tiXfT OF • *4 PLANE AND SOLID GEOMETRY BY C. A. HART INSTRUCTOR OF MATHEMATICS, WADLEIGH HIGH SCHOOL, NEW YORK CITY AND DANIEL D. FELDMAN HEAD OF DEPARTMENT OF MATHEMATICS, ERASMUS HALL HIGH SCHOOL, BROOKLYN WITH THE EDITORIAL COOPERATION OF J. H. TANNER AND VIRGIL SNYDER PROFESSORS OF MATHEMATICS IN CORNELL UNIVERSITY NEW YORK •:• CINCINNATI •:• CHICAGO AMERICAN BOOK COMPANY H-3 Copyright, 1911, 1912, by AMERICAN BOOK COMPANY. Entered at Stationers' Hall, London. H.-F. PLANK AND SOLID GEOMETRY. W. P. I v" PREFACE This book is the outgrowth of an experience of many years in the teaching of mathematics in secondary schools. The text has been used by many different teachers, in classes of all stages of development, and under varying conditions of sec- ondary school teaching. The proofs have had the benefit of the criticisms of hundreds of experienced teachers of mathe- matics throughout the country. The book in its present form is therefore the combined product of experience, classroom test, and severe criticism. The following are some of the leading features of the book : Tlie student is rapidly initiated into the subject. Definitions are given only as needed. The selection and arrangement of theorems is such as to meet the general demand of teachers, as expressed through the Mathe- matical Associations of the country. Most of the proofs have been given in full. In the Plane Geometry, proofs of some of the easier theorems and construc- tions are left as exercises for the student, or are given in an incomplete form. In the Solid Geometry, more proofs and parts of proofs are thus left to the student ; but in every case in which the proof is not complete, the incompleteness is specifically stated. The indirect method of proof is consistently applied. The usual method of proving such propositions, for example, as Arts. 189 and 415, is confusing to the student. The method used here is convincing and clear. The exercises are carefully selected. In choosing exercises, each of the following groups has been given due importance : (a) Concrete exercises, including numerical problems and problems of construction. (6) So-called practical problems, such as indirect measure- ments of heights and distances by means of equal and similar triangles, drawing to scale as an application of similar figures, ..problems from physics, from design, etc, 304026 iv PREFACE (c) Traditional exercises of a more or less abstract nature. The arrangement of the exercises is pedagogical Compara- tively easy exercises are placed immediately after the theorems of which they are applications, instead of being grouped to- gether without regard to the principles involved in them. For the benefit of the brighter pupils, however, and for review classes, long lists of more or less difficult exercises are grouped at the end of each book. The definitions of closed figures are unique. The student's natural conception of a plane closed figure, for example, is not the boundary line only, nor the plane only, but the whole figure composed of the boundary line and the plane bounded. All definitions of closed figures involve this idea, which is entirely consistent with the higher mathematics. The numerical treatment of magyiitudes is explicit, the funda- mental principles being definitely assumed (Art. 336, proof in Appendix, Art. 595). This novel procedure furnishes a logical, as well as a teachable, method of dealing with incommensurables. The area of a rectangle is introduced by actually measuring it, thereby obtaining its measure-number. This method permits the same order of theorems and corollaries as is used in the parallelogram and the triangle. The correlation with arithmetic in this connection is valuable. A similar method is employed for introducing the volume of a parallelopiped. Proofs of the superposition theorems and the concurrent line theorems will be found exceptionally accurate and complete. The many historical notes ivill add life and interest to the work. The carefully arranged summaries throughout the book, the collection of formulas of Plane Geometry, and the collection of formulas of Solid Geometry, it is hoped, will be found helpful to teacher and student alike. Argument and reasons are arranged in parallel form. This arrangement gives a definite model for proving exercises, ren- ders the careless omission of the reasons in a demonstration impossible, leads to accurate thinking, and greatly lightens the labor of reading papers. PREFACE v Every construction figure contains all necessary construction lines. This method keeps constantly before the student a model for his construction work, and distinguishes between a figure for a construction and a figure for a theorem. TJie mechanical arrangement is such as to give the student every possible aid in comprehending the subject matter. The following are some of the special features of the Solid Geometry : The vital relation of the Solid Geometry to the Plane Geometry is emphasized at every point. (See Arts. 703, 786, 794, 813, 853, 924, 951, 955, 961, etc.) The student is given every possible aid in forming his early space concepts. In the early work in Solid Geometry, the average student experiences difficulty in fully comprehending space relations, that is, in seeing geometric figures in space. The student is aided in overcoming this difficulty by the intro- duction of many easy and practical questions and exercises, as well as by being encouraged to make his figures. (See § 605.) As a further aid in this direction, reproductions of models made by students themselves are shown in a group (p. 302) and at various points throughout Book VI. Tlie student's knowledge of the things about him is constantly draivn upon. Especially is this true of the work on the sphere, where the student's knowledge of mathematical geography has been appealed to in making clear the terms and the relations of figures connected with the sphere. The same logical rigor that characterizes the demonstrations in the Plane Geometry is used throughout the Solid. The treatment of the polyhedral angle (p. 336), of the prism (p. 345), and of the pyramid (p. 350) is similar to that of the cylinder and of the cone. This is in accordance with the recommendations of the leading Mathematical Associations throughout the country. The grateful acknowledgment of the authors is due to many friends for helpful suggestions; especially to Miss Grace A. Bruce, of the Wadleigh High School, New York; to Mr. Edward B. Parsons, of the Boys' High School, Brooklyn ; and to Professor McMahon, of Cornell University. CONTENTS PLANE GEOMETRY PAGE SYMBOLS AND ABBREVIATIONS viii INTRODUCTION 1 BOOK I. RECTILINEAR FIGURES 19 Necessity for Proof 21 Polygons. Triangles 22 Superposition 25 Measurement of Distances by Means of Triangles ... 32 Loci 42 Parallel Lines 65 Quadrilaterals. Parallelograms 84 Concurrent Line Theorems 100 Construction of Triangles 106 Directions for the Solution of Exercises 110 Miscellaneous Exercises Ill BOOK II. THE CIRCLE 113 Two Circles . 131 Measurement 133 Miscellaneous Exercises 156 BOOK III. PROPORTION AND SIMILAR FIGURES . . 161 Similar Polygons 176 Drawing to Scale 188 Miscellaneous Exercises 206 BOOK IV. AREAS OF POLYGONS 209 Transformation of Figures 223 Miscellaneous Exercises . . . . .... 239 BOOK V. REGULAR POLYGONS. MEASUREMENT OF THE CIRCLE 245 Measurement of the Circumference and of the Circle . . 258 Miscellaneous Exercises 274 vi CONTENTS vii PAGE FORMULAS OF PLANE GEOMETRY . . . . .282 APPENDIX TO PLANE GEOMETRY 284 Maxima and Minima 284 Variables and Limits. Theorems 291 Miscellaneous Theorems ........ 294-298 SOLID GEOMETRY BOOK VI. LINES, PLANES, AND ANGLES IN SPACE . 299 Lines and Planes . * . . 301 Dihedral Angles . 322 Polyhedral Angles * . .. .336 BOOK VII. POLYHEDRONS . . ... . .343 Prisms 345 Pyramids 350 Mensuration of the Prism and Pyramid 354 Areas 354 Volumes . . 358 Miscellaneous Exercises 381 BOOK VIII. CYLINDERS AND CONES 383 Cylinders 383 Cones • 388 Mensuration of the Cylinder and Cone 392 Areas 392 Volumes 406 Miscellaneous Exercises 414 BOOK IX. THE SPHERE 417 Lines and Planes Tangent to a Sphere 424 Spherical Polygons 429 Mensuration of the Sphere 444 Areas . . , 444 Volumes 457 Miscellaneous Exercises on Solid Geometry .... 467 FORMULAS OF SOLID GEOMETRY . . . . .471 APPENDIX TO SOLID GEOMETRY 474 Spherical Segments 474 The Prismatoid 475 Similar Polyhedrons 477 .INDEX 481 SYMBOLS AND ABBREVIATIONS = equals, equal to, is equal to. =fc does not equal. > greater than, is greater than. < less than, is less than. =0= equivalent, equivalent to, is equiva- lent to. ~ similar, similar to, is similar to. S^ is measured by. JL perpendicular, perpendicular to, is perpendicular to. Js perpendiculars. || parallel, parallel to, is parallel to. ||s parallels. . . . and so on (sign of continuation). v since. .-, therefore. /-n arc ; AB, arc AB. £7, OD parallelogram, parallelograms. O, (D circle, circles. Z, A angle, angles. A, A triangle, triangles. q.e.d. Quod erat demonstrandum, which was to be proved. q.e.f. Quod erat faciendum, which was to be done. The signs +, — , X , + have the same meanings as in algebra. rt. right. str. straight. ext. exterior. int. interior. alt. alternate. def. definition. ax. axiom. post. postulate. hyp- hypothesis. prop. proposition. prob. problem. th. theorem. cor. corollary. cons. construction. ex. exercise. fig. figure. iden. identity. comp. complementary. sup. supplementary. adj. adjacent. homol . homologous. viii PLANE GEOMETRY INTRODUCTION 1. The Subject Matter of Geometry. In geometry, although we shall continue the use of arithmetic and algebra, our main work will be a study of what will later be defined (§ 13) as geometric figures. The student is already familiar with the physical objects about him, such as a ball or a block of wood. By a careful study of the following exercise, he may be led to see the relation of such physical solids to the geometric figures with which he must become familiar. Exercise. Look at a block of wood (or a chalk box). Has it weight ? color ? taste ? shape ? size ? These are called properties of the solid. What do we call such a solid ? A physical solid. Can you think of the properties of this jj , solid apart from the block of wood ? Imagine the block removed. Can you imagine the space which it occupied ? What name would you give to this space ? A geometric solid. What properties has it that the block possessed ? Shape and size. What is it that separates this geometric solid from surrounding space ? How thick is this surface ? How many surfaces has the block ? Where do they intersect ? How many intersections are there ? How wide are the intersections ? how long ? What is their name ? They are lines. Do these lines intersect ? where ? How wide are these intersections ? how thick ? how long ? Can you say where this one is and so distinguish from where that one is ? What is its name ? It is a point. If you move the block through space, what will it generate as it moves ? What will the surfaces of the block generate ? all of them ? Can you move a surface so that it will not generate a solid ? Yes, by moving it along itself. What will the edges of the block generate ? Can you move an edge so that it will not generate a surface ? What will the corners generate ? Can you move a point so that it will not generate a line ? PLANE GEOMETRY FOUR FUNDAMENTAL GEOMETRIC CONCEPTS 2. The space in which we live, although boundless and unlimited in extent, may be thought of as divided into parts. A physical solid occupies a limited portion of space. The portion of space occupied by a physical solid is called a geometric solid. 3. A geometric solid has length, breadth, and thickness. It may also be divided into parts. The boundary of a solid is called a surface. 4. A surface is no part of a solid. It has length and breadth, but no thickness. It may also be divided into parts. The boundary of a surface is called a line. 5. A line is no part of a surface. It has length only. It may also be divided into parts. The boundary or extremity of a line is called a point. A point is no part of a line. It has neither length, nor breadth, nor thickness. It cannot be divided into parts. It is position only. THE FOUR CONCEPTS IN REVERSE ORDER 6. As we have considered geometric solid independently of surface, line, and point, so we may consider point indepen- dently, and from it build up to the solid. A small dot made with a sharp pencil on a sheet of paper represents approximately a geometric point. 7. If a point is allowed to move in space, the path in which it moves will be a line. A piece of fine wire, or. a line drawn on paper with a sharp pencil, represents approximately a geometric line. This, how- ever fine it may be, has some thickness and is not therefore an ideal, or geometric, line. 8. If a line is allowed to move in space, its path in general will be a surface. INTRODUCTION 3 9. If a surface is allowed to move in space, its path in general will be a geometric solid. 10. A solid has threefold extent and so is said to have three dimensions ; a surface has twofold extent and is said to have two dimensions ; a line has onefold extent or one dimension ; a point has no extent and has therefore no dimensions. 11. The following may be used as working definitions of these four fundamental concepts : A geometric solid is a limited portion of space. A surface is that which bounds a solid or separates it from an adjoining solid or from the surrounding space. A line is that which has length only. A point is position only. DEFINITIONS AND ASSUMPTIONS 12. The primary object of elementary geometry is to deter- mine, by a definite process of reasoning that will be introduced and developed later, the properties of geometric figures. In all logical arguments of this kind, just as in a debate, certain fundamental principles are agreed upon at the outset, and upon these as a foundation the argument is built. In ele- mentary geometry these fundamental principles are called definitions and assumptions. The assumptions here mentioned are divided into two classes, axioms and postulates. These, as well as the definitions, will be given throughout the book as occasion for them arises. 13. Def. A geometric figure is a point, line, surface, or solid, or a combination of any or all of these. 14. Def. Geometry is the science which treats of the properties of geometric figures. 15. Def. A postulate may be defined as the assumption of the possibility of performing a certain geometric operation. Before giving the next definition, it will be necessary to introduce a nostulate. 4 PLANE GEOMETRY 16. Transference postulate. Any geometric figure may be moved from one position to another without change of size or shape. 17. Def. Two geometric figures are said to coincide if, when either is placed upon the other, each point of one lies upon some point of the other. 18. Def. Two geometric figures are equal if they can be made to coincide. 19. Def. The process of placing one figure upon another so that the two shall coincide is called superposition. This is an imaginary operation, no actual movement taking place. LINES 20. A line is usually designated by two capital letters, as line AB. It may be designated . „ also by a small letter placed somewhere on the line, as line a. , Fig. 1. 21. Straight Lines. In § 7 we learned that a piece of fine wire or a line drawn on a sheet of paper represented approximately a geometric line. So also a geometric straight line may be represented approximately by a string stretched taut between two points, or by the line made by placing a ruler (also called a straightedge) on a flat surface and drawing a sharp pencil along its edge. 22. Questions. How does a gardener test the straightness of the edge of a flower bed ? How does he get his plants set out in straight rows ? How could you test the straightness of a wire ? Can you think of a wire not straight, but of such shape that you could cut out a piece of it and slip it along the wire so that it would always fit ? If you re- versed this piece, so that its ends changed places, would it still fit along the entire length of the wire ? If you turned it over, would it fit ? Would the piece cut out fit under these various conditions if the wire were straight ? INTRODUCTION 23. Def. A straight line is a line such that, if any portion of it is placed with its ends in the line, the entire portion so placed will lie in the line, how r ever it may be applied. Thus, if AB is a straight line, and if any portion of AB, as CD, is placed on any other part of AB, with its ends in AB, A S 5 9 every point of CD will lie in AB. FlG - 2 - A straight line is called also a right line. The word line, unqualified, is understood to mean straight line. 24. Straight line postulate. A straight line may be drawn from any one point to any other. 25. Draw a straight line AB. Can you draw a second straight line from A to B ? If so, where will every point of the second line lie (§ 23) ? It then follows that : Only one straight line can be drawn between two points; i.e. a straight line is determined by two points. 26. Draw two straight lines AB and CD intersecting in point P. Show that AB and CD cannot have a second point in com- mon (§ 23). It then follows that: Two intersecting straight lines can have only one point in com- mon; i.e. two intersecting straight lines determine a point. 27. Def. A limited portion of a straight line is called a line segment, or simply a line, or a segment. Thus, in Fig. 2, AC, CD, and DB are line segments. 28. Def. Two line segments which lie in the same straight line are said to be collinear segments. 29. Def. A curved line (or curve) is a line no portion of which is straight, as GH. 30. Def. A broken line is a line made up of different succes- sive straight lines, as KL. * 4 Fig. 3. 6 TLANE GEOMETRY 31. Use of Instruments. Only two instruments are permitted in the constructions of plane geometry: the ruler or straight- edge for drawing a straight line, already spoken of in § 21 ; and the compasses, for constructing circles or arcs of circles, and for transferring line segments from one position to another. Thus to add two lines, as AB and CD, draw, with a ruler, a straight line OX. Place one leg of the compasses at A and the other at B. Next place one leg at and cut off segment OM equal to AB. In a similar manner lay off MN equal to CD. Then AB + CD — 0M+ MN = ON. Show how to subtract AB from ON. What is the remainder? Ex. 1. Can a straight line move so that its path will not be a surface ? If so, how ? Ex. 2. Can a curved line move in space so that its path will not be a surface ? If so, how ? Ex. 3. Can a broken line move in space so that its path will not be a surface ? Ex. 4. Draw three lines as AB, CD, and ^ EF. Construct the sum of AB and CD ; of AB and EF; of AB, CD, and EF. C Ex. 5. Construct : («) the difference be- E tween AB and CD ; (6) the difference between FlG 5 CD and EF; (c) the difference between AB and EF. Add the results obtained for (a) and (6) and see whether the sum is the result obtained for (c). Ex. 6. Draw a line twice as long as AB (the sum of AB and AB) ; Jhree times as long as AB. INTRODUCTION SURFACES 32. Plane Surface. It is well known that the carpenter's straightedge is applied to surfaces to test whether they are flat and even. If, no matter where the straightedge is placed on the surface, it always fits, the surface is called a plane. Now if we should use a powerful magnifier, we should doubt- less discover that in certain places the straightedge did not exactly fit the surface on which it was placed. A sheet of fine plate glass more nearly approaches the ideal. 33. Questions. Test the surface of the blackboard with a ruler to see whether it is a plane. How many times must you apply the ruler ? Can you think of a surface such that the ruler would fit in some positions (a great many) but not in all ? Can you think of a surface not plane but such that a piece of it could be cut out and slipped along the rest so that it would fit ? Would it fit if turned over (inside out) ? Fig. 6. 34. Def. A plane surface (or plane) is a surface of unlimited extent such that whatever two of its points are taken, a straight line joining them will lie wholly in the surface. 35. Def. A curved surface is a surface no portion of which is plane. 36. Def. A plane figure is a geometric figure all of whose points lie in one plane. Plane Geometry treats of plane figures. 37. Def. A rectilinear figure is a plane figure all the lines of which are straight lines. Ex. 7. How can a plane move in space so that its path will not be a solid ? Ex. 8. Can a curved surface move in space so that its path will not be a solid ? If so, how ? PLANE GEOMETRY ANGLES 38. Def. An angle is the figure formed by two straight lines which diverge from a point. The point is the vertex of the angle and the lines are its sides. 39. An angle may be designated by a number placed within it, as angle 1 and angle 2 in Fig. 7, and angle 3 in Fig. 8. Or L M H Fig. 9. three letters may be used, one on each side and one at the vertex, the last being read between the other two; thus in Fig. 7, angle 1 may be read angle ABC, and angle 2, angle CBD. An angle is often designated also by the single letter at its vertex, when no other angle has the same vertex, as angle F in Fig. 8. 40. Revolution postulate. A straight line may revolve in a plane, about a point as a pivot, and when it does revolve contin- , uously from one position to another, it passes once and only once through every intermediate position. 41. A clear notion of the magnitude of an angle may be obtained by imagining that its two sides were at first collinear, and that one of them has revolved about a point common to the two. Thus in Fig. 8. we may imagine FG first to have been in the position FE and then to have revolved about F as a pivot to the position FG. 42. Def. Two angles are adjacent if they have a common vertex and a common side which lies between them ; thus in Fig. 7, angle 1 and angle 2 are adjacent; also in Fig. 9, angle JIMK and angle KML are adjacent. INTRODUCTION 9 43. Two angles are added by placing them so that they are adjacent. Their sum is the angle formed by the two sides that are not common ; thus in Fig. 10, the sum of angle 1 and angle 2 is angle ABC. C Fig. 10. 44. The difference between two angles is found by placing them so that they have a vertex and a side in common but with the common side not between the other two. If the other two sides then happen to be collinear, the difference between the angles is zero and the angles are equal. If the other sides are not collinear, the angle which they form is the difference between the two angles compared ; thus in Fig. 11, the difference between angle 1 and angle 2 is angle ABC. Fig. 11. 45. Def. . If one straight line meets another so as to make two adjacent angles equal, each of these angles is a right angle, and the lines are said to be perpendicular to each other. Thus, if DC meets AB so that angle BCD and angle DCA are equal angles, each is a right angle, and lines AB and CD are said to be 4 ~C~ B perpendicular to each other. Fig. 12. 10 PLANE GEOMETRY 46. Def. If two lines meet, hut are not perpendicular to each other, they are said to be oblique to each other. 47. Def. An acute angle is an angle that is less than a right angle ; as angle 1, Fig. 13. 48. Def. An obtuse angle is an angle that is greater than a right angle and less than two right angles ; as angle 2, Fig. 14. 49. Def. A reflex angle is an angle that is greater than two right angles and less than four right angles; as angle 2, Fig. 15. 50. Note. Two lines diverging from the same point, as BA and BC, Fig. 15, always form two positive angles, as the acute angle 1 and the reflex angle 2. Angle 1 may be thought of as formed by the revolution of a line counter- clockwise from the position BA to the position BC, and should be read angle ABC. Angle 2 may be thought of as formed by the revolution of a line counter-clockwise from the position BC to the position BA, and should be read angle CBA. 51. Def. Acute, obtuse, and reflex angles are sometimes called oblique angles. Ex. 9. (a) In Fig. 16, if angle 1 equals angle 2, what kind of angles are they ? (b) Make a statement with regard to the lines AB and CD. (c) If angle 3 does not equal angle 4, what kind of angles are they ? (eZ) Make a statement with regard to the lines AB and CE. D Fig. 16. 7? Ex. 10. A plumb line is suspended from the top of the blackboard. What kind of angles does it make with a horizontal line drawn on the blackboard ? with a line on the blackboard neither horizontal nor vertical ? INTRODUCTION 11 Ex. 11. Suppose the minute hand of a clock is at twelve. Where may the hour hand be so that the two hands make with each other : (a) an acute angle ? (£>) a right angle ? (c) an obtuse angle ? Ex. 12. Draw: (a) a pair of adjacent angles; (6) a pair of non- adjacent angles. Ex. 13. Draw two adjacent angles such that : (a) each is an acute angle ; (&) each is a right angle ; (c) each is an obtuse angle ; (e?) one is acute and the other right ; (e) one is acute and the other obtuse. Ex. 14. In Fig. 17, angle 1 -f angle 2 = ? angle 3 + angle 4 = ? angle BAD + angle DAF = ? angle DAF — angle 3 = ? angle 2 + angle DAF — angle 4 = ? angle 4 + angle BAE — angle 1 = ? Ex. 15. Name six pairs of adjacent angles in Fig. 17. Ex. 16. Draw two non-adjacent angles that have : (a) a common vertex ; (&) a common side ; (c) a common vertex and a common side. 52. Def. A line is said to be bisected if it is divided into two equal parts. FlG ' 17 ' 53. Def. The bisector of an angle is the line which divides the angle into two equal angles.* Ex. 17. Draw a line AB, neither horizontal nor vertical, (a) Draw freehand a line perpendicular to AB and not bisecting it : (6) a line bisecting AB and not perpendicular to it. ASSUMPTIONS 54. 1. Things equal to the same thing, or to equal things, are equal to each other. 2. If equals are added to equals, the sums are equal. 3. If equals are subtracted from equals, the remainders are equal. 4. If equals are added to unequals, the sums are unequal in the same order. 5. If equals are subtracted from unequals, the remainders are unequal in the same order. ' * The proof that every angle has but one bisector will be found in the Appendix, § 599, 12 PLANE GEOMETRY 6. If unequals are subtracted from equals, the remainders are unequal in the reverse order. 7. (a) If equals are multiplied by equals, the products are equal; (6) if unequals are multiplied by equals, the products are unequal in the same order. 8. (a) If equals are divided by equals, the quotients are equal ; (b) if unequals are divided by equals, the quotients are unequal in the same order. 9. If unequals are added to unequals, the less to the less and the greater to the greater, the sums are unequal in the same order. .10. If three magnitudes of the same kind are so related that the first is greater than the second and the second greater than the third, then the first is greater than the third. 11. The whole is equal to the sum of all its parts. 12. The whole is greater than any of its parts. 13. Like powers of equal numbers are equal, and like roots of equal numbers are equal. 14. Transference postulate. Any geometric figure may be moved from one position to another without change of size or shape. (See § 16.) 15. Straight line postulate I. A straight line may be drawn from any one point to any other. (See § 24.) 16. Straight line postulate H. A line segment may be pro- longed indefinitely at either end. 17. Revolution postulate. A straight line may revolve in a plane, about a point as a pivot, and when it does revolve continu- ously from one position to another, it passes once and only once through every intermediate position. (See § 40.) 55. A ssumptions 1-13 are usually called axioms. That is, an axiom may be denned as a statement whose truth is as- sumed.* Ex. 18. Illustrate the first five assumptions above by using arithmeti- cal numbers only. Ex. 19. Illustrate the next five by using general numbers (letters) only * See Appendix, § 600. INTRODUCTION 13 DEMONSTRATIONS 56. It has been stated (§ 12) that the fundamental princi- ples agreed upon at the outset as forming the basis of the logi- cal arguments in geometry are called definitions, axioms, and postulates. Every new proposition advanced, whether it is a statement of a truth or a statement of something to be per- formed, must by a process of reasoning be shown to depend upon these fundamental principles. This process of reasoning is called a proof or demonstration. After the truth of a state- ment has thus been established, it in turn may be used to establish new truths. The propositions here mentioned are divided into two classes, theorems and problems. 57. Def. A theorem is a statement whose truth is required to be proved or demonstrated. For example, " If two angles of a triangle are equal, the sides opposite are equal" is a theorem. There are two parts to every theorem : the hypothesis, or the conditional part ; and the conclusion, or the part to be proved. In the theorem just quoted, " If two a7igles of a tri- angle are equal " is the hypothesis ; and " the sides opposite are equal " is the conclusion. . Ex. 20. Write out carefully the hypothesis and the conclusion of each of the following : (a) If you do your duty at all times, you will be rewarded. (6) If you try to memorize your proofs, you will never learn geometry. (c) You must suffer if you disobey a law of nature. (d) Things equal to the same thing are equal to each other. (e) All right angles are equal. 58. Def. A corollary is a statement of a truth easily de- duced from another truth. Its correctness, like that of a theorem, must be proved. 59. Def. A problem, in general, is a question to be solved. As applied to geometry, problems are of two kinds, namely, problems of construction and problems of computation. 14 PLANE GEOMETRY 60. From the revolution postulate (§ 40) and from the definition of a perpendicular (§ 45) we may deduce the follow- ing corollaries : 61. Cor. I. At every point in a straight line there exists a perpendicular to the line. Given line AB and point C in AB. To prove that there exists a JL to AB at a as en. D ,2-' ••• S E ! ,♦'* 1 c Fig. 18. B Let CE (Fig. 18) meet AB at O, so that Z 1 < Z 2. Then if CE is revolved about C as a pivot toward position CA, Z 1 will continuously increase and Z 2 will continuously decrease (§40)- .-. there must be one position of CE, as CD, in which the two A formed with AB are equal. In this position CD±AB (§45). 62. Cor. II. At every point in a straight line there exists only one perpendicular to the line. Given CD A. AB at C, i.e. Z BCD = /.DCA. To prove CD the only JL to AB at C. E J) i \ » \ \ C Fig. 19. B Let CE (Fig 19) be any line from C A other than CD. Let C# fall between CD and (M* Then ZBCE>ZBCD; i.e.> ZDCA. And ZECA , and angle B = angle E ; repeat the proof by superposition, marking the lines with colored crayon as soon as their positions are determined. Ex. • 54. Place polygon I upon polygon II so that some part of I shall fall upon its equal in II. Discuss the resulting positions of the remaining parts of the figure. Ex. 55. If two quadrilaterals have three sides and the included angles of one equal respectively to three sides and the included angles of the other, and arranged in the same order, are the quadrilaterals equal ? Prove. 106. Note. The method of superposition should be used for proving fundamental propositions only. In proving other propositions it is neces- sary to show merely that certain conditions are present and to quote theorems, previously proved, which state conclusions regarding such conditions. Ex. 56. If at any point in the bisector of an angle a perpendicular to the bisector is drawn meeting the sides of the angle, the two triangles thus formed will be equal. Ex. 57. If equal segments, measured from the point of intersection of two lines, are laid off on one of the lines, and if perpendiculars to this line are drawn at the ends of these segments, two • equal triangles will be formed. Ex. 58. If at the ends of a straight line per- pendiculars to it are drawn, these perpendiculars will cut off equal segments upon any line which bisects the given line and is not perpendicular to it. Fig. 1. Fig. Ex. 59. If two angles of a triangle are equal, the bisectors of these angles are equal (Fig. 1). Ex. 60. If two triangles are equal, the bisector of any angle of one is eqjjal to the bisector of the corresponding angle of the other (Fig. 2), 28 PLANE GEOMETRY Proposition III. Theorem 107. Two triangles are equal if two sides and the in- cluded angle of one are equal respectively to two sides %nd the included angle of the other. Given A ABC and MNO, AB = MN, AG = MO, and Z A = Z M. To prove A ABC = A MNO. Argument 1. Place A ABC upon A MNO so that AC shall fall upon its equal MO, A upon M, C upon 0. 2. Then AB will become col- linear with MN. 3. Point B will fall on point N. 4. . * . B C will coincide with NO. 5. A ABC=A MNO. Q.E.D. Reasons 1. Any geometric figure may be moved from one posi- tion to another without change of size or shape. § 54, 14. 2. ZA = ZM, by hyp. 3. AB = MN, by hyp. 4. Only one str. line can be drawn between two points. § 25. 5. Two geometric figures are equal if they can be made to coincide. § 18. 108. Cor. Two right triangles are equal if the two sides including the right angle of one are equal respectively to the two sides including the right angle of the other. Ex. 61. Prove Prop. HI by placing AB upon MN, BOOK I 29 109. Def. In equal figures, the points, lines, and angles in one which, when superposed, coincide respectively with points, lines, and angles in the other, are called homologous parts. Hence : 110. Homologous parts of equal figures are equal. Ex. 62. If two straight lines bisect each other, the lines joining their extremities are equal in pairs. Hint. To prove two lines or two angles equal, try to find two triangles, each containing one of the lines or one of the angles. If the triangles can be proved equal, and the two lines or two angles are homologous parts of the triangles, then the lines or angles are equal. The parts given equal may be more easily remembered by marking them with the same symbol, or with colored crayon. Ex. 63. In case the lines in Ex. 62 are perpendicular to each other, what additional statement can you make ? Prove its correctness. Ex. 64. If equal segments measured from the vertex are laid off on the sides of an angle, and if their extremities are joined to any point in the bisector of the angle, two equal triangles will be formed. Ex. 65. If two medians of a triangle are perpendicular to the sides to which they are drawn, the triangle is equilateral. Ex. 66. If equal segments measured from the vertex are laid off on the arms of an isosceles triangle, the lines joining the ends of these segments to the opposite ends of the base will be equal (Fig. 1). Fig. 1. Fie. 2. Ex. 67. Extend Ex. 66 to the case in which the equal segments are laid off on the arms prolonged through the vertex (Fig. 2). 30 PLANE GEOMETRY Proposition IV. Theorem 111. The base angles of an isosceles triangle are equal. B Given isosceles A ABC, with AB and BC its equal sides. To prove Z A = Z C. Argument Reasons 1. Let BD bisect Z ABC. 1. Every Z has but one bi- sector. § 53. o In A ABB and .DSC, XB = BC. 2. By hyp. 3. BD = BD. 3. By iden. 4. Z1 = Z2. 4. By cons. 5. .-. Aabd = Adbc. 5. Two A are equal if two sides and the included Z of one are equal re- spectively to two sides and the included Z of the other. § 107. 6. .\Za=Zc. 6. Homol. parts of equal fig- Q.E.D. ures are equal. § 110. 112. Cor. I. The bisector of the angle at the vertex of an isosceles triangle is perpendicular to the base and bisects it. 113. Cor. n. An equilateral triangle is also equi- angular. Ex. 68. The bisectors of the base angles of an isosceles triangle are eaual. BOOK I 31 114. Historical Note. Exercise 69 is known as the pons asi- norum, or bridge of asses, since it has proved difficult to many beginners in geometry. This proposition and the proof here suggested are due to Euclid, a great mathematician who wrote the first systematic text-book on geometry. In this work, known as Euclid's Elements, the exer- cise here given is the fifth prop- osition in Book I. Of the life of Euclid there is but little known except that he was gentle and modest and " was a Greek who lived and taught in Alexandria about 300 b.c." To him is attributed the saying, " There is no royal road to geometry. ' ' His appreciation of the culture value of geometry is shown in a story related by Stobaeus (which is probably authentic). "A lad who had just begun geometry asked, *■ What do I gain by learning all this stuff ? ' Euclid called his slave and said, ' Give this boy some coppers, since he must make a profit out of what he learns.' " Euclid Ex. 69. By using the accompanying diagram prove that the base angles of an isosceles triangle are equal. Hint. Prove A ABE = A BBC. Then prove A ACE = ABAC. Ex. 70. (a) If equal segments measured from the vertex are laid off on the arms of an isosceles triangle, the lines drawn from the ends of the segments to the foot of the bisector of the vertex angle will be equal. (6) Extend (a) to the case in which the equal segments are laid off on the arms prolonged through the vertex. Ex. 71. (a) If equal segments measured from the ends of the base are laid off on the arms of an isosceles triangle, the lines drawn from the ends of the segments to the foot of the bisector of the vertex angle will be equal. (6) Extend (a) to the case in which the equal segments are laid off on the arms prolonged through the ends of the base. 32 PLANE GEOMETRY Ex. 72. (a) If equal segments measured from the ends of the base are laid off on the base of an isosceles triangle, the lines joining the vertex of the triangle to the ends of the segments will be equal. (6) Extend (a) to the case in which the equal segments are laid off on the base prolonged (Fig. 1) . Fig. 1. Fig. 2. Ex. 73. (a) If equal segments measured from the ends of the base are laid off on the arms of an isoscles triangle, the lines drawn from the ends of the segments to the opposite ends of the base will be equal. (6) Extend (a) to the case in which the equal segments are laid off on the arms prolonged through the ends of the base (Fig. 2). Ex. 74. Triangle ABC is equilateral, and AE = BF = CD. Prove triangle EFD equi- lateral. 115. Measurement of Distances by Means of Triangles. The theorems which prove triangles equal are applied practically in measuring distances on the surface of the earth. Thus, if it is desired to find the distance between two places, A and B, which are separated by a pond or other obstruction, place a stake at some point accessible to both A and B, as F. Measure the distances FA and FB ; then, keeping in line with F and B, measure CF equal to FB, and, in line with F and A, measure FE equal to FA. Lastly measure CE, and the distance from A to B is thus obtained, since AB is equal to CE. Can this method be used when A and B are on opposite sides of a hill and each is invisible from the other ? BOOK I 33 Ex. 75. Show how to find the distance across a river by taking the following measurements. Measure a convenient dis- tance along the bank, as BT, and fix a stake at its mid-point, F. Proceed at right angles to BT from T to the point P, where F, S, and P are in line ; measure PP. Ex. 76. An army engineer wished to obtain quickly the approximate distance across a river, and had no instruments with which to make measure- ments. He stood on the bank of the river, as at A, and sighted the opposite bank, or B. Then without raising or lowering his eyes, he faced about, and his line of sight struck the ground at C. He paced the distance, AC\ and gave this as the distance across the river. Explain his method. Ex. 77. Tell what measurements to make to obtain the distance between two inaccessible points, B and S (Fig. 1). Ex. 78. The fact that a triangle is deter- mined if its base and its base angles are given was used as early as the time of Thales (640 b.c.) to find, the distance of a ship at sea; the base of the triangle was usually a lighthouse tower and the base angles were found by ob- servation. Draw a figure and explain. Fig. 1. Ex. 79. Explain the following method of finding SB (Fig. 2). Place a stake at S, and another at a convenient p point, Q, in line with S and B. From a con- venient point, as P, measure TS and TQ. Pro- long QT, and make. TF equal to QT. Prolong ST, and make TB equal to ST. Then keep in line with F and B, until a point is reached, as G, where T and B come into line. Then BO is equal to the required distance, BS. Ex. 80. In an equilateral triangle, if two lines are drawn from the ends of the base, making equal angles with the base, the lines are equal. Is this true of every isosceles triangle? Fig. 2. 34 PLANE GEOMETRY Proposition V. Theorem 116. Two triangles are equal if the three sides of one are equal respectively to the three sides of the other. ■ Given A ABC and RST, AB=RS, BC = ST, and CA = TR. To prove A ABC = A RST. 2. Argument Place A RST so that the longest side RT shall fall upon its equal AC, R upon A, T upon C, and so that S shall fall oppo- site B. Draw BS. 3, A ABS is isosceles. 3. 4. .\Z1 = Z2. 4. 5. A BCS is isosceles. 5. G. .-.Z3 = Z4. 6. 7. Z1 + Z3 = Z2 + Z4. 7. 1. 2. Reasons Any geometric figure may- be moved from one posi- tion to another without change of size or shape. § 54, 14. A str. line may be drawn from any one point to any other. § 54, 15. AB = RS, by hyp. The base A of an isosceles A are equal. § 111. BC—ST, by hyp. Same reason as 4. If equals are added to equals, the sums are equal. § 54 ? 2. BOOK I 35 Argument 8. .*. Z ABC=/-CSA. 9. .'. A ABC = A CSA; i.e. A ABC = A RST. Q.E.D. Reasons 8. The whole = the sum of all its parts. § 54, 11. 9. Two A are equal if two sides and the included Z of one are equal respec- tively to two sides and the included Z of the other. § 107. Ex. 81. (a) Prove Prop. V, using two obtuse triangles and applying the shortest side of one to the shortest side of the other. (b) Prove Prop. V, using two right triangles and applying the shortest side of one to the shortest side of the other. 117. Question. Why is not Prop. V proved by superposition? SUMMARY OF CONDITIONS FOR EQUALITY OF TRIANGLES a side and the two adjacent angles two sides and the included angle three sides a side and the two adjacent angles two sides and the included angle three sides 118. Two triangles are equal if of one are equal respectively to of the other. . Ex. 82. The median to the base of an isosceles triangle bisects the angle at the vertex and is perpendicular to the base. Ex. 83. In a certain quadrilateral two adjacent sides are equal ; the other two sides are also equal. Find a pair of triangles which you can prove equal. 36 PLANE GEOMETRY Ex. 84. If the opposite sides of a quadrilateral are equal, the opposite angles also are equal. Ex. 85. If two isosceles triangles have the same base, the line joining their vertices bisects each vertex angle and is perpendicular to the -com- mon base. (Two cases.) Ex. 86. In what triangles are the three medians equal ? Ex. 87. In what triangles are two medians equal ? Ex. 88. If three rods of different lengths are put together to form a triangle, can a different triangle be formed by arranging the rods in a different order ? Will the angles opposite the same rods always be the same ? Ex. 89. If two sides of one triangle are equal respectively to two sides of another, and the median drawn to one of these sides in the first is equal to the median drawn to the corresponding side in the second, the triangles are equal. 119. Def. A circle is a plane closed figure whose boundary is a curve such that all straight lines to it from a fixed point within are equal. The curve which forms the boundary of a circle is called the circumference. The fixed point within is called the center, and a line joining the center to any point on the circumference is a radius. 120. It follows from the definition of a circle that : All radii of the same circle are equal. 121. Def. Any portion of a circumference is called an arc. 122. Assumption 18. Circle postulate. A circle may be con- structed having any point as center, and having a radius equal to any finite line. 123. The solution of a problem of construction consists of three distinct steps : (1) Tlie construction, i.e. the process of drawing the required figure with ruler and compasses. (2) The proof a demonstration that the figure constructed fulfills the given conditions. (3) The discussion, i.e. a statement of the conditions under which there may be no solution, one solution, or more than one. BOOK I 37 Proposition VI. Problem 124. To construct an equilateral triangle, with a given line as side. A Given line BC. To construct an equilateral triangle on BC. I. Construction 1. With B as center and BC as radius, construct circle CAB. 2. With C as center and BC as radius, construct circle BMA. 3. Connect point A, at which the circumferences intersect, with B and C. 4. A ABC is the required triangle. Argument 1. AB = BC 3Hid CA — BC. 2. .'. AB = BC=CA. II. Proof 2. 3. .'.A ABC is equilateral. Q.E.D. Reasons All radii of the same circle are equal. § 120. Things equal to the same thing are equal to each other. § 54, 1. A A having its three sides equal is equilateral. § 95. III. Discussion This construction is always possible, and there is only one solution. (See § 116.) 38 PLANE GEOMETRY Proposition VII. Problem 125. With a given vertex and a given side, to construct an angle equal to a given angle. Given vertex A, side AB, and Z CDE. To construct an Z equal to Z CDE and having A as vertex and AB as side. I. Construction 1. With D as center, and with any convenient radius, de- scribe an arc intersecting the sides of Z D at F and G, respec- tively. 2. With A as center, and with the same radius, describe the indefinite arc IH, cutting AB at /. 3. With I as center, and with a radius equal to str. line FG, describe an arc intersecting the arc IH at K. 4. Draw AK. 5. Z BAK = Z CDE, and is the Z required. II. Proof Argument 1. Draw FG and IK. 2. In A FDG and IAK, DF—AI. 3. DG — AK. 4. FG = IK. Reasons 1. A str. line may be drawn from any one point to any other. § 54, 15. 2. By cons. 3. By cons. 4. By cons. BOOK I 39 Argument 5. .* . A FDG = A IAK. 6. .• . ZBAK = Z.CDE. Q.E.D. Eeasons 5. Two A are equal if the three sides of one are equal respectively to the three sides of the other. §116. 6. Homol. parts of equal figures are equal. § 110. III. Discussion This construction is always possible, and there is only one solution. Ex. 90. Construct a triangle, given two sides and the included angle. Ex. 91. Construct an isosceles triangle, given the vertex angle and Ex. 92. Construct a triangle, given a side and the two adjacent angles. Ex. 93. Construct an isosceles triangle, given an arm and one of the equal angles. Ex. 94. How many parts determine a triangle ? Do three angles determine it ? Explain. Ex. 95. Construct an isosceles triangle, given the base and an arm. Ex. 96. Construct- a scalene triangle^ given the three sides. 126. Def. The bisector of an angle of a triangle is the line from the vertex of the angle bisecting the angle and limited by the opposite side of the triangle. a Ex. 97. In what triangles are the three bisectors equal ? Ex. 98. In what triangles are two bisectors, and only two, equal ? Ex. 99. In what triangles are the medians, the bisectors, and the altitudes identical ? 40 PLANE GEOMETRY Proposition VIII. Problem 127. To construct the bisector of a given angle. Given Z.ABC. To construct the bisector of Z ABC. I. Construction 1. With B as center, and with any convenient radius ; describe an arc intersecting BA at E and BC at D. 2. With D and E as centers, and with equal radii, describe arcs intersecting at F. 3. Draw BF. 4. BF is the bisector of Z ABC. II. Proof Argument Reasons 1. In AEBF and ^J5Z>, £.0 as £D. 1. By cons. 2. EF = DF. 2. By cons. 3. BF = BF. 3. By iden. 4. .'. AEBF = AFBD. 4. Two A are equal if the three sides of one are equal respectively to the three sides of the other. §116. 5. .'. Z EBF — Z.FBD. 5. Homol. parts of equal .figures are equal. § 110. BOOK I 41 Argument 6. .'. BF is the bisector of Z ABC. Q.E.D. Reasons 6. The bisector of an Z. is the line which divides the Z. into two eqnal A § 53. III. Discussion This construction is always possible, and there is only one -solution. Draw an obtuse angle and divide it into: (a) four (6) eight equal angles. Construct the bisector of the vertex angle of an isosceles Ex. 100. equal angles Ex. 101. triangle. Ex. 102. Draw two intersecting lines and construct the bisectors of the four angles formed. Ex. 103. Bisect an angle between two bisectors in Ex. 102, and find the number of degrees in each angle. Ex. 104. Construct the bisector of an exterior angle at the base of an isosceles triangle. Ex. 105. Construct the bisectors of the three angles of any triangle. What can you infer about them ? Can the correctness of this inference be proved by making a careful construction ? 128. Def. The distance between two points is the leugth of the straight line joining them. Thus if three points, A, B, and C, are so located that AB = AC, A is said to be equidistant from B and C. Ex. 106. Find all the points on the blackboard which are one foot from a fixed point, P, on the blackboard. Ex. 107. Draw a line, AB, on the blackboard and mark some point near the line, as P. Find all the points in AB that are a foot from P. Ex. 108. Mark a point, Q, on the blackboard. Find all the points on the blackboard which are : (a) ten inches from Q ; (6) four inches from Q. How far are the points of (6) from the points of (a), if the distance is measured on a line through Q ? 42 PLANE GEOMETRY LOCI 129. In many geometric problems it is necessary to locate all points which satisfy certain prescribed conditions, or to determine the path traced by a point which moves according to certain ijxed laws. Thus, the points in a plane two inches from a given point are in the circumference of a circle whose center is the given point and whose radius is two inches. Again, let it be required to find all points in a plane two inches from one fixed point and three inches from another. All points two inches from the fixed point P are in the cir- cumference of the circle LMS, having P for center and having a radius equal to two inches. All points three inches from the fixed point Q are in the cir- cumference of the circle LET, having Q for center and having a radius equal to three inches. If the two circles are wholly outside of each other, there will be no points satisfying the two prescribed conditions ; if the two circumferences touch, but do not intersect, there will be one point ; if the two circum- ferences intersect, there will be two points. It will be proved later (§ 324) that there cannot be more than two points which satisfy both of the given conditions. 130. Def. A figure is the locus of all points which satisfy one or more given conditions, if all points in the figure satisfy the given conditions and if these conditions are satisfied by no other points. A locus, then, is an assemblage of points which obey one or more definite laws. It is often convenient to locate these points by thinking of them as the path traced by a moving point the motion oi which is controlled by certain fixed laws. BOOK I 43 131. In plane geometry a locus may be composed of one or more points or of one or more lines, or of any combination of points and lines. 132. Questions.* — What is the locus of all points in space two inches from a given point ? What is the locus of all points in space two inches from a given plane ? What is the locus of all points* in space such that perpendiculars from them to a given plane shall be equal to a given line ? What is the locus of all points on the surface of the earth midway between the north and south poles ? 23|° from the equator ? 23£° from the north pole ? 00° from the equator ? What is the locus of a gas jet four feet from the ceiling of this room ? four feet from the ceiling and five feet from a side wall ? four feet from the ceiling, five feet from a side wall, and six feet from an end wall ? Ex. 109. Given an unlimited line AB and a point P. Find all points in AB which are also : (a) three inches from P; (b) at a given distance, a, from P. Ex. 110. Given a circle with center and radius six inches. State, without proof, the locus : (a) of all points four inches from ; (&) of all points five inches from the circumference of the circle, measured on the radius or radius prolonged. Ex. 111. Given the base and one adjacent angle of a triangle, what is the locus of the vertex of the angle opposite the base ? (State without proof.) Ex. 112. Given the base and one other side of a triangle, what is the locus of the vertex of the angle opposite the base ? (State without proof.) Ex. 113. Given the base and the other two sides of a triangle, what is the locus of the vertex of the angle opposite the base ? Ex. 114. Given the base of a triangle and the median to the base, what is the locus of the end of the median which is remote from the base ? Ex. 115. Given the base of a triangle, one other side, and the median to the base, what is the locus of the vertex of the angle opposite the base ? 133. Question. In which of the exercises above was a triangle determined ? * In order to develop the imagination of the student the authors deem it advisable in this article to introduce questions involving loci in space. It should be noted that no proofs of answers to these questions are demanded. v * 44 PLANE GEOMETRY Proposition IX. Theorem 134. Every point in the perpendicular bisector of a line is equidistant from the ends of that line. A C B Given line AB, its ± bisector CD, and P any point in CD. To prove PA PB. Argument 1. In A APC and CPB, AC as CB. 2. PC=PC. 3. Z PC A = Z BCP. 4. .*.A APC = A CPB. 5. .\PA = PB. Q. E. D. REASON8 1. By hyp. 2. By iden. 3. All rt. A are equal. § 64. 4. Two A are equal if two sides and the included Z of one are equal re- spectively to two sides and the included Z of the other. § 107. 5. Homol. parts of equal fig- ures are equal. § 110. Ex. 116. Four villages are so located that B is 25 miles east of A, C 20 miles north of A, and D 20 miles south of A. Prove that B is as far from C as it is from D. Ex. 117. In a given circumference, find the points equidistant from two given points. A and B. BOOK I 45 135. Def. One theorem is the converse of another when the conclusion of the first is the hypothesis of the second, and the hypothesis of the first is the conclusion of the second. The converse of a truth is not always true ; thus, " All men are bipeds " is true, but the converse, " All bipeds are men," is false. "All right angles are equal" is true, but "All equal angles are right angles" is false. 136. Def. One theorem is the opposite of another when the hypothesis of the first is the contradiction of the hypothesis of the second, and the conclusion of the first is the contradic- tion of the conclusion of the second. The opposite of a truth is not always true ; thus, " If a man lives in the city of New York, he lives in New York State," is true, but the opposite, " If a man does not live in the city of New York, he does not live in Neiv York State," is false. 137. Note. If the converse of a proposition is true, the opposite also is true ; so, too, if the opposite of a proposition is true, the converse also is true. This may be evident to the student after a consideration of the fol- lowing type forms : (1) Direct (2) Converse (8) Opposite If ^lis.B, If CisD, If A is not B, Then C is D. Then A is B. Then C is not D. If (2) is true, then (3) must be true. Again, if (3) is true, then (2) must be true. 138. A necessary and sufficient test of the completeness of a definition is that its converse shall also be true. Hence a definition may be quoted as the reason for a converse or for an opposite as well as for a direct statement in an argument. Ex. 118. State the converse of the definition for equal figures ; straight line ; plane surface. Ex. 119. State the converse of : If one straight line meets another straight line, the sum of the two adjacent angles is two right angles. Ex. 120. State the converse and opposite of Prop. IX. Ex. 121. State the converse of Prop. I. Is it true? 46 PLANE GEOMETRY Proposition X. Theorem (Converse of Prop. IX) 139. Every point equidistant from the ends of a line lies in the perpendicular bisector of that line. Given line RS, and point Q such, that QR = QS, To prove that Q lies in the _L bisector of RS. Argument 1. Let QT bisect Z RQS. 2. QR = QS. 3. .'.A RQS is isosceles. 4. .'. QT is the _L bisector of RS. 5. .'. Q lies in the _L bisector of RS. Q.E.D. 5. Reasons Every Z has but one bi- sector. § 53. By hyp. A A having two sides equal is an isosceles A. § 94. The bisector of the Z at the vertex of an isosceles A is ± to the base and bisects it. § 112. By proof. 140. Cor. I. Every point not in the perpendicular bi- sector of a line is not equidistant from the ends of the line. Hint. Use § 137, or contradict the conclusion and tell why the con- tradiction is false. BOOK I 47 141. Cor. n. The locus of all points equidistant from the ends of a given line is the perpendicular bisector of that line. (For model proofs of locus theorems, see pp. 297 and 298.) 142. Cor. in. Two points each equidistant from the ends of a line determine the perpendicular bisector of the line. Hint. Use § 139 and § 25. 143. In order to prove that a locus problem is solved it is necessary and sufficient to show two things : (1) That every point in the proposed locus satisfies the prescribed conditions. (2) That every point outside of the proposed locus does not satisfy the prescribed conditions. Instead of proving (2), it may frequently be more convenient to prove : (2') That every point which satisfies the prescribed con- ditions lies in the proposed locus. 144. Note. In exercises in which the student is asked to "Find a locus, " it must be understood that he has not found a locus until he has given a proof with regard to it as outlined above. The proof must be based upon a direct proposition and its opposite ; or, upon a direct propo- sition and its converse. Ex. 122. Find the locus of all points equidistant from two given points A and B. Ex. 123. In a given unlimited line AB, find a point equidistant from two given points G and D not on this line. Ex. 124. Given a circle with center 0, also a point P. Find all points which lie in the circumference of circle O, and which are also (a) two inches from P ; (6) a distance of d from P. Ex. 125. Find all points at a distance of d from a given point P, and at the same time at a distance of m from a given point Q. Ex. 126. Given a circle with radius r. Find the locus of the mid- points of the radii of the circle. Ex. 127. Given two circles having the same center. State, without proof, the locus of a point equidistant from their circumferences. Ex. 128. The perpendicular bisector of the base of an isosceles tri- angle passes through the vertex, - 48 PLANE GEOMETRY Proposition XI. Problem 145. To construct the perpendicular bisector of a given straight line. V I I \ 'H IF x Fig. 1. B A Fig. 2. Given line AB (Fig. 1). To construct the perpendicular bisector of AB. I. Construction 1. With A as center, and with a convenient radius greater than half AB, describe the arc EF. 2. With B as center, and with the same radius, describe the arc HK. 3. Let c and D be the points of intersection of these two arcs. 4. Connect points C and D. 5. CD is the _L bisector of AB. II. The proof and discussion are left as an exercise for the student. Hint. Apply § 142. 146. Note. The construction given in Fig. 2 may be used when the position of the given line makes it more convenient. 147. Question. Is it necessary that CA shall equal CB? that CA ghall equal DA (Fig. 1) ? Give the equations that must hold. BOOK I 49 Proposition XII. Problem 148. To construct a perpendicular to a given straight line at a given point in the line. \ A D\ C te B Given line AB and point C in the line. To construct a _L to AB at C I. Construction 1. With C as center, and with any convenient radius, draw arcs cutting AB on each side of C, as at D and E. 2. Then with D and E as centers, and with a longer radius, draw two arcs intersecting each other at F. 3. Drawee. 4. FC is J_ to AB at a II. The proof and discussion are left as an exercise for the student. , Ex. 129. Construct the perpendicular bisector of a line given at the bottom of a page or of a blackboard. Ex. 130. Divide a given line into four equal parts. Ex. 131. Construct the three medians of a triangle. Ex. 132. Construct a perpendicular to a line at a given point when the given point is one end of the line. (Hint. Prolong the line.) Ex. 133. Construct a right triangle, given the two arms a and b. Ex. 134. Construct a right triangle, given the hypotenuse and an arm. Ex. 135. Construct the complement of a given angle. Ex. 136. Construct an angle of 45°; of 135°. Ex. 137. Construct a quadrilateral, given four sides and the angle between two of them. * 4 50 PLANE GEOMETRY Proposition XIII. Problem 149. From a point outside a line to construct a per- pendicular to the line. *£- B Xq Given line AB and point P outside of AB. To construct a _L from P to AB. I. Construction 1. With. P as center, and with a radius of sufficient length, describe an arc cutting AB at points C and D. 2. With c and D as centers, and with any convenient radius, describe arcs intersecting at Q. 3. Draw PQ. 4. PQ is a _L from P to AB. , II. The proof is left as an exercise for the student. The discussion will be given in § 154. 150. Question. Must P and Q be on opposite sides of AB? Is it necessary that PC=QG? Ex. 138. Construct the three altitudes of an acute triangle. Do they seem to meet ? where ? Ex. 139. Construct the three altitudes of a right triangle. Where do they seem to meet ? Ex. 140. Construct the three altitudes of an obtuse triangle. Where do they seem to meet ? Ex. 141. Construct a triangle ABC, given two sides and the median drawn to one of them. Abbreviate thus : given a, b r and m^. BOOK I 51 I 151. Analysis of a problem of construction. In the more difficult problems of construction a course of reasoning is some- times necessary to enable the student to discover the process of drawing the required figure. This course of reasoning is called the analysis of the problem. It is illustrated in § 152 and is more fully treated in the exercises following § 274. 152. Note. * In such problems as Ex. 141, it is well first to imagine the problem solved and to sketch a figure to represent the desired con- struction. Then mark (with colored crayon, if convenient) the parts supposed to be given. By studying carefully the relation of the given parts to the whole figure, try to find some part of the figure that you can construct. This will generally be a triangle. After this part is constructed it is usually an easy matter to complete the required figure. Thus : Problem. Let it be required to con- struct an isosceles triangle, given an arm and the altitude upon it. By studying the figure with the given parts marked (heavy or with colored crayon), it will be seen that the solution of the problem depends in this case upon the construc- tion of a right triangle, given the hypotenuse and one arm. The right triangle ABB may now be constructed, and it will be readily seen that to complete the construction it is only necessary to prolong BD to C, making BC = AB, and to connect A and C. Ex. 142. Construct a triangle ABC, given two sides and an altitude to one of the given sides. Abbreviate thus : given a, b, and hi,. Ex. 143. Construct a triangle ABC, given a side, an adjacent angle, and the altitude to the side opposite the given angle. Abbreviate thus : given a, B, and ft*. Ex. 144. Construct an isosceles triangle, given an arm and the me- dian to it. Ex. 145. Construct an isosceles triangle, given an arm and the angle which the median to it makes with it. Ex. 146. Construct a triangle, given two sides and the angle which a median to one side makes : (a) with that side ; (&) with the other side. Ex. 147. Construct a triangle, given a side, an adjacent angle, and its bisector. 52 PLANE GEOMETRY t Proposition XIV. Theorem 153. If one side of a triangle is prolonged, the exterior angle formed is greater than either of the remote in- terior angles. B F Given A ABC with AC prolonged to D, making exterior Z DCB. To prove Z DCB > Z ABCov Z CAB. Argument 1. Let E be the mid-point of BCj draw AE, and pro- long it to F, making EF = AE. Draw CF. 2. In A ABE and EFC y BE = EC. 3. AE = EF. 4. Z BE A = Z CEF. 5. .'. A ABE = A EFC. .-.Zb = Zfce. Reasons 1. A str. line may be drawn from any one point to any other. § 54, 15. 2. By cons. E is the mid- point of BC. 3. By cons. 4. If two str. lines inter- sect, the vertical A are equal. § 77. 5. Two A are equal if two sides and the included Z of one are equal re- spectively to two sides and the included Z of the other. § 107. 6. Homol. parts of equal fig- ures are equal. § 110. BOOK I 53 7. Argument Z DCB > Z FCE. 8. .-. Z DC£ > Z 5. 9. Likewise, if BC is pro- longed to G, Z.ACG> Z CAB. 10. But Z DCB = Z .4CG. 11. .*. Z Z)C5 > Z CAB. 12. Reasons 7. The whole > any of its parts. § 54, 12. 8. Substituting Z.B for its equal Z i^C^. 9. By bisecting line AC, and by steps similar to 1-8. 10. Same reason as 4. 11. Substituting A DCB for its equal Z.ACG. 12. By proof. ZDCB> A ABC or Z CAB. Q.E.D 154. Cor. From a point outside a line there exists only one perpendicular to the line. Hint. If there exists a second _L to AB from A 7^ — \f B P, as PC, then A PC A and Z. PKA are both rt. A and are therefore equal. But this is impossible by § 153- 155. §§ 149 and 154 may be combined in one statement as follows : From a point outside a line there exists one and only one per- pendicular to the line, Ex. 148. A triangle cannot contain two right angles. Ex. 149. In the figure of Prop. XIV, is angle DCB necessarily greater than angle BCA ? than angle B ? j± Ex. 150. In Fig. 1, prove that : (1) Angle 1 is greater than angle CAE or angle AEC ; (2) Angle 5 is greater than angle CBA or angle BAE ; (3) Angle EDA is greater than angle 3 ; (4) Angle 4 is greater than angle DAE. Ex. 151. In Fig. 2, show that angle 6 is greater than angle 7 ; also that angle 9 is greater than angle A. Fig. 2. sA K *-> ^aA V B C I Fig. 1. A X2 ) E i£-r^ ^6 54 PLANE GEOMETRY Proposition XV. Theorem 156. If two sides of a triangle are unequal, the angle opposite the greater side is greater than the angle opposite the less side. Given A ABC with BC > BA. To prove Z CAB > Z C. Argument Reasons 1. On BC lay off BD = AB. 1. Circle post. §§ 122, 157. • 2. Draw AD. 2. Str. line post. I. § 54, 15. 3. Then Z. =Z2. 3. The base A of an isosceles A are equal. § 111. 4. If one side of a A is pro- longed, the ext. Z formed > either of the remote int. A. § 153. 5. Substituting Z 1 for its equal Z 2. 6. The whole > any of its parts. § 54, 12. 7. If three magnitudes of the same kind are so related that the first > the sec- ond and the second > the third, then the first q.e.d. > the third. § 54, 10. 157. Note. Hereafter the student will not be required to state postulates and definitions in full unless requested to do so by the teacher. 4. NowZ2 > Z.C. 5. r.Zl > Zc. (5. 7. But Z CAB > Z 1. /. ZCAB > ZC. BOOK I 55 158. Note. When two magnitudes are given unequal, the laying off of the less upon the greater will often serve as the initial step in de- veloping a proof. Ex. 152. Given the isosceles triangle EST, with ST the base and BT prolonged any length, as to K. Prove angle KSR greater than angle K. Ex. 153. If two adjacent sides of a quadrilateral are greater respec- tively than the other two sides, the angle included between the two shorter sides is greater than the angle between the two greater sides. Ex. 154. If from a point within a triangle lines are drawn to the ends of one of its sides, the angle between these lines is greater than the angle between the other two sides of the triangle. (See Ex. 151.) 159. The indirect method, or proof by exclusion, consists in contradicting the conclusion of a proposition, then showing the contradiction to be false. The conclusion of the proposition is thus established. This process requires an examination of every possible contradiction of the conclusion. For example, to prove indirectly that A equals B it would be necessary to consider the only three suppositions that are admissible in this case, viz. : (i) A > B} (2) A < B, \ (3) A = B. \ By proving (1) and (2) false, the truth of (3) is established, i.e. A = B. This method of reasoning is called reductio ad absurdum. It enables us to establish a conclusion by showing that every contradiction of it leads to an absurdity. Props. XVI and XVII will be proved by the indirect method. 160. Question. Would it be possible to base a proof upon a contra- diction of the hypothesis ? 161. (a) In the use of the indirect method the student should give, as argument 1, all the suppositions of which the case he is considering admits, including the conclusion. As reason 1 the number of such possi- ble suppositions should be cited. (&) As a reason for the last step in the argument he should state which of these suppositions have been proved false. 56 PLANE GEOMETRY Proposition XVI. Theorem (Converse of Prop. IV) 162. If two angles of a triangle are equal, the sides opposite are equal. Given A RST with Z.R — Z.T. To prove r = t . Argument 1. r > tj rt\ then Zi2>Zr. 3. This is impossible. 4. Next suppose r the Z opposite the less side. §156. By hyp., Zs = ZT. Same reason as 2. Same reason as 3. The two suppositions, r > t and r < t, have been proved false. 163. Cor. An equiangular triangle is also equilateral. Ex. 155. The bisectors of the base angles of an isosceles triangle form an isosceles triangle. BOOK I 57 Proposition XVII. Theorem (Converse of Prop. XV) 164. If two angles of a triangle are unequal, the side opposite the ir eater angle is greater than the stole opposite the less angle. A b Given A ABC with ZA>ZC, To prove a>c. Argument 1. a < c, a — c, or a > c 2. First suppose a < c; then Z.Ac. Q.E.D. Reasons 1. In this case only three sup- positions are admissible. 2. If two sides of a A are un- equal, the Z opposite the greater side > the Z opposite the less side. §156. 3. By hyp., ZA>Zc. 4. The base A of an isosceles A are equal. § 111 5. Same reason as 3. 