every circle. This constant value is denoted by the symbol 7r; then, C _,. (1) It is shown by methods of higher mathematics that the ratio 7r is incommensurable; its numerical value can only be obtained approximately. Its value to the nearest fourth decimal place is 3.1416. 334. Equation (1) of ~ 333 gives C=7rD. REGULAR POLYGONS 167 That is, the circumference of a circle is equal to its diameter multiplied by 7r. We also have C- 2 wrR. That is, the circumference of a circle is equal to its radius multiplied by 2 7r. 335. Def. In circles of different radii, similar arcs, similar segments, and similar sectors are those which correspond to equal central angles. PROP. XIV. THEOREM 336. The area of a circle is equal to one-half the product of its circumference and radius. Given R the radius, C the circumference, and S the area of a O. To Prove S= C x R. (1) Proof. 1. Circumscribe a regular polygon about the 0; let P denote its perimeter, and Kits area. 2. We have KJ= I P x R. (~ 317) 3. If number of sides of polygon be indefinitely increased, K approaches limit X, and I P x R the limit I C x R. (~ 329) 4. Equating limits (?) gives equation (1). 337. We have C= 2 7rR; then, S = rR x R = 7rR2. That is, the area of a circle is equal to the square of its radius multiplied by 7r. Again, S = 4 7r X 4 R2 = 7r X (2 R)' If D denote the diameter of the 0, S = v rD2. That is, the area of a circle is equal to the square of its diameter multiplied by;,r. 168 PLANE GEOMETRY BOOK V 338. Let S and S' denote the areas of two ~ whose radii are R and, and diamete and R, nd and D', respectively. S 7'R2 R2 Then, St- 7rR12 -- R'" and S ^r T~~jD""_ 0rD2 D2 and -S (~ 337) That is, the areas of two circles are to each other as the squares of their radii, or as the squares of their diameters. 339. Let s be the area, and c the arc, of a sector of a 0, whose area is S, circumference C, and radius R. Since a sector is the same part of the 0 that its arc is of the circumference, s c S =C' Ors=c x C. Then, by ~ 336, s=c x R= c x R. Hence, the area of a sector equals one-half the product of its arc and radius. Since similar sectors are like parts of the ( to which they belong (~ 335), it follows that Similar sectors are to each other as the squares of their radii. Ex. 22. The area of a circle is 64 r; find the circumference. Ex. 23. Give geometric method for cutting the largest possible octagon from a board one foot square. Ex. 24. The area of a circle is 154, find its circumference and radius. Ex. 25. The area of a circle is 1447r. Find the area of an inscribed equilateral triangle. Ex. 26. The radius of a circle is 14; find its area. If the radius were doubled, how would the area be affected? Ex. 27. What is the area of the largest circle which can be cut from a triangular piece of cardboard whose edges are 5, 13, and 12 inches, respectively? Ex. 28. The ratio of the areas of two circles is —, one circumference is 37.6992; find the other circumference. Will more than one circumference satisfy the given conditions? Why? (r = 3.1416.) REGULAR POLYGONS 169 PROP. XV. PROBLEM 340. Given p and P, the perimeters of a regular inscribed and of a regular circumscribed polygon of the same number of sides, to find p' and P', the perimeters of a regular inscribed and of a regular circumscribed polygon having double the number of sides. A' M F Ni B A \, \\ /, o Solution. 1. Let AB be side of polygon whose perimeter is p. 2. Let radius OF bisect arc AB at F; let radii OA and OB extended cut tangent at F at A' and B', respectively; then A'B' is side of polygon whose perimeter is P. (~ 309) 3. Draw chords AF, BF, and tangents AMlV, Bn, meeting A'B' at MA and N, respectively; then AF and MN are sides of polygons whose perimeters are p' and P'.' (~ 312) 4. If n denotes number of sides of polygons whose perimeters are p and P, and 2 n number of sides of polygons whose perimeters are p' and P', AB = 9, A'B' =-P AF-, MN P (1) n n 2 ni 2n 5. Line OM bisects Z A'OF; whence, M= MF OF' (~~ 175, 230) _P OA' 6. Again, -=. (~ 315) p OF 7. Then, AI and Pp=A (?) p iMF, p MF +P _- - A A'7' P P' P 8. Then, P+ A'F = - _ 2 p 2 MF JMN 2 ' 2 n P' 170 PLANE GEOMETRY-BOOK V 9. Then, P' (P+p)=2 PXp, orP' = 2 P+P (2) P+p 10. Again, since Z ABF= Z AFM, A ABF, AFM are similar. (~~ 192, 196, 236) A F AIF 11. Then A AB AF' AB,, jT,, or AF2 =AB x MF. (?) 12. Thenby(1),42 X4- 42 4?2 n 4n 4?Z2 13. Then, p'2 =p x P', and p'= p x f'. (3) 341. We will now show how to compute an approximate value of Xr (~ 333). If the diameter of a 0 is 1, the side of an inscribed square is 1- V2 (~ 319); hence, its perimeter is 2 V2. Again, the side of a circumscribed square is equal to the diameter of the 0; hence, its perimeter is 4. We then put in equation (2), ~ 340, P = 4, and p = 2 V2 = 2.82843. Then, Pt = 2 P x p = 3.31371. P.+p We then put in equation (3), ~ 340, p = 2.82843, and P' = 3.31371. Then, p' = Vp x ' = 3.06147. These are the perimeters of the regular circumscribed and inscribed octagons, respectively. Repeating the operation with these values, we put in (2), P = 3.31371, and p = 3.06147. Then, P = 2P —xP = 3.18260. P+p We then put in (3), p = 3.06147 and P' = 3.18260. Then, ' = Vp x ' = 3.12145. These are, respectively, the perimeters of the regular circumscribed and inscribed polygons of sixteen sides. REGULAR POLYGONS 171 In this way, we form the following table: No. OF PERIMETER OF PERIMETER OF SIDES REG. CIRC. POLYGON REG. INSC. POLYGON 4 4. 2.82843 8 3.31371 3.06147 16 3.18260 3.12146 32 3.15172 3.13665 64 3.14412 3.14033 128 3.14222 3.14128 256 3.14175 3.14151 512 3.14163 3.14157 The last result shows that the circumference of a ( whose diameter is 1 is > 3.14157, and < 3.14163. Hence, an approximate value of rr is 3.1416, correct to the fourth decimal place. Note. The value of wr to fourteen decimal places is 3.14159265358979. Ex. 29. The radius of a circle is 12. What is the radius of a circle having twice the area? Ex. 30. The area of the equilateral triangle inscribed in a circle is one-half the area of the regular hexagon inscribed in the same circle. Ex. 31. Two circumferences are in the ratio 3 to 4. What is the ratio of their radii? their diameters? their areas? Ex. 32. The areas of two circles are in the ratio 1 to 2. What is the ratio of their radii? of their diameters? Ex. 33. The areas of two circles are 836 and 616, respectively. If the radius of the second is 257, what is the radius of the first? Ex. 34. One side of an inscribed equilateral triangle is 10; find the radius of the circle. Ex. 35. What is the area of a sector of a circle whose radius is 12, if the angle of the sector is 45~? Ex. 36. Find the ratio of the.area of a circle to the area of its circumscribed square. Ex. 37. Find the ratio of the area of a circle to the area of its circumscribed equilateral triangle. 172 PLANE GEOMETRY-BOOK V Ex. 38. Find the ratio of the perimeters of the circumscribed and inscribed equilateral triangles. Ex. 39. A wheel revolves 55 times in travelling 1045 r ft. What is 4 its diameter in inches? If r represents the radius, a the apothem, s the side, and k the area, prove that Ex. 40. In a regular octagon, s = r12 - /2, a = i r/2 + V2, and k = 2 r2 V2. Ex. 41. In a regular dodecagon, s = r/2 - V, a = ~ r/2 + /3, and k = 3 r2. Ex. 42. In a regular octagon, s = 2 a(2 - 1), r = a4 - 2 /2, and k = 8 a2(V2 - 1). Ex. 43. In a regular dodecagon, s = 2 a(2 - /3), r = 2 aV2 - V, and k = 12 a2(2 - 3). Ex. 44. In a regular decagon, a = ~ r/10 + 2 /5. (Find the apothem.) Ex. 45. Find the number of degrees in a radian, an arc whose length is equal to that of the radius of the circle. (7r = 3.1416.) Ex. 46. A circular grass plot whose diameter is 52 feet is surrounded by a walk of uniform width whose area is 753.984. Find the width of the walk. (r =3.1416.) 04 8 16 hNumber of sides 64 Ex. 47. Find the areas of regular inscribed polygons of 4, 8, 16, 32, 64, 128 sides, and of the circumscribing circle (~ 341). Using area for vertical measurements and number of sides for horizontal measurements, make a graph of the data showing the approach of the area of the polygon to that of the circle as the number of sides increases. REGULAR POLYGONS 173 LOCI 342. Def. If a series of points, all of which satisfy a certain condition, lie in a certain line, and every point in this line satisfies the given condition, the line is said to be the locus of the points. For example, all points which satisfy the condition of being equally distant from the extremities of a straight line, lie in the perpendicular erected at the middle point of the line (~ 56). Also, every point in the perpendicular erected at the middle point of a line satisfies the condition of being equally distant from the extremities of the line (~ 55). Hence, the perpendicular erected at the middle point of a straight line is the Locus of points which are equally distant from the extremities of the line. Again, all points which satisfy the condition of being within an angle, and equally distant from its sides, lie in the bisector of the angle (~ 99). Also, every point in the bisector of an angle satisfies the condition of being equally distant from its sides (~ 98). Hence, the bisector of an angle is the locus of points which are within the angle, and equally distant from its sides. Ex. 48. Find the locus of points in a plane three inches from a fixed point in the plane. Construct the locus. Ex. 49. What is the locus of points in a plane equally distant from all points in the circumference of a circle? Ex. 50. Find the locus of the vertices of triangles having a fixed base and the same altitude. Construct the locus. Ex. 51. Two equal chords intersect; find the locus of their intersections. Ex. 52. The length of the hypotenuse of a right triangle is fixed; find the locus of the centres of the inscribed circles; of the circumscribed circles. Ex. 53. Find the locus of the middle points of chords drawn through a fixed point within a circle. 174 PLANE GEOMETRY-BOOK V Ex. 54. Construct the locus of points equidistant from two given lines. How many cases are there? Ex. 55. Draw the approximate locus of the middle points of the hypotenuses of all right triangles having a hypotenuse of fixed length and the vertex of the right angle fixed. Ex. 56. Construct the locus of the centres of circles tangent to two lines which intersect at right angles. Is there more than one system of these circles? Ex. 57. Find the locus of the centres of circumferences which pass through a fixed point. NEW SOLID GEOMETRY SOLID GEOMETRY --- BOOK VI LINES AND PLANES IN SPACE. DIEDRAL ANGLES. POLYEDRAL ANGLES 343. Def. A plane is said to be determined by certain lines or points when one plane, and only one, can be drawn through these lines or points. PROP. I. THEOREM 344. A plane is determined by a straight line and a point wvithout the line. A B Given point C without str. line AB. To Prove that a plane is determined by AB and C. Proof. 1. If any plane MN be drawn through AB, it may be revolved about AB as an axis until it contains point C. 2. Hence, a plane can be drawn through AB and C; and it is evident that but one such plane can be drawn. 345. A plane is determined by three points not in the same straight line. For if two of the points be joined by / *C 7 a straight line, one plane, and but one, / B can be drawn through this line and the remaining point (~ 344). 175 SOLID GEOMETRY- BOOK VI 346. A plane is determined by two intersecting straight lines. For, by ~ 344, one plane, and but one, can be drawn through AB and C; and B this plane contains BC by ~ 6. / 347. A plane is determined by two parallel straight lines. By the definition of ~ 68, two 11s lie in the same plane; and by ~ 344, but /A B one plane can be drawn through A and /C D/ any point C of CD. PROP. II. THEOREM 348. The intersection of two planes is a straight line. B A Q Given line AB the intersection of planes MN and PQ. To Prove AB a str. line. Proof. 1. Draw a str. line between points A and B. 2. By ~ 6, this str. line lies in plane MN, and also in plane PQ. 3. Then it must be the intersection of planes MN and PQ. 4. Hence, the line of intersection AB is a str. line. Ex. 1. Can a triangle be a warped surface, that is, must all parts of it lie in the same plane? Why? Is this true of any other rectilinear figure? Ex. 2. How many planes determine a line? How many planes are suficient to determine a point? A solid? Is the number of points necessary to determine these bounding planes the same as the number of planes? Ex. 3. Fold a sheet of paper. Is the fold a curved line? A broken line? A straight line? Why? LINES AND PLANES IN SPACE 177 349. Defs. If a straight line meets a plane, the point of intersection is called the foot of the line. A straight line is said to be perpendicular to a plane when it is perpendicular to every straight line drawn in the plane through its foot. A straight line is said to be parallel to a plane when it cannot meet the plane however far they may be extended. A straight line which is neither perpendicular nor parallel to a plane, is said to be oblique to it. Two planes are said to be parallel to each other when they cannot meet however far they may be extended. 350. The following is given for reference: A perpendicular to a plane is perpendicular to every straight line drawn in the plane through its foot. PRO.P. III. THEOREM 351. At a given point in a plane, but one perpendicular to the plane can be drawn. T A/l M Given line PQ 1 to plane MN at P. To Prove PQ the only I which can be drawn to MN at P. Proof. 1. If possible, let PT be another I. to MNV at P. 2. Let the plane determined by PQ and PT intersect 1~N in line IHK; then, both PQ and PT are I II(K. [A I to a plane is I to every str. line drawn in the plane through its foot.] (~ 350) 3. But this is impossible; for, in plane HKT, only one -_ can be drawn to HKI at P. [At a given point in a str. line, but one __ to the line can be drawn.] (~ 15) 178 SOLID GEOMETRY-BOOK VI 4. Then only one L can be drawn to MN at P. Note. As a rigorous proof of the theorem that, at a given point in a plane, a perpendicular to the plane can be drawn, is too difficult for the average pupil, it is omitted. PROP. IV. THEOREM 352. A straight line perpendicular to each of two straight lines at their point of intersection is perpendicular to their plane. 'M\ ' " F VI Given line AB I lines AC and AD, and plane MN drawn through AC and AD. To Prove AB I plane MNfV. Proof. 1. Let AE be any other str. line drawn through A in MN; draw line CD intersecting AC at C, AE at E, AD at D. 2. Prolong BA to F, making AF = AB, and draw lines BC, BE, BD, FC, FE, and FD. 3. In A BCD and CDF, CD = CD. 4. Since AC and AD are _L to BF at its middle point, BC = CE, and BD = D. [If a I be erected at the middle point of a str. line, any point in the I is equally distant from the extremities of the line.] (~ 54, I) 5. Then, A BCD = A CDF. [Two A are equal when the three sides of one are equal respectively to the three sides of the other.] (~ 52) 6. Now revolve A BCD about CD as an axis until it coincides with A COD. i"~~~~ ~~J~~~r~ ~i~iiiiiiiiiiiiiiiiiiiiiiiiiiiiA:::t~:i ~ ~ ~ ~ llj1jj 41:::::::::::: /~~ -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~1 Z~~4. LINES AND PLANES IN SPACE 179 7. Then, point B will fall on point F, and line BE will coincide with line EF; that is, BE = EF. 8. Hence, since points A and E are each equally distant from B and F, line AE is L BF. [Two points, each equally distant from the extremities of a str. line, determine a I at its middle point.] (~ 56) 9. Since AE is any str. line, AB is I_ to every str. line drawn through A in lMN, and is therefore I to plane MN. [A str. line is said to be I to a plane when it is I to every str. line drawn in the plane through its foot.] (~ 349) PROP. V. THEOREM 353. All the perpTendiculars to a straight line at a given point lie in c plane perpendicular to the line. B z c D Given AC, AD, and AE any three Js to line AB at A. To Prove that they lie in a plane _L to AB. Proof. 1. Let MN be the plane determined by AC' and AD; then, plane,N is 1 AB. [A str. line I to each of two str. lines at their point of intersection is I to their plane.] (~ 352) 2. Let the plane determined by AB and AE intersect 7MN in line AE'; then, AB _L AE'. [A _ to a plane is I to every str. line drawn in the plane through its foot.] (~ 350) 3. In plane ABE, only one J_ can be drawn to AB at A. [At a given point in a str. line, but one I to the line can be drawn.] (~ 15) 4. Then, AE' and AE coincide, and AE lies in plane MN. 5. Then, all the Js to AB at A lie in a plane I AB. 180 SOLID GEOMETRYI- BOOK VI 354. It follows from ~ 353 that Through a given point in a straight line, a plane can be drawn perpendcicular to the line, and but one. 355. Through a given point without a straight line, a plane can be draiwn perpendicular to the line, and but one. If C is the given point without line A AB, draw line CB AB, and BD any other I to AB at B. / Then, the plane drawn through BC / B and BD will be 1 AB. [A str. line I to each of two str. lines at their point of intersection is I to their plane.] (~ 352) Again, every plane through C 1_ AB must intersect the plane determined by AB and BC in a line from C 1 AB. [A I to a plane is I to every str. line drawn in the plane through its foot.] (~ 350) But only one _ can be drawn from C to AB. [From a given point without a str. line, but one I can be drawn to the line.] (~ 59) Then, every plane through CIL AB must contain BC, and be l to AB at B. But only one plane can be drawn through B 1 AB. [Through a given point in a str. line, but one plane can be drawn I to the line.] (~ 354) Hence, but one plane can be drawn through CJ- AB. Ex. 4. If we have five points, four of which are in same plane, how many planes do they determine? If no four of the five were in the same plane, would the number of planes change? PROP. VI. THEOREM 356c If from a point in a perpendicular to a plane, oblique lines be drlawn to the plane, I. Tico oblique lines cutting off equal distances from the foot of the pe)lendicular are eqtal. LINES AND PLANES IN SPACE 181 II. Of two oblique lines cutting off unequal distcmaces from the foot of the perpendicular, the more remote is the greater. Given line AB L plane MN at B; AC and AD oblique lines meeting IMNV at equal distances from B, and AE an oblique line meeting MNY at a greater distance from B than AC. To Prove AC= AD, and AE > AC. Proof. 1. Draw lines BC, BD, and BE. 2. In A ABC, ABD, AB = AB; and by hyp., BC= BD. 3. Also, Z ABC-Z ABD. [A I to a plane is L to every str. line drawn in the plane through its foot. ] (~ 350) 4. Then, A ABC= A ABD. [Two A are equal when two sides and the included Z of one are equal respectively to two sides and the included Z of the other.] (~ 46) 5. Then, AC= AD. [In equal figures, the homologous parts are equal.] (~ 48) 6. Again, on BE take BF= BC, and draw line AF; then, since BF= BC, AF= AC. [If from a point in a I to a plane, oblique lines be drawn to the plane, two oblique lines cutting off equal distances from the foot of the I are equal. ] (~ 356, I) 7. But, AB L BE. [A L to a plane is L to every str. line drawn in the plane through its foot.] (~ 350) 8. Then, AE > AF, or AC. [If oblique lines be drawn from a point to a str. line, of two oblique lines cutting off unequal distances from the foot of the I from the point to the line, the more remote is the greater.] (~ 64, II) 182 SOLID GEOMETRY -BOOK VI PROP. VII. THEOREMI 357. (Converse of Prop. VI.) If from a point in a perpendictlar to a plane, oblique lines be (draicwn to the plane, I. Two equal oblique lines cut off equal distances from the foot of the perpendicular. II. Of two unequal oblique lines, the greater cuts off the greater distance fr-om the foot of the perpendicular. Given line AB _L to plane IMN at B; AC, AD, and AE oblique lines from A to MIN, AC and AD being equal and AE > AC; also, lines BC, BD, and BE. (Fig. of Prop. VI.) To Prove BC= BD, and BE > BC. (The proof is left to the pupil.) PROP. VIII. THEOREM 358. If through the foot of a perpendicular to a plane a line be drarcn att right angles to any line in the plane, the line drawn. from its intersection with this line to any point in the peipendicular will be perpendicular to the line in the plane. B Given line AB _ to plane MIN at A, line AE J_ to any line CD in ALV, and line BE from E to any point B in AB. To Prove BE _L CD. Proof. 