Digitized by tine Internet Archive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsofplanesOObowsrich THE ELEMENTS OF ^ Plane and Solid Geometey. WITH NUMEEOUS EXERCISES. BY EDWARD A. BOWSER, LL.D, Professor of Mathematics and Ekginekring in Rutgers College. ( vT' NEW YORK: D. VAN NOSTRAND COMPANY, Publishers, 23 MuRiiAY AND 27 WARiiEif Stueets, 1890. CopyRiOHT, 1890, By E. a. Bowsk^. PREFACE. "TN the present work on Elementary Geometry, or what is com- -*- monly known as the Euclidian Geometry, it is aimed to com- bine the excellencies of Euclid with those of the best modern writers, especially of Legendre, and Rouche and Comberousse. Many of the demonstrations are those of Euclid, with minor changes frequently introduced, and the syllogistic form is retained through- out ; but the arrangement is quite different. Many objections have been made against Euclid. His definitions are not all of them the best, nor are they in their proper places. His treatment of angles is deficient. His arrangement of the propo- sitions is often poor, mixing straight lines, angles, triangles, etc., without any regular classification ; and his demonstrations are sometimes cumbersome and prolix. Nevertheless, although numerous attempts have been made to improve upon Euclid, it still remains the great model, the unrivalled original, on which is founded the whole system of elementary Geometry. Perhaps a more finished specimen of exact logic has never been produced than that of the old Greek Geometer. In the present treatise it is desired to effect two objects: (1) to teach geometric truths ; (2) to discipline and invigorate the mind, to train it to habits of clear and consecutive reasoning. Accord- ingly, more numerous propositions have been given, and the demonstrations made more complete, than either object alone would seem to demand. In each proposition is a distinct statement, of what is given, of what is required, and of the proof. Each assertion in the proof begins a new line, and is accompanied by a reference to the preced- ing principle on which the assertion depends. These references are quoted two or'^three times in small type, and afterwards referred to only by number. The student should alicays be ready, if required, to quote the proper reference, and to show its application. The text is so arranged that the enunciation, figure, and proof of each proposition 183623 "' IV PREFACE. are in view together ; and notes are directly appended to the propo- sitions to which they refer. A few symbols and abbreviations of words have been freely used, but only such as have long been employed by mathematicians, and are recognized by the majority of teachers. In the figures the given lines are represented by full lines, those which are added to aid in the demonstration, by sfwrt-dotted lines, and the remlting lines by dashed, or long-dotted lines. In solid geometry the dotted lines represent those which a solid body would conceal. The propositions marked with a star may be omitted without interfering with the continuity of the work, but the omission is not recommended. The exercises distributed through the text are quite easy, and may all be worked out by the average student ; those at the end of each book are more advanced, and have been carefully graded, with hints appended to many of the more diflScult ones. It is only in original demonstration that the student can acquire mental power. More discipline is gained in working out one demonstration, without aid, than by learning a number of them that are given by others. A student can never really compreliend a subject if he only tries to understand and remember what the book says. The subject can become known to him only by his thinking upon it. To develop the power of independent thought in the student, is the most important part of the teacher's work, and it is the most difficult. In preparing this work I have consulted some of the best Ameri- can and English books on Euclidian Geometry, and am especially indebted to the text-book recently published bj' the English Asso- ciation for the Improvement of Geometrical Teaching. I have also derived assistance from a number of French works, especially from those of Catalan, Briot, and Legendre. while the Traite de Geomelrie by Rouche and De Comberousse has been my constant companion. It remains for me to express my thanks to Prof. R. W. Prentiss, of the Nautical Almanac Office, for reading the MS., and to Mr. I. S. Upson, the College Librarian, for reading the proof sheets. E. A. B, Rutgers College, ) New Brunswick, N. J., May, 1890- \ TABLE OF CONTENTS. INTRODUCTION. PAGE Definitions » 1 Straight Lines 4 Plane Angles 5 Angular Measure 8 Superposition 10 Postulates and Axioms 12 Symbols and Abbreviations = o 13 PLANE GEOMETRY, BOOK I. RECTILINEAR FIGURES. Perpendicular and Oblique Lines 14 Parallel Lines : , 23 Triangles 33 Quadrilaterals 46 Polygons 52 Miscellaneous Theorems 56 Symmetry 64 Exercises. 67 BOOK 11. THE CIRCLE. Definitions 74 Arcs and Chords 78 Relative Position of Two Circles. ... 88 Measurement of Angles 90 Quadrilaterals , 102 Problems of Construction. 104 Exercises. Theorems 123 Loci 127 Problems 129 Y VI TABLE OF C0NTENT8. BOOK III. RATIO AND PROPORTION : SIMILAR FIGURES. PAGE Ratio and Proportion 134 Proportional Lines 143 Similar Figures 149 Similar Triangles 150 Similar Polygons 155 Numerical Relations between the Different Parts of a Triangle 158 Problems of Construction 168 Applications 173 Exercises. Theorems 175 Problems 178 BOOK IV. AREAS OF POLYGONS. Measurement of Areas 18q Comparison of Areas 187 Problems of Construction jgi Applications 200 Exercises. Theorems 201 Problems 206 BOOK V. REGULAR POLYGONS. THE CIRCLE. MAXIMA AND MINIMA. Regular Polygons 208 The Measure of the Circle , 215 Principle of Limits 216 Problems of Construction 224 Problems of Computation 229 Value of Tt. Method of Perimeters 231 Maxima and Minima 234 Exercises. Theorems , 243 Numerical Exercises 245 Problems _ . 247 Exercises in Maxima and Miniina 248 TABLE OF CONTENTS. VU SOLID GEOMETRY, BOOK VI. PLANE AND SOLID ANGLES. PAGE Definitions 250 Lines and Planes 252 Parallel Planes 258 Diedral Angles. Definitions 264 Polyedral Angles. Definition'^ 276 Exercises. Theorems 282 Loci. 284 Problems 285 BOOK vn. rOLYEDRONS. Prisms and Parallelopipeds. Definitions 286 Pyramids 302 Similar Polyedrons 316 Regnlar Polyedrons 319 General Properties of Polyedrons 325 Exercises. Theorems 327 Numerical Exercises 329 Problems 333 BOOK VIII. THE SPHERE. Circles of the Sphere and Tangent Planes 334 Spherical Triangles and Polygons 342 Relation of a Spherical Polygon to a Polyedral Angle ... 344 Symmetrical Spherical Triangles 345 Polar Triangles 349 Relative Areas of Spherical Figures 354 Exercises. Theorems 360 Numerical Exercises 361 Problems 363 BOOK IX. THE THREE ROUND BODIES. The Cylinder 366 The Cone 371 The Sphere 878 Exercises. Theorems. 386 Numerical Exercises 388 OF fhi ELEMENTARY GEOMETRY. INTEODUCTION. Definitions. 1. Space is indefinite extension in every direction. All material bodies occupy limited portions of space, and have length, hreadth, thick nes'i, for m^ smd positj^on. The mate- rial body occupying any portion of space is called a physi- cal solid. The part of space which is or may be occupied by a material body is called a geometric solid. A physical solid is therefore a 7-eal body, while a geometric solid is only the fonii of a physical solid, and is the one treated of in Geometry. The term solid will be used for brevity to de- note a geometric solid. 2. A solid is a limited portion of space, and has length, hreadth, and thickness. Length, hreadth, and thickness are called the three dimensions of the solid. 3. A surface is the limit or boundary of a solid, and has only two dimensions, length and breadth. A surface has no thickness, for if it had any, however small, it would form part of the solid, and would be space of three dimen- sions. 4. A li7ie is the limit or boundary of a surface, and has only one dimension, namely, length. A line has no breadth, for if it had any, however small, it would form part of the surface, and would be space of two dimensions; and if in addition it had any thickness, it would be space of three dimensions; hence a line has neither breadth nor thickness. 1 2 OEOMETRT. 5. A point is the limit or extremity of a line, and has position, but neither length, breadth, nor thickness. A point has no length, for if it had any, however small, it would form part of the line of which it is the extremity; and it can have neither breadth nor thickness because the line has none. 6. If we suppose a solid to be divided into two parts which touch each other, the division between the two parts is a surface. This surface can have no thickness, for if it had a thickness, however small, it would be a part either of the one solid or the other, and would therefore be a solid and not a surface. Again, if we suppose a surface cut into two parts which touch each other, the division between the two parts is a line. This line can have no thickness, because the surface has none, and it can have no breadth, for it forms no part of either surface. If we suppose a line cut into two parts which touch each other, the division between the two parts is a point. This point can have neither breadth nor thickness, because the line has none, and it can have no length, for it forms no part of either line. Euclid regarded a point merely as a mark of position, and he at- tached to it no idea of size and shape. Similarly, he considered that the properties of a line arise only from its length and position, without reference to that minute breadth which every line must really have if actually drawn, even though the most perfect instruments are used. We cannot make the points, lines, and surfaces of Geometry. A dot, made on paper or on the blackboard, will have length, breadth, and thickness, and hence will not be a real point. Yet the dot may be taken as an imperfect representation of the real point. So also a line, drawn on paper or on the blackboard, will have breadth and thickness, and hence will not be a real line. Yet the line which we draw may be taken to represent the real line. 7. We have considered a surface as the boundary of a, solid, a line as the boundary of a surface, and a point as the limit of a line. On the other hand, inversely, we may re- INTRODUCTION'. 3 gard a line as generated by the motion of a point, a surface as generated by the motion of a line, and a solid as generated by the motion of a surface. Again, each of these may be re- garded in a purely abstract manner, distinct from each other. Thus, we may suppose a surface to exist in space sepa- rately from the solid whose boundary it forms, and to be of iDilimited extent. Similarly, we may suppose a line to exist in space sepa- rately from the surface whose boundary it forms, and to be of tinlimited lengtli. Likewise we may suppose a point to exist in space sepa- rately from the line, and to have only position. The points, lines, surfaces, and solids of Geometry are called geometric points, lines, surfaces, and solids. 8. A draight line, or rigid line, is one which has the same direction at A - B every point, as the line AB. 9. A curved line is one no part of which is straight, but changes its C^^^^' D direction at every point, as the line CD. 10. A broken line is a line made up of different successive straight ^ p. , lines, as the line EF. The word line, used alone, signifies a straight line; and the word curve, a curved line. 11. K plane surface, or, simply, a plane, is a surface in which the right line joining any two points in it lies wholly in the surface. 12. A curved surface is one no part of which is plane. 13. A figure is any definite combination of points, lines, surfaces, or solids. A plane figure is one formed of points and lines in a plane. If the figure is formed of right lines only, it is called a rectilinear, or right-lined, figure. The figure of a solid depends upon the relative position 4 GEOMETRY. of the points in its surface. Lines, surfaces, and solids are the geometric figures. When the extent of lines, surfaces, and solids is considered they are called magnitudes, but when their /o/-m or sJiape is considered they are csiWed figures. 14. Geometry is the science which treats of magnitude, form, and position. Thus it is the province of Geometry to investigate the properties of solids, of surfaces, and of the figures constructed on surfaces. Plane Geometry treats of plane figures. Solid Geometry, called also Geometry of Space and Geom- etry of TJiree Dimensions, treats of solids, of curved sur- faces, and of the figures described on curved surfaces. Straight Lines. 15. K finite straight line is a straight line contained be- tween two definite points which are its extremities. When a straight line is produced indefinitely it is called an in- definite straight line. Any finite straight line may be sup- posed at anytime to be produced into an indefinite straight line. Two finite straight lines are said to be equal, or of equal length, when the extremities of the one line can be made to coincide respectively with the extremities of the other. If any line, as OB, be produced through to A, the parts OB and OA A^ -^ B are said to have opposite dii'ections Fig. 2 from the common point 0. Every straight line AB has two opposite directions, the one from A toward B, expressed by *' the line AB,'' and the other from B toward A, expressed by "the line BA.'' If a line BC is to be produced toward I), we should ex- press this by saying that " BC is to _ be produced "\ but if it is to be pro- B C duced toward A, we should express ^*^' ^ this by saying that '^ CB is to be produced." Straight lines are added together by ])lacing them one INTRODUCTION. O after anotlier in succession in the same straight line so that one extremity of each newly added line coincides with one extremity of the last added line, and so that no part of any newly added line coincides with any part of the last added line. Thus, AB, BO, and CD, Fig. 3, are added together and form the straight line AD. AB, BO, and CD are called the jmrts of AD, and AD is called the sum of AB, BC, and CD. Plane Akgles. 16. K plane ancjle is the opening between two straight lines drawn from the same point. ^B The straight lines are called the arms or sides of the angle, and the common point is called its vertex. Thus the q^ A lines OA, OB are said to contain, or ^'9' * include^ or form the angle at 0. When there are several angles at one point, any one of them is expressed by three letters, putting the letter at the vertex between the other two. Thus, if the straight lines OA, OB, OC meet at the point 0, the angle contained by the lines OA, OB is named the angle AOB or BOA; the angle contained by the lines OA, OC is named the angle AOC or COA; ' '"'9- ^ and the angle contained by OB, OC is named the angle BOO or COB. When there is only one angle at a point, it may be de- noted either by the single letter at that point, or by three letters as above. Thus in Fig. 4 the angle at the point may be denoted either by the angle or by AOB or by BOA. 17. Adjacent angles are angles which have a common vertex and one common arm, their non-coincident arms 6 GEOMETRY. being on opposite sides of the common arm. Thus the angles A OB and COB (Fig. 5) are adjacent angles, of which OB is the common arm. Of the two straight lines OB, 00 (Fig. 5) it is easily seen that the opening between OA and OC is greater than the opening between OA and OB. This we express by saying that the angle AOC is greater than the angle AOB. The 7nagnitude or size of an angle depends entirely upon the extent of opening between its sides, and is not altered by changing the length of its sides. 