6. The two suppositions, a BC. Let bisectors of angles A and B meet at 0. Prove AO > BO. Ex. 159. A line drawn from the vertex of an isosceles triangle to any point in the base is less than one of the equal sides of the triangle. Ex. 160. If ABC and ABB are two triangles on the same base and on the same side of it such that AC — BD and AD = BC, and if AC and BD intersect at 0, prove triangle AOB isosceles. Ex. 161. Prove Prop. XVI by using the figure and method of Ex. 69. Ex. 162. Upon a given base is constructed a triangle one of the base angles of which is double the other. The bisector of the.larger base angle meets the opposite side at the point P. Find the locus of P. Ex. 163. If the four sides of a quadrilateral are equal, its diagonals bisect each other. Ex. 164. The diagonals of an equilateral quadrilateral are perpen- dicular to each other, and they bisect the angles of the quadrilateral. Ex. 165. If two adjacent sides of a quadrilateral are equal and the other two sides are equal, one diagonal is the perpendicular bisector of the other. Tell which one is the bisector and prove the correctness of your answer. Ex. 166. If, from a point in a perpendicular to a line, oblique lines are drawn cutting off equal segments from the foot of the perpendicular, the oblique lines are equal. Ex. 167. State and prove the converse of Ex. 166. Ex. 168. If, from a point in a perpendicular to a line, oblique lines are drawn cutting off unequal segments from the foot of the perpendicular, the oblique lines are unequal. Prove by laying off the less segment upon the greater. Then use Ex. 166, § 153, and § 164. Ex. 169. If, from any point in a perpendicular to a line, two unequal oblique lines are drawn to the line, the oblique lines will cut off unequal segments from the foot of the perpendicular. Prove by the indirect method. Ex. 170. By means of Prop. XIV, prove that the sum of any two angles of a triangle is less than two right angles. Ex. 171. Construct a triangle ABC, given two sides, a and 6, and the altitude to the third side, c. (See § 152.) BOOK I 59 Proposition XVIII. Theorem 167. The sum of any two sides of a triangle is greater than the third side Given A ABC. To prove a + c > b. Argument Reasons 1. Prolong c through B prolongation BD = until a. 1. Str. line post. II. § 54, 16. 2. Draw CD. 2. Str. line post. I. § 54, 15. 3. In isosceles A BDC, /.D = ZDCB. 3. The base A of an isosceles A are equal. § 111. 4. But Zdca > Zdcb. 4. The whole > any of its parts. § 54, 12. 5. .-. Adca > Zd. 5. Substituting Z.D for its equal, Z.DCB. 6. .'. in A ADC, AD> b. 6. If two A of a A are unequal, the side opposite the greater Z is > the side opposite the less angle. §164. 7. .: a + c> b. q.e.d. 7. Substituting a -f c for AD. 168. Cor. I. Any side of a triangle is less than the sum and greater than the difference of the other two. 169. Cor. II. Any straight line is less than the sum of the.parts of a broken line having the same extremities. 60 PLANE GEOMETRY 170. Note to Teacher. Teachers who prefer to assume that " a straight line is the shortest line between two points " may omit Prop. XVIII entirely. Then Prop. XVII may be proved by a method similar to that used in Prop. XV. (See Ex. 172.) Ex. 172. Prove Prop. XVII by using the hint contained in § 158. Ex. 173. If two sides of a triangle are 14 and 9, between what limit- ing values must the third side be ? Ex. 174. If the opposite ends of any two non-intersecting line seg- ments are joined, the sum of the joining lines is greater than the sum of the other two lines. ft Ex. 175. Given two points, P and R, and a line AB not passing through either. To find a point O, on AB, such that PO + OB shall be as small as possible. This exercise illustrates the law by which light is reflected from a mirror. The light from the object, P, is reflected and appears to come from L, as far behind the mirror as P is in front of it. Ex. 176. If from any point within a triangle lines are drawn to the extremities of any side of the triangle, the sum of these lines is less than the sum of the other two sides of the triangle. Hint. Let ABC be the given triangle, D the point within. Prolong AD until it intersects BG at E. Apply Prop. XVIII. 171. Note to Teacher. Up to this point all proofs given have been complete, including argument and reasons. In written work it is frequently convenient, however, to have students give the argument only. These two forms will be distinguished by calling the former a complete demonstration and the latter, which is illustrated in Prop. XIX, argument only. It is often a sufficient test of a student's understanding of a theorem to have him state merely the main points involved in a proof. This may be given in enumerated steps, as in Prop. XXXV, or in the form of a paragraph, as in Prop. XLIV. This form will be called outline of proof. BOOK I 61 Proposition XIX. Theorem 172. If two triangles have two sides of one equal respec- tively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second. Given two A ABC and DEF with AB = DE y BO = EF, but /.ABC> /-F. To prove AC > DF. Argument Only 1. Place A DEF on A ABC so that BE shall fall upon its equal AB, D upon A, E upon B. 2. EF will then fall between AB and BC. Denote A DEF in its new position by ABF. 3. Draw BK bisecting Z FBC and meeting AC at K. 4. Draw KF. 5. In A FBK and KBC, BC = BF. 6. BK = BK. 7. Zfbk = Zkbc. 8. .'. A FBK = A KBC. 9. .'.KF=KC. 10. AK+ KF>AF. 11. .'. AK-\-KC>AF. 12. That is, AC> AF. 13. ..'. AC>DF. Q.E.D. 62 PLANE GEOMETRY Ex. 177. (a) Draw a figure and discuss the case for Prop. XIX when F falls on AC; when F falls within triangle ABC. (&) Discuss Prop. XIX, taking AC (the base) = DF, AB=DE, and angle CAB greater than angle D. B ~ Ex. 178. Prove Prop. XIX by using Fig. 1. Hint. In A A CE, A CEA > A BE A. ;. Z. CEA > Z. EAB >ZEAC. FlG . i Ex. 179. Given triangle ABC with AB greater than BC, and let point P be taken on AB and point Q on CB, so that AP — CQ. Prove ^4^ greater than CP. Ex. 180. Would the conclusion of Ex. 179 be true if P were taken on AB pro- B longed and Q on CB prolonged ? Prove. Ex. 181. In quadrilateral ABCD if AB = CD, and angle OZM is greater than angle DAB, prove AC greater than BD. Proposition XX. Theorem (Converse of Prop. XIX) 173. If two triangles have two sides of one equal respec- tively to two sides of the other, but the third side of the first greater than the third side of the second, then the angle opposite the third side of the first is greater than the angls opposite the third side of the second. C D Given A ABC and DEF with AB = DE, BC = EF, but AC > DF. To prove /-B > Z.E. BOOK I 63 Argument 1. ZbZe. 2. First suppose Zb Ze. Q.E.D. REASON8 1. In this case only three sup- positions are admissible. 2. If two A have two sides of one equal respectively to two sides of the other, but the included Z of the first > the included Z of the second, then the third side of the first > the third side of the second. § 172. 3. By hyp., AC > DF. 4. Two A are equal if two sides and the included Z of one are equal respec- tively to two sides and the included Z of the other. § 107. 5. Homol. parts of equal fig- ures are equal. § 110. 6. Same reason as 3. 7. The two suppositions, Zb < ZE and Zb = Ze, have been proved false. Ex. 182. If two opposite sides of a quadrilateral are equal, but its diagonals are unequal, then one angle opposite the* greater diagonal is greater than one angle opposite the less diagonal. Ex. 183. If two sides of a triangle are unequal, the median drawn to the third side makes unequal angles with the third side. Ex. 184. If from the vertex S of an isosceles triangle EST a line is drawn to point P in the base BT so that HP is greater than PZ 7 , then angle BSP is greater than angle PST. Ex. 185. If one angle of a triangle is equal to the sum of the other tw», the triangle can be divided into two isosceles triangles. 64 PLANE GEOMETRY SUMMARY OP THEOREMS FOR PROVING ANGLES UNEQUAL 174. (a) When the angles are in the same triangle : If two sides of a triangle are unequal, the angle opposite the greater side is greater than the angle opposite the less side. (6) When the angles are in different triangles: If two triangles have two sides of one equal respectively to two sides of the other, but the third side of the first greater than the third side of the second, then the angle opposite the third side of the first is greater than the angle opposite the third side of the second. (c) An exterior angle of a triangle is greater than either re- mote interior angle. SUMMARY OF THEOREMS FOR PROVING LINES UNEQUAL 175. (a) When the lines are in the same triangle: The sum of any two sides of a triangle is greater than the third side. Any side of a triangle is less than the sum and greater than the difference of the other two^ If two angles of a triangle are unequal, the side opposite the greater angle is greater than the side opposite the less angle. (b) When the lines are in different triangles: If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second. (c) Every point not in the perpendicular bisector of a line is not equidistant from the ends of the line. The perpendicular is the shortest straight line from a point to a line. Any straight line is less than the sum of the parts of a broken line having the same extremities. In some texts Exs. 168 and 169 are given as theorems. BOOK I 65 176. Note. The student should note that the proof of Prop. XIX depends upon the device of substituting a broken line for an equal straight line. This method is further illustrated by the following exercises. Ex. 186. Prove by the method discussed above that lines drawn to the ends of a line from any point not in its perpendicular bisector are unequal. Ex. 187. Construct an isosceles triangle, given the base and the sum of the altitude and a side. Solution. Let b be the required base and let a + s be the sum of the altitude and a side. Imagine the problem solved, giving A ABC. By marking the given lines and studying the figure, the following procedure will be found to be possible : On an unlimited line CE lay off CB = b ; next draw the J_ bisector of CB and upon it lay off FD = a + s. It is now necessary to break off a part of a + s to form AB. The fact that AB must equal AD should suggest an isosceles A of which BD will be the base and of which it is required to find the vertex. But the _L bisector of the base of an isosceles A passes through the vertex. Therefore, the solution is completed by drawing the ± bisector of BD, which determines A, and by drawing AB and AC. Ex. 188. Construct a triangle, given the base, an adjacent angle, and the sum of the other two sides. Ex. 189. Construct a right triangle, given one arm and the sum of the hypotenuse and the other arm. . \ \ « \ i + I \ / " 8 \ .:?*< t A Jr^ -' >*6ry \ \ * ' "/ \ \ /N'"\ \ /'A\ \\ / F \\ C b B E b a + 8 PARALLEL LINES 177. Def. Two lines are parallel if they lie in the same plane and do not meet however far they are prolonged either way. 178. Assumption 19. Parallel line postulate. Two intersect- ing straight lines cannot both be parallel to the same straight line. 179. The following form of this postulate is sometimes more convenient to quote : Through a given point there exists only one line, parallel to a given line. 66 PLANE GEOMETRY Proposition XXI. Theorem 180. If two straight lines are parallel to a third straight line, they are parallel to each other. D Given lines a and b, each II c. To prove a II b. Argument Reasons 1. a and b are either II or 1. In this case only two sup- not II. positions are admissible. 2. Suppose that a is not II b ; then they will meet at 2. By def. of II lines. § 177. some point as D. ♦ 3 This is impossible. 3. Parallel line post. § 178. 4. .-. a II b. 4. The supposition that a and b are not II has been Q.E.D. proved false. 181. Def. A transversal is a line that intersects two or more other lines. 182. Defs. If two straight lines are cut by a transversal, of the eight angles formed, 3, 4, 5, 6 are interior angles ; 1, 2, 7, 8 are exterior angles ; 4 and 5, 3 and 6, are alternate interior angles ; 1 and 8, 2 and 7, are alternate exterior angles ; 1 and 5, 3 and 7, 2 and 6, 4 and 8, are corresponding angles (called also exterior interior angles) BOOK I 67 Proposition XXII. Theorem 183. If two straight lines are cut by a transversal making a pair of alternate interior angles equal, the lines are parallel. M N >*P Q B< Given two str. lines MN and OQ cut by the transversal AB in points C and D, making Z MOD = ZQDC. To prove MN II OQ. Argument 1. MN and OQ are either II or not II. 2. Suppose that if JV is not II OQ; then they will meet at some point as P, forming, with line DC, A PDC. 3. Then /.MOD > ZQDC. 4. This is impossible. 5. /. MN II OQ. Q.E.D. 3. Reasons 1. In this case only two sup- positions are admissible. 2. By def. of II lines. § 177. If one side of a A is pro- longed, the ext. Z formed > either of the remote int. A. § 153. Z.MCD = ZQDC, by hyp. The supposition that MN and OQ are not II has been proved false. 184. Cor. I. If two straight lines are cut by a trans- versal making a pair of corresponding angles equal, the lines are parallel. Bint. Prove a pair of alt. int. A equal, and apply the theorem. 68 PLANE GEOMETRY 185. Cor. II. If two straight lines are cut by a trans- versal making a pair of alternate exterior angles e-qual, the lines are parallel. (Hint. Prove a pair of alt. int. A equal.) 186. Cor. HI. If two straight lines are cut by a trans- versal making the sum of the two interior angles on the same side of the transversal equal to two right angles, the lines are parallel. ^ Given lines m and n cut by the J m transversal t making Zl + Z2 = 2rt.A jt To prove m || n. Argument 1. Zl + Z2 = 2rt.A 2. Zl + Z3 = 2rt.Zs. 3. .♦. Zl + Z2 = Zl-fZ3. 4. .-. Z2 = Z3. 5. .*. m II n. Q.E.D. Reasons 1. By hyp. 2. If one str. line meets an- other str. line, the sum of the two adj. A is 2 rt. A. § 65. 3. Things equal to the same thing are equal to each other. § 54, 1. 4. If equals are subtracted from equals, the remain- ders are equal. § 54, 3. 5. If two str. lines are cut by a transversal making a pair of alt. int. A equal, the lines are II. § 183. 187. Cor. IV. If two straight lines are perpendicular to a third straight line, they are parallel to each other. Ex. 190. If two straight lines are cut by a transversal making the sum of the two exterior angles on the same side of the transversal equal to two right angles, the lines are parallel. BOOK I 69 Ex. 191. In the annexed diagram, if angle » BGE = angle CHF, are AB and CD parallel ? V Trove. A _^£ R Ex. 192. In the same diagram, if angle HGB Q — *»£ D = angle GHC, prove that the bisectors of these \.tt angles are parallel. Ex. 193. If two straight lines bisect each other, the lines joining their extremities are parallel in pairs. Ex. 194. In the diagram for Ex. 191, if angle BGE and angle FHD are supplementary, prove AB parallel to CD. Ex. 195. If two adjacent angles of any quadrilateral are supplemen- tary, two sides of the quadrilateral will be parallel. Proposition XXIII. Problem 188. Through a given point to construct a line parallel to a given line. A A ¥—Y /R K D Given line AB and point P. To construct, through, point P, a line II AB, I. Construction 1. Draw a line through P cutting AB at some point, as R. 2. With P as vertex and PS as side, construct Z. YPS = ZBRP. §125. 3. XY will be II AB. II. The proof and discussion are left as an exercise for the student. Ex 196. Through a given point construct a parallel to a given line by using : (a) § 183 ; (6) § 185 ; (c) § 187. 70 PLANE GEOMETRY Proposition XXIV. Theorem (Converse of Prop. XXII) 189. If two parallel lines are cut by a transversal, the alternate interior angles are equal. Given || lines AB and CD cut by the transversal EF at points G and H. To prove Z A GH = Z DHG. Argument 1. Either Z A GH = Z DHG, or ZAGHJ= ZDHG. 2. Suppose ZAGH^ZDHG, but that line XY, through G, makes ZXGH= ZDHG. 3. Then XY || CD. 4. But AB || CD. 5. It is impossible that AB and XY both are || CD. 6. .'.ZAGH= ZDHG. Q.E.D. Reasons 1. In this case only two sup- positions are admissible. 2. With a given vertex and a given side, an Z may be constructed equal to a given Z. § 125. 3. If two str. lines are cut by a transversal making a pair of alt. int. A equal, the lines are || . §183. 4. By hyp. 5. Parallel line post. § 178. 6. The supposition that ZAGH^Z DHG has been proved false. BOOK I 71 ^ 190. Cor. I. (Converse of Cor. I of Prop. XXII). // two parallel lines are cut by a transversal, the corresponding angles are equal. A* Given two || lines XY and MN out by the transversal AB, form- ing corresponding A 1 and 2. ^ To prove Z 1 = Z2. Hint. Z 3 = A 2 by § 189. \£ 191. Cor. n. (Converse of Cor. II of Prop. XXII). If two parallel lines are cut by a transversal, the alter- nate exterior angles are equal. 192. Cor. m. (Converse of Cor. Ill of Prop. XXII). If two parallel lines are cut by a transversal, the sum of the two interior angles on the same side of the trans- versal is two right angles. * Given two II lines XY and MN X_ cut by the transversal AB, form- ing int. A 1 and 2. jf To prove Z 1 + Z 2 = 2 rt. A. Argument Only 1. Zl + Z3 = 2rt. A. 2. Z3 = Z2. 3. .-. Z1 + Z2 = 2rt. A Q.E.D. 193. Cor. IV. A straight line perpendicular to one of two parallels is perpendicular to the other also. 194. Cor. V. (Opposite of Cor. Ill of Prop. XXII). If two straight lines are cut by a transversal making the sum of the two interior angles on the same side of the transversal not equal to two right angles, the lines are not parallel. (Hint. Apply § 137, or use the indirect method.) 72 PLANE GEOMETRY 195. Cor. VI. Two lines perpendicular respectively to two intersecting lines also intersect. Given two intersecting lines AB A and CD, and BE _L AB, CF ± CD. To prove that BE and CF also intersect. Argument Only 1. Draw CB. 2. Zfcd is a rt. Z. 3. .-. Zl.< art.Z. 4. Likewise Z2 < a rt. Z. 5. .-. Z1 + Z2 < 2rt. A. 6. .-. BE and CF also intersect. q.e.d. 196. Def. Two or more lines are said to be concurrent if they intersect at a common point. Ex. 197. If two parallels are cut by a transversal so that one of the angles formed is 45°, how many degrees are there in each of the other seven angles ? Ex. 198. If a quadrilateral has two of its sides parallel, two pairs of its angles will be supplementary. „ -, Ex. 199. In the annexed diagram AB is % parallel to CD, . and EF is parallel to OH. Prove (a) angle 1 — angle 2 ; \ \S (&) angle 1 + angle 3 = 2 right angles. F H Ex. 200. If a line is drawn through any point in the bisector of an angle, parallel to one of the sides of the angle, an isosceles triangle will be formed. Ex. 201. Draw a line parallel to the base of a triangle, cutting the sides so that the sum of the segments adjacent to the base shall equal the parallel line. Ex. 202. If two parallel lines are cut by a transversal, the bisectors of a pair of corresponding angles are parallel. BOOK I 73 Ex. 203. State and prove the converse of Ex. 190. Ex. 204. State and prove the converse of Ex. 202. Ex. 205. If a line is drawn through the vertex of- an isosceles triangle parallel to the base, this line bisects the exterior angle at the vertex. Ex. 206. State and prove the converse of Ex. 206. Ex. 207.' If a line joining two parallels is bisected, any other line through the point of bisection and limited by the parallels is bisected. Ex. 208. If through the vertex of one of the acute angles of a right triangle a line is drawn parallel to the opposite side, the line forms with the hypotenuse an angle equal to the other acute angle of the triangle. Ex. 209. The acute angles of a right triangle are complementary. Ex. 210. If two sides of a triangle are prolonged their own lengths through the common vertex, the line joining their ends is parallel to the third side of the triangle. Ex. 211. In an isosceles triangle, if equal segments measured from the vertex are laid off on the arms, the line joining the ends of the seg- ments is parallel to the base of the triangle. (Hint. Draw the bisector of the vertex Z. of the A.) Ex. 212. Extend Ex. 211 to the case where the segments are external to the triangle. SUMMARY OF THEOREMS FOR PROVING LINES PARALLEL 197. (1) If two straight lines are cut by a trans- i versal making the alternate interior angles equal, alternate exterior angles equal, corresponding angles equal, interior angles on one side of the transversal supplementary, the lines are parallel. (2) Two straight lines perpendicular to a third straight line are parallel to each other. (3) Two straight lines parallel to a third straight line are parallel to each other. (4) After parallelograms have been studied, the fact that the opposite sides of a parallelogram are parallel may be used (§ 220). 74 PLANE GEOMETRY Proposition XXV. Theorem 198. Two angles whose sides are parallel, each to each, are either equal or supplementary. Given Zl and the A at E, with AB II EF and A C II DE. To prove Zl Z5 = 2rt.Zs. Z2=Z3, and Zl + Z4 = 2 rt. 4 Zl + Argument 1. Prolong £.4 and Z>J£ until they intersect at some point as H. 2. Z1 = Z6. 3. Z2 = Z6. 4. .-. Z1 = Z2. 5. Z3=Z2. 6 .-. Z1 = Z3. 7. Z2 + Z4 = 2rt.A 8. .-. Z 1 + Z 4 = 2 rt. Zs. Reasons 1. Str.linepost.il. §54,16. 2. Corresponding A of || lines are equal. § 190. 3. Same reason as 2. 4. Things equal to the same thing are equal to each other. § 54, 1. 5. If two str. lines intersect, the vertical A are equal. §77. 6. Same reason as 4. 7.. If one str. line meets an- other str. line, the sum of the two adj. A is 2 rt. A. § 65. 8. Substituting Z 1 for its equal Z 2. BOOK 1 75 Argument 9. Z5=Z4. 10. .-. Zl + Z5 = 2rt.A. Q.E.D. Reasons 9. Same reason as 5. 10. Substituting Z5 for its equal Z4. 199. Note. Every angle viewed from its vertex has a right and a left side ; thus, in the annexed diagram, the right side of ZA is r and the left side, I ; the right side of Z.B is r and the left, I. Fig. 2, Fig. 3. In the diagram of Prop. XXV, Zl and Z2, whose sides are || right to right (ABto EF) and left to left (AC to EK) are equal ; while Zl and Z4, whose sides are || right to left (AB to EF) and left to right {AG to ED) are supplementary. Hence : 200. (a) If two angles have their sides parallel right to right and left to left, they are equal (b) If two angles have their sides parallel right to left and left to right, they are supplementary. Ex. 213. In Fig. 3, above, show which is the right side of each of the four angles about O. Ex. 214. If a quadrilateral has its opposite sides parallel, its opposite angles are equal. Ex. 215. Given two equal angles having a side of one parallel to a side of the other, are the other sides necessarily parallel ? Prove. Ex. 216. If two sides of a triangle are parallel respectively to two homologous sides of an equal triangle, the third side of the first is parallel to the third side of the second. Ex. 217. Construct a triangle, given an angle and its bisector anil th« 'altitude drawn from the vertex of the given angle. 76 PLANE GEOMETRY Proposition XXVI. Theorem 201. Two angles whose sides are perpendicular, each to each, are either equal or supplementary. H K C D Given Zl and the A at P, with OA _L EC and OB J_ HD. To prove Z1=Z2=Z3, Zl+Z4=2rt. Z§,Z1+Z5=2 rt. A. Hint. Draw OE II CK and OFW DH. Prove OE ± OA and OF± OB. Prove Z 7 = Z 1, and prove Z 7 = Z 2. 202. Note. It will be seen that Z\ and Z2, whose sides are ± right to right (OA to PIT) and left to left (OB to PIT), are equal ; while Zl and Z4, whose sides are ± right to left (0.4 to PC) and left to right (OB to P^T), are supplementary. Hence : 203. (a) If two angles have their sides perpendicular right to right and left to left, they are equal. (b) If two angles have their sides perpendicular right to left and left to right, they are supplementary. Ex. 218. If from a point outside of an angle perpendiculars are drawn to the sides of the angle, an angle is formed which is equal to the given angle. Ex. 219. In a right triangle if a perpendicular is drawn from the vertex of the right angle to the hypotenuse, the right angie is divided into two angles which are equal respectively to the acute angles of the triangle. Ex. 220. If from the end of the bisector of the vertex angle of an isosceles triangle a perpendicular is dropped upon one of the arms, the perpendicular forms with the base an angle equal to half the vertex angle. BOOK I 77 Proposition XXVII. Theorem 204. The sum of the angles of any triangle is two right angles. Given A ABC. To prove A A + /.ABC + AC Argument Through B draw BE Al + /ABC+ A 2 = 2 rt. A AC. 3. Zl=Zi. = 2 rt. A. Reasons 1. Parallel line post. §179. 2. The sum of all the ^ about a point on one side of a str. line passing through that point = 2 rt. A. §66. 3. Alt. int. A of II lines are equal. § 189. 4. Same reason as 3. 5. Substituting for A 1 and 2 their equals, z£ ^1 and C, respectively. 205. Cor. I. In a right triangle the two acute angles are complementary. 206. Cor. II. In a triangle there can be but one right angle or one obtuse angle. 207. Cor. in. If two angles of one triangle are equal respectively to two angles of another, then the third angle of the first is equal to the third angle of the second. 208. Cor. IV. If two right triangles have an acute angle of one equal to an acute angle of the other, the othgr acute angles are equal. 4. Z2 = Za 5. .-. Aa + Aabc + Ac = 2 rt. A. Q.E.D. 78 PLANE GEOMETRY 209. Cor. V. Two right triangles are equal if the hy- potenuse and an acute angle of one are equal respectively to the hypotenuse and an acute angle of the other. 210. Cor. VI. Two right triangles are equal if a side and an acute angle of one are equal respectively to a side and the homologous acute angle of the other. 211. Cor. VII. Two right triangles are equal if the hypotenuse and a side of one are equal respectively to the hypotenuse and a side of tJie other. Given rt. A ABC and BEF y with AB = BE and BC=EF. To prove A AB C = A BEF. Hint. Prove DFK a str. line ; then A DEK is isosceles. 212. Cor. VIII. The altitude upon the base of an isosceles triangle bisects the base and also the vertex angle. 213. Cor. IX. Each angle of an equilateral triangle is one third of two right angles, or 60°. 214. Question. Why is the word homologous used in Cor. VI but not in Cor. V ? Ex. 221. If any angle of an isosceles triangle is 60°, what is the value of each of the two remaining angles ? Ex. 222. If the vertex angle of an isosceles triangle is 20°, find the angle included by the bisectors of the base angles. Ex. 223. Find each angle of a triangle if the second angle equals twice the first and the third equals three times the second. Ex. 224. If one angle of a triangle is m° and another angle Z°, write an expression for the third angle. Ex. 225. If the vertex angle of an isosceles triangle is a°, write an expression for each base angle. BOOK I 79 Ex. 226. Construct an angle of 60° ; 120° ; 30° ; 15°. Ex. 227. Construct a right triangle having one of its acute angles 60°. How large is the other acute angle ? Ex. 228. Construct an angle of 160°. Ex. 229. above. Ex. 230. Ex. 231. Ex. 232. Prove Prop. XXVII by using each of the diagrams given Given two angles of a triangle, construct the third angle. Find the sura of the angles of a quadrilateral. The angle between the bisectors of two adjacent angles of a quadrilateral is equal to half the .sum of the two remaining angles. Ex. 233. The angle between the bisectors of the base angles of an isos- celes triangle is equal to the exterior angle formed by prolonging the base. Ex. 234. If two straight lines are cut by a transversal and the bisectors of two interior angles on the same side of the transversal are perpendicular to each other, the lines are parallel. Ex. 235. If in an isosceles triangle each of the base angles is one fourth the angle at the vertex, a line drawn perpendicular to the base at one of its ends and meeting the opposite side prolonged will form with the adjacent side and the exterior portion of the opposite side an equilateral triangle. Ex. 236. Two angles whose sides are perpendicular each to each are either equal or supple- mentary. (Prove by using the annexed diagram.) Ex. 237. If at the ends of the hypotenuse of a right triangle per- pendiculars to the hypotenuse are drawn meeting the other two sides of the triangle prolonged, then the figure contains five triangles which are mutually equiangular. Ex. 238. If one angle of a triangle is 50° and another angle is 70°, find the other interior angle of the triangle ; also the exterior angles of the triangle. What relation is there between an exterior angle and the two remote interior angles of the triangle ? 80 PLANE GEOMETRY Proposition XXVIII. Theorem 215. An exterior angle of a triangle is equal to the sum of the two remote interior angles. B C Given A ABC with Z.DCB an exterior Z. To prove A BOB = A A -f AB. The proof is left as an exercise for the student. Hint. Draw CE II AB. Ex. 239. The bisector of an exterior angle at the vertex of an isosceles triangle is parallel to the base. Ex. 240. If the sura of two exterior angles of a triangle is equal to three right angles, the triangle is a right triangle. Ex. 241. The sum of the three exterior angles of a triangle is four right angles. Ex. 242. What is the sum of the exterior angles of a quadrilateral ? Ex. 243. If the two exterior angles at the base of any triangle are bisected, the angle between these bisectors is equal to half the sum of the interior base angles of the triangle. Ex. 244. If BE bisects angle B of triangle ABC, and AE bisects the exterior angle DA C, angle E is equal to one half angle C. Ex. 245. D is any point in the base BC of isosceles triangle ABC. The side AC is prolonged through C to E so that CE = CD, and DE is drawn meeting AB at F. Prove angle EFA equal to three times angle AEF. Ex. 246. The mid-point of the hypotenuse of a right triangle is equidistant from the three vertices. Prove by laying off on the right angle either acute angle. Z>7 £o) - *(*0) Prc 81 'R0P0SITI0N XXIX. Theorem 216. The sum of all the angles of any polygon is twice as many right angles as the -polygon has sides, less four right angles. c A F Given polygon ABODE • • •, any polygon having n sides. To prove the sum of its A — 2 n rt. A — 4 rt. A. Argument 1. From any point within the polygon such as 0, draw lines to the vertices. 2. There will be formed n A. . 3. The sum of the A of each A thus formed = 2 rt. A. 4. .-. the sum of the A of the n A thus formed =2w rt. A. 5. The sum of all the A about = 4 rt. A, 6. .*. the sum of all the A of the polygon = 2 n it. A - 4 rt. A. Q.E.D. Reasons Straight line post I. § 54, 15. 2. 3. 4. Each side of the polygon will become the base of a A. The sum of the A of any A is 2 rt. A. § 204. If equals are multiplied by equals, the products are equal. § 54, 7 a. 5. The sum of all the A about a point = 4 rt. A. § 67. The sum of all the A of the polygon = the sum of the A of the n A — the sum of all the A about 0. 6. 217. Cor. Each angle of an equiangular polygon of sides is equal to * 4 2(n - 2) right angles. 82 PLANE GEOMETRY Proposition XXX. Theorem 218. If the sides of any polygon are prolonged in succes- sion one way, no two adjacent sides being prolonged through the same vertex, the sum, of the exterior angles thus formed is four right angles. 4/D Given polygon P with Zl, Z2, Z3, Z4, ..-its successive ex- terior angles. To prove Z 1 + Z 2 + Z3 + Z4 + ••• = 4 rt. A. Argument 1. Zl+^6 = 2 r t. 4 Z2 + Z7 = 2 rt. Z, and so on ; i.e. the sum of the int. Z and the ext. Z at one vertex = 2 rt. A, 2. .-. the sum of the int. and ext. zs at the n vertices = 2 n rt. A. 3. Denote the sum of all the interior A by J and the sum of all the ext. A by E\ then E+I=2 n rt. Z. 4. But/=2nrt.Z-4rt.Z. 5. .-..£ = 4 rt. Z; t\e. Zl + Z2 + Z3+Z4+ ... = 4rt.Z. q.e.d. 1. 2. Reasons If one str. line meets an- other str. line, the sum of the two adj. Z is 2 rt. Z. § 65. If equals are multiplied by equals, the products are equal. § 54, 7 a. 3. Arg. 2. 4. The sum of all the Z of any polygon = 2 n rt. A - 4 rt. Z. § 216. 5. If equals are subtracted from equals, the remain- ders are equal. § 54, 3. BOOK I 83 219. Note. The formula 2 n rt. A — 4 rt. A (§ 216) is sometimes more useful iu the form (;i — 2) 2 rt. A. "fix. 247. Find the sum of the angles of a polygon of 7 sides ; of 8 sides ; of 10 sides. Ex. 248. Prove Prop. XXIX by drawing as many diagonals as possible from one vertex. Ex. 249. How many diagonals can be drawn from one vertex in a polygon of 8 sides ? of 50 sides ? of n sides ? Show that the greatest num- ber of diagonals possible in a polygon of n sides (using all vertices) is n(n - 3) , 2 Ex. 250. How many degrees are there in each angle of an equi- angular quadrilateral ? in each angle of an equiangular pentagon ? Ex. 251. How many sides has a polygon the sum of whose angles is 14 right angles ? 20 right angles ? 540° ? Ex. 252. How many sides has a polygon the sum of whose interior angles is double the sum of its exterior angles ? Ex. 253. Is it possible for an exterior angle of an equiangular poly- gon to be 70°? 72°? 140°? 144°? Ex. 254. How many sides has a polygon each of whose exterior angles equals 12° ? Ex. 255. How many sides has a polygon each of whose exterior angles is one eleventh of its adjacent interior angle ? Ex. 256. How many sides has a polygon the sum of whose interior angles is six times the sum of its exterior angles ? Ex. 257. How many sides has an equiangular polygon if the sum of three of its exterior angles is 180° ? Ex. 258. Tell what equiangular polygons can be put together to make a pavement. How many equiangular triangles must be placed with a common vertex to fill the angular magnitude around a point ? 84 PLANE GEOMETRY QUADRILATERALS. PARALLELOGRAMS QUADRILATERALS CLASSIFIED WITH RESPECT TO PARALLELISM 220. Def. A parallelogram is a quadrilateral whose oppo- site sides are parallel. 221. Def. A trapezoid is a quadrilateral having two of its opposite sides parallel and the other two not parallel. 222. Def. A trapezium is a quadrilateral having no two of its sides parallel. PARALLELOGRAMS CLASSIFIED WITH RESPECT TO ANGLES 223. Def. A rectangle is a parallelogram having one right angle. \ It is shown later that all the angles of a rectangle are right angles. 224. Def. A rhomboid is a parallelogram having an oblique angle. It is shown later that all the angles of a rhomboid are oblique. 225. Def. A rectangle having two adjacent sides equal is a square. It is shown later that all the sides of a square are equal. 226. Def. A rhomboid having two adjacent sides equal is a rhombus. It is shown later that all the sides of a rhombus are equal. 227. Def. A trapezoid having its two non-parallel sides equal is an isosceles trapezoid. 228. Def. Any side of a parallelogram may be regarded as its base, and the line drawn perpendicular to the base from any point in the opposite side is then the altitude. 229. Def. The bases of a trapezoid are its parallel sides, and its altitude is a line drawn from any point in one base per- pendicular to the other. BOOK I 85 Proposition XXXI. Theorem 230. Any two opposite angles of a parallelogram are equal, and any two consecutive angles are supplementary. Given O ABCD. To prove : (a) ZA = Zc, and ZB = Zd ; (6) any two consecutive A, as A and B, sup. Argument 1. ZA = Z C and Zb=Zd. 2. A A and B are sup. Q.E.D. Reasons If two A have their sides II right to right and left to left, they are equal. § 200, a. If two II- lines are cut by a transversal, the sum of the two int. A on the same side of the transversal is two rt. A. § 192. 231. Cor. All the angles of a rectangle are right angles, and all the angles of a rhomboid are oblique angles. Ex. 259. If the opposite angles of a quadrilateral are equal, the figure is a parallelogram. Ex. 260. If an angle of one parallelogram is equal to an angle of another, the remaining angles are equal each to each. Ex. 261. The bisectors of the angles of a parallelogram (not a rhom- bus op a square) inclose a rectangle. 86 PLANE GEOMETRY Proposition XXXII. Theorem 232. The opposite sides of a parallelogram are equal. Given O ABCD. To prove AB = CD and BC=AD. The proof is left as an exercise for the student. 233. Cor. I. All the sides of a square are equal, and all the sides of a rhombus are equal. 234. Cor. II. Parallel lines intercepted between the same parallel lines are equal. 235. Cor. III. The perpendiculars drawn to one of two parallel lines from any two points in the other are equal. 236. Cor. rv. A diagonal of a parallelogram divides it into two equal triangles. Ex. 262. The perpendiculars drawn to a diagonal of a parallelogram from the opposite vertices are equal. Ex. 263. The diagonals of a rhombus are perpendicular to each other and so are the diagonals of a square. Ex. 264. The diagonals of a rectangle are equal. Ex. 265. The diagonals of a rhomboid are unequal. Ex. 266. If the diagonals of a parallelogram are equal, the figure is a rectangle. Ex. 267. If the diagonals of a parallelogram are not equal, the figure is a rhomboid. Ex. 268. Draw a line parallel to the base of a triangle so that the portion intercepted between the sides may be equal to a given line. Ex. 269. Explain the statement : Parallel lines are everywhere equi- distant. Has this been proved ? BOOK I 87 Ex. 270. Find the locus of a point that is equidistant from two given parallel lines. Ex. 271. Find the locus of a point: (a) one inch above a given horizontal line ; (&) two inches below the given line. Ex. 272. Find the locus of a point : (a) one inch to the right of a given vertical line ; (6) one inch to the left of the given line. Ex. 273. Given a horizontal line OX and a line OY perpendicular to OX. Find the locus of a point three inches above OX and two inches to the right of OY. 237. Historical Note. Rene" Descartes (1596-1650) was the first to observe the importance of the fact that the position of a point in a plane is determined if its distances, say x and y, from two fixed lines in the plane, perpendic- ular to each other, are known. He showed that geometric fig- ures can be represented by algebraic equations, and de- veloped the subject of analytic geometry, which is known by his name as Cartesian geome- try. Descartes was born near Tours in France, and was sent at eight years of age to the famous Jesuit school at La Fl§che. He was of good fam- ily, and since, at that time, most men of position entered either the church or the army, he chose the latter, and joined the army of the Prince of Orange. One day, while walking in a street in a Holland town, he saw a placard which challenged every one who read it to solve a certain geometric problem. Descartes solved it with little difficulty and is said to have realized then that he had no taste for military life. He soon resigned his commission and spent five years in travel and study. After this he lived a short time in Paris, but soon retired to Holland, where he lived for twenty years, devoting his time to mathematics, philosophy, astronomy, and physics. His work in philosophy was of such importance as to give him the name of the Father of Modern Philosophy. Descartes 88 PLANE GEOMETRY Proposition XXXIII. Theorem 238. The diagonals of a parallelogram bisect each other. Given O A BCD with its diagonals AC and BB intersecting at 0. To prove AO = OC and BO = OB. Hint. Prove A OBC = A OB A. .. AO = OC and BO = OD. Ex. 274. If through the vertices of a triangle lines are drawn parallel to the opposite sides of the triangle, the lines which join the vertices of the triangle thus formed to the opposite vertices of the given triangle are bisected by the sides of the given triangle. Ex. 275. A line terminated by the sides of a parallelogram and passing through the point of intersection of its diagonals is bisected at that point. Ex. 276. How many parallelograms can be constructed having a given base and altitude ? What is the locus of the point of intersection of the diagonals of all these parallelograms ? Ex. 277. If the diagonals of a parallelogram are perpendicular to each other, the figure is a rhombus or a square. Ex. 278. If the diagonals of a parallelogram bisect the angles of the parallelogram, the figure is a rhombus or a square. Ex. 279. Find on one side of a triangle the point from which straight lines drawn parallel to the other two sides, and terminated by those sides, are equal. (See § 232.) Ex. 280. Find the locus of a point at a given distance^ from a given finite line AB. Ex. 281. Find the locus of a point at a given distance from a given line and also equidistant from the ends of another given line. Ex. 282. Construct a parallelogram, given a side, a diagonal, and the altitude upon the given side. BOOK I 89 Proposition XXXIV. Theorem (Converse of Prop. XXXII) 239. If the opposite sides of a quadrilateral are equal, tlie figure is a parallelogram. A D Given quadrilateral ABCD, with AB = CD, and BC=AD. To prove ABCD a O. Argument Keasons 1. Draw the diagonal BD. 1. Str. line post. I. § 54, 15. 2. In A ABD and BCD, AB = CD. 2. By hyp. 3. BC = AD. 3. By hyp. 4. BD = BD. 4. By iden. 5. .'. A ABD = A BCD, 5. Two A are equal if the three sides of one are equal respectively to the three sides of the other. § 116. 6. .: A ABD = ZCDB. 6. Homol. parts, of equal fig- ures are equal. § 110. 7. .'. AB II CD. 7. If two str. lines are cut by a transversal making a pair of alt. int. A equal, the lines are II. § 183. 8. Likewise Z.BDA = /.DBC. 8. Same reason as 6. 9. .'. BC II AD. 9. Same reason as 7. 0. .'. ABCD is a O. q.e.d. 10. By def. of a O. § 220. Ex. 283. Construct a parallelogram, given two adjacent sides and th» ^eluded angle. 90 PLANE GEOMETRY Ex. 284. Construct a rectangle, given the base and the altitude. Ex. 285. Construct a square, given a side. Ex. 286. Through a given point construct a parallel to a given line by means of Prop. XXXIV. Ex. 287. Construct a median of a triangle by means of a parallelo- gram, (1) using §§ 239 and 238 ; (2) using §§ 220 and 238. Ex. 288. An angle of a triangle is right, acute, or obtuse according as the median drawn from its vertex is equal to, greater than, or less than half the side it bisects. Proposition XXXV. Theorem 240. If two opposite sides of a quadrilateral are equal and parallel, the figure is a parallelogram. A D Given quadrilateral ABCD, with BC both equal and II to AD. To prove ABCD a O. Outline of Proof 1. Draw diagonal BD. 2. Prove A BCD — A ABD. 3. Then Zcdb == /.ABD and AB II CD. 4. .'.ABCD is a O. Ex. 289. If the mid-points of two opposite sides of a parallelogram are joined to a pair of opposite vertices, a parallelogram will be formed. Ex. 290. Construct a parallelogram, having given a base, an adjacent angle, and the altitude, making your construction depend upon § 240. Ex. 291. If the perpendiculars to a line from any two points in an- other line are equal, then the lines are parallel. Ex. 292. If two parallelograms have two vertices and a diagonal in common, the lines joining the other four vertices form a parallelogram. BOOK I 91 Proposition XXXVI. Theorem (Converse of Prop. XXXIII) 241. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. B A D Given quadrilateral ABCD with its diagonals AC and BD in- tersecting at so that AO = CO and BO = 1)0. To prove ABCD a O. Argument Only 1. In A OBC and ODA, BO = DO. 2. C0 = AO. 3. Z3 = Z4. 4. .-.A OBC=A OBA. 5. .'. BC=AB. 6. AlsoZl = Z2. 7. .*. BC\\ AD. 8. .-. ABCD is a O. q.e.d. Ex. 293. In parallelogram ABCD, let diagonal AC be prolonged through A and C to X and Y, respectively, making AX = CY. Prove XB YD a parallelogram. Ex. 294. If each half of each diagonal of a parallelogram is bisected, the lines joining the points of bisection form a parallelogram. Ex. 295. Lines drawn from the vertices of two angles of a triangle and terminating in the opposite sides cannot bisect each other. Ex. 296. State four independent hypotheses which would lead to the conclusion, "the quadrilateral is a parallelogram." Ex. 297. Construct a parallelogram, given its diagonals and an angle between them. 92 PLANE GEOMETRY Proposition XXXVII. Theorem 242. Two parallelograms are equal if two sides and the included angle of one are equal respectively to two sides and the included angle of the other. II Given ZU I and II with AB = EF, AD = EH, and Z.A= /.E. To prove OI = OII. 7. 8. Argument Place O I uponO II so that AB shall fall upon its equal EF, A upon E, B upon F. Then AD will become col- linear with EH. Point D will fall on H. Now DC II AB, and HG II EF. .-. DC and HG are both II AB. .: DC will become collinear with HG, and C will fall somewhere on HG or its prolongation. Likewise BC will become collinear with FG, and C will fall somewhere on FG or its prolongation. .*. point C must fall on point G. 9. .-.01 = 011. Q.E.D. Reasons 1. Transference post. §54,14. 2. /.A = ZE, by hyp. 3. AD = EH, by hyp. 4. By def . of a O. § 220. 5. AB and ^i* coincide, Arg. 1. 6. Parallel line post. §179. 7. By steps similar to 4, 5, and 6. 8. Two intersecting str. lines can have only one point in common. § 26. 9. By def. of equal figures. § 18. BOOK I 93 243. Quadrilaterals QUADRILATERALS CLASSIFIED 1. Opposite sides II : Parallelogram. (a) Right-angled: Rectangle. Two adj. sides equal: Square. (b) Oblique-angled: Khomboid. Two adj. sides equal : Rhombus. 2. Two sides II, other two non-ll : Trap- ezoid. (a) Two non-ll sides equal : Isosceles trapezoid. 3. No two sides II : Trapezium. Ex. 298. If it is required to prove a given quadrilateral a rectangle, show by reference to § 243 that the logical steps are to prove first that it is a parallelogram ; then that it has one right angle. Ex. 299. If a given quadrilateral is to be proved a square, show that the only additional step after those in Ex. 298 is to prove two adjacent sides equal. Ex. 300. If a given quadrilateral is to be proved a rhombus, what are the three steps corresponding to those given in Ex. 298 and Ex. 299 ? Ex. 301. Since rectangles and rhomboids are parallelograms, they possess all the general properties of parallelograms. What property dif- ferentiates rectangles from rhomboids (1) by definition ? (2) by proof ? (See Ex. 266 and Ex. 267.) Ex. 302. (a) What two properties that have been proved distinguish squares from other rectangles ? (See Ex. 277 and Ex. 278.) (b) What two properties that have been proved distinguish rhombuses from other rhomboids? (See Ex. 277 and Ex. 278.) (c) Show that the two properties which distinguish squares and rhom- buses from the other members of their class are due to the common prop- erty possessed by squares and rhombuses by definition. Ex. 303. The mid-point of the hypotenuse of a right triangle is equi- distant from the three vertices. Ex. 304. If a line AB of given length is moved so that its ends always touch the sides of a given right angle, what is the locus of the mid-point of %B? 94 PLANE GEOMETRY Proposition XXXVIII. Theorem 244. If three or more parallel lines intercept equal seg- ments on one transversal, they intercept equal segments on any other transversal. Given II lines AG, BH, CJ, DK, etc., which intercept the equal segments AB, BC, CD, etc., on transversal AF, and which intercept segments GH, HJ, JK, etc., on transversal GL. To prove GH = HJ = JK, etc. Argument 1. Draw GM, HN, JR, etc. II AF. 2. Now AG MB, BHNC, CJRD, etc., are HJ. 3. .'.GM = AB, HN = BC, JR = CD, etc. 4. And AB = BC = CD, etc. 5. .-. GM = HN = JR, etc. 6. Again GM, HN, JR, etc., are II to each other. 7. .-. Zl = Z2 = Z3, etc. 8. And Z4 = Z5 = Z6, etc. Reasons 1. Parallel line post. § 179. 2. By def. of a O. § 220. 7. The opposite sides of a O are equal. § 232. By hyp. Ax. 1. § 54, 1. If two str. lines are II to a third str. line, they are II to each other. § 180. Corresponding A of II lines are equal. § 190. If two A have their sides II right to right and left to left, they are equal. § 200, a. BOOK I 95 9. .-. A GHM = A HJN = AJKE, etc. 10. .'. GH= HJ = JK, etc. Q.E.D. 9. Two A are equal if a side and the two adj. A of one are equal respectively to a side and the two adj. A of the other. § 105. 10. Homol. parts of equal figures are equal. § 110. 245. Question. Are the segments that the parallels intercept on one transversal equal to the segments that they intercept on another trans- versal ? Illustrate. 246. Cor. I. The line bisecting one of the non-paral- lel sides of a trapezoid and parallel to the bases bisects the other of the non-parallel sides also. 247. Cor. n. The line joining the mid-points of the non-parallel sides of a trapezoid is (a) parallel to the bases; and (b) equal to one half their sum. Hint, (a) Prove EF WBC and AD by the indirect method. (6) Draw GH II CD. Prove AH=GB; then prove EF = GC = % (GC + HD) = KBC+AD). 248. Cor. m. The line bisect- ing one side of a triangle and parallel to another side bisects the third side. G B c V 3/ \, A \ \ H 249. Cor. IV. The line joining the mid-points of two sides of a triangle is parallel to the third side and equal to one half the third side. . Hint. Draw CF II BA. Prove CF=AE = EB. 96 PLANE GEOMETRY Proposition XXXIX. Problem 250. To divide a given straight line into any number of equal parts. S \! A"' V \ \ KS •'' \L \ s* \ \ \ \ \ A H B Given straight line AB. To divide AB into n equal parts. I. Construction 1. Draw the unlimited line AX. 2. Take any convenient segment, as AR, and, beginning at A, lay it off n times on AX.' 3. Connect the nth point of division, as K, with B, 4. Through the preceding point of division, as P, draw a line PH II KB. § 188. 5. Then HB is one nth of AB. 6. .-. HB, if laid off successively on AB, will divide AB into n equal parts. II. The proof and discussion are left as an exercise for the student. Ex. 305. Divide a straight line into 7 equal parts. Ex. 306. Construct an equilateral triangle, having given the perim- eter. Ex. 307. Construct a square, having given the perimeter. 251. Def. The line joining the mid-points of the non- parallel sides of a trapezoid is called the median of the trape- zoid. Ex. 308. Show, by generalizing, that Cor. Ill, Prop. XXXVIII, may be obtained from Cor. I -and Cor. IV from Cor. II. BOOK I 97 Ex. 309. The lines joining the mid-points of the sides of a quadri- lateral taken in order form a parallelogram. •Rx, 310. What additional statement can you make if the quadri-' lateral in Ex. 309 is an isosceles trapezoid ? a rectangle ? a rhombus ? a square ? Ex. 311. lines drawn from the mid-point of the base of an isosceles triangle to the mid-points of its equal sides form a rhombus or a square. When is the figure a rhombus ? when a square ? Ex. 312. The mid-points of the sides of a quadrilateral and the mid- points of its two diagonals are the vertices of three parallelograms whose diagonals are concurrent. Ex. 313. What is the perimeter of each parallelogram in Ex. 312 ? Ex. 314. Construct a triangle, given the mid-points of its sides. Ex. 315. Through a given point within an angle construct a line, limited by the two sides of the angle, and bisected at the given point. Ex. 316. Eveiy diagonal of a parallelogram is trisected by the lines joining the other two vertices with the mid-points of the opposite sides. Ex. 317. If a triangle inscribed in another triangle has its sides parallel respectively to the sides of the latter, its vertices are the mid- points of the sides of the latter. Ex. 318. If the lower base KT of trapezoid BSTK is double the upper base BS, and the diagonals intersect at O, prove OK double OS, and Or double OB. Ex. 319. Construct a trapezoid, given the two bases, one diagonal, and one of the non-parallel sides. In the two following exercises prove the properties which require proof, state those which follow by definition, and those which have been proved in the text : Ex. 320. Properties possessed by all trapezoids : (a) Two sides of a trapezoid are parallel. (b) The two angles adjacent to either of the non-parallel sides are supplementary. (c) The median of a trapezoid is parallel to the bases and equal to one half their sum. Ex. 321. In an isosceles trapezoid : (a) The two non-parallel sides are equal. (b) The angles at each base are equal and the opposite angles are supplementary. * (.c) The diagonals are equal. 98 PLANE GEOMETRY Proposition XL. Theorem 252. The two perpendiculars to the sides of an angle from any point in its bisector are equal. B D A Given Z ABC', P any point in BR, the bisector of ZABC; PD and PE, the is from P to BA and BC respectively. To prove PD = PE. Argument Only 1. In rt. A DBP and PBE, PB = PB. 2. Z DBP — Z PBE. 3. .-. ADBP = APBE. 4. .'. PD =PE. Q.E.D. 253. Prop. XL may be stated as follows : Every point in the bisector of an angle is equidistant from the sides of the angle. Ex. 322. Find a point in one side of a triangle which is equidistant from the other two sides of the triangle. Ex. 323. Find a point equidistant from two given intersecting lines and also at a given distance from a fixed third line. Ex. 324. Find a point equidistant from two given intersecting lines and also equidistant from two given parallel lines. Ex. 325. Find a point equidistant from the four sides of a rhombus. Ex. 326. The two altitudes of a'rhombus are equal. Prove. Ex. 327. Construct the locus of the center of a circle of given radius, which rolls so that it always touches the sides of a given angle. Do not prove, BOOK I 99 Proposition XLI. Theorem (Opposite of Prop. XL) 254. The two perpendiculars to the sides of an angle from any point not in its bisector are unequal. C B G D A Given /.ABC; P any point not in BR, the bisector of /ABC; PD and PE, J§ from P to BA and BC respectively. To prove PD =£ PE. Outline of Proof Draw FG J_ BA ; draw PG. Then FE = FG. Now PF-\-FG> PG. .'.PE>PG. But PG > PD. .'. PE > PD. 255. Prop. XLI may be stated as follows : Every point not in the bisector of an angle is not equidistant from the sides of the angle. 256. Cor. I. (Converse of Prop. XL). Every point equi- distant from the sides of an angle lies in the bisector of the angle. Hint. Prove directly, using the figure for § 262, or apply § 137. 257. Cor. II. The bisector of an angle is the locus of all points equidistant from the sides of the angle. Ex. 328. What is the locus of all points that are equidistant from a pair of intersecting lines ? 100 PLANE GEOMETRY .CONCURRENT LINE THEOREMS Proposition XLII. Theorem 258. The bisectors of the angles of a triangle are con- current in a point which is equidistant from the three sides of the triangle. Given A AB O with AF, BE, CD the bisectors of A A, B, and C respectively. To prove: (a) AF, BE, CD concurrent in some point as 0; (b) the point equidistant from AB, BC, and CA. 1. Argument BE and CD will intersect at some point as 0. 2. 3. Draw OL, OH, and OG, J§ from to AB, BC, and CA respectively. V is in BE, OL = OH. is in CD, OG=OH. OL = OG; Reasons 1. If two str. lines are cut by a transversal making the sum of the two int. A on the same side of the transversal not equal to 2 rt. A, the lines are not II. § 194. 2. From a point outside a line there exists one and only one J_ to the line. § 155. 3. The two Js to the sides of an Z from any point in its bisector are equal. §252. 4. Same reason as 3. 5. Things equal to the same thing are equal to each other. § 54 ; 1. book i aoi Argument 6. .* . AF, the bisector of Z.CAB, passes through O. 7. .-. AF, BE, and CD are con- current in 0. 8. Also is equidistant from AB, BC, and CA. q.e.d. Reasons 6. Every point equidistant from the sides of an Z lies in the bisector of the Z. § 256. 7. By def . of concurrent lines. §196. 8. By proof, OL = OH= OG. 259. Cor. The point of intersection of the bisectors of the three angles of a triangle is the locus of all points equidistant from the three sides of the triangle. Ex. 329. Is it always possible to find a point equidistant from three given straight lines ? from four given straight lines ? Ex. 330. Find a point such that the perpendiculars from it to three sides of a quadrilateral shall be equal. (Give geometric construction.) Ex. 331. Prove that if the sides AB and AG of a triangle ABC are prolonged to E and F, respectively, the bisectors of the three angles BAC, EBC, and BCF all pass through a point which is equally distant from the three lines AE, AF, and BC. Is any other point in the bisector of the angle BAC equally distant from these three lines? Give reason for your answer. Ex. 332. Through a given point P draw a straight line such that perpendiculars to it from two fixed points Q and B shall cut off on it equal segments from P. (Hint. See § 246.) Ex. 333. Construct the locus of the center of a circle of given radius, which rolls so that it always touches the sides of a given triangle. Do not prove. Ex. 334. Find the locus of a point in one side of a parallelogram and equidistant from two other sides. In what parallelograms is this locus a vertex of the parallelogram ? Ex. 335. Find the locus of a point in one side of a parallelogram and equidistant from two of the vertices of the parallelogram. In what class of parallelograms is this locus a vertex of the parallelogram ? Ex. 336. Construct the locus of the center of a circle of given radius which rolls so that it constantly touches a given circumference. Do not prove. PLANE GEOMETRY Proposition XLIII. Theorem 260. The perpendicular bisectors of the sides of a tri- angle are concurrent in a point which is equidistant from the three vertices of the triangle. Given A ABC with FG, HK, ED, the _L bisectors of AB, BC, CA. To prove : (a) FG, HK, ED concurrent in some point as ; (&) the point equidistant from A, B, and C. Argument 1. FG and ED will intersect at some point as 0. 2. Draw OA, OB, and OC. 3. v is in FG, the _L bisector of AB, OB = OA; and v is in DE, the _L bi- sector of CA, 0C= OA. 4. .-. OB = OC. 5. .-. HK, the _L bisector of BC, passes through 0. 6. .*. FG, HK, and ED are con- current in 0. J. Also is equidistant from A, B, and C. q.e.d. Reasons 1. Two lines X respectively to two intersecting lines also intersect. § 195. 2. Str. line post. I. § 54, 15. 3. Every point in the _L bi- sector of a line is equi- distant from the ends of that line. § 134. 4 Ax. 1. § 54, 1. 5. Every point equidistant from the ends of a line lies in the _L bisector of that line. § 139. 6. By def . of concurrent lines §196. 7o By proof, OA=OB = OC. BOOK I 103 261. Cor. The point of intersection of the perpendic- ular bisectors of the three sides of a triangle is the locus of all points equidistant from the three vertices of the triangle. Ex. 337. Is it always possible to find a point equidistant from three given points ? from four given points ? Ex. 338. Construct the perpendicular bisectors of two sides of an acute triangle, and then construct a circle whose circumference shall pass through the vertices of the triangle. Ex. 339. Construct a circle whose circumference shall pass through the vertices of a right triangle. Ex. 340. Construct a circle whose circumference shall pass through the vertices of an obtuse triangle. Proposition XLIV. Theorem 262. The altitudes of a triangle are concurrent. Given A ABC with its altitudes AD, BE, and CF, To prove AD, BE, and CF concurrent. Outline of Proof Through the vertices A, B, and C, of triangle ABC, draw lines || BC, AC, and AB, respectively. Then prove, by means of /I7, that AD, BE, and CF are the _L bisectors, respectively, of the sides of the auxiliary A HKL. Then, by Prop. XLIII, AD, BE^sad CF are concurrent. q.e.d. 104 PLANE GEOMETRY Proposition XLV. Theorem 263. Any two medians of a triangle intersect each other in a trisection point of each. Given A ABC with AD and CE any two of its medians. To prove that AD and CE intersect in a point O such that OD = \AD and OE = i CE. Outline of Proof 1. AD and CE will intersect at some point as 0. § 194. 2. Let R and S be the mid-points of AO and CO respectively. 3. Quadrilateral REDS is a O. 4. .-. AR= RO= OD and CS = SO = OE. 5. That is, OD = ± AD and 0^ = J- CB. q.e.d. 264. Cor. T/^e £&ree medians of a triangle are con- current. 265. Def. The point of intersection of the medians of a triangle is called the median center of the triangle. It is also called the centroid of the triangle. This point is the center of mass or center of gravity of the triangle. Ex. 341. Draw a triangle whose altitudes will intersect on one of its sides, and repeat the proof for Prop. XLIV. Ex. 342. Draw a triangle whose altitudes will intersect outside of the triangle, and repeat the proof for Prop. XLIV. Ex. 343. Prove Prop. XLV by prolonging OD its own length and drawing lines to B and C from the end of the prolongation. Ex. 344. Construct a triangle, given two of its medians and the angle between them. BOOK I 105 Proposition XLVI. Theorem 266. If one acute angle of a right triangle is double the other, the hypotenuse is double the shorter side Given rt. A ABC with Z A double Z B. To prove AB — 2 AC. Argument Only 1. Draw CE, making Z 1 = A A. 2. ZA = 60°. 3. .'. Zl=60°. 4. .\Z AEC = 60°. 5. .'. AE= AC= EC. 6. In A EBC, Z2 = 30°. 7. Z B = 30°. 8. .*. EB =EC. 9. ,\EB = AE=AC. 10. .'. AB = 2 AC. Q.E.D. 267. Prop. XLVI is sometimes stated : In a thirty-sixty de- gree right triangle, the hypotenuse is double the shorter side. 268. Historical Note. Such a triangle (a 30°-60° right triangle) is spoken of by Plato as "the most beautiful right-angled scalene triangle." Ejc. 345. Prove Prop. XLVI by drawing CE = CA. Ex. 346. Prove Prop. XLVI by drawing lines through the ends of the, hypotenuse parallel to the other two sides, thus forming a rectangle. 106 PLANE GEOMETRY CONSTRUCTION OF TRIANGLES 269. A triangle is determined, in general, when three parts are given, provided that at least one of the given parts is ar line. The three sides and the three angles of a triangle are called its parts ; but there are also many indirect parts ; as the three medians, the three altitudes, the three bisectors, and the parts into which both the sides and the angles are divided by these lines. A A 270. The notation given in the annexed figures may be used for brevity : A, B, C, the angles of the triangle ; in a right triangle, angle C is the right angle. a, b, c, the sides of the triangle; in a right triangle, c is the hypotenuse. m ai m b , m c , the medians to a, b, and c respectively. h a , h bi h c , the altitudes to a, b, arid c respectively. t a , t b , t c , the bisectors of A, B, and C respectively. l u , d a , the segments of a made by the altitude to a, s a , r a , the segments of a made by the bisector of angle A. 271. The student should review the chief cases of construc- tion of triangles already given : viz. a, B, c ; A, b, C ; a,b,c: right triangles : a, b ; b, c ; b, B ; b, A ; c, A ; review also § 152. Ex. 347. State in words the first eight cases given in § 271. BOOK I 107 Proposition XLVII. Problem 272. To construct a triangle having two of its sides equal respectively to two given lines, and the angle oppo- site one of these lines equal to a given angle. Y B Fig. 1. Given lines a and b, and Z.B. To construct A ABC. I. Construction 1. Draw any line, as BX. 2. At any point in BX, as B, construct Z XBY = to the given A B. § 125. 3. On .BFlay off BC = a. 4. With C as center and with b as radius, describe an arc cutting BX at A. 5. A ABC is the required A. II. The proof is left as an exercise for the student. III. Discussion (1) b may be greater than a; (2) b may equal a ; (3) b may be less than a. (1) If b > a, there will be one solution, i.e. one A and only one can be constructed which shall contain the given parts. This case is shown in Fig. 1. (2) If b = a, the A will be isosceles. The construction will be the same as for case (1). 108 PLANE GEOMETRY A B (3) If b < a. Make the construction as for case (1). If b > the _L from C to BX, there will be two solutions, since A ABC and BBC, Fig. 2, both contain the required parts. If b equals the _L from Cto BX there will be one solution. The A will be a rt. A. If b < the _L from C to BX, there will be no solution. In the cases thus far considered, the given Z was acute. The discussion of the cases in which the given Z is a rt. Z and in which it is an obtuse Z is left to the student. 273. Question. Why is (1) the only case possible when the given angle is either right or obtuse ? 274. The following exercises are given to j I illustrate analysis of problems and to show the use of auxiliary triangles in constructions. Ex. 348. Construct a triangle, given a, h a , m a . Analysis. Imagine the problem solved as in Fig. 1, and mark the given parts with heavy lines. The triangle AHM is determined and may be made the basis of the construction. Ex. 349. Construct a triangle, given 6, m a , m c Analysis. From Fig. 2 it will be seen that tri- angle AOC may be constructed. Its three sides are known, since AO — \m a and CO — f m c . Ex. 350. Construct a triangle, given a, h a , h c . Analysis. In Fig. 3, right triangle CHB is determined. The locus of vertex A is a line parallel to CB, so that the distance between it and CB is equal to hw BOOK I 109 Ex. 351. Construct atriangle, given b ,c, m a . Analysis. Triangle ABK, Fig. 4, is deter- mined by three sides, b, c, and 2ra a . Since ABKC is a parallelogram, AK bisects CB. Ex. 352.. Construct a triangle, given a, »n a , \ / ^"' and the angle between m a and a. *£ Analysis. Triangle AM G is determined, as shown in Fig. 5. A ~?B K Fig. 4. rr~v M d S « N Fig. 6. Ex. 353. Construct a trapezoid, given its four sides. Analysis. Triangle EST, Fig. 6, is determined. Ex. 354. Construct a parallelogram, given q jj the perimeter, one base angle, and the altitude. 2£37 K M Construction. The two parallels, CH and AJE, Fig. 7, may be drawn so that the distance A B- E between them equals the altitude ; at any point FlG - •• B construct angle EBC equal to the given R M base angle ; draw CA, bisecting angle FOB ; measure AE equal to half the given perimeter ; complete parallelogram CHEB. f S N Ex. 355. Construct a trapezoid, given the Fig. 8. two non-parallel sides and the difference between the bases. (See Fig. 8.) Construct a triangle, given : Ex. 356. A, h a , t a . Ex. 360. B, h a b. Ex. 357. h a} l a , d a . Ex. 361. a«, t a , h a . Ex. 358. h a , m a , l a . Ex. 362. a, m b , C. Ex. 359. a, U, C. Ex. 363. a, U, B. Construct an isosceles trapezoid, given : Ex. 364. The two bases and the altitude. Ex. 365. One base, the altitude, and a diagonal. Ex. 366. A base, a diagonal, and the angle between them. 110 PLANE GEOMETRY DIRECTIONS FOR THE SOLUTION OF EXERCISES 275. I. Make the figures clear, neat, accurate, and general in character. II. Fix firmly in mind hypothesis and conclusion with ref- erence to the given figure. III. Recall fundamental propositions related to the propo- sition in question. \ IV. If you can find no theorem which helps you, contradict the conclusion in every possible way {reductio ad absurdum) and try to show the absurdity of the contradiction. V. Make frequent use of the method of analysis, which con- sists in assuming the proposition proved, seeing what results follow until a known truth is reached, and then retracing the steps taken. VI. If it is required to find a point which fulfills two conditions, it is often convenient to find the point by the Intersection of Loci. By finding the locus of a point which satisfies each condition separately, it is possible to find the points in which the two loci intersect; i.e. the points which satisfy both conditions at the same time. VII. See § 152 and exercises following § 274 for method of attacking problems of construction. The method just described under V is a shifting of an uncertain issue to a certain one. It is sometimes called the Method of Successive /Substitutions. It may be illustrated thus : 1. A is true if .B.is true. 3. But C is true. 2. B is true if C is true. 4. .-. A is true. This is also called the Analytic Method of proof. The proofs of the theorems are put in what is called the Synthetic form. But these were first thought through analytically, then rearranged in the form in which we find them. BOOK I 111 MISCELLANEOUS EXERCISES Ex. 367. The perpendiculars drawn from the extremities of one side of a triangle to the median upon that side are equal. Ex. 368. Construct an angle of 75° ; of 97^°. Ex. 369. Upon a given line find a point such that perpendicufars from it to the sides of an angle shall be equal. Ex. 370. Construct a triangle, given its perimeter and two of its angles. Ex. 371. Construct a parallelogram, given the base, one base angle, and the bisector of the base angle. Ex. 372. Given two lines that would meet if sufficiently prolonged. Construct the bisector of their angle, without prolonging the lines. Ex. 373. Construct a triangle, having given one angle, one adjacent side, and the difference of the other two sides. Case 1 : The side oppo- site the given angle less than the other unknown side. Case 2 : The side opposite the given angle greater than the other unknown side. Ex. 374. The difference between two adjacent angles of a parallelo- gram is 90° ; find all the angles. Ex. 375. A straight railway passes 2 miles from a certain town. A place is described as 4 miles from the town and 1 mile from the railway. Represent the town by a point and find by construction how many places answer the description. Ex. 376. Describe a circle through two given points which lie out- side a given line, the center of the circle to be in that line. Show when no solution is possible. Ex. 377. Construct a right triangle, given the hypotenuse and the difference of the other two sides. Ex. 378. If two sides of a triangle are unequal, the median through their intersection makes the greater angle with the lesser side. Ex. 379. Two trapezoids are equal if their sides taken in order are equal, each to each. Ex. 380. Construct a right triangle, having given its perimeter and an acute angle. Ex. 381. Draw a line such that its segment intercepted between two given indefinite lines shall be equal and parallel to a given finite line. Ex. 382. One angle of a parallelogram is given in position and the point of intersection of the diagonals is given ; construct the parallelo- gram. 112 PLANE GEOMETRY Ex. 383. third side. Construct a triangle, given two sides and the median to the Ex. 384. If from any point within a triangle lines are drawn to the three vertices of the triangle, the sum of these lines is less than the sum of the, sides of the triangle, and greater than half their sum. Ex. 385. Repeat the proof of Prop. XIX for two cases at once, using Figs. 1 and 2. Ex. 386. If the angle at the vertex of an isosceles triangle is four times each base angle, the perpendicular to the base at one end of the base forms Fig. 1. Fig. 2. with one side of the triangle, and the prolongation of the other side through the vertex, an equilateral triangle. Ex. 387. The bisector of the angle G of a triangle ABC meets AB in D, and DE is drawn parallel to AG meeting BO in E and the bisector of the exterior angle at G in F. Prove DE = EF. Ex. 388. Define a locus. Find the locus of the mid-points of all the lines drawn from a given point to a given line not passing through the point. Construct an isosceles trapezoid, given the bases and one Construct a square, given the sum of a diagonal and one Ex. 389. angle. Ex. 390. side, Ex. 391. The difference of the distances from any point in the base prolonged of an isosceles triangle to the equal sides of the triangle is constant. Ex. 392. Find a point X equidistant from two intersecting lines and at a given distance from a given point. Ex. 393. When two lines are met by a transversal, the difference of two corresponding angles is equal to the angle between the two lines. % Construct a triangle, given •, Ex. 394. A, h a , l a . Ex. 398. Ex. 395. A, t a , s a . Ex. 399. Ex. 396. a, h a , l a . Ex. 400. Ex. 397. a, 6 + c, A Ex. 401. a, m. m c , h CJ B. 6, c, A, B, b -TcT BOOK II THE CIRCLE 276. Def. A circle is a plane closed figure whose boundary is a curve such that all straight lines to it from a fixed point within are equal. 