1. On CD take EC = ED. 2. Draw lines AC, AD, BC, and BD; then, AC = AD. [If a I be erected at the middle point of a str. line, any point in the I_ is equally distant from the extremities of the line.] (~ 54, I) 3. Then, BC = BD. [If from a point in a I to a plane, oblique lines be drawn to the plane, two oblique lines cutting off equal distances from the foot of the 1- are equal.] (~ 356, I) LINES AND PLANES IN SPACE 183 4. Then, since points B and E are equally distant from C and D, BE I CD. [Two points, each equally distant from the extremities of a str. line, determine a I at its middle point.] (~ 56) PROP. IX. THEOREM 359. From a given point without a plane, one perpendicular to the plane can be drcawn, and but one. N\ N Given point A without plane MN. To Prove that a 1L can be drawn from A to MNV, and but one. Proof. 1. Let DE be any line in MN; draw line AlFI DE, line BF in plane JVMN1 DE, line AB ~ BF, and line BE. 2. Now, EF is 1_ to plane ABF. [A str. line 1 to each of two str. lines at their point of intersection is I to their plane.] (~ 352) 3. Then, since BF is drawn through the foot of EF, I_ to line AB in plane ABF, BE is IL AB. [If through the foot of a I to a plane a line be drawn at rt. I to any line in the plane, the line drawn from its intersection with this line to any point in the I will be I to the line in the plane.] (~ 358) 4. Then AB, being IJ to BE and BF, is I MN. [A str. line I to each of two str. lines at their point of intersection is L to their plane.] (~ 352) 5. If possible, let AC be another I from A to MN; then A ABC will have two rt. As, which is impossible. [A 1L to a plane is I to every str. line drawn in the plane through its foot.] (~ 350) 6. Then, but one I can be drawn from A to MV. 184 SOLID GEOMETRY-BOOK VI 360. If AB is the I, and AC any other line, from A to MN (Fig. of Prop. IX), and line BC be drawn, AB 1_ BC. [A L to a plane is - to every str. line drawn in the plane through its foot.] (~ 350) Then, AB < AC. [The I is the shortest line that can be drawn from a point to a str. line.] (~ 95) Then, the peependicular is the shortest line that can be draown from a point to a plane. Note. The distance of a point from a plane signifies the length of the perpendicular from the point to the plane. PROP. X. THEOREM 361. If two straight lines are parallel, a plane drazn through one of them, not coinciding with the plane of the parallels, is parallel to the other. iil N Given line a II to line b, and plane IMN drawn through b, not coinciding with the plafie of the 11s. To Prove a I1 MrN. Proof. 1. Since a and b lie in a plane which intersects M_~V in b, if a meets AIV, it must be somewhere in b. 2. But since a is II b, it cannot meet it (~ 68). 3. Then, a and AlN cannot meet, and are II (~ 349). Ex. 5. If the circumference of a circle be drawn in a plane and at its centre a perpendicular to the plane be erected, any point in this perpendicular is equidistant from the circumference. Ex. 6. The plane which bisects a straight line at right angles is the locus of all points equidistant from the extremities of the line. LINES AND PLANES IN SPACE 185 PROP. XI. THEOREM 362. If a straight line is parallel to a plane, the intersection of the plane with any plane drawn through the line is parallel to the line..Ma Given line a II to plane MN; and b the intersection of MN with any plane drawn through a. To Prove a II b. (a and b lie in the same plane, and cannot meet.) 363. If a line and plane are parallel, a parallel to the line through any point of the plane lies in the plane. Let line a be 11 to plane MN (Fig. of Prop. XI), and line b drawn II to a through any point Ai in MN. The plane drawn through a and A intersects ViNV in a II to a. [If a str. line is 11 to a plane, the intersection of the plane with any plane drawn through the line is II to the line.] (~ 362) But through A only one 11 can be drawn to a. [But one str. line can be drawn through a given point II to a given str. line.] Whence, b lies in MN. (~ 69) PROP. XII. THEOREM 364. If two parallel planes are cut by a third plane, the intersections are parallel. M Given II planes zMNI and PQ cut by plane AB in lines a and b, respectively. To Prove a II b. (a and b lie in the same plane, and cannot meet.) 186 SOLID GEOMETRY - BOOK VI 365. If c and d are II lines included between II planes MN and PQ (Fig. of Prop. XII), the figure bounded by a, c, b, and d is a,, and c = d. [In any 07, the opposite sides are equal.] (~ 104) Then, parcallel lines included between parallel planes are equal. PROP. XIII. THEOREM 366. Through a given straight line, a plane can be drawnz parcallel to any other straight line. Given lines a and b. To Prove that a plane can be drawn through b II a. (Draw line c II a; then use ~ 361.) Note. If a is II b, an indefinitely great number of planes can be drawn through b 1I a (~ 361); otherwise, but one such plane can be drawn, for every plane through b II a must contain c (~ 363), and but one plane can be drawn through b and c. PROP. XIV. THEOREM 367. Through a given point a plane can be drazwn parallel to any two straight lines in space. a -/A]- — B, Given point A- and lines and ---b. Given point A and lines a and b. To Prove that a plane can be drawn through A li to a and b. (The proof is left to the pupil; see ~ 361.) Note. If a and b are II, an indefinitely great number of planes can be drawn through A II to a and b (~ 361); otherwise, but one such plane can be drawn. LINES AND PLANES IN SPACE 187 PROP. XV. THEOREM 368. Two pependciculars to the same plane are parallel. Given lines AB and CD 1_ to plane J1TN at B and D, respectively. To Prove AB II CD. Proof. 1. Let A be any point of AB, and draw line AD. 2. Also, draw line BD, and line DF in plane MILV 1_ BD. 3. Then, CD _ DF. [A I to a plane is 1 to every str. line drawn in the plane through its foot. ] (~ 30) 4. Also, AD DF. [If through the foot of a I to a plane a line be drawn at rt. A to any line in the plane, the line drawn from its intersection with this line to any point in the I will be I to the line in the plane.] (~ 358) 5. Then, CD, AD, and BD lie in the same plane. [All the Is to a str. line at a given point lie in a plane 1_ to the line.] (~ 353) 6. Since points A and B lie in the plane of the lines AD, BD, and CD, AB lies in this plane. [A plane is a surface such that the str. line joining any two of its points lies entirely in the surface.] (~ 6) 7. That is, AB and CD lie in the same plane. 8. Again, AB and CD are 1 BD. [A L to a plane is I to every str. line drawn in the plane through its foot. ] (~ 350) 9. Then, AB 11 CD. [Two Is to the same str. line are 11.] (~ 70) 188 SOLID GEOMETRY-BOOK VI 369. If one of tzwo parallel lines is perpendicular to a plane, the other is also perpendicular to the plane. A c Let lines AB and CD be 11, and AB 1 31 to plane V. / Since two TS to the same plane are II / (~ 368), a L from C to MN is II AB. But through C, only one II can be drawn to line AB (~ 69), whence CD LI MN. 370. If each of two straight lines is parallel to a third straight line, they are parallel to each other. a M b c Let lines a and b be 11 line c. Drawing plane MN J_ c, it is IJ to a and b (~ 369); then, a. 1 b (~ 368). Ex. 7. Find the locus of points three inches fron a given plane. PROP. XVI. THEOREM 371. Two planes perpendicular to the same straight line are parallel. a P Given planes 1MN and PQ I to line a. To Prove MN I1 PQ. (Prove as in ~ 70; by ~ 355, but one plane can be drawn through a given point I to a given str. line.) LINES AND PLANES IN SPACE 189 PROP. XVII. THEOREM 372. If each of two intersecting lines is parallel to a plane, their plane is parallel to the given plane. i l _ /l P~ —, --— E Given lines AB and in plane N, II to plane P Given lines AB and AC, in plane MN, 11 to plane PQ. To Prove MNZ II PQ. Proof. 1. Draw line AD PQ, line DE 11 AB, and line DF l AC; then, DE and DF lie in PQ. [If a line and a plane are II, a II to the line through any point of the plane lies in the plane.] (~ 363) 2. Whence, AD is I to DE and DF. [A i to a plane is I to every str. line drawn in the plane through its foot.] (~ 350) 3. Therefore, AD is I to AB and AC. [A str. line I to one of two 11s is I to the other.] (~ 72) 4. Then, AD I MN. [A str. line _ to each of two str. lines at their point of intersection is. to their plane.] (~ 352) 5. Then, MN II PQ. [Two planes I to the same str. line are II.] (~ 371) Ex. 8. Find locus of points equidistant from all points in the circumference of a circle. Ex. 9. A set of points are always equidistant from the vertices of a triangle; find their locus. SOLID GEOMETRY-BOOK VI PROP. XVIII. THEOREMn 373. A straight line pe2rendicular to one of two parallel planes is perpendicular to the other also. P - Q Given planes MN and PQ II, and line a I PQ. To Prove a MNV. Proof. 1. Let two planes through a intersect MN in lines b and c, and PQ in lines d and e, respectively. 2. Then, b 11 d, and c II e. [If two 11 planes are cut by a third plane, the intersections are II.] (~ 364) 3. But ais to d and e. [A I to a plane is I to every str. line drawn in the plane through its foot.] (~ 350) 4. Then, a is I to b and c. [A str. line I to one of two Ils is I to the other.] (~ 72) 5. Then, a I MVN. [A str. line I to each of two str. lines at their point of intersection is I to their plane.] (~ 352) 374. Two parallel planes are everywhere equally distant. (Note, p. 184.) If planes if and PQ (Fig. of Prop. XVIII) are II, all lines from points in lMNIV to PQ are II. [Two I to the same plane are 11.] (~ 368) Therefore, these lines are all equal. [ II lines included between II planes are equal.] (~ 365) LINES AND PLANES IN SPACE 191 375. Throu7gh a given point a plane can be drawn parallel to a given plane, and but one. Let A be any point without plane PQ; / draw line AiB I PQ, and through A draw / j plane NIN t AIB; then MIN II PQ. p [Two planes I to the same str. line are 11.] B/ (~ 371) ------ If another plane could be drawn through A 11 PQ, it would have to be L- AB. [A str. line I to one of two 1I planes is I to the other also.] (~ 373) It would then coincide with MIV. [Through a given point in a str. line, but one plane can be drawn _ to the line.] (~ 354) Then but one plane can be drawn through A II PQ. PROP. XIX. THEOREM 376. If two angles not in the same plane have their sides parallel and extending in the same direction, they are equal, and their planes care parallel. _/,V Q Given As BAC and B'A'C' in planes MN and PQ, respectively, with AB and AC II, respectively, to A'B' and A'C', and extending in the same direction. To Prove Z BAC = L B'A'C', and MN II PQ. Proof. 1. Lay off AB = A'B' and AC = A'C'; draw lines AA', BB', CC', BC, and B'C'. SOLID GEOMETRY-BOOK VI 2. Since AB is equal and II to A'B', ABB'A' is a 17. [If two sides of a quadrilateral are equal and II, the figure is a L7.] (~ 107) /AG'C 3. Whence, AA' is equal and II to BB'. [The opposite sides of a 0D are equal.] A ' O (~104) 4. Similarly, ACC'A' is a C7, and AA' is equal and II to CC'. 5. Then, BB' is equal and it to CC'. [If each of two str. lines is 11 to a third str. line, they are 11 to each other.] (~ 370) 6. Whence, BB'C'C is a ~7, and BC= B'C'. 7. Then, AABC = AA'B'C'. [Two A are equal when the three sides of one are equal respectively to the three sides of the other.] (~ 52) 8. Then, Z BAC = / B'A'C'. [In equal figures, the homologous parts are equal.] (~ 48) 9. Again, lines AB and AC are 11 to plane PQ. [If two str. lines are II, a plane drawn through one of them, not coinciding with the plane of the Ils, is II to the other.] (~ 361) 10. Then, MN II PQ. [If each of two intersecting lines is II to a plane, their plane is 11 to the given plane.] (~ 372) A Ex. 10. Are two straight lines which / are perpendicular to the same straight line always parallel? Prove your conclusion. Ex. 11. If astraight line be perpendicular \ to each of two parallel planes and through its \! middle point Xa plane be drawn intersecting the two parallel planes, then any line drawn Q in this plane through X and terminating in _ the parallel planes is bisected at X. LINES AND PLANES IN SPACE 193 PROP. XX. THEOREM 377. If two straight lines are cut by three parallel planes, the corresponding segments are proportional. M P, B.....D B Given 11 planes MN, PQ, and RS intersecting lines AC and A'C' in points A, B, C, and A', B', C', respectively. To Prove BC BC BC B' C Proof. 1. Draw line AC'; let plane through AC and AC' intersect PQ in line BD, and RS in line CC'; then BD I! CC'. [If two II planes are cut by a third plane, the intersections are 1I.] (~ 364) AB AD 2. Then, BC DC(1) BC DC' [A II to one side of a A divides the other two sides proportionally.] (~ 227) 3. In like manner, AD A'B (2) DC' B'C' ABAIBI 4. From (1) and (2), AC B A' [Things which are equal to the same thing are equal to each other.] (Ax. 1) Ex. 12. If two lines which have their extremities in two parallel planes be cut by a third plane parallel to these planes, then one of these lines is to each part cut off as the other line is to the cor- responding part. Ex. 13. If two planes are parallel to the same / "plane, they are parallel to each other. 194 SOLID GEOMETRY- BOOK VI Ex. 14. Three parallel planes are so placed that the ratio of the distance from the first plane to the second to the distance from the second to the third is 3. A line makes an angle of 45~ with one of these planes, and the distance between the first and second plane is 20; find the segments of the line made by these planes. Ex. 15. A line parallel to each of two intersecting planes is parallel to their intersection. DIEDRAL ANGLES DEFINITIONS 378. If two planes meet in a straight line, A the figure formed is called a diedral angle. B The line of intersection of the planes is called the edge of the diedral angle, and the planes the faces. D Thus, in the diedral angle formed by planes -E F BD, BF, BE is the edge, and BD, BF the faces. A diedral angle may be designated by two letters on its edge; or, if several diedral angles have a common edge, by four letters, one in each face and two on the edge, the letters on the edge being named between the other two. Thus, we may read the above diedral angle A BE, or ABEF. B C We call two diedral angles cdjacent when they have the same edge, and a common face between them; as, ABEC and CBED. We call two diedral angles vertical when E the faces of one are the extensions of the faces of the other. 379. A plane angle of a diedral angle is the angle between two straight lines drawn one in each face, perpendicular to the edge at the same D point. A Thus, if lines AB and AC be drawn in faces DE and DF, respectively, of diedral angle A' DG, perpendicular to DG at A, Z BAC is a G - plane angle of the diedral angle. F LINES AND PLANES IN SPACE 195 380. If BAG and BI'A'C' (Fig. of ~ 379) are plane /s of diedral I DG, AB I A'B' and AC iI AC'. (~ 70) Then, / BAG = / B'A'C'. (~ 376) That is, all plane angles of a diedral angle are equal. 381. Since a plane perpendicular to the edge of a diedral angle intersects the faces in lines perpendicular to the edge (~ 350), a plane perpencdicnlar to the edge of a diecal angle intersects the faces in lines which form the plane cogle of the diecral angle (~ 379). 382. Two diedral angles are eqytal when their faces may be made to coincide. PROP. XXI. THEOREM 383. Tw?,vo diedracl a)gles are equal if their plane angles are equal. A A B~ I B Given ABC and A'B'C' plane /s of diedral /s BD and B'D', respectively, and Z ABC /1 AB'I. To Prove diedral / BD = diedral / B'D'. Proof. 1. Place- diedrals so that A'B' shall coincide with AB, and BC' with BC. 2. Then, BD I plane ABC, B'D' I plane A'B'C'. (~ 352) 3. Then, B'D' and BD coincide (~ 351); whence faces of diedrals coincide, and diedrals are equal. (~ 346) 384. It follows froin ~ 383 that If two planies intersect, their verticalc diecral angles are equal. For their plane Z are equal. (~ 44) 196 SOLID GEOMETRY-BOOK VI PROP. XXII. THEOREM 385. If two diedral angles are equal, their plane angles are equal. (Converse of Prop. XXI.) Given ABC and A'B'C' plane As of diedral As BD and B'D', respectively, and BD = B'D'. (Fig. of Prop. XXI.) To Prove / ABC = A'BC'. (Apply B'D' to BD so that face A'D' shall coincide with AD, and C'D' with CD, point B' falling at B.) 386. Defs. If a plane meets another in such a way as to make the adjacent diedral angles equal, each is called a right diedral angle, and the planes are said to P be perpendicular. Thus, if plane PQ meets plane MN _ R in such a way as to make diedral As PRQM and PRQN equal, each of these is a right diedral Z, and MNI PQ. Q It may be shown as in ~ 15 that throgh a given line in a plane, a plane can be drawn perpendicular to the given plane, and but one. PROP. XXIII. THEOREM 387. 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.,P M Q -N Given planes PQ and MN -, intersecting in line QR, and line AB in plane PQ _ QR. To Prove AB -L MM. Proof. 1. Draw line C'BC in plane MNI QR; ABC and ABC' are plane A of diedral Asformed by planes. (~ 379) LINES AND PLANES IN SPACE 197 2. Then, diecral /s PRQN and PRQM are equal. (~ 386) 3. Then, Z ABC = Z ABC', or Z ABCOs a rt. Z. (~ 385) 4. Then, AB L 3JIN, being 1 to BC and BQ. (~ 352) 388. If two planes are perpendictlcar, a perpendicular to one of them at any point of their intersection lies in the other. Let planes PQ and MN (Fig. of Prop. XXIII) be J, and intersect in line QR, and line AB be drawn from any point B of QR I MN. A line drawn in PQ from B I QR is _ JMN. (~ 387) From B but one I- can be drawn to MN (~ 351); whence AB lies in PQ. 389. If two planes are perpendicular, a perpendicular to one of themn fromn any point of the other lies in the other. (The proof is left to the pupil; use figure of Prop. XXIII.) PROP. XXIV. THEOREM 390. If a straight line is perpendicular to a plane, every plane drawn through the line is perpendicular to the plane. P I -. ---. — Q Given line AB 1 plane IxV, and PQ any plane drawn through AB, intersecting MIN in QR. To Prove PQ I MN. Proof.. 1. Draw line C'BC in MN I QR. 2. Since ABI BQ (~ 350), ABC and ABC' are plane As of diedral As PQ RIV and PR QM, respectively. 3. Then, A ABC and ABC' are equal, being rt. As. (~ 350) 4. Then, diedral A PRQNV, PRQMZ are equal (~ 383), or PQ 1_ 1/N. 198 SOLID GEOMETRY- BOOK VI PROP. XXV. THEOREM 391. A plane perpendicular to each of two intersecting planes is perpendicular to their intersection. Given planes PQ, RS 1 MIN, intersecting in line AB. To Prove AB I MN. (By ~ 388, a ~1 to MN at B lies in both PQ and RS.) PROP. XXVI. THEOREM 392. Every point in the bisecting plane of a diedral angle is equally distant from its faces. Given P any point in bisecting plane BE of diedral Z ABDC, and lines PM and PL I to AD and CD, respectively. To Prove PM = PN. Proof. 1. Let the plane of PM and PN intersect planes AD, BE, and CD in lines FM, FP, and FN, respectively. 2. We have plane PFMNV IAD and CD, or 1_BD. (~~ 390, 391) 3. Then, /PFM= Z PFN; being plane As of diedral As ABDE and CBDE, respectively. (~~ 381, 385) 4. Also, AsPMF and PNF are rt. /. (~ 350) 5. By ~ 60, rt. A PLM, PVNF are equal (PF = PF), and PM= PN. LINES AND PLANES IN SPACE 199 PROP. XXVII. THEOREM 393. (Converse of Prop. XXVI.) Any point which is within a diedral angle, and equally distant from its faces, lies in the bisecting plane of the diedral angle. Given point P within diedral Z ABDC, equally distant from AD and CD, and plane BE determined by BD and P. (Fig. of Prop. XXVI.) To Prove that BE bisects diedral Z ABDC. (Prove A PFM = A PF, and Z PFM = / PFN; then use ~ 383.) 394. It follows from ~~ 392 and 393 that The locus of points in space equally distant fromt the faces of a diedrca alngle is the plane bisecting the diedral angle. PROP. XXVIII. THEOREM 395. Through a given straight line without a plane, a plane can be drawn perpecndicular to the given plane, and but one, unless the line is perpendicular to the plane. A Given line AB without plane /YIN, not I_ MN. To Prove that a plane can be drawn through AB IL MN, and but one. Proof. 1. Draw AC_ M/N; then ACDB L MN by ~ 390. 2. If more than one plane could be drawn, their common intersection, AB, would be I. MN (~ 391), which is impossible. [If AB is L MNV, an indefinitely great number of planes can be drawn through AB I 1I: Y, by ~ 390.] SOLID GEOMETRY-BOOK VI 200 396. Defs. The projection of a point on a plane is the foot of the perpendicular drawn from the point to the plane. The projection of a line on a plane is the line which contains the projections of all its points. 397. The projection of a straight line on a plane is a straight line. Let line CD (Fig. of Prop. XXVIII) be the projection of str. line AB on plane MN; draw a plane through AB IL 7MV. By ~ 389, the Is to dMN from all points of AB will lie in this plane; whence, CD is a str. line. (~ 348) Ex. 16. Two parallel planes are cut by a third plane. Prove that the corresponding diedral angles are equal. PROP. XXIX. THEOREM 398. The angle between a straight line and its projection on a plane is the least angle which it makes with cay line drawn in the plane through its foot. N Given line BC the projection of line AB on plane MNV, and BD any other line drawn through B in ~MNV. To Prove Z ABC < ZABD. Proof. 