GIFT OF V %*~*?r^ vOl^S , NEW PLANE AND SOLID GEOMETRY. BY a WOOSTER WOODRUFF BEMAX H PROFESSOR OF MATHEMATICS IX THE UNIVERSITY OF MICHIGAN DAYID EUGENE SMITH PROFESSOR OF MATHEMATICS IX TEACHERS COLLEGE COLUMBIA UNIVERSITY, NEW YORK BOSTON, U.S.A. GTXX & COMPANY, PUBLISHERS C&e ^tfjenarum press 190.°, Copyright, 1895, 1899, 1900, by Wooster Woodruff Beman and David Eugene Smith ALL RIGHTS RESERVED PREFACE. Ix presenting a revision of their " Plane and Solid Geom- etry " (Boston, 1895), the authors feel that an explanation of its distinctive features may be of service to the teacher. It is sometimes asserted that we should break away from the formal proofs of Euclid and Legendre and lead the student to independent discovery, and so we find text-books that give no proofs, others that give hints of the demonstrations, and still others that draw out the demonstration by a series of questions which, being capable of answer in only one way, merely conceal the Euclidean proof. But, after all, the experience of the world has been that the best results are secured by setting forth a minimum of formal proofs as models, and a maximum of unsolved or unproved propositions as exercises. This plan has been followed by the authors, and the success of the first edition has abundantly justified their action. There is a growing belief among teachers that such of the notions of modern geometry as materially simplify the ancient should find place in our elementary text-books. With this belief the authors are entirely in sympathy. Accordingly they have not hesitated to introduce the ideas of one-to-one correspondence, of anti-parallels, of negative magnitudes, of general figures, of prismatic space, of similarity of point systems, and such other concepts as are of real value in the early study of the science. All this has been done in a con- iii iv rUEFACE. servative way, and such material as the first edition showed to be at all questionable has been omitted from the present revision. Within comparatively recent years the question of methods of attack has interested several leading writers. Whatever has been found to be usable in elementary work the authors have inserted where it will prove of most value. To allow the student to grope in the dark in his efforts to discover a proof, is such a pedagogical mistake that this innovation in American text-books has been generally welcomed. Upon this point the authors have freely drawn from the works of Petersen of Denmark, and of Eouche and de Comberousse of France, and from the excellent treatise recently published by Hadamard (Paris, 1898). With this introduction of modern concepts has necessarily come the use of certain terms and symbols which may not generally be recognized by teachers. These have, however, been chosen only after most conservative thought. None is new in the mathematical world, and all are recognized by the leading writers of the present time. They certainly deserve place in our elementary treatises on the ground of exactness, of simplicity, and of their general usage in mathematical literature. The historical notes of the first edition have been retained, it being the general consensus of opinion that they add materially to the interest in the work. For teachers who desire a brief but scholarly treatment of the subject the authors refer to their translation of Fink's "History of Elementary Mathematics" (Chicago, The Open Court Pub- lishing Co., 1899). For the limitations of elementary geom- etry, the impossibility of trisecting an angle, squaring a circle, etc., teachers should read the authors' translation of PREFA CE. V Klein's valuable work, "Famous Problems of Elementarv Geometry " (Boston. Ginn & Company). It is impossible to make complete acknowledgment of the helps that have been used. The leading European text-books have been constantly at hand. Special reference, however, is due to such standard works as those of Henrici and Treutlein, " Lehrbuch der Elementar-Geometrie," the French writers already mentioned, and the noteworthy contributions of the recent Italian school represented by Faifofer, by Socci and Tolomei, and by Lazzeri and Bassani. Teachers are urged to consider the following suggestions in using the book : 1. Make haste slowly at the beginning of plane and of solid geometry. 2. Never attempt to give all of the exercises to any class. Two or three hundred, selected by the teacher, should suffice. 3. Eequire frequent written work, thus training the eye, the hand, and the logical faculty together. The authors' Geometry Tablet (Ginn & Company) is recommended for this work. W. W. BEMAN, Anx Arbor, Mich. D. E. SMITH, Brockport, N. Y. June 15, 1899. CONTENTS. INTRODUCTION. PAGE 1. Elementary Definitions ....... 1 2. The Demonstrations of Geometry ..... 9 3. Preliminary Propositions ....... 13 PLAXE GEOMETRY. Book I. — Rectilixear Figures. 1. Triangles .21 2. Parallels and Parallelograms ...... 43 3. Problems ........... 67 4. Loci of Points 80 Book II. — Equality of Polygons. 1. Theorems 90 2. Problems . . ■ 109 3. Practical Mensuration 112 Book III. — Circles. Definitions 114 1. Central Angles ......... 116 2. Chords and Tangents 119 3. Angles formed by Chords, Secants, and Tangents . . 128 4. Inscribed and Circumscribed Triangles and Quadrilat- erals . 136 5. Two Circles .......... 143 6. Problems 146 vii viii CONTENTS. Appendix to Book III. PAGE Methods 152 Book IV. — Ratio and Proportion. 1. Fundamental Properties 150 2. The Theory of Limits ........ 167 3. A Pencil of Lines Cut by Parallels 170 4. A Pencil Cut by Antiparallels or by a Circumference . 177 5. Similar Figures ......... 182 6. Problems ........... 104 Book V. — Mensuration of Plane Figures. Regular Polygons and the Circle. 1. The Mensuration of Plane Figures 100 2. The Partition of the Perigon 205 3. Regular Polygons 200 4. The Mensuration of the Circle 216 Appendix to Plane Geometry. 1. Supplementary Theorems in Mensuration .... 226 2. Maxima and Minima 220 3. Concurrence and Collinearity 238 SOLID GEOMETRY. Book VI. — Lines and Planes in Spacf 1. The Position of a Plane in Space. The Straight Li the Intersection of two Planes 2. The Relative Position of a Line and a Plank 3. Pencil of Planes ...... 4. Polyhedral Angles ..... 5. Problems ........ NK AS 244 251 265 274 280 COX TENTS. Book VII. POLYHEDRA. 1. General and Regular Polyhedra . 283 2. Parallelepipeds ........ 288 3. Prismatic and Pyramidal Space. Prisms and Pyramids . 291 4. The Mensuration of the Prism 298 5. The Mensuration of the Pyramid . Book VIII. — The Cylinder, Cone, and Sphere Similar Solids. 308 1. The Cylinder .......... 317 2. The Cone 321 ■j. The Sphere 326 4. The Mensuration of the Sphere ...... 355 5. Similar Solids 364 TABLES. Numerical Tables Biographical Table Table of Etymologies . Index . 371 372 375 379 PLANE AND SOLID GEOMETKY. 3>^C PLANE GEOMETRY. INTRODUCTION. 1, ELEMENTARY DEFINITIONS. 1. In Arithmetic the student has considered the science of numbers, and has found, for example, that a number which ends in 5 or is divisible by 5. In Algebra he has studied, among other things, the equation, and has found that if J x — 1 — 5, x must equal 12. In Geometry he is to study form, and he will find, for example, that two triangles must necessarily be equal if the three sides of the one are respectively equal to the three sides of the other. Before beginning the subject, however, there are certain terms which, although familiar, are used with such exactness as to require careful explanation. These terms are solid, sur- face, line, angle (with various kinds of each), and point. As with most elementary mathematical terms, such as number, space, etc., it is difficult to give them simple and satisfactory definition. Explanations can, however, be given which will lead the student to a reasonable understanding of them. 2. The space with which we are familiar and in which we live is evidently divisible. Any limited portion of space is called a solid. In geometry no attention is given to the substance of which the solid is composed. It may be water, or iron, or air, or 2 PLANE GEOMETRY. [Intr. wood, or it m?y be a vacuum. Indeed, geometry considers only the space occupied by the substance. This space is called a geometric solid, or simply a solid, while the substance is called a physical solid. Thus, a ball is a physical solid ; the space which the ball occupies is a geometric solid. 3. That which separates one part of space from an adjoin- ing part is called a surface. So we speak of the surface of a ball, the surface of the earth, etc. 4. Every surface is divisible. That which separates one part of a surface from an adjoining part is called a line. 5. Every line is divisible. That which separates one part of a line from an adjoining part is called a point. A point is not divisible. Thus, in the figure the surface of the block separates the space occu- pied by the block from all the rest of space. This surface is divisible in many ways ; for example, it is divided into two parts by the line passing from A through § B and C and back to A. This line is divisible \ in many ways ; for example, it is separated \ into three parts by the points A, B, C. In the case of a line that returns into itself, — i.e. a closed line, like the one just mentioned, — two points are necessary com- pletely to separate one part from the other. It is impossible to draw mechanically a geometric line. A chalk mark, a thread, a fine wire, an ink mark, are all very thin physical solids used to represent lines ; for this purpose they are very helpful. So, too, a dot may be used to represent a point, and a sheet of paper may be used to represent a surface, although each is really a physical solid. 6. The preceding definitions start from the solid and take the surface, line, and point in order. It is also possible to start with the point and proceed in reverse order. The point is the simplest geometric concept ; it lias position, but not magnitude. Sec. 7.] ELEMENTARY DEFINITIONS. 3 A moving point describes a line. This may be represented by a pencil point moving on a piece of paper. A moving line describes, in general, a surface. This may be represented by a crayon lying flat against the blackboard, and moving sidewise. How may a line move so as not to describe a surface ? A moving surface describes, in general, a solid. Thus, the surface of a glass of water, as it moves upward, may be said to describe a solid. How may a surface move so as not to describe a solid ? 7. Through two points any number of lines may be imagined to pass. For example, through the points Pi, I (read " P-one, P-two") the lines q, r, may be imagined to pass. A straight line is a line which is determined by any two of its points. In the figure, s represents a straight line, for, given the points P u P 2 on the line, its position is fixed ; it is determined. But q and r do not represent straight lines, because Pi and P 2 do not determine them. The word line, used alone, is to be understood to refer to a straight line. The expression straight line is used to mean both an unlim- ited straight line and a portion of such a line. In case of doubt, line-segment, or merely segment, is used to mean a limited straight line. As has been seen, a point is usually named by some capital letter. A segment is usually A B C named by naming its end points, ! ■ 1 ■ ■ — or by a single small letter. In the annexed figure, AB, AC, BC, and o are marked off. Two segments are said to be equal when they can be made to coincide. 4 PLANE GEOMETRY. [Intr. 8. If three points, A, B, C, are taken in order on a line, as in the preceding figure, then the line-segment AC is called the sum of the line-segments AB and BC, and AB is called the difference between AC and BC. 9. If a point divides a line-segment into two equal seg- ments, it is said to bisect the line-segment and p is called its mid-point. A line is easily bisected by the use of a straight- edge and compasses, thus : With centers A and B, and equal radii, describe arcs intersecting at P and P'. A~ Draw PP'. This bisects AB. The proof of this fact is given later. 10. If a segment is drawn out to greater length, it is said to be produced. 'p^ To produce AB means to extend it through B, toward C, in the second figure in § 7. To produce BA means to extend it through A, away from B. 11. A line not straight, but made up of straight lines, is called a broken line. 12. Through three points, not in a straight line, any num- ber of surfaces may be imagined to pass. For example, through the points A, B, C the surfaces P and S may be imagined to pass. A plane surface (also called a plane) is a surface which is determined by any three of its points not in a straight line. In the figure, P represents a plane, for it is determined by the points A, B, C. But S does not represent such a surface. A plane is indefinite in extent unless the contrary is stated. To produce it means to extend it in length or breadth. Secs. 13, 14.] ELEMENTARY DEFINITIONS. 5 13. If two lines proceed from a point, they are said to form an angle, the lines being called the arms, and the point the vertex, of that angle. The size of the angle is independent of the length of the arms ; the size depends merely upon the amount of turning necessary to pass from one arm to the other. The methods of naming an angle will be seen from the annexed figures. It is convenient to letter an angle around the vertex, as indicated by the arrows, that is, opposite to the course of clock-hands, or counter-clockwise. C -A \ ^ Angle m. Angle 0. Angle A OB. Angle a b. Angle AOB. A line proceeding from the vertex, turning about it counter- clockwise from the first arm to the second, is said to turn through the angle, the angle being greater as the amount of turning is greater. 14. If the two arms of an angle lie in the same straight line on opposite sides of the vertex, a straight angle is said to be formed. If the angle still further increases, until the moving arm has performed a complete revolution, thus passing through two straight angles, a perigon is said to be formed. For practical purposes angles are measured in degrees, min- utes, and seconds. A perigon is said to con- b £ tain 360°. ° A In general, if tWO lines A0B ' a stT&i S^t angle. A perigon, or angle _ BOA, a straight angle. of 360°. are drawn irom O, two angles, each less than a perigon, are formed. Of these the smaller is always to be understood if "the angle at 0" is mentioned, unless the contrary is stated. 6 PLANE GEOMETRY. [Lntr. 15. If a line turns through an angle, all points or line- segments through which it passes in its turning, except the vertex, are said to be within the angle. Other points or lines are either on the arms or without the angle. 16. Two angles, ab, a'b', are said to be equal when, without changing the relative position of a and b, angle ab may be placed so that a lies along a', and b along b'. This equality is tested by placing one angle on the other, the vertices coinciding. Then if the arms can he made to coincide, the angles are equal, otherwise not. 17. If three lines, OA, OB, OC, proceed from a common point 0, OB lying within the angle AOC, then angles A OB and BO C are called ad- jacent angles. Angle AOC is called the sum of the angles A OB, BOC. Either of the adjacent angles is called the difference be- tween angle AOC and the other of the adjacent angles. As two angles may be added, so several may be added. 18. If a line divides an angle into two equal angles, it is said to bisect the angle and is called its bisector. In the annexed figure, if angle A OY equals angle YOB, then OY is the bisector of angle A OB. And, in general, to bisect any magnitude means to divide it into two equal parts. An angle is easily bisected by the use of a straight-edge and compasses, thus : If AOB is the given angle, mark off with the compasses OC equal to OD. Then with C and D as centers and CD as a radius draw two arcs intersecting at P and P'. The line joining PorF with is the required bisector. The proof of this fact is given later. Secs. 19-22.] ELEMENTARY DEFINITIONS 7 19. A right angle is half of a straight angle, It follows from this definition that the sum of two rigid angles is a straight angle; and from the definitions of a straight angle and of a perigon, that the sum of two straight an- gles, or of four right angles, is a perigon. C It also follows that a straight angle contains 180° and a right angle contains 90°. 20. If two lines meet and form a right angle, each line is said to be perpendicular to the other. Each is also spoken of as a perpendicular to the other. Thus, in the preceding figure, BO is perpendicular to CA, or is a perpendicular to CA. The segment PO is called the per- pendicular from P to CA, since it will presently be proved that it is unique ; that is, that there is one and only one perpendic- ular. is called the foot of that perpendicular. The word unique, meaning one and only one, is frequently used in mathematics. A line is easily drawn perpendicular to another line by the use of a straight-edge and compasses. This is seen in the figure in § 9, where PP" is perpendicular to A B. 21. An angle less than a right angle is said to be acute ; one greater than a right angle but less than a straight angle is said to be obtuse ; one greater than a straight angle but less than a perigon is said to be reflex or convex. 22. Two lines which form an acute, obtuse, or reflex angle are said to be oblique to each other. Acute, obtuse, and reflex angles are classed under the gen- eral term oblique angles. The meaning of the expressions oblique lines, an oblique, foot of an oblique, will be understood from § 20. Draw a figure representing acute, obtuse, and reflex angles, oblique lines, an oblique from P to CA, the foot of an oblique. 8 PLANE GEOMETRY. [Intr. 23. Two angles are said to be complements of each other if their sum is a right angle. Two angles are said to be supple- ments of each other if their sum is a straight angle. Two angles are said to be conjugates of each other if their sum is a perigon. If one angle is the complement of another, the two angles are said to be complemental or comple- mentary. Similarly, if one angle is the supplement of another, the two angles are c _ / a said to be supplemental or supplementary. In the annexed figure, angles AOB and BOC are supplemental, also angles BOC and COD, D etc. 24. If two lines, CA, DB, intersect at 0, as in the above figure, the angles A OB and COD are called vertical or opposite angles; also the angles BOC and DO A. Exercises. 1. How many degrees in a right angle ? How many minutes ? How many seconds ? 2. What is the complement of one-half of a right angle ? of one- fourth ? 3. How many degrees in the supplement of an angle of (a) 75° ? (b) 90°? (c) 150°? (d) 179°? 4. Also in the complement of an angle of (a) 75° ? (b) 1° ? (c) 89° ? (d)45°? (e) 90°? (f)0°? 5. Also in the conjugate of an angle of (a) 270° ? (b) 180° ? (c) 359° ? (d) 90°? (e) 1°? (f) 360°? 6. Draw a figure showing that two straight lines determine one point ; also one showing that three straight lines determine, in general, three points. 7. How many degrees in each of the two conjugate angles which the hour and minute hands of a clock form at 4 o'clock ? 8. If six lines, proceeding from a point, divide a perigon into six equal angles, express one of those angles (a) in degrees, (b) as a fraction of a right angle, (c) as a fraction of a straight angle. Secs. 25-27.1 DEMONSTRATIONS OF GEOMETRY. 2. THE DEMONSTRATIONS OF GEOMETRY. 25. The object of geometry is the investigation of truths con- cerning combinations of lines and points, and of the methods of making certain constructions from lines and points. 26. A proposition is a statement of either a truth to be demonstrated or a construction to be made. For example, geometry investigates this proposition : If two lines intersect, the vertical angles are equal. It also investigates the methods of drawing a line perpendicular to another line, and various other propo- sitions requiring some construction. Propositions are divided into two classes — theorems and problems. A theorem is a statement of a geometric truth to be demon- strated. A problem is a statement of a geometric construction to be made. For example : Theorem, If two lines intersect, the vertical angles are equal. — Problem, Required through a point in a line to draw a perpen- dicular to that line. 27. There are a few geometric statements so obvious that the truth of them may be taken for granted, and a few geometric operations so simple that it may be assumed that they can be performed. Such a statement, or the claim to perform such an operation, is called a postulate. The geometric operations thus assumed require the use of the straight-edge and compasses. The straight-edge and the compasses are the only instruments recognized in elementary geometry. The postulates used in this work are set forth from time to time as required. At present three general classes suffice, as follows : 10 PLANE GEOMETRY. [Intr. 28. Postulates of the Straight Line. 1. Two points determine a straight line. This follows from the definition. 2. Two straight lines in a plane determine a point. 3. A straight line may be drawn and revolved about one of its points as a center so as to include any assigned point in space. 4. A straight line-segment may be produced. 5. A straight line is divided into two parts by any one of its points. 29. Postulates of the Plane. 1. Three points not in a straight line determine a plane. This follows from the definition. 2. A straight line through two points in a plane lies wholly in the plane. Thus, if part of a straight line lies in an unlimited plane blackboard, the whole line lies in the blackboard. 3. A plane may be passed through a straight line and re- volved about it so as to include any assigned point in space. 4. A portion of a plane may be produced. 5. A plane is divided into two parts by any one of its straight lines, and space is divided into two funis by any plane. 30. Postulate of Angles. All straight angles are equal. 31. There arc also a number of simple statements, of a general nature, so obvious that- the truth of them may be taken for granted. These are called axioms. The following arc the axioms mosl frequently used in geometry, and they arc so important that they should be learned by number. Sec. 32.] DEMONSTRATIONS OF GEOMETRY. 11 32. Axioms. 1. Tilings which are equal to the same thing, or to equal things, are equal to each other. That is, (1) if A = JJ, and C = B, then 4 = C. Or, (2) if J. = B, and 7> = C, and C = D, then J. = Z). 2. If equals are added to equals, the sums are equal. That is, if A = B, and if C = B. then A + C = B + D. f>. If equals are subtracted from equals, the remainders are equal. That is, if A = B, and if C = 1), then A - C = B - D. ^4. If equals ore added to unequals, the sums are unequal in the same sense. That is, if A = B. and if C is greater than D, then A + C is greater than B ~ I). ■h^ If equals are subtracted from unequals, the remainders are unequal in the same sense. That is, if ^4 = B, and if C is greater than D, then C — A is greater than D - B. 6. If equals are multiplied by equals, the products are equal. That is, if A = B. and m is any number, then mA = mB. s 7. If equals are divided by equals, the quotients are equal. A B That is, as in axiom 6, — =— • It will be seen that axiom 6 covers m m axiom 7. for m may be a fraction. 8. The whole is greater than any of its parts, and equals the sum of all its parts. The latter part of this axiom is merely the definition of whole. 9. If three magnitudes are so related that the first is greater than the second, while the second is greater than, or equal to, the third, then the first is g renter than the third. E.g. if A is greater than 7?. and if B is greater than, or equal to, C, then A is greater than C. 12 PLANE GEOMETRY. [Intr. 33. Symbols and Abbreviations. The following are used in this work, and are inserted here merely for reference, and not for memorizing : e.g. Latin, exempli gratia, for example. i.e. Latin, id est, that is. since. therefore, pt., pts. point, points, rt. right, st. straight. ax. axiom, post. postulate, def. definition, prop. proposition, th. theorem, pr. problem, cor. corollary, subst. substitution, prel. preliminary, const. construction, ppd. parallelepiped. ^ arc. O, (s) circle, circles. A, A triangle, triangles. □ , Q[] square, squares. I I. \j~\ rectangle, rectangles. O, UJ parallelogram, parallelo- grams. Z., A angle, angles. + pins, increased by. — minus, diminished by. X , • , and absence of sign, de- note multiplication. /. de- > < > < and fractional form, note division. is equal, or equivalent, to. is identical with, as AB = AB, or coincides with. is congruent to. is similar to. approaches as a limit. is greater than. is less than. is not equal to, i.e. > or < . is not greater than, i.e. = or < . is not less than, i.e. = or >. is perpendicular to, or a per- pendicular. is parallel to, or a parallel. and so on. The above take the plural also; thus, = means are equal, as well as is equal. The manner of reading some of the familiar symbols is suggested, as follows : P', P-prime; P", P-second ; P'", P-third, etc. P 1? P-one ; P 2 , P-two, etc. A'B', A-prime B-prime, etc. A{, A-one-prime, etc. References to preceding propositions are made by book and proposi- tion thus, I, prop. IV ; if the first Roman numeral is omitted, the prop- osition is in the current book. Section references are also used. Other simple abbreviations are occasionally used, but they will be easily understood. Prop. I.] PRELIMINARY PROPOSITIONS. 13 3. PRELIMINARY PROPOSITIONS. 34. The following theorems are designed to show to the beginner the nature of a geometric proof, and to lead him by easy steps to appreciate the logic of geometry. Some of them might properly have been incorporated in Book I, and others might have been omitted altogether ; but they form a group of simple propositions which lead the student up to the more diffi- cult work of geometry, and for that reason they are inserted here. The student and the teacher are advised to proceed slowly until the logic of the subject is understood, and under no circumstances to allow mere memorizing of the proofs. Proposition I. 35. Theorem, All right angles are equal. Suggestion, The only angles of whose equality we are thus far assured are straight angles. Hence in some way we must base our proof of this theorem on the postulate of angles, which asserts this fact. We then consider how a right angle is related to a straight angle, and the proof is at once suggested. any two right angles, r, r\ that r = r'. Given To prove Proof, 1. r and r are halves of straight angles. Def . rt. Z (§ 19. A right angle is half of a straight angle.) 2. All straight angles are equal. § 30 3. .'. all right angles, and hence r and ?•', are equal. Ax. 7 (If equals are divided by equals, the quotients are equal.) 14 PLANE GEOMETRY. [Intr. Proposition II. 36. Theorem. At a given point in a given line not more than one perpendicular can be drawn to that line in the same plane. Y ,.7 V Given YY f _L XX at 0. To prove that no other perpendicular can be drawn to XX\ at 0, in the same plane. Proof. 1. Suppose that another _L, ZZ', could be drawn. 2. Then ZXOZ would be a rt. Z. Def . ± (If two lines meet and form a rt. Z, each is said to be ± to the other. ) 3. But Z XOY is a rt. Z. Given ; def. _L § 20 (For it is given that YY' _L XX', and the def. of a _L is given in step 2.) 4. .-. Z XO Y would equal Z XOZ. Prop. I (All right angles are equal.) 5. J>ut this is impossible. Ax. 8 (The whole is greater than any of its parts, etc.) 6. .'. the supposition of step 1 is absurd, and a second perpendicular is impossible. Note. In prop. I we proved directly from the definition of straight angle that all right angles are equal. In prop. II a different method of proof is followed. We have here supposed that the theorem is false and have shown that this supposition is absurd. Such proofs have long been known by the name " reductio ad ahs/tnlnm," a reduction to an absurd- ity. They are also called indirect proofs. Props. Ill, IV.] PRELIMINARY PROPOSITIONS. 15 Proposition III. 37. Theorem. The complements of equal angles are equal. Suggestion*. Three lines of proof may present themselves. We may base our proof on the equality of straight angles, as we did in prop. I, or we may take an indirect proof as in prop. II, beginning by supposing the theorem false and showing the absurdity of this supposition, or we may base the proof on prop. I. Since the complements suggest right angles, which of the three methods would it probably be best to follow ? A 0' A' Given two equal A, AOB, A'O'B', and their complements. BOC, B'O'C, respectively. To prove that ABOC = A B'O'C. Proof. 1. A AOC and A'O'C are rt. A. Def. compl. (§ 23. Two A are said to be complements if their sum is a rt. Z.) 2. .'. Z AOC = Z A'O'C. Prop. I (All right angles are equal.) 3. But Z AOB = Z A'O'B'. Given 4. .\ A BOC = A B'O'C Ax. 3 (If equals are subtracted from equals, the remainders are equal.) Proposition IV. 38. Theorem. The supplements of equal angles are equal. Let the student draw the figure and give the proof after the manner of prop. III. Use only four steps in the proof. Given To prove Proof. 16 PLANE GEOMETRY. [Intr. Proposition V. 39. Theorem. The conjugates of equal angles are equal. Given two equal angles, ab, a'b'. To prove that Aba — A b'a'. Proof. 1. The given A may be so placed that a lies along a', and b along b'. Def. equal A (§ 16. Two A, ab, a'b', are said to be equal when Z ab can be placed so that a lies along a', and 6 along b'.) 2. But then A ba must equal A b'a'. Def. equal A Proposition VI. 40. Theorem. If two lines cut each other, the vertical angles are equal. Suggestion. After examining the figure the student might say that because Z a + Z b = st. Z, and Z b + Z a' = st. Z, .-. Z a + Z 6 = Z 6 + Z a', and then subtract Z 6 from these equals ; or he might say that Za — Za' because each is the supplement of Z b. He should always feel encouraged to try various proofs, selecting the shortest and the clearest. Does the following proof meet these requirements ? Given two lines cutting each oilier, forming two pairs of opposite angles, a, a', and b, b'. To prove that A a — A a '. Proof. 1. Aa and A a' are supplements of Ab. Def. suppl. (§ 23. Two A are said to be supplements if their sum is a st. Z.) 2. .\Aa = A«'. Prop. IV (The supplements <>1" equal angles are equal.) Props. VII, VIII.] PRELIMINARY PROPOSITIONS. 17 Proposition VII. 41. Theorem. A line-segment can be bisected in only one point. P A ST B Given a line-segment AB, bisected at M. To prove that there is no other point of bisection. Proof. 1. Suppose another point of bisection exists, as P. be- tween M and B. 2. Then since AM and AP are both halves of A P. they are equal. Ax. 7 (State ax. 7.) 3. But this is impossible, for AM is part of AP. Ax. 8 (State ax. 8.) 4. .'. the supposition that there is a second point of bisection is absurd. (Another reductio ad absurdum, as in prop. II.) Proposition VIII. 42. Theorem. An angle can be bisected by only one line. (The student may prove this after the manner of prop. VII.) Exercises. 9. Of two supplemental angles, a and b, (a) suppose a — 2b, how many degrees in each ? (b) suppose a = 3 6, how many ? 10. How many straight lines are, in general, determined by three points? by four? (The points in the same plane.) 11. If of five angles, a, b. c, d, e. whose sum is a perigon, a = 20°, b = 30°, c — 40°, d = 50°, how many degrees in e ? 12. Of three angles whose sum is a perigon, the first is twice the sec- ond, and the second three times the third; how many degrees in each ? 18 PLANE GEOMETRY. [Intr. Proposition IX. 43. Theorem. The bisectors of two adjacent angles formed by one line cutting another are perpendicular to each other. Suggestion. Considering the figure, we see that to prove OA _L OB we must show that Z A OB is a rt. Z. Now the only way that we have as yet of showing an angle to be a right angle is to show that it is half of a straight angle. But evidently Z A OY is half of Z XOY, because Z XOY is bisected ; similarly, Z YOB is half of Z YOX', and this suggests the following proof. Given two lines, XX', YY', cutting at ; also OA, OB, bisecting A XOY, YOX'. respectively. that OA _L OB. AAOY = \AXOY. Z YOB = ±A YOX'. To prove Proof. 1. 2. 3. ;.ZAOB = \AXOX'. (If equals are added to equals, the sums are equal.) 4. .-. Z AOB = i of a st. Z. Def. st. Z (§ 14. If the two arms of an Z lie in the same st. line on opposite sides of the vertex, a st. Z is said to be formed.) Given; § 18 (liven; § 18 Ax. 2 5. .-. Z AOB = ait. Z. (§ 19. A rt. Z is half of a st. Z.) 6. .-. 0A± OB. Def. rt. Z Def. _L (§ 20. If two lines meet and form a rt. Z, each line is said to be ± to the other.) Prop. X.] PRELIMINARY PROPOSITIONS. 10 Proposition X. 44. Theorem. The bisectors of the four angles which two intersecting lines make with each other form two straight lines. B Given XX' intersecting YY' at O, OA bisecting Z XOY, OB bisecting AYOX'. 00 bisecting AX'OY' and OD bisecting Z Y'OX. To prove that CO A and DOB are straight lines. Proof. 1. A A OB and BOO are rt. A. Prop. IX (State prop. IX.) 2. .'.the two together form a st. angle. Def. it. Z (§ 19. State the definition.) 3. .*. 00 A is a st. line. Def. st. Z (§ 14. State the definition.) 4. Similarl v for DOB. 45. The nature of a logical proof should now be understood. Before continuing, however, the following points should be emphasized : a. Every statement in a proof must be based upon a postu- late, an axiom, a definition, or some proposition previously considered of which the student is prepared to give the proof again when he refers to it. 20 PLANE GEOMETRY. [Intr. b. No statement is true simply because it appears to be true from a figure which the student may have drawn, no matter how carefully. Many cases will be found, for example, where angles appear equal when they are not so. c. The arrangement of the discussion of a theorem is as follows : Given. Here is stated, with reference to the figure which accompanies the proof, whatever is given by the theorem. To prove. Here is stated the exact conclusion to be de- rived from what is given. Proof. Here are set forth, in concise steps, the statements to prove the conclusion just asserted. If the proof is written on the blackboard, the steps should be numbered for convenient reference by class and teacher. The teacher will state how much in the way of written or indicated authorities shall be required after each step. Corollarv. A corollary is a proposition so connected with another as not to require separate treatment. The proof is usually simple, but it must be given with the same accuracy as that of the proposition to which it is attached. It is usually sufficient to say, This is proved in step 4 ; or, This follows from steps 2 and 5 by axiom 3, etc. In every case the stu- dent should (1) clearly prove the corollary, but (2) do so as concisely as possible. A corollary may also follow from a definition; thus, from the definitions of Proposition and Theo- rem the following might be stated as a corollary : Every theorem is a proposition, but not every proposition is a theo- rem; and as a part of our definition of a Perigon we incor- porated the corollary (the term then being undefined) that a perigon equals two straight angles. Note. Any item of interest may be inserted under this head. Exercises. 13. Of the proofs of the preliminary theorems, state which are direct and which indirect. (See note on p. 14.) 14. How can you form a right angle by paper folding ? Prove it. BOOK I. — RECTILINEAR FIGURES. 1. TRIANGLES. 46. A figure is any combination of lines and points formed under given conditions. E.g. an angle is a figure, for it is a combination of two lines and one point formed under the condition that the two lines proceed from the point. 47. A rectilinear figure is a figure of which all the lines are straight. Plane geometry treats of figures in one plane, — plane figures. Hence in plane geometry, which in this work extends through Books I to V inclusive, the word figure used alone denotes a plane figure, and all propositions and definitions refer to such figures placed in one plane. 48. If the two end-points of a broken line coincide, the fig- ure obtained is called a polygon, and the broken line its perim- eter. The vertices of the angles made by the segments of the perimeter are called the vertices of the polygon, and the seg- ments between the vertices are called the sides of the polygon. 49. The perimeter of a polygon divides the plane into two parts, one finite (the part inclosed), the other infinite. The finite part is called the surface of the polygon. or for brevity simply the polygon. A point is said to be within or without the polygon according as it lies within or without this finite part. A polygon. The figure ABCBE is a polygon (the sides being produced for a sub- sequent definition). 21 22 PLANE GEOMETRY. [Bk. 50. In passing counter-clockwise around the perimeter of a polygon the angles on the left are called the interior angles of the polygon, or for brevity simply the angles of the polygon. Such are the angles CBA, DCB, EDC, in the figure on p. 21. 51. If the sides of a polygon are produced in the same order, the angles between the sides produced and the following sides are called the exterior angles of the polygon. Such are the angles XBC, YCD, in the figure on p. 21. They are the angles through which one would turn, at the successive corners, in walking around the polygon. 52. A line joining the vertices of any two angles of a poly- gon which have not a common arm, is called a diagonal. Such a line would be the one joining A and C in the figure on p. 21. The sides, angles, and diagonals of a polygon are often called its paints. 53. A polygon which has all of its sides equal is called equilateral. 54. Two polygons are said to be mutually equilateral, or one is said to be equilateral to the other, when the sides of the one are respectively equal to the sides of the other. A polygon which has all of its angles equal is called equi- angular. Two polygons are said to be mutually equiangular, or one is said to be equiangular to the other, when the angles of the one are respectively equal to the angles of the other. 55. A polygon of three sides is called a triangle; one of four sides, a quadrilateral. Secs. 56-59.] TRIANGLES. 23 56. Any side of a polygon may be called its base, the side on which the hgnre appears to stand being usually so called, as AB in the figure on p. 21. In the case of a triangle, the vertex of the angle opposite the base is called the vertex of the triangle, the angle itself being called the vertical angle of the triangle, and the other two angles the base angles. Thus, in the first triangle on p. 25, C is the vertex of the triangle, Z C 57. Two figures which may be made to coincide in all their parts by being placed one upon the other are said to be con- gruent. For example, two line-segments may be congruent, or two angles, or two triangles, etc. 58. The operation of placing one figure upon the other so that the two shall coincide is called superposition, and the figures are sometimes called superposable (a synonym of con- gruent). This is illustrated in prop. I. Superposition is an imaginary operation. It is assumed as a postulate (§ 61) that figures may be moved about in space with no other change than that of position. The actual move- ment is, however, left for the imagination. 59. It will hereafter be explained and defined that polygons of the same shape are called similar, the symbol of similarity being ^, and that those of the same area are called equal or equivalent, the symbol being =. Congruent figures are both similar and equal, and hence the symbol for congruence is =, a symbol used in modified form by the great mathematician Leibnitz. The symbol ~~ is derived from the letter S, the initial of the Latin si mills, similar. 24 PLANE GEOMETRY. [Bk. I. Many writers use equal for congruent, and equivalent for equal, as above defined. But because of the various meanings of the word equal, and its general use as a synonym for Equality. Similarity. Congruence. equivalent, the more exact word congruent with its suggestive symbol is coming to be employed. The student should be familiar with this other use of the words equal and equivalent. 60. It is customary to designate the sides of a triangle by the small letters correspond- ing to the capital letters which designate the opposite vertices. A Thus, in the figure, side a is opposite vertex A, etc. 61. It now becomes necessary to assume three other pos- tulates. Postulates of Motion. 1. A figure may he moved about in space with no other change than that of position, and so that any one of its points may be made to coincide with any assigned point in space. That is, we may pick up one polygon and place it on another without changing its shape or size. 2. A figure may be moved about in space while one of its points remains fixed. Such movement is called rotation about a center, the center being the fixed point. 3. A figure may be moved about in space while two of its points remain fixed. Such movement is called rotation about an axis, the axis being the line determined by the two points. Prop. I. TRIANGLES. 25 Proposition I. 62. Theorem. If two triangles have two sides and the in- cluded angle of the one respectively equal to two sides and the included angle of the other, the triangles are congruent. Given A c b A' the A ABC, A'B'C such that c = c', b = b', and ZA = ZA'. To prove that A ABC ^ A A'B'C. Proof. 1. Place A A'B'C on A ABC so that A' falls on A, and c' coincides with its eqnal c. 2. Then b' may be caused to fall on b, because Z A' = Z A. 3. Then C will fall at C, because b' = b. 4. will coincide with a, §61,1 § 61, 2 Given ; § 61, 3 Given; § 57 §28,1 5. (Two points determine a straight line.) AABC = AA'B'C, by definition of congruence. § 57 Notes. This is a proof by superposition. The theorem may be stated, A triangle is determined when two sides and the included angle are given. In the exercises hereafter given, the proofs are to be given in full ; when a question is asked, a proof of the answer is to be given ; when a theorem is suggested, it is to be completely stated and then proved. 26 PLANE GEOMETRY. [Bk. Proposition II. 63. Theorem. If two triangles have two angles and the included side of the one respectively equal to two angles and the included side of the other, the triangles are congruent. K B B Given the A ABC and A'B'C such that Z.C = ZC, ZB = Z B', and a = a'. To prove that A ABC ^ A A'B'C Proof. 1. Place A A'B'C on A ABC so that a' falls on a and Z C coincides with its equal Z C. 2. Then B' will fall on B because a' = a. 3. Then c' will fall on c because Z-B' ' = /LB. 4. .*. A' will coincide with A. (Two straight lines determine a point.) 5. .'. A ABC ^ A A'B'C, by definition of congruence. § 57 Note. Prop. II, and prop. Ill following, are attributed to Thales. § 61 Given Given § 28, 2 Exercises. 15. In the figure on p. 19, given that OA bisects angle XOY, and that OB is perpendicular to OA, prove that OB bisects angle YOX'. 16. Show that the distance BA across a lake may be measured by setting up a stake at 0, A' sighting across it to fix the lines A'B and B'A, laying off OA' — OA, and OB' — OB, and then measuring B'A' Sacs. 64, 65.] TRIANGLES. 27 64. Reciprocal Theorems. The student will notice that prop- ositions I and II have a certain similarity. Indeed, if the words side and angle are interchanged in prop. I. it becomes prop. II, and if interchanged in prop. II that becomes prop. I. Theorems of this kind are called reciprocal. The relation is more clearly seen by resorting to parallel columns. Prop. I. If two triangles have Prop. II. If two triangles Lave two sides and the included angle of two angles and the included side of the one respectively equal to two the one respectively equal to two sides and the included angle of the angles and the included side of the other, the triangles are congruent. other, the triangles are congruent. Moreover, if small letters and capitals are interchanged in the proof of prop. I, the proof becomes that of prop. II. 65. The principle involved is called the Principle of Reci- procity, and is extensively used in geometry. But the student must not suppose that because a theorem is true its reciprocal theorem is also true ; in elementary geometry, involving measurements, the reciprocal is often false. The principle is, however, of great value even here, for it leads the student to see the relation between propositions, and it often suggests new possible theorems for investigation. For these purposes we shall use it. At present it is sufficient to say that for many theorems of plane geometry reciprocal theorems may be formed by re- placing the words point by line, line by point, (ingles of a triangle by (opposite^ sides of a triangle, sides of a triangle by (opposite) angles of a triangle. Exercises. 17. F.xplain this statement and tell why it is true : Any two sides and the included angle of a triangle determine the remaining parts. 18. State the reciprocal of ex. 17 and tell whether it is true, and why. 28 PLANE GEOMETRY. [Bk. I. Proposition III. 66. Theorem. If two sides of a trionigle are equal, the angles opposite those sides are equal. Given the A ABC with AC = BC. To prove that AA=Z.B. Proof. 1. Suppose m to bisect Z. ba. 2. Then v b = a, Given and Z. bin — A ma, and m = m, 3. . ' . A AM C £ A BMC, Prop. I (State prop. I.) and AA—/-B, by definition of congruence. § 57 Corollary. If a triangle is equilateral, it is also equi- angular. For by the theorem the angles opposite the equal sides are equal. 67. Definitions. The line from any vertex of a triangle to the mid-point of the opposite side is called the median to that side. In the above figure, CM is the median to AB. If a triangle has two equal sides, it is called an isosceles triangle. Sec. 68.] TRIANGLES. 29 The third side is called the base of the isosceles triangle, and the equal sides are called the sides. A triangle which has no two sides equal is called a scalene triangle. The distance from one point to another is the length of the straight line-segment joining them. The distance from a point to a line is the length of the per- pendicular from that point to that line. That this perpendicular is unique will be proved later. This is the meaning of the word distance in plane geometry. In speaking of points on a curved surface (for example, the earth's surface), distance may be measured on a curved line. 68. In the figure of prop. Ill, A AMC ^ A BMC, as proved. .\AM=MB, and Z CMA = Z BMC, and hence each is a right angle. In cases of this kind the points A and B are said to be symmetric with respect to an axis. Hence, in the figure, CM is called an axis of symmetry. And, in general, two systems of points, A x , B x , C\, , A 2 , B 2 , C 2 , , are said to be symmetric vitlt respect to an axis when all lines, A X A 2 , B X B 2 , , are bisected at right angles by that axis. Also, two figures are said to be symmetric with respect to an axis when their systems of points are symmetric. A single figure, like that of prop. Ill, is said to be sym- metric with respect to an axis when this axis divides it into two svmmetric figures. Exercises. 19. If four lines go out from a point making four angles of which the first and third are equal, and the second and fourth are equal, prove that the four lines form two intersecting straight lines. 20. In the figure on p. 10, if a line passes through and bisects angle XOA, prove that it also bisects angle X'OC 30 PLANE GEOMETRY. [Bk. I. Proposition IV. 69. Theorem. If two angles of a triangle are equal, the sides opposite those angles are equal. A B Given the A ABC with Z A = Z B. To prove that a = b. Proof. 1. Suppose that a =£ b, and that a > b. 2. Then let BX, a part of a, equal &, and join A and A' 3. Then '.'ZB=Z.BAC, Given and J£ = AB, .'.AABC^A BAX. Why ? 4. .'.the supposition leads to an absurdity, for A ABO A BAX, Ax. 8 (State ax. 8.) and . ' . a~jj> b. In the same way it may be shown that a <£ b, and . ' . a = b. Corollary. If a triangle is equiangular, it is a /so equi- lateral. (Why ?) Exercise. 21. If four points, A, B, C, D, are placed in order on a line, and if AC = BD, prove that AB = CD. Prop. V. TRIANGLES. 31 Proposition V. 70. Theorem. If any side of a triangle is produced, the exterior angle is greater than either of the interior angles not adjacent to it. C A B Given the A ABC, with AB produced to X. To prove that Z XBC > Z C, and also > Z BA C '. Proof. 1. Suppose BC bisected at 31. AM drawn and produced to P so that 3IP = AM, and BP drawn. Why ? Why ? § 57 Why ? Why ? Similarly, by producing CB, bisecting AB at X, producing CX, etc.. it can be shown that an angle equal to Z XBC is greater than ABAC. 2. Then ■ .• Z BMP = Z CJ/J. r.ABPJI^A CA3I, and ZPB3T=ZC. 3. But Z.XBO Z.PB3L 4. ..Z.XBOZ C Exercises. 22. Show that, in the figure of prop. V, Z XBC > Z 1L4 C by following out in full the proof suggested in step 5. • 23. In the figure of prop. V, join C to any point in the segment A B and prove that Z CBA + ZBAC< 180°. 24. If a diagonal of a quadrilateral bisects two angles, the quadrilateral has two pairs of equal sides. 25. How many equal lines can be drawn from a given point to a given line ? Show that if another is supposed to be drawn, an absurdity results. 32 PLANE GEOMETRY. [Bk. I. Proposition VI. 71. Theorem. If two sides of a triangle are unequal, the opposite angles are unequal and the greater side has the greater angle opposite. Given the A ABC, with a > b. To prove that ZA>ZB. Proof. 1. Suppose Z C bisected by YY' cutting AB at D, CA' made equal to CA, and DA' drawn. 2. Then A. ID C ^ A. 4 'DC, and ZA = Z CA 'D. Why? 3. But Z CA'D > AB. Prop. V (§ 70. If any side of a A is produced, the exterior angle is greater than either of the int. A not adjacent to it.) 4. ..ZA>ZB. Subst. 2 in 3 Exercises. 26. State, without proof, the reciprocal of prop. VI. 27. Can a scalene triangle have two, equal angles ? Proof. 28. Prove prop. VI by drawing AA' instead of DA', and proving that Z A > Z A' AC = Z CA'A > Z B. 29. ABCD is a quadrilateral of which DA is the longest side and BC the shortest. Which is greater, ZB or ZD? Prove it. (Suggestion : Draw BD.) Also Z C or Z A ? Prove it. 30. How many perpendiculars can be drawn to a given line from a point outside that line? Show that any other supposition violates prop. V. 31. ABC is a triangle having Z B = twice ZA; Z B is bisected by a line meeting b at D ; prove that AD — BD. Prop. VII.] TRIANGLES. 33 Proposition VII. 72. Theorem. If two angles of a triangle are unequal, the opposite sides are unequal and the greater angle has the greater side opposite. Given the A ABC with Z A > Z B. To prove that a > b. Proof. 1. a 4- b, for if a = b, then Z A = Z B. Why ? 2. a < b, for if a < b, then Z .i< Z J5. Prop. VI. State it. 3. .'.a must be greater than b. Note. It must not be inferred from props. VI, VII that, because one angle of a triangle is twice as large as another, one side is twice as long as another. Exercises. 32. Prove that if the bisector of any angle of a triangle is perpendicular to the opposite side, the triangle is isosceles. 33. Suppose any point taken on the perpendicular bisector of a line ; is it equally or unequally distant from the ends of the line ? Give the proof in full. 34 a. Prove that in an isosceles triangle ABC, where a =b, the bisector of Z C, produced to c, bisects side c. 34 b. Prove that in an isosceles triangle abc, where ZA = ZB, the bisector of side c, joined to C, bisects Z C. 35. After reading § 73, state the converse of each of the following : (a) prop. Ill ; (b) prop. IV ; (c) prop. VI ; (d) prop. VII ; (e) this state- ment, If the animal is a horse, then the animal has two eyes. Of these converses, how many are true ? 36. What kind of a triangle is formed by joining the mid-points of the sides of an equilateral triangle ? Prove it. 34 PL A NE GE OME TR Y. [B k . I. 73. The Law of Converse. Two theorems are said to be the converse, each of the other, when what is given in the one is what is to be proved in the other, and vice versa. E.g. props. VI and VII. The converse of a theorem must not be con- fused with its reciprocal. Props. I and II are reciprocal, but not converse. Because a theorem is true its converse is not necessarily true. For example, prel. prop. I may be stated thus : Given that A r and r / are rt. A, to prove that Z r = Z r' ; the converse is, Given that Zr = Zr', to prove that they are rt. A. This converse is evidently false, for Z r could equal Zr' without their being rt. A. But there is one important class of converse theorems, illus- trated by props. IV and VII, that should be mentioned. When- ever three theorems have the following relations, their converses must be true : 1. If it has been proved that when A> B, then X > Y, and 2. " " " A = B, " X = T, " 3. " " " A < B, « X < Y, then the converse of each of these is true. For 1'. If X > Y, then A can neither be equal to nor less than B, without violating 2 or 3 ; .. A > B. (Converse of 1.) 2'. If A' = Y, then A can neither be greater nor less than B, without violating 1 or 3 ; .'. A = B. (Converse of 2.) 3'. If X < Y, then A can neither be greater than nor equal to B, without violating 1 or 2 ; .'. A < B. (Converse of 3.) The law just proved will hereafter be referred to as the Law of Converse. By its use the proof of the converse of many theorems, where true, is made very simple. The student should not proceed further unless the Law of Converse is thoroughly understood, and its proof mastered. Prop. VII may now be proved by the Law of Converse, thus : If a > b, then Z A > Z B. Prop. VI If a = b y " Z A = Z B. " III Ifa Z B, then a > b. Sec. 74.] TRIANGLES. 