277. Def. The curve which forms the boundary of a circle is called the circumference. Q \R 278. Def. The fixed point within is called the center, and a line joining the center to any point on the circum- ference is called a radius, as QR. 279. From the above definitions and of equal figures, § 18, it follows that : (a) All radii of the same circle are equal. (b) All radii of equal circles are equal. (c) All circles having equal radii are equal 280. Def. Any portion of a circumference is called an arc, as DF, FC, etc. 281. Def. A chord is any having its ex- the circumfer- i 282. Def. A diameter is a chord which passes through the center, as BG. 283. All U3 from the definition straight line tremities on ence, as DF. \. ^incB^, ny diameter is twice a radius, it follows that of a circle are equal. 114 PLANE GEOMETRY Proposition I. Theorem 284. Every diameter of a circle bisects the circumfer- ence and the circle. M Given circle AMBN with center O, and AB, any diameter. To prove: (a) that AB bisects circumference AMBN) (b) that AB bisects circle AMBN. Argument Turn figure AMB on AB as an axis until it falls upon the plane of ANB. Arc AMB will coincide with arc ANB. .-. arc AMB = arc ANB ; i.e. AB bisects circumference AMBN. 4. Also figure AMB will coincide with figure ANB. 5. .'.figure AMB = figure ANB', i.e. AB bisects circle AMBN. Q.E.D. Reasons 1. § 54, 14. 2. § 279, a. 3. § 18. 4. § 279, a. 5. § 18. Ex. 402. A semicircle is described upon each of the diagonals of a rectangle as diameters. Prove the semicircles equal. Ex. 403. Two diameters perpendicular to each other divide a circum- ference into four equal arcs. Prove by superposition. Ex. 404. Construct a circle which shall pass through two given points. Ex. 405. Construct a circle having a given radius r, and passing through two given points A and B BOOK II 115 285. Def. A secant of a cir- cle is a straight line which cuts the circumference in two points, but is not terminated by the circumference, as MN. 286. Def. A straight line is tangent to, or touches, a cir- cle if, however far prolonged, it meets the circumference in but one point. This point is called the point of tangency. HK is tangent to circle at point T, and T is the point of tangency. 287. Def. A sector of a circle is a plane closed figure whose boundary is composed of two radii and their intercepted arc, as sector SOB. 288. Def. A segment of a circle is a plane closed figure whose boundary is composed of an arc and the chord joining its extremi- ties, as segment DCE. 289. Def. A segment which is one half of a circle is called a semi- circle, as segment AMB. 290. Def. Ad arc which is half of a circumference is called a semicircumference, as arc AMB. 291. Def. An arc greater than a semicircumference is called a major arc, as arc DME\ an arc less than a semicir- cumference is called a minor arc, as arc DCE. 292. Def. A central angle, or angle at the center, is an angle whose vertex is at the center of a circle and whose sides are radii. Ex. 406. The line joining the centers of two circles is 6, the radii are 8 and 10, respectively. What are the relative positions of the two circles ? ..Ex. 407. A circle can have only one center. • 116 PLANE GEOMETRY Proposition II. Theorem 293. In equal circles, or in the same circle, if two cen- tral angles are equal, they intercept equal arcs on the cir- cumference; conversely, if two arcs are equal, the central angles that intercept them are equal. I. Given equal circles ABM and CDN, and equal central A G and Q, intercepting arcs AB and CD, respectively. To prove £b = CD. Argument 1. Place circle ABM upon circle CDN so that center O shall fall upon center Q, and QA shall be collinear with QC. 2. A will fall upon C. 3. OB will become collinear with QD. 4. .-. B will fall upon D. 5. .-. AB will coincide with CD. 6. .'. AB = CD. Q.E.D. Reasons 1. §54,14. 2. § 279, b. 3. By hyp. 4. § 279, 6. 5. § 279, b. 6. §18. II. Conversely : Given equal circles ABM and CDN, and equal arcs AB and CD, intercepted by A and Q, respectively. To prove Zo = ZQ. BOOK II 117 2. Argument Place circle ABM upon circle CDN so that center shall fall upon center Q. Kotate circle ABM upon O as a pivot until AB falls upon its equal CD, A upon C, B upon D. OA will coincide with QCand OJ? with QD. .-. Z = Z Q. Q.E.D. Reasons 1. §54,14. 2. §54,14. 3. §17. 4. §18. 294. Cor. In equal circles, or in the sams circle, if two central angles are unequal, the greater angle inter- cepts the greater arc; conversely, if two arcs are unequal, the central angle that intercepts the greater arc is the greater. (Hint. Lay off the smaller central angle upon the greater.) 295. Def. A fourth part of a circumference is called a quadrant. From Prop. II it is evident that a right angle at the center intercepts a quadrant on the cir- cumference. Thus, two ± diam- eters AB and CD divide the circumference into four quadrants, AC, CB, BD, and DA. 296. Def. A degree of arc, or an arc degree, is the arc intercepted by a central angle of one degree. 297. A right angle contains ninety angle degrees (§ 71) ; therefore, since equal central angles intercept equal arcs on the circumference, a quadrant contains ninety arc degrees. Again, four right angles contain 360 angle degrees, and four right angles at the center of a circle intercept a complete cir- cumference ; therefore, a circumference contains 360 arc degrees. Hence, a semicircumference contains 180 arc degrees. Ex. 408. Divide a given circumference into eight equal arcs ; sixteen equaJ .arcs. 118 PLANE GEOMETRY Ex. 409. Divide a given circumference into six equal arcs ; three equal arcs ; twelve equal arcs. Ex. 410. A diameter and a secant perpendicular to it divide a cir- cumference into two pairs of equal arcs. Ex. 411. Construct a circle which shall pass through two given points A and B and shall have its center in a given line c. Ex. 412. If a diameter and another chord are drawn from a point in a circumference, the arc intercepted by the angle between them will be bisected by a diameter drawn parallel to the chord. Ex. 413. If a diameter and another chord are drawn from a point in a circumference, the diameter which bisects their intercepted arc will be parallel to the chord. Proposition III. Theorem 298. In equal circles, or in the same circle, if two chords are equal, they subtend equal arcs; conversely, if two arcs are equal, the chords that subtend them are equal. I. Given equal circles and Q, with equal chords AB and CD To prove AB = CD. Argument 1. Draw radii OA, OB, QC, QD. 2. In A OAB and QCD, AB = CD. 3. OA = QC and OB = QD. 4. .-.A OAB = A QCD. 5. .-. Z0 = Z Q. 6. .'. AB= CD. Q.E.D. Reasons 1. § 54, 15. 2. By hyp. 3. §279, b. 4. §116. 5. §110. 6. § 293, I. BOOK II 119 II. Conversely : Given equal circles and Q, and equal arcs AB and CD. To prove chord AB = chord CD. Argument 1. Draw radii OA, OB, QC, QD. 2. AB = CD. 3. .*. Zboa = Zdqc. 4. OA = QC and OB = QD. 5. .: A OAB = A QG'D. 6. .-. chord ^4i? = chord CD. Q.E.D. Keasons § 54, 15. By hyp. § 293, II. § 279, b. §107. §110. Ex. 414. If a circumference is divided into any number of equal arcs, the chords joining the points of division will be equal. Ex. 415. A parallelogram inscribed in a circle is a rectangle. Ex. 416, If two of the opposite sides of an inscribed quadrilateral are equal, its diagonals are equal. Ex. 417. State and prove the converse of Ex. 416. 299. Def. A polygon is inscribed in a circle if all its vertices are on the circumference. Thus, polygon ABODE is an inscribed polygon. 300. Def. If a polygon is in- scribed in a circle, the circle is said to be circumscribed about the poly- gon. _^ Ex. 418. Inscribe an equilateral hexagon in a circle ; an equilateral triangle. Ex. 419. The diagonals of an inscribed equilateral pentagon are equal. Ex. 420. If the extremities of any two intersecting diameters are joined, an inscribed rectangle will be formed. Under what conditions will the rectangle be a square ? Ex. 421. State the theorems which may be used in proving arcs equal. StafceVthe theorems which may be used in proving chords equal. 120 PLANE GEOMETRY Proposition IV. Theorem 301. In equal circles, or in the same circle, if two chords are unequal, the greater chord subtends the greater minor arc ; conversely, if two minor arcs are unequal, the chord that subtends the greater arc is tlie greater. Reasons 1. § 54, 15. " 2. § 279, a. 3. By hyp. 4. § 173. 5. § 294. I. Given circle 0, with chord AB > chord CD. To prove AB > CD. Argument 1. Draw radii OA, OB, OC, OD 2. In A OAB and OCD, OA = 3. Chord AB > chord CD. 4. .•. Zl > Z2. 5. .*. AB > CD. Q.E.D. II. Conversely : Given circle 0, with AB > CD. To prove chord AB > chord CD. Argument Reasons 1. Draw radii OA, OB, OC, OD. 1. § 54, 15. 2. In A OAB and OCD, OA = OC, OB — OD. 2. § 279, a. 3. AB > / SU^ 3. By hyp. 4. .\^C> Z2. 4. §294. 5. .-. chord AB > chord CD. q.e.d. 5. § 172. Ex. 422. Prove the converse of Prop. IV by the indirect method BOOK II 121 Proposition V. Theorem 302. The diameter perpendicular to a chord bisects the chord and also its subtended arcs. J A f n E \ V ° Given chord AB and diameter CD _L AB at E. To prove AE = EB, AC= CB, and AB = DB. Argument 1. Draw radii OA and OB. 2. In A OAB, OA = OB. 3. .-.A OAB is an isosceles A. 4. .-. OE bisects AB, and AE = EB. 5. Also OE bisects Z BOA, and Zl 6. .'. Z.AOD = /.DOB. 7. .'. AC= CB and AD = DB. Z2. Q.E.D. Reasons 1. § 54, 15. 2. § 279, a. 3. § 94. 4. § 212. 5. § 212. 6. § 75. 7. § 293, I, 303. Cor. I. The perpendicular bisector of a chord passes through the center of the circle. 304. Cor. II. The locus of the centers of all circles which pass through two given points is the perpendicu- lar bisector of the line which joins the points. 305. Cor. ni. The locus of the mid-points of all chords of a circle parallel to a given line is the diameter per- pendicular to the line. (For complete proof, see p. 298.) Ex. 423. If the diagonals of an inscribed quadrilateral are unequal, its opposite sides are unequal. 122 PLANE GEOMETRY Ex. 424. Through a given point within a circle construct a chord which shall be bisected at the point. Ex. 425. Given a line fulfilling any two of the five following condi- tions, prove that it fulfills the remaining three: 1. A diameter. 2. A perpendicular to a chord. 3. A bisector of a chord. 4. A bisector of the major arc of a chord. 5. A bisector of the minor arc of a chord. Ex. 426. Any two chords of a circle are given in position and magni- tude ; find the center of the circle. Ex. 427. The line passing through the middle points of two parallel chords passes through the center of the circle. Ex. 428. Given an arc of a circle, find the center of the circle. Proposition VI. Problem 306. To bisect a given arc. A Given AB, an arc of any circle. To bisect AB. The construction, proof, and discussion are left as an exercise for the student. Ex. 429. Construct -an arc of 45° ; of 30°. Construct an arc of 30°, using a radius twice as long as the one previously used. Are these two 30° arcs equal ? Ex. 430. Distinguish between finding the " mid-point of an arc " and the " center of an arc." BOOK II 123 Proposition VII. Theorem 307. In equal circles, or in the same circle, if two chords are equal, they are equally distant from the center; con- versely, if two chords are equally distant from the center, they are equal. I. Given circle O with chord AB = chord CD, and let OE and OF be the distances of AB and CD from center O, respectively. To prove OE = OF. Argument 1. Draw radii OB and OC. 2. E and F are the mid-points of AB and CD, respectively. .*. in rt. A OEB and OCF, EB = CF. OB = OC. .'. A OEB = A OCF. .'. OE— OF. Q.E.D. II. Conversely : Given circle with OE, the distance of chord AB from center 0, equal to OF, the distance of chord CD from center 0. To prove chord AB = chord CZ). Hint. Prove A OEB = A OCJl Reasons 1. § 54, 15. 2. § 302. 3. § 54, 8 a. 4. § 279, a. 5. § 211. 6. § 110. Ex. 431. If perpendiculars from the center of a circle to the sides of an inscribed polygon are equal, the polygon is equilateral. Ex. 432. If through any point in a diameter two chords are drawn making equal angles with the diameter, the two chords are equal. 124 PLANE GEOMETRY Proposition VIII. Theorem 308. In equal circles, or in the same circle, if two chords are unequal, the greater chord is at the less dis- tance from the center. Given circle with chord AB > chord CD, and let OF and OH be the distances of AB and CD from center 0, respectively. To prove OF < OH. Argument 1. From A draw a chord AE, equal to DC. 2. From draw OG±AE. 3. Draw FG. 4. AB > CD. 5. .'. AB > AE. 6. F and G are the mid-points of AB and AE, respectively. 7. .-. ^F > AG. 8. .-. Zl >Z2. 9. Z ^0 = Z 0£4. 10. .-. Z3 chord CD. The proof is left as an exercise for the student. Hint. Begin with A OGF. 311. Cor. I. A diameter is greater than any other chord. 312. Cor. n. The locus of the mid-points of all chords of a circle equal to a given chord is the circumference hav- ing the same center as the given circle, and having for ra- dius the perpendicular from the center to the given chord. Ex. 433. Prove Prop. IX by the indirect method. Ex. 434. Through a given point within a circle construct the mini- mum chord. Ex. 435. If two chords are drawn from one extremity of a diameter, making unequal angles with it, the chords are unequal. Ex. 436. The perpendicular from the center of a circle to a side of an inscribed equilateral triangle is less than the perpendicular from the center of the circle to a side of an inscribed square. (See § 308.) 126 PLANE GEOMETRY Proposition X. Theorem 313. A tangent to a circle is perpendicular to the radius drawn to the point of tangency. A M T B Given line AB, tangent to circle O at T, and OT, a radius drawn to the point of tangency. To prove AB ± OT. Argument Reasons 1. Let M be any point on AB other than T; 1. § 286. then M is outside the circumference. 2. Draw OM, intersecting the circumference 2. § 54, 15. at s. 3. OS < om. 3. § 54, 12. 4. OS = OT. 4. § 279, a. 5. .-. OT< OM. 5. § 309. 6. .-. OT is the shortest line that can be 6. Arg. 5. drawn from to AB. 7. /. OT±AB ; i.e. AB± OT. Q.E.D. 7. §165. 314. Cor. I. (Converse of Prop. X). A straight line perpendicular to a radius at its outer extremity is tangent to the circle. Hint. Prove by the indirect method. In the figure for Prop. X, suppose that AB is not tangent to circle O at point T; then draw CD through T, tangent to circle O. Apply § 63. 315. Cor. II. A perpendicular to a tangent at the point of tangency passes through the center of the circle. BOOK II 127 316. Cor. m. A line drawn from the center of a circle perpendicular to a tangent passes through the point of tangency. 317. Def. A polygon is circum- scribed about a circle if each side of the polygon is tangent to the circle. In the same figure the circle is said to be inscribed in the polygon. Ex. 437. The perpendiculars to the sides of a circumscribed polygon at their points of tangency pass through a com- mon point. Ex. 438. The line drawn from any vertex of a circumscribed polygon to the center of the circle bisects the angle at that vertex and also the angle between radii drawn to the adjacent points of tangency. Ex. 439. If two tangents are drawn from a point to a circle, the bisector of the angle between them passes through the center of the circle. Ex. 440. The bisectors of the angles of a circumscribed quadrilateral pass through a common point. Ex. 441. Tangents to a circle at the extremities of a diameter are parallel. Proposition XI. Problem 318. To construct a tangent to a circle at any given point in the circumference. The construction, proof, and discussion are left as an exer- cise for the student. (See § 314.) Ex. 442. Construct a quadrilateral which shall be circumscribed about a circle. What kinds of quadrilaterals are circumscriptible ? Ex. 443. Construct a parallelogram which shall be inscribed in a circle. What kinds of parallelograms are inscriptible ? Ex. 444. Construct a line which shall be tangent to a given circle and parallel to a given line. Ex. 445. Construct a line which shall be tangent to a given circle and 'perpendicular to a given line. 128 PLANE GEOMETRY 319. Def. The length of a tangent is the length of the seg- ment included between the point of tangency and the point from which the tangent is drawn j as TP in the following figure. Proposition XII. Theorem 320. If two tangents are drawn from any given point to a cirele, these tangents are equal. Given PT and PS, two tangents from point P to circle 0. To prove PT=PS. The proof is left as an exercise for the student. Ex. 446. The sum of two opposite sides of a circumscribed quadrilat- eral is equal to the sum of the other two sides. Ex. 447. The median of a circumscribed trapezoid is one fourth the perimeter of the trapezoid. Ex. 448. A parallelogram circumscribed about a circle is either a rhombus or a square. Ex. 449. The hypotenuse of a right triangle circumscribed about a circle is equal to the sum of the other two sides minus a diameter of the circle. Ex. 450. If a circle is inscribed in any triangle, and if three triangles are cut from the given triangle by drawing tangents to the circle, then the sum of the perimeters of the three triangles will equal the perimeter of the given triangle. BOOK II 129 Proposition XIII. Problem 321. To inscribe a circle in a given triangle. Given A ABC. To inscribe a circle in A ABC. I. Construction 1. Construct AE and CD, bisecting A CAB and BCA, respec- tively. § 127. 2. AE and CD will intersect at some point as 0. § 194. 3. From draw OF A. AC. § 149. 4. With as center and OF as radius construct circle FGH. 5. Circle FGH is inscribed in A ABC. II. The proof and discussion are left for the student. 322. Def. A circle which is tangent to one side of a triangle and to the other two sides prolonged is said to be escribed to the triangle. ,_ Ex. 451. Problem. To escribe a circle to a given triangle. Ex. 452. (a) Prove that if the lines that bi- sect three angles of a quadrilateral meet at a com- mon point P, then the line that bisects the remaining angle of the quadri- lateral passes through P. (6) Tell why a circle can be inscribed in this particular quadrilateral. Ex. 453. In triangle ABC, draw XY parallel to PC so that XY+BC=BX + CY. , Dx. 454. Inscribe a circle in a given rhombus. 130 PLANE GEOMETRY Proposition XIV. Problem 323. To circumscribe a circle about a given triangle. Given A ABC. To circumscribe a circle about A ABC. The construction, proof, and discussion are left as an exercise for the student. 324. Cor. Three points not in the same straight line determine a circle. Ex. 455. Discuss the position of the center of a circle circumscribed about an acute triangle ; a right triangle ; an obtuse triangle. Ex. 456. Circumscribe a circle about an isosceles trapezoid. Ex. 457. Given the base of an isosceles triangle and the radius of the circumscribed circle, to construct the triangle. Ex. 458. The inscribed and circumscribed circles of an equilateral triangle are concentric. Ex. 459. If upon the sides of any triangle equilateral triangles are drawn, and circles circumscribed about the three triangles, these circles will intersect at a common point. Ex. 460. The two segments of a secant which are between two con- centric circumferences are equal. Ex. 461. The perpendicular bisectors of the sides of an inscribed quadrilateral pass through a common point. Ex. 462. The bisector of an arc of a circle is determined by the center of the circle and another point equidistant from the extremities of the chord of the arc. Ex. 463. If two chords of a circle are equal, the lines which connect their mid-points with the center of the circle are equal. BOOK II 131 TWO CIKCLES 325. Def. The line determined by the centers of two circles is called their line of centers or center-line. 326. Def. Concentric circles are circles which have the same center. Proposition XV. Theorem 327. If two circumferences meet at a -point which is not on their line of centers, they also meet in one other point, M N Given circumferences M and N meeting at P, a point not on their line of centers OQ. To prove that the circumferences meet at one other point, as R. Argument 1. Draw OP and QP. Rotate AOPQ about OQ as an axis until it falls in the position QRO. OR = OP = a radius of circle M. .-. R is on circumference M. Also QR = QP = a radius of circle N. .\ R is on circumference N. .\ R is on both circumference M and circumference N: i.e. circumferences M and N meet at R. Q.E.D. Reasons 1. §54,15. 2. § 54, 14. 3. By cons. 4. § 279, a. 5. By cons. 6. § 279, a. 7. Args. 4 and 6. 328. Cor. I. If two circumferences intersect, their line of venters bisects their common chord at right angles. 132 PLANE GEOMETRY 329. Cor. II. If two circumferences meet at one point only, that point is on their line of centers. Hint. If they meet at a point which is not on their line of eenters, they also meet in another point (§ 327). This contradicts the hypothesis. 330. Def. Two circles are said to touch or be tangent to each other if they have one and only one point in common. They are tangent internally or externally according as one circle lies within or outside of the other. Tangent externally. Tangent internally. Concentric. 331. From § 330, Cor. II may be stated as follows : If two circles are tangent to each other, their common point lies on their line of centers. 332. Cor. III. If two circles are tangent to each other, they have a common tangent line at their point of contact. Hint. Apply § 314. 333. Def. A line touching two circles is called an external common tangent if both circles lie on the same side of it ; the line is called an internal common tangent if the two circles lie on opposite sides of it. Fig. 1. Fig. 2. Thus a belt connecting two wheels as in Fig. 1 is an illus- tration of external common tangents, while a belt arranged as in Fig. 2 illustrates internal common tangents. BOOK II 133 334. Questions. In case two circles are tangent internally how many common tangents can be drawn ? in case their circumferences intersect ? in case they are tangent externally ? in case they are wholly outside of each other ? in case one is wholly within the other ? Ex. 464. If two circles intersect, their line of centers bisects the angles between the radii drawn to the points of intersection. Ex. 465. If the radii of two intersecting circles are 5 inches and 8 inches, what may be the length of the line joining their centers ? Ex. 466. If two circles are tangent externally, tangents drawn to them from any point in their common internal tangent are equal. Ex. 467. Two circles are tangent to each other. Construct their com- mon tangent at their point of contact. Ex. 468. Construct a circle passing through a given point and tan- gent to a given circle at another given point. Ex. 469. Find the locus of the centers of all circles tangent to a given circle at a given point. MEASUREMENT 335. Def. To measure a quantity is to find how many times it contains another quantity of the same kind. The result of the measurement is a number and is called the numerical measure, or measure-number, of the quantity which is measured. The measure employed is called the unit of measure. Thus, the length or breadth of a room is measured by find- ing how many feet there are in it ; i.e. tiow many times it con- tains a foot as a measure. 336. It can be shown that to every geometric, magnitude there corresponds a definite number called its measure-number. The proof that to every straight line segment there belongs a measure-number is found in the Appendix, § 595. The method of proof there used shows that operations with measure-numbers follow the ordinary laws of algebra. 337. Def. Two quantities are commensurable if there exists a measure that is contained an integral number of times in each. Such a measure is called a common measure of. the two quantities. 134 PLANE GEOMETRY Thus, a yard and a foot are commensurable, each containing an inch a whole number of times ; so, too, 12£ inches and 18f inches are commensurable, each containing a fourth of an inch a whole number of times. 338. Questions. If two quantities have a common measure, how many common measures have they ? Name some common measures of 12£ inches and 18| inches. What is their greatest common measure ? What is their least common measure ? 339. Def. Two quantities are incommensurable if there exists no measure that is contained an integral number of times in each. It will be shown later that a diagonal and a side of the same square cannot be measured by the same unit, without a re- mainder; and that the diagonal is equal to V2 times the numerical measure of the side. Now V2 can be expressed only approximately as a simple fraction or as a decimal. It lies between 1.4 and 1.5, for (1.4) 2 = 1.96, and (1.5) 2 = 2.25. Again, it lies between 1.41 and 1.42,* between 1.414 and 1.415, between 1.4142 and 1.4143, and so on. By repeated trials values may be found approximating more and more closely to V2, but no decimal number can be obtained that, taken twice as a factor, will give exactly 2. 340. When we speak of the ratio of one quantity to another, we have in mind their relative sizes. By this is meant not the difference between the two, but how many times one con- tains the other or some aliquot part of it. In algebra the ratio of two numbers has been defined as the indicated quotient of the first divided by the second. Since to each geometric mag- nitude there corresponds a number called its measure-number (§ 336), therefore : 341. Def. The ratio of two geometric magnitudes may be defined as the quotient of their measure-numbers, when the same measure is applied to each. * The student should multiply to get the successive approximations. BOOK II 135 Thus, if the length of a room is 36 feet and the width 27 feet, the ratio of the length to the width is said to be the ratio of 36 to 27 ; i.e. f f , which is equal to f . The ratio of the width to the length is |-J, which is equal to j. The term ratio is never applied to two magnitudes that are unlike. 342. Def. If the two magnitudes compared are commen- surable, the ratio is called a commensurable ratio and can always be expressed as a simple fraction. 343. Def. If the two magnitudes compared are incommen- surable, the ratio is called an incommensurable ratio and can be expressed only approximately as a simple fraction. Closer and closer approximations to an incommensurable ratio may be obtained by repeatedly using smaller and smaller units as measures of the two magnitudes to be compared and by finding the quotient of the numbers thus obtained. Two magnitudes, e.g. two line segments, taken at random are usually incommensurable, commensurability being compara- tively rare. 344. Historical Note. The discovery of incommensurable magni- tudes is ascribed to Pythagoras, whose followers for a long time kept the discovery a secret. It is believed that Pythagoras was the first to prove that the side and diagonal of a square are incommensurable. A more complete account of the work of Pythagoras will be found in § 510. Ex. 470. What is the greatest common measure of 48 inches and 18 inches ? Will it divide 48 inches — 18 inches ? 48 inches — 2 x 18 inches ? Ex. 471. Draw any two line segments which have a common meas- ure. Find the sum of these lines and, by laying off the common measure, show that it is a measure of the sum of the lines. Ex. 472. Given two lines, 5 inches and 4 inches long, respectively. Show by a diagram that any common measure of 5 inches and 4 inches is also a measure of 15 inches plus or minus 8 inches. Ex. 473. Find the greatest common divisor of 728 and 844 by division and^ point out the similarity of the process to that used in Prop. XVI 136 PLANE GEOMETRY Proposition XVI. Problem 345. To determine whether two given lines are commen- surable or not; and if they are commensurable, to find their common measure and their ratio. A F B i i i-i-j C E G D i i — i — i — i_i Given lines AB and CD. To determine : (a) whether AB and CD are commensurable and if so, (6) what is their common measure; and (c) what is the ratio of AB to CD. I. Construction 1. Measure off AB on CD as many times as possible. Sup- pose it is contained once, with a remainder ED. 2. Measure off ED on AB as many times as possible. Sup- pose it is contained twice, with a remainder FB. 3. Measure off FB on ED as many times as possible. Sup- pose it is contained three times, with a remainder GD. 4. Measure off GD on FB as many times as possible, and so on. 5. It is evident that this process will terminate only when a remainder is obtained which is a measure of the remainder immediately preceding. 6. If this process terminates, then the two given lines are commensurable, and the last remainder is their greatest com- mon measure. 7. For example, if GD is a measure of FB, then AB and CD are commensurable, GD is their greatest common measure, and the ratio of AB to CD can be found. II. Proof Argument 1. Suppose FB = 2 GD. 2. ED = EG+ GD. "Reasons 1. See I, 7. 2. §54,11. BOOK II 137 3. 4. 5. 6. 7. 8. 9. LO. .-. ED = SFB + GD = 1 GD. AB — 2ED-\- FB. .'. AB=14:GD + 2 CD = 16 GD. CD = AB + ED. .'. CD = 16 GD + 7 GD = 23 GD. .-. ^4D and CD are commensurable. Also, GD is a common measure of AB and CD. The ratio of AB to CD = the ratio of 16 GD to 23 CD = if. Q.E.D. 3. 4. 5. 6. 7. 8. 9. 10. §309. § 54, 11. §309. § 54, 11. §309. §337. Args. 5 7. §341. and III. A full discussion of this problem will be found in the Appendix, § 598. CONSTANTS AND VARIABLES. LIMITS 346. Consider an isosceles triangle ABC, whose base is AC and whose altitude is LB. Keeping the base AC the same (con- stant) i suppose the altitude to change (vary). If LB increases, what will be the effect upon the lengths of AB and CB ? what the effect upon the base angles ? upon the vertex angle ? Will the base angles always be equal to each other? What limiting value have they ? Is the base angle related to half the vertex angle or are the two independent ? Whal relation is there ? Is this relation constant or does it change ? Imagine the altitude of the triangle to diminish. Repeat the questions given above, considering the altitude as decreasing. What is now the limiting value for the altitude ? what for the length of one of the equal sides ? for the base angles ? for the angle at the vertex ? Ex. 474. Consider an isosceles triangle with a constant altitude and a^ variable base. Repeat the questions given above. 138 PLANE GEOMETRY Ex. 475. Consider an isosceles triangle with constant base angles, but variable base. Tell what other constants and what other variables there would be in this case. Ex. 476. If through any point in the base of an isosceles triangle lines are drawn parallel to the equal sides of the triangle, a parallelo- gram will be formed whose perimeter will be constant ; i.e. the perimeter will be independent of the position of the point. 347. Def. A magnitude is constant if it does not change throughout a discussion. 348. Def. A magnitude is variable if it takes a series of different successive values during a discussion. 349. Def. If a variable approaches a constant in such a way that the difference between the variable and the constant may be made to become and remain smaller than any fixed number previously assigned, however small, the constant is called the limit of the variable. 350. The variable is said to approach its limit as it becomes more and more nearly equal to it. Thus, suppose a point to move from A toward B, by successive steps, under the A £ ^ F ? restriction that at each step it must go over one half the segment between it and B. At the first step it reaches C, whereupon there remains the segment CB to be traveled over ; at the next step it reaches D, and there remains an equal segment to be covered. Whatever the number of steps taken, there must always remain a segment equal to the segment last covered. But the segment between A and the moving point may be made to differ from AB by as little as we ptease, i.e. by less than any previously assigned value. For assign some value, say, half an inch. Then the point, continuing to move under its governing law, may approach B until there remains a segment less' than half an inch. Whatever be the value assigned, the variable segment from A to the moving point may be made to differ from the constant segment AB by less than the assigned value. BOOK II 139 Again, the numbers in the series 4, 2, 1, ^, \, \, etc., in which each term is one half of the preceding term, approach as a limit as the number of terms in the series is increased. For if we assign any value, as y-g-^-g-, it is evident that a term of the series may be found which is less than y-oxro-g-; it is also evident that no term of the series can become 0. 351. In elementary geometry the variables that approach limits are usually such that they cannot attain their limits. There are, however, variables that do attain their limits. The limiting values of algebraic expressions are frequently of this kind ; thus, the expression — — - approaches lasa; approaches X ~\~ -L 0, and has the limit 1 when x becomes zero. 352. Def. Two variables are said to be related when one depends upon the other so that, if the value of one variable is known, the value of the other can be obtained. For example, the diagonal and the area of a square are re- lated variables, for there is a value for the area for any value which may be given to the diagonal, and vice versa. 353. Questions. On the floor is a bushel of sand. If we keep adding to this pile forever, how large will it become ? Does it depend upon the law governing our additions ? If we add one quart each hour, how large will it become ? If we add one quart the first hour, a half quart the second hour, a fourth quart the third hour, etc., each hour adding one half as much as the preceding hour, how large will the pile become ? 354. Historical Note. Achilles and the Tortoise. One of the early Greek schools of mathematics, founded during the fifth century b.c, at Elea, Italy, and known as the Eleatic School, was famous for its investigations of problems involving infinite series. Zeno, one of the most prominent members, proposed this question: He "argued that if Achilles ran ten times as fast as a tortoise, yet if the tortoise had (say) 1000 yards start, it could never be overtaken : for, when Achilles had gone the 1000 yards, the tortdise would still be 100 yards in front of him ; by the time he had covered these 100 yards, it would still be 10 yards in front of him ; and so on forever : thus Achilles would get nearer and nearer to the tortoise but never overtake it." Was Zeno right ? If not, ca%you find the fallacy in his argument ? 140 PLANE GEOMETRY Proposition XVII. Theorem 355. If two variables are always equal, and if each ap- proaches a limit, then their limits are equal. Given two variables, V and V', which are always equal and which approach as limits L and L\ respectively. To prove L = L\ Argument 1. Either L = L', or L=£ L\ 2. Suppose that one limit is greater than the other, say L > L' ; then V, in ap- proaching L, may assume a value be- tween L' and L j i.e. V may assume a value > L\ 3. But v' cannot assume a value > L'. 4. .-. V may become > V'. 5. But this is impossible, since V and v' are always equal. 6. .*. L = L\ q.e.d. Ke a sons 1. § 161, a. 2. § 349. 3. § 349. 4. Args. 2 and 3. 5. By hyp. 6. § 161, 6. 356. Question. In the above proof are V and V f increasing or de- creasing variables ? The student may adapt the argument above to the case in which V and V are decreasing variables. Ex. 477. Apply Prop. XVII to the accompanying figure, where variable V is represented by the line AB, variable V by the line CD, limit L by line AE, and limit V by line CF. B H 357. Note. It will be seen that, in the application of Prop. XVII, there are three distinct things to be considered : (1) Two variables that are always equal; (2) The limits of these two variables ; (3) The equality of these limits themselves. BOOK II 141 Proposition XVIII. Theorem 358. An angle at the center of a circle is measured by its intercepted arc. Given central ZAOB and AB intercepted by it; let ZCOE be any unit Z (e. g. a degree), and let CE, intercepted by the unit Z, be the unit arc. To prove the measure-number of ZAOB, referred to ZCOE, equal to the measure-number of AB, referred to CE. I. If ZAOB and ZCOE are commensurable. (a) Suppose that ZCOE is contained in ZAOB an integral number of times. Argument 1. Apply Z COE to Zaob as a measure. Suppose that ZCOE is contained in Z AOB r times. 2. Then r is the measure-number of Z AOB referred to ZCOE as a unit. -°). Now the r equal central A which com- pose ZAOB intercept r equal arcs on the circumference, each equal to GE. 4. .\ r is the measure-number of AB re- ferred to CE as a unit. 5. .*. the measure-number of ZAOB, re- ferred to ZCOE as a unit, equals the measure-number of AB, referred to ■* 4 CE as a unit. q.e.d. Reasons 1. § 335. 2. § 335. 3. § 293, 1. 4. § 335. 5. §54,1. 142 PLANE GEOMETRY I. If Z AOB and Z COE are commensurable. (b) Suppose that Z COE is not contained in Z AOB an integral number of times. The proof is left as an exercise for the student. Hint. . Some aliquot part of ZCOE must be a measure of /.AOB. (Why ?) Try i ZCOE, \ ZCOE, etc. II. If Z AOB and Z COE are incommensurable. Argument Let Z 1 be a measure of Z COE. Ap- ply Z 1 to Z AOB as many times as possible. There will then be a re- mainder, Z.FOB, less than Zl. Z AOF and Z COS are commensurable. .-. the measure-number of A AOF, re- ferred to Z CO.E as a unit, equals the measure-number of AF } referred to CE as a unit. Reasons 1. §339. 2. §337. 3. § 358, I. BOOK II 143 Reasons 4. §335. 5. § 294. 6. Args.4and5. Argument 4. Now take a smaller measure of /.COE. No matter how small a measure of Z COE is taken, when it is applied as a measure to /AOB, the remainder, Z FOB, will be smaller than the Z taken as a measure. 5. Also FB will be smaller than the arc in- tercepted by the Z taken as a measure. 6. .*. the difference between /AOF and Z AOB may be made to become and remain less than any previously as- signed Z, however small; and like- wise the difference between AF and AB, less than the arc intercepted by the assigned Z. 7. .*. Z AOF approaches Z AOB as a limit, and AF approaches Xb as a limit. 8. Hence the measure-number of Z .40^ approaches the measure-number of /AOB as a limit, and the measure- number of AF approaches the meas- ure-number of AB as a limit. ' 9. But the measure-number of Z AOF is always equal to the measure-number of AF. 10. .*. the measure-number of /.AOB, re- ferred to Z COE as a unit, equals the measure-number of AB, referred to CE as a unit. q.e.d. 359. If a magnitude is variable and approaches a limit, then, as the magnitude varies, the successive measure-numbers of the variable approach as their limit the measure-number of the limit of the magnitude. (This theorem will be found in the Appendix, § 597.) 7. §349. 8. §359. 9. Arg. 3. 10. §355. 144 PLANE GEOMETRY 360. Cor. In equal circles, or in the same circle, two angles at the center have the same ratio as their inter- cepted arcs. Hint. The measure-numbers of the angles are equal respectively to the measure- numbers of their intercepted arcs. Therefore the ratio of the angles is equal to the ratio of the arcs. Ex. 478. Construct a secant which shall cut off two thirds of a given circumference. Ex. 479. Is the ratio of two chords in the same circle equal to the ratio of the arcs which they subtend ? Illustrate your answer, using a semicircumference and a quadrant. 361. The symbol oc will be used for is measured by. oc is the symbol of variation, and the macron (— ) means long or length' Hence oc suggests varies as the length of. 362. From § 336 it follows directly that : (a) In equal circles, or in the same circle, equal a7igles are measured * by equal arcs ; conversely, equal arcs measure equal a,ngles. (sum \p) The measure of the ( stii/m. the [difference of two angles is equal to (b) The measure of the sum difference (c) The measure of any multiple of an angle is equal to that same multiple of the measure of the angle. \ of the measures of the angles. 363. Def. An angle is said to be inscribed in a circle if its vertex lies on the circumference and its sides are chords. 364. Def. An angle is said to be inscribed in a segment of a circle if its vertex lies on the arc of the segment and its sides pass through the extremities of that arc. * It is, of course, inaccurate to speak of measuring one magnitude by another magnitude of a different kind.; but, in this case, it has become a convention so general that the student needs to become familiar with it. More accurately, in Prop. XVII I, the measure' number of an angle at the center, referred to any unit angle, is the same as the measure- number of its intercepted arc when the unit arc is the arc intercepted by the unit angle. BOOK II 145 Proposition XIX. Theorem 365. An inscribed angle is measured by one half its intercepted arc, B Given inscribed Z ABC, To prove that Z ABC cc \ AC. I. Let one side of Z ABC, as AB y pass through the center oi the circle (Fig. 1). Argument 1. Draw radius OC. 2. OB = OC. 3. .-. Z1 = Z2. 4. Z1 + Z2 = Z3. 5. .\2Z1 = Z3. 6. 7. ButZ3oc^tc. 8. „\ i Z 3 or Z 1 oc \ AC; I e. Z ABC radius QN. Then, with O as center and with OL — OM — $7Vas radius, construct circle LPK. 3. Construct tangent QP from point Q to circle LPK. § 373. 4. Draw OP and prolong it to meet circumference MTB at T. 5. Draw QVWOT. § 188. 6. Draw TV. 7. TV is tangent to circles MTB and NVS. II. The proof and discussion are left as an exercise for the student. Construct a second common external tangent to circles MTB by same Ex. 497. and NVS method. How is the method modified if the two circles are equal ? Ex. 498. Construct a common internal tangent to two circles. Hint. Follow steps of Ex. 496 except step 2. Make OL = OM + QN. Ex. 499. By moving Q toward in the precedin figure, show when there are four common tangents ; when only three ; when only two ; when only one ; when none. BOOK II 151 Proposition XXII. Problem 374. With a given line as chord, to construct a segment of a circle capable of containing a given angle. 4 B *xC VI / Given line AB and Z M. C \^ ^*"*^„^^ \ / To construct, with AB as ^ S^XH chord, a segment of a circle capable of containing Z Jf. ]& I. Construction 1. Construct an Z, as ZCDE, equal to the given Z M. § 125. 2. With H, any convenient point on DE, as center and with AB as radius, describe an arc cutting DC at some point, as K. 3. Draw UK. 4. Circumscribe a circle about AKDH. § 323. 5. Segment KDH is the segment capable of containing Z-M. II. The proof and discussion are left to the student. 375. Questions. Without moving AB, can you construct a A with AB as base and a vertex Z = Z M? What is the sum of the base A ? 376. Cor. The locus of the vertices of all triangles hav- ing a given base and a given angle at their vertices is the arc which forms, with the given base, a segment capable of containing the given angle. Ex. 500. On a given line construct a segment that shall contain an angle of 105° ; of 135°. Ex. 501. Find the locus of the vertices of all triangles having a com- mon base 2 inches long and having their vertex angles equal to 60°. Ex. 502. Construct a triangle, having given &, hi, and B. 152 PLANE GEOMETRY Proposition XXIII. Theorem 377. An angle formed by two chords which intersect within a circle is measured by one half the sum of the arc intercepted between its sides and the arc intercepted be- tween tJie sides of its vertical angle. tL B Given two chords AB and CD, intersecting at E. To prove that Zl«i (BD + AC). Argument Draw AD. Z1=Z2 + Z3. Z_2*\Bb. Z 3 * \ 2b. .'. Z 1 oc i (BD + AC). Q.E.D. Reasons § 54, 15. §215. §365. §365. § 362, &. §309. Ex. 503. One angle formed by two intersecting chords intercepts an arc of 40°. Its vertical angle intercepts an arc of 60°. How large is the angle ? Ex. 504. If an angle of two intersecting chords is 40° and its inter- cepted arc is 30°, how large is the opposite arc ? Ex. 505. If two chords intersect at right angles within a circumfer- ence, the sum of two opposite intercepted arcs is equal to a semicircum- ference. Ex. 506. If M is the center of a circle inscribed in triangle ABC and if AM is prolonged to meet the circumference of the circumscribed circle at D, prove that BD ss DM = DC. BOOK II 153 Proposition XXIV. Theorem 378. An angle formed by a tangent and a chord is measured by one half its intercepted arc. Given A ABC formed by tangent AB and chord BC, To prove that AABC<£\BC. Argument Reasons 1. Draw diameter BD. 1. § 64, 15. 2. Aabd is a rt. Z. 2. § 313. 3. .-. A ABB * \ a semicircumf erence ; i.e. A ABB cc \ arc BOB. 3. §297. 4. A CBD QCi8. 4. §365. 5. .-. A ABB — Z C#D « | (arc BCB — CB). 5. § 362, b. G. .'. A ABC QC J 2K7. Q.E.D. 6. §309. Ex. 507. In the figure of § 378, if arc BC = 100°, find the number of degrees in angle ABC; in angle CBD ; in angle CBE. Ex. 508. If tangents are drawn at the extremities of a chord which subtends an arc of 120°, what kind of triangle is formed ? Ex. 509. If a tangent is drawn to a circle at the extremity of a chord, the mid-point of the subtended arc is equidistant from the chord and the tangent. Ex. 510. Solve Prop. XXII by means of Prop. XXIV. Hint. Observe that in the figure for Prop. XXIV any angle inscribed in »egpnent BBC would be equal to angle ABC. 154 PLANE GEOMETRY Proposition XXV. Theorem 379. An angle formed by two secants intersecting out- side of a circumference, an angle formed by a secant and a tangent, and an angle formed by two tangents are each measured by one half the difference of the intercepted arcs. Fig. 2. Fig. 3. I. An angle formed by two secants (Fig. 1). Given two secants BA and BC, forming Z 1. To prove that Z 1 cc A (AC — BE). Argument Draw CD. 2. Z1 + Z2 = Z3. 3. .-. Z 1 = Z 3 - Z 2. 4. Z 3 cc -i- ac, and Z 2 « \ BE. 5. Reasons § 54, 15. §215. § 54, 3. § 365. 5. §362,6. 6. § 309. .-. Z3-Z2 <* \{ac-be). 6. .-. Z 1 QC 1 (AC - BE). Q.E.D. II. An angle formed by a secant and a tangent (Fig. 2). III. An angle formed by two tangents (Fig. 3). The proofs of II and III are left to the student. 380. Note. In the preceding theorems the vertex of the angle may be : (1) within the circle ; (2) on the circumference ; (3) outside the circle. Ex. 511. Tell how to measure an angle having its vertex in each of the three possible positions with regard to the circumference. BOOK II 155 Proposition XXVI. Theorem 381. Parallel lines intercept equal arcs on a circum- ference. n „ H Fig. 1. Fig. 2. I. If the two II lines are secants or chords (Fig. 1). Given II chords AB and CZ), intercepting arcs AC and BD. To prove AC = BD. Argument 1. Draw CB. 2. Z1 = Z2. 3. Z.\^\ac, andZ2ocij£b. 4. .-. \AC=\BB. 5. .'. AC = i£d. Q.E.D. Reasons 1. § 54, 15. 2. § 189. 3. § 365. 4. § 362, a. 5. § 54, 7 a. II. If one of the II lines is a secant and the other a tangent (Fig. 2). III. If the two II lines are tangents (Fig. 3). The proofs of II and III are left as exercises for the student. Ex. 512. In Fig. 1, Prop. XXV, if angle 1 equals 42° and arc BE equals 40°, how many degrees are there in angle 3 ? in arc AC? Ex. 513. In Fig. 2, if arc DC equals 60° and angle 3 equals 100°, find the number of degrees in angle 1. Ex. 514. In Fig. 3, if angle 1 equals 65°, find angle 2 and angle 3. Ex. 515. An angle formed by two tangents is measured by 180° minus the intercepted arc. Ex. 516. If two tangents to a circle meet at an angle of 40°, how nxaftjr degrees of arc do they intercept ? 156 PLANE GEOMETRY Ex. 517. Is the converse of Prop. XXVI true ? Prove your answer. Ex. 518. If two sides of an inscribed quadrilateral are parallel, the other two sides are equal. Ex. 519. Is the converse of Ex. 518 true ? Prove your answer. Ex. 520. If, through the point where the bisector of an inscribed angle cuts the circumference, a chord is drawn parallel to one side of the angle, this chord will equal the other side of the angle. Ex. 521. If through the points of intersection of two circumferences parallels are drawn terminating in the circumferences, these parallels will be equal. Ex. 522. Prove that a trapezoid inscribed in a circle is isosceles. Ex. 523. If two pairs of sides of an inscribed hexagon are parallel, the other two sides are equal. Ex. 524. If the sides AB and BC of an inscribed quadrilateral ABGD subtend arcs of 60° and 130°, respectively, and if angle AED, formed by the diagonals AC and BD intersecting at E, is 75°, how many degrees are there in arcs AD and DC ? how many degrees in each angle of the quad- rilateral ? MISCELLANEOUS EXERCISES Ex. 525. Equal chords of a circle whose center is C intersect at E. Prove that CE bisects the angle formed by the chords. Ex. 526. A common tangent to two unequal circles intersects their line of centers at a point P ; from P a second tangent is drawn to one of the circles. Prove that it is also tangent to the other. Ex. 527. Through one of the points of intersection of two circum- ferences draw a chord of one that shall be bisected by the other circumference. Ex. 528. The angle ABC is any inscribed angle in a given segment of a circle ; AC is prolonged to P, making CP equal to CB. Find the locus of P. Ex. 529. Given two points P and Q, and a straight line through Q. Find the locus of the foot of the perpendicular from P to the given line, as the latter revolves around Q. Ex. 530. If two circles touch each other and a line is drawn through the point of contact and terminated by the circumferences, the tangents at its ends are parallel. Ex. 531. If two circles touch each other and two lines are drawn through the point of contact terminated by the circumferences, the chords joining the ends of these lines are parallel. BOOK II 157 Ex. 532. If one arm of a right triangle is the diameter of a circle, the tangent at the point where the circumference cuts the hypotenuse bisects the other arm. Ex. 533. Two fixed circles touch each other externally and a circle of variable radius touches both externally. Show that the difference of the distances from the center of the variable circle to the centers of the fixed circles is constant. Ex. 534. If two circles are tangent externally, their common internal tangent bisects their common external tangent. Ex. 535. If two circles are tangent externally and if their common external tangent is drawn, lines drawn from the point of contact of the circles to the points of contact of the external tangent are perpendicular to each other. Ex. 536. The two common external tangents to two circles meet their line of centers at a common point. Also the two common internal tan- gents meet the line of centers at a common point. Ex. 537. Two circles whose radii are 17 and 10 inches, respectively, are tangent externally. How long is the line joining their centers ? how long if the same circles are tangent internally ? Ex. 538. If a right angle at the center of a circle is trisected, is the intercepted arc also trisected ? Is the chord which subtends the arc tri- sected ? Ex. 539. Draw a line intersecting two given circumferences in such a way that the chords intercepted by the two circumferences shall equal two given lines. What restriction is there on the lengths of the given lines ? Ex. 540. Construct a triangle, given its base, the vertex angle, and the median to the base. Under what conditions will there be no solution ? Ex. 541. In the same circle, or in equal circles, two inscribed triangles are equal, if two sides of one are equal respectively to two sides of the other. Ex. 542. If through the points of intersection of two circumferences two lines are drawn terminating in the circumferences, the chords which join their extremities are parallel. Ex. 543. The tangents drawn through the vertices of an inscribed rectangle, which is not a square, form a rhombus. Ex. 544. The line joining the center of the square described upon the hypotenuse of a right triangle, to the vertex of the right angle, bisects the righf^angle. 158 PLANE GEOMETRY Ex. 545. If two common external tangents or two common internal tangents are drawn to two circles, the segments of these tangents inter- cepted between the points of contact are equal. Ex. 546. Through two given points draw two parallel lines at a given distance apart. Ex. 547. In a given circle inscribe a chord of given length which pro- longed shall be tangent to another given circle. Ex. 548. Find the locus of the middle point of a chord drawn from a given point in a given circumference. Ex. 549. The locus of the intersections of the altitudes of triangles having a given base and a given angle at the vertex is the arc which forms with the base a segment capable of containing an angle equal to the supplement of the given angle at the vertex. Hint. Let ABO be one of the A and O the intersection of the altitudes. In quadrilateral FOEC, /.EOF is the supplement of Z FOE. .-. Z EOF is the supplement of Z BOA. Ex. 550. In a circle, prove that any chord which bisects a radius at right angles subtends an angle of 120° at the center. Ex. 551. Construct an equilateral triangle, having given the radius of the inscribed circle. Ex. 552. The two circles described upon two sides of a triangle as diameters intersect upon the third side. Ex. 553. All triangles circumscribed about the same circle and mutually equiangular are equal. Ex. 554. If two cir- cumferences meet on their line of centers, the circles are tangent to each other. Fig. 1. Fig. 2. M between and Q OP+PQ>OQ. OP= OM. .'.PQ>MQ. .'. P is not on circumference &. Outline of Proof Fig. 1. II. M not between O and Q, Fig. 2. 0Q+ QP>OP. .*. OQ+ QP>OM. .-. QP> QM. .-. P is not on circumference S. BOOK II 159 Ex. 555. If two circumferences intersect, neither point of intersection is on their line of centers. Ex. 556. In any right triangle ABG, right-angled at G, the radius of the inscribed circle equals \{a + b — c) and the radius of the escribed circle tangent to c equals |(a + b + c). Ex. 557. In the accom- . T B P panying figure a, b, c, are the ^^^~ jT ~"\ s*y ^N. sides of triangle AB G. Prove: ^v„( 1$T \ 1. a = TP; Sr^f \ / 2. AP=l(a + b + c); ^*^-_^/ 3. TB = l(a + c-b). «"\ Ex. 558. Trisect a quadrant ; a semicircumference ; a circumference. Ex. 559. Describe circles about the vertices of a given triangle as centers, so that each shall touch the other two. Ex. 560. Construct within a given circle three equal circles, so that each shall touch the other two and also the given circle. Construct a triangle, having given : Ex. 561. h a , h c , a Ex. 562. A, B, and B, the radius of the circumscribed circle. Ex. 563. a, B, B. Ex. 564. C and the segments of c made by t c . Ex. 565. C and the segments of c made by h c . Ex. 566. r, the radius of the inscribed circle, and the segments of c made by t c > Construct a right triangle ABC, right-angled at C, having given : Ex. 567. c, h c . Ex. 568. c and one segment of c made by h c . Ex. 569. The segments of c made by h c . Ex. 570. The segments of c made by t c . Ex. 571. c and a line I, in which the vertex of G must lie. Ex. 572. c and the perpendicular from vertex G to a line I. Ex. 573. c and the distance from G to a point P. Construct a square, having given : Ex. 574. The perimeter. Ex. 575. A diagonal. Ex. 576. The sum of a diagonal and a side. 160 PLANE GEOMETRY Construct a rectangle, having given : Ex. 577. Two non-parallel sides. Ex. 578. A side and a diagonal. Ex. 579. The perimeter and a diagonal. Ex. 580. A diagonal and an angle formed by the diagonals. Ex. 581. A side and an angle formed by the diagonals. Ex. 582. The perimeter and an angle formed by the diagonals. Construct a rhombus, having given : Ex. 583. A side and a diagonal. Ex. 584. The perimeter and a diagonal. Ex. 585. One angle and a diagonal. Ex. 586. The two diagonals. Ex. 587. A side and the sum of the diagonals. Ex. 588. A side and the difference of the diagonals. Construct a parallelogram, having given : Ex. 589. Two non-parallel sides and an angle. Ex. 590. Two non-parallel sides and a diagonal. Ex. 591. One side and the two diagonals. Ex. 592. The diagonals and an angle formed by them. Construct an isosceles trapezoid, having given : Ex. 593. The bases and a diagonal. Ex. 594. The longer base, the altitude, and one of the equal sides. Ex. 595. The shorter base, the altitude, and one of the equal sides. Ex. 596. Two sides and their included angle. Construct a trapezoid, having given : Ex. 597. The bases and the angles adjacent to one base. Ex. 598. The bases, the altitude, and an angle. Ex. 599. One base, the two diagonals, and their included angle. Ex. 600. The bases, a diagonal, and the angle between the diagonals. Construct a circle which shall : Ex. 601. Touch a given circle at P and pass through a given point Q Ex. 602. Touch a given line I at P. Ex. 603. Touch three given lines two of which are parallel. Ex. 604. Touch a given line I at P and also touch another line m, Ex. 605. Have its center in line Z, cut I at P, and touch a circle K BOOK III PROPORTION AND SIMILAR FIGURES 382. Def. A proportion is the expression of the equality of two ratios. Example. If the ratio — is equal to the ratio -, then the equation b d -=- is a proportion. This proportion may also be written a : b = c : d or b d a :b : : c :d, and is read a is to b as c is to d. 383. Def. The four numbers a, 6, c, d are called the terms of the proportion. 384. Def. The first term of a ratio is called its antecedent and the second term its consequent ; therefore : The first and third terms of a proportion are called ante- cedents, and the second and fourth terms, consequents. 385. Def. The second and third terms of a proportion are called its means, and the first and fourth terms, its extremes. •386. Def. If the two means of a proportion are equal, this common mean is called the mean proportional between the two extremes, and the last term of the proportion is called the third proportional to the first and second terms taken in order; thus, in the proportion a : 6 = 6 : c, 6 is the mean proportional between a and c, and c is the third proportional to a and 6. 387. Def. The fourth proportional to three given numbers is the fourth term of a proportion the first three terms of which are the three given numbers taken in order; thus, if a : 6 = c : d, d is called the fourth proportional to a, 6, and c. m 162 PLANE GEOMETRY Proposition I. Theorem 388. If four numbers are in proportion, the product of the extremes is equal to the product of the means. Given a : b = c : d. To prove ad = be. Argument 1 - — £ b~ ' d 2. bd = bd. 3. .'.ad = bc. q.e.d. Reasons 1. By hyp.* 2. Byiden. 3. §54, la. 389. Note. The student should observe that the process used here' is merely the algebraic "clearing of fractions," and that as fractional equations in algebra are usually simplified by this process, so, also, pro- portions may be simplified by placing the product of the means equal to the product of the extremes. 390. Cor. I. The mean proportional between two num- bers is equal to the square root of their product. 391. Cor. II. If two proportions have any three terms of one equal respectively to the three corresponding terms of the other, then the remaining term of the first is equal to the remaining term of the second. Ex. 606. Given the equation m : r = d : c ; solve (1) for d, (2) for r, (3) for m, (4) for c. Ex. 607. Find the fourth proportional to 4, 6, and 10 j to 4, 10, and 6; to 10, 6, and 4. Ex. 608. Find the mean proportional between 9 and 144 ; between 144 and 9. Ex. 609. Find the third proportional to f and f ; to f and f. 392. Questions. What rearrangement of numbers can be made in Ex. 607 without affecting the required term ? in Ex. 608 ? in Ex. 609 ? Ex. 610. Find the third proportional to a 2 — 6 2 and a — b. Ex. 611. Find the fourth proportional to a 2 — b 2 , a— b, and a + 6. * See § 382 for the three wdys of writing a proportion. BOOK III 163 Ex. 612. If in any proportion the antecedents are equal, then the consequents are equal and conversely. Ex. 613. If a : b = c : d, prove that ma : kb = mc : kd. Ex. 614. If I : k = b : f», prove that Ir : kr = be : mc. Ex. 615. If x : y = b : c, prove that dx:y = bd:c. Ex. 616. If x : y = 6 : c, is (fcc : y = 6 : cd a true proportion ? Proposition II. Theorem (Converse of Prop. I) ♦ 393 If the product of two numbers is equal to the prod- uct of two other numbers, either pair may be made the means and the other pair the extremes of a proportion. Given ad = be. To prove a:b = c:d. Reasons 1. By hyp. 2. By iden. 3. § 54, 8 a. Argument 1. ad — be. 2. bd = bd. a c t i - = - ; i.e. a:o = c:d. q.e.d. b d The proof that a and d may be made the means and b and c the extremes is left as an exercise for the student. 394. Note. The pupil should observe that the divisor in Arg. 2 above must be chosen so as to give the desired quotient in the first mem- h k ber of the equation : thus, if hi = kf, and we wish to prove that - = -, we must divide by fl ; then M = K i.e. ? =-. fl fl f I Ex. 617. Given pt = cr. Prove p : r = c : t ; also, c :p = t: r. Ex. 618. From the equation rs = Im, derive the following eight pro- portions: ,77 7 r:l = m:s, s:l = m:r, I :r = s:m, m:r = s:l r :m = l :s, s :m = l:r, I: s = r : m, m:s = r :l. Ex. 619. Form a proportion from 7 x 4 = 3 x a ; from ft = gb. How can the proportions obtained be verified ? 164 PLANE GEOMETRY Ex. 620. Form a proportion from (a + c) (a — b) = de. Ex. 621. Form a proportion from m 2 — 2 mn + n 2 = ab. Ex. 622. Form a proportion from c 2 + 2 cd + d 2 = a + &. Ex. 623. Form a proportion from (a + b)(a— 6) =4x, making a; (1) an extreme ; (2) a mean. Ex. 624.' If 7 a + 3 y : 12 = 2a + y : 3, find the ratio a : y. Proposition III. Theorem 395. If four numbers are in proportion, they are in proportion by inversion; that is, the second term is to the first as tlxe fourth is to the third. Given a:b=c:d. To prove b : a = d : c. Argument Reasons 1. a:b = c:d. 1. By hyp. 2. .•. ad = be. 2. § 388. 3. \ b : a = d : c. Proposition IV. Q.E.D. Theore 3. M § 393. 396. If four numbers are in proportion, they are in proportion by alternation; that is, the first term is to the third as the second is to the fourth. Given a:b = c: d. To prove a:c — b:d. Argument Reasons 1. a:b = c:d. 1. By hyp. 2. .\ ad = be. 2. §388. 3. .•. a:c = b:d. Q.E.D. 3. §393. Ex. 625. If x : - 1 / = y : |, what is the value of the ratio x : y ? Ex. 626. Transform a : x = 4 : 3 so that x shall occupy in turn every place in the proportion. book in 165 397. Many transformations may be easily brought about by the following method: (1) Reduce the conclusion to an equation in its simplest form (§ 388), then from this derive the hypothesis. (2) Begin with the hypothesis and reverse the steps of (1). This method is illustrated in the analysis of Prop. V. Proposition V. Theorem 398. If four numbers are in proportion, they are in proportion by composition; that is, the sum of the first two terms is to the first {or second) term as the sum> of the last two terms is to the third {or fourth) term. Given a:b = c:d. To prove : {a) a -+- b : a , = c 4- d : c ; (6) a -f b : b = c -f d : d. I. Analysis (1) The conclusions required above, when reduced to equa- tions in their simplest forms, are as follows : (a) (b) 1. ac +ad = ac + bc. 1. be + bd = ad + bd. 2. Whence ad = be. 2. Whence be = ad. 3. .*. a: b — c:d. 3. .-. a:b = c:d. . (2) Now begin with the hypothesis and reverse the steps. II. Proof {a) Argument 1. a:b = c:d. 2. .-. ad = be. 3. ac = ac. 4. .-.ac-\-ad=ac-\-bc; i.e. a{c + d)= c{a-\- b). 6. .-.a +b:a = c + d: c. q.e.d Reasons 1. By hyp. 2. § 388. 3. By iden. 4. §54,2. 5. §393. (fy The proof of (6) is left as an exercise for the student. 166 PLANE GEOMETRY 627. If* = ?,find*-+Z; ^±1. y 5 x y Proposition VI. Theorem 399. iy four numbers are in proportion, they are in proportion by division; that is, the difference of the first two terms is to tlw first (or second) term as the differ- ence of the last two terms is to the third (or fourth) term. Given a:b = c:d. To prove : (a) a — b:a = c — dic\ (b) a — bib = c — did. I. The analysis is left as an exercise for the student. Reasons 1. By hyp. 2. §388. 3. By iden. 4. §54,3. II. Proof (a) Argument 1. a ib = cid. 2. \ ad = be. 3. ac = ac. 4. .* . ac — ad m ac — be, i.e. a(c — d) = c (a — b). 5. .* a — bia — c — die. Q.E.D. 5. §393. (b) The proof of (b) is left as an exercise for the student. Ex. 628. If - = -, find ^^ ; X -^UL. y 3 x y Proposition VII. Theorem 400. If four numbers are in proportion, they are in proportion by composition and division; that is, the sum of the first two terms is to their difference as the sum of the last two terms is to their difference. Given a ib = ci d. To prove a -\- bia— b = c-\-d:c — d. BOOK III Argument 1. a:b = = c:d. 2. a + b_ c + d a c 3. And a — b a c — d c 4. a + b_ ,5 167 i.e. a + b:a — b = c + die — d. q.e.d. Reasons 1. By hyp. 2. §398. 3. §399. 4. §54, 8 a. Proposition VIII. Theorem 401. In a series of equal ratios the sum of any number of antecedents is to the sum of the corresponding conse- quents as any antecedent is to its consequent. Given a:b = c:d = e:f = g :h. To prove a -f- c + e -f g : b + d + / -J- h = a: b. I. Analysis Simplifying the conclusion above, we have : ab -|- be + be + bg = ab -f- ad -f af + ah. The terms required for the first member of this equation, and their equivalents for the second member, may be obtained from the hypothesis. II. Proof Argument Reasons 1. ab = ab. 1. By iden. 2. be = ad. 2. §388. 3. be = af. 3. §388. 4. bg = ah. 4. §388. 5. .'. ab + bc + be + bg = ab-\-ad+af-\- ah ; i . e . b (a + c + e + g) = a (b + d + /+ h) . 5. § 54, 2. 6. .'. a+ c -f e + g : b + d + f+h = a :b. ^'4 Q.E.D. 6. §393. 168 PLANE GEOMETRY Ex. 629. Ex. 630. With the hypothesis of Prop. VIII, prove a + c + e:b + d+f=g:h. If ffl = « = ft = .g prove ffl-« + *-J> = g. r t g q r-t+g—q q Ex. 631. If *±ll - 1 find 5. Hint. Use Prop. VI. 2y 4 y Ex. 632. If .a : b = c : d, show that & — a:& + a = cl — c:c? + c. Ex. 633. Given the proportion a : 6 = 11 : 6. Write the proportions that result from taking the terms (1) by inversion ; (2) by alternation ; (3) by composition ; (4) by division ; (5) by composition and division. Proposition IX. Theorem 402. The products of the corresponding terms of any number of proportions form a proportion. Given a : b = c : d, e:f=g:h, i:j — k:l. To prove aei : bfj = cgk : did. 2. Argument b~d' e _g . i _k aei __ cgk m i.e. aei : bfj = cgk : did. Q.E.D. Reasons By hyp. 2. §54, 7 c 403. Cor. If four numbers are in proportion, equi- multiples of the first two and equimultiples of the last two are also in proportion. Hint. Given a :b = x:y. To prove am :bm = nx :yiy. Ex. 634. If r : s = m i t, is fr : qs = qm -.ft ? Prove your answer. Ex. 635. If a : b = 3 : 4 and x : y — 8 ;9, find the value of ax : by. BOOK III 169 Ex. 636. If four numbers are in proportion, equimultiples of the antecedents and equimultiples of the consequents are also in proportion. Ex. 637. If r : s = t : m, prove 3 s -j- 2 m : 4 s = 3 r + 2 1 : 4 r, Ex. 638. If to : a; = y : g, prove are + 60 : ex 4- tte = aw + 6y : cw + r \ X, X, 1 51 1 * c c/ R « \M b \L S Given lines a, b, and c. To construct the fourth proportional to a, b, and c. I. Construction 1. From any point, as R, draw two indefinite lines RS and RT. 2. On RS lay off RM = a and ML = b. 3. On RT lay off RF = c. 4. Draw MF. 5. Through L construct LG II MF. § 188. 6. FG is the fourth proportional to a, b, and c. II. The proof and discussion are left as an exercise for the student. 412. Question. Could the segments a, b, and c be laid off in any other order ? 413. Cor. I. To construct the third proportional to two given lines. 414. Cor. n. To divide a given line into segments pro- portional to two or more given lines. Ex. 654. Divide a given line into segments in the ratio of 3 to 5. Ex. 655. Divide a given line into segments proportional to 2, 3, and 4. BOOK III 175 Ex. 656. Construct two lines, given their sum and their ratio. Ex. 657. Construct two lines, given their difference and their ratio. Ex. 658. If a, b, and c are three given lines, construct x so that: (a) x:a = b:c, (&) x = — • b Proposition XIII. Theorem (Converse of Prop. XI) 415. If a straight line divides two sides of a triangle proportionally, it is parallel to the third side. A F E C Given A ABC, and BE so drawn that — To prove BE II BC. AB AE Argument 1. BE and BC are either II or not II. 2. Suppose that BE is not II BC, but that some other line through B, as BF, is II BC. AB _ AC AB ~ AF AB _ AC AB ~ AE' 3. Then 4. But 5. .-. AF= AE. 6. This is impossible. 7. .'. BE II BC. Q.E.D. Reasons 1. § 161, a. 2. §179. 3. §409. By hyp. §391. § 54, 12. § 161, b. 416. Cor. If a straight line divides two sides of a tri- angle into segments which are proportional, it is parallel to.the third side. Thus, if AB-.BB = AE: EC, BE is II BC. 176 PLANE GEOMETRY Ex. 659. In the diagram at the right, if AB = 15, AC = 12, AD = 10, and AE = 8, prove D,£ parallel to BC. Ex. 660. If AB = 50, #£ = 15, AE = 28, and EC = 12, is Z># parallel to BC ? Prove. Ex. 661. If D^ || BC, AB = 25, Di? = 5, and.4C = 20, find AE. Ex. 662. If BE || .BC, AB= 30, ^li> = 25, and EC = 4, find .4C. SIMILAR POLYGONS 417. Def. If the angles of one polygon, taken in order, are equal respectively to those of another, taken in order, the poly- gons are said to be mutually equiangular. The pairs of equal angles in the two polygons, taken in order, are called homolo- gous angles of the two polygons. 418. Def. If the sides of one polygon, taken in order as antecedents, form a series of equal ratios with the sides of another polygon, taken in order as consequents, the polygons are said to have their sides proportional. Thus, in the accom- panying figure, if a : I = b : m = c : n = d : o = e : p, the two polygons have their sides proportional. The lines forming any ratio are called homologous lines of the two polygons, and the ratio of two such lines is called the ratio of similitude of the polygons. 419. Def. Two polygons are similar if they are mutually equiangular and if their sides are proportional. BOOK III 177 Proposition XIV. Theorem 420. Two triangles which are mutually equiangular are similar. A K C D Given A ABC and DEF with Za = Zd, Zb = Ze, and Zc = ZF. To prove A ABC ~ A DEF. Argument Ebasons 1. Place A DEF on A AB C so that Z coincide with Z A, DE falling D shall on ^5 1. § 54, 14. and DF on AC. Represent A DEF in its new position by A AHK. 2. ZB = Z E = Z AHK. 2. By hyp. 3. .'. HK II BC. 3. §184. 4. AB _AC " AH~~ AK 4. § 409. 5. . AB _AC t " DE~ DF 5. §309. 6. By placing A DEF on A ABC so that 6. By steps sim Ze shall coincide with Zb, it may ilar to 1-5. be shown that AB _BC DE ~~ ^' 7. AB _BC_AC ' DE EF DF' 7. § 54, 1. 8. ,\ A ABC ~ A DEF. Q.E.D. 8. §419. 421. Cor. I. If tivo triangles have tivo angles of one equal respectively to two angles of the other, the triangles art 'similar. 178 PLANE GEOMETRY 422. Cor. II. Two right triangles are similar if an acute angle of one is equal to an acute angle of the other. 