1. Take BD BC; draw lines AC, AD. 2. In A ABC, ABD, AB AB, and BC = BD. 3. We have AC < AD (~ 360); the result follows by ~ 101 We call Z ABC the angle between line AB and plane MN. PROP. XXX. THEOREM 399. Two straight lines, not in the samne plane, have one common perpendicular, and but one; and this line is the shortest lile that can be drawn between them. LINES AND PLANES IN SPACE 201 A K B -D Given lines AB and CD, not in the same plane. To Prove that one common _L to AB and CD can be drawn, and but one; and that this line is the shortest line that can be drawn between AB and CD. Proof. 1. Draw plane MN through CD II AB. (~ 366) 2. Draw plane AH through ABI IMN (~ 395); let their intersection meet CD at G, and draw AG in plane AHT1 GH. 3. We have AG I MN (~ 387); then, AG L CD. (?) 4. Now AB II GH (~ 362); then, AG I AB. (?) 5. If possible, let EK be another I to AB and CD; draw EF 11 AB, and KL in plane AHI GH; then EF lies in MN. (~ 363) 6. Then EII ED and EF (~ 72); and EK i MN. (?) 7. Since KL I _MN (~ 387), we have two Is from K to MN, which is impossible; then only one common I can be drawn. 8. Again, EK > KL, or > AG (~~ 360, 96); this proves last statement of theorem. Ex. 17. Two parallel lines make equal C angles with a plane, and their projections on A\ the plane have the same ratio as the lines. Ex. 18. Two parallel lines, AB and CD, are projected on a plane QR. The ratio of R I the projections is 2. The projection of CD E' —\B is 9. AB is 14. Find CD. F D Ex. 19. A line AB makes an angle of A C, 30~ with the plane QR. The distances fromn A and B to QR are 16 and 8, respectively. -— _ l CD is parallel to AB, and its projection on ' QR is 14. Find AB and CD. 202 SOLID GEOMETRY-BOOK VI Ex. 20. An equilateral ABC, whose area is 25V3, has its side BC parallel to a plane QR. The plane of A ABC makes an angle of 45~ with plane QR. Find the area of the projection of A ABC on plane RQ. Ex. 21. The length of a rectangle ABCD is 10, and its width 8. Side 10 is parallel to plane QR. Side 8 makes an angle of 30~ with QB. Find area of the projection of /7 ABCD on plane QR, correct to three decimal places. A. C I. I ---R --- B C PROP. XXXI. THEOREM 400. Two diedral angles are to each other as their plane angles. Case I. tWhen the plane angles are commensurable..B1 ---- A.B'_ ---l: - 2-)' - Given ABC and A'B'C', plane As of diedral As ABDC and A'B'D'C', respectively, and commensurable. ABDC Z ABC To Prove ABDC ZAB (1) A'B'D' C - A'Bf' Proof. 1. Let / ABE be contained 4 times in Z/ABC, 3 times in Z A'B'C'; find ratio Z ABC to / A'B'C'. 2. Pass planes through edges of ciedrals, and division lines of plane As; subdivisions of diedrals are all equal. (~ 383) 3. Find ratio diedral ABDC to diedral A'B'D'C'. 4. From results (1) and (3), we have equation (1). LINES AND PLANES IN SPACE 203 Case II. When the plane angles are incommensurable. B A lB A' Given ABC and A'B'C' plane As of diedral AsABDC and A'B'D'C', respectively, and incommensurable. To Prove ABDC ABC(2) A'B'D'C' ZA BA' BC Proof. 1. Divide ZABC into any number of equal parts. 2. Let one of the parts be contained exactly in- ZA'B'E, with a remainder Z C'B'E < one of the parts. 3. Pass plane through B'D' and B'E. ABDC / ZABC 4. By Case I, ABDC - AB A'B'D'E, /AIB'E 5. If number of subdivisions of Z ABC be indefinitely inABD)C ABC creased, find limits ABDC and ABC A'B'D'E ZA'B'E 6. We have equation (2) by equating the limits. (~ 187) By ~ 400, the plane angle may be taken as the measure of the diedral angle. POLYEDRAL ANGLES DEFINITIONS 401. A polyedral angle is a figure consisting of three or more triangles, called faces, having for their bases '0 the sides of a polygon, and for their common vertex a point without its plane; as O-ABCD. We call 0 the vertex of the polyedral, and polygon ABCD the base; we call s A OB, A — A4 BOC, etc., the face angles, and their sides AO, OB, etc., the edges. 204 204 ~SOLID GEOMETRY - BOOK VI The polyedral angle is not regarded as limited by the base; thus, the face A OB means the plane between OAl and OB extended indefinitely. A triecdral angle is a polyedral angle of thr'ee faces. Two polyedral angles are called vertical when the edges of one are the prolongations of the edges of the other. 402. A polyedral angle, is called convex when its base is a convex polygon (~ 11-8). 403. Two polyedral angles are equal when they. can be applied to each other so that their faces shall coincide. 404. Two polyedral angles are said to be symmetrical when the face and diedral angles of one are equal respectively to 0 0' tbe homologous face and diedral angles of the other, if the equal parts occur in the rever'se A --- c c' -- Al order. Thus, if face As AOB, 1300, B and OQA are equal respectively to face /s A'0'B', B'0'C', and C'O'A', and diedral A OA, 013, and 00 to diedral A QA'f, 0'B', and 00C', triedral A 0-ABC and 0'-A'B'C' are symmetrical. Puop. XXXII. THEOREM 405. Two vertical polyeclral angles are symmetrical. '0 ~0 FIG. 1. FIn. 2. Given 0-ABC and 0-A'B'C' (Fig. 1) vertical triedral As. POLYEDRAL ANGLES 205 To Prove O-ABC and O-A'B'C' symmetrical. Proof. 1. The face s of the vertical triedrals are equal. (~ 44) 2. Diedrals OA and OA' are vertical (~ 378); for faces AOB and A'OB' are in one plane, as also are faces AOC and A'OC. 3. Then diedrals OA, OA' are equal. (~ 384) 4. Similarly, diedrals OB and OB', as also OC and OC' are equal. 5. Since equal parts are in reverse order, triedrals are symmetrical. (~ 404) That the equal parts of the triedrals occur in reverse order may be seen by moving O-A'B'C' II to itself to the right, and then revolving it, as shown in Fig. 2, about an axis passing through 0, until face OA' C' comes into the same plane as before; edge OB' being on this side of, instead of beyond, plane OA' C'. In like manner, the theorem may be proved for two polyedral A, having any number of faces. PROP. XXXIII. THEOREM 406. The sum of any two face angles of a triedral angle is greater than the third. The theorem requires proof only in the case where the third face angle is greater than either of the others. 0 -A s B Given in triedral Z O-ABC, face / AOC > face / AOB or face Z BOO. To Prove / AOB + Z BOC > Z AOC. (1) 206 SOLID GEOMETRY-BOOK VI Proof. 1. In the face AOC take line OD= OB, making Z AOD = AOB; pass plane through B, D cutting faces in AB, BC, CA. 2. We have A AOB = A AOD (~ 49); whence AB = AD. 3. ByAx. 7, AB+BC >AD+ DC. 4. From results (2) and, (3), BC > DC. 5. From results (2), (4), and OB = OD, Z BOC > Z COD. (~ 101) A. 6. Adding AOB to first member of this, and equal Z AOD to second member, and putting Z AOC for Z AOD + Z COD, gives inequality (1). PROP. XXXIV. THEOREM 407. Th7e sum of the face angles of any convex polyedral angle is less than four right angles. 0 A B Given O-ABCDE a convex polyedral L. To Prove Z AOB + / BOC + etc. < 4 rt. A. Proof. 1. Draw lines from any point 0', in base of polyedral, to vertices A, B, etc. 2. We have Z OAE + Z OAB > Z O'AE + Z O'AB. (~ 406) 3. Similarly Z OBA + Z OBC > Z O'BA + Z O'BC, etc. 4. Adding inequalities, sum of base A of A OAB, OBC, etc., is > sum of base As of A O'AB, O'BC, etc. 5. Since sum of all /s of A OAB, OBC, etc., equals sum of all A of A O'AB, O'BC, etc. (~ 84), sum of As at 0 is < sum of A at 0', which proves theorem. POLYEDRAL ANGLES 27 207 PROP. XXXV. THEO-REM 408. ITf twvo triedral cangles have the face angles of one equal respectively to the face angles of the other, their homologous diedral angles are ecjuali ~~~~ -A A B FIG. 1. FIG. 2. FIG. 3. Given, in triedral A 0-ABC and O'-A'B'C',1 Z A OB = / AV0B, I/ BOO = Z B'O'C'2 and Z OQA =/ C'O'A'. To Prove diedral / QA = diedral Z Q'A'. Proof. 1. Lay off OA, OB, OC', O'A', O'B', 00C' all equal; draw linies AB, BC, OCA, AIBR, BRC', C'A1. 2. We have A QAB =A O'A'B' (~ 46); then AB - A'B'. (9) 3. In like mnanner BC = B'C'. CA,= C'A'. 4. Then A ABC = A A'B'C' ( 52); and Z/EAF-=Z E'A'F. 5. Take AD = A'D'; and draw DE in face QAB 1I QA, meeting AB at E (it will meet, since A QAB is isosceles, and / QAB acute). 6. Draw DF in face QAC I OAl, meeting AC at F; and D'E' and DE'F in faces O'A'B', O'A'C', I O'A', meeting A'B' at El, and A'C' at F'; and draw linies EF, E'F'. 7T. In rt. A ADE, A'DE', AD = A'D'; and from equal A QAB, QAfB', Z DAE = Z D'A'E'. (?) 8. Then A ADE =A A'D'E' (~ 89); whence, AE =AE' and DE = D'E'. 9. In like manner, AF =A'F', DF =D'F'. 208 SOLID GEOMETRY - BOOK VI 10. From results (4), (8) and (9), AAEF; = AA lE'F', and EF= E'F'. 11. From results (8), (9) and (10), A DEF = A D'E'F'. (~ 52) 12. Then, / EDF = Z E'D'F' (?); these are plane A of diedrals OA and 0 A (~ 379), whence the diedrals are equal by ~ 383. The above proof holds for Fig. 3 as well as for Fig. 2; in Figs. 1 and 2, the equal parts occur in the same order, and in Figs. 1 and 3 in the reverse order. 409. It follows from ~ 408 that if two triedral angles have the face angles of one equal respectively to the face angles of the other, 1. They are equal if the equal parts occur in the same order. For if triedral Z O'-A'B'C' (Fig. 2) be applied to O-ABC so that diedral A O'A' and OA coincide, point 0' falling at 0, then since / A'O'C' = Z AOC, and / A'O'B' = Z AOB, O'B' will coincide with OB, and O'C' with OC. 2. Thiey are symmetrical if the equal parts occur in the reverse order. M A. D -V ' I Ex. 22. What is the locus of points 'X X a'< equidistant from two intersecting lines, I and also equidistant from two given points? C BOOK VII POLYEDRONS DEFINITIONS 410. A polyedron is a solid bounded by polygons. We call the bounding polygons the faces of the polyedron, their sides the edges, and their vertices the vertices; a diagonal is a straight line, joining two vertices not in the same face. The least number of polygons which can bound a polyedron is four. A polyedron of four faces is called a tetraedron; of six faces, a hexaedron; of eight faces, an octaedron; of twelve faces, a dodecaedron; of twenty faces, an icosaedron. 411. A polyedron is called convex when the section made by any plane is a convex polygon (~ 118). All polyedrons considered hereafter will be understood to be convex. 412. Any two solids are called equivalent when they have the same volume. PRISMS AND PARALLELOPIPEDS 413. A prism is a polyedron, two of whose faces are equal polygons lying in parallel planes, having their homologous sides parallel, the other faces be- ing parallelograms (~ 107). We call the equal and parallel faces the bases of the prism, and the others the lateral faces; we call the edges which are not sides of the bases the lateral edyes, and the sum of the areas of the lateral faces the lateral area. 209 SOLID GEOMETRY -BOOK VII The altitude is the perpendicular distance between the planes of the bases. 414. The following is given for reference: The bases of a prism are equal. 415. It follows from ~ 413 that the lateral edges of a prism are equal and parallel. (~ 104) 416. A prism is called triangular, quadrangular, etc., according as its base is a triangle, quadrilateral, etc. 417. A right prism is a prism whose lateral edges are perpendicular to its bases. The lateral faces are rectangles (~ 350). An oblique prism is a prism whose lateral edges are not perpendicular to its bases. 418. A regular prism is a right prism whose base is a regular polygon. 419. A truncated prism is a portion of a prism included between the base, and a plane, not parallel to the base, cutting all the lateral edges. The base of the prism and the section made by the plane are called the bases of the truncated prism. 420. A right section of a prism is a section made by a plane cutting all the lateral edges, and perpendicular to them. 421. A pcarallelopiped is a prism whose bases are parallelograms; that is, all the faces are parallelograms. / _ — A right parallelopiped is a parallelopiped whose lateral edges are perpendicular to its bases. A rectangular parallelopiped is a right parallelopiped whose bases are rectangles; that is, all the faces are rectangles. A cube is a rectangular parallelopiped whose six faces are all squares. POLYEDIRONS 211 PROP. I. THEOREM B 422. The sections of a prism made by two parallel planes which cat all the lat- - eral edges, are equalpolygons. D F Given II planes OF and C'F' cutting U ur- E all the lateral edges of prism AB. D F To Prove A section ODEEG = C'D'E'IF'G'. A Proof. 1. The homologous sides CD, D', etc., are I (~ 364) 2. Then, the homologous sides and As are equal (~~ 105, 1, 376); then use ~ 123. PROP. II. THEOREM 423. Two -prismns are equal when the faces including a triedr al angle of one are equal respectively to the jftces including a triedral angle of the other-, and similarly placed. K K' L H Gie, A B A' B' Giein prisms AlT and A'H', faces ABODE, AG, and AL equal respectively to faces A'B'O'D'E', A', and A '; the equal parts being similarly placed. To Prove prism AH=- prism A'H. Proof. 1 Since Z EAB = EAB'l EAF= LEAF, / FAIB ZF'AB' (~ 48) triedral /s A-BEE, AL-B'E'F' are equal. (~ 409, t) 2. Apply prisms so that A'B'C'D'E', A'G', and A'L' shall coincide with ABODE, AG, and AL, respectively..3. Then edges OJI', Df-Y' fall on OH, DJY and vertices H', K' on H, K. (~ 69, 415) 4. Then faces B'H', OK', D'L', II'L' coincide with BH, OK, DL, HL (~~ 346, 347); and prisms are equal. 212 SOLID GEOMETRY- BOOK VII 424. Two right prisms are equal when they have equal bases and equal altitudes; for by inverting one of the prisms if necessary, the equal faces will be similarly placed. 425. Note. The proof in ~423 holds for two truncatedprisms. PROP. III. THEOREM 426. An oblique prism is equivalent to a right prism, having for its base a right section of the oblique prism, and for its altitu.de a lateral edge of the oblique prism..F Lr A' uB C F r K Given FI' a right prism, having for its base FI a right section of oblique prism AD', and its altitude FF' equal to AA', a lateral edge of AD'. To Prove AD' FK. Proof. 1. In truncated prisms AI, A'IC' faces AD, A'D' are equal (?); then, AB A'B'. 2. By ~ 422, FK, F'K' are equal; then, FG = F'G'. 3. Since FF' = AA', AF= A'F'; similarly, BG B'G'. 4. Also homologous A of AG, A'G' are equal. (~~ 413, 76) 5. Then AG = A'G' (~ 123); similarly, BH = B'H'. 6. Then AlK= A'K'. (~~ 423, 425) 7. Taking AK from solid AK', we get FI', and taking A'K1, we get AD'; then AD' = Fi'. POLYEDRONS 213 PROP. IV. THEOREM 427. The opposite lateral faces of a parallelopiped are equal and parallel. Given AC and A'C' the bases of parallelopiped AC'. To Prove AB' and DC', also \ AD' and BC', equal and II. \, — Proof. 1. Since AB is equal and II to DC, and AA' to DD', B Z A'AB = Z D'DC, and AB' II DC'. (~ 376) 2. Then AB'= DC' (~ 110); similarly for AD', BC'. PROP. V. THEOREM 428. The plane passed through two diagonally opposite edges of a parallelopiped divides it into two equivalent triangular prisms. D'.A A 1 Given plane AC' passing through edges AA' and CC' of parallelopiped A'C. To Prove prism ABC-A' = prism ACD-A'. (1) Proof. 1. Let rt. section EFGH intersect plane AC' in EG. 2. Faces AB', DC' are 11, and EF II GH. (~~ 427, 364) 3. Also, EH II FG; EFG H is a D, and A EFG = A EGH. (?) 4. By ~ 426, ABC-A' is a rt. prism of base EFG and altitude AA', and ACD-A' to a rt. prism of base EGH and altitude AA'; also, the rt. prisms are equal, having equal bases and altitudes (~ 424), which proves (1). SOLID GEOMETRY BOOK VII PROP. VI. THEOREM 429. The lateral area of a prism is equal to the perimeter of a right section multiplied by a lateral edge. B C Given DEFGH a rt. section of prism AC'. To Prove lat. area AC' = (DE + EF + etc.) x AA'. Proof. 1. We have DE l AA', EFL BB', etc. (~ 350) 2. By ~~ 282, 415, area AB' + area BC' + etc. = DE x AA' + EF x AA', etc. = (DE + EF + etc.) X AA'. 430. It follows from ~ 429 that the lateral area of a right prism is equal to the perimeter of the base multipled by the altitude. PROP. VII. THEOREM 431. Two rectangular parallelopipeds having equal bases are to each other as their altitudes. The phrase " rectangular parallelopiped " in the above statement signifies the volume of the rectangular parallelopiped. Case I. IVhen the altitudes are commensurable. P Q cle uA\~ ------ V,/ll'a / B. ---- Given P and Q reet. parallelepipeds, with equal bases, and commensurable altitudes, AA' and BB'. POLYEDRONS 215 PAA' To Prove ' (1) Proof. 1. Let AC be contained 4 times in AA', 3 times in BB'; find ratio AA' to BB'. 2. Pass planes through points of division of AA', BB', L to AA', BB'; subdivisions of P and Q are all equal. (~ 424) 3. Find ratio P to Q. 4. From results (1) and (3), we have equation (1). Case II. When the altitudes are incommensurable. P D AI --- — BLV Given P and Q rect. parallelopipeds, with equal bases, and incommensurable altitudes, AA' and BB'. - AA' To Prove _ (2) Q BB' Proof. 1. Divide AA' into any number of equal parts; let one of the parts be contained a certain number of times in BB', with a remainder CB' < one of the parts. 2. Pass plane CD _L BB', forming rect. parallelopiped BD. P AA' 3. By Case I, _ D B-C 4. If number of subdivisions of AA' be indefinitely increased, P AA' find limits of and --. BD BC 5. We have equation (2) by equating limits. (~ 187) Ex. 1. The base of a right prism is a square whose side is 6; if the lateral edge of this prism is 10, what is the lateral area? If the diagonal of the base is 6, what is the lateral area? 216 SOLID GEOMETRY-BOOK VII 432. Note. The dimensions of a rectangular parallelopiped are the three edges meeting at any vertex. Then, the theorem of ~ 431 may be expressed: If two rectangular parallelopipeds have two dimensions of one equal respectively to two dimensions of the other, they are to each other as their third dimensions. Ex. 2. If the base of a right prism is an equilateral triangle whose side is 4, and the lateral edge of the prism is 10, what is the lateral area and total area? If the altitude of the base is 4, what are the above areas? PRop. VIII. THEOREM 433. Two rectangular parallelopipeds having equal altitudes are to each other as their bases. Q ~P ~ ~ ~ ~ I a a a ' Given P and Q rectangular parallelopipeds, with altitude c, and dimensions of the bases a, b, and a', b', respectively. To Prove a x (~ 280). (1) Q a' x b' Proof. 1. If R is a rectangular parallelopiped with altitude c, and base dimensions a', b, P and R have each the dimensions b, c and are as their third dimensions (~ 432). 2. Then P a (2) R cat 3. Since R and Q have dimensions a', c, R b =MltpRy ( b (3) Q b' 4. Multiplying (2) and (3), and cancelling R. gives (1). POLYEDRONS 217 434. Note. The theorem of ~ 433 may be expressed: Two rectangular parallelopipeds having a dimension of one equal to a dimension of the other, are to each other as the products of their other two dimensions. PROP. IX. THEOREM 435. Any two rectangular parallelopipeds are to each other as the products of their three dimensions. P Q i — A. Cl-C' C a c / '". " p cl' _ / ca. a' a' Given P and Q reet. parallelopipeds with the dimensions a, b, c, and a', b', c', respectively. P axbxc To Prove Q a' X b' x c' (If R is a rect. parallelopiped with dimensions a', b', c, find P R values of - and - by ~~ 434, 432.) R Q Ex. 3. Find the lateral area of a right prism whose base is a regular hexagon, if one of the lateral faces has the base 10, and altitude 8. Ex. 4. The edges of two rectangular parallelopipeds are 9, 12, 15, and 15, 12, 18, respectively. Find the ratio of their volumes. 436. Def. The volhwe of a solid is its ratio to another solid, called the unit of volume, adopted arbitrarily as the unit of measure (~ 180). The usual unit of volume is a cube (~ 421) whose edge is some linear unit; for example, a cubic inch or a cubic foot, 218 SOLID GEOMETRY —BOOK VII PROP. X. THEOREM 437. If the unit of volume is the cube whose edge is the linear unit, the volume of a rectangular parallelopiped is equal to the product of its three dimensions. P a 1 Given a, b, and c the dimensions of rect. parallelopiped P, and Q the unit of volume; that is, a cube whose edge is 1. To Prove vol. P = a x b x c. Proof. 1. We have -a x b X (~435) Q I X 1 2. Since Q is unit of volume, - is volume P. (~ 436) Q 438. Note. In all succeeding theorems relating to volumes, it is understood that the unit of volume is the cube whose edge is the linear unit, and the unit of strface the square whose side is the linear unit. The following are direct consequences of ~ 437: 439. The volume of a cube equals the cube of its edge. 440. The voltume of a rectangular parallelopiped equals the product of its base and altitude. Ex. 5. The volume of a cube is 64 cubic inches. Using proportion, find the edge of a cube having eight times the volume. Ex. 6. A cubical tank holds 30 gallons of water; find an edge of the tank in feet. Ex. 7. Find the ratio of a diagonal of a cube to an edge. Is this ratio commensurable? Ex. 8. The volume of a cube is 16 3. Find the diagonal. Ex. 9. The diagonal of a cube is D\/3. Find the volume. ,,, ~ )lil i ~~lllllllll iiii i iijl~ ~iiiiiiii 4'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~'::I::::::::: A ~ ~ ~ ~ ~ ~ ~ ~ ~ cBl~liiilllll.5iijjiii ll~ii 7.:''" '':'~ '''": ":::::::::::::::::::: POLYEDRONS 219 PROP. XI. THEor.E M 441. The volume of any parallelopip~ec is equcd' to thte Product of its base aci-t altitude. Dr U'K' I A ~F1, —D II11 A Be' L Given AE the altitude of parallelolpiped AC'. To Prove vol. A'-area ABCD x A,-E. Proof. 