18. Angles are equal when they can be placed one upon the other so that the vertex and sides of the one can be made to coincide with the vertex and sides of the other. Thus the angles ABO and DEF are equal if ABO can be placed upon DEF so that while BA coincides with ED, BO shall also coincide with EF. 19. The angle formed by joining two or more angles together is called their sum. Angles are added together by placing them so as to be adjacent to each other. Thus the sum of the two angles ABC, PQK, is the angle ABE, formed by applying the side QP to the side BO so that the vertex Q shall fall on the vertex B, and the side QR on the opposite side of BO from BA. If the angles ABC, PQR are equal to each other, the angle ABR is doiihle either of them, and the common side BC is said to hisect the angle between the non-coincident sides BA and BR. 20. When a straight line standing on another makes the adjacent angles equal to each other, each of the angles is Fig. 6 INTRODUCTION. Fig. 8 called a right angle; and the straight line which stands on the other is said to be perpendicu- lar or at right angles to it. Thus, if the adjacent angles AOC and BOO are equal to each other, each is a right angle, and the line CO is per- pendicular to AB. The point is called the foot of the perpendicular. 21. A straight angle has its arms extending in opposite directions so as to be in the same straight line. Thus, if the arms OA, OB are in the same straight line, the angle formed by them is called a straight angle. Since the sum of the two right angles AOC and BOC (Fig. 8) is the angle AOB (19)*, a right angle is half a straight angle. 22. An acute angle is an angle which is less than a right angle, as ^^ the angle A. 23. An obtuse angle is an angle which is greater than a right angle, and less than a straight angle, as the angle BAC. 24. When the sum of two -angles is a right angle, each is called the A ^ complement of the other, and the ^'9- " two are called complementary angles. Thus, if the angle BAG is a right angle, the angles BAD, J)AC are complements of each other. 25. When the sura of two angles is a straight angle, each is called the supplement of the other, and the two are called sup- plementary adjacent angles. Thus, if the angle AOB (Fig. 9) is a straight angle, Fig. 12 the angles BOC, CO A are supplements of each other. * An Arabic uumenil in parenthesis refers to an article. 8 OEOMETBY. Hence, when one line stands on another, the two adjacent angles are supplements of each other. Hence a right angle is equal to its supplement. The supplement of an acute angle is obtuse, and, con- versely, the supplement of an obtuse angle is acute. 26. A reflex angle is an angle which is greater than a straight angle, and less than two straight angles, as the angle 0. Acute, obtuse, and reflex angles are called oblique angles, in distinction from right and straight angles ; and intersecting lines which are not per- pendicular to each other are called oblique lines. 27. Where two angles are contained between two inter- secting lines on opposite sides of the A>^ ^D vertex, they are called opposite or ver- tical angles. Thus, AOC and BOD are 6pposite or vertical angles, as also AOD and COB. Fig. 14 An^gular Measure. 28. A right line drawn from the vertex and turning about it in the plane of the angle from the position of co- incidence with one side of the angle to that of coincidence with the other side, is said to turn through the angle, and the angle is the greater as the quantity of turning is greater. Thus, suppose that the right line OP (Fig, 15) is capable of revolv- ing about the point 0, like the hands of a watch, but in the oppo- site direction, and that it has passed successively from the po- sition OA to the positions occupied by OB, 00, OE, etc. Then it is clear that the line must have done F"'g. 's more turning in passing from OA to 00 than in passing INTRODUCTION. 9 from OA to OB; and consequently the angle AOC is said to be greater than the angle AOB. When the revolving line OP turns from coincidence with OA to the position OB it is said to describe or to generate the angle AOB. When the revolving line has turned from the position OA to the position OD, perpendicular to OA, it has generated the right angle AOD ; when it has turned to the position OA', it has generated the straight angle AOA'; when it has turned to the position OF, it has gen- erated the reflex angle AOF; when it has turned entirely around to the position OA, it has generated two straight angles. Hence, the whole angle which a line must turn through, about a point in a plane, to take it around to its first position, is two straight angles, or four right angles. Again, since the revolving line may turn from one po- sition to the other in either of two b^ directions, two angles are formed by two lines drawn from a point. Thus, if OA, OB be the sides of an angle, a line may turn from the po- sition OA to the position OB about the point in either of the two directions indicated by the arrows, giving the obtuse angle AOB (marked a), and the reflex angle AOB (marked h). Angles are generally measured in degrees, minutes, and seconds. A degree is the ninetieth part of a right angle, or the three hundred and sixtieth part of four right angles. A minute is the sixtieth part of a degree; and a second is the sixtieth part of a minute. Degi-ees, minutes, and seconds are denoted by the symbols °, ', ". Thus 7 degrees, 24 minutes, and 38 seconds is written, 7° 24' 38". Hence, when the revolving line OP (Fig. 15) has turned through one-fourth of a revolution, it has generated a right angle, or 90° ; when it has made half a revolution, it has generated a straight angle, or 180° ; and when it has made \ 10 GEOMETRY. a luliole revolution, it has generated two straight angles, or 360°. Superposition. 29. The placing of one geometric magnitude on another, such as a line on a line, an angle on an angle, etc., is called superposition. The superposition employed in Geometry is only mental, that is, we conceive one magni- tude to he taken up and laid down upon the other; and then, if we can prove that they coincide, we infer that they are equal. This is the ultimate test of the equality of two geometric magnitudes. Thus, if two straight lines are to be compared, we con- ceive one to be taken up and placed on the other, and find whether their two ends can be made to coincide. If so, they are equal ; if not, they are unequal. If two angles are to be compared, we conceive one to be taken up and applied to the other. If they can be so placed that their vertices coincide in position and their sides in direction, the angles are equal (18). Superposition involves the principle* that '*any figure may be taken up, transferred from one position to another, and laid down again without change of form or size." Magnitudes which coincide with one another throughout their whole extent are said to be equal. Definitions of Terms. 30. A theorem is a truth which requires proof. 31. A prohlem is a question which requires solution, such as a particular line to be drawn, or a required figure to be constructed. 32. An axiom is a self-evident truth, which is admitted without proof. 33. K post-date assumes the possibility of solving a cer- tain problem. * Euclid makes frequeiit use of this principle, witJiQut explicitly gtatin^ it. — Casey. INTRODUCTION, 11 34. A jn'oposition is a general term for a theorem, problem, axiom, or postulate. 35. A demonstration is the course of reasoning by which we prove a theorem to be true. 36. A corollary is a conclusion which folloAvs imme- diately from a theorem. 37. A lemma is an auxiliary theorem required in the demonstration of a principal theorem. 38. A scJiolium is a remark upon one or more propo- sitions. 39. An hypothesis is a supposition made either in the enunciation of a proposition or in the course of a demon- stration. 40. A solution of a problem is the method of construc- tion which accomplishes the required end. 41. A construction is the drawing of such lines and curves as may be required to prove the truth of a theorem, or to solve a problem. 42. The enunciation of a theorem consists of two parts : the hypothesis, or that which is assumed ; and the conclusion, or that which is asserted to follow therefrom. Thus, in the typical theorem. If A is B, then C is D, the hypothesis is that A is B, and the conclusion that is D. 43. The emcnciafion of a problem consists of two parts: the data, or things supposed to be given; and the qiimsita, or things required to be done. 44. Two theorems are said to be converse, each of the other, whenj;he hypothesis of each is the conclusion of the other. Thus, If C is D, then A is B, is the converse of the typical theorem in (42). 12 GEOMETBT, Postulates. 45. Let it be granted — 1. That a straight line may be drawn from any one point to any oth'er point. 2. That a terminated straight line may be produced to any distance in a straight line. 3. That a circle may be described with any centre, at any distance from that centre. Axioms. 46. 1. Things which are equal to the same thing are equal to each other. 2. If equals be added to equals the sums will be equal. 3. If equals be taken from equals the remainders will be equal. 4. If equals be added to unequals the sums will be un- equal. 5. If equals be taken from unequals the remainders will be unequal. 6. Things which are double the same thing, or equal things, are equal to one another. 7. Things which are halves of the same thing, or of equal things, are equal to one another. 8. The whole is greater than any of its parts. 9. The whole is equal to the sum of all its parts. Geometric Axioms. 10. A straight line is the shortest distance between any two points. 11. If two straight lines have two points in comm on, they will coincide throughout their whole length, and form but one straight line. 12. Through a given point only one straight line can be drawn parallel to a given straight line. INTRODUCTION. 13 Symbols and Abbreviations. 47. The following are some of the principal symbols and abbreviations that are commonly used in Colleges and Public Schools. These symbols and abbreviations are gen- erally employed in the present book. They are recom- mended to the student for writing out the propositions and demonstrations on the blackboard or in exercise books, as their use will greatly shorten the work. -|- plus, or together with. adj adjacent. — minus, or diminished by. alt alternate. X multiplied by. ax axiom. -^ divided by. cons construction. .'. therefore. cor corollary. _ j equals, cyl cylinder. ~~ ( or is (or are) equal to. def definition. > is (or are) greater than. ext exterior. < is (or are) less than. fig figure. Z angle. hyp hypothesis. Zs angles. int interior. A triangle. opp opposite. As triangles. post postulate. £17 parallelogram. prob problem. ZZ7s parallelograms. prop proposition. ± perpendicular. pt point. ±s perpendiculars. rt right. 1 1 parallel. sim similar. I Is parallels. sq square. O circle. st straight. Os circles. sup supplementary. Oce circumference. 0ces circumferences. The words ''join AB^' are used as an abbreviation for *'draw a straight line from A to B.^' The initial letters Q. e. d., placed at the end of a theorem, stand for the Latin words duod erat Demonstrandum, meaning loliich tvas to be proved. The letters Q. E. F., placed at the end of a problem, stand for duod erat Faciendum, which was to be done. PLANE GEOMETRY. Book I. EECTILINEAR FIGURES. Perpen^dicular and Oblique Lines. Proposition 1 . Theorem. 48. All straight angles are equal to one another.* Hyp. Let AB, AC be the arms of a st. Z. whose vertex is A, and ^ J 'C DE, DF the arms of another st. Z whose vertex is D. ^ ^ ^ To prove / BAG = Z EDF. Proof. Because the Z BAG is a st. Z ^ .'. BA and AG are in the same st. line BAG. (21) Because the Z EDF is a st. Z , .-. ED and DF are in the same st. line EDF. (21) Kow if the Z BAG be applied to the Z EDF so that the vertex A shall coincide with the vertex D and the arm AB with the arm DE, then the arm AG will coincide with the arm DF, because, if two st. lines have tivo p)oints in common they will coincide throughoiit their ivhole lengthy and form hut one st. line (Ax. 11). .-. Z BAG = Z EDF. q.e.d. 49. Cor. 1. All right angles are equal to one another. \ (21), also (Ax. 7). 50. Cor. 2. The complements of equal angles are equal to each other, and the siq^plemeiits of equal angles are equal to each other. 51. Cor. ^. At a given j^oint in a given straight line only one perpendicular can he drawn to that line. * The Elements of Plane Geometry, by Association for the Improvement of Geometric Teaching, p, 20a. t This is Euclid's 11th axiom. 14 BOOK L— PERPENDICULAR AND OBLiqUE LINES. 15 Proposition 2. Theorem. 52. If one sir ai gilt line meet another straight line, the sum of the two adjacent angles ii equal to two right angles. Hyp. Let the st. line DO meet the st. line AB at C. To prove Z ACD + Z DOB = 2 rt. Zs. ,^ Proof From C draw CE at rt. Zs ^ c ^ to AB. Then Z ACD = Z ACE + Z ECD. Add Z DCB to each. .-. Z ACD+ Z DCB= Z ACE+ z ECD + Z DCB. ijf equals be added to equals, the sums will be equal {Ax. 3). Again, Z ECB ^ Z ECD + Z DCB. Add Z ACE to each. .-. Z ACE+ Z ECB= Z ACE+ Z ECD + Z DCB. If equals be added to equals, the sums will be equal {Ax. 2). Hence Z ACE+ Z ECB =r z ACD + Z DCB. (Ax. 1)* But Z ACE + Z ECB := 2 rt. Z s. . (Cons. ) .-. Z ACD+ zDCB = 2rt.Zs. q.e.d. ScH. The angles ACD, DCB are supplementary adjacent angles (25). 53. Cor. If one of the angles ACD, DCB is a right angle, the other is also a right angle. The sum of all the angles on the same side of a straight li7ie, at a cdmbmo^i point, is equal to two right angles ; for their sum is equal to the sum of the two adjacent angles ACD, DCB. ♦ Tbe student should quote every reference in full. W PLANE GEOMETRY. Proposition 3. Theorem. 54. Conversely, if the sum of two adjacent angles is equal to two right angles, their exterior sides are in oiw straight line. -B Hyp. Let Z ACD + Z DCB = 2rt. Zs. To prove that AC and CB are in one st. line. Proof. If CB be not in the same st. line as AC, let CE be in the same st. line as AC. ThenZ ACD + Z DCE = 2 rt. Zs. being sup. adj. I s (52). But z ACD+ Z DCB = 2i»t. Zs. (Hyp.) .