35 74. Suggestions as to the Treatment of the Exercises. Tims far the student has been left to his own ingenuity in treating the exercises. A few suggestions should now be given. 1. In attacking a. theorem take the most general figure possible. E.g. if a theorem relates to a triangle, draw a scalene triangle ; an equilateral or an isosceles triangle often deceives the eye, and leads away from the demonstration. Draw all figures accurately ; an accurate figure often suggests the demonstration. But the student who relies too much upon the accuracy of the figure in the demonstration itself is liable to go astray. 2. Be certain that what is given and what is to be proved arc clearly stated, with reference to the letters of the figure. This has been done in all of the theorems thus far proved. The neglect to do so in the exercises is one of the most fruitful sources of failure. 3. Then begin by assuming the theorem true .- see what fol- lows from that assumption • then see if this can be proved true without the assumption ; if so, try to reverse the process. E.g. suppose PO _L X'X, and PB, PA two obliques cutting off OA > OB, as in the figure, and that it is required to prove PA > PB. Assume it true ; then Z b > Z a. Now see if Zb> Z a without the assumption ; Zb>Zc, which = Z d. which > Z a, by prop. V ; .-. Zb> Z a, without the assumption. Xow re- * o B A verse the process ; v Z 6 > Za, .-. PA > PB by prop. VII. 4. Or begin by assuming the theorem false, and endeavor to show the absurdity of the assumption. \ lied actio ad absurdum.) 5. To secure a clearer understanding of the theorem it is often well to follow Pascal's advice and substitute the defini- tion for the name of the thing defined. E.g. suppose it is to be proved that the median to the base of an isos- celes triangle is perpendicular to the base. Instead of saying : '• Given CAT the median to the base of the isosceles triangle ABC" (see figure on p. 28), it is often better to say : "Given A ABC, with AC = B<7, and M taken on AB so that AM - MB,'' for then the facts stand out prominently without any confusing terms. 36 PLANE GEOMETRY. [Bk. I. Proposition VIII. 75. Theorem. The sum of any two sides of a triangle is greater than the third side. X Given the A ABC. To prove that a + b > c. Proof. 1. Suppose Z C bisected by CD. Then Z CD A >ZDCB. Prop. V. State it 2. And • . ' Z A CD = ZDCB, Step 1 .•.ZC f Di>ZiC r i). .*. o > AD. Prop. VII. State it Similarly, a > DB. 3. .'.a + b>c. Corollary. The difference of any two sides of a triangle is less than the third side. For if a + b > c, and c > 6, then a > c — &, by ax. 5. Exercises. 37. Two equal lines, ^4C and AD, are drawn on oppo- site sides of a line AB and making equal angles with it ; BC and BD are drawn. Show that BC and BD also make equal angles with AB. 38. P, Q, E are points on the sides AB, BC, CA, respectively, of an equilateral triangle ABC, such that AP = i?Q = Ctf ; joining P, Q, and R, prove that A PQE is equilateral. (Notice that ex. 36 is merely a special case of this one.) 39 a. The bisectors of the equal angles of an isosceles triangle form, with the base, an isosceles triangle. 39 b. The mid-points of the equal sides of an isosceles triangle form, with the vertex, the vertices of an isosceles triangle. Peop. IX. TRIANGLES. 37 Proposition IX. 76. Theorem. If from the ends of a side of a triangle two lines are drawn to a point within the triangle, their sum is less than the sum of the other two sides of the triangle, hut they contain a greater angle. C Given the A ABC, P a point within, and BP and PA drawn. To prove that (1) BP + PA < a + b, (2) Z APB > Z C. Proof. 1. Produce AP to meet a at X. Then XP + PA = XA < XC + b, Ax. 8 ; prop. VIII (State ax. 8 and prop. VIII.) and BP < BX + XP. Prop. VIII 2. .-.BP + XP + PA < BX + XC + XP + b. 3. .'. BP + PA < a + h. which proves (1). Why ? 4. Also, Z APB > Z PXB > Z C which proves (2). Why ? Exercises. 40 a. If the equal 40 6. If the equal angles of an sides of an isosceles triangle are isosceles triangle are bisected, the bisected, the lines joining the points angles formed by the lines of bi- of bisection with the vertices of the section and the equal sides are equal angles are equal. equal. 41. The perimeter of a quadrilateral is less than twice the sum of its two diagonals. 38 PLANE GEOMETRY. [Bk. I. Proposition X. 77. Theorem. If two triangles have two sides of the one respectively equal to two sides of the other, but the included angles unequal, then the third sides are unequal, the greater side being opposite the greater angle. Fig. 2. Fig. 3. Given the A A X B X C X and A 2 B 2 C 2 , with a x = a 2 , b x = b 2 , but Z.C X >Z. C 2 . To prove that c x > c 2 . Proof. 1. Suppose A A 2 B 2 C 2 placed on AA X B X C X so that b 2 and b x , being equal, coincide. § 61 Then ' .' Z C\ > Z C 2 , side a 2 must fall within Z C x , as in Fig. 3. 2. Suppose CM drawn bisecting Z B 2 CB X , and B 2 M drawn. 3. Then in A B X MC, B 2 3IC, CB X = CB 2 , Given CM = CM, Z MCB X = Z B 2 CM. Step 2 4. .\AB X MC^AB 2 MC, and MB, = MB 2 . Prop. I 5. But AM + MB 2 > All,. Prop. VIII .'. AM ' + MB X > AB 2 , or c x > c 2 . The proof is the same when B 2 falls above A X B X . Prop. XL] TRIANGLES. 39 Proposition XL 78. Theorem. If two triangles have two sides of the one respectively equal to two sides of the other, but the third sides unequal, then the included angles are unequal, the greater angle being opposite the greater third side. Given A A l B l C l and A 2 B 2 C 2 . with a, = c 2 . To prove that Z. C\ > Z C 2 . Proof. 1. It lias been shown that if a x = a 2 , b x = b 2 , and if Z C, > Z 6*,, then c, > c 2 . Prop. X 2. And if Z C\ = « « « = " Prop. I 3. « " Z d < " « « < " Prop. X 4. .'. the converses are true, which proves the theorem. § 73. Law of Converse (Explain the Law of Converse. Since this law is so often used, it should be reviewed frequently.) Exercises. 42. Are props. X and XI reciprocals ? converses ? 43. In A ABC, suppose CA > AB, and that points P, Q are taken on AB, CA respectively, so that PB = CQ. Prove that BQ < CP. 44. Investigate ex. 43 when P is taken on AB produced, and Q on AC prod Heed. 45. The equal sides, AC, BC, of an isosceles triangle ABC are pro- duced through the vertex to P and Q respectively, so that AP = BQ. Prove that BP = AQ. 46. Prove that the straight line joining any two points is less than any broken line joining them. 47. Prove that the perimeter of a triangle is less than twice the sum of the three medians. 48. In a quadrilateral, prove that the sum of either pair of opposite sides is less than the sum of its two diagonals. 49. If the perpendicular from any vertex of a triangle to the opposite side divides that side into two segments, how does each of these segments compare in length with its adjacent side of the triangle ? Prove it. 40 PLANE GEOMETRY. [Bk. 1. Proposition XII. 79. Theorem. If two triangles have the three sides of the one respectively equal to the three sides of the other, the tri- angles are congruent. C Given A ABC, AB'C, with AB = AB', BC = B'C, and AC = AC. To prove that A ABC ^ A AB'C. Proof. 1. Suppose no side longer than AC. Then the A, being mutually equilateral, may be placed with AC in common, and on opposite sides of AC. Draw BB'. 2. Then Z CBB' = Z BB'C, Prop. Ill and Z B'BA = Z AB'B. Why ? 3. .-. Z CBA = Z AB'C. Why ? 4. • .'.A ABC ^ A AB'C. Why? ^4C is evidently an axis of symmetry (§ 6S) in the figure. Exercises. 50. Suppose three sticks to be hinged together to form a triangle, could the sides be moved so as to change the angles ? On what theorem does the answer depend ? How would it be with a hinged quadrilateral ? 51. Ascertain and prove whether or not a quadrilateral is determined when the four sides and either diagonal are given in fixed order. 52. Also when the four sides and one angle are given in fixed order. 53. How many braces would it take to stiffen a three-sided plane figure ? four-sided ? five-sided ? Prop. XIII.] TRIANGLES. 41 Proposition XIII. 80. Theorem. If two triangles have two angles of the one respectively equal to tivo angles of the other, and the sides opposite one pair of equal angles equal, the triangles are congruent. X B Y Given A AB C, A'B' C, with Z A = Z A', Z B = Z B\ b = b'. To prove that A ABC^ A A'B'C. Proof. 1. Place A A'B'C on A ABC so that A' falls at A, A'B' lies along AB, and C and C both lie on the same side of AB. 61 2. Then because Z A = Z A', and b = b', b' coincides with b, and C ' with C. 3. Now B' cannot fall between A and B, as at X, for then Z CXJ, which = Z B', would be greater than Z B. Prop. V. State it 4. Neither can i?' fall on AB produced, as at Y, for then Z Y, which = Z I?', would be less than Z B. Prop. V 5. .*. B' must fall at 2?, and the A are congruent. § 57 Exercises. 54. If YO meets X'X at O, and YA, YB are drawn meet- ing X'X dXA,B; and if YA = YB, and AO^OB, which is the greater, A A YO or Z Y7i ? 55. Consider the diagonals of an equilateral quadrilateral, (a) as to their bisecting each other, (6) as to the kind of angles they make with each other. State the theorems which you discover and prove them. 42 PLANE GEOMETRY. [Bk. I. Proposition XIV. 81. Theorem. If tivo triangles have tivo sides of the one respectively equal to two sides of the other, and the angles opposite one pair of equal sides equal, then the angles opposite the other pair of equal sides are either equal or supplemental, and if equal the triangles are congruent. C' C' A^ ~~X B U "BT X' ^B' Given A ABC, A'B'C, with a = a',b = b', Z B = Z B'. To prove that either (1) Z A = Z A' and A ABC ^ A A'B'C, or (2) ZA + ZA' = st. Z. Proof. 1. Place A A'B'C on Ai^Cso that B' falls at B, a' coincides with its equal a, and A' and A fall on the same side of a. § 61 2. Then v Z.B = AB', B'A' lies along BA. 3. Then either A' falls at ^4, the A are congruent and Z^4 = Z^4'; or else A' falls at some other point on BA, as at X, and A A'B'C ^ A XJ5C. 4. But v CX=b' = b, .'. Z.A = Z. CXA. Prop. III. State it 5. And v Z CXA + Z J5X Z a', and that Q is drawn as in the figure, making an Z equal to Z.a'. 2. Then Q would be parallel to P' . Why ? 3. But this would be impossible, V P II P\ § 85 (Two intersecting straight lines cannot both be parallel to the same straight line.) 4. Similarly, it is absurd to suppose that Z Ac. Corollaries. 1. A line perpendicular to one of two paral- lels is perpendicular to the other also. For it cuts the other (§ 85, cor.) and the alternate angles are equal right angles. Prop. XVII.] PARALLELS A XI) PARALLELOGRAMS. 47 2. A line cutting two parallels makes corresponding angles equal, and the interior, or the exterior, angles on the same side of the transversal supplemental. For the alternate angles are equal (prop. XVII), and hence prop. XV applies. 3. If the alternate or the corresponding angles are unequal, or if the interior angles on the same side of the transversal are not supplemental, then the Jims are not parallel, but meet on that side of the transversal on which the sum of the inte- rim' angles is less than a straight angle. For the lines cannot be parallel, by prop. XVII and cor. 2. Further, suppose Z c + Z b' < st. Z ; then v Za'-f //)'= st. Z, it follows that Zc < Za'. .-. P and P / cannot meet towards P\ for then Z c would be greater than Z a', prop. V. Let the student give the proof in fidl furni, in steps. 4. Two lines respectively perpendicular to two intersecting lines cannot be parallel. For, in the annexed figure, let AB _L X. CD _L Y; y^ join A and C. Then Z EAC < rt. Z, and ZACD < rt. Z ; .-. their sum is < st. Z ; .-. cor. 3 applies. A/~ Give proof in full form in steps. \p /q 5. If the arms of one angle are parallel or perpendicular to the arms of another, the angles are equal or supplemental. The proof is left to the student. Exercises. 63. In the figure of prop. XV, suppose a = c' — 120° 30', how large is each of the other angles ? 64. In the same figure, suppose a + d' = st. Z, and a — 2 d. how large is each of the other angles ? 65. If a transversal cuts two lines making the sum of the two interior angles on the same side of the transversal a straight angle, one of them being 30° 27', how large is each of the other angles ? 48 PLANE GEOMETRY. [Bk. I. Proposition XVIII. 87. Theorem. Lines parallel to the same line are paral- lel to each other. Given A II M, and B II M. To prove that A II B. Proof. 1. Suppose T a transversal, making corresponding A a, m, b, with A, 31, B, respectively. 2. Then v A\\M, .-.Aa = A m. Prop. XVII, cor. 2 3. And v B\\ M, .\Z.b = Z.m. 4. .\/.a = £b. . Why? 5. .'. AWB. Prop. XVI, cor. 1. State it Exercises. 66. In prop. XVIII, if T cuts A, must it necessarily cut M ? Why ? If it cuts M, must it necessarily cut B ? Why ? 67. Prove that a line parallel to the base of an isosceles triangle makes equal angles with the sides or the sides produced. (The line may pass above, through, or below the triangle, or through the vertex.) 68. If through any point equidistant from two parallels, two transver- sals are drawn, prove that they will cut off equal segments of the parallels. 69. ABC is a triangle, and through P, the point of intersection of the bisectors of Z B and ZC,a line is drawn parallel to BC, meeting AB at M, and CA at N. Prove that MN = MB + CN. 70. Through the mid-point of the segment of a transversal cut off by two parallels, a straight line passes, terminated by the parallels. Prove that this line is bisected by the transversal. Prop. XIX.] PARALLELS AND PARALLELOGRAMS. 49 Proposition XIX. 88. Theorem. In any triangle, (1) any exterior angle equals the sum of the two interior non-adjacent angles ; (2) the sum of the three interior angles is a straight angle. A B Given A ABC, with AB produced to X. To prove that (1) AXBC = AA + A C; (2) AA + AB + AC = st. Z. Proof. 1. Suppose BY II AC, and A named as in the figure. 2. Then Z x = Z a, Why ? and Ay = Ac. Why ? 3. .'.Ax + Ay, or Z XBC, = Z « + Ac, which proves (1). Ax. 2 4. But Ax + Ay + Ab = st. A. Def. st. Z 5. .'. Aa + Ab + Ac = st. Z, by substituting 3 in 4, which proves (2). Notes. 1. Prop. XIX, (2), is attributed to Pythagoras. 2. The theorem is one of the most important of geometry. To it and to its corollaries (p. 50) frequent reference is hereafter made. Exercises. 71. PQR is a triangle having PQ = PR ; RP is produced to S so that PS = RP; QS is drawn. Prove that QS _L RQ. 72. Prove prop. XIX, (2), by drawing through C, in the figure given, aline II AB. 73. Also by assuming any point P on AB, drawing PC, and showing that Z BPC + Z CPA = st. Z , and also equals the sum of the interior angles. 74. State the reciprocal of prop. VIII, and prove or disprove it. 50 PLANE GEOMETRY. [Bk. 1. Corollaries to prop. XIX. 1. If a triangle has one rigid angle, or one obtuse angle, the other angles are acute. For the sum of all three is a straight angle. 2. Every triangle has at least two acute angles. V For if it had none or only one, the sum of the others would equal or exceed what kind of an angle, and thus violate what theorem ? 3. From a point outside a given line not more than one perpendicular can be drawn to that line. For if two could be drawn, a triangle could be formed having how many right angles, thus violating what corollary ? 4. If a triangle has a right angle, the two acute angles are complement al. For the sum of all three must equal two right angles ; therefore, etc. 5. If tivo triangles have tivo sides of the one respectively equal to two sides of the other, and the angles opposite one pair of equal sides right angles, or equal obtuse angles, the triangles are congruent. I For prop. XIV then applies ; the oblique angles cannot be supplemental. 6. If two angles of one triangle equal two angles of another, the third angles are equal. (Why ?) 7. Two triangles are congruent if two angles and any side of the one are respectively equal to the corresponding ptarts of the other. (Why ?) 8. Each angle of an equilateral triangle is one-third of a straight angle. (Why ?) 89. Definitions. A triangle, one of whose angles is a right angle, is called a right-angled triangle. A triangle, one of whose angles is an obtuse angle, is called an obtuse-angled triangle. A triangle, all of whose angles are acute, is called an acute- angled triangle. The side opposite the right angle of a right-angled triangle is called the hypotenuse. Sec. 90.] PARALLELS AND PARALLELOGRAMS. 51 90. Summary of Propositions concerning Congruent Triangles. Two triangles are congruent if the following parts of the one are equal to the corresponding parts of the other : 1. Two sides and the included angle. Prop. I 2. Two angles and the included side. Prop. II 3. Three sides. Prop. XII 4. Two angles and the side opposite one, Prop. XIII or, more generally, two angles and a side. Prop. XIX, cor. 7 5. Two sides and the angle opposite one, provided that angle Js not acute. Prop. XIV, and prop. XIX, cor. 5 If the angle is acute, then from two sides and the acute angle opposite one of them two different triangles may be possible. This is therefore known as the ambiguous case. If the side opposite the acute angle is not less than the given adjacent side, the case is not ambiguous. Why ? Draw the figures illustrating the ambiguous case. These propositions can be summarized in one general propo- sition : A triangle is determined 'when any three independent parts are given, except in the ambiguous case. It should be noted that the three angles are not three independent parts, since when any two of them are given the third is determined. (Prop. XIX.) Exercises. 75. In a right-angled triangle, the mid-point of the hypote- nuse is equidistant from the three vertices. (Suppose a line drawn from the vertex C of the right angle making with a an angle equal to Z B.) 76. In a right-angled triangle, a perpendicular let fall from the vertex of the right angle, upon the hypotenuse, cuts off two triangles mutually equiangular to the original triangle. 77. If a JL x and b ± y, and x intersects y, then Zab — Z xy. 78. In the annexed figure, Z aai = Z bb x . Prove that (1) Z aa\ — Z ab + Z ba-i ; (2) Z bb x = Z 6a i + Z ai&i. b,l 79. How many degrees in each angle of an isosceles /"b right-angled triangle ? also of an isosceles triangle whose vertical angle is 72° ? 178°? 60°? 52 PLANE GEOMETRY. [Bk. I. Proposition XX. 91. Theorem. Of all lines drawn to a given line from a given external point, the perpendicular is the shortest; of others, those making equal angles with the perpendicular are equal ; and of two others, that which makes the greater angle with the perpendicular is the greater. Given PO±XX'; FA, FA', FB, oblique to XX 1 , with Z A'FO = Z OF A < Z OFB. To prove that (1) PO < PA, (2) PA' = PA, (3) FB > PA, or FA'. Proof. 1. ZPAO< ZtAOP. Prop. XIX, cor. 1 2. .'. PO < PA, which proves (1). Prop. VII 3. Z AOP = Z POA', Prel. prop. I £A'PO = £OPA, AVhy? and PO = PO. 4. .-. AAOF^AA'OP, and PA' = PA, which proves (2). Why ? 5. Z BAP is obtuse, v it > Z AOP, Prop. V and Z PB is acute, ' . * Z />' OP is rt. Why ? 6. .. PB> PA, or its equal PJ', by step 4, which proves (3). Prop. VII Prop. XX.] PARALLELS AND PARALLELOGRAMS. 53 Corollaries. 1. From a given external point tit ere can be two, and only two, equal obliques of given length to a given line. Prove it by a reductio ad absurdum. 2. If from a point not on a perpendicular drawn t<> a line at its mid-point, lines are drawn to the ends of the line, these lines are unequal and the one cutting the perpendicular is the greater. , Let Z be the point, not on OP, in the figure. Suppose ZA' to cut OP at Y. Then ZA' = ZY+YA>ZA. 3. The converse of cor. 2 is true. For if ZA' cuts OP, then ZA' > ZA, by cor. 2. " ZA " " " " <• » '• " " " Z is on " " " = " (Why ?) .-. the Law of Converse (§ 73) evidently applies to this case. 4. Of two obliques from a point to a line, that which meets the line at the greater distance from the foot of the perpen- dicular is the greater. For if OR>OA, then Z OPB > Z OP A. (Why ?) .-. prop. XX applies. 5. Two obliques from a point to a line, meeting that line at equal distances from the foot of the perpendicular, are equal, make equal angles with this line and also with the perpen- dicular. Give the proof in full. 6. Two equal obliques from a point to a line cut off equal segments from the foot of the perpendicular. Draw the figure. It will then be seen that prop. XIX, cor. 5, applies. The _L is evidently an axis of symmetry (§ 68). Exercises. 80. - A line perpendicular to the bisector of any angle of a triangle makes an angle with either arm of that angle equal to half the sum of the other two angles ; and, unless parallel to the base, it makes an angle with the line of the base equal to half the difference of those angles. 81. In an isosceles triangle, the perpendicular from the vertex, the median to the base, and the bisector of the vertical angle all coincide. 54 PLANE GEOMETRY. [Bk. I. 92. Definitions. A polygon is said to be convex when no side produced cuts the surface of the polygon. A polygon is said to be concave when a side produced cuts the surface of the polygon. A polygon is said to be cross when the perimeter crosses itself. The word polygon is understood, in elementary geometry, to refer to a convex or concave polygon unless the contrary is stated. Convex. Concave. Cross. A general quadrilateral. If all of the sides of a polygon are indefinitely produced, the figure is called a general polygon. If a polygon is both equi- angular and equilateral, it is said to be regular. By the term regular polygon, a regular convex polygon is under- stood unless the contrary is stated. A polygon is called a tri- angle, quadrilateral, penta- gon, hexagon, heptagon, octagon, nonagon, decagon, dodecagon, pentedecagon, n-gon has 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, ... The student, even if unacquainted with Latin or Greek, should under- stand the derivation of these common terms. From the Latin are derived the words and prefix tri-angle (three-angle), quadri-lateral (four-side), nona- (nine) ; from the Greek are derived poly-gon (many-angle), penta- (five), hexa- (six), hepta- (seven), octo- (eight), deca- (ten), dodeca- (twelve), Much light will be thrown on the meaning of various geometric terms by consulting the Table of Etymologies in the Appendix. Regular convex polygon. Regular cross polygon. according as it ?i-sides. Prop. XXI. 1 PA RALLELS A N I) PA II A LL E L OG RA MS. 55 Proposition XXI. 93. Theorem. The sum of the interior angles of an n-gon is (n — 2) straight angles. P Given 1\ a polygon of n sides. To prove that the sum of the interior angles is (u — 2) straight angles. Proof. 1. P may be divided into (n — 2) A by diagonals which do not cross ; for, (a) A 4-gon (quadrilateral) is a A + a A, .-. 'J A, or (4-2) A. (b) A 5-gon (pentagon) is a 4-gon + a A, .-. 3 A, or (5-2) A. (c) A 6-gon (hexagon) is a 5-gon + a A, .-. 4 A. or (6-2) A. (d) And every addition of 1 side adds 1 A. (e) .*. for an n-gon there are (n — 2) A. 2. The sum of the A of each A is a st. Z. Prop. XIX 3. .*. the sum of the interior A of an w-gon is (// — 2) st. A, because these equal the sum of the A of the A. Corollary. If each of two angles of a quadrilateral is a right angle, the other two angles are supplemental. (Why ?) Exercise. 82. How many diagonals in a common convex pentagon ? hexagon ? 56 PLANE GEOMETRY. [Bk. I. 94. Generalization of Figures. If a thermometer registers 70° above zero, it is ordinarily stated, in scientific works, that it registers + 70°, while 10° below zero is indicated by — 10°, the sign changing from + to — as the temperature decreases through zero. Similarly, west longitude is represented by the sign -f-, while longitude on the other side of 0° (i.e. east) is represented by the sign — , the longitude changing its sign in passing through zero. So in speaking of temperature it is said that 10° + (— 10°) = 0, meaning thereby that if the tem- perature rises 10° from 0, and then falls 10°, the result of the two movements is the original temperature, 0. This custom holds in geometry. Thus, in this figure, if the segment between B and C is thought of as extending from B to C, it would be named BC ; and, as is usually done in geometry with , ~ lines thought of as extending to the right, it would be considered Sijyositive line. But if it is thought of as extending from C to B, it would be named CB, and con- sidered a negative line. Hence it is said that BC + CB = 0, an expression borrowed from algebra, where it would appear in a form like x + (— x) = 0. Similarly, with regard to angles : the turning of an arm in a sense opposite to that of a clock-hand, counter-clockwise, is considered positive, while the turning in the opposite sense is considered nega- tive. Thus, Z XOA is considered posi- tive, but the acute Z AOX is considered negative, and this is indicated by the state- ment, — Z.XOA — acute Z A OX. Hence, as in the case of lines, Z XOA + (- Z XOA) = Z XOA + acute Z AOX = zero. On this account we pay special attention to the manner of lettering angles, distinguishing between Z XOA and Z A OX. It is only recently that negative angles have been considered in elementary geometry, and hence the older works paid no attention to the order of the naming of the arms. Sec. 95.] PARALLELS AND PARALLELOGRAMS. 57 95. These considerations enable us to generalize man)- fig- ures, with interesting results. Thus, prop. XXI is true for a cross polygon as well as for the simple cases usually consid- ered. If, in Fig. 1, P is moved through AB to the position C Fig. 1. Fig. 2. shown in Fig. 2, we shall still have Z A (which has passed through and has become negative) + Zi?-fZC + ZP (which is now reflex) = 2 st. angles. Exercises. 83. Prove the last statement made above. 84. How many points of intersection, at most, of the sides of a gen- eral quadrilateral ? pentagon ? 85. How many diagonals, at most, has a general quadrilateral ? 86. Prove prop. XXI by connecting each vertex with a point within the figure, thus forming n &, giving n st. A, and then subtracting the two around 0. 87. In an isosceles triangle, the perpendiculars from the ends of the base to the opposite sides are equal. 88. H the bisector of the vertical angle of a triangle also bisects the base, the triangle is isosceles. 89. H the base AB of A ABC is produced to X, and if the bisectors of Z XBC and ABAC meet at P, what fractional part is A P of A C ? 90. Given two parallels and a transversal, what angle do the bisectors of the interior angles on the same side of the transversal make with each other ? 91. H one angle of an isosceles triangle is given, and it is known whether it is the vertical angle or not, then the other two angles are determined. 5$ PLANE GEOMETRY. [Bk. Proposition XXII. 96. Theorem. The sum of the exterior angles of any poly- gon is a perigon. Given P and Q, two ^-gons. To prove that the sum of the exterior A = 360° in each »-gon. Proof. 1. In P, each interior Z -f its adjacent exterior Z = 180°. § 14, def. st. Z 2. /. sum of int. and ext. A— n -180 3. But sum of int. A = (n - 2) • 180°. Ax. 6 Why ? Ax. 3 4. .-. sum of ext. A = 2 • 180° = 360°. The proof for Q is the same, if Z a is considered negative. Exercises. 92. Each exterior angle of an equilateral triangle equals how many times each interior angle ? 93. Each exterior angle of a regular heptagon equals what fractional part of each interior angle ? 94. Each exterior angle of a regular n-gon equals what fractional part of each interior angle ? See if the result found is true if n = 3, or 4. 95. Is it possible for the exterior angle of a regular polygon to be 70° ? 72°? 75°? 120°? 96. Prove prop. XXII independently of prop. XXI by taking a point anywhere in the plane of the figure (inside or outside the polygon, or on the perimeter) and drawing par- allels to the sides from that point, and showing that the sum of the exterior angles equals the perigon about that point. Sec 97.] PARALLELS AND PARALLELOGRAMS. 59 97. Definitions. A quadrilateral whose opposite sides are parallel is called a parallelogram. A quadrilateral that lias one pair of opposite sides parallel is called a trapezoid. Trapezium is a term often applied to a quadrilateral no two of whose sides are parallel. By the definition of trapezoid here given it will be seen that the paral- lelogram may be considered a special form of the trapezoid. The parallel sides of a trapezoid are called its bases, and are distin- guished as upper and lower. If the two opposite non-parallel sides of a trapezoid are equal, the trapezoid is said to be isosceles. Q Upper Base c Lower Base D^ Upper Base „ . / Lower Base \ „ A b Parallelogram. Trapezoid. Isosceles trapezoid. Iii the above figures, angles A, 7>, or £, C, or C, D, or D, A, are called consecutive angles. Angles A, C, or B, I), are called opposite angles. V Exercises. 97. If the student has proved ex. 96, let him prove prop. XXI from that. \. Prove that the quadrilateral formed by the bisectors of the angles of any quadrilateral has its opposite angles supplemental. 99. Show that in ex. 98 the angles bisected may be either the four interior or the four exterior angles. 100. If from the ends of the base of an isosceles triangle perpendicu- lars are drawn to the opposite sides, a new isosceles triangle is formed, each of its base angles being half the vertical angle of the original triangle. 101. The hypotenuse is greater than either of the other sides of a right- angled triangle. 102. From the vertex of the right angle of a right-angled triangle, is it possible to draw, to the hypotenuse, a line longer than the hypotenuse ? Proof. 103. A line from the vertex of an isosceles triangle to any point on the base produced is greater than either side. Is this also true for a scalene triangle ? 60 PLANE GEOMETRY [Bk. I. Proposition XXIII. 98. Theorem. Any two consecutive angles of a parallelo- gram are supplemental, and any two opposite angles are equal. D 7 Given O ABCD. To prove that . (1) Z A + Z B = st. Z, (2) AA = AC. Proof. 1. Z ^ + Z 7? = st. Z, which proves (1). Prop. XVII, cor. 2 2. Z B + Z C = st. Z. Why ? 3. .\ZA + ZB = ZB + ZC. Why ? 4. .-. Z .4 = Z C, which proves (2). Why ? Corollary. 7/ o«e angle of a parallelogram is a right angle, all of its angles are right angles. (Why ?) 99. Definitions. If one angle of a parallelogram is a right angle, the parallelogram is called a rectangle. By the corollary, all angles of a rectangle are right angles. A parallelogram that has two adjacent sides equal is called a rhombus. It is shown in prop. XXIV, cor. 1, that all of its sides are equal. A rectangle that has two adjacent sides equal is called a square. It is shown in prop. XXIV, cor. 1, that all of its sides are equal. A square is thus seen to be a special form of a rhombus. Rhombus. Square. Prop. XXIV.] PARALLELS AND PARALLELOGRAMS. 61 Proposition XXIV. 100. Theorem. In any varallelogram, (1) either* diagonal divides it into two congruent triangles, (2) the opposite sides are equal. B Given O ABCD. To prove that (1) A ABC s? A CD A, (2) AB = DC. Proof. 1. In the figure, Zx = Z.x', Ay — Z-y 1 , and AC = AC. Prop. (?) 2. .-. A ABC^A CD A, which proves (1). Prop. II 3. .'. AB = DC, which proves (2). § 57 Similarly for diagonal BD, and sides BC and AD. Corollaries. 1. If two adjacent sides of a parallelogram are equal, all of its sides are equal. For by step 3 the other sides are equal to these. Hence, as stated in § 99, all of the sides of a rhombus are equal. 2. The diagonals of a parallelogram bisect each other. For if diagonal BD cuts AC at O, then, by prop. II, A ABO = A CDO, whence AO = OC, and BO = OB. In the annexed figure, if a and a' are perpendicular to P and P' , two parallels (prop. XVII, cor. 1), they are parallel (prop. XVI, cor. 3). Hence a = a', by prop. P XXIV. This fact is usually expressed by p— saying, 3. Tic o parallel lines are everywhere equidistant from each other. 62 PLANE GEOMETRY [Bk. Proposition XXV. 101. Theorem. If a convex quadrilateral has tivo opposite sides equal and parallel, it is a parallelogram. Given a convex quadrilateral ABCD, with AB — DC, and AB II DC. To prove that ABCD is a parallelogram. Proof. 1. In the figure Z x = Z x', BD = BD, and AB = DC. 2. .\A ABD 5£ A GOB, and Zy = Z y'. 3. .-. #C II JlA 4. .'. ABCD is a O by definition. Prop. (?) Given Prop. (?) Prop. XVI Exercises. 104. It is shown in Physics that if two forces are pulling from the point J5, and the first force is represented (see fig. to prop. XXV) by BA, and the second by BC, the resultant (resulting force) will be rep- resented by the diagonal BD. Show that, if the two forces do not pull in the same line, the resultant is always less than the sum of the two forces. 105. If two equal lines bisect each other at right angles, what figure is formed by joining the ends ? 106. If the diagonals of a rectangle are perpendicular to each other, prove that the rectangle is a square. 107. On the diagonal BD of O ABCD, P and Q are so taken that BP = QD. Show that APCQ is a parallelogram. Suppose P is on DB produced, and Q on BD produced. 108. Prove that the diagonals of a rectangle are equal. Prove that the diagonals of a rhombus arc perpendicular to each other and bisect the angles of the rhombus. 1 Prop. XXVI.J PARALLELS AND PARALLELOGRAMS. 63 Proposition XXVI. 102. Theorem. If two parallelograms have two adjacent *ides and any angle of the one respectively equal to the corresponding parts of the other, they are congruent. B' A A B Given LU ABCD, A'B'CD', in which AB = A'B\ AD = A'D'. andZD = ZD'. To prove that O ABCD ^ O A'B'CD'. Proof. 1. ZA = ZA'.ZB = Z B\ Z C = Z C, for they are equal or supplemental to D or D'. Prop. XXIII 2. CD = CD', BC = B'C, for they are equal to sides that are known to be equal. Pro}). XXIV 3. Apply lj ABCD to O A'B'CD' so that AB coin- cides with its equal A'B', A falling on A'. Then AD can be placed on AD' because Z A = Z .!'. Then D will fall on D', because AD = A'D'. Similarly, C will fall on C\ and C£ on C'£\ Corollaries. 1. Two rectangles are congruent if two ad- jacent sides of the one are equal to any two adjacent sides of the other. (Why ?) 2. Two squares are congruent if a side of the one equals >< side of the other. (Why?) Exercises. 109. Is a parallelogram determined when any two sides and either diagonal are given ? when two adjacent sides and either diag- onal are given ? 110. The angle- at either base of an isosceles trapezoid are equal. 64 PLANE GEOMETRY. [Bk. I. Proposition XXVII. 103. Theorem. If there are two pairs of lines, all of which are parallel, and if the segments cut off by each pair on any transversal are equal, then the segments cut off on any other transversal are equal also. Given four parallels, of which JP X , P 2 cut off a segment a, and P 3 j P 4 cut off an equal segment b, on a trans- versal T, and cut off segments a', b', respectively, on transversal T. To prove that v a' = b'. Proof. 1. Suppose x and y II T as in the figure. 2. Then, in the figure, Z wx = Z P 2 T = Z P 4 T = Z zy. Prop. XVII, cor. 2 3. And Z a'w = Z b'z. Why ? 4. And x = a = b = y. Prop. XXIV 5. .'.A wa'sc = A zb'y, and a' = b'. Prop. XIX, cor. 7 Corollaries. 1. 7^ «- system of paral- lels cuts off equal segments on one trans- versal, it does on every transversal. For if a = 6i or b- 2 , a' = b\ or 0%, respectively, and similarly for the other transversals. 2. The line through the mid-point of one side of a triangle^ parallel to another side, bisects the third side. Draw a third parallel through the vertex. Then cor. 1 proves it. Prop. XXVII.] PARALLELS AND PARALLELOGRAMS. 65 3. The line joining the mid-points of two sides of a triangle is parallel to the third side. For if not, suppose through the mid-point of one of those sides a line is drawn parallel to the base ; then this must bisect the other side, by cor. 2 ; .-.it must coincide with the line joining the mid-points, or else a side would be bisected at two different points. (This is the converse of cor. 2. Draw the figure.) 4. The line joining the mid-points of two sides of a triangle equals half the third side. (Prove it.) Exercises. 111. The line joining the mid-points of the non-parallel sides of a trapezoid is parallel to the bases. 112. In a right-angled triangle the mid-point of the hypotenuse is equidistant from the three vertices. (This exercise has been given before, and will be repeated, since it is important and admits of divers proofs. It is here easily proved by prop. XXVII, cor. 2 ; for if a = b, then a' — b' \ but p II e, .-. p ± a'. .-. x = b = a.) a ' b ' 113. The lines joining the mid-points of the sides of a triangle divide it into four congruent triangles. 114. If one of the equal sides CB of an isosceles triangle ABC is pro- duced through the base, and if a segment BD is laid off on the produced part, and an equal segment AE is laid off on the other equal side, then the line joining D and E is bisected by the base. (Consider the cases in which BD< CB, BD = CB, BD>CB.) 115. If the mid-points of the adjacent sides of any quadrilateral are joined, the figure thus formed is a parallelogram. (Consider this theorem for cases of concave, convex, and cross quadrilaterals, and for the special case of an interior angle of 180°.) 116. The lines joining the mid-points of the opposite sides of a quadri- lateral bisect each other. Consider for the special cases mentioned in ex. 115. 117. The line joining the mid-points of the diagonals of a quadri- lateral, and the lines joining the mid-points of its opposite sides, pass through the same point. 118. P and Q are the mid-points of the sides AB and CD of the parallelogram ABCD. Prove that PD and BQ trisect (divide into three equal segments) the diagonal A C. 66 PLANE GEOMETRY. [Bk. I. EXERCISES. 119. "What is the sum of the interior angles of a polygon of 20 sides ? of 30 sides ? 120. How many degrees in each angle of a regular polygon of 12 sides ? of 20 sides ? 121. How many sides has a polygon the sum of whose interior angles is 48 right angles ? 122. The vertical angle of a certain isosceles triangle is 11° lo' 20"; how large are the base angles ? 123. The exterior angle of a certain triangle is 140°, and one of the interior non-adjacent angles is a right angle ; how many degrees in each of the other two interior angles ? 124. Each exterior angle of a certain regular polygon is 10° ; how many sides has the polygon ? 125. If P is any point on the side BC of A ABC, then the greater of the sides AB, AC, is greater than AP. 126. If the diagonals of a quadrilateral bisect each other, prove that the quadrilateral is a parallelogram. Of what corollary is this the converse ? Prove that the diagonals of an isosceles trapezoid are equal. 127. Conversely, prove that if the diagonals of a trapezoid are equal, the trapezoid is isosceles. 128. Is a parallelogram determined when its two diagonals are given ? when its two diagonals and their angle are given ? 129. ABC is a triangle ; AC is bisected at M; BM is bisected at N; AN meets BC at P; MQ is drawn parallel to AP to meet BC at Q. Prove that BC is trisected (see ex. 118) by P and Q. 130. A, C are points on the same side of XX' ; B is the mid-point of AC; through A, B, C parallels are drawn cutting XX in A', B', C* Prove that AA' + CC = 2 BB'. 131. A straight line drawn perpendicular to the base AB of an isos- celes triangle ABC cuts the side CA at B and BC produced at E ; prove that CEB is an isosceles triangle. 132. ABC is a triangle, and the exterior angles at B and C are bisected by the straight lines BB, CB respectively, meeting at B ; prove that Z. CBB + jZi = a right angle. 133. In the triangle ABC the side BC is bisected at E, and AB at G; AE is produced to F so that EF = AE, and CG is produced to H so that Gil = CG. Prove that F, B, iJare in one straight line. PROBLEMS. 67 3. PROBLEMS. 104. Definitions. A curve is a line no part of which is straight. C ire«2£/W£ % 105. A circle is the finite portion of a plane bounded by a curve, which is called the circumference, and is such that all points on that line are equidistant from a point within the figure called the center of the circle. A circle is evidently described by a line-seg- ment making a complete rotation in a plane, about a fixed point (the center). 106. A straight line terminated by the center and the cir- cumference is called a radius, and a straight line through the center terminated both ways by the circumference is called a diameter of the circle. 107. A part of a circumference is called an arc. Note. The above definitions are substantially those usually met in elementary geometries. The student will find, after leaving this subject, that the word circle is often used for circumference. Indeed, there is good authority for so using the word even in elementary geometry. 108. From the above definitions the following corollaries may be accepted without further proof : 1. A diameter of a circle is equal to the sum of two radii of that circle. 2. Circles having the some rodii ore congruent. 3. A point is within a circle, on its circumference, or outside the circle, according as the distance from that point to the center is less than, equal to, or greater than, the radius. 68 PLANE GEOMETRY. [Bk. I. 109. It now becomes necessary to assume certain postulates relating to the circle. Postulates of the Circle. 1. All radii of the same circle are equal, and hence all diameters of the same circle are equal. 2. If an unlimited straight line passes through a point within a circle, it must cut the circumference at least twice, and so for any closed figure. That it cannot cut the circumference more than twice is proved in III, prop. VI, cor. 3. If one circumference intersects another once, it intersects It again. 4. A circle has but one center. 5. A circle may be constructed with any center, and with a radius equal to any given line-segment. This postulate requires the use of the compasses. As has been stated, the only instruments allowed in elementary geometry are the compasses and the straight-edge, a limitation due to Plato. In the more advanced geometry, where other curves than the circle are studied, other instru- ments are permitted. 110. Order to be observed in the solution of problems : Given. For example, the angle A. Required. For example, to bisect that angle. Construction. A statement of the process of solving, using only the straight-edge and compasses in drawing the figure described. Proof. A proof that the construction has fulfilled the requirements. % Discussion. Any consideration of special cases, of the limitations of the problem, etc. If a problem has but a single solution, as that an angle may be bisected but once, the solution is said to be unique. Prop. XXVIII.] PROBLEMS. 69 Proposition XXVIII. 111. Problem. To bisect a given angle. Given the A AOB. Required to bisect it. Construction. 1. With center describe an arc cutting AO&t C, and OB at D. § 109, post, of O 2. Draw DC. § 28. post. st. line 3. Describe arcs with centers D, C, and radius DC. Post. (?) 4. Join their intersection P, with 0. Post. (?) Then Z AOB is bisected, YY' being an axis of symmetry (§ 68). Proof. 1. Draw DP, CP '; then OD = DC, DP = DC, DC = CP. ..DP = CP. OP = OP. .'.A OCP^A ODP, and Z COP = ZPOD. § 109, 1 Ax. (?) But COROJ Prop. XII An angle may be divided into 2, Jf, 8, 16, £", equal angles. (How ?) 70 PLANE GEOMETRY. [Bk. I. 112. Note on Assumed Constructions. It has been assumed, up to prop. XXVIII, that all constructions were made as required for the theorems. Thus an equilateral triangle has been frequently mentioned, although the method of constructing one has not yet been indicated ; a regular heptagon has been mentioned in ex. 93, and reference might be made to certain results following from the trisection of an angle, although the solutions of the problems, to construct a regular heptagon, and to trisect any angle, are impossible by elementary geometry. But the possibility of solving such problems has nothing to do with the logical sequence of the theorems ; one may know that each angle of a regular heptagon is 5 • ISO , whether the regular heptagon admits of construction or not. Nevertheless, an important part of geometry con- cerns itself with the construction of certain figures — a part of utmost practical value and of much interest to the student of mathematics. 113. Suggestions on the Solution of Problems. The methods of logically undertaking the solution of problems will be dis- cussed at the close of Book III. But at present one method, already suggested on p. 35, should be repeated : In attempting the solution of a problem, assume that the solution has been accomplished ; then analyze the figure and see what results follow; then reverse the process, making these results precede the solution. For example, in prop. XXVIII, assume that ZAOB has been bisected by YY' ; if that were done, and if any point, P, on YY' were joined to points equidistant from O, on the arms, say C and D, then A OCP would be congruent to A ODP ; now reverse the process and attempt to make A OCP congruent to A ODP ; this can be done if OD can be made equal to OC, and PD to PC, because OP= OP ; but this can be done by § 109, 5. This method of attacking a problem, without which the student will grope in the dark, is called Geometrical Analysis. Exercises. 134. Give the solution of prop. XXVIII, using P' instead of P. Why is P better than P' for practical purposes ? In what case would the construction fail for the point P' ? In that case how many degrees in Z A OB ? 135. In prop. XXVIII, in what case would P' fall below O ? Give the solution in that case, after connecting P' and and producing P'O. Prop. XXIX., PROBLEMS. 71 Proposition X XIX. 114. Problem. To draw a perpendicular to a given line from a given internal point. y \ o \c Xp Y Solution. This is merely a special case of prop. XXVIII, the case in which Z AOB is a straight angle. (Why ?) The construction and. proof are identical with those of prop. XXVIII, and the student should give them to satisfy himself of this fact. Exercises. 136. What kind of a quadrilateral is CPDP' ? Prove it. 137. Prove that any point on BA is equidistant from P and P'. Also that any point on YY' is equidistant from D and C. 138. In step 3 of the construction of prop. XXVIII might the radius equal two times DC ? If so, complete the solution. Is there any limit to the length of the radius in that step ? 139. In the figure of prop. XXVIII, suppose ZPCO = 130°. Find the number of degrees in the various other angles, not reflex, of the figure. 140. In the figure of prop. XXVIII, prove that the reflex angle BOA is bisected by YY', that is, by PO produced. 141. Also prove that YY' is the perpendicular bisector of DC. 142. Also prove that if is connected with P and with P', OP* will fall on OP. (Prel. prop. VIII.) 72 PLANE GEOMETRY. [Bk. I. Proposition XXX. 115. Problem. To draw a perpendicular to a given line from a given external point. A s / / A s — "\ Given the line XX' and the external pt. P. Required to draw a perpendicular from P to XX'. Construction. 1. Draw PR cutting XX'. § 28 2. With center P and radius PR const, a O. § 109, post, of O 3. Join A and A', where the circumference cuts XX', with P. § 28, post, of st. line 4. Bisect Z A' PA. Prop. XXVIII The bisector, PO, is the required perpendicular. Proof. 1. PA = PA', § 109, 1 Z OPA = ZA'PO, Const., 4 and PO = PO. 2. .-.AAPO^AA'PO, Prop. I andZiOP = ZPO# 3. .-. Z ,4 OP is a rt. Z, and PO _L XX'. §§ 19, 20 Note. The solution of this problem is attributed to (Enopides. Exercises. 143. Find in a given line a point equidistant from two given points A and B, the mid-point of AB being also given. 144. Find a point equidistant from three given points A, B, C, the mid-points of A B and BC being also given. Prop. XXXI.] PROBLEMS. 73 Proposition XXX I. 116. Problem. To bisect a given line. Given the line AB. Required to bisect it. Construction. 1. With, centers A. B, and equal radii describe arcs intersecting at P and P'. Post. (?) 2. Draw PP'. Post. (?) 3. Then PP' bisects AB. Proof. (Let the student give it. Draw AP\ P'B, BP, PA.) Exercises. 145. Through a given point t<:> draw a line making equal angles with the arms of a given angle. Discuss for various relative posi- tions of the point. 146. To draw a perpendicular to a line from one of its extremities, when the line cannot be produced. (Ex. 112 suggests a plan.) 147. Through two given points on opposite sides of a given line draw two lines which shall meet in the given line and include an angle which is bisected by that line. 148. If two isosceles triangles have a common base, the straight line through their vertices is a perpendicular bisector of the base. 74 PLANE GEOMETRY. [Bk. I. Proposition XXXII. 117. Problem. From a given point in a given line to draw a line making with the given line a given angle. C B Given the line AB, the point P in it, and the angle 0. Required from P to draw a line making with AB an angle equal to Z 0. Construction. 1. On the arms of ZO lay off OC — OB by describing an arc with center O and any radius OC. § 109, post, of O 2. Draw CD. § 28, post, of st. line 3. With center P and radius OC, describe a circum- ference cutting PB in C. Post. (?) 4. With center C and radius CD, describe an arc cut- ting the circumference in D'. Post. (?) Draw PD, and this is the required line. Proof. Draw CD'; then, A PCD' and A OCD being mutually equilateral, Why ? A PCD ^ A OCD, and Z CPD' = Z COD. Prop. XII Exercises. 149. Prove that, the circumferences must cut at D' as stated in step 4. 150. See if the solution of prop. XXXII is general enough to cover the cases where the Z O is straight, reflex, a perigon. 151. From a given point in a given line to draw a line making an angle supplemental to a given angle. Prop. XXXIIL] PROBLEMS. 75 Proposition XXXIII. 118. Problem. Through a given point to draw a line paral- lel to a given line. V Given the line AB and the point P. Required through P to draw a line parallel to AB. Construction. 1. Join P with any point. 0, on AB. § 28. post, of st. line 2. From P draw PC making Z OPC = Z POA. (?) Then PC is the required line. Proof. PCWAB. Why? Discussion. The solution fails if P is on the unlimited line AB. Exercises. 152. Through a given point to draw a line making a given angle with a given line. Notice that the solution is not unique. 153. Through a given point to draw a transversal of two parallels, from which the parallels shall cut off a given segment. Discussion should show when there are two solutions, when only one, when none. 154. To construct a polygon (say a hexagon) congruent to a given polygon. 155. Through two given points to draw two lines forming with a given unlimited line an equilateral triangle. 156. Three given lines meet in a point ; draw a transversal such that the two segments of it, intercepted between the given lines, may be equal. Is the solution unique ? 157. From P, the intersection of the bisectors of two angles of an equi- lateral triangle, draw parallels to two sides of the triangle, and show that these parallels trisect (see ex. 118) the third side. fl 76 PLANE GEOMETRY. [Bk. I. Proposition XXXIV. 119. Problem. To construct a triangle, given the three ,. Given a, b, c, three sides of a triangle. Required to construct the triangle. Construction. 1. With the ends of b as centers, and with radii a, c, describe circumferences. § 109 2. Connect either point of intersection of these circum- ferences with the ends of b. § 28 Then is the required A constructed. Proof. It was constructed on b, and the other sides equal a, c. § 109, 1 Discussion. If the two circumferences do not intersect, a solution is impossible, for then either a > b + c, a = b + c, a = c — b, or a < c — b, and in none of . these cases is a triangle possible. Prop. VIII and cor. Corollary. To construct an equilateral triangle on a given line-segment. The first proposition of Euclid's "Elements of Geometry." Euclid proceeded upon the principle of logical sequence of propositions, with no attempt at grouping the theorems and the problems separately. He found this corollary (a problem) the best proposition with which to begin his system. Exercise. 158. In a given triangle inscribe a rhombus, having one of its angles coincident with a given angle of the triangle, and the other three vertices on the three sides of the triangle. Prop. XXXV.] PROBLEMS. 77 Propositiox XXXV. 120. Problem. To construct a triangle, given two sides and the included angle. b Given the sides a, b, and the included angle k. Required to construct the triangle. Construction. 1. From either end of b draw a line making with b the angle k. Prop. XXXII 2. On that line mark off a by describing an arc of radius a. § 109 3. Join the point thus determined with the other end of b. § 28 Then the triangle is constructed. Proof. By step 2 the line marked off equals a, and by step 1 Z-b = Z. k, and it is constructed on b. Exercises. 159. To trisect a right angle. (Construct an equilateral triangle on one arm.) 160. On the side AC of A ABC to find the point P such that the par- allel to AB, from P, meeting BC at Z>, shall have PD = A P. 161. To construct a triangle, having given two angles and the perpen- dicular from the vertex of the third angle to the opposite side. 162. Draw a line parallel to a given line, so that the segment inter- cepted between two other given lines may equal a given segment. 163. Given the three mid-points of the sides of a triangle, to construct the triangle. 164. Through a given point P in an angle AOB to draw a line, termi- nated by OA and 07?, and bisected at P. (Through P draw a II to BO cutting OA in X ; on XA lay off XY = OX ; draw YP.) '0/li-o* 78 PLANE GEOMETRY. [Bk. I. Proposition XXXVI. 