423. Cor. in. If a line is drawn parallel to any side of a triangle, this line, with the other two sides, forms a triangle which is similar to the given triangle. Ex. 663. Upon a given line as base construct a triangle similar to a given triangle. Ex. 664. Draw a triangle ABC. Estimate the lengths of its sides. Draw a second triangle DEF similar to ABC and having DE equal to two thirds of AB. Compute DF and EF. Ex. 665. Any two altitudes of a triangle are to each other inversely as the sides to which they are drawn. Ex. 666. At a cer- tain hour of the day a tree, BC, casts a shadow, CA. At the same time a vertical pole, ED, casts a shadow, DF. What measurements arc necessary to determine the height of the tree ? E ^ Ex. 667. If CA is found to be 64 feet ; DF, 16 feet what is the height of the tree ? Ex. 668. To find the distance across a river from A to B, a point C was located so that BC was perpendicular to AB at B. CD was then measured off 100 feet in length and perpendicular to BC at C. The line of sight from D to A intersected BC at E. By measurement CE was found to be 90 feet and EB 210 feet. What was the distance across the river ? BOOK III 179 Ex. 669. Two isosceles triangles are similar if the vertex angle of one equals the vertex angle of the other, or if a base angle of one equals a base angle of the other. 424. It follows from the definition of similar polygons, § 419, and from Prop. XIV that: (1) Homologous angles of similar triangles are equal. (2) Homologous sides of similar triangles are proportional. (3) Homologous sides of similar Mangles are the sides opposite equal angles. 425. Note. In case, therefore, it is desired to prove four lines pro- portional, try to find a pair of triangles each having two of the given lines as sides. If, then, these triangles can be proved similar, their homologous sides will be proportional. By marking xoith colored crayon the lines required in the proportion, the triangles can readily be found. If it is desired to prove the product of two lines equal to the product of two other lines, prove the four lines proportional by the method just sug- gested, then put the product of the extremes equal to the product of the means* 426. Def. The length of a secant from an external point to a circle is the length of the segment included between the point and the second point of intersection of the secant and the circumference. Ex. 670. If two chords intersect within a circle, es- tablish a proportionality among the segments of the chords. Place the product of the extremes equal to the product of the means, and state your result as a theorem. Ex. 671. If two secants are drawn from any given point to a circle, what are the segments of the secants? Does the theorem of Ex. 670 still hold with regard to them? Ex. 672. Rotate one of the secants of Ex. 671 about fjg^ 88 «n. the point of intersection of the two until the rotating fC \ secant becomes a tangent. What are the segments of the f \ | secant which has become a tangent ? Does the theorem of \ \ J Ex. 670 still hold ? Prove. ^\_ *S *By the product of two lines is meant the product of their measure-numbers. This wiH \p discussed aerain in Book IV. 180 PLANE GEOMETRY Proposition XV. Theorem 427. Two triangles which have their sides proportional are similar. E Given A BEF and ABC such that — = — = AB BO AG To prove A DEF ~ A ABC. Argument 1. On BE lay off DH — AB, and on DF lay off BK=AC. 2. Draw ££ ' o BE _BF f AB~ AC . BE BF 4. .•. = — • BH BK 5. .\ HK II EF. 6. .-. A DEF ~ A BHK. It remains to prove A BHK = A ^£(7. „ BE _EF BH~ HK But BE EF I.e. BE BH EF BC AB BC 9. .'.HK=BC. 10. Now BH= AB and DK=AC 11. .\ A BHK=A ABC. 12. But A BEF~ A DiMr. 13. .-. A BEF ~ A ^BC. Q.E.D. Reasons 1. §54,14. 2. §54,15. 3. By hyp. 4. §309. 5. §415. 6. §423. 7. §424,2. 8. By hyp. 9. §391. 10. Arg. 1. 11. §116. 12. Arg. 6. 13. §309. Ex. 673. If the sides of two triangles are 9, 12, 15, and 6, 8, 10, respectively, are the triangles similar ? Explain. BOOK III 181 Ex. 674. Construct a triangle that shall have a given perimeter and shall be similar to a given triangle. Ex. 675. Construct a trapezoid, given the two bases and the two diagonals. Hint. How do the diagonals of a trapezoid divide each other ? Proposition XVI. Theorem 428. If two triangles have an angle of one equal to an angle of tlie other, and the including sides propor- tional, the triangles are similar. Given A ABO and DEF with Z A = Z D and — = A To prove A AB C ~ A DEF. DE BF Argument Reasons 1. Place A DEF on A 4BC so that Z coincide with Z A, DE falling and DF on AC. Kepresent in its new position by A ASK. D shall on AB, ADEF 1. § 54, 14. ?. AB _AC m DE~ DF' 2. By hyp. 3. AB_AO t AH~~ AK 3. §309. 4. .'. HK || BC. 4. §415. 5. .: A ABC ~ A AHK. 5. §423. 6. But A AHK is A DEF transferred to a 6. Arg. 1. different position. 7. .'. A ABC ~ A DEF. Q.E.D. 7. § 309. * Ex. 676. Two triangles are similar if two sides and the median drawn to one of these sides in one triangle are proportional to two sides and the corKe'|ponding median in the other triangle. 182 PLANE GEOMETRY a B E Fig. 2. Ex. 677. In triangles AB C and DBC, Fig. \,AB = AC and BD = BC. Prove triangle ABC similar to triangle DBG. Ex.678. In Fig. 2, AB . AG = AE : AD. Prove triangle AB G simi- lar to triangle ADE. Ex. 679. If in triangle ABC, Fig. 3, GA = BC, and if 7) is a point such that GA : AB = AB : .4Z>, prove ^1Z? = BD. Ex. 680. Construct a triangle similar to a given triangle and having the sum of two sides equal to a given line. Proposition XVII. Theorem 429. Two triangles that have their sides parallel each to each, or perpendicular each to each, are similar. Fig. 1. Given A ABC and A ! B'c', with AS, BC, and CA II (Fig. 1) or _L (Fig. 2) respectively to a'b', b'c', and c'a'. To prove A AB C ~ A A'b'c'. Argument 1. AB, BC, and CA are II or _L respectively to A'B\ B'c', and C'A'. 2. .*. A A, B, and C are equal respectively or are sup. respectively to A A', B ', and c\ Reasons 1. By hyp. 2. §§198,201. BOOK III 183 Argument 3. Three suppositions may be made, there- fore, as follows : (1) Za + Za'=2tLA,Zb+Zb' = 2 rt. A, Zc + Zc' = 2 rt. A. (2) Za=Za',Zb + Zb' = 2i±A, Z C + Z c" = 2 rt. A. (3) Z A = Z A', Zb = Zb'; hence, also, Za= Z &. 4. According to (1) and (2) the sum of the A of the two A is more than four rt. A. 5. But this is impossible. 6. .-. (3) is the only supposition admissible; i.e. the two A are mutually equi- angular. 7. .-. AABC~AA'B'c'. q.e.d. Reasons 3. §161, a. 4. §54,2. 5. §204. 6. 161, b. 7. §420. 430. Question. Can one pair of angles in Prop. XVII he supple- mentary and the other two pairs equal ? SUMMARY OP CONDITIONS FOR SIMILARITY OF TRIANGLES 431. I. Two triangles are similar if they are mutually equi- angular. (a) Two triangles are similar if two angles of one are equal respectively to two angles of the other. (6) Two right triangles are similar if an acute angle of one is equal to an acute angle of the other. (c) If a line is drawn parallel to any side of a triangle, this line, with the other two sides, forms a triangle which is similar to the given triangle. II. Two triangles are similar if their sides are proportional. III. Two triangles are similar if they have an angle of one equal to an angle of the other, and the including sides pro- portional. IV. Two triangles are similar if their sides are parallel each to ea^h, or perpendicular each to each. 184 PLANE GEOMETRY Ex. 681. Inscribe a triangle in a circle and circumscribe about the circle a triangle similar to the inscribed triangle. Ex. 682. Circumscribe a triangle about a circle and inscribe in the circle a similar triangle. Ex. 683. The lines joining the mid-points of the sides of a triangle form a second triangle similar to the given triangle. Ex. 684. ABC is a triangle inscribed in a circle. A line is drawn from A to P, any point of BC, and a chord is drawn from B to a point Q in arc BC so that angle ABQ equals angle ABC. Prove AB x AC = AQ x AB. Proposition XVIII. Theorem 432. The bisector of an angle of a triangle divides the opposite side into segments which are proportional to the other two sides. Given A ABC with BP the bisector of Z ABC. To prove AP:PC= AB:BC. Argument Reasons 1. Through C draw CE II PB, meeting AB 1. §179. prolonged at E. 2. In A AEC, AP : PC — AB : BE. 2. §410. 3. Now Z 3 = Z 1. 3. §190. 4. AndZ4 = Z2. 4. §189. 5. But Z 1 = Z 2. 5. By hyp. 6. .-. Z3 = Z4. 6. § 54, 1. 7. .-. BC = BE. 7. §162. 8. .\ AP : PC = AB : BC. Q. E.D. 8. §309. BOOK III 185 Ex. 685. The sides of a triangle are 8, 12, and 15. Find the seg- ments of side 8 made by the bisector of the opposite angle. Ex. 686o In the triangle of Ex. 685, find the segments of sides 12 and 15 made by the bisectors of the angles opposite. Proposition XIX. Theorem 433. The bisector of an exterior angle of a triangle di- vides the opposite side externally into segments which are proportional to the other two sides. Given A ABC, with BP the bisector of exterior Z CBF. To prove AP : PC = AB : BC. Argument Reasons 1. Through C draw CE 11 PB, meeting AB 1. §179. at E. 2. Then in A ABP, AP : PC = AB : BE. 2. §409. 3. Now Z3 = Z1. 3. §190. 4. And Z4 = Z2. 4. §189. 5. But Z1=Z2. 5. By hyp. G. .-. Z3 = Z4. 6. § 54, 1. 7. ,\ BC= BE. 7. § 162. - 8. ,\ AP: PC=AB : BC. 8. §309. Q.E.D. Ex. 687. Compare the lettering of the figures for Props. XVIII and XIX, and also the steps in the argument. Could one argument serve for the two cases ? Ex. 688. The sides of a triangle are 9, 12, and 16. Find the segments of side 9 made by the bisector of the exterior angle at the opposite vertex. 186 PLANE GEOMETRY 434. Def. A line is divided harmonically if it is divided internally and externally into segments whose ratios are numeric cally equal ; thus, if line AC is divided internally at P and ex- ternally at Q so that the ratio of | f \ AP to PC is numerically equal to ^ PC Q the ratio of AQ to QC, AC is said to be divided harmonically. Ex. 689. The bisectors of the interior and exterior angles at any vertex of a triangle divide the opposite side harmonically. Ex. 690. Divide a given straight line harmonically in the ratio of 3 to 6 ; in the ratio of a to 6, where a and b are given straight lines. Proposition XX. Theorem 435. In two similar triangles any two homologous alti- tudes have the same ratio as any two homologous sides. Given two similar A ABC and DEF f with two corresponding altitudes AH and DK. To prove AH DK _AB ' DE _BC _ ~ EF~ CA ' FD Argument 4. But^ In rt. A ABH and DEK, Z B = AE. .-. AABH~ A DEK. , AH, oppo site Z B _ A B, opposite Z.BHA ' ' DK, opposite Z.E DE,o])ipositeZ.EKD BC _ CA DE EF FD ' AH = AB = BC _ CA DK~ DE EF FD Q.E.D. Reasons 1. §424,1. 2. §422. 3. § 424, 2. 4. §424,2. 5. §54,1. book in 187 Proposition XXI. Theorem 436. If three or more straight lines drawn through a common point intersect two parallels, the corresponding segments of the parallels are proportional. Given lines PA, PB, PC, PD drawn through a common point P and intersecting the II lines AB and A'd' at points A, B, C, D and A\ B', C', d', respectively. _ AB BC CD To prove —_==_—- = -—-. a'b' b'c' c'd' Argument Reasons 1. AD II A'D'. 1. % hyp. 2. .'. A APB ~ A A'PB f . 2. §423. 3. AB _PB A'B'~ PB 1 ' 3. § 424, 2. 4. Likewise A BPC ~ A B f P&, 4. Args. 1-2. 5. And ^=^L. b'c' pb' 5. § 424, 2. 6. AB _ BC " a'b' ~ b'c'' 6. § 54, 1. 7. Likewise it can be proved that 7. By steps sim- BC _ CD B'C'~ C'D'' ilar to 1-6. 8. AB BC CD .'• = = . Q.E.D. A'B' B'C' CD' 8. § 54, 1. Ex. 691. If three or more non-parallel straight lines intercept pro- portional segments on two parallels, they pass through a common point. 188 PLANE GEOMETRY Ex. 692. A man is riding in an automobile at the uniform rate of 30 miles an hour on one side of a road, while on a footpath on the other side a man is walking in the opposite direction. If the distance between the footpath and the auto track is 44 feet, and a tree 4 feet from the footpath continually hides the chauffeur from the pedestrian, does the pedestrian walk at a uniform rate ? If so, at what rate does he walk ? Ex. 693. Two sides of a triangle are 8 and 11, and the altitude upon the third side is 6. A similar triangle has the side homologous to 8 equal to 12. Compute as many parts of the second triangle as you can. Ex. 694. In two similar triangles, any two homologous bisectors are in the same ratio as any two homologous sides. Ex. 695. In two similar triangles, any two homologous medians are in the same ratio as any two homologous sides. 437. Drawing to Scale. Measure the top of your desk. Make a drawing on paper in which each line is -^ as long as the corresponding line of your desk. Check your work by measuring the diagonal of your drawing, and the corresponding line of your desk. This is called drawing to scale. Map draw- ing is a common illustration of this principle. The scale of the drawing may be represented : (1) by saying, " Scale, -j^" or " Scale, 1 inch to 12 inches " ; (2) by actually drawing the scale as indicated. B 12 24 36 Ex. 696. Using the scale above, draw lines on paper to represent 24 inches ; 3 feet 3 inches. Ex. 697. On the black- board draw, to the scale above, a circle whose diame- ter is 28 feet. Ex. 698. The figure represents a farm drawn to the scale indicated. Find the cost of putting a fence around the farm, if the fenc- ing costs $2.50 per rod. B 40 160 BOOK III 189 Proposition XXII. Theorem 438. If two polygons are composed of the same number of triangles, similar each to each and similarly placed, the polygons are similar. C E J Given polygons ABODE and FGHIJ with A ABC ~ A FGH f AACD ~ A FHI, A ABE ~ A FIJ. To prove polygon ABODE ~ polygon FGHIJ. Argument Reasons 1. In A ABC and FGH, Z.B = Z G. 1. § 424, 1. 2. Also Z 1 = Z 2. 2. § 424, 1. 3. In A ACD and i?W/, Z 3 = Z4. 3. § 424, 1. 4. .-.Z1 + Z3=Z2 + Z4. 4. § 54, 2. 5. .-. Z .BCD = Z (?#7. 5. § 309. 6. Likewise Z CDE = Z flTJ", Z.E = /.J, 6. By steps sim- and Zeab = Zjfg. ilar to 1-5. 7. .-. polygons abcde and FGHIJ are mutually equiangular. 7. By proof. 8. In A ^£C and TOff. 4? = *?= ^. .FG G# iZF 8. § 424, 2. 9. In A Sandra/, ^ = ^ = ^. JZF HI IF 9. § 424, 2. 10. And in A ADE and FIJ, — = — = ^. TF I J JF 10. § 424, 2. 11. . AB _BC _CD _DE _EA FG ~~ &#~ #/ "~ IJ~~JF' 11. § 54, 1. 12. .-. polygon ABCDE ~ polygon FGHIJ. 12. §419 Q.E.D. 190 PLANE GEOMETRY 439. Cor. Any two similar polygons may be divided into the same number of triangles similar each to each and similarly placed. Proposition XXIII. Problem 440. Upon a line homologous to a side of a given poly- gon, to construct a polygon similar to the given polygon. C R Given polygon AD and line MQ homol. to side AE. To construct, on MQ, a polygon ~ polygon AD. I. Construction 1. Draw all possible diagonals from A, as AC and AD. 2. At M, beginning with MQ as a side, construct A 7, 8, and 9 equal respectively to A 1, 2, and 3. § 125. 3. At Q, with MQ as a side, construct Z 10 equal to A 4, and prolong side QP until it meets ML at P. § 125. 4. At P, with PM as a side, construct Zll equal to Z5, and prolong side PO until it meets MR at 0. § 125. 5. At 0, with OM as a side, construct Z12 equal to A6, and prolong side ON until it meets MF at N. § 125. 6. MP is the polygon required. II. Proof Argument 1. A ADE ~ A MPQ, A ACD ~ A MOP, and A ABC ~ A MNO. 2. .-. polygon MP ~ polygon AD. q.e.d. Reasons 1. §421. 2. § 438. book in 191 III. The discussion is left as an exercise for the student. Proposition XXIV. Theorem 441. The perimeters of two similar polygons are to each other as any two homologous sides. Given ~ polygons P and Q, with sides a, b, c, d, e, and / homol. respectively to sides k, I, m, n, o, andjp. _ perimeter of P a To prove *■ — = -. perimeter of Q k Argument a _^_ !L— -— f — / ^ I m n o p k 2. IV U IIV IV \J JJ a + b+ c +d+e+f = a k + I +m + ?i + o-|-/ ) & Thati S ,P erimeterofP = ° perimeter of Q k Q.E.D. Reasons 1. §419. 2. § 401. 3. § 309. Ex. 699. The perimeters of two similar polygons are 152 and 138 ; a side of the first is 8. Find the homologous side of the second. Ex. 700. The perimeters of two similar polygons are to each other as any two homologous diagonals. Ex. 701. The perimeters of two similar triangles are to each other as any two homologous medians. Ex. 702. If perpendiculars are drawn to the hypotenuse of a right triangle at its extremities, and if the other two sides of the triangle are prolonged to meet these perpendiculars, the figure thus formed contains fivf, triangles each of which # is similar to any one of the others. 192 PLANE GEOMETRY Proposition XXV. Theorem 442. In a right triangle, if the altitude upon the hypotenuse is drawn: I. The triangles thus formed are similar to the given triangle and to each other. II. The altitude is a mean proportional between the segments of the hypotenuse. III. Either side is a mean proportional between the whole hypotenuse and the segment of the hypotenuse adjacent to that side. Given rt. A ABC and the altitude CD upon the hypotenuse. To prove : I. A BCD ~ A ABC, A ADC ~ A ABC, and A BCD ~ A ADC. II. BD: CD= CD : DA. III. AB : BC=BC:DB and AB: AC= AC: AD. I. Argument 1. In rt. A B CD and ABC, Z.B = /.B. 2. .'. A BCD ~ A ABC. 3. In rt. A ADC and ABC, Za = A A. 4. .-. A ADC ~ A ABC. 5. .:Z1 = Zb. 6. .'. A BCD ~ A ADC. Q.E.D. Reasons 1. By iden. 2. §422. 3. By iden. 4. § 422. 5. § 424, 1. 6. §422. II. The proof of II is left as an exercise for the student Hint. Mark (with colored chalk, if convenient) the lines required in the proportion. Decide which triangles will furnish these lines, and use the fact that homologous sides of similar triangles are proportional. BOOK III 193 III. The proof of III is left as an exercise for the student. Hint. Use the same method as for II. 443. Cor. I. In a right triangle, if the altitude upon the hypotenuse is drawn: I. The square of the altitude is equal to the product of the segments of the hypotenuse. Thus, CD 2 = BD • DA. (See § 442, II.) II. The square of either side is equal to the product of the whole hypotenuse and the segment of the hypote- nuse adjacent to that side. Thus, BC 2 = AB • DB } and AX?— AB-AD. (See § 442,111.) 444. Cor. II. If from any point in the circumference of a circle a perpendicular to a diameter is drawn, and if chords are drawn from the point to the ends of the diameter : I. The perpendicular is a mean proportional between the segments of the diameter. II. Either chord is a mean proportional between the whole diameter and the segment of the diameter adjacent to the chord. Hint. A APB is a rt. A. Apply Prop. XXV, II and III. Ex. 703. In a right triangle, the squares of the two sides are pro- portional to the segments of the hypotenuse made by the altitude upon it. Hint. Apply § 443, II. Ex. 704. The sides of a right triangle are 9, 12, and 15. (1) Compute the segments of the hypotenuse made by the altitude upon it. (2) Compute the length of the altitude. Ex. 705. If the two arms of a right triangle are 6 and 8, compute the length of the perpendicular from the vertex of the right angle to the hypotenuse. 194 PLANE GEOMETRY Proposition XXVI. Problem 445. To construct a mean proportional between two given lines. / t' N f I !0 3 rr^rr^ Given two lines, m and w. To construct a line Z so that m:l = l:n. I. Construction 1. On any indefinite str. line, as AB, lay off AG = m and CZ) = n. 2. With 0, the mid-point of AD, as center and with a radius equal to OB, describe a semicircumference. 3. At G construct GEWAD, meeting the semicircumference at E. § 148. 4. CE is the required line I. II. The proof and discussion are left as an exercise for the student. Ex. 706. By means of § 444, II, construct a mean proportional be- tween two given lines by a method different from that given in Prop. XXVI. Ex. 707. Use the method of Prop. XXVI to construct a line equal to V3 ab, a and b being given lines. Analysis. Let x = the required line. Then x = V3ab. .-. x* = 3 ab. ,\ 3 a : x = x : b. Ex. 708. Construct a line equal to a V3, where a is a given line. Ex. 709. Using a line one inch long as a unit, construct a line equal to VS ; V5 ; V6. Choosing your own unit, construct a line equal to 3 a/2, 2\% 5\/5. book in 195 Proposition XXVII. Theorem 443. In any right triangle the square of the hypotenuse is equal to the sum of the squares of the othe? ■ two sides. B cd 4 Given rt. A ABC, with its rt. Z. at C. To prove a 2 -\- b 2 = c 2 . Argument Reasons 1. From C draw CD _L AB forming seg- 1. §155. ments I and d. 2. a 2 = cL 2. § 443, II. 3. b 2 = c • d. 3. § 443, II. 4. .\a 2 + b 2 =c(l+d). 4. § 54, 2. 5. .*. a 2 + b 2 = c • c = c 2 . q.e.d. 5. §309. 447. Cor. I. The square of either side of a right tri- angle is equal to the square of the hypotenuse minus the square of the other side. 448. Cor. n. The diagonal of a square is equal to its side multiplied by the square root of two. Outline of Proof 1. d 2 =s 2 + s 2 =2s 2 . 2. .\ d = s-y/2. q.e.d. 449. Historical Note. The property of the right triangle stated in Prop. XXVII was known at a very early date, the ancient Egyptians, 2000 b.c, having made a right triangle by stretching around three pegs a cor* measured off into 3, 4, and 5 units. See Note, Book IV, § 510. 196 PLANE GEOMETRY Ex. 710. By means of § 448, construct a line equal to V2 inches. Ex. 711. If a side of a square is 6 inches, find its diagonal. Ex. 712. The hypotenuse of a right triangle is 15 and one arm is 9. Find the other arm and the segments of the hypotenuse made by the perpendicular from the vertex of the right angle. Ex. 713. Find the altitude of an equilateral triangle whose side is 6 inches. Ex. 714. Find a side of an equilateral triangle whose altitude is 8 inches. Ex. 715. Divide a line into segments which shall be in the ratio of 1 to V2. Ex. 716. The radius of a circle is 10 inches. Find the length of a chord 6 inches from the center ; 4 inches from the center. Ex. 717. The radius of a circle is 20 inches. How far from the cen- ter is a chord 'whose length is 32 inches ? whose length is 28 inches ? Ex. 718. Tn a circle a chord 24 inches long is 5 inches from the center. How far from the center is a chord whose length is 12 inches ? 450. Def. The projection of a point upon a line is the foot of the perpendicular from the point to the line. 451. Def. The projection of a line segment upon a line is the segment of the second line included between the projections of the extremities of the first line upon the second. Thus, C is the projection of A upon MN, D is the projection of B upon MN } and CD is the projection of AB upon MN. B B 9 A. M~C Irk M C D N M V * A Ex. 719. In the figures above, under what condition will the projec- tion of AB on M2f be a maximum ? a minimum ? Will the projection CD ever be equal to AB ? greater than AB ? Will the projection ever be a point ? Ex. 720. In a right isosceles triangle the hypotenuse of which is 10 inches, find the length of the projection of either arm upon the hypotenuse. Ex. 721. Find the projection of one side of an equilateral triangle upon another if each side is 6 inches. BOOK III 197 Ex. 722. Draw the projections of the shortest side of a triangle upon each of the other sides : (1) in an acute triangle ; (2) in a right triangle ; (3) in an obtuse triangle. Draw the projections of the longest side in each case. Ex. 723. Two sides of a triangle are 8 and 12 inches and their included angle is 60°. Find the projection of the shorter upon the longer. Ex. 724. In Ex. 723, find the projection of the shorter side upon the longer if the included angle is 30° ; 45°. Ex. 725. Parallel lines that have equal projections on the same line are equal. Proposition XXVIII. Problem 452. In any triangle to find the value of the square of the side opposite an acute angle in terms of the other two sides and of the projection of either of these sides upon the other. A X D Fig. 1. Given ABAX, with Z.X acute; and p, the projection of b upon a. To find the value of a 2 in terms of a, b, and p. Argument 1. In rt. A BAD, a? = AD 2 + DB 2 . 2. But AD 2 = b 2 - p 2 . 3. And DB = a—p (Fig. 1) or p — a (Fig. 2) . 4. .-. DB* = a 2 — 2 ap + p 2 . 5. .-. a,* 2 = b 2 — p 2 -f a 2 — 2 ap -f p 2 ; i.e. x? = a 2 + b 2 — 2 ap. q.e.f. 453. Question. Why is it not necessary to include here the figure aad^discussion for a right triangle ? Reasons 1. §446. 2. §447. 3. § 54, 11. 4. § 54, 13. 5. §309. 198 PLANE GEOMETRY 454. Prop. XXVIII may be stated in the form of a theorem as follows : In any triangle, the square of the side opposite an acute^ angle is equal to the sum of the squares of the other two sides dimin- ished by twice the product of one of these sides and the projec- tion of the other side upon it. Ex. 726. If the sides of a triangle are 7, 8, and 10, is the angle opposite 10 obtuse, right, or acute ? Why ? Ex. 727. Apply the statement of Prop. XXVIII to the square of an arm of a right triangle. Ex. 728. Find x (in the figure for Prop. XXVIII) in terms of a and b and the projection of a upon b. Ex. 729. If the sides of a triangle are 13, 14, and 15, find the pro- jection of the first side upon the second. Ex. 730. If two sides of a triangle are 4 and 12 and the projection of the first side upon the second is 2, find the third side of the triangle Proposition XXIX. Theorem 455. In any obtuse triangle, the square of the side oppo- site the obtuse angle is equal to the sum of the squares of the other two sides increased by twice the product of one of these sides and the projection of the other side upon it. Given A BAX with Z.X obtuse, and p, the projection of b upon a. To prove (c 2 = a 2 + b 2 + 2 ap BOOK 111 199 Argument 1. In rt. A BAD, a?=Al? + Bb\ 2. But AB 2 = b 2 -p\ 3. And DB = a+p. 4. .\ BB 2 = a 2 + 2 ap +p\ 5 .-. ^ = 6 2 -p 2 +a 2 + 2ap+P 2 ; i. e. x 1 = a 2 + b 2 -f 2 ap. q.e.d. Reasons 1. §446. 2. §447. 3. §54,11. 4. § 54, 13. 5. §309. 456. From Props. XXVIII and XXIX, we may derive the following formulas for computing the projection of one side of a triangle upon another ; thus if a, 6, and c represent the sides of a triangle : From Prop. XXVIII, p = ^ + b * "~ ^ (1) 2a From Prop. XXIX, - p = a * + & 2 ~ g ( 2 ) It is seen that the second members of these two equations are identical and that the first members differ only in sign. Hence, formula (1) may always be used for computing the length of a projection. It need only be remembered that if p is positive in any calculation, it indicates that the angle oppo- site c is acute; while if p is negative, the angle opposite c is obtuse. It can likewise be shown (see Prop. XXVII) that if p = 0, the angle opposite c is a right angle. Ex. 731. Write the formula for the projection of a upon b. Ex. 732. In triangle ABC, a = 15, 6 = 20, c = 25; find the projec- tion of b upon c. Is angle A acute, right, or obtuse ? Ex. 733. In the triangle of Ex. 732, find the projection of a upon b. Is angle C acute, right, or obtuse ? Ex. 734. The sides of a triangle are 8, 14, and 20. Is the angle opposite the side 20 acute, right, or obtuse ? Ex. 735. If two sides of a triangle are 10 and 12, and their included angle is 120°, what is the value of the third side ? Ex. 736. If two sides of a triangle are 12 and 16, and their included angle is 45°, find the third side. 200 PLANE GEOMETRY Ex. 737. If in triangle ABC, angle C = 120°, prove that AS 2 = BC 2 + AC 2 + AC- BC. Ex. 738. If a line is drawn from the vertex C of an isosceles tri- angle ABC, meeting base AB prolonged at D, prove that dl> 2 -CB i = AB. BD. Proposition XXX. Theorem 457. In any triangle, the sum of the squares of any two sides is equal to twice the square of half the third side increased by twice the square of the median upon that side. Given A ABC with m a , the median to side a. To prove b 2 + c 2 = 2 (-Y + 2 m a \ Argument 1. Suppose 6 > c ; then Z ADC is obtuse and Z BDA is acute. 2. Let p be the projection of m a upon a. 3. Then from A A D C, 4. And from A ABD, & + m a 2 (2) " """ 5. .-. 6 2 -hc 2 =2^J+2m a 2 . p. Q.E.D. Reasons 1. § 173. 2. §451. 3. §455. 4. §454. 5. §54,2. BOOK III 201 458. Cor. The difference of the squares of two sides of any triangle is equal to twice the product of the third side and the projection of the median upon the third side. Hint. Subtract the equation in Arg. 4 from that in Arg. 3 (§ 457) member from member. Ex. 739. Write the formula involving the median to b ; to c. Ex. 740. Apply Prop. XXX to a triangle right-angled at A ; at B ; at C. 459. It will be seen that the formula of Prop. XXX con- tains the three sides of a triangle and a median to one of these sides. Hence, if the three sides of a triangle are given, the median to any one of them can be found ; also, if two sides and any median are given, the third side can be found. Ex. 741. If the sides of a triangle ABC are 5, 7, and 8, find the lengths of the three medians. Ex. 742. If the sides of a triangle are 12, 16, and 20, find the median to side 20. How does it compare in length with the side to which it is drawn ? Why ? Ex. 743. In triangle ABC, a = 16, b = 22, and m c = 17. Find c. Ex. 744. In a right triangle, right-angled at C, m c = 8£ ; what is c ? Find one pair of values for a and b that will satisfy the conditions of the problem. Ex. 745. The sum of the squares of the four sides of any parallel- ogram is equal to the sum of the squares of its diagonals. Ex. 746. The sum of the squares of the four sides of any quadrilateral is equal to the sum of the squares of its diagonals increased by four times the square of the line joining their mid-points. Ex. 747. Construct a triangle ABC, given 6, c, and m a . Ex. 748. Construct a triangle ABC, given a, b, and the projection of b upon a. Ex. 749. Compute the side of a rhombus whose diagonals are 12 and 16. Ex. 750. If the square of the longest side of a triangle is greater than the sum of the squares of the other two sides, is the triangle obtuse, right* or acute ? Why ? 202 PLANE GEOMETRY Proposition XXXI. Theorem 460. If through a point within a circle two chords are drawn, the product of the two segments of one of these chords is equal to the product of the two segments of the other. Given P, a point within circle 0, and AB and CD, any two chords drawn through P. To prove PA • PB = PC • PD. The proof is left as an exercise for the student. Hint. Prove A APC ~ A PDB. Ex. 751. In the figure for Prop. XXXI, if PA = 5, PB = 12, and PD = 6, find PC. Ex. 752. In the same figure, if PC = 10, PD = 8, and AB = 21, find PA and PB. Ex. 753. In the same figure, if PC = 6, DC = 22, and AB = 20, find AP and PB. Ex. 754. In the same figure, if PA = m, PC = n, and PD — r, find PB. Ex. 755. If two chords intersecting within a circle are of lengths 8 and 10, and the second bisects the first, what are the segments of the second ? Ex. 756. By means of Prop. XXXI construct a mean proportional between two given lines. Ex. 757. If two chords intersect within a circle and the segments of one chord are a and b inches, while the second chord measures d inches, construct the segments of the second chord. Hint. Find the locus of the mid-points of chords equal to d. BOOK III 203 Ex. 758. If two lines AB and CD intersect at E so that AE ■ EB = CE • ED, then a circumference can be passed through the four points A, B, C, D. Proposition XXXII. Theorem 461. // a tangent and a secant are drawn from any given point to a circle, the tangent is a mean propor- tional between the whole secant and its external segment. Given tangent PT and secant PB drawn from the point P to the circle 0. m PB To prove — - : PT PT PC Argument 1. Draw crand BT. . 2. In A PBT and CTP, Zp = Zp. 3. Z2 = Z2'. 4. .-. APBT~ A CTP. 5 . PB, opposite Z 3 in A PBT ' PT, opposite Z 3' in A CTP _ PT, opposite Z 2 in A PBT ~ PC, opposite Z 2' in A CTP Q.E.D. Reasons 1. §54,15. 2. Byiden. 3. § 362, a. 4. §421. 5. §424,2. 462. Cor. I. If a tangent and a secant are drawn from any given point to a circle, the square of the tangent is equal to the product of the whole secant and its external segment. 204 PLANE GEOMETRY 463. Cor. II If two or more secants are drawn from any given point to a circle, the product of any secant and its external segment is constant. T^r' Given secants PD, PE, PF, drawn from point P to circle O, P and let their external segments be denoted by PA, PB, PC, respectively. To prove PD • PA = PE • PB =PF • PC, Argument 1. From P draw a tangent to circle 0, as PT. 2. PT 2 = PD • PA; PT 2 = PE • PB; PT 2 = PF • PC. 3. .'. PD- PA = PE>PB = PF- PC. Q.E.D. 1. F Reasons §286. 2. §462. 3. § 54, 1. Ex. 759. If a tangent and a secant drawn from the same point to a circle measure 6 and 18 inches, respectively, how long is the external segment of the secant ? Ex. 760. Two secants are drawn to a circle from an outside point. If their external segments are 12 and 9, and the internal segment of the first secant is 8, what is the length of the second secant ? Ex. 761. The tangents to two intersecting circles from any point in their common chord (prolonged) are equal. Ex. 762. If two circumferences intersect, their common chord (pro- longed) bisects their common tangents. 464. Def. A line is said to be divided in extreme and mean ratio if it is divided into two parts such that one part is a mean proportional between the whole line and the other part. Thus, AB is divided in extreme and mean ratio at P if AB : AP = ■ AP : PB. This division is known as the golden section. BOOK III 205 Proposition XXXIII. Pi-oblem 465. To divide a line internally in extreme and mean ratio. x •«w i ' t X V ? I I I I I f I / A P B Given line AB. To find, in AB f a point P such that AB:AP = AP: PB. I. Construction 1. From B draw 50 _L ^4 5 and — \AB. § 148. 2. With O as center and with BO as radius describe a cir- cumference. 3. From A draw a secant through 6, cutting the circumfer- ence at C and D. 4. With A as center and with AC as radius draw CP, cutting AB at P. 5. AB : ^IP = AP : PP. II. Proof Argument 1. AB is tangent to circle O. 2. .-. AD : AB = AB : AC. 3. .-. AD — ^1P: AB = AB — AC: AC. 4. .-. AD— CD : AB = AB— AP : AP. 5. .'. AP : AB = PB : ^P. 6. .'. ^1P: AP = JP: PB. Q.E.D. Reasons 1. §314. 2. §461. 3. § 399. 4! § 309. 5. § 309. 6. § 395. HJ. The discussion is left as an exercise for the student. 206 PLANE GEOMETRY Ex. 763. Divide a line AB externally in extreme and mean ratio. Hint. In the figure for Prop. XXXIII prolong BA to P', making FA = AD. Then prove AB : P'^l = PA : P'B. Ex. 764. If the line I is divided internally in extreme and mean ratio, and if s is the greater segment, find the value of s in terms of I. Hint. I : s = s : I — s. Ex. 765. A line 10 inches long is divided internally in extreme and mean ratio. Find the lengths of the two segments. Ex. 766. A line 8 inches long is divided externally in extreme and mean ratio. Find the length of the longer segment. MISCELLANEOUS EXERCISES Ex. 767. Explain how the accompanying figure can be used to find the distance from A to B on opposite sides of a hill. CE = %BC, CD = \AC. ED is found by measurement to be 125 feet. What is the distance AB ? Ex. 768. A little boy wished to obtain the height of a tree in his yard. He set up a ver- tical pole 6 feet high and watched until the shadow of the pole measured exactly 6 feet. He then measured quickly the length of the tree's shadow and called this the height of the tree. Was his answer correct ? Draw figures and explain. Use this method for measuring the height of your school building and flag pole. Ex. 769. If light from a tree, as AB, is allowed to pass through a small aperture O, in a window shutter W, and strike a white screen or wall, an inverted image of the tree, as CD, is formed on the screen. If the dis- tance OE= 30 feet, OF =8 feet, and the length of the tree AB=%5 feet, find th3 length of the image CD. Under what condition will the length of the image equal the length of the tree ? This exercise illustrates the principle of the photographer's camera. Ex. 770. By means of Prop. XXXII construct a mean proportional between two given lines. Ex. 771. In a certain circle a chord 5V5 inches from the center is 20 inches in length. Find the length of a chord 9 inches from the center. BOOK III 207 Ex. 772. Compute the length of : (1) the common external tangent, (2) the common internal tangent, to two circles whose radii are 8 and 6, respectively, and the distance between whose centers is 20. Ex. 773. If the hypotenuse of an isosceles right triangle is 16 inches, what is the length of each arm ? Ex. 774. If from a point a tangent and a secant are drawn and the segments of the secant are 4 and 12, how long is the tangent ? Ex. 775. Given the equation VOl+JIl - ?_5» ; so lve for x. X + c c Ex. 776. Find a mean proportional between a 2 + 2 ab + b 2 and a 2 - 2 ab + b 2 . Ex. 777. The mean proportional between two unequal lines is less than half their sum, Ex. 778. The diagonals of a trapezoid divide each other into segments which are proportional. Ex. 779. ABC is an isosceles triangle inscribed in a circle. Chord BD is drawn from the vertex B, cutting A y the base in any point, as E. Prove BD : AB—AB : BE. Ex. 780. In a triangle ABC the side AB is 305 feet. If a line parallel to BC divides AC in the ratio of 2 to 3, what are the lengths of the segments into which it divides AB ? Ex. 781. Construct, in one figure, four lines whose lengths shall be that of a given unit multiplied by V2, V3, 2, V5, respectively. Ex. 782. Two sides of a triangle are 12 and 18 inches, and the perpen- dicular upon the first from the opposite vertex is 9 inches. What is the length of the altitude upon the second side ? Ex. 783. If a : b = c : d, show that K«) ! K 6 )=K«)=K*> Also translate this fact into a verbal statement. Ex. 784. If a constant is added to or subtracted from each term of a proportion, will the resulting numbers be in proportion ? Give proof. Ex. 785. li r : s = t : q, is3r + -:s=7t:2q? Prove your answer. u Ex. 786. One segment of a chord drawn through a point 7 units from the center of a circle is 4 units long. If the diameter of the circle is 15 units, what is the length of the other segment ? Ex. 787. The non-parallel sides of a trapezoid and the line joining the mid-points of the parallel sides, if prolonged, are concurrent. 208 PLANE GEOMETRY Ex. 788. Construct a circle which shall pass through two given points and be tangent to a given straight line. Ex. 789. The sides of a triangle are 10, 12, 15. Compute the lengths of the two segments into which the least side is divided by the bisector of the opposite angle. „ Ex. 790. AB is a chord of a circle, and CE is any A ^^%^\. R chord drawn through the middle point C of arc AB, cut- ting chord AB at D. Prove CE : CA = CA : CD. Ex. 791. Construct a right triangle, given its perime- ter and an acute angle. ^- & Ex. 792. The base of an isosceles triangle is a, and the perpendicular let fall from an extremity of the base to the opposite side is b. Find the lengths of the equal sides. Ex. 793. AD and BE are two altitudes of triangle CAB. Prove that AD:BE= CA : BC. Ex. 794. If two circles touch each other, their common external tangent is a mean proportional between their diameters. Ex. 795. If two circles are tangent internally, all chords of the greater circle drawn from the point of contact are divided proportionally by the circumference of the smaller circle. Ex. 796. If three circles intersect each other, their common chords pass through a common point. Ex. 797. The square of the bisector of an angle of a triangle is equal to the product of the sides of this angle diminished by the product of the segments of the third side made by the bisector. Given A ABC with t, the bisector of ZB, dividing side b into the two segments s and r. To prove t 2 = ac — rs. Outline of Proof 1. Prove that a : t = t+ m : c. 2. Then ac = P + tm = t 2 + rs. 3. .*. t 2 = ac — rs. Ex. 798. In any triangle the product of two sides is equal to the product of the altitude upon the third side and the diameter of the circumscribed circle. Hint. Prove A ABB ~AEBC. Then prove ac = Jut. BOOK IV AREAS OF POLYGONS 466. A surface may be measured by finding how many times it contains a unit of surface. The unit of surface most fre- quently chosen is a square whose side is of unit length. If the unit length is an inch, the unit of surface is a square whose side is an inch. Such a unit is called a square inch. If the unit length is a foot, the unit of surface is a square whose side is a foot, and the unit is called a square foot. 467. Def. The result of the measurement is a number, which is called the measure-number, or numerical measure, or area of the surface. B C U Fig. 1. Rectangle A C = 15 U. 468. Thus, if the square IT is contained in the rectangle ABCD (Fig. 1) 15 times, then the measure-number or area of rectangle ABCD, in terms of U, is 15. If the given square is not -contained in the given rectangle an integral number of 209 210 PLANE GEOMETRY G U E H Fig. 2. Rectangle EG = 8 U+. times without a remainder (see Fig. 2), then by taking a square which is an aliquot part of U, as one fourth of U, and applying M E H Fig. 3. Rectangle EG = *£ U+= U\ U+. it as a measure to the rectangle (see Fig. 3) a number will be obtained which, divided by four,* will give another (and G 1 -u- 1 Fig. 4. Rectangle AO = ^U+= 111 U+. usually closer) approximate area of the given rectangle. By proceeding in this way (see Fig. 4), closer and closer approxi- mations of the true area may be obtained. * It takes four of the small squares to make the unit itself. BOOK IV 211 469. If the sides of the given rectangle and of the unit square are commensurable, a square may be found which is an aliquot part of U, and which is contained in the rectangle also an integral number of times. 470. If the sides of the given rectangle and of the unit square are incommensurable, then closer and closer approxima- tions to the area may be obtained, but no square which is an aliquot part of U will be also an aliquot part of the rectangle (by definition of incommensurable magnitudes). There is, how- ever, a definite limit which is approached more and more closely by the approximations obtained by using smaller and smaller subdivisions of the unit square, as these subdivisions approach zero as a limit.* 471. Def. The measure-number, or area, of a rectangle which is incommensurable with the chosen unit square is the limit which successive approximate measure-numbers of the rec- tangle approach as the subdivisions of the unit square approach zero as a limit. For brevity the expression, the area of a figure, is used to mean the measure-number of the surface of the figure with respect to a chosen unit. 472. Def. The ratio of any two surfaces is the ratio of their measure-numbers (based on the same unit). 473. Def. Equivalent figures are figures which have the same area. The student should note that : Equal figures have the same shape and size; such figures can be made to coincide. Similar figures have the same shape. Equivalent figures have the same size. Ex. 799. Draw two equivalent figures that are not equal. * For none of these approximations can exceed a certain fixed number, for example (h + \)(b + 1), where the measure applied is contained in the altitude h times with a re- mainder less than the measure, and in the base b times with a remainder less than the 212 PLANE GEOMETRY Ex. 800. Draw two equal figures on the blackboard or cut them out of paper, and show that equal figures may be added to them in such a way that the resulting figures are not equal. Are they the same size ? Ex. 801. Draw figures to show that axioms 2, 7, and 8, when applied to equal figures, do not give results which satisfy the test for equal figures. 1 T t ff X f JLL > 7 ^ t B ~"---.° r TT T ■*• ±± TV t-i. Fig. 1. Fig. 2. Ex. 802. Fig. 1 above represents a card containing 64 small squares, cut into four pieces, I, II, III, and IV. Fig. 2 represents these four pieces placed together in different positions forming, as it would seem, a rectangle containing 65 of these small squares. By your knowledge of similar triangles, try to explain the fallacy in the construction. 474. Historical Note. Geometry is supposed to have had its origin in land surveying, and the earliest traditions state that it had its beginning in Egypt and Babylon. The records of Babylon were made on clay tablets, and give methods for finding the approximate areas of several rectilinear figures, and also of the circle. The Egyptian records were made on papyrus. Herodotus states that the fact that the inundations of the Nile caused changes in the amount of taxable land, rendered it neces- sary to devise accurate land measurements. This work was done by the Egyptian priests, and the earliest manu- script extant is that of Ahmes, who lived about 1700 b.c. This manu- script, known as the Rhind papyrus, is preserved in the British Museum. It is called "Directions for knowing all dark things," and is thought to be a copy of an older manuscript, dating about 3400 b.c. In addition to problems in arithmetic it contains a discussion of areas. Problems on pyramids follow, which show some knowledge of the properties of similar figures and of trigonometry, and which give dimensions, agreeing closely with those of the great pyramids of Egypt. The geometry of the Egyptians was concrete and practical, unlike that of the Greeks, which was logical and deductive, even from its beginning. BOOK IV 213 Proposition I. Theorem 475. The area of a rectangle is equal to the product of its base and its altitude. (See § 476.) B C U A D u Given rectangle ABCD, with base AD and altitude AB, and let U be the chosen unit of surface, whose side is u. To prove the area of ABCD = AD • AB. I. If AD and AB are each commensurable with u. (a) Suppose that u is contained in AD and AB each an in- tegral number of times. Argument 1. Lay off u upon AD and AB, respectively. Suppose that u is contained in AD r times, and in AB s times. 2. At the points of division on AD and on A B erect Js to AD and AB, respectively. 3. Then rectangle ABCD is divided into unit squares. 4. There are r of these unit squares in a row on AD, and s rows of these squares in rectangle ABCD. 5. .-. the area of ABCD = r «s. 6. But r and s are the measure-numbers of AD and AB, respectively, referred to the linear unit u. 7. .-'.*>fche area of A BCD = AD • AB. q.e.d. 1. Reasons §335. 2. § 63. 3. §466. 4. Arg. 1. 5. §467. 6. Arg. 1. 7. § 309. 214 PLANE GEOMETRY (b) If u is not a measure of AD and AB, respectively, but if some aliquot part of u is such a measure. The proof is left to the student. U A QD u II. If AD and AB are each incommensura ble with u. Argument 1. Let m be a measure of u. Apply m as a measure to AD and iB as many times as possible. There will be a remainder, as QD, on AD, and a re- mainder, as PB, on AB, each less than m. 2. Through Q draw W_L ^-D, and through P draw PiW J_ AB. 3. Now J. Q and ^LP are each commensu- rable with the measure m, and hence commensurable with u. 4. .*. the area of rectangle APMQ=A Q- A P. 5. Now take a smaller measure of u. No matter how small a measure of u is taken, when it is applied as a measure to AD and AB, the remainders, QD and PP, will be smaller than the measure taken. 6. .-. the difference between AQ and AD may be made to become and remain less than any previously assigned segment, however small. Reasons 1. § 339. 2. § 63. 3. § 337. 4. § 475, I. 5. § 335. 6. Arg. 5. BOOK IV 215 Argument 7. Likewise the difference between AF and AB may be made to become and remain less than any previously as- signed segment, however small. 8. .-. A Q approaches AD as a limit, and AP approaches AB as a limit. 9. .-. AQ-AP approaches AD-AB as a limit. 10. Again, the difference between APMQ and A BCD may be made to become . and remain less than any previously assigned area, however small. 11. .-. APMQ approaches ABCD as a limit. 1 2. But the area of APMQ is always equal to AQ • AP. 13. .-. the area of ABCD = AD-AB. Q.E.D. Reasons 7. Arg. 5. 8. § 349. 9. § 477. 10. Arg. 5. 11. § 349. 12. Arg. 4. 13. § 355. III. If AD is commensurable with u but AB incommensu- rable with u, The proof is left as an exercise for the student. 476. Note. By the product of two lines is meant the product of the measure-numbers of the lines. The proof that to every straight line segment there belongs a measure-number is given in § 595. 477. If each of any finite number of variables approaches a finite limit, not zero, then the limit of their product is equal to the product of their limits. (See § 593.) 478. Cor. I. The area of a square is equal to the square of its side. 479. Cor. n. Any two rectangles are to each other as the products of their bases and their altitudes. Outline of Proof. Denote the two rectangles by R and R\ their bases by b and b', and their altitudes by h and h', re- spectively. Then R = b • h and R' = b '• h'. R Bf b-h b' • h'' 216 PLANE GEOMETRY 480. Cor. III. (a) Two rectangles having equal bases are to each other as their altitudes, and (b) two rectangles having equal altitudes are to each other as their bases. Outline op Proof (a\ — — ^ ' ^ — ^L (1\ ;?_ — fr • j _ & Ex. 803. Draw a rectangle whose base is 7 units and whose altitude is 4 units and show how many unit squares it contains. Ex. 804. Find the area of a rectangle whose base is 12 inches and whose altitude is 5 inches. Ex. 805. Find the area of a rectangle whose diagonal is 10 inches, and one of whose sides is 6 inches. Ex. 806. If the area of a rectangle is 60 square feet, and the base, 5 inches, what is the altitude ? Ex. 807. If the base and altitude of a rectangle are 2\ inches and \\ inches, "respectively, find the area of the rectangle. Ex. 808. Find the area of a square whose diagonal is 8 V2 inches. Ex. 809. Find the successive approximations to the area of a rec- tangle if its sides are V 10 and V 5, respectively, using 3 times 2 for the first approximation, taking the square roots to tenths for the next, to hundredths for the next, etc. Ex. 810. Compare two rectangles if a diagonal and a side of one are d and s, respectively, while a diagonal and side of the other are d' and s'. Ex. 811. Construct a rectangle whose area shall be three times that of a given rectangle. Ex. 812. Construct a rectangle which shall be to a given rectangle in the ratio of two given lines, m and n. Ex. 813. Compare two rectangles whose altitudes are equal, but whose bases are 15 inches and 3 inches, respectively. Ex. 814. From a given rectangle cut off a rectangle whose area is two thirds that of the given one. Ex. 815. If the base and altitude of a certain rectangle are 12 inches and 8 inches, respectively, and the base and altitude of a second rectangle are 6 inches and 4 inches, respectively, compare their areas. BOOK IV 217 Proposition II. Theorem 481. The area of a parallelogram equals the product of its base and its altitude. R b C Given O ARCD, with base b and altitude h. To prove area of ARCD, = b-h. Argument 1. Draw the rectangle ERCF, having b as base and h as altitude. 2. In rt. A DCF and ARE, DC = AR. 3. Also CF = RE. 4. .*. A DCF = A ARE. 5. Now figure ARCF = figure ARCF. G. .-. area of ARCD = area of ERCF. 7. But area of ERCF =b-h. 8. .-. area of ARCD = b-h. q.e.d. 482. Cor. I. Parallelograms having equal bases and equal altitudes are equivalent. 483. Cor. n. Any two parallelograms are to each other as the products of their bases and their altitudes. Hint. Give a proof similar to that of § 479. 484. Cor. m. (a) Two parallelograms having equal bases are to each other as their altitudes, and (b) two par- allelograms having equal altitudes are to each other as their bases. (Hint. Give a proof similar to that of § 480.) Reasons 1. §223. 2. §232. 3. §232. 4. §211. 5. By iden. (5. § 54, 3. 7. §475. 8. § 54, 1. 218 PLANE GEOMETRY Proposition III. Theorem 485. The area of a triangle equals one half the product of its base and its altitude. X B 1. Reasons and Let 1. §179. E.D. 2. 3. 4. 5. §220. §236. §481. § 54, 8 a. A b C Given A ABC, with base b and altitude h. To prove area of A ABC =. ±b-h. Argument Through A draw a line II CB, through B draw a line II CA. these lines intersect at X. 2. Then AXBC is a O. 3. /. A ABC = |D AXBC. 4. But area of AXBC = b>h. 5. .-. area of A ABC = J b * h. < 486. Cor. I. Triangles having equal bases and equal altitudes are equivalent. 487. Cor. II. Any two triangles are to each other as the products of their bases and their altitudes. Outline op Proof Denote the two A by T and T', their bases by b and b', and their altitudes by h and h', respectively. 1* \V .h'~ Then T \ b - h and f £&' ///. b' ■ h' 488. Cor. III. (a) Two triangles having equal bases are to each other as their altitudes, and (b) two triangles having equal altitudes are to each as their bases. BOOK IV 219 Outline of Proof / \ T b-h h , b << T ^ b-h b W T'~b-h' ft" K ; t b'-h b r Cor. IV. A triangle is equivalent to one half of a parallelogram having the same base and altitude. Ex. 816. Draw four equivalent parallelograms on the same base. Ex. 817. Find the area of a parallelogram having two sides 8 inches and 12 inches, respectively, and the included angle 60°. Find the area if the included angle is 45°. Ex. 818. Find the ratio of two rhombuses whose perimeters are 24 inches and 16 inches, respectively, and whose smaller base angles are 30°. Ex. 819. Find the area of an equilateral triangle having a side equal to 6 inches. Ex. 820. Find the area of an equilateral triangle whose altitude is 8 inches. Ex. 821. Construct three or more equivalent triangles on the same base. Ex. 822. Find the locus of the vertices of all triangles equivalent to a given triangle and standing on the same base. Ex. 823. Construct a triangle equivalent to a given triangle and hav- ing one of its sides equal to a given line. Ex. 824. Construct a triangle equivalent to a given triangle and hav- ing one of its angles equal to a given angle. Ex. 825. , Construct a triangle equivalent to a given triangle and hav- ing two of its sides equal, respectively, to two given lines. Ex. 826. Divide a triangle into three equivalent triangles by drawing lines through one of its vertices. Ex. 827. ' Construct a triangle equivalent to § of a given triangle ; | of a given triangle. Ex. 828. Construct a triangle equivalent to a given square. Ex. 829. The area of a rhombus is equal to one half the product of its diagonals. Ex. 830. If from any point in a diagonal of a parallelogram lines are drawn to the opposite vertices, two pairs of equivalent triangles are formed. Ex. 831. Two lines joining the mid-point of the diagonal of a quadri- lateral to the opposite vertices divide the figure into two equivalent parts. Ex. 832. Find the area of a triangle if two of its sides are 6. inches and.9 inches, respectively, and the included angle is 60°, 220 PLANE GEOMETRY Proposition IV. Problem 490. To derive a formula for the area of a triangle in terms of its sides. C a B D C Given A ABC, with sides a, b, and c. To derive a formula for the area of A ABC in terms of a, b } and c. Argument Reasons 1. Let h a denote the altitude upon a, p the pro- 1. § 485. jection of b upon a, and T the area of A ABC. Then T=-ah a = -h a . 2 ' 2 a 2. h a 2 =b 2 -p 2 = (b+p)(b-p). 3. Butp = a2 + 52 - c2 (Fig. 1), or -<* + *- _ t. 6. Now let a + 6 + c = 2 s. Then a + &-c = 2s- 2c = 2(s — c); a-& + c = 2s-2& = 2(s — 6); 6 + c-a = 2s-2a = 2(s-a). 2. § 447. 3. § 456. 4. § 309. 5. § 54, 13. 6. § 54, 3. BOOK IV 221 Argument '. Then K=yj 2 s 2(s-a) 2(8-6) 2(s-c) 4 ft 2 = - Vs(s — d)(s — b)(s — c). ft 8. /. T=- • - Vs(s - ft)(s - b)(s - c) = V«(s — ft)(s — &)(s — c). Q.E.F. Reasons 7. § 309. 8. § 309. Ex. 833. Find the area of a triangle whose sides are 7, 10, and 13. Ex. 834. If the sides of a triangle are a, 6, and c, write the formula for the altitude upon b ; upon c. (See Prop. IV, Arg. 7.) Ex. 835. In triangle ABC, a = 8, b = 12, c = 16 ; find the area of triangle ABC ; the altitude upon b ; the altitude upon c. Proposition V. Theorem 491. The area of a triangle is equal to one half the product of its perimeter and the radius of the inscribed circle. A b Given A ABC, with area T, sides a, b, and c, and radius of inscribed circle r. To prove T= 1 (ft + b -f- c) r. Outline of Proof 1. Area of A OBC = \ a • r\ area of A OCA = i 6 . r; area of A OAB—^c-r. 2. .'. r = i(ft-r-&-fc)r, q.e.d. 492. Cor. The area of any polygon circumscribed about a circle is -equal to one half its perimeter multi- pU'&l by the radius of the inscribed circle. 222 PLANE GEOMETRY Ex. 836. If the area of a triangle is 15 V3 square inches and its sides are 3, 5, and 7 inches, find the radius of the inscribed circle. Ex. 837. Derive a formula for the radius of a circle inscribed in a triangle in terms of the sides of the triangle. Outline of Solution 1. T=±(a + b + c)r = $ (2 s)r = sr. Ex. 838. Derive a formula for the radius of a circle circumscribed about a triangle, in terms of the sides of the triangle. Outline of Solution (See figure for Ex. 798.) 1. dh = ac; i.e. d or 2 i2 = — . h 2. ,'.M= = Q.E.F. 2 h 4 V s(8 -a) (s - b) - c) Ex. 839. If the sides of a triangle are 9, 10, and 11, find the radius of the inscribed circle ; the radius of the circumscribed circle. Ex.840. The sides of a triangle are 3a, 4 a, and 5 a. Find the radius of the inscribed circle ; the radius of the circumscribed circle. What kind of a triangle is it ? Verify your answer by comparing the radius of the circumscribed circle with the longest side. Ex. 841. Derive formulas for the bisectors of the angles of a triangle in terms of the sides of the triangle. Outline of Solution (See figure for Ex. 797.) 1. t b 2 =ac — rs. 2. But a :c=r:s. 3. .-. a -\- c : a =b : r. 4. .-. r = -5*-. 5. Likewise 8 = -^-. 6. .-. tf = ac- ahH a + c a + c (a-\-c) 2 ac(a + b + c)(a-b+c) = 4acs{s-b K 7< ... % = _^_ V acs(s _ 6) . (a + cy (a + cy ■ a + c K J 8. Likewise t a = y/bcs(s — a) and t c = Vabs(s — c) • b+c a + 6 K J Q- E - F - Find r, B, T, h a , m a , and t a , having given : Ex.842, a = 11, b = 9, c= 16. What kind of an angle is C? Ex. 848. a = 13, b = 15, c = 20. What kind of an angle is C? Ex. 8,44. a = 24, b = 10, c = 26. What kind of an angle is C ? BOOK IV 223 TRANSFORMATION OF FIGURES 493. Def. To transform a figure means to find another figure which is equivalent to it. Proposition VI. Problem 494. To construct a triangle equivalent to a given polygon. Given polygon ABGDEF. To construct a A =o= polygon ABCDEF. (a) Construct a polygon =c= ABCDEF, but having one side less. I. Construction 1. Join any two alternate vertices, as C and A. 2. Construct BG II CA, meeting FA prolonged at G. § 188. 3. Draw CG. 4. Polygon GCDEF =o polygon ABCDEF and has one side less. 1. II. Proof Argument A AGC and ABC have the same base CA, and the same altitude, the J_ be- tween the lis CA and BG. .-. A AGC =c= A ABC. But polygon A CDEF — poly gon ACDEF. .-. polygon GCDEF o= polygon AB CDEF. Q.E.P. Reasons 1. §235. 2. §486. 3. By iden. 4. §54,2. 224 PLANE GEOMETRY (6) In like manner, reduce the number of sides of the new polygon GCDEF until A DHK is obtained. The construction, proof, and discussion are left as an exercise for the student. Ex. 845. Transform a scalene triangle into an isosceles triangle. Ex. 846. Transform a trapezoid into a right triangle. Ex. 847. Transform a parallelogram into a trapezoid. Ex. 848. Transform a pentagon into an isosceles triangle. Ex. 849. Construct a triangle equivalent to § of a given trapezium. Ex. 850. Transform $ of a given pentagon into a triangle. Ex. 851. Construct a rhomboid and a rhombus which are equivalent, and which have a common diagonal. Proposition VII. Theorem 495. The area of a trapezoid equals the product of its altitude and one half the sum of its bases. V A b D Given trapezoid AFED, with altitude h and bases b and b 1 . To prove area of AFED = I- (h 4- b^h. BOOK IV 225 1. 2. Argument Draw the diagonal AE. The altitude of A AED, considering b as base, is equal to the altitude of A AFE, considering b' as base, each being equal to the altitude of the trapezoid, h. .-. area of A AED = \ b • h. Area of A AFE = \b' • h. .*. area of trapezoid AFED = ^(b + b')h. Q.E.D. Reasons 1. § 54, 15. 2. § 235. 3. § 485. 4. § 485. 5. §54,2. 496. Cor. The area of a trapezoid equals the product of its altitude and its median. 497. Question. The ancient Egyptians, in attempting to find the area of a field in the shape of a trapezoid, multiplied one half the sum of the parallel sides by one of the other sides. For what figure would this method be correct ? Ex. 852. Find the area of a trapezoid whose bases are 7 inches and 9 inches, respectively, and whose altitude is 5 inches. Ex. 853. Find the area of a trapezoid whose median is 10 inches and whose altitude is 6 inches. « Ex. 854. Through a given point in one side of a given parallelogram draw a line which shall divide the parallelogram into two equivalent parts. Will these parts be equal ? Ex. 855. Through a given point within a parallelogram draw a line which shall divide the parallelogram into two equivalent parts. Will these parts be equal ? Ex. 856. If the mid-point of one of the non-parallel sides of a trape- zoid is joined to the extremities of the other of the non-parallel sides, the area of the triangle formed is equal to one half the area of the trapezoid. Ex. 857. Find the area of a trapezoid whose bases are b and b' and whose other sides are each equal to s. Ex. 858. If the sides of any quadrilateral are bisected and the points of bisection joined, the included figure will be a parallelogram equal in area to -half the original figure. 226 PLANE GEOMETRY Proposition VIII. Theorem 498. Two triangles which have an angle of one equal to an angle of the other are to each other as the products of tTie sides including the equal angles. A G Given A ABC and BEF, with Z A = Z.B. A ABC AC • AB To prove A BEF BF • BE Argument 1. Let h be the altitude of A ABC upon side AC, and h' the altitude of A BEF upon side BF. Then, A ABC AC-h AC m h_ BF h r * 2. 3. 4. 5. * A BEF BF • h' In rt. A ABG and BEK, ZA = ZB. .'. A ABG ~ A BEK. h _AB^ h'~~ BE' A ABC AC AB _ AC ♦ AB BF BE~ A BEF BF -BE Q.E.D. Reasons 1. § 487. 2. By hyp. 3. § 422. 4. § 424,2. 5. § 309. Ex. 859. Draw two triangles upon the blackboard, so that an angle of one shall equal an angle of the other. Give a rough estimate in inches of the sides including the equal angles in the two triangles, and compute the numerical ratio of the triangles. Ex. 860. If two triangles have an angle of one supplementary to an angle of the other, the triangles are to each other as the products of the sides including the supplementary angles. BOOK IV 227 Proposition IX. Problem 499. To construct a square equivalent to a given tri- angle Given A ABC, with base b and altitude h. To construct a square =0= A ABC. I. Analysis 1. Let # = the side of the required square; then h. \b : x = x : h. 4. 5. .*. the side of the required square will be a mean pro- portional between i b and h. II. Construction 1. Construct a mean proportional between J b and h. Call it x. § 445. 2. On x, as base, construct a square, S. 3. . §544. 4. § 168. 5. § 59, 10. §447. I, Arg. 3. Argument , CB, which is ^ AB, can be made less than any previously assigned value, however small. 4. But OB — OC < CB. 5. .-. OB — OC, being always less than CB, can be made less than any previously assigned value, however small. Q.E.D. II. 1. Again, OB 2 — OC 2 = CB 2 . 2. But CB can be made less than any pre- viously assigned value, however small. 3. .*. CB can be made less than any previ- ously assigned value, however small. 4. .♦. 0~B" — OC , being always equal to CB , can be made less than any previously assigned value, however small. Q.E.D. 544. If a variable can be made less than any assigned value, the quotient of the variable by any constant, except zero, can be made less than any assigned value. 545. If a variable can be made less than any assigned value, the square of that variable can be made less than any assigned value, (For proofs of these theorems see Appendix, §§ 586 and 589.) 3. § 545. 4. §309. Ex. 1001. Construct the following designs: (1) on the blackboard, making each line 12 times as long as in the figure ; (2) on paper, making each line 4 times as long : 262 PLANE GEOMETRY Proposition XI. Theorem 546. By repeatedly doubling the number of sides of regular circumscribed and inscribed polygons vf the same number of sides, and making the polygons always regular : I. TJveir perimeters approach a common limit. II. Their areas approach a common limit. Given P and p the perimeters, R and r the apothems, and K and k the areas respectively of regular circumscribed and inscribed polygons of the same number of sides. To prove that by repeatedly doubling the number of sides of the polygons, and making the polygons always regular : I. P and p approach a common limit. II. K and k approach a common limit. I. Argument 1. Since the two regular polygons have the P_R - p r ■p _ R — r same number of sides, P—p = P K But by repeatedly doubling the number of sides of the polygons, and making them always regular, R — r can be L Reasons §538. 2. § 399. §54, 7 a. § 543, I. BOOK V 263 Argument made less than any previously as- signed value, however small. 5. .-. ~~ r can be made less than any pre- R viously assigned value, however small. 6. .-. P ~ r can be made less than any R previously assigned value, however small, P being a decreasing variable. D n. 7 -\ P—p, being always equal to P , R can be made less than any previously assigned value, however small. 8. .\ P andp approach a common limit. Q.E.D. Reasons 5. § 544. G. § 547. 7. § 309. 8. § 548. II. The proof of II is left as an exercise for the student. Hint. Since the two regular polygons have the same number of sides, E- El (§ 539). The rest of the proof is similar to steps 2-8, § 546, I. k r 2 547. If a variable can be made less than any assigned value, the product of that variable and a decreasing value may be made less than any assigned value. 548. If tioo related variables are such that one is always greater than the other, arid if the greater continually decreases while the less continually increases, so that the difference between the two may be made as small as ive please, then the tivo variables have a common limit ivhich lies between them. (For proofs of these theorems see Appendix, §§ 587 and 594.) 549. Note. The above proof is limited to regular polygons, but it can be shown that the limit of the perimeter of any inscribed (or circum- scribed) polygon is the same by whatever method the number of its sides is successively increased, provided that each side approaches zero as a limit. 264 PLANE GEOMETRY 550. Def. The length of a circumference is the common limit which the successive perimeters of inscribed and circum- scribed regular polygons (of 3, 4, 5, etc., sides) approach as the number of sides is successively increased and each side ap- proaches zero as a limit. The term "circumference" is frequently used for "the length of a circumference." (See Prop. XII.) 551. The length of an arc of a circumference is such a par< of the length of the circumference as the central angle which intercepts the arc is of 360°. (See § 360.) 552. The approximate length of a circumference is found in elementary geometry by computing the perimeters of a series of regular inscribed and circumscribed polygons which are ob- tained by repeatedly doubling the number of their sides. The perimeters of these inscribed and circumscribed polygons, since they approach a common limit, may be made to agree to as many decimal places as we please, according to the number of times we double the number of sides of the polygons. Proposition XII. Theorem 553. The ratio of the circumference of a circle to its diameter is the same for all circles. Given any two circles with circumferences Cand (f, and with radii R and R', respectively. c_ = &_ 2R 2r' To prove —— = - — , BOOK V 265 Argument 1. Inscribe in the given circles regular polygons of the same number of sides, and call their perimeters P and P\ o rp, P R 2R 2. Then — = — = -—. P' R' 2R' 3. P _ P' 2R~2R r 4. As the number of sides of the two regu- lar polygons is repeatedly doubled, P approaches C as a limit, and P 1 approaches C' as a limit. 5. \ — approaches — as a limit. 2R ™ 2R 6. Also — • approaches — . as a limit. 2R' ff 2r' p P* 7. But — is always equal to - — -. 1 R ZR 8. 2R 2r' Q.E.D. Reasons 1. § 517, a. 2. §538. 3. § 396. 4. § 550. 5. § 408, b. 6. § 408, b. 7. Arg. 3. 8. §355. 554. Def. This constant ratio of the circumference of a circle to its diameter is usually represented by the Greek let- ter ir. It will be shown (§ 568) that its value is approxi- mately 3| ; or, more accurately, 3.1416. 555. Cor. I. TTie circumference of a circle is equal to 2 irR. 556. Cor. n. Any two circumferences are to each other as their radii. Ex. 1002. If the radius of a wheel is 4 feet, how far does it roll in two revolutions ? Ex. 1003. How many revolutions are made by a wheel whose radius is 3 feet in rolling 44 yards ? Ex. 1004. (a) Find the width of the ring between two concentric circumferences whose lengths are 2 feet and 3 feet, respectively. (&) Assuming the earth's equator to be 25,000 miles, find the width of the ring between it and a concentric circumference 1 foot longer. (f\ Write your inference in the form of a general statement. 266 PLANE GEOMETRY Proposition XIII. Theorem 557. The area of a regular polygon is equal to one half the product of its perimeter and its apothem.- C \Pr. Given regular polygon ABCD apothem. To prove area of ABCD • Argument 1. In polygon ABCD • • • , inscribe a circle. 2. Then r, the apothem of regular polygon ABCD • • • , is the radius of circle 0. 3. .-. area of ABCD ...■■jPr, q.e.d. , P its perimeter, and r its Reasons 1. § 530. 2. § 533. 3. §492. Ex. 1005. Find the area of a regular hexagon whose side is 6 inches. Ex. 1006. The area of an inscribed regular hex- agon is a mean proportional between the areas of the inscribed and circumscribed equilateral triangles. Ex. 1007. The figure represents a flower bed drawn to the scale of 1 inch to 20 feet. Find the number of square feet in the flower bed. 558. Def. The area of a circle is the common limit which the successive areas of inscribed and circumscribed regular polygons approach as the number of sides is successively in- creased and each side approaches zero as a limit. BOOK V 267 Proposition XIY. Theorem 559. The area of a circle is equal to one half the prod- uct of its circumference and its radius. Given circle 0, with radius R, circumference C, and area K. To prove K=\CR. Argument Circumscribe about circle O a regular polygon. Call its perimeter P and its area S. Then S — ^pr. As the number of sides of the regular circumscribed polygon is repeatedly doubled, P approaches C as a limit. .*. i PR approaches \ CR as a limit. Also s approaches Zasa limit. But S is always equal to J PR. .'. K—\ CR. Q.E.D. Reasons 1. §517,6. 2. § 557. 3. §550. 4. § 561. 5. § 558. 6. Arg. 2. 7. § 355. 560. The product of a variable and a constant is a variable. 561. TJie limit of the product of a variable OMd a constant, not zero, is the limit of the variable multiplied by the constant. (Proofs of these theorems will be found in the Appendix, §§ 585 and 590.) 562. Cor. I. The area of a circle is equal to irR 2 . Hint. K = \ C R = I • 2 ttR . R = iri? 2 . 268 PLANE GEOMETRY 563. Cor. II. The areas of two circles are to each other as the squares of their radii, or as the squares of their diameters. 564. Cor. III. The area of a sector whose angle is a° is (See § 551.) 