1. On AB prolonged take EU ==AB; draw planes FIX', UH' 1I EU, meeting AB', DC', DC prolonged, forming rt. parallelopiped Eli'- AC'. (~ 426) 2. On HTG prolonged, take NAM= fIG; draw planes NP', ML' I NYM meeting 'H'G', K'F', KF prolonged, forming rt. parallelopiped LX' ElH' (~ 426), and hence = AC'. 3. Since EU, is I Gil', planes LIL, MH' are I. (~390) 4. Since MIM' is IL MIV;M it is I plane LHU. (~ 38T) 5. Then LIMM'V is a rt. 4 and LM'11 a rectangle (~ ~ 350, 77); whence LN' is a rect. parallelopiped. 6. Then vol. LN' (or AC')=- area LMNYP x MM'. (~440) 7. Nowv rect. LNY= rect. Eli, having equal bases MNV, UH, and same altitude (~ 111); also, rect. FL[=ctT:AC', having equal bases EU, AB, and same altitude (~ 283); then, LN~ct AC. 8. Suhstitnte in result of (6) area ABC'D for area LMN-P, aad put MM' - AE (~ 374). Ex. 1.0. If the edge of a cube equals a, find the diagonal of a face, also the diagonal of the cube. Ex. 11. The diagonal of a cube is 12. Find its edge. 220 SOLID GEOMETRY - BOOK VII Ex. 12. The diagonal of a cube is d. Find its volume. Ex. 13. The area of the surface of a cube is 1 square foot. Find its edge and the edge of a cube the area of whose surface is 2 square feet. Ex. 14. A bin in the form of a rectangular parallelopiped must contain 3281 bushels and is 6 feet in height. Find number of square feet of material necessary for construction. Will a change in the form of the base make any change in the amount of material used? Ex. 15. The diagonal of a cubical box is 18x/3 inches. How many bushels of grain will it contain? Ex. 16. Pine lumber, green, weighs about 830 Kg. per cubic meter, kiln dried, 460 Kg. per cubic meter. Find the weight of a pile of 16-foot kiln-dried lumber 8 ft. wide and 12 ft. high, allowing one-tenth of the volume for air spaces. Ex. 17. If the diagonals of a rectangular parallelopiped meet in a point, any line through this point terminating in the surface of the parallelopiped is bisected by this point. Is this true of any parallelopiped? PROP. XII. THEOREM 442. Thie volume of a triangular prism is equal to the product of its base and altitude. -A B - D a Given AE the altitude of triangular prism AB-C-. To Prove vol. ABC-C' = area ABC' x AE. Proof. If edges of parallelopiped ABCD-D' be II AB, BC, BB', respectively, by ~ 428, ABC-C' I ABCD-D'; by ~ 441, vol. ABC-C' = I area, ABCD x AE = area ABC X AE. (?) POLYEDRONS 221 PROP. XIII. THEOREM 443. The volume of any prism is equal to the product of its base and altitude. Given any prism. To Prove its volume equal to the product of its base and altitude. Proof. 1. Divide the prism into triangular prisms by planes through a lateral edge and the diagonals of the base. 2. Since volume of each triangular prism equals product of its base and altitude (~ 442), the sum of the volumes of the triangular prisms equals the sum of their bases times their common altitude, which proves the theorem. 444. It follows from ~ 443 that 1. Prisms having equivalent bases and equal altitudes are equivclent. 2. Two prisms having equal altitudes are to each other as their bases. 3. Two prisms having equivalent bases are to each other as their altitudes. 4. Any two prisms are to each other as the products of their bases by their altitudes. Ex. 18. The volume of a right prism whose base is a regular hexagon is 600. One side of the base is 10; find the altitude. Ex. 19. The base of a right prism is an isosceles right triangle whose hypotenuse is c. The volume of the prism is v; find the lateral area. Ex. 20. The sides of a right section of an oblique triangular prism are 4, 5, and 7. The lateral edge is 14. Find the volume. Ex. 21. The lateral edges of a prism whose base is a regular hexagon make an angle of 60~ with the base. A perpendicular from the centre of the upper base meets the lower base at a point 10 feet from its centre. If a side of the lower base is 6, find the volume of the prism. 222 SOLID GEOMETRY-BOOK VII Ex. 22. A ditch is a right prism in form; its length is 100 feet. Its ends are isosceles trapezoids whose bases are 10 feet and 3 feet, respectively. Its depth is 4 feet. Find its volume. Ex. 23. The volume of a triangular prism is 288. The base is an equilateral triangle whose side is 8. Find the altitude of the prism. PYRAMIDS DEFINITIONS 445. A pyramid is a polyedron bounded by a polygon, called the base, and a series of triangles having a common vertex. The common vertex of the triangular faces is called the vertex of the pyramid. The triangular faces are called the lateral faces, the edges terminating at the vertex the lateral edges, and the sum of the areas of the lateral faces the lateral area. The altitude is the perpendicular distance from the vertex to the plane of the base. 446. A pyramid is called triangular, quadrangzlar, etc., according as its base is a triangle, quadrilateral, etc. A regular pyramid is a pyramid whose base is a regular polygon, and whose vertex lies in the perpendicular to the base, at its centre. 447. A truncated pyramid is a portion of a pyramid included between the base and a plane cutting all the lateral edges. The base of the pyramid and the section made by the plane are called the bases of the truncated pyramid. 448. A frustum of a pyramCid is a truncated pyramid whose bases are parallel. The altitude is the perpendicular distance between the planes of the bases. POLYEDRONS 223 PROP. XIV. THEOREM 449. In a regular pyramid, the lateral edges are equal, and the lateral faces are equal isosceles triangles. Given regular pyramid O-ABCDEF. o To Prove OA = OB, etc., and A OAB= A OBC, etc. (Prove by ~~ 356, I, and 52.) 450. Def. The slant height of a regular pyramid is the altitude of any lateral face. / '_ Or, it is the line drawn from the vertex \ D of the pyramid to the middle point of any B side of the base. (~ 91) PROP. XV. THEOREM 451. The lateral faces of a frustum of a regular pyramid are equal trapezoids. 0 A' B C Given AC' a frustum of regular pyramid O-ABCDE. To Prove faces AB', BC', etc., equal trapezoids. Proof. 1. Since A OAB = A OBC (~ 449), we may apply A OAB to A OBC so that OB, OA, AB shall coincide with OB, OC, BC, respectively; also A'B' II AB, B'C' 1I BC. (?) 2. Then A'B', B'C' coincide (~ 69); this proves theorem. 452. It follows from ~ 451 that the lateral edges of a frutstum of a regular pyramid are equal. 224 SOLID GEOMETRY - BOOK VII PROP. XVI. THEOREM 453. The lateral area of a regular pyramid is equal to the perimeter of its base multiplied by one-half its slant height. 0 Given slant height OH of regular pyramid O-ABCDE. To Prove lat. area O-ABCDE = (AB + BC + etc.) X 1- OH. (By ~ 450, OH is the altitude of each lateral face.) PROP. XVII. THEOREM 454. The lateral area of a.rustum of a regytlar pyramid is equal to one-half the sum of the perimeters of its bases, multipflied, by its slant height. E' B C Given slant height HH' of the frustum of a regular'pyramid AD'. To Prove lat. areaAD' = (AB + AB' + BC+ B'C' + etc.) x HH', (HH1' is the altitude of each lateral face.) POLYEDRONS 9. r PROP. XVIII. THEOREM 455. if a pyramid be cut by a plane _parallel to the base, the lateral edges and cdtitude are divided proportionally, and the section is similar to the base. B',O n OA OB O ' Bi Given plane A'BC'DII to base of pyramid 0-ABD, cutting faces OAR, OBO, OCD, and ODA B in lines A 'B', B'C', C'D', anti D'A', respectively, anti altitude OP at P'. OB' OC' O O' AFB To Prove A OB 0',etc. (1 anti o A'B'C'D' siimilar to ABCD. Proof. 1. To prove (1), pass plane JhIN through 0 II ABOD, atnd use ~ 377. 2. Again, since A'B' II ABl, B'CY' II BC, etc. (?), homologous ~c of Al'B'C'D' andi ABOD are eqlual. (~ 376) OA' A'B' 3. Also, ~ OA'B', OAlB are similar (~ 238), andOA R OA AB 4. Also B etc. whence by (1), A'B' B' etc. OB BC AB BC' 5. From results (2) and (4), A'B'C'D', ABCD are similar. (~ 233) Ex. 24. All the faces of a pyramid are equal equilateral triangles. The altitude of the pyramid is \G. Find the ed-e. 226 SOLID GEOMETRY-BOOK VII 456. Since A'B'C'D' and ABCD are similar, area A'B'C'D' A'B'2 2 area ABCD AB2 But from ~ 455, A'B' ' O OP' area A'B'C'D' OP'2 = _ whence AB OA OP ' area ABCD opF Hence, the area of a section of a pyramid, parallel to the base, is to the area of the base as the square of its distance friom the vertex is to the square of the altitude of the pyramid. 457. If two pyramids have equal altitudes and equivalent bases, sections parallel to the 0 0' bases equally distant from the vertices are equivalent. / Let pyramids O-ABC, D -— \F O'-A'B'C' have c bases, and same altitude H; let DEF, ------- A A, A D'E'F' be sections 1I to bases, at distance h from O, 0'. B B1 area DE h2 and area D'E'F' h2 By ~ 456, and area ABC -H' area ABIC' - Then, area DEF area D'E'F' area ABC area A'B'C' Since ABC A'B'C', area DEF = area D'E'F'. Ex. 25. The base of a triangular pyramid is an equilateral triangle, whose side is 16. Each lateral edge of the pyramid makes an angle of 60~ with the base; find the altitude of the pyramid. Ex. 26. The locus of all points equidistant from the faces of a triedral angle is the intersection of the planes which bisect its diedrals. Ex. 27. Prove that the lateral area of a regular pyramid is greater than the area of its base. Ex. 28. The base of a regular pyramid is a square whose side is 8. The slant height makes an angle 45~ with the base; find the lateral area. POLYEDRONS 227 PROP. XIX. THEOREM 458. Two triacngular pyramids having equal altitudes and equivalent bases are equivalent. Q r ~ 0 0f d- a- --—; —5' b b'. Given o-abc and o'-a'b'c' triangular pyramids with equal altitudes and - bases. To Prove vol. o-abc = vol. o'-a'b'c'. Proof. 1. Place pyramids with bases in same plane; let PQ be common altitude, and divide it into three equal parts. 2. Through points of division pass planes II to base, cutting o-abc in def, ghkc, and o'-a'b'c' in d'ef', g'h'k'; then def= d'ef', ghk l g'h'k'. (~ 457) 3. With abc, def, ghk as lower bases, construct prisms X, Y, Z, with lateral edges = and II to ad; with d'ef', g'h'k' as upper bases, construct prisms Y', Z', with lateral edges = and II to a'd'; then prism Y == Y', also Z Z'. (~ 442) 4. Then, sum of prisms circumscribed about o-abc exceeds sum of prisms inscribed in o'-a'b'c' by vol. X. 5. o-abc is < sum of X, Y, Z; and > sum of prisms = Y', Z', with def, gihk as upper bases, and lateral edges = and II to ad. 6. o'-a'b'c' is > (Y' + Z'); and < sum of prisms X, Y, Z, with Cb'c', dc'ef', g'h'k' as lower bases, and lateral edges = and 11 to a'd'. 7. Then volumes of pyramids differ by < the difference of the volumes of the two systems of prisms; or by < vol. X, by result (4). 228 SOLID GEOMETRY -BOOK VII 8. By increasing number of divisions of PQ, vol. X can be made < any assigned volume, however small. 9. Then, volumes of pyramids cannot differ; or vol. o-abh = vol. o'-a'b'c'. 459. Since vol. o'-a'b'c' is > (vol. Y' + vol. Z'), which = (vol. Y+ vol. Z), and < (vol. X+ vol. Y+ vol. Z), vol. o'a'b'c' must be < vol. X; and hence approaches the limit 0 when the number of subdivisions is indefinitely increased. PROP. XX. THEOREM 460. A triangular pyramid is equivalent to one-third of a triangular prism having the same base and altitude. df \ ^ " 1.^ c ' e D F e\ c7 ac a,- c. b Given triangular pyramid E-ABC, and prism ABC-DEF with same base and altitude. To Prove vol. E-ABC = vol. ABC-DEF. Proof. 1. Passing plane through C, D, E, prism is divided into triangular pyramids E-ABC, C-DEF, E-ACD. 2. Since E-ABC, C-DEF have base ABC = DEFand same altitude (~~ 414, 374), volumes are equal (~ 458). 3. Since C-DEF (or E-DCF), and E-ACD, have base DCF= ACD (~ 105), and same altitude, volumes are equal. (?) 4. Since volumes of pyramids are equal, vol, E-ABC = n vol. ABC-DEF, POLYEDRONS 229 461. By ~ 460, the volume of a triangular pyramid is equal to one-third the product of its base and altitude. (~ 442) PROP. XXI. THEOREM 462. Thie volume of any pyramid is equal to one-third the product of its base and altitude. Given any pyramid. To Prove its volume equal to one-third the product of its base and altitude. (Prove as in ~ 443.) 463. It follows from ~ 462 that 1. Pyramids having equivalent bases and equal altitudes qre equivalent. 2. Two pyramids having equal altitudes are to each other as their bases. 3. Two pyramids having equivalent bases are to each other as their altitudes. 4. Any two pyramids are to each other as the products of their bases by their altitudes. Ex. 29. The altitude of a pyramid is a, the base is b. A plane, parallel to the base, bisects one edge of the pyramid; find the volume of the pyramid cut off by the plane. Ex. 30. The base of a pyramid is an equilateral triangle. The line from the vertex to the intersection of the altitudes of the triangular base makes an angle of 60~ with the base of the pyramid, and the projection of the vertex of the pyramid on the base meets this plane at a point 16 from the intersection of the altitudes of the base. The altitude of the base is 8 3; find the volume of the pyramid. 230 SOLID GEOMETRY - BOOK VII PROP. XXII. THEOREM 464. Twvo tetraedr-ons having a triedral angle of ommi #r a triedral angle of the other-, are to each other as the the edges including the equal triedral angles. 01 B A A Given V and V' the volumes of tetraedrons 0-ABC and O-A'B'C', respectively, having common triedral Z 0. V -OA x OB x OC To Prove V 4 B 0 VI OA1' x OBI x OC' Proof. 1. Draw lines CP, C"P' 1. face OAB; let their plane intersect OA'1' in line OFF'. 2. Since OAB, OA'B' are the bases, and CP, C'P' the altitudes, of pyramids 0-QAB, C'-OA'B', V area QAB x OP area QAB OF V'area OA'B' x C'P' ) area OAR' C'P' 3. N ow area QAB - OA x OR90 area OA'B' OA' x OB' CP OC 4. In similar rt. A OOP, OO'P' (~ ~352, 237), - _ (?) O'P' OCI 5. Substitute results (3) and (4) in (1). Ex. 31. The diagonals and edges of a cube B determine triangles, which form the faces of six equal quadrangnlar pyramids. Ex. 32. The volume of a right triangular prism is equal to the product of the area of a face, F ABCD, and one-half the altitude, EF, of the base. C B POLYEDRONS 231 PROP. XXIII. THEOREM 465. The volume of a frustum of a pyramid is equal to the sum of its bases and the mean proportional between its bases, multiplied by one-third its altitude. 0 A D --- —^_^D 4Q B - B Given area of lower base B, area of upper base b, and altitude It, of AC', a frustrum of any pyramid O-AC. To Prove vol. AC' = (B +b + VB x b) x ~ H (~ 217). Proof. 1. Let altitude OP cut A'C' at Q. 2. Since AC' is difference of pyramids O-AC, O-A'C', vol. AC'= B x (H + OQ) - b x OQ (~ 462) =B x + B x 1OQ-b x OQ = B x 1 U+ (B-b) x oQ. (1) 3. By ~ 456, B - ((H+ OQ)2 or BH t+ OQ 09 oQ' OQ 4. Clear of fractions, OQ /B = H Vb + OQ V/b, or OQ (VB - V) = /V6b. 5. Multiply both sides by V/B + V/b, OQ(B-b) = HV/B x b+ H x b. 6. Substitute this in (1). Ex. 33. The altitude of a pyramid whose base is a square is 9; the area of the base is 400: (a) find the lateral area; (b) the volume; (c) the distance from the vertex to the section parallel to the base and having an area 100. 232 SOLID GEOMETRY - BOOK VII PRoP. XXIV. THEOREM 466. The volume of a truncated triangular prism is equal to the product of a right section by one-third the sum of the lateral edges. D Given GHC and DKL rt. sections of truncated triangular prism ABO-DEF. To Prove vol. ABC-DEF = area GIIC x ~ (AD ~ BE + CE). Proof. 1. In rt. section DKL, draw DM I 1 KL; truncated prism ABC-DEF consists of rt. triangular prism GHC-DK-L, and pyramids D-EKILE, C-ABHG. 2. Since lateral edges of a prism arc equal (?), vol. GHC-DKYL = area GHC x GD (?) = area GHC x N (GD + HA-+ CL). 3. Since DY is altitude of pyramid D-EKLF (~ 387), vol. D-EKLF = area EKLF x I DYM (?. 4. Since KL1 is altitude of trapezoid EKLE (~ 350), vol. D-EKLE - I (KE + LF) x KL x DM (?) vol.D-EIF =2 3 - (.KL x DV) x N (KE~+ LEF) 2 = area DKL x N (HE — LE), or area GHC x N(KE~LF). (? 5. Similarly, vol. C-ABJIG = area GIIC x 1 (AG ~BH). POLYEDRONS 233 6. Adding results (2), (4), (5), vol. A-BCDEF equals area GHC x '(AG + GD + BH + HK+ KE + CL + LF). 467. By ~ 466, the volume of a truncated right triangular prism is equal to the product of its base by one-third the sum of the lateral edges. Ex. 34. Given any tetraedron and the medians of the faces which form a triedral, and planes passed through these medians in pairs, then these planes and the base form a second tetraedron which is equivalent to one-fourth the given tetraedron. Ex. 35. A side of the lower base of a frustum of a pyramid whose base is a square is 12 feet, a side of the upper base is 8 feet; the altitude of the frustum is 18 feet; find its volume. Ex. 36. A pyramid has a triangular base whose area is 144 square feet; its altitude is 14 feet. At what distance from the vertex must a plane be passed parallel to the base that the area of the section made by the plane may be 96 square feet? Ex. 37. A frustum of a pyramid is cut by a plane, parallel to and halfway between the bases; find the ratio of the perimeter of this section of the frustum to the sum of the perimeters of the bases. Do the areas of the bases and the section have the same ratio as their perimeters? Ex. 38. A concrete stack is in the form of a frustum of a rectangular pyramid. The lower base is 10 feet long and 8 feet wide, the upper base 5 feet long and 4 feet wide, the height, 50 feet -all outside measurements. The thickness of the wall is one foot. How much Portland cement was needed for construction, if the mixture is one part cement to nine parts sand and gravel? Ex. 39. The altitude of a triangular pyramid is 14. The sides of the base are 6, 8, and 10; find the volume of the frustum formed by passing a plane through the pyramid parallel to the base and 7 from the vertex. Ex. 40. The face angles at the vertex of a regular triangular pyramid are right angles. Find an expression for the volume in terms of an edge. 234 SOLID GEOMETRY -BOOK VII SIMILAR POLYEDRONS 468. Def. Two polyedrons are said to be similar when they have the same number of faces similar each to each and simnilarly placed, and have their homologous polyedral angles equal. PROP. XXV. THEOREM 469. The ratio of any two homologous edges of two similar polyedrons is equal to the ratio of any other two homologous edges. A C Given, in similar polyedrons AF and A'F', edge AB homologous to edge A'B', and edge EF to edge E'F'; and faces AC and DF similar to faces A'C' and D'F', respectively. AB EF To Prove A' AB' E'F' By ~ 234, 2, AB' = CD' 470. It follows from ~ 469 that any two homologous faces of two similar polyedrons are to each other as the squares of any two homologous edges. - area ~ABC%) AB~2 EF2 For by ~ 291, area ABCD AB - F area A'B'C'D' — 2 -F2' 471. By ~ 470, the entire su:faces of two similar polyedrons are to each other as the squares of any two homologous edges. Fr by 224 area ABCD + area CDEF, etc. - EF2 For by area 2ODE, etc area AFB'C'D~' + area C'D'E'Fl., etc. 2 POLYEDRONS 235 PROP. XXVI. THEOREM 472. Two tetraedrons are similar when the faces including a triedral angle of one are similar, respectively, to the faces including a triedral angle of the other, and similarly placed. A B AD ~4-D C' Given, in tetraedrons ABCD, A'B'C'D', face ABC similar to A'B'C', ACD to A'C'D', and ADB to A'D'B'. To Prove ABCD, A'B'C'D' similar. BC AC CD AD BD Proof. 1. We have B — = CD A'D' (?) BC' A'C'- C'D' AD' B'D' 2. Then BCD, B'C'D' are similar; and As BAC, CAD, DAB equal, respectively, to As BA'C', C'A'D', D'A'B'. (?) 3. Then triedral A A-BCD, A'-B'C'D' are equal. (~ 409, 1) 4. Similarly, any two homologous triedrals are equal; and ABCD, A'B'C'D' similar. (?) PROP. XXVII. THEOREM 473. Two tetraedrons are similar when a diedral angle of one is equal to a diedral angle of the other, and the faces including the equal diedral angles similar each to each, and similarly placed. B 236 SOLID GEOMETRY - BOOK VII Given, in tetraedrons ABCD, A'B'C'D', diedral Z AB equal to diedral / A'B', face ABC similar A to A'B'C', and face ABD similar to A' A'B'D'. To Prove ABCD, A'B''D' similar. D, Proof. 1. Apply A'B'C'D' to ABCD so that diedral A'B' shall coincide with diedral AB, and point A1' with A. 2. Since / B'A'C' = BAC and Z B'A'D' = BAD, A'C' will coincide with AC, and A'D' with AD, and Z C'A'D'=Z CAD. A'C' A'1B A'.3. Since A — = A- B A D (?), A C'A'D', CAD are similar. 