121. Problem. To construct a triangle, given two sides and the angle opposite one of them. Given two sides of a triangle, a, b, and Z k opposite a. Required to construct the triangle. Construction. 1. At either end of b draw a line making with b an angle equal to Z k. Prop. (?) 2. With the other end of b as a center, and a radius a, describe a circumference. Post. (?) 3. Join the points where the circumference cuts the line of step 1, with the center. Post. (?) Then the triangle is constructed. Proof. For it has the given side b, and the given Z k, and the lines of step 3 equal a. § 109, 1 Discussion. If the circumference cuts the line twice, two solu- tions are possible, and the triangle is ambiguous (see prop. XIV). If it touches the line without cutting it, what about the solution ? If it does not meet the line, no solution is possible. If Z k is right or obtuse, or if a<£b, only one solution is possible (prop. XIX, cor. 5), Draw a figure for each of these cases, and show from the drawings that the statements made in the discussion are true. Exercise. 165. XX% YY', are two given lines through 0, and P is a given point ; through P to draw a line to XX', which shall he hisected by YY'. Investigate for various positions of P. as where P is within the ZXOY. the Z YOX\ on OY, or on OX. Prop. XXXVII.] PROBLEMS. 79 Proposition XXXVII. 122. Problem. To construct a triangle, given two angles and the included side. X A A B B A B Given two angles, A, B, and the included side AB. Required to construct the triangle. Construction. 1. From A draw A X making with AB an angle equal to Z A. Prop. XXXI 1 2. Similarly, from B draw BY, making an angle equal to Z B." Prop. XXXII C being the intersection of AX, BY, then ABC is the required A. Proof. (Let the student give it.) Discussion. If AX, BY, do not intersect, what follows ? Proposition XXXVIII. 123. Problem. To construct a triangle, given two angles and a side opposite one of them. Solution. Subtract the sum of the angles (found by prop. XXXII) from 180° and thus find the third angle (prop. XIX). The problem then reduces to prop. XXXVII. Proposition XXXIX. 124. Problem. To construct a square on a given line as a side. 80 PLANE GEOMETRY. [Bk. I. 4. LOCI OF POINTS. 125. The place of all points satisfying a given condition is called the locus of points satisfying that condition. Indeed, the word locus (Latin) means simply place (English, locality, locate, etc.) ; the plural is loci. For example, if points are on this page and are one inch from the left edge, their locus is evidently a straight line parallel to the edge. Furthermore, the locus of points at a given distance r from a fixed point is the circumference described about with a radius r. This statement, although very evident, is made a theorem (prop. XL) because of the frequent reference to it. Of course in this discussion, as elsewhere in Books I— V, the points are all supposed to be confined to one plane. In Plane Geometry the loci considered will be found to con- sist of one or more straight or curved lines. It is a common mistake to assume that a locus, which one is trying to discover, consists of a single line. It may consist of two lines, as in prop. XLII. 126. In proving a theorem concerning the locus of points it is necessary and sufficient to prove two things : 1. That any point on the supposed locus satisfies the condition; 2. That any point not on the supposed locus does not satisfy the condition. For if only the first were proved, there might be some other line in the locus ; and if only the second were proved, the sup- posed locus might not be the correct one. Exercise. 166. State, without proof, what is (1) the locus of points \ in. from a given straight line ; (2) the locus of points equidistant from two parallel lines. k Prop. XL.] LOCI OF POINTS. 81 Proposition XL. 127. Theorem. The locus of points at a given distance from a given point is the circumference described about that point as a center, with a radius equal to the given distance. Given the point 0, the line r. and the circumference C described about with radius /■. To prove that C is the locus of points r distant from 0. Proof. 1. Let P x , P 2 , P 3 be points on C\ within the circle, and without the circle, respectively. Let OP 2 produced meet C in B. and OP 3 meet Cm A. 2. Then OP 1 = OB = OA = r, § 109, 1 and OP, < OB, and OP z > OA. Ax. 8 3. .*. any point on C is r distant from 0, and any point not on C is not r distant from 0. Exercises. 167. Has it been proved in prop. XL that the required locus may not be merely the arc cut off by r and OP\ ? If so, where ? 168. What is the locus of points at a distance of i in. from the above circumference, the distance being measured on a line through 0? 169. Lighthouses on two islands are 10 miles apart ; show that there are two points at sea which are exactly 12 miles from each. 170. How would you find, by the intersection of two loci, a point on this page 1 in. from in the above figure, and 3 in. from the right edge of the paper ? 82 PLANE GEOMETRY. [Bk. I. Proposition XLI. 128. Theorem. The locus of points equidistant from two given points is the perpendicular bisector of the line joining them. Y Given two points Xand X\ and IT' _L XX' at the mid- point 0. To prove that YY' is the locus of points equidistant from X and X\ Proof. 1. Let P be any point on YY', and P' be any point not on YY'. Draw PX, PX', P'X, P'X'. 2. Then PX= PX', Prop. XX, cor. 5 and P'X' > P'X. Prop. XX, cor. 2 3. Hence any point on YY' is equidistant from X and X', but any point not on YY' is unequally distant from X and A"'. .-. YY' is the locus. § 125, def. Exercises. 171. Required to find a point which is 1 in. from X and | in. from X' in the above figure. Is there more than one such point ? 172. Required to find a point which is equidistant from X and X' in the above figure, and 1 in. from 0. Is there more than one such point ? Prop XLII. LOCI OF POINTS. 83 Proposition XLII. 129. Theorem. The locus of points equidistant from two given lines consists of the- bisectors of their included angles. Given OA and OB, two lines intersecting at 0, and XX' and YY' the bisectors of the angles at 0. To prove that XX' and YY' form the locus of points equidis- tant from OA and OB. Proof. 1. Let Q be any point on neither XX' nor YY'; let QB _L OB, QA _L OA, QA cut OX in P, PA' _L 0#. Draw QA'. Since $ may be moved, P may be considered as any point on OX. 2. Then A OAP ^ A OA'P, and JP = A' P. Prop. XIX, cor. 7 3. Also, A'P + PQ> A'Q > BQ. Why ? 4. .-. AQ,ovAP + PQ> BQ. Why? 5. .\ any point P on XX' (or on 77') is equidistant from OA and 0^, but any point Q on neither XX i nor 77' is unequally distant from OA and OB. 84 PLANE GEOMETRY. [Bk. I. Corollaries. 1. If the given lines are parallel, the locus is a parallel midway between them. (Prove it.) The student should imagine the effect of keeping points A, A' fixed, and moving farther to the left. YY' moves with 0, but XX' keeps its position as the lines approach the condition of being parallel. 2. The locus of points at a given distance from a given line consists of a pair of parallels at that distance, one on each side- of the fixed line. (Prove it.) 130. Definitions. Three or more lines which Three or more points which meet in a point are said to be lie in a line are said to be concurrent. collinear. Proposition XLIII. 131. Theorem. The perpendicular bisectors of the three sides of a triangle are concurrent. Given a triangle of sides a, b, c, and x, y, z their respective perpendicular bisectors. To prove that x, y, z are concurrent. Proof. 1. x and y must meet as at P. Prop. XVII, cor. 4 2. Then P is equidistant from B and C, and C and A. Prop. XLI 3. .'. P is on the perpendicular bisector of c ; Why ? i.e. z passes through P. Prop. XLIIL] LOCI OF POINTS. 85 Corollaries. 1. The point equidistant from three non- coUinear points is the intersection of the perpendicular bisectors of any two of the lines joining them. Step 2. 2. There Is one circle, and only one, whose circumference passes through three non-collinear points. Let A, B, C be the three points. Then by step 2 they are equidistant from P, the intersection of x and y. And v x and y contain all points equidistant from A, B, and C, and can intersect but once, there is only one point P. And v there is only one center and one radius, there is one and only one circle. 3. Circumferences having three points in common are iden- tical. Otherwise cor. 2 would be violated. 4. If from a point more than two lines to a circumference are equal, that point Is the center of the circle. For suppose a circumference through A, B. C, and suppose PA = PB = PC. Now with center P and radius PA a circumference can be described through A, B, C, because it is given that PA=PB = PC. § 108, cor. 3 And this is identical with the given circumference. Prop. XLIIL cor. 3 .-. its center must be identical with the given center, since a O cannot have two centers. § 109, 4 Exercises. 173. The proof of prop. XLIII is, of course, the same if the triangle is right-angled or obtuse-angled. The figures, however, show interesting positions for P ; consider them. 174. Required to find a point at a given distance d from a fixed point 0, and equidistant from two given intersecting lines. How many such points can be found in general ? 175. Required to find a point equidistant from two given intersecting lines, and equidistant from two given points. How many such points can be found in general ? 86 PLANE GEOMETRY. [Bk. I. Proposition XL IV. 132. Theorem. The bisectors of the interior and exterior angles of a triangle are concurrent four times by threes. .-'/^ F >-' A. XR Given the A ABC, and the bisectors of the interior and exterior angles, lettered as in the figure. To prove that these bisectors are concurrent four times by threes ; that is, 3 meet at P 1? 3 at P 2 , etc. Proof. 1. v Z CAM > Z CBM, .'. Z GAM > Z HBM. Prop. V 2. .-. AG and BR meet as at P 3 . Prop. XVII, cor. 3 3. Z HBM+ Z P^ < Z 5 + Z A < 180°. Prop. XIX .'. BH and .^P meet as at P v Prop. XVII, cor. 3 4. PP_L PiT, and AG _L JP, Prel. prop. IX .*. PP and AG meet as at P 4 . Prop. XVII, cor. 4 5. Also, P x is equidistant from a and c, from c and b, and .'. from a and &, Prop. XLII ,\P X lies on CT. Similarly for P 4 . Prop. XLII 6. Similarly, P 2 and P 3 lie on CN. .'. the four points P x , P 2 , P 3 , P 4 , are points of concurrence of the bisectors. Prop. XLV.] LOCI OF POINTS. 87 Proposition XLV. 133. Theorem. The perpendiculars from the vertices of a triangle to the opposite sides are concurrent. A\ Z X' Given the A ABC. To prove that the perpendiculars from A, B, C, to a, b, c, respectively, are concurrent. Proof. 1. Through A, B, C, respectively, suppose B'C II CB, A'C II CA, A'B' II BA. 2. Then ABCB' and ABAC are UJ. Def. O 3. .-. B'C = AB= CA'. and C is the mid-point of B'A\ Prop. XXIV; ax. 1 4. Similarly, A and B are mid-points of B'C\ CA'. 5. If AX, BY, CZ±B'C, CA', A'B', respectively, they are concurrent, as at 0. Prop. XLIII 6. And they are also the perpendiculars from A, B, C to a, b, c. Prop. XVII, cor. 1 Note. The theorem is due to Archimedes. 134. Definition. To trisect a magnitude is to cut it into three equal parts. Exercise. 176. In prop. XLIV suppose C moves down to the side c. What becomes of Pi, P 2 , P 3 , P 4 ? 88 PLANE GEOMETRY [Bk. I. Proposition XLVI. 135. Theorem. The medians of a triangle are concurrent in a trisection point of each. Given the A ABC and the medians BY, AX, intersecting at 0. To prove that (1) the median from C must pass through 0, (2) OX = I AX, OY=iBY, etc Proof. 1. Suppose CO drawn, and produced indefinitely, cutting AB at Z. 2. Suppose AP II OB ; CO must cut AP, as at P. § 85 3. Draw PB. Then V CY= YA, .'. CO = OP. Prop. XXVII, cor. 2 4. And V CO = OP, and CX= XB, .'. OX II PB. Prop. XXVII, cor. 3 5. .'. APBO is &EJ, AZ = ZB. and OZ = ZP. Prop. XXIV, cor. 2 6. . ' . OZ is a median, and it passes through 0. 7. And v 0Z= £ OP, .'. O^ = £ CO, or J CZ. Simi- larly for OF and OA: Exercise. 177. The sum of the three medians of a triangle is greater than three-fourths of its perimeter. Sacs. 13&-139.] LOCI OF POINTS. 89 136. Definitions. The point of concurrence of the perpen- dicular bisectors of the sides of a triangle is called the cir- cumcenter of the triangle. (Prop. XLIII.) The reason will appear later when it is shown that this point is the center of the circum-scvihed circle. (See Table of Etymologies.) 137. The point of concurrence of the bisectors of the interior angles of a triangle is called the in-center of the triangle ; the points of concurrence of the bisectors of two exterior angles and one interior are called the ex-centers of the triangle. (Prop. XLIV.) It will presently be proved that the in-center is the center of a circle, in-side the triangle, just touching the sides ; and that the ecc-centers are centers of circles, out-side the triangle, just touching the three lines of which the sides of the triangle are segments. Hence the names in-center and ex-center. 138. The point of concurrence of the three perpendiculars from the vertices to the opposite sides is called the orthocenter of the triangle. (Prop. XLY.) 139. The point of concurrence of the three medians of a tri- angle is called the centroid of that triangle. (Prop. XLYI.) It is shown in Physics that this point is also the center of mass, or center of gravity of the plane surface of the triangle. It is, therefore, sometimes called bv those names. Exercises. 178. If a triangle is acute-angled, prove that both the circumcenter and the orthocenter lie within the triangle. 179. In prop. XLVI, if X, Y, Z be joined, prove that the A XYZ will be equiangular with the A ABC. 180. Is there any kind of a triangle in which the in-center, circum- center. orthocenter, and centroid coincide ? If so, what is it ? Prove it. 181. In the figure of prop. XLVI, connect X, F, Z, and prove that is also the centroid of A XYZ. 182. In ex. 179, prove that if the mid-points of the sides of A XYZ are joined, is also the centroid of that triangle ; and so on*. BOOK II. — EQUALITY OF POLYGONS. 1. THEOREMS. 140. Definitions. Two polygons are said to be adjacent if they have a segment of their perimeters in common. 141. Suppressing the common segment of the perimeters of two adjacent polygons, a polygon results which is called the sum of the two polygons. Simi- larly for the sum of several polygons, and for the difference of two overlapping polygons. 142. Surfaces which may be divided into the same number of parts respectively congruent, or which are the differences between congruent surfaces, are said to be equal. This property is often designated by the expressions equivalent, equal in area, of equal content, etc.; but the use of the word congruent, for identically equal, renders the word equal sufficient. The definition is more broadly treated in Book V. 143. The altitude of a trapezoid is the perpendicular distance between the base lines. Hence a trapezoid can have but one alti- tude, a, unless it becomes a parallelogram. 144. The altitude of a triangle with reference to a given side as the base, is the distance from the opposite ver- tex to the base line. Hence a triangle can have three distinct altitudes, viz. a u a 2 , a 3 , in the figure. 90 Prop. I.j EQUALITY OF POLYGONS. 91 Proposition I. 145. Theorem. Parallelograms on the same base or on equal bases and between the same parallels are equal. Given UJ ABCD, ABC'D', on the same base AB, and between the same parallels P. P'. To prove that O ABCD = O A BCD'. Proof. 1. AD = BC, AD' = BC. DC = AB = D'C. Why ? 2. In Fig. 1, adding CD'. DD' = CC. Ax. 2 3. .'.ABC'C^AAD'D. Why? 4. But ABC'D = ABC'D. . ' . O AB CD = O AB C ! D. Ax. 3 Similarly for Figs. 2 and o. In Fig. 2, CZ)' has become zero ; in Fig. 3, it has become negative. The meaning of •• between the same parallels " is apparent. Corollaries. 1. A parallelogram equals a rectangle of the same base and the same altitude. (Why ?) 2. Parallelograms hawing equal bases and equal altitudes are equal. (Why ?) 3. Of two parallelograms having equal altitudes, that is the greater ivhich has the greater base ; and of two having equal bases, that is the greater which has the greater altitude. (Why ?) 4. Equal parallelograms on the same base or on equal bases have equal altitudes. Law of Converse, § 73, after cors. 2 and 3. Give it in full. 92 PLANE GEOMETRY. Proposition II. [Be. il 146. Theorem. Triangles on the same base or on equal bases and between the same parallels are equal. Given A ABC, ABC on the base AB, and between the same parallels AB, C'C. To prove that A ABC = A ABC. Proof. 1. In the figure, suppose AD II BC, BD' II AC. Then ABCD, ABD'C are equal UJ. Why ? 2. And, since A ABC, ABC are their halves, I, prop. XXIV .\AABC=AABC. Ax. 7 Corollaries. 1. A triangle equals half of a parallelogram, or half of a rectangle, of the same base and the same altitude as the triangle. By step 2, and prop. I, cor. 1. 2. Triangles having equal bases and equal altitudes are eq ual. 3. Of two triangles having equal altitudes, that is the greater which has the greater base ; and of two having equal bases, that is the g?*eater ivhich has the greater altitude. (Why ?) 4. Equal triangles on the same base or on equal bases have equal altitudes. (Why ?) Note. In props. I and II if the figures are on equal bases they can evidently be placed on the same base. Hence the proofs given are sufficient. Prop. III.] EQUALITY OF POLYGOXS. 93 Proposition III. 147. Theorem. A trapezoid is equal to half of the rect- angle whose base is the sum of the two parallel sides, and whose altitude is the altitude of the trapezoid. b \i L B D Given the trapezoid ABCD. To prove that ABCD equals half of a rectangle with the same altitude, and with base equal to AB -f- DC. Proof. 1 . About 0, the mid-point of B C. revolve AB CD through 180° to the position A'CBD', leaving its original trace. 2. Then, '■' Zc' = Z c, and Zb + Zc = st. Z, .-. Z b + Z c' = Z b + Z c = st. Z, and . • . ABD' is a st. line. § 14, clef. st. Z . Similarly, DC A' is a st. line. 3. Also, v ZD' = ZD, and ZA + ZD=st. Z. .'.ZA + ZD' = ZA + ZD = st. Z, .'. D'A' li AD, I, prop. XVI, cor. 2 and .-. ADA'D is a O. § 97, def. O 4. The base of the O = AB + DC, and the O = 2 • ^CD. Why ? 5. .'. ^LBOZ> = £ O = £ required □. Prop. I, cor. 1 Exercises. 183. P is any point within O ABCD. Prove that A PAB + A PCD = £ O ABCD. Suppose P is outside of LJ ABCD. 184. A quadrilateral equals a triangle of which two sides equal the a \ diagonals of the quadrilateral, and the included angle of those sides 2 x-y C r A B JL For, in the above figure, AP = x 2 . BH = y 2 , PD = xy, CM=xy, and AC = (x - y) 2 . But A C = AP + BH - PD- CH, yy + //- - 2 xy. The truth of the corollary is, however, evident from prop. VI, if the agreement as to signs is considered ; for if y becomes 0, then 2 xy = 0, and y 2 = ; and as y passes through and becomes negative, 2xy also becomes negative, but y 2 remains positive because it is the rectangle of - y and - y. Exercises. 198. If ABC MNA is the perimeter of any polygon, and P is any point in the plane, then A PAB + APBC + + A PMN + A PNA is constant. 199. If A, B. C, I) are four collinear points, in order, then A B • CI) -f AD • BC = AC • Bl). (Euler.) Investigate when B moves to and through C. 100 PLANE GEOMETRY. [Bk. II. Proposition VII. 154. Theorem. The difference of the squares on two lines equals the rectangle of the sum and difference of those lines. x-y x O» F y y 2 y y Given ABCD, a square on x, and AEFG, a square on y. To prove that !/ 2 = ( x + U) ( x - y)- Proof. 1. Suppose the squares placed as in the figure, and GF produced to BC. Then the fs in the figure are all Def. □ equal, as also the sides of x 2 . 2. .'. the (x — y)'s are equal. 3. But or 4. .-.x 2 x-y* X' Ax. 3 Ax. 8 Prop. V Ax. 3 f 2 + x(x-y)+ y(x-y), y 2 + (x + tj)(x-y). f= (x+y)(x-y)- Corollary. If a point divides a line internally or externally into two segments, the rectangle of the segments equals the differ- ence of the square on half the line and the square on the seg^ ment between the mid-point of the line and the point of division. 1. If AB is the line, P (either P x or P 2 ) the point of divi- sion, and M the mid-point, it is to be proved that 1— , , A p B P a AP ■ PB = AM 2 - MP 2 . 2. Let AB = y, AP = x, then PB = y - x, AM = \ y, and MP = x-\y. ' 3. But x(y ->=(!)•-( , by prop. VII. Sec. 155.] EQUALITY OF P0LY90XS. 101 155. Reciprocity between Algebra and Geometry. From props. V. VI, VII, it is evident that a reciprocity exists between algebra and geometry which is likely to be of great advantage to each. This reciprocity will be more clearly seen by resort- ing to parallel columns. Geometric Theorems. If x, y, are line-segments, and xy, xz, represent the rectangles of x and y, x and z, , and x(y-\-z) represents the rectangle of x and y + z, and x 2 represents the square on x, then 1. x(y + z)= xy-\-xz. Prop. V 2. (x + yf=x 2 + y 2 + 2xy. Prop. VI Algebraic Theorems. If a, b, are numbers, and ab, ac represent the products of a and b, a and r, , and a(b + e) represents the product of a and b + e, and a 2 represents the second power of a, then 1. a (b + e) = ab + ac. 2. (a + by- = « 2 + b 2 + 2ab. 3. x 2 = ± Prop. VI, cor. 1 4. (x — y) 2 = x 2 -f y 2 — 2 xy. Prop. VI, cor. 2 5. x 2 — y 2 =(x-\-y)(x — y) . Prop. VII "-= <;. Const. 2 Ax. 1 Exs. 215-2:30.] PROBLEMS. HI EXERCISES. 215. If one angle of a triangle is two-thirds of a straight angle, show that the square on the opposite side equals the sum of the squares on the other two sides, together with their rectangle. 216. Prove prop. XI for a concave quadrilateral. 217. If ZP = 180° and SP = PQ. show that prop. XI reduces to a previous theorem. 218. Prove prop. XI. cor. directly from prop. X without reference to "% prop. XL 219. If ABCD is any quadrilateral, and the mid-points of the diagonals ^ "are joined by a line bisected at -If, and if P is any point, then PA 2 + PB 2 + PC 2 - PD- = MA* + MB 2 + M& + ML 2 + 4 PM*. 220. To construct a parallelogram equal to a given triangle, and having one of its C D E "angles equal to a given angle. V^v / / 221. To construct a parallelogram equal \ \^s/ / / to a given square, on the same base and V___y_^Ny X n having an angle equal to half the angle of A M B the square. 222. To construct an isosceles triangle equal to a given triangle, and on the same base. 223. To construct a triangle equal to a given parallelogram, and having one of its angles equal to a given angle. 224. To construct a parallelogram equal to a given triangle, and having its perimeter equal to that of the triangle. (In the figure of ex. 220 how must 3ID compare with BC + CA ?) 225. To construct a square equal to the sum of two given squares. (Apply prop. VIII.) 226. On a given line to construct a rectangle equal to a given rectangle. 227. On one side of a triangle as a diagonal to construct a rhombus equal to the given triangle. 228. Prove that in any triangle three times the sum of the squares on the sides equals four times the sum of the squares on the three medians. 229. Also that three times the sum of the squares on the lines joining the centroid to the vertices equals the sum of the squares on the sides. 230. If one angle of a triangle is one-third of a straight angle, show that the square on the opposite side equals the sum of the squares on the other two sides less their rectangle. 112 PLANE GEOMETRY. [Bk. II, 3. PRACTICAL MENSURATION. 164. For practical purposes a surface is measured as follows : 1. A square unit is defined as a square which is one linear unit long and one linear unit wide. That is, a square inch is a square that is 1 in. long and 1 in. wide ; a square meter is a square that is 1 m. long and 1 m. wide, etc. In the figure the shaded square is considered as a square unit. 2. If two sides of a rectangle are 3 in. and 5 in. respec- tively, then, in the figure, the area of the strip AB is 5 X 1 sq. in., and the total area is 3 X 5 X 1 sq. in., or 15 sq. in. ^ B Theoretically, a rectangle rarely has sides Jj~ — — — — both of which exactly contain any linear unit, however small. Such cases are discussed in Book IV. But for practical purposes the above method is approximate to any required degree. At present it is necessary for the student to learn that geometry gives him an instrument for practical work. It will accordingly be assumed that the measurements can be made to any degree of approximation, and that the expressions area, measure, etc. , are understood in their ordinary sense. It has already been explained that the rectangle of two lines corresponds to the product of two numbers ; hence, in practice, lines ar be subtended by their com- mon chord. In the figure. BI) and I)B are each said to be subtended by chord BI). The word suhtend is variously used in geometry. It means to extend under or to be opposite to. Hence in a triangle a side is said to subtend an opposite angle, a chord is said to subtend an arc, etc. 177. A portion of a circle cut off by an arc and two radii drawn to its extremities is called a sector, and the central angle standing on that arc is called the angle of the sector. In the figure. OAB is a sector, and LAOB is its angle. 116 PLANE GEOMETRY. [Bk. III. 1. CENTRAL ANGLES. Proposition I. 178. Theorem. In the sa?ne circle or in equal circles, if two central angles are equal, the arcs on which they stand are equal also, and of two unequal central angles the greater stands on the greater arc. Given M, M', two equal circles, and central angles AOB = A'O'B', AOOA'O'B'. To prove that AB = A^B', and AC > jVB'. Proof. 1. Place O M ' on O M so that Z A'O'B' coincides with its equal Z A OB. Then A' coincides with A, and B' with B. § 165, def. 2. Then A'B' coincides with AB, because its points are equidistant from 0. § 165, def. O 3. Also, v ZAOOZ A'O'B', .-.ZAOO ZAOB. 4. .*. C is not in Z A OB, and AC > IB. Ax. 8 5. And v AB = A T B', .'. AC > A r B'. Ax. 9 The proof is essentially the same for a single circle, and so in general when equal circles are involved. Prop. I.] CENTRAL ANGLES. 117 Corollaries. 1. Sectors of the same circL or of equal circles, which have equal angles,' are equal. For, by steps 1, 2, they coincide.. 2. Sectors of the same circle, or of equal circles, which have unequal angles, are unequal, the greater having the greater angle. This is proved by superposition in steps 3. 4. 5, of the proposition. 3. The two arcs into which the circumference is divided by a diameter are equal . For their central angles are straight angles, and these being equal the arcs are equal by the proposition. 4. The two figures into which a circle is divided by a diam- eter are equal. For their central angles are straight angles. Hence by cor. 1 they are equal. This corollary is attributed to Thales. 179. Definition. The figure formed by a semicircumference and the diameter joining its extremities is called a semicircle. It is proved (cor. 4) that all semicircles, cut from the same circle, are equal. Hence the name, semi- meaning half. 180. Since the 360 equal angles, into which the perigon at the center of a circle is imagined to be divided, stand on equal arcs by prop. I, the ordinary mensuration of angles by degrees is also used for arcs. Similarly for minutes, seconds, and other measurements. Hence the common expression, an angle at the center is measured by the subtended arc. The expression is not strictly correct ; we do not measure an angle by an arc, but the angle and arc have the same numerical measure, as will be proved in § 254. We might as truly say that an arc is measured by its central angle. But the expression is so commonly used, and has found its way into so many text-books and examination papers, that the student needs to become familiar with it. 118 PLANE GEOMETRY. [Bk. III. Proposition II. 181. Theorem. In the same circle or in equal circles, if two arcs are equal, the central angles which they subtend are equal also, and of two unequal arcs the greater subtends the greater central angle. Proof. If and 0' are two central angles, and A, A' are the arcs on which they stand, it has been proved in prop. I that If > 0', then A > A', « = 0', " A = A', « < 0', « A< A'. Hence the converses are true, by the Law of Converse, § 73. Corollaries. 1. In the same circle or in equal circles, equal sectors have equal angles; and of two unequal sectors, the greater has the greater angle. Law of Converse, § 73, from prop. I, cors. 1, 2. 2. A central angle is greater than, equal to, or less than, a right angle, according as the arc on which, it stands is greater than, equal to, or less than, a quadrant. (Why ?) Exercises. 248. If two lines drawn to a circumference, from a point within the circle, are equal, they subtend equal central angles. 249. Prove the converse of ex. 248. 250. Two circumferences cannot bisect each other. 251. Suppose from the point P on a circumference two equal chords, PA, PB, are drawn. Prove (1) that these chords subtend equal central angles, (2) that they subtend equal arcs. 252. The arc AB is bisected by the point M, and MC is a diameter ; prove that chord AC = chord BC. 253. How many degrees in the central angle standing on a third of a circumference? a fourth ? a fifth ? Prop. III.] CHORDS AND TANGENTS. 119 2. CHORDS AND TANGENTS. Proposition III. 182. Theorem. In the same circle or in equal circles, if two arcs are equal they are subtended by equal chords, and of two unequal minor arcs the greater is subtended by the greater chord. Given two equal circles, M, M' ; two equal arcs, K, K' ; and two unequal minor arcs, K > K". To prove that, as lettered in the figure, chords AB = A'B', AB > CA'. Proof. 1. Draw the radii OA, OB, O'A', O'B', O'C. Then v K= K', .'. Z AOB = Z A'O'B'. Prop. II 2. But v OM=OM', .-. OA = OB= O'A' = O'B' = O'C. 3. .'.A OAB ^ A O'A'B', and AB = A'B'. Why ? 4. Also, V K> K", r.A AOB > Z CO' A', Prop. II .-. AB > CA'. I, prop. X Corollary. In the same circle or in equal circles, of two unequal major arcs, the greater is subtended by the less chord. 120 PLANE GEOMETRY. [Bk. III. Proposition IV. 183. Theorem. In the same circle or in equal circles, if two chords are equal they subtend equal major and equal minor arcs ; and of two unequal chords the greater subtends the greater minor and the less major arc. Proof. Let C, C be two chords of the same circle or of equal circles ; JVJ N' their corresponding minor arcs ; J } J' " " major arcs. From prop. Ill, if N > N', or if J < J\ then C > C, '< N=N', " « J=J', « C = C, " JYJ', " C Z P l OM, .'.AO>P 1 0, I, prop. XX and .-. P x is within the O. § 108, def. O, cor. 3 3. And v Z MOP, > Z MOB, .-. P 2 0> BO, I, prop. XX and .*. P 2 is without the O. § 108, def. O, cor. 3 Corollary. A straight line cannot meet a circumference in more than two points. For every other point on that line must be either between or not between those two points, and hence must lie either within or without the circle. Exercises. 262. Prove that, in general, two chords of a circle can- not bisect each other. What is the exception ? 263. What is the locus of the mid-points of a pencil of parallel chords of a circle ? Why ? Prop. VII. CHORDS AND TANGENTS. 123 Proposition VII. 186. Theorem. In the same circle or in equal circles, equal chords are equidistant from the center; and of tiro unequal chords the greater is nearer the center. Given two equal © M. M', with chords AB = A'B", AE > A'B', and OC, 01), O'C _L's from center to AB, AE. and from center O 1 to A'B'. To prove that (1) OC=0'C, (2) OD < O'C Proof. 1. C, C bisect AB, A'B', Prop. V AC = A'C", being halves of equal chords. Ax. 7 OA = O'A' Draw OA, O'A'; then and ZC = Z C", Prel. prop. I .'. A A CO ^ A A' CO', I. prop. XIX, cor. 5 and OC = O'C, which proves (1). 3. And v AE > A'B', then AE > AB, which equals A'B'. Ax. 9 4. .'. minor AEE > AFB, so that E does not lie on AEB. Prop. IV 5. And V and AB are on opposite sides of AE, .-. OC cuts AE, as at G, and OD < OG. I, prop. XX 6. And v OG < OC, r. OD < OC. Ax. 9 7. And V OC = O'C, .-. 0/> < O'C". Ax. 9 124 PLANE GEOMETRY. [Bk. III. Proposition VIII. 187. Theorem. In the same circle or in equal circles, chords that are equidistant from the center are equal ; and of two chords unequally distant, the one nearer the center is the greater. Proof. If c, c' are two chords of the same circle or of equal circles, and d, d' are the respective perpendiculars from the center upon them ; then from prop. VII, If c > c', then d < d', " c = c', " d = d', " c < c', " d > d'. Hence the converses are true by the Law of Converse, § 73. Corollary. The diameter is the greatest chord in a circle. For its distance from the center is zero. Exercises. 264. AB is a fixed chord of a circle, and XY is any other chord having its mid-point P on AB. What is the greatest and what is the least, length that XY can have ? 265. What is the locus of the mid-points of equal chords of a circle ? 266. Two parallel chords of a circle are 6 inches and 8 inches, respec- tively, and the distance between them is 1 inch. Find the radius. 267. Two chords are drawn through a point on a circumference so as to make equal angles with the radius drawn to that point. Prove that the chords subtend equal arcs. 268. If from the extremities of any diameter perpendiculars to any secant are drawn, the segments between the feet of the perpendiculars and the circumference will be equal. Draw the various figures. 269. If two equal chords of a circle intersect, the segments of the one are equal respectively to the segments of the other. 270. Find the shortest chord which can be drawn through a given point in a circle. 271. The circumference of a circle whose center lies on the bisector of an angle cuts equal chords, if any. from the arms. Prop. IX.] CHORDS AND TANGENTS. 125 Proposition IX. 188. Theorem. Of all lines passing through a point on a circumference, the perpendicular to the radius drawn to that point is the only one that does not meet the circumference again. Given point P on the circumference of a O with center 0, and AB, PC, respectively perpendicular and oblique to OP at P. To prove that AB does not meet the circumference again, but that PC does. Proof 1. Let 031 A. PC, and OX be any oblique to AB. Then 031 < OP, I, prop. XX and .. 31 is within the O, and PC cuts the circum- ference again. §§ 108, 109 2. Also, OX > OP, Why ? and .". X, any point except P on AB. is without the O. § 108, def. O, cor. .3 3. .'. the perpendicular does not meet the circumference again, but an oblique does. 189. Definitions. The unlimited straight line which meets the circumference of a circle in but one point is said to touch, or be tangent to, the circle at that point. The point is called the point of contact, or point of tangency, and the line is called a tangent. 126 PLANE GEOMETRY. [Bk. III. A tangent from a point to a circle is to be understood as the segment of the tangent between the point and the circle. If the two points in which a secant cuts a circumference continually approach, the secant approaches the condition of tangency. Hence the tangent is sometimes spoken of as a secant in its limiting position. Corollaries. 1. One, and only one, tangent can be drawn to a circle at a given point on the circumference. For the tangent is perpendicular to the radius at that point, and there # is only one such perpendicular. (Has this been proved ?) 2. Any tangent is perpendicular to the radius drawn to the point of contact. (Why?) 3. A line perpendicular to a radius at its extremity on the circumference is tangent to the circle. (Why?) 4. The center of a circle lies on the perpendicular to any tangent at the point of contact. For the radius to that point is perpendicular to the tangent, and as there is only one such perpendicular at that point (prel. prop. II), that perpendicular must be the radius. 5. The perpendicular from the center to a tangent meets it at the point of contact. For the radius to that point is perpendicular to the tangent, and there is only one perpendicular from the center to the tangent. Exercises. 272. Show that of these three properties of a line, (1) the passing through the center of a circle, (2) the being perpendicular to a given chord, (3) the bisecting of that chord, any two in general necessi- tate the third. In what special case is there an exception ? 273. If a chord is bisected by a second chord, and the second by a third, and the third by a fourth, and so on, the points of bisection approach nearer and nearer the center. 274. Tangents drawn to a circle from the extremities of a diameter are parallel. 275. The diameter of a circle bisects all chords which are parallel to the tangent at either extremity. Prop. X.] CHORDS AND TANGENTS. 127 Proposition X. 190. Theorem. An unlimited straight line cuts a circum- ference, touches the circle, or does not meet the circle, accord- ing as its distance from the center of the circle is less than, equal to, or greater than, the radius. Given OA, OB, OC, the perpendiculars from center of O M, to lines S, T, X, and respectively less than, equal to, greater than, the radius. To prove that S is a secant, T a tangent, N a line not meet- ing M. Proof. 1. A, B, C are respectively within the O, on the circum- ference, or without the O. § 108, def. O, cor. 3 *2. .'. Sis a secant. § 109, 2 3. And T is a tangent. Prop. IX, cor. 3 4. N II T. I, prop. XVI, cor. 3 5. And .". jV cannot meet O M because it cannot cross T. Corollary. The converses are true. Let the student state this corollary in full, and show that the Law of Converse (§ 73) applies. Exercises. 276. What is the locus of the extremities of equal tan- gents drawn from points on a circumference ? 277. Two tangents meet at a point the length of a diameter distant from the center of the circle. How many degrees in their included angle ? * 128 PLANE GEOMETRY. [Bk. III. 3. ANGLES FORMED BY CHORDS, SECANTS, AND TANGENTS. 191. Definitions. A segment of a circle is either of the two portions into which the circle is cut by a chord. If a segment is not a semicircle, it is called a major or a minor segment according as its arc is a major or minor arc. E.g. BBC is a minor segment, and BDE is a major segment. The fact that the word segment is used to mean a part of a line, and also a part of a circle, will not present any difficulty, since the latter use is rare, and the sense in which the word is used is always evident. It means "a part cut off," and is therefore applicable to both cases. 192. The angle, not reflex, formed by two chords which meet on the circumference is called an inscribed angle, and is said to stand upon, or be subtended by, the arc which lies within the angle and is cut off by the arms. It is also called an angle inscribed in, or simply an angle in, the segment whose arc is the conjugate of the arc on which it stands. /.ABB is an inscribed angle, standing on AB ; it is also an angle in the segment BCBEA. Similarly. ABBA is in the segment BAG and stands on BA. 193. Points lying on the same circumference are called coney clic. Exercises. 278. If from the extremities of any chord perpendiculars to that chord are drawn, they will cut off equal segments measured from the extremities of any diameter. (Draw a perpendicular from the center to the chord.) 279. If a tangent from a point B on a circumference meets two tan- gents from A, C, on the circumference, in points X, Y ; and if the lines joining the center to A, X, Y, C, are a, x, ?/, c, respectively, then Zxy = /ax -\- / ye, and XY = AX + YC. Prop. XI.] CHORDS AND TANGENTS. 129 Proposition XI. 194. Theorem. An inscribed angle equals half the central an; jle standing on the same arc. Fig. 2. Fio. 3. Given AVB an inscribed angle, and AOB the central angle on the same arc AB. To prove that Z AVB = £ Z A OB. Proof. 1. Suppose VO drawn through center 0, and produced to meet the circumference at X. Then Z XVB = Z VBO. I, prop. Ill 2. And Z XOB = Z XVB + Z FJ50, Why ? = 2 Z XTO. Step 1 3. r.A.XVB = \£XOB. Ax. 7 4. Similarly Z .! FA = JZ A OX (each = zero in Fig. 2), and .•.ZJF5 = iZJ05. Ax. 2 The proof holds for all three figures, point A having moved to X (Pig. 2), and then through X (Pig. 3). 195. The theorem is often stated thus : An inscribed angle is measured by half its intercepted arc This expression, like that mentioned in § 180 is not strictly correct. The angle and the arc simply have the same numerical measure as proved later in § 254. 130 PLANE GEOMETRY. [Bk. IT1. Corollaries. 1. Angles in the same segment, or in equal segments, of a circle are equal. (Why ?) 2. If from a point on the same side of a chord as a given segment, lines are drawn to the ends of that chord, the angle included by those lines is greater than, equal to, or less than, an angle in that segment, according as the point is within, on the arc of, or without, the segment. This follows from cor. 1 and from I, prop. IX. Draw the figure and prove. 3. The converse of cor. 2 is true by the Law of Converse. Hence the locus of the vertex of a constant angle whose arms pass through two fixed points is an arc. Let the student state the converse in full, and give the proof. Exercises. 280. In the figures on p. 129, prove that if P is taken ' anywhere on BY, then Z PBV + Z BVP is constant. 281. In Fig. 3, p. 129, if BO is produced to meet the circumference at W, and the point of intersection of BW and AV is called F, prove that A YVB and "MM Fare mutually equiangular. 282. What is the locus of the vertex of a triangle on a given base and with a given vertical angle ? Prove it. 283. In Fig. 1, p. 129, suppose A to move freely on the arc VAXB, and suppose A AVB, VBA bisected by lines meeting at P. Show that the locus of P is a constant arc. 284. If the vertices of a hexagon are concyclic, the sum of any three alternate interior angles is a perigon. (That is, the sum of three angles, taking every other one.) 285. Two equal chords with a common extremity are symmetric with respect to the diameter through that extremity, as an axis ; so also are their corresponding arcs. 286. If from any point P. on the diameter AB, PX and PY are drawn to the circumference on the same side of AB and making ZXP^l = ZBPY. then & APX and YPB are mutually equiangular. 287. If any number of triangles on the same base and on the same side of it have equal vertical angles, the bisectors of the angles are concurrent. 288. Prove that two chords perpendicular to a third chord at its extremities are equal. Prop. XII.] CHORDS AND TANGENTS. 131 Proposition XII. 196. Theorem. An angle in a segment is greater than, equal to, or less than, a right angle, according as the segment is less than, equal to, or greater than, a semicircle. Given the segments ABE, ACE, ABE of a circle with center 0, respectively less than, equal to. greater than, a semicircle. To prove that A AEB, AEC, AEB are respectively greater than, equal to, less than, a right angle. Proof. 1. Draw OB, OB. Then v Z AEB = \ reflex AAOB, Prop. XI .'./.AEB > rt. Z. ' Z AEC = i st. Z AOC, Why ? ..ZAEC= it. Z. Z A EB = \ oblique ZAOB, Why ? .-.A AEB < rt, Z. 2. And And Corollary. A segment is less than, equal to, or greater than, a semicircle, according as the angle in it is greater than, equal to, or less than, a right angle. From prop. XII, by the Law of Converse, § 73. Let the student write out the proof. Note. The discovery that an angle in a semicircle is a right angle is attributed to Thales, who, tradition asserts, sacrificed an ox to the gods in honor of the event. 132 PLANE GEOMETRY. [Bk. III. Propositiox XIII: 197. Theorem. An angle formed by a tangent and a chord of a circle equals half of the central angle standing on the intercepted arc. C Given AB a chord, XX' a tangent through A, and the center of the circle. To prove that Z XAB = i Z A OB, and ZBAX' = iZBOA. Proof. 1. Produce AO to meet the circumference at C. Then A XAC =\AAOC, = \ st. Z. Why ? 2. And •.' ZBAC =±ZBOC, Why? .-. Z XAB = i Z AOB. Ax. 3 3. Also, V Z C^X' = | Z COJ, Why ? .-. Z #4X' = }Z 1*0.4. Ax. 2 Corollary. Tangents to a circle from the same external jjoint are equal. For, connect the points of tangency, and two angles of the triangle are equal by this theorem. 198. The theorem is often stated thus: An angle formed by a tangent and a chord of a circle is measured by half its inter- cepted arc. See § 195. &Y Prop. XIV.] CHORDS AND TANGENTS. L33 Proposition XIV. 199. Theorem. An angle formed by two unlimited inter- secting lines which meet the circumference equals either the sum or the difference of half the central angles on the inter- cepted arcs, according as the point of intersection is within or without the circle. Fig. 1. Fig. 2. Given two lines XX', YY' meeting a circumference at A, A' and B, B', respectively, and intersecting at P. To prove that Z A'PB' equals half the central angle on A'B' plus or minus half that on AB, according as P is within or without the circle. Proof. Suppose AB' drawn. Then Z A'PB' = Z A'AB' ± Z AB'B. § 88 == J cent, Z on AB' ± | cent. Z on AB. Prop. XI The theorem is thus re-stated for two of the special cases : Corollaries. 1. An angle formed by two chords equals the sum, of half the central angles on the intercepted arcs. See Fig. 1. (State this as suggested in § 195.) 2. An angle formed by two secants intersecting without tin- circle equals the difference of half the central angles on the intercepted arcs. See Fig. 2. (State this as suggested in § 195.) 134 PLANE GEOMETRY. [Bk. III. Prop. XIV is, of course, true for tangents as well as chords and secants. The following figures represent special cases. Fig. 3. Fig. 4. Fig. 5. Fig. 6. Fig. 7. Fig. 3 is a special case where P is at 0, and merely affirms that a central angle equals itself. Fig. 4 shows that prop. XI is a special case of prop. XIV. Fig. 6 shows the same for prop. XIII. Corollary. 3. An angle formed by a secant and a tan- gent, or by two tangents, equals the difference of half the central angles on the intercepted arcs. See Figs. 7 and 8. Prop. XV.] CHORDS AND TANGENTS. 135 Proposition XV. 200. Theorem. If two parallel liar* intercept arcs on a circumference, those arcs are equal. Given two parallel lines. I and II. intercepting arcs AB. ' ' I>. on the circumference of a circle with center 0. To prove that AB = CI). Proof. 1. Suppose YOY _L I. and to cut BC at 31 Fig. 1. Then YY' ± II. I. prop. XVII, cor- 1 2. And BM = MC, and AM= JIB. Prop. V 3. .'.AB=CD. Ax. 3 Note. The proof is the same for Figs. 2. 3 ; in Fig. 2. BC equals zero; and in Fig. 3, DA also equals zero. It should be noticed that Figs. 1, 2, 3, respectively, may be considered as special, or at least as limiting cases of Figs. 2. 7. and 8 of prop. XIV. In prop. XIV as P moves farther and farther to the right the lines come nearer and nearer to being parallel, the angle A PB approaches nearer and nearer zero, and hence the cen- tral angles on arcs BA, A'B' approach nearer and nearer equality. It might therefore be inferred that when the lines become parallel, the arcs become equal, as proved in prop. XV. Exercise. 289. The chords which join the extremities of two equal arcs are either parallel, or else they intersect and are equal and cut off equal segments from each other. 136 PLANE GEOMETRY. [Bk. III. INSCRIBED AND CIRCUMSCRIBED TRIANGLES AND QUADRILATERALS. 201. Definitions. If the ver- tices of the angles of a poly- gon lie on a circumference, the polygon is said to be in- scribed in the circle, and the circle is called a circumscribed circle. Inscribed quadrilateral. Circumscribed circle. If the lines of the sides of a polygon are tangent to a circle, the polygon is said to be circumscribed about the cir- cle, and the circle is called an inscribed or escribed circle, according as it lies within or without the polygon. Circumscribed quadrilateral. Inscribed circle. Inscribed cross quadrilateral. Circumscribed circle. Circumscribed quadrilateral. Escribed circle. The words inscriptible, circumscriptible, escriptible mean capable of being inscribed in, circumscribed about, escribed to, a circle. Exercise. 290. Tf any two chords cut within the circle, at right angles, the sum of the squares on their segments equals the square on the diameter. Prop. XVI.] VIBCLE8 AND POLYGONS. L37 Proposition XVI. 202. Theorem. A circumference can be described to pass through the three vertices of any triangle. (Circumscribed circle.) c Given the points A, B, C, the vertices of A ABC. To prove that a circumference can be described to pass through A, B, C. Proof. 1. There is such a circumference. $ 131, cor. 2 2. And the center of the O can be found. § 131, cor. 1 Note. The relation between prop. XVI and prop. XVII should be noticed. Similarly for props. XVIII and XIX, and for XX and XXI. Exercises, 291. Prove from prop. XVI and prop. XI that the sum of the interior angles of any triangle equals a straight angle. 292. If the hypotenuse of a right-angled triangle is the diameter of a circle, the circumference passes through the vertex of the right angle. (Corollary. The median from the vertex of the right angle of a right- angled triangle equals half of the hypotenuse.) 293. A line-segment of constant length slides so as to have its extremi- ties constantly resting on two lines perpendicular to each other. Find the locus of its mid-point. 294. If a circle is described on the line joining the orthocenter to any vertex, as a diameter, prove that the circumference passes through the feet of the perpendiculars from the other vertices to the opposite sides. 295. Prove that the perpendiculars from the vertices of a triangle to the opposite sides bisect the angles of the triangle formed by joining their feet ; the so-called Pedal Triangle. 138 PLANE GEOMETRY. [Bk III. Proposition XVII. 203. Theorem. A circle can be described tangent to the three lines of any triangle. {Inscribed and escribed circles.) t Given the lines a, b, c, forming a A ABC. To prove that a circle can be described tangent to a, b, c. Proof. 1. Let be the in-center, O x , 2 , 3 the ex-centers. Let OP, OQ, OB± a, b, c. Then A AB 5£ A A Q 0, and ABBO^ABPO. I, prop. XIX, cor. 7 2. .-. OQ = OB= OP. Why ? 3. .'. P, Q, B are concyclic § 108, def. O, cor. 3 4. And V AB _L OB, AB is a tangent. Prop. IX, cor. 3 Similarly, a, b, c, are tangent to the other three (D. Corollary. A circle can be described tangent to three lines not all parallel nor concurrent. Prop. XVIII.] CIRCLES AND POLYGONS. 139 Proposition XVIII. 204. Theorem. In an inscribed quadrilateral the swn or difference of two opposite angles equals the sum or differ- ence of the other two opposite angles, according as the quad- rilateral is convex or cross. Given the inscribed convex quadrilateral ABCD. To prove that in Fig. l,Zi + ZC = Z5 + ZD. Proof for Fig. i. 1. Z A + Z C = \ central Z on 1W + DB = st. Z. Prop. XI 2. Similarly, Z B + Z i> = st. Z. 3. ,'.Zi + Z(J = Z5 + ZJ). § 30 Proof for Fig. 2. If the quadrilateral is cross, Z. C — Z. A = Z.D — /LB, since each equals zero. Why ? Corollaries. 1. A parallelogram inscribed in a circle has all of its angles equal, and is therefore a rectangle. (Why ?) 2. The opposite angles of an inscribed convex quadrilateral are supplemental. Exercises. 296. In the figure of prop. XIII, if P is the mid-point of arc AB, prove that P is equidistant from AX and AB. Suppose the arc BCA is taken, instead of AB. 297. If a circle is described on one side of a triangle as a diameter, prove that the circumference passes through the feet of the perpendiculars drawn to the other two sides from the opposite vertices. 140 PLANE GEOMETRY [Bk. III. Proposition XIX. 205. Theorem. In a circumscribed quadrilateral the sum or difference of two opposite sides equals the sum or differ- ence of the other two opposite sides, according as the quadri- lateral is convex or cross. b, Fig. 1. Fig. 2. Given the circumscribed convex quadrilateral abed. To prove that in Fig. 1, a + c = b + d. Proof for Fig. z, as lettered. 1. a x = d 2 , a 2 = b x , c x = b 2 , c 2 = d x . Prop. XIII, cor. 2. .'. a t + a 2 + c x + c 2 — b x + b 2 + r/ x + d 2 . Ax. 2 .3. Or, a'-\-c = b + d. Ax. 8 Proof for Fig. 2. If the quadrilateral is cross, c — a = d — b. 1. v c x = b 2 , and c 2 = d x , .'. c = />. : + r^. 2. v a x = r/ 2 , and a 2 = b x , .'. a = b x + d 2 . 3. .' . r — a = d — b. Ax. 3 Corollary. ^4 parallelogram circumscribed about a circle has all of its sides equal, and is therefore a rhombus. (Why?) Exercises. 298. The bisector of an angle formed by a tangent and chord bisects the intercepted arc. 299. Given two pairs of parallel chords, ABWA'IV, and BCW B'C ; prove that AC' WA'C. Prop. XX.J CIRCLES AND POLYGONS. 141 Proposition XX. 206. Theorem. If the sum of two opposite angles of a quadrilateral equals the sum of the other two opposite angles, the quadrilateral is inscriptible. Given the quadrilateral ABCD such that Z.A + Z.C = £B + £D. To prove that ABCD is iuscriptible. Proof. 1. Suppose the circumference determined by A, B, C not to pass through D, but to cut CD at E. Prop. XVI Draw AE. Then £B + £ AEC = ZC + Z BAE. Prop. XVI I [ 2. But Z B + Z2> = Z C + A A, and .-. Z AEC - Z D = Z ^J^ - Z /I, or, Z ^JZ> = - Z £\m I, prop. XIX But this is absurd : hence step 1 is absurd. The proof is the same for D'. Corollary. If two opposite angles of a quadrilateral are supplemental, the quadrilateral is inscriptible. Exercises, 300. A square is inscriptible. 301. Every equiangular quadrilateral is inscriptible. 302. The intersection of the diagonals of an equiangular quadrilateral is the center of the circumscribed circle. 303. A circle is described on one of the equal sides of an isosceles triangle as a diameter. Prove that the circumference bisects the base. 142 PLANE GEOMETRY. [Bk. III. Proposition XXI. 207. Theorem. If the sum of two opposite sides of a quadrilateral equals the sum of the other two opposite sides, the quadrilateral is circumscriptible. D E p / A "B Given the quadrilateral ABCD such that AB + CD = BC + DA. To prove that ABCD is circumscriptible. Proof. 