360^ Ex. 1008. Find the area of a circle whose radius is 3 inches. Ex. 1009. Find the area of a sector the angle of which is (a) 45°, (&) 120°, (c) 17°, and the radius, 5 inches. Find the area of the segment corresponding to (6). Hint. Segment = sector — triangle. Ex. 1010. If the area of one circle is four times that of another, and the radius of the first is 6 inches, what is the radius of the second ? Ex. 1011. Find the area of the ring included between the circumfer- ences of two concentric circles whose radii are 6 inches and 8 inches. Ex. 1012. Find the radius of a circle whose area is equal to the sum of the areas of two circles with radii 3 and 4, respectively. Ex. 1013. In the figure the diameter AB = 2B, AC = \AB, AD = \ AB, and E is any point on AB. (1) Find arc AC + arc CB ; arc AD + arc DB ; arc AE + arc EB. (2) Com- pare each result with semicircumference AB. 565. Def. In different circles similar arcs, similar sectors, and similar segments are arcs, sectors, and segments that cor- respond to equal central angles. Ex. 1014. Ex. 1015. radii. Ex. 1016. ments are to the squares of their radii. Hint. Prove s ector AOB _ A AOB sector CPD A CPD apply §§ 396 and 399. Similar arcs are to each other as their radii. Similar sectors are to each other as the squares of their Similar seg- each other as Then BOOK V 269 Proposition XV. Problem 566. Given a circle of unit diameter and the side of a regular inscribed polygon of n sides, to find the side of a regular inscribed polygon of 2n sides. Given circle ABF of unit diameter, AB the side of a regular inscribed polygon of n sides, aud CB the side of a re'gular in- scribed polygon of 2 n sides ; denote AB by s and CB by x. To find x in terms of s. Argument 1. Draw diameter CF ; draw BO and BF. 2. Z CBF is a rt. Z. 3. Also CF is the _L bisector of AB. L .: CB 2 = CF- CK. 5. Now CF = 1, .50 1 CO 6. .% C5 =#2 1 1 2' ' 2 1 . CK= CK=C0 — KO = --K0. 7. .-. x 2 = - — -JW — ir# -Mtjr Vl-s 2 /rr Q.E.F. Reasons 1. §54,15. 2. §367. 3. §142. 4. § 443, II. 5. By cons. 6. §309. 7. §447. 8. §54,13. 270 PLANE GEOMETRY Proposition XVI. Problem 567. Given a circle of unit diameter and the side of a regular circumscribed polygon of n sides, to find the side of a regular circumscribed polygon of 2n sides. Given circle O of unit diameter, AB half the side of a reg- ular circumscribed polygon of n sides, and CB half the side of a regular circumscribed polygon of 2 n sides ; denote AB by | and CB by |. To find x in terms of s. 2. Argument Draw CO and AO. A BOC = Z BOA. 3. .-. in A OAB, AC : CB = AO : BO. 4. But AC = AB — CB. 5. And 6. .'. AB — CB : CB »->/ AB + J50 AB -[-BO : BO. 7. Substituting | for ^J5, - for C£, and - for 'BO, ■HS vr s — x : x = yV + I : 1. Reasons 1. §54,15. 2. §517, 6. 3. §432. 4. § 54, 11. 5. §446. 6. §309. 7. §309. 8. §403 ? BOOK V 271 9. 10. Argument . s — x = x Vs 2 + 1. .\ X = 1 + V s 2 + 1 Q.E.F. Reasons 9. §388. 10. Solving for x. Ex. 1017. Given a circle of unit diameter and an inscribed and a circumscribed square ; compute the side of the regular inscribed and the regular circumscribed octagon Proposition XVII. Problem 568. To compute the approximate value of the circum- ference of a circle in terms of its diameter ; i.e. to compute the value of ir. Given circle A BCD, with unit diameter. To compute approximately the circumference of circle ABCD in terms of its' diameter ; i.e. to compute the value of ir. Argument The ratio of tne circumference of a circle to its diameter is the same for all circles. Since the diameter of the given circle is unity, the side of an inscribed square I will be | V2. Reasons 1. §553. 2. §522. 272 PLANE GEOMETRY Argument 3. By using the formula x = \^—- 4. VI- the sides of regular inscribed polygons of 8, 16, 32, etc., sides may be com- puted ; and by multiplying the length of one side by the number of sides, the length of the perimeter of each polygon may be obtained. The re- sults are given in the table below. Likewise if the diameter of the given circle is unity, the side of a circum- scribed square will be 1. By using the foimula x = 1 + Vi^TT the sides of regular circumscribed polygons of 8, 16, 32, etc., sides may be computed ; and by multiplying the length of one side by the number of sides, the length of the perimeter of each polygon may be obtained. The results are given in the following table. Reasons 3 §566. 4. §522. 5. §567. NlTMBEB OF BIDES Perimeter of inscribed POLYGON Perimeter of circumscribed polygon 4 2.828427. 4.000000 8 3.061467 3.313708 16 3.121445 3.182597 32 3.136548 3.151724 64 3.140331 3.144118 128 3.141277 3.142223 256 3.141513 3.141750 512 3.141572 3.141632 . 1024 3.141587 3.141602 2048 3.141591 3.141595 4096 3.141592 3.141593 BOOK V 273 These successive perimeters will be closer and closer approxi- mations of the length of the circumference. By continuing to double the number of sides of the inscribed and circumscribed polygons, perimeters may be obtained which agree to as many orders of decimals as desired. The last numbers in the"table show that the length of a cir- cumference of unit diameter lies between 3.141592 and 3.141593. .-. tt, the ratio of any circumference to its diameter, to five deci- mal places is 3.14159. The value commonly used is 3.1416. 569. Historical Note. The earliest known attempt to find the area of a circle was made by Ahmes, an Egyptian priest, as early as 1700 b.c. His method gave for ir the equivalent of 3. 1604. His manuscript is pre- served in the British Museum. Archimedes (250 b.c.) gave the value y, his method being similar to that given in the text. Hero of Alexandria gave 3 and 3}. Ptolemy (about 150 b.c) gave 3.1417. Metius of Holland (1600 a.d.) gave fff, which is correct to six places. Lambert (1750 a.d.) proved w an irrational number, and Lindemann (1882) proved it transcendental, i.e. not expressible as a root of an alge- braic equation. By methods of the calculus the value of ir has been computed to sev- eral hundred places. Richter carried it to 500 decimal places, and Shanks, in 1873, gave 707 places. It is impossible to " square the circle," i.e. to obtain by accurate con- struction, with the use of ruler and compasses only, a square equivalent to the area of a circle. Ex. 1018. Find the area of a circle whose radius is 5 inches. (Let «■ = 8.1416.) | Ex. 1019. The side of an inscribed square is 4 inches. What is the area of the circle ? Ex. 1020. What is the area of a circle inscribed in a square whose side is 6 inches ? Ex. 1021. The figure represents a circular grass plot drawn to the scale of 1 inch to 24 feet. Measure carefully the radius of the circle and find the number of square feet in the grass plot. 274 PLANE GEOMETRY Ex. 1022. The figure represents a belt from drive wheel to wheel Q. The diameter of wheel is 2 feet and of wheel Q 16 inches. If the drive wheel makes 75 revo- lutions per minute, how many rev- olutions per minute will the smaller wheel make ? Ex. 1023. The radius of a circle is 6 inches. What is the area of a segment whose arc is 60° ? Ex. 1024. The radius of a circle is 8 inches. What is the area of the segment subtended by the side of an inscribed equilateral triangle ? Ex. 1025. The diagonals of a rhombus are 16 and 30 ; find the area of the circle inscribed in the rhombus. MISCELLANEOUS EXERCISES Ex. 1026. An equiangular polygon inscribed in a circle is regular if the number of sides is odd. Ex. 1027. An equilateral polygon circumscribed about a circle is regular if the number of sides is odd. Ex. 1028. Find the apothem and area in terms of the radius in an equilateral triangle ; in a square ; in a regular hexagon. Ex. 1029. The lines joining the mid-points of the apothems of a regu- lar pentagon form a regular pentagon. Find the ratio of its area to the area of the original pentagon. Ex. 1030. Within a circular grass plot of radius 6 feet, a flower bed in the form of an equilateral triangle is inscribed. How many square feet of turf remain ? Ex. 1031. The area of a regular hexagon inscribed in a circle is 24 V3. What is the area of the circle ? Ex. 1032. From a circle of radius 6 is cut a sector whose central angle is 105°. Find the area and perimeter of the sector. (ir = 2 r 2 .) Ex. 1033. Prove that the following method of _ inscribing a regular pentagon and a regular decagon .in a circle is correct. Draw diameter CD perpen- dicular to diameter AB ; bisect OA and join its mid- point to D ; take EF = ED and draw FD. FD will be the side of the required pentagon, and OF the side of the required decagon. BOOK V 275 Ex. 1034. Divide a given circle into two segments such that any angle inscribed in one segment is twice an angle inscribed in the other ; so that an angle inscribed in one segment is three times an angle inscribed in the other ; seven times. Ex. 1035. Show how to cut off the corners of an equilateral triangle so as to leave a regular hexagon ; of a square to leave a regular octagon. Ex. 1036. The diagonals of a regular pentagon form a regular penta- gon. Ex. 1037. The diagonals joining alternate vertices of a regular hexa- gon form a regular hexagon one third as large as the original one. Ex. 1038. The area of a regular inscribed octagon is equal to the product of the side of an inscribed square and the diameter. Ex. 1039. If a is the side of a regular pentagon inscribed in a circle whose radius is 2?, then a = -5- v 10 — 2V5* 29 Ex. 1040. The area of a regular inscribed dodecagon is equal to three times the square of the radius. Ex. 1041. Construct an angle of 9°. Ex. 1042. Construct a regular pentagon, given one of the diagonals. Ex. 1043. Through a given point construct a line which shall divide a given circumference into two parts in the ratio of 3 to 7 ; in the ratio of 3 to 5. Can the given point lie within the circle? Ex. 1044. Transform a given regular octagon into a square. Ex. 1045. Construct a circumference equal to the sum of two given circumferences. Ex. 1046. Divide a given circle by concentric circumferences into four equivalent parts. Ex. 1047. In a given sector whose angle is a right angle inscribe a square. Ex. 1048. In a given sector inscribe a circle. Ex. 1049. If two chords of a circle are perpendicular to each other, the sum of the four circles having the four segments as diameters is equivalent to the given circle. Ex. 1050. The area of a ring between two concentric circumferences whose radii are B and R r respectively is tt(B 2 — B' 2 ). Ex. 1051. The area of the surface between two concentric circles is equal to twice the area of the smaller circle. Find the ratio between theij radii. 276 PLANE GEOMETRY MISCELLANEOUS EXERCISES ON PLANE GEOMETRY Ex. 1052. If equilateral triangles are constructed on the sides of any given triangle, the lines joining the vertices of the given triangle to the outer vertices of the opposite equilateral triangles are equal. Ex. 1053. If, on the arms of a right triangle as diameters, semicircles are drawn so as to lie outside of the triangle, and if, on the hypotenuse as a diameter, a semicircle is drawn passing through the vertex of the right angle, the sum of the areas of the two crescents included between the semicircles is equal to the area of the given triangle. Ex. 1054. The area of the regular inscribed triangle is half that of the regular inscribed hexagon. Ex. 1055. From a given point draw a secant to a circle such that its internal and external segments.shall be equal. Ex. 1056. Show that the diagonals of any quadrilateral inscribed in a circle divide the quadrilateral into four triangles which are similar, two and two. Ex. 1057. Through a point P, outside of a circle, construct a secant VAB so that JJ? 2 = PAx PB. Ex. 1058. The radius of a circle is 6 feet. What are the radii of the circles concentric with it whose circumferences divide its area into three equivalent parts ? B a ^ . Ex. 1059. Given parallelogram ABCD, 2? the mid-point of BC; prove 0T=\ TC. Ex. 1060. Given PTa tangent to a circle at point T, and two other tangents parallel to each other cutting PT at A and B respectively ; prove that the radius of the circle is a mean proportional between AT and TB. Ex. 1061. Show that a mean proportional between two unequal lines is less than half their sum. Ex. 1062. Given two similar triangles, construct a triangle equiva- lent to their sum. Ex. 1063. The square of the side of an inscribed equilateral triangle is equal to the sum of the squares of the sides of the inscribed square and of the inscribed regular hexagon. Ex. 1064. Prove that the area of a circular ring is equal to the area of a circle whose diameter equals a chord of the outer circumference which is tangent to the inner* 0. BOOK V 277 Ex. 1065. It two chords drawn from a common point P on the cir- cumference of a circle are cut by a line parallel to the tangent through P, the chords and the segments of the chords between the two parallel lines are inversely proportional. Ex. 1066. Construct a segment of a circle similar to two given simi- lar segments and equivalent to their sum. Ex. 1067. The distance between two parallels is a, and the distance between two points A and B in one parallel is 2 b. Find the radius of the circle which passes through A and P, and is tangent to the other parallel. Ex. 1068. Tangents are drawn through a point 6 inches from the circumference of a circle whose radius is 9 inches. Find the length of the tangents and also the length of the chord joining the points of contact. Ex. 1069. If the perimeter of each of the figures, equilateral triangle, square, and circle, is 396 feet, what is the area of each figure ? Ex. 1070. The lengths of two sides of a triangle are 13 and 15 inches, and the altitude on the third side is 12 inches. Find the third side, and also the area of the triangle. (Give one solution only.) Ex. 1071. If the diameter of a circle is 3 inches, what is the length of an arc of 80° ? Ex. 1072. AD and BC are the parallel sides of a trapezoid ABCD, whose diagonals intersect at E. If F is the mid-point of BC, prove that FE prolonged bisects AD. Ex. 1073. Given a square ABCD. Let E be the mid-point of CD, and draw BE. A line is drawn parallel to BE and cutting the square. Let P be the mid-point of the segment of this line within the square. Find the locus of P when the line moves, always remaining parallel to BE. Describe the locus exactly, and prove the correctness of your answer. Ex. 1074. Let ABCD be any parallelogram, and from any point P in the diagonal AC draw the straight line PM cutting AB in M, BC in jV, CD in P, and AD in K. Prove that PM • PN= PK • PL. Ex. 1075. Find the area of a segment of a circle whose 'height is 4 inches and chord 8V3 inches. Ex. 1076. A square, whose side is 5 inches long, has its corners cut off in such a way as to make it into a regular octagon. Find the area and the perimeter of the octagon. Ex. 1077. Into what numbers of arcs less than 15 can the circumfer- enQe* of a circle be divided with ruler and compasses only ? 278 PLANE GEOMETRY Ex. 1078. Through a point A on the circumference of a circle chords are drawn. On each one of these chords a point is taken one third of the distance from A to the other end of the chord. Find the locus of these points, and prove that your answer is correct. Ex. 1079. In what class of triangles do the altitudes meet within the triangle ? on the boundary ? outside the triangle ? Prove. Ex. 1080. Given a triangle ABC and a fixed point D on side AC ; draw the line through D which divides the triangle into two parts of equal area. Ex. 1081. The sides of a triangle are 5, 12, 13. Find the radius of the circle whose area is equal to that of the triangle. Ex. 1082. In a triangle ABC the angle C is a right angle, and the lengths of AC and BC are 5 and 12 respectively ; the hypotenuse BA is prolonged through A to a point D so that the length of AD is 4 ; CA is prolonged through A to E so that the triangles AED and ABC have equal areas. What is the length of AE? Ex. 1083. Given three points A, B, and C, not in the same straight line ; through A draw a straight line such that the distances of B and C from the line shall be equal. Ex. 1084. Given two straight lines that cut each other ; draw four circles of given radius that shall be tangent to both of these lines. Ex. 1085. Construct two straight lines whose lengths are in the ratio of the areas of two given polygons. Ex. 1086. The radius of a regular inscribed polygon is a mean pro- portional between its apothem and the radius of the similar circumscribed polygon. Ex. 1087. Draw a circumference which shall pass through two given points and bisect a given circumference. Ex. 1088. A parallelogram is constructed having its sides equal and parallel to the diagonals of a given parallelogram. Show that its diagonals are parallel to the sides of the given parallelogram. Hint. Look for similar triangles. Ex. 1089. If two cnords are divided in the same ratio at their point of intersection, the chords are equal. Ex. 1090. The sides AB and iO of a triangle ABC are bisected in D and E respectively. Prove that the area of the triangle BBC is twice that of the triangle DEB. Ex. 1091. Two circles touch externally. How many common tan- gents have they ? Give a construction for the common tangents. BOOK V 279 Ex. 1092. Prove that the tangents at the extremities of a chord of a circle are equally inclined to the chord. Ex. 1093. Two unequal circles touch externally at P ; line AB touches the circles at A and B respectively. Prove angle APB a right angle. Ex. 1094. Find a point within a triangle such that the lines joining this point to the vertices shall divide the triangle into three equivalent parts. Ex. 1095. A triangle ABC is inscribed in a circle. The angle B is equal to 50° and the angle C is equal to 60°. What angle does a tangent at A make with BC prolonged to meet it ? Ex. 1096. The bases of a trapezoid are 8 and 12, and the altitude is 6. Find the altitudes of the two triangles formed by prolonging the non- parallel sides until they intersect. Ex. 1097. The circumferences of two circles intersect in the points A and B. Through A a diameter of each circle is drawn, viz. AC and AD. Prove that the straight line joining C and D passes through B. Ex. 1098. How many lines can be drawn through a given point in a plane so as to form in each case an isosceles triangle with two given lines in the plane ? Ex. 1099. The lengths of two chords drawn from the same point in the circumference of a circle to the extremities of a diameter are 5 feet and 12 feet respectively. Find the area of the circle. Ex. 1100. Through a point 21 inches from the center of a circle whose radius is 15 inches a secant is drawn. Find the product of the whole secant and its external segment. Ex. 1101. The diagonals of a rhombus are 24 feet and 40 feet re- spectively. Compute its area. Ex. 1102. On the sides AB, BC, CA of an equilateral triangle ABC measure off segments AD, BE, CF, respectively, each equal to one third the length of a side ; draw triangle DEF ; prove that the sides- of triangle DEF are perpendicular respectively to the sides of triangle ABC. Ex. 1103. Construct as if (a) - = -; (6) x = a V£ x 3 Ex. 1104. Find the area included between a circumference of radius 7 and an inscribed square. Ex. 1105. What is the locus of the center of a circle of given radius whose circumference cuts at right angles a given circumference ? Ex. 1106. Two chords of a certain circle bisect each other. One of them js 10 inches long ; how far is it from the center of the circle ? 280 PLANE GEOMETRY Ex. 1107. Show how to find on a given straight line of indefinite length a point O which shall be equidistant from two given points A and B in the plane. If A and B lie on a straight line which cuts the given line at an angle of 45° at a point 7 inches distant from A and 17 inches from jB, show that OA will be 13 inches. Ex. 1108. A variable chord passes, when prolonged, through a fixed point outside of a given circle. What is the locus of the mid-point of the chord ? Ex. 1109. A certain parallelogram inscribed in a circle has two sides 20 feet in length and two sides 15 feet in length. What are the lengths of the diagonals ? Ex. 1110. Upon a given base is constructed a triangle one of the base angles of which is double the other. The bisector of the larger base angle meets the opposite side at the point P. Find the locus of P. Ex. 1111. What is the locus of the point of contact of tangents drawn from a fixed point to the different members of a system of con- centric circles ? Ex. 1112. Find the locus of all points, the perpendicular distances of which from two intersecting lines are to each other as 3 to 2. Ex. 1113. The sides of a triangle are a, 6, c. Find the lengths of the three medians. Ex. 1114. Given two triangles; construct a square equivalent to their sum. Ex. 1115. In a circle whose radius is 10 feet, two parallel chords are drawn, each equal to the radius. Find the area of the portion be- tween these chords. Ex. 1116. A has a circular garden and B one that is square. The distance around each is the same, namely, 120 rods. Which has the more land, A or B ? How much more has he ? Ex. 1117. Prove that the sum of the angles of a pentagram (a five- pointed star) is equal to two right angles. Ex. 1118. AB and A'B' are any two chords of the outer of two con- centric circles ; these chords intersect the circumference of the inner circle in points P, Q and P', Q f respectively : prove that^lP- PB=A'P' • P'B'. Ex. 1119. A running track consists of two parallel straight por- tions joined together at the ends by semicircles. The extreme length of the plot inclosed by the track is 176 yards. If the inside line of the track is a quarter of a mile in length, find the cost of seeding this plot at | cent a square yard, (tt = - 2 f a .) BOOK V 281 Ex. 1120. If two similar triangles, ABC and DEF, have their homologous sides parallel, the lines AD, BE, and OF, which join their homologous vertices, meet in a point. Ex. 1121. In an acute triangle side AB = 10, AC =7, and the pro- jection of AC on AB is 3.4. Construct the triangle and compute the third side BC. Ex. 1122. Divide the circumference of a circle into three parts that shall be in the ratio of 1 to 2 to 3. Ex. 1123. The circles having two sides of a triangle as diameters intersect on the third side. Ex. 1124. Construct a circle equivalent to the sum of two given circles. Ex. 1125. Assuming that the areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles, prove that the bisector of an angle of a triangle divides the opposite side into segments propor- tional to the adjacent sides. Ex. 1126. In a circle of radius 5 a regular hexagon is inscribed. Determine (a) the area of one of the segments of the circle which are exterior to the hexagon ; (6) the area of a triangle whose vertices are three successive vertices of the hexagon ; (c) the area of the ring bounded by the circumference of the given circle and that of the circle inscribed in the hexagon. Ex. 1127. Find the locus of the extremities of tangents to a given circle, which have a given length. Ex. 1128. A ladder rests with one end against a vertical wall and the other end upon a horizontal floor. If the ladder falls by sliding along the floor, what is the locus of its middle point ? Ex. 1129. An angle moves so that its magnitude remains constant and its sides pass through two fixed points. Find the locus of the vertex. Ex. 1130. The lines joining the feet of the altitudes of a triangle form a triangle whose angles are bisected by the altitudes. Ex. 1131. Construct a triangle, given the feet of the three altitudes. Ex. 1132. If the radius of a sector »s 2, what is the area of a sector whose central angle is 152° ? Ex. 1133. The rectangle of two lines is a mean proportional between the squares on the lines. Ex. 1134. Show how to inscribe in a given circle a regular polygon simtter to a given regular polygon. 282 PLANE GEOMETRY FORMULAS OF PLANE GEOMETRY 570. In addition to the notation given in § 270, the follow- ing will be used : a = side of polygon in general. b = base of a plane figure. b' = bases of a trapezoid. C = circumference of a circle. D = diameter of a. circle. E = sum of exterior angles of a polygon. h = altitude of a plane figure. I = sum of interior angles of a polygon. K = area of a figure in general. I = line in general. P = perimeter of polygon in general. p = projection of b upon a. B = radius of circle, or radius of regular polygon. r = apothem of regular polygon, or radius of inscribed circle. s — the longer of two segments of a line ; or s = %(a + b + c). X = angle in general. x a = side of a triangle opposite an acute angle. x = side of a triangle opposite an obtuse angle. FIGURE FORMULA REFERENCE Any triangle. A + B + C = 180°. §204. Polygon. /= (n- 2)180°. §§ 216, 219. E = 4 rt. A. §218. Central angle. X 9? intercepted arc. §358. Inscribed angle. X ?c i intercepted arc. ' §365. Angle formed by two X«| sum of arcs. §377. chords. Angle formed by tangent X<* | intercepted arc. §378. and chord. Angle formed by two X ?c | difference of arcs. § 379. secants. Angle formed by secant Xcc i difference of arcs. §379. and tangent. Angle formed by two tan- X^l difference of arcs. §379. gents. Similar polygons. P _a P> a'' §441. Right triangle. c 2 = a 2 + b*. §446. Any triangle. Xc ? - a? + V* - 2 ap. §452. Obtuse triangle. x< ? = a°- + & 2 + 2 ap. §455. FORMULAS 283 FIGURE FORM Any triangle. &Hc 2 =2^y Line divided in extreme l : s — s : I — and mean ratio. Rectangle. K=b-h. Square. K=a\ Parallelogram, K=b-h. Triangle. K=\b-h. Polygon. Circle inscribed in triangle. DLA REFERENCE + 2 m a \ § 457. s. § 465 and Ex. 763. §475. §478. §481. §485. h a = - vs(s — a) (i a b)(s-c). K = y/s(s-a)(s-b)(i K=l(a + b + c)r. K=\Pr. c). §490. §490. §491. §492. _ \/g(s-q)(s-6)(8-c) t Ex> 837> Circle circumscribed about triangle. Bisector of angle of tri- angle. Trapezoid. Regular polygons of same number of sides. Circles. Regular polygon. Circle. B abc Circles. Sector. Segment. 4Vs(s— a)(s— 6)(s— c) 2 b + c K=\(b + b')h P _B _r P' B' r' ' K' JK' 2 r'* C = C 2B 2B'' C = 2tB. a b 1 ' K=lPr. K=\C-B. K=ttB\ K' B'* D' 2 ' ~_ central Z _ y/bcsis— a). B*. 360= K= sector ^ triangle. Ex. 838. Ex. 841. §495. §538. §539. § 553. § 556. §557. §559. §562. §563. §564. Ex. 1009. APPENDIX TO PLANE GEOMETRY MAXIMA AND MINIMA 571. Def. Of all geometric magnitudes that satisfy given conditions, the greatest is called the maximum, and the least is called the minimum.* 572. Def. Isoperimetric figures are figures which have the same perimeter. Proposition I. Theorem 573. Of all triangles having two given sides, that in which these sides include a right angle is the maximum. Given A ABC and AEC, with AB and AC equal to AE and AC respectively. Let Z CAB be a rt. Z and Z CAE an oblique Z. To prove A ABC> A AEC. Draw the altitude EF. A ABC and AEC have the same base, AC. Altitude AB > altitude EF. .'. A ABC > A AEC. Q.E.D. 574. Cor. I. Conversely, if two sides are given, and if the triangle is a maximum, then the given sides include a right angle. Hint. Prove by reductio ad absurdum. * In later mathematics a somewhat broader use will be made of these terms. 284 APPENDIX 285 575. Cor. II. Of all parallelograms having given sides, the one that is rectangular is a maximum, and con- versely. Ex. 1135. Construct the maximum parallelogram having two lines of given lengths as diagonals. Ex. 1136. What is the minimum line from a given point to a given line? Ex. 1137. Of all triangles having the same base and altitude, that which is isosceles has the minimum perimeter. Proposition II. Theorem 576. Of all equivalent triangles having the same base, that which is isosceles has the least perimeter. nCD A C Given equivalent A ABC and AEC with the same base AC, and let AB= BC and AE^EC. To prove AB + BC + CA < AE + EC + CA. Draw CF J_ AC and let CF meet the prolongation of AB at G. Draw EG and BE and prolong BE to meet GC at F. BF II AC. Z.CBF=Z.FBG. BF bisects CG and is X CG. .-. BC = BG and EC = EG. AB + BG < AE-\-EG. .'. AB+BC < AE + EC. ^ 'j. AB + BC+CA < AE + EC 4- CA. Q.E.D. 286 PLANE GEOMETRY Ex. 1138. Of all equivalent triangles having the same base, that which has the least perimeter is isosceles. (Prove by reductio ad absur- dum.) Ex. 1139. Of all equivalent triangles, the one that has the minimum perimeter is equilateral. Ex. 1140. State and prove the converse of Ex. 1139. * Proposition III. Theorem 577. Of all isoperimetric triangles on the same base, the isosceles triangle is tJis maximum. AH C Given isosceles A ABC and any other A as AEC having the same base and the same perimeter as A ABC. To prove A ABC > A AEC. Draw BH± AC, EF from E \\ AC, and draw AF and FC. A AFC is isosceles. ,\ perimeter of A AFC < perimeter of A AEC. .*. perimeter of A AFC < perimeter of A ABC. .'. AF < AB. .-. FH< BH. .'. A AFC < A ABC. .: A AEC A AEC. Q.e.d. Ex. 1141. Of all triangles having a given perimeter and a given base, the one that has the maximum area is isosceles. Ex. 1142. What is the maximum chord of a circle ? What is the maximum and what the minimum line that can be drawn from a given exterior point to a given circumference ? APPENDIX 287 Ex. 1143. Of all triangles having a given perimeter, the one that has the maximum area is equilateral. Proposition IV. Theorem 578. Of all polygons having all their sides but one equal, respectively, to given lines taken in order, the maximum can be inscribed in a semicircle having the undetermined side as diameter. C A H Given polygon ABCEFH, the maximum of all polygons sub- ject to the condition that AB, BC, CE, EF, FH, are equal respec- tively to given lines taken in order. To prove that the semicircumference described with AH as diameter passes through B, C, E, and F. Suppose that the semicircumference with AH as diameter does not pass through some vertex, as E. Draw AE and EH. Then Z AEH is not a rt. Z. Then if the figures ABCE&nd EFH&re revolved about E until AEH becomes a rt. Z, A AEH will be increased in area. .-. polygon ABCEFH can be increased in area without chang- ing any of the given sides. But this contradicts the hypothesis that polygon ABCEFH is a maximum. .*. the supposition that vertex E is not on the semicircumfer- ence is false. .-. the semicircumference passes through E. In the same way it may be proved that every vertex of the polygon lies on the semicircumference. q.e.d. Ex. 1144. Given the hase and the vertex angle of a triangle, con- «%ict the triangle so that its area shall be a maximum. 288 PLANE GEOMETRY Ex. 1145. Find the point in a given straight line such that the tan- gents drawn from it to a given circle contain a maximum angle. Proposition V. Theorem 579. Of all polygons that have their sides equal, re- spectively, to given lines taken in order, the polygon that can be circumscribed by a circle is a maximum. Given polygon ABCD which is circumscribed by a O, and polygon A'b'c'd' which cannot be circumscribed by a O, with AB — A'B', BC=B'C', CD = &D', and DA = D'A'. To prove ABCD > A'b'&D'. From any vertex as A draw diameter AE; draw E C and ED. On C'D', which equals CD, construct Ad'c'e' equal to ADCE; draw A'e'. The circle whose diameter is A'E' does not pass through all the points B', C', D'. (Hyp.) .*. either ABCE or EDA or both must be greater, and neither can be less, than the corresponding part of polygon A'b'c'e'd' (§ 578). .-. ABCED > A'B'C'E'D'. But A DCE = A D'C'e\ .'. ABCD > A'B'C'D' Q.E.D. Ex. 1146. In a given semicircle inscribe a trapezoid whose area is a maximum. Ex. 1147. Of all equilateral polygons having a given side and a given number of sides, the one that is regular is a maximum. APPENDIX 289 Proposition VI. Theorem 580. Of all isoperimetric polygons of the same number of sides, the maximum is equilateral. Given polygon ABODE the maximum of all isoperimetric polygons of the same number of sides. To prove AB = BC = CD = DE = EA. Suppose, if possible, BC > CD. On BD as base construct an isosceles A BFD isoperimetric with A BCD. A BFD > A BCD. .'. polygon ABFDE > polygon ABCDE. But this contradicts the hypothesis that ABCDE is the maxi- mum of all isoperimetric polygons having the same number of sides. .'. BC=CD. In like manner any two adjacent sides may be proved equal. .-. AB = BC = CD = DE = EA. Q.E.D. 581. Cor. Of all isoperimetric polygons of the same number of sides, the maximum is regular. Ex. 1148. In a given segment inscribe a triangle whose perimeter is a maximum. 290 PLANE GEOMETRY Proposition VIT. Theorem 582. Of two isoperimetric regular polygons, that which has the greater number of sides has the greater area. Given the isoperimetric polygons P and Q, and let P have one more side than Q. To prove P > Q. In one side of Q take any point as H. EFHG may be considered as an irregular polygon having the same number of sides as P. .*. P > EFHG ; i.e. P > Q. Q.E.D. Proposition VIII. Theorem 583. Of two equivalent regular polygons, that which has the greater number of sides has the smaller pe- rimeter. H Given square S =o= regular hexagon H. To prove perimeter of S > perimeter of H. Construct square R isoperimetric with H. APPENDIX 291 Area of H > area of R ; i.e. area of S > area of R. .-. perimeter of S > perimeter of R. .*. perimeter of S > perimeter of H. q.e.d. 584. Cor. Of all polygons having a given number of sides and a given area, that which has a minimum perimeter is regular. Ex. 1149. Among the triangles inscribed in a given circle, the one that has a maximum perimeter is equilateral. Ex. 1150. Of all polygons having a given number of sides and in- scribed in a given circle, the one that has a maximum perimeter is regular. VARIABLES AND LIMITS. THEOREMS Proposition I. Theorem 585. If a variable can be made less than any assigned value, the product of the variable and any constant can be made less than any assigned value. Given a variable V, which can be made less than any previ- ously assigned value, however small, and let K be any constant. To prove that v • K may be made as small as we please, i.e. less than any assigned value. Assign any value, as a, no matter how small. Now a value for V may be found as small as we please. Take v < ~. Then v- K< a; i.e. V- K may be made less than any assigned value. q.e.d. 586. Cor. I. If a variable can be made less than any assigned value, the quotient of the variable by any con- stant, except zero, can be made less than any assigned value. Hint. — ;= — • V, which is the product of the variable and a constant. A K 587. Cor. II. If a variable can be made less than any assigned value, the product of that variable and a de- t 292 PLANE GEOMETRY creasing value may be made less than any assigned value. Hint. Apply the preceding theorem, using as K a value greater than any value of the decreasing multiplier. 588. Cor. III. The product of a variable and a variable may be a constant or a variable. 589. Cor. IV. If a variable can be made less than any assigned value, the square of that variable can be made less than any assigned value. (Apply Cor. II.) Ex. 1151. Which of the corollaries under Prop. I is illustrated by the theorem: " The product of the segments of a chord drawn through a fixed point within a circle is constant " ? Proposition II. Theorem 590. The limit of the product of a variable and a con- stant, not zero, is the limit of the variable multiplied by the constant. Given any variable V which, approaches the finite limit i, and let K be any constant not zero. To prove the limit of K • V= K • L. Let R = L — V; then V—L — R. .\ K- V=K- L — K- R. But the limit of K . R = 0. .-. the limit of K . V= the limit of (K • L — K • R) = K • L. Q.E.D. 591. Cor. The limit of the quotient of a variable by a constant is the limit of the variable divided by the constant. Hint. — = — • F, which is the product of the variable and a constant. K jBl Proposition III. Theorem 592. If two variables approach finite limits, not zero, then the limit of their product is equal to the product of their limits. APPENDIX 293 Given variables V and V 1 which approach the finite limits L and L\ respectively. To prove the limit of V • V' == L • L 1 . Let R = L—V and R' = L' — r'. Then V= L — R and F' = £' — R ! . .-. V • V' = L • L' —(L' • R + L > R' — R - R^. But the limit of (jJ • R + £ • i?' - R • i?') = 0. .-. the limit of F • V' = the limit of [L • i' — (l' . i? + i • R' — R.R')] = L-L\ .-. the limit of F • V' = £ • i'. q.e.d. 593. Cor. 7/ eac/z/ of any finite number of variables approaches a finite limit, not zero, then the limit of their product is equal to the product of their limits. Proposition IV. Theorem 594. If two related variables are such that one is always greater than the other, and if the greater con- tinually decreases while the less continually increases, so that the difference between the two may be made as small as we please, then the two variables have a com- mon limit which lies between them. A P Q R L R' Q' P' Given the two related variables AP and AP', AP' greater than AP, and let AP and AP' be such that as AP increases AP' shall decrease, so that the difference between AP and AP' shall ap- proach zero as a limit. To prove that AP and AP 1 have a common limit, as AL, which lies between AP and AP'. Denote successive values of AP by AQ, AR, etc., and denote the corresponding values of AP' by AQ', AR', etc. Since every value which AP assumes" is less than any value which AP' assumes (Hyp.) .-. AP < AR'. ^ *j$ut AP is continually increasing. 294 PLANE GEOMETRY A P Q R L R' Q' P' Hence AP has some limit. (By def. of a limit, § 349.). Since any value which AP* assumes is greater than every value which AP assumes (Hyp.) .*. AP' > AR. But AP 1 is continually decreasing. Hence AP 1 has some limit. (By def. of a limit, § 349.) Suppose the limit of AP =£ the limit of AP f . Then let the limit of AP be AK, while that of AP' is AK*. Then AK and AK 1 have some finite values, as m and ra', and their difference is a finite value, as d. But the difference between some value of AP and the cor- responding value of AP' cannot be less than the difference of the two limits AK and AK'. This contradicts the hypothesis that the difference between AP and AP 1 shall approach zero as a limit. .*. the limit of AP = the limit of AP' and lies between AP and ap\ as AL. q.e.d. 595. Theorem. With every straight line segment there is associated a number which may be called its measure- number. For line segments commensurable with the unit this theo- rem was considered in §§ 335 and 336; we shall now consider the case where the segment is incommensurable with the chosen unit. Given the straight line segment a and the unit segment u; to express a in terms of u. Apply u (as a measure) to a as many times as possible, sup- pose t times, then v ' t.u tu ••• times p 2 p s p* respectively, then h ^ ^to + 1 U ^ ^t 3 + l - 2 - -. u< a<^— • u, J. . u < a < -?J-- -u, pi pi p6 pi Now the infinite series of increasing numbers t, J , — , •••, p p 2 none of which exceeds the finite number t + 1, defines a num- ber n (the limit of this series) which we shall call the measure- number of a with respect to u. Moreover, this number n is unique, i.e. independent of p (the number of parts into which . the unit was divided), for if m is any number such that m < n, then m ' u < a, and if m > n, then m • u > a ; we are therefore justified in associating the number n with a, and in saying that n • u = a. 596. Note. Manifestly, the above procedure may be applied to any geometric magnitude whatever, i.e. every geometric magnitude has a unique measure-number. 597. Cor. If a magnitude is variable and approaches a limit, then, as the magnitude varies, the successive measure-numbers of the variable approach as their limit the measure-number of the limit of the magnitude. 598. Discussion of the problem : To determine wlwther two given lines are commen- surable or not; and if they are commensurable, to find their common measure and their ratio (§ 345). Moreover, GD is the greatest common measure of AB and CD. For every measure of AB is a measure of its multiple GE. Hence, every common measure of AB and GD is a common measure of GE and CD and therefore a measure of their differ- 296 PLANE GEOMETRY ence ED, and therefore of AF, which is a multiple of ED Hence, every common measure of AB and CD is a common measure of AB and AF and therefore a measure of their differ- ence FB. Again, every common measure of ED and FB is a common measure of ED and EG (a multiple of FB) and hence of their difference GD. Hence, no common measure of AB and CD can exceed GD. Therefore, GD is the greatest common measure of AB and CD. Now, if AB and CD are commensurable, the process must ter- minate ; for any common measure of AB and CD is a measure of each remainder, and every segment applied as a measure is less than the preceding remainder. Now, if the process did not terminate, a remainder could be reached which would be less than any assigned value, however small, and therefore less than the greatest common measure, which is absurd. If AB and CD are incommensurable, the process will not ter- minate ; for, if it did, the last remainder obtained would be a common measure of AB and CD, as shown above. .C 599. Theorem. An angle can be bisected by only one line. Given Z ABC, bisected by BT. ^>^^ To prove that no other ^ j, bisector of Z ABC exists. ^ ' "~ „ Suppose that another Z?-"^-^^^^^ bisector of Z ABC exists, ■ 'A e.g. BF. Then Z ABF = Z.ABT. This is impossible. .-. no other bisector of A ABC exists. q.e.d. 600. Note on Axioms. The thirteen axioms (§ 54) refer to num- bers and may be used when referring to the measure-numbers of geometric magnitudes. Axioms 2-9 are not applicable always to equal figures. (See Exs. 800 and 801.) Axioms 7 and 8 hold for positive numbers only, but do not hold for negative numbers, for zero, nor for infinity ; axioms 11 and 12 hold only when the number of parts is finite. APPENDIX 297 601. The methods of proving locus theorems are illustrated by the following proofs of § 141 and § 305. (a) The locus of all points equidistant from the ends of a given line is tJie perpendicular bisector of that line. F P 4 4>" F ]P ( Fk r b 3. 1. C Fig. 2. Given line AB and its _L bisector, CF. To prove CF the locus of all points equidistant from A and B. 3. First Method (Fig. l) Argument Let P be any point in CF. Then P is equi- distant from A and B, i.e. every point in CF satisfies the prescribed condition. Let Q be any point not in CF. Then Q is unequally distant from A and B, i.e. no point outside of CF satisfies the prescribed condition. /. CF is the locus of all points equi- distant from A and B. q.e.d. Second Method (Fig. 2) Argument Same as Arg. 1, above. Let R be any point such that RA =RB. Then R lies in CF } i.e. every point which satisfies the prescribed condi- tion lies in CF. Same as Arg. 3, above. Reasons 1. § 134. 2. § 140. 3. § 130. Reasons 2. § 139. 298 PLANE GEOMETRY (b) The locus of the mid-points of all chords of a circle parallel to a given line is the diameter perpendicular to the line. ' R f(- £~ ( ° a\j"" ~rp Given circle 0, line AB, and diameter RS _L AB. To prove RS the locus of the mid-points of all chords of circle that are II AB. Argument Let P be any point in diameter RS. Through P draw FG II AB. Now RS J. AB. .'. RS _L FG. .-. P is the mid-point of FG, a chord II AB. Let Q be any point not in diameter RS. Through Q draw II K II AB, intersecting RS in T. Then AS J_ HK. .-. T is the mid-point of HK, i.e. Q is not the mid-point of HK, a chord II AB. .'. RS is the locus of the mid-points of all chords of circle that are II AB. Q.E.D. Reasons 1. §179. 2. 3. 4. 5. By hyp. § 193. § 302. §179. 6. 7. §193. § 302. 8. § 130 SOLID GEOMETRY BOOK VI LINES, PLANES, AND ANGLES IN SPACE 602. Def. Solid geometry or the geometry of space treats of figures whose parts are not all in the same plane. (For defini- tion of plane or plane surface, see § 34.) 603. From the definition of a plane it follows that : (a) If two points of a straight line lie in a plane, the whole line lies in that plane. (b) A straight line can intersect a plane in not more than one point. 604. Since a plane is unlimited in its two dimensions (length and breadth) only a portion of it can be shown in a figure. This is usually repre- sented by a quad- rilateral drawn as a parallelogram. Thus MN represents a plane. Sometimes, however, conditions make it necessary to represent a plane by a figure other than a parallelogram, as in § 617. Ex. 1152. Draw a rectangle freehand which is supposed to lie : (a) in a vertical plane ; (b) in a horizontal plane. May the four angles of the rectangle of (a) be drawn equal ? those of the rectangle of (5) ? 299 300 SOLID GEOMETRY 605. Note. In the figures in solid geometry dashed lines will be used to represent all auxiliary lines and lines that are not supposed to be visible but which, for purposes of proof, are represented in the figure. All other lines will be continuous. In the earlier work in solid geometry the stu- dent may experience difficulty in imagining the figures. If so, he may find it a great help, for a time at least, to make the figures. By using pasteboard to represent planes, thin sticks of wood or stiff wires to repre- sent lines perpendicular to a plane, and strings to represent oblique lines, any figure may be actually made with a comparatively small expenditure of time and with practically no expense. For reproductions of models actually made by high school students, see group on p. 302 ; also §§ 622, 633, 678, 756, 762, 770, 797. 606. Assumption 20. Revolution postulate. A plane may revolve about a line in it as an axis, and as it does so revolve, it can contain any particular point in space in one and only one position. 607. From the revolution postulate it follows that : Through a given straight line any number of planes may be passed. For, as plane MN revolves about AB as an axis (§ 606) it may occupy an unlimited num- ber of positions each of which will represent a different plane through AB. 608. Def. A plane is said to be determined by given condi- tions if that plane and no other plane fulfills those conditions. 609. From §§ 607 and 608 it is seen that: A straight line does not determine a plane. Ex. 1153. How many planes may be passed through any two points in space ? why ? Ex. 1154. At a point P in a given straight line AB in space, con- struct a line perpendicular to AB. How many such lines can be drawn ? BOOK VI /^ 301 LINES AND PLANES Proposition I. Theorem 610. A plane is determined by a straight line and a point not in the line. ^ 1* N A-±— --* /_ ^ / f /j' ' Given line AB and P, a point not in AB. To prove that AB and P determine a plane. Argument 1. Through AB pass any plane, as MN. 2. Revolve plane MN about AB as an axis until it contains point P. Call the plane in this position RS. 3. Then plane RS contains line AB and point P. 4. Furthermore, in no other position can plane MN, in its rotation about AB, contain point P. 5. .-. RS is the only plane that can contain AB and P. 6. .'. AB and P determine a plane. q.e.d. 611. Cor. I. A plane is determined by three points not in the same straight line. Hint. Let A, J5, and C be the three given point??. Join A and B by a straight line, and apply § 610. 612. Cor. II. A plane is determined by two intersec- ting^ straight lines. Reasons 1. §607. 2. § 606. 8. Arg. 2. 4. § 606. 5. Arg. 4. ('). § 608. 302 SOLID GEOMETRY 613. Cor. III. A plane is determined by two parallel straight lines. Ex. 1155. Given line AB in space, and P a point not in AB\ Con- struct, through P, a line perpendicular to AB. Ex. 1156. Hold two pencils so that a plane can he passed through them. In how many ways can this be done, assuming that the pencils are lines ? why ? # Ex. 1157. Can two pencils be held so that no plane can be passed through them ? If so, how ? Ex. 1158. In measuring wheat with a half bushel measure, the meas- ure is first heaped, then a straightedge is drawn across the top. Why is the measure then even full ? Ex. 1159. Why is a surveyor's transit or a photographer's camera always supported on three legs rather than on two or four ? Ex. 1160. How many planes are determined by four straight lines, no three of which lie in the same plane, if the four lines intersect : (1) at a common point ? (2) at four different points ? 614. Def. The intersection of two surfaces is the locus of all points common to the two surfaces. 615. Assumption 21. Postulate. Two planes having one point in common also have another point in common. Reproduced from Models made by High School Students BOOK VI 303 Proposition IT. Theorem 616. If two -planes intersect, their intersection is a straight line- S N Given intersecting planes MN and RS. To prove the intersection of MN and RS a str. line. Argument 1. Let A and B be any two points common to the two planes MN and RS. Draw str. line AB. Since both A and B lie in plane MN, str. line AB lies in plane MN. Likewise str. line AB lies in plane RS. Furthermore no point outside of AB can lie in both planes. .-. AB is the intersection of planes MN and RS. But AB is a str. line. .-. the intersection of MN and RS is a str. line. q.e.d. Reasons 1. § 615. 2. § 54, 15. 3. § 603, a. 4. § 603, a. 5. § 610. 6. § 614. 7. Arg. 2. 8. Args. 6 and 7. Ex. 1161. Is it possible for more than two planes to intersect in a straight line ? Explain. Ex. 1162. By referring to §§ 26 and 608, give the meaning of the expression, "Two planes determine a straight line." Ex. 1163. Is the statement in Ex. 11G2 always true ? Give reasons for.your answer. 304 SOLID GEOMETRY Proposition III. Theorem 617. If three planes, not passing through the same line, intersect each other, their three lines of intersection are concurrent, or else they are parallel, each to each. M M N M Fig. 1 Fig. 2. Given planes MQ, PS, and RN intersecting each other in lines MN, PQ, and RS-, also : I. Given MN and RS intersecting at (Fig. 1). To prove MN, PQ, and RS concurrent. Argument .• is in line MN, it lies in plane MQ. .• O is in line RS, it lies in plane PS. •. O, lying in planes MQ and PS, must lie in their intersection, PQ. \ PQ passes through 0; i.e. MN, PQ, and RS are concurrent in 0. q.e.d. II. Given MN II RS (Fig. 2). To prove PQ II MN and RS. Argument 1. PQ and MN are either || or not II. 2. Suppose that PQ intersects MN ; then MN also intersects RS. 3. But this is impossible, for MN II RS. 4. .-. PQ Reasons 1. § 603, a. 2. § 003, a. 3. § 614. 4. Arg. 3. MN. 5. Likewise PQ II RS. Q.E.D. Reasons § 161, a. § 617, I. 3. By hyp. 4. § 161, b. 5. By steps sim- ilar to 1-4. BOOK VI 305 618. Cor. If two straight lines are parallel to a third straight line, tJiey are parallel to each other. I E F Given lines AB and CD, each II EF. To prove AB II CD. Argument AB and CD are either || or not II. Through AB and EF pass plane AF, and through CD and EF pass plane CF. Pass a third plane through AB and point C, as plane BC. Suppose that AB is not II CD ; then plane BC will intersect plane CF in some line other than CD, as CH. Then CH II EF. But CD II EF. .-. CH and C2), two straight lines in plane CF, are both II EF. This is impossible. .'. AB li CD. Q.E.D. Reasons 1. § 161, a. 2. § 613. « 3. § 610. 4. § 613. 5. § 617, II. 6. By hyp. 7. Args. 5 and 6. 8. § 178. 9. § 161, b. 619. Def. A straight line is perpendicular to a plane if it is perpendicular to every straight line in .the plane passing through the point of intersection of the given line and plane. 620. Def. A plane is perpendicular to a straight line if the line is perpendicular to the plane. 621. Def. If a line is perpendicular to a plane, its point of intersection with the plane is called the foot of the perpendicular. 306 SOLID GEOMETRY * Proposition IV. Theorem 622. If a straight line is perpendicular to each of two intersecting straight lines at their point of inter- section, it is perpendicular to the plane of those lines. Given str. line FB _L AB and to BC at B, and plane MN con- taining AB and BC. To prove FB J_ plane MN. Outline of Proof 1. In plane MN draw AC; through J? draw any line, as BH y meeting AC at H. 2. Prolong FB to E SO that BE=FB', draw AF, HF, CF, AE, HE, CE. 3. AB and BC are then _L bisectors of FE ; i.e. FA = AE, FC = CE. ) 4. Prove A AFC = A EAC ; then Z HAF = Z EAH. 5. Prove A HAF = A EAH ; then HF = ##. 6. .-.bhA-FE; i.e. FB±BH, any line in plane 3fiV passing through B. 7. .'. FB±MN. 623. Cor. All the perpendiculars that can be drawn to a straight line at a given point in the line lie in a plane perpendicular to the line at the given point. BOOK VI 307 Proposition V. Problem 624. Through a given point to construct a plane per- pendicular to a given line. Fig. 1 Given point P and line AB. To construct, through P, a plane _L AB, I. Construction 1. Through line AB and point P pass a plane, as APD (in Fig. 1, any plane through AB). §§ 607, 610. 2. In plane APD construct PD, through P, J_ AB. §§ 148, 149. 3. Through AB pass a second plane, as ABC. § 607. 4. In plane ylPC, through the foot of PD, construct a _L to AB (PC in Fig. 1, DC in Fig. 2). § 148. 5. Plane MN, determined by C, D, and P, is the plane required. II. The proof is left as an exercise for the student. Hint. Apply § 623. III. The discussion will be given in § 625. Ex. 1164. Tell how to test whether or not a flag pole is erect. Ex. 1165. Lines AB and CD are each perpendicular to line EF. Are AB and CD necessarily parallel ? Explain. Do they necessarily lie jh the same plane ? why or why not ? 308 SOLID GEOMETRY Proposition VI. Theorem 625. Through a given point there exists only one plane perpendicular to a given line. ^8 S^ :::: y /r—^^p/ M £ 1 * M*- 1 ' \B \B Fig. 1. Fig. 2. Given plane MN, through P, _L AB. To prove MN the only plane through P ± AB. Argument Only 1. Either MN is the only plane through P ± AB or it is not. 2. In MN draw a line through P intersecting line AB, as PR. 3. Let plane determined by AB and PR be denoted by APR. 4. Suppose that there exists another plane through P A. AB; let this second plane intersect plane APR in line PS. 5. Then AB _L PR and also PS; i.e. PR and PS are _L AB. 6. This is impossible. 7. .*. MN is the only plane through P_L AB. q.e.d. 626. Question. In Fig. 2, explain why AB JL PS. 627. §§ 624 and 625 may be combined in one statement: Through a given point there exists one and only one plane per- pendicular to a given line. 628. Cor. I. The locus of all points in space equidis- tant from the extremities of a straight line segment is the plane perpendicular to the segment at its mid-point. 629. Def. A straight line is parallel to a plane if the straight line and the plane cannot meet. 630. Def. A straight line is oblique to a plane if it is neither perpendicular nor parallel to the plane. 631. Def. Two planes are parallel if they cannot meet. BOOK VI 309 Proposition VII. Theorem 632. Two planes perpendicular to the same straight line are parallel. a M V N 7 * Given planes MN and RS, each ± line AB. To prove MN II RS. Hint. Use indirect proof. Compare with § 187. Proposition VIII. Theorem 633. If a plane intersects two parallel planes, the lines of intersection are parallel. Given II planes MN and RS, and any plane PQ intersecting MN and RS in AB and CD, respectively. To prove AB II CD. Hint. Show that AB and CD cannot meet. 634. Cor. I. Parallel lines intercepted between the same parallel planes are equal. (Hint. Compare with § 234.) .Ex. 1166. State the converse of Prop. VIII. Is it true? 310 SOLID GEOMETRY Proposition IX. Theorem 635. If two angles, not in the same plane, have their sides parallel respectively, and lying on the same sidle of the line joining their vertices, they are equal-* Given Z ABC in plane MN and Z DEF in plane RS with BA and BC II respectively to ED and EF, and lying on the same side of line BE. To prove Z ABC= Z BEF. Argument Reasons 1. Measure off BA as ED and BC = EF. 1. § 122. 2. Draw AD, CF, AC, and DF. 2. § 54, 15. 3. BA II ED and BC II EF. 3. By hyp. 4. Then ^DFjB and C^J^.B are DD. 4. §240. 5. .-. AD = BE and CF as BE. 5. § 232. 6. .'. AD= CF. 6. § 54, 1. 7. Also AD II 5JE and CF II £J£. 7. § 220. 8. .*. AD II CF. 8. § 618. 9. .-. ^CF2) is a O. 9. § 240. 10. .'. AC=DF. 10. § 232. 11. But £vl = FZ) and BC—EF. 11. Arg. 1. 12. .'. A ABC = A DFF. 12. § 116. 13. .'. Z ABC= Z DEF. Q.E.D. 13. §110. Ex. 1167. Prove Prop. IX if the angles lie on opposite sides of BE. * It will also be seen (§ 645) that the planes of these angles are parallel. BOOK VI 311 Proposition X. Theorem 636. If one of two -parallel lines is perpendicular to a plane, the other also is perpendicular to the plane. A B^ ^E Given AB II CD and AB _L plane MN. To prove CD _L plane MN. Argument Reasons 1. Through D draw any line in plane MN, 1. § 54, 15. as DF. 2. Through B draw BE in plane MN II DF. 2. § 179. 3. Then Z ABE = Z CDF. 3. § 635. 4. But Z XB# is a rt. Z. 4. § 619. 5. .'. Z CD.F is a rt. Z; i.e. CD _L D.F, any 5. § 54, 1. line in plane MN through D. 6. .\ CD X plane MN. Q- E - D - 6. §619. Ex. 1168. In the accompanying diagram AB and CD lie in the same plane. Angle CBA = S5°, angle BCD = 35°, angle ABE= 90°, .RE lying in plane Jf2V. Is CD necessarily perpendicular to plane MN? Prove your answer. Ex. 1169. Can a line be perpendicular to each of two intersecting planes ? Prove. Ex. 1170. If one of two planes is per- pendicular to a given line, but the other is not, the planes are not parallel. Ex. 1171. If a straight line and a plane are each perpendicular to the »s*j,me straight line, they are parallel to each other. 812 SOLID GEOMETRY Proposition XL Problem 637. Through a given point to construct a line per- pendicular to a given plane. R M ■------"1 rN r ... 4"/ 'F i / projec- tion OA. To prove : I. PB = PA; II. PC > PA. The proof is left as an exercise for the student. 659. Cor. I. (Converse of Prop. XIX). Of all oblique lines drawn from a point to a plane : I. Equal oblique lines have equal projections. II. Unequal oblique lines have unequal projections, and the longer line has the greater projection. 660. Cor. II. The locus of a point in space equidistant from all points in the circumference of a circle is a straight line perpendicular to the plane of the circle and passing through its center. 661. Cor. III. The shortest line from a point to a given plane is the perpendicular from that point to the plane. 662. Def. The distance from a point to a plane is the length of the perpendicular from the point to the plane. BOOK VI 321 663. Cor. IV. Two parallel planes are everywhere equally distant. (Hint. See § 634.) 664. Cor. V. If a line is parallel to a plane, all points of the line are equally distant from the plane. Ex. 1178. In the figure of § 658, if PO = 12 inches, PA = 15 inches, and PC = 20 inches, find OA and CA' . Ex. 1179. Find the locus of all points in a given plane which are at a given distance from a point outside of the plane. Ex. 1180. By applying § 660, suggest a practical method of con- structing a line perpendicular to a plane : (a) Through a point in the plane ; (6) Through a point not in the plane. Ex. 1181. Find a point in a plane equidistant from all points in the circumference of a circle not lying in the plane. Ex. 1182. Find the locus of all points equidistant from two parallel planes. Ex. 1183. Find the locus of all points at a given distance d from a given plane MN. Ex. 1184. Find the locus of all points in space equidistant from two parallel planes and equidistant from two fixed points. Ex. 1185. A line and its projection upon a plane always lie in the same plane. Ex. 1186. (a) The acute angle that a straight line makes with its own projection upon a plane is the least angle that it makes with any line passing through its foot in the plane. (b) With what line passing through its foot and lying in the plane does it make the greatest angle ? Hint, (a) Measure off BD = BC. Which is greater, AD or AC? By means of § 173, prove A ABC < A ABB. 665. Def. The acute angle that a straight line, not perpen- dicular to a given plane, makes with its own projection upon the plane, is called the inclination of the line to the plane. Ex. 1187. Find the projection of a line 12 inches long upon a plane, ikth,e inclination of the line to the plane is 30° ; 45° ; 60°. 322 SOLID GEOMETRY DIHEDRAL ANGLES 666. Defs. A dihedral angle is the figure formed by two planes that diverge from a line. The planes forming a di- hedral angle are called its faces, and the intersection of these planes, its edge. 667. A dihedral angle may be designated by reading in order the two planes forming the angle; thus, an angle formed by planes AB and CD is angle AB-CD, and is usually written angle A-BC-D. If there is no other dihedral angle having the same edge, the line forming the edge is a sufficient designation, as dihedral angle BC. 668. Def. Points, lines, or planes lying in the same plane are said to be coplanar. 669. A clear notion of the magnitude * of a dihedral angle may be obtained by imagining that its two faces, considered as finite portions of planes, were at first coplanar and that one of them has revolved about a line com- mon to the two. Thus in the figure we may imagine face CD first to have been in the position of face AB and then to have revolved about 5Cas an axis to the position of face CD. 670. Def. The plane angle of a dihedral angle is the angle formed by two straight lines, one in each face of the dihedral angle, perpendicular to its edge at the same point. Thus if EF, in face AB, is _L BC at F, and FH, in face CD, is J_ BC at F, then Z EFH is the plane Z of the dihedral Z A-BC-D. Ex. 1188. All plane angles of a dihedral angle are equal. Ex. 1189. Is the plane of angle EFH (§ 667) perpendicular to the edge BC? Prove. State your result in the form of a theorem. Ex. 1190. Is Ex. 309 true if the quadrilateral is a quadrilateral in space, i.e. if the vertices of the quadrilateral are not all in the same plane ? Prove. BOOK VI 323 671. Def. Two dihedral angles are adjacent if they have a common edge and a common face which lies between them ; thus Z A-BC-D and Z B-CB-E are adj. dihe- D* dral A. 672. Def. If one plane meets another so as to make two adjacent dihe- dral angles equal, each of these angles is a right dihedral angle, and , the planes are said to be perpendicular to each other. Thus if plane HP meets plane LM so that dihedral A H-KL-M and M-LK-N are equal, each Z is a rt. dihedral Z, and planes HP and LM are _L to each other. Ex. 1191. By comparison with the definitions of the corresponding terms in plane geometry, frame exact definitions of the following terms : acute dihedral angle ; obtuse dihedral angle ; reflex dihedral angle ; oblique dihedral angle ; vertical dihedral angles ; complementary di- hedral angles ; supplementary dihedral angles ; bisector of a dihedral angle ; alternate interior dihedral angles ; corresponding dihedral angles. Illustrate as many of these as you can with an open book. Ex. 1192. If one plane meets another plane, the sum of the two adjacent dihedral angles is two right dihedral angles. Hint. See proof of § 65. Ex. 1193. If the sum of two adjacent dihedral angles is equal to two right dihedral angles, their exterior faces are coplanar. Hint. See proof of § 76. Ex. 1194. If two planes intersect, the vertical dihedral angles are equal. „ 'Hint. See proof of § 77. 324 SOLID GEOMETRY Proposition XX. Theorem 673. If two dihedral angles are equal, their -plane angles are equal. Given two equal dihedral A BO and b'o' whose plane A are A MNO and M'n'o', respectively. To prove Z MNO = Z M'N'O'. Argument Reasons 1. Superpose dihedral Z BO upon its equal, 1. § 54, 14. dihedral Z B'o', so that point N of edge BO shall fall upon point N' of edge B'O'. 2. Then MN and m'n', two lines in plane 2. § 670. AB, are _L 7iC at point N. 3. .'. MN and J/V are collinear. 3. § 62. 4. Likewise NO and JV'O' are collinear. 4. §§ 670, 62. 5. .'.A MNO = Z M'N'O'. q.e.o. 5. §18. 674. Cor. I. The plane angle of a right dihedral angle is a right angle. 675. Cor. II. If two intersecting planes are each per- pendicular to a third plane, their intersections with the third plane intersect each other. BOOK VI 325 Given planes AB and CD _L plane MN and intersecting each other in line DB ; also let AE and FC be the intersections of planes AB and CD with plane MN. To prove that AE and FC intersect each other. Argument 1. Either AE II FC or AE and FC intersect each other. 2. Suppose AE II FC. Then through 77, any point in DB, pass a plane J/fi7> _L 7A7, intersecting FC in if and AE in 2,. 3. Then plane HKL is _L ;(# also. 4. .*. Z 7/£X is the plane Z of dihedral Z .FC, and Z ifLTJ is the plane Z of di- hedral Z „4#. 5. But dihedral A FC and AE are rt. dihe- dral A. .'.A HKL and JTL7J are rt. A. .'.A HKL contains two rt. A. But this is impossible. AE and FC intersect each other, q.e.d. Reasons 1. 161, a. 2. § 627. § 636. § 670. 5. § 672. §674. Arg. 6. §206. § 161, b. Ex. 1195. Find the locus of all points equidistant from two given points in space. Ex. 1196. Find the locus of all points equidistant from three given points in space. Ex. 1197. Are the supplements of equal dihedral angles equal? complements ? Prove your answer. Ex. 1198. If two planes are each perpendicular to a third plane, can they be parallel to each other? Explain. If they are parallel to ea'qh other, prove their intersections with the third plane parallel. 326 SOLID GEOMETRY Proposition XXI. Theorem (Converse of Prop. XX) 676. If the -plane angles of two dihedral angles are equal, the dihedral angles are equal. Given two dihedral A BC and b'c' whose plane A MNO and M'N'O' are equal. To prove dihedral A BC = dihedral Z B'o'. Argument 1. Place dihedral A BC upon dihedral Ab'c' so that plane A MNO shall be superposed upon its equal, plane Am'n'O'. 2. Then BC and B'c' are both _L MN and NO at N. 3. .-. BC and B'c' are both _L plane MNO at N. 4. .-. BC and ^'c' are collinear. 5. .-. planes AB and jl'JJ', determined by MN and J5C, are coplanar ; also planes CD and C'D', determined by BC and NO, are coplanar. 6. .-. dihedral A BC = dihedral Z #'c'. Q.E.D. Reasons 1. § 54, 14. 2. § 670. 3. § 622. 4. § 638. 5. § 612. 6. §18. BOOK VI 327 677. Cor. If the plane angle of a dihedral angle is a right angle, the dihedral angle is a right dihedral angle. Ex. 1199. Prove Ex. 1194 by applying § 676. Ex. 1200. If two parallel planes are cut by a transversal plane, the alternate interior dihedral angles are equal. Hint. Let Z ABC be the plane Z of jj dihedral Z V-WX-Y. Let the plane XL determined by AB and BO intersect plane f JW MN in CD. Then AB and CD lie in / j&- 47 the same plane and are II (§ 633). Prove / JmL / y that Z DCB is the plane Z of dihedral _- XI / ly Z M-Z T-X. */ J jS V Ex. 1201. State the converse of Ex. /D f/C / 1200, and prove it by the indirect method. zf y 1 Ex. 1202. If two parallel planes are 1/ cut by a transversal plane, the correspond- ing dihedral angles are equal. (Hint. See proof of § 190.) Ex. 1203. State the converse of Ex. 1202, and prove it by the indirect method. Ex. 1204. If two parallel planes are cut by a transversal plane, the sum of the two interior dihedral angles on the same side of the transversal plane is two right dihedral angles. (Hint. See proof of § 192.) Ex. 1205. Two dihedral angles whose faces are parallel, each to each, are either equal or supplementary dihedral angles. (Hint. See proof of § 198.) Ex. 1206. A dihedral angle has the same numerical measure as its plane angle. (Hint. Proof similar to that of § 358.) Ex. 1207. Two dihedral angles have the same ratio as their plane angles. Ex. 1208. Find a point in a plane equidistant from three given points not lying in the plane. Ex. 1209. If a straight line intersects one of two parallel planes, it must, if sufficiently prolonged, intersect the other also. (Hint. Use the indirect method and apply §§ 663 and 664.) Ex. 1210. If a plane intersects one of two parallel planes, it must, if sufficiently extended, intersect the other also. (Hint. Use the indirect lfte'^iod and apply § 652. ) 328 SOLID GEOMETRY Proposition XXII. Theorem 678. If a straight line is perpendicular to a plane, every plane containing this line is perpendicular to the given plane. El M Given str. line AB _L plane MN and plane PQ containing line AB and intersecting plane MN in CD. To prove plane PQ _L plane MN. Argument 1. AB _L CD. 2. Through B, in plane MN, draw BE _L CD. 3. Then Z ABE is the plane Z of dihedral Z Q-CD-M. 4. But Z ABE is a rt. Z. 5. .-. dihedral Z Q-CD-M is a rt. dihedral Z, and plane PQJL plane ifJV. q.e.d. Eeasons 1. §619. 2. §63. 3. §670. §619. §677. Ex. 1211. If from the foot of a per- pendicular to a plane a line is drawn at right angles to any line in the plane, the line drawn from the point of intersec- tion so formed to any point in the per- pendicular is perpendicular %o the line of the plane. Hint. Make KE = EH. Prove AK = AH, and apply § Ex. 1212. In the figure of Ex. 1211, if AB = 20, BE JEK= 10, find AK. 142. 4V11, and BOOK VI 329 Proposition XXIII. Theorem 679. If two planes are perpendicular to each other, any line in one of them, perpendicular to their intersec- tion, is perpendicular to the other. M ■ j tf- '■"£ ':.'■' :~\ / iskH .. •: -^ "--';"■". .-. B El M Given plane PQ _L plane MN, CD their line of intersection, and AB, in plane PQ, _L CD. To prove AB _L plane MN, Argument 1. Through B, in plane MN, draw BE _L CD. 2. Then Z ABE is the plane Z of the rt. dihedral Z Q-CD-M. 3. .*. Z J£# is a rt. Z, and AB J_ M. 4. But AB J_ CD. 5. .-. AB _L plane JfJT. q.e.d. Reasons 1. §63. 2. §670. 3. §674. 4. By hyp. 5. §622. 680. Cor. If two planes are perpendicular to each other, a line perpendicular to one of them at any point in their line of intersection, lies in the other. Hint. Apply the indirect method, using §§ 679 and 638. Ex. 1213. If a plane is perpendicular to the edge of a dihedral angle, is it perpendicular to each of the faces of the dihedral angle ? Prove your answer. Ex. 1214. The plane containing a straight line and its projection upon a plane is perpendicular to the given plane. Ex. 1215. If two planes are perpendicular to each other, a line per- pendicular to one of them from any point in the other lies in the other p,la,ne. 330 SOLID GEOMETRY Proposition XXIV. Theorem 681. If each of two intersecting planes is perpendicu- lar to a third plane ; I. Their line of intersection intersects the third plane. II. Their line of intersection is perpendicular to the third plane. Given planes PQ and RS _L plane MN and intersecting each other in line AB. To prove : I. That AB intersects plane MN. II. AB J_ plane MN. I. Argument 1. Let planes PQ and RS intersect plane MN in lines PD and RE, respectively. 2. Then PD and RE intersect in a point as C. 3. .*. AB passes through C; i.e. AB intersects plane MN. q.e.d. II. Argument 1. Either AB _L plane MN or it is not. 2. Suppose AB is not _L plane MN, but that some other line through C, the point common to the three planes, is _L plane MN, as line OF. 3. Then CF lies in plane PQ, also in plane RS. 4. .-. CF is the intersection of planes PQ and RS. 5. .*. planes PQ and RS intersect in two str. lines, which is impossible. 6. .-. AB 1. plane MN. q.e.d. Reasons 1. §616. 2. §675. 3. § 617, L Reasons 1. § 161, a 2. §639. 3. §680. 4. §614; 5. §616. 6. § 161, b. BOOK VI 331 Proposition XXV. Problem 682. Through any straight line, not perpendicular to a plane, to construct a plane perpendicular to the given plane. ^,5 Given line AB not _L plane MN. To construct, through AB, a plane _L plane MN. The construction, proof, and discussion are left as an exer- cise for the student. Hint. Apply § 678. For discussion, see § 683. 