-. Z ACD + Z DCE = Z ACD + Z DCB. (Ax. 1) Take away the common Z ACD. .-. Z DCE= ZDCB, (Ax. 3) which is impossible (Ax. 8), unless CE coincides with CB. .'. AC and CB are in one st. line. Q.E.D. EXERCISES. 1. Find the number of degrees in an angle if it is three times its complement. 2. Find the number of degrees in an angle if its com- plement and supplement are together equal to U0°. -^ ^ " BOOK I.—PEUPENI)ICULAU AND OBLiqUE LINES. 17 Proposition 4- Tlieorem. 55. If two atraiglit lines mtersect each other, the vertical angles are equal. -B Hyp, Let the st. lines AB and CD cut at E. To prove Z AEC = Z BED. Proof. Z CEA + Z CEB = 2 rt. Zs, being sup. adj. Z« (52). Z AEC + Z AED = 2 rt. Zs, being sup. adj . Z« (52). .-. Z CEA + Z CEB = z AEC + Z AED. ^^^ rt. Z« are egw«^ to one another (49). Take away the common Z AEC. .-. Z CEB = Z AED. (Ax. 3) In the same way it may be proved that Z AEC = Z BED. Q.E.D. 56. Cor. 1. If two straight lines intersect each other, the four angles which they make at the point of intersection, are together equal to four right angles. If one of the four angles is a right angle, the other three are right angles, and the lines are mutually perpendicular to each other. 57. Cor. 2. If any numher of straight lines meet at a point, the sjim of all the angles made by consecutive lines, in a plane, is equal to four right angles. 18 PLANE GEOMETRY. I Proposition 5. Theorem. 58. From a i^oint without a straight line only one 'per- pendicular can he drawn to that line; and this perpen- dicular is the shortest distance from the point to the line. Hyp, Let AB be the given st, line, P the point without the line, PC a ± from P to AB, and PD any other line from P to AB. (1) To prove that PD is not ± to AB. Proof. Let the part of the plane above the line AB, and which con- tains P, be revolved about AB as \' an axis until the point P comes P into the position P'. Then, since Z ACP is a rt. Z , Z ACP' is also a rt. Z . (29) .-. PCP' is a St. line. (54) If Z ADP is a rt. Z , Z ADP' is also a rt. Z . (29) .-. PDP' is a St. line; (54) that is, between two points there are two straight lines, which is impossible. (Ax. 11) .-. Z ADP is not a rt. Z , and .-. PD is not ± to AB. (20) .*. PC is the only ± to AB from the point P. (2) To prove PC < PD. Proof. Since in the revolution of CPD about AB as an axis, the point P comes into the position P', .-. PC = P'C, and PD = P'D; .-. PC + P'C = 2 PC, and PD + DP' = 2 PD. PC + P'C < PD + DP'. .•.2PC<2PD, and.-. PC < PD. But (29) (Ax. 2) (Ax. 2) (Ax. 10) Q.E.D. 59. ScH. By the distance of a point from a line is meant the shortest distance, that is, the length of the perpendicu- lar from the point to the line. BOOK I.-PERPENDICULAR AND OBLIQUE LINES. 19 Proposition 6. Theorem. 60. Two oblique lines drawn from a j)oint to a straight line, cutting off equal distances from tlie foot of the perpen- dicular, are equaL Hyp. Let CD be the ± from C to the line EF, and CA and CB two oblique lines so that AD = DB. Tojirove CA = CB. Proof, Let the part CAD be revolved about CD as an axis until it comes into the plane of CDB. Because Z CDA ^ Z CDB = rt. Z , .*. DE will take the direction of DF. Because DA = DB, the point A will fall upon the point B. .•.CA = CB. (Hyp.) (Hyp.) (Ax. 11) Q.E.D. 61. Cqr. Two equal ohlique lines draion from a point to a straight line, make equal angles with that line, and cut off equal distances from the foot of the perpendicular^ 20 PLANE GEOMETRY. Proposition 7. Theorem. 62. If from any point two lines he drawn to the ends of a straight line, their sum will he greater than the sum of tiuo other lines similarly draion, hut included hy them. ' Hyp. Let AB, AC be two lines drawn from the point A to the ends of the line BC, and DB, DC two lines similarly drawn, but included by AB, AC. To prove AB + AC > DB + DC. Proof Produce BD to meet AC at E. Then BA + AE > BE. (Ax. 10) Add EC to each. .-. BA + AC > BE + EC. (Ax. 2) Again, DE + EC> DC. (Ax. 10) Add BD to each. . • . BE + EC > BD + DC. (Ax. 2) Much more . • . BA + AC > BD + DC. q.e.d. EXERCISES. 1. Write out in full the proof of the second part of Prop. 4, that the angle AEC is equal to the angle BED. 2. Prove that the straight line which bisects the angle AEC, in Prop. 4, bisects also the vertical angle BED. BOOK L-PERPENDICULAB AND OBLIQUE LINES. 21 Proposition 8. Theorem. 63. Of two oblique lines draiun from the same point to the sa7ne straight line, that lohich meets the line at the greater distance from the foot of the perpendicular is the greater. P Hyp. Let PC be _L from P to AB, and PE, PD two oblique lines so that CE > CD. Toj^rove PE > PD. Proof Produce PC to P', making CP' = PC. Join DP', EP'. Because AB is ± to PP' at its middle point, .-. PD := DP', and PE == EP'. (60) But PE + EP' > PD + DP'. (62) .•.2PE>2PD, .-. PE>PD. If tliQ two oblique lines are on opposite sides of PC, as PE and PD', and if CE > CD', take CD = CD', and join PD. Then PD = PD'. (60) But PE > PD, as just proved. .-. PE > PD'. Q.E.D. 64. CoR. 1. Only two equal straiglit lines can he draion from a point- to a straight line, 65. Cor. 2. Of two unequal oblique lines, the greater exits off the greater distance from the foot of the perpen- dicnlar. 22 PLANE GEOMETRY. Proposition 9. Tiieorem. 1^ 66. 1. Every ijoint m the perpendicular erected at the middle of a straight line is equally distant from the e.c- t7'emities of the line. \ 2. Every point without the perpendicular is unequal'ty \ distant from the extremities of the line. Hyp. Let DO be 1 to AB at its middle point G, P any point in DO, and any point without DO. Draw PA, PB, and OA, OB. (1) To prove PA = PB. Proof Because DO 1 to AB, and AO = OB, . • . PA = PB. (2) To prove OA > OB. Proof. OA will cut the 1 DO at E; join EB. Then OB < OE + EB. But EB = EA. .-. OB AB - AC. q.e.d. BOOK L—TBIAN0LE8. Proposition 20. Theorem. 97. The sum of the three angles of any triangle is equal to two right angles. Hyp, Let ABO be any A. C ,,£ To prove ZA + zABC+ZC = 2rt.Zs. Proof. Produce AB to T), and draw BE |) to AC. Then since BE is I| to AC, . • . ext. Z EBD = int. Z A, (77) and ZCBE = alt. ZC (72) Add ZABCtoeach. . • . Z ABC+ Z CBE+ Z EBD = Z ABC+ Z C+ Z A. (Ax. 2) But ZABC + ZCBE + ZEBD = 2rt. Zs. The sum of all the /_son the same side of a st. line at a point = 2rt. Zs (53). .-. ZABC+ZC+ZA = 2rt. Zs. (Ax. 1) Q.E.D. 98. CoE. 1. Since Z EBD = Z A, and Z CBE = Z C, .•.ZCBD=ZA-hZC. Hence^ the exterior angle of a triangle is equal to the sum of the two opjjosite interior angles. 99. CoE. 2. If two a7igles of a triangle are given, or merely their sum, the third angle can he found hy subtract- ing this sum from two right angles. 100. CoE. 3. If two triangles have iivo angles of the one equal to two angles of the other, the third angles are equal. 101. CoE. 4. A triangle can have hut one right angle, or hit one obtuse angle. 102. Coe". 5. In any right-angled triangle the two acute angles are complementary. 103. CoE. 6. Each angle of an equiangular triarigle t> two-thirds of a right angle. 36 PLANE GEOMETRY. Proposition 21. TFieorem. 104. Two triangles are equal when two sides and the included angle of the one are equal respectively to two sides and the included angle of the other. Hyp, Let ABC, DEF be two AS, having AB = DE, AC = DF. ZA = ZD. To prove A ABC = A DEF. Proof, Apply the A ABC to the A DEF (29) so that the point A shall fall on D, and the side AB on DE. Then because AB = DE (Hyp.), . • . pt. B must fall on E. Because Z A = Z D (Hyp.). . • . AC must lie along DF. And because AC = DF (Hyp.), . * . pt. C must fall on F. Now, since B falls on E, and C on F, . •. BC must coincide with EF. (Ax. 11) Hence, since BC coincides with EF, . • . BC = EF, Z B = Z E, Z C = Z F. Therefore the two As coincide in all their parts, and hence are equal (29). ' Q.E.D. Note.— Wlien all the parts of one triangle are respectively equal to all the parts of another triangle, the triangles are said to be "equal in all respects." Such triangles are said to be identically equal, or congruent. This proposition is proved by " the method of superposition," i.e., it is shown that one of the triangles could be placed on the other so as to cover it exactly without overlapping; and then, since all their parts would coincide, the two tri- angles are equal by definition (29). BOOK I.— TRIANGLES. 37 Proposition 22. Theorem. 105. Ttvo triangles are equal when a side and tlie two 1 1 adjacent angles of the one are eqnal respectively to a side 11 and the two adjacent angles of the other, til Hyp, Let ABC, DEF be two ' AS having ZA=:ZD, ZB = Z E, AB = DE. To prove A ABC = A DEF. Proof. Apply the A ABC to the A DEF so that the point A shall fall on D, and the side AB onDE. Then, (Hyp.) Because Because (Hyp.) (Hyp.) because AB = DE, . • . pt. B must fall on E. Z A = Z D, . • . AC must lie along DF. Z B = Z E, .•. BC must lie along EF. Because AC and BC fall upon DF and EF, respectively, . • . the point C, falling upon both lines DF and EF, must fall at their point of intersection F. . • . the two A s coincide in all their parts, and are equal. Q.E.D. 106. Cor. 1. Tiuo right-angled triangles are equal luhen the hypotemise a7id. a7i acute angle of the one are equal respectively to the hypotenuse and an acute angle of the other, 107. Cor. 2. Two right-angled triangles are equal when a side and an acute angle of the one are equal respectively to a side ayid homologous acute a7igle of the other. 38 PLANE GEOMETRY, Proposition 23. Theorem. 108. Two triangles are equal if the three sides of the me are equal respectively to the three sides of the other, Hijjp, Let ABC, DEF be two AS having AB = DE, AC = DE, BC = EF. To prove A ABC = A DEE. Proof Apply the A ABC to the A DEE so that the side AB shall coincide with its equal DE, and the vertex C fall at F' on the opposite side of DE from F. Join EF' which cuts DE at H. Because DF = DE', and EF = EF', (Hyp.) .*. points D and E are equally distant from F and F'. .-. DE is _L to EF' at its middle pt. H. (67) Now revolve the A DEF about DE as an axis till it comes into the plane of the A DEE'. Because / DHF = /_ DUE' = a rt. Z (67) and HF = HE', .-. pt. F will fall on F', . • . the two A s coincide in all their parts, and are equal. Q.E.D. 109. ScH. When two triangles are equal, the equal angles lie opposite the equal sides; and conversely, the equal sides lie opposite the equal angles. Thus, the angles A and D are equal or homologous angles (95). Also the angles C and F are homologous; likewise the angles I^ and E. ^Krhypotenuse and a side of the one are equal respectively to the hypotenuse and a side of the other. BOOK L— TRIANGLES. 39 Proposition 24. Theorem. I Hyp, Let ABC, DEF be two rt. A s having hyp. AC = DF, and BC = EF. To prove A ABC = A DEF. Proof Apply the A ABC to the A DEF so that BC shall coincide with its equal EF. Then, since Z B = Z E = a rt. Z , .*. BA will lie along ED, and pt, A will fall somewhere on ED. But the equal oblique lines CA and FD cut off on ED equal distances from the foot of the J_ EF (61). .-. pt. A will fall at D. .-. the two A s coincide in all their parts, and are equal. Q.E.D. EXERCISES, 1. If BC, the base of an isosceles triangle, ABC, is pro- duced to any point D, show that AD is greater than either of the equal sides. 2. Prove that the sum of the distances of any point from the three vertices of a triangle is greater than half its perimeter. 3. If one of the acute angles of a right triangle is 40° 14' 48", what is the value of the other acute angle ? 40 PLANE GEOMETRY. Proposition 25. Theorem. 111. In an isosceles triangle the angles opjjosite the equal sides are equal. Hyp. Let ABC be an isosceles A having AC = BC. To prove Z A = Z B. Proof. Draw the line CD from the vertex C, to the middle pt. D of the base. Then, in the As ADC and BDC, because AC = BC, AD = DB, (Hyp.) (Cons.) CD = CD, (being common to both As,) .-. A ADC = A BDC. Two AS are equal if the three sides are equal each to each (108). { .'. Z A = Z B, being opposite equal sides (109). Q.E.D. 112. Cor. 1. Since A ADC = A BDC, we have Z ACD = Z BCD, and Z ADC = Z BDC. Therefore, the straight line which joins the vertex to the middle jjoint of the hase of an isosceles triangle, is at right a?igles to the base, and bisects the vertical angle. Hence, also, the bisector of the vertical angle of an isosceles triangle bisects the base at right angles. 113. Cor. 2. TJie j)erpendicular from the vertex to the base of an isosceles triangle bisects the base and the angle at the vertex. An equilateral triangle is also equiangular. BOOK L-T1UANGLE8. 41 Proposition 26. Theorem. 114. Conversely, if two angles of a triangle are equal, the sides opposite the equal angles are equal, and the triangle IS isoceles. Htjp. Let ABC be a A having ZA=ZB. To prove AC = BC. Proof. Draw CD, bisecting the Z ACB. Then, in As ACD, BCD, Also, ZA=. ZB. (Hyp.) ZACD:= ZBCD. (Cons.) .-. ZADC= ZBDC. (100) CD is common. .-. aACD= a BCD, having a side and the two adjacent angles equal, each to eoAili (105). . • . AC == BC, Jyeing homologous sides of equal as (109). Q.E.D. 115. Coil. Au equiangular triangle is also equilateral. EXERCISES. 1. ABC is an isosceles triangle, with AB = AC. The bisectors of the angles B and C meet at 0. Prove that CO = BO. - 2. ABC is a triangle ; BA is produced to D so that AD = AC, and DC is joined. Prove that Z BCD > Z BDO. 3. The angle C is twice as large as either of the angles A and B : how many degrees are there in each angle ? 42 PLANE GEOMETRY. Proposition 27. Theorem. 116. If one side of a triangle he greater than another, the angle opposite the greater side is greater than the ayigle opposite the less, c Hyp, Let ABC be a A having the side AC>BC. To prove ZABC>ZA. Proof From CA cut off CD = CB. Join BD, Then, since CD = CB, .-. ZCDB=: zCBD. (Ill) But ext. ZCDB > opposite int. /DAB of aADB. (98) .-. also ZCBD>ZDAB. But Z CBA > Z CBD. (Ax. 8) Much more .*. ZCBA>ZDAB. q.e.d. EXERCISES. 1. Prove Prop. 27 by producing CB to E, making CE = CA. 2. ABO is a triangle in which OB, 00 bisect the angles B, 0, respectively; show that, if AB is greater than AC;, then OB is greater than 00. BOOK I.-TRIANOLES. 43 Proposition 28. Theorem. 117. Conversely, if one angle of a triangle he greater than another, the side opposite the greater angle is greater than the side opposite the less. Hyjy. Let ABC be a A having Z ABO > Z BAG. To prove AC>BC. Proof. Draw BD so that ZABD= ZBAD. Then AD = BD, being opposite equal Zs (114). « But in A BCD, BD + DC > BC. Mtlier side of a A is < the sum of the oilier tioo (96). And, since AD = BD, . • . BD + DC = AC. .-. AC>BC. Q.E.D. Note. — These two propositions may be stated briefly as follows: In every triangle, the greater side is opposite the greater angle, and conversely, the gi eater angle is opposite the greater side. 118. Cor. The hypotenuse is the greatest side of a right-angled triangle. EXERCISES. 