1. Suppose the O tangent to AB, BC, CD not to be tangent to DA, but to be tangent to EA. Prop. XVII Then AB + CE = BC + EA. Prop. XIX 2. But AB + CD = BC + DA, Given and .'. CD - CE, or ED, = DA- EA. Ax. 3 But this is absurd ; hence step 1 is absurd. I, prop. VIII, cor. The proof is the same for D'. Exercises. 304. A square is circumscriptible. (Notice the relation between exs. 300-302 and exs. 304-306.) 305. Every equilateral quadrilateral is circumscriptible. 306. The intersection of the diagonals of an equilateral quadrilateral is the center of the inscribed circle. 307. A', B" are the feet of perpendiculars from A . B on a. b in A ABC ; M is the mid-point of AB. Prove that Z B'A'M - Z MB' A' = AC. Sec. 208.T TW0 CIRCLES. 14^ 5. TWO CIRCLES. 208. Definitions. Two circles are said to touch or to be tan- gent when their circumferences have one, and only one, point in common. They are said to be internally or externally tangent according as one circle lies within or without the other. The more accurate expression, a tangent circumference, is often used instead of a tangent circle. The line determined by the centers of two circles is called their center-line ; the segment of the center-line, between the centers, is called their center-segment. If two circles have a common center, they are said to be concentric. The expression concentric circumferences is also used. Exercises. 308. A triangle is inscribed in a circle. Prove that the sum of three angles, one in each segment of the circle, exterior to the triangle, equals a perigon. 309. Prove that a perpendicular from the orthocenter of a triangle to a side, produced to the circumference of the circumscribed circle, is bisected by that side. 310. Prove that the bisectors of any angle of an inscribed quadri- lateral and the opposite exterior angle meet on the circumference. 311. If the diagonals of an inscribed quadrilateral bisect each other, what kind of a quadrilateral is it ? 312. Prove that if two consecutive sides of a convex hexagon inscribed in a circle are respectively parallel to their opposite sides, the remaining sides are parallel to each other. 313. Prove that the bisectors of the angles formed by producing the opposite sides of an inscribed quadrilateral to meet, are perpendicular to each other. (A proof may be based on cors. 1 and 2 of prop. XIV.) 314. Prove that if the diagonals of an inscribed quadrilateral are perpendicular to each other, the line through their intersection perpen- dicular to any side bisects the opposite side. (Brahmagupta's theorem.) 144 PLANE GEOMETRY. [Bk. III. Proposition XXII. 209. Theorem. If two circumferences meet in a 'point which is not on their center-line, then (1) they meet in one other point, (2) their center-line is the perpendicular bisector of their common chord, (3) their center-segment is greater than the difference and less than the sum of the radii. Fig. 1. Fig. 2. Given 31 and iV, two circumferences with centers A, B, meeting at P not on AB. To prove that (1) they meet again, as at P'; (2) AB _L PP' and bisects it, as at C\ (3) AB > the difference between AP and BP and < AP + BP. Proof. 1. In Fig. 1, suppose AABP revolved about AB as an axis of symmetry, thus determining A AP'B. Then V AP 1 = AP, and BP' = BP, .'. P' is on both M and N, which proves (1). § 108, def. O, cor. 3 2. In Fig. 2, V AP = AP, and BP = BP, § 109, 1 3. .'. A and B lie on the _L bisector of PP', which proves (2). I, prop. XLI 4. AB > the difference between AP and BP and < AP + BP, which proves (3). § 75 and cor. Corollary. If two circumferences meet at one point only^ that point is on their center-line. \ Why ?) Prop. XXIIL] TWO CIRCLE*. 145 Proposition XXIII. 210. Theorem. If two circles meet on their center-line, they are tangent. P 0' A Given and O', the centers of two circles with radii OA, 0'A } which meet on their center-line at A. To prove that the circles are tangent. Proof. 1. Let P be any point, other than A, on circumference with center 0, and draw OP, OP. Then 00' + O'P > OP or its equal OA. I, prop. VIII 2. And v 00'= OA - O'A, .-. OA- 0'A+ OP> OA, or OP > OA, by adding O'A and subtracting OA. Axs. 4. 5 3. .". P is without the circle with center 0'. § 108. def. O, cor. 3 4. And ".' the © have only one point in common, .'. they are tangent. § 208 Corollaries. 1. If two circumferences intersect, neither 'point of intersection is on the center-line. (Why?) 2. If two circles touch, they have a common tangent-line at the 'point of contact. For a perpendicular to their center-line at that point is tangent to both. (Why ?) Exercise. 315. Find the locus of the centers of all circles tangent to a given circle at a given point. y. 146 PLANE GEOMETRY. [Bk. 111. 6. PROBLEMS. Proposition XXIV. 211. Problem. To bisect a given arc. Solution. 1. Draw its chord AB. 2. Draw PCI AB at its rnid-point. Then PC bisects the arc. § 28 §§ 114, 116 Prop. V, cor. 2 Proposition XXV. 212. Problem. To find the center of a circle, given its circumference or any arc. Given a circumference, or an arc ABC. Required to find the center of the circle. Solution. 1. Draw two chords from B, as BA, BC. § 28 2. Draw their _L bisectors DD\ EE\ §§ 114, 116 intersecting at the center O. § 131, cors. 1, 4 Note. Hereafter it will be assumed that the center is known if an arc is known, for it may always be found by this problem. Prop. XXVI] TWO CIRCLES. 147 Proposition XXVI. 213. Problem. To draw a tangent to a given circle from a given point. 1. If the point is on the circumference. Solution. 1. At the given point erect a perpendicular to the radius drawn to the point. I, prop. XXIX 2. This is the required tangent, and the solution is unique. Prop. IX, cors. 3, 1 2. If t tie point is without tlie circle. Given a circle PP'B, with center 0; also an external point A. Required from A to draw a tangent to O PPB. Construction. 1. Draw AO. §28 2. Bisect AO at M. I, prop. XXXI 3. Describe a O with center 3L radius MO. § 109 4. Join A to intersections of circumferences. § 28 Then these lines, AP, AP', are the required tangents. Proof. 1. The circumferences will have two points in common, and only two. Prop. XXII; I. prop. XLIIL cor. 3 2. And v A APO, OP' A are rt. A, Why ? .'. AP, AP' are tangents. Why ? (Would this solution hold for case 1 ?) 148 PLANE GEOMETRY. [Bk. III. Proposition XXVII. 214. Problem. To draw a common tangent to two given circles. Fig. 1, Fig. 2. Given two circles A, B, with radii r, r' (r > ?•'), and centers 0, 0', respectively. Required to draw a common tangent to them. Construction. 1. Describe © A x , A % (Figs. 1 and 2), with centers 0, and radii /• — r' and r + >•', respectively. § 109 2. From 0' draw tangents 0'C\, 0'C SJ to © A x , A*. Prop. XXVI 3. Draw OC\, OC 2 , cutting circumferences A at E x , E 2 . § 28 4. Draw 0'D X II OE x , and O'D, II ^ a O. § IIS 5. Draw ^7)^ E 2 D 2 ; they are the tangents. Proof. 1. Zs C 1? C 2 are rt. A Why? 2. In Fig. 1, V C X E X = 0^ - 0C X = /• - (;• - r') = r', .'. C X E X = and II 0'D X . Const. 1, 4 3. .'. C x O'D x E x is a □, and A E x , 1\ are rt. A, I, props. XXV. XXIII. cor. and .'. D X E X is tangenl to CD A. /»'. Prop. IX. cor. 3 Prop. XXVII.] TWO CIRCLES. 149 Similarly, in Fig. 2, E 2 C 2 = and II D 2 0', and E 2 D 2 0'C 2 is a □ , and D 2 E 2 is a tangent. In both figures a second tangent can evidently be drawn, the solution being analogous to that above given. Hence there are four tangents in general. Note. The following special cases are of interest. Fig. 3. Fig. 4. (3 Fig. 5. Fig. In Fig. 3 the two circles have moved to external tangency, and the two interior tangents have closed up into one. In Fig. 4 the circumfer- ences intersect and the interior tangents have vanished. In Fig. 5 the circles have become internally tangent and the two exterior tangents have closed up into one. In Fig. 6 the circle B lies wholly within the circle A, and the tangents have all vanished. In all cases the center-line is evi- dently an axis of symmetry. Exercises. 316. All tangents drawn from points on the outer of two concentric circumferences to the inner are equal. 317. Find the locus of the centers of all circles touching two intersect- ing lines. (Show that it is a pair of perpendiculars.) Suppose the two lines were parallel instead of intersecting. 150 PLANE GEOMETRY. [Bk. III. Proposition XXVIII. 215. Problem. On a given line-segment as a chord to con- struct a segment of a circle containing a given angle. c )C m ^y \ 0' r Given the line-segment AB and the A N. Required on AB to construct a segment of a circle, containing AN. Construction. 1. Draw BD and AC, making A ABD, BAC equal to A N. I, prop. XXXII 2. Draw YY', the X bisector of AB. I, prop. XXXI 3. Draw J_'s to ^C, BD, from J, #. I, prop. XXIX These _L's will intersect YY' at the centers of the © whose segments on AB are required. Proof. 1. The two J_'s from A, B, meet YY', as at 0', 0. I, prop. XVII, cor. 4 2. is the center of O with chord AB and tangent BD. Prop. V, cor. 2 ; prop. IX, cor. 4 3. .'.A ABD, or Z N, = £ central Z on AEB, Prop. XIII = Z in segment AB Y. Prop. XI Similarly for segment Y'BA, where ABAC = £ central Z on the intercepted arc. Sec 216. TWO CIRCLES. 151 216. Definitions. Two intersecting arcs are said to form an angle, meaning thereby the angle included by their respective tangents at the point of intersection. An arc and a secant are said to form an angle, meaning thereby the angle included by the secant and the tangent to the arc at the point of meeting. E.g. in the figure of prop. XXVIII, OB is said to make a right angle. with the circumference EBA, because it is perpendicular to the tangent at B. Exercises. 318. The bisectors of the interior and the exterior verti- cal angles of a triangle meet the circumscribed circumference in the mid-points of the arcs into which the base divides that circumference, and the line joining those points is the diameter which bisects the base. 319. A triangle whose angles are, respectively, 30°, 50°, 100° is inscribed in a circle ; the bisectors of the angles meet the circumference in A, B, C. Find the number of degrees in the angles of A ABC. 320. The three sides of A ABC are, respectively, 412 in., 506 in., 514 in.; required the lengths of the six segments formed by the three points of tangency of the inscribed circle. 321. The radii of two concentric circles are 29 in. and 36 in., respec- -Ac- tively. In the larger circle a chord is drawn tangent to the smaller; required its length. 322. Two circumferences of circles of radii 0.5 ft. and 1.2 ft. intersect so that the tangents drawn at their point of intersection are perpendicu- lar to each other. Required the distance between the centers. 323. The distance between the centers of two circles of radii 7 in. and 4 in., respectively, is 8 in. Required the length of their common tan- gent, between the points of tangency. Is there more than one answer ? 324. The distance between the centers of two circles of radii 327 in. and 115 in., respectively, is 729 in. Required the length of their com- mon exterior tangent, between the points of tangency. 325. The distance between the centers of two circles is 165 in. ; the radii are 62 in. and 48 in., respectively. Calculate, correct to 0.001, the length of the longest line parallel to the center-line and 30 in. from it, limited by the circumferences. 326. Through the point A, 6 in. from the center of a circle of radius 4.5 in., two tangents, AT, A T ', are drawn. Calculate the length of the chord TT' and its distance from the center. X APPENDIX TO BOOK III. — METHODS. 217. The student has already been informed of three im- portant methods of attacking a proposition : (1) By Analysis (§ 113). (2) By Intersection of Loci (I, props. XLIII, XLIY). (3) By Reductio ad Absurdum (§ 74). He is now prepared to discuss these somewhat more fully. 218. I. Method of Analysis. This method, first found in Euclid's Geometry, though attributed to Plato, may be thus described : Analysis is a kind of inverted solution; it assume* the 'proposition proved, considers what results follow, and con- tinues to trace these results until a known proposition is reached. It then seeks to reverse the process and to give the usual, or Synthetic, proof. A more modern form of analysis is sometimes known as the Method of Successive Substitutions. In this the student sub- stitutes in place of the given proposition another upon which the given one depends, and so on until a familiar one is reached. The student reasons somewhat as follows : 1. I can solve A if I can solve B. 2. And I can solve B if I can solve C. 3. But I can solve C. Or he reasons thus: 1. A is true if B is true. 2. And B is true if C is true. 3. But C is true. 4. Hence A and B are true. 162 Sec. 218.] METHODS OF ATTACK. 153 Illustrative Exercises. 1. Through a given point to draw a line to make equal angles with two Intersecting lines. Analysis. Suppose x, y the lines, P the point, and I the required line ; then, in the figure, Z. c = Z. a ■ + Z b ; but "•' Z a is to equal Ah. :.Z' , = 2Za; .'.if Zeis bisected, and a line is drawn through P parallel to this bisector, the con- struction is effected. Now that the method is discovered, give the solution in the ordinary way. 2. Tli rough a given point to draw a line such that the seg- ments intercepted, by the perpendiculars let fall upon it from two given points shall be equal. B \P x D \ D X A' Analysis. Suppose P the given point through which the line x is to be drawn, and A and B the other given points ; then, in the figure, AD and BD' _L x, and DP is to equal PD'. Further, if AP is produced to meet BD' produced at A', then A DP A 5£ A D'PA'j and .'. AP = PA'. But V A and P are given, AP can be drawn, and PA' found ; .'.A' can be found, and .*. A'B; then from Pal can be drawn to A'B, and the problem is solved. Always give the solution in the ordinary way. 154 PLANE GEOMETRY. [Bk. III. 3. If two circles are tangent, any secant drawn through their point of contact cuts off segments from one that contain angles equal to the angles in the segments of the other. Analysis. 1. Let CD be the common tangent to © 0, 0' at their point of contact P. Ill, prop. XXIII, cor. 2 2. Then an Z in segment A = an Z in segment A', if Za = Z a'. Ill, prop. XIII 3. But Za = Z a'. Prel. prop. VI . C V d^T 7\ Y Exercises. 327. To construct a trap- ezoid, given the four sides. Analysts. Assume the figure drawn. Then if d is moved parallel to itself and between c and a, to the position FZ, the A XYZ can easily be constructed (I, prop. XXXIV). The process may now be reversed and the trapezoid constructed. 328. To place a line so that its extremities shall rest upon two given circumferences, the line being equal and parallel to another line. Analysis. If and O' are the given circles, &nd>AB the given line, and if O 0' is moved along a line parallel and equal to AB, then either XY or X'Y' answers the conditions. Hence the process may be reversed ; first describe O 0", and then from T, T draw YX and Y'X' = and II BA. Exs. 329-339.1 METHODS OF ATTACK, 155 t EXERCISES. 329. Given two parallels, AT. X Y'. with a transversal WZ limited by AT and X'Y' ; also two points A, B. not between the parallels, and on opposite sides of them. Required to join A and B by the shortest broken line which shall have MN, the intercept between AFand X'Y\ parallel to WZ. A nalysis. If any MN in the figure is moved along NB parallel to its original position, until N coincides with B and M is at P, then AM X P < AM 2 P or AM S P (I, prop. VIII) ; hence AM\N\B is the shortest broken line. Hence the process may be reversed ; first draw BP II and = ZW; then join A and P, thus fixing M\ : and then draw M\N\ II WZ. 330. Through one of the two points of inter- section of two circumferences to draw a line from which the two circumferences cut off chords having a. given difference. (The projection of the center- segment on the required line equals half the given difference ; hence move this projection to the position OA ; the right-angled A OCA can now be constructed, and the required line will be parallel to OA.) 331. In ex. 330, show that if the two chords lie on opposite sides of P. the sum replaces the difference. 332. In a given circle to draw a chord equal and parallel to a given line. 333. From a ship two known points are seen under a given angle; the ship sails a given distance in a given direction, and now the same two points are seen under another known angle. Find the positions of the ship. (On the line joining the known points, construct segments to contain the given angles; the problem then reduces to ex. 328.) 334. Construct a trapezoid, given the diagonals, their included angle, and the sum of two adjacent sides. 335. To construct a triangle given a and the orthocenter. /% 336. Also, given a and the centroid. * xv $A 337. To draw a tangent to a given circle, perpendicular to a given line. 338. To construct a triangle, ABC, having given c. Z C, and the foot of the perpendicular from C to c. 339. Find the locus of the points of contact of tangents drawn from a fixed point to a system of concentric circles. * 156 PLANE GEOMETRY. [Bk. III. 219. II. Method of Intersection of Loci. This method, adapted chiefly to the solution of problems, has already been used in Book I (props. XLIII, XLIV). So long as it is known merely that a point is on one line, its position is not definitely known ; but if it is known that the point is also on another line, its position may be uniquely determined. For example, if it is known that a point is on each of two intersecting lines, the point is uniquely determined as their point of intersection ; but if the point is on a straight line and a circumference which the line intersects, it may be either of the two points of inter- section. For convenience of reference the following theorems are stated, and will be referred to by the letters prefixed : a. The locus of points at a given distance from a given point is the circumference described about that point as a center, with a radius equal to the given distance. (§ 127.) b. The locus of points at a given distance from a given line consists of a pair of parallels at that distance, one on each side of the fixed line. (§ 129, cor. 2.) c. The locus of points equidistant from two given points is the perpjendicular bisector of the line joining them. (§ 128.) d. The locus of points equidistant from two given lines con- sists of the bisectors of their included angles ; if the lines are parallel, it is a parallel midway between them. (§ 129.) e. The locus of points from which a given line subtends a given angle is an arc subtended by that chord. (§ 195, cor. 3.) Abbreviations. The following abbreviations will be used : In the triangle ABC the altitudes on the sides a } b, c will be designated by h n , h h , h c , respectively ; the corresponding medians by m a , m b , m c ; the corresponding angle-bisectors terminated by a, b, c, by v a , v b , v c ; the radii of the inscribed and circumscribed circles by r, B, respectively ; the radius of the escribed circle touching a, and touching b and c produced, by r a , and similarly for r b , r c . Sec. 220.] METHODS OF ATTACK. 157 220. Definition. A triangle is said to be inscribed in another when its vertices lie respectively on the sides of the other. Exercises. 340. To describe a circumference with a given radius, and * (1) Passing through two given points. (Combine a and c.) X (2) Passing through one given point and touching a given line, (a, b.) (3) Passing through one given point and touching a given circle, (a. ) (4) Touching a given line and a given circle, (a, 6.) (5) Touching two given circles, (a.) 341. In a given triangle to inscribe a triangle with two of its sides given, and the vertex of their included angle given, (a.) -V- 342. To describe a circumference passing through a given point and touching a given line, or a given circle, in a given point, (c.) -X 343. On a given circumference to find a point having a given distance from a given line. '(6.) f344. On a given line, not necessarily straight, to find a point equi distant from two given points, (c.) t345. Describe a circumference touching two parallel lines and passing through a given point, (d, a.) 346. Find a point from which two given line-segments are seen under cf (or subtend) given angles, (e.) (Pothenot's problem.) 347. Construct the triangle ABC. given o, //,„ m a . 348. Also, given ZA, a, h (l . — — 349. Also, given ZA, a, m a . * 350. Also, given a, h b , h c . I X 351. Also, given Z A, h a , v a . (First construct the right-angled tri- angle with side h a and hypotenuse v a .) 352. Also, given h a , m a , B. (First construct the right-angled triangle with side h n and hypotenuse m a ; then find the circumcenter by a, c.) 353. Also, given a, R. h b . (First construct the right-angled triangle with side h b and hypotenuse a ; then find the circumcenter by a.) 354. Also, given c, r, Z A = 90° ; c, r,. Z A = 90° ; 6. r. ZA = 90° ; or 6, r c , Z A = 90°. 355. Describe two circles of given radii /v r 2 , to touch one another, and to touch a given line on the same side of it. 4 158 PLANE GEOMETRY. [Bk. III. MISCELLANEOUS EXERCISES. 356. If two circumferences intersect, any two parallel lines drawn through the points of intersection and terminated by the respective circumferences are equal. 357. If the center-segment of two circles is (1) greater than, (2) equal to, the sum of the two radii, the circumferences (1) do not meet, (2) are tangent. 358. The greatest of all lines joining two points, one on each of two given circumferences, is greater than the center-segment by the sum of the radii. 359. If two circles, whose centers are O, (7, are tangent at P, and a line through P cuts the circumferences at A, A', prove that OA II (/A '. Two cases ; external and internal tangency. Show that the proposition is true for any number of circles. 360. Through a vertex of a triangle to draw a straight line equally distant from the other vertices. 361. Describe a circle of given radius to touch two given lines. Show that a solution is, in general, impossible if the lines are parallel, but that otherwise there are four solutions. 362. From what two points in the plane are two circles seen under equal angles ? 363. Given an equilateral triangle, ABC, find a point P such that the circles circumscribing 4 PBC, PC A, PAB are all equal. 364. To divide a circle into two segments so that the angle contained in one shall be double that contained in the other. 365. From two given points to draw lines meeting a given line in a point and making equal angles with that line, the points being on (1) the same side of the given line, (2) opposite sides of the given line. 366. To draw, through a given point, a secant from which two equal circumferences shall cut off equal chords. Discuss the number of solu- tions for various positions of the given point. 367. Through one of the points of intersection of two circumferences to draw a chord of one circle which shall be bisected by the circumference of the other. 368. Two opposite vertices of a given square move on two lines at right angles to each other. Find the locus of the intersection of the diagonals. 369. Find the locus of the intersection of two lines passing through two fixed points on a circumference and intercepting an arc of constant length. BOOK IV. — RATIO AND PROPORTION. 1. FUNDAMENTAL PROPERTIES. 221. Introductory Xote. The inference was drawn in Book II (§ 155) that a relation exists between algebra and geometry with the following correspondence : Geometry. Algebra. A line-segment. A number. The rectangle of two line-segments. The product of two numbers. And as it was assumed that a straight line may be represented by a number, so it may be assumed that any other geometric magnitude, such as an arc, an angle, a surface, etc., may be represented by a number. With these assumptions, the fun- damental properties of Batio and Proportion may be proved either "by algebra or by geometry, as may be most convenient, the proof being valid for both of these subjects. The purely geometric treatment is too difficult for the beginner. 222. Definitions. To measure a magnitude is to find how many times it contains another magnitude of the same kind, called the unit of measure. 223. A ratio is the quotient of the numerical measure of one magnitude divided by the numerical measure of another magnitude of the same kind. Tor example, the ratio of a line 8 ft. long to one 16 ft. long is T 8 ? , or \ ; that of one 16 ft. long to one 8 ft. long is 2. The ratio of a to b is expressed by the symbols - • a : 6, a/5, or a -4- b. If the ratio r = r, then a = r • b. b 159 160 PLANE GEOMETRY. [Bk. IV. 224. The practical method of finding the ratio of two magnitudes is to measure them, and to divide the numerical result of one measurement by that of the other. But if two line-segments have a common measure, their ratio and their common measure may be found by the following process : Let AB and CD be the two lines. A« ■ ■ — ►-.b Apply CD as often as pos- F , , , p sible to AB, and suppose that AB = 2 CD + EB, EB < CD. Similarly, apply EB to CD, and suppose that CD = 2EB + FD, FD < EB. Similarly, apply FD to EB, and suppose that EB = FD+ GB, GB < FD. Similarly, apply GB to FD, and suppose that FD = 3 GB with no remainder. Then FD = 3 GB. EB = FD + GB= 4 GB. CD = 2EB + FD= 8 GB + 3 GB = 11 GB. AB = 2 CD + EB = 22 GB + 4 GB = 26 ££\ .*. Gi? is a common measure, and the ratio of CZ> to AB is i£ by definition of ratio. 225. Definitions. Two magnitudes that have a common measure are said to be commensurable; if they have no com- mon measure they are said to be incommensurable. For example, two surfaces having areas 10 sq. in. and 15 sq. in. are said to be commensurable, there being the common measures 5 sq. in., 1 sq. in., 2.5 sq. in., etc. But if the length of one line is represented by V2, and the length of another by 1, then there is no common measure, and the lines are said to be incommensurable. A ratio may therefore be an integer, or a fraction, or an irrational number such as V2. Prop. I.] RATIO AND PROPORTION. 161 For practical purposes all magnitudes may be looked upon as commensurable, since a unit of measure can be so taken that the remainder may be made as small as we wish. 226. In the ratio a : b, a and b are called the terms of the ratio, — the former, a, being called the antecedent, and the latter, b, the consequent. 227. If the ratio a : b equals the ratio c : d, the four terms are said to form a proportion. The four terms are also said to be in proportion. The terms a and b are also said to he proportional to c and d. (I c This equality of ratios is indicated by the symbol =, e.g., t=^' a : b = c : d, or a/b = c/d, read "a is to b as c is to cZ.' 1 Instead of the parallel bars ( = ), the double colon (: :) is also used in this connection as a sign of equality, the proportion being written a : b : : c : d. The double colon is not. however, as extensively used as formerly. 228. The first and last terms of a proportion are called the extremes, and the other terms the means. Thus in the proportion 2:3 = 6: 9, 3 and 6 are the means and 2 and 9 are the extremes. Proposition I. 229. Theorem. If a : b = c : d, then ad = be. (Jb C Proof. From j- = ;v it folloAvs, by multiplying equals by bd, that ad = be. Ax. 6 Hence If four numbers are in pro- If four lines are in propor- portion, the product of the tion, the rectangle of the means equals the product of means equals the rectangle of the extremes. the extremes. 1G2 PLANE GEOMETRY. [Bk. IV, Proposition II. 230. Theorem. If ad = be, then a : b = c : d. Proof. Divide the given equals by bd. Hence If the product of two num- bers equals the product of two other numbers, either two may be made the means and the other two the extremes of a proportion. Ax. If the rectangle of two lines equals the rectangle of two other lines, either two may be made the means and the other two the extremes of a proportion. Proposition III. -* 231. Theorem. If a : b = c : d, then (1) a : c = b : d, (2) d : b = c : a, and (3) b : a = d : c. Proof. 1. and and 2. And 3. Hence If four numbers are in pro- portion, the following inter- changes may be made : (1) the means, (2) the extremes, (3) each antecedent and its cor- responding consequent. Prop. I Prop. II Prop. II Stepl Prop. II If four magnitudes are in proportion, the following in- terchanges may be made: (1) the means, (2) the extremes, (3) each antecedent and its corresponding consequent. ad - = be, a c b %7 which proves (i), d b = 6 = —j a which proves (2)- be = = ad b a d — — j c which proves (3). Prop. IV.] RATIO AND PROPORTION. 163 Definitions. The proportion a : c = b : d is often spoken of as the proportion a : b = e : d taken by alternation. The proportion b : a = d : e is also spoken of as the propor- tion a:b = c:d taken by inversion. Hence prop. Ill may be stated: If four magnitudes are in proportion they are in proportion by alternation and also by inversion. But to take a proportion by alternation, the magnitudes must be similar. Thus 1 2 :$4 = $8 :| 16, therefore, by alternation, 82 : 88 = 84 : $16. But the proportion 82:84 = 8 days : 16 days cannot be taken by alter- nation, for $2:8 days = 84 : 16 days means nothing, $2:8 days not being a ratio (§ 223). Proposition IV. 232. Theorem. If a : b = c : d, then (1) ka : b = kc : d, and (2) a : kb = c : kd. 01 G Proof. From - = -> it follows, by multiplying by k, that — = -7 j which proves (1). Ax. 6 b d v 7 a c And also, — = — > by dividing by /.', which proves (2). nO KCL Ax. 7 Hence if four magnitudes are in proportion, and both ante- cedents or both consequents are multiplied by the same num- ber, the magnitudes are still in proportion. Corollary. If four magnitudes are in proportion, and all are multiplied by the same number, the results are in proportion. Note. The number k may be integral, fractional, or irrational. 164 PLANE GEOMETRY. [Bk. IV. Proposition V<. 233. Theorem. If a : b = c : d, then (1) a±b:b = c±d:d, (2) a ± b : a = c ± cl : c, and (3) a±b:a=f:b = c±d:cq=d. Proof. 1. From t = v it follows 6 a Axs. 2, 3 which proves (1). Prop. Ill Axs. 2, 3 that or 6 at 2. It is also true that b_d a c and a c or U±b c±d which proves (2). to V 3. Or, by subtracting first, a a c qp d c a ±b e ± d Axs. 3, 2 and .'. =- ; bv dividing in the last two a if b c ifd J equations. Ax. 7 The proportion a + b : b = c + d : d is often spoken of as the proportion a : b = c : d taken by composition, and a — b : b = c — d : d as the same proportion taken by division, and a±b:azpb = c±dicqpd as the same proportion taken by composition and division. Props. VI, VII.] RATIO AND PROPORTION. 165 Proposition* VI. 234. Theorem. If ^ = J- 2 - = ^ = , the terms all beinq \ b 2 b 3 y magnitudes of the same kind, then a-. a 9 -± or — or bj + b 2 + bj b Proof. 1. ^i = |l, Prop. Ill aud ...fd^s^-iiJs. p v «2 ^2 2. ...^±?=f S Prop. Ill #1 + 6 2 #2 3. Then, as in steps 1, 2, a x + a 2 + a 3 a 3 Given &1 + & 2 + &3 l>% and so on, however many ratios there may be. Proposition VII. 235. Theorem. If a : b = c : d, then a 2 : b 2 = c 2 : d 2 . _ a c a z c Proof. ,._ = _,..._ = _ Ax . 6 Hence If four numbers are in pro- If four lines are in propor- portion, their squares are also tion, the squares on those lines in proportion. are also in proportion. Corollary. If a : b = c : d, and m : n = x : y, then am : bn = ex : dy. 166 PLANE GEOMETRY. [Bk. IV. Proposition VIII. 236. Theorem, a : b = ka : kb. Proof. kab = kab, or a • kb = b • ka. .'. a : b = ka : kb. Prop. 11 Note. The number k may be integral, fractional, or irrational. 237. Definitions. If a : b = c : x, x is called the fourth pro- portional to a, b, c. Corollaries. 1. By three of the four terms of a proportion the other is determined. For if a :'b = c : x, or x : b = c : a, or c :x = d : 6, etc., it follows that ax ■= be, whence x = bc/a, a fixed number. 2. If a : b = a : x, then b = x. For if, in the proof of cor. 1, c = a, then 6 = x. 238. If, in a proportion, the two means are equal, as in a : x = x : b, this common mean is called the mean proportional, or geometric mean, between the two extremes. Corollaries. The mean proportional be- The geometric mean be- tween two numbers equals the tween tivo lines equals the side square root of their product. of that square which equals their rectangle. Because the number representing the square units of area of a rectangle is the product of the two numbers representing the linear units in two adjacent sides, the expression product of two lines is often used for rectangle of two lines. Exercises. 370. Find a mean proportional between 2 and 32. 371. Find a fourth proportional to 3, 7, 15. 372. What number must be added to each of the numbers 2, 1, 5, 3, to have the results in proportion ? Secs. 239, 240.] THE THEORY OF LIMITS. 167 2. THE THEORY OF LIMITS. 239. Definitions. A quantity is called a variable if, in the course of the same investigation, it may take indefinitely many values ; on the other hand, a quantity is called a con- stant if. in the course of the same investigation, it keeps the same value. M, M 2 M 3 B E.g. if a line AB is bisected at M t , and M X B at M 2 , and M 2 B at 3f 3 , and so on, and if x represents the line from A to any of the points Jfj , M 2 , then £ is a variable, but AB is a constant. It is customary, as in algebra, to represent variables by the last letters of the alphabet, and constants by the first letters. 240. If a variable x approaches nearer and nearer a con- stant a. so that the difference between x and a can become and remain smaller than any quantity that may be assigned. then a is called the limit of x. E.g. in the above figure. AB is the limit of x. But if the point M simply slides along the line, passing through B, then, although the difference between AM and AB. or x and a, can become smaller than any quantity which may be assigned, it does not remain smaller, for when M passes through B this difference increases. Hence AB is not then the limit of x. That " x approaches as its limit a n is indicated by the symbol x = a. Corollary. If x = a, then a — x is a variable whose limit is zero : thai is, a — x = 0. Although the variable has been taken, in this discussion, as increasing towards its limit, it may also be taken as decreasing. Thus if we bisect a line, bisect its half, and continue to bisect indefinitely, the variable segment is evidently approaching a limit zero. 168 PLANE GEOMETRY. [Bk. IV. Proposition IX. 241. Theorem. If, while approaching their respective limits, tivo variables have a constant ratio, their limits have that same ratio. B C _j j. U- D Given X and X', two variables, such that as they increase they approach their respective limits AB, or L, and AC, or L', and have a constant ratio r. To prove that L : V = r, or that X : X' = L : X'. Proof. If the ratio X: X' is not equal to the ratio L : L', then (1) it must equal the ratio of L to something less than L', or (2) it must equal the ratio of L to something greater than V. It will be shown that both of these suppositions are absurd. I. To show that (1) is absurd. 1. Suppose X:X' = L:L'-DC. Then '.' X:X'= r, .'.X=rX', and L = r(V - DC). § 223, def. ratio 2. Then L- X= r(L' -DC- X'). Ax. 3 3. But v X'=L) .'. L' — X' can become as small as we please. " " " less than DC, and .'. r(L' — X' — DC) can become negative. 4. But V X> Lj .'. L — X cannot become negative. 5. .*. step 2 is absurd, for a negative quantity cannot equal one not negative. Prop. IX.] THE THEORY OF LIMITS. 169 II. To show that (2) is absurd. 1. Suppose X: X' = L : V + CD'. Then £ - X = /• (X' + CD' - X'), as in step 2, p. 168. 2. But r(D' 4- CD' — X') cannot become less than r • CD'. 3. And L — X— 0, because L is the limit of X. 4. .'.if step 1 were true, a quantity, L — X, which can become as small as we please, would equal a quan- tity not less than r • CD', which is absurd. The proof would be substantially the same if the two variables were supposed to decrease toward a limit. Corollaries. 1. If, while approaching their respective limits, two variables are always equal, their limits are equal. For their ratio is always 1. 2. If, while approaching their respective limits, two variables have a constant ratio, and one of them is always greater than the other, the limit of the first is greater than the limit of the second. For the limits have the same ratio as the variables. Exercises. 373. If a : b — c : d, prove that a 2 bd + b~c + be = ab' 2 c + abd + ad. 1. Since be = ad, Prop. I the equation is true if a-bd + b' 2 c = ab 2 c + abd. 2. Now if in place of each ad we put be, we see that the equation is true if ab' 2 c -f b 2 c — ab 2 c + b 2 c. But this is an identity. Hence the proof is complete. 374. lia:b = c:d, prove that a — c : b — d = V a- + c- : V& 2 + d 2 . Also that Va 2 + c 2 : \lb 2 + d 2 = A/ac + - : \Jbd + — ■ Also that a + mb • a — nb = c. + md : c — nd. 170 PLANE GEOMETRY. [Bk. IV 3. A PENCIL OF LINES CUT BY PARALLELS. 242. Definitions. Through a point any number of lines can be passed. Such lines are said to form a pencil of lines. The point through which a pencil of lines passes is called the vertex of the pencil. A A pencil of three lines. A pencil of four parallels. The annexed pencil of three lines is named "V — ABC." To conform to the idea of a general figure, set forth in §§ 94, 95, the word pencil is also applied to parallel lines, the vertex being spoken of as " at infinity." 243. Definition. Two lines are said to be divided proportion- ally when the segments of the one have the same ratio as the corresponding segments of the other. Exercises. 375. If a : b = c : d, prove that (1) a + b + c + d:b + d = c + d:d. (2) in (a + mb) :n(a — nb) = m(c + md) : n(c — nd). (3) a (a + b + c + d) = (a + 6) (a + c). (4) a?c + ad 2 : b 2 d + bd 2 = (a + c) 3 : (b + d)*. 376. If b is a mean proportional between a and c, prove that a 2 - b' 2 + c 2 1 1 1 P ~ P + <* b*. 377. Show that there is no finite number which, when added to each of four unequal numbers in proportion, will make the resulting sums in proportion. 378. If a :b = c :d, and u :v = x :y, prove that au + bv : au — bv = ex -f dy -. ex — dy. Prop. X. PENCILS OF LINES. 171 Proposition X. 244. Theorem. The segments of a transversal of a pencil of parallels are 'proportional to the corresponding segments of any other transversal of the same pencil. I 1 ,/H 1== n m nil/. t— !/ , / (I \ bJ/-~. n m Given the pencil of parallels P, cutting from two trans- versals T and T the segments A, B and C, £>, respectively. To prove that A : B = C : D. Proof. 1. Suppose A and B divided into equal segments I, and that A = reZ, while B = ?i7. (In the figure, n = 6, n' = 4.) Then if ll's to P are drawn from the points of divi- sion, C is the sum of n equal segments m, and D is the sum of n' equal segments w. I, prop. XXVII, cor. 1 A _ wZ _ ?i ?iw. _ 6' '"" ~B = n T l = n' = n^~i = ~D' Whv ? Note. The preceding proof assumes that -4. and B are commensurable. The following proof is valid if A and B are incommensurable. 172 PLANE GEOMETRY 245. Proof for incommensurable case. [Bk. IV 1. Suppose A divided into equal segments I, and that A — nl, while B = n'l + some remainder, x, such that x < /. Then if ll's to P are drawn from the points of divi- sion, C is the sum of n equal segments m, and D is the sum of n' equal segments m, + a remainder y such that y < m. 2. Then Z^ lies between n'l and (w' + 1) /. Step 1 3. .'. — lies between — - and -r— A nl nl 4jt in the figure, between — and (>/ ) D n'm (n[ + 1) m while — lies between and L ( nm, nm 4. .'. — and — both lie between — and A C n n and .'. they differ by less than (In the figure, by less than \.) 1 Prop. X.] PENCILS OF LINES. 173 5. And v - can be made smaller than any assumed n difference, by increasing n } .'.to assume any differ- ence leads to an absurdity. a . B D A C JJT b. ■'A = r 0T ^ == D' PropI11 Corollaries. 1, A line parallel to one side of a triangle divides the other two sides proportionally. For in the figure, if BCO is the triangle, the lines OB. OC are cut by parallels. Hence BB X : B x O = CC X : C x O. 2. The corresponding segments of the lines of a pencil cut off (from the vertex) by parallel transversals are proportional. In the above figure, 0.4 : 0A X = OB : 0B X = OC : 0C\ = by prop. X. 3. The segments of the lines of a pencil cut off {from the vertex) by parallel transversals are proportional to the corre- sponding segments of the transversals. To prove that, in the above figure, AB : A X B X = OA :OAi= OB : 0B V Draw through A x a line II to OB cutting AB at X. Then AB : XB = OAiOA x = OB : 0B X . Prop. X But XB - AiB v I, prop. XXIV 4. Parallel transversals are divided proportionally by the lines of a pencil. To prove that, in the above figure, AB \BC = A\B\\ B X C X . By cor. 3, AB : A X B X = BO : B x = BC : B X C X . Hence, etc. 174 PLANE GEOMETRY. [Bk. IV. Proposition XI. 246. Theorem. A line can be divided, internally or ex- ternally, into segments having a given ratio, except that if it is divided externally the ratio cannot be unity. Given Fig. l. Fig. 2. the line AB, and two lines s 1} s 2 having a given ratio. To prove that AB can be divided in the ratio Sj : s 2 , except that in the case of external division s x cannot equal s 2 . Proof. 1. Suppose AM drawn making, with AB, an angle < 180°; that AC be taken = s 1} and CD = s 2 ; that DB be drawn, and CP II DB. 2. Then AP : PB = s x : s 2 , as required. Prop. X, cor. 1 3. In Pig. 2, if s t = s 2 , where does D fall ? What is then the relation of CP to AB ? Hence show that the division is impossible in this case. Corollary. The point of internal division is unique ; like- wise the point of external division. From step 2, AB : PB = Si + s 2 : s 2 , AB, Si -f s 2 , and s 2 , all being constants ; but by three terms of a proportion the fourth is determined. (§ 237, def. of 4th proportional, cor. 1.) 247. Note. Instead of saying that the external division, if the ratio is unity, is impossible, it is often said that the point of division, P, is at infinity. In the case of internal division, the ratios AP : PB and AC : CD are evidently positive ; but in the case of external division each ratio is evidently negative because PB and CD are negative. In both cases step 2 is evidently true. Prop. XII. 1 PENCILS OF LINES. 175 Proposition XII. 248. Theorem. A line which divides two sides of a triangle proportionally is parallel to the third. Given the triangle ABC, and DE so drawn that AD : DC = BE : EC. To prove that DE II AB. Proof. 1. Suppose DE not ii AB, but that DX II AB. Then BX : XC = AD : D C. Prop. X, cor. 1 2. But this is impossible, for the division of BC in the ratio AD : DC is unique. Prop. XI, cor. 3. .'. DX must be identical with DE, and DE II AB. The proof is the same for all of the figures. Exercises. 379. In the above figures, if AD : DC = BE : EC = m : n, ^ { and if the line through A and E cuts the line through B and D at P, then prove that J.P : PE = BP : PD = m + n : n. 380. If ex. 379 has been proved, show from it that the centroid of a triangle divides the medians in the ratio of 2:1. 381. Prove prop. XI on the following figures : n 176 PLANE GEOMETRY. [Bk. IV. Proposition XIIL 249. Theorem. If any angle of a triangle is bisected, in- ternally or externally, by a line cutting the opposite side, then the opposite side is divided, internally or externally, respectively, in the ratio of the other sides of the triangle. Given A ABC, the bisector of Z C meeting AB at P. To prove that AP : PB = AC : BC. Proof. 1. Let BE II PC, meeting AC produced at E, in Fig. 1. Then Z EBC = Z PCB = Z ACP = Z CEB. Given ; I. prop. XVII, cor. 2 2. .'.BC=CE. Why? 3. But in A A BE, AP : PB = A C : CE, Prop. X, cor. 1 and .'. AP : PB = AC : BC Why ? The proof for Pig. 2 is the same if step 1 is changed to Z CBE = ZBCP = Z PCX = Z BEC 250. Definition. When a line is divided internally and externally into segments having the same ratio, it is said to be divided harmonically. If the internal and external points of division of AB, in prop. XIII, are P and P' then AB is divided harmonically by P and P'. Exercise. 382. The hypotenuse of a ri^ht-ancled triangle is divided harmonically by any pair of lines through the vertex of the right angle, making equal angles with one of its arms. Sec. 251.] PENCILS OF LINES. 177 4, A PENCIL CUT BY ANTIPARALLELS OR BY A CIRCUMFERENCE. 251. Definitions. If a pencil of two lines — XY is cut by two parallel lines AB, MX, and if MX revolves, through a straight angle, about the bisector of Z XO Y as an axis, falling in the position A V B U then AB and A X B X are said to be anti- parallel to each other. OA and 0A X are called corresponding segments of the pencil, as are also OB and 0B X . A and A x are called corresponding l>oints, as are also B and B x . Corollary. If Z A — Z A x , in the above figure, then AB and A X B X are antiparallel to each other. Exercises. 383. From P, a given point in the side AB of A ABC, draw a line to AC produced so that it will be bisected by BC. 384. Investigate ex. 383 when P is on AB produced. 385. If the vertices of A XYZ lie on the sides of A abc so that x II a, y II b, z II c, then X, F, Z bisect «, b, c. 386. In prop. XIII, suppose Z B = Z A ; also, suppose ZB! = OA : OA OB-.OB x . (Prop. X, cor. 3, etc.) 390. In the above figures, if ^4i coincides with B. and if OB = b, OA = a, OB x = b x , then 6 2 = ab x . 391. If from the vertex of a right-angled triangle a perpendicular p is drawn cutting the hypotenuse c into two segments x, y, adjacent to sides a, 6, respectively, then (1) a and p are antiparallels of the pencil 6, c ; (2) a is a mean proportional between c and x ; (3) p is a mean propor- tional between x and y ; (4) 6 2 = cy, a 2 = ex, and .-. a 2 + b 2 = c (x + y) = c 2 . (Thus a new proof is found for the Pythagorean proposition.) ,= * Prop. XV.] PENCILS OF LINES. 179 Proposition XV. 253. Theorem. If a pencil of lines cuts a circumference, the product of the two segments from the vertex is constant, whichever line is taken. A, A, Fig. 1.— The point O on the chord. Fig. 2. — The point O on the chord produced. Given AB X and A X B, two chords, each divided at into two segments. To prove that AO • OB x = A x O • OB. Proof. 1. Suppose AB, A X B X drawn. Then £A = £ A v Why ? 2. .'. AB and A X B X are antiparallel, § 251, cor. and .'. AO ■ OB x = A x O ■ OB. Prop. XIV, cor. The proposition is entirely general and . should be proved for the fol- lowing cases. Fig. 3. — The point O at the end of the chord. FIG. 4. — Chord A X B becomes zero. Fig. 5. —Chord AB Y also becomes zero. Corollary. The tangent from the vertex of a pencil to a circumference is a mean proportional between the two segments of any other line of the pencil . In Fig. 4, AO ■ B x O=A x O • BO=B0 2 . Therefore AO : BO=BO : B x O. W' 180 PLANE GEOMETRY. [Bk. IV. Proposition XVI. 254. Theorem. In the same circle or in equal circles central angles are proportional to the arcs on which they stand. Given A and B, two central angles standing on arcs C and D, respectively. To prove that A: B = C : D. Proof. 1. If A and B are in different circles, they may be placed in the relative positions shown in the figure. § 108, def. O, cor. 2 Suppose A and B divided into equal A x, and suppose A = nx, and B = n'x. (In the figure, n = 6, nf = 4.) 2. Then C is divided into n equal arcs y, " y. Ill, prop. I and D « »' 3. ^4 _ 7103 w ny _ 6' . a c " B I) Why ? Why ? Note. The above proof assumes that A and B are commensurable, and hence that they can be divided into equal angles x. The proof on p. 181 is valid if A and B are incommensurable. Skcs. 255, 256.] PENCILS OF LINES. 255. Proof for incommensurable case. 181 1. Suppose A divided into equal A x, and suppose A = nx, while B = n'x + some remainder w, such that w < x. Then C is divided into u equal arcs y, and D is the sum of n' equal arcs y + a remainder .?, such that z < y. 2. Then B lies between n'x and (»*' + 1) x, and D lies between n'y and (V + 1) y. B , 2> v ^ '. , «' '»' + 1 3. .*..— and — both he between — and AC n n (In the figure, between | and f.) J ) T) 1 And .*. — and — differ by less than -■ AC- n Why ? Why ? Why ? Why ? 4, And '.' - can be made smaller than any assumed n difference, by increasing n. .'. to assume any difference leads to an absurdity. B D . AC 5. .-._ = -, whence - = -• Corollary. In the same circle or in equal circles sectors are proportion"! to their angles or to their arcs. 256. This proposition is often stated, A central angle is measured by its intercepted arc. See § 180. ,v- v *3 182 PLANE GEOMETRY. [Bk. IV. 5. SIMILAR FIGURES. 257. Definitions. We have (§ 59) roughly defined similar figures as figures having the same shape. But this is unsatis- factory because the word shape is not defined. We therefore proceed scientifically to define 1. Similar systems of jioints, and then 2. Similar figures. Two systems of points, A x , B x , C 1} and A 2 , B 2 , C 2 , , are said to be similar when they can be so placed that all lines, A X A 2 , B^B 2 , C x C 2 , , joining corresponding points form a pencil whose vertex, 0, divides each line into segments having a constant ratio r. In the figure, OA x : OA 2 = OB x : OB 2 = 258. Two figures are said to be similar when their systems of points are similar. The symbol of similarity w*, already mentioned, is due to Leibnitz. It is derived from the letter S. The following are illustrations of similar figures involving circles: Concentric circles. Any circles. Secs. 259-261.] SIMILAR FIGURES. 183 The following are illustrations of similar rectilinear figures A 2 v Any line-segments. Four similar triangles. Three similar quadrilaterals. 259. When two similar figures are so placed that lines through their corresponding points form a pencil, they are said to be placed in perspective, and the vertex of that pencil is called their center of similitude. The figures above and on p. 182 are placed in perspective, and in each case is the center of similitude. Two similar figures may evidently be so placed that the center of similitude will fall within both, or between them, or on the same side of both, as is seen in the above illustrations. 260. Two systems of points, A 1} B lf C x , and A. 2 , B 2 , C. 2 , , are said to be symmetric with respect to a center when all lines, A^4 2 , B X B 2 , C X C 2 , , are bisected by 0. 261. Two figures are said to be symmetric with respect to a center when their systems of points are symmetric -with respect to that center. E.g. in the figure, A A x BiCi, A 2 B 2 C 2 are symmetric with respect to 0. 184 PLANE GEOMETRY. [Bk. IV. 262. In similar figures, if the ratio, r, known as the ratio of similitude, is 1, the figures are evidently symmetric with respect to a center. Hence Central Symmetry is a special case of Similar Figures in Perspective. The term Center of Similitude is due to Euler. Corollaries. 1. Congruent figures are similar. For if made to coincide, any point in their plane is evidently a center of similitude, the ratio of similitude being 1. Or, they may be placed in a position of central symmetry. 2. To any ])oint in a system there is one and only one corre- sponding point of a similar system with respect to a given center. * If A\ and A % are corresponding points in two similar systems in per- spective, and is the center of simili- tude, then every point Pi on OAi has a unique corresponding point P 2 on OA 2 . For OA l :OA 2 = OP^ : OP 2 , .-. 0P 2 is unique. § 237, cor. 1 Exercises. 392. What is the limit of 1/x as x increases indefinitely ? of 1/(1 + x) as x = ? as x = 1 ? 393. In A ABC, P is any point in AB, and Q is such a point in CA that CQ — PB ; if PQ and BC, produced if necessary, meet at X, prove that CA : AB = PX : QX. (From P draw a line \\ AC.) 394. In the annexed figure of a "Diagonal Scale," AB is 1 centimeter. Show how, by means of the scale and a pair of dividers, to lay off 1 millimeter, 0.5 millimeter, 0.3 millimeter, etc. On what proposition or corollary does this measurement of fractions of a millimeter depend ? /" 395. A BCD is a parallelogram ; from A a line is drawn cutting BD I in E, BC in F, and DC produced in G. Prove that AE is a mean pro- j portional between EF and EG. 396. ABC is a triangle, and through D, any point in c, DE is drawn II a to meet b in E ; through C, CF is drawn II EB to meet c produced in F. Prove that AB is a mean proportional between AD and A F. V 1 6 ■ 4 ' .1 ' :• l - Prop. XVII.] SIMILAR FIGURES. 185 Proposition XVII. 263. Theorem. Tivo triangles are similar if they have two angles of the one equal to two angles of the other, respectively. Given the A A X B X C X , A 2 B 2 C 2 , with ZA l = Z.A 2 , ZC 1 = ZC 2 . To prove that A A X B X C 1 ^ A A 2 B 2 C. Proof. 1. Place one A on the other so that Z C\ coincides with Z C 2 , as at 0, OA 2 falling on OA x . Let OX 2 X x be any line through 0, cutting A 2 B 2 at X 2 , and A X B X at X x . . Then \- Z.A 2 = ZA X , .'. A 2 B 2 II A X B X . I, prop. XVI, cor. 1 2. .'. OA x : OA 2 = OB x : OB 2 = OX x : OX 2 = = r. Prop. X, cor. 2 V 3. And all points on OA x and OB x have their corre- sponding points on OA 2 and OB 2 , respectively. V § 262, cor. 2 4. .'. the A are similar, being the center of simili- tude. § 258 Corollaries. 1. Mutually equiangular triangles are similar. 2. If two triangles have the sides of the one respectively parallel or perpendicular to the sides of the other, they are similar. For by § 86, cor. 5, they can be proved to be mutually equiangular. y 186 PLANE GEOMETRY. [Bk. IV. Proposition XVIII. 