4. Then ABCD, A'B'C'D' are similar by ~ 472. PROP. XXVIII. THEOREM 474. Two similar polyedrons may be decomposed into the sacme number of tetraedrons, similar each to each, and similarly placed.:E A!X~f BLOX'F Given AF and A'F' similar polyedrons, vertices A and A' being homologous. To Prove that they may be decomposed into the same number of tetraedrons, similar each to each, and similarly placed. Proof. 1. Divide all faces of AF, except the ones having A as a vertex, into A, by drawing diagonals from one vertex. 2. Divide all faces of A'F', except those having A' as a vertex, into A similar to those in AF, and similarly placed. (~ 247) 3. Passing planes through A and A' and the sides of the A, AF and AF' are decomposed into same number of tetraedrons, similarly placed. POLYEDRONS 237 4. If ABCF, A'B'C'F' are homologous tetraedrons, A ABC is similar to A'B'C', and BCF to B'C'F'. (~ 247) 5. Since AF and A'F' are similar, diedrals BC and B'C' are equal (?); then ABCF, A'B'C'F' are similar. (~ 473) 6. Similarly, we may prove any two homologous tetraedrons similar. PROP. XXIX. THEOREM 475. Two similar tetraedrons are to each other as the cubes of their homologous edges. A B D o C C' Given V, V' the volumes of similar tetraedrons ABCD, A'B'C'D', vertices A, A' being homologous. V AR' To Prove AB V' A'Bi' Proof. 1. Since triedrals at A, A' are equal (?), V AB x AC x AD AB AC AD (~ 464) = (1) V' A'B' x AC' x AD' 464 B' A-C1 AD' 2. Express AC and AD in terms of AB, A'B'. (~ 469) A C' A'D' 3. Substitute this in (1). 476. Any two similar polyedrons are to each other as the cubes of their homologous edges. For any two similar polyedrons may be decomposed into the same number of tetraedrons, similar each to each (~ 474); and any two homologous tetraedrons are to each other as the cubes of their homologous edges. (~ 475) 238 SOLID GEOMETRY-BOOK VII REGULAR POLYEDRONS 477. Def. A regular polyedron is a polyedron whose faces are equal regular polygons, and whose polyedral angles are all equal. PROP. XXX. THEOREM 478. It is not possible to construct more than five regular convex polyedrons. Proof. 1. A convex polyedral / must have at least three faces, and the sum of its face A < 360~. (~ 407) 2. Since the Z of an equilateral A is 60~, we can form convex polyedrl As by combining 3, 4, or 5 equilateral A, but not more than 5; then not more than 3 regular convex polyedrons can be bounded by equilateral A. 3. Since the / of a square is 90~, we can form a convex polyedral Z by combining 3 squares, but not more than 3; then not more than one regular convex polyedron can be bounded by squares. 4. Since the Z of a regular pentagon is 108~, we can form a convex polyedral Z by combining 3 regular pentagons, but not more than 3; then not more than. one regular convex polyedron can be bounded by regular pentagons. 5. Since the Z of a regular hexagon is 120~, and of a regular polygon of more than 6 sides > 120~, no convex polyedral Z can be formed by combining regular hexagons, or regular polygons of more than 6 sides. 479. Note. The regular polyedrons are the regular tetraedron, octaedron, and icosaedron, bounded respectively by four, eight, and twenty equilateral triangles; the regular hexaedron, or cube, bounded by six squares; and the regular dodecaedron, bounded by twelve regular pentagons. To construct them, draw the following figures on cardboard; cut them out entire, and on the interior lines cut the cardboard half through; the edges may then be brought together to form the respective solids. POLYEDRONS 239 _I.. X\ v Tetraedron Hexaedron Octaedron Dodecaedron Icosaedron Ex. 41. One edge of a regular tetraedron is a; find the volume. Ex. 42. In any regular tetraedron planes passed through the bisectors, in pairs, of the face angles of any triedral angle, enclose, with the base, a tetraedron whose volume is I that of the given tetraedron. Ex. 43. One edge of a regular tetraedron is 6: find the area of the entire surface and the volume. Ex. 44. Find the volume of a regular tetraedron whose edge is one foot. Compare this volume with that of a cube having the same edge. Ex. 45. In each face of a regular tetraedron draw lines joining the middle points of the edges. I)o the triangles thus formed form the faces of a regular tetraedron? BOOK VIII THE CYLINDER AND CONE DEFINITIONS 480. A cylindrical surface is a surface generated by a moving straight line, which constantly intersects a given plane curve, and in all its positions is parallel to a given straight line, not C in the plane of the curve. N B E Thus, if line AB moves so as constantly to intersect plane curve AD, - and is constantly 11 line MN,' not in A B/ the plane of the curve, it generates a cylindrical surface. We call the curve the directrix, the moving line the generatrix, and any position of it, EF, an element of the surface. A cylinder is a solid bounded by a cylindrical surface, called the lateral surface, and two parallel planes called the bases. The altitude of a cylinder is the perpendicular distance between the planes of its bases. A right cylinder is a cylinder the elements of whose lateral surface are perpendicular to its bases. A circular cylinder is a cylinder whose base is a circle. 240 THE CYLINDER 241 A right circular cylinder is called a cylinder of revolution because it may be generated by the revolution of a rectangle about one of its sides as an axis (~ 371). Similar cylinders of revolution are cylinders by the revolution of similar rectangles about homologous sides as axes. 481. It follows from the definitions of ~ 480 that the elements of the lateral surface of a cylinder are equal and parallel. (~ 365) PROP. I. THEOREM 482. A section of a cylinder made by a plane passing through cia element of the lateral surface is a parallelogram. D a se e A Given ABCD a section of cylinder AF, made by a plane passing through AB, an element of the lateral surface. To Prove section ABCD a C7. Proof. 1. We have AD, BC II str. lines. (~~ 348, 364) 2. If a str. line be drawn from C in plane ACII AB, it is an element. (~~ 481, 71) 3. Then, it must be the intersection of plane AC and the cylindrical surface and coincide with CD; and CD II AB. 483. It follows from ~ 482 that a section of a right cylinder made by a plane perpendicular to its base is a rectangle. 242 SOLID GEOMETRY-BOOK VIII PROP. II. THEOREM 484. The bases of a cylinder are equal. C' E A' "G Given cylinder AB'. To Prove base A'B'= base AB. Proof. 1. Take E', F', (' any three points in perimeter A'B'; draw elements E'E, F'F G'G, and lines EF, FG, GE, E'F', F'G', G'E'. 2. We have EE', FF' equal and II (?); whence EE'F'F is a 7. (?) 3. Then E'F'= EF. (?) Similarly E'G' = EG, F'G' = FG. 4. Then A E'F'G' A EFG (?); and base A'B' may be applied to AB so that E' shall fall at E, F' at F, G' at G. 5. Then, all points in perimeter A'B' will fall in perimeter AB, and bases are equal. The following are consequences of ~ 484. 485. The sections of a cylinder made by two parallel planes cutting all the elements are equal. For they are the bases of a cylinder. 486. The section of a cylinder made by a plane parallel to the base is equal to the base. THE CYLINDER 243 MEASUREMENT OF CYLINDERS DEFINITIONS 487. The lateral area of a cylinder is the area of its lateral surface. A right section of a cylinder is a section made by a plane perpendicular to the elements of its lateral surface. A prism is said to be inscribed in a cylinder when its lateral edges are elements of the cylindrical surface, and its bases inscribed in the bases of the cylinder. A plane is said to be tangent to a cylinder when it contains one, and only one, element of the lateral surface. A prism is said to be circumscribed about a cylinder when its lateral faces are tangent to the cylinder, and its bases circumscribed about the bases of the cylinder. 488. If a prism whose base is a regular polygon be inscribed in, or circumscribed about, a circular cylinder (~ 480), and the number of sides of its base be indefinitely increased, 1. The lateral area of the prism approaches the lateral area of the cylinder as a limit. 2. The volume of the prism approaches the volume of the cylinder as a limit. IIr I:I I / I II 1 I, I II, -1 1/ -. II _ _ 3. The perimeter of a right section of the prism approaches the perimeter of a right section of the cylinder as a limit.* Ex. 1. Find area of the section made by a plane drawn through an element of the lateral surface of a cylinder of revolution, whose altitude is 12, and radius of base 6, if the distance from the centre of the base to the cutting plane is 3. * Rigorous proofs of these statements are beyond the scope of the present treatise. SOLID GEOMETRY - BOOK VIII 244 PROP. III. THEOREM 489. The lateral area of a circular cylinder is equal to the perimeter of a right section multiplied by an element of the lateral surface. Given S the lateral area, P the perime- ter of a rt. section, E an element of the lateral surface, of a circular cylinder. -— ' To Prove S = P x E. (1) / Proof. 1. Inscribe in the cylinder a prism whose base is a regular polygon; let S' denote its lateral area, and P' the perimeter of a rt. section. 2. We have ' = P' x E. (~ 429) 3. If number of sides of base be indefinitely increased, find limits of S' and P' x E. (~ 488) 4. We have (1) by Theorem of Limits. (~ 187) 490. By ~ 489, the lateral area of a cylinder of revolution is equal to the circumference of its base multiplied by its altitude. 491. If S denotes the lateral area, T the total area, H the altitude, R the radius of the base, of a cylinder of revolution, S — 2r RH. (~ 334) And, T= 2 rRHIE 2 7 R (~ 337) = 27rR (+ R). PROP. IV. THEOREM 492. The volume of a circular cylinder is equal to the product of its base and altitude. Given V the volume, B the area of the base, and H the altitude, of a circular cylinder. To Prove V=B x H. (1) Proof. 1. Inscribe in cylinder a prism whose base is a regular polygon; let V' denote its volume, B' the area of its base. THE CYLINDER 245 2. We have V'= B' x H. (~ 443) 3. If number of sides of base of prism be indefinitely increased, find limits V' and B' x H. (~~ 488, 329) 4. Apply Theorem of Limits. 493. If V denotes the volume, H the altitude, and R the radius of the base, of a circular cylinder, V== 7R2H. (?) Ex. 2. The radius of the base of a right circular cylinder is 9. The altitude is 8. Find the lateral area. If this cylinder were oblique, would a right section be a circle? Ex. 3. The lateral area of a cylinder of revolution is 120 7r. The area of the base is 36 7r. Find the altitude. Ex. 4. A point in space moves so that it is always at a constant distance from a fixed straight line; find its locus. Ex. 5. The radius of the base of a cylinder of revolution is 12. The altitude is 13. Find the volume. Ex. 6. The radius of the base of an oblique circular cylinder is 6. The distance between the planes of the bases is 10; find the volume. Ex. 7. The radius of the base of a circular cylinder is 8. The elements of the lateral surface make A an angle of 30~ with the plane of the base. The D length of an element is 25. Find volume of K — \ - i cylinder. \ -'Ex. 8. The line which joins any two opposite points in the diagonally opposite edges \ of a parallelopiped is bisected by the plane I7 passed through the other two edges. D Ex. 9. If from any two points in the R diagonally opposite base edges of a parallelo- -! A piped perpendiculars be drawn to the plane A,'-iF passed through the other two edges, these p" perpendiculars are equal. 7. ML 246 SOLID GEOMETRY-BOOK VIII PROP. V. THEOREM 494. The lateral or total areas of two similar cylinders of revolution (~ 480) are to each other as the squares of their altitudes, or as the squares of the radii of their bases; and their volumes are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases. Given S and s the lateral areas, T and t the total areas, V and v the volumes, H and h the altitudes, and R and r the radii of the bases, of two similar cylinders of revolution. T H2 _R2 V H3 R3 To Prove S=T 12 R and - = s t h2 r2' v hr3 r3 Proof. 1. Since rectangles are similar, =_ I= Rh (?) h r h + r 2. By ~ 491, S 2RH =-x R R s- 2 rrh r r rr2 h2' and T_ 2 7rR(H + R) R R = R2_ H2 t 2 rr(h + r) r r r2 h' 3. ~493V 7rR2H R R R3 Hx 3. By 493, X-= -= — V 7vrr2h r2 r r3 h3 Ex. 10. Water weighs 1000 Kg. per cubic meter. The average weight of the same amount of milk is 1033 Kg. A cylindrical milk-can contains 81 gallons of milk. How many pounds of milk will it contain? If the diameter of its base is 10 inches, what is its altitude? Ex. 11. An oil tank is a cylinder of revolution, whose length is 30 feet, and radius of base 3.5 feet. The average weight of petroleum is 795 Kg. per cubic meter. What weight of oil will the tank contain? THE CONE 247 THE CONE DEFINITIONS 495. A conical surface is a surface generated by a moving straight line, which constantly intersects a given plane curve, and passes through a given point not in the plane of the curve. Thus, if line OA moves so as con- B' stantly to intersect plane curve ABC, and c constantly passes through point 0, not in the plane of the curve, it generates a coni- cal surface. We call the moving line the generatrix, A C the curve the directrix, the given point the vertex, and any position of the generatrix, B OB, an element of the surface. If the generatrix be supposed indefinite in length, it will generate two conical surfaces of indefinite extent, O-A'B'C', O-ABC, called the upper and lower nappes, respectively. A cone is a solid bounded by a conical surface, and a plane cutting all its elements. We call the plane the base of the cone, and the conical surface the lateral surface; the altitude is the perpendicular distance from the vertex to the plane of the base. A circular cone is a cone whose base is a circle. The axis of a circular cone is a straight line drawn from the vertex to the centre of the base. A right circular cone is a circular cone whose axis is perpendicular to its base. A right circular cone is called a cone of revolution because it may be generated by the revolution of a right triangle about one of its legs as an axis. 248 SOLID GEOMETRY-BOOK VIII Similar cones of revolution are generated by the revolution of similar right triangles about homologous legs as axes. A frustum of a cone is a portion of a cone included between the base and a plane parallel to the base. The base of the cone is called the lower base, and the section made by the plane the upper base, of the frustum. The altitude is the perpendicular distance between the planes of the bases. A plane is said to be tangent to a cone, or frustum of a cone, when it contains one, and only one, element of the lateral surface. PROP. VI. THEOREM 496. A section qf a cone made by a plane passing through the vertex is a triangle. 0 Given OCD a section of cone OAB made by a plane passing through vertex 0. To Prove section OCD a A. Proof. 1. We have CD a str. line. (?) 2. Str. lines OC, OD, drawn in plane 0CD2 are elements (~ 495), and hence the intersections of the plane with the conical surface; then section OCD is bounded by str. lines. THE CONE 249 PROP. VII. THEOREM 497. A section of a circular cone made by a plane parallel to the base is a circle. S A B'7 Xt1 t B Given A'B'C' a section of circular cone S-ABC, made by a plane II to the base. To Prove AB'C' a (. Proof. 1. Take A', B' any two points in perimeter A'B'C'. 2. Let planes determined by A', B', and axis OS intersect base in radii OA, OB, the section in O'A', O'B', and the lateral surface in str. lines SA'A, SB'B. (~ 496) 3. Then O'A' 1I OA and O'B' II OB (?), and A SOA, SOB are similar to A SO'A', SOB', respectively. (?) 4. Then, O' and OB (? OA SO OB SO O'A' O'B' Then, (?) OA OB 5. Since OA= OB (?) O'A'= O'B'; A', B' being any two points in perimeter, section A'B'C' is a (. Ex. 12. Two circular cylinders have equal altitudes, but the radius of the base of the one is double the radius of the base of the other. Compare (a) their lateral areas; (b) their volumes. Ex. 13. Two circular cylinders have equal altitudes, but the radius of the base of the first is 1 the radius of the base of the second. Compare (a) their lateral areas; (b) their volumes. 250 SOLID GEOMETRY- BOOK VIII MEASUREMENT OF CONES DEFINITIONS 498. The lateral area of a cone, or frustum of a cone, is the area of its lateral surface. The slanzt height of a cone of revolution is the straight line from the vertex to any point in the circumference of the base. The slant height of a frustum of a cone of revolution is that portion of the slant height of the cone included between the bases of the frustum. 499. A pyramid is said to be inscribed in a cone when its vertex coincides with the vertex of the cone, and its base is inscribed in the base of the cone. A pyramid is said to be circumscribed about a cone when its vertex coincides with the vertex of the cone, and its base is circumscribed about the base of the cone. A frustum of a pyramid is said to be inscribed in a frustum of a cone when its bases are inscribed in the bases of the frustum of the cone. A frustum of a pyramid is said to be circumscribed about a frustum of a. cone when its bases are circumscribed about the bases of the frustum of the cone. 500. If a pyramid whose base is a regular polygon be inscribed in, or circumscribed about, a circular cone (~ 495), and the number of sides of its base be indefinitely increased, 1. The lateral area of the pyramid approaches the lateral area of the cone as a /- ~^ limit. 2. The volume of the pyramid approaches the volume of the cone as a limit.* Rigorous proofs of these statements are beyond the scope of the present treatise. THE CONE 251 501. If a firusteum of a pyramid vwhose base is a regular polygon be inscribed in, or circumscribed about, a frustum of a circular cone, and the number of sides of its base be indefinitely increased, 1. The lateral area of the frustum of the pyramid approaches the lateral area of the frustum of the cone as a limit. 2. The volume of the frustum of the pyramid approaches the volume of the frustum of the cone as a limit.* PRop. VIII. THEOREM 502. The lateral area of a cone of revolution is equal to the circumference of its base, multiplied by one-half its slant height. Given S the lateral area, C the circumference of the base, and L the slant height, / of a cone of revolution. / To Prove S= CxL. /-\ Proof. 1. Circumscribe about the cone a regular pyramid; let S' denote its lateral area, and C' the perimeter of its base. 2. Since sides of base of pyramid are bisected at points of contact with base of cone (~ 174), the slant heights of pyramid and cone are equal, and S' = C' x ~ L. (~~ 450, 453) 3. If number of sides of base of pyramid be indefinitely increased, find limits S' and C' X I L. (~~ 500, 329) 4. Apply Theorem of Limits. 503. If S denotes lateral area, T total area, L the slant height, and R the radius of the base, of a cone of revolution, S= 2-rR x IL (?) = 7rRL. And, T = 7rRL + 7rR2 (?) = 7R(L +?). Ex. 14. The diameter of the base of a right circular cone is equal to' the altitude. Find the lateral and total areas. Ex. 15. The edge of a regular tetraedron is 8: find area of the lateral surface of the circumscribed cone. * Rigorous proofs of these statements are beyond the scope of the present treatise. 252 SOLID GEOMETRY-BOOK VIII PROP. IX. THEOREM 504. The volume of a circular cone is equal to the area of its base, multiplied by one-third its altitude. Given V the volume, B the area of the base, and H the altitude, of a circular cone. To Prove V= B X 1 It. (Inscribe a pyramid whose base is a regular polygon.) 505. If V denotes the volume, H the altitude, and R the radius of the base, of a circular cone, V= -7rR2H. (?) PROP. X. THEOREM 506. The lateral or total areas of two similar cones of revolution are to each other as the squares of their slant heights, or as the squares of their altitudes, or as the squares of the radii of their bases; and their volumes are to each other as the cubes of their slant heights, or as the cubes of their altitudes, or as the cubes of the radii of their bases. Given S and.s the lateral areas, T and t the total areas, Vand v the volumes, L and I the slant heights, H and h the altitudes, and R and r the radii of the bases, of two similar cones of revolution (~ 495). S T L9 3 H' B' d LV H' B' To Prove T L = = Rand = s t 12 hl2 r2 v 1 h3 43 (The proof is left to the pupil; compare ~ 494.) THE CONE 253 PROP. XI. THEOREM 507. The lateral area of a firustzum of a cone of revolution is eqcual to the sum of the circumferences of its bases, multiplied by one-half its slant height. Given S the lateral area, C, c the circumferences of the bases, L the slant height, of a frustum of a cone of revolution. To Prove S = (C+ c) x - L. Proof. 