1. In a A ABC, it AC is not greater than AB, show that any straight l^e drawn through the vertex A and termi- nated by the base BC, is less than AB. 2. Any two sides of a triangle are together greater than twice the straight line drawn from the vertex to the middle point of the third side. 44 PLANE GEOMETRY. Proposition 29. Theorem. 119. i/* two triangles have two sides of the one equal respectively to two sides of the other, but the iiicluded angle 171 the first triangle greater than the included angle in the second, then the third side of the first triangle is greater than the third side of the second. Hyp, Let ABC, DEF be two A s, having AB = DE, AC = DF, but ZBAC>ZEDF. To prove BC>EF. Proof. Apply the A ABC to the A DEF so that AB shall coincide with DE, and the pt. C fall at H. Join FH. Then, . since DH = DF, .-. ZDHF= ZDFH. But ZDHF>ZEHF. The wlwU 18 greater than any of its parts (Ax. 8). .-. also zr)FH>zEHF. But ZEFH>zDFH. Much more . • . Z EFH > Z EHF. .-. EH>EF. TTie greater side is opposite the greater angle (117). But EH = BC. . • . BC > EF. q.e.d. (Hyp.) (Ill) (Ax. 8) BOOK L— TRIANGLES, 45 Proposition 30. Theorem. 120. Conversely, if tivo triangles have two sides of the one equal respectively to two sides of the other, hut the third side of the first triangle greater than the third side of the second, then the included angle of the first triangle is greater than the included angle of the secotid. Hyp. Let ABC, DEF be two As having AB = DE, AC = DF, but BC > EF. To prove Z A > Z D. Proof If Z A is not > iT), then either (1) Z A = ZD, or (2) ZA< ZD. (1) If ZA= zD, then BC = EF, Immng two sides and the included Z equal, each to each (104). But this is contrary to the hypothesis. (2) If ZA ZD. Q.E.D. Note.— Prop. 30 is here proved by an indirect method. This method is sometimes called the reductio ad absurdum. It consists in assuming that the conchision to be proved is not tnie, and showing tliat this assumption leads to an absurdity, or to a result inconsistent with the hypothesis. / 46 PLANE GEOMETRY. Quadrilaterals. 121. A quadrilateral is a plane figure bounded by four straight lines, which are called its sides. The straight lines which join opposite angles of a quadri- lateral are called the diagonah. 122. A trapezium is a quadrilateral which has no two of its sides parallel. 123. A trapezoid is a quadrilateral which has two of its sides parallel. The parallel sides of a trapezoid are called the bases, and the perpendicular distance between them is called the altitude. The line joining the middle points of the non-parallel sides is called the middle parallel*' of the trapezoid. 124. A parallelogram is a quadri- lateral which has its opposite sides parallel. . The bases of a pafallelogram are the side on which it^ands and the opposite side. The perpendicular distance be- tween the bases is called the altitude. 125. A rectajigle is a parallelogram whose angles' are right angleXf^ 126. A square is a rectangle whose sides are all equal. J 127. A rhomboid is a parallelogram whose angles are oblique and whose ad- jacent sides are unequal. 128. A rhombus, or lozenge, is a par- allelogram whose sides are all equal. § * Called also the median. t Called also a right-angled parallelogram. X Callt^d also an equilateral rectangle. § Called also equilateral rhomboid. BOOK I— QUADRILATERALS. 47 Proposition 3 1 . Theorem. 129. 1)1 every parallelogram, the op2)osite sides are eqirnl, and the opposite angles are eqicah Hyp. LetABCDbea^. D __^ To prove DC = AB, AD = BO, \ ^> ^ Z.K^ IQ, ZD=.ZB. \o^^^ \ Proof. Draw the diagonal AC. ^ B Because DC is || to AB, (Hyp.) and AD is II to BO, (Hyp.) .-. ZDCA= ZBAC, and ^ ZACB=ZCAD, being alt 4nt. Is (73). Hence the whole Z DCB = Z BAD. (Ax. 9) Now in the As ACD, ACB, because (ZDCA= ZBAC,) (ZCAD=ZACB,[ (,l"st proved) and AC is common, .-. A ACD = A ACB, having a side and the two adjacent Z,s equal, each to each (105) .-. DC = AB, AD =BC, being liomologous sides of equal as (109). and Z D = Z B. q.e. d. *> 130. Cor. 1. A diagonal of a parallelogram divides it into two equal triangles. 131. Cor. 2. Two parallels included between tivo other parallels are equal. 48 PLANE GEOMETRY. Proposition 32. Theorem. 132. If the O'piiosite sides of a quadrilateral are equal, the figure is a parallelografu. Hyp. Let ABCD be a quadrilat- eral having AB = CD, AD = BC. To prove ABCD is a CJ. Proof. Draw the diagonal AC. In the As ACD, ACB, because AD = BC, (Hyp.) AB = CD, (Hyp.) AC is common, .-. aACD= a ACB. (108) .-. ZACD = ZCAB, and Z CAD = Z ACB, being homologous /s of equal as (109). .*. ABisli to CD, and AD is II to BC, tJie alt. -int. Zs being equal (75). . • . ABCD is a fU by definition. (124) Q.E.D. EXERCISES. 1. If one angle of a parallelogram is a right angle, prove that all its angles are right angles. 2. Prove that two parallels are everywhere equally distant. 3. If, in the figure of Prop. 32, BE be drawn parallel to AC and meeting DA produced to E, prove that the paral- lelogram EBCA will be equal to the parallelogram ABCD. BOOK L-QUADRILATERAL8. 49 Proposition 33. Theorem. 133. If tivo opposite sides of a qttadrilaterdl are equcd ayid parallel, the figure is a parallelogram. Hyp, Let ABCD be a quadrilat- D_ eral, having AB = and || to DC. To prove ABCD is a m. Proof. Draw the diagonal AC. In the A s ACD, ACB, because AB = CD, (Hyp.) AC is common, ZACD= ZCAB, being alt. -int. /. 8 {12), .-. aACD= aACB, Jiamng two sides and the included A equal, each to each (104). . • . AD = BC, being homologous sides of equal as (109). . * . the figure ABCD is a CD, Jiamng its opposite sides equal (132). q.e.d. EXERCISES. 1. Prove that the straight lines which bisect two adjacent angles of a parallelogram cut each other at right angles. 2. AB, CD, EF are three equal and parallel straight lines ; prove that the triangle ACE is equal to the triangle BDF. 50 PLANE GEOMETRY, Proposition 34. Theorem. 134. The diagonals of a parallelogram Used each other. Hyp. Let ABCD be a /Z7, whose diagonals intersect at 0. To prove AO = OC, DO = OB. Proof. lu the As AOB, COD, " ^ AB = DC, hdnig opp. sides of the OJ (129), ZABO= ZCDO, and ZBAO= ZDCO, being alt. -int. As (72). .-. aAOB= aCOD, having a side and the two adj. Zs equal, each to each (105). . • . AO = OC, and DO = OB. (109) Q.E.I). EXERCISES. 1. If the opposite angles of a quadrilateral are equal, the figure is a parallelogram. 2. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. 3. If the diagonals of a parallelogram are equal, the fig- ure is a rectangle ; if they also intersect at right angles, it is a square. 4. The straight lines joining the middle points of the opposite sides of any quadrilateral bisect each other. 5. The diagonals of a rhombus bisect each other at right angles. G. The diagonals of a rectangle are equal. BOOK L^qUADRILATEIlAL8, 51 Proposition 35. Theorem. 135. Two parallelograms are equal when two adjacent sides and the included angle of the one are equal respectively to two adjacent sides and the included angle of the other. Hyp, Let ABCD, A'B'C'D' be two ^^s, having AB = A'B', AD = A'D', V \ ZA = ZA'. \ \ To prove ru ABCD =diA'B'G'D\ jy c' Proof Apply ZZ7ABCD to \ \ ^A'B'C'D', so that Z A shall coin- \ \^ cide with Z A', and the side AB a' b' with the equal side A'B'. Because Z A = Z A', and AD = A'D', (Hyp. ) .-. AD will fall on A'D', and pt. D on pt. D'. Because DC is || to AB, and D'C is || to A'D', (124) .-. DC will fall on D'C, and pt. C somewhere on D'C. Through a given pt. only one st. I. can be drawn \\ to a given st. I. (Ax. 12) Also, because BC is || to AD, and B'C is || to A'D', (124) .♦. BC will fall on B'C, and pt. C somewhere on B'C. (Ax. 12) .*. since pt. C falls on D'C and B'C, it must fall at their pt. of intersection C. .'. the two£I7s coincide throughout, and are equal, q.e.d. 136. COK. Two rectangles are equal ivhen they have two adjacent sides eqxial, each to each. 52 PLANE GEOMETRY, POLYGOl^S. 137. A polygon is a plane figure bounded by straight lines; as ABODE. The straight lines are called the sides of the polygon; and their sum is called the perimeter of the polygon. The angles of the polygon are the angles formed by the adjacent sides with each other ; and the vertices of these angles are also called the vertices of the polygon. 138. The angles of the polygon measured on the side of the enclosed surface are called interior angles. An exterior angle of a polygon is an angle between any side and the continuation of an adjacent side. A diagonal is a line joining any two vertices that are not consecutive; as AD. 139. Polygons are named from the number of their sides, as follows : A polygon of three sides is a triangle; one of four sides, a quadrilateral; one of five sides, a pentagon; one of six sides, a hexagon; one of seven sides, a heptagon; one of eight sides, an octagon; one of nine sides, a nonagon; one of ten sides, a decagon; one of twelve sides, a dodecagon; one of fifteen sides, a quindecagon. 140. An equilateral polygon is one which has all its sides equal. An equiangular polygon is one which has all its angles equal. A re^i^Z^Mi- polygon is one which is both equilateral and equiangular. 141. A cow_ye^_polygon is one each of whose interior angles is less than a straight angle; as ABODE. BOOK L— POLYGONS. t)6 142. A concave polygon is one in which at least one of the interior angles is reflex (26), as GHKMNO, in which the interior angle KMN is reflex.* The polygons considered in this g^ work will be understood to be convex, unless otherwise stated. It is evident that a polygon has as many angles as sides. 143. Two polygons are mutually equilateral when the sides of the one are equal respectively to the sides of the other, taken in the same order; as the polygons ABCD, A'B'C'D', in which AB = A'B', BC = B'C, etc. 144. Two polygons are mutually equiangular when the angles of the one are equal respectively to the angles of the other, taken in the same order; as the polygons PQRS, P'Q'R'S', in which Z P = Z P', ZQ = ZQ', etc. 145. In polygons which are mutually equilateral or mu- tually equiangular, any two corresponding sides or angles are called homologous (95). Except in the case of triangles, two polygons may be mu- tually equilateral without being mutually equiangular, and mutually equiangular without being mutually equilateral. 146. A polygon may be divided into triangles by draw- ing diagonals from one of its vertices ; and the number of triangles into which any polygon can thus be divided is evidently equal to the number of its sides, less two. When two po^gons can be divided by diagonals into the same number of tri- angles, equal each to each, and similarly placed, the polygons are equal ; for they can be applied one to the other, and the corresponding triangles will evi- dently coincide, and therefore the polygons w ill coincide throughout. When two polygons are both mutually equilateral and mutually equiangular. they are equal, for they can be applied one to the other so as to coincide. * Called also a reentrant angle. 54 PLANE GEOMETRY, Proposition 36. Theorem. 147. The sum of the interior angles of a polygon is equal to two right angles taken as many times less two as the polygon has sides, F E ,/.•(*■''' ^' — ->» ]j^^-^'- Hyp, Let ABCDEF be a polygon having n sides. To prove ZA+zB+zC +etc. = 2 rt. Zs {n-2). Proof From any vertex A, draw the diagonals AC, AD, AE. The polygon will be divided into as many triangles less two as it has sides (146). Thus, there are (n — 2) triangles, whose angles make the interior angles of the polygon. Now, the sum of the Z s of a A = 2 rt. Z s. (97) .*. the sum of the Z s in the polygon = 2 rt. Z s (/^ — 2). Q.E.D. 148. Cor. 1. 2 rt. Zs (/^ - 2) - 2?^ rt. Zs -4 rt. Zs. Therefore, the sum of the angles of a polygon is also equal to tivice as many right angles as the figure has sides, less four right angles. 149. Cor. 2. Tlie sum of the angles of a quadrilateral is equal to two right angles taken (A — 2) times, i.e., four right angles. TJie sum of the angles of a pentagon is equal to two right angles talcen (5 — 2) times, i.e., six right a7igles, etc. 150. Cor. 3. Each ayigle of an equiangular polygon of fi, sides is ^ — - right angles. BOOK I.^POLYGONS. 65 Proposition 37. Theorem. 151, The sum of the exterior angles of a polygon, made hy producing each of its sides in the smne direction, is equal to four right angles. Hyp. Let ABODE be a polygon having its n sides pro- duced in the same direction. To prove Zci + Zl) + Z.c+ etc. = 4 rt. Z s. Proof Any int. Z A + its adj. -ext. Z «^ = 3 rt. Z s, (52) . • . all the int. Z s + all the ext. Zs = 2nrtZ s. But all the int. Z s = 27^ rt. Z s - 4 rt. Z s. (148) . • . all the ext. Z s = 4 rt. Z s. (Ax. 3) Q.E.D. EXERCISES. 1. Express in terms of a right angle, and also in degrees, the magnitude of each interior angle of (1) a regular hexa- gon, (2) a regular octagon, and (3) a regular decagon. 2. If one side of a regular hexagon is produced, show that the exterior angle is equal to the angle of an equilateral triangle. 3. The exterior angle of a regular polygon is one-fifth of a right angle : find the number of sides in the polygon. 4. The interior angle of a regular polygon is five-thirds of a right angle: find the number of sides.in the polygon. 5. The side AB of the triangle ABO is produced to D, so that BD is equal to BO : prove that the angle ABO is double the angle ADO. 6. How many sides has a polygon, the sum of whose in- terior angles is four times that of its exterior angles ? / O 56 PLANE GEOMETBY. I M Miscellaneous Theorems. Proposition 38. Theorem. 152. Jf three or more parallels intercept equal lengths on any transversal, they intercept equal lengths on every transversal. Hyp. Let the || s AE, BF, CG, DH, cut the two transversals MN and OP at the pts. A, B, 0, D, and E, F, G, H, so that AB = BC = CD = etc. To prove EF = FG = GH == etc. Proof. From E, F, G, draw EK, FL, GS, all II to MN. Then A \e B P\f ■= K \ J I\h EK = AB, FL = BO, GS = CD, hdng opp. sides of a r'7(129). (Ax. 1) Also and .•.EK = FL=GS = etc. ZFEK=ZGFL = ZHGS, Z EFK = Z FGL = Z GHS, being ext.-int. AsofW lines (77). .-. aEFK = aFGL=aGHS, Mmng a side and the two adj. As equal, each to each (100, 105). .•.EF = FG = GH, being homologous sides of equal as (109). q.e.d. 153. Cor. Since FK = GL = HS, .•.BF-AE=:FK, and CG - BF = GL = FK. Therefore, the intercepfted p^drt of each jMrallel ivill differ in length from the next intercept hy the same amount. I BOOK L-MISCELLANEOVS THEOllEMS. 57 Proposition 39. Theorem. 154. The straight line drawn through the middle point of a side of a triangle parallel to the base, bisects the re- maining side, and is equal to half the base. Hyp. Let ABC be a A, D the middle poiut of AB, and DE a line || to BC, meet- A ing AC at E. /\ To prove (1) AE = EC ; o/- \e and (2) DE = iBC. / \ Proof (1) Through A draw a line || b C to BC. . ~ Then, because AD = DB, ^4^ '^ jli^(Hyp.) ...AE^EC. ^^^^^ €C if lis intercept equal lengths on one transversal, they intercept equal lengths on every transversal (152). (2) The intercepted part BC exceeds DE as much as DE exceeds the intercepted part at A (153). But this latter intercept = 0. .•.BC-DE = DE. .