264. Theorem. If two triangles have one angle of the one equal to one angle of the other, and the including sides pro- portional, the triangles are similar. Given A A X B X C^ A 2 B 2 C 2 , such that Z C\ = Z C 2 and a x : a 2 = h : b 2 . To prove that A A X B X C\ v-~ A A 2 B 2 C 2 . Proof. 1. v Z C 2 = Z Cu AA 2 B 2 C 2 can be placed on AA 1 B 1 C 1 so that C 2 falls at C 1? i? 2 on a 1? and J 2 on Z> P 2. Then V C^ 2 : C X A X = C\B 2 : C^ l5 .-. A 2 B 2 II J^. Props. XII, V 3. .'.A A l B 1 C l and A A 2 B 2 C 1 are mutually equiangular. I, prop. XVII, cor. 2 4. .'. AA^iCi ^ AA 2 B 2 C\ and its congruent AA 2 B 2 C 2 . Prop. XVII, cor. 1 Exercises. 397. ABC, DBA are two triangles with a common side AB. If P is any point on AB, and PX II AC, and PF II AD, meeting BC and PD in X and F, respectively, prove that A YBX -~ A DPC. 398. ABCD is a quadrilateral. Prove that if the bisectors of A A, C meet on diagonal BD, then the bisectors of AB, D will meet on diago- nal AC. 399. Construct a triangle, having given the base, the vertical angle, and the ratio of the remaining sides. (Intersection of loci and prop. XIII.) 400. In A ABC, CM is a median ; A BMC, CM A are bisected by lines meeting a and b in B and Q, respectively. Prove that QR II AB. Prop. XIX. SIMILAR FIGURES. 18' Proposition XIX. 265. Theorem. If tivo triangles have their sides propoi tional, they are similar. C, C s Given AA-^Bi^C^ A 2 B 2 C 2 such that a x : a, — b x : I> 2 = c x : c 2 ° To prove that A A l B l C 1 ^ A A 2 B 2 C 2 . Proof. 1. On C X A X , C\B X lay off C\X = b 2 , and C X Y = a 2 , and draw XY. Then o 3. But and 4. and 5. and '.' a 1 :a 2 = b 1 : b 2 , and Z. C\ = Z. C\, '. A XYC\ ^ A A X B X C X . Prop. XVIII a x : a 2 = Cj : r 27 Given a x : a 2 = c, : A'3~. Prop. X, cor. 3 .'. Cl :r 2 = c, :J7, Why? A A X B X d — A A 2 B 2 C 2 . I. prop. XII Steps 2, 5 Exercises. 401. The product of the two segments of any chord drawn through a given point within a circle equals the square of half the shortest chord that can be drawn through that point. 402. If P is a point on AB produced, the tangents fiom P to all circumferences through A and B are equal, and hence such points are concyclic. 403. If from any point P on the side CA of a right-angled triangle ABC, PQ is drawn perpendicular to the hypotenuse AB at Q, then AP • A C = AQ ■ AB. Suppose P to be taken (1) at C ; (2) at A ; (3) on A C produced. h' 188 PLANE GEOMETRY. [Bk. IV. Proposition XX. 266. Theorem. Similar triangles have their corresponding sides proportional and their corresponding angles equal. Given two similar triangles, A X B X C X , A 2 B 2 C 2 , A x correspond- ing to A 2 , B x to B 2 , C x to C 2 . To prove that A X B X : A 2 B 2 = B X C X : B 2 C 2 = and that ZB X = Z-B 2 Proof. 1. Suppose the A placed in perspective. Then OA x : OA 2 = OB x : OB 2 = OC x : OC 2 . § 258 2. .". A X B X II A 2 B 2 , and so for other sides. Props. XII, V 3. .-. Z OB x A x = ^ OB 2 A 2 , and Z C x B x O = Z C 2 B 2 0. I, prop. XVII, cor. 2 4. .'./]>! = Z7> 2 , and so for other angles. Ax. 2 5. Also, OB, : OB 2 = A X B X : A 2 B 2 , = B X C X : B 2 C 2 . Prop. X, cor. 3 6. .*. A X B X : A 2 B 2 = B X C X : B 2 C 2 , and so for other sides. Note. This is the converse of props. XVII, XIX. Corollaries. 1. The corresponding altitudes of two similar triangles have the same ratio as any two corresj»> mling sides. Why? 2. The corresponding sides of similar triangles are opposite the equal angles. In what step is this proved ? Secs. 267, 268.] SIMILAR FIGURES. 189 267. Summary of Propositions concerning Similar Triangles. Two triangles are similar if 1. (a) Two angles of the one equal two angles of the other. Prop. XVII (b) They are mutually equiangular. Prop. XVII, cor. 1 (c) The sides of the one are parallel to the sides of the other. Prop. XVII, cor. 2 (d) The sides of the one are perpendicular to the sides of the other. Prop. XVII, cor. 2 2. One angle of the one equals one angle of the other and the including sides are proportional. Prop. XVIII 3. Their corresponding sides are proportional. Prop. XIX If two triangles are similar, 1. They are mutually equiangular. Prop. XX 2. Their corresponding sides are proportional. Prop. XX 3. Their corresponding altitudes are proportional to their corresponding sides. Prop. XX, cor. 1 268. It should further be observed that, in general, Three conditions determine congruence. (See § 90.) Two conditions determine similarity. For these conditions are 1. Two angles equal. (Prop. XVII.) 2. One angle and one ratio. (Prop. XVIII.) 3. Two ratios ; for if the sides are a, b, c, and a', b', c', then if -==—-, and - = — , the A are similar, since - must also equal — * b b' c c' c c' Exercises. 404. If X is any point in the side a, or a produced, of A ABC, and if r& and r c are the radii of circles circumscribed about A ABX and A AXC, respectively, then r h :r c — c\b. (Join the centers and prove two triangles similar.) 405. If one of the parallel sides of a trapezoid is double the other, prove that the diagonals intersect one another in a point of trisection. 190 PLANE GEOMETRY. [Bk. IV. Proposition XXI. 269. Theorem. If two polygons are mutually equiangular and have their corresponding sides proportional, they are similar. B,Z ^Z ^^> Given two polygons, A X B X C X and A 2 B 2 C 2 , such that ZA X = ZA 2 , ZB X = ZB 2 , , and A X B X : A 2 B 2 = B X C X : B 2 C 2 = To prove that J^Ci ^ A, B 2 C 2 Proof. 1. Place A 2 B 2 II A X B X . Then v the A of one polygon == the corresponding A of the other, the remaining sides may be made parallel respectively. I, prop. XVII, cor. 5 2. If A X B X > A 2 B 2 , then B X C X > B 2 C 2 , , because the ratios are equal. 3. Draw A X A 2 , B X B 2 , Then A X B X B 2 A 2 is not a O; also B X C\C 2 B 2 , etc.; and A X A 2 meets B X B 2 as at 0, B X B 2 meets C^CVas at 0', etc. I, prop. XXIV 4. But B x 0' : B 2 0' = B X C X : B 2 C 2 = A X B X : A 2 B 2 = B x O: B 2 0, » Prop. X, cor. 3 which/is impossible unless and 0' coincide. Prop. XI, cor. 5. .'. the two figures are similar, and (> is the center of similitude. § 258 Prop. XXL] SIMILAR FIGURES. 191 In step 2, if A X B X = A 2 B 2 , then B X C X = B 2 C 2 , , and the polygons are congruent and therefore similar. § 262. cor. 1 Corollaries. 1. If two polygons are similar, titer/ are mutually equiangular and their corresponding sides are pro- portional. For if placed in perspective as on p. 190, 1. OA x : OA 2 = OB x : OB 2 . § 258 2. .-. A X B X II A 2 B 2 , and so for other sides. Prop. XII 3. .-. ZL'i = Z B 2 , and so for other angles. I, prop. XVII, cor. 5 4. Also A X B X : A 2 B 2 = B x O : B 2 = B X C X : B 2 C 2 = Prop. X, cor. 3 2. Polygons similar to the same polygon are similar to each other. For they have angles equal to those of the third polygon, and the ratios of their sides equal the ratios of the sides of the third polygon. 3. The perimeters of similar polygons have the same ratio as the corresponding sides. For by cor. 1, A X B X : A 2 B 2 = B X C X : B 2 C 2 = = r. .-. A X B X + B X C X + :A 2 B 2 + B 2 C 2 + = r. (Why ?) 4. Two similar polygons can be divided into the same num- ber of triangles similar each to each, and similarly placed. For and (X coincide, and the figures can be placed having within each. The triangles A x OB x , A 2 OB 2 are then similar, by prop. XVII. Exercises. 406. If from a point outside a circle a pair of tangents and a secant are drawn, the quadrilateral formed by joining in succes- sion the four points thus determined on the circumference has the rect- angles of its opposite sides equal. 407. AB is a diameter, and from A a line is drawn to cut the circum- ference in C and the tangent from B in D. Prove that the diameter is the mean proportional between A C and AD. 408. In O ABCD. P, Q are points in a line parallel to AB ; PA and QB meet at R, and PB and QC meet at S. Prove that RS II AB. 409. Chords AB, CB are produced to meet at P, and PF is drawn parallel to BA to meet CB produced in F. Prove that PF is the mean proportional between FB and FC. 192 PLANE GEOMETRY. [Bk. IV. Proposition XXII. 270. Theorem. In a right-angled triangle the perpendicu- lar from the vertex of the right angle to the hypotenuse divides the triangle into two triangles which are similar to the whole and to each other. D Given A ABC, with Z C a right angle, and CD _L AB. To prove that (l)AiCD^A ABC. (2) A CBD ~ A ABC. (3) AACD^A CBD. Proof. 1. ' . • Z CD A = ZACB, Why ? and Z A = Z A, .'.AACD^AABC, which proves (1). Prop. XVII 2. Similarly A CBD ~ A ABC, which proves (2). Prop. XVII 3. .*. A ACD ~ A CBD, which proves (3). Prop. XXI, cor. 2 Corollaries. 1. Either side of a right-angled triangle is the mean proportional between the hypotenuse and its segment adjacent to that side. For from step 1, AB : AC = AC : AD ; and from 2, AB.BC = BC-.DB. 2. The perpendicular from the vertex of the right angle to the hypotenuse is the mean proportional between the segments of the hypotenuse. For from step 3, AD : CD = CD : DB. Exs. 410-422.] SIMILAR FIGURES. 193 EXERCISES. 410. Prove the converse of prop. XXII : If the perpendicular drawn from the vertex of a triangle to the base is the mean proportional i between the segments of the base, the triangle is right-angled. Jfflr\ 411. Prove that any chord of a circle is the mean proportional I between its projection on the diameter from one of its extremities, and the diameter itself. 412. In the figure on p. 192, if AD represents three units, and DB represents one unit, what number is represented by CD ? 413. Prove that if a perpendicular is let fall from any point on a cir- cumference, to any diameter, it is the mean proportional between the segments into which it divides that diameter. 414. Prove that if two fixed parallel tangents are cut by a variable IV tangent, the rectangle of the segments of the latter is constant. 415. Through any point in the common chord of two intersecting j£. circumferences two chords are drawn, one in each circle. Prove that the four extremities of these chords are concyclic. 416. If the bisectors of the interior and exterior angles at B, in the figure of prop. XXII, meet b at F and E, respectively, prove that BC is the mean proportional between FC and CE. i/ 417. Calculate each of the segments into which the bisectors of the y\ angles of a triangle divide the opposite sides, the lengths of the sides being 9 in., 12 in., and 15 in., respectively. J& 418. From the points A, B, on a line AB, 25 in. long, perpendiculars AC, BD are erected such that AC = 13 in., BD = 7 in. On AB the point is taken such that ZBOD = ZCOA. Calculate the distances AO, OB. 419. Given a trapezoid ABCD, with the non-parallel sides AD, BC divided at E, F, respectively, in the ratio of 2 to 3, to calculate the length of EF, knowing that AB = 12.45 in., and DC = 38.5 in. 420. Calculate the sides of a right triangle, knowing that their respec- tive projections on the hypotenuse are 2.88 in. and 5.12 in. 421. The two sides of a right triangle are respectively 10 in. and 24 in. Required the lengths of their projections on the hypotenuse, and the distance of the vertex of the right angle from the hypotenuse. (To 0.001.) 422. The two sides of a right triangle are respectively 3.128 in. and 4.275 in. Required the lengths of the two segments into which the bisector of the right angle divides the hypotenuse. (To 0.001.) 194 PLANE GEOMETRY. [Bk. IV. 6. PROBLEMS. Proposition XXIII. 271. Problem. To divide a line-segment into parts propor- tional to the segments of a given line. Given the line OX', and the line OX divided into segments OA, AB, Required to divide OX' into segments proportional to OA, AB, Construction. 1. Placing the lines oblique to each other at a common end-point 0, draw XX' . § 28 2. From A, B, draw lines II XX', cutting OX' at A', B', I, prop. XXXIII Then OX' is divided as required. Proof. '.' OX, OX' are two transversals of a pencil of ll's, the corresponding segments are in proportion. Prop. X Corollaries. 1. A given line can be divided into parts proportional to any number of given lines. For that number of given lines may be laid off as OA, AB, BC, on OX. 2. A line can be divided into any number of equal parts. Note. While a straight line can be divided into any number of equal parts, by means of the straight edge and the compasses, a circumference cannot be divided into 7, 9, 11, 13, and, in general, any prime number of equal parts beyond 5. The exceptions are noted in Book V. Prop. XXIV.] PROBLEMS. . 1 ( J5 Proposition XXIV. 272. Problem. To find the fourth proportional to three given lines. Given three lines, a, b, c. Required to find x such that a : b = c : x. Construction. 1. From the vertex of a pencil of two lines, with the compasses lay off a, b, in order, on one line, and c on the other line. 2. Join the end-points of a, c, remote from the vertex, by I § 28 3. From the end-point of b, remote from a. draw a line parallel to I. I, prop. XXXIII This will cut off x, the line required. Proof. a : b — c : x. Prop. X, cor. 1 273. Definition. If a : b = b : x, x is called the third pro- portional to a and b. Corollary. The third proportional to two given lines van be found. For to find x such that a : b = b : x. make c = b in the above solution. Exercises. 423. The problem admits of a considerable variation of the figure, as suggested by the figure given in ex. 383. Invent another solution from this suggestion. 424. How many inches in the fourth proportional to lines respectively 2 in.. 3 in.. 5 in. long? In the third proportional to lines respectively 2 in. , 7 in. long ? 196 PLANE GEOMETRY. [Bk. IV. Proposition XXV. 274. Problem. To find the mean proportional between two given D B Given two lines, AD, DB. Required to find the mean proportional between them. Construction. 1. Placing AD, DB end to end in the same line, bisect AB at 0. I, prop. XXXI 2. With center and radius OB, describe a circle. § 109 3. Prom D draw DC J- AB, to meet circumference at C. I, prop. XXIX Then CD is the mean proportional. Proof. AD : CD = CD : DB. Prop. XXII, cor. 2, and § 238 275. Definition. A line is said to be divided in extreme and mean ratio by a point when one of the segments is the mean proportional between the whole line and the other segment. Thus, AB is divided internally in extreme and mean ratio at P, if AB : AP = AP : PB ; and externally in f p such ratio at P', if AB : AP* = AP' : P'B. E 1 To say that AB : AP = AP : PB is A B merely to say that AP 2 = AB • PB. This division is often known as the Golden Section or the Median Section. If the student understands quadratic equations he will see that if the length of AB is 6, and if AP = x, then PB = G — x, and v AP 2 = AB • PB, ... x 2 = 6 (6 - x), or x 2 + 6x - 36 - 0. Solving, x = - 3 =fc 3 VE. Prop. XXVI.] PROBLEMS. 197 Proposition XXVI. 276. Problem. To divide a line i)i extreme and mean ratio. Given the line AB. Required to divide AB in extreme and mean ratio ; i.e. to find P such that AB • PB = AP 2 . Construction. 1. Draw CB J_ AB and —\AB. 2. Describe a O with center C and radius CB. 3. Draw AC cutting the circumference in A' and Y. 4. Describe two arcs with center A and radii AX and A Y, thus fixing points P, P'. These are the required points. Proof for point P. Proof for point P'. AB 2 = AX- AY = AP(AX+XY) = AP (AP + AB) = AP 2 + AP • AB. .' . AB (AB - AP) = AP 2 . .'.AB'PB = AP 2 . AB 2 = AY- AX = P'A (A Y - XY) = P'A (P'A - AB) = P'A 2 - AB • P'A. AB (AB + P'^) = P'A 2 . .'.AB-P'B = P'A 2 . .'. AB is divided internally at P and externally at P' in Golden Section. It should be noticed that if the sense of the lines as positive or nega- tive is considered (that is, considering AP — — PA), the above solutions would be identical if X and Y were interchanged, and P' substituted for P. 198 PLANE GEOMETRY [Bk. IV. Proposition XXVII. 277. Problem. On a given line-segment as a side corre- sponding to a given side of a given polygon, to construct a polygon similar to that polygon. D ^C Fig. 2. Given the polygon ABCD and the line-segment A'B'. Required to construct on A'B' as a side corresponding to A B, a polygon A'B' CD' ~ ABCD. Construction. 1. In Fig. 1, place A'B' II AB. I, prop. XXXIII Draw AA', BB', meeting at 0; draw OC, OD. § 28 Draw B'C II BC, CD' II CD. I, prop. XXXIII Draw D'A'. Then A'B'C'D' — ABCD. Proof. 1. v OA:OA'=OB:OB'=OC:OC'=OD:OD', §244 .-. D'A' II DA. Prop. XII 2. v A'B':AB= OB': OB = B'C:BC = » Prop. X, cor. 3 and Z C'B'A' = Z CBA, and so for the other A, I, prop. XVII, cor. 2; ax. 3 . • . A'B' CD' ~ AB CD. Prop. XXI If A'B' = AB, as in Fig. 2, draw from C, D IPs to AA'; otherwise the construction is as above. It is left to the student to prove D'A' II DA, and A'B'C'D' ^ ABCD. .'. A'B'C'D' ~~ ABCD by § 262, cor. 1. BOOK V. — MENSURATION OF PLANE FIGURES. REGULAR POLYGONS AND THE CIRCLE. 1. MENSURATION OF PLANE FIGURES. Proposition I. 278. Theorem. Tivo rectangles having equal altitudes are proportional to their bases. , Given two rectangles R and R', with altitude a, and with bases b, b', respectively. To prove that R : R' = b:b'. Proof. 1. Suppose b and b' divided into equal segments, /, and suppose b = nl, and b' = n'l. (In the figures, n = 6, n' = 4.) Then if _k are erected from the points of division, R = n congruent rectangles al, and R' = ?i' " " « R R' n - al u' -al Why ? Note. The above proof assumes that b and b' are commensurable, and hence that they can be divided into equal segments I. The proposi- tion is, however, entirely general. The proof on p. 200 is valid if b and b' are incommensurable. 199 200 PLANE GEOMETRY. 279. Proof for incommensurable case. R' [Bk. V. ■ 1. Suppose b divided into equal segments I, and suppose b = nl, while b' = ?i'l + some remainder x, such that x < I. Then if _L's are erected from the points of division, R = n congruent rectangles al, and 22'= ? such that ax < al. al + a remainder ax, Why ? 2. Then b' lies between n'l and (V + 1) I, (In the figure, between 4 1 and 5 Z. ) and B' lies between n' • PA, .' . OP < PA, and .'.MP < PA. .-.incuts MY. Prop. VII.] PARTITION OF THE PEIUGON. 207 2. v A OPB is isosceles, OB = PB = PJ, the radius. 3. Also, 0A a + OB 1 = AB 2 + 2 0Jf • OA II, prop. IX, cor. 1 And •.-2 03I=OP, .'. 04 1 + OP 2 = ^P 2 + OP ■ OA. 4. And V OB 2 = PA 2 = OP • OJ, .-. OA 2 + OP 2 = ^4P 2 + OB 2 , from step 3, .-. OA 2 = AB 2 , and .-. 0.1 = AB. 5. .-. Z 0P.4 = Z0 = Z BPO, I, prop. Ill = AA + Z. PBA. I, prop. XIX And v £A = £ PBA, I, prop. Ill .\Z0 = 2ZJ. 6. .'. Z is f of a st. Z, or £ of a perigon. I, prop. XIX Corollary, yl perigon can be divided into 5 • ,J" equal parts. For if n = 0, then 5 • 2 n = 5 • 1 = 5, so that the corollary reduces to the problem itself. If n = 1, then 5 • 2 n = 5 • 2 = 10, and by bisecting Z A OB, the resulting angle is jL f a perigon. Similarly, by bisecting again, ^ f a perigon is formed, and so on. Exercises. 433. In the figure on p. 200, let OP= x, PA = r ; then show that x--(VE — 1). (Omit exs. 433, 434 if the student has not had quadratic equations.) 434. In the same figure, if OP = x and OA = a, show that x = |(S-V6). 435. On the sides a, b, c. of an equilateral triangle, points X, Y, Z are so taken that BX : XG - CY : YA = AZ : ZB = 2 : 1. Find the ratio of A XYZ to A .4£C. 208 PLANE GEOMETRY. [Bk. V. Proposition VIII. 287. Problem. To divide a perigon into fifteen equal angles. D\ \ C /B A Solution. 1. Make Z. AOB — \ of a perigon. Prop. VII 2. Make Z ^40D = ^ of a perigon. Prop. VI 3. Bisect Z BOD. I, prop. XXVIII 4. Then /. BOC '=£(£ — £) perigon = ^ of a perigon. Corollary. ^4 perigon can be divided into 15 • #* equal angles. Explain. 288. Note. That a perigon could be divided into 2", 3 ■ 2", 5 ■ 2 W , 15 • 2 n equal angles, was known as early as Euclid's time. By the use of the compasses and straight edge no other partitions were deemed possible. In 1796 Gauss found, and published the fact in 1801, that a perigon could be divided into 17, and hence into 17 • 2 n equal angles ; furthermore, that it could be divided into 2 m -f 1 equal angles if 2 m + 1 was a prime number ; and, in general, that it could be divided into a number of equal angles represented by the product of different prime numbers of the form 2 m -f 1. Hence it follows that a perigon can be divided into a number of equal angles represented by the product of 2 n and one or more different prime numbers of the form 2 m -f 1. It is shown in the Theory of Num- bers that if 2 m + 1 is prime, m must equal 2p ; hence the general form for the prime numbers mentioned is 2 2P + 1. Gauss's proof is only semi- geometric, and is not adapted to elementary geometry. Exercises. 436. Including the divisions of a perigon suggested by Gauss, there are 25 possible divisions below 100. What are they ? 437. As in ex. 43G, there are 13 possible divisions between 100 and 300. What are they ? Prop. IX.] REGULAR POLYGONS. 209 3. REGULAR POLYGONS. Proposition IX. 289. Problem. To inscribe in a circle a regular 'polygon having a given number of sides. Given a circle with center and radius OA. Required to inscribe in the circle a regular ?i-gon. Construction. 1. Divide the perigon into n equal parts (n being limited as in props. V-VIII and cors.) as AOB, BOC, COD, , B, C, D, lying on the circum- ference. 2. Draw AB, BC, CD § 28 Then ABCD is an inscribed regular %-gon. Proof. 1. AAOB^ABOC^ACOD^ , and AB = BC = CD = I, prop. I 2. .'. AB = BC= CD= Ill, prop. IV 3. .\ZDCB = Z. CBA = , '.' each staDds on (n — 2) arcs equal to AB. Ill, prop. XI, cor. 1 4. .". ABCD is an inscribed regular polygon. §§ 92, 201 210 PLANE GEOMETRY. [Bk. V. Corollaries. 1. The side of an inscribed regular hexagon equals the radius of the circle. Then Z AOB = \ of 360° = GO ; .-. Z BA 0, which = Z OB A = 60°. .-. A ABO is equilateral. 2. An inscribed equilateral polygon is regular. For by step 3 of the proof it is also equiangular ; and being both equi- lateral and equiangular, it is regular. Proposition X. 290. Problem. To circumscribe about a circle a regular polygon having a given number of Given a circle with center and radius OA. Required to circumscribe about this circle a regular ?i-gon. Construction. 1. Divide the perigon O into n equal parts {ii being limited as in props. V— VIII and cors.) by lines OW, OX, OY, 2. Bisect A WOX } XO Y, by radii to A, Ii, I, prop. XXVIII 3. From A, B, C, draw tangents to meet OJJ'at 7>, OX at E, Ill, prop. XXVI Then DEFG is the required polygon. Prop. X.] REG ULA R POL YG OXS. 211 Proof. 1. v DE± OA, EF± OB, , III, prop. IX, cor. 2 .-.A OAE^A OBE, and AE = BE, OE = OK I, prop. II 2. .'. the tangents from A and 7> meet X at the same point, E. 3. And v Z.DOE = A EOF, and Z OEZ) = Z i^O, .'. A DOE^A EOF, and Z># = ## Const. 1 Step 1 Why ? 4. Also, v Z GEE = A FED, each being the supple- ment of an Z equal to Z WOX, (Xame the Z.) I, prop. XXI, cor. .'. DEFG is a circumscribed regular polygon. §§ 92, 201 Corollaries. 1. The side of a regular hexagon circum- scribed about a circle of diameter 1, is 1/V3, or i v3. For it is (as in prop. IX, cor. 1) the side of an equilateral A whose altitude is ^. This is easily shown to be 1/Vs. (Show it.) 2. A circumscribed equiangular polygon is regular. Prove that any two adjacent sides are equal. Exercises. 438. In a right-angled triangle, any polygon on the hypote- nuse equals the sum of two similar polygons described on the sides as corresponding sides of those polygons. (Suggestion : P 2 : P z =z b 2 : c 2 ; .-. P 2 + P 3 :P 3 = &a + c 2 : C 2 = a 2 : c 2 = P 1 :P Z ; .-. P 2 + P 3 : Pi = P 3 : Ps = 1. This is one s of the generalized forms of the Pythago- rean theorem.) 439. If r is the radius of the circle, and 8 is the side of the inscribed equilateral triangle, then s = r V3. 212 PLANE GEOMETRY. [Bk. V. Proposition XI. 291. Problem. To circumscribe a circle about a given regular polygon. E D C B Given the regular polygon A BCD Required to circumscribe a circle about it. Construction. Bisect A DCB, CBA, the bisectors meeting at 0. Then is the center and OB the radius. Proof. 1. Draw OA, OD, OE, Then V A OCB, CBO are halves of oblique A, each is less than a rt. Z. 2. .'. CO and BO cannot be II, and they meet as at 0. 3. And v ZCBO = Z OB A, Const, and AB = BC, § 92, clef. reg. pol. 4. .'.A ABO £ A C50, and 6U = OC. Why ? Similarly each of the lines OB, OD, = OC. 5. .*. is the center, and OA, OB, are radii. § 108 292. Note. The inscription and circumscription of regular polygons are seen to depend upon the partition of the perigon. Elementary geometry is thus limited to the inscription and circumscription of regular polygons of 2", 3-2", 5 • 2", 15 • 2 n sides ; or, since the discovery by Gauss, to poly- gons the number of whose sides is represented by the product of 2 n and one or more different prime numbers of the form 2 m + 1. In addition to regular convex polygons, cross polygons can also be regular, the common five-pointed star being an example. Prop. XII.] REGULAR POLYGONS. 213 Proposition XII. 293. Problem. To inscribe a circle in a given regular polygon. Given a regular polygon WXY Required to inscribe a circle in it. Construction. 1. Circumscribe a circle about it. Prop. XI 2. From center of tins O draw OA _L WX. I, prop. XXX With center 0, and radius OA, a O may be inscribed. Proof. 1. Draw OB, OC, _L XY, YZ, Then v OA bisects WX, .'.A lies between W and X, and so for B, C, Ill, prop. V 2. And*.' WX=XY= , .\ OA = OB = Ill, prop. VII 3. .'.if with center and radius OA a O is described, then WX, XY, will be tangent to the O, III, prop. IX, cor. 3 and .'. the O is inscribed in the polygon. § 201, def. inscr. O Exercises. 440. Solve prop. XI by bisecting the sides AB, BC by perpendiculars, thus determining 0. 441. Inscribe a regular cross pentagon in a circle. (The regular cross pentagon, the pentagram, was the badge of the Pythagorean school.) 214 PLANE GEOMETRY. [Bk. V. Corollaries. 1. The inscribed and circumscribed circles of a regular polygon are concentric. For from step 2 of the construction and step 2 o^f: the proof, is the center of both circles. 2. The bisectors of the angles of a regular polygon meet in the common in- and circumcenter. For by the proof of prop. XI they meet in O, and by cor. 1 is the common in- and circumcenter. 3. The perpendicular bisectors of the sides of a regular polygon meet in the common in- and circumcenter. (Why ?) 294. Definitions. The radius of the circumscribed circle is called the radius of a regular polygon ; the radius of the inscribed circle, the apothem of that polygon ; the common center of the two circles, the center of that polygon. E.g. in the figure below, r is the radius, m the apothem, and the center of the regular polygon, part of which is shown as inscribed in the circle. Proposition XIII. 295. Theorem. The area of a regular polygon equals half the product of the apothem and perimeter. Given an inscribed regular polygon, of area a, perimeter p, apothem m. To prove that a = ^ mp. Prop. XIII.] REGULAR POLYGONS. 215 Proof. Let be the center and /• the radius of the circum- scribed circle. Let t be one of the A formed by joining to two consecutive vertices, and s a side of the polygon. Then area t equals -J- ms. "Why ? .'. the area of the polygon equals the sum of the areas of the triangles = \m X the sum of the sides = \ mp. Ax. 2 Corollaries. 1. The areas of regular polygons of the same number of sides are proportional to the squares of their %pothemSj of their radii, or of their sides. For - = i^L = J^L ; and from similar & and IV, prop. XX, a $ mp mp m r s p . . .j. a m 2 r 2 s 2 — = — = — = —• .-.by substitution — = —^ = — = — • m r s p a m' 2 r 2 s- 2. The 'perimeters of regular polygons of the same number of sides are proportional to their apothems, their radii, or their sides. Proved with cor. 1. Exercises. 442. The distance from the center to a side of the inscribed equilateral triangle equals r/2. 443. Draw a diameter AB of a circle with center ; then with center A and radius A draw an arc cutting the circumference in C, D ; draw CD. DB, BC, and prove A BCD equilateral. 444. The area of an inscribed regular hexagon is a mean propor- tional between the areas of the inscribed and circumscribed equilateral triangles. 445. Show how, with compasses alone, to divide a circumference into six equal arcs. 446. Prove that if AB, CD, two diameters of a circle, are perpen- dicular to each other, then ACBD is an inscribed square. 447. Let OX be the perpendicular bisector of line-segment AB at ; lay off on OX, OD = AO ; and, on DX, lay off DC = DB ; then prove that C is the center of the O circumscribed about the regular octagon of which AB is a side. 216 PLANE GEOMETRY. [Bk. V. 4. THE MENSURATION OF THE CIRCLE. 296. Postulate of Limits. The circle and its circumference are the respective limits which the inscribed and circum- scribed regular polygons and their perimeters approach, if the number of their sides increases indefinitely. The following may be read by the student in connection with the postulate, although it does not constitute a proof : 1. In the figure, suppose an in- and circumscribed regular ?i-gon rep- 360° resented. Then each exterior angle equals in each figure. 360° 180° 2. .-. each interior angle equals 180° > and .-. Z a = 90° ° n n 3. .-. if n increases indefinitely, Za = 90°, and p = r. 4. .-. the inscribed polygon = the circle, and its perimeter circumference. Similarly for the circumscribed polygon. the Coeoll aries. 1. The circumscribed regular polygon and its perimeter are respectively greater than the circle and its circumference ; the inscribed, and its perimeter, less. 2. If, on any finite closed curve, n points are assumed equidistant from each other, and each connected with the succeeding point by a straight line, then the curve is the limit which the broken line approaches if n increases indefinitely. Prop. XIV.] MENSURATION OF THE CIRCLE. 217 Proposition XIV. 297. Theorem. The ratio of the circumference to the diameter of a circle is constant. Proof. 1. Suppose any two circles, of circumferences c, c', radii r, r', and diameters d, d', respectively, to have similar regular polygons inscribed in them, of perimeters p, p\ respectively. Then p :p' = r : >•', Prop. XIII, cor. 2 = 2r:2r' = d: d'. IV, prop. VIII 2. And v >\ ?•', d, d' do not change when the number of sides of the polygons is doubled, quadrupled, , § 294, def. radius polyg. and '.' p = c, and p* = e', § 296, post, of limits r .c:c' = d:d'. IV, prop. IX 3. .' . c : d = c' : d' = the same for any (D. IV, prop. Ill Note. This constant ratio c : d is designated by the symbol it (pi), the initial letter of the Greek word for circumference (periphereia). The value of it is discussed in prop. XVII. Corollaries. 1. c = 7rd, or 2tti\ For if - = 7t, then c = nd. d 2. If the radius of a, circle is 1, then Q, = 2ir, or a semi- circumference equals it. 3. The circumferences of two circles are proportional to their radii. _, c 2 nr r For - = = — • c / 2 itr' r' Exercises. 448. Find, in terms of the radius of the circle, r, the side, apothem, and area of the inscribed and circumscribed equilateral triangle. 449. Also of the inscribed and circumscribed square. 218 PLANE GEOMETRY. [Bk. V. Proposition XV. 298. Problem. Griven the sides of the regular inscribed and circumscribed n-gons, to find the side of the regidar circumscribed 2 n-gon. A' M' C Solution. 1. In the figure, let AB = a side of the regular inscribed w-gon, i n ; " A'B' = " « " circumscribed " e n . Then BM ' = " " " inscribed 2?i-gon, i 2n ; and BC = J " " " circumscribed " c 2n . 2. But v OC bisects Z M'OB', Why ? .'. CB':M'C= OB': OM' (= OB), IV, prop. XIII = A'B' : .47?, IV, prop. X, cor. 3 = c « ■ k> 3. .'. CB' + M'C : J/'C = c n + s : i n , IV, prop. V or M'B':M'C = c n + i n : /„. 4. . ' . 2 M'B' :2M'C = c n + i n : i n . IV, prop. VIII 5. .-. , c B :c 2B = c fI + /„:/,„ or Hn n + '» Prop. XVI.] MENSURATION OF THE CIRCLE. 219 Proposition XVI. 299. Problem. Given the sides of the regular inscribed n-gon and the regular circumscribed 2 n-gon, to find the side of the regular inscribed 2 n-gon. Solution. 1. In the figure on p. 218, A M'BM - A CM'D. . Why ? .\ M'B : BM = CM' : JI'I). Why ? 2- Or h. :*<. = **.:*»,,, Why? * .•.^ = iV2^./,, Why? Corollaries. 1. //'p n . p 2n > P n ? ^n represent the perimeters of the polygons with sides i n , i 2n , c n , c 2n , respectively, then, (!) P ^» = ¥irt' and ( 2 ) ft- = v iv^- r n I p n For c 2n = Pi>„/2?i, c w = P„/w, /*„ =p 2 «/-«, and i„ = p n /n; substi- tute these in the tinal steps of props. XV, XVI. From prop. XV, P-j» _ P„/n-p„/n _ Pn-Pn/n 2n P n /n + p n / » P« + P» 2 " P»+P» From prop. XVI, P2 2ra » = I /JT 5J? . Pn i 2 \ 2n n P2n= ^Pn-P'2i 2. c n = — =? where r ?s f/^p radius. Vr 2 -ii n 2 >2f = f : Jf 2 - (| V = r : Vr^ - i tf. /' l " by multiplying by in. 220 PLANE GEOMETRY. [Bk. V. Proposition XVII. 300. Theorem. The approximate value of it is 3.14159 +. Proof. 1. In a regular hexagon inscribed in a circle of diam- eter 1, i G = J, and .'.^ 6 = 3. Prop. IX, cor. 1 2. Of the regular hexagon circumscribed about that O, c 6 = l/V3. Prop. X, cor. 1 3. .\P a = 6-c 6 = 3.4641016 4. From jh and P 6 can be found p^ and P 12 . Props. XV, XVI 5. From p>\2 and P 12 can be found ^> 24 and P 24 , and so on. Props. XV, XVI If the process were continued to a 1536-gon, ^ 1536 would be found to be 3.1415904, and P 1536 would be found to be 3.1415970. 6. And '.' c, or ird, which equals it • 1 or ir, lies between p n and P„, however large n may be, § 296, post, of limits, cor. 1 .". 7T lies between 3.1415904 and 3.1415970, and is, ' therefore, approximately 3.14159 +. Exercises. 450. The diagonals of a regular pentagon cut each other in extreme and mean ratio. 451. If ABODE is a regular pentagon, and AD cuts BE at P, prove that AP : AE = AE : AD. 452. To construct a regular pentagon equal to the sum of two given regular pentagons. 453. Find, in terms of the radius of the circle, r, the side of the inscribed regular pentagon. (Omit unless ex. 433 was taken.) 454. Also of the inscribed and circumscribed regular hexagon. 455. Also of the inscribed and circumscribed regular dodecagon. 456. Also of the inscribed regular decagon. (Depends on ex. 453.) Secs. 301, 302.] MENSURATION OF THE CIRCLE. 221 301. Notes. The computation in prop. XVII, which the student is not expected to make, is as follows : So. of sides 6 p 3. p 3.4641016 12 3.1058285 3.2153903 24 48 3.1326286 3.1393502 3.1596599 3.1460862 96 3.1410319 3.1427146 192 3.1414524 3.1418730 384 768 3.1415576 3.1415838 3.1416627 3.1416101 1536 3.1415904 3.1415970 302. The following historical notes on it are inserted to show the student how the subject of the mensuration of the circle has grown. The early approximation for 7T, in use among the ancient people, was 3. See I Kings, vii, 23; II Chron. iv, 2. "What is three hand-breadths around is one hand-breadth through. " — The Talmud. Ahmes, however, gave the equivalent of 3.1604. Archimedes seems to have been the first to employ geometric methods similar to that of props. XV, XVI for approximating it. He announced, "The circumference of a circle exceeds 3 times the diameter by a part which is less than i, but more than i^, of the diameter." Hero of Alexandria used both 3 and 3i. Ptolemy of Alexandria gave 3 T yo- Aryabhatta found 3.1416, by a method similar to that of prop. XVII. Brahmagupta used the values of Archimedes; also f§§£ and ||ib tne last being only another form for Ptolemy's. Metius gave the easily remembered value 355/113. Ludolph van Ceulen computed it to the equivalent of over 30 decimal places (the decimal fraction was not yet invented), and wished it engraved on his tomb at Leyden. On this account it is often called in Germany, "the Ludolphian number." Vega carried it to 140 decimal places. Dase carried it to 200 decimal places. Kichter carried it to 500 decimal places. More recently Shanks carried it to 707 decimal places. The symbol it is first used in this sense in Jones's " Synopsis Palma- riorum Matheseos," London, 1706. 222 PLANE GEOMETRY. [Bk. V. 303. Definition. It is now necessary fro extend our idea of equal surfaces. The definition at the beginning of Book II, § 142, is true, and it suffices for the cases there under con- sideration. But when curvilinear figures are compared with rectilinear, it is impossible to cut the surfaces into parts respectively congruent. Hence, we enlarge the definition, thus : Two surfaces are said to be equal if they have the same numerical measure in terms of a common unit. Thus, a circle having an area of 2 m 2 would equal a rectangle 2 m long by 1 m broad, even though they could not be cut into parts respectively congruent. 304. Table of Values. The following table of values of expressions involving ir will be found useful in computations concerning the circle, sphere, cylinder, cone, etc. : it = 3.14159 Vtt = 1.77245 180°/ tt = 57°. 29578 ;r/4 = 0.78540 1/Vtt = 0.56419 tt/180 = 0.01745 1 / it = 0. 31831 it V2 = 4. 44288 Approximate values : 7T 2 = 9.86960 V^72 = 1.25331 % = - 2 T 2 - = 3}, fff. The table is repeated, with other tables of value in numerical compu- tations, at the end of this work. 305. Radian Measure of Angles and Arcs. Since if A = any central angle and a = its arc, Aist.Z. = a: semi#ircumf. = a : wr. .'. A : st. Z./ir = a:r, or A: 180° /ir =a:r, or ^:57°.29+ = a : r. That is, the ratio of a central angle to st. Z./nr equals the ratio of its arc to an arc of the same length as the radius. Just as the "degree" is the unit for both angle and arc measure, it being understood to be ^^ of a perigon in the one case and ^^ of a circumference in the other, so a special name is given to st. Z /ir and to an arc which equals a radius in length ; this name is radian. In other words, a radian > Y Sec. 306.] MENSURATION OF THE CIRCLE. 223 is — of a st. Z., in angle measure, and — of a semicircumfer- IT IT ence, or an arc equal to a radius in length, in arc measure. Since r = — 180°, .*. r = 57°.29 + , where r stands for radian. IT irr it Also v 180° = 7T/-, .". 1° = — — - or — — of a radian, or .01745.33 of a radian. In most work in advanced mathematics the radian measure is used exclusively. In common measurements the degree is used. It is necessary in this work to use both. It is customary to express an angle in radians by the Greek letters a (alpha), (3 (beta), y (gamma), , the first letters of that alphabet. 306. Corollary. The length of an arc equals the product of the radius by the angle i In radians. For if a = length of arc, and a = its Z in radians, then - = — , 9 ^v C 2 7T .-. a = a = a ■ r. 27t Exercises. 457. Express the following in radians : 10°, 21° 20', 57°, 58°, 90°. 458. Express the following in degrees: 1.3090 r, .8058 r, .3636 r, .1687 r, .0029 r. 459. Express the following in radians : 100°, 180°, 270°. 460. Express the following in degrees : 3.4907 r, 5.2359 r, 0.2832 r, nr. 461. Find the lengths of arcs of 47° 50', 61° 20', 75° 40', the radius being 10. 462. Given the lengths of the following arcs, to find the radii of the various circles : 75° 10', 131.19 ; 32° 20', 2.822 ; 4°, .0698. 463. Show that the perimeters of the inscribed and circumscribed squares, the diameter of the circle being 1, are respectively 2.8284271 and 4 ; hence, find the perimeters of the inscribed and circumscribed regular octagons, and thus show that the value of it may be approxi- mated in this way. 464. The circumferences of certain © are 43.9823, 84.8230, 128.8053, 185.5340, 204.2035 ; find the diameters. 224 PLANE GEOMETRY. [Bk. V. Proposition XVIII. 307. Theorem. The area of a circle equals half the prod- uct of its circumference and radius. Given a, c, r, the area, circumference, and radius of a circle. To prove that a = ^ cr. Proof. 1. If a', p represent the area and perimeter of a cir- cumscribed regular polygon, then the apothem of that polygon is r. § 294 2. And a' = \pr. Prop. XIII 3. But a' = a, and \ pr = \ cr. § 296, post, of limits 4. .'. a = \ cr. IV, prop. IX, cor. 1 Corollaries. 1. a = ttv 2 . For c = 2 7tr. 2-« = £- (Why?) . 3. If s represents the area of a sector, and a its angle in radians, then s = v 2 a/2. For s : itr 2 = a : 2 it. (IV, prop. XVI, cor.) 4. Of two unequal circles, the greater has the greater circum- ference. c 2 For, by cor. 2, a = 4 it .-. 4 na = c 2 . .-. as the area increases, the circumference increases also. 5. The areas of two circles are proportional to the squares of their radii. „ a itr 2 r 2 For - = — - = — a itr 2 r" 2 Sec. 308.] MENSURATION OF THE CIRCLE. 225 308. Historical Note on Quadrature of the Circle. The ex- pression, " to square the circle," means to find the side of a square whose area equals that of a given circle. The solution of this problem by elementary geometry has been proved to be impossible. It nevertheless occupied the attention of many mathematicians before this impossibility was shown, and many ignorant people still attempt it. Some of the Pythagorean school claimed to have solved it, Anaxagoras (died 428 b.c.) wrote upon it, and hundreds of writers since then have discussed the sub- ject. It is closely related to finding a straight line equal to a given circum- ference ("to rectify the circumference"), and the two depend upon finding the value of it exactly. That it cannot be expressed exactly, nor as the root of a rational algebraic equation, was shown by Lindemann in 1882. For the mathematical discussion, see Klein's "Famous Problems of Elementary Geometry," translated by the authors. (Boston, Ginn & Co.) Exercises. 465. What is the radius of that circle of which the number of square units of area equals the number of linear units of circumference ? 466. Also, of which the number of square units of area equals the number of linear units of radius ? 467. Give a formula for a in terms of d, and the constant it. 468. A circle equals a triangle of which the base equals the circumfer- ence and the altitude equals the radius. 469. Find the areas of circles with radii 5, 7, 21, 35, 47, 50. (In these computations, for uniformity let it = 3.1416.) j 470. Also with diameters 2, 8, 11, 31, 42, 97. ^ 471. Find the radii of circles of areas 78.5398, 2042.8206, 4536.4598. 472. Also the diameters of circles of areas 2123.7166, 3318.3072, 56.745017. 473. Also the circumferences of circles of areas 95.0332, 452.3893. \y 474. Also the areas of circles of circumferences 267.0354, 191.6372. 475. The area of the ring formed between the circumferences of two * concentric circles of radii rj., r 2 , where T\ > r 2 , is it (r x + r 2 ) (j\ — r 2 ). 476. The area of that portion of the ring of ex. 475 cut off by the arms of the central angle a radians is \a{r x i- r 2 ) (ri — r 2 ) ; or, if ai, a 2 are arcs bounding that portion, the area = I (cii + a 2 ) (ri — r 2 ). Note. The remainder of the work may be omitted without destroying the integrity of the course. APPENDIX TO PLANE GEOMETRY. 1. SUPPLEMENTARY THEOREMS IN MENSURATION. Proposition XIX. 309. Theorem. If the sides of a triangle are a, b, c, and if s = i (a + b 4- c), s x = s — a, s 2 = s — b, s 3 = s — <3, then the area equals V s • 8 1 - s 2 • s 3 . Fig. b Fig. 2. Proof. 1. a 2 = b 2 + c 2 ^2 be', 2 be' taking the sign - for Fig. 1, + for Fig. 3, and being for Fig. 2. § 159 .*. c' = ± (b 2 + c 2 — a 2 ) /2 b, by solving the above equation for c'. Axs. 2, 7 2. But h 2 = c 2 - c' 2 = (c + c') (c - c') § 154 = [e + (b 2 + c 2 - a 2 ) /2 ft] [c - (6 2 + c 2 - a 2 ) /2 ft], by substituting the value of c' given in step 1. .-. K 1 = (2 be + £ 2 + c 2 - a 2 ) (2 6c - 6 2 - c 2 + a 2 ) /4 ft 2 , by removing parentheses and simplifying. 3. .'. 4 b 2 h 2 = [<7, + c) 2 - a 2 ] [a 2 - (b - c) 2 ], by multi- plying by 4 & 2 and factoring. .-.4 ^A 2 = (b + c -f A ABC V Proof. Suppose C\D J_ AB. Then AC\ > DC l9 I, prop. XX and .'. its equal AC 2 > BC\. .'.AABC 2 > AABC 19 II, prop. II, cor. 3 since they have the same bases but different altitudes. Exercises. 482. Find in radians the angle a: of a sector of a circle of radius r, such that the number of square units of its area equals the number of linear units of its entire perimeter. 483. Interpret the result of ex. 482 for r = 2 ( 1 + - Y Discuss it for r <£ 2. Discuss it for r < 2 (l + -Y 484. In the Sulvasutras, early semi-theological writings of the Hindus, it is said: "Divide the diameter into 15 parts and take away 2; the remainder is approximately the side of the square equal to the circle." From this compute their value of it. 485. On AB describe a semicircle, and in it inscribe the isosceles tri- angle ABC ; on BC and CA describe semicircles opposite the A ABC. Show that A ABC = the sum of the two limes thus formed. (The limes of Hippocrates.) 486. Six lights are placed regularly on the circumference of a circle of radius 21 ft.; what are the distances of each from each of the others ? (To 0.01.) Prop. XXIII.] MAXIMA AX1) MIX IMA. 231 Proposition XXIII. 315. Theorem. Of all isoperimetric triangles on the same base the isosceles is the maximum. ^B' Given two isoperiuietric A ABC and ABX, A ABC being isosceles, with AC — BC. To prove that AABC> A ABX. Proof. 1. On AC produced, let CB' = AC; draw B'B, B'X; suppose CD II AB. Then v AC = CB', .-. BD = DB'. I, prop. XXVII, cor. 2 2. And v CB = AC, .-. CB= CB', and CDA.BB'. Why ? Ax. 1 3. . • . A C + CJB = ^ C + CT' < AX + X£'. Ax. 2 ; I, prop. VIII 4. * . ' .IX -f- X£ = A C + G#, Why ? .'. AT + X£ < AX + X#', and .-. XB lie on the same side of CD, I, prop. XX, cor. 3 and .'.A ABC > A ABX. II, prop. II, cor. 3 Corollary. Of all isoperimetric triangles, that which is equilateral is the maximum. (Why ?) 232 PLANE GEOMETRY. [Bk. V. Proposition XXIV. 316. Theorem. Of all triangles having the same base and area, the isosceles has the minimum perimeter. Given the A AB C and ABX having the same base and area, with AC = BC. To prove that perimeter ABC < perimeter ABX. Proof. 1. Suppose CYWAB; AC produced so that CB' = AC; B'X drawn ; and B'B drawn cutting C Y at D. Then v A ABC = A ABX, .'. CY passes through X. II, prop. II, cor. 4 2. And v AC= CB', ,\BD = DB'. I, prop. XXVII, cor. 2 3. And v A BBC ^ A B'DC, I, prop. XII .-.CDA.BB', Why? and .'. XB = XB'. I, prop. XX 4. But AC + CB' < AX+ XB', I, prop. VIII and .'. AC + CB < AX + XB. 5. .-. perim. ABC < perim. ABX. Why ? Corollary. Of all equal triangles, that which is equi- lateral has the minimum perimeter. For whatever side is taken as the base, the perimeter is less if the other two sides are equal. Prop. XXV.] MAXIMA AND MINIMA. 233 Proposition XXV. 317. Theorem. If the ends of a line of given length are joined by a straight line, and the area of the figure enclosed is a maximum, it takes the form of a semicircle. P __ Given a line APB (the curve in the figure), of given length, and AB joining its end-points. To prove that, if the area of the figure ABP is a maximum, ABP is a semicircle. Proof. 1. Let P be any point on the line; then joining A and P, B and P, let the segments cut off by AP, BP be called s 1} s 2 , and A ABP called t, as in the figure. Then Z P is a right angle ; for if not, without changing s u s 2 , the area of t could be increased by making Z P right. Prop. XXII 2. But this is impossible if ABP is a maximum, and similarly for any other point on APB. Why ? 3. .'. the area enclosed is a maximum when the line connecting A and B subtends a right angle at every point on the curve. Note. It will be seen that examples of maxima or minima involve also the idea of symmetry (§ 68). This fact is of value in solving problems in maxima and minima. Exercise. 487. Given the points A, B, on the same side of line X'X, to find on X'X a point P such that Z X'PA = Z BPX. Prove that AP + PB is the shortest path from A to X'X and back to B. (Reflected ray of light.) 234 PLANE GEOMETRY. [Bk. V. Proposition XXYI. 318. Theorem. Of all isoperimetric plane figures the maximum is a circle. Proof. Suppose A, B points bisecting the given perimeter, AB cutting the figure into two segments, s u s 2 . Then are maxima when they are semicircles, and AB is a diameter, Why ? Proposition XXVII. 319. Theorem. Of all equal plane figures the circle has the minimum perimeter. Given circle C — plane figure P. To prove that circumference C < perimeter P. Proof. 1. Suppose X a circle of circumference equal to perim- eter P. Then P < X, Prop. XXVI and .'. C < X. Subst. 2. .*. circumference C < circumference A', § 307, cor. 4 and . * . circumference C < perimeter P. Subst. Prop. XXVIII. MAXIMA AND MINIMA. 235 Proposition XXVIII. 320. Theorem. A polygon with given sides is a maximum when it is inscriptible. Given two polygons, F and P', with given sides a, b, c , F being inscribed in a circle, and F' not inscriptible. To prove that F > F'. Proof. 1. Name the circular segments on a, b, (opposite P), A, B, ; suppose congruent segments constructed on a, b, (opposite P'). Then F + A + B + > F' + A + B + Prop. XXVI 2. .\P> P' Why? Exercises. 488. If the diagonals of a parallelogram are given, its area is a maximum when it is a rhombus. 489. What is the minimum line from a given point to a given line ? Where has this been proved ? 490. Into what two parts must a given number be divided so that the product of those parts shall be a maximum ? (Compare ex. 479.) 491. As a corollary to ex. 479, show that of isoperimetric rectangles the square is the maximum. 492. Find the point in a given straight line such that the tangents drawn from it to a given circle contain the maximum angle. 493. A straight ruler, 1 foot long, slips between the two edges of the floor (the edges making a right angle). Find the position of the ruler when the triangle formed by the edges and ruler is a maximum; also the area of that triangle. 236 PLANE GEOMETRY. [Bk. V. Proposition XXIX. 321. Theorem. Of all isojjerimetric polygons of a given number of sides, the maximum is regular. B X Given P, the maximum polygon of a given perimeter and a given number of sides. To prove that P is regular. Proof. 1. Any two adjacent sides, AB, BC, must be equal. For if unequal, as AX, XC, then A AXC could be replaced by A ABC, having AB = BC, thus enlarging P without changing the perimeter. But this is im- possible because P is a maximum. Prop. XXIII 2. And hence P is inscriptible because its sides are given. Prop. XXVIII 3. .". P is regular. V, prop. IX, cor. 2 Exercises. 494. Considering only the relation of space enclosed to amount of wall, what would be the most economical form for the ground plan of a house ? 495. Of all triangles in a given circle, what is the shape of the one having the greatest area ? Prove it. 496. Through a point of intersection of two circumferences draw the maximum line terminated by the two circumferences. 497. Of all triangles of a given base and area, the isosceles has the greatest vertical angle. 498. Draw the minimum straight line between two non-intersecting circumferences. Prop. XXX.] MAXIMA AXD MINIMA. 237 Proposition XXX. 322. Theorem. Of two isoperimetric regular polygons, that having the greater number of sides is the greater. Proof. 1. Let ABCD be a square, P a point on DA, A PCX isoperimetric with A PCD and having CX= PX. Then A PCX > A PCD, Prop. XXIII and .-. pentagon ABCXP > □ ABCD. Ax. 4 2. But pentagon ABCXP would, with the same perim- eter, be greater if it were regular. Prop. XXIX 3. .'. a regular pentagon is greater than an isoperi- metric square. Similarly, a regular hexagon would be greater than an isoperimetric regular pentagon, and so on. Exercises. 499. A cross-section of a bee's cell is a regular hexagon. Show that this is the best form for securing the greatest capacity with a given amount of wax (perimeter). 500. Find the maximum rectangle inscribed in a given semicircle. 501. Find the minimum square inscribed in a given square. 502. Draw the minimum tangent from a variable point in a given line to a given circle. 503. What is the area of the largest triangle that can be inscribed in a circle of radius 5 ? 504. Given a square of area 1. Find the area of an isoperimetric (1) equilateral triangle, (2) regular hexagon, (3) circle. 238 PLANE GEOMETRY. [Bk. V. 3. CONCURRENCE AND COLLINEARITY. Proposition XXXI. 323. Theorem. If X, Y, Z are three points on the sides a- b, c, respectively, of a triangle ABC, such that the pterpen- diculars to the sides at these points are concurrent, then (BX 2 - XC 2 ) + (CY 2 - YA 2 ) + (AZ 2 - ZB 2 ) = ; and conversely. Proof. Let P be the point of concurrence, and draw PA, PB, PC. Then (BX 2 - XC 2 ) + (CY 2 - YA 2 ) + (AZ 2 - ZB 2 ) = PB 2 - PC 2 + PC 2 - PA 2 + PA 2 - PB 2 = 0, for BX 2 - XC 2 = (BP 2 - PX 2 ) - (PC 2 - PX 2 ) = BP 2 -PC 2 , and so for the rest. Conversely: 1. Suppose the _b from X, Y, to meet at P; and suppose PZ' _L c. Then as above, (BX 2 - XC 2 ) + (CY 2 - YA 2 ) + (AZ' 2 - ZB 2 ) = 0. 2. But (BX 2 - XC 2 ) + (CY 2 - YA 2 ) + (AZ 2 - ZB 2 ) = 0, and AZ' 2 - Z'B 2 = AZ 2 - ZB 2 . Why ? 3. .'. AZ' 2 - AZ 2 = Z'B 2 - ZB 2 - but those differences have opposite signs and cannot be equal unless each is zero. 4. .*. Z must coincide with Z'. Prop. XXXII.] CONCURRENCE AND COLLINEARITY. 239 Proposition XXXII. 324. Theorem. If three lines, x, y, z, drawn from the vertices of triangle ABC to meet a, b, c in X, Y, Z, are con- current, then AZ EX CI' ZB XC — — = 1 ; and conversely. Proof. 1. Let P be the point of concurrence. Then v A APC, PBC have the base PC, they are proportional to their altitudes, and .'.to AZ, ZB. Why ? AZ A .IPC 2. and and ZP A PPG' BXABPA XC~ A APC 9 CY APBC YA~ ABBA _ . AZ BX CY 3 - ' ZBXC'YA = 1 - Ax ' 6 Conversely : Let CP meet c in Z' ; then as above, .4Z' PA' CT_ ZP * XC ' YA ~ K _ . ^ px cr °- But zpxg'T^ = l Glven 6. ' ' Z?B~ ZB 7. .'. Z' must coincide with Z. IV, prop. XI, cor. Note. Ceva's theorem. 240 PLANE GEOMETRY. [Bk. V. Proposition XXXIII. 325. Theorem. If three points, X, Y, Z, lying respectively on the three sides a, b, c of triangle ABC, are collinear, then and conversely. AZ BX CY ZB ' XC ' YA -1 2. And similarly, and 3. Ax. 6 Proof. 