1. Circumscribe about the frustun of the cone a frustum of a regular \ pyramid; let S' denote its lateral area. / C', c' the perimeters of its bases. 2. As in ~ 502, the slant heights of the frustums are equal. 3. Then, S' = (C' + c') X - L. (~ 454) 4. If number of sides of bases of frustum of pyramid be indefinitely increased, find limits S' and (C' + c') x - L. (~~ 501, 329) 5. Apply Theorem of Limits. 508. If S denotes the lateral area, L the slant height, R, r the radii of the bases, of a frustum of a cone of revolution, S = (27R + 2 7r) x - L(?) = ((R + r) L. 509. We may write the first result of ~ 508 S=2 r(x q (I? + ) XL. But, 2 r x 1 (?R + r) is the circumference of a section equally distant from the bases. (~ 132) Whence, the lateral area of afrustum of a cone of revolution is equal to the circumference of a section equally distant from its bases, multiplied by its slant height. Ex. 16. In two similar cones of revolution, the slant height of the first is 16 and its lateral area is 128 7r. The lateral area of the second is 50 r: find the radius of the base of the second. Ex. 17. The area of the entire surface of a cylinder of revolution is 96 7r. Its altitude and diameter are equal. Find radius of base. 254 SOLID GEOMETRY — BOOK VIII PROP. XII. THEOREM 510. The volume of a frustum of a circular cone is equal to the sum of its bases and a mean, proportional between its bases, multplied by one-third its altitude. Given V the volume, B, b the areas of the bases, H the altitude, of a frustum of a circular cone. To Prove T=(B + b + VB x b) x H. (Inscribe a frustum of a pyramid whose base is a regular polygon; then apply ~ 465.) 511. If Vdenotes the volume, H the altitude, and R and r the radii of the bases, of a frustum of a circular cone, B = rR, b = rr2, and B X b = Vr2R2r2 = rRr. (?) Then, V= (rR2 + rr2 + 7rRr) X ~ H= 1 7r(R2 + r2 + Rr)H. Ex. 18. The radius of the base of a circular cylinder is a feet. The distance between the bases is equal to the distance from the centre of the lower base to the projection of the centre of the upper base upon the plane of the lower base: find the volume. Is there more than one such cylinder? Ex. 19. The radius of the base of a cone is 8. A line joining the vertex and the centre of the base makes an angle of 30~ with the base. The distance from the centre of the base to the projection of the vertex on the plane of the base is 12. Find volume of cone. Ex. 20. In Ex. 19, if a plane be passed through the middle point of THE CONE 255 the altitude and parallel to the base of the cone, find the volume of the frustum formed. Ex. 21. The lateral area of a right circular cone is 16 V/5 r. The altitude of the cone and the diameter of its base are equal. Find the area of the entire surface. Ex. 22. The volumes of two similar polyedrons are 27 and 512 cubic feet, respectively. Compare their homologous edges. If the area of the surface of the first is 36 square feet, find the area of the surface of the second. Ex. 23. If four straight lines not in the same plane enclose a warped surface, the lines which join the middle points of these lines in order enclose a parallelogram. Ex. 24. A portion of the Chicago Drainage Canal is in the form of a right prism, whose base is an isosceles trapezoid. Width at water-line 162 feet. Width at bottom 160 feet. Depth of water 22 feet. In one minute 300,000 cubic feet of water pass a given point. Find the rate per hour of the current. Ex. 25. Three metal pipes are each 30 inches long. The first two are in the form of a right prism, with a square and a regular hexagon, respectively, as bases; the third in the form of a cylinder of revolution, the radius of whose base is 10. The bases of the three pipes have equal perimeters. Find their volumes. Ex. 26. A silo is in the form of a cylinder of revolution 18 feet in altitude surmounted by a roof in the form of a conical surface whose elements are inclined 45~ to the upper base of the cylinder. The diameter of the base of the cylinder is 10 feet. Find the number of square feet of tin necessary to roof the silo, also the volume of ensilage the cylindrical part will contain. Ex. 27. A cistern made of concrete is in the form of a cylinder of revolution with the frustums of two cones of revolution. Diameter of base of cylinder 12 feet; altitude of cylinder 8 feet; altitude of lower frustum 2 feet; diameter of bottom of cistern 10 feet; altitude of upper frustum 6 feet; diameter of opening 5 feet- all outside measurements. Inside diameter of base of cylinder, 8 feet. Inside diameter of upper base of lower frustum 8 feet, of bottom 6 feet. Inside diameter of lower base of upper frustum 8 feet, of upper base 3 feet. Using one part cement to nine parts sand and gravel, find amount of cement necessary for construction. BOOK IX THE SPHERE DEFINITIONS 512. A sphere is a solid bounded by a surface, all points of which are equally distant from a point within called the centre. A radius of a sphere is a straight line drawn from the centre to the surface; a diameter is a straight line drawn through the centre, having its-extremities in the surface. 513. By the definition of ~ 512, all radii of a sphere are equal. Also, all diameters are equal, since each is the sum of two radii. 514. Two spheres are equal when their radii are equal. For they can be applied one to the other so that their surfaces shall coincide throughout. Conversely, the radii of equal spheres are equal. 515. A line (or plane) is said to be tangent fo a sphere when it has one, and only one, point in common with the surface; the common point is called the point of contact. A polyedron is said to be inscribed in a sphere when all its vertices lie in the surface of the sphere; in this case the sphere is said to be circumscribed about the polyedron. A polyedron is said to be circumscribed about a sphere when all its faces are tangent to the sphere; in this case the sphere is said to be inscribed in the polyedron. 516. A sphere may be generated by the revolution of a semicircle about its diameter as an axis. 256 THE SPHERE 257 For all points of such a surface are equally distant from the centre of the 0. (?) PROP. I. THEOREM 517. A section of a sphere made by a plane is a circle. P Given ABC a section of sphere APC made by a plane. Given ABC a section of sphere APC made by a plane. To Prove ABC a 0. Proof. 1. If 0 is centre of sphere, draw 00' -L plane ABC. 2. Take A, B any two points in perimeter ABC; draw lines OA, OB, O'A, O'B. 3. Since OA = OB (?), O'A = O'B. (~ 357) 4. Since A, B are any two points in perimeter, ABC is a 0. 518. Defs. A great circle of a sphere is a section made by a plane passing through the centre; as ABC. A small circle is a section made by a plane which does not pass through the A centre. The diameter perpendicular to a circle of a sphere is called the axis of the circle; its extremities are called the poles. The following are immediate consequences of ~~ 517, 518. 519. The axis of a circle of a sphere passes through the centre of the circle. 520. All great circles of a sphere are equal. For their radii are radii of the sphere. 258 SOLID GEOMETRY BOOK IX 521. Every great circle bisects the sphere and its surface. For if the portions of the sphere formed by the plane of the great ( be separated, and placed so that their plane surfaces coincide, the spherical surfaces falling on the same side of this plane, the two spherical surfaces will coincide throughout; for all points of either surface are equally distant from the centre. 522. Any two great circles bisect each other. For the intersection of their planes is a diameter of the sphere, and therefore a diameter of each 0. (~ 153) 523. Between any two points on the sc-:face of a sphere, not the extremities of a diameter, an arc of a great circle, less than a semi-circumference, can be drawn, and but one. For the two points, with the centre of the sphere, determine a plane which intersects the surface of the sphere in the required are. If the points are the extremities of a diameter, an indefinitely great number of arcs of great ~ can be drawn between them; for an indefinitely great number of planes can be drawn through the diameter. 524. Def. The distance between two points on the surface of a sphere, not at the extremities of a diameter, is the arc of a great circle, less than a semi-circumference, B drawn between them. Thus, the distance between points C and D \ is arc CED, and not arc CAFBD. 525. An arc of a circle may be drawn through any three points on the surface of a sphere. For the three points determine a plane which intersects the surface of the sphere in the required are. Ex. 1. The radius of a sphere is 18 inches. Find area of circle made by a plane passing through the sphere two inches from the centre. Ex. 2. Find the locus of points in space three inches from a fixed point. THE SPHERE 259 PROP. II. THEOREM 526. All points in the circumference of a circle of a sphere are equally distant from each of its poles. P Pt Given P, P' the poles of 0 ABC of sphere APC; A, B any two points in circumference ABC, and great 0 arcs PA, PB. To Prove arc PA = arc PB (~ 524). Proof. 1. Let axis PP' intersect plane ABC at O0 draw lines OA, OB, and chords PA, PB. 2. Since 0 is centre of 0 ABC (~ 519), OA = OB. (?) 3. Then chord PA =chord PB (?), and arc PA= arc PB. (?) 4. In like manner, all points in circumference ABC may be proved equally distant from P'. 527. Def. The polar distance of a circle of a sphere is the distance (~ 524) from the nearer of its poles, or from either pole if they are equally near, to the circumference. Thus, in above figure, polar distance of 0 ABC is arc PA. 528. If, in the figure of Prop. II, ABC is a great 0, and 0 the centre of the sphere, arc PA is a quadrant. (~ 190) Hence, the polar distance of a great circle is a quadrant. An arc of a circle may be drawn on the surface of a sphere by placing one foot of the compasses at the nearer pole of the circle, the distance between the feet being equal to the chord of the polar distance. Ex. 3. Find the locus of points on the surface of a sphere equidistant from the circumferences of two great circles of the sphere. Ex. 4. If three great circles on a sphere intersect, how many points have their circumferences in common? Is number of points the same if 260 SOLID GEOMETRY- BOOK IX circles are equal small circles? Find greatest possible number of intersection points common to the circumferences of three unequal small circles. Ex. 5. Mount Rainier is 14,400 feet in height. Considering the earth a sphere whose radius is 4000 miles, at what distance can the peak of this mountain be seen? PROP. III. THEOREM 529. If a point on the su:tface of a sphere lies at a quadrant's distance from each of two points in the arc of a great circle, it is a pole of that arc. The term quadrant, in Spherical Geometry, usually signifies a quadrant of a great circle. P Given point P on surface of sphere APC, AB an are of great o ABC, and PA and PB quadrants. To Prove P a pole of are AB. (PO is I to OA and OB; then use ~ 352.) Ex. 6. The circumference of a small circle, whose radius is 16, bisects the arcs of great circles drawn from a given great circle to one of its poles. Find the diameter of the sphere. Ex. 7. A cylinder of revolution is inscribed in a sphere whose radius is 18. Lines are drawn from the centre of the sphere to all points in the intersections of the cylinder and the surface of the sphere. If a plane be passed through the axis of the conical surface thus formed, the elements of the two nappes lying in this plane make an angle of 90~. Find volume of solid included between the conical surface of two nappes and the surface of the cylinder. Ex. 8. In Ex. 7 find the volume of the cylinder, if the radius of the sphere is 12 and the elements make an angle of 60~. Does this statement admit of more than one solution? THE SPHERE 261 PROP. IV. THEOREM 530. The intersection of two spheres is a circle, whose centre is in the straight line joining the centres of the spheres, and whose plane is perpendicular to that line. Given two intersecting spheres. To Prove their intersection a 0, whose centre is in line joining centres of spheres, and whose plane is I to this line. Proof. 1. Let 0, O' be centres of two (; draw line 00', intersecting common chord AB at C; 00' bisects AB at rt. /s. (?) 2. If entire figure be revolved about 00', the ( generate spheres whose centres are 0, 0' (~ 516), and AC a 0 1 00'. (?) PROP. V. THEOREM 531. A plane perpendicular to a radius of a sphere at its extremity is tangent to the sphere.,' B A Given plane MN I to radius OA of sphere AC at A. To Prove JMN tangent to sphere. (Prove as in ~ 169.) 262 SOLID GEOMETRY -BOOK IX PROP. VI. THEOREM 532. A plane targent to a sphere is perpendicular to the radius drawn to the 2potit'- ontact. (Converse of Prop. V.) Given plane MN tangent to sphere AC at A (Fig. of Prop. V) and radius OA. To Prove MN I OA. (Prove as in ~ 170.) Ex. 9. The radii of two spheres are 9 and 12, respectively. The distances between their centres is 15. Find the area of their circle of intersection. Ex. 10. If two spheres are tangent to the same plane at the same point, the line joining their centres passes through the point of contact. PROP. VII. THEOREM 533. A sphere can be circumscribed about any tetraedron. A _B D Given tetraedron ABCD. c To Prove that a sphere can be circumscribed about it. Proof. 1. Draw lines KE, KF in faces ACD, BCD, bisecting CD at rt. Zs, and let E be centre of circumscribed 0 of ACD, and F of BCD. (~ 206) 2. We have plane ELKF I CD (?), and to faces ACD, BCD. (~ 390) 3. Draw EG I ACAD, FH BCD; EG, FI lie in EKF. (~ 388) 4. EG, FH must meet at 0, unless II, which cannot be unless ACD, BCD are in same plane (~ 368); which is contrary to hyp. THE SPHERE 263 5. 0, being in EG, is equally distant from A, C, D; and being in Flt, equally distant from B, C, D. (~ 356, I) 6. Then, a spherical surface with 0 as centre, and OA as radius, will pass through A, B, C, D. Ex. 11. Given two planes the plane angle of whose diedral is 60~. Describe a sphere of radius a to which each of these planes shall be tangent. Is there more than one such sphere? Ex. 12. Through a point at a distance 12 from the centre of a sphere of radius 6 draw a plane tangent to the sphere. Ex. 13. What is the locus of the centres of spheres which are tangent to each of two parallel planes? Ex. 14. Three lines not in the same plane intersect at a point. Find the locus of the centres of spheres tangent to the three lines. Ex. 15. What is the locus of the centres of spheres which have a given radius and whose surfaces pass through a fixed point? PROP. VIII. THEOREM 534. A sphere can be inscribed in any tetraedron. A.B - C D Given tetraedron ABCD. To Prove that a sphere can be inscribed in it. Proof. If bisecting planes of diedral As BC, BD, CD meet at point 0, this point is equally distant from faces of tetraedron. (~ 392) 264 SOLID GEOMETRY-BOOK IX Ex. 16. A light C is equally distant from all points in the circumference of an opaque circle whose radius is AB. The ray of light, line CBD, makes an angle of 60~ with the plane QR, parallel to B circle A. AB is 8, and the area of shadow on R QR is 576 7r. Find volume of frustum of cone whose bases are circle A and the shadow. Q Ex. 17. A system of spheres is tangent to the faces of a triedral. Each face angle of the triedral is 90~. Find locus of the centres of the spheres. 535. Defs. The angle formed by two intersecting curves is the angle formed by tangents to the curves at their point of intersection. A spherical angle is the angle formed by two intersecting arcs of great circles. PROP. IX. THEOREM 536. A spherical angle is measured by an arc of a great circle having its vertex as a pole, included between its sides produced if necessary. A Given AB, AB' arcs of great ( on sphere AC; AD, AD' tangent to AB, AB', respectively, and BB' an arc of great ( with A as pole, included between AB, AB'. To Prove Z DAD' measured by arc BB'. Proof. 1. Draw radii OA, OB, OB'; since arcs AB, AB' are quadrants (~ 528), As AOB, AOB' are rt. As. (?) 2. Then OB II AD, OB' II AD' (?); and Z DAD' = Z BOB'. (~ 376) 3. Z BOB' is measured by arc BB' (?); then Z DAD' is. THE SPHERE 265 537. Since plane BOB' (Fig. of Prop. IX) is 1 OA (~ 352), planes AOB and BOB' are 1. (~ 390) Now a tangent to arc AB at B is I BOB'. (~ 388) Then it is 1 to a tangent to arc BB' at B. (~ 350) Then, spherical / ABB' is a rt. Z. (~ 535) That is, an arc of a great circle drawn from the pole of a great circle is perpendicular to its circumference. 538. By ~ 536, the angle formed by two arcs of great circles is the plane angle of the diedral angle between their planes. (~ 379) SPHERICAL POLYGONS AND SPHERICAL PYRAMIDS DEFINITIONS 539. A spherical polygon is a portion of B the surface of a sphere bounded by three or. more arcs of great circles; as ABCD. / The bounding arcs are called the sides of the spherical polygon, and are usually measured in degrees. The angles of the spherical polygon are the spherical angles (~ 535) formed by the adjacent sides, and their vertices are called the vertices of the spherical polygon. A diagonal of a spherical polygon is an arc of a great circle joining any two vertices which are not consecutive. A spherical triangle is a spherical polygon of three sides. A spherical triangle is called isosceles when it has two sides equal; equilateral when all its sides are equal; and right when it has a right angle. 540. The planes of the sides of a spherical polygon form a polyedral angle, whose vertex is the centre of the sphere, and whose face angles are measured by the sides of the spherical polygon (~ 191). 266 SOLID GEOMETRY -BOOK IX Thus, in figure ~ 539, the planes of the sides of the spherical polygon form a polyedral angle, O-ABCD, whose face As AOB, BOC, etc., are measured by arcs AB, BC, etc., respectively. A spherical polygon is called convex when the polyedral angle formed by the planes of its sides is convex (~ 402). 541. A spherical pyramid is a solid bounded by a spherical polygon and the planes of its sides; as O-ABCD, figure ~ 539. The centre of the sphere is called the vertex of the spherical pyramid, and the spherical polygon the base. 542. Two spherical pyramids are equal when their bases are equal. For they can be applied one to the other so as to coincide throughout. 543. If circumferences of great circles be drawn with the vertices of a spherical triangle as poles, they divide the surface of the sphere into eight spherical triangles. Thus, if circumference B'C'B" be drawn a1' with vertex A of spherical A ABC as pole, circumference A'C'A" with B as pole, and circumference A'B"A"B' with B' / C as pole, the surface of the sphere is \/ divided into eight spherical A; A'B'C', A'B"C', A"B'C', and A"B"C' on the A" hemisphere represented in the figure, the others on the opposite hemisphere. Of these eight spherical A, one is called the polar triangle of ABC, and is determined as follows: Of the intersections, A' and A", of circumferences drawn with B and C as poles, that which is less than a quadrant's distance from A, i.e. A', is a vertex of the polar triangle; and similarly for the other intersections. Thus, A'B'C' is the polar A of ABC. THE SPHERE 267 544. Two spherical polygons, on the same or equal spheres, are said to be symmetrical when the sides and angles of one are equal, respectively, to the sides and angles of the other, if the equal parts occur in the reverse order. Thus, if spherical A ABC and A A' A'B'C', on the same or equal spheres, have sides AB, BC, CA equal, re- spectively, to sides A'B', B'C', C'A', B C' and As A, B, C to As A', B', C', and the equal parts occur in the reverse order, the A are symmetrical. It is evident that, in general, two symmetrical spherical polygons cannot be placed so as to coincide throughout. PROP. X. THEOREM 545. If one spherical triangle is the polar triangle of another, the second spherical triangle is the polar triangle of the first. A ' BC Given A'B'C' the polar A of spherical A ABC; A, B, 0 being the poles of arcs B'C', C'A', A'B', respectively. To Prove ABC the polar A of spherical A A'B'C'. Proof. 1. Since B is the pole of AC', and C of A'B', A' is a quadrant's distance fron B and C. (~ 528) 2. Then A' is pole of arc BC.. (~ 529) 3. Similarly, B' is pole of CA, and C' of AB. 4. Then, ABC is polar A of A'B'C'; for of the two intersections of circumferences with B', C' as poles, A is < a quadrant from A'; and similarly for other vertices. (~ 543) Two spherical triangles, each of which is the polar triangle of the other, are called polar triangles. 