-. DE^^BC. Q.E.D. 155. Cor. 1. The line joini7ig the middle points of two sides of a triangle is parallel to the third side, and equal to half of it, 156. Cor. 2. Let ABCD be a trape- Dy ^C zoid,' AG =: CD, and GH |1 to AB. ^ Then BH = HC, for, since AB, GH, and DC are parallels intercept- A''^ ^B ing equal lengths AG and GD on the transversal AD, they intercept equal lengths on every transversal (1^^) Also - AB - GH = GH -DC. (153) .•.GH==:i(AB + DC). Therefore, the line drawn through the middle point of one of the non-parallel sides of a trapezoid parallel to the bases, bisects the opposite side, and is equal to half their sum. 58 PLANE GEOMETRY, The Locus of a Poij^^t. 157. The locus of a point is the line, or system of lines, which contains all the points, and only those points, that satisfy a given condition. A locus is sometimes defined as the path traced out by a point which moves in accordance with a given law. Thus, the locus of a point which is always at a given dis- tance from a given straight line, is a pair of straight lines. 158. In order to infer that a certain line, or system of lines, is the locus of a point under a given condition, it is necessary to prove, (1) that any point which fulfils the given condition is on the supposed locus ; and (2) that every point on the supposed locus satisfies the give?i condition, 159. T7ie locus of a point tvhich is always equidistant from two given points, is the perpendic^dar bisector of the line joining them (67). EXERCISE. Find the locus of the middle point of a straight line drawn from a given point to meet a given straight line of unlimited length. Let A be the given pt., and BC the ^ ! p — /b x'E C given st. line. _±LL__Pi'jQrl!^ (1) Let AD be any straight line i^^^"'' drawn from A to meet BC, and let P p^ be its mid. pt. Draw AF J. to BC, and bisect AF at H. Join HP, and produce it indefinitely. Then HP is || to FD. (155) . • . P is on the st. line which passes through the fixed pt. Hand is II to BC. (2) Every pt. in HP, or HP produced, satisfies the re- quired condition. For in this st. line take any pt. Q. Join AQ, and produce it to meet BC in E. Then Q is the mid. pt. of AE. (154) Hence the required locus is the st. line || to BC and pass- ing through the mid. pt. of the J_ from A to BO. BOOK L— MISCELLANEOUS THEOREMS. 59 Proposition 40. Theorem. 160. (1) Every point in the bisector of an angle is equally 11 distant from the sides of the angle; and (2) conversely, every point luithin an angle, and equally distant from its sides, is in the bisector of the angle. ^C (1) Hyp, Let AE be the bisector of the Z BAG ; P any point in AE ; and /^^^ ^^^E PD, PH J_s to AB, AC. To prove PD = PH. (59) f Proof Because in the rt. A s APD^ APH, ZDAP = ZHAP, (Hyp.) and AP is common, .-. aAPD=a APH, having the hypotenuse and an acute Z equal in each (106). .•.PD = PH. being homologous sides of equal as (109). (2) Hyp, Let PDi.to AB = PH _L to AC. To prove that P is on the bisector of Z BAO. Proof Join PA. Then A APD = A APH, having the hypotenuse and a side equal in each (110). .-. Z PAD= Z PAH, teing homologous As of equal As (109). . • . P is on the bisector of the angle BAG. q.e.d. 161. GoR. ^very point within the angle, but not on the bisector, is unequally distant from the tivo sides, 162. ScH. The bisector of an angle is the locus of all the points situated within the angle^ which are eq^ially distant from its sides, y' 60 PLANE GEOMETRY. The Concuerekce of Straight Lines iif a TRIAiq^GLE. 163. Three or more straight lines are said to be concur- rent when they meet in a point. 164. Three or more points are said to be coUitiear when they lie upon one straight line. 165. A straight line from any vertex of a triangle to the middle. point of the opposite side is called a, medial, or a 7nedian, of the triangle. Proposition 41 . Theorem. 166. The bisectors of the three angles of a triangle are concurrent. Hyp. Let ABC be a A ; A, OB, DC, the bisectors of the Z s A, B, and 0. To prove that these bisectors meet in a pt. Proof Let the bisectors OA, OB meet at 0. Draw the _Ls OL, OM, ON. Because is on the bisector of / BAC, . • . OL = OK (160) Because is on the bisector of Z ABC, .•.OL=:OM. (160) .•.OK = OM. (Ax. 1) .*. is on the bisector of the Z ACB, heing equally distant from its sides [160 (2)], that is, must lie on the bisector CD. . • . the bisectors of the three Z s meet at the point 0. Q.E.D. 167. Cor. TJie point is equalUj distant from the three sides of the triangle. BOOK L— MISCELLANEOUS THEOBEMS. 61 Proposition 42. Theorem. 168. TJie perpendicular Msectors of the three sides of a triangle are concurreiit. Hyp, Let ABC be a A ; D, E, F, the middle pts. of its sides ; and DO, EO, EH, the i_s erected at D, E, F. To prove that these J_s meet in a pt. Proof The two ±s OD, OE, since they cannot be 1| , will meet at the pt. 0. Join OA, OB, OC. Because is in the _L bisector OD, .•.OA = OB. (66) Because is in the _L bisector OE, . -.06 = 00. {m) .-.OArrOC. (Ax. 1) .-. is in the 1 bisector of AC; (67) that is, must lie on the J_ bisector HE. . • . the three J. bisectors meet at the pt. 0. q.e.d. 169. Cor. The point of intersection of the perpendicular bisectors, is equally distant from the three vertices of the triangle. EXERCISES. 1. If the 1;riangle in Prop. 41 is equilateral, find the value of the angle AOB. 2. If the same triangle is isosceles, and the angle is three times as great as either of the angles A and B, find the value of the angle AOB. 62 PLANE GEOMETRY, Proposition 43. Theorem. 170. The peiyendiculars from the vertices of a triangle to the opposite sides are concurrent. Hyp. Let ABC be a A, and p^- -^ -^^q AD, BE, CF the three J_s from A, B, C to the opp. sides. To provg that these J_s meet ill a pt. Proof. Through -A, B, 0, draw RQ, RP, PQ || respectively to BC, AC, AB. Then, since ABPC is a OJ, ^ A \ v: (124) .-. BP = AC. (139) And since ARBC is a CD, (124) .-. RB = AC. (129) .-. RB = BP. (Ax.l) Similarly, RA = AQ, and PC = CQ; that is, A, B, C, are the mid. pts. of QR, RP, PQ. Since AC is |1 to PR, and BE is J_ to AC, .-. BE is also i. to PR. A line 1 to one of two Wsis i. to the other (71). Similarly, AD and CF are JL to RQ and PQ. . * . these three J_s meet in a pt., leing tlie JL bisectors oftlie three sides oftJie A PQR (168). q.e.d. 171. Def. The intersection of the perpendiculars from the vertices of a triangle to the opposite sides is called its orthocentre. The triangle formed by joining the feet of the perpen- diculars is called the pedal or orthoQentric triangle. BOOK 1. -MISCELLANEOUS THEOREMS 63 -^ Proposition 44. Theorem. 172. The three medial lines of a triangle are concurrent in a point of trisection, * the greater segment in each being toiuards the angular point. Hijp. Let ABO be a A ; D, E, F, the mid. pts. of BO, AO, AB; and AD, BE, OF, the three medial lines. To prove that these lines meet in the pt. that trisects them. Proof, Let the two medials BE and OF meet in 0. Bisect BO in H, and 00 in K. Join HK, KE, EF, FH. In the A BOO, because H and K are the mid. pts. of BO and 00, .-. HK is II and = to J BO. (155) Also, in thfe A ABO, because E and F are the mid. pts. of AO and AB, .-. EF is II and = to I BO. (155) .-. EFHK is a OJ, having two opp. sides equal and \\ (133). .-. is the mid. pt. of Ell and FK. The diagonals ofacj bisect each other (134). .-. EO = OH = HB, and FO = OK = KO; or EO --= J- EB, and FO = i FO. That is, the medial BE cuts the medial OF at a pt. 0, one- third the way from F to 0. In the same way it may be proved that the medial AD cuts the medial OF at a pt. one-tliird the way from F to 0; that is, at the same pt. 0. .'. The three^medials are concurrent in the point that tri- sects them. q.p:.d. Note.— The point of intersection of the three medians of a triangle is called the centroid. * When a line is divided into three equal parts it is said to be trisected, 64 PLANE GEOMETRY. Symmetry. Symmetry with ref^pect to an axis. \P IvT 173. Two points are said to be symmetrical with respect to a straight line, when the straight line bisects at right angles tli^ straight line joining the two points. . Thus, the two points P and P' are symmetrical with respect to the line MN, if MN bisects PP' at right angles. The straight line MN is called the axis of symmetry. If we take the plane containing the pt. P, and turn it about the axis MN, until the upper part is brought down on the part below MN, the line AP will take the direction AP', and the point P will coincide with the point P'. Thus, when two points are symmetrical with re- spect to an axis, if one of the parts of the plane be revolved about the axis to bring ]t down on the other part, the symmetrical points coincide. C ip' 1 74. Two figures are said to be symmetrical with re- spect to an axis, when every point in one figure has its symmetrical point in the other. Thus, the figures ABC, A'B'C are symmetrical with respect to the axis MN, if every point in the figure ABC has a symmetrical point in A'B'C with respect to MN. In all cases, two figures that are symmetrical with re- BOOK I.—SYMMETliY. 65 II I spect to an axis, can be applied one to the other, by revolv- ing either about the axis; consequently they are equal. The corresponding symmetrical lines of symmetrical fig- res are called Jionioloc/ous lines. Thus, in the symmetric- 1 figures, ABC, A'B'C, the homologous lines are AB ind A'B', BC and B'C, AC and A'C Symmetry luith respect to a point. 175. Two points are said to \>q symmetrical vf\i\\YQ^^eci to a third point, when this third point bisects the straight line joining the two points. Thus, P and P' are symmetrical with respect to A, if the straight line PP' is P A P bisected at A. The point A is called the centre of symmetry, 116. Two figures are said to be symmetrical with re- spect to a centre, when every point in one figure has its symmetrical point in the other. Thus, the figures ABC, A'B'C are symmetrical with respect to the centre 0, if every point in the figure ABC has a symmetrical point in A'B'C. 177. A figure is symmetrical with respect to an axis, when it can be divided by that axis into two figures symmetrical with respect to the axis. A figure is symmetrical with re- spect to a centre, when every straight line drawn through that centre cuts the figure in two points symmetrical with respect to this centre. 66 PLANE GEOMETRY. Proposition 45. Theorem. 178. If a figure is symmetrical with respect to two axes at right angles to each other, it is also symmetrical with respect to their intersection as a centre. P G Y F _. f^ - -^->9 H ^^^ — -y E ^" X P B Y'° Hyp, Let the figure ABCDEFGH be symmetrical with respect to the two _L axes XX', YY', which intersect at 0. To prove that is the centre of symmetry of the figure. Proof. Let P be any pt. in the perimeter of the figure. Draw PRP' J_ to XX', and PSQ J. to YY'. Join RS, OP', and OQ. Then, because the figure is symmetrical with respect to XX', .-. PR := P'R. And since PR = OS, .-. P'R = OS. .-. RP'0SisaZl7, Jiatring two opp. sides = and || (133). .-. RSis = and || to P'O, being opp. sides of a CJ (124). Also, since the figure is symmetrical with respect to YY', .-.PS^SQ. And since PS = OR, .-. SQ = OR. . •. SROQ is a /=7; . • . RS is = and 1| to OQ. Because both P'O and OQ are = and || to RS, . • . the points P', 0, Q are in the same st. line which is bisected at 0. .*. any st. line P'OQ, drawn through 0, is bisected at 0. . • . the figure is symmetrical with respect to as a cen- tre (177). 9.E.P, BOOK L— EXERCISES. 67 EXERCISES. The following theorems are given for the exercise of the student, that he may work out his own demonstrations. The student can make no solid acquisitions in geometry, without frequent practice in the application of the princi- ples he has acquired. lie should not merely learn the demonstration of a number of theorems, but he should ac- quire the power of grasping and demonstrating geometric theorems for himself, and this power can never be gained by memorizing demonstrations. He should understand that the ability to investigate, to reason for himself, is the chief object for the attainment of which he should strive. Diligent application, systematic practice in devising proofs of new propositions, is indispensable. In the process of finding a demonstration the student should first construct a diagram, and state the hypothesis, including in the statement not only what the theorem says, but what it implies. He should also examine the conclu- sion, and see what it says and what it implies, and discover the relation between the hypothesis and the conclusion. A correct diagram is most useful in suggesting the steps by which a theorem is to be demonstrated. If the student will ask himself why he takes any particular step, he may avoid the habit of random guessing, and with more cer- tainty discover the correct and direct process for effecting the demonstration. Sometimes it will be necessary to draw additional lines in the diagram, and to call to mind the different theorems which apply to the figure thus formed. The demonsti-ation must be framed in the simplest manner, but without omitting any logical step. This is a matter of practice, in which no general rule can be given. The student should express each step of the demonstra- tion completely and fully. Tlie most common fault is that of passing over steps in the demonstration because the con- clusion seems to be obvious, 68 PLANE QEOMETllT. It is often the case that the clearness of intuition acquired by a practised geometrician will make him impatient of the successive steps of detailed reasoning, and he will be eager to conduct his pupil to the desired end by a shorter and an easier road. But it is a great misfortune to the learner if he is deprived of that useful disciplme in geometric reason- ing which, however tedious it may seem in its application to short and easy propositions, is indispensable to the in- vestigation of more lengthy and difficult ones. One of the great objects of the study of geometry is to cultivate the habit of examining the logical foundations of those conclusions which are accepted with- out critical examination. The feeling of security that a conclusion is right before its foundation has been examined is a most fruitful source of erroneous opinions, and the person who neglects the habit of inquiring into what appears obvious is liable to pass over things which, had they been carefully examined, would have changed the conclusion.* 1. If the angles ABC and ACB at the base of an isosceles triangle be bisected by the lines BD, CD, show that DBC will be an isosceles triangle. 2. BAC is a triangle having the angle B double the angle A. If BD bisects the angle B and meets AC at D, show that BD is equal to AD. 3. In the triangle ABC, the angle A = 50°, the angle B = 70°. What angle will the bisectors of these two angles make with each other? 4. In the preceding triangle, what will be the values of the three exterior angles ? 5. A given angle BAG is bisected ; if CA is produced to G, and the angle BAG bisected, prove that the two bisect- ing lines are at right angles to each other. 