683. Cor. Through a straight line, not perpendicular to a plane, there exists only one plane perpendicular to the given plane. Hint. Suppose there should exist another plane through AB ± plane MN. What would you know about AB ? 684. §§ 682 and 683 may be combined in one statement as follows : Through a straight line, not perpendicular to a plane, there exists one and only one plane perpendicular to the given plane. Ex. 1216. Apply the truth of Prop. XXIV : (a) to the planes that intersect at the corner of a room ; (6) to the planes formed by an open book placed perpendicular to the top of the desk. Ex. 1217. If a plane is perpendicular to each of two intersecting plaaes, it is perpendicular to their intersection. 332 SOLID GEOMETRY T^ Proposition XXVI. Problem 685. To construct a common perpendicular to any two straight lines in space. K X Given AB and CD, any two str. lines in space. To construct a line _L both to AB and to CD. I. Construction 1. Through CD construct plane MN II AB. § 647. 2. Through AB construct plane AF _L plane MN intersecting MN in EF, and CD in H. § 682. 3. Through H construct HK, in plane AF, _L EF. § 148. 4. HK is J_ to both ^5 and CD and is the line required. II. Proof Argument 1. AB II plane JfJT* 2. .-. EF II J£. 3. But HK _L #F. 4. .*. HK±AB. 5. Also #ir _L plane JOT. 6. .-. HKl_ CD. 7. .-. HK _L to both ^i£ and CD. q.e.d. Reasons 1. By cons. §641. By cons. §193. §679. § 619. 7. Args. 4 and 6. III. The discussion will be given in § 686. 686. Cor. Between two straight lines in space {not in the same plane) there exists only one common per- pendicular. BOOK VI 333 Hint. Suppose JI, in figure of § 685, a second JL to AB and CD. Through Y draw ZW || AB. What is the relation of XY to AB? to ZW? to CD? to plane MN? Through X draw XB ± EF. What is the relation of XB to plane MN ? Complete the proof. 687. §§ 685 and 686 may be combined in one statement as follows : Between two straight lines in space (not in the same plane) there exists one and only one common perpendicular. Ex. 1218. A room is 20 feet long, 15 feet wide, and 10 feet high. Find the length of the shortest line that can be drawn on floor and walls from a lower corner to the diagonally opposite corner. Find the length of the line that extends diagonally across the floor, then along the intersec- tion of two walls to the ceiling. Ex 1219. If two equal lines are drawn from a given point to a given plane, the inclinations of these lines to the given plane are equal. If two unequal lines are thus drawn, which has the greater inclination ? Prove. Ex. 1220. The two planes determined by two parallel lines and a point not in their plane, intersect in a line which is parallel to each of the given parallels. Ex. 1221. If two lines are parallel, their projections on a plane are either the same line, or parallel lines. Ex. 1222. If each of three planes is perpendicular to the other two : (a) the intersection of any two of the planes is perpendicular to the third plane ; (6) each of the three lines of intersection is perpendicular to the other two. Find an illustration of this exercise in the classroom. Ex. 1223. If two planes are parallel, no line in the one can meet any line in the other. Ex. 1224. Find all points equidistant from two parallel planes and equidistant from three points : (a) if the points lie in neither plane ; (b) if the points lie in one of the planes. Ex. 1225. Find all points equidistant from two given points, equi- distant from two parallel planes, and at a given distance d from a third plane. Ex. 1226. If each of two intersecting planes is parallel to a given line, the intersection of the planes is parallel to the line. Ex. 1227. Construct, through a point in space, a straight line that shall be parallel to two intersecting planes. 334 SOLID GEOMETRY Proposition XXVII. Theorem 688. Every point in the plane that bisects a dihedral angle is equidistant from the faces of the angle' Given plane BE bisecting the dihedral Z formed by planes AC and CD ; also PH and PK Js from P, any point in plane BE, to faces AC and CD, respectively. To prove PH= PK. Argument 1. Through PH and PK pass plane MN intersecting plane AC in CH, plane CD in Off, plane BE in PC, and edge BC in C. 2. Then plane MN _L planes AC and CD; i.e. planes ^ C and CD are J_ plane MN. /%. .\ BC _L plane UK, 4. .-. BC± CH, CP, and C^. 5. .-. Apch and jBTCP are the plane z§ of the dihedral A E-BC-A and D-CB-E. 6. But dihedral Z E-BC-A = dihedral Z D-CB-E. 7. .-. Zpch=Zkcp. 8. Also PC = PC. 9. .*. rt. A PC#= rt. A KCP. 10. .'. PH= PK. Q.E.D. Reasons 1. § 612, 616. 2. § 678. 3. §681,11. 4. § 619. 5. § 670. 6. By hyp. 7. § 673. 8. Byiden. 9. § 209. 10. § 110. BOOK VI 335 689. Cor. I. Every point equidistant from the two faces of a dihedral angle lies in the plane bisecting the angle. 690. Cor. n. The plane bisecting a dihedral angle is the locus of all points in space equidistant from the faces of the angle. 691. Cor. m. Problem. To construct the bisector of a given dihedral angle. Ex. 1228. Prove that a dihedral angle can be bisected by only one plane. Hint. See proof of § 599. Ex. 1229. Find the locus of all points equidistant from two inter- secting planes. Of how many planes does this locus consist ? Ex. 1230. Find the locus of all points in space equidistant from two intersecting lines. Of how many planes does this locus consist ? Ex. 1231. Find the locus of all points in space equidistant from two parallel lines. Ex. 1232. Find the locus of all points in space equidistant from two intersecting planes and equidistant from all points in the circumference of a circle. Ex. 1233. Find the locus of all points in space equidistant from two intersecting planes and equidistant from two fixed points. Ex. 1234. Find the locus of all points in space equidistant from two intersecting planes, equidistant from two parallel planes, and equidistant from two fixed points. Ex. 1235. If from any point within a dihedral angle lines are drawn perpendicular to the faces of the angle, the angle formed by the perpen- diculars is supplementary to the plane angle of the dihedral angle. Ex. 1236. Given two points, P and #, one in each of two intersecting planes, M and N. Find a point Xin the intersection of planes M and N such that PX+XQ is a minimum. Ex. 1237. Given two points, P and Q, on one side of a given plane MN. Find a point X in plane MN such that PX + XQ shall be a minimum. «fyNT. See Ex. 175. 336 SOLID GEOMETRY POLYHEDRAL ANGLES 692. Def. A polyhedral angle is the figure generated by a moving straight line segment that continually intersects the boundary of a fixed polygon and one extremity of which is a fixed point not in the plane of the given polygon. A poly- hedral angle is sometimes called a solid angle. 693. Defs. The moving line is called the v generatrix, as VA ; the fixed polygon is called a the directrix, as polygon ABODE; the fixed point is called the vertex of the polyhedral , angle, as V. I 694. Defs. The generatrix in any position 4/ J is an element of the polyhedral angle; the /\/ \ Af elements through the vertices of the poly- Bf" tt( gon are the edges, as VA, VB, etc. ; the ( \ portions of the planes determined by the edges of the polyhedral angle, and limited by them are the faces, as AVB, BVC, etc.; the angles formed by the edges are the face angles, as A AVB, BVC, etc.; the dihedral angles formed by the faces are called the dihedral angles of the poly- hedral angle, as dihedral A VA, VB, etc. 695. Def. The face angles and the dihedral angles taken together are sometimes called the parts of a polyhedral angle. 696. A polyhedral angle may be designated by a letter at the vertex and one on each edge, as V-ABCDE. If there is no other polyhedral angle having the same vertex, the letter at the vertex is a sufficient designation, as V. 697. Def. A convex polyhedral angle is a polyhedral angle whose directrix is a convex polygon, i.e. a polygon no side of which, if prolonged, will enter the polygon; as V-ABCDE. In this text only convex polyhedral angles will be considered. 698. Defs. A trihedral angle is a polyhedral angle whose directrix is a triangle (tri-gon) ; a tetrahedral angle, a polyhe- dral angle whose directrix is a quadrilateral (tetra-gon) ; etc. BOOK VI 337 699. Defs. A trihedral angle is called a rectangular, birec- tangular, or trirectangular trihedral angle according as it contains one, two, or three right dihedral angles. 700. Def. An isosceles trihedral angle is a trihedral angle having two face angles equal. Ex. 1238. By holding an open book perpendicular to the desk, illus- trate birectangular and trirectangular trihedral angles. By placing one face of the open book on top of the desk and the other face along the side of the desk against the edge, illustrate a rectangular trihedral angle. Ex. 1239. Is every birectangular trihedral angle isosceles ? Is every isosceles trihedral angle birectangular ? 701. From the general definition of equal geometric figures (§ 18) it follows that: Two polyhedral angles are equal if they can be made to coincide. Proposition XXVIII. Theorem 702. . Two trihedral angles are equal : I. If a face angle and the two adjacent dihedral an- gles of one are equal respectively to a face angle and tlie two adjacent dihedral angles of the other; II. If two face angles and the included dihedral angle of one are equal respectively to two face angles and the included dihedral angle of the other: provided the equal parts are arranged in the same order. The proofs are left as exercises for the student. 703. Questions. Compare care- fully the wording of I above and the accompanying figures with the wording and figures of § 105. What in I takes the place of A in § 105? side? adj. A? What, in the accompanying figure, cor- responds to A ABC in the proof of §105? A DEF? AC? DF? point A? point C? If these and similar changes are made in the proof of § 1Q5^ will it serve as a proof of I above ? Compare II above with § 107. 338 SOLID GEOMETRY Proposition XXIX. Theorem 704. Two trihedral angles are equal if the three face angles of one are equal respectively to the three face angles of the other, and the equal parts are arranged in the same order. Given trihedral A V-ABC and V'-A'B'C', Z AVB = Z A'V'b' , Z B VC=Z B'v'c', Z CVA = Z c'v'a', and the equal face angles arranged in the same order. To prove trihedral Z V-ABC= trihedral Z v'-A'b'c'. 1. Outline of Proof Since, by hyp., any two face A of V-ABC, as A AVB and BVC, are equal, respectively, to the two corresponding face A of v'-A'b'c', it remains only to prove the included dihe- dral A VB and v'b' equal. § 702, II. (See also § 705.) 2. Let face A AVB and BVC be oblique A\ then from any point E in VB, draw ED and EF, in planes AVB and BVC, respec- tively, and _L VB. 3. Since A AVB and BVC are oblique A, ED and EF will meet VA and VC in D and F, respectively. Draw FD. 4. Similarly, lay off V*E' = VE and draw A D'e'f'. Prove rt. A DVE=vt. A D'V'E' ; then VD— V'd', ED=E'd'. Prove rt. A EVF= rt. A E'v'f'-, then VF= v'f', EF=E , f'. Prove A FVD = A f'v'd' ; then FD = f'd'. . A DEF = A D'E'F' : then Z DEF = Z D'E'F' BOOK VI 339 9. But A DEF and d'e'f' are the plane A of dihedral A VB and r'B', respectively. 10. .-. dihedral Z VB = dihedral Z v'b'. 11. .-. trihedral Z F-XBC== trihedral Z F'-^'.B'C'. q.e.d. 705. Note. If all the face A are rt. A, show that all the dihedral A are rt. dihedral A and hence that all are equal. If two face A of a trihe- dral Z are rt. A, show that the third face Z is the plane Z of the included dihedral Z, and hence that two homologous dihedral A, as VB and V'B', are equal. It remains to prove that Prop. XXIX is true if only one face Z of the first trihedral Z and its homologous face Z of the other are rt. A, or if all face A are oblique. 706. Questions. State the proposition in Bk. I that corresponds to § 704. What was the main step in the proof of that proposition ? Did that correspond to proving dihedral Z VB of § 704 = dihedral Z V'B' ? 707. Def. Two polyhedral angles are said to be sym- metrical if their corresponding parts are equal but arranged in reverse order. By making symmetrical polyhedral angles and comparing them, the student can easily satisfy himself that in general they cannot be made to coincide. 708. Def. Two polyhedral angles are said to be vertical if the edges of each are the prolongations of the edges of the other. It will be seen that two vertical, like two symmetrical, poly- hedral angles have their corresponding parts equal but arranged in reverse order. Two Equal Polyhedral Two Vertical Poly- Two Symmetrical Poly- .„ • Angles hedral Angles hedral Angles 340 SOLID GEOMETRY Proposition XXX. Theorem 709. Two trihedral angles are symmetrical : I. If a face angle and the two adjacent dihedral an- gles of one are equal respectively to a face angle and the two adjacent dihedral angles of the other; II. If two face angles and the included dihedral angle of one are equal respectively to two face angles and the included dihedral angle of the other ; III. If the three face angles of one are equal respec- tively to the three face angles of the other : provided the equal parts are arranged in reverse order. V" A '» . /» \ i * \ ' * \ ' * \ / # \ / * v / » \ / » \ / * \ / * \ / ' x / • x The proofs are left as exercises for the student. Hint. Let V and V' be the two trihedral A with parts equal but arranged in reverse order. Construct trihedral Z V" symmetrical to V. Then what will be the relation of V" to V ? of V to V? Ex. 1240. Can two polyhedral angles be symmetrical and equal ? vertical and equal ? symmetrical and vertical ? If two polyhedral angles are vertical, are they necessarily symmetrical ? if symmetrical, are they necessarily vertical ? Ex. 1241. Are two trirectangular trihedral angles necessarily equal? Are two birectangular trihedral angles equal ? Prove your answers. Ex. 1242. If two trihedral angles have three face angles of one equal respectively to three face angles of the other, the dihedral angles of the first are equal respectively to the dihedral angles of the second. BOOK VI 341 / Proposition XXXI. Theorem 710. The sum, of any two face angles of a trihedral angle is greater than the third face angle- V c Given trihedral Z V-ABC in which the greatest face Z is AVB. To prove Z BVC + Z CVA > Z AVB. Outline of Proof 1. Iii face AVB draw VD making Z DVB = A BVC, and through D draw any line intersecting va in E and VB in F. 2. On VC lay off VG = VD and draw FG and GE. 3. Prove A FVG = ADVF; then ^(? = FD. 4. But FG + GE > FD + DE; .: GE > D#. 5. In A GVE and £FD, prove Z GVE > Z J£FD. 6. But Z FVG = Z.DVF. 7. .'. Z J^FG +Z (?FE >Z#F.Z) + ZZ)^; i.e. ZBVC + Z CTJ. > Z ^FS. Q.E.D. 711. Question. State the theorem in Bk. I that corresponds to Prop. XXXI. Can that theorem be proved by a method similar to the one used here ? If so, give the proof. Ex. 1243. If, in trihedral angle V-ABC, angle BVC = 60°, and angle CVA = 80°, make a statement as to the number of degrees in angle AVB. Ex. 1244. Any face angle of a trihedral angle is greater than the difference of the other two. 342 SOLID GEOMETRY Proposition XXXII. Theorem 712. The sum of all the face angles of any convex, poly- hedral angle is less than four right angles- V Given polyhedral Z V with n faces. To prove the sum of the face A at V Z 4 rt. A. Hint. Let a plane intersect the edges of the polyhedral Z in A, B, C, etc. From O, any point in polygon ABC . . ., draw OA, OB, OC, etc. How many A have their vertices at V? at ? What is the sum of all the A of all the A with vertices at V? at O ? Which is the greater, Z ABV + Z FBO or Z ABO + Z 07? O ? Then which is the greater, the sum of the base A of A with vertices at F, or the sum of the base A of A with vertices at O ? Then which is greater, the sum of the face A about F, or the sum of the A about ? 713. Question. Is there a proposition in plane geometry corre- sponding to Prop. XXXII ? If so, state it. If not, state the one that most nearly corresponds to it. Ex. 1245. Can a polyhedral angle have for its faces three equi- lateral triangles ? four ? five ? six ? Ex. 1246. Can a polyhedral angle have for its faces three squares ? four ? Ex. 1247. Can a polyhedral angle have for its faces three regular pentagons ? four ? Ex. 1248. Show that the greatest number of polyhedral angles that can possibly be formed with regular polygons as faces is five. Ex. 1249. Can a trihedral angle have for its faces a regular decagon and two equilateral triangles? a regular decagon, an equilateral tri- angle, and a square ? two regular octagons and a square ? BOOK VII POLYHEDRONS 714. Def. A surface is said to be closed if it separates a finite portion of space from the remaining space. 715. Def. A solid closed figure is a figure in space composed of a closed surface and the finite portion of space bounded by it. 716. Def. A polyhedron is a solid closed figure whose bounding surface is composed of planes only. 717. Defs. The intersections of the bounding planes are called the edges ; //\ >i the intersections of the edges, the /wl™ffl*^ vertices; and the portions of the /$lM bounding planes bounded by the edges, jSfflll II lllllfllfc'^ the faces, of the polyhedron. jM^WtSSm^ 718. Def. A diagonal of a polyhe- B ^^M^^Mmm^W dron is a straight line joining any two ^|["W vertices not in the same face, as AB. >|pi« 11 719. Defs. A polyhedron of four ° y faces is called a tetrahedron; one of six faces, a hexahedron; one of eight faces, an octahedron; one of twelve faces, a do- decahedron ; one of twenty faces, an icosahedron ; etc. Ex. 1250. How many diagonals has a tetrahedron ? a hexahedron ? Ex. 1251. What is the least number of faces that a polyhedron can have ? edges ? vertices ? Ex. 1252. How many edges has a tetrahedron ? a hexahedron ? an octahedron ? Ex. 1253. How many vertices has a tetrahedron? a hexahedron? an octahedron ? Ex. 1254. If E represents the number of edges, F the number of faces, and Fthe number of vertices in each of the polyhedrons mentioned in Exs. 1252 and 1253, show that in each case E+2 = V+F. This result is known as Euler's theorem. 343 344 SOLID GEOMETRY Ex. 1255. Show that in a tetrahedron 8 = (V—2) 4 right angles, where 8 is the sum of the face angles and Fis the number of vertices. ' Ex. 1256. Does the formula, 8= (V—2) 4 right angles, hold for a hexahedron ? an octahedron ? a dodecahedron ? 720. Def. A regular polyhedron is a polyhedron all of whose faces are equal regular polygons, and all of whose polyhe- dral angles are equal. 721. Questions. How many equilateral triangles can meet to form a polyhedral angle (§ 712) ? Then what is the greatest number of regular polyhedrons possible having equilateral triangles as faces ? What is the greatest number of regular polyhedrons possible having squares as faces ? having regular pentagons as faces ? Can a regular polyhedron have as faces regular polygons of more than five sides ? why ? What, then, is the maximum number of kinds of regular polyhedrons possible ? 722. From the questions in § 721, the student has doubtless drawn the conclusion that not more than five kinds of regular polyhedrons exist. He should convince himself that these five are possible by actually making them from cardboard as indicated below : Tetrahedron Hexahedron Octahedron Dodecahedron Icosahedron BOOK VII 345 723. Historical Note. The Pythagoreans knew that there were five regular polyhedrons, but it was Euclid who proved that there can be only five. Ilippasus (aire. 470 B.C.), who discovered the dodecahedron, is said to have been drowned for announcing his discovery, as the Pythago- reans were pledged to refer the glory of any new discovery " back to the founder." PRISMS * 724. Def. A prismatic surface is a surface generated by a moving straight line that continually intersects a fixed broken line and remains parallel to a fixed straight line not coplanar with the given broken line. £j B C Prism Prismatic Surface 725. Defs. By referring to § 693, the student may give the definitions of generatrix and directrix of a prismatic surface. Point these out in the figure. 726. Def. A prism is a polyhedron whose boundary consists- of a prismatic surface and two parallel planes cutting the generatrix in each of its positions. 727. Defs. The two parallel plane sections are the bases of the prism, as ABODE and FGHKL ; the faces forming the pris- matic surface are the lateral faces, as AG, BH, etc.; the inter- sections of the lateral faces are the lateral edges, as AF, BG, etc. In this 'text only prisms whose bases are convex polygons will be considered. * This treatment of prisms and pyramids is given because of its similarity to the treat- ment of cylinders and cones given in §§ 819-822 and 8o7-S4l). 346 SOLID GEOMETRY 728. Def. A right section of a prism is a section formed by a plane which is perpendicular to a lateral edge of the prism and which cuts the lateral edges or the edges prolonged. rA \ \ Right Prism Regular Prism Oblique Prism 729. Def. A right prism is a prism whose lateral edges are perpendicular to the bases. 730. Def. A regular prism is a right prism whose bases are regular polygons. 731. Def. An oblique prism is a prism whose lateral edges are oblique to the bases. 732. Defs. A prism is triangular, quadrangular, etc., accord- ing as its bases are triangles, quadrilaterals, etc. 733. Def. The altitude of a prism is the perpendicular from any point in the plane of one base to the plane of the other base. 734. The following are some of the properties of a prism ; the student should prove the correctness of each : (a) Any two lateral edges of a •prism are parallel. (b) The lateral edges of a prism are equal. (c) Any lateral edge of a right prism is equal to the altitude. (d) TJie lateral faces of a prism are parallelograms. (e) The lateral faces of a right prism are rectangles. (f) The bases of a prism are equal polygons. (g) TJie sections of a prism made by two parallel planes cutting all the lateral edges are equal polygons. (h) Every section of a 2^'ism made by a plane parallel to the base is equal to the base. BOOK VII 347 Proposition I. Theorem 735. Two prisms are equal if three faces including a trihedral angle of one are equal respectively, and simi- larly placed, to three faces including a trihedral angle of the other A B A B Given prisms AI and A'l', face AJ = face A'j', face AG = face A'G', face AD = face A'd'. To prove prism AI= prism A*f. Argument 1. A BAF, FAE, and BAE are equal, respec- tively, to A b'a'f', f'a'e', and b'a'e'. 2. . \ trihedral Z A = trihedral Z A 1 . 3. Place prism AI upon prism A'l 1 so that trihedral Z A shall be superposed upon its equal, trihedral Z A'. 4. Faces A J, AG, and AD are equal, re- spectively, to faces A'j', A'G 1 , and A'D'. 5. .*. J, F, and G will fall upon J\ F 1 , and G', respectively. G. CH and c'h' are both || BG. 7. .*. CiJand C'h' are collinear. 8. .-. H will fall upon#'. 9. Likewise 2" will fall upon l\ 10.^.-. prism AI— prism A'l'. Q.E.D. 5. 10. Reasons § no. §704. § 54, 14. By hyp. §18. § 734, a. § 179. § 603, b. By steps sim- ilar to 6-8. § 18. 348 SOLID GEOMETKY 736. Def. A truncated prism is the portion of a prism in- cluded between the base and a section of the prism made by a plane oblique to the base, but which cuts all the edges of the prism. 737. Cor. I. Two truncated prisms are equal if three faces including a trihedral angle of one are equal re- spectively to three faces including a trihedral angle of the other, and the faces are similarly placed. 738. Cor. H. Two right prisms are equal if tliey have equal bases and equal altitudes. Ex. 1257. Two triangular prisms are equal if their lateral faces are equal, each to each. Ex. 1258. Classify the polyhedrons whose faces are : (a) four tri- angles ; (6) two triangles and three parallelograms ; (c) two quadri- laterals and four parallelograms ; (d) two quadrilaterals and four rectangles ; (e) two squares and four rectangles. Ex. 1259. Find the sum of the plane angles of the dihedral angles whose edges are the lateral edges of a triangular prism ; a quadrangular prism. (Hint. Draw a rt. section of the prism.) Ex. 1260. Every section of a prism made by a plane parallel to a lateral edge is a parallelogram. Ex. 1261. Every section of a prism made by a plane parallel to a lateral face is a parallelogram. Ex. 1262. The section of a parallelopiped made by a plane passing through two diagonally opposite edges is a parallelogram. §3 iiiiiiiiiiiiiiF Oblique Parallelopiped Right Parallelopiped Rectangular Cube Parallelopiped 739. Def. A parallelopiped is a prism whose bases are parallelograms. BOOK VII 349 740. Def. A right parallelopiped is a parallelopiped whose lateral edges are perpendicular to the bases. 741. Def. A rectangular parallelopiped is a right parallelo- piped whose bases are rectangles. 742. Def. A cube (i.e. a regular hexahedron) is a rectangu- lar parallelopiped whose edges are all equal. 743. The following are some of the properties of a parallel- opiped ; the student should prove the correctness of each : (a) All the faces of a parallelopiped are parallelograms. (b) All the faces of a rectangular parallelopiped are rectangles. (c) All the faces of a cube are squares. (d) Any two opposite faces of a parallelopiped are equal and parallel. (e) Any two opposite faces of a parallelopiped may be taken as the bases. Ex. 1263. Classify the polyhedrons whose faces are : (a) six paral- lelograms ; (&) six rectangles ; (c) six squares ; (d) two parallelograms and four rectangles ; (e) two rectangles and four parallelograms ; (/) two squares and four rectangles. Ex. 1264. Find the sum of all the face angles of a parallelopiped. Ex. 1265. Find the diagonal of a cube whose edge is 8 ; 12 ; e. Ex. 1266. Find the diagonal of a rectangular parallelopiped whose edges are (5, 8, and 12 ; whose edges are a, &, and c. Ex. 1267. The edge of a cube : the diagonal of a face : the diagonal of the cube =l:x:y ; find x and y. Ex. 1268. Find the edge of a cube whose diagonal is 20 V3 ; d. Ex. 1269. The diagonals of a rectangular parallelopiped are equal. Ex. 1270. The diagonals of a parallelopiped bisect each other. Ex. 1271. The diagonals of a parallelopiped meet in a point. This point is sometimes called the center of the parallelopiped. Ex. 1272. Any straight line through the center of a parallelopiped, with its extremities in the surface, is bisected at the center. Ex. 1273. The sum of the squares of the four diagonals of a rectan- gular parallelopiped is equal to the sum of the squares of the twelve edges. Ex. 1274. Is the statement in Ex. 1273 true for any parallelopiped ? 350 SOLID GEOMETRY PYRAMIDS 744. Def. A pyramidal surface is a surface generated by a mov- ing straight line that continually intersects a fixed broken line and that passes through a fixed point not in the plane of the broken line. 745. Defs. By re- ferring to §§ 693 and 694, give the defini- tions of generatrix, directrix, vertex, and element of a pyramidal surface. Point these out in the figure. 746. Def. A pyram- idal surface consists of two parts lying on opposite sides of the vertex, called the up- per and lower nappes. 747. Def. A pyramid is a polyhedron whose boundary con- sists of the portion of a pyramidal surface extending from its vertex to a plane cutting all its elements, and the section formed by this plane. 748. Defs. By referring to § 727, the student may give the definitions of base, lateral faces, and lateral edges of a pyramid. The vertex of the pyramidal surface is called the vertex of the pyramid, as V. Point these out in the figure. In this text only pyramids whose bases are convex polygons will be considered. BOOK VII 351 749. Defs. A pyramid is triangular, quadrangular, etc., ac- cording as its base is a triangle, a quadrilateral, etc. 750. Questions. How many faces has a triangular pyramid ? a tet- rahedron ? Can these terms be used interchangeably ? How many dif- ferent bases may a triangular pyramid have ? 751. Def. The altitude of a pyramid is the perpendicular from the vertex to the plane of the base, as VO in the figure below, and in the figure on preceding page. V Regular Pyramid Truncated Frustum of Frustum of Pyramid Triangular Pyramid Regular Pyramid 752. Def. A regular pyramid is a pyramid whose base is a regular polygon, and whose vertex lies in the perpendicular erected to the base at its center. 753. Def. A truncated pyramid is the portion of a pyramid included between the base and a section of the pyramid made by a plane cutting all the edges. 754. Def. A frustum of a pyramid is the portion of a pyra- mid included between the base and a section of the pyramid made by a plane parallel to the. base. 755. The following are some of the properties of a pyramid; the student should prove the correctness of each : (a) TJie lateral edges of a regular pyramid are equal. (b) TJie lateral edges of a frustum of a regular pyramid are equal. (c) The lateral faces of a regular pyramid are equal isosceles triangles. (d) The lateral faces of a frustum of a regular pyramid are equal. .isosceles trapezoids. 352 SOLID GEOMETRY Proposition II. Theorem 756. If a -pyramid is cut by a plane parallel to the base : I. The edges and altitude are divided proportionally. II. The section is a polygon similar to the base. Given pyramid V— ABODE and plane MN || base AD cutting the lateral edges in F, G, H, I, and J and the altitude in P. VD_VC_ _ VQ t VF VG~ VH~ ~ VP II. FGHIJ~ ABODE. To prove : i. ™ I. Argument 1. Through V pass plane RS II plane KL 2. Then plane RS II plane MN. VA _ VB VB _ VO VO VF~ VO' VG~ VH' VH VA_VB _VC _ VF~~ VG~ VH~ 3. 4. Reasons ane KL. 1. § 652. 2. § 654. VO , ■ — , etc. VP 3. § 650. VO — . Q.E VP T>. 4. § 54, 1. II. The proof of II is left as an exercise for the student. 757. Cor I. Any section of a pyramid parallel to the base is to the base as the square of its distance from the vertex is to the square of the altitude of the pyramid. Hint. Prove AB 1 VB 2 VO 1 BOOK VII 353 758. Cor. n. If two -pyramids having equal altitudes are eut by planes parallel to their bases, and at equal dis- tances from their vertices, the sections have the same ratio as the bases. Hint. Apply § 757 to each pyramid. 759. Cor. III. If two pyramids have equiva- E lent bases and equal al- titudes, sections made by planes parallel to the bases, and at equal distances from the vertices, are equivalent. Ex. 1275. Is every truncated pyramid a frustum of a pyramid ? Is every frustum of a pyramid a truncated pyramid? What is the lower base of a frustum of a pyramid ? the upper base ? the altitude ? Ex. 1276. Classify the figures whose faces are as indicated below : (a) one quadrilateral and four triangles ; (6) one square and four equal isosceles triangles ; (c) one pentagon and five triangles ; (d) two pentagons and five trapezoids ; (c) two squares and four equal isosceles trapezoids ; (/) two regular hexagons and six rectangles. Ex. 1277. In the figure of § 758, if VP = 12, PO = 8, VA = 28, and VB = 25, find VF and VG. Ex. 1278. The base of a pyramid, whose altitude is 2 decimeters, contains 200 square centimeters. Find the area of a section 6 centimeters from the vertex ; 10 centimeters from the vertex. Ex. 1279. The altitude of a pyramid with square base is 16 inches ; the area of a section parallel to the base and 10 inches from the vertex is 56 i square inches. Find the area of the base. Ex. 1280. The altitude of a pyramid is H. At what distance from the vertex must a plane be passed parallel to the base so that the section formed shall be : (a) one half as large as the base ? (6) one third ? (c) op^ nth ? 354 SOLID GEOMETRY Ex. 1281. Prove that parallel sections of a pyramid are to each other as the squares of their distances from the vertex of the pyramid. Do the results obtained in Ex. 1280 fulfill this condition ? Ex. 1282. Each side of the base of a regular hexagonal pyramid is 6 ; the altitude is 15. How far -from the vertex must a plane be passed parallel to the base to form a section whose area is 12 V3 ? Ex. 1283. The areas of the bases of a frustum of a pyramid are 288 square feet and 450 square feet ; the altitude of the frustum is 3 feet. Eind the altitude of the pyramid of which the given figure is a frustum. Ex. 1284. The bases of a frustum of a regular pyramid are equi- lateral triangles whose sides are 10 inches and 18 inches, respectively ; the altitude of the frustum is 8 inches. Find the alti- tude of the pyramid of which the given figure is a frustum. Ex. 1285. The sum of the lateral faces of any pyramid is greater than the base. Hint. In the figure, let VE be the altitude of face VAD and VO the altitude of the pyramid. Which is the greater, VE or OE ? MENSURATION OF THE PRISM AND PYRAMID Areas 760. turn of 761. lowing a, 6, c = B = b = E = H = h = L = a pyramid, or a frus- of its lateral faces. Def. The lateral area of a prism, a pyramid is the sum of the areas In the mensuration of the prism and pyramid the fol- notation will be used : dimensions of a rectangu- lar parallelopiped. area of base in general or of lower base of a frustum. area of upper base of a frustum. lateral edge, or element, or edge of a tetrahedron in general. altitude of a solid. altitude of a surface. slant height. S = T = V = V lt V2 ... = vertex of a pyramid. perimeter of right sec- tion or of the lower base of a frustum. perimeter of upper base of a frustum. lateral area. total area. volume in general. volumes of smaller sol- ids into which a larger solid is divided. BOOK VII 355 Proposition III. Theorem 762. The lateral area of a prism is equal to the product of the perimeter of a right section and a lateral edge. Given prism AK with MQ a rt. section, E a lateral edge, S the lateral area, and P the perimeter of rt. section MQ. To prove S = P • E. Argument Reasons 1. Rt. section MQ _L Al, CJ, etc. 1. § 728. 2. .'.MN±AI; NQ±.CJ\ etc. 2. §619. 3. .-. MN is the altitude of O AJ\ NQ is 3. § 228. the altitude of O CK; etc. 4. .-. area of O AJ = JW- AI= MN- E-, area of CJ CK = NQ • CJ= NQ > E-, etc. 5. CJAJ+CJCK-] =s(JCir + JTQ +—)*• 6. .*. 8 =c P • #. 763. Cor. The lateral area of a right prism is equal to the product of the perimeter of its base and its altitude. Hint. Thus, if P = perimeter of base and H = altitude, S = P • H. 764. Def. The slant height of a regular pyramid is the alti- tude of any one of its triangular faces. 765. Def. The slant height of a frustum of a regular pyra- mid is the altitude of any one of its trapezoidal faces. 4. §481. 5. §54,2. 6. §309. 356 SOLID GEOMETRY Proposition IV. Theorem 766. The lateral area of a regular pyramid is equal to one half the product of the perimeter of its base and its slant height. C D Given regular pyramid O-ACD • •• with the perimeter of its base denoted by P, its slant height by L, and its lateral area by S. To prove S = £- P . L. Argument Reasons 1. Area of A AOC = £ AC % OR = -J- AC • L; area of A COD = \CB • L-, etc. 1. §485. 2. .-. AAOC+ ACOD+ ... = ±(AC+CD-\ )L. 2. § 54, 2. 3. .'. S = ± P • L. Q.E.D. 3. § 309. 767. Cor. The lateral area of a frustum of a regular pyramid is equal to one half the product of the sum of the perimeters of its bases and its slant height. Hint. Prove 8 = \ ( P + p) L. Ex. 1286. Find the lateral area and the total area of a regular pyramid each side of whose square base is 24 inches, and whose altitude is 16 inches. Ex. 1287. The sides of the bases of a frustum of a regular octagonal pyramid are 15 centimeters and 24 centimeters, respectively, and the slant height is 30 centimeters. Find the number of square decimeters in the lateral area of the frustum. BOOK VII 357 Ex. 1288. Find the lateral area of a prism whose right section is a quadrilateral with sides 5, 7, 9, and 13 inches, and whose lateral edge is 15 inches. Ex. 1289.- Find the lateral area of a right prism whose altitude is 16 inches and whose base is a triangle with sides 8, 11, and 13 inches. Ex. 1290. The perimeter of a right section of a prism is 45 deci- meters ; its altitude is 10 V3 decimeters ; and a lateral edge makes with the base an angle of 60°. Find the lateral area. Ex. 1291. Find the altitude of a regular prism, one side of whose triangular base is 5 inches and whose lateral area is 195 square inches. Ex. 1292. Find the total area of a regular hexagonal prism whose altitude is 20 inches and one side of whose base is 10 inches. Ex. 1293. Find the total area of a cube whose diagonal is 8 VS. Ex. 1294. Find the edge of a cube if its total area is 294 square centimeters ; if its total area is T. Ex. 1295. Find the total area of a regular tetrahedron whose edge is 6 inches. Ex. 1296. Find the lateral area and total area of a regular tetra- hedron whose slant height is 8 inches. Ex. 1297. Find the lateral area and total area of a regular hexagonal pyramid, a side of whose base is 6 inches and whose altitude is 10 inches. Ex. 1298. Find the total area of a rectangular parallelopiped whose edges are 6, 8, and 12 ; whose edges are a, b, and c. Ex. 1299. Find the total area of a right parallelopiped, one side of whose square base is 4 inches, and whose altitude is 6 inches. Ex. 1300. The balcony of a theater is supported by four columns whose bases are regular hexagons. Find the cost, at 2 cents a square foot, of painting the columns if they are 20 feet high and the apothems of the bases are 10 inches. Ex. 1301. In a frustum of a regular triangular pyramid, the sides of the bases are 8 and 4 inches, respectively, and the altitude is 10 inches. Find the slant height and a lateral edge. Ex. 1302. In a frustum of a regular hexangular pyramid, the sides of the bases are 12 and 8, respectively, and the altitude is 16. Find the lateral area. Ex. 1303. In a regular triangular pyramid, the side of the base is 8 inches, and the altitude is 12 inches ; a lateral face makes with the base an angle of 60°. Find the lateral area. 358 SOLID GEOMETRY Volumes 768. Note. The student should compare carefully §§ 769-776 with the corresponding discussion of the rectangle, §§ 466-473. 769. A solid may be measured by finding how many times it contains a solid unit. The solid unit most frequently chosen is a cube whose edge is of unit length. If the unit length is an inch, the solid unit is a cube whose edge is an inch. Such a unit is called a cubic inch. If the unit length is a foot, the solid unit is a cube whose edge is a foot, and the unit is called a cubic foot. Fig. 1. Rectangular Parallelopiped AD = 60 77. 770. Def. The result of the measurement is a number, which is called the measure-number, or numerical measure, or volume of the solid. 771. Thus, if the unit cube U is contained in the rectangular D K Fio. 2. Rectangular Parallelopiped AD = 24 IT- BOOK VII 359 parallelopiped AD (Fig. 1) 60 times, then the measure-number or volume of rectangular parallelopiped AD, in terms of U, is 60. If the given unit cube is not contained in the given rectangu- lar parallelopiped an integral number of times without a re- mainder (Fig. 2), then by taking a cube that is an aliquot part of U, as one eighth of U, and applying it as a measure to the rectangular parallelopiped (Fig. 3), a number will be obtained Fig. 3. Rectangular Parallelopiped AD = &±* U+ = 39| U+. which, divided by 8,* will give another (and usually closer) approximate volume of the given rectangular parallelopiped. By proceeding in this way (Fig. 4), closer and closer approxi- mations to the true volume may be obtained. Fig. 4. Rectangular Parallelopiped AD = *%%& U+ = 41 T 9 3 TJ+. > I * It takes eight of the small cubes to make the unit cube itself. 360 SOLID GEOMETRY 772. If the edges of the given rectangular parallelopiped and the edge' of the unit cube are commensurable, a cube may be found which is an aliquot part of U, and which will be con- tained in the rectangular parallelopiped an integral number of times. 773. If the edges of the given rectangular parallelopiped and the edge of the unit cube are incommensurable, then closer and closer approximations to the volume may be obtained, but no cube which is an aliquot part of U will be also an aliquot part of the rectangular parallelopiped (by definition of incom- mensurable magnitudes). There is, however, a definite limit which is approached more and more closely by the approximations obtained by using smaller and smaller subdivisions of the unit cube, as these subdivisions approach zero as a limit. 774. Def. The volume of a rectangular parallelopiped which is incommensurable with the chosen unit cube is the limit which . successive approximate volumes of the rectangular parallelopiped approach as the subdivisions of the unit cube approach zero as a limit. For brevity the expression the volume of a solid, or simply the solid, is used to mean the volume of the solid with respect to a chosen unit. 775. Def. The ratio of any two solids is the ratio of their measure-numbers, or volumes (based on the same unit). 776. Def. Two solids are equivalent if their volumes are equal. 777. Historical Note. The determination of the volumes of polyhedrons is found in a document as ancient as the Rhind papyrus, which is thought to be a copy of a manuscript dating back possibly as far as 3400 b.c. (See § 474.) In this manuscript Ahmes calculates the con- sents of an Egyptian barn by means of the formula, V— a • b • (c + \ c), where a, b, and c are supposed to be linear dimensions of the barn. But unfortunately the exact shape of these barns is unknown, so that the accuracy of the formula cannot be tested. BOOK VII 361 Proposition V. Theorem 778. The volume of a rectangular parallelopiped is equal to the product of its three dimensions. JL u AT W Given rectangular parallelopiped AD, with dimensions AC, AF, and A G ; and U the chosen unit of volume, whose edge is u. To prove the volume of AD = AC • AF • AG. I. If AC, AF, and AG are each commensurable with u. (a) Suppose that u is contained in AC, AF, and AG each an integral number of times. Argument 1. Lay off u upon AC, AF, and AG. Sup- pose that u is contained in AC r times, in AF s times, and in AG t times. 2. At the points of division on AC, AF, and AG draw planes _L AC, AF, and AG. 3. Then AD is divided into unit cubes. 4. There are r of these unit cubes in a row along AC, s of these rows in parallelopiped AK, and t such paral- lelopipeds in parallelopiped AD. 5. .*. the volume of AD = r • s • t. 6. But r, s, and £ are the measure-numbers of AC, AF, and AG, respectively, re- ferred to the linear unit u. 7. 'the volume of AD=AC-AF- AG. q.e.d. Reasons §335. 2. § 627. 7. § 769. Arg. 1. § 770. Arg. 1. § 309. 362 SOLID GEOMETRY (b) Suppose that u is not a measure of AC, AF, and AG, re- spectively, but that some aliquot part of u is such a measure. The proof is left as an exercise for the student. II. If AC, AF, and AG are each incommensurable with u. Argument 1. Let m be a measure of u. Apply m as a measure to AC, AF, and AG, respec- tively, as many times as possible. There will be remainders, as MC, NF, and QG, each less than m. 2. Through M draw plane MK ± AC, through. N draw plane NK _L AF, and through Q draw plane QK _L AG. 3. Now AM, AN, and AQ are each com- mensurable with the measure m, and hence with u, the linear unit. 4. .-. the volume of rectangular parallel o- piped AK = AM • AN • AQ. 5. Now take a smaller measure of u. No matter how small a measure of u is taken, when it is applied as a measure to AC, AF, and AG, the remainders, MC, NF, and QG, will be smaller than the measure taken. Reasons 1. §339. 2. §627. 3. §337. 4. § 778, I. 5. §335. BOOK VII 363 Argument 6. .-. the difference between AM and AC, the difference between AN and AF, and the difference between AQ and AG, may each be made to become and remain less than any previously assigned segment, however small. 7. .-. AM approaches AC as a limit, AN approaches AF as a limit, and AQ approaches AG as a limit. 8. .-. AM • AN- AQ approaches AC- AF - AG as a limit. 9. Again, the difference between rectan- gular parallelopiped AK and rec- tangular parallelopiped AD may be made to become and remain less than any previously assigned volume, however small. 10. .*. the volume of rectangular parallelo- piped AK approaches the volume of rectangular parallelopiped AD as a limit. 11. But the volume of AK is always equal to AM- AN • AQ. 12. .-. the volume of AD=AC-AF>AG. q.e.d. Reasons 6. Arg. 5. 7. §349. 8. §593. Arg. 5. 10. §349. 11. Arg. 4. 12. §355. III. If AC is commensurable with u but AF and AG are in- commensurable with u. IV. If AC and AF are commensurable with u but AG is in- commensurable with u. The proofs of III and IV are left as exercises for the student. 779. Cor I. The volume of a cube is equal to the cube of its edge. H*N£. Compare with § 478. 364 SOLID GEOMETRY 780. Cor., II. Any two rectangular parallelopipeds are to each other as the products of their three di- mensions. (Hint. Compare with § 479. ) 781. Note. By the product of a surface and a line is meant the product of the measure-numbers of the surface and the line. 782. Cor. III. The volume of a rectangular parallelo- piped is equal to the product of its base and its altitude. 783. Cor. IV. Any two rectangular parallelopipeds are to each other as the products of their bases and their altitudes. (Hint. Compare with § 479.) 784. Cor. V. (a) Two rectangular parallelopipeds having equivalent bases are to each other as their altitudes; (b) two rectangular parallelopipeds having equal altitudes are to each other as their bases. (Hint. Compare with § 480.) 785. Cor. VI. (a) Two rectangular parallelopipeds hav- ing two dimensions in common are to each other as their third dimensions, and (b) two rectangular parallelopi- peds having one dimension in common are to each other as the products of their other two dimensions. 786. Questions. What is it in Book IV that corresponds to volume in Book VII ? to rectangular parallelopiped ? State the theorem and corol- laries in Book IV that correspond to §§ 778, 779, 782, 783, and 784. Will the proofs given there, with the corresponding changes in terms, apply- here ? Compare the entire discussion of §§ 466-480 with §§ 769-785. Ex. 1304. Find the volume of a cube whose diagonal is 5\/3 ; d. Ex. 1305. The volume of a rectangular parallelopiped is V; each side of the square base is one third the altitude of the parallelopiped. Find the side of the base. Find the side of the base if V = 192 cubic feet. Ex. 1306. The dimensions of two rectangular parallelopipeds are 6, 8, 10 and 5, 12, 16, respectively. Find the ratio of their volumes. Ex. 1307 . The total area of a cube is 300 square inches ; find its volume. Ex. 1308. The volume of a certain cube is V; find the volume of a cube whose edge is twice that of the given cube. Ex. 1309. The edge of a cube is a ; find the edge of a cube twice as large ; {. e. containing twice the volume of the given cube. BOOK VII 365 787. Historical Note. Plato (429-348 b.c.) was one of the first to discover a solution to that famous problem of antiquity, the duplication of a cube, i.e. the finding of the edge of a cube whose volume is double that of a given cube. There are two legends as to the origin of the problem. The one is that an old tragic poet rep- resented King Minos as wishing to erect a tomb for his son Glau- cus. The king being dissatis- fied with the dimensions (100 feet each way) proposed by his architect, exclaimed : " The in- closure is too small for a royal tomb ; double it, but fail not in the cubical form." The other legend asserts that the Athenians, who were suf- Plato fering from a plague of typhoid fever, consulted the oracle at Delos as to how to stop the plague. Apollo replied that the Delians would have to double the size of his altar, which was in the form of a cube. A new altar was constructed having its edge twice as long as that of the old one. The pestilence became worse than before, whereupon the Delians ap- pealed to Plato. It is therefore known as the Delian problem. Plato was born in Athens, and for eight years was a pupil of Socrates. Plato possessed considerable wealth, and after the death of Socrates in 399 b.c. he spent some years in traveling and in the study of mathe- matics. It was during this time that he became acquainted with the members of the Pythagorean School, especially with Archytas, who was then its head. No doubt it was his association with these people that gave him his passion for mathematics. About 380 b.c he returned to his native city, where he established a school. Over the entrance to his school was this inscription: "Let none ignorant of geometry enter my door." Later an applicant who knew no geometry was actually turned away with the statement : u Depart, for thou hast not the grip of philosophy." Plato is noted as a teacher, rather than an original discoverer, and his contributions to geometry are improvements in its method rather than additions to its matter. He valued geometry mainly as a "means of ed- ucation in right seeing and thinking and in the conception of imaginary processes." It is stated on good authority that "Plato was almost as imporfcrat as Pythagoras to the advance of Greek geometry." 366 SOLID GEOMETRY Proposition VI. Theorem 788. An oblique prism is equivalent to a right prism, whose base is a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism. o' B C Given oblique prism AD f ; also rt. prism EN' with base EN a rt. section of ad', and with EE', LL', etc., lateral edges of EN', equal to AA', bb', etc., lateral edges of Ad\ To prove oblique prism AD* 'o rt. prism EN'. Outline of Proof 1. In truncated prisms AN and A'N', prove the A of face BK equal, respectively, to the A of face b'k'. 2. Prove the sides of face BK equal, respectively, to the sides of face b'k'. 3. .-. face BK= face b'k'. 4. Similarly face £ilf=face B'm', and face BE— face b'e'. 5. .-. truncated prism AN = truncated prism a'n' (§ 737). 6. But truncated prism A' If = truncated prism A'N. 7. .-. oblique prism AD'=ort. prism EN'. q.e.d. 789. Question. Is there a theorem in Book IV that corresponds to Prop. VI ? If not, formulate one and see if you can prove it true. BOOK VII 367 Proposition VII. Theorem 790. The volume of any parallelopiped is equal to the product of its base and its altitude. \2*M m S:K*j£ n Given parallelopiped I with its volume denoted by V, its base by B, and its altitude by H. To prove V = B • H. Argument 1. Prolong edge AC and all edges of I II AC. 2. On the prolongation of AC take DF = AC, and through D and F pass planes _L AF, forming rt. parallelopiped 77. 3. Then I =^ //. 4. Prolong edge FK and all edges of II II FK. 5. On the prolongation of FK take MN = FK, and through M and N pass planes _L FN , forming rectangular parallelo- piped III. 6. Then II o III. 7. .-.IoIII. 8. Again, BoB' = b". 9. Also H, the altitude of 1,= the altitude of ///. 10. But the volume of 111= b" . H. 1-1* .-. v= B • H. Q.E.D. Reasons § 54, 16. § 627. § 788. § 54, 16. § 627. 6. § 788. 7. §54,1 8. § 482. 9. § 663. 10. § 782. 11. §309. 368 SOLID GEOMETRY 791. Cor* I. Parallelopipeds having equivalent bases and equal altitudes are equivalent. 792. Cor. II. J_ny two parallelopipeds are to' each other as the products of their bases and their altitudes. 793. Cor. III. (a) Two parallelopipeds having equiva- lent bases are to each other as their altitudes, and (b) two parallelopipeds having equal altitudes are to each other as their bases. 794. Questions. What expression in Book IV corresponds to vol- ume of a parallelopiped ? Quote the theorem and corollaries in Book IV that correspond to §§ 790-793. Will the proofs given there, with the corresponding changes, apply here ? Ex. 1310. Prove Prop. VI by subtracting the equal truncated prisms of Arg. 5 from the entire figure. Ex. 1311. The base of a parallelopiped is a parallelogram two adja- cent sides of which are 8 and 15, respectively, and they include an angle of 30°. If the altitude of the parallelopiped is 10, find its volume. Ex. 1312. Four parallelopipeds have equivalent bases and equal lat- eral edges. In the first the lateral edge makes with the base an angle of 30° ; in the second an angle of 45° ; in the third an angle of 60° ; and in the fourth an angle of 90°. Find the ratio of the volumes of the four parallelopipeds. Ex. 1313. Find the edge of a cube equivalent to a rectangular par- allelopiped whose edges are 6, 10, and 15 ; whose edges are a, b, and c. Ex. 1314. Find the diagonal of a cube whose volume is 512 cubic inches ; a cubic inches. Ex. 1315. The edge of a cube is a. Find the area of a section made by a plane through two diagonally opposite edges. Ex. 1316. How many cubic feet of cement will be needed to make a box, including lid, if the inside dimensions of the box are 2 feet 6 inches, 3 feet, and 4 feet 6 inches, if the cement is 3 inches thick ? Hint. In a problem of this kind, always find the volume of the whole solid, and the volume of the inside solid, then subtract. Ex. 1317. The volume of a rectangular parallelopiped is 2430 cubic inches, and its edges are in the ratio of 3, 5, and 6. Find its edges. Ex. 1318. In a certain cube the area of the surface and the volume have the same numerical value. Find the volume of the cube. BOOK VII 369 Proposition VIII. Theorem 795. The plane passed through two diagonally opposite edges of a parallelopiped divides it into two equivalent triangular prisms. jj Given plane AG passed through edges AE and CG of paral- lelopiped BH dividing it into the two triangular prisms ABC-F and CDA-H. To prove prism ABC-F =c= prism CDA-H. Argument Only 1. Let MNOP be a rt. section of parallelopiped BH, cutting the plane AG in line MO. 2. Face AF II face DG and face AH II face BG. 3. .'. MN\\ PO and MP II NO. 4. .-. MNOP is a O. 5. . \ A MNO = A OPM. 6. Now triangular prism ABC-F on rt. prism whose base is A MNO, a rt. section of prism ABC-F, and whose altitude is AE, a lateral edge of prism ABC-F. 7. Likewise triangular prism CDA-H =c=& rt. prism whose base is A OPM and whose altitude is AE. 8. But two such prisms are equivalent. 9. .-. prism ABC-F o prism CDA-H. q.e.d. 796. Questions. Is there a theorem in Book I that corresponds to Prop. VIII? If so, state it. Could an oblique prism exist such that a right section, as MNOP, might intersect either base ? If so, draw a figuf 1 to illustrate. 370 SOLID GEOMETRY Proposition IX. Theorem 797. The volume of a triangular prism is equal to the product of its base and its altitude. Given triangular prism A CD-X with its volume denoted by V, its base by B, and its altitude by H. To prove V= B - H. The proof is left as an exercise for the student. 798. Questions. What proposition in Book IV corresponds to Prop. IX above ? Can you apply the proof there given ? What is the name of the figure CZ in § 797 ? What is its volume ? What part of CZ is A CD-X (§ 795) ? Ex. 1319. The volume of a triangular prism is equal to one half the product of any lateral face and the perpendicular from any point in the opposite edge to that face. Hint. The triangular prism is one half of a certain parallelopiped (§ 795). Ex. 1320. The base of a coal bin which is 8 feet deep is a triangle with sides 10 feet, 15 feet, and 20 feet, respectively. How many tons of coal will the bin hold considering 35 cubic feet of coal to a ton ? Ex. 1321. One face of a triangular prism contains 45 square inches ; the perpendicular to this face from a point in the opposite edge is 6 inches. Find the volume of the prism. Ex. 1322. During a rainfall of | inch, how many barrels of water will fall upon a ten-acre field, counting 7£ gallons to a cubic foot and 31| gallons to a barrel ? Ex. 1323. The inside dimensions of an open tank before lining are 6 feet, 2 feet 6 inches, and 2 feet, respectively, the latter being the height. Find the number of pounds of zinc required to line the tank with a coat- ing \ inch thick, a cubic foot of zinc weighing 6860 ounces. book vn 371 Proposition X. Theorem 799. The volume of any prism is equal to the product of its base and its altitude. ^ AC Given prism AM with its volume denoted by V, its base by B, and its altitude by H. To prove V = B . H. Argument 1. From any vertex of the lower base, as A, draw diagonals AD, AF, etc. 2. Through edge AI and these diagonals pass planes AK, AM, etc. 3. Prism AM is thus divided into triangu- lar prisms. 4. Denote the volume and base of trian- gular prism ACD-J by v l and b x ; of ADF-K by v 2 and b 2 ; etc. Then Vj = b x H ; v 2 = b 2 H; etc. 5. .-. ^ + ^2+ ... = (b 1 + b 2 +-~)H. 6. .-. Fs^.a ■ , 1. Reasons § 54, 15. 2. § 612. 3. Q.E.D. equivalent § 732. §797. § 54, 2. § 309. bases and 800. Cor. I. Prisms having equal altitudes are equivalent. 801. Cor. n. Any two prisms are to each other as the products of their bases and their altitudes. 802. Cor. m. (a) Two prisms having equivalent bases are to each other as their altitudes: (b) two prisms having equal altitudes are to each other as their bases. 372 SOLID GEOMETRY * Proposition XI. Theorem 803. Two triangular pyramids having equivalent bases and equal altitudes are equivalent. & /A / wN Given triangular pyramids O-ACD and O'-a'c'd' with base ACD =c= base A'c'd', with altitudes each equal to QR, and with volumes denoted by V and V', respectively. To prove V = V' . Argument 1. V = V*, V < V f , or V > V'. 2. Suppose V < V', so that v' — v=k, a constant. For convenience, place the two pyramids so that their bases are in the same plane, MN. 3. Divide the common altitude QR into n equal parts, as QX, XY, etc., and through the several points of division pass planes II plane MN. 4. Then section FGU o section f'g'u', section JEW o section j'k' w', etc. 5. On FGU, JEW, etc., as upper bases, con- struct prisms with edges II DO and with altitudes = QX. Denote these prisms by II, III, etc. G. On A'c'D', F'g'u', etc., as lower bases, construct prisms with edges II D'o' and with altitudes = QX. Denote these prisms by I', II' ? etc. Reasons 1. 161, a. 2. § 54, 14. 3. §653. 4. §759. 5. §726. 6. §726. BOOK VII 373 Argument 7. Then prism II =c= prism II', prism III =c= prism III', etc. 8. Now denote the sum of the volumes of prisms II, III, etc., by 5; the sum of the volumes of prisms I', II', III', etc., by S' ; and the volume of prism I' by v'. Then S' — s = v'. 9. But V 1 K-ACD. Q.E.D. 1. Reasons §611. 2. 3. §646. §664. 4. 5. §806. §§749,750. 0. 7. 8. 9. §309. By steps sim- ilar to 3-7 § 54, 1. §309. 380 SOLID GEOMETRY 817. Cor. I. The volume of a truncated right triangu- lar prism is equal to one third the product of its base and the sum of its lateral edges. ^/I^ 818. Cor. n. The volume of any -^ I truncated triangular prism is equal I II to one third the product of a right sec- /V^-,Y / tion and the sum of its lateral edges. I jk/^ Hint. Rt. section ACD divides truncated / / / triangular prism QR into two truncated right Qi^--- J--J triangular prisms. ^^>^^ Ex. 1334. The base of a truncated right triangular prism has for its sides 13, 14, and 15 inches ; its lateral edges are 8, 11, and 13 inches. Find its volume. Ex. 1335. In the formula of §815: (1) put 6 = and compare result with formula of § 805 ; (2) put b = B and compare result with formula of § 799. Ex. 1336. A frustum of a square pyramid has an altitude of 13 inches ; the edges of the bases are 2| inches and 4 inches, respectively. Find the volume. Ex. 1337. The edges of the bases of a frustum of a square pyramid are 3 inches and 5 inches, respectively, and the volume of the frustum is 204£ cubic inches. Find the altitude of the frustum. Ex. 1338. The base of a pyramid contains 144 square inches, and its altitude is 10 inches. A section of the pyramid parallel to the base divides the altitude into two equal parts. Find : (a) the area of the section ; (6) the volume of the frustum formed. Ex. 1339. A section of a pyramid parallel to the base cuts off a pyra- mid similar to the given pyramid. Ex. 1340. The total areas of two similar tetrahedrons are to each other as the squares of any two homologous edges. Ex. 1341. The altitude of a pyramid is 6 inches. A plane parallel to the base cuts the pyramid into two equivalent parts. Find the altitude of the frustum thus formed. Ex. 1342. Two wheat bins are similar in shape ; the one holds 1000 bushels, and the other 800 bushels. If the first is 15 feet deep, how deep is the second ? Ex 1343. A plane is passed parallel to the base of a pyramid cut- ting the altitude into two equal parts. Find : (a) the ratio of the section to the base ; (&) the ratio of the pyramid cut off to the whole pyramid. BOOK VII 381 MISCELLANEOUS EXERCISES Ex. 1344. Find the locus of all points equidistant from the three edges of a trihedral angle. Ex. 1345. Find the locus of all points equidistant from the three faces of a trihedral angle. Ex. 1346. (a) Find the ratio of the volumes and the ratio of the total areas of two similar tetrahedrons whose homologous edges are in the ratio of 2 to 5. (6) Find the ratio of their homologous edges and the ratio of their total areas if their volumes are in the ratio of 1 to 27. Ex. 1347. (a) Construct three or more equivalent pyramids on the same base. (6) Find the locus of the vertices of all pyramids equivalent to a given pyramid and standing on the same base. Hint. Compare with Exs. 821 and 822. Ex. 1348. The altitude of a pyramid is 12 inches. Its base is a regular hexagon whose side is 5 inches. Find the area of a section paral- lel to the base and 4 inches from the base ; 4 inches from the vertex. Ex. 1349. A farmer has a corncrib 20 feet long, a cross section of which is represented in the figure, the numbers denoting feet. If the crib is entirely filled with corn in the ear, how many bushels of corn will it contain, counting 2 bushels of corn in the ear for 1 bushel of shelled corn. (Use the approxi- mation, 1 bushel = 1\ cubic feet. For the exact volume of a bushel, see Ex. 1439.) Ex. 1350. A wheat elevator in the form of a frus- tum of a square pyramid is 30 feet high ; the edges of its bases are 12 feet and 6 feet, respectively. How many bushels of wheat will it hold ? (Use the approximation given in Ex. 1349.) Ex. 1351. A frustum of a regular square pyramid has an altitude of 12 inches, and the edges of its bases are 4 inches and 10 inches, respec- tively. Find the volume of the pyramid of which the frustum is a part. Ex. 1352. In a frustum of a regular quadrangular pyramid, the sides of the bases are 10 and 6, respectively, and the slant height is 14. Find the volume. Ex. 1353. Find the lateral area of a regular triangular pyramid whose altitude is 8 inches, and each side of whose base is 6 inches. Ex. 1354. The edge of a cube is a. Find the edge of a cube 3 times as large ; n times as large. 382 SOLID GEOMETRY Ex. 1355. A berry box supposed to contain a quart of berries is in the form of a frustum of a pyramid 5 inches square at the top, 4| inches square at the bottom, and 2$ inches deep. The United States djry quart contains 67.2 cubic inches. Does the box contain more or less than a quart ? Ex. 1356. The space left in a basement for a coal bin is a rectangle 8 x 10 feet. How deep must the bin be made to hold 10 tons of coal ? Ex. 1357. The figure represents a barn, the numbers denoting the dimensions in feet. Find the number of cubic feet in the barn. Ex. 1358. Let AB, BC, and BD, the dimensions of the barn in Ex. 1357, be denoted by a, 6, and c, respectively. Substitute the values of a, 6, and c in Ahmes' formula given in § 777. Compare your result with the result obtained in Ex. 1357. Would Ahmes' formula have been correct if the Egyptian barns had been similar in shape to the barn in Ex. 1357? Ex. 1359. How much will it cost to paint the barn in Ex. 1357 at 1 cent per square foot for lateral surfaces and 2 cents per square foot for the roof ? Ex. 1360. The barn in Ex. 1357 has a stone foundation 18 inches wide and 3 feet deep. Find the number of cubic feet of masonry if the outer surfaces of the walls are in the same planes as the sides of the barn. Ex. 1361. The volume of a regular tetrahedron is \ 6 -V2. Find its edge, slant height, and altitude. Ex. 1362. The edge of a regular octahedron is a. Prove that the volume equals— V2. Ex. 1363. The planes determined by the diagonals of a parallelo- piped divide the parallelopiped into six equivalent pyramids. Ex. 1364. A dam across a stream is 40 feet long, 12 feet high, 7 feet wide at the bottom, and 4 feet wide at the top. How many cubic feet of material are there in the dam ? how many loads, counting 1 cubic yard to a load ? Give the name of the geometrical solid represented by the dam. Ex. 1365. Given S the lateral area, and H the altitude, of a regular square pyramid, find the volume. Ex. 1366. Find the volume V, of a regular square pyramid, if its total surface is T, and one edge of the base is a. BOOK VIII CYLINDERS AND CONES CYLINDERS 819. Def. A cylindrical surface is a surface generated by a moving straight line that continually intersects a fixed curve and remains parallel to a fixed straight line not coplanar with the given curve. CD K Fig. 1. Cylindrical Surface Fig. 2. Cylinder 820. Defs. By referring to §§ 693 and 694, the student may give the definitions of generatrix, directrix, and element of a cylindrical surface. Point these out in the figure. The student should note that by changing the directrix from a broken line to a curved line, a prismatic surface becomes a cylindrical surface. 821. Def. A cylinder is a solid closed figure whose boun- dary consists of a cylindrical surface and two parallel planes cutting the generatrix in each of its positions, as DC. 383 384 SOLID GEOMETRY 822. Defs. The two parallel plane sections are called the bases of the cylinder, as AC and DF (Fig. 4) ; the portion of the cylindrical surface between the bases is the lateral -surface of the cylinder ; and the portion of an element of the cylin- drical surface included between the bases is an element of the cylinder, as MN. 823. Def . A right cylinder is a cylinder whose elements are perpendicular to the bases. Fig. 3. Right Cylinder Fig. 4. Oblique Cylinder 824. Def. An oblique cylinder is a cylinder whose elements are not perpendicular to the bases. 825. Def. The altitude of a cylinder is the perpendicular from any point in the plane of one base to the plane of the other base, as HE in Figs. 3 and 4. 826. The following are some of the properties of a cylinder ; the student should prove the correctness of each : (a) Any two elements of a cylinder are parallel and equal (§§ 618 and 634). (b) Any element of a right cylinder is equal to its altitude. (c) A line drawn through any point in the lateral surface of a cylinder parallel to an element, and limited. by the bases, is itself an element (§§ 822 and 179). BOOK VIII 385 Proposition I. Theorem 827. The bases of a cylinder are equal. Given cylinder AD with bases AB and CD. To prove base AB = base CD. Argument Only 1. Through any three points in the perimeter of base AB, as E, F, and G, draw elements EH, FK, and GL. 2. Draw EF, FG, GE, HK, KL, and LH. 3. EH is II and = FK; .-. EK is a O. 4. .: EF= HK; likewise FG = KL and GLE =£#. 5. .-. AEFG = Ahkl. 6. .-. base AB may be placed upon base CD so that X, F, and G'will fall upon H, K, and £, respectively. 7. But .E, .F, and G are any three points in'the perimeter of base AB ; i.e. every point in the perimeter of base AB will fall upon a corresponding point in the perimeter of base CD. 8. Likewise it can be shown that every point in the perim- eter of base CD will fall upon a corresponding point in the perimeter of base AB. 9. .*. base AB may be made to coincide with base CD. 10. .*. base AB = base CD. q.e.d. 828. Cor. I. The sections of a cylinder made by two parallel planes cutting all the eleinents are equal. 829. Cor. n. Every section of a cylinder made by a plcttle parallel to its base is equal to the base. 386 SOLID GEOMETRY Ex. 1367. Every section of a cylinder made by a plane parallel to its base is a circle, if the base is a circle. Ex. 1368. If a line joins the centers of the bases of a cylinder, this line passes through the center of every section of the cylinder parallel to the bases, if the bases are circles. ' 830. Def. A right section of a cylinder is a section formed by a plane perpendicular to an element, as section EF. Fig. 1. Cylinder with Circular Base AB WB Fig. 2. Circular Cylinder Fig. 3. Right Cir- cular Cylinder 831. Def. A circular cylinder is a cylinder in which a right section is a circle ; thus, in Fig. 2, if rt. section EF is a ©, cylinder AD is a circular cylinder. 832. Def. A right circular cylinder is a right cylinder whose base is a circle (Fig. 3). 833. Questions. In Fig. 1, is rt. section EF a O ? In Fig. 2, is base AB a O ? In Fig. 3, would a rt. section be a O ? 834. Note. The theorems and exercises on the cylinder that follow will be limited to cases in which the bases of the cylinders are circles. When the term cylinder is used, therefore, it must be understood to mean a cylinder with circular bases. See also § 846. Ex. 1369. Find the locus of all points at a distance of 6 inches from a straight line 2 feet long. Ex. 1370. Find the locus of all points : (a) 2 inches from the lateral surface of a right circular cylinder whose altitude is 12 inches and the radius of whose base is 5 inches ; (&) 2 inches from the entire surface. Ex. 1371. A log is 20 feet long and 30 inches in diameter at the smaller end. Find the dimensions of the largest piece of square timber, the same size at each end, that can be cut from the log. BOOK VIII Proposition II. Theorem 387 835. Every section of a cylinder made by a plane pass- ing through an element is a parallelogram. (See § 834.) E Given cylinder AB with base AK, and CDEF a section made by a plane through element CF and some point, as D, not in CF, but in the circumference of the base. To prove CDEF a O. Argument 1. Through D draw a line in plane DF II CF. 2. Then the line so drawn is an element; i.e. it lies in the cylindrical surface. 3. But this line lies also in plane DF. 4. Reasons § 179. § 826, c. Arg. 1. §614. it is the intersection of plane DF with the cylindrical surface, and coin- cides with DE. 5. .'. DE is a str. line and is II and= CF. 5. § 826, a. 6. Also CD and EF are str. lines. 6. § 616. 7. .-.CDEF is a O. • q.e.d. 7. §240. 836. Cor. Every section of a right circular cyliivder made by a plane passing through an element is a rec- tangle. ■ Ex. 1372. In the figure of Prop. II, the radius of the base is 4 inches, element CF is 12 inches, CD is 1 inch from the center of the base, and CF makes with CD an angle of 60°. Find the area of section CDEF. Ex. 1373. Every section of a cylinder, parallel to an element, is a parallelogram. How is the base of this cylinder restricted ? (See § 834.) 388 SOLID GEOMETRY CONES Conical Surface 837. Def. A conical surface is a surface generated by a moving straight line that continually intersects a fixed curve and that passes through a fixed point not in the plane of the curve. 838. Defs. By referring to §§ 693, 694, and 746, the stu- dent may give the definitions of generatrix, directrix, vertex, element, and upper and lower nappes of a conical surface. Point these out in the figure. The student should observe that by changing the directrix from a broken line to a curved line, a pyramidal surface becomes a conical surface. 