1. Let I, m, n be perpendiculars from A, B, C onlF. Then by similar A, ZB being here negative, AZ I ZB~ -m BX _m XC~ n CY _n YA~ I AZ BX CY *'' ZB' XC' YA Conversely : Let XY meet AB in Z' ; then as above, AZ' BX CY Z'B' XC YA AZ BX CY ZB XC YA AZ' AZ •'* Z'B~ ZB 7. .'. Z' must coincide with Z. IV, prop. XI, cor. Note. Menelaus's theorem. 4. But 1. -1. Given Prop. XXXIV.] CONCURRENCE AND COLLINEARITT. 241 Proposition XXXIV. 326. Theorem. If a circumference intersects the sides, a, b, c, of a triangle ABC, in the points Ai and A 2 , Bi and B 2 , Ci and C 2 , respectively, then ACi BAi CBi AC2 BA2 CB2 _ 1 CiB ' AiC " B X A " C 2 B A 2 C ' B 2 A ~~ * Proof. 1. AC 1 -AC 2 = B l A-B 2 A, Why? and BA X • BA 2 = C\B • C 2 B, and CB l - CB 2 = A X C - A 2 C. 2. .'. by axs. 6 and 7, the above result follows. Note. This theorem, known as Carnot's theorem, is not a proposi- tion in concurrence or collinearity. It is introduced as leading to the proof of the very celebrated theorem following, one commonly known as the Mystic Hexagram, discovered by Pascal at the age of 16. The theorem is also easily proved when the triangle is inscribed or circumscribed. Exercises. 505. By means of Ceva's theorem, prove that the three medians of a triangle are concurrent. 506. Also, that the bisectors of the three interior angles of a triangle are concurrent. 507. Also, that the bisectors of two exterior and of the other interior angles of a triangle are concurrent. 508. Also, that the perpendiculars from the vertices of a triangle to the opposite side are concurrent. 242 PLANE GEOMETRY. [Bk. V. Proposition XXXV. 327. Theorem. If the opposite sides of an inscribed hex- agon intersect, they determine three collinear points. Given an inscribed hexagon, ABCDEF, such that BA and DE meet at P, CD and AF at Q, BC and FE at R. To prove that P, Q, R are collinear. Proof. 1. Call the A determined by AB, CD, and EF, LMX. as in the figure. Then from Menelaus's theorem, LP MD NE and and PM DN MQ NF QN NR RL FL LB BM EL LA AM MC CN = -1, = -1, = -1. 2. .'. By multiplying and recalling Cam of s theorem, LP MQ NR PM ' QN ' RL~ 3. .'. by Menelaus's theorem, P, Q, R are collinear. Exs. 509-521.] CONCURRENCE AXD COLLINEARITY. 243 MISCELLANEOUS EXERCISES. 509. Show that the following is a special case of prop. XXXI: The perpendicular bisectors of the sides of a triangle are concurrent. 510. Also, the perpendiculars from the vertices of a triangle to the opposite sides are concurrent. 511. If three circumferences intersect in pairs, the common chords are concurrent. 512. By means of Menelaus's theorem, prove that the points in which the three bisectors of the exterior angles of a triangle meet the opposite sides are collinear. 513. Also, that the points in which the two bisectors of two interior angles of a triangle and the other exterior angle meet the opposite sides are collinear. 514. The orthocenter, 0, of A ABC is determined by the perpendicu- lars AD, BE. Prove that AO • 01) = BO ■ OE. 515. Draw a circle with a central right angle ^4 05. A and B being on the circumference; bisect ZAOB by 031, meeting AB at 31; draw MP __ OA ; then see if the following is true in general : AB = chord AB + PA. (Consider special cases, AB = 120°, 180°, 360°.) 516. Given the base and the vertical angle of a triangle ; construct it so that its area shall be a maximum. 517. AB is a diameter of a circle of center O ; from any point P on the circumference, PC is drawn perpendicular to AB) from C a perpen- dicular CE is drawn to OP. Prove that PC is a mean proportional between OA and PE. 518. On side a of A ABC, point P is taken such that Z PAC = Z B. Prove that CP . CB = AP 2 : AB~. Investigate for three cases, ZA<, = ,>ZB. 519. ABC is a triangle right-angled at C ; CD±c. Prove that AD-.LB = CA--.BC 2 . 520. If O, C are the centers of two fixed circles, such that the cir- cumference of C passes through O. and if a tangent to circumference of at T cuts circumference of 0" at X, Y. then OX ■ OY is constant. (If the center-line meets the circumference of C at A, A XTO <-" A AYO.) 521. If is the orthocenter of triangle ABC, and A', B' ', C are the mid-points of a.b,c; 3I a . 3T h . 3I C are the mid-points of AO, BO, CO; P a , P}j, P c are the feet of the perpendiculars from A. B, C to a. b. c ; prove that A\ B' . C", M a , 3I b . 3E. P a , P b , P c are coney clic. (The •• Nine Points Circle.") SOLID GEOMETRY. BOOK VI. — LINES AND PLANES IN SPACE. 1. THE POSITION OF A PLANE IN SPACE. THE STRAIGHT LINE AS THE INTERSECTION OF TWO PLANES. 328. Definitions. Through three points, not in a straight line, any number of surfaces may be imagined to pass. For example, through the points A, B, _/_ A^ C the surfaces P and S may be imagined to 329. A plane surface (also called & plane) is a surface which is determined by any three of its points not in a straight line. In the figure, P represents a plane, for it is determined by the points A, B, C. But S does not represent such a surface. A plane is, of course, supposed to be indefinite in extent. This definition, and the following postulates, are repeated, for con- venience, from the Plane Geometry. In drawing a figure it should be remembered that a plane, like a line, has no thickness, and that it is indefinite in extent. Nevertheless, it aids the eye in understanding the figure, if we represent the plane as a rectangle, lying in perspective, and having a slight thickness. Exercises. 522. Show that if there are given four points in space, no three being collinear, the number of distinct straight lines determined by them is six ; if there are five points, the number of lines is ten. 523. Hold two pencils in such a way as to show that a plane cannot, in general, contain two straight lines taken at random in space. 244 Prop. I.] LINES AND PLANES IN SPACE. 245 330. Postulates of the Plane. (See § 29.) 1. Three points not in a straight line determine a plane. 2. A straight line through two points in a plane lies wholly in the plane. 3. A plane may be pjassed through a straight line and revolved about it so as to include any assigned point in space. 4. A portion of a plane may be produced. 5. A plane is divided into two parts by any one of its straight lines, and space is divided into two parts by any plane. 331. Solid Geometry treats of figures whose parts are not all in one plane. Proposition I. 332. Theorem. A plane is determined by a straight line and a point not in that line. Given the line AB, and the point P not in that line. To prove that AB and P determine a plane. Proof. 1. Only one plane contains pts. A, B, and P. § 330, 1 (§ 330, 1. Three points not in a straight line determine a plane.) 2. And that plane contains line AB. § 330, 2 (§ 330, 2. A straight line joining two points in a plane lies wholly in the plane.) 3. .*. only one plane contains AB and P. 333. Definition. Lines or points which lie in the same plane are said to be coplanar. 246 SOLID GEOMETRY. [Bk. VI. Corollaries. 1. A plane is determined by two intersecting lines. Let the lines AB, CD intersect at 0. Then only one plane contains AB and C. Prop. I And that plane contains the point 0, for lies in the line AB. § 330, 2 And since that plane contains C and O, it contains CD. § 330, 2 2. A plane is determined by tiuo parallel lines. For the parallels lie in one plane, by definition (§ 82). And only one plane can contain these parallels, since a plane is deter- mined by either line and any point of the other. Draw the figure. 3. If a plane contains one of two ])arallel lines and any point of the other, it contains both parallel lines. For it must be identical with the plane determined by the two paral- lels ; otherwise more than one plane could contain either parallel and any point in the other. Proposition II. 334. Theorem. The intersection of two planes is a straight line. /N Given two intersecting planes, M, N. To prove that their intersection is a straight line. Prop. II.] LINES AND PLANES IN SPACE. 247 Proof. 1. Let P be a point common to M and N. Then a pencil of lines through P, in the plane X, must lie partly on one side of 21 and partly on the other, because 21 divides space into two parts. § 330, 5 2. Hence, in general, a line connecting a point in the pencil on one side of 21, with a point on the other side, must cut 21 at some other point than P, — say at Q. 3. Then 21 and X have two points in common. 4. Then every point in the straight line through P and Q lies in plane 21, § 330, 2 and also in plane X, for the same reason. 5. .'. the straight line PQ is common to both planes. 6. If there were any point not in PQ, common to 21 and X, the planes would coincide. Prop. I Corollary. A point common to two planes lies in their line of intersection. Proved in step 6. Exercises. 524. State the four methods, already mentioned, of determining a plane. 525. Is it possible for three planes to have a straight line in common ? Draw a figure to illustrate. 526. If two planes have three points in common, will they necessarily coincide ? 527. Four planes, no three containing the same line, intersect in pairs ; how many straight lines do they determine by their intersections ? 528. What is the only rectilinear polygon that is necessarily plane ? Why? 529. Prove that all transversals of two parallel lines are coplanar with the parallels. 530. What is the reason that a three-legged chair is always stable on the floor while a four-legged one may not be ? 248 SOLID GEOMETRY. [Bk. VI. Proposition III. 335. Theorem. If three planes, not containing the same line, intersect in pairs, the three lines of intersection are either concurrent or parallel. Given planes AD, CF, EB, intersecting in AB, CD, EF. To prove that AB, CD, EF are either concurrent or parallel. Proof. Case I. If CD meets AB, as at 0, to show the three lines concurrent. 1. v is in AB, it is in plane EB. § 330, 2 2. Similarly, V is in CD, it is in plane CF. Why ? 3. V is in planes EB and CF, EF passes through 0. Prop. II, cor. 4. .'. AB, CD, EF are concurrent in 0. Case II. If CD II AB, to show the three lines parallel. 1. If AB were not II EF, CD would pass through their common point. Case I 2. But this is impossible, for CD II AB. Given 3. If CD were not II EF, AB would pass through their common point. Case I 4. But this is impossible, for CD II AB. . Given 5. .*. as no two can meet, and as each pair is coplanar, they are parallel. Def. II lines Corollary. If tivo intersectl» APYO. To prove that PA PA. I, prop. XX Conversely : Given PO ± MX, PY and PA two obliques such that PA < PY. To prove that Z PAD > Z P YO. Proof. 1. Suppose X taken on OA so that OX = Y. 2. Then A POX ^ A PO Y, Z PXO =ZPYO, and PX=PY. Why? 3. ..PA< PX, '.' PA PX. Why? Prop. XII.] LINES AND PLANES IN SPACE. 259 6. .'. X is on OA produced, and .'.ZPAO> Z PXO. I, prop. V 7. .'. Z PJ > Z P FO. Subst. 2. Given PO J_ MN, PY and P.4 two obliques such that OA < OY. To prove that PAlnne. For if QfP II QP, then QP = Q'P' < QfR. Prop. XI, 1 355. Definition. The length of the common perpendicular from one line to another is called the distance between those lines. K Sbcs. 350-358.] PENCIL OF PLANES. 265 3. PENCIL OF PLANES. 356. Definitions. Any number of planes containing the same line are said to form a pencil of planes ; the line is called its axis. 357. Any two planes of a pencil are said to form a dihedral angle. LMN, a pencil of planes ; A B, the axis of the pencil. Dihedral angles formed by the planes M and ^V. Dihedral angle MN measured by plane angle BOC. AOthe edge of the dihe- dral angle. The two planes of a dihedral angle are called the faces, and the axis of the pencil is called the edge of the dihedral angle. Two intersecting planes form more than one dihedral angle, just as two intersecting lines form more than one plane angle, the latter term now being used to designate an angle made by lines in a plane. 358. A plane of a pencil turning about the axis from one face of a dihedral angle to the other is said to turn through the angle, the angle being greater as the amount of turning is greater. Since the size of a dihedral angle depends only upon the amount of turning just mentioned, it is independent of the extent of the faces. 266 SOLID GEOMETRY. [Bk. VI. 359. If perpendiculars are erected from any point in the edge of a dihedral angle, one in each face, the size of the plane angle thus formed evidently varies as the size of the dihedral angle. Hence a dihedral angle is said to be measured by that plane angle, or, strictly, to have the same numerical measure. 360. A dihedral angle is said to be acute, right, obtuse, oblique, reflex, straight, according as the measuring plane angle is so, and it is usually named by its measuring plane angle, or merely by its faces in counter-clockwise order. The terms adjacent angles, bisector, sum and difference of dihedral angles, point within or without the angle, complement, supplement, conjugate, and vertical angles will readily be understood from the corresponding terms in plane geometry. As with plane angles the smallest angle made by two intersecting lines is, in general, to be understood unless the contrary is stated, so with dihedral angles. If a dihedral angle is right, the planes are said to be perpen- dicular to each other. E.g. in the following figure, ZY ± MN.. Proposition XVIII. 361. Theorem. If a line is perpendicular to a plane, any jolane containing this line is also perpendicular to that plane. Given OY perpendicular to the plane MN, and ZY any plane containing OY. To prove that ZY± MN. Prop. XIX.] PENCIL OF PLANES. 267 Proof. 1. Suppose OX, in MN, _L OZ, the intersection of MN and ZY. Then A YOZ, XO Y are right A. Why ? 2. But v Z Z07 fixes the measure of the dihedral Z, §359 .'. Zr_L JlfiV. Def. Proposition XIX. 362. Theorem. If two planes are perpendicular to each other, any line in one of them, perpendicular to their inter- section, is perpendicxdar to the other. Given the planes ZYA.MN, OZ their intersection, and OX, in MX, _L OZ. To prove that OX 1. ZY. Proof. 1. Let OY, in ZY, he ± to OZ at 0. Then Z A^O Y is the measuring angle. § 359 * 2. .-. A XOY is right. Def. _L planes 3. But ' Z ZOX is also right, Why ? .'. OXA.ZY. Why? Corollaries. 1. If two planes are pern endicular to each other, a line from any point in their line of intersection, per- pendicular to either, lies in the other. By the theorem, OY ± MN, and it lies in ZY; and by prop. VII, cor. 2, only one perpendicular to MN can be drawn from O. 268 SOLID GEOMETRY [Bk. VI. 2. Through a point without a line not more than one plane can pass perpendicular to that line. For if through Y another plane could pass _L OX, it would pass through O, v A XOY = rt. Z, and only one _L can be drawn from Y to OX. But the plane would also include line OZ, else there would be two Js from to OX in the plane MN. Proposition XX. 363. Theorem. If each of two intersecting planes is per- pendicular to a third plane, their line of intersection is also perpendicular to that plane. Given two planes, Q, R, intersecting in OP, and each per- pendicular to plane M. To prove that OP _L M. Proof. 1. A _L to Jf from lies in Q and in R. Prop. XIX, cor. 1 2. .'. it coincides with OP, the only line common to Q and 22. .\ OP J_ M. Exercises. 564. To construct a plane containing a given lino, and parallel to another given line. (Assumed in step 1 of prop. XVII.) 565. Prove that vertical dihedral angles are equal. 566. How many degrees in the measure of the dihedral angle between the plane of the earth's equator and the ecliptic ? 567. Prove that the edge of a dihedral angle is perpendicular to the plane of the measuring angle. 568. Prove that a line and its projection on a plane determine a second plane perpendicular to the first. Prop. XXI.] PENCIL OB PLANES. 269 Proposition XXI. 364. Theorem. Any point in a plane which bisects a dihedral angle is equidistant from the faces of the angle. Given a dihedral angle, with faces Q, B, and edge CD, bisected by plane B; P, any point in B, with PX± Q, PY±B. To prove that PX=PY. Proof. 1. Let M be the plane of PX, PF, and D its intersec- tion with CD. 2. Then vPJlft .'. MA. Q. Why? 3. Similarly, M _1_ R. Why? 4. .'.MA. CD. Prop. XX 5. . ' . CD _L DX, D Y, DP, whose A therefore measure the dihedral A. § 359 6. .'.AXDP = APDY. And v A X= A Y= rt. Z, and DP = DP, 7. .'.A DXP ^ADYP, and PX = PY. § 88, cor. 7 Corollary. TAe focz^s of points that are equidistant from two intersecting planes is the pair of planes bisecting their dihedral angles. 270 SOLID GEOMETRY. [Bk. VI. Proposition XXII. 365. Theorem. If from any point lines are drawn per- pendicular to two intersecting planes, the angle formed by these perpendiculars has a measure equal or supplemental to that of the dihedral angle of the planes. Given the planes M, Q, intersecting in i ; lines PX _L M, PY J_ Q-, and plane PYX cutting i at A. To prove that Z YPX is equal or supplemental to the dihe- dral angle MQ. Proof. 1. Plane YXP _L M, also _L Q. Prop. XVIII 2. .-. plane YXP _L i. Prop. XX 3. .'. XA and YA J_ i. § 339 4. .-. Z XA Y measures dihedral Z MQ. § 359 5. But vZZ=Zr=rt. Z, Given .*. Z YPX is supplemental to AX AY, or dihedral Z Jf©. I, prop. XXI, cor. Corollary. If the point is within the dihedral angle, the angles are supplemental. Definition. If two planes do not meet, however far pro- duced, they are said to be parallel. The term "pencil of parallels is applied to planes as well as to lines. Prop. XXIII.] PENCIL OF PLANES. 271 Proposition XXIII. 366. Theorem. Planes perpendicular to the same straight line are parallel I x 1 M \ r~^\ Given two planes, M, X, J_ line XY, at X, Y, respectively. To prove that 31 II X. Proof. If M and X should meet, as at P, then two planes would pass through P _L XY, which is impossible. Prop. XIX, cor. 2 Exercises. 569. Prove that through a point without a plane any number of lines can pass parallel to the plane. 570. Problem : To bisect a dihedral angle. 571. To find the locus of points equidistant from two fixed planes, and equidistant from two fixed points. I 572. To find a point equidistant from two given planes, equidistant from two given points, and also at a given distance from a third plane. 573. Prove prop. XXII for the case in which the point P is taken in plane M. 574. In the figure on p. 270, as Z XA Y increases from zero to a straight angle, what change does Z YPX undergo ? 575. Also, suppose Z XAY = 120°; what angle will PY make with plane M , if produced through Q to M ? 576. Given two points, F, W, in two intersecting planes, If, Q, respectively. Find Z in the line of intersection of M and Q, such that VZ + ZW shall be a minimum. 577. If from two points on a line parallel to a plane, parallel lines are drawn to that plane, a parallelogram is formed. 272 SOLID GEOMETRY. [Bk. VI. Proposition XXIV. 367. Theorem. The lines in which two p>arallel planes intersect a third plane are parallel. Given two parallel planes, M, N, intersected by a third plane, T, in lines a, b. To prove that a II b. Proof. 1. a and b are in the same plane T. 2. And they cannot meet, because they are in M and N respectively, and M II N. 3. .'. they are parallel by definition. Corollary. A line perpendicular to one of two parallel planes is perpendicular to the other. Pass two planes through that line and apply prop. XXIV and the def. of a plane ± to a line. Exercises. 578. Through a given point only one plane can pass parallel to a given plane. 579. If two parallel planes intersect two other parallel planes, the four lines of intersection are parallel. 580. Parallel lines have parallel projections on any plane. (Suppose, as a special case, that the lines are perpendicular to the plane.) 581. If two lines are at right angles, are their projections on any plane also at right angles ? 582. If two planes are perpendicular to each other, any line perpen- dicular to one of them is either parallel to or lies in the other. Prop. XXV.] PENCIL OF PLANES. 273 Proposition XXV. 368. Theorem. If two straight lines are cut by parallel planes, the corresponding segments are proportional. Given ABC, DEF, two lines cut by planes P, Q, R, in points A, B, C and 1), E, F. To prove that AB : BC = BE : EF. Proof. 1. Suppose the line GIIF, drawn through F, II ABC, cutting P, Q at G, H, respectively. Then AC, GF determine a plane; also DF, GF. Prop. I, cors. 2, 1 2. .-. AGWBH II CF, and DG II EH. Why ? 3. .-. AB = GH, and BC = HF. I, prop. XXIV 4. But v GH: HF=DE: EF, TV, prop. X, cor. 1 .'. AB:BC = I>E: EF. Subst. 3 in 4 Exercises. 583. In a gymnasium swimming tank the water is 5 ft. deep, and the ceiling is 9 ft. above the water; a pole 18 ft. long rests obliquely on the bottom of the tank and touches the ceiling. How much of the pole is in the water ? 584. In the figure of prop. XXV, connect C and D, and prove the theorem without using the line FG. 274 SOLID GEOMETRY. |Bk. VI. 4. POLYHEDRAL ANGLES. 369. Definitions. When a portion of space is separated from the rest by three or more planes which meet in but one point, the planes are said to form, or to include, a polyhedral angle. A polyhedral angle is also called a solid angle. As two intersecting lines form an infinite number of plane angles, but the smallest is considered unless the contrary is stated, and similarly with two intersecting planes, V so three or more intersecting planes form an infi- nite number of polyhedral angles, but, as with plane and dihedral angles, only the smallest is considered. VAB, VBC, , the faces. The lines of intersection of the planes of a polyhedral angle, each with the next, are called the edges of the polyhedral angle. ilP^i^^B C On account of the complexity of the general Apolyhedral angle . figure, the planes which form a polyhedral angle v-ABCB. r,thever- are considered as cut off by the edges, as in the tex ; VA, VB, VC, above figure. So also the edges, which may be ^'J^^jf s; pla " es produced indefinitely, are considered as cut off by the vertex unless the contrary is stated. The portions of the planes which form a polyhedral angle, limited by the edges, are called the faces of the angle. 370. Polyhedral angles contained by 3, 4, , n planes are termed respectively trihedral, tetrahedral, /7-hedral angles. A polyhedral angle is specifically designated by a letter at its vertex, or by J,hat letter followed by a hyphen, and letters on the successive edges. 371. Congruent polyhedral angles are such as have their dihedral angles equal, and the plane angles of their faces also equal and arranged in the same order. Secs. 372, 373.] POLYHEDRAL ANGLES. 275 372. Symmetric polyhedral angles are such as have their dihe- dral angles equal, and the plane angles of their faces also equal, but arranged in reverse order. A X C B B Symmetric polyhedral angles. Opposite polyhedral angles. Thus, in the above figure, V and V are symmetric trihedral angles, the letters showing the reverse arrangement. Some idea of this reverse arrangement may be obtained by thinking of two gloves, fitting the right and left hands respectively. As two such gloves are not congruent, so, in general, two symmetric polyhedral angles are not congruent. 373. Opposite polyhedral angles are such that each is formed by producing the edges and faces of the other through the vertex. Exercises. 585. How many edges in an n-hedral angle ? How many dihedral angles ? How many plane face angles ? How many vertices ? 586. If a straight line is oblique to one of two parallel planes, it is to the other. 587. If a plane intersects all the faces of a tetrahedral angle, what kind of a plane figure is formed by the lines of intersection ? What, in the case of a trihedral angle ? 588. Does the magnitude of a polyhedral angle depend upon the lengths of the edges ? « 589. Construct from stiff paper two symmetric trihedral angles, with face angles of about 30°, 60°, 45°, and see if they are congruent. (No proof required.) 590. If each of two intersecting lines is parallel to a plane, so is the plane of those lines. 276 SOLID GEOMETRY. TBk. VI. Proposition XXVI. 374. Theorem. Opposite polyhedral angles are symmetric. B' Given V-ABCD, any polyhedral angle, and V-A'B'C'D', its opposite polyhedral angle. To prove that V-ABCD and V-A'B'C'D' are symmetric. Proof. 1. Z.AVB = AA'VB', ZBVC = ZB' VC, Prel. prop. VI 2. Dihedral A with edges VB, VB', being formed by the same planes, have equal (vertical) measuring angles. Prel. prop. VI 3. So for the other dihedral A. But the order of arrangement in the one is reversed in the other. .*. the polyhedral A are symmetric. Note. That the order of the angles is reversed appears more clearly to the eye by making two opposite trihedral angles of pasteboard. It is also seen by tipping the upper angle over, as has been done in the figure to the right. Exercises. 591. If the edges of one polyhedral angle are respectively perpendicular to the faces of a second polyhedral angle, then the edges of the latter are respectively perpendicular to the faces of the former. 592. Two parallel planes intersecting two parallel lines cut off equal segments. Prop. XXVII.] POLYHEDRAL ANGLES. 277 Proposition XXYII. 375. Theorem. In any trihedral angle the sum of any two face-angles is greater than the third. Given the trihedral angle V-XYZ. To prove that ZLYVZ + ZZ VX > Z XV Y. Proof. 1. If Z XVY> either Z YVZ or Z ZVX, no proof is necessary. Why not ? 2. If ZA7T > either Z FFZ or Z ZVX, suppose it > A ZVX. 3. Then in plane VXY suppose VW drawn, making Z XVW = Z ZVX. Suppose VC taken on VZ, equal to VP on VW, and a plane passed through C, P, and any point A of FX. Let this plane intersect VY at P. 4. Then A ^ FP ^ A ^ TC, and A C = AP. I, prop. I 5. But v AC+CB> AB, or AP + PB, Why ? .-. CP > PB. Why? 6. .'. in APF^and CVB, ZBVO ZPVB. I, prop. XI 7. . • . Z C FJ + Z £ J r C> Z A VP + Z P VB, or Z yl FP. Or Z TFZ -f- Z Z VX > Z IFF. Ax. 4 278 SOLID GEOMETRY. [Bk. VI. Corollaries. 1. In any trihedral angle the difference of any two face-angles is less than the third. For if the face-angles are a, 6, c, then since a + 6 > c, .-.a>c — b. 2. In any 'polyhedral angle any face-angle is less than the sum of all the other face-angles. For the polyhedral angle may be divided into a number of trihedral angles, and prop. XXVII repeatedly applied. Note. Prop. XXVII and corollaries suppose that each face-angle is less than a straight angle. This is in accordance with the note under the definition of a polyhedral angle. 376. Definition. A polyhedral angle is said to be convex when any polygon formed by a plane cutting every face, is convex ; otherwise it is said to be concave. Proposition XXVIII. p* 377. Theorem. In any convex polyhedral angle the sum of the face-angles is less than a perigon. Given any convex polyhedral angle, V-ABC To prove that Z A VB + Z B VC + Z C VI) + < perigon. Proof. 1. Let the faces of the angle be cut by a plane. This will form a convex polygon of n sides (n = 5 in the figure), abc Def. convex polyh. Z Prop. XXV1II.J POLYHEDRAL ANGLES. 279 Let S v = sum of plane A a Vb, bVc, , at the vertex ; S b = sum of plane AbaV, Vba, cbV, , at the bases of the A ; and ^ = sum of plane A cba, deb, , of the polygon. 2. Then S p = (n - 2) st. A, or S p + 2 st. A = n st. A. I, prop. XXI 3. And S v + S b = n st. A, since there is a st. Z for each A. I, prop. XIX 4. .". 8 V + S b = ^ + perigon. Steps 2 and 3 ; ax. 1 5. And V S b > S p , Prop. XXVII .'. S v < perigon. Exercises. 593. The three planes which bisect the three dihedral angles of a trihedral angle intersect in a common line whose points are equidistant from the three faces. (See prop. XXI, cor., and I, prop. XLIV.) 594. Suppose a polyhedral angle formed by three, four, five equilateral triangles. What is the sum of the face-angles at the vertex ? 595. If lines through any point and the vertices, A, B, C, , of a polygon, cut a plane parallel to the plane of that polygon in A', B', C", , prove that A'B'C *>~ ABC and that the ratio of similitude is that of OA' to OA. 596. In ex. 595, the more remote is from the planes ABC , A'B'C' , the more nearly do AA', BB', CC become parallel; suppose they become parallel, state and prove the resulting theorem. 597. In ex. 595, if plane A'B'C were not parallel to plane ABC , prove that the corresponding sides, AB, A'B', and BC, B'C, and CD, CD', , would, in general, meet in points on the intersection of the two planes. 598. Two planes, each parallel to a third plane, are parallel to each other. 599. Ex. 598 is analogous to I, prop. XVIII. State the theorem and corollaries analogous to I, prop. XVII and its corollaries, and prove them. 280 SOLID GEOMETRY. [Bk. VI. 5. PROBLEMS. Proposition XXIX. 378. Problem. Through a given point to pass a plane perpendicular to a given line : (1) the point being without the line, (2) the point being on the line. 1. Given the line YY', and point P without. Required through P to pass a plane _L YY'. Construction. 1. From P draw PO _L YY'. I, prop. XXX 2. From draw another line OX A. YY'. I, prop. XXIX Then MN, the plane of OP, OX, is the required plane. Proof. V YY'±OP, and YY' _L OX, .'. YY'±MX. §339 2. Given the line YY', and the point upon it. Required through to pass a plane _L IT. Construction and Proof. Draw OP and 0X1. YY'. This can be done because the three lines are not required to be coplanar. Then the plane XOP 1 V V. § 339 Props. XXX, XXXL] PROBLEMS. 281 Proposition XXX. 379. Problem. Through a given point to pass a plane parallel to a given plane. Solution. Draw two intersecting lines in the given plane. Through the given point draw two lines parallel to these lines, thus determining the required plane. Proposition XXXI. 380. Problem. Through a given point to draw a line per- pendicular to a given plane : (1) the point being without the plane, (2) the point being in the plane. 1. Given the plane MN, and the point P without. Required to draw a perpendicular from P to MN. Construction. 1. Draw PC ±AB, any line in MN. I, prop. XXX 2. In MN draw CE _L AB. I, prop. XXIX Draw PP' _L CE. I, prop. XXX Then PP' is the required perpendicular. Proof. 1. CA _L plane CPP'. Prop. VI, cor. 1 2. Draw P'D II CA ; then P'D _L plane CPP'. Prop: IX 3. .-. Z DP'P is right, and PP' X P'D. § 339 4. But PP' _L CP'. and .'. PP' ± MX. Prop. VI, cor. 1 282 SOLID GEOMETRY. 2. Given the plane MN, and the point R within it. [Bk. VI. Required through R to draw a perpendicular to MN. Construction. 1. From any external point S draw ST JL MN. Case 1 2. From R draw RQ II TS. I, prop. XXXIII Proof. Then RQ is the required perpendicular. Why ? Exercises. 600. From the point of intersection of two lines to draw a line perpendicular to each of them. > 601. To determine the point whose distances from the three faces of a given trihedral angle are given. Is it unique ? 602. From the vertex of a trihedral angle to draw a line making equal angles with the three edges. f 603. The three planes, through the bisectors of the face-angles of a trihedral angle, perpendicular to those faces, intersect in a common line wdiose points are equidistant from the edges. (See I, prop. XLIII.) 604. In how many ways can a polyhedral angle be formed with equi- lateral triangles and squares ? 605. Prove that a straight line makes equal angles with parallel planes. 606. If each of two intersecting planes is parallel to a given line, prove that their intersection is coplanar with that line. 607. Frove that parallel lines make equal angles with parallel planes. 608. Are planes perpendicular to the same plane parallel ? 609. In the figure of prop. XXV, without drawing FG, draw CD and AF; then show that the four lines CD, CA, FD, FA intersect plane Q in the vertices of a parallelogram. 610. Given two lines, not coplanar, and a plane not containing either line, required to draw a straight line which shall cut both given lines and shall be perpendicular to the plane. (Project both lines on the plane.) BOOK VII. — POLWEDBA. 1. GENERAL AND REGULAR POLYHEDRA. 381. Definitions. A solid whose bounding surface consists entirely of planes is called a polyhedron; the polygons which bound it are called its faces ; the sides of those polygons, its edges ; and the points where the edges meet, its vertices. 382. If a polyhedron is such that no straight line can be drawn to cut its surface more than twice, it is said to be convex; otherwise it is said to be concave. Unless the contrary is stated, the word polyhedron means convex poly- hedron. The word convex will, however, be used wherever necessary for special emphasis. 383. If the faces of a polyhedron are congruent and regular polygons, and the polyhedral angles are all congruent, the polyhedron is said to be regular. Exercises. 611. Draw a figure of a polyhedron of four faces. Count the edges, faces, and vertices, and show that the number of edges plus two equals the number of faces plus the number of vertices. 612. Do the same for a polyhedron of five faces ; also of six faces. 613. Take a piece of chalk, apple, or potato, and see if a seven-edged polyhedron can be cut from it. 614. What is the locus of points on the surface of a polyhedron equi- distant from two given vertices ? (The distances are to be taken as usual on a straight line, and not necessarily on the surface.) 615. What is the locus of points equidistant from two given non- parallel faces of a given polyhedron ? 616. To find a point equidistant from two given vertices of a polyhe- dron, and from two given non-parallel faces. 283 284 SOLID GEOMETRY [Bk. VII. Proposition I. 384. Theorem. If a convex polyhedron has e edges, v ver- tices, and f faces, then e + 2 = f + v. Given To prove Proof. 1 ABC Z, a convex polyhedron of e edges, v ver- tices, / faces. that e + 2=f+v. Imagine ABC Z formed by adding adjacent faces, beginning with any face as ABCD of a sides, then adding face M, of b sides, and so on. (It is advisable to build up a rectangular box of paste-board while reading the proof.) Let e r = the number of edges, and v r = the number of vertices, after r faces have been put together. (Thus when we put 2 rectangles together in building up the box, we have located 7 edges and 6 vertices; i.e. e» = 7, v 2 = C, in this case.) 2. Then, since the first face had a sides, .*. e x = a and v l = a. (In the box, e x = 4, vi = 4.) 3. v adding an adjacent face 71/, of b sides, gives only (b — 1) new edges, and (b — 2) new vertices (Why ?), (In the box, adding a second rectangle to the first gives only 3 new edges because we have 1 in common with the first face, and 2 new vertices because we have 2 in common with the first face.) .'. e 2 = a -f- b — 1, v 2 = a + b — 2, so that e a — v 2 = 1. (In the box, e 2 = 4 + 4 - 1 = 7, v 2 = 4 + 4 - 2 = C, so that e 2 — «a = 1.) Prop. I.] GENERAL AND REGULAR POLYIIEDRA. 285 4. Therefore we have e 2 = *>a + 1. Now while the number of edges common to two successive open surfaces will vary according to the way in which the additions are made, the addition of a new face evidently increases e by one more unit than it increases v. .-.es^Vs + 2, €i = r 4 + 3, and, in general, e r = v r + r - 1, or e r — v r = r — 1. 5. But the addition of the last, or /th face, as XYZ, after all the others have been put together, gives no new edges or vertices, •'• e f - l 'f = e /-i - v f-i =f~ 2 ' (In the box, adding the last face merely puts on the cover, adding no new edges or vertices. .-. e 6 — »e = e 5 — v 5 = 4, which is evidently true, because 12, the number of edges, minus 8, the number of vertices, is 4.) 6. That is, e - v =f- 2, so that e + 2 =/+ u; for e f = e, and v f = v. Corollary. For every polyhedron there is another which, with the 8, VI, prop. V .-. O AC ^ CD EG, which proves (1). I, prop. XXVI Similarly for other opposite faces. 4. And v PQ II SB, and P,S' II QE, XI, prop. XXIV .-. Pi? is a O, which proves (2). § 97. def. O Corollary. A parallelepiped has three sets of pta.rallel and equal edges, four in each set. 388. Definition. If the faces of a parallelepiped are all rectangles, it is called a rectangular parallelepiped. It will be noticed that as axes of symmetry enter into the study of plane figures (§ 68), and especially of regular figures, so planes of sym- metry and axes of symmetry enter into the study of solids. A plane of symmetry divides the solid into halves, related to each other as a figure is related to its image in a mirror. Planes of symmetry play an important part in the study of crystals. The term axis of symmetry will be under- stood from Plane Geometiy. Exercises. 618. Prepare a table showing the number (1) of faces, (2) of edges, (3) of vertices, (4) of sides in each face, (5) of plane angles at each vertex, of all of the five regular polyhedra. 619. How many degrees in the sum of the face-angles at one vertex of a regular tetrahedron? hexahedron? octahedron? dodecahedron? icosahedron ? 620. The perpendiculars to the faces, through their centers, of a ^ regular tetrahedron are concurrent in n point equidistant from all of the vertices, from all of the faces, and from all of the edges. 621. Prove that no polyhedron can have less than six edges. 622. In a regular tetrahedron three times the square on an altitude -W* equals twice the square on an edge. 623. Certain crystals have their corners cut off, that is, the vertices of their polyhedral angles replaced by planes. Suppose a regular hexa- hedral crystal has its trihedral angles replaced by planes, how many faces has the new crystal ? How many edges ? vertices ? Is Euler's theorem satisfied ? 624. How many planes of symmetry and how many axes of sym- metry has a regular hexahedron ? octahedron ? f 290 SOLID GEOMETRY. [Bk. VII. Pkopositiox IV. 389. Theorem. In any parallelepiped, 1. The four diagonals are concurrent in the mid-point of each. 2. The sum of the squares on the four diagonals equals the sum of the squares on the twelve edges. A B Given a parallelepiped with diagonals AG, BH, CE, DF. To prove that (1) the diagonals are concurrent at 0, the mid- point of each ; (2) AG 2 + BH 2 + CE 2 + BE 2 = AB 2 + BC 2 + Proof. 1. v BF= and II BR, Why ? .-.DBFH is a. CJ. Why? 2. .'. ED and BH bisect each other at 0. § 100, cor. 2 Similarly, BH and CE, CE and A G, bisect each other. 3. And '.' there is only one point of bisection of BH and CE, § 41 . * . BH, CE, A G, and DF are concurrent at 0. 4. And v AG 2 + CE 2 = AC 2 + CG 2 -f- GE 2 + EA* } and DF 2 + BH 2 = BF 2 + FH 2 h HB 2 + BB 2 , II, prop. XI, cor. 5. .'. by adding, and noting that AC 2 + 7>/> >2 = AB 2 + BC 2 + CD 2 + i>vi' 2 , etc., the theorem is proved. Secs. 390-393.] PRISMATIC AND PYRAMIDAL SPACE. 291 3. PRISMATIC AND PYRAMIDAL SPACE. PRISMS AND PYRAMIDS. 390. Definitions. A prismatic surface is a surface made up of portions of planes, the intersections of which are all parallel to one another. 391. If, counting from any plane of a prismatic surface as the first, each plane intersects its succeeding plane, and the last one intersects the first, the sur- face is said to enclose a prismatic space. The lines of intersec- tion are called the edges, and the portions of the planes between the edges, the faces, of the prismatic space A prismatic sur- face. A portion of a prismatic space, quadrangular and convex. ABCD, a right section. The edges and the faces are supposed to be unlimited in length. It will be readily seen that a prismatic space is related to entire space as a plane polygon is to its entire plane. It will therefore be inferred that theorems relating to polygons have corresponding theorems relating to prismatic spaces. 392. A section of a prismatic space, made by a plane cutting its edges, is called a transverse section. If it is perpendicular to the edges, it is called a right section. 393. A prismatic space is said to be triangular, quadrangular, rectangular, pentagonal, , n-gonal, according as a transverse section is a triangle, quadrilateral, rectangle, pentagon, , n-gon, and to be convex or concave according as a transverse section is a convex or a concave polygon. 292 SOLID GEOMETRY. [Bk. VII. Prismatic spaces may be such that transverse sections are convex, con- cave, or cross polygons. All theorems not involving mensuration will at once be seen to apply to each class. But on account of the complexity of the figures, the third form (cross) is not considered in this work. 394. The portion of a prismatic space included between tw r o parallel transverse sections is called a prism, the two transverse sections being called the bases of the prism. Thus in the figure on p. 293 the portion of the prismatic space P, between S and S', is a prism. S and S' are the basis. The signification of the terms edges, faces, and prismatic surface of a prism, upper and lower bases of a prism, triangular prisms, etc., will be inferred from the above definitions. By transverse and right sections of a prism are to be understood the transverse and right sections of its prismatic space. The sides of the bases of a prism are also called edges ; where con- fusion is liable to arise these are called base edges, and the edges of the prismatic space are called lateral edges. :# Exercises. 625. In the figure of prop. IV, prove that Oi, 0, 2 are collinear. 626. Also that X = EA/2. 627. Also that if AG is a rectangular parallelepiped, 0\0 is perpen- dicular to line EG. 628. Also that if the diagonals of all the faces are drawn, and the points of intersection of the diagonals of the opposite faces are con- nected, these connecting lines are concurrent at O, the mid-point of each. 629. Prove that the square on a diagonal of a rectangular parallele- piped equals the sum of the squares on three concurrent edges. 630. If the edge of a cube is represented by Vs, find the diagonal. 631. Prove that the four diagonals of a rectangular parallelepiped are equal. 632. Show that the edge, diagonal of a face, and diagonal, of a cube, are proportional to 1, V2, Vij. 633. If the plane PR, of prop. Ill, were also to cut the faces UF and 1)1>, what would be the plane figure resulting? What would be the relation of its opposite sides? Prop. Y.] PRISMATIC AND PYRAMIDAL SPACE. 293 Proposition V. 395. Theorem. Parallel transverse sections of a prismatic space are congruent polygons. Given the prismatic space P, with S, S', two parallel trans- verse sections. To prove that S ^ S'. Proof. 1. '•' the sides of S II sides of S', respectively, VI, prop. XXIV .\ A of £ = A of S', respectively. VI, prop. V 2. And '•* the sides of >S' also equal the sides of S', respectively, I, prop. XXIV .'. by superposition, S is evidently congruent to S'. Corollaries. 1. The bases of a prism are congruent poly- gons. 2. The faces of a j^rism are parallelograms. 3. The lateral edges of a prism are equal. Exercise. 634. Suppose in the figure of prop. Ill, another plane II to PR, cutting the same faces as PR, hut not the other faces. Prove that it would cut out a parallelogram congruent to PR. 294 SOLID GEOMETRY. [Bk. VII. 396. Definitions. A pyramidal surface is a surface made up of portions of planes which have but one point in common. A pyramidal surface. A portion of a pyramidal space, quad- rangular and convex. V, the vertex; ABCD, a transverse section; V-ABCD, a pyramid, ABCD being its base. 397. If, counting from any plane of a pyramidal surface as the first, each plane intersects its succeeding plane, and the last one intersects the first, the surface is said to contain a pyramidal space. Unlike a prismatic space, a pyramidal space is double, its parts lying on opposite sides of the common point. The lines of intersection of the planes are called the edges, the portions of the planes between the edges, the faces, and the point of intersection of the edges, the vertex, of the pyramidal space. The edges and faces are supposed to be unlimited in length. 398. A section of a pyramidal space, made by a plane cutting all of its edges on the same side of the vertex, is called a transverse section. 399. The terms triangular, , n-gonal, concave, convex pyramidal space are defined as the like terms for prismatic space. Secs. 400, 401.] PRISMATIC AND PYRAMIDAL SPACE. 295 400. The portion of a pyramidal space included between the vertex and a transverse section is called a pyramid, the transverse section being called its base, and the vertex of the space, the vertex of the pyramid. Thus the figure V-ABC below is a pyramid, ABC being the base and V the vertex. The distance from the vertex of a pyramid to the plane of its base is called the altitude of the pyramid. Thus in the figure below, YV is the altitude of pyramid V-ABC. The signification of the terms edges, faces, transverse section, base edges, etc. , of a pyramid can be inferred from the preceding definitions. 401. The portion of a pyramidal space included between two transverse sections on the same side of the vertex is called a truncated pyramid ; if the transverse sections are parallel, it is called a frustum of a pyramid, the two sections being called the bases of the frustum. A frustum of a pyramid is therefore a special form of a truncated pyramid. A pyramid is also a special case of a truncated pyramid, the upper base being zero. The distance from any point in one base of a frustum of a pyramid to the plane of the other base is called the altitude of the frustum. T, a truncated pyramid ; ABCXYZ, a frustum of the pyramid V-ABC ; VV, the altitude of the pyramid ; ABC, XYZ, the lower and upper bases of the frustum ; XX , the altitude of the frustum. 296 SOLID GEOMETRY. [Bk. VII. Proposition VI. 402. Theorem. Parallel transverse sections of a pyrami- dal space are similar polygons, whose areas are proportional to the squares of the distances from the vertex to the cutting planes. Given P, a pyramidal space with vertex V, cut by two parallel planes, B, B', making transverse sections ABC = S, A'B'C = S', respectively ; VXA. B, VX' _L B'. To prove that (1) S~ S', (2) S:S'= VX 2 : VX' 2 . Proof. 1. v the sides of S are II to the sides of S', VI, prop. XXIV .*. VA: VA'= VB: VB' = and so on for other points. IV, prop. X, cor. 2 2. .'. S~ S', which proves (1). § 258, def. sim. figs. 3. And v AB : A'B' = VA : VA' = VX : VX', IV, prop. X, cors. 1, 3 and 8 : S' = AJi 2 : A'V><\ V, prop. IV .'. S: S' = VX* : VX". IV, prop. VII Prop. VI.] PRISMATIC AND PYRAMIDAL SPACE. 297 Note. The definition of similar figures, given in Book IV, § 258, is general ; the center of similitude and the given figures may or may not be in the same plane. Thus in the figure on p. 296, V is the center of similitude of the triangles ABC and A'B'C, and in the figure on p. 295, V is the center of similitude of the triangles XYZ and ABC. Neither is the definition limited to plane figures ; we may have similar solids as well. Thus two balls are similar, or two cubes, or two regular tetra- hedra, etc. Corollaries. 1. If a pyramid is cut by a plane parallel to the base, (1) the edges and altitude are divided proportion- ally, (2) the section is similar to the base. If the planes in the proof on p. 296 are on the same side of F, step 3 proves (1), and step 2 proves (2). Or, in the figure on p. 295, VV'-.XX' = VA :XA, and A ABC ^ A XYZ. 2. In pyramids having equal bases and equal altitudes, transverse sections parallel to the bases, and equidistant from them, are equal ; if the bases are congruent, so are the sections. 1. Let s, s' be the areas of the sections, b, b' the areas of the bases, d the distance of the section from the vertex, and a the altitude. 2. Then from prop. VI, s : b = d 2 : a 2 , and s' : b' = d 2 : a 2 . .-. s:b = s' : b'. Ax. 1 3. But b = b\ .-. s = s'. 4. And if the bases are congruent, so are the sections, since they are both similar and equal to the bases. 3. The bases of a frustum of a pyramid are similar figures. For they are parallel transverse sections of a pyramidal space. Hence step 2, p. 296, proves the corollary. Exercise. 635. In the figure on p. 296, suppose Z. ABC a right angle, AB = 3 in., AC = 5 in., VB = 10 in., and the area of A A'B'C' = 12 sq. in. ; find the length of VB'. 298 SOLID GEOMETRY. [Bk. VII. 4. THE MENSURATION OF THE PRISM. 403. Definition. The area of the prismatic surface of a prism is called the lateral area of the prism. Similarly for the pyramid, and for the cylinder and cone, to be defined hereafter. Proposition VII. 404. Theorem. The lateral area of a prism equals the product of an edge and the perimeter of a right section. Given the prism P; a right section R with sides s u s 2 , ; f,f 2 , the faces of the prism; e, an edge. To prove that the lateral area of P is e • (s t + s 2 + ). Proof. 1. ' • ' by definition of right section, B _L e, . ' . s x _L e. § 339 2. V/u f*> are UJ, Prop. V, cor. 2 .'. area/! = e-s x . V, prop. II, cor. 3 3. And V the edges are equal, Prop. V, cor. 3 .'. area f2 — e-s 2 , and so for the other faces. 4. .*. lateral area= e\«i + e- s 2 + == e • (s t + s 2 + )• Ax. 2 Sec. 405.] THE MENSURATION OF THE PRISM. 299 405. Definitions. A prism whose edges are perpendicular to the base is called a right prism ; if the edges are oblique to the base it is called an oblique prism. E.g. on p. 301 R is a right prism and is an oblique prism. So a cube is a special kind of a right prism, and the parallelepiped illustrated on p. 290 is an oblique prism. The distance from any point in one base of a prism to the plane of the other base is called the altitude of the prism. Similarly for a parallelepiped, which is a special kind of prism. Corollary. The lateral area of a right prism equals the product of the altitude and the perimeter of the base. For the altitude here equals the edge. Exercises. 636. Required the lateral area of a prism of edge 3 in., the right section being an equilateral triangle of area i V3 sq. in. Also the lateral area of one of edge 3 in. , the right section being a square of diagonal V2 in. 637. Required the lateral area of a right prism whose base is a square of area 9 sq. in., and whose altitude equals the diagonal of the base. Also required the total area. 638. Required the total area of a right prism whose base is an equi- lateral triangle of area ± V3, and whose altitude equals a base edge. 639. Required the total area of a right prism whose base is a regular hexagon whose side is 1 in., the altitude of the prism being equal to the diameter of the circle circumscribing the base. 640. Required the lateral area of a prism of edge |, the right section being a regular hexagon of area § Vs. 641. Required the total area of the prism mentioned in ex. 640, supposing it to be a right prism. 642. A converse of prop. VI is as follows : If two similar polygons have their corresponding sides parallel, and lie in different planes, the lines through their corresponding vertices are concurrent. Prove it. (A generalization of the idea of similar figures in perspective ; see the definition of similar figures § 258, and the note at the top of p. 297.) 643. Investigate and prove whether or not any three faces of a tetra- hedron are together greater than the fourth. 300 SOLID GEOMETRY. [Bk. VII. Proposition VIII. 406. Theorem. Prisms cut from the same prismatic space and having equal edges are equal. Given two prisms, P, P', cut from the same prismatic space S, and having equal edges e. To prove that P = P. Proof. 1. If 7T=the portion of the prismatic space between P and P, then by adding e to the edges of K, each edge of P + K= an edge of K -f- P. Ax. 2 2. Then '•' P + K can evidently slide along in the prismatic space and occupy the position of K -f P', .'.P + K*zK+P'. §57 3. ..P = P. Ax. 3 Corollaries. 1. Right jyrisms having equal altitudes and congruent bases are congruent. 2. An oblique prism is equal to a right prism whose base and altitude are respectively a right section and edge of the oblique j)ris7ti. Prop. IX.J THE MENSURATION OF THE PRISM. 301 Proposition IX. 407. Theorem. The two triangular prisms into which any parallelepiped is divided by a plane through two opposite edges are equal. Given and E, parallelepipeds with equal edges, cut from a prismatic space, B being right ; also, P, a plane through two opposite edges of that space, cutting R, into two triangular prisms, T x and T 2 , X x and X 2 , respectively. To prove that (1) T x = T 2 , (2) A\ = X 2 . Proof. 1. The base of T l ^ the base of T % . I, prop. XXIV 2. '•' they have the same altitude, .". T x = T 2 . Why ? 3. Xi = -Tu and X 2 = T 2 . Prop. VIII 4. And v T x = T 2 , .'. X, = X 2 . Ax. 1 Corollary. A trianrjular prism is half of a parallele- piped of the same altitude, whose base is the parallelogram of which one side of the triangular base is the diagonal and the other two are the sides. (Why ?) 302 SOLID GEOMETRY. [Bk. VII. Proposition X. 408. Theorem. Any parallelepiped is equal to a rectan- gular parallelepiped of equal base and equal altitude. Given a parallelepiped, III. To prove that III equals a rectangular parallelepiped as I, of equal base and equal altitude. Proof. 1. Let II be a parallelepiped on the same base, />, as III, formed by a rectangular prismatic space, B, cutting the prismatic space S of the figure. Let I be a rectangular parallelepiped cut from B, with a base B' = B, and a base edge e' consequently equal to base edge e of II. II, prop. I, cor. 4 2. Then III = II, being part of S and having a com- mon edge. Prop. VIII 3. And I = II, being part of B and having an equal edge. Prop. VIII 4. Ill = I. Ax. 1 Prop. XI.] THE MENSURATION OF THE PRISM. 303 Proposition XI. 409. Theorem. Two rectangular parallelepipeds having congruent bases are proportional to their altitudes. R' pj^ Given two rectangular parallelepipeds, R and R', with alti- tudes a and a' respectively, and with bases b. To prove that R : R' = a : a'. Proof. 1. Suppose a and a' divided into equal segments, I, and suppose a = nl, and a' = n'l. (In the figures, n — 6, n' = 4.) Then if planes pass through the points of division, parallel to the bases, R = n congruent rectangular ppds. bl, and R' = n' " " " « 7i - bl n _a ~a r 2. .