6. ACB, ADB are two triangles on the same side of AB, such that AC = BD, and AD = BC, and AD and BC in- tersect at : prove that the triangle AOB is isosceles. 7. ABC is a triangle, and the angle A is bisected by a line which meets BC at D: show that BA is greater than BD, and CA is greater than CD. * Newcomb. BOOK L— EXERCISES. 69 8. If one angle of a triangle is equal to the sum of the ( other two, the triangle can be divided into two isosceles triangles. 9. If the angle C of a triangle is equal to the sum of the angles A and B, the side AB is equal to twice the line join- ing C to the middle point of AB. 10. A line bisects the angle A of a triangle ABC ; from B a perpendicular is drawn to this bisecting line, meeting it at D, and BD is produced to meet AC or AC produced at E : show that BD = DE. 11. A line is drawn terminated by two parallel lines; through its middle point any line is drawn and terminated by the parallel lines. Show that the second line is bisected at the middle point of the first. 12. If through any point equidistant from two parallel lines, two lines be drawn cutting the parallel lines, they will intercept equal portions of these parallel lines. 13. If the line bisecting the exterior angle of a triangle be parallel to the base, show that the triangle is isosceles. 14. If a line be drawn bisecting one of the angles of a triangle to meet the opposite side, the lines drawn from the point of intersection parallel to the other sides, and termi- nated by these sides, will be equal. Let ABC be the a ; let a line be drawn bisecting z A and meeting BC at D, etc. 15. The side BC of a triangle ABC is produced to a point D; the angle ACB is bisected by the line CE which meets AB at E. A line is drawn through E parallel to BC, meet- ing AC at F, and the line bisecting the exterior angle ACD at G. Show that EF = FG. 16. A line drawn at right angles to BC, the base of an isosceles triangle ABC, cuts the side AB at D and CA pro- duced at E: show that AED is an isosceles triangle. From A draw a line bisecting ZBAC, and meeting BC at F, etc. 70 PLANS! omMETRT. 17. If the lines bisecting the angles at the base of an isosceles triangle be produced to meet, they will contain an angle equal to an exterior angle of the triangle. ''18. A is the vertex of an isosceles triangle ABC, and BA i^ produced to D, so that AD = BA; and DC is drawn: show that BCD is a right angle. •^OtMl9. ABC is a triangle, and the exterior angles at B and ..^. are bisected by the lines BD, CD respectively, meeting at , p.; show that the angle BDC together with half the angle BAC make up a right angle. 20. In the triangle ABC the side BC is bisected at E, and AB at G; AE is produced to F so that EF =: AE, and CG is produced to H so that GH == CG: show that FB and HB are in one straight line. 21.' Lines are drawn through the extremities of the base of an isosceles triangle, making angles with it on the side remote from the vertex, each equal to one-third of one of the equal angles of the triangle and meeting the sides 'pro- duced : show that the three triangles thus formed are isosceles. 22. AEB, CED are two lines intersecting at E ; lines AC, DB are drawn forming two triangles ACE, BED; the angles ACE, DBE are bisected by the lines CF, BF, meet- ing at F. Show that the angle CFB is equal to half the sum of the angles EAC, EDB. I AEC = I ECB + I EBC, etc. 23. If a quadrilateral have two of its opposite sides parallel, and the other two equal but not parallel, any two of its opposite angles are together equal to two right angles. 24. On the sides AB, BC, and CD of a parallelogram ABCD three equilateral triangles are described, that on BC towards the same parts as the parallelogram, and those on AB, CD towards the opposite parts: show that the dis- tances of the vertices of the trianorles on AB, CD from that BOOK I.-EXERCI8E8. 71 on BC are respectively equal to the two diagonals of the panillelogram. 25. A, B, C are three points in a straight line, such that AB = BC: show that the sum of the perpendiculars from A and C on any straight line which does not pass between A and is double the perpendicular from B on the same line. 26. In a right triangle, show that the medial drawn from the right angle is equal to half the hypotenuse. 27. If P and Q are the feet of the perpendiculars from A on the lines bisecting the angles B and C of a triangle ABC ; show that PQ is parallel to BC. Produce AP, AQ to meet BC in D, E, etc. 28. AD, BE, CF, the perpendiculars from the vertices of a triangle ABC, intersect in ; prove that the angle BOF = the angle BAC, the angle FO A = the angle ABC, and the angle BOD = the angle BCA. 29. What is the magnitude of each angle of the follow- ing regular figures ? a pentagon, an octagon, a decagon. 30. From the angle A of a triangle ABC a perpendicu- lar is drawn to the opposite side, meeting it, produced if necessary, at D; from the angle B a j)erpendicular is drawn to the. opposite side, meeting it, produced if necessary, at E: show-that the lines which join D and E to the middle point of AB are equal. 31. From the angles at the base of a triangle perpen- diculars are drawn to the opposite sides, produced if neces- sary: show that the line joining the points of intersection will be bisected by a perpendicular drawn to it from the middle point of the base. 32. The lines which join the middle points of adjacent sides of any quadrilateral, form a parallelogram. 33. The sides AB, AC of the triang'e ABC are bisected in D and E respectively; BE and CD are produced to F 72 PLANE GEOMETRY. and G, so that EF = BE and DG = DO : prove that F, A, G are collinear. 34. ABODE is a regular pentagon; AO, AD are joined: prove that each of the angles AOD, ADO is double the angle OAD. Draw DF II to AE, F being in AC, etc. 35. Two medians of a triangle are equal : prove (with- out assuming that they trisect each other) that the triangle is isosceles. 36. ABCD, BAOE are parallelograms on the same baso AB, such that the diagonals BD, AM are equal : prove that the other diagonals AO, BO are equal. 37. ABOD is a parallelogram; E is the middle point of BO; AB and DE produced meet in F: prove that the triangle DBF is half the parallelogram ABOD. 38. If two right triangles ABO, ABD be on the same hypotenuse AB, and the vertices and D be joined, the pair of angles subtended by any side of the quadrilateral thus formed are equal. 39. The angles made with the base of an isosceles tri- angle by perpendiculars from its extremities on the equal sides are each equal to half the vertical angle. 40. The angle included between the internal bisector of one base angle of a triangle and the external bisector of the other base angle is equal to half the vertical angle. 41. The sum of the distances of any point m the base of an isosceles triangle from the equal sides is equal to the dis- tance of either extremity of the base from the opposite side. 42. The sum of the perpendiculars from any point in the interior of an equilateral triangle is equal to the per- pendicular from any vertex on the opposite side. 43. AD and BO are two parallel lines cut obliquely by AB, and perpendicularly by AO ; and between these lines we draw BED, cutting AO in E, such that ED = 2AB ; prove that the angle DBO is one-third of ABO. 44. If be the point of concurrence of the bisectors of BOOK I— EXERCISES. 73 the angles of the triangle ABC, and if AO produced meet BC in D, and from 0, OE be drawn perpendicular to BC ; prove that the angle BOD = the angle COE. 45. The bisectors of the external angles of a quadrilat- eral form a circumscribed quadrilateral the sum of whose opposite angles is equal to two right angles. 4G. The locus of a point equidistant from two given in- tersecting lines is two lines at right angles to each other. Bisect the adj. Zs between the given Unes, etc. 47. Any line is drawn cutting two fixed intersecting lines, nd the two angles on the same side of it are bisected : find -the locus of the point of intersection of the bisecting lines. 48. The leaf of a book is turned down so that the corner always lies on the same line of printing : find the locus of the foot of the perpendicular from the corner to the crease. 49. ABO is a triangle, D is the middle point of BO, E the middle point of AD ; let BE produced meet AO in F : prove that AO is trisected in F. Draw DG II to BF meeting AC in G, etc. 50. D, E, F are the middle points of the sides of a tri- angle, N is the foot of the perpendicular from the angle opposite D to the side which D bisects : prove that z EDF = Z ENF. Let D, E, F lie in BC, CA, AB respectively; ED is II to AB,and DF II to AC, etc. 51. If A, B, denote the angles of. a A, prove that i(A + B), i(B + 0), i(0 + A) will be the angles of a A formed by any side and the bisectors of the external angles between that side and the other sides produced. 52. If the exterior angles of a A be bisected, the three external As formed on the sides of the original A are equi- angular. 53. If a hexagon have its opposite sides equal and paral- lel, the three straight lines joining the opposite angles are concurrent. 54. Show that a parallelogram is symmetrical with respect to its centre. Book II. THE CIKCLE. Definitions. 179. A circle is a plane figure bounded by a curved line called the circiimfereilce, every point of which is equally distant from a point within called the centre. A radius is a straight line drawn from the centre to the circumference. A diameter is a straight line drawn through the centre, and terminated both ways by the circumference. Thus, in the figure, ABODE is the circumference; the space included within the circumference is the circle; is the centre; OA, OB, 00, are radii; AOO is a diameter. From the definition of a circle, it is evident that all its radii are equal; and also that all its diameters are equal, and each double the radius. 180. An arc of a circle is any part of the circumfer- ence, as AED. A semi'Circumference is an arc equal to one-half the cir- cumference. 181. A chord is the straight line which joins any two points on the circumference,* as AD. * The arc is sometimes said to be subtended by its chord, 74 BOOK 1L-DEFINITI0N8. 75 Chords of a circle are said to be equally distant from the centre, when the perpendiculars drawn to them from the centre are equal. The chord on which the greater perpen- dicular falls is said to he further from the centre, ROTE. — Every chord subtends two arcs whose sum is the circumference. A rd which does not pass through the centre, subtends two unequal arcs. Thus, AD subtends both the arc AED and the arc ABCD; of these, the greater is called the major arc, and the less the minor arc. Thus, the major arc is greater, and the minor are less than the semi-circumference. The major and minor arcs, into which a circumference is divided by a chord, are said to be conjugate to each other. When an arc and its chord are spoken of, the minor arc is always meant, unless otherwise stated. 182. A secant of a circle is a straight line of indefinite length which cuts the circum- ference in two points, as AB. A tangent to a circle is any straight line which meets the circumference, but, being pro- duced, does not cut it, as CD. Such a line is said to touch the circle at a point, and the point is called the ji^om^ of contact, or point of tangency, A secant may be considered as a chord produced, and a chord may be con- sidered as the part of the secant that is within the circle. If a secant, which cuts a circle at the points P and Q. be gradually turned about P as fixed, the point Q will ultimately approach the fixed point P. When the secant PQB reaches this limiting position, it be- comes the tangent to the circle at the point P. 183. Two circles are tangent to each other when they are tangent to the same straight line at the same point. They are said to have internal con- tact or external contact, according as one circle is entirely within or en- tirely without the other. 76 PLANE OEOMETRT. Circles that have the same centre are said to be con- centric. 184. A segment of a circle is the figure included between a chord and its arc, as ACB, or AFB. A segment is called a major or minor segment, according as its arc is a major or mmor arc. The chord of the segment is sometimes called the base of the segment. A semicircle is the segment included between a diameter and a semi-circumference, as DFE. 185. A straight line is inscribed in a circle when its extremities are on the cir- cumference, as AC. 186. An inscribed angle is one whose vertex is in the circumference, and whose sides are chords of the circle, as Z ACB. The /_ AGE, whose vertex is at the centre 0, is called the angle at the centre. 187. An angle in a segment is the angle formed by two straight lines drawn from any point in the arc of the seg- ment to the extremities of the base of the segment, as Z ACB or ZADB. 188. A sector of a circle is the figure bounded by two radii and the arc inter- cepted between them, as AOB, or BOC. 189. A polygon is said to be inscribed in a circle, when all its vertices are on the circumference, as ABCD. 190. A circle is said to be circum- scribed about a polygon, when the circum- ference passes through each vertex of the polygon. BOOK IL— DEFINITIONS. Tl Vm 191. A polygon is said to be circumscribed about a circle , when each of its sides is tangent to the circumference, as ABCD. The circle is then said to be inscribed in the polygon. 192. From the definition of a circle it follows; (1) The distance of a point from the centre of a circle is less than, equal to, or greater than the radius, according as the 2^oint isicithin, on, or without the circumference. Hence (157), the locus of a point, which is always at a given distance from a given point, is the circumference of a circle, of which the given point is the centre, and the given distance is the radius, (2) Circles of equal radii are identically equal. For, if one circle be applied to the other so that their centres co- incide, their circumferences will coincide, since all the points of both are at the same distance from the centre (179). (3) A straight line cannot meet the circumference of a circle in more than two points. For, if it could meet it in three points, these three points would be equally distant from the centre (179). There would then be three equal straight lines drawn from the same point to the same straight line, w^ich is impossible (64). 78 PLANE GEOMETRY. Arcs and Chords. Proposition 1 . Theorem. 193. Every diameter divides the circle and the circum- ference into two equal parts, and is greater than any other chord. Hyp, Let AB be the diameter of the ACB, and AC any chord not passing through the centre. To prove that AB bisects the O and the Oce, and that AB > AC. Proof. Join OC. Then Jet the plane containing the arc ACB be turned about AB as an axis until it falls on the plane of AEB. The arc ACB will coincide with the arc AEB, since every point in ACB is at the same distance from the centre as every point in AEB (179). Hence the arc ACB = the arc AEB, and the surface ACB = the surface AEB. (29) Also, AC < AO + OC, i.e., AO + OC > AC. But AO + OC = AO + OB, since OC and OB are radii of the same © (179). ,• . AO + OB > AC, or AB > AC. q.e.d. BOOK II.