839. Def. A cone is a solid closed figure whose boundary consists of the portion of a conical surface extending from its vertex to a plane cutting all its elements, and the section formed by this plane. 840. Defs. By referring to §§ 748 and 822, the student may give the definitions of vertex, base, lateral surface, and element of a cone. Point these out in the figure. 841. Def. A circular cone is a cone containing a circular section such that a line joining the vertex of the cone to the center of the section is perpendicular to the section. Thus in Fig. 4, if section AB of cone V-CD is a O with center 0, such that vo is J_ the section, cone V-CD is a circular cone. 842. Def. The altitude of a cone is the perpendicular from its vertex to the plane of its base, as VC in Fig. 3 and VO in Fig. 5\ Fio. 2. Cone BOOK VIII 389 843. Defs. In a cone with a circular base, if the line join- ing its vertex to the center of its base is perpendicular to the Fig. 3. Cone with Circular Base Fig. 4. Circular Cone V Fig. 5. Right Circu- lar Cone plane of the base, the cone is a right cir- cular cone (Fig. 5). If such a line is not perpendicular to the plane of the base, the cone is called an oblique cone (Fig. 3). 844. Def. The axis of a right circular cone is the line joining its vertex to the center of its base, as vo, Fig. 5. 845. The following are some of the properties of a cone; the student should prove the correct- ness of each : (a) The elements of a right circular cone are equal. (b) TJie axis of a right circular cone is equal to its altitude. (c) A straight line drawn from the vertex of a cone to any point in the perimeter of its base is an element. 846. Note. The theorems and exercises on the cone that follow will be limited to cases in which the bases of the cones are circles, though not necessarily to circular cones. When the term cone is used, therefore, it must be understood to mean a cone with circular base. See also § 834. Ex. 1374. What is the locus of all points 2 inches from the lateral surface, and 2 inches from the base, of a right circular cone whose alti- tudes 12 inches and the radius of whose base is 5 inches ? 390 SOLID GEOMETRY Proposition III. Theorem 847. Every section of a cone made by a plane passing through its vertex is a triangle. V Given cone V-AB with base AB and section VCD made by a plane through V. To prove VCD a A. Argument From V draw str. lines to C and D. Then the lines so drawn are elements ; i.e. they lie in the conical surface. But these lines lie also in plane VCD. .-. they are the intersections of plane VCD with the conical surface, and coincide with VC and VD, respectively. Also CD is a str. line. .*. VC, VD, and CD are str. lines and VCD is a A. q.e.d. Reasons 1. § 54, 15. 2. § 845, c. ' 3. § 603, a. 4. §614. 5. § 616. 6. §92. Ex. 1375. What kind of triangle in general is the section of a cone through the vertex, if the cone is oblique ? if the cone is a right circular cone ? Can any section of an oblique cone be perpendicular to the base of the cone? of a right circular cone ? Explain. Ex. 1376. Find the locus of all straight lines making a given angle with a given straight line, at a given point in the line. What will this locus be if the given angle is 90° ? Ex. 1377. Find the locus of all straight lines making a given angle with a given plane at a given point. What will the locus be if the given angle is 90° ? BOOK VIII 391 Proposition IV. Theorem 848. Every section of a cone made by a plane parallel to its base is a circle. (See § 846.) V Given CD a section of cone V-AB made by a plane II base AB. To prove section CD a O. Outline of Proof 1. Let R and S be any two points on the boundary of section CD-, pass planes through OF and points R and S. 2. Prove A VOM ~ A VPR and A VON ~ A VPS. 3. Then °£. VO A ON — and — VO OM i.e. — = ■ ON PR VP PS VP ' PR PS 4. But OM — ON-, .-. PS = PR-, i.e. P is equidistant from any two points on the boundary of section CD. 5. .-. section CD is a O. q.e.d. 849. Cor. Any section of a cone parallel to its base is to the base as the square of its distance from the vertex is to the square of the altitude of the cone. V By § 563, Prove Then Outline of Proof section CD PR base AB ~OM 2 PR _ 0M~ VP _ ' vo~ VE VF section CD ~vt£ base AB VF 2 "F$r applications of §§ 848 and 849, see Exs. 1380-1384, 392 SOLID GEOMETRY MENSURATION OF THE CYLINDER AND CONE Areas 850. Def. A plane is tangent to a cylinder if* it contains an element, but no other point, of the cylinder. 851. Def. A prism is inscribed in a cylinder if its lateral edges are elements of the cylinder, and the bases of the two figures lie in the same plane. 852. Def. A prism is circumscribed about a cylinder if its lateral faces are all tangent to the cylinder, and the bases of the two figures lie in the same plane. Ex. 1378. How many planes can be tangent to a cylinder? If -two of these planes intersect, the line of intersection is parallel to an element. How are the bases of the cylinders in §§ 850-852 restricted ? (See § 834.) 853. Before proceeding further it might be well for the student to review the more important steps in the develop- ment of the area of a circle. In that development it was shown that : (1) The area of a regular polygon circumscribed about a circle is greater, and the area of a regular polygon inscribed in a circle less, than the area of the regular circumscribed or in- scribed polygon of twice as many sides (§ 541). (2) By repeatedly doubling the number of sides of regular circumscribed and inscribed polygons of the same number of sides, and making the polygons always regular, their areas approach a common limit (§ 546). (3) This common limit is defined as the area of the circle (§ 558). (4) Finally follows the theorem for the area of the circle (§ 559). It will be observed that precisely the same method is used throughout the mensuration of the cylinder and the cone. Compare carefully the four articles just cited with §§ 854, 855, 857, and 858. BOOK VIII 393 Proposition V. Theorem 854. I. The lateral area of a regular prism circum- scribed about a right circular cylinder is greater than the lateral area of the regular circumscribed prism whose base has twice as many sides. II. The lateral area of a regular prism inscribed in a right circular cylinder is less than the lateral area of the regular inscribed prism whose base has twice as many sides. The proof is left as an exercise for the student. Hint. See § 541. Let the given figures represent the bases of the actual figures. Ex. 1379. A regular quadrangular and a regular octangular prism are inscribed in a right circular cylinder with altitude 25 inches and radius of base 10 inches. Find the difference between their lateral areas. Ex. 1380. The line joining the vertex of a cone to the center of the base, passes through the center of every section parallel to the base. Ex. 1381. Sections of a cone made by planes parallel to the base are to each other as the squares of their distances from the vertex. Ex. 1382. The base of a cone contains 144 square inches and the altitude is 10 inches. Find the area of a section of the cone 3 inches from the vertex ; 5 inches from the vertex. Ex. 1383. The altitude of a cone is 12 inches. How far from the vertex must a plane be passed parallel to the base so that the section shall be one half as large as the base ? one third ? one nth ? Ex. 1384. The altitude of a cone is 20 inches ; the area of the section parallel to the base and 12 inches from the vertex is 90 square inches. Find 'Jie area of the base. 394 SOLID GEOMETRY Proposition VI. Theorem 855. By repeatedly doubling the number of sides of the bases of regular prisms circumscribed about, and in- scribed in, a right circular cylinder, and making the bases always regular polygons, their lateral areas approach a common limit. Given H the common attitude, R and r the apothems of the bases, P and p the perimeters of the bases, and S and s the lateral areas, respectively, of regular circumscribed and in- scribed prisms whose bases have the same number of sides. Let the given figure represent the base of the actual figure. To prove that by repeatedly doubling the number of sides of the bases of the prisms, and making them always regular polygons, S and s approach a common limit. Argument 1. S = P • H and s = p- H. 2. S_P s p' 3. But?-* p r 4. S_R s r 5. S — s R— r S R Reasons § 763. 2. § 54, 8 a. 3. § 538. 4. § 54, 1. 5. § 399. BOOK VIII 395 Argument 6. 9. s— s = s But by repeatedly doubling the num- ber of sides of the bases of the prisms, and making them always regular polygons, R — r can be made less than any previously assigned value, however small. .*. can be made less than any previously assigned value, however small. R s R can be made less than any previously assigned value, however small, S being a decreasing variable. 10. .-. S — s, being always equal to S ~ , R can be made less than any previously assigned value, however small. 11. .-. S and s approach a common limit. Q.E.D. Reasons 6. § 54, 7 a. 7. § 543, I. 8. § 586. 9. § 587. 10. §309. 11. § 594. 856. Note. The above proof is limited to regular prisms, but it can be shown that the limit of the lateral area of any inscribed (or circum- scribed) prism is the same by whatever method the number of the sides of its base is successively increased, provided that each side approaches zero as a limit. (See also § 549.) Compare the proof of § 855 with that of § 546, I. 857. Def. The lateral area of a right circular cylinder is the common limit which the successive lateral areas of circum- scribed and inscribed regular prisms (having bases containing 3, 4, 5, etc., sides) approach as the number of sides of the bases is successively increased and each side approaches zero as a limit. 396 SOLID GEOMETRY Proposition VII. Theorem 858. The lateral area of a right circular cylinder is equal to the product of the circumference of its base and its altitude. Given a rt. circular cylinder with its lateral area denoted by S, the circumference of its base by C, and its altitude by //. To prove S = C • H. Argument Circumscribe about the rt. circular cylinder a regular prism. Denote its lateral area by S', the perimeter of its base by P, and its altitude by h. Then S' = P > H. As the number of sides of the base of the regular circumscribed prism is repeatedly doubled, P approaches O as a limit. .-. P • H approaches C • H as a limit. Also S' approaches S as a limit. But s' is always equal to P • H. .'. S=C • H. Q.E.D. Reasons 1. § 852. 2. § 763. 3. § 550. 4. § 590. 5. § 857. 6. Arg. 2. 7. § 355. 859. Cor. If S denotes the lateral area, T the total area, II the altitude, and B the radius of the base, of a right circular cylinder, BOOK VIII 397 S=2ttRH', T= 2 ttRH + 2ttR 2 = 2 ttR{H + r). 860. Note. Since the lateral area of an oblique prism is equal to the product of the perimeter of a right section and a lateral edge (§ 762), the student would naturally infer that the lateral area of an oblique cylinder with circular bases is equal to the product of the perimeter of a right sec- tion and an element. This statement is true. But the right section of an oblique cylinder with circular base is not a circle. And since the only curve dealt with in elementary geometry is the circle, this theorem and its applications have been omitted here. Ex. 1385. Find the lateral area and total area of a right circular cylinder whose altitude is 20 centimeters and the diameter of whose base is 10 centimeters. Ex. 1386. How many square inches of tin will be required to make an open cylindrical pail 8 inches in diameter and 10 inches deep, making no allowance for waste ? Ex. 1387. In a right circular cylinder, find the ratio of the lateral area to the sum of the two bases. What is this ratio if the altitude and the radius of base are equal ? Ex. 1388. Find the altitude of a right circular cylinder if its lateral area is S and the radius of its base B. Ex. 1389. Find the radius of the base of a right circular cylinder if its total area is T and its altitude is H. 861. Def. Because it may be generated by a rectangle re- volving about one of its sides as an axis, a right circular cylinder is sometimes called a cylinder of rev- olution. 862. Questions. What part of the cylinder will side CD, opposite the axis AB, generate? What will AD and BC generate ? What will the plane AC gen- erate ? What might CD in any one of its positions be called ? f^ZZ^wT"*^ 863. Def. Similar cylinders of revolution are cylinders generated by similar rectangles revolving about homologous sides as axes. 398 SOLID GEOMETRY Proposition VIII. Theorem 864. The lateral areas, and the total areas, of two similar cylinders of revolution are to each other as the squares of their altitudes, and as the squares of the radii of their bases. H Given two similar cylinders of revolution with their lateral areas denoted by S and S 1 , their total areas by T and T', their altitudes by Hand H 1 , and the radii of their bases by i? and R', respectively. , N 8 H 2 R 2 To prove : (a) — = — - = — - • S' IV T* H 12 R 12 Argument 1. S = 2ttRH and S' = 2 7ri?'#'. 2. S 2 7ri2H RH R H ' S'~2 ttR'h' ~~ R'H' ~ R' H' a But rectangle RH ~ rectangle R'H' . 4. 5. S _ # H _H 2 " S'~ H' ' H'~~ H' 2 ' 6. A1 S i? i? U 2 7. Again, r= 2 7ri2(H+ i?) and 7-' = 2 ,rtf '(If + *')• Reasons 1. § 859. 2. § 54,8 a. 3. § 863. 4. § 419. 5. § 309. 6. § 309. 7. § 859. BOOK VIII 399 Argument « 2L — R ( H + R ) ._ J*. . H+R " T'~ R\H' + R') ~ R' ' H' + R' 9. But, from Arg. 4, H + R II H' + R' H f R! 10. 11. Also —= R H_ t H^__H^_ B* ' H'~ H' 2 ' R! Rf R" Q.E.D. Reasons 8. § 54,8 a. 9. § 401. 10. § 309. 11. § 309. Ex. 1390. The altitudes of two similar cylinders of revolution are 5 inches and 7 inches, respectively, and the total area of the first is 676 square inches. Find the total area of the second. Ex. 1391. The lateral areas of two similar cylinders of revolution are 320 square inches and 500 square inches, and the radius of the base of the larger is 10 inches. Find the radius of the base of the smaller. Ex. 1392. Two adjacent sides of a rectangle are a and b; find the lateral area of the cylinder generated by revolving the rectangle : (1) about a. as an axis ; (2) about b as an axis. Put the results in the form of a general statement. Have you proved this general statement ? 865. Def. The slant height of a right circular cone is a straight line joining its vertex to any point in the circumference of its base. Thus any element of such a cone is its slant height. 866. Def. A plane is tangent to a cone if it contains an element, but no other point, of the cone. 867. Def. A pyramid is inscribed in a cone if its base is inscribed in the base of the cone and its vertex coincides with the vertex of the cone. 868. Def. A pyramid is circumscribed about a cone if its base is circumscribed about the base of the cone and its vertex coincides with the vertex of the cone. 869. The student may state and prove the theorems on the right circular cone corresponding to those mentioned in § 854. Ex. 1393. How many planes can be tangent to a cone? Through what point must each of these planes pass ? Prove. How are the bases of th* cones in §§ 866-868 restricted ? (See § 846.) 400 SOLID GEOMETRY Proposition IX. Theorem 870. By repeatedly doubling the number of sides of the bases of regular pyramids circumscribed about, and in- scribed in, a right circular cone, and making the bases always regular polygons, tl%eir lateral areas approach a common limit. Given H the common altitude, L and I the slant heights, R and r the apothems of the bases, P and p the perimeters of the bases, and S and s the lateral areas, respectively, of regular circumscribed and inscribed pyramids whose bases have the same number of sides. Let the given figure represent the base of the actual figure. To prove that by repeatedly doubling the number of sides of the bases of the pyramids, and making them always regular polygons, S and s approach a common limit. Reasons 1. § 766. 2. § 54, 8 a. 3. § 538. 4. § 309. 5. § 399. Argument 1. S = 1 PL and s = ^ pi. 2. S_PL_P ■ L s pi p I 3. Butf = * p r 4. S_R L_RL s r I rl 5. S — s RL — rl RL book: VIII 401 = s Argument RL — rl 10. 11. 12. 13. 14. 15. 16. 17. RL Now L — lI2 S' ir \ ) T l H » L f% R I2' The proof is left as an exercise for the student. Hint. Apply the method of proof used in Prop. VIII. 879. Def. A frustum of a cone is the portion of the cone included between the base and a section of the cone made by a plane parallel to the base. 880. Questions. What are the upper and lower bases of a frustum of a cone ? the altitude ? What kind of a figure is the upper base of a frustum of a right circular cone (§ 848) ? 881. Def. The slant height of a frustum of a right circular cone is the length of that portion of an element of the cone included between the bases of the frustum. Ex. 1397. Every section of a frustum of a cone, made hy a plane passing through an element, is a trapezoid. 404 SOLID GEOMETRY Proposition XII. Theorem 882. The lateral area of a frustum of a right circular cone is equal to one half the product of the sum of the circumferences of its bases and its slant height. Given frustum AM, of right circular cone lateral area denoted by S, the circumferences C and c, the radii of its bases by R and r, and byz. To prove S = i(C+c)L. Argument 1.5 = lateral area of cone O-AF minus lateral area of cone O-DM. 2. Let V denote the slant height of cone -DM. Then S=±c(L + L') ±CL' = $CL + |2/(C-c). It now remains to find the value of l! . c r But A OKD ~ A OQA. • R_OA_ L + L' r~ OD~ L* . c L-\-L' c L': L' CL C-c S = ±CL + ± cL C—c ((7-c)=l(C+c)X. Q.E.D. O-AF, with its of its bases by its slant height Reasons . § 54, 11. 2. § 873. 3. § 556. §422. § 424, 2. § 54, 1. Solving for L\ § 309. BOOK VIII 405 883. Cor. I. If s denotes the lateral area, T the total area, L the slant height, and R and r the radii of the bases, of a frustum, of a right circular cone, S = ttL(R + r) ; T = ttL(R + r) + tt^R 2 + r 2 ). 884. Cor. n. The lateral area of a frustum of a right circular cone is equal to the product of its slant height and the circumference of a section midway between its bases. Ex. 1398. If 8 denotes the lateral area, L the slant height, and C the circumference of a section midway between the bases, of a frustum of a right circular cone, then S = CL. Ex. 1399. In the formulas of § 883 : (a) make r= and compare results with formulas of § 875 ; (&) make b = B and compare results with formulas of § 859. Ex. 1400. The altitude of a frustum of a right circular cone is 16 inches, and the diameters of its bases are 20 inches and 30 inches, respectively. Find its lateral area and also its total area. Ex. 1401. In the figure, AB and CD are arcs of circles ; OA = 2 inches, OB = 5 inches, and ZBOC= 120°. Cut figure ABCB out of paper and form it into a frustum of a cone. Find its lateral area and also its total area. Ex. 1402. A frustum of a right circular cone whose altitude is 4 inches and radii of bases 4 inches and 7 inches, respectively, is made as in- dicated in Ex. 1401. Find the radius of the circle from which it must be cut. Ex. 1403. The sum of the total areas of two similar cylinders of rev- olution is 216 square inches, and one altitude is f of the other. Find the total area of each cylinder. Ex. 1404. A regular triangular and a regular hexangular pyramid are inscribed in a right circular cone with altitude 20 inches and with radius of base 4 inches. Find the difference between their lateral areas. Ex. 1405. Cut out of paper a semicircle whose radius is 4 inches, and find its area. Form a cone with this semicircle and find its lateral area by § 875. Do the two results agree ? Ex. 1406. The slant height, and the diameter of the base, of a right circular cone are each equal to L. Find the total area. 406 SOLID GEOMETRY VOLUMES Proposition XIII. Theorem 885. I. The volume of a prism whose base is a regular polygon and which is circumscribed about a cylinder is greater than the volume of the circumscribed prism whose, base is a regular polygon with twice as many sides. II. The volume of a prism whose base is a regular poly- gon and which is inscribed in a cylinder is less than the volume of the inscribed prism whose base is a regular polygon with twice as many sides. The figures and proofs are left as exercises for the student. Proposition XIV. Theorem 886. By repeatedly doubling the number of sides of the bases of prisms circumscribed about, and inscribed in, a cylinder, and making the bases always regular poly- gons, their volumes approach a common limit. Given H the common altitude, B and b the areas of the bases, and V and v the volumes, respectively, of circumscribed and inscribed prisms whose bases are regular and have the same number of sides. Let the given figure represent the base of the actual figure. BOOK VIII 40T To prove that by repeatedly doubling the number of sides of the bases of the prisms, and making them always regular poly- gons, V and v approach a common limit. Argument V= B • H and v = b - H. . v — B ' H — B " v~ b • H~ b' V — v B — b V V—V = V 5. Similar to Arg. 7, § 855. 6. Similar to Arg. 8, § 855. 7. Similar to Arg. 9, § 855. 8. Similar to Arg. 10, § 855. 9. .*. V and v approach a common limit. Q.E.D. 887. Note. The proof of § 886 is limited in the same manner as the proof of § 855. Read § 856. 1. Reasons § 799. 2. § 54,8 a. 3. § 399. 4. § 54,7 a. 5. § 546, II. 6. § 586. 7. § 587. 8. § 309. 9. § 594. Ex. 1407. The total area of a right circular cone whose altitude is 10 inches is 280 square inches. Find the total area of the cone cut off hy a plane parallel to the base and 6 inches from the base. Ex. 1408. The altitude of a right circular cone is 12 inches. What part of the lateral surface is cut off by a plane parallel to the base and 6 inches from the vertex ? Ex. 1409 . The altitude of a right circular cone is II. How far from the vertex must a plane be passed parallel to the base so that the lateral area and the total area of the cone cut off shall be one half that of the original cone ? one third ? one nth ? 888. Def. The volume of a cylinder is the common limit which the successive volumes of circumscribed and inscribed prisms approach as the number of sides of the bases is suc- cessively increased, and each side approaches zero as a limit. 408 SOLID GEOMETRY Proposition XV. Theorem The volume of a cylinder is equal to the product of its base and its altitude. Given a cylinder, with its volume denoted by V, its base by B, and its altitude by H. To prove V=B • H. Argument 1. Circumscribe about the cylinder a prism whose base is a regular poly- gon. Denote its volume by V* and the area of its base by B'. 2. Then V' = B' . H. 3. As the number of sides of the base of the circumscribed prism is re- peatedly doubled, B' approaches B as a limit. 4. .-. B' • H approaches B • H as a limit. 5. Also v' approaches V as a limit. 6. But V f is always equal to B' • H. 7. .*. V—B • H. Q.E.D. Reasons 1. § 852. 2. § 799. 3. §558. 4. § 590. 5. § 888. 6. Arg. 2. 7. § 355. 890. Cor. // v denotes the volume, H the altitude, and R the radius of the base, of a cylinder, V=*R 2 H. BOOK VIII 409 Proposition XVI. Theorem 891. The volumes of two similar cylinders of revolution are to each other as the cubes of their altitudes, and as the cubes of the radii of their bases. 1 1 1 k i i 1 .i i_,/y j < rtt> Given two similar cylinders of revolution with their volumes denoted by V and r', their altitudes by H and H', and the radii of their bases by R and R', respectively. V H s R s To prove V' H' Argument V = ttR^H and F* = ttR ,2 H' . V __ TrtfH __ R 2 H _ R 2 H_ " V'~7rR' 2 H'~ R' 2 !? II' But rectangle RH ~ rectangle R'H 1 . " H'~ R r '*' r'~ R' 2 ' R'~ R' 3 ' But, from Arg. 4, |^ = ^. ii J V H* i.e. — = — - V' H' R 3 Q. E. D. Reasons 1. §890. 2. § 54, 8 a. 3. § 863. 4. § 419. 5. § 309. 6. § 54, 13. 7. § 309. Ex. 1410. The volumes of two similar cylinders of revolution are 135 cubic inches and 1715 cubic inches, respectively, and the altitude of the nr&t is 3 inches. Find the altitude of the second. 410 SOLID GEOMETRY Ex. 1411. A cylinder of revolution has an altitude of 12 inches and a base with a radius of 5 inches. Find the total area of a similar cylinder whose volume is 8 times that of the given cylinder. Ex. 1412. The dimensions of a rectangle are 6 inches and 8 inches, respectively. Find the volume of the solid generated by revolving the rectangle : (a) about its longer side as an axis ; (b) about its shorter side. Compare the ratio of these volumes with the ratio of the sides of the rectangle. Ex. 1413. Cylinders having equal bases and equal altitudes are equivalent. Ex. 1414. Any two cylinders are to each other as the products of their bases and their altitudes. Ex. 1415. (a) Two cylinders having equal bases are to each other as their altitudes, and (6) having equal altitudes are to each other as their bases. Ex. 1416. The volume of a right circular cylinder is equal to the product of its lateral area and one half the radius of its base. Ex. 1417. Cut out a rectangular piece of paper 6x9 inches. Roll this into a right circular cylinder and find its volume (two answers). Ex. 1418. A cistern in the form of a right circular cylinder is to be 20 feet deep and 8 feet in diameter. How much will it cost to dig it at 5 cents a cubic foot ? Ex. 1419. Find the altitude of a right circular cylinder if its volume is V and the radius of its base R. Ex. 1420. In a certain right circular cylinder the lateral area and the volume have the same numerical value, (a) Find the radius of the base. (6) Find the volume if the altitude is equal to the diameter of the base. Ex. 1421. A cylinder is inscribed in a cube whose edge is 10 inches. Find : (a) the volume of each ; (ft) the ratio of the cylinder to the cube. Ex. 1422. A cylindrical tin tomato can is 4 T % inches high, and the diameter of its base is 4 inches. Does it hold more or less than a liquid quart, i.e. *f* cubic inches ? 892. The student may : (a) State and prove the theorems on the cone corresponding to those given in §§ 885 and 886. (6) State, by aid of § 888, the definition of the volume of a BOOK VIII 411 Proposition XVII. Theorem 893. The volume of a cone is equal to one third the product of its base and its altitude. Given a cone with its volume denoted by V, its base by B, and its altitude by H. To prove V=\B • H. The proof is left as an exercise for the student. 894. Question. What changes must be made in the proof of Prop. XV to make it the proof of Prop. XVII ? 895. Cor. If v denotes the volume, H the altitude, and R the radius of the base of a cone, V=\ttE*H. Ex. 1423. Any two cones are to each other as the products of their bases and altitudes. Ex. 1424. The slant height of a right circular cone is 18 inches and makes with the base an angle of 60° ; the radius of the base is 8 inches. Find the volume of the cone. Ex. 1425. The base of a cone has a radius of 12 inches ; an element of the cone is 24 inches long and makes with the base an angle of 30°. Find the volume of the cone. Ex. 1426. The hypotenuse of a right triangle is 17 inches and one side is 15 inches. Find the volume of the solid generated by revolving the triangle about its shortest side as an axis. Ex. 1427. A cone and a cylinder have equal bases and equal alti- tudes. Find the ratio of their volumes. 896. Historical Note. To Eudoxus is credited the proof of the proposition that "every cone is the third part of a cylinder on the same bage and with the same altitude." Proofs of this proposition were also give*n later by Archimedes and Brahmagupta. (Compare with § 809.) 412 SOLID GEOMETRY Proposition XVIII. Theorem 897. The volumes of two similar cones of revolution are to each other as the cubes of their altitudes, as the cubes of their slant heights, and as the cubes of the ra- dii of their bases. Given two similar cones of revolution with their volumes denoted by V and V 1 , their altitudes by H and H\ their slant heights by L and L\ and the radii of their bases by R and R\ respectively. „ V H 3 L 3 R* To prove - = _ = _ = _. The proof is left as an exercise for the student. Hint. Apply the method of proof used in Prop. XVI. Ex. 1428. If the altitude of a cone of revolution is three fourths that of a similar cone, what other fact follows by definition ? Compare the circumferences of the two bases ; their areas. Compare the total areas of the two cones ; their volumes. Ex. 1429. If the lateral area of a right circular cone is 1 T \ times that of a similar cone, what is the ratio of their volumes ? of their altitudes ? Ex. 1430. Through a given cone X two planes are passed parallel to the base ; let Y denote the cone cut off by the upper plane, and Z the entire cone cut off by the lower plane. Prove that Y and Z are to each other as the cubes of the distances of the planes from the vertex of the given cone X. Hint. See Ex. 1381. Ex. 1431. Show that § 897 is a special case of Ex. 1430. Ex. 1432. The lateral area of a cone of revolution is 144 square inches and the total area 240 square inches. Find the volume. BOOK VIII 413 Proposition XIX. Theorem 898. The volume of a frustum of a cone is equal to one third the product of its altitude and the sum of its lower base, its, upper base, and the mean proportional between its two bases. O Given frustum AM, of cone O-AF, with its volume denoted by F, its lower base by B, its upper base by b, and its altitude by H. To prove V=\H(B + b + V^6). The proof is left as an exercise for the student. Hint. In the proof of § 815, change "pyramid" to "cone." 899. Cor. If v denotes the volume, H the altitude, and R and r the radii of the bases of a frustum of a cone, Ex. 1433. Make a frustum of a right circular cone as indicated in Ex. 1401, and of the same dimensions. Find its volume. Ex. 1434. A tin pail is in the form of a frustum of a cone ; the diameter of its upper base is 12 inches, of its lower base 10 inches. How high must the pail be to hold 2| gallons of water? (See Ex. 1422.) Ex. 1435. A cone 6 feet high is cut by a plane parallel to the base and 4 feet from the vertex ; the volume of the frustum formed is 456 cubic inches. Find the volume of the entire cone. . Ex. 1436. Find the ratio of the base to the lateral area of a right circular cone whose altitude is equal to the diameter of its base. 414 SOLID GEOMETRY MISCELLANEOUS EXERCISES Ex. 1437. The base of a cylinder is inscribed in a face of -a cube whose edge is 10 inches. Find the altitude of the cylinder if its volume is equal to the volume of the cube. ' Ex. 1438. A block of marble in the form of a regular prism is 10 feet long and 2 feet 6 inches square at the base. Find the volume of the largest cylindrical pillar that can be cut from it. Ex. 1439. The Winchester bushel, formerly used in England, was the volume of a right circular cylinder 18| inches in internal diameter and 8 inches in depth. Is this the same volume as the bushel used in the United States (2150.42 cubic inches) ? Ex. 1440. To determine the volume of an irregular body, it was immersed in a vessel containing water. The vessel was in the form of a right circular cylinder the radius of whose base was 8 inches. On placing the body in the cylinder, the surface of the water was raised 10| inches. Find the volume of the irregular solid. Ex. 1441. In draining a certain pond a 4-inch tiling (i.e. a tiling whose inside diameter was 4 inches) was used. In draining another pond, supposed to contain half as much water, a 2-inch tiling was laid. It could not drain the pond. What was the error made ? Ex. 1442. A grain elevator in the form of a frustum of a right cir- cular cone is 25 feet high ; the radii of its bases are 10 feet and 5 feet, respectively ; how many bushels of wheat will it hold, counting \\ cubic feet to a bushel ? Ex. 1443. The altitude of a cone with circular base is 16 inches. At what distance from the vertex must a plane be passed parallel to the base to cut the cone into two equivalent parts ? Ex. 1444. Two sides of a triangle including an angle of 120° are 10 and 20, respectively. Find the volume of the solid generated by revolv- ing the triangle about side 10 as an axis. Ex. 1445. Find the volume of the solid generated by revolving the triangle of Ex. 1444 about side 20 as an axis. Ex. 1446. Find the volume of the solid generated by revolving the triangle of Ex. 1444 about its longest side as an axis. Ex. 1447. The slant height of a right circular cone is 20 inches, and the circumference of its base 4tt inches. A plane parallel to the base cuts off a cone whose slant height is 8 inches. Find the lateral area and the volume of the frustum remaining. BOOK VIII 415 Ex. 1448. A cone has an altitude of 12.5 feet and a base whose radius is 8.16 feet ; the base of a cylinder having the same volume as the cone has a radius of 6.25 feet. Find the altitude of the cylinder. Ex. 1449. A log 20 feet long is 3 feet in diameter at the top end and 4 feet in diameter at the butt end. (a) How many cubic feet of wood does the log contain ? (b) How many cubic feet are there in the largest piece of square timber that can be cut from the log ? (c) How many cubic feet in the largest piece of square timber the same size throughout its whole length ? (d) How many board feet does the piece of timber in (c) contain, a board foot being equivalent to a board 1 foot square and 1 inch thick ? Hint. In (&) the larger end is square ABCD. What is the smaller end ? In (c) one end is square EFGH. What is the other end ? Ex. 1450. The base of a cone has a radius of 16 inches. A section of the cone through the vertex, through the center of the base, and per- pendicular 'to the base, is a triangle two of whose sides are 20 inches and 24 inches, respectively. Find the volume of the cone. Ex. 1451. The hypotenuse of a right triangle is 10 inches and one side 8 inches ; find the area of the surface generated by revolving the tri- angle about its hypotenuse as an axis. Ex. 1452. A tin pail in the form of a frustum of a right circular cone is 8 inches deep ; the diameters of its bases are 8^ inches and 10| inches, respectively. How many gallons of water will it hold ? (One liquid gallon contains 231 cubic inches.) Ex. 1453. The altitude of a cone is 12 inches. At what distances from the vertex must planes be passed parallel to the base to divide the cone into four equivalent parts ? Hint. See Ex. 1430. Ex. 1454. Find the volume of the solid generated by an equilateral triangle, whose side is a, revolving about one of its sides as an axis. Ex. 1455. Regular hexagonal prisms are inscribed in and circum- scribed about a right circular cylinder. Find (a) the ratio of the lateral areas of the three solids ; (6) the ratio of their total areas ; (c) the ratio of'tfreir volumes. 416 SOLID GEOMETRY Ex. 1456. How many miles of platinum wire fo of an inch in diam- eter can be made from 1 cubic foot of platinum ? . Ex. 1457. A tank in the form of a right circular cylinder is 5 feet long and the radius of its base is 8 inches. If placed so that its axis is horizontal and filled with gasoline to a depth of 12 inches, how many gallons of gasoline will it contain ? Hint. See Ex. 1024. Ex. 1458. Find the weight in pounds of an iron pipe 10 feet long, if the iron is | inch thick and the outer diameter of the pipe is 4 inches. (1 cubic foot of bar iron weighs 7780 ounces.) Ex. 1459. In a certain right circular cone whose altitude and radius of base are equal, the total surface and the volume have the same numeri- cal value. Find the volume of the cone. Ex. 1460. Two cones of revolution lie on opposite sides of a common base. Their slant heights are 12 and 5, respectively, and the sum of their altitudes is 13. Find the radius of the common base. Ex. 1461. The radii of the lower and upper bases of a frustum of a right circular cone are R and R', respectively. Show that the area of a section midway between them is Zl — ± L . Ex. 1462. A plane parallel to the base of a right circular cone leaves three fourths of the cone's volume. How far from the vertex is this plane ? How far from the vertex is the plane if it cuts off half the volume ? Answer the same questions for a cylinder. Ex. 1463. Is every cone cut from a right circular cone by a plane par- allel to its base necessarily similar to the original cone ? why ? How is it with a cylinder ? why ? Ex. 1464. Water is carried from a spring to a house, a distance of | mile, in a cylindrical pipe whose inside diameter is 2 inches. How many gallons of water are contained in the pipe ? Ex. 1465. A square whose side is 6 inches is revolved about one of its diagonals as an axis. Find the surface and the volume of the solid generated. Can you rind the volume of the solid generated by revolving a cube about one of its diagonals as an axis ? Hint. Make a cube of convenient size from pasteboard, pass a hat- pin through two diagonally opposite vertices, and revolve the cube rapidly. Ex. 1466. Given V the volume, and R the radius of the base, of a right circular cylinder. Find the lateral area and total area. Ex. 1467. Given the total area T, and the altitude H, of a right circular cylinder. Find the volume. BOOK IX THE SPHERE 900. Def. A sphere is a solid closed figure whose boundary is a curved surface such that all straight lines to it from a fixed point within are equal. 901. Defs. The fixed point within the sphere is called its center; a straight line joining the center to any point on the surface is a radius; a straight line passing through the center and having its extremities on the surface is a diameter. 902. From the above definitions and from the definition of equal figures, § 18, it follows that : (a) All radii of the same sphere, or of equal spheres, are equal. (b) All diameters of the same sphere, or of equal spheres, are equal. (c) Spheres having equal radii, or equal diameters, are equal. (d) A sphere may be generated by the revolution of a circle about a diameter as an axis. Ex. 1468. Find the locus of all points that are 3 inches from the sur- face of a sphere whose radius is 7 inches. Ex. 1469. The three edges of a trihedral angle pierce the surface of a sphere. Find the locus of all points of the sphere that are : (a) Equidistant from the three edges of the trihedral angle. (b) Equidistant from the three faces of the trihedral angle. Ex. 1470. Find a point in a plane equidistant from three given points in space. Ex. 1471. Find the locus of all points in space equidistant from the threfc sides of a given triangle. 417 418 SOLID GEOMETRY Proposition I. Theorem 903. Every section of a sphere made by a plane is a circle. Given AMBN a section of sphere made by a plane. To prove section AMBN a O. Argument 1. From draw OQ JL section AMBN. 2. Join Q to C and B, any two points on the perimeter of section AMBN. Draw OC and OB. 3. In rt. A OQC and OQB, OQ = OQ. 4. OC — OB. 5. .'. A OQC = A OQB. 6. .'. QC = QB -, i.e. any two points on the perimeter of section AMBN are equi- distant from Q. 7. .'. section AMBN is a O. q.e.d. Reasons 1. § 639. 2. § 54, 15. 3. By iden. 4. § 902, a. 5. §211. 6. § 110. 276. 904. Def. A great circle of a sphere is a section made by a plane S^ m —-- ^ which passes through the center of A the sphere, as O CRBS. c/-"-"" ffjl 905. Def. A small circle of a \^**** ^J>_ sphere is a section made by a plane \ £ which does not pass through the center of the sphere, as O AMBN. BOOK IX 419 906. Def. The axis of a circle of a sphere is the diameter of the sphere which is perpendicular to the plane of the circle. 907. Def. The poles of a circle of a sphere are the extremi- ties of the axis of the circle. Ex. 1472. Considering the earth as a sphere, what kind of circles are the parallels of latitude ? the equator ? the meridian circles ? What is the axis of the equator ? of the parallels of latitude ? What are the poles of the equator ? of the parallels of latitude ? Ex. 1473. The radius of a sphere is 17 inches. Find the area of a section made by a plane 8 inches from the center. Ex. 1474. The area of a section of. a sphere 45 centimeters from the center is 784 ir square centimeters. Find the radius of the sphere. Ex. 1475. The area of a section of a sphere 7 inches from the center is 576 ir. Find the area of a section 8 inches from the center. 908. The following are some of the properties of a sphere ; the student should prove the correctness of each : (a) In equal sjjheres, or in the same sphere, if two sections are equal, they are equally distant from the center, and conversely. Hint. Compare with § 307. (b) In equal spheres, or in the same sphere, if two sections are unequal, the greater section is at the less distance from the center, and conversely. (Hint. See §§308, 310.) (c) In equal spheres, or in the same sphere, all great circles are equal. (Hint. See § 279, c.) ' (d) The axis of a small circle of a sphere passes through the center of the circle, and conversely. (e) Any two great circles of a sphere bisect each other. (/) Every great circle of a sphere bisects the surface and the sphere. (g) Through two points on the surface of a sphere, not the ex- tremities of a diameter, there exists one and only one great circle. (h) Through three points on the surface of a sphere there exists one and only one circle 909. Def. The distance between two points on the surface of a sphere is the length of the minor arc of the great circle joining them. 420 SOLID GEOMETRY Proposition II. Theorem 910. All points on the circumference of a circle of a sphere are equidistant from either pole of the circle. =^Q Given C and D any two points on the circumference, and P and R the poles, of O AMBK. To prove arc PC= arc PD and arc RC = arc RD. The proof is left as an exercise for the student. Hint. Apply § 298. 911. Def. The polar distance of a circle of a sphere is the distance between any point on its circumference and the nearer pole of the circle. 912. Cor. I. The polar distance of a great circle is a quadrant. 913. Cor. II. Jn equal spheres, or in the same sphere, the polar distances of equal circles are equal. Ex. 1476. What is the locus of all points on the surface of the earth at a quadrant's distance from the north pole ? from the south pole ? from the equator ? from a point P on the equator ? at a distance of 23 \° from the south pole ? 23|° from the equator ? 180° from the north pole ? Ex. 1477. Considering the earth as a sphere with a radius of 4000 miles, calculate in miles the polar distance of : (a) the Arctic Circle ; (6) the Tropic of Cancer ; (c) the equator. Ex. 1478. State a postulate for the construction of a circle on the surface of a sphere corresponding to § 122, the postulate for the construc- tion of a circle in a plane. BOOK IX Proposition III. Theorem 421 914. A -point on the surface of a sphere at the distance of a quadrant from each of tivo other points {not the extrem- ities of the same diameter) on the surface, is the pole of the great circle passing through these two points. Given PC and PD quadrants of great © of sphere 0, and ACDB a great O passing through points C and D. To prove P the pole of great O ACDB. Argument Draw OC, OD, and OP. PC = 90° and PD = 90°. .'.A POC and POD are rt. A\ i.e. OP J_ OC and OD .'. OP _L the plane of O ACDB. .-. OP is the axis of O ACDB. .'. P is the pole of great O ACDB. q.e.d. Reasons 1. § 54, 15. 2. By hyp 3. § 358. 4. § 622. 5. § 906. 6. § 907. Ex. 1479. Assuming the chord of a quadrant of a great circle of a sphere to be given, construct with compasses an arc of a great circle through two given points on the surface of the sphere. Ez. 1480. If the planes of two great circles are perpendicular to each other, each passes through the poles of the other. Ex. 1481. Find the locus of all points in space equidistant from two given points and at a given distance d from a third given point. Ex. 1482. Find the locus of all points in space at a distance d from a given point and at a distance m from a given plane. Ex. 1483. Find the locus of all points in space equidistant from two given*points and also equidistant from two given parallel lines. 422 SOLID GEOMETRY Proposition IV. Theorem 915. The intersection of two spherical surfaces is the circwrriference of a circle. Given two spherical surfaces generated by intersecting cir- cumferences and Q revolving about line MKf as an axis. To prove the intersection of the two spherical surfaces the circumference of a 0. Outline op Proof 1. Show that MN _L AB at its mid-point (§ 328). 2. Show that AC, revolving about axis MN, generates a plane. 3. Show that A generates the circumference of a O. 4. The locus of A is the intersection of what (§ 614) ? 5. .*. the intersection of the two spherical surfaces is the cir- cumference of a O. Q.E.D. Ex. 1484. Find the locus of all points in space 6 inches from a given point P and 10 inches from another given point Q. Ex. 1485. The radii of two intersecting spheres are 12 inches and 16 inches, respectively. The line joining their centers is 24 inches. Find the circumference and area of their circle of intersection. 916. Def. An angle formed by two intersecting arcs of circles is the angle formed by tangents to the two arcs at their point of intersection; thus the Z formed A* by arcs AB and AC is plane A DAE. \sT\~ — — "^-F 917. Def. A spherical angle is an angle formed by two intersecting arcs of great circles of a sphere.* g * A different meaning is sometimes attached to the expression "spherical angle" in advanced mathematics. BOOK IX Proposition V. Theorem 423 918. A spherical angle is measured by the arc of a great circle having the vertex of the angle as a pole and intercepted by the sides of the angle, prolonged if nec- essary. P Given spherical Z APB, with CD an arc of a great O whose pole is P and which is intercepted by sides PA and PB of Z APB. To prove that Z APB oc CD. Outline of Proof 1. Draw radii OP, OC, and OB. 2. From P draw PR tangent to PA and PT tangent to PB. 3. Prove OC and OD each _L OP. 4. Prove OC II PR, OD II PT, and hence Z COD = Z RPT. 5. But Z COD oc CD; .*. Z iZPT, i.e. Z ARB QC CD. Q.E.D. 919. Cor. I. A spherical angle is equal to the plane angle of the dihedral angle formed by the planes of the sides of the angle. 920. Cor. II. The sum of all the spherical angles about a point on the surface of a sphere equals four right angles. Ex. 1486. By comparison with the definitions of the corresponding terms in plane geometry, frame exact definitions of the following classes of spherical angles : acute, right, obtuse, adjacent, complementary, supple- mentary, vertical. Ex. 1487. Any two vertical spherical angles are equal. Ex. 1488. If one great circle passes through the pole of another great-circle, the circles are perpendicular to each other. 424 SOLID GEOMETRY LINES AND PLANES TANGENT TO A SPHERE 921. Def. A straight line or a plane is tangent to a sphere if, however far extended, it meets the sphere in one and only one point. 922. Def. Two spheres are tangent to each other if they have one and only one point in common. They are tangent internally if one sphere lies within the other, and externally if neither sphere lies within the other. Proposition VI. Theorem 923. A plane tangent to a sphere is perpendicular to the radius drawn to the point of tangency. Given plane AB tangent to sphere at T, and or a radius drawn to the point of tangency. To prove plane AB A. OT. The proof is left as an exercise for the student. 924. Question. What changes are necessary in the proof of § 313 to make it the proof of § 923 ? 925. Cor. I. (Converse of Prop. VI). A plane perpen- dicular to a radius of a sphere at its outer extremity is tangent to the sphere. Hint. See § 314. ' Ex. 1489. A straight line tangent to a sphere is perpendicular to the radius drawn to the point of tangency. Ex. 1490. State and prove the converse of Ex. 1489, BOOK IX 425 Ex. 1491. Two lines tangent to a sphere at the same point determine a plane tangent to the sphere at that point. Ex. 1492. Given a point P on the surface of sphere O. Explain how to construct : (a) a line tangent to sphere O at P; (6) a plane tangent to sphere O at P. Ex. 1493. Given a point B outside of sphere Q. Explain how to con- struct : (a) a line through M tangent to sphere Q ; (6) a plane through B tangent to sphere Q. Hint. Compare with § 373. Ex. 1494. Two planes tangent to a sphere at the extremities of a diameter are parallel. Ex. 1495. If the straight line joining the centers of two spheres is equal to the sum of their radii, the spheres are tangent to each other. Hint. Show that the radius of the O of intersection of the two spheres (§ 915) is zero. 926. Def. A polyhedron is circumscribed about a sphere if each face of the polyhedron is tangent to the sphere. 927. Def. If a polyhedron is circumscribed about a sphere, the sphere is said to be inscribed in the polyhedron. 928. Def. A polyhedron is inscribed in a sphere if all its vertices are on the surface of the sphere. 929. Def. If a polyhedron is inscribed in a sphere, the sphere is said to be circumscribed about the polyhedron. Ex. 1496. Find the edge of a cube inscribed in a sphere whose radius is 10 inches. Ex. 1497. Find the volume of a cube: (a) circumscribed about a sphere whose radius is 8 inches ; (&) inscribed in a sphere whose radius is 8 inches. Ex. 1498. A right circular cylinder whose altitude is 8 inches is in- 3cribed in a sphere whose radius is 6 inches. Find the volume of the cylinder. Ex. 1499. A right circular cone, the radius of whose base is 8 inches, is inscribed in a sphere with radius 12 inches. Find the volume of the cone. Ex. 1500. Find the volume of a right circular cone circumscribed aboflt fe, regular tetrahedron whose edge is a. 426 SOLID GEOMETRY Proposition VII. Problem 930. To inscribe a sphere in a given tetrahedron. V Given tetrahedron V-ABC. To inscribe a sphere in tetrahedron V-ABC. I. Construction 1. Construct planes BABS, SBCT, and TCAB bisecting dihedral A whose edges are AB, BC, and CA, respectively. § 691. 2. From 0, the point of intersection of the three planes, construct OF 1. plane ABC. § 637. 3. The sphere constructed with as center and OF as radius will be inscribed in tetrahedron V-ABC. II. Proof Argument 1. Plane RABS, the bisector of dihedral Z AB, lies between planes ABV and ABC; i.e. it intersects edge VC in some point as D. 2. .-. plane RABS intersects plane BCV in line BD and plane ACV in line AD. 3. Plane SBCT lies between planes BCV and ABC; i.e. it intersects plane RASB in a line through B between BA and BD, as BS. 4. Similarly plane TCAR intersects plane RABS in a line through A between Reasons By cons. § 616. By cons. By steps ilar to sim- 1-3. BOOK IX 427 Argument AB and AD, as AR ; and plane SBCT intersects plane TCAR in a line through G as CT. 5. But AR and £S pass through the in- terior of A ABB. 6. .-. AR and BS intersect in some point as 0, within A ABB. 7. .-. AR, BS, and CT are concurrent in point 0. 8. From draw OH, OK, and OL _L planes VAB, VBC, and FCM, respectively. 9. v is in plane OAB, 0F = OH. 10. v is in plane OB c, OF= OK. 11. v is in plane OCA, OF = OL. 12. .-. OF = OH = OK = OL. 13. .-. each of the four faces of the tetra- hedron is tangent to sphere 0. 14. .-. sphere O is inscribed in tetrahedron V-ABC. Q.E.D; Reasons 5. Args. 3 & 4. 6. § 194. 7. § 617, I. 8. § 639. 9. § 688. 10. § 688. 11. § 688. 12. §54,1. 13. § 925. 14. §§ 926, 927. III. Discussion The discussion is left as an exercise for the student. Ex. 1501. The six planes bisecting the dihedral angles of a tetrahe- dron meet in a point which is equidistant from the four faces of the tetrahedron. Hint. Compare with § 258. Ex. 1502 . Inscribe a sphere in a given cube. Ex. 1503. The volume of any tetrahedron is equal to the product of its surface and one third the radius of the inscribed sphere. Hint. See §§ 491 and 492. Ex. 1504. Find a point within a triangular pyramid such that the planes determined by the lines joining this point to the vertices shall di- vide the pyramid into four equivalent parts. 'rfiNT. Compare with Ex. 1094. 428 SOLID GEOMETRY Proposition VIII. Problem 931. To circumscribe a sphere about a given tetrahe- dron. V Given tetrahedron V-ABC. To circumscribe a sphere about tetrahedron V-ABC. I. Construction 1. Through D, the mid-point of AB, construct plane DO A. AB ; through E, the mid-point of BC, construct plane EOA.BC; and through F, the mid-point of VB, construct plane FO J_ VB. 2. Join O, the point of intersection of the three planes, to A. 3. The sphere constructed with as center and OA as radius will be circumscribed about tetrahedron V-ABC. II. Outline of Proof 1. Prove that the three planes OB, OE, and OF intersect each other in three lines. 2. Prove that these three lines of intersection meet in a point, as O. 3. Prove that OA = OB = OC = OV. 4. .*. sphere is circumscribed about tetrahedron V-ABC. Q.E.D. 932. Cor. Four points not in the same plane determine a sphere. 933. Questions. Are the methods used in §§ 930 and 931 similar to those used in §§ 321 and 323 ? In § 930 could the lines forming the edges of the dihedral angles bisected be three lines meeting in one vertex ? In § 931 could the three edges bisected be three lines lying in the same face ? BOOK IX 429 Ex. 1505. The six planes perpendicular to the edges of a tetrahedron at their mid-points meet in a point which is equidistant from the four ver- tices of the tetrahedron. Ex. 1506. The four lines perpendicular to the faces of a tetrahedron, and erected at their centers, meet in a point which is equidistant from the four vertices of the tetrahedron. Ex. 1507. Circumscribe a sphere about a given cube. Ex. 1508. Circumscribe a sphere about a given rectangular paral- lelopiped. Can a sphere be inscribed in any rectangular parallelopiped ? Explain. Ex. 1509. Find a point equidistant from four points in space not all in the same plane. SPHERICAL POLYGONS 934. Def. A line on the surface of a sphere is said to be closed if it separates a portion of the surface from the remain- ing portion. 935. Def. A closed figure on the surface of a sphere is a figure composed of a portion of the surface of the sphere and its bounding line or lines. 936. Defs. A spherical polygon is a closed figure on the surface of a sphere whose boundary is composed of three or more arcs of great circles, as ABCD. The bounding arcs are called the sides of the polygon, the points of intersection of the arcs are the vertices of the poly- gon, and the spherical angles formed by the sides are the ari'gtes of the polygon. 430 SOLID GEOMETRY' 937. Def. A diagonal of a spherical polygon is an arc of a great circle joining any two non-adjacent vertices. 938. Def. A spherical triangle is a spherical polygon "having three sides. Ex. 1510. By comparison with the definitions of the corresponding terms in plane geometry, frame exact definitions of the following classes of spherical triangles : scalene, isosceles, equilateral, acute, right, obtuse, and equiangular. Ex. 1511. With a given arc as one side, construct an equilateral spherical triangle. Hint. Compare the cons., step by step, with § 124. Ex. 1512. With three given arcs as sides, construct a scalene spheri- cal triangle. 939. Since each side of a spherical polygon is an arc of a great circle (§ 936), the planes of these arcs meet at the center of the sphere and form at that point a polyhedral angle, as polyhedral Z O-ABCD. This polyhedral angle and the spheri- cal polygon are very closely related. The following are some of the more important relations; the student should prove the correctness of each : 940. (a) The sides of a spherical polygon have the same measures as the corresponding face angles of the poly- hedral angle. (b) The angles of a spherical polygon have the same measures as the corresponding dihedral angles of the polyhedral angle. Thus, sides AB, BC, etc., of spherical polygon ABCD have the same measures as face AAOB, BOC, etc., of polyhedral Z O-ABCD ; and spherical A ABC, BCD, etc., have the same measures as the dihedral A whose edges are OB, OC, etc. These relations make it possible to establish certain prop- erties of spherical polygons from the corresponding known properties of the polyhedral angle, as in §§ 941 and 942. BOOK IX Proposition IX. Theorem 431 941. The sum of any two sides of a spherical triangle is greater than the third side. Given spherical A ABC. To prove AB + BC > CA. Argument 1. Z AOB + Z BOC > Z COA. 2. Z ^05 oc ^Z?, Z BOCKBC, Z CO^ QC (3. 3. .-. ^i? + £C> CM. Q.E.D. Reasons 1. §710. 2. § 940, a. 3. § 362, b. Proposition X. Theorem 942. The sum of the sides of any spherical polygon is less than 360°. Given spherical polygon ABC ••• with n sides. To prove AB + BC + ••• < 360°. • 'Hint. See §§ 712 and 940, a. 432 SOLID GEOMETRY Ex. 1513. In spherical triangle ABC, arc AB — 40° and arc BC = 80°. Between what limits must arc CA lie ? Ex. 1514. Any side of a spherical polygon is less than the sum of the remaining sides. Ex. 1515. If arc AB is the perpendicular bisector of arc CD, every point on the surface of the sphere and not in arc AB is unequally distant from C and D. Ex. 1516. In a spherical quadrilateral, between what limits must the fourth side lie if three sides are 60°, 70°, and 80°? if three sides are 40°, 50°, and 70° ? Ex. 1517. Any side of a spherical polygon is less than 180°. Hint. See Ex. 1514. 943. If, with A, B, and C the vertices of any spherical tri- angle as poles, three great circles are constructed, as B'c'ED, C'A'd, and EA'b', the surface of the sphere will be divided into eight spherical triangles, four of which are seen on the hemi- sphere represented in the figure. Of these eight spherical tri- angles, ^'.B'C'is the one and only one that is so situated that A and A 1 lie on the same side of BC, B and B' on the same side of AC, and C and C' on the same side of AB. This particular tri- angle a'b'c' is called the polar triangle of triangle ABC. 944. Questions. In the figure above, A A'B'C, the polar of A ABC, is entirely outside of A ABC. Can the two A be so constructed that : (a) A'B'C is entirely within ABC ? (&) A'B'C is partly outside of and partly within ABC? Ex. 1518. What is the polar triangle of a spherical triangle all of whose sides are quadrants ? BOOK IX 433 Proposition XT. Theorem 945. If one spherical triangle is the polar of another, then the second is the polar of the first. Given A A'b'c' the polar of A ABC. To prove A ABC the polar of A A'b'c'. Argument Reasons 1. § 912. 2. § 912. 3. § 914. 4. By steps simi- lar to 1- 3. 5. § 943. 1. A is the pole of B'c' ; i.e. AC' is a quad- rant. 2. B is the pole of A'C' ; i.e. BC' is a quad- rant. 3. .'. C' is the pole of AB. 4. Likewise b' is the pole of AC, and A' is the pole of BC. 5. .'.A ABC is the polar of A a'b'c'. q.e.d. 946. Historical Note. The properties of polar triangles were dis- covered about 1626 a.d. by Albert Girard, a Dutch mathematician, born in Lorraine about 1595. They were also discovered independently and about the same time by Snell, an "infant prodigy," who at the age of twelve was familiar with the standard mathematical works of that time and who is remembered as the discoverer of the well-known law of refraction of light. Ex. 1519. Determine the polar triangle of a spherical triangle having two of its sides quadrants and the third side equal to 70° ; 110° ; (90 — a)° ; (9G^a)°. 434 SOLID GEOMETRY Proposition XII. Theorem 947. In two polar triangles each angle of one and that side of the other of which its vertex is the pole are together equal, numerically, to 180°- E a Given polar A ABC and A'b'c', with sides denoted by a, b, c, and a', b', c', respectively. To prove: (a) Z ^4-a' = 180°, Z B+b' = l$0°, Z c+c'= 180°; (6) ZA'+a=180°, ZB'+b=180°, Z C'+c=180°. (a) Argument Only 1. Let arcs AB and AC (prolonged if necessary) intersect arc B'C' at D and E, respectively ; then c'D = 90° and EB' = 90°. 2. .-. c'D + EB' = 180°. 3. . • . C'E + ED + EB + ZXB' = 180° ; i.e. ED + a' = 180°. 4. But ED is the measure of Z A. 5. .\Z^ + a' = 180°. 6. Likewise Z 5 + &' = 180°, and Z C + c' = 180°. q.e.d. (6) The proof of (6) is left as an exercise for the student. Hint. Let BC prolonged meet A'B' at .ffand A'C at if. 948. Question. In the history of mathematics, why are polar tri- angles frequently spoken of as supplemental triangles ? Ex. 1520. The angles of a spherical triangle are 75°, 85°, and 145°. Find the sides of its polar triangle. Ex. 1521. If a spherical triangle is equilateral, its polar triangle is equiangular ; and conversely. BOOK IX 435 Proposition XIII. Theorem 949. The sum of tiie angles of a spherical triangle is greater than 180° and less than 540°. Given spherical A ABC with sides denoted by a, b, and c. To prove ZA + Zb + Zc> 180° and < 540°. Argument 1. Let A A'b'c', with sides denoted by a', b', and c', be the polar of A ABC. 2. Then Z A + a' = 180°, Z B +V ' = 180°, Zc+c'= 180°. 3. .:ZA + ZB+Zc + (a'+b'+c') = 54:0 o . 4. But a' + b' + c' <360°. 5. .:ZA + ZB+Z <7>180°. 6. Again, a 1 + b' + c' > 0°. 7. .-. Z^ + Z^B + Z c< 540°. q.e.d. Reasons 1. § 943. 2. § 947. 3. § 54, 2. 4. § 942. 5. § 54, 6. 6. § 938. 7. § 54, 6. 950. Cor. In a spherical triangle there can be one, two, or three right angles ; there cajv be one, two, or three obtuse angles. 951. Note. Throughout the Solid Geometry the student's attention has constantly been called to the relations between definitions and theorems of solid geometry and the corresponding definitions and theorems of plane geometry. In the remaining portion of the geometry of the sphere there will likewise be many of these comparisons, but here the student must be particularly on his guard for contrasts as well as comparisons. For ex- 436 SOLID GEOMETRY ample, while the sum of the angles of a plane triangle is equal to exactly 180°, he has learned (§ 949) that the sum of the angles of a spherical tri- angle may be any number from 180° to 540° ; while a plane triangle can have but one right or one obtuse angle, a spherical triangle may have one, two, or three right angles or one, two, or three obtuse angles (§ 950). If the student will recall that, considering the earth as a sphere, the north and south poles of the earth are the poles of the equator, and that all meridian circles are great circles perpendicular to the equator, it will make his thinking about spherical triangles more definite. 952. Def. A spherical triangle containing two right angles is called a birectangular spherical triangle. 953. Def. A spherical triangle having all of its angles right angles is called a trirectangular spherical triangle. Thus two meridians, as NA and NB, making at the north pole an acute or an obtuse Z, form with the equator a birectangular spherical A. If the Z between NA and NB is made a rt. Z, A ANB becomes a trirectangular spherical A. Ex. 1522. What kind of arcs are NA and NB ? Then what arc measures spherical angle ANB ? Are two sides of any birectangular spherical triangle quadrants? What is each side of a trirectangular spherical triangle ? Ex. 1523. If two sides of a spherical triangle are quadrants, the triangle is birectangular. (Hint. Apply § 947.) Ex. 1524. What is the polar triangle of a trirectangular spherical triangle ? Ex. 1525. An exterior angle of a spherical triangle is less than the sum of the two remote interior angles. Compare this exercise with § 215. Make this new fact clear by applying it to a birectangular spherical triangle whose third angle is : (a) acute; (6) right; (c) obtuse. BOOK IX 437 Proposition XIV. Theorem 954. In equal spheres, or in the same sphere, two spher- ical triangles are equal : I. If a side and the two adjacent angles of one are equal respectively to a side and the two adjacent angles of the other; II. If two sides and the included angle of one are equal respectively to two sides and tlie included angle of the other; III. If the three sides of one are equal respectively to the three sides of the other : provided the equal parts are arranged in the same order. The proofs are left as exercises for the student. Hint. In each of the above cases prove the corresponding trihedral A equal (§§ 702, 704) ; and thus show that the spherical A are equal. 955. Questions. Compare Prop. XIV, I and II, with § 702, I and II, and with §§ 105 and 107. Could the methods used there be employed in § 954 ? Is the method here suggested preferable ? why ? 956. Def. Two spherical polygons are symmetrical if the corresponding polyhedral angles are symmetrical. ^> 957. The following are some of the properties of symmet- rical spherical triangles; the student should prove the correct- ness of each : (a) Symmetrical spherical triangles have their parts respectively equal, but arranged in reverse order. (b) Two isosceles symmetrical spherical triangles are equal. Hint. Prove (6) by superposition or by showing that the correspond- ing trihedral A are equal. 438 SOLID GEOMETRY Proposition XV. Theorem 958. In equal spheres, or in the same sphere, two sym- metrical spherical triangles are equivalent. A A' Given symmetrical spherical A ABC and A'B'c' in equal spheres O and O'. To prove spherical A ABC=^ spherical A A'b'C*. Argument 1. Let P and P' be poles of small © through A, B, C, and A', B', C', respectively. 2. Arcs AB, BC, CA are equal, respectively, to arcs A'B', B'C', C'A'. 3. .-. chords AB, BC, CA, are equal, respec- tively, to chords A'b', b'c', c'a'. 4. .-. plane A ABC = plane A a'b'c*. 5. .-. O ABC = O A'B'C'. 6. Draw arcs of great (D PA, PB, PC, p'a', p'b', and P'c'. . 7. Then PA = PB = PC= P'A'= p'b'= p'c'. 8. .-. isosceles spherical A APB and a' p'b' are symmetrical. 9. .'. A APB == A A' P'B'. 10. Likewise A BPC = A b'p'c' and A CPA = A C'P'A'. 11. .-. spherical A ABCo spherical Aa'b'c'. Q.E.D. Reasons 1. § 908, h. 2. § 957, a. 3. § 298, II. 4. §116. 5. § 324. 6. § 908, g. 7. § 913. 8. §956. 9. §957, b. 10. By steps simi- lar to 8-9. 11. § 54, 2. BOOK IX 439 Proposition XVI. Theorem 959. In equal spheres, or in the same sphere, two spheri- cal triangles are symmetrical, and therefore equivalent : I. If a side and the two adjacent angles of one are equal respectively to a side and the two adjacent angles of the other ; II. // two sides and the included angle of one are equal respectively to two sides and the included angle of the other; III. If the three sides of one are equal respectively to the three sides of the other : provided the equal parts are arranged in reverse order. The proofs are left as exercises for the student. Hint. In each of the above cases prove the corresponding trihedral A symmetrical (§ 709) ; and thus show that the spherical & are symmetrical. •Ex. 1526. The bisector of the angle at the vertex of an isosceles spherical triangle is perpendicular to the base and bisects it. Ex. 1527. The arc drawn from the vertex of an isosceles spherical triangle to the mid-point of the base bisects the vertex angle and is per- pendicular to the base. Ex. 1528. State and prove the theorems on the sphere correspond- ing to the following theorems on the plane : (1) Every point in the perpendicular bisector of a line is equidistant from the ends of that line (§ 134). (2) Every point equidistant from the ends of a line lies in the perpen- dicular bisector of that line (§ 139). (3) Every point in the bisector of an angle is equidistant from the sides of the angle (§ 253). Hint. In the figure for (3), corresponding to the figure of § 252, draw PD ± AB and lay off BE = BD. Ex. 1529. The diagonals of an equilateral spherical quadrilateral are perpendicular to each other. Prove. State the theorem in plane geome- try that corresponds to this exercise. Ex. 1530. In equal spheres, or in the same sphere, if two spherical triangles are mutually equilateral, their polar triangles are mutually equi- angular ; and conversely. 440 SOLID GEOMETRY Proposition XVII. Theorem 960. In equal spheres, or in the same sphere, if two spherical triangles are mutually equiangular, they are mutually equilateral, and are either equal or symmet- rical. Given spherical A T and T* in equal spheres, or in the same sphere, and mutually equiangular.. To prove T and T f mutually equilateral and either equal or symmetrical. Argument Let A P and P ' be the polars of A T and T', respectively. T and T' are mutually equiangular. Then P and P' are mutually equi- lateral. .*. P and P' are either equal or sym- metrical, and hence are mutually equiangular. .*. T and T' are mutually equilateral. .-. T and T' are either equal or sym- metrical. Q.E.D. Reasons §943. By hyp. §947. 4. §§ 954, III, and 959, III. §947. Same as 4. reason Ex. 1531. In plane geometry, if two triangles are mutually equi- angular, what can be said of them ? Are they equal ? equivalent ? Ex. 1532. Find the locus of all points of a sphere that are equidis- tant from two given points on the surface of the sphere ; from two given points in space, not on the surface of the sphere. BOOK IX 441 Proposition XVIII. Theorem 961. The base angles of an isosceles spherical triangle are equal. b Given isosceles spherical A ABC, with side AB = side BC. To prove Z A = Z C. Hint. Compare with § 111. Proposition XIX. Theorem 962. If two angles of a splxerical triangle are equal, the sides opposite are equal, g- >f Given spherical A RST with Ar = Z T. To prove r = t. Argument Let A R's'T* be the polar of A RST. AR — At. .-. r' = t'. .'. &=!*. «'^ r = t. Q.E.D. Keasons 1. § 943. 2. By hyp. 3. § 947. 4. § 961. 5. § 947. 442 SOLID GEOMETRY Proposition XX. Theorem 963. If two angles of a spherical triangle are unequal, the side opposite the greater angle is greater than the side opposite the less angle. Given spherical A ABC witli Z A > Z C. To prove BC> AB. Argument 1. Draw an arc of a great O AD making Z1=Z a 2. Then Td = 'dc. 3. But bd + a1>> AB. 4. . \ BD + DC > AB ; i.e. BO > AB. Q.E.D. 1. Reasons § 908, g. 2. § 962. 3. § 941. 4. § 309. Ex. 1533. In a birectangular spherical triangle the side included by the two right angles is less than, equal to, or greater than, either of the other two sides, according as the angle opposite is less than, equal to, or greater than 90°. Ex. 1534. An equilateral spherical triangle is also equiangular. Ex. 1535. If two face angles of a trihedral angle are equal, the dihedral angles opposite are equal. Ex. 1536. State and prove the converse of Ex. 1534. Ex. 1537. State and prove the converse of Ex. 1535. Ex. 1538. The arcs bisecting the base angles of an isosceles spheri- cal triangle form an isosceles spherical triangle. Ex. 1539. The bases of an isosceles trapezoid are 14 inches and 6 inches and the altitude 3 inches ; find the total area and volume of the solid generated by revolving the trapezoid about its longer base as an axis. Ex. 1540. Find the total area and volume of the solid generated by revolving the trapezoid of Ex. 1530 about its shorter base as an axis. BOOK IX 443 Proposition XXI. Theorem 964. If two sides of a spherical triangle are unequal, the angle opposite the greater side is greater than the angle opposite the less side. B Given spherical A ABC with BC > AB. To prove Z A > Z C. The proof is left as an exercise for the student. Hint. Prove by the indirect method. 