268 SOLID GEOMETRY-BOOK IX PROP. XI. THEOREM 546. In two polar triangles, each angle of one is measured by the supplement of that side of the other of which it is the pole. A' a D a' e Given A, B, 0, A', B', and C' the As, in degrees, of polar A ABC, A'B'C'. Let A be the pole of B'C', B of C'A', C of A'B', A' of BC, B' of CA, and C' of AB. Let BC expressed in degrees be denoted by a, CA by b, AB by c, BC' by a', C'A' by b', A'B' by c'. To Prove A = 180 - a', B = 180 -b', C = 180 - c', A'= 180 - a, B' = 180 - b, C' = 180~ - c. Proof. 1. Extend arcs AB, AC to meet B'C' at D and E. 2. Arcs &E, C'.D are quadrants. (~ 528) 3. Then'B'E + C'D = 180~, or DE + B'C' = 180~. 4. Since A is pole of B'C', arc DE measures Z A. (~ 536) 5. Then result (3) becomes A + a' = 180~, or A = 180 - a'. 6. Similarly we prove theorem for any Z of either A. 547. Def. Two spherical polygons on the same or equal spheres are said to be mutually equilateral, or mutually equiangular, when the sides or angles of one are equal, respectively, to the homologous sides or angles of the other, whether taken in the same or in the reverse order. 548. It follows from ~ 546 that if two spherical triangles on THE SPHERE the same or equal spheres are mutually equeiangular, their polar triangles are mutually equilateral. For any two homologous sides of the polar A are supplements of equal As, and therefore equal. PRoP. XII. THEOREM 549. Any side of a spherical triangle is less than the sum of the other two sides. Given AB any side of spherical A ABC. To Prove AB < AC + BC. (By ~ 406, / AOB < / AOC + Z BOO; these As are measured by AB, AC, BC, respectively.) PROP. XIII. THEOREM 550. The sum of the sides of a convex spherical polygon is less than 360~. B Given convex spherical polygon ABCD. To Prove AB + BC + CD +DA< 360~. (By ~ 407, sum of / AOB, BOC, COD, and DOA is < 360~.) 270 SOLID GEOMETRY -BOOK IX PROP. XIV. THIEOREM 551. The sum of the angles of a spherical triangle is greater than two, and less than six, right angles. A' ct bt 'C^f ' at Given spherical A ABC. To Prove A + B + C > 180, and < 540~. Proof. 1. Let A'B'C' be polar A of ABC; a', b', c' being sides having A, B, C, respectively, as poles. 2. By ~ 546, A= 180~-a', B=180~ -b', C= 180 - c'. 3. Adding, A + B + C= 540- (a'+ b' + c'); (1) therefore, A + B + C < 540~. -4. Since a' + b' + c' is < 360~ (~ 550), by (1), A+B + C> 180~. 552. By ~ 551, a spherical A may have one, two, or three rt. A, or one, two, or three obtuse As. 553. Def. We call a spherical triangle having two right angles a bi-rectangular triangle, and one having three right angles a tri-rectanglar triangle. Ex. 18 Each side of a spherical triangle is 90~. Find angles and sides of its polar triangle. Ex. 19. The angles of a spherical triangle are 93~, 128~, and 117~, respectively. Find number of degrees in each side of its polar triangle. THE SPHERE 271 PROP. XV. THEOREM 554. If two spherical triangles on the same or equal spheres have two sides and the included angle of one equal, respectively, to two sides and the included angle of the other, I. They are equal if the equal parts occur in the same order. II. They are symmetrical if the equal parts occur in the reverse order. A~/' D D) I. Given ABC, DEF spherical A on the same or equal spheres, having AB = DE, AC= DF, and Z A = Z D; the equal parts occurring in the same order. To Prove A ABC = A DEF. Proof. Superpose A ABC upon A DEF so that AB shall fall on DE, and AC on DF; BC falls on EF by ~ 523. II. Given ABC, D'E'F' spherical A on the same or equal spheres, having AB = D'E', AC = D'', and / A = Z D'; the equal parts occurring in the reverse order. To Prove ABC and D'E'F' symmetrical. Proof. 1. Make A DEF, on the same or an equal sphere symmetrical to D'E'F', having DE = D'E', DF = D'F', and L D = Z D'; the equal parts occurring in the reverse order. 2. Then AB = DE, AC = DF, and Z A = Z D (?);.whence A ABC- A DEF (~ 554, I) and is symmetrical to D'E'F'. 272 SOLID GEOMETRY- BOOK IX PROP. XVI. THEOREM 555. If two spherical triangles on the same or equal spheres have a side and two adjacent angles of one equal, respectively, to a side and two adjacent angles of the other, I. They are equal if the equal parts occur in the same order. II. They are symmetrical if the equal parts occur in the reverse order. (The proof is left to the pupil; use Fig. of Prop. XV.) PROP. XVII. THEOREM 556. If two spherical triangles on the same or equal spheres are mutually equilateral, I. They are equal if the equal parts occur in the same order. II. They are symmetrical if the equal parts occur in the reverse order. A D ]s -.,..\o 2E '0o Given ABC, DEF mutually equilateral spherical A on equal spheres with side BC equal to EF. To Prove ABC and DEF equal or symmetrical according as the equal parts occur in the same or in the reverse order. Proof. 1. Draw radii of spheres OA, OB, OC, O'D, O'E, O'F. 2. Since triedrals O-ABC, O'-DEF have homologous face As equal (?), diedrals OA, O'D are equal. (~ 408) 3. Then plane A of OA, O'D are equal. (~ 385) 4. Then, Z BAC = EDF (~ 538); and theorem is proved by ~ 554. The theorem may be proved in a similar manner when given spherical A are on same sphere. THE SPHERE PROP. XVIII. THEOREM 557. If two spherical triangles on the same or equal spheres are mutually equiangular, I. They are equal if the equal parts occur in the same order. II. They are symmetrical if the equal parts occur in the reverse order. A' D' / z /J I ''',!/ AC\ B.i ~a F' Given ABC, DEF mutually equiangular spherical A on the same or equal spheres. To Prove ABC and DEF equal or symmetrical according as the equal parts occur in the same or in the reverse order. Proof. 1. Let A'B'C', D'E'F' be polar A of ABC, DEF. 2. A A'B'C', D'E'F' are mutually equilateral (~ 548); then, homologous s of A'B'C', D'E'F' are equal (~ 556). 3. Since ABC is polar A of A'B'C', and DEF of D'E'F' (?), ABC, DEF are mutually equilateral (~ 548); and theorem follows by ~ 556. 558. If three diameters of a sphere be B so drawn that each is perpendicular to the other two, the planes determined by them C ----'-E — divide the surface of the sphere into eight.A equal tri-rectangular triangles. For by ~ 537, each Z of each spherical A is a rt. Z, and the A are equal by ~ 557. 559. By ~ 558, the surface of a sphere is eight times the surface of one of its tri-rectangular triangles. 274 SOLID GEOMETRY-BOOK IX PROP. XIX. THEOREM 560. In an isosceles spherical triangle the angles opposite the equal sides are equal. Given, in spherical A ABC, AB = AC. 4 To Prove Z B = / C. Proof. 1. Draw AD an arc of a great 0, bisecting BC at D. 2. By ~ 556, II, A BD, ACD are sym- metrical; and Z B= Z C. a 561. An isosceles spherical triangle is equal to the spherical triangle which is symmetrical to it. For the equal parts occur in the same order. PROP. XX. THEOREM 562. If two angles of a spherical triangle are equal, the sides opposite are equal. (Converse of Prop. XIX.) Given in spherical A ABC, Z B = Z C. To Prove AB= AC. / Proof. 1. Let A'B'C' be polar A of ABC; B being pole of A'C', and C of A'B'. / \ 2. Since A'C' is sup. of Z B, and A'B' of B - -Xc Z C (?), A'C' = A'B' (?), and B'= / C'. (~ 560) 3. Since AB is sup. of Z C', and AC of Z B' (~ 545), AB = AC. PROP. XXI. THEOREM 563. If two angles of a spherical triangle are unequal, the sides opposite are unequal, and the greater A side lies opposite the greater angle. / Given, in spherical AABC, Z ABC >Z C. To Prove AC > AB. (Compare ~ 95. Draw BD an arc of a B, great 0 meeting AC at D, and making Z CBD equal to L C.) SPHERICAL POLYGONS AND SPHERICAL PYRAMIDS 275 PROP. XXII. THEOREM 564. If tawo sides of a spherical triangle are unequal, the angles opposite are unequal, aned the greater angle lies opposite the greater side. (Converse of Prop. XXI.) (Prove by Reductio ad absurdumn.) PROP. XXIII. THEOREM 565. The shortest line on the surface of a sphere between two given points is the arc of a great circle, not greater than a semicircumference, which joins the points. Given points A and B on the sur- / face of a sphere, and AB an arc of a / great 0, not > a semi-circumference. To Prove AB the shortest line on I\ the surface of the sphere between A and B. Proof. 1. Let C be any point on arc AB; and DCF, ECG arcs of small ~ with A, B as poles, and AC, BC as polar distances. 2. Arcs DCF, ECG have only point C common; for if F is any other point in DCF, and AF, BF arcs of great (, AF=AC. (~526) 3. By ~ 549, AF + BF > AC + BC; subtracting AF from first member) and AC from second, we have BF > BC (or BG), and F is without 0 ECG. 4. To prove shortest line on surface passes through C. let any other line on surface, ADEB, not passing through C, cut arc DCF at D, and arc ECG at E. 5. Whatever the nature of AD, an equal line can be drawn from A to C; and a line equal to BE from B to C. 6. Then a line can be drawn from A to B through C, equal to AD + BE, and hence < ADEB by line DE. 7. Then no line which does not not pass through C can be the shortest line from A to B; and C being any point in AB, the shortest line passes through every point of AB. 276 SOLID GEOMETRY- BOOK IX Ex. 20. A right cone is divided into two equivalent parts by a plane which is drawn parallel to its base. In what ratio is the altitude of the cone divided by the plane? MEASUREMENT OF SPHERICAL POLYGONS 566. Defs. A lune is a portion of the sur- A face of a sphere bounded by two semi-circumferences of great circles; as ACBD. The angle of the lune is the angle formed ( D by its bounding arcs. A spherical wedge is a solid bounded by a lune and the planes of its bounding arcs. B The lune is called the base of the spherical wedge. 567. It is evident that two lunes on the same sphere, or equal spheres, are equal when their angles are equal. Also, two spherical wedges in the same sphere, or equal spheres, are equal when the angles of the lunes which form their bases are equal. PROP. XXIV. THEOREM 568. The spherical triangles corresponding to a pair of vertical triedral angles are symmetrical. B CB' Given AOA', BOB', COC' diameters of sphere AC; also, planes determined by them, intersecting the surface ii arcs AB, BC, CA, A'B', B'C', C'A'. To Prove spherical A ABC, A'B'C' symmetrical. Proof. 1. Since Z AOB = Z A'OB' (?), AB = A'B'. (?) 2. Similarly, BC = B'C', CA = C'A'; now use ~ 556, 2. MEASUREMENT OF SPHERICAL POLYGONS 277 PROP. XXV. THEOREM 569. Two spherical triangles corresponding to a pair of vertical triedral angles are equivalent. a' Given AOA', BOB', COC' diameters of sphere AB; also, planes determined by them, intersecting the surface in arcs AB, BC, CA, A'B', B'C', C'A'. To Prove area ABC = area A'B'C'. Proof. 1. Let P be pole of small 0 passing through A, B, C; draw diameter of sphere PP', and arcs of great ~ PA, PB, PC, P'A', P'B', P'C'; then PA = PB = PC. (~ 526) 2. A PAB, P'A'B' are symmetrical (~ 568); whence A PAB = P'A'B'. (~ 561) 3. Similarly, A PBC = A P'B'C', A PCA = A P'C'A'. 4. A PAB, PBC, PCA compose A ABC, and A P'A'B', P'B'C', P'C'A' compose A A'B'C'; which proves theorem. If P and Pt fall without spherical A ABC and A'B'C', we should take the sum of two isosceles spherical A, diminished by the third. 570. Two symmetrical spherical triangles are equivalent. 571. Since spherical pyramids O-APB, O-BPC, O-CPA equal, respectively, spherical pyramids O-A'P'B', O-B'P'C', O-C'P'A' (~ 542), vol. O-ABC= vol. O-A'B'C'. Whence, the spherical pyramids corresponding to a pair of vertical triedral angles are -equivalent. 278 SOLID GEOMETRY -BOOK IX PROP. XXVI. THEOREM 572. Two lunes on the same sphere, or equal spheres, are to each other as their angles. The word " lzne,' in the above statement, signifies the area of the lune. Case I. When the angles are commensurable. B Given ACBD, ACBE lunes on sphere AB, having their As CAD, CAE commensurable. ACBD Z CAD To ProveACB AD ACBE =Z CAE ' Proof. 1. Let Z CAa be contained 5 times in / CAD and 3 times in L CAE; find ratio / CAD to Z CAE. 2. Extend arcs of division of Z CAD to B; subdivisions of ACBD, ACBE are all equal. (~ 567) 3. Find ratio ACBD to ACBE. 4. From results (1) and (3), we have equation (1). The theorem may be proved in a similar manner when the given lunes are on equal spheres. Case II. When the angles are incommensurable. A Given ACBD, ACBE lunes on sphere AB, having As CAD, CAE incommensurable. MEASUREMENT OF SPHERICAL POLYGONS 279 To Prove A BD — _ Z CAD ACBE / CAE (Prove as in ~~ 188 or 227. Let Z CAD be divided into any number of equal parts, and apply one of these parts to Z CAE as a unit of measure.) 573. The surface of a lune is to the surface of the sphere as the angle of the lune is to four right angles. For the surface of a sphere may be regarded as a lune whose Z is equal to 4 rt. As. 574. If the unit of measure for angles is the right angle, the area of a lune is equal to twice its angle, multiplied by the area of a tri-rectangular triangle. Let L be area of lune, A the numerical measure of its Z referred to a rt. Z as unit of measure, T the area of a trirectangular triangle; since area surface of sphere is 8 T (~ 559), =A (~ 572); and L = x x8T=2Ax T. 8T 4 4 Let it be required, for example, to find the area of a lune whose Z is 50~, on a sphere the area of whose surface is 72. The / of the lune referred to a rt. / as the unit of measure is 5; and Tis of 72, or 9. Then the area of the lune is 2 x 5 X 9, or 10. 575. Def. A tri-rectangular pyramid is a spherical pyramid whose base is a tri-rectangular triangle. 576. It may be proved, as in ~ 572, that Two spherical wedges in the same sphere, or equal spheres, are to each other as the angles of the lunes which form their bases. 577. It may be proved that if the unit of measure for angles is the right angle, the volume of a spherical wedge is equal to twice the angle of the lane which forms its base, multiplied by the volume of a tri-rectangular pyramid. 578. Def. The spherical excess of a spherical triangle is the excess of the sum of its angles above 180~ (~ 551). Thus, if the ZA of a spherical A are 65~, 80~, and 95~, its spherical excess is 65~ + 80~ + 95~ - 180~, or 60~. 280 SOLID GEOMETRY BOOK IX PROP. XXVII. THEOREM 579. If the unit of measure for angles is the right angle, the area of a spherical triacgle is equal to its spherical excess, multiplied by the area of a tri-rectangular triangle. A B / --- B..a / Given A, B, C the numerical measures of the As of spherical A ABC, referred to a rt. Z as the unit, and T the area of a trirectangular A. To Prove area ABC = (A + B + C- 2) x T. (1) Proof. 1. Draw diameters AA', BB', CC', and circumferences ABA'B', ACAC', BCB'C'; ABA'B' is a lune whose Z is A. 2. By ~ 574, area ABC + area A'BC= 2 A x T. (2) 3. Similarly, area ABC + area AB'C = 2 B x T. (3) 4. Lune ACBC', whose Z is C, is composed of A ABC, ABC'; and area ABC' = area A'B'C. (~ 569) 5. Then, area ABC + area A'BC = 2 C x T. (4) 6. Add (2), (3), (4); since A ABC, A'BC, AB'C, A'B'C form surface of hemisphere whose area is 4 T (~ 559), 2 area ABC + 4 T= (2A + 2 B + 2 C) x T. 7. Transposing 4 T, and dividing by 2, we obtain equation (1). Let it be required, for example, to find the area of a spherical A whose A are 105~, 80~, and 96~, on a sphere the area of whose surface is 144. The spherical excess of the spherical A is 100~, or 1 - referred to a rt. Z as the unit of measure; and the area of a tri-rectangular A is - of 144, or 18. Then the area of the spherical A is Io — x 18, or 20. MEASUREMENT OF SPHERICAL POLYGONS 281 580. It may be proved, as in ~ 579, that If the unit of measure for angles is the right angle, the volume of a triangular spherical pyramid is equal to the spherical excess of its base, multiplied by the volume of a tri-rectangular pyramid. PROP. XXVIII. THEOREM 581. If the unit of measure for angles is the right angle, the area of any spherical polygon is equal to the sum of its angles, diminished by as many times two right angles as the figure has sides less two, multiplied by the area of a tri-rectanygular triangle. Given K the area of any spherical polygon, n the number of its sides, s the sum of its As referred to a rt. Z as the unit, and T the area of a tri-rectangular A. To Prove K= [s - 2 (n - 2)] X T. (1) Proof. 1. Divide spherical polygon into n - 2 spherical A by drawing diagonals froml any vertex. 2. By ~ 579, area each A equals sum of its s, minus 2 rt. s, multiplied by T. 3. Then, sum of areas of A equals sum of their /s (or sum of A of polygon), minus n - 2 times 2 rt. As, multiplied by T; which proves equation (1). 582. It may be proved, as in ~ 581, that If the unit of measure for angles is the right angle, the volume of any spherical pyramid is equal to the sum of the angles of its base, diminished by as many times two right angles as the base has sides less two, multiplied by the volume of a tri-rectangular pyramid. Ex. 21. The sides of a spherical triangle, on a sphere the area of whose surface is 156, are 44~, 63~, and 97~, respectively. Find the area of its polar triangle. Ex. 22. Find the area of a spherical hexagon whose angles are 120~, 139~, 148~, 155~, 162~, and 167~, on a sphere the area of whose surface is 280. 282 SOLID GEOMETRY-BOOK IX Ex. 23. Given an opaque sphere whose radius OB is 16 v2 feet. A light is placed at C, a point so chosen that Z AOB = 45~, AB being the radius of the small circle which inter- cepts the light. The shadow cast by the sphere upon the plane QR is a circle with centre at 31, and has an area of / O 9216 wr square feet. Find the distance, R N CM, from the light to the plane QR. / If M were not the centre of the shadow and the shadow were not a circle, would the circle whose centre is A and the plane QR be parallel? Ex. 24. Find the volume of a pentagonal spherical pyramid the angles of whose base are 109~, 128~, 137~, 153~, and 158~, respectively; the volume of the sphere being 180. MEASUREMENT OF THE SPHERE DEFINITIONS 583. A zone is a portion of the surface of a sphere included between two parallel planes. The circumferences of the circles which bound the zone are called the bases, and the perpendicular distance between their planes the altitude. A zone of one base is a zone one of whose bounding planes is tangent to the sphere. A spherical segment is a portion of a sphere included between two parallel planes. The circles which bound it are called the bases, and the perpendicular distance between them the altitude. A spherical segment of one base is a spherical segment one of whose bounding planes is tangent to the sphere. 584. If semicircle ACEB be revolved A about diameter AB as an axis, and CD c_ and EF are lines I AB, arc CE generates E i a zone whose altitude is DF, CEFD a spherical segment whose altitude is DF, arc AC a zone of one base, and ACD a spherical segment of one base. MEASUREMENT OF TEIE SPHERE 283 585. If a semicircle be revolved about its diameter as an axis, the solid generated by any sector of A the semicircle is called a spherical sector. Thus, if semicircle ACDB be revolved D -/ about diameter AB as an axis, sector OCD generates a spherical sector. The zone generated by the arc of the sector is called the base of the spherical B sector. PROP. XXIX. THEOREM 586. The area of the surface generated by the revolution of a straight line about a straight line in its plane, not parallel to and not intersecting it, as an axis, is equal to its projection on the axis, multiplied by the circumference of a circle, whose radius is the perpendicular erected at the middle point of the line and terminating in the axis. M -— 4 - B - ------- D F Given str. line AB revolved about str. line FM in its plane, not 11 to and not intersecting it, as an axis; lines AC and BD I FM, and EF the L erected at the middle point of AB terminating in FM. To Prove area AB* = CD x 2 7rEF (~~ 254, 334). (1) Proof. 1. Draw line AG I BD, line EH _ CD; AB generates lateral surface of a frustum of a cone of revolution. 2. By ~ 509, area AB -- AB x 2 7rEH. (2) 3. Since A ABG, EFH are similar (?), A = EF. (? 'AGrC EH ( 4. Then, AB x EH= AG x EF= CD x EF. (?) 5. Substituting in (2), gives equation (1). * The expression " area AB' is used to denote the area of the surface generated by AB. 284 SOLID GEOMETRY - BOOK IX PROP. XXX. THEOREM 587. The area of a zone is equal to its altitude multiplied by the circumference of a great circle. A A' B BD' BGi ar AX B' 0 Given are AB revolved about diameter OM as an axis, lines AA', BB' 1 OM, and R the radius of the arc. To Prove area zone generated by AB = A'B' x 2 rR. (1) Proof. 1. Divide AB into equal arcs AC, CD, DB; draw chords AC, CD, DB, lines CC', DD' 0M, and OE I AC. 2. By ~ 586, area AC AC' x 2 T-OE, area CD= CD' x 2 7rOE, etc. 3. Adding, area generated by broken line ACDB = (A'C' + CD' + etc.) x 2 rOE = A'B' x 2 7rOE. 4. If subdivisions of AB be bisected indefinitely, area generated by broken line ACDB approaches area generated by arc AB as a limit,* and A'B' x 2 wrOE approaches A'B' x 2 7rR as a limit. (~ 330t) 5. Prove equation (1) by Theorem of Limits. 