—ARCS AND CHORDS, 79 Proposition 2. Theorem. 194. In the smne circle, or in equal circles, equal arcs have equal chords, and suMend equal angles at the centre, Hijp, Let ABO, DEF be equal ^ Os, and AB, DE equal arcs. To prove I C^ / V F chord AB = chord DE, and Z ACB = Z DFE. Proof, Apply the O ABC to the O DEF, so that the centre C may fall on the centre F, and the radius CA on the radius FD. Then, because the Os are equal, (Hyp.) .*. the pt. A will fall on the pt. D, and their Oces will coincide, since all t/ie pis. of both are at iJie same distance from tJie centre (179). Because arc AB = arc DE, (Hyp.) . • . the pt. B will fall on the pt. E, and the chord AB will coincide with chord DE. (Ax. 11) .-. chord AB = chord DE. and Z ACB = Z DFE. Q.E.p. Note.— A line, straight or curved, is said to subtend a certain angle from a certain point when the lines drawn from the point to the ends of the line form that angle. Thus, the chord AB or the arc AB subtends the angle ACB frpm the point C. 195. Cor. Sectors of eqtcal arcs in the same or equal circles are equal. 80 PLANE GEOMETRY. Proposition 3. Theorem. 196. Li the same circle, or in equal circles, equal angles at the centre intercept equal chords and eqtial arcs on the circumference. Hyp, Let ABO, DEF be equal A^m^ D^rTTr^E OS, and let Z C = Z F. (\/| To prove \ ^ / V ^ chord AB = chord DE, V^_^ and arc AB = arc DE. Proof. Apply the O ABO to the O DEF, so that the centre may fall on the centre F, and the radius OA on the radius FD. Then, because the Os are equal, (Hyp.) .*. the pt. A will fall on the pt. D, and their ©ces will coincide. (179) Because Z = Z F, (Hyp.) .-. OB must fall on FE; and since OB = FE, (Hyp.) . ». the pt. B will fall on the pt. E. !N"ow since A coincides with D, and B with E, and the Oce of the O ABO with the ©ce of the O DEF, . • . chord AB must coincide with chord DE, (Ax. 11) and the arc AB must coincide with the arc DE. .• . chord AB = chord DE, and arc AB = arc DE. q.e.d. 197. Cor. 1. Sectors of equal angles at the centre in the same or eqtial circles are eqiiah 198. OoR. 2. In the same or equal circles equal chords subtend equal arcs and equal ajigles at the centre. BOOK IL-^ARCS AND CHORDS. 81 Proposition 4. Theorem. I 199. In the same circle, or in equal circles, of two un- equal minor arcs, the greater is subtended by the greater chord. D Hyp. In the ©ABO, let arc AB > arc AD. To prove chord AB > chord AD. Proof, Draw the radii CA, CD, CB. Because the arc AB > the arc AD, (Hyp.) . • . the pt. D must be between the pts. A and B, and .-. ZACB>ZACD. Then in the A s ABC, ADC, we have CB = CD, CA = CA, ZACB>ZACD. . • . chord AB > chord AD. since the As ham two sides equal each to eacJi, and the included As unequal (119). q.e.d. 200. Cor. Conversely: If the chord AB > the chord AD, the arc AB > the arc AD. For, if the arc AB were equal to the Jirc AD, the chord AB would be equal to the chord AD (194); and if the arc AB were less than the arc AD, the chord AB would be less than the chord AD (199). Til en, since the arc AB cannot be equal to, nor less than the arc AD, it must be greater. Hence, the greater chord subtends the greater minor arc* 82 PLANE GEOMETRY. Proposition 5. Theorem. 201. The radius which is perpendicular to a chord bisects the chord and/ the arc which it suUends, Hyp. In the O ABC let the radius OD be _L to the chord AB at E. To prove AE = EB, and arc AD = arc DB. Proof. Join OA, OB. Then, since OA = OB, (Radii) OE is common, rt. zAEO^rt.ZBEO, .-. aAOE= aBOE. (Hyp.) Two rt. AS are equal if they have the hypotenuse and a side equal each to each (110). and .-.AE^^EB, ZAOE= ZBOE. .•. arc AD = arc DB, since equal ^sat tJie centre intercept equal arcs on the Qce (196). q.e.d. 202. Cor. 1. TJie perpendicidar Usector of a chord passes through the centre of the circle, and bisects both the arc and the angle at the centre subtended by the chord. 203. Cor. 2. A radius which bisects a chord is i^erpen- dicular to the chord and bisects the subtended arc. 204. Cor. 3. A radius ichich bisects an arc is perpen- dicular to the chord of that arc at its middle point, and bisects the angle at the centre which the arc subtends. 205. Cor. 4. The locus of the middle point of parallel chords in a circle is the dia?mter perpendicular to these chordsy BOOK II.— ARCS AND CHORDS. 83 Proposition 6. Theorem. . 206. In the smne circle, or in equal circles , equal chords (I are equally distant from the centre; and, conversely, chords 11 which are equally distant from the centre are equal to one I another. I B (1) Hyp. Let AB, CD be equal chords in the O ABDO, and let OF,OE be J. to AB, CD. To prove OF = OE. Proof Join OA, OC. Then, since OF, OE are i. to AB, CD, /. AB, CD are bisected at F, E. . (201) But AB = CD. (Hyp.) .•.AP = CE. (Ax. 7) Also, AO = CO. (Kadii) .-. rt. A AOF = rt. A COE. (110) .-. OF = OE. (2) Hyp. Let the i.s OF, OE be equal. To prove AB = CD. Proof. Since OF = OE, (Hyp.) and *. OA = OC, (Eadii) . •. rt. A AOF = rt. A COE. (110) .-. AF = CE, and . • . AB = CD. (201 and Ax. 6) Q.E.D. > 84 PLANE GEOMETRY. Proposition 7. Theorem. 207. In the same circle , or in equal circles, of two un- equal chords, the less is at the greater distance from the centre. Hyj). In the O ABD, let the chord CD < the chord AB, and let the J_s OE, OF be drawn from the centre to CD, AB. To prove OE > OF. Proof I Since chord AB > chord CD, .-.arc AB >arcCD. (200) On arc AGB take arc AG = arc CD. Draw the chord AG, and the J_ OH. Since arc AG = arc CD, . • . chord AG = chord CD. Equal arcs have equal chords (194). And .-.OH^OE. Equal chords are equally distant from tJie centre (206). Because arc AG < arc AB, the pt. G must fall within the arc AB. .* . _L OH will cut the chord AB at some pt. K, Now, OH > OK. The whole is > any of its parts (Ax. 8). And OK > OF. T/ie ± is the shortest distance from a pt. to a line (58). .-. a fortiori, OH > OF. .•.OE>OF. Q.E.D 208. Cor. Conversely, of two chords unequally distant from, the centre, the one which is at the greater distance is the less, EXERCISE. If two chords of a circle cut each other, and make equal angles with the sti*aight line which joins their point of iutersection to the centre, prove that the chords are equal., BOOK II.—THE CIHGLE. TANGENTS, 85 Proposition 8. Tiieorem. 7 209. A straight line 'perpendicular to a radius at its extremity is a tangent to the circle. Hyp. Let be the centre of a O, OA the radius, and BC a line J. to OA at A. To prove BC tangent to the O. Proof, In BC take any pt. D, other than A; join OD. Then, since OA is _L to BC, (Hyp.) .-. OA < OD. Tlie L is the shortest distance from a pt. to a line (58). and .'. the pt. D is without the circle. [192. (1)] BC has every pt. except A without the O, . • . BC is a tangent to the O at A. (182) Q.E.D. 210. Cor. 1. Conversely, a tangent to a circle at any point is perpendicular to the radius drawn to that point, 211. Cor. 2. The perpendicular to a ta^igent at the point of tangency passes though the centre of the circle, 212. Cor. 3. The straight line drawn from the centre perpendicular to the tangent meets it in the point of con- tact, 213. Cor. 4. Ofily one tangent can he drawn to a circle at a given point on the circumference. 86 PLANE GEOMETRY. i Proposition 9. Theorem. 214. Two parallel lines intercept equal arcs on the cir- cumference. There may he three cases. Hyp. Let AB, CD be the || s. Case I. When AB, CD are secants. To prove arc AC = arc BD. Proof. Draw the radius OE J_ to one of the II s. It will then be J. to the other || . A St. line ± to one of two \\sis ± to the other (71). .-. arc AE = arc BE, and arc CE = arc DE, (201) . • . arc AC = arc BD. (Ax. 3) Case II. When AB is a tangent and CD is a secant. To prove arc CE = arc DE. Proof. Draw the radius OE to the pt. of contact E. Then OE is J. to AB (210), and to its || CD (71). .-. arcCE = arcDE, (201). Case III. When AB, CD are tangents. To prove arc EMH = arc ENH. Proof. Draw the secant MN || to AB. Then arc ME = arc NE, and arc MH = arc NH. Adding, arc EMH = arc ENH. q.e.d. 215. Cor. 1. Conversely, if the arcs intercepted hy tico secants are equal, the secants are parallel. 216. Cor. 2. Tlte straight line joining the points of coiir tact of tioo parallel taiigents is a diameter. l]^^ ase II) ¥ BOOK II.-THE CIRCLE. 87 Proposition lO. Thieorem. 217. Tliroxigh three given points not in the saine straight line, one circumference, and only one, can be draion, I ^^^^. ;^ I \ "\ I?/ B Hyp, Let A, B^ bo the three given pts. not in a st. line. To prove that one Oce, and only one, can be drawn through A, B, 0. Proof, Join AB, BO. Bisect AB, BC by the _Ls DF, EG. Since AB, BC are not in the same st. line, (Hyp-) .*. the JLs DF, EG must meet at some pt. 0. (76) Because G is in the J_ DF, .'. it is equidistant from A and B. (66) And because G is in the J_ EG, . *. it is equidistant from B and 0. (66) Then, because G is equidistant from A, B, C, .-. the Oce described with centre G and radius GA will pass through A, B, C. Again, only one Oce can be so described. For if any Oce pass through A, B, C, its centre will be at once in the J_ bisectors DF, EG, and .*. at their pt. of intersection. But two st. lines cannot intersect in more than one pt. .*. there is only one Oce that can pass through A, B, and C. Q.E.D. 218. Cor. 1. Tivo circumferences cannot intersect in more than two points, 219. Cor. 2. Two circumferences ivhicli have three points cojnmon coincide. PLANE GEOMETRY. Relative Position of Two Circles. Proposition 1 1 . Theorem. 220. If Uoo circumfeveiices intersect each other, the right line joining their centres bisects their common chord at right angles. Hyp, Let 0, 0' be the centres of two Oces which intersect each other; and A, B their pts. of inter- section. To 2Jrove that the line 00' bi- sects AB at rt. Z s. Proof. Because and 0' are each equally distant from AandB, (179) .•. the line 00' bisects AB at rt Z s. (67) 221. CoR. 1. Conversely, the perpendicular bisector of a common chord passes through the centres of both circles, 222. Cor. 2. If we suppose the circles to be moved so that the point A approaches the line 00', tlie pt. B will also approach the line; and since the line 00' is always perpendicular to the mid- dle of AB (220),, the two points A and B will ultimately come together on the line 00', and be united in a single point common to the two circles. The common chord AB will then be a common tangent to the two circumferences at their point of contact. Hence, when tivo circumferences are tangent to each other, their point of contact is in the straight line joining their centres ; and the perpe7idicular at this 2)oint is a common tangent to the two circumferences. 1 BOOK II.—TWO CIRCLES. Proposition 12. Theorem. 223. If two circumferences intersect each other, the dis- tance between their centres is less than the sum and greater Vian the difference of the radii. \ Hyp. Let the Os with centres 0, 0' intersect at A. Join 00', AO, AO'. To yrove 00' < OA + AO', and > OA - AO'. Proof. In the aOAO' 00' < OA + AO', and 00' > OA - AO'. Eiilier side of a t^ < the sum and > ilie difference of the other two sides (96). q.e.d. 224. Cor. 1. If tlie distance of the centres of two circles is greater than the sum of their radii, they are tvholly exterior to each other. 2t2ib. Cor. 2. If the distance of the centres of two circles is equal to the sum of the radii, they are tangent exter- nallij. 226. Cor. 3. If the distance of the centres is less than the sum and greater than the difference of the radii, the circles intersect. 227. Cor. 4. If the distarice of the centres is equal to the diff^erence of the radii, the circles are tangent internally. 228. Cor. 5. If the distance of the centres is less than the diff'erencel^of the i^adii, one circle is wholly luithin the other. ScH. If two circles intersect and the radius of either circle drawn to a point of section touches the other circle, the circles intersect orthogonally, i.e., at right angles. 90 PLANE GEOMETRY. EXERCISES. 1. Through a given point P either Jnside or outside a given circle whose centre is 0, two straight lines PAB, PCD are drawn making equal angles with OP, and cut- ting the circle in A, B, C, D: prove that AB = CD, and PA = PC. 2. P is a point inside a circle whose centre is : prove that the chord which is at right angles to OP is the shortest chord that can be drawn through P. Let APB be ± to OP, CPD any other chord through P: draw OE ± to CD, etc. 3. If two circles cut each other, any two parallel straight lines drawn through the points of intersection to cut the circles are equal. 4. Two circles whose centres are A and B intersect at C; through C two chords DCE, FCG are drawn equally in- clined to AB and terminated by the circles: show that DE = FG. 5. Prove that the two tangents drawn to a circle from an external point are equal and equally inclined to the straight line joining the point to the centre of the circle. 6. A is a point outside a given circle whose centre is 0; with centre A and radius AO a circle is described, and with centre and radius equal to the diameter of the given circle another circle is discribed cutting the last in B; OB is joined cutting the given circle in E: prove that AE is tangent to the given circle. The Measukement of Angles. 229. To measure a quantity is to find how many times it contains another quantity of the same kind taken as a standard of comparison. This standard is called the irnit. Thus, if we wish to measure a line, we must take a unit of lengtlif and see how many times it is contained in the line to be measured. BOOK IL—TIIE CIRCLE. 91 The number which shows how many times a quantity contains the unit, is called the numerical measure of that quantity. 230. The rehitive magnitude of two quantities, meas- ured by the number of times which the first contains the second, is called their ratio. Thus, the ratio of A to B is y , or A : B. Since the ratio of two quantities is found by dividing the first by the second, therefore the ratio of two quantities is the number which would express the measure of the first, if the second were taken as unity. The ratio of two quantities is the same as the ratio of their numerical measures. Thus, if A contains the unit rn 28 times, and B contains ., ^ ^. , A 28w 28 it 9 times, we have ^- = -7: — = -7^ . B 9m 9 231. When a quantity is contained an exact number of times in two quantities of its kind, it is called their com- mon measure. Two quantities are commensurable when they have a common measure. The ratio of two commensurable quantities can be ex- pressed by a whole number or by a fraction. Thus, if each of the two lines A and B contains some line an exact number of times. A' ^ ^ ■■ • as for example if A contains it 5 ^^___^___^__^___^ times and B contains it 4 times, the two lines A and B are com- C' • mensurable, and the line C is their common measure. Their ratio is expressed by the fraction f . If is not contained an exact number of times in A and B, but if there be a common measure which is contained, say, 25 times in A and 19 times in B, then the ratio of A to B is the fraction f f . 