965. Questions. Could Prop. XXI have been proved by the method used in § 156 ? Does reason 4 of that proof hold in spherical A ? See Ex. 1525. Ex. 1541. If two adjacent sides of a spherical quadrilateral are greater, respectively, than the other two sides, the spherical angle included between the two shorter sides is greater than the spherical angle between the two greater sides. Hint. Compare with Ex. 153. Ex. 1542. Find the total area and volume of the solid generated by revolving the trapezoid of Ex. 1539 about the perpendicular bisector of its bases as an axis. Ex. 1543. Pind the radius of the sphere in- scribed in a regular tetrahedron whose edge is a. Hint. Let be the center of the sphere, A the center of face VED, and B the center of face EDF. Then OA = radius of inscribed sphere. Show that rt. A VAO and VBC are similar. Then VO : VC = VA : VB. VC, VA, and VB can be found (Ex. 1328). Find VO, then OA. Ex. 1544. Pind the radius of the sphere circumscribed about a regu- lar tejtrahedron whose edge is a. Hin*. In the figure of Ex. 1543, draw AD and OD. 444 SOLID GEOMETRY MENSURATION OF THE SPHERE Areas Proposition XXII. Theorem 966. If an isosceles triangle is revolved about a straight line lying in its plane and passing through its vertex but not intersecting its surface, the area of the surf ace gener- ated by the base of the triangle is equal to the product of its projection on the axis and the circumference of a circle whose radius is the altitude of the triangle- x X X c OH D Y Y Y Fig. 1. Fig. 2. Fig. 3. Given isosceles A AOB with base AB and altitude OE, a str. line XY lying in the plane of A AOB passing through O and. not intersecting the surface of A AOB, and CD the projection of AB on J7; let the area of the surface generated by AB be denoted by area AB. To prove area AB = CD • 2 irOE. I. If AB is not II XY and does not meet XY (Fig. 1). Argument Only 1. From E draw EH A. XY. 2. Since the surface generated by AB is the surface of a frustum of a rt. circular cone, area AB = AB • 2 -n-EH. 3. From A draw AK± BD. 4. Then in rt. A BAK and OEH, Z BAR = Z OEH. AC H KJ) A E B, K BOOK IX 445 5. .-. ABAK ~ AOEH. 6. .*. AB.AK— OE: EH) i.e. AB • EH = AK • OE. 7. But AK= CD; .'. AB -EH= CD • OE. 8. .-. area AB — CD -2 -nOE. q.e.d. II. If AB is not II XY and point A lies in XY (Fig. 2). III. If ^B II XT (Fig. 8). The proofs of II and III are left as exercises for the student. Hint. See if the proof given for I will apply to Figs. 2 and 3. 967. Cor. I. If half of a regular polygon with an even number of sides is circumscribed about a semicircle, the area of the surface generated by its semiperimeter as it revolves about the diameter of the semicircle as an axis, is equal to the product of the diameter of the regular polygon and the circumference of a circle whose radius is R, the radius of the given semicircle. Outline of Proof 1. Area A'b' = A'f' -2ttR; area B'c' = f'o' • 2 ttR; etc. 2. .-. area a'b'c'-- = (a'f' + f'o' + ) 2 ttR = A'E' • 2 ttR. 968. Cor. n. If half of a regular polygon with an even number of sides is inscribed in a semicircle, the area of the surface generated by its semi- perimeter as it revolves about the diameter of the semicircle as an axis, is equal to the product of the diameter of the semicircle and the circumference of a circle whose radius is the apothem of the regular polygon. ' if int. Prove area ABC • • • = AE • 2 ira. 446 SOLID GEOMETRY 969. Cor. m. If halves of regular polygons with the same even number of sides are circumscribed about, and inscribed in, a semicircle, then by repeatedly doubling the number of sides of these polygons, and making the poly- gons always regular, the surfaces generated by the semi- perimeters of the polygons as they revolve about the diameter of the semicircle as an axis approach a com- mon limit. Outline of Proof 1. If S and s denote the areas of the surfaces generated by the semi- perimeters A'b'c'--- and ABC--- as they revolve about a'e' as an axis, then 6' = A'E' • 2 *-# (§ 967) ; and s = AE • 2 ttcl (§ 968). 2. .-. - = S A'E 1 • 2 ttR 8 AE • 2 7ra But polygon a'b'c'-- A'E' a'b' A'E' AE polygon ABC-- • (§ 438). = -(§§419,435). AE AB a V S s a R 2 a 2 S R z 7. .-. by steps similar to Args. 6-11 (§ 855), S and s ap- proach a common limit. q.e.d. 970. Def. The surface of a sphere is the common limit Which the successive surfaces generated by halves of regular polygons with the same even number of sides approach, if these semipolygons fulfill the following conditions : (1) They must be circumscribed about, and inscribed in, a great semicircle of the sphere; (2) The number of sides must be successively increased, each side approaching zero as a limit. BOOK IX 447 Proposition XXIII. Theorem 971. The area of the surface of a sphere is equal to four times the area of a great circle of the spliere. A 1. Reasons § 517, a. Given sphere with its radius denoted by R, and the area of its surface denoted by S. To prove S = 4 -irR 2 . Argument 1. In the semicircle ACE inscribe ABCDE, half of a regular polygon with an even number of sides. Denote its apothem by a, and the area of the surface generated by the semiperime- ter as it revolves about AE as an axis bys'. 2. Then S' = AE • 2 ira ; i.e. S' = 2 R • 2 ttcl = 4 irRa. 3. As the number of sides of the regular polygon, of which ABCDE is half, is repeatedly doubled, S' approaches S as a limit. 4. Also a approaches R as a limit. 5. .-. 4 nR • a approaches 4 irR • R, i.e. 4 irR\ as a limit. 6. But S' is always equal to 4 7ri? • a. 1." V- 5 = 4 irR 2 . Q.E.D. 2. § 968. §970. § 543, I. § 590. Arg. 2. § 355. 448 SOLID GEOMETRY 972. Cor. I. The areas of the surfaces of two spheres are to each other as the squares of their radii and as the squares of their diameters. Outline of Proof 1. S = 4 7 ri? 2 andS' = 4 7 ri2' 2 ; .-.?- = i^- = £. ' 8' 4tt2*' 2 R 12 2 But R2 = 4 R2 = ( 2 R ^ = D * • • - = — ' U R ,2 ~ ±R< 2 ~ (2rJ~ D' 2 ' '"' S'~D' 2 ' 973. Historical Note. Prop. XXIII is given as Prop. XXXV in the treatise entitled Sphere and Cylinder by Archimedes, already spoken of in S 809. Ex. 1545. Find the surface of a sphere whose diameter is 10 inches. Ex. 1546. What will it cost to gild the surface of a globe whose radius is 1| decimeters, at an average cost of § of a cent per square centimeter ? Ex. 1547. The area of a section of a sphere made by a plane 11 inches from the center is 3000 ir square inches. Find the surface of the sphere. Ex. 1548. Find the surface of a sphere circumscribed about a cube whose edge is 12 inches. Ex. 1549. The radius of a sphere is R. Find the radius of a sphere whose surface is twice the surface of the given sphere ; one half ; one nth. Ex. 1550. Find the surface of a sphere whose diameter is 2 R, and the total surface of a right circular cylinder whose altitude and diameter are each equal to 2 R. Ex. 1551. From the results of Ex. 1550 state, in the form of a theorem, the relation of the surface of a sphere to the total surface of the circumscribed cylinder. Ex. 1552. Show that, in Ex. 1550, the surface of the sphere is exactly equal to the lateral surface of the cylinder. 974. Historical Note. The discovery of the remarkable property that the surface of a sphere is two thirds of the surface of the circum- scribed cylinder (Exs. 1550 and 1551) is due again to Archimedes. The discovery of this proposition, and the discovery of the corresponding proposition for volumes (§ 1001), were the philosopher's chief pride, and he therefore asked that a figure of this proposition be inscribed on his tomb. His wishes were carried out by his friend Marcellus. (For a further account of Archimedes, read also § 542.) BOOK IX 449 975. Defs. A zone is a closed figure on the surface of a sphere whose boundary is composed of the circumferences of two circles whose planes are parallel. The circumferences forming the boundary of a zone are its bases. Thus, if semicircle NES is revolved about NS as an axis, arc AB will gen- erate a zone, while points A and B will generate the bases of the zone. 976. Def. The altitude of a zone is the perpendicular from any point in the plane of one base to the plane of the other base. 977. Def. If the plane of one of the bases of a zone is tan- gent to the sphere, the zone is called a zone of one base. Thus, arc NA or arc RS will generate a zone of one base. 978. Questions. Is the term " zone " used in exactly the same sense here as it is in the geography ? Name the geographical zones of one base ; of two bases. Name the five circles whose circumferences form the bases of the six geographical zones. Which of these are great circles ? * 979. Cor. n. The area of a zone is equal to the product of its altitude and the circumference of a great circle. Outline *of Proof Let S denote the area generated by broken line a'b'c', s by broken line ABC, and Z by arc ABC-, let DE, the altitude of the. zone, be denoted by H. Then S = D'E' . 2 ttR ; B S = DE • 2 7T<2. .-. ?=£• (See Args. 2-5, s a 2 § 969.) Then by steps similar to §§ 969-971, z = h.2ttR. % The student will observe that the projection used in this figure is different from that usedXn the other figures. 450 SOLID GEOMETRY 980. Cor. III. In equal spheres, or in the same sphere, the areas of two zones are to each other as their altitudes. 981. Question. In general, surfaces are to each other as the prod- ucts of two lines. Is § 980 an exception to this rule ? Explain. Ex. 1553. The area of a zone of one base is equal to the area of a circle whose radius is the chord of the arc generating the zone. Hint. Use §§ 979 and 444, II. Ex. 1554. Show that the formula of § 971 is a special case of § 979. Ex. 1555. Find the area of the surface of a zone if the distance between its bases is 8 inches and the radius of the sphere is 6 inches. Ex. 1556. The diameter of a sphere is 16 inches. Three parallel planes divide this diameter into four equal parts. Find the area of each of the four zones thus formed. Ex. 1557. Prove that one half of the earth's surface lies within 30° of the equator. Ex. 1558. Considering the earth as a sphere with radius B, find the 7? 2 7? area of the zone adjoining the north pole, whose altitude is — ; Is the one area twice the other ? Ex. 1559. Considering the earth as a sphere with radius B, find the area of the zone extending 30° from the north pole ; 60° from the north pole. Is the one area twice the other ? Ex. 1560. Considering the earth as a sphere with radius i?, find the area of the zone whose bases are parallels of latitude: (a) 30° and 45 c from the north pole ; (6) 30° and 45° from the equator. Are the two areas equal ? Explain your answer. Ex. 1561. How far from the center of a sphere whose radius is B must the eye of an observer be so that one sixth of the surface of the sphere is visible ? Hint. Let E be the eye of the observer. Then AB must = — . Find OB, then use § 443, II. 3 » w , Ex. 1562. What portion of the surface of a sphere can be seen if the distance of the eye of the observer from the center of the sphere is 2B ? SB? nB? Ex. 1563. The radii of two concentric spheres are 6 inches and 10 inches. A plane is passed tangent to the inner sphere. Find : (a) the area of the section of the outer sphere made by the plane ; (&) the area of the surface cut off of the outer sphere by the plane, BOOK IX 451 982. Def. A lune is a closed figure on the surface of a sphere whose boundary is composed of two semicircumferences of great circles, as NASB. 983. Defs. The two semicircumfer- ences are called the sides of the lune, as NAS and NBS; the points of inter- section of the sides are called the vertices of the lune, as N and S-, the spherical angles formed at the vertices by the sides of the lune are called the angles of the lune, as A ANB and BSA. 984. Prove, by superposition, the following property of lunes : In equal spheres, or in the same sphere, two lunes are equal if their angles are equal. 985. So far, the surfaces considered in connection with the sphere have been measured in terms of square units, i.e. square inches, square feet, etc. For example, if the radius of a sphere is 6 inches, the surface of the sphere is 4<7r6 2 , i.e. 144 ir square inches. But, as the sides and angles of a lune and a spherical polygon are given in degrees and not in linear units, it will be necessary to introduce some new unit for determining the areas of these figures. For this purpose the entire surface of a sphere is thought of as being divided into 720 equal parts, and each one of these parts is called a spherical degree. Hence : 986. Def. A spherical degree is yj-g- of the surface of a sphere. Now if the area of a lune or of a spherical triangle can be ob- tained in spherical degrees, the area can easily be changed to square units. For example, if it is found that the area of a spherical triangle is 80 spherical degrees, its area is -f^, i.e. ^ of the entire surface of the sphere. On the sphere whose radius is 6 inches, the area of the given triangle will be -J- of 144 7r square inches, i.e. 16 ■* square inches. The following theorems are for the purpose of determining the areas of figures on- tye surface of a sphere in terms of spherical degrees. 452 SOLID GEOMETRY Proposition XXIV. Theorem 987. The area of a lune is to the area of the surface of tl%e sphere as the number of degrees in the angle of the lune is to 360. N N Q E Given lune NASB with the number of degrees in its Z denoted by N f its area denoted by L, and the area of the surface of the sphere denoted by S ; let O EQ be the great O whose pole is N. m L N To prove - = 8 360 I. If arc AB and circumference EQ are commensurable (Fig. 1). Argument Reasons 1. Let m be a common measure of arc AB 1. § 335. and circumference EQ, and suppose that m is contained in arc AB r times and in circumference EQ t times. arc AB r ~t' Then 4. circumference EQ Through the several points of division on circumference EQ pass semicir- cumferences of great circles from N to S. Then lune NASB is divided into r lunes and the surface of the sphere into t lunes, each equal to lune NCSB. § 341. § 908, h. 4. §984. BOOK IX Argument Reasons 5. L — L " S~t 5. §341. a L arc AB 6. § 54, 1. S circumference EQ 7. But arc AB is the measure it contains N degrees. of Ztf; i.e. 7. §918. 8. And circumference EQ contains 360° 8. §297. 9. L_ N S~S60 Q.] E.D. 9. §309. 453 II. If arc AB and circumference EQ are incommensurable (Fig. 2). The proof is left as an exercise for the student. Hint. The proof is similar to that of § 409, II. 988. Cor. I. The area of a lune, expressed in spherical degrees, is equal to twice the number of degrees in Us angle. L = JL (§987). 8 360 V J Outline of Proof L N ** 720 360* L = 2n. Ex. 1564. Find the area of a lune in spherical degrees if its angle is 35°. What part is the lune of the entire surface of the sphere ? Ex. 1565. Find the area of a lune in square inches if its angle is 42° and the radius of the sphere is 8 inches. (Use tr = - 2 /.) Ex. 1566. In equal spheres, or in the same sphere, two lunes are to each other as their angles. Ex. 1567. Two lunes in unequal spheres, but with equal angles, are to each other as the squares of the radii of their spheres. Ex. 1568. In a sphere whose radius is R, find the altitude of a zone equivalent to a lune whose angle is 45°. Ex. 1569. Considering the earth as a sphere with radius B, rind the area of the zone visible from a point at a height h above the surface of the earth. 989. Def. The spherical excess of a spherical triangle is thre^excess of the sum of its angles over 180°. 454 SOLID GEOMETRY Proposition XXV. Theorem 990. The area of a spherical triangle, expressed in spherical degrees, is equal to its spherical excess. B i A r / i J -— A / / \ : / / A Given spherical A ABC with its spherical excess denoted by E. To prove area of A ABC = E spherical degrees. Argument 1. Complete the circumferences of which AB, BC, and CA are arcs. 2. Aab'c' and A'BC are symmetrical. 3. .-. AAB'C' oAA'BC 4. .-. A ABC + A AB'C'o A ABC -{-A A'BC. 5. .-., expressed in spherical degrees, A ABC + A AB'C' f©= lune A = 2 A ; A ABC + A AB'C = lune B =2 B-, A ABC + A ABC' = lune C = 2 C. .-. 2A^LBC-f-(A^BC+AAB'(7' + A^LB , C -f A ABC')= 2(A + B+ C). But A ABC+ A AB'C'+ A AB'C+AABC' = surface of a hemisphere = 360. .-. 2 A ABC+ 360 = 2 (A + 5 + c). .-. A ^BC + 180 = ^ + 5 + a . •. A ^z?c = (i+B + 6')- 180, i.e. ^ spherical degrees.' q.e.d. 6. 8. 9. 10. Reasons § 908, g. § 956. § 958. § 54, 2. § 988. 6. § 54, 2. 7. §985. 8. § 309. 9. § 54, 8 a. 10. § 54, 3. BOOK IX 455 991. In § 949 it was proved that the sum of the angles of a spherical triangle is greater than 180° and less than 540°. Hence the spherical excess of a spherical triangle may vary from 0° to 360°, from which it follows (§ 990) that the area of a spherical triangle may vary from yf^ to ff£ of the entire surface ; i.e. the area of a spherical triangle may vary from nothing to \ the surface of the sphere. Thus in a spherical triangle whose angles are 70°, 80°, and 100°, respectively, the spherical excess is (70° + 80° + 100°) - 180° = 70° ; i.e. the area of the given triangle is -f^ of the surface of the sphere. 992. Historical Note. Menelaus of Alexandria (circ. 98 a.d.) wrote a treatise in which he describes the properties of spherical triangles, although there is no attempt at their solution. The expression for the area of a spherical triangle, as stated in § 990, was first given about 1626 a.d. by Girard. (See also § 946.) This theorem was also discovered in- dependently by Cavalieri, a prominent Italian mathematician. Ex. 1570. If three great circles are drawn, each perpendicular to the other two, into how many trirectangular spherical triangles is the surface divided ? Then what is the area of a trirectangular spherical triangle in spherical degrees ? Test your answer by applying Prop. XXV. Ex. 1571. Find the area in spherical degrees of a birectangular spherical triangle one of whose angles is 70° ; of an equilateral spherical triangle one of whose angles is 80°. What part of the surface of the sphere is each triangle ? Ex. 1572. The angles of a spherical triangle in a sphere whose sur- face has an area of 216 square feet are 95°, 105°, and 130°. Find the number of square feet in the area of the triangle. Ex. 1573. In a sphere whose diameter is 16 inches, find the area of a triangle whose angles are 70°, 86°, and 120°. Ex. 1574. The angles of a spherical triangle are 60°, 120°, and 160°, and its area is lOOf square inches. Find the radius of the sphere. (Use Ex. 1575. The area of a spherical triangle is 90 spherical degrees, and the angles are in the ratio of 2, 3, and 5. Find the angles. Ex. 1576. Find the angle (1) of an equilateral spherical triangle, (2) of a lune, each equivalent to one third the surface of a sphere. Ex. 1577. Find the angle of a lune equivalent to an equilateral spherical triangle one of whose angles is 84°. 456 SOLID GEOMETRY Proposition XXVI. Theorem 993. The area of a spherical polygon, expressed in spherical degrees, is equal to the sum of its angles dimin- ished by 180° taken as many times less two as the poly- gon has sides. c Given spherical polygon ABCD • • • with n sides ; denote the sum of its angles by T. To prove area of polygon ABCD • ••, expressed in spherical degrees, = T— (n — 2)180. 1. 2. Argument From any vertex such as A, draw all possible diagonals of the polygon, forming n — 2 spherical A, I, II, etc. Then, expressed in spherical degrees, AI = (Z 1 + Z2 + Z3)-180; A 11= (Z 4 + Z 5 + Z 6) -180; etc. .-. A I + AII + ... = r-(n- 2)180. .*. area of polygon ABCD ••• = T-(n-2) 180. q.e.d. Reasons § 937. 2. § 990. § 54, 2. § 309. Ex. 1578. Prove Prop. XXVI by using a figure similar to that used in § 216. Ex. 1579. Find the area of a spherical polygon whose angles are 80°, 92°, 120°, and 140°, in a sphere whose radius is 8 inches. Ex. 1580. Find the angle of an equilateral spherical triangle equiva- lent to a spherical pentagon whose angles are 90°, 100°, 110°, 130°, and 140°. Ex. 1581. Find one angle of an equiangular spherical hexagon equivalent to six equilateral spherical triangles each with angles of 70°. BOOK IX 457 4f£c§E Xo M r v: y .N Ex. 1582. The area of a section of a sphere 63 inches from the cen- ter is 256 ir square inches. Find the surface of the sphere. Ex. 1583. The figure represents a sphere inscribed in a cylinder, and two cones with the bases of the cylinder as their bases and the center of the sphere as their vertices. Any plane, as BS, is passed through the figure parallel to MN, the plane of* the base. Prove that the ring between section AB of the cylinder, and section CD of the cone, is always equivalent to the section of the sphere. Ex. 1584. Find the volume of a barrel 30 inches high, 54 inches in circumference at the top and bottom, and 64 inches in circumference at the middle. Hint. Consider the barrel as the sum of two frustums of cones. Ex. 1585. Given T the total area, and B the radius of the base, of a right circular cylinder. Find the altitude. Ex. 1586. Given #the lateral area, and B the radius of 'the base, of a right circular cone. Find the volume. Ex. 1587. Given /S'the lateral area, and T the total area, of aright circular cone. Find the radius and the altitude. Volumes 994. Note. The student should not fail to observe the striking similarity in the figures and theorems, as well as in the definitions, relat- ing to the areas and volumes connected with the measurement of the sphere. A careful comparison of the following articles will emphasize this similarity : AREAS VOLUMES AREAS VOLUMES §966 § 995 § 979 §§ 1004, 1005 §967, 968 §996, a §982 § 1006 §969 § 996, b §984 § 1007 §970 § 996, c §987 § 1008 §971 §997 §988 § 1009 § 972 §999 §936 § 1010 $975 § 1002 §990 § 1012 §•5077 §1003 § 993 § 1013 458 SOLID GEOMETRY Proposition XXYII. Theorem 995. If an isosceles triangle is revolved about a straight line lying in its plane and passing through its vertex but not intersecting its surface, the volume of the solid gener- ated is equal to the product of the surface generated by the base of the triangle and one third of its altitude. Y Fig. 2. Fig. 3. Given isosceles A AOB with altitude OE, and a str. line XY lying in the plane of A AOB, passing through O and not inter- secting the surface of A AOB-, let the volume of the solid gen- erated by AAOB revolving about XF as an axis be denoted by volume AOB. To prove volume AOB = area AB • 1 OE. I. If AB is not II XY and does not meet XY (Fig. 1). Argument Only 1. Draw AC and BD _L XY. 2. Prolong BA to meet XY at F. 3. Then volume AOB = volume FOB — volume FOA. 4. Volume FOB — volume FDB + volume BOB. 5. .'. volume FOB =\tt ~BD 2 • FD + \ 7T BD 2 • DO = 1 7T BD Z (FD + DO) = \ttBD- BD > FO. BD • FO = twice area of A FOB — BF ■ OE. 6. But 7. .-. volume FOB = ^ tt BD • BF 8. But ttBD • BF as area FB. 9. .-. volume FOB = area FB OE = ttBD BF • I OE. iOK BOOK IX 459 10. Likewise volume FOA = area FA • \ OE. 11. .-. volume AOB = area FB • i OE — area FA • = (area FB — area FA) i Q# = area AB • J OE. iOE Q.E.D. II. If AB is not II XY and point A lies in XY (Fig. 2). The proof is left as an exercise for the student. Hint. See Arg. 9 or Arg. 10 of § 995, I. III. If AB II XY (Fig. 3). The proof is left as an exercise for the student. Hint. Volume AOB = volume ACDB — twice volume CO A. 996. The student may : . (a) State and prove the corollaries on the volume of a sphere corresponding to §§ 967 and 968. (b) State and prove the theorem on the volume of a sphere corresponding to § 969. 1. Outline of Proof V= (area A'b'c' *-)$$ = A'E' • 2 ttR • 1 R = %7rR 2 >A'E' (§§ 996, a and 967). v = (area AB C • • • )\ a = AE • 2 -n-a • -J- a = f 7ra 2 XE (§§ 996, a and 968). F_ |ir^ > »^J , _JZ a f A'E' = i? 3 2 * XE ~ a 3 TTcr • ^1J£ 4. Proceed as in § 855, observing that since the limit of a = R (§ 543, I), the limit of a 3 = R 3 (§ 593) ; i.e. R^^-a 3 may be made less than any previously assigned value, however small. (c) State, by aid of § 970, the definition of the volume of a 460 SOLID GEOMETRY Proposition XXVIII. Theorem 997. The volume of a sphere is equal to the product of the area of its surface and one third its radius. Given sphere with its radius denoted by R, the area of its surface by S, and its volume by V. To prove V= S • ± R. Argument 1. In the semicircle ACE inscribe ABODE, half of a regular polygon with an even number of sides. Denote its apothem by a, the area of the surface generated by the semiperimeter as it revolves about AE as an axis by S', and the vol- ume of the solid generated by semi- polygon ABODE by V'. Then r = 8' \R. 1005. Cor. IV. If v denotes the volume of a spherical sector, z the area of the zone forming its base, H the alti- tude of the zone, and R the radius of the sphere, V=z-\R (§1004) = {H-2ttR)\R (§979) = %vR'*H. pole and whose altitude is Ex. 1600. Considering the earth as a sphere with radius B, find the volume of the spherical sector whose base is a zone adjoining the north 7? 2 Tt — ; — • Is the one volume twice the other? 3 3 Compare your results with those of Ex. 1558. Ex. 1601. Considering the earth as a sphere with radius R, find the volume of the spherical sector whose base is a zone extending : (a) 30° from the north pole ; (ft) 60° from the north pole. Is the one volume twijie the other ? Compare your results with those of Ex. 1559. 464 SOLID GEOMETRY Ex. 1602. Considering the earth as a sphere with radius i?, find the volume of the spherical sector whose base is a zone lying between the parallels of latitude : (a) 30° and 45° from the north pole ; (&) 30° and 45° from the equator. Are the two volumes equal ? Compare your results with those of Ex. 1560. Ex. 1603. Considering the earth as a sphere with radius i?, find the area of the zone whose bases are the circumferences of small circles, one 30° north of the equator, the other 30° south of the equator. What part of the entire surface is this zone ? Ex. 1604. What part of the entire volume of the earth is that por- tion included between the planes of the bases of the zone in Ex. 1603 ? • Hint. This volume consists of two pyramids and a spherical sector. Ex. 1605. A spherical shell 2 inches in thickness contains the same amount of material as a sphere whose radius is 6 inches. Find the radius of the outer surface of the shell. Ex. 1606. A spherical shell 3 inches thick has an outer diameter of 16 inches. Find the volume of the shell. • Ex. 1607. Find the volume of a sphere circumscribed about a rec- tangular parallelopiped whose edges are 3, 4, and 12. Ex. 1608. Find the volume of a sphere inscribed in a cube whose volume is 686 cubic centimeters. Ex. 1609. The surface of a sphere and the surface of a cube are each equal to &. Find the ratio of their volumes. Which is the greater ? Ex. 1610. In a certain sphere the volume and the circumference of a great circle have the same numerical value. Find the surface and the vol- ume of the sphere. Ex. 1611. How many bullets \ of an inch in diameter can be made from a sphere of lead 10 inches in diameter ? from a cube of lead whose edge is 10 inches ? 1006. Defs. A spherical wedge is a solid closed figure whose bounding surface consists of a lune and the planes of the sides of the lune. The lune is called the base of the spherical wedge, and the angle of the lune the angle of the spherical wedge. 1007. Prove, by superposition, the following property of wedges : In equal spheres, or in the same sphere, two spherical wedges are equal if their angles are equal. BOOK IX 465 Proposition XXIX. Theorem 1008. The volume of a spherical wedge is to the volume of the sphere as the number of degrees in the angle of the spherical wedge is to 360. N N Q E S S Given spherical wedge NASB with the number of degrees in its Z denoted by N, its volume denoted by W, and the volume of the sphere denoted by V; let O EQ be the great O whose pole is N. m W N To prove — = . P v 360 The proof is left as an exercise for the student. Hint. The proof is similar to that of § 987. 1009. Cor. I. The volume of a spherical wedge is equal to the product of its base and one third the radius of the sphere. Outline of Proof — = -^- (§1008). v 360 v } : W S \R = L \ R, where S represents 360 360 the area of the surface of the sphere, and L the area of the lune, i.e. the area of the base of the spherical wedge. fix. 1612. In a sphere whose radius is 16 inches, find the volume of a spherical wedge whose angle is 40°. 466 SOLID GEOMETRY 1010. Defs. A spherical pyramid is a solid closed figure whose bounding surface consists of a spherical polygon and the planes of the sides of the spherical polygon. The spherical polygon is the base, and the center of the sphere the vertex, of the spherical pyramid. 1011. By comparison with § 957, b and § 958, prove the following prop- erty of spherical pyramids : In equal spheres, or in the same sphere, two triangular spherical pyramids whose bases are symmetrical spherical triangles are equivalent. 1012. Cor. n. The volume of a spherical triangular pyramid is equal to the product of its base and one third the radius of the sphere. Outline of Proof 1. Pyramid O-AB'c' =c= pyramid (y-A'BC (§ 1011). 2. .-. pyramid O-ABC + pyramid O-AB'C 1 =c= wedge A = 2 A . i R (§ 1009) ; pyramid O-ABC + pyra- mid O-AB'C = wedge B = 2 B • ^ R; pyramid O-ABC + pyramid O-ABC' — wedge C = 2 c • \ R. 3. .-. twice pyramid O-ABC + hemispnere = 2 (A + B -f C) -| R. 4. .-. twice pyramid O-ABC + 360 • i R = 2 (A + B -f c) i R. 5. .-. pyramid O-ABC = (A+ B + C— 180) J R = AABC >\R = K - \R. q.e.d. 1013. Cor. in. The volume of any spherical pyramid is equal to the product of its base and one third the ra- dius of the sphere. (Hint. Compare with §805.) Ex. 1613. Show that the formula of § 997 is a special case of §§ 1004, 1009 and 1013. Ex. 1614. In a sphere whose radius is 12 inches, find the volume of a spherical pyramid whose base is a triangle with angles 70°, 80°, and 90°. BOOK IX 467 MISCELLANEOUS EXERCISES ON SOLID GEOMETRY Ex. 1615. A spherical pyramid whose base is an equiangular penta- gon is equivalent to a wedge whose angle is 30°. Find an angle of the base of the pyramid. Ex. 1616. The volume of a spherical pyramid whose base is an equi- angular spherical triangle with angles of 105° is 128 ir cubic inches. Find the radius of the sphere. Ex. 1617. In a sphere whose radius is 10 inches, find the angle of a spherical wedge equivalent to a spherical sector whose base has an alti- tude of 12 inches. Ex. 1618. Find the depth of a cubical tank that will hold 100 gallons of water. Ex. 1619. The altitude of a pyramid is H. At what distance from the vertex must a plane be passed parallel to the base so that the part cut off is one half of the whole pyramid ? one third ? one nth ? Ex. 1620. Allowing 550 pounds of copper to a cubic foot, find the weight of a copper wire \ of an inch in diameter and 2 miles long. Ex. 1621. Disregarding quality, and considering oranges as spheres, i.e. as similar solids, determine which is the better bargain, oranges averaging 2f inches in diameter at 15 cents per dozen, or oranges averaging 3^ inches in diameter at 30 cents per dozen. Ex. 1622. In the figure, B, C, and D are the mid-points of the edges of the cube meet- ing at A. What part of the whole cube is the pyramid cut off by plane BCD ? Hint. Consider ABC as the base and D as the vertex of the pyramid. Ex. 1623. Is the result of Ex. 1622 the same if the figure is a rectangular parallelopiped ? Ex. 1624. It is proved in calculus that in order that a cylindrical tin can closed at the top and having a given capacity may require the small- est possible amount of tin for its construction, the diameter of the base must equal the height of the can. Find the dimensions of such a can holding 1 quart ; 2 gallons. Ex. 1625. A cylindrical tin can holding 2 gallons has its height equal to the diameter of its base. Another cylindrical tin can with the same capacity has its height equal to twice the diameter of its base. Find the ratio of the amount of tin required for making the two cans. Is your answer consistent with the fact contained in Ex. 1624 ? any parallelopiped ? 468 SOLID GEOMETRY Ex. 1626. A cannon ball 12 inches in diameter is melted, and the lead is cast in the form of a cube. Find the edge of the cube. Ex. 1627. The cube of Ex. 1626 is melted, and the lead is cast in the form of a cone, the diameter of whose base is 12 inches. Find the altitude of the cone. Ex. 1628. Find the weight of the cannon ball in Ex. 1626 if a cubic foot of iron weighs 450 pounds. Ex. 1629. The planes determined by the diagonals of a cube divide the cube into six equal pyramids. Ex. 1630. Let D, E, F, and G be the mid-points of VA, AB, BC, and CV, respectively, of triangular pyramid V-ABC. Prove DEFG a parallelogram. Ex. 1631. In the figure, is plane DEFQ par- yt allel to edge AC? to edge VB? Prove that any r yr l\ section of a triangular pyramid made by a plane Si~~"\ A parallel to two opposite edges is a parallelogram. yS \ I \ Ex. 1632. The three lines joining the mid- ^^^Ijr*"* ,"7 \tr points of the opposite edges of a tetrahedron bisect F^****JrE each other and hence meet in a point. • Hint. Draw DF and EG. Are these two of the required lines ? Ex. 1633. In a White Mountain two-quart ice cream freezer, the can is 4f inches in diameter and 6| inches high ; the tub is 6| inches in diameter at the bottom, 8 inches at the top, and 9f inches high, inside measurements, (a) Does the can actually hold 2 quarts ? (6) How many cubic inches of ice can be packed about the can ? Ex. 1634. Find the total area of a regular tetrahedron whose alti- tude is a centimeters. Ex. 1635. The lateral faces of a triangular pyramid are equilateral triangles, and the altitude of the pyramid is 6 inches. Find the total area. Ex. 1636. In the foundation work of the Woolworth Building, a 55- story building on Broadway, New York City, it was necessary, in order to penetrate the sand and quicksand to bed rock, to sink the caissons that contain the huge shafts of concrete to a depth, in some instances, of 131 feet. If the largest circular caisson, 19 feet in diameter, is 130 feet deep and was filled with concrete to within 30 feet of the surface, how many loads of concrete were required, considering 1 cubic yard to a load ? Ex. 1637. From A draw a line meeting line XFin B ; let G be the mid-point of AB. Find the locus of C as B moves in line XY. Ex. 1638. In Ex. 1637, let XY be a plane. Find the locus of C as B moves arbitrarily in plane XY. BOOK IX 469 Ex. 1639. A granite shaft in the form of a frustum of a square pyramid contains 16 If cubic feet of granite ; the edges of the bases are 4 feet and \\ feet, respectively. Find the height of the shaft. Ex. 1640. The volume of a regular square pyramid is 42§ cubic feet ; its altitude is twice one side of the base, (a) Find the total surface of the pyramid ; (6) find the area of a section made by a plane parallel to the base and one foot from the base. Ex. 1641. Allowing'l cubic yard to a load, find the number of loads of earth in a railway cut | mile in length, the average dimensions of a cross section being as represented in the figure, the numbers denoting feet. Give the name of the geometrical solid represented by the cut. 12 "7" Why is it not a frustum of a pyramid ? 16 Ex. 1642. For protection against fire, a tank in the form of a frustum of a right circular cone was placed in the tower room of a certain public building. The tank is 16 feet in diameter at the bottom, 12 feet in di- ameter at the top, and 16 feet deep. If the water in the tank is never allowed to get less than 14 feet deep, how many cubic feet of water would be available in case of an emergency ? how many barrels, counting i\ cubic feet to a barrel ? Ex. 1643. A sphere with radius B is inscribed in a cylinder, and the cylinder is inscribed in a cube. Find : (a) the ratio of the volume of the sphere to that of the cylinder ; (b) the ratio of the cylinder to the cube ; (c) the ratio of the sphere to the cube. Ex. 1644. A cone has the same base and altitude as the cylinder in Ex. 1643. Find the ratio of the cone : (a) to the sphere ; (&) to the cylinder; (c) to the cube. Ex. 1645. In a steam-heated house the heat for a room was supplied by a series of 10 radiators each 3 feet high. gl The average cross section of a radiator is s\ TTTN shown in the figure, the numbers denoting yj 2 £L/ inches. It consists of a rectangle with a ®* semicircle at each end. Find the total radiating surface in the room. Ex. 1646. A coffee pot is 5 inches deep, 4J inches in diameter at the top, and 5^ inches in diameter at the bottom. How many cups of coffee will it hold, allowing 6 cups to a quart ? (Answer to nearest whole number.) Ex. 1647. Any plane passing through the center of a parallelopiped divides it into two equivalent solids. Are these solids equal ? Ex. 1648. From two points, P and B, on the same side of plane AB, two lines are drawn to point in plane AB, making equal angles with the^plane. Find the locus of point O. (Hint. See Ex. 1237.) 470 SOLID GEOMETRY Ex. 1649. A factory chimney is in the form of a frustum of a regu- lar square pyramid. The chimney is 120 feet high, and the edges of its bases are 12 feet and 8 feet, respectively. The flue is 6 feet square throughout. How many cubic feet of material does the chimney contain? Ex. 1650. Find the edge of the largest cube that can be cut from a regular square pyramid whose altitude is 10 inches and one side of whose base is 8 inches, if one face of the cube lies in the base of the pyramid. Ex. 1651. Fig. 1 represents a granite monument, the numbers denoting inches. The main part of the stone is 5 feet high, the total height of the stone being 5 feet 6 inches. Fig. 2 represents a view of 10 \ B 16 Bie A 36 B 10 / B \ 9 36 Fig. 2. 9 16 ?r^ ,8 20 8 Fig. 3. the main part of the stone looking directly from above. Fig. 3 repre- sents a view of the top of the stone looking directly from above. Calcu- late the volume of the stone. Hint. From Fig. 2 it is seen that the main part of the stone con- sists of a rectangular parallelopiped A, four right triangular prisms B, and a rectangular pyramid at each corner. Fig. 3 shows that the top con- sists of a right triangular prism and two rectangular pyramids. Ex. 1652. The monument in Ex. 1651 was cut from a solid rock in the form of a rectangular parallelopiped. How many cubic feet of granite were wasted in the cutting ? Ex. 1653. In the monument of Ex. 1651 the two ends of the main part, and the top, have a rock finish, the front and rear surfaces of the main part being polished. Find the number of square feet of rock finish and of polished surface. Ex. 1654. The base of a regular pyramid is a triangle inscribed in a circle whose radius is B, and the altitude of the pyramid is 2B. Find the lateral area of the pyramid. Ex. 1655. Find the weight in pounds of the water required to fill the tank in Ex. 1323, if a cubic foot of water weighs 1000 ounces. BOOK IX 471 Ex. 1656. By using the formula obtained in Ex. 1543, find the vol- ume of the sphere inscribed in a regular tetrahedron whose edge is 12. Ex. 1657. By using the formula obtained in Ex. 1544, find the volume of the sphere circumscribed about a regular tetrahedron whose edge is 12. Ex. 1658. A hole 6 inches in diameter was bored through a sphere 10 inches in diameter. Find the vol- ume of the part cut out. Hint. The part cut out consists of two spherical cones and the solid generated by revolving isosceles A BOC about X7as an axis. Ex. 1659. Check your result for Ex. 1658 by finding the volume of the part left. Ex. 1660. Find the area of the spherical surface left in Ex. 1658. Ex. 1661. Four spheres, each with a radius of 6 inches, are placed on a plane surface in a triangular pile, each one being tangent to each of the others. Find the total height of the triangular pile. Ex. 1662. Find the total height of a triangular pile of spheres, each with radius of 6 inches, if there are three layers; four layers; n layers. FORMULAS OF SOLID GEOMETRY 1014. In addition to the notation given in § 761, the follow- ing will be used : A, B, C, ••• = number of degrees in the angles of a spherical polygon . a, b, c, ••• = sides of a spherical polygon. B = base of spherical sec- tor, wedge, and pyramid. C = circumference of base in general or of lower base of frus- tum of cone, c = circumference of up- per base of frustum of cone. D = diameter of a sphere. E = spherical excess of a spherical triangle. H = K = L = N = S: T W: Z : altitude of zone or spherical sector, area of a spherical triangle or spherical polygon, area of lune. number of degrees in the angle of a lune or wedge, radius of base in general, of lower base of frustum of cone, or of. sphere, radius of upper base of frustum of cone, area of surface of a sphere, sum of the angles of a spheri- cal polygon, volume of a wedge, area of a zone. 472 SOLID GEOMETRY FIGURE I Prism. Right prism. Regular pyramid. Frustum of regular pyramid. Rectangular parallelopiped. Cube. Rectangular parallelopipeds. Rectangular parallelopiped. Rectangular parallelopipeds. Any parallelopiped. Parallelopipeds. Triangular prism. Any prism. Prisms. Triangular pyramid. Any pyramid. Pyramids. Similar tetrahedrons. Frustum of any pyramid. Truncated right triangular prism. Right circular cylinder. Similar cylinders of revolution. Right circular cone. Similar cones of revolution. FORMULA 8 = P E. 8=P- H. S=%P.L. V=a-b-c. V=E\ V a-b-c V a'- b' • c' v= B- H. V _ B ■H V B' H' v= B- H. V B -H V B< W v = B- H. v= B- H. V B H V B' H' v= \* H. v= 1* H. v__ B ■H V B' . H i V_ = E^ V E' s ' V=}H(B + b+y/Tn>). V= \B(E + E' + E"). 8 = C-H. 8 = 2wBH. T=2ttB(H+B). S_ = H* _B* T = BT* _B* T' H' 2 B' 2 ' 8 = \C-L. 8 = w BL. T=ttB(L + B). & __ H* _ Z2 ._ B*_ 8' H' 2 ~ L' 2 ~ B' 2 ' REFERENCE §762. ' § 763. §766. §767. §778. §779. §780. §782. §783. §790. §792. §797. §799. §801. §804. §805. §807. §812. §815. §817. §858. § 859. §859. §864. §864. §873. §875. §875. §878. BOOK IX 473 FIGURE FORMULA REFERENCE Similar cones of revolution. T _ H 2 _ L 2 _ B 2 T> H' 2 L' 2 B' 2 ' §878. Frustum of right circular cone. S = %(C+c)L. §882. S = irL(B + r). §883. T = «-£(JB+r)+T(JP+f«). §883. Cylinder with circular bases. V=BH. §889. V=irB 2 H. §890. Similar cylinders of revolution, V _H 3 _ B 3 V H' 3 B' 3 ' §891. Cone with circular base. V=\B-H. §893. V = \ irB 2 • H. §895. Similar cones of revolution. V _ H* _ L* _ B 3 V H' 3 V 3 B' 3 §897. Frustum of cone with circular base. V = \ H(B + b -f VB • b). §898. V=%irH(B 2 + r 2 + B ■r). §899. Spherical triangle. a + b > c. §941. Spherical polygon. a + 6 + c+-.-<360°. §942. Polar triangles. .4 + a' = 180°, B+b< = 180°, §947. Spherical triangle. A + B+ G> 180° and < 540 §949. Sphere. S=4:7rB 2 . §971. Spheres. S _B 2 _D 2 8' B' 2 D' 2 ' §972. Zone. Z = H'2tB. §979. Zones. Z _H Z' H>' §980. Lune. L_ N 8 360 ' §987. L = 2N. §988. Spherical triangle. K=(A+B+C) -180° = E. §990. Spherical polygon. K=T- (n- 2)180. §993. Sphere. V=S.IB. §997. V=i*&=\T&. §998. Spheres. V _B 3 _D 3 V B' 3 D' 3 ' §999. Spherical sector. V=Z-\B. §1004. V=*ttB 2 H. §1005. Spherical wedge. W _ N V 360* § 1008. W=L\R. §1009. Spherical triangular pyramid. V=K\B. § 1012. Ar,y spherical pyramid. V=K.\B. § 1013. 474 SOLID GEOMETRY APPENDIX TO SOLID GEOMETRY SPHERICAL SEGMENTS 1015. Defs. A spherical segment is a solid closed figure whose bounding surface consists of a zone and two parallel planes. Spherical Segment of Two Bases Spherical Segment of One Base The sections of the sphere formed by the two parallel planes are called the bases of the spherical segment. 1016. Defs. State, by aid of §§ 976 and 977, definitions of: (a) Altitude of a spherical segment. (6) Segment of one base. Proposition I. Problem 1017. To derive a formula for the volume of a spher- ical segment in terms of the radii of its bases and its altitude. E Given spherical segment generated by ABCD revolving about EF as an axis, with its volume denoted by r, its altitude by h, and the radii of its bases by r x and r 2 , respectively. To derive a formula for V in terms of i\, r 2 , and h. APPENDIX 477 . \ pyramid P-ADC = J H • m x + } II • m x + \H • mj = \ H • 4 j%. .*. the volume of all lateral pyramids = \ H>4tM. .'. V=\H>B + \H-b+\H.±M=±rH(B+b + 4:M). Q.E.F. Rs. 1665. By substituting in the prismatoid formula, derive the formula for : (a) the volume of a prism (§ 799) ; (6) the volume of a pyramid (§ 805) ; (c) the volume of a frustum of a pyramid (§ 815). Ex. 1666. Solve Ex. 1651 by applying the prismatoid formula to each part of the monument. SIMILAR POLYHEDRONS* 1022. The student should prove the following : (a) Any two homologous edges of two similar polyhedrons have the same ratio as any other two homologous edges. (b) Any two homologous faces of two similar polyhedrons have the same ratio as the squares of any two homologous edges. (c) The total surfaces of two similar polyhedrons have the same ratio as the squares of any two homologous edges. 1023. Def. The ratio of similitude of two similar polyhe- drons is the ratio of any two homologous edges. 1024. Def. If two polyhedrons AB CD ••• and a'b'c'd' ... are so situated that lines from a point to A', b', C', d' } etc., are divided by points g' C' A, B, C, D, etc., in such ^^^7\--^/\ a manner that ^"^^^/\ ! \ oa[ _ q& _ q& _ qif _ ^ oa ~ ob ~ oc ~ od ~ E' the two polyhedrons are said to be radially placed. Ex. 1667. Construct two polyhedrons radially placed and so that point lies between the two polyhedrons ; within the two polyhedrons. * See § 811. In this discussion only convex polyhedrons will be considered. 478 SOLID GEOMETRY Proposition III. Theorem 1025. Any two radially placed polyhedrons are similar. (See Fig. 2 below.) Given polyhedrons EC and 'E'c' radially placed with respect to point 0. To prove polyhedron EC ~ polyhedron E'c'. AB, BC, CD, and DA are II respectively to a'b', B'c', c'd\ and D'A'. § 415. .-. ABCD II A'B'C'D', and is similar to it. § 756, II. Likewise each face of polyhedron EC is ~ to the correspond- ing face of polyhedron E'c', and the faces are similarly placed. Again, face All II face a'h', and face AF II face A'F'. .-. dihedral Z AE = dihedral Z A'e'. Likewise each dihedral Z of polyhedron EC is equal to its corresponding dihedral Z of polyhedron E'c'. .-. each polyhedral Z of polyhedron EC is equal to its corre- sponding polyhedral Z of polyhedron E'c'. § 18. .-. polyhedron EC ~ polyhedron E'c'. §811. q.e.d. Proposition IV. Theorem 1026. Any two similar polyhedrons may be radially placed. q' Fig. i. Fig. 2. Given two similar polyhedrons XM and E'&. To prove that XM and E'c' may be radially placed. APPENDIX 479 Outline of Proof 1. Take any point within polyhedron E'C? and construct polyhedron EC so that it is radially placed with respect to E'C 1 and so that OA' : 0A = OB' : OB = •• = A'B' : KL. 2. Then polyhedron EC ~ polyhedron E'c'. § 1025. 3. Prove that the dihedral A of polyhedron EC are equal, respectively, to the dihedral A of polyhedron XM, each being equal, respectively, to the dihedral A of polyhedron E'c'. 4. Prove that the faces of polyhedron EC are equal, respec- tively, t6 the faces of polyhedron XM. 5. Prove, by superposition, that polyhedron EC = XM. 6. .-. polyhedron XM may be placed in the position of EC. 7. But EC and E'c' are radially placed. 8. .-. XM and E'c' may be radially placed. q.e.d. Proposition V. Theorem 1027. If a pyramid is cut by a plane parallel to its base : I. The pyramid cut off is similar to the given pyra- mid. II. The two pyramids are to each other as the cubes of any two homologous edges. O The proofs are left as exercises for the student. Hint. For the proof of II, pass planes through OB' and diagonals B'D', B'E', etc., dividing each of the pyramids into triangular pyramids. Then pyramid O-BCD ~ pyramid O-B'C'D' ; pyramid O-EBD ~ pyra- mid O-E'B'D', etc. Use § 812 and a method similar to that used in § 505. 480 SOLID GEOMETRY Proposition VI. Theorem. 1028. Two similar polyhedrons are to each other as the cubes of any two homologous edges. C Given two similar poly- hedrons XM and E'c', with their volumes denoted by V and V', respectively, and with KL and A'b' two homol. edges. :3 To prove KL A'B 1 Place XM in position EC, so that XM and E'C } are radially placed with respect to point within both polyhedrons. § 1026. Denote the volumes of pyramids O-ABCD, O-AEFB, etc., by v x , v 2 , etc., and the volumes of pyramids O-A'b'c'd', O-A'e'f'b', etc., by v-iy v 2 ', etc. Then °i _ V But AB A'B'~ -^ ; ^ = = 3 ; etc - §1027,11. j%,= - (§1022, a); AB A^B' A'E' 3 ; i _ •k' AB IB 1 Vi + V 4- AB A r B' §401. polyhedron EC __ AB 3 . V __ KL 3 polyhedron E'C' ^B 13 ' V' J^ 3 Q.E.D. 1029. Note. Since § 1028 was assumed early in the text (see §814), the teacher will find plenty of exercises throughout Books VII, VIII, and IX illustrating this principle. INDEX (The numbers refer to articles.) Art. Acute triangle 98 Adjacent angles 42 Adjacent dihedral angles . . .671 Alternation 396 Altitude, of cone 842 of cylinder 825 of parallelogram .... 228 of prism 733 of prismatoid .... 1020 of pyramid 751 of spherical segment . . 1016 of trapezoid 229 of triangle 100 of zone 976 Analysis, of a proportion . . . 397 of constructions .... 151, 152, 176 (Ex. 187) , 274, 499 of exercises 274 Angle 38 acute 47 at center of circle .... 292 at center of regular polygon 534 bisector of . . . 53, 127, 599 central 292 degree of 71 dihedral 666 inscribed in circle .... 363 inscribed in segment . . . 364 magnitude of ... 41, 669 obtuse 48 of lune 983 of one degree 71 of spherical wedge . . . 1006 of two intersecting arcs . 916 polyhedral .692 reflex 49 right 45, 71 right and left side of . . .199 sides of 38 solid 692 spherical 917 straight 69 tetrahedral 698 trihedral 698 vertex of 38 Angles 38 added 43 adjacent 42 alternate exterior .... 182 ' alternate interior . . . .182 Art. Angles, complementary ... 68 corresponding 182 designation of . 39, 50, 667, 696 difference of 44 equal 44 exterior 87, 182 exterior interior 182 homologous . . . 417, 424 interior 86, 182 oblique 51 of a polygon 86 subtracted ...... 44 sum of 43 supplementary 69 supplementary-adjacent • • 72 vertical 70 Antecedents 384 Apothem 533 Application 79 Arc 121, 280 degree of 296 length of 551 major 291 minor 291 Arcs, similar 565 Area, of a figure 471 of circle 558 of rectangle .... 468-471 of surface 467 Argument only 171 Arms, of isosceles triangle . . 94 of right triangle 96 Assumptions 12, 54 Axiom 55 Axioms 54 note on 600 Axis, of circle of sphere . . . 906 of right circular cone . . . 844 Base, of cone 840 of isosceles triangle .... 94 of polygon 99 of pyramid 748 of spherical pyramid . . 1010 of spherical sector . . . 1002 Bases, of cylinder 822 of prism 727 of spherical segment . . . 1015 of zone 975 Birectangular spherical triangle . 952 481 482 INDEX Art. Bisector, of angle . . .53, 127, 599 of line . , . . . . 52, 145 Bisectors of angles of triangle . 126 Center, of circle .... 119, 278 of regular polygon . . . .531 of sphere 901 Center-line 325 Central angle . . ... . . 292 Centroid 265 Chord 281 Circle 119, 276 arc of 121, 280 area of 558 center of 119, 278 chord of 281 circumference of . 119, 277, 550 circumscribed 300 determined 324 diameter of 282 escribed 322 inscribed 317 radius of ..... 119, 278 secant of 285 sector of 287, 564 segment of 288 tangent to 286 Circles, concentric 326 tangent, externally . . . 330 tangent, internally . . . 330 Circular cone 841 Circular cylinder 831 Circumference . . 119, 277, 550 length of 550, 552 measurement of .... 540 Circumscribed circle .... 300 Circumscribed polygon . . . 317 Circumscribed polyhedron . . 926 Circumscribed prism .... 852 Circumscribed pyramid . . . 868 Circumscribed sphere .... 929 Clockwise motion ..... 86 Closed figure . 83, 714, 715, 934, 935 Closed line 82 Coincidence 17 Collinear segments .... 28 Commensurable quantities 337, 342, 469 Common measure 337 Compasses, use of 31 Complementary* angles ... 68 Complete demonstration . . .171 Composition 398 Composition and division . . 400 Concentric circles 326 Conclusion 57 Concurrent lines 196 Cone 839 altitude of 842 Art. Cone, base of 840 circular 841 element of 840 frustum of 879 lateral surface of .... 840 oblique 843 of revolution 876 plane tangent to 866 right circular 843 spherical 1003 vertex of 840 volume of 892,6 Cones, similar 877 Conical surface 837 directrix of 838 element of 838 generatrix of 838 lower nappe of . . . . . 838 upper nappe of 838 vertex of 838 Consequents 384 Constant 346, 347 Construction 123 of triangles 269 Continued proportion .... 405 Converse theorem 135 Convex polyhedral angle . . . 697 Coplanar points, lines, planes . 668 Corollary 58 Counter-clockwise motion . . 50 Cube 742 Cylinder 821 altitude of 825 bases of 822 circular 831 element of 822 lateral surface of .... 822 oblique 824 of revolution 861 plane tangent to 850 right _ . - 823 right circular 832 right section of 830 volume of 888 Cylinders, similar 863 Cylindrical surface 819 directrix of .820 element of 820 generatrix of 820 Definition, test of 138 Degree, of angle 71 of arc 296 spherical 986 Demonstration 56, 79 complete 171 Determined circle 324 Determined line and point . 25, 26 Determined plane 608 INDEX 483 Art. Determined triangle . . 133, 269 Diagonal 88 of polyhedron 718 Diameter, of circle 282 of sphere 901 Dihedral angle 666 " ;e of 666 faces of . 666 plane angle of 670 right 672 Dihedral angles, adjacent . .671 Dimensions 10 Directrix, of conical surface . . 838 of cylindrical surface . . . 820 of polyhedral angle . . . . 693 of prismatic surface . . . 725 of pyramidal surface . . . 745 Discussion ....... 123 Distance, between two points . 128 from point to line . . . .166 from point to plane . . . 662 on surface of sphere . . . 909 polar 911 Division 399 Dodecahedron 719 Drawing to scale 437 Edge of dihedral angle .... 666 Edges, of polyhedral angle . . 694 of polyhedron 717 Element, of cone 840 of conical surface .... 838 of cylinder 822 of cylindrical surface . • . 820 of polyhedral angle . . . . 694 of pyramidal surface . . . 745 Equal figures 18, 473 homologous parts of . 109,110 Equiangular polygon .... 90 Equidistance 128 Equilateral polygon .... 89 Equilateral triangle 95 Equivalent figures 473 Equivalent solids 776 Escribed circle 322 Euclid's Elements 114 Excess, spherical 989 Exterior angle of polygon . . 87 Extreme and mean ratio . . . 464, 465 (Ex. 763), 526 Extremes 385 Face angles of polyhedral angle . 694 Faces, of dihedral angle . . . 666 of polyhedral angle .... 694 of polyhedron 717 Figure, closed 83 ' J ^"geometric 13 Art. Figure, plane . . . , . 36 rectilinear .... . 37 Figures, equal . . . , 18, 473 equivalent .... . 473 isoperimetric . 572 similar . . .419. 473 transformation of . . . 493 Foot of perpendicular . 621 Formulas, of plane geometry . 570 of solid geometry . .1014 Fourth proportional . . . 387 Frustum of cone . . . 879 slant height of . . . . 881 Frustum of pyramid . 754 lateral area of ... . . 760 slant height of . . . . 765 Generatrix, of conical surface . 838 of cylindrical surface . . . 820 of polyhedral angle .... 693 of prismatic surface . . . 725 of pyramidal surface . . .-745 Geometric figure 13 Geometric figures, equal ... 18 Geometric solid . . . . 2, 9, 11 Geometry 14 of space 602 plane 36 solid 602 subject matter of ... . 1 Golden section 464 Great circle 904 Harmonical division .... 434 Hexagon 92 Hexahedron 719 Historical notes Achilles and the tortoise . 354 Ahmes .... 474, 569, 777 Archimedes, 542, 569, 809, 896 973, 974, 1001 Archytas 787 Area of a circle 569 Athenians 787 Brahmagupta . . . 809, 896 Cavalieri 992 Cicero 542 Descartes 237 Division of circumference . 516 Egyptians 371, 449, 474, 497, 569, 777 Euclid . . . 114, 510, 516, 723 Eudoxus . . . 809, 896, 1000 Gauss 516, 520 Girard, Albert . . .946, 992 Hero of Alexandria . . . 569 Herodotus 474 Hippasus 723 KingHiero .542 484 INDEX Art. Historical notes Lambert .. . 569 Lindemann 569 Menelaus of Alexandria . . 992 Metius of Holland . . . 569 Morse, S. F. B 520 Newton 542 Origin of geometry .... 474 Plato .... 268, 542, 787 Plutarch 371 Pons asinorum 114 Ptolemy . 569 Pythagoras 344,510,514,723,787 Richter 569 Seven Wise Men . . . .371 Shanks 569 Snell 946 Socrates 787 Squaring the circle . . . 569 Stobseus 114 Thales 371, 510 Zeno 354 Homologous parts .... 109, 110, 417,418,424 Hypotenuse 96 Hypothesis . .. 57 Icosahedron 719 Inclination of line to plane . . 665 Incommensurable quantities . . 339, 343, 344, 470 Indirect proof 159, 161 Inscribed angle 363 Inscribed circle 317 Inscribed polygon 299 Inscribed polyhedron . . . .928 Inscribed prism 851 Inscribed pyramid 867 Inscribed sphere 927 Instruments, use of 31 Intersection of two surfaces . . 614 Inversion 395 Isoperimetric figures .... 572 Isosceles trapezoid 227 Isosceles triangle 94 arms of 94 base of 94 sides of 94 vertex angle of 94 Lateral area, of frustum . . . 760 of prism 760 of pyramid 760 of right circular cone . . . 872 of right circular cylinder . 857 Lateral edges, of prism . . . 727 of pyramid 748 Lateral faces, of prism .... 727 of pyramid ..,,,, 748 Art. Lateral surface, of cone . . . 840 of cylinder 822 of frustum of pyramid , m .760 Length, of arc 551 of circumference . . . 550, 552 of perpendicular 166 of secant 426 of tangent 319 Limits . . 346, 349-351, 590-594 Line 4, 7, 11, 27 bisected 52, 145 broken 30 closed 82 curved 29 determined 25 divided externally .... 406 divided harmonically . . . 434 divided in extreme and mean ratio 464, 465 (Ex. 763), 526 divided internally .... 406 inclination of 665 oblique to plane 630 of centers 325 parallel to plane 629 perpendicular to plane . . . 619 projection of 656 right 23 segments of ... . 27, 406 straight 21, 23 tangent to sphere . . . .921 Line segment 27 Lines 20 concurrent 196 difference of 31 divided proportionally . . 407 homologous 109, 110, 418, 424 oblique 46 parallel 177 perpendicular 45 product of . 425, 475, 476, 511 rectangle of 425, 475, 476, 511 Locus .... 129, 130, 131, 601 as an assemblage . . . .130 as a path 130 finding of a 144 of a point 130 of all points 130 problem, solution of . 143, 60,1 Lune 982 angle of 983 sides of 983 vertices of 983 Magnitude of an angle . . 41, 669 Major arc 291 Maximum 571 Mean proportional . . . . .386 Means -385 INDEX 485 Art. Measure, common 337 numerical . . . 335, 467, 770 unit of 335, 466 Measurement 335 of angle 358 of arc 358 of circle . . . . . . 540, 558 of circumference . . . 540, 552 of distances by means of triangles 115 of line 335, 595 of rectangle .... 468-471 of surface 466 Measure-number 335, 467, 471, 595, 770 Median, of trapeze id . . . .251 of triangle 102 Median center of triangle . . . 265 Methods of attack 103, 104, 106, 110 (Ex. 62), 115, 151, 152, 158, 159, 176 (Ex. 187), 269-274, 275, 397, 398, 425, 499, 502, 513 Minimum 571 Minor arc 291 Nappes, upper and lower . 746, 838 Numerical measure . 335, 467, 770 Oblique angles 51 Oblique cone 843 Oblique cylinder 824 Oblique lines 46 Obtuse triangle 97 Octahedron 719 One to one correspondence 595, 596 Opposite theorem 136 Optical illusions 81 Outline of proof 171 Parallel lines 177 Parallel planes 631 Parallelogram 220 altitude of 228 base of 228 Parallelograms classified . 223, 243 Parallelopiped 739 rectangular 741 right 740 Pentagon 92 Perimeter of polygon .... 85 Perpendicular 45, 166, 619, 620, 672 foot of 621 Perpendicular planes .... 672 Pi , 554 evaluation of 568 Plane 32, 33, 34 determined 608 ' '* perpendicular to line . . • 620 Art. Plane, tangent to cone .... 866 tangent to cylinder .... 850 tangent to sphere . . . .921 Plane angle of dihedral angle . 670 Plane figure 36 Plane geometry 36 Planes, parallel 631 perpendicular 672 postulate of 615 Point 5, 6, 11 determined 26 of tangency 286 Polar distance of circle . . .911 Polar triangle 943 Poles of circle 907 Polygon 84 angles of 86 base of 99 circumscribed 317 diagonal of 88 equiangular 90 equilateral . - 89 exterior angles of .... 87 inscribed 299 perimeter of 85 regular 91, 515 sides of 84 spherical_-^_-.v__!__i_, • • .936 ^.-vertices of . . . . "" . . . 84 "Polygons, mutually equiangular 417 sides proportional . . . .418 similar 419 Polyhedral angle 692 convex 697 dihedral angles of ... . 694 edges of 694 element of 694 face angles of ..... 694 faces of ....... 694 parts of : 695 vertex of . 693 Polyhedral angles, symmetrical 707 vertical 708 Polyhedron 716 circumscribed about sphere . 926 diagonal of 718 edges of 717 faces of 717 inscribed in sphere . . . 928 regular 720 vertices of 717 Polyhedrons, radially placed . 1024 similar 811, 1022 Pons asinorum 114 Portraits Archimedes 542 Descartes 237 Euclid 114 Gauss 520 486 INDEX Art. Portraits Plato 787 Pythagoras 510 Thales . 371 Postulate 15 circle 122 of planes 615 parallel line .... 178, 179 revolution ... 40, 54, 606 straight line 24, 54 transference 16, 54 Prism 726 altitude of 733 bases of 727 circumscribed about cylinder 852 inscribed in cylinder . . .851 lateral area of 760 lateral edges of 727 lateral faces of 727 oblique 731 quadrangular 732 regular 730 right 729 right section of 728 triangular 732 truncated . 736 Prismatic surface 724 Prismatoid 1019 altitude of 1020 Problem defined 59 Problems of computation ... 59 Problems of construction . . . 59 analysis of . . 151, 152, 176, (Ex. 187), 274, 499 discussion of 123 proof of 123 solution of 123 Product of lines 425, 475, 476, 511 Projection, of line . . . 451, 656 of point 450, 655 Proof 79, 123 analytic method of ... . 275 by exclusion .... 159, 161 by successive substitutions 275 indirect . . . . . 159, 161 necessity for 81 outline of 171 reductio ad absurdum . 159, 161 synthetic method of . . . 275 Proportion 382 analysis of 397 antecedents of 384 by alternation 396 by composition 398 by composition and division 400 by division 399 by inversion 395 consequents of 384 continued .,.,,, 405 Art. Proportion, extremes of . . . 385 means of 385 terms of ....... 383 ways of writing .... * 382 Proportional, fourth .... 387 mean 386 third 386 Proportions simplified . . 389, 397 Proposition 56 Pyramid 747 altitude of 751 base of 748 circumscribed about cone . 868 frustum of 754 inscribed in cone .... 867 lateral area of 760 lateral edges of 748 lateral faces of 748 quadrangular 749 regular . 752 slant height of 764 spherical 1010 triangular 749, 750 truncated . 753 vertex of . 748 Pyramidal surface 744 Quadrangular prism .... 732 Quadrangular pyramid . . . 749 Quadrant 295 Quadrilateral 92 Quadrilaterals classified . 220, 243 Radially placed polyhedrons . 1024 Radius, of circle . . . 119, 278 of regular polygon .... 532 of sphere 901 Ratio .... 340-343, 382, 384 antecedent of 384 consequent of 384 extreme and mean .... 464, 465 (Ex. 763), 526 of any two surfaces . . .472 of similitude .... 418, 1023 of two commensurables . . 342 of two incommensurables . 343 of two magnitudes . . . .341 of two solids 775 Rectangle 223 area of 468-471 of two lines 425, 475, 476, 511 Rectangular parallel opiped . -741 Rectilinear figure 37 Reductio ad absurdum . 159, 161 Regular polygon . . . . 91, 515 angle at center of ... . 534 apothem of 533 center of 531 radius of 532 Regular polyhedron , . , , 720 INDEX 487 Art. Regular prism 730 Regular pyramid 752 Related variables . . . 352, 594 Rhomboid 224 Rhombus 226 Right circular cone 843 axis of 844 lateral area of 872 slant height of 865 Right circular cylinder . . . 832 lateral area of 857 Right cylinder 823 Right parallelopiped .... 740 Right prism 729 Right section, of cylinder . . 830 of prism . 728 Right triangle 96 arms of 96 hypotenuse of 96 sides of 96 Ruler, use of ...... 31 Scale, drawing to 437 Scalene triangle 93 Secant 285 length of 426 Sector 287, 564 spherical 1002 Segment, of circle 288 of line 27 of sphere 1015 Segments, added 31 collinear 28 difference of 31 of a line 406 similar 565 subtracted 31 sum of 31 Semicircle 289 Semicircumference .... 290 Series of equal ratios . . 401, 405 Sides, of angle 38 of polygon 84 Similar arcs . 565 Similar cones of revolution . . 877 Similar cylinders of revolution . 863 Similar polygons 419 Similar polyhedrons . .811, 1022 Similar sectors 565 Similar segments 565 Similarity of triangles . . . .431 Similitude, ratio of . . . 418, 1023 Slant height, of frustum of cone . 881 of frustum of pyramid . . 765 of pyramid 764 of right circular cone . . . 865 Small circle of sphere .... 905 •Solid, geometric . . . . 2, 9, 11 $olid angle 692 Art. Solid geometry 602 Solution, of exercises .... 275 of locus problems .... 143 of problems 123 of theorems 79 Sphere 900 center of 901 circumscribed about poly- hedron 929 diameter of 901 great circle of 904 inscribed in polyhedron . . 927 line tangent to 921 plane tangent to .... 921 radius of 901 small circle of 905 surface of 970 volume of 996,c Spheres, tangent externally . . 922 tangent internally .... 922 tangent to each other . . . 922 Spherical angle 917 Spherical cone 1003 Spherical degree 983 Spherical excess 989 Spherical polygon 936 angles of 936 diagonal of 937 sides of 936 vertices of 936 Spherical polygons, symmetrical 956 Spherical pyramid .... 1010 base of 1010 vertex of 1010 Spherical sector 1002 base of 1002 Spherical segment . . . . 1015 altitude of 1016 bases of 1015 of one base 1016 Spherical surface 970 Spherical triangle 938 birectangular 952 trirectangular 953 Spherical wedge 1006 angle of 1006 base of 1006 Square 225 Squaring the circle 569 Straight angle 69 Straight line 21, 23 determined 25 oblique to plane 630 parallel to plane .... 629 perpendicular to plane . . 619 Straightedge 21 Summary, of divisions of cir- cumference 529 of equal triangle theorems . 118 488 INDEX Akt. Summary, of formulas of plane geometry 570 of parallel line theorems . . 197 of quadrilaterals .... 243 of similar triangle theorems . 431 of trapezoids . . Exs. 320, 321 of unequal angle theorems . 174 of unequal line theorems . . 175 Superposition .... 19, 103, 106 Supplemental triangles . . . 948 Supplementary angles .... 69 Supplementary-adjacent angles . 72 Surface 3, 8, 11 closed 714 conica 837 curved 35 cylindrical 819 of sphere 970 plane 32, 33, 34 prismatic 724 pyramidal 744 Surfaces 32 Symmetrical polyhedral angles . 707 Symmetrical spherical polygons . 956 Tangent 286 external common .... 333 internal common .... 333 length of 319 Tetrahedral angle 698 Tetrahedron 719, 750 Theorem 57 converse of 135 opposite of 136 solution of 79 Third proportional . . . . . 386 Transformation . . ... . . 493 Transversal 181 Trapezium 222 Trapezoid . . ... . . . .221 altitude of 229 bases of 229 isosceles 227 median of 251 Triangle 92 acute 98 altitudes of 100 bisectors of angles of . . .126 centroid of 265 construction of 269 determined .... 133, 269 equilateral 95 isosceles 94 median center of .... 265 medians of 102 notation of 270 obtuse 97 parts of 269 Art. Triangle, polar 943 right 96 scalene 93 spherical - . 938 vertex angle of .... 94, 99 vertex of 99 Triangles, classified .... 93-98 similarity of 431 supplemental 948 Triangular prism 732 Triangular pyramid . . . 749, 750 Trihedral angle 698 birectangular 699 isosceles 700 rectangular 699 trirectangular 699 Trirectangular spherical triangle 953 Truncated prism 736 Truncated pyramid .... 753 Unit, of measure .... 335, 466 of volume 769 Variable . . . 346, 348, 585-594 approaches its limit . . . 350 independent 346 limit of 349 reaches its limit 351 related 352, 594 Vertex, of angle 38 of cone 840 of conical surface .... 838 of polyhedral angle . . . 693 of pyramid 748 of pyramidal surface . . . 745 of spherical pyramid . . 1010 of triangle 99 Vertex angle of triangle . . 94, 99 Vertical angles 70_ Vertical polyhedral angles . . 708 Vertices, of lune 983 of polygon 84 of polyhe ron 717 of spherical polygon . . . 936 Volume, of cone 892, b of cylinder . . . . . .888 of rectangular parallelopiped 774 of solid 769, 770 of sphere 996, c unit of . 769 Wedge, spherical 1006 Zone 975 altitude of 976 bast of 975 of one base 977 MILNE'S STANDARD. ALGEBRA By WILLIAM J. MILNE, Ph.D., LL.D., President of the New York State Normal College, Albany, N. Y. $I.OO THE Standard Algebra conforms to the most recent courses of study. The inductive method of presentation is followed, but declarative statements and observations are used, instead of questions. Added to this kind of unfold- ing and development of the subject are illustrative problems and explanations to bring out specific points, the whole being driven home by varied and abundant practice. ^f The problems are fresh in character, and besides the tradi- tional problems include a large number drawn from physics, geometry, and commercial life. They are classified accord- ing to the nature of the equations involved, not according to subject matter. The statement of necessary definitions and of principles is clear and concise, but the proofs of principles, except some important ones, are left for the maturer years of the pupil. ^[ Accuracy and self-reliance are encouraged by the use of numerous checks and tests, and by the requirement that re- sults be verified. The subject of graphs is treated after simple equations, introduced by some of their simple uses in repre- senting statistics, and in picturing two related quantities in the process of change, and again after quadratics. Later they are utilized in discussing the values of quadratic expressions. Factoring receives particular attention. Not only are the usual cases given fully and completely with plenty of practice, but the factor theorem is taught. ^f The helpful and frequent reviews are made up of pointed oral questions, abstract exercises, problems, and recent college entrance examination questions. The book is unusually handy in size and convenient for the pocket. The page size is small. ROBBINS'S PLANE TRIGONOMETRY By EDWARD R. ROBBINS, Senior Mathematical Mas- ter, William Perm Charter School, Philadelphia, Pa. $0.6o THIS book is intended for beginners. It aims to give a thorough familiarity with the essential truths, and a satisfactory skill in operating with those processes. It is illustrated in the usual maimer, but the diagrams are more than usually clear-cut and elucidating. ^[ The work is sound and teachable, and is written in clear and concise language, in a style that makes it easily under- stood. Immediately after each principle has been proved, it is applied first in illustrative examples, and then further im- pressed by numerous exercises. Accuracy and rigor of treat- ment are shown in every detail, and all irrelevant and ex- traneous matter is excluded, thus giving greater prominence to universal rules and formulas. •fj The references to Plane Geometry preceding the first chapter are invaluable. A knowledge of the principles of geometry needed in trigonometry is, as a rule, too freely taken for granted. The author gives at the beginning of the book a statement of the applied principles, with reference to the sections of his Geometry, where such theorems are proved in full. Cross references in the text of the Trigonometry to those theorems make it easy for the pupil to review or to supplement imperfect knowledge. ^| Due emphasis is given to the theoretical as well as to the practical applications of the science. The number of ex- amples, both concrete and abstract, is far in excess of those in other books on the market. This book contains four times as many exercises as most books, and twice as many as that having the next lowest number. AMERICAN BOOK COMPANY (313) f ^- UNIVERSITY OF CALIFORNIA LIBRARY This book is DUE on the last date stamped below. OCT 7 1947 5Mar4oHJ 31 May 49 rp OCT 6 SEP 7 1980 BY 1980 •WAr/o*j r^ W LD 21-100m-12,*46(A2012sl6)4120 it c. ^tiB^s c ti3 5 0<