-.=: = hi Why Note. The student should notice the resemblance between this theorem and Bk. V, prop. X. The above proof assumes that a and a' are commensurable, and hence that they can be divided into equal segments I. The proposition is, however, entirely general. The proof on p. 304 is valid if a and a' are incommensurable. Exercise. 644. Given the diagonals, a, 6, c, of three unequal faces "»-., of a rectangular parallelepiped, to compute the edges. 304 SOLID GEOMETRY. [Bk. VII. 410. Proof for incommensurable case. R a\\\ 1. Suppose a divided into equal segments /, and suppose a — nl, while a' = n'l + some remainder x, such that x < I. Then if planes pass through the points of division, parallel to the bases, R = n congruent rectangular ppds. bl, and R' = 71' « « " " + a re- mainder bx such that bx < bl. 2. Then a' lies between n'l and (V + 1) /, and H' lies between ?i' • bl and (n' + 1) bl. 3. .'. — lies between — and > a n n while -=- lies between — and li n n Why ? Why? a' li' 1 4. .*. — and — differ by less than -■ a R J n Why ? 5. And v — can be made smaller than any assumed difference, by increasing n, .'.to assume any difference leads to an absurdity. a' R' . R a 6. .'. — = — 5 whence — = — :■ a R R' a' Prop. XII.] THE MENSURATION OF THE PRISM. 301 Proposition XII. 411. Theorem. Two rectangular parallelepipeds of equal altitudes are proportional to their bases. P P' a. "I a a Given » two rectangular parallelepipeds, P, P', having alti- tudes a, and bases be and yz, respectively. To prove that P : P' = bc: yz. Proof. 1. Suppose a rectangular parallelepiped Q to have an altitude a and a base ?/c. 2. Then v ac may be considered the base of P and Q, . ^_^ Q c 3. And similarly, -^ = — P P' Prop. XI Why ? Why ? 412. Definition. The length, breadth, and thickness of a rectangular parallelepiped are called its three dimensions. Exercise. 645. If through a point on a diagonal plane of a parallele- piped planes are passed parallel to the two pairs of faces not intersected by the diagonal, the parallelepipeds on opposite sides of that diagonal plane are equal. (See II, prop. IV.) 306 SOLID GEOMETRY. [Bk. VII. Proposition XIII. m 413. Theorem. Two rectangular parallelepipeds are pro- portional to the products of their three dimensions. Given two rectangular parallelepipeds, P, P', of dimen- sions a, b, c, and a', b', c', respectively. To prove that P : P' = abc : a'b'c'. Proof. 1. Suppose a rectangular parallelepiped Q to have the three dimensions a', b, c'. 2. Then and 3. Corollaries. 1. The volume of a rectangular parallele- piped equals the product of its three dimensions. This means that the number which represents the volume is the product of the three numbers representing the dimensions. That is, the number of times the unit of volume is contained in the given parallelepiped, is the product of the numbers of times the unit of length is contained in three concurrent edges. If T" is a cube, of edges 1, 1, 1; then P' is the unit of measure of volume. But P : P' then becomes 7* : 1 , and abc s 1 • 1 • 1 theu becomes abc : 1. .«. P: 1 = abc : 1, or P = abc. P ac Q~ a'c r Prop. XII Q _b P'~ b'' Prop. XI P abc P' ~ a'b'c' Why ? Prop. XIII.] THE MENSURATION OE THE PRISM. 307 2. The volume of any parallelepiped equals the product of its base and altitude. For (prop. X) it equals a rectangular parallelepiped of equal base and equal altitude. 3. The volume of a, triangular prism equals the product of its base and altitude. Cor. 2 with prop. IX, cor. Let the student give the proof in detail. 4. The volume of any prism equals the product of its base and altitude. For it can be cut into triangular prisms by diagonal planes through a lateral edge, the sum of the bases of the triangular prisms being the base of the given prism. .-. cor. 3 applies. Let the student draw the figure and give the proof in detail. 5. Any prism equals a rectangular parallelepiped of equal base and equal altitude. Cors. 4, 2. 6. The volume of an oblique prism equals the product of an edge and a, right section. Cor. 4 with prop. VIII, cor. 2. 7. Prisms having equal bases are proportional to their alti- tudes. For if a is the altitude and b the base, then P = ab, and P' = a'b'. If b — &', then P' = a'b. Hence P : P' — ab : a'b — a-.a'. 8. Prisms having equal altitudes are proportional to their bases. Prisms having equal bases and equal altitudes are equal. Let the student give the proof. Exercises. 646. What is the edge of the cube whose volume equals that of a rectangular parallelepiped with edges 2.4 m, 0.9 m, 0.8 m ? 647. From the given edge e of a cube, compute (1) the cube's entire surface, (2) its diagonal, (3) its volume. 648. Draw a figure illustrating geometrically the formula (a + bf = a 3 + & 3 + 3 a 2 b + 3 ab 2 . 308 SOLID GEOMETRY [Bk. VII. 5. MENSURATION OF THE PYRAMID. 414. Definitions. A regular pyramid is a pyramid whose base is a regular convex polygon, the perpendicular to which, at its center, passes through the vertex of the pyramid. 415. The slant height of a regular pyramid is the distance from the vertex to any side of the base. E.g. VB in the annexed figure. 416. The portion of the slant height of a regular pyramid cut off by the bases of a frustum is called the slant height of the frustum. Corollary. The slant height of a regular pyramid, or of a frustum of a regular pyramid, is the same on ivhat ever face it is measured. Let the student show that the faces are all congruent ; hence that the slant heights are equal. Exercises. 649. To pass a plane through a given pyramid parallel to the base, so that the section shall equal half the base. 650. The edges of a rectangular parallelepiped are 3, 4, 5 ; required the total area of the faces, the areas of its diagonal planes, the length of its diagonal line, andjthe lengths of the diagonals of its faces. Similarly for a cube of edge V2. 651. If a cubic block of sandstone at a temperature of 0° Centigrade has an edge 1 m long, and if for every 1° Centigrade increase of tempera- ture the edge increases 0.000012 of its length at 0°, find the volume at 40° Centigrade. 652. A brick has the dimensions 25 cm, 12 cm, 6 cm, but on account of sbrinkage in baking, the mold is 27.5 cm long, and proportionally wide and deep. What per cent does the volume of the brick decrease in baking ? Prop. XIV.] MENSURATION OF TUE PYRAMID. 309 Proposition XIV. 417. Theorem. The lateral area of the frustum of a regular pyramid equals half the product of its slant height and the sum of the perimeters of its bases. Given BB', a frustum of a regular pyramid, h its slant height, s a side of base B and p its perimeter, s' a side of base B' and p' its perimeter, I the lateral area. To prove that I = \ h (jp + p*). Proof. 1. The area of each face = \h (s + $')- V, prop. IT, cor. 5 2. Adding all the faces, and remembering that p is the sum of the sides s, and p' of the sides s', we have I = ih(p +p r ). Corollary. The lateral area of a regular pyramid equals half the product of its slant height and the perimeter of its base. For in the above theorem, let B' = ; then s' and p' = ; .-. I = } hp. Exercises. 653. Prove the above corollary independently of the theorem. 654. What is the lateral area of a regular pyramid whose base is a a triangle of altitude % V 3, and whose slant height is a ? 655. What is the total area of a frustum of a regular hexagonal pyramid whose base edges are respectively 3 — v3 and 3 + v3, and whose slant height is 10 ? 310 SOLID GEOMETRY. [Bk. VII. Proposition XV. 418. Theorem. Pyramids having equal bases and equal altitudes are equal. V Given pyramids VAZ, V'A'Z', having equal bases, and having equal altitudes h. To prove that pyramid VAZ = pyramid YA'ZK Proof. 1. Suppose their bases in the same plane M, and their vertices on the same side of M. Suppose their altitude It, divided into n equal parts and planes passed through the division-points par- allel to 31. Then these planes will make equal corresponding transverse sections because the bases are equal. Prop. VI, cor. 2 2. Suppose planes passed through the sides of these sections parallel to an edge of the pyramid, making a set of prisms in each pyramid, A, B, and A', B>, 3. Then V A = A', and B = B', Prop. XIII, cor. 8 .'. A + B + = A' + B'+ Ax. 2 Prop. XV.] MENSURATION OF THE PYRAMID. 311 4. But if n increases indefinitely, A + B + =pyr. VAZ, and A' + B' + = pyr. VA'Z*. 5. /. pyr. VAZ = pyr. VAZ'. IV, prop. IX, cor. 1 Corollaries. 1. A pyramid having a parallelogram for its base is divided into equal pyramids by a plane through its vertex and two opposite vertices of the base. For the two pyramids have equal bases and a common altitude. 2. A pyramid having a parallelogram for its base equals twice a triangular pyramid of the same altitude, tvhose base equals half that parallelogram. (Why?) 3. A triangular pyramid can be constructed equal to any given n -gonad pyramid. II, prop. XII, and this theorem. Exercises. 656. Find the area of the entire surface of a regular tetrahedron of altitude h. 657. Find the altitude of a regular tetrahedron of total area a. 658. Find, by § 417, the total area of a cube of edge e. 659. What is the length of the base edge of a regular triangular pyramid which is equal to a regular hexagonal pyramid of the same altitude, the base edge being 1 ? 660. In prop. XIV, cor., B' was supposed to decrease to ; supposing^ instead, that B' increases until it equals B, show that step 2 of the theorem gives the usual formula for the lateral area of a prism. f 661. Prove that frustums of pyramids having equal bases and equal altitudes, which themselves have equal altitudes, are equal. 662. A pyramid has for its base a regular hexagon with its shorter diagonal Vs ; the altitude equals the longer diagonal; required the lateral area of the pyramid. 663. Find the total area of the pyramid mentioned in ex. 662. 664. The lower base of a frustum of a regular pyramid is a square of area s 2 ; the area of the upper base is half that of the lower one ; the slant height is s ; required the lateral area. 312 SOLID GEOMETRY. [Bk. VII. Proposition XVI. 419. Theorem. A triangular prism can be divided into three equal triangular pyramids. D F E D D F ^s^/ 1 ^ ^ 1 k< 4g >^ NA / A^ /C A C A Given ABCDEF, a triangular prism. To prove that ABCDEF can be divided into three equal tri- angular pyramids. Proof. 1. v A, E, C determine a plane, also C, D, E, § 330, 1 .*. ABCDEF = three triangular pyramids, viz., E-ABC, E-ACD, C-DEF. Ax. 8 2. But v AABC^ADEF, . ' . E-AB C = C-DEF. Prop. XV 3. And C-DEF = E-DCF = E-ACD, because they have a common altitude from E to plane ACFD, and equal bases. Prop. XV 4. . ' . E-AB C = C-DEF = E-A CD. Ax. 1 Corollaries. 1. A triangular pyramid is one-third of a triangular prism of the same base and same, altitude. For the prism is three times the pyramid. 2. Any pyramid is one-third of a prism of the same base and same altitude. For, dividing the base into triangles by drawing diagonals, the pyra- mid may be considered as made up of triangular pyramids, each of which is a third of a triangular prism of the same base and same alti- tude ; hence the sum of the triangular pyramids, or the given pyramid, equals one-third the sum of the triangular prisms, or one-third of a prism of the same base and same altitude. Sec. 420.] MENSURATION OF THE PYRAMID. 313 3. The volume of a pyramid equals one-third the product of its base and altitude. Cor. 2, and prop. XIII, cor. 4. 4. Pyramids having equal bases are proportional to their altitudes; having equal altitudes, to their l>ases. -^ , t t / , ,w i V %ab ab For if p = | ao, and p =|ao, then — = = p' % a'b' a'h' , ■<■ , t/ , ab a And it 6 = o r then = — • a'6' a' Or if a = a , then = — . a'b' b' Or if a. — a' and b = b\ then — = 1, orp = p', as stated in prop. XV. P' 420. Definitions. A polyhedron which has for bases any two polygons in parallel planes, and for lateral faces triangles or trapezoids which have one side in common with one base and the opposite vertex or side in common with the other base, is called a prismatoid. The altitude of a prismatoid is the perpendicidar distance between the planes of its bases. Exercises. 665. Find the volume of the pyramid mentioned in ex. 662. 666. A church-tower is capped by a regular octagonal pyramid whose height is 55.5 m, and whose base edge is 4.9 m. Required the volume. 667. A pentagonal pyramid has equal lateral and base edges, 1 in. Find the lateral area. 668. Find the volume of a cube the diagonal of whose face is a V2. 669. Each face of a given triangular pyramid is an equilateral triangle whose side is 3. Find the total area. 670. Find the volume of the tetrahedron mentioned in ex. 656. 671. Also of the pyramid mentioned in ex. 667. 672. An edge of a regular octahedron is 1 in. Find the volume 673. A pyramid stands on a square base of edge 1 m ; the lateral edge of the pyramid is also 1 m. Find the lateral area and volume. 314 ■SOLID GEOMETRY. [Bk. VII. Proposition XVII. 421. Theorem. The volume of a prismatoid of bases b and 1/, altitude h, and transverse section m midway between the bases, is expressed by the formula v = - {b + b' + 4?>i). Fig. 1. Fig. 2. Fig. 3. Proof. 1. If any face, ABFE, of the prismatoid P, is a trape- zoid, divide it into two triangles by a diagonal EB. Let V be any point in m ; join F to the vertices of P ; then P will be divided into two pyramids (Fig. 2) of bases b, V and vertex V, and also pyramids of vertex V and triangular bases ABE, etc. (Pig. 3.) Let EB meet m at D; call A TDC m v (Fig. 3.) 2. Then the volume of r-b = ib-t and V-b' = j-6'. -■ Prop. XVI, cor. 3 This completes Fig. 2. 3. Pyramid V-ABE = E-CVD + B-C1P) + V-ABC. Ax. 8 4. Of these, and Prop. XVI, cor. 3 Prop. XVII.] MENSURATION OF THE PYRAMID. 315 5. But F-ABC = twice V-CBD (or B-CVD), v AABC = twice A C£D, having edge AB = 2 • CD, and a common altitude. Prop. XV I. cor. 4 6. .-. V-ABC = | /«!•-/ 7 . . • . py ram id V- ABE = - • 4 % A x s . 2, 8 and .". the sum of the pyramids of the form of V-ABE = ^±m. Axs. 2 3 8 8. .-.P = |(ft' + ft' + 4i»). Axs. 2. 8 u Note. The Prismatoid Formula, o = - (6 + // + 4 ???), is of great value G in the mensuration of solids. From it can be derived formulae for the volumes of all of the solids of elementary geometry. f^.- Coeollary. The volume of the frustum of a pyramid, of bases b, b'. and altitude h, is -(b + b' 4- Vbb'). o Fur if e, e' are corresponding sides of 6, b\ then | (e -f e') is the corre- sponding side of to. (Why ?) e %/ & , e ' ^ •'■ -n - = —7= i and — 7- = -7= • , prop. IV.) , and .-. 2 Vwi = V 6 + v&'. i (e + e') Vm .-. 4to = 6 + 6' + 2 VW, which may be substituted in the Prismatoid Formula. Exercises. 674. By letting (1) b' = 0, and (2) b' — b, show that (1) prop. XVI, cor. 3, and (2) prop. XIII, cor. 4, follow, as special cases, from the Prismatoid Formula. 675. Calling a prismatoid whose lower base b is a rectangle of length I and width t*>, and whose upper base b' is a line e parallel to a base edge, and whose altitude is h, a icedge, find a formula for the volume of a wedge. 316 SOLID GEOMETRY. [Bk. VII. EXERCISES. 676. The base of a wedge is 4 by 6, the altitude is 5, and the edge e is 3. Find the volume. (See ex. 675 ) Also, when e = 0. 677. The altitude of a pyramid is divided into five equal parts by planes parallel to the base. Find the ratios of the various frustums to one another and to the whole pyramid. 678. Two pyramids, P, P', have square bases, and are such that the altitude of P equals twice the altitude of P', but the base edge of P is half as long as the base edge of P'. Find the ratio of their volumes. 679. Find the volume of a cube whose diagonal is Vo. 680. A frustum of a pyramid has for its bases squares whose sides are respectively 0.6 m, 0.5 m; the altitude of the frustum is 0.9 m. Find the volume. 681. Given the volume v, and the bases 6, &', of a frustum of a pyramid, to find a formula for (1) its altitude, (2) the altitude of the whole pyramid. 682. A granite monument is in the form of a frustum of a square pyramid, surmounted by a pyramid ; the sides of the bases of the frustum are 1 m and 0.8 m, and the altitude of the frustum is 1.8 m ; the altitude of the pyramidal top is 0.45 m. A cubic meter of water weighs a metric ton, and granite is three times as heavy as water. Find the weight of the monument. 683. An excavation 1.5 m deep, rectangular at top and bottom, and in the form of a frustum of a pyramid, has its upper base 10 m wide and 16 m long, and the lower base 7.5 m wide. How many cubic meters of earth would it take to fill it to a depth of 0.75 m ? 684. The volume of a cube is six times that of the regular octahedron formed by joining the centers of the faces of the cube. 685. Find the volume of a prismatoid of altitude 3.5 cm, the bases being rectangles whose corresponding dimensions are 3 cm by 2 cm, and 3.5 c^n by 5 cm. 686. It is usual to find the volume of a pile of broken stones by taking the product of the altitude and the area of a transverse mid-section. Compare this with the Prismatoid Formula and find what relation it assumes between m and b -f b'. Is this relation true in the case of a pyramid ? 687. The volume of a pyramid equals the product of the altitude and a transverse section (parallel to the base) how far from the vertex ? BOOK VIII. — THE CYLINDER, CONE, AND SPHERE. SIMILAR SOLIDS. 1. THE CYLINDER. 422. Definitions. A curved surface is a surface no part of which is plane. The number of kinds of curved surfaces is unlimited, just as the number of kinds of curves in a plane is unlimited. But as among plane curves the circumference is the best known, so there are certain curved surfaces which are better known than others, and these are treated in this book. 423. A cylindrical surface is a surface generated by a straight line, called the generatrix, which moves so as constantly to pass through a given curve, called the directrix, and to remain parallel to its original position. The surface of a piece of straight pipe, or the surface of the paper in a roll, is an example. 424. A straight line in any position of the generatrix is called an element of the cylindrical surface. 425. If the directrix is a closed curve, the cylindrical surface incloses a space of unlimited length, called a cylindrical space. 426. A section of a cylindrical space, made by a plane cutting its elements, is called a transverse section. If it is perpen- dicular to the elements, it is called a right section. 317 One form of a cylindric- al surface. AB CBD, the directrix ; BB', an element ; BCB',a. por- tion of a cylindrical space. 318 SOLID GEOMETRY. [Bk. VIII. As a transverse section of a prismatic space may be a convex, con- cave, or cross polygon, so a transverse section of a cylindrical space may be a curve of any shape if only its end-points meet. All theorems, if the signs are properly considered, will be seen to apply to each of the three forms of transverse section, corresponding to convex, concave, and cross polygons. The third is, however, too complex for treatment in elementary works. 427. The portion of a cylindrical space included between two parallel transverse sections is called a cylinder. E.g. the portion between planes P and P' in the figure on p. 310. • The terms bases and altitude of a cylinder will be understood, without further definition, from the corresponding definitions under the prism. The student should, throughout this section, notice the relation of cylin- drical spaces to prismatic spaces. A cylinder is considered as having the same directrix as its cylindrical space, and as having for elements the segments of the elements of the cylindrical surface included between its bases. A cylinder is said to be right or oblique according as its elements are perpendicular or oblique to the bases. If the base of a cylinder is a circle, the cylinder is said to be circular. 428. Postulate of the Cylinder. A cylindrical surface may be constructed with any directrix and with any original posi- tion of the generatrix. In solid geometry constructions are allowed which require other instruments than the compasses and straight-edge. For example, this postulate requires the generatrix to move constantly parallel to its origi- nal position, a construction manifestly impossible by the use of merely these two instruments. Exercises. 688. Draw a figure of a convex cylinder; a concave cylinder ; a cross cylinder. 689. Prove that if a transverse section of a cylindrical space is per- pendicular to one element it is a right section. Prop. I.] THE CYLINDER. 319 Proposition I. 429. Theorem. Parallel transverse sections of a cylin- drical space are congruent. Given a cylindrical space S, cut by two parallel planes, P, P', so as to form two transverse sections, L, L'. To prove that L = L'. Proof. 1. Let AA', BB\ CC be segments of elements between P and P\ any point in P, and 00' II A A', meeting P' at 0' ; join to A, B, C, and 0' to A', B', C". 2. Then 00', A A' determine a plane. VI, prop. I, cor. 2 3. And v OA I! O'A', OB II 0'5', OC II O'C, § 367 .-.Z^O# = Z^'0'.£' 3 ZAOC = ZA'0'C, §337 4. Also, OA = 0'^', 0^ = O'B', T, prop. XXIV .'.if L is placed on V so that falls on 0' and (L4 lies on O'A', A will fall on A', B on £', etc. 5. Similarly, for every point of L there is a single corresponding point of L' on which it will fall. .'. the figures are congruent. § 57, def. congruence Corollaries. 1. The bases of a cylinder are congruent. • 2. The elements of a cylinder are equal. (Why ?) 320 SOLID GEOMETRY. [Bk. VIII. Proposition II. 430. Theorem. Cylinders cut from the same cylindrical space, and having equal elements, are equal. Given two cylinders, AD, A'D', cut from the same cylin- drical space S, and having equal elements AC, A'C To prove that AD = A'D'. Proof. 1. ■.'AC = A'C' J and A'C = A'C, .:AA'=CC'. Ax. 3 2. Similarly for BB' and DD', and for all other seg- ments of the same elements, included between AB, A'B', and CD, CD'. 3. And v CD^AB, and C'D'^A'B', Prop. I .*. solid CD' can be made to slide along in S and coincide with solid AB', since they are equal in all their parts. 4." Adding the common part A'D, AD = A'D'. Ax. 2 Corollary. The cylindrical surfaces of two cylinders cut from the same cylindrical space, and having equal elements, are equal. For it is proved, in step 3, that they can be made to coincide. Secs. 431-434.] THE CONE. 321 2. THE CONE. 431. Definitions. A conical surface is a surface generated by a straight line which moves so as constantly to pass through a given curve and contain a given point called the vertex. The terms generatrix, directrix, elements will be understood from §§ 423, 424. 432. The portions of the conical surface on opposite sides of the vertex are called the nappes, and are usually distinguished as upper and lower. 433. If the directrix is a closed curve, the conical surface incloses a double space, on opposite sides of the vertex, known as a conical space. A section of a conical space made by a plane cutting all of its elements on the same side of the vertex is called a transverse section. A conical surface. DX, the directrix; V, the vertex ; N. X', the lower and upper nappes; V-JJX, a cone, with base the closed figure DX. 434. The portion of a conical space included between the vertex and a transverse section is called a cone, the transverse section being called its base. A cone is considered as having the same directrix and vertex as its conical space, and the segments of the elements between the vertex and base are called the elements of the cone. The distance from the vertex of a cone to the plane of the base is called the altitude of the cone. If the base of a cone is a circle, the cone is said to be circular. In that case the line determined by the vertex and the center of the base is called the axis of the cone. If this axis is perpendicular to the base, the cone is called a right circular cone ; if oblique, an oblique circular cone. 322 SOLID GEOMETRY. [Bk. VIII. A right circular cone is often called a cone of revolution, because it can be generated by the revolution of a right-angled triangle about one of its shorter sides. A right circular cylinder is often called a cylinder of revolution. (Why ?) 435. Postulate of the Cone. A conical surface may be con- structed with any directrix and any vertex. 436. Relation of Cone and Pyramid. If points A, B, C, are taken on the perimeter of the base of a cone, and joined to the vertex V, and if planes be passed through VA and VB, VB and VC, , a pyramid will be formed, called an inscribed pyramid. If the base of the cone is bounded by a convex curve, the base of the pyramid will be a polygon inscribed in it. But whether the base is convex or not, the pyramid is called an inscribed pyramid. Pyramids inscribed in cones. The first figure is a right circular cone. The inscribed pyramids are indicated by dotted lines. It, the altitude. 437. If the base of the cone is a circle, and a regular polygon is circumscribed about it, the planes determined by the sides of the polygon and the vertex of the cone form, with the polygon, a pyramid which is said to be circumscribed about the circular cone. There are other forms of circumscribed pyramids, but the one here mentioned is the only one that is necessary for this work. The slant height of a right circular cone is denned as the slant height of the circumscribed pyramid. (Why V) Secs. 438-440.] THE CONE. 323 438. If a pyramid is inscribed in or circumscribed aLoui a cone, a transverse section of the pyramid and cone cuts off, toward the base, a frustum of a cone and an inscribed or circum- scribed frustum of a pyramid. A ABC, a circumscribed frustum of a pyramid ; A'B'C, an inscribed frustum of s, regular pyramid ; s, the slant height of the frustum of the cone. The terms bases, altitude, and lateral surface will be understood from the terms used with the pyramid and the frustum of a pyramid. 439. From the above definitions it is evident that, if the inscribed or circumscribed frustum of a pyramid has equilateral bases, then if the number of lateral faces increases indefinitely, the frustum of the pyramid, its bases, and its lateral surface, approach as their respective limits the frustum of the cone, its bases, and its lateral surface, but that the altitude does not vary. If a frustum of a right pyramid be circumscribed about the frustum of a right circular cone, the slant height of the frustum of the pyramid may be called the slant height of the frustum of the cone. Hence the following 440. Corollary. If F is the frustum of a cone, and F' the inscribed or circumscribed frustum of a pyramid, of equilateral bases, and if b 1? b 2 , 1, v are the bases, lateral surface, and volume, respectively, of F, and b/, b 2 ', l f , v' the bases, lateral surface, and volume, respectively, of ' F', then if the number of faces of F' increases indefinitely, V^bx, b 2 ' = l),, l' = l, v' = v. 324 SOLID GEOMETRY. [Bk. VIII. Proposition III. 441. Theorem. The lateral area of a frustum of a right circular cone equals one-half the product of the slant height and the sum of the circumferences of its bases. Given a frustum of a right circular cone, I its lateral area, c x and c 2 the circumferences of its upper and lower bases, respectively, and 5 its slant height. To prove that I — J s(c x + c 2 ). Proof. 1. Let I', pi, 2h, s be the lateral area, the perimeters of the upper and lower bases, and the slant height, respectively, of the circumscribed frustum F of a regular pyramid. 2. Then V = %s( Pl + p 2 ). VII, prop. XIV 3. But if the number of faces of F increases indefi- nitely, V = I, jh == ( 'i ? 2h == c 2, while the slant height is the same. § 440 4. .'. I = £s (c x + c 2 ). IV, prop. IX, cor. 1 Corollaries. 1. If the radii of the upper and lower bases are r 1? r 2 , respectively, then 1 = 7rs (r L -f- r 2 ). 2. If r 3 = the radius of the circle midway between the bases of the frustum, then 1 = 2 7rr 3 s. For r 3 = (r 1 + r 2 )/2. Why? J 3. The lateral area of a right circular cone equals half the product of its slant height and the circumference of the base. If the upper base of a frustum of a cone decreases to zero, what does the frustum become ? At the same time what does C\ of step 4 become ? 4. The lateral area of a right circular cylinder equals the product of its altitude and the circumference of the base. If, in step 4, c x = c 2 , what does / equal ? What does s equal ? Prop. IV.] THE CONE. 325 Proposition IV. 442. Theorem. The volume of the frustum of a cone of bases h v b 2 and altitude h is expressed by the formula ▼ = 3 (\ + b 2 + Vb^). Proof. 1. Let v', h, V> b 2 be the volume, altitude, and bases, respectively, of an inscribed frustum of a pyramid with an equilateral base. h 2. Then v' = - (V + bj + VVV). VII, prop. XVII o 3. But if the number of faces of v' increases indefinitely, v' = v, bx — b u b. 2 ' — b 2 , while h is constant. § 440 4. .". v = - (b x + b 2 + ^s/b x b 2 ). IV, prop. IX, cor. 1 o Corollaries. 1. If the frustum is circular, and the radii of b : , b 2 are i\, r 2 , respectively, then v = \ 7rh (i\ 2 + r 2 2 + r 1 r 2 ). 2. If v 3 = the radius of the circle midway between the bases of a frustum of a circular cone, and if h is the altitude, and x x , r 2 are the radii of the bases, then v = i 7rh (i^ 2 + r 2 2 -f- 4 r 3 2 ). See prop. Ill, cor. 2. 3. The volume of a cone of base b and altitude h is expressed by the formula v = \ hb. Let b 2 — in prop. IV. 4. The volume of a circular cone, the radius of whose base is r, is expressed by the formula v = \ 7rr 2 h. 5. The volume of a cylinder of base b and altitude h is expressed by the formula v = hb. Let &i = 6 2 . 6. The volume of a cylinder of altitude h and base radius r is expressed by the formula v = 7rr 2 h. 326 SOLID GEOMETRY. [Bk. VIII. 3. THE SPHERE. 443. Definitions. A sphere is the finite portion of space bounded by a surface, which is called a spherical surface and is such that all points upon it are equidistant from a point within called the center of the sphere. A straight line terminated by the center and the spherical surface is called a radius, and a straight line through the center, terminated both ways by the spherical surface, is called a diameter of the sphere. A section of a sphere made by a plane is called a plane section. A sphere. O, the center. OA,OB, radii. AB, a diameter. 444. Corollaries. 1. A diameter of a sphere is equal to the sum of tivo radii of that sphere. 2. Spheres having the same radii are congruent, and con- versely. 3. A point is within a sphere, on its surface, or outside the sphere, according as the distance from that point to the center is less than, equal to, or greater than, the radius. 445. Postulates of the Sphere. (Compare § 109.) 1. All radii of the same sphere are equal, and hence all ilia meters of the same sphere are equal. 2. If an unlimited straight line passes through a point within a sphere, it must cut the surface at least twice. 3. If an unlimited plane, or if a spherical surface, intersects a spherical surface, it must intersect it in a closed line. 4. A sphere has but one center. 5. A sphere may be constructed with any center, and with a radius equal to any given line segment. Prop. V, THE SPHERE. 327 Proposition V. 446. Theorem. A plane section of a sphere is a circle. Given a sphere with center 0, and a section ABDC made by a plane M. To prove that ABDC is a circle. Proof. 1. M intersects the sphere in a closed line. § 445, 3 2. Suppose joined to two points AB on that line, and OC±3I; draw CA, CB. 3. Then v A OCB, OCA are rt, and OC = OC, and OB = OA, .'. A CBO ^ A CAO, and CB = CA. § 88, cor. 5 So for any other points on the closed line. 4. .'. ABDC is a circle and C is its center. § 165, def. O 447. Definitions. A great circle of a sphere is a circle passing through its center ; a small circle, one not passing through its center. Corollaries. 1. The line determined by the center of a sphere and the center of any small circle of that sphere is perpendicular to that circle. For the line OC from the center of the sphere perpendicular to the circle has been proved to coincide with the line determined by the center of the circle and the center of the sphere, and there is only one line from the center of the sphere perpendicular to the circle. 328 SOLID GEOMETRY. [Bk. VIII. 2. Of two circles of a sphere, the first is greater than, equal to, or less than, the second, according as its distance from the center is less than, equal to, or greater than, that of the second. For AC 2 - f 2 — OC 2 ; .-. the smaller OC, the greater AC, etc. 3. A great circle has the same center and radius as the sphere itself ; hence all great circles of a given sphere are equal. 4. A great circle bisects the sphere and the spherical surface. For if the two parts are applied one to the other, they will coincide ; if they did not, the definition of sphere would be violated. 5. Two great circles bisect each other. They have the same center, and hence a common diameter. 448. The student should notice the relation between the sphere and circle. Thus in prop. V and its corollaries : The Circle. The Sphere. A portion of a line cut off by a A portion of a plane cut off by circumference is a chord. a spherical surface is a circle. The greater a chord, the less its The greater a circle, the less its distance from the center. distance from the center. A diameter (great chord) bisects A great circle bisects the sphere the circle and the circumference. and the spherical surface. Two diameters (great chords) Two great circles bisect each bisect each other. other. Hence may be anticipated a line of theorems on the sphere, derived from those on the circle, by making the following substitutions: 1. Circle, 2. circumference, 1. Sphere, 2. spherical surface, 3. line, 4. chord, 5. diameter. 3. plane, 4. circle, 6. great circle. 449. Definitions. The diameter of a sphere, perpendicular to a circle of that sphere, is called the axis of that circle, and its extremities are called the poles of that circle. The two equal parts into which a great circle divides a sphere are called hemispheres, their curved surfaces being called hemispherical surfaces. Corollary. The axis of a circle pusses through its center. Prop. VI.] THE SPHERE. 329 Proposition VI. 450. Theorem. The straight lines joining any tivo points on the circumference of a circle of a sphere to one of the poles of that circle are equal. Given the circle ABC, and its poles P, P'; PA, PB con- necting P with any two points on the circumference. To prove that PA = PB. Proof. 1. v OP ±Q ABC at C, .'. OP±AC and BC. Why? 2. And V PC = PC, and CA = CB, § 109 '.'. A ACP ^ A BCP, and PA = PB. Why ? Corollary. Great-circle arcs from a pole of a circle to points on the circumference of that circle are equal. (Why ?) 451. Definitions. The length of the great-circle arc joining a pole to any point on the circumference of a circle is called the polar distance of the circle. The shorter polar distance of small circles is to be understood. A fourth of the circumference of a great circle is called a quadrant. Corollaries. 1. Circles of the same sphere, having equal polar distances, are equal. (Why ?) 2. The polar distance of a great circle is a quadrant. (Why ?) 330 SOLID GEOMETRY. [Bk. VIII. Proposition VII. 452. Theorem, If on a spherical surface, each of the great-circle arcs joining a point to two other points {not the extremities of a diameter of the sphere) is a quadrant, then that point is a pole of the great circle through those points. Given P, A, B, three points on a spherical surface, and such that PA = PB = a quadrant; A, B are not extremities of a diameter ; is the center. To prove that P is the pole of the great circle ABO. Proof. 1. '•' PA = PB = a quadrant, .*. Z POA = Z POB = a rt. Z. Ill, prop. II, cor. 2 2. .-. PO _L O ABO. VI, prop. VI, cor. 1 3. ,\ P is a pole of O ABO. § 449, def. pole Exercises. 690. How many points on a spherical surface determine a small circle ? How many, in general, determine a great circle ? 691. Prove that parallel circles of a sphere have the same poles. 692. In the theorem : A diameter which is perpendicular to a chord bisects it, make the substitutions suggested in § 448, and prove the result- ing proposition. 693. Similarly for III, prop. VI. 694. What is the locus of points at a given distance r from a fixed point C ? Prop. VIII.] THE SPHERE. 331 Proposition YIII. 453. Theorem. Of all planes through a point on a sphere the plane perpendicular to the radius drawn to that point is the only one that does not meet the sphere again. Given point P on a sphere with center 0, and M, N, two planes respectively perpendicular and oblique to OP at P. To prove that M does not meet the spherical surface again, but that N does. Proof. 1. Let OB _L X, and OA be any oblique to M. Then v OP is oblique to N } Why ? .-. OB < OP. VI, prop. XI 2. And v OP ± M t .-. OA> OP. VI, prop. XI 3. .". B is within, and ^1 without, the sphere. § 444, cor. 3 4. .'. N meets the spherical surface in more than one point. § 445, 3 5. And v A is any point in M 3 except P, Step 3 .'. M meets the surface onlv at P. 332 SOLID GEOMETRY. [Bk. VIII. 454. Definitions. A plane (or line) which, meeting a spher- ical surface in one point, does not meet it again, is said to be tangent to the sphere at that point. The point is called the point of tangency, or point of contact, and the plane (or line) is called a tangent plane (or line). Corollaries. 1. One and only one plane can be passed through a given point on a sphere, tangent to that sphere. (Why?) 2. Any tangent plane is perpendicular to the radius at the point of contact. For it cannot be oblique and be a tangent plane. Step 4. 3. A plane perpendicular to a radius at its extremity on the spherical surface is tangent to the sphere. k Exercises. 695. To find a point in a given plane, equidistant from two fixed points in that plane, and at a given distance ci! from a point C not in that plane. Discuss for 0, 1, 2 solutions. 696. Prove that the lateral area of any right cylinder equals the product of its altitude and the perimeter of the base. (Inscribe a prism and apply the theorem of limits.) 697. How many square feet in the surface of a cylindrical water tank, open at the top, its height being 40 ft., and its diameter 40 ft. ? 698. Considering the moon as a circle of diameter 21G0.6 miles whose center is 234,820 miles from the eye, what is the volume of the cone whose vertex is the eye and whose base is the full moon ? 699. Find a point whose distance from a fixed point is d and whose distance from each of two intersecting planes is cV. Discuss the solution as to impossible cases, and the number of such points for possible cases. 700. Find the locus of points equidistant from two given points, and at a given distance d from a given point. 701. To determine a plane which shall pass (1) through a given line and be (2) through a given point and be at a given distance from a given at a given distance from a given point. line. Prop. IX.] THE SPHERE. 333 Proposition IX. 455. Theorem. Four jjoints, not coplanar, determine a spherical surface. Given four points, A, B, C, D, not coplanar. To prove that A, B, C, D determine a spherical surface. Proof. 1. Draw AB, BC, CD, DA, AC. Let E be the circumcenter of A ACD, F of A ABC, EH 1. ACD, FJ±ABC. 2. Then F, F are on the _L bisectors of AC; call these _L bisectors GF, OF. I, prop. XLI 3. And v FA = EC = ED (Why ?), .'. any point on EH is equidistant from A, C, D. VI, prop. XI, 3 Similarly, any point on FJ is equidistant from A, B, C. 4. But CA ± plane EOF. Why ? 5. .'. planes ABC, ACD _1_ EOF. Why ? G. .'. both EH and FJ lie in plane EOF. VI, prop. XIX, cor. 1 7. And V FJ meets EH, uniquely, as at P, I, prop. XVII, cor. 4 .*. P is the center of a sphere whose surface passes through A, B, C, D, and there is only one such sphere. 334 SOLID GEOMETRY. [Bk. VIII. 456. Definitions. A sphere is said to be circumscribed about a polyhedron if the vertices of the polyhedron all lie on the spherical surface ; the polyhedron is then said to be inscribed in the sphere. Corollaries. 1. Two spherical surfaces having four common points, not coplanar, coincide. For by step 7 they have the same center, P, and the same radius. 2. The perpendiculars to the four faces of a tetrahedron, through the circumcenters of those faces, are concurrent. For each of these perpendiculars passes through P, the center of the sphere whose surface is determined by the four vertices. 3. A sphere can be circumscribed about any tetrahedron. 457. The angle between two great-circle arcs is defined as being the plane angle between tangents to those arcs at their point of meeting. P E.g. the angle made by arcs AP, BP is defined as the plane angle A'PB'. 458. From this definition follow these corollaries : 1. The angle made by two arcs has the same numerical measure as the dihedral angle of their planes. (§ 359.) 2. An angle made by tivo arcs has the same numerical meas- ure as the arc which these arcs intercept on the circumference of the great circle of wh ich the vertex is the pole. That is, Z APB = /.A'PB'- ZAOB, which has the same numerical measure as AB. Secs. 459-463.] THE SPHERE. 335 459. A spherical polygon is a portion of a spherical surface bounded by arcs of great circles. The words sides, angles, vertices, etc., are used as with plane polygons. 460. A spherical polygon is said to be convex when each side produced leaves the entire polygon on the same hemi- sphere ; otherwise it is said to be concave. In the figure, ABP is a convex polygon, for if any side, as PB, is produced it leaves the entire polygon on the hemisphere to the left of PB. But QRS T is concave, because side SB, or QR, produced, leaves part of the polygon on one hemisphere thus formed, and part on the other. 461. Corollary. No side of a convex spherical polygon is greater than a semicircumference. For if AP> semicircumference, suppose XP = a semicircumference. Then v great circles bisect each other (prop. V, cor. 5), PB must pass through X ; but then PB produced would leave part of the polygon on one hemisphere and part on the other, so that it could not be convex. 462. A lune is a portion of a spherical surface bounded by the semicircumferences of two great circles. The angle of a lune is that angle tow r ard the lune made by the bounding arcs. In the figure, PAP'B is a lune, and Z. APB, or /. BP'A, is its angle. The limiting cases of a lune are evidently a semicircumference, when the angle is zero, and a spherical surface, when the angle is 360°. 463. Corollary. Limes on the same sphere, and having the same angle, are congruent. For one can evidently be made to coincide with the other. 336 SOLID GEOMETRY. [Bk. VIII. Proposition X. 464. Theorem. On the same sphere or on equal spheres lunes are proportional to their angles. (In this figure the eye is supposed to be looking down on the sphere from above the angle of the lune, as on the North Pole of the earth. This allows only half of each lune to be seen.) Given two lunes, C and D, with angles A and B respec- tively. To prove that A : B = C : D. Proof. 1. If C and D are on different spheres, they can be placed in the relative positions shown in the figure. § 444, cor. 2 Suppose A and B divided into equal A, x, and sup- pose A = nx, and B = n'x. (In the figure n = 6, n' = 4.) 2. Then C is divided into n congruent lunes, y, and D " " n' " « § 463 3. m * m A = 2L=*=3L = l. Why? B n'x n' n'y D y ' Exercise. 702. The six planes perpendicular to the six edges of a tetrahedron at the mid-points of its edges, meet in a point, (Is this point the center of a particular sphere ?) Prop. X.] THE SPHERE. 337 465. Proof for incommensurable case. (Compare § 410.) 1. Suppose A divided into equal A, x, and suppose A = nx, while B = n'x + some remainder w, such that w < x. Then C is divided into n congruent lunes, y, and D is the sum of n' congruent lunes, y, + a remain- der z, such that z < y. 2. Then B lies between n'x and (n* + 1) x, Why ? and D " " n'y " (%' + 1) */. Why ? 3. .". — lies between — and — , Why? A nx nx ... 2> v . ra'v , (V + 1W while — lies between — — and — • C ny ny 4. . ' . — and — both lie between — and A 6 n n 7? T) 1 5. .*. — and — differ by less than -■ Why ? j\. o n> 6. And v - can be made smaller than any assumed n J difference, by increasing n, .*.to assume any difference leads to an absurdity. B D . AC 7. /. 2 = -, whence -==-• 338 SOLID GEOMETRY. [Bk. VIII. 466. Definition. The solid bounded by a lune and two semi-circles is called a spherical wedge. The angle of the lune is called the angle of the wedge. The word iingula is sometimes used for spherical wedge. Corollaries. 1. A lune is to the spherical surface on which it lies as the angle of the lune is to a perigon. For the spherical surface may be considered as a lune whose angle is a perigon. 2. A spherical ivedge is to the sphere of which it is a part as the angle of the wedge is to a perigon. In the proof of prop. X, if we should substitute the word wedge for the word lune, and consider the sphere as a wedge whose angle is a perigon, the corollary would evidently be proved. * Exercises. 703. To draw a plane tangent to a given sphere, from a point on the sphere. (See III, prop. XXVI.) Also, to draw one from an external point. 704. To find the locus of centers of spheres whose surfaces (1) pass through two given points ; (2) are tangent to two given coplanar lines ; (3) are tangent to two given planes. (As special cases, the lines may be parallel and the planes may be parallel.) 705. What is the locus of the centers of spheres whose surfaces (1) pass through the vertices of a given triangle ? (2) are tangent to the sides of a given triangle ? 706. To find the center of a sphere whose surface includes both a given circumference and a point not in the plane of that circumference. (As a special case, suppose the point is on the perpendicular to the plane of the given circle through the center.) 707. In the figure of prop. IX show that E, G, F, P are concyclic. Hence show that six circumferences intersect by threes in the ciivum- centers of the faces of a tetrahedron, and all intersect in the center of the circumscribed sphere. 708. To construct a sphere of given radius whose surface shall contain three given points. 709. Also, of given radius whose surface shall contain two given points and be tangent to a given plane. Secs. 467, 468.] THE SPHERE. 339 The Relation of Spherical Polygons to Polyhedral Angles. 467. If the center of a sphere is at the vertex of a pyramidal space, the pyramidal surface cuts from the spherical surface two spherical polygons. In the above figure the two polygons are ABCD, A'B'C'D'. These polygons have their like-lettered angles and sides equal respectively. For example. /. A = Z A', since they have the same numerical measure as the opposite dihedral angles of planes ADOD'A' and ABOB'A'. Also, AB = A'B\ since the central angles BOA and B'OA' are equal. 468. But the equal elements of these polygons are arranged in reverse order. And as the polyhedral angles are called opposite and are proved (VI, prop. XXVI) symmetric, so the spherical polygons are called opposite spherical polygons. And since these have just been shown to have their corresponding elements equal but arranged in reverse order, they are called symmetric spherical polygons. Thus all opposite polygons are symmetric ; but since polygons can slide around on the sphere, it follows that symmetric polygons are not necessarily opposite, although they are congruent to opposite polygons. 340 SOLID GEOMETRY. [Bk. VIII. 469. Since the dihedral angles of the polyhedral angles have the same numerical measure as the angles of the spherical polygons, and the face angles of the former have the same numerical measure as the sides of the latter, it is evident that to each property of a polyhedral angle corresponds a reciprocal property of a spherical polygon, and vice versa. This relation appears by making the following substitutions : Polyhedral Angles. Spherical Polygons. a. Vertex. a. Center of Sphere. b. Edges. b. Vertices of Polygon. c. Dihedral Angles. c. Angles of Polygon. d. Pace Angles. d. Sides. 470. In addition to the correspondences between poPyhedral angles and spherical polygons, it will be observed that a relation exists between a straight line in a plane and a great- circle arc on a sphere. Thus, to a plane triangle corresponds a spherical triangle, to a straight line perpendicular to a straight line corresponds a great-circle arc perpendicular to a great-circle arc, etc. The word arc is always understood to mean great-circle arc, in the geometry of the sphere, unless the contrary is stated. It is very desirable that every school have a spherical blackboard, with large wooden compasses for the drawing of both great and small circles. It is only by the use of such helps that students come to a clear knowl- edge of spherical geometry. If such a blackboard is at hand, it is recom- mended that many problems and exercises of Book I be investigated on the sphere. E.g. the problem, To bisect a given arc, corresponds to I, prop. XXXI, and the solutions are quite similar. Likewise the prob- lems, To bisect a given angle, To draw a perpendicular to a given line from a given internal point, etc., have their corresponding problems in spherical geometry. Exercise. 710. State without proof the proposition in the geometry of the sphere corresponding to the following : Every face angle of a con- vex polyhedral angle is less than a straight angle. Props. XI-XIII. THE SPHERE. 341 Proposition XI. 471. Theorem. (a) In any trihedral angle, each face angle being less than a straight angle, the sum of any tic o face angles is (a') In any spherical tri- angle, each side being less than a semicircumference, the sum of any two sides is greater, and their difference greater, and their difference less, than the third angle. less, than the third side. Proof. In VI, prop. XXVII, with its cor. 1, (a) has been proved. Hence (a') is also proved. § 469 Proposition XII. 472. Theorem. (a) In any polyhedral angle, each face angle being less than a straight angle, any face an- gle is less than the sum of the remaining face angles. (a') In any spherical poly- gon, each side being less than a semicircumference, any side is less than the sum of the remaining sides. Proof. In VI, prop. XXVII, cor. 2, (a) has been proved. Hence (a') is also proved. § 469 Proposition XIII. 473. Theorem. (a) In any convex polyhe- (a') In any convex spherical dral angle the sum of the face polygon the sum of the sides angles is less than a perigon. is less than a circumference. Proof. In VI, prop. XXVIII, (a) has been proved. Hence (a') is also proved. § 469 342 -SOLID GEOMETRY. [Bk. VIII. Proposition XIV. 474. Theorem. (a) No face angle of a convex (a') No side of a convex p>olyhedral angle is greater spherical polygon is greater than a straight angle. than a semicircumference. Proof. From § 461, (a') is true. Hence (a) is also proved. § 469 475. Definitions. If ABC If O-ABC is a trihedral is a spherical triangle, and A', angle, and OA', OB', OC are z? ^r B', C are the poles of a, b, c, respectively, and if A and A', B and B', C and C lie on the same side of a, b, c, respec- tively, then A A'B'C is called the polar triangle of ABC. perpendiculars to a, b, c, the faces opposite A, B, C, respec- tively, and if A and A', B and B', C and C lie on the same side of a, b, c, respectively, then trihedral Z O-A'B'C is called the polar trihedral angle of O-ABC. In referring to polar triangles ABC, A'B'C', the above arrangement of elements will always be intended. Also, in referring to symmetric spherical triangles, ABC and A'B'C, it will always be understood that Z.A = /. A', etc., and AB — A'B' , etc. The polar triangle of ABC is often called the polar of ABC. It is evident from the one-to-one correspondence of § 475, that to every proposition concerning polar triangles corresponds a proposition concerning polar trihedral angles, and vice versa. Prop. XV.] THE SPHERE. 343 PROrOSTTTOX XV. 476. Theorem. If one spherical triangle is the polar of a second, then the second is also the polar of the first. Given a spherical triangle, ABC, and A'B'C its pol^r. To prove that A ABC is the polar of A A'B'C. Proof. 1. In the figure suppose AC, AB', drawn. 2. Then, v B 1 is a pole of b, .'. AB' is a quadrant. Prop. VI, cor. 2 Similarly, v C is a pole of c, .'. AC is a quadrant. 3. .'.A is a pole of a'. Prop. VII Similarly, B and C are poles of J' and c', respectively. 4. And v A, A' are on the same side of a', and so for the other vertices and sides, .'.A ABC is the polar of A A'B'C. § 475 Corollary. If one trihedral angle is the polar of a second, then the second is also the polar of the first. For from the one-to-one correspondence of § 475, the proof is evi- dently identical with the above. Note. One triangle may fall entirely within or entirely without its polar ; or one may be partly within and partly without the other. Simi- larly, one trihedral angle may fall entirely within or entirely without its polar trihedral angle, or may be partly within and partly without the latter. 344 SOLID GEOMETRY. [Bk. VIII. Proposition XVI. 477. Theorem. Any angle of a spherical triangle has the same numerical measure as the supplement of the opposite side of its polar. Given ABC, a spherical triangle, and A'B'C its polar. To prove that the numerical measure of any angle C is 180° -c'; of 6", 180° - c. Proof. 1. Suppose a, b to cut c' in E\ D\ respectively, and a\ V to cut c in E, D, respectively. 2. Measure of Z C = that of tTe\ § 458, 2 But D% = AE' + IVB' - AB< = 90 o 4-90 o -i 7 fe' = 180°-c'. Why 90°? 3. Similarly for Z C, substituting A, B, D, E, for A, B', D\ E\ and vice versa, in the above proof. Corollaries. 