92 PLANE GEOMETRY. Generally, if there be a common measure which is con- tained m times in A and 7i times in B, their ratio will be m n ' 232. Two quantities are incommensuralle when they have no common measure. The ratio of such quantities is ^ ^^jcalM an incornmcnsurable ratio. This ratio cannot be ex- ^^^^actlyexpressed in figures; but its numerical value can be obtained approximately as near as we please. Thus, suppose A and B are two lines whose ratio is V2, We cannot find any fraction which is exactly equal to V"2', but by taking a sufficient number of decimals, we may find V^ to any required degree of approximation. Thus, V^= 1.4142135 . . . . , and therefore 1^2 > 1.414213 and < 1.414214. That is, the ratio of A to B lies between iHo f o o ^^^ To-o MM y ^^^d therefore differs from either of these ratios by less than one-millionth. And since the decimals may be continued without end in extracting the square root of 2, it is evident that this ratio can be expressed as a fraction with an error less than any assignable quantity. In general, when A and B are incommensurable, divide B into n equal parts each equal to x, so that B = nx, where n is an integer. Also let A > mx but <(y;i + 1)^;; tlien A ^ mx T ^ Un + l)x ^ > — and < ^^ ' — '- : B nx nx that is, :jT lies between — and -; so that ^ differs B n n n m . 1 from — by a quantity less than — . And since n can be taken as great as we please, - may be made as small as we n BOOK IL—MEASUnEMENT OF ANGLES. 93 please until it becomes less than any assignable value, though it can never reach absolute zero. We say, therefore, that zero is the limit of - as n is in- creased indefinitely. Hence, Uvo integers can he found whose ratio loill express the ratio of two incominensurable quantities to any re- quired degree of accuracy. 233. Theorem. Two incommensuraUe ratios are equal, if their approximate numerical values always remain equal while the common measure is iiidefi^iitely diminished. Hyp, Let A:B and A': B' be two incommensurable m ratios, whose true values always differ from — by less than — . n To prove ' A:B = A':B'. Proof Since each of the ratios ^, :^, differs from - by less than — , A A' 1 the difference between ^r and vr> must be < - . B B' n But, by diminishing the common measure, n can be made as great as we please, and therefore — may be made as small as we please, i, e., less than any assignable quantity, however small. Hence, the difference between A : B and A': B' must be less than any assignable quantity, however small. .-. A:B = A':B'. 94 PLANE GEOMETRY. Proposition 13. Theorem. 234. In the same circle, or in equal circles, angles at the centre are in the same ratio as their intercepted arcs. Hyp, Let AOB, AOC be any two Z s at the centres of two equal Os, /^ ^\ /^^^^\ and AB, AC their intercepted [ O \( o j arcs. \ / ' ' ''/V/ / 'i \/ • ' ^* ^\/d ZAOB arcAB ^^-i-^Ll /nLL>^^ To prove ^^^ = -_^_. „ Case I. When the arcs are commenstiraUe, Proof. Take M, any common measure of AB and AC, and suppose it to be contained five times in AB and four times in AC. rrn arc AB 5 Then ^-^ =z - (1) arc AC 4 ^ ^ Draw radii to tlie several pts. of division of tlie arcs AB, AC, dividing the Z AOB into five / s and / AOC into four Zs. These Z s are all equal. In the same © or in equal Qs equal arcs subtend equal /.s at the ceirtre (194). ZAOB 5 (2) (Ax. 1) ' * ZA0C""4' Therefore, from (1) and (2), ZAOB _ arc AB Z AOC "arc AC' Case II. When the arcs are incomme^isurahle. Proof. In this case we know (232) that we may always find an arc AD as nearly equal as 2veplease to AC, and such that AB, AD are commensurable. Join OD; then Z AOB arc AB ,^ ^, "^XrvTk = tt^' (Case J ) Z AOD arc AD '^ ^ BOOK n.-MEASUREMENT OF ANGLES. 95 Now, these two ratios being always equal while the com- mon measure is indefinitely diminished, they will be equal when D moves up to and as nearly as we please coincides with 0. (233) I Z AOB arc AB ZAOG~arc AG* ^ 235. CoE. Li the same circle, or in equal circles, sectors are in the same ratio as their arcs; for sectors are equal when their arcs are equal. (195) 236. ScH. Since the angle at the centre of a circle, and the arc intercepted by its sides increase and decrease in the same ratio (234), the numerical measure of the angle is the same as that of the arc. This theorem, being of very frequent use, is expressed briefly by saying that an angle at the centre is measured hy its interce'pted arc.^ This means simply that an angle at the centre is the same part of the whole angular magnitude about the centre that its inter- cepted arc is of the whole circumference. 237. The circumference is divided, like the angular magnitude about the centre (28), into 360 equal parts called degrees. The degree is divided into 60 equal parts called minutes, and the minute into 60 equal parts called seconds. Hence the unit of angle and the unit of arc are both called a degree. When the angle becomes a right angle (20), the arc becomes a quarter of the circumference, or a q\iadrant. When the angle becomes a straight angle (21), the arc be- comes a semi-circumference, and so on. A right angle and a quadrant are both expressed by 90". Two right angles and a semi-circumference are both expressed by 180°. Four right angles and a circumference are both expressed by 360°. *Boucb6 et Comberousse, p. 64. 96 PLANE GEOMETUY. I Proposition 14. Theorem. 238. Ail inscribed angle is measured hij one-half the arc intercepted between its sides. Hyp. Let BAG be an Z inscribed in tlie OABO and intercepting the arc BO. To prove Z BAG is measured by ^ arc BG. Gase I. When tlie centre is loithin the /BAG. Proof. Join AO, produce it to D, and join OB, OG. In aAOB, Z BOD = Z OAB + Z OBA. The ext. /. of a A equals the sum of the opp. int. Z« (98). But, since OA OB, .-. Z OAB = Z OBA, being opp. equal sides (111). .-. Z BOD = 21 OAB. Similarly, Z GOD = 2Z OAG. .-.whole ZBOG :=2zBAG. But Z BOG is measured by arc BC. The Z at the centre is measured by the intercepted arc (236) .*. 2 Z BAG is measured by arc BO. .*. Z BAG is measured by ^ arc BG. Gase II. Wlien the centre is ivithoiit the ZBAG. Proof. Join AO, produce it to D, and join OB, OG. Then, Z DOB = 2 Z DAB, (Gase I) and Z DOG = 2 Z DAG. (Gase I) .•.ZD0G-ZD0B = 2ZDAG -2ZDAB. .-.Z BOG ==2 ZBAG. (Radii) (Ax. 2) f BOOK TI.~MEASUREMENT OF ANQLE8. 97 But Z BOC is measured by arc EC. (236) .-. Z BAG is measured by ^ arc BC. q.e.d. Note.— This theorem is equally true when the angle at the centre is greater than two right angles, as the student may show. 239. Cor. 1. All migles inscribed q in the same segment are equal; for each is measured by one-half the same arc AFB. 240. Cor. 2. Every angle AHB, inscribed in a semicircle, is a right angle; for it is measured by one-half a semi-circumference, or by a quad- A rant (237). 241. Cor. 3. Evertj angle BAC, inscribed in a segment greater than a semicircle, is an acute angle; for it is measured by one-half the arc BDC, which is less than a quadrant. Every a7igle BDC, inscribed in a segment less than a semicircle, is an obtuse aiigle; for it is measured by one- half the arc BAC, which is greater than a quadrant. 242. Cor. 4. The opposite angles of an inscribed quad- rilateral are supplementary ; for the sum of the Zs A and ,._-- D isjn easured by one-h alf_t he Qce, wh ich is the measure '^ ^ritwo right Z s (237); therefore the Zs are supplement- ary (25). §B PLANE GEOMETRY. Proposition 1 5. Tiieorem. 243. An angle formed ly a tangent and a chord from the point of contact is measured iy one-half the intercepted arc. Hyp, Let AC be a tangent to the ^ OBHE at B, and BD any chord of the ^-^ Ofrom B. / To prove Z CBD is measured by / i arc BHD. ( Proof. From B draw BE at rt. Z s \ to AC.' \^^ BE is a diameter of the O. ^ ^ C TJie 1 to a tangent at the pt. of contact passes tJiroiigh the centre of the O (311). Then, rt. Z CBE is measured by J semiOce BHE, and Z DBE is measured by \ arc DE. (238) Therefore, subtracting, Z CBD is measured by \ arc BHD. Similarly, Z ABD is measured by ^ arc BED. Q.E.D. 244. ScH. This proposition is a particular case of Prop. 14. Thus, let the side BD remain fixed, while the side BH turns about B, as in (28), until it becomes the tangent BC at the point B. In every position of the chord BH, the inscribed angle HBD is measured by half the intercepted arc HD. Therefore, when the chord BH becomes the tan- gent BC, the angle CBH is measured by half the arc BHD. EXERCISES. 1. If the angle BAC at the circumference of a circle be half that of an equilateral triangle, prove that BC is equal to the radius of the circle. 2. If a hexagon be inscribed in a circle, show that the sum of any three alternate angles is four right angles. 3. If two circles intersect in the points A, B, and any two lines ACD, BFE, be drawn through A and B, cutting one of the circles in the points C, E, and the other in the points D, F, the line CE is parallel to DF. BOOK II.-MEASUREMENT OF ANGLES. 99 Proposition 1 6. Theorem. 245. An angle formed ly two cJiords wliich intersect witlmi a circlcy is measured by one-half the sum of the arcs intercepted hetweenits sides and between its sides 2)roduced. Hyp, Let BD, CE be two chords intersecting at A within the O BODE. To prove Z BAG is measured by ^ (arc BO + ^rc DE). Proof Join BE. ZBAC = ZAEB + ZABE. TJie ext. /_ of a t. equals tJie sum of the opp. int. Zs (98). But Z AEB is measured by J arc BO, and Z ABE is measured by J arc DE. (238) adding, Z BAO is measured by J (arc BO + arc DE). Q.E.D. EXERCISES. 1. If arc BO = 84° and Z CAD is a rt. I, how many de- grees are there in the arc-DE ? 2. The side^ of a quadrilateral touch a circle, and the straight lines drawn from the centre of the circle to the vertices cut the circumference in A, B, 0, D : show that AC, BD, which intersect inside the circle, are at right angles to each other. 100 PLANE GEOMETRY. Proposition 1 7. Theorem. 246. A7i angle formed hy Uvo secants which intersect loithout a circle, is measured by one-half the difference of the intercepted arcs. Hyp. Let AC, AB be two secants intersecting at A without the O BCED. To prove Z A is measured by \ (arc BC - arc DE). Proof, Join BE. ZBEO=ZA+ZB (98) .•.ZA= ZBEC- ZB. But Z BEC is measured by ^ arc BC. and Z B is measured by \ arc DE. (238) .*. Z A is measured by | (arc BC — arc DE). q.e.d. 247. ScH. Prop. 14 may be considered as a special case of Props. J 6 and 17 by conceiving AB in (245) and (246) to move parallel to its present position until D reaches E. When D reaches E, the arc DE becomes zero, and BAC be- comes an inscribed angle, measured by half its intercepted arc. EXERCISES. 1. If arcBC = 80° and zB degrees in the angle A. / , 14°, find the number of w 2. A, B, C are three points on the circumference of a circle, the bisectors of the angles A, B, C meet in D, and AD produced meets the circle in E : prove that ED = EC. 3. If a quadrilateral be described about a circle, the angles at the centre subtended by the opposite sides are supplemental. BOOK IL-MEASUliEMENT OF ANGLES, 101 Proposition 18. Theorem. 248. An angle formed hy a tangent and a secant is measured hy one-half the difference of the intercepted arcs. C H Hy]?. Let AC, AB be a tangent and a secant intersect- ing at A. To 2Jrove Z A is measured by \ (arc BHE — arc DE). Proof Join BE. ZBEC = ZA+ZB (98) .-. ZA=: ZBEO-ZB. But Z BEG is measured by | arc BHE, (238) and Z B is measured by ^ arc DE. (238) .'. Z A is measured by \ (arc BHE — arc DE). Q.E.D. 249. Cor. The angle formed hy ttoo tangents is meas- ured hy one-half the difference of the intercepted arcs, EXERCISES. 1. Two tangents AB, AC are drawn to a circle; D is any point on the circumference outside the triangle ABC: show that the sum of the angles ABD and ACD is constant. 2. If a variable tangent meets two parallel tangents it subtends a right angle at the centre. 102 PLANE GEOMETRY. Quadrilaterals. Proposition 1 9. Theorem. 250. If the opposite angles of a quadrilateral are sup- plementary, the quadrilateral may he inscribed in a circle. Hyp. Let ABCD be a quadrilateral in wliich ZB+ ZD=2rt. /s. To prove the pts. A, B, C, D are in the same O. Proof Through the three pts. A, B, C describe a O. If this O does not pass through D, it "" '" will cut AD, or AD produced, at some other pt. than D. Let E be this pt. Join EC. Because the quadrilateral ABCE is inscribed in a O, .-. ZABC+ zAEC = 2rt. Zs. T?ie opp. Z.sof an inscribed quad, are supplementary (242). But Z ABC + Z ADC = 2 rt. Z s. (Hyp.) .-. Z ABC + Z AEC = Z ABC + Z ADC. (Ax. 1) .-. ZAEC= ZADC; (Ax. 3) that is, an ext. Z of a A == an int. opp. Z , which is im- possible. (98) .*. the circle which passes through A, B, C, must pass through D. q.e.d. 251. Def. Points which lie on the circumference of a circle are called concyclie. A cyclic quadrilateral is one which is inscribed in a circle. EXERCISE. If two opposite sides of a cyclic quadrilateral be pro- duced to meet, and a perpendicular be let fall on the bi- sector of the angle between them from the point of inter- section of the diagonals : prove that this perpendicular will bisect the angle between the diagonals, BOOK IL—qUADlULATEHALS, 103 V Proposition 20. Theorem. 252. hi any quadrilateral circiunscribing a circle, the um of one pair of opposite sides is equal to the sum of the ther pair. Hyp, Let ABCD be a quadri- lateral circumscribing a O. To prove AB + CD = AD + BO. Proof. From the centre draw the radii to the pts. of contact E, Y, G, H, and draw OB. Thenrt.AOBE=rt.AOBF,(no) ^ .-. EB = rB. Similarly, EA = HA, GD = HD, GO = FO. Adding these four equations, we have EB + EA + GD + GC = FB + HA + HD + FO, or AB + CD = BC + AD. q.e.d. EXERCISES. 1. The Ime joining the middle points of two parallel chords of a circle passes through the centre. 2. The chords that join the extremities of two equal arcs of a circle towards the same parts are parallel. 3. The sum of the angles _su btended a t the centre of a circle by two opposite sides of a circumscribed quadrilateral is equal to two right angles. 104 PLANE GEOMETRY. 4. An isosceles triangle has its vertical angle equal to the exterior angle of an equilateral triangle. Prove that the radius of the circumscribing circle is equal to one of the equal sides of the given triangle. . 5. Prove that the radius of the circle inscribed in an ^(.equilateral triangle is equal to one- third of the altitude of I \he triangle. 6. The quadrilateral ABCD is inscribed in a circle; AB, DC produced meet in E: prove that the triangles ACE, BDE, and also the triangles ADE, BCE, are mutu- ally equiangular. 7. The quadrilateral ABCD is inscribed in a circle; AB, DC produced meet in E, and BC, AD produced meet in F; the sides AB, BC, CD subtend arcs of 