588. The proof of ~ 587 holds for any zone which lies entirely on the surface of a hemisphere; for, in that case, no chord is 1I 01M, and ~ 586 is applicable. Since a zone not entirely on the surface of a hemisphere is the sum of two zones, each of which does lie entirely on the surface of a hemisphere, the theorem of ~ 587 holds for any zone. * A rigorous proof of this is beyond the scope of the present treatise. t The theorem of ~ 330 is evidently true when, instead of the perimeter of a regular inscribed polygon, we have a broken line consisting of equal straight lines inscribed in an arc. MEASUREMENT OF THE SPHERE 285 589. If S denotes the area of a zone, h its altitude, R the radius of the sphere, S = 2 7rRh. 590. Since the surface of a sphere may be regarded as a zone whose altitude is a diameter of the sphere, The area of the surface of a sphere is equal to its diameter multiplied by the circumference of a great circle. 591. Let S denote the area of the surface of a sphere, R its radius, and D its diameter. Then, S = 2 R x 2 rR = 4 7rR2. That is, the area of the surface of a sphere is equal to the square of its radius multiplied by 4 Tr. Again, S = 7r x (2 R)2 = rD2. That is, the area of the surface of a sphere is equal to the square of its diameter multiplied by 7r. 592. Since 7rR2 is the area of a great 0 (?), the surface of a sphere is equivalent to four great circles. 593. The areas of the surfaces of two spheres are to each other as the squares of their radii, or as the squares of their diameters. (The proof is left to the pupil; compare ~ 338.) Ex. 25. Find the area of the surface of the earth included in the north temperate zone. (Call earth a sphere whose radius is 4000 miles and the altitude of the zone 2164.8 miles.) Ex. 26. A sphere is 32 feet in diameter. Find the area of the surface visible to an observer standing 32 feet from the centre of the sphere. Ex. 27. The radii of two spheres are in the ratio i. What is the ratio of the areas of their surfaces? Ex. 28. The area of the surface of a sphere is three times the area of the surface of another. Find the ratio of their radii. If the area of one is M times that of the other, find ratio of radii. Ex. 29. Find the area of a spherical triangle whose angles are 75~, 105~, and 100~, respectively, on a sphere whose radius is 8 inches. Ex. 30. A plane bisects a radius of a sphere. Find the ratio of the areas of the surfaces into which the surface of the sphere is divided. SOLID GEOMETRY- BOOK IX PROP. XXXI. THEOREM 594. If an isosceles triangle be revolved about a straight line in its plane, not parallel to its base, as an axis, which passes through its vertex without intersecting its surface, the volume of the solid generated is equal to the area of the surface generated by the base, multiplied by one-third the altitude. A 'F e/*^", \, --- ED 0 Given isosceles A OAB revolved about str. line OF in its plane, not 11 to base AB, as an axis; and line OC_1AB. To Prove vol. OAB* = area AB x 1 OC. (1) Proof. 1. Draw AD, BE I OF; extend BA to meet OF at F. 2. vol. OBF = vol. OBE + vol. BEF = TrBE2 x OE + 7rBE2 x EF (~505) = rBE2 x (OE + EF)= rBE x BE x OF. 3. Since BE OF = OC x BF, each being twice area OBF, vol. OBF== 1 7rBE X OC X BF = 7rBE X BF X I OC. 4. Then (~ 503), vol. OBF= area generated by BF X OC0. (2) 5. Similarly, vol. OAF=area generated by AF x ~ C0. (3) 6. Subtracting (3) from (2) gives equation (1). Ex. 31. The area of the surface of a sphere is 900 7r. Find volume of an inscribed cone whose axis coincides with a fixed diameter and the radius of whose base is 22. How many cones satisfy the condition? * The expression "vol. OAB" is used to denote the volume of the solid generated by OAB. MEASUREMENT OF THE SPHERE 287 PROP. XXXII. THEOREM 595. The volume of a spherical sector is equal to the area of the zone which forms its base, multiplied by one-third the radius of the sphere. M D. n B,o Given sector OAB revolved about diameter OMf as an axis, and R the radius of the arc. To Prove volume of spherical sector generated by OAB equal to area of zone generated by AB x ~ R. Proof. 1. Divide AB into equal arcs, AC, CD, DB; draw chords AC, CD, DB, also lines OC, OD, and line OE 1 AC. 2. By ~ 594, vol. OAC = area AC x i OE, vol. OCD = area CD x Q OE, etc. 3. Adding, volume generated by polygon OACDB = (area AC0+ area CD, etc.) x 3 OE = area ACDB x ~ OE. 4. If subdivisions of AB be bisected indefinitely, volume generated by polygon OACDB approaches volume generated by sector OAB as a limit, and area ACDB x ~ OE approaches area generated by arc AB x R as a limit. (~~ 329, 330t) 5. Prove theorem by Theorem of Limits. It is evident, as in ~ 588, that ~ 595 holds for any spherical sector. Ex. 32. The angles of a spherical quadrilateral are 95~, 115~, 135~, and 110~, respectively. What is the area of the quadrilateral if the radius of the sphere is 7 inches? * A rigorous proof of this is beyond the scope of the present treatise. t See note, foot p. 284. 288 SOLID GEOMETRY-BOOK IX 596. If Vdenotes the volume of a spherical sector, h the altitude of the zone which forms its base, R the radius of the sphere, V= 2 7rRh x R (~ 589) = rR2h. 597. Since a sphere may be regarded as a spherical sector whose base is the surface of the sphere, The volume of a sphere is equal to the area of its surface multiplied by one-third its radius. 598. Let V denote the volume of a sphere, R its radius, and D its diameter. Then, V= 4 7rR2 x R (~ 591) = 4 7rR3. That is, the volume of a sphere is equal to the cube of its radius multiplied by 4 1r. Again, V= rD2 x 1 D (~ 591) - 1 7rD3. That is, the volume of a sphere is equal to the cube of its diameter multiplied by I 7r. 599. The volumes of two spheres are to each other as the cubes of their radii, or as the cubes of their diameters. (The proof is left to the pupil.) 600. The volume of a spherical pyramid is equal to the area of its base multiplied by one-third the radius of the sphere. Let P denote volume of spherical pyramid, K the area of its base, R the radius of the sphere, n the number of sides of the base, s the sum of its As referred to a rt. Z as the unit of measure, T the area of a tri-rectangular A, T' the volume of a trirectangular pyramid, S the area of the surface of the sphere, and Vits volume. Then, P 2 (58 - 2)] X T1 _ (~~ 51, 582) K [s - 2(n - 2) X T - MEASUREMENT OF THE SPHERE 289 V 8 T' T' Also, = ST T =T'9) Then, = 7 (~ ~ 591, 598) = 1R; and therefore K S 4 7R2 3 P=K'x 1R. Ex. 33. A tank in the form of a cylinder of revolution is 2 meters in altitude, and the diameter of its base is 1 meter. How many gallons of water will it contain? Ex. 34. A bar of iron in the form of a rectangular parallelopiped is 12 inches long, 8 inches wide, and 4 inches thick. The bar is melted and cast into a spherical ball. Find the radius of the ball. Ex. 35. A brass rod is in the form of a cylinder of revolution, whose length is 4 feet and diameter of base 3 inches, capped at each end by a hemisphere whose diameter is 3 inches. The rod and hemispheres have a uniform thickness of I inch. Find its'weight. (One cubic foot of water weighs 621 pounds, and brass is 8.2 times as heavy as water.) Ex. 36. A spherical metal shell is ~ inch in thickness., The outside diameter of the shell is 10 inches. Find the amount of metal in the shell. Ex. 37. The diameter of a sphere, the altitude and diameter of a cone of revolution and of a cylinder of revolution are each a inches. Their volumes are in what ratio? Their entire surfaces are in what ratio? Ex. 38. Find ratio of the volumes of a sphere to its circumscribed tube. Ex. 39. Find the ratio of the volume of a sphere to its circumscribed regular tetraedron. Ex. 40. The volume of a spherical wedge is 75. The volume of the sphere is 325. Find the angle of the base of the spherical wedge. Ex. 41. Find the ratio of the volume of a sphere to its circumscribed cylinder of revolution. Could an oblique circular cylinder circumscribe the sphere? Why? Ex. 42. Find the ratio of the volume of a sphere to the volume of the inscribed cube. 290 290 ~SOLID GEOMETRY - BOOK IX PROP. XXXIII. PROBLEM 601. Given the radii of the bases, and the altitude, of a spherical segment, to find its vol ame. A,M A' r Given 0 the centre of arc ADB, lines AA', BB' I to diameter 0OX, AA' =r', BB' = r, and A'B' = h. Required to express volume of spherical segment generated by the revolution of ADBR'A' about OXf as an axis in terms of r, r', h. Solution. 1. Draw lines OA, OB, AB; also, line 00.1I AR, and line AE 1I BR'; denote radius OA by B. 2. vol. ADBB'A' = vol. ACJBD + vol. ARR'A'. (1) 3. vol. AC1DB = vol. OADB -- vol. OAR. 4. By ~ 596, vol. 0ADB = 2 7rB'h. 5. By ~ 594, vol. OAR = area AR x OC0 = h x 2 irOC x 1 00 (~ 586) = 2 rOO'h. 6. Then, vol. AODB = ~ —Rh - 2w00'h 6C )(R~02h. 7'. By ~ 253 ' 0 (AI' (~ARB() AB'. 8. Then, vol. ACDBR= 2 -X {AB Xh rABh = I wr(BE' AEif)h = 7r( - r')2 ~ h2]h. 9. Substitute in (1), by ~ 511, vol. ADBB'A' -1 7r[(r - r')2 ~ h2]h + 1- 7(2 r-' + 2 r" + 2 rr')h -L 7w(r2 - 2 rr' + r"2 + h2 ~_ 2 r2 + 2 r"2 + 2 rr')h j- I '(3 r.2 + 3 9,"2h ~ 1 rrh' =3 7 w(12 + r"2) h ~ _I 7r/h3. 602. If r denotes radius of base, h the altitude, of a spherical segment of one base, its volume is I irT'2h + 1- 7rh'. MEASUREMENT OF THE SPHERE 2 291 Ex. 43. Find the volume of a cone of revolution inscribed in a sphere whose volume is 4500 7r. The axis of the cone is 15 and coincides with a fixed diameter. Cal there be more than one such cone? Ex. 44. A plane is drawn through the axis of a cone of revolution. The intersection of the plane and the cone is an equilateral triangle whose side is 12. Find the area of the surface of a sphere inscribed in the cone. Ex. 45. A plane is passed through the axis of a cone of revolution. The intersection of the plane and the cone is an equilateral triangle. Find the ratio of the volume of the sphere inscribed in the cone to the volume of the cone. Ex. 46. The legs of a right triangle are 18 and 12, respectively. The triangle revolves about the hypotenuse as an axis. Find the area of the surface generated. Ex. 47. A hollow iron column in the shape of a frustum of a cone of revolution is 10 feet long; one base is 10 inches in outer diameter, and 8 inches in inner, and the other base is 6 inches in outer diameter, and 4 inches in inner. The column is melted and formed into a bar in the shape of a rectangular parallelopiped, 4 inches wide and 2 inches thick. Find length of bar. Ex. 48. A 10-pound spherical iron cannon ball is 3 inches in diameter. Find the diameter of a 20-pounder. Ex. 49. Measure the circumference of a great circle of any sphere of known diameter. Using the measurement as given, compute the volume of the sphere. Ex. 50. Given a semicircle of radius r, and a chord parallel to the diameter of the semicircle. The radii drawn from the middle point of the diameter to the extremities of the chord form an angle of 60~. The semicircle is revolved about its diameter as an axis. Find the volume of the solid generated by the segment between the chord and the arc. Ex. 51. Given an equilateral triangle ABD, C and a light placed at point C, equidistant from A, B and D, and at a distance 12 from plane A cg-B ABD. A side of ABD is 8 and plane ABD is / parallel to plane QR. AtB'D' is the shadow of ABD upon QR, and CO' is perpendicular to /A 4 A'B'D', meeting ABD at O. The distance from / - / C to QR is 64. Find area A'B'D'. Q INDEX TO DEFINITIONS. Acute angle, ~ 37. Adjacent angles, ~ 13. diedral angles, ~ 378. Alternate-exterior angles, ~ 74. -interior angles, ~ 74. Altitude of a cone, ~ 495. of a cylinder, ~ 480. of a frustum of a cone, ~ 495. of afrustum of a pyramid, ~ 448. of a parallelogram, ~ 103. of a prism, ~ 413. of a pyramid, ~ 445. of a spherical segment, ~ 583. of a trapezoid, ~ 103. of a triangle, ~ 73. of a zone, ~ 583. Angle, ~ 10. at the centre of a regular polygon, ~ 308. between two intersecting curves, ~ 535. inscribed in a segment, ~ 149. of a lune, ~ 566. Angles of a polygon, ~ 115. of a quadrilateral, ~ 102. of a spherical polygon, ~ 539. of a triangle, ~ 21. Angular degree, ~ 39. Antecedents of a proportion, ~ 213. Apothem of a regular polygon, ~ 308. Arc of a circle, ~ 23. Area of a surface, ~ 278. Axiom, ~ 32. Axis of a circle of a sphere, ~ 518. of a circular cone, ~ 495. Base of a cone, ~ 495. of a polyedral angle, ~ 401. of a pyramid, ~ 445. of a spherical pyramid, ~ 541. of a spherical sector, ~ 585. of a spherical wedge, ~ 566. of a triangle, ~ 73. Bases of a cylinder, ~ 480. of a parallelogram, ~ 103. of a prism, ~ 413. of a spherical segment, ~ 583. of a trapezoid, ~ 103. of a truncated prism, ~ 419. of a truncated pyramid, ~ 447. of a zone, ~ 583. Bi-rectangular triangle, ~ 553. Broken line, ~ 5. Central angle, ~ 149. Centre of a parallelogram, ~ 108. of a regular polygon, ~ 308. Chord of a circle, ~ 147. Circle, ~ 23. circumscribed about a polygon, ~ 152. 293 294 INDEX TO DEFINITIONS. Circle inscribed in a polygon, ~ 152. Circles tangent externally, ~ 151. tangent internally, ~ 151. Circular cone, ~ 495. cylinder, ~ 480. Circumference, ~ 23. Commensurable magnitudes, ~ 181. Common measure, ~ 181. tangent, ~ 151. Complement of an angle, ~ 40. of an arc, ~ 189. Complementary angles, ~ 40. Concave polygon, ~ 119. Concentric circles, ~ 146. Cone, ~ 495. of revolution, ~ 495. Conical surface, ~ 495. Consequents of a proportion, ~ 213. Constant, ~ 184. Converse of a proposition, ~ 67. Convex polyedral angle, ~ 402. polyedron, ~ 411. polygon, ~ 118. spherical polygon, ~ 540. Corollary, ~ 32. Corresponding angles, ~ 74. Cube, ~ 421. Curve, ~ 5. Cylinder, ~ 480. of revolution, ~ 480. Cylindrical surface, ~ 480. Decagon, ~ 116. Degree of arc, ~ 189. Determination of a plane, ~ 343. of a straight line, ~ 35. Diagonal of a polyedron, ~ 410. of a polygon, ~ 115. of a quadrilateral, ~ 102. of a spherical polygon, ~ 539. Diameter of a circle, ~ 142. of a sphere, ~ 512. Diedral angle, ~ 378. Dimensions of a rectangle, ~ 279. of a rectangular parallelopiped, ~ 432. Directrix of a conical surface, ~ 495. of a cylindrical surface, ~ 480. Distance between two points on the surface of a sphere, ~ 524. of a point from a line, ~ 96. of a point from a plane, ~ 360. Dodecaedron, ~ 410. Dodecagon, ~ 116. Edge of a diedral angle, ~ 378. Edges of a polyedral angle, ~ 401. of a polyedron, ~ 410. Element of a conical surface, ~ 495. of a cylindrical surface, ~ 480. Enneagon, ~ 116. Equal figures, ~ 12. polyedral angles, ~ 403. Equiangular polygon, ~ 117. triangle, ~ 22. Equilateral polygon, ~ 117. spherical triangle, ~ 539. triangle, ~ 22. Equivalent solids, ~ 412. surfaces, ~ 279. Exterior angles, ~ 74. of a triangle, ~ 73. Extremes of a proportion, ~ 213. Face angles of a polyedral angle, ~ 401. Faces of a diedral angle, ~ 378. of a polyedral angle, ~ 401. of a polyedron, ~ 410. Foot of a line, ~ 349. Fourth proportional, ~ 215. INDEX TO DEFINITIONS. 295 Frustum of a cone, ~ 495. of a pyramid, ~ 448. of a pyramid circumscribed about a frustum of a cone, ~ 499. of a pyramid inscribed in a frustum of a cone, ~ 499. Generatrix of a conical surface, ~ 495. of a cylindrical surface, ~ 480. Geometrical figure, ~ 8. Geometry, ~ 9. Great circle of a sphere, ~ 518. Lateral faces of a pyramid, ~ 445. Lateral surface of a cone, ~ 495. of a cylinder, ~ 480. Legs of a right triangle, ~ 22. Limit of a variable, ~ 185. Line, ~ 2. Locus of a series of points, ~ 342. Lower base of a frustum of a cone, ~ 495. nappe of a conical surface, ~ 495. Lune, ~ 566. I Hendecagon, ~ 116, Heptagon, ~ 116. Hexaedron, ~ 410. Hexagon, ~ 116. Homologous, ~ 48. Hypotenuse of a ~ 22. Hypothesis, ~ 45. right triangle, Icosaedron, ~ 410. Incommensurable magnitudes, ~ 181. Indirect method of proof, ~ 66. Inscribed angle, ~ 149. Inscriptible polygon, ~ 152. Interior angles, ~ 74. Isosceles spherical triangle, ~ 539. trapezoid, ~ 103. triangle, ~ 22. Lateral area of a cone, ~ 498. of a cylinder, ~ 487. of a frustum of a cone, ~ 498. of a prism, ~ 413. of a pyramid, ~ 445. Lateral edges of a prism, ~ 413. of a pyramid, ~ 445. Lateral faces of a prism, ~ 413. Material body, ~ 1. Mean proportional, ~ 214. Means of a proportion, ~ 213. Measure of a magnitude, ~ 180. of an angle, ~ 39. Median of a triangle, ~ 138. Mutually equiangular polygons, ~ 121. equiangular spherical polygons, ~ 547. equilateral polygons, ~ 120. equilateral spherical polygons, ~ 547. Numerical measure, ~ 180. Oblique angles, ~ 37. lines, ~ 37. prism, ~ 417. Obtuse angle, ~ 37. Octaedron, ~ 410. Octagon, ~ 116. Parallel lines, ~ 68. planes, ~ 349. Parallelogram, ~ 103. Parallelopiped, ~ 421. Pentagon, ~ 116. Perimeter of a polygon, ~ 115. Perpendicular lines, ~ 14. planes, ~ 386. Plane, ~ 6. 296 INDEX TO DEFINITIONS. Plane angle of a diedral angle, ~ 379. figure, ~ 8. geometry, ~ 9. tangent to a cone, ~ 495. tangent to a cylinder, ~ 487. tangent to a frustum of a cone, ~ 495. tangent to a sphere, ~ 515. Point, ~ 3. of contact of a line tangent to a circle, ~ 150. of contact of a line tangent to a sphere, ~ 515. of contact of a plane tangent to a sphere, ~ 515. Polar distance of a circle of a sphere, ~ 527. triangle of a spherical triangle, ~ 543. Poles of a circle of a sphere, ~ 518. Polyedral angle, ~ 401. Polyedron, ~ 410. circumscribed about a sphere, ~ 515. inscribed in a sphere, ~ 515. Polygon, ~ 115. circumscribed about a circle, ~ 152. inscribed in a circle, ~ 152. Postulate, ~ 32. Prism, ~ 413. circumscribed about a cylinder, ~ 487. inscribed in a cylinder, ~ 487. Problem, ~ 32. Projection of a line on a line, ~ 254. of a line on a plane, ~ 396. of a point on a line, ~ 254. of a point on a plane, ~ 396. Proportion, ~ 210. Proposition, ~ 32. Pyramid, ~ 445. circumscribed about a cone, ~ 499. inscribed in a cone, ~ 499. Quadrangular prism, ~ 416. pyramid, ~ 446. Quadrant, ~ 154. Quadrilateral, ~ 102. Radius of a circle, ~ 23. of a regular polygon, ~ 308. of a sphere, ~ 512. Ratio, ~ 180. Reciprocally proportional magnitudes, ~ 258. Rectangle, ~ 103. Rectangular parallelopiped, ~ 421. Rectilinear figure, ~ 8. Re-entrant angle, ~ 119. Regular polyedron, ~ 477. polygon, ~ 305. prism, ~ 417. pyramid, ~ 446. Rhomboid, ~ 103. Rhombus, ~ 103. Right angle, ~ 14. angled spherical triangle, ~ 539. circular cone, ~ 495. cylinder, ~ 480. diedral angle, ~ 386. parallelopiped, ~ 421. prism, ~ 417. section of a cylinder, ~ 487. section of a prism, ~ 420. triangle, ~ 22. Scalene triangle, ~ 22. Secant, ~ 150. Sector of a circle, ~ 147. Segment of a circle, ~ 147. Segments of a line by a point, ~ 231. Semicircle, ~ 154. INDEX TO DEFINITIONS. 297 Semi-circumference, ~ 154. Sides of a polygon, ~ 115. of a quadrilateral, ~ 102. of a spherical polygon, ~ 539. of a triangle, ~ 21. of an angle, ~ 10. Similar arcs, ~ 335. cones of revolution, ~ 495. cylinders of revolution, ~ 480. polyedrons, ~ 468. polygons, ~ 233. sectors, ~ 335. segments, ~ 335. Slant height of a cone of revolution, ~ 498. of a frustum of a cone of revolution, ~ 498. of a frustum of a regular pyramid, ~ 450. of a regular pyramid, ~ 450. Small circle of a sphere, ~ 518. Solid, ~ 1. geometry, ~ 9. Sphere, ~ 512. circumscribed about a polyedron, ~ 515. inscribed in a polyedron, ~ 515. Spherical angle, ~ 535. excess of a spherical triangle, ~ 578. polygon, ~ 539. pyramid, ~ 541. sector, ~ 586. segment, ~ 583. segment of one base, ~ 583. triangle, ~ 539. wedge, ~ 566. Square, ~ 103. Straight line, ~ 5. divided in extreme and mean ratio, ~ 272. Straight line oblique to a plane, ~ 349. parallel to a plane, ~ 349. perpendicular to a plane, ~ 349. tangent to a circle, ~ 150. tangent to a sphere, ~ 515. Straight lines divided proportionally, ~ 226. Subtended arc, ~ 147. Supplement of an angle, ~ 40. of an arc, ~ 189. Supplementary-adjacent. angles, ~42. angles, ~ 42. Surface, ~ 1. Surface of a material body, ~ 1. of a solid, ~ 1. Symmetrical polyedral angles, ~ 404. spherical polygons, ~ 544. Tangent circles, ~ 151. Tetraedron, ~ 410. Theorem, ~ 32. Third proportional, ~ 214. Transversal, ~ 74. Trapezium, ~ 103. Trapezoid, ~ 103. Triangle, ~ 21. Triangular prism, ~ 416. pyramid, ~ 446. Triedral angle, ~ 401. Tri-rectangular triangle, ~ 553. pyramid, ~ 575. Truncated prism, ~ 419. pyramid, ~ 447. Unit of measure, ~ 180. of surface, ~ 278. of volume, ~ 410. 298 INDEX TO DEFINITIONS. Upper base of a frustum of a cone, ~ 495. Upper nappe of a conical surface, ~ 495. Variable, ~ 184. Vertex of a cone, ~ 495. of a conical surface, ~ 495. of a polyedral angle, ~ 401. of a pyramid, ~ 445. of a spherical pyramid, ~ 541. of a triangle, ~ 73. of an angle, ~ 10. Vertical angle of a triangle, ~ 73. angles, ~ 37. diedral angles, ~ 378. polyedral angles, ~ 401. Vertices of a polyedron, ~ 410. of a polygon, ~ 115. of a quadrilateral, ~ 102. of a spherical polygon, ~ 539. of a triangle, ~ 21. Volume of a solid, ~ 410. Zone, ~ 583. of one base, ~ 583.