1. If two spherical triangles arc mutually equiangular, their polars are mat mi 11;/ equilateral ; if mutually equilateral, their polars arc mutually equiangular. 2. The sum of the angles of 2'. The sum of the dihedral a spherical triangle is greater angles of a trihedral angle is than one and less than three greater than one and less than straight angles. three straight angles. Prop. XVII. THE SPHERE. 345 For by prop. XIII (a'), < a' + b' + c' < 360°. .-. by subtracting from 3 ■ 180°, 3 • 180°> (180°- a') + (180°- b') + (180°- c') > 180 c .-. by prop. XVI, 3 ■ 180° >Z A + Z B + ZC >180 c 478. Definitions. If ABCD A' is a spherical polygon, and A\ B', C, D', are the poles of XA, AB, BC, CD, , respectively, and if A', B', lie on the same side of XA, AB, that the polygon does, then A'B'C'D' If O-ABCD X is a poly- hedral A, and OA', OB', OC, OD', are _b to planes OXA, OAB, OBC, , respectively, and if A', B', lie on the same side of planes OXA, OAB, OBC, that the polyhedral angle does, then O-A'B'C'D' is called the polar polygon is called the polar poly- of ABCD hedral angle of O-ABCD Polar trihedral angles are also called supplemental trihedral angles. 479. A spherical triangle is said to be birectangular if it has two right angles, trirectangular if it has three. Proposition XVII. 480. Theorem. (a) Two opposite or two symmetric trihedral angles are congruent if each has two equal dihedral angles, or two equal face angles. (a') Two opposite or two symmetric spherical triangles are congruent if each has tiro equal angles or two equal sides. Proof for (a'). 1. Their sides and angles are respectively equal but arranged in reverse order. § 468 2. But if to the order ABC corresponds B'A'C, and if B 1 = A', then B' and A' may be interchanged. 3. Then to the order ABC will correspond A'B'C, and the A are congruent by superposition. 346 SOLID GEOMETRY. [Bk. VIIL Proposition XVIII. 481. Theorem. Tivo symmetric spherical triangles on the same sphere or on equal spheres are equal. Given two symmetric spherical triangles, ABC, A'B'C, on the same sphere. To prove that A ABC = A A'B'C. Proof. 1. The plane of A, B, C determines a small circle. 2. Let be the pole of the O, and similarly for 0' and spherical A A'B'C. 3. Then V side AB = side A!B\ .'. chord AB = chord A'B'. (In the figure they are not drawn because AB is so nearly straight.) Ill, prog. 1 1 L Similarly for chords BC, B'C, and CA, CA'. 4. .'. plane A ABC ^ plane A A'B'C. I, prop. XII 5. .'. O ABC — O A'B'C, being circumscribed about, congruent plane A. Why ? 6. .-. 0?1 = 0?i = 6C = OA' = (fji' = CC'. Why ? 7. .". spherical A JO/7 £ .-l'O'T?', BOC £ 5'0'C, C04 ^ C"0'4'. §§ 468, 480 (a') 8. .'. A .!/;(/ = A A'B'C. Ax. 2 Prop. XIX.] THE SPFfERE. 347 Proposition XIX. 482. Theorem. Two triangles on the same sphere, or on equal spheres, are either congruent or symmetric and equal if (a) two sides and the in- (b) two angles and the in- cluded angle eluded side of the one figure are equal to the corresponding parts of the other. Proof. If the parts are arranged in the same order, the triangles can be brought into coincidence, as in I, props. I, II. If they are arranged in reverse order, then one triangle is congruent to the triangle symmetric to the other. Why? Corollary. Two triliedral angles are either congruent or symmetric and equal if (a) two face angles and the (b) two dihedral angles and the included dihedral angle included face angle of the one figure are equal to the corresponding parts of the other. For from the one-to-one correspondence of § 475, the proof is evidently identical with the above, without the labor of drawing the figures. 348 SOLID GEOMETRY. [Bk. VIII. Proposition XX. 483. Theorem. (a) If a trihedral angle has (a') If a spherical triangle two dihedral angles equal to has two angles equal to each each other, the opposite face other, the opposite sides are angles are equal. equal. Given the A ABC, with ZA = ZB. To prove that a = b. Proof. 1. Let A A'B'C be symmetric to A ABC, so that a = a', b = b', etc. 2. Then v ZA = ZB, .'. ZA' must equal ZB', and the A are congruent and a' = b. Prop. XIX 3. But '•' a = a\ and a' = b, .'. a = b, which proves (a'). Ax. 1 Hence (a) is also proved. § 469 Corollaries. 1. (a) If a trihedral angle (a') ^4?? equiangular spher- has its three dihedral angles ical triangle is equilateral, equal, it has also its three face angles equal. In A ABC, if Z A = Z B, then a = b ; and if Z C also equals Z B, c also equals b. .: if Z A = Z B = Z C, a = b = c. Ax. 1 2. (a) If a trihedral angle (a') If a spherical triangle has two face angles equal to has two sides equal to each each other, the opposite dihe- other, the opposite angles are dral angles are equal. equal. The proof is almost identical with that of I, prop. Ill, and hence is left for the student. Prop. XXL] THE SPUE UK. 349 Proposition XXI. 484. Theorem. Two triangles on the same sphere, or on equal spheres, are either congruent or symmetric and equal if (a) the three sides (b) the three angles of the one figure are equal to the corresponding parts of the other. (a) Given A ABC, A'B'C, mutually equilateral, the sides being arranged in the same order ; also A ACX symmetric to A A'B'C. To prove that A ABC = A A'B'C, A ABC is symmetric to A ACX. Proof. 1. Place A ACX as in the figure ; draw BX. Then Z BXC = Z CBX, and Z AXB = Z XBA. Prop. XX, cor. 2 2. .\ZAXC = ZCBA, i.e. ZB =ZX = ZB'. Similarly with the other angles. 3. .\ A A'B'C ^ A ABC. 4. And A ABC is symmetric to A ACX Ax. 3 Why ? Why t 350 SOLID GEOMETRY. [Bk. VIII. (b) Given (Let the student state it.) To prove (Let the student state it.) Proof. 1. Their polars are mutually equilateral. Why ? 2. .'. their polars are congruent or symmetric. Why ? 3. .*. A ABC and A A'B'C are mutually equilateral. Props. XV, XVI, cor. 1 4. .'.AABC*£ or symmetric to A A'B' 6". Prop. XXI (a) Corollary. Two trihedral angles are either congruent or symmetric and equal if (a) the three face angles (b) the three dihedral angles of the one figure are equal to the corresponding parts of the other. Exercises. 711. A plane isosceles triangle can have its equal sides of any length. Discuss as to a spherical isosceles triangle on a given sphere. 712. As with plane triangles, the pole (circuracenter) may fall outside the triangle, or on a side. Prove theorem 481 for those cases. 713. Prove I, prop. XII (a corresponding theorem of Plane Geometry) by the method of prop. XVIII. 714. Draw the figure of a spherical quadrilateral and its polar; also of a four-faced polyhedral angle and its polar. 715. Prove that if one spherical polygon is the polar of another, then the second is the polar of the first. State the reciprocal theorem for polyhedral angles. (The special case of the quadrilateral may be taken.) Prop. XXII.] THE SPHERE. 351 Proposition XXII. 485. Theorem. For a great circle to be perpendicular to a small circle, it is necessary and it is sufficient that the circum- ference of the former pass through a pole of the latter. Given a small circle S, with P and P' its poles, G a great circle, and the center of the sphere. To prove that for G to be J_ to S it is necessary and it is sufficient that its circumference pass through P. Proof. 1. PP' _L 8. Def. pole 2. And if G passes through P it passes through PP', and G ± S. VI, prop. XVIII 3. .*. it is sufficient that G contain P. 4. Furthermore it is necessary; for if G _L S, then PP lies in G or else PP' II G. Why ? 5. But PP' is not II to G, for each contains 0. Why ? 6. .'. it is necessary, and it is sufficient, that 6r contain P. Corollaries. 1. Through a poi?it X, within or on a, sphere, it is possible to pass one great circle perpendicular to a given circle S, and only one unless X is on the straight line through the poles of S. For PP' passes through the center 0, and PP' and X determine a great circle _L S, unless X is on PP'. (Why ?) May X be without the sphere ? 352 SOLID GEOMETRY. [Bk. VIII. 2. If the circumferences of two great circles are drawn per- pendicular to a third circumference, they will intersect at the poles of the circle of that third circumference. 486. Definition. If a great circle is perpendicular to a small circle, their circumferences are said to be perpendicular to each other. Proposition XXIII. 487. Theorem. If from a point on a sphere arcs of great circles both perpendicular and oblique, are drawn to any circumference, then, 1. The shorter perpendicular is less than any oblique; 2. Two obliques cutting off equal arcs from the foot of this perpendicular are equal ; 3. Of two obliques cutting off unequal arcs from the foot of this perpendicular, the one cutting off the greater arc is the greater. Given S, any circle of a sphere ; P any point on the spherical surface; minor arcs, PC J_ circumference S, PA, PB, PD obliques ; PC = CD, and AC > CD or its equal BC. To prove that (1) PC < PB ; (2) PB = PD ; (3) Rk > PD or its equal PB. Prop. XXIII.] THE SPHERE. 353 Proof. 1. Suppose N the pole of S, on the same side of S as P; draw NA, NB, ND. 2. CP produced passes through X. Prop. XXII, cor. 2 3. NB, or its equal NC, < NT + PB. Prop. XI, (a') 4. .'.PC PB, and pQ > 7^P. Why ? 10. .'. PE+ PA, or PA, > PB. 488. Definition. The excess of the sum of the angles of a spherical n-gon over (ii — 2) straight angles is called the spherical excess of the ?i-gon. Hence the spherical excess of a 2-gon (lime), 3-gon (triangle), 4-gon, is the excess of the sum of its angles over 0, 1, 2, straight angles. Exercises. 716. Prove that if in prop. XVI the word polygon is sub- stituted for triangle, the resulting theorem is true, and state the corollary that follows from it, analogous to corollary 1 of prop. XVI. 717. What is meant by the spherical excess of a spherical decagon ? What is the spherical excess, in degrees, of a triangle whose angles are 75°, 90°, 100°? 718. What is the spherical excess, in radians, of a triangle whose angles are 80°, 90°, 100°? Also of a triangle whose angles are 1, 2, and 3 radians, respectively ? 354 SOLID GEOMETRY. [Bk. VIII. Proposition XXIV. 489. Theorem. A spherical triangle equals a lune icJwse angle is half the spherical excess of the triangle. B' Given T, a spherical triangle, with angles a, b, c. To prove that T = a lime whose angle is \ (a + b -f c — st. Z). Proof. 1. Let A, B, C = limes of A a, b, c, respectively (in the figure they are AA', BB', CC), and S= surface of sphere. 2. A AB'C and ABC are mutually equilateral, for AC + CA = semicircumference = AC + CA ; hence AC = AC, and so for the other sides. Ax. 3 3. .'. A AB'C = A ABC, so that T + A AB'C = hme A. Prop. XXI 4. .\A + (B- T) + (C-T) =$S, Ax. 8 or T=±(A + B + C-iS). Axs. 3, 7 5. But '.* i S = a lime whose Z is a st. Z, § 402 .'. T = a lune whose Z is J- (a + /> + c — st. Z). Corollary. A spherical polygon equals a lune whose angle is half the spherical excess of the polygon. For the polygon can he cut into triangles as in Plane Geometry. The practical method of measuring a spherical polygon is given in § 493, cor. 3. Prop. XXV.] THE MENSURATION OF THE SPHERE. 355 4. THE MENSURATION OF THE SPHERE. Proposition XXV. 490. Theorem. The area of the surface of a sphere of radius r is -I 7rr 2 . B A' M' B'O Proof. 1. A semicircle cut off by a diameter XX, revolving about X'X as an axis, generates a sphere. 2. Let AB be one of a number of chords inscribed in arc XBX', forming half of a regular polygon. Let 031 ± AB, thus bisecting it ; III, prop.V let A A', MM, BB', all be _L to X'X, and AC II X'X. 3. Then AB, revolving about axis X'X, generates the surface I = 2 it - AB ■ M'M Prop. Ill, cor. 2 4. But V A A CB — A J/J/' 0, Why ? .-. 03f : J/' J/ = AB : J C = A B : .!'£'. 5. .'. ^LB • IT If = A'B' • OK IV. pr0 p. I 6. .'. Z = 2it • .!'£' • OM. Subst. in 3 7. Summing for all frustums, including two cones, the sum of the lateral surfaces = 2 7r • OM • (X'A 1 + A'B' + ) = 2tt • OM -2 r. Axs. 2, 8 8. But if the number of sides increases indefinitely, the sum of the lateral surfaces == surface of sphere, s, and OM = r ; .'. * = 2ir • r • 2r = 4 7rr\ IV, prop. IX, cor. 1 356 SOLID GEOMETRY. [Bk. VIII. 491. Definitions. That part of a spherical surface which is included between two parallel planes which cut or touch the surface, is called a zone. / — — \ The solid bounded by the zone and the two parallel planes is x r Zones and spherical segments. In first figure, Called a Spherical seg- lower base is zero. ment. The distance between the two parallel planes determining a zone and a spherical segment is called the altitude of the zone and the segment. The circumferences in which the planes intersect the spher- ical surface are called the bases of the zone, and the circles are called the bases of the segment. In case of tangent planes the bases may one or both reduce to zero. If one base only reduces to zero, the zone, or segment, is said to have one base. 492. Definition. As a plane ' angle is often said to be measured by the ratio of the intercepted arc to the whole circumference (§ 256), so a polyhedral angle is said to be measured by the ratio of the intercepted spherical polygon to the whole spherical surface. The practical method of measuring a polyhedral angle is given in § 493, cor. 4. 493. Corollaries. 1. The area of a zone of altitude a, on a sphere of radius r, is 2 7rra. For, prop. XXV, step 7, the sum of the lateral surfaces may approach zs their limit a zone, in which case X'A' + A'B' + = a, and OM == r. 2. The area of a lune of angle a (expressed in radian measure) on a sphere of radius r, is 2 a? 2 . By prop. X, I : 4 7tr 2 = a : 2 it. 3. The area of a spherical polygon of spherical excess a (expressed in radian measure) is av 2 . Prop. XXV.] THE MENSURATION OF THE SPHERE. 357 For by prop. XXIV, the polygon equals the lune whose angle is a/2. .-. the area = 2 • — • r- = ar 2 , by cor. 2. 4. The measure of a polyhedral angle whose intercepted spherical polygon has a spherical excess a is For by definition, § 490, it is a 4-7T ar' 2 4 7tr 2 5. The area generated by a chord of a circle revolving about a diameter which does not cut it, equals 2tt times the product of its projection on that diameter, and the distance from the center to the chord. (Why ?) G. The areas of two spheres are propjortional to the squares of their radii. _ a 4 7T/-2 r °- For — = = — ■ a 4 tit 1 r 2 Exercises. 719. What is the area of a spherical triangle the sura of whose angles is 4 radians, on a sphere of radius 1 ft. ? 720. Also of one the sum of whose angles is 270°, r being 2 ft. ? 721. Also of one the sum of whose angles is 180°, on any sphere ? 722. Also of one the sum of whose angles is 237° 29', r being 10 in. ? 723. Also of a spherical quadrilateral the sum of whose angles is 417° 29', on a sphere of radius 2 in. ? 724. Also of a spherical pentagon the sum of whose angles is 4 straight angles, on a sphere of radius 5 in. ? 725. What is the measure in radians of a polyhedral angle the spherical excess of whose intercepted spherical polygon is 8 it ? 726. What is the ratio of a trihedral angle the sum of the angles of whose intercepted spherical triangle is 1.5 7r radians, to a tetrahedral angle the sum of the angles of whose intercepted spherical quadrilateral is 2.5 it radians ? 727. What is the area of a spherical triangle whose angles are 70°, 80°, 90°, on a sphere of diameter 20 in. ? 728. Show that a trirectangular triangle is its own polar. 729. The locus of points on a sphere, from which great-circle arcs perpendicular to the arms of an angle are equal, is the great-circle arc bisecting that angle. 358 SOLID GEOMETRY. [Bk. VIII. Proposition XXVI. 494. Theorem. Two solids lying between two parallel planes, and such that the ttvo sections made by any plane parallel to the given planes are equal, are themselves equal. Given two solids, S, S', lying between parallel planes, M, N, and such that the two sections A, A\ or B, B\ , made by any plane Q, or R, , are equal, i.e. A = A'B = B\ To prove that s=sr Proof. 1. Let K, K' be two segments of S, S', lying between the sections A and B, and A' and B' ; let the alti- tude of K, K' be 1/n of the altitude h of S and S'. 2. Suppose two straight lines to move so as always to be perpendicular to Q, and to touch the perimeters of A, A', thus generating two cylinders (or prisms, or combinations of cylinders and prisms) of altitude h/n as in Fig. 2. As the volumes of both prisms and cylinders are expressed by the same formula, v = bh } we may speak of these solids as cylinders, C, C. 3. Then C = C, since they have equal bases and alti- tudes ; and so for other pairs of cylinders described in the Same wav. with altitude h/n. Prop. XXVI.] THE MENSURATION OF THE SPHERE. 359 4. .'. the sum of the solids like C = the sum of the solids like C", whatever n equals. 5. But if n increases indefinitely, h/n decreases in- definitely, and it is evident that the sum of the solids like C = S, while the sum of the solids like 6" = S\ 6. .'. S= S'. IV, prop. IX, cor. 1 495. This important proposition is known as Cavalieri's theorem. It will be seen that VII, prop. XV, is merely a special case of this proposition. We shall base the mensura- tion of the volume of the sphere upon it. Solids of this kind are often called Cavalieri bodies. Exercises. 730. A spherical triangle is to the surface of the sphere as the spherical excess is to eight right angles. 731. The locus of points on a sphere, from which great-circle arcs to two fixed points on the sphere are equal, is the circumference of a great circle perpendicular to the arc joining those points at its mid-point. 732. There is evidently a proposition of plane geometry analogous to Cavalieri's theorem, beginning, "Two plane surfaces lying between two parallel lines " State this proposition and prove it. 733. From ex. 732 prove that triangles having equal bases and equal altitudes are equal. 734. What is the ratio of the surface of a sphere to the entire surface of its hemisphere ? 735. Prove that the areas of zones on equal spheres are proportional to their altitudes. 736. Find the ratio of the surfaces of two spheres, in terms of their radii, i\ and r 2 . 737. What is the ratio of the area of a great circle of a sphere to the area of its spherical surface ? 738. If a meter is 0.0000001 of a quadrant of the earth's circumfer- ence, and the earth is assumed to be a sphere, how many square myria- meters of surface has the earth ? 739. What is the radius of the sphere whose area is 1 square unit ? Answer to 0.001. 360 SOLID GEOMETRY. [Bk. VIII. Proposition XXVII. 496. Theorem. The volume of a sphere of radius r is expressed by the formula v = j 77-r 3 . Proof. 1. Suppose the sphere circumscribed by a cylinder, and suppose two cones formed with the bases of the cylinder as their bases, and their vertices at the center of the sphere. Suppose the solid to be cut by a plane Q, parallel to the bases, and x distant from the center of the sphere. 2. Then since x also equals the radius of the O cut — ^ from the cone, because the altitude of the cone equals the radius of its base, .*. area of ring CD between cone and cylinder = 7r(r 2 -a- 2 ). 3. But the area of the O AB cut from the sphere is also it (r 2 — x 2 ), because its radius is V r* — .>■'-. 4. .'. the sphere and the difference between the cone and cylinder are two Cavalieri bodies, and .'. they are equal. Prop. XXVI 5. .'. v = tt>- 2 • 2r - Trr 2 ■ ",'* = i tt/- 8 . Why ? Prop. XXVII.] THE MENSURATION OF THE SPHERE. 361 Corollaries. 1. The volume of a sphere equals two- thirds the volume of the circumscribed cylinder. (Arehimedes's theorem.) / The volume of the circumscribed cylinder is evidently itr- • 2 r, or 2 7tr\ 2. The volume of a sphere equals the product of its surface by one-third of its radius. For the surface is -4 7rr' 2 , prop. XXV ; and f itf 1 r • 4 7tr~. 3. The volumes of two spheres are proportional to the cubes of their radii. 4. The volume of a spherical segment of one hasp, of altitude a, is expressed by the formula v = \ 7ra 2 (3 r — a). For, as in the theorem, it equals the difference between a circular cylinder of radius r and altitude a, and the frustum of a cone, of the same altitude and with bases of radii r and (r — a). .-. v = itr-a — \na \r- + (r — a)' 2 + r (r — a)] Prop. IV, cor. 1 = i7ra 2 (3r-a). 497. Definitions. A spherical sector is the portion of a sphere generated by the revolution of a circular sector about any diameter of its circle as an axis. The base of the spherical sector is the zone generated by the arc of the circular sector, and the altitude is the altitude of that zone. If the base of the spherical sector is a zone of one base only, the spherical sector is called a spherical cone. Spherical sectors. The upper one a spherical cone. 362 SOLID GEOMETRY. [Bk. VIII. Corollaries. 1. The volume of a spherical cone, whose base b has an altitude a, is expressed by the formula v = J 7rr 2 a, or v = i br. For it evidently equals the sum of a cone and a spherical segment of one base. What does the latter equal, by § 496, cor. 4 ? Show that the cone = \ it (r — a) [r 2 — (r — a) 2 ]. Then add the results, and show that the sum is § itr 2 a. But 6 = 2 nra. (Why ?) 2. The volume of a spherical sector, whose base is b and alti- tude a, is expressed by the formula v = §7rr 2 a, or v = ± br. For it equals the difference between two spherical cones. Suppose these to have altitudes a\ , « 2 , and bases b { , b 2 , and volumes Vi , v 2 , respectively. Then v = Vi — v 2 = j nr 2 a\ — § 7tr 2 a 2 — § nr 2 (<3i — a 2 ). But ci! - a 2 = a. (Why ?) .-. u = 1 7rr 2 a. Now show that v = \br.- Exercises. 740. Show that if the directrix of a cylinder is the cir- cumference of a great circle of a sphere, and the generatrix is perpen- dicular to that circle, and the bases of the cylinder are circles tangent to the sphere, then the cylinder may be said to be circumscribed about the sphere. 741. After considering ex. 740, show that the surface of a sphere is two-thirds the entire surface of the circumscribed cylinder. (Archimedes. ) 742. Find the ratio of a spherical surface to the cylindrical surface of the circumscribed cylinder. 743. What is the radius of that sphere the number of square units of whose area equals the number of linear units in the circumference of one of its great circles ? 744. What is the ratio of the entire surface of a cylinder circum- scribed about a sphere to the entire surface of its hemisphere ? 745. What is the area of the entire surface of a spherical segment the radii of whose bases are n, r 2 , the radius of the sphere being r ? 746. A cone has for its base a great circle of a sphere, and for its vertex a pole of that circle. Find the ratio of the curved surfaces of the cone and hemisphere ; of the entire surfaces. 747. Show that the area of a zone of one base (the other base is zero) equals that of a circle whose radius is the chord of the generating arc. Prop. XXVIII.] THE MENSURATION OF THE SPHERE. 3G3 Proposition XXVIII. 498. Theorem. The volume of a spherical segment of altitude a, whose bases have radii r v r 2 , is expressed by the formula v = i vra [3 (i-! 2 + r 2 2 ) + a 2 ] , or v = 1 7ra (i*! 2 + r 2 2 ) + § 7ra 3 . Proof. 1. Let the above figure represent a segment cut from the figure of prop. XXVII. 2. Then if the distances of the circles of radii i\ and r 2 , from the center 0, are x x and x 2 , respectively, the radii of the bases of the frustum of the cone are x x and x 2 . ^Vhy ? 3. v = cylinder — frustum, Prop. XXVII, step 4 = iri^a — i Tra (x t * + x 2 + x l x 2 ) Prop. IV, cors. 1, 6 = i ira (6 >» 2 - 2 a^ 2 - 2 a^ - 2 ay 2 ) = i tt« [3 (>- 2 - x, 2 ) + 3 (r 2 - x 2 2 ) + («, - z 2 ) 2 ]. 4. But '•' olyhedral angles are either congruent or symmetric, and their corresponding edges are in proportion, the constant ratio being the ratio of similitude. D, Given two similar polyhedra, J l B l C l and A.Ji.J\ , or A X B X C X and A S B 9 C B Prop. XXIX.] SIMILAR SOLIDS. 365 To prove that (1) Z B l A 1 I) 1 = Z B % AJ> % or Z B 3 A Z D 3 ; (2) dihedral angle with edge A X B X = dihedral angle with edge A 2 B 2 , or with edge A 3 B 3 ; (3) polyhedral angle A x = polyhedral angle A 2 and is symmetric to polyhedral angle A 3 as arranged in the figure ; and (4) A X B X : A 2 B 2 = the ratio of similitude. Proof. 1. Let the polyhedra be placed in perspective (§ 259), the center of similitude, AyB x C x and A 2 B 2 C 2 on one side of 0, and A 3 B 3 C 3 on the other. 2. Then as in IV. prop. XX, A X B X II A 2 B 2 II A 3 B 3 and ZV^i II A4 II I) Z A 3 . 3. .'. Z B 1 A l I) 1 = Z B 2 AJ) 2 = Z B 3 A 3 D 3 , and similarly for other face angles, which proves (1). VI, prop. V 4. The trihedral Z A x = Z X because the face Z are respectively equal and similarly placed, and is sym- metric to Z A 3 because the face angles are respec- tively equal and placed in reverse order. Prop. XXI, cor. 5. So for the other trihedral Z. And V polyhedral Z, as D x , D 2 , D 3 , can be cut into congruent or symmetric trihedral A similarly placed, as by the planes A l C\D 1 , A 2 C 2 D 2 , A 3 C 3 D 3 , they too are con- gruent or symmetric. 6. .'. the dihedral Z are equal, which proves (2), and the corresponding polyhedral Z are congruent or symmetric, which proves (3). 7. The corresponding edges, as A l B l , A 2 B 2 , A 3 B 3 , being corresponding sides of similar A OA x B x , OA 2 B 2 , OA 3 B 3 , have the ratio of similitude, which proves (4). 366 SOLID GEOMETRY. [Bk. VIII. Corollaries. 1. If the ratio of similitude is 1, the poly- hedra are either congruent or symmetric. 2. Corresponding faces of similar polyhedra are propor- tional to the squares of any two corresponding edges. Step 7, and V, prop. IV. Proposition XXX. 501. Theorem. Tivo similar polyhedra can be divided into the same number of tetrahedra similar each to each and similarly placed. Proof. 1. In the figure below, the plane of A lf C lf D l and the plane of A 2 , C 2 , D 2 cut off tetrahedra A l B i C l D 1 , 2. Any point P 1 in the one has a corresponding point P 2 in the other, such that OP x : OP 2 = the ratio of similitude. Why ? 3. Hence the tetrahedra are similar. Proposition XXXI. 502. Theorem. The volumes of similar polyhedra are to each other as the cubes of their corresponding edges. Given the similar polyhedra A l B l C 1 , A 2 I>.,(\, , A t B s Cg having volumes v u v.,, v s , respec- tively. Prop. XXXI.] SIMILAR SOLIDS. 367 To prove that >\ : r 2 = A X B? : J.,/;,' ; . >\ : r 3 = .I^ 3 : A % B£. Proof. 1. Place the polyhedra in perspective, as in the figure, the center of similitude being 0. Divide the polyhedra into similar tetrahedra, simi- larly placed, A X B X C X D X , A 2 B 2 C 2 D 2 , A 3 B 3 C 3 D 3 , being corresponding tetrahedra. Prop. XXX Let t x , t. 2 , t s represent the volumes of these respec- tive tetrahedra, p u p 2} jh their altitudes from D x , D 2 , JJ 3 , and a lf a 2 , a 3 the areas of AA 1 B l C 1 , A 2 B 2 C 2 , A 3 B 3 C 3 . 2. Then v* l = Jjp l a 1 , and t. 2 = i 2 ? 2 a 2 > .\t l :t 2 = jh«i -Ptfh' 3. But a, : a, = A X B X 2 : A 2 B. 2 2 , V, prop. IV and p x : i> 2 = D 1 A 1 : D 2 A 2 = A X B X : A 2 B 2 . IV, prop. XX 4. .'. p x a x '■ lh a 2 = A X B X Z : A 2 B.? . IV, prop. VII, cor. 5. .'. t v : t 2 = A X B X 3 : A 2 B}. From steps 2, 4 Similarly the other tetrahedra are proportional to the cubes of their corresponding edges, which edges are proportional to the particular edges A X B X and A 2 B 2 . 6. .'. the sum of the tetrahedra making up the polyhe- dron A 1 B 1 C\ has the same ratio to the sum of the tetrahedra making up the polyhedron A 2 B 2 C 2 as A X B X Z has to A 2 B.?, or v t : v 2 = A X B X 3 : A 2 B 3 . Similarly v x : r 3 = A X B X 3 : A 3 B 3 3 , v 2 '. v 3 = A 2 £> 2 '• A 3 u 3 , = Jj 2 i^ 2 '. Jj 3 C 3 , = C 2 D 3 : C 3 D 3 , etc. 368 SOLID GEOMETRY. [Bk. VIIL Proposition XXXII. 503. Theorem. Any two spheres are similar. Proof. Let the spheres be placed in concentric position. Then V the ratio of their radii is constant any point on the surface of the one has on the surface of the other its similar point, with respect to the center, the ratio being i\ : r 2 . .'. the spheres are similar. Proposition XXXIII. 504. Theorem. Two right circular cylinders are similar if their elements have the same ratio as the radii of the ir bases. Proof. 1. Let the cylinders have the radii r 19 r. 2 , and the alti- tudes h lt h 2 , respectively, and be placed with their axes in the same line, their mid-points coincid- ing at 0. Let the semi-altitudes be OA 1} OJ. 2 , and let aline from cut the bases in B lf B 2 , not necessarily on the circumferences, and one from cut the cylindrical surf aces in C lt C 2 , respectively. 2. Then '•' the altitudes are proportional to the radii, Prop. XXXIV.] SIMILAR SOLIDS. 369 .'. OAj : OA 2 = r x :r 2 . And ••' A X B X \\A. 2 B,, Why? .-.OB,: OB,, = OA 1 : OA, = r, : r 2 . IV, prop. X 3. And '•' the axes coincide .\ the elements are parallel, and .*. OC,: OC, = r x :r 2 . .'. the points of the respective cylinders are similar with respect to as a center. Corollaries. 1. The areas of the cylindrical surfaces of two similar cylinders are proportional to the squares of their altitudes. For «i = 2 7tri7ii and a 2 = 2 7tr 2 h 2 . ai _ rdii ' ' a 2 foho But v ^ = ^ by prop. XXXIII, ' ' a 2 h 2 2 2. The volumes of two similar rigid circular cylinders are proportional to the cubes of their altitudes. Proposition XXXIV. 505. Theorem. Two right circular cones are similar if their altitudes have the same ratio as the radii of their bases. Place the bases in concentric position. The proof is then so similar to that of prop. XXXIII that it is left for the student. Corollaries. 1. The areas of the surfaces of two similar right circular cones are proportional to the squares of their altitudes. 2. The volumes of two similar right circular cones are pro- portional to the cubes of their altitudes. 370 SOLID GEOMETRY. [Bk. VIII.] EXERCISES. 750. The mean radii of the earth and moon are respectively 3956 miles, 1080.3 miles. Show that their volumes are as 49 to 1, nearly. 751. The mean diameter of the planet Jupiter being 80,657 miles, find the ratio of its volume to that of the earth. 752. The sun's diameter is about 109 times the earth's. Find the ratio of their volumes. 753. What is the radius of that sphere whose number of square units of surface equals the number of cubic units of volume ? 754. Also of that whose number of cubic units of volume equals the number of square units of area of one of its great circles. 755. Also of that whose number of cubic units of volume equals the number of linear units of the circumference of a great circle. 756. Two planes cut a sphere of radius 1 m, at distances 0.8 m and 0.5 m from the center. Find (1) the area of the zone between them, (2) the volume of the corresponding spherical segment. 757. A solid cylinder 20 cm long and 2 cm in diameter is terminated by two hemispheres. The solid is melted and molded into a sphere. Find the diameter of the sphere. 758. A meter was originally intended to be 0.0000001 of a quadrant of the circumference of the earth. Assuming it to be such, and the earth to be a sphere, find its radius in kilometers. 759. A cone, a sphere, and a cylinder have the same altitudes and diameters. Show that their volumes are in arithmetical progression. 760. Given a sphere of radius 10. How far from its center must the eye be in order to see one-fourth of its surface ? 761. If a tetrahedron is cut by a plane parallel to one of its faces, the tetrahedron cut off is similar to the first. 762. The areas of the surfaces of two similar polyhedra are proportional to the squares of their corresponding edges. NUMERICAL TABLES. 371 506. NUMERICAL TABLES. Formula or Mensuration. The numbers refer to the pages. Ab- breviations : b, base ; h, altitude ; r, radius ; a, area ; c, circumference ; p. perimeter; s, slant height; », volume; m, mid-section. Parallelogram, 202, a = bh. Circle, 217, 224, c = 27rr. Triangle, 202, a = \ bh. a = nr 2 . Trapezoid, 202, a = $ (b + b') h. Arc, 223, = a • r. Parallelepiped, 307, v = bh. Prism, 307, v — bh. Lateral area, right prism, 298, a = ph. Prismatoid, 314, v = \h{b + 6' + 4 m). Pyramid, 313, v = %bh. Lateral area, regular pyramid, 309, a = ^s. Frustum of pyramid, 315, v = i- h (b + b' + V&&'). Lateral area, frustum of regular pyramid, 309, a = £ (p + p') s. Right circular cylinder, 324, 325, v = bh = xr-h. Lateral a = ch = 2 nrh. Right circular cone, 324, 325, v = \bh = J 7rr 2 /i. Lateral a = £ cs = 7frs. Frust. of rt. circ. cone, 325, v — ^ith (i\ 2 + r 2 2 + rir 2 ). Sphere, 355, 360, v = 1 7rr 3 . a = 4 tzT 2 . Lune, 356, a = 2 ar 2 . Spherical polygon, 356, a = ar 2 . Zone, 356, a = 2 ;rWL Spherical segment, 363, v = \nh [3 (n« + r 2 2 ) + ft 2 ]. Spherical sector, 362, v = f 7Tr 2 ft = £&r. Most Important Expressions involving 7T. * = 3.141693. 1/7T = 0.31830989. 180°/ ?r = 57°.29578. 7T/4 = 0.785398. tt 2 = 9.86960440. tt/180 = 0.01745. ?r/3 = 1.047198. V# = 1.77245385. Approximate values ; | tt = 4.188790. 1/Var = 0.56418958. Certain Numerical Results frequently used. V2 = 1.4142. VlO = 3.1623. VE = 1.7100. V3= 1.7821. VJ =0.7071. V6 =1.8171. V5 =2.2361. V2 =1.2599. V7 =1.9129. V6 = 2.4495. V3 = 1.4422. V9 = 2.0801. 3 " V7 =2.6458. VI =1.5874. VlO 2.1544. 372 BIOGRAPHICAL TABLE. 507. BIOGRAPHICAL TABLE. The following table includes only those names mentioned in this work, although numerous others might profitably be considered by the student. The history of geometry may be said to begin in Egypt, the work of Ahmes, copied from a treatise of about 2500 B.C., containing numerous geometric formulae. The scientific study of the subject did not begin, however, until Thales visited that country, and carried the learning of the Egyptians back to Greece. The period of about four hundred years from Thales to Archimedes may be called the golden age of geometry. The contributions of the latter to the mensuration of the circle and of certain solids practically closed the scientific study of the subject in ancient times. Only a few contributors, such as Hero, Ptolemy, and Menelaus, added anything of importance during the eighteen hundred years which preceded the opening of the seventeenth century. Within the past three hundred years several important propositions and numerous improvements in method have been added, but the great body of ele- mentary plane geometry is quite as Euclid left it. In recent times a new department has been created, known as Modern Geometry, involving an extensive study of loci, collinearity, concurrence, and other subjects beyond the present range of the student's knowledge. The pronunciations here given are those of the Century Cyclopedia of Names. The first date indicates the year of birth, the second the year of death. All dates are a.d. unless the contrary is indicated by the sign — . The letter c. stands for circa, about, b. for born, d. for died. Numbers after the biographical note refer to pages in this work. Key. L. Latin, G. Greek, dim. diminutive, fern, feminine, a fat, a fate, a far, a fall, a ask, a fare, e met, e mete, e her, i pin, i pine, o )i<>/. o note, 6 move, 6 nor, u tub, u mute, u pull. h French nasalizing n. ch German ch. s as in leisure. t as in nature. A single dot under a vowel indicates its abbreviation. A double dot under a vowel indicates that the vowel approaches the short sound of u, as in put. BIOGRAPHICAL TABLE. 373 Ahmes (a'mes). c. — 1700. Egyptian priest Wrote the oldest extant work on mathematics 221 Anaxagoras (an-aks-ag'0-ras). - 499, - -428. Greek philosopher and mathematician 225 Archimedes (ar-M-mS'dk). -287, -212. Syracuse, Sicily. The greatest mathematician, and physicist of antiquity . 87. 221, 353, 354 Aryabhatta (ar-ya-bha'ta). b. 470. One of the earliest Hindu mathematicians. Wrote on Algebra and Geometry 221 Bhaskara (bhas'ka-ra). 12th cent. Hindu mathematician . . . 104 Brahmagupta (brah-ma-gop'ta). b. 508. Hindu mathematician. One of the earliest Indian writers 143, 221 Carnot (kar-no'), Lazare Nicholas Marguerite. 1753, 1823. French physicist and mathematician. Contributed to Modern Geom- etry 241. 242 Oavalieri (ka-va-le-a'iv). Bonaventnra. 1598, 1647. Prominent Ital- ian mathematician 351 Ceulen (koilen). Ludolph van. 1540. 1010. Dutch geometrician . 221 Ceva (cha'va), Giovanni. 1048, c. 1737. Italian geometrician, 239. 241 Dase (da'ze), Zacharias. 1824, 1861. Famous German computer . 221 Descartes (da-kart'), Kene. 1596, 1650. Eminent French mathe- matician, physicist, and philosopher. Founder of the science of Analytic Geometry 285 Euclid (u'klid). c. — 300. Eminent writer on Geometry in the Alexandrian School, at Alexandria, Egypt, His "Elements," the first scientific text-book on the subject, is still the standard in the schools of England 76, 152, 208 Euler (oiler), Leonhard. 1707, 1783. Swiss. One of the greatest mathematicians of modern times 99, 108, 285, 289 Gauss (gous), Karl Friedrich. 1777. 1855. German. One of the greatest mathematicians of modern times 208, 212 Hero (he'ro)of Alexandria. More properly Heron (he'ron). c. — 110. Celebrated Greek surveyor and mechanician 221. 227 Hippocrates (hi-pok'ra-tez) of Chios, b. c. — 470. Author of the first elementary text-book on Geometry 230 Jones (jonz), William. 1675-1749. English teacher 221 Klein (kiln), Christian Felix. 1849. Professor at Gottingen . . 225 Leibnitz (lib'nits), Gottfried Wilhelm. 1646, 1716. Equally cele- brated as a philosopher and a mathematician. One of the founders of the science of the Calculus 23, 182 Lindemann (lin'de-man), Ferdinand, b. 1852. German professor 225 374 BIOGRAPHICAL TABLE. Meister (mis'ter), Albrecht. 1724-1788. German mathematician . 98 Menelaus (men-e-la/us). c. 100. Greek mathematician and astrono- mer. One of the early writers on Trigonometry . . 240, 242, 243 Metius (metius). Anthonisz, Adrisen. Called Metius from Metz, his birthplace. 1527-1607 221 Monge (mohzh), Gaspard. 1746, 1818. French. Founder of the science of Descriptive Geometry. One of the founders of the Polytechnic School of Paris 97 CBnopides (e-nop'i-dez). c. - 465. Early Greek Geometer ... 72 Pascal (pas-kal'), Blaise. 1623, 1662. Celebrated French mathemati- cian, physicist, and philosopher 241 Plato (pla'to). c. — 429, — 348. Greek philosopher and founder of a school that contributed extensively to Geometry, 68, 106, 152, 286 Pothenot (po-te-no'), Laurent, d. 1732. French professor . . .157 Ptolemy (tore-mi). Claudius Ptolemseus. 87, 165. One of the greatest of astronomers, geographers, and geometers of the later Greeks 221, 228 Pythagoras (pi-thag'o-ras). c. — 580, c. — 501. Founder of a cele- brated school in Lower Italy. One of the foremost of the early mathematicians 49, 103, 286 Richter (rich'ter). 1854. German computer 221 Thales (tha'lez). - 640, - 548. One of the Seven Wise Men of Greece. Introduced the study of Geometry from Egypt, 26, 117, 131 Vega, Georg, Freiherr von. 1756-1802. Professor of mathe- matics at Vienna 221 TABLE OF ETYMOLOGIES. This table includes such of the pronunciations and etymologies of the more common terms of Geometry as are of greatest value to the student. The equivalent foreign word is not always given, but rather the primitive root as being more helpful. The pronunciations and etymologies are those of the Century Dictionary. See Biographical Table, p. 372. Abscissa (ab-sis'a). L. cut off. Acute (a-kut'). L. acutus, sharp. Adjacent (a-ja'sent). L. ad, to, + jacere, lie. Angle (ang'gl). L. angulus, a cor- ner, an angle ; G. ankylos, bent. Antecedent (an-te-se'dent). L. ante, before, + ceclere, go. Bisect (bi-sekf). L. 6*'-, two-, + sectus, cut. Center (sender). L. centrum, center; G. kentron, from kentein, to prick. Centroid (sen'troid). G. kentron, center, + eiclos, form. Chord (kord). G. chorde, string. Circle (sir'kl). L. circulus, dim. of circus (G. kirkos), a ring. Circumference (ser-kum'f e-rens) . L. circum, around (see Circle), + ferre, to bear. Collinear (ko-lin'e-ar). L. com-, together, + linea, line. Commensurable (ko-men'su-ra-bl) , L. com-, together, + mensurare, measure. Complement (kom'ple-ment). L. complementum, that which fills, from com- (intensive) + plere, fill. Concave (kon'-kav). L. com- (in- tensive) + cavus, hollow. Concentric (kon-sen'trik). L. con-, together, + centrum, center. Concurrent (kon-kur'ent). L. con-, together, + currere, run. Concyclic (kon-sik'lik). L. con-, together, + cyclicus, from G. kyklikos, from kyklos, a circle, related to kyliein, roll (compare Cylinder). Congruent (kong / gro-ent). L. con- gruere, to agree. Consequent (kon'se-kwent). L. con-, together, + sequi, follow. Constant (kon'stant). L. con-, to- gether, + stare, stand. Converse (kon'vers). L. con-, to- gether, + vertere, turn. Convex (kon'veks). L. convexus, vaulted, from con-, together, -f vehere, carry. Corollary (kor'o-la-ri). L. corolla- Hum, a gift, money paid for a garland of flowers, from corolla, dim. of corona, a crown. Cylinder (sirin-der). G. kyllndros, from kyliein, roll. 375 376 TABLE OF ETYMOLOGIES. Decagon (dek'a-gon). G. deka, ten, + gonia, an angle. Degree (de-gre'). L. de, down, 4- gradus, step. Diagonal (dl-ag'o-nal). G. dia, through, + gonia, a corner, an angle. Diameter (di-am'e-ter). G. dia, through, 4- metron, a measure. Dihedral (dl-he'dral). G. di-, two, + hedra, a seat. Dimension (di-men'shpn). L. dis-, apart, 4- metiri, measure. See Measure. Directrix (di-rek'triks). L. fem. of director, from directus, direct. Dodecahedron (do"dek-a-he'drpn). G. dodeka, twelve, + hedra, a seat. Equal (e'kwal). L. cequalis, equal, from cequus, plain. Equiangular (e-kwi-ang'gu-liir). L. cequus, equal, + angulus, angle. Equilateral (e-kwi-lat'e-ral). L. cequus, equal, + latus, side. Equivalent (e-kwiv'a-lent). L. cequus, equal, 4- valere, be strong. Escribed (es-krlbd')- L. e, out, + scribere, write. Excess (ek-ses'). L. ex, out, -f cedere, go ; i.e. to pass beyond. Frustum (frus'turn). L. a piece. Generatrix (jen'e-ra-triks). L. fem. of generator, from generare, beget, from genus, a race. Geometry (je-om'e-tri). G. ge, the earth, + metron, a measure, -gon, a termination, G. gonia, an angle. Harmonic (har-mon'ik). G. har- monia, a concord, related to har- mos, a joining. A line divided internally and externally in the ratio 2:1, is cut into segments representing 1, f, £. Pythago- ras first discovered that a vibrat- ing string stopped at half its length gave the octave of the original note, and stopped at two-thirds of its length gave the fifth. Hence the expression "harmonic division " of a line. Hemisphere (hem'i-sfer). G. hemi-, half, + sphaira, a sphere. -hedron, a termination, G. hedra, a seat. Hepta-, in combination, G. seven. Hexa-, in combination, G. six. Hexagram (hek'sa-gram). G. hex, six, 4- gramma, a line. Hypotenuse (hi-pot'e-nus). G. hypo, under, 4- teinein, stretch. Inclination (in-kli-na'shon). L. in, on, 4- clinare, lean. Incommensurable (in-ko-men'su- ra-bl). L. in-, not, 4- com-, to- gether, 4- mensurare, measure. Infinity (in-fin'i.-ti). L. in-, not, 4 finitus, bounded. Inscribed (in-skribd 7 ). L. in, in, 4- scribere, write. Isosceles (i-sos'e-lez). G. isos, equal, -f skelos, leg. Lateral (lat'e-ral). L. latus, a side. Locus (lo'kus). L. a place. Com- pare locality. Lune (lun). L. luna, the moon. Major (ma/jor). L. greater, com- parative of magnus, great. Maximum (mak'si-mum). L. great- est, superlative of magnus, great. Mean (men). L. medius, middle. TABLE OF ETYMOLOGIES. ;;:: Measure (mezh'iir). L. mensura, a measuring. See Dimension. Median (me'di-au). See Mean. Mensuration (men-su-ra'shon). See Measure. Minimum (min'i-nium). L. least. Minor (mi 'nor). L. less. Nappe (nap). French, a cloth, sheet, surface. Oblique (ob-lek 7 or ob-Hk'). L. ob, before, + liquis, slanting. Obtuse (ob-tus'). L. obtusus, blunt, from ob, upon, + twidere, strike. Octo-, octa-, in combination, L. and G., eight. Opposite (op'o-zit). L. ob, before, against, + ponere, put, set. Ordinate (or'di-nat). L. ordo (or- din-), a row. Orthocenter (Or'tho-sen-ter). G. ortho-, straight, -j- kentron, center. Orthogonal (or-thog'o-nal). G. orthos, right, + gonia, an angle. Parallel (par'a-lel). G. para, be- side, + allelon, one another. Parallelepiped (par-a-lel-e-pip'ed or -pl'ped). Gr. parallelos, parallel, + epipedon, a plane surface, from epi, on, -f pedon, ground. Parallelogram (par-a-lel'S-grani) . G. parallelos, parallel, + gram ma, line. Pedal (ped'al or pe'dal). L. peda- lis. pertaining to the foot, from pes (ped-). foot. Pencil (pen'sil). L. penic ilium, a painters 1 pencil, a brush. Perigon (per'i-gon). G. peri, around, + gonia, a corner, angle. Perimeter (pe-rim'e-ter). G. peri, around, + metron, measure. Perpendicular (per-pen-dik u-lar). L. perpendiculum , a plumb-line, from per, through, -f pcndere. hang. Perspective (per-spek'tiv). L. per, through, + specere, see. it (pi). Initial of G. periphereia, periphery, circumference. Pole (pol). G. polos, a pivot, hinge, axis, pole. Polygon (pol'i-gon). G. polys, many, + gonia, corner, angle. Polyhedron, (pol-i-he'drpn) G. polys, many, + hedra, seat. Postulate (pos'tu-lat). L. postula- tum, a demand, from poscere, ask. Prism (prizm). G. prisma, some- thing sawed, from priein. saw. Prismatoid (priz'ma-toid). G. pris- ma (t~), + eidos, form. Projection (pro-jek'skpn). L. pro, forth, + jacere, throw. Pyramid (pir'a-mid). G. pyramis, a pyramid, perhaps from Egyp- tian pir-em-us, the slanting edge of a pyramid. Quadrant (kwod'rant). L. quad- rants, a fourth part. See Quadrilateral. Quadrilateral (kwod-ri-lat'e-ral). L. quattuor (quadri-), four, -f latus, (later-), side. Radius (ra'di-us). L. rod, spoke of a wheel. Ratio (ra'shio). L. a reckoning, calculation, from reri, think, estimate. Reciprocal (re-sip'ro-kal). L. re- ciprocus. returning, from re-, back, and pro. forward, with two adjective terminations. Rectangle (rek'tang-gl). L. rectus, 378 TABLE OF ETYMOLOGIES. right, + angulus, an angle. See Angle. Rectilinear (rek-ti-lin'-ar). L. rec- tus, right, + linea, a line. Reflex (re'fleks or re-fleks'). L. re-, back, -f flectere, bend. Regular (reg / u-lar). L. regula, a rule, from regere, rule, govern.. Rhombus (rom'bus). G. rhombos, a spinning top. Scalene (ska-len'). G. skalenos, uneven, unequal ; related to skel- los, crooked-legged. Secant (se'kant). L. secare, cut, as also Sector, Section, Seg- ment. Segment (seg'ment). See Secant. Semicircle (sem'i-ser-kl). L. semi-, half, + circulus, circle. Similar (sim'i-lar). L. similis, like. Solid (sol'id). L. solidus, firm, compact. Sphere (sfer). G. sphaira, a ball. Square (skwar). L. quadra, a square, from quattuor, four. Straight (strat). Anglo-Saxon, streht, from streccan, stretch. Subtend (sub-tend'). L. sub, under, -f tender e, stretch. Successive (suk-ses'iv). L. sub, un- der, + cedere, go. Sum (sum). L. summa, highest part. Compare Summit. Superposition (su'per-po-zish'qn). L. super, over, + ponere, lay. Supplement (sup'le-ment). L. sub., under, + plere, fill ; to fill up. Surface (ser-fas). L. superficies, surface, from super, above, + fades, form, figure, face. Symbol (sim'bpl). G. symbolos, a sign by which one infers some- thing, from sun, together, + bal- lein, put. Tangent (tan'j.ent). L. tangere, touch. Tetrahedron (tet-ra-he'drpn). G. tetra-, four, -|- hedra, seat. Theorem (the'o-rem). G. theorema, a sight, a principle contemplated. Transversal (trans-ver'sal). L. trans, across, -f vertere, turn. Trapezium (tra-pe'zi-um). G. tra- pezion, a table, dim. of trapeza, a table, from tetra, four, + pous, foot. Trapezoid (tra-pe'zoid). G. trapeza, table, + eidos, form. Tri-, in composition, L. tres (tri-), G. treis (tri-), three. See Secant, -hedron, Angle, for meaning of trisect, trihedral, triangle. Truncate (trung'kat). L. truncare, cut off, from Old L. troncus, cut off, mutilated. Unique (u-nek'). L. unicus, from unus, one. Vertex (ver'teks). L. vertere, turn. Zone (zon). G. zone, a girdle, belt. INDEX. PAGE Abbreviations 12 Acute angle 6, 266 Acute-angled triangle ... 50 Adjacent angles .... 6, 266 " polygons .... 90 Alternate angles 43 Alternation 163 Altitude 90, 295, 313, 321, 350, 361 Ambiguous case 51 Analysis 70, 152 Angle . 5, 115, 151, 256, 261, 265, 334, 335 Antecedent 161 Antiparallel 177 Apothem 214 Arc 67, 114 Area 112, 202 Arm 5 Assumed constructions ... 70 Axioms 11 Axis .... 29, 265, 321, 328 Base, 23, 29, 59, 292, 295, 321, 350, 361 Birectangular triangle . . . 345 Bisect 4, 5, 266 Broken line - . 4 Cavalieri's theorem .... 359 Center 67, 114, 326 " line 143 " of mass (gravity) . . 89 " of polygon .... 214 " of similitude .... 183 PAGF Center of symmetry . . .183 " segment 143 Central angle .115 " symmetry . . . .183 Centroid 89 Chord 114 Circle 67, 114. 327 Circular cone . . . .321, 322 " cylinder 318 Circumcenter 89 Circumference ... 67, 114 Circumscribed, 136, 212, 322, 334 Collinear 84, 238 Commensurable 160 Complement . . . 8, 115, 266 Composition 164 Concave 54, 283, 335 Concentric 143 Concurrent 84, 238 Coney clic 128 Cone 321, 322, 361 " of revolution .... 322 Congruent 23, DO Conical surface, space . . .321 Conjugate . . . . 8, 115, 266 Consecutive angles .... 59 Consequent 161 Constant 167 Contact 125, 332 Converse 34 Convex . . . . 7, 54, 283, 335 Coplanar 245 379 380 INDEX. Corollary 20 Corresponding angles ... 43 " segments, points, 179, 182 Counter-clockwise Cross polygons Curve .... Curved surface Cylinder . . . " of revolution . Cylindrical surface, space Degree . . Determined Diagonal Diagonal scale Diameter . Difference . Dihedral Dimensions Directrix Distance Distributive law Division . . . Division, internal o . 54 . 67 . 317 . 318 . 322 . 317 5, 222 . 3, 25, 51 . 22, 288 . . .184 67, 114, 326 90, 115, 266 . . . 265 . . . 305 . 317, 321 29, 259, 264 ... 96 . 164, 170 and external, 95, 176 Edges . 265, 274, 283, 291, 294 Element 317, 321 Equal . . 6, 24, 90, 114, 222 Equiangular 22 Equilateral 22 Equivalent 24, 90 Escribed ....... 136 Ex-center 89 Excess, Spherical . . . .353 Exterior angles . . . . 22, 43 Extreme and mean ratio . . 196 Extremes 161 Faces . 265, 274, 283, 291, 294 Figure 21 Fourth proportional .... 166 Frustum 295, 322 PAGE General polygon 54 Generalization of figures . . 56 Generatrix 317, 321 Geometry 1, 9, 21 Geometric mean 166 Golden section . . . 196, 197 Great circle 327 Harmonic division .... 176 Hemisphere 328 Hero's formula 227 Hexagram, Mystic .... 241 Hypotenuse 50 In-center 89 Inclination 256 Incommensurable . . . .160 Indirect proof 14 Infinity 170, 174 Inscribed, 128, 136, 157, 212, 322, 334 Instruments of geometry . . 208 Interior angles . . . . 22, 43 Inversion 163 Isoperimetric 229 Isosceles 28, 59 Lateral area 298 Law of Converse 34 Limit .... 167, 216, 323 Line 2, 3 Locus 80, 156 Ludolphian number . : . .221 Lune 335 Major 115, 128 Maximum 229 Mean proportional .... 166 Means 161 Measure, 112, 117, 129, 159, 202, 266, 356 Median 28 " section 196 Mensuration, 112, 226, 298, 356 Methods of attack . 35,70,152 INDEX. 381 PAGE Mid-point 4 Minimum 229 Minor 115, 128 Motion ........ 24 Mutually equiangular, etc. . 22 Mystic hexagram .... 241 Nappes 321 Negative .... 56, 97, 98 Nine points circle .... 243 Number, 101, 159, 166, 201, 306 Oblique 7, 252, 266 " circular cone . . . 321 " cylinder 318 " prism 291 Obtuse angle 7, 266 Obtuse-angled triangle ... 50 One-to-one correspondence, 101, 159 Opposite ... 8, 59, 275, 339 Orthocenter 89 Orthogonal 104 Parallel . . . 43, 46, 262, 270 Parallelepiped 288 Parallelogram 59 Parts of polygon 22 Pedal triangle 137 Pencil .... 170, 265, 270 Perigon 5 Perimeter 21 Perpendicular . 7, 252, 266, 352 Perspective 183 tc 217, 220-222 Plane 4, 244 " figure, Geometry ... 21 " of symmetry .... 289 " section 326 Platonic bodies 286 Point 2 Polar distance 329 " triangle, polyhedral angle . . . .342, 345 PAGE Pole 328 Polygon . . . 21, 54, 335, 339 Polyhedral angle . . . 271, 339 Polyhedron 283 Positive and negative . . 56, 97 Postulates, 9, 10, 24, 46, 68, 216, 245, 318, 322, 326 Preliminary propositions . . 13 Prism 292 Prismatic surface, space . . 291 Prismatoid 313 Problem 9 Produced 4 Product of lines . 166, 201, 306 Projection . . . 104, 255, 256 Proof 19, 35 Proportion 161 Proposition 9 Pyramid 295 Pyramidal surface, space . . 294 Pythagorean Theorem, 102, 103, 211 Quadrant 115, 329 Quadrature of the circle . . 225 Quadrilateral 22 Radian ........ 222 Radius ... 67, 114, 214, 326 Ratio 159, 184 Reciprocal Theorems, 27, 101, 340 Rectangle 60, 94 Rectangular parallelepiped . 289 Rectilinear figure 21 Reductio ad absurdum . 14, 152 Reflex angle 7 Regular polygon 54 " polyhedron . . 283, 286 " pyramid 308 Rhombus 60 Right angle 7, 266 Right-angled triangle ... 50 Right circular cone .... 321 382 INDEX. PAGE Right section .... 291, 317 Rotation 24 Scalene 29 Secant 114 Sector 115, 361 Segment ... 3, 95, 128. 356 Semicircle 117 Semicircumference .... 115 Sense of lines and surfaces, 56, 97 Sheaf of lines 364 Side 21 Similar systems of points, figures ... 23, 182, 364 Slant height . . 308, 322, 323 Small circle 327 Solid 1-3 " angle 274 " Geometry 245 Sphere 326 Spherical cone 361 " excess 353 " polygon .... 335 " sector 361 " segment .... 356 " surface 326 wedge 338 Square 60, 94 Squaring the circle .... 225 Straight angle 5, 266 " line 3 Subtend 115, 128 PAGE Sum ... 4, 5, 90, 115, 266 Superposition 23 Supplement . . . 8, 115, 266 Surface, 3, 21, 273, 291, 292, 294, 317, 321, 326 Symbols ........ 12 Symmetry, 29, 183, 233, 275, 339 Tables 371 Tangent .... 125, 143, 332 Terms of a proportion . . . 161 Tetrahedral 274 Theorem 9 Third proportional . . . .195 Touch 125, 143 Transversal 43 Transverse section, 291, 294, 317, 321 Trapezium .59 Trapezoid 59 Triangle 22, 342 Trihedral 274 Trisect 87 Truncated pyramid .... 295 Unique 7 Unit 112, 159, 202 Variable 167 Vertex, 5, 21, 23, 170, 283, 294, 295, 321 Vertical 8, 266 Wedge 315, 338 Zone 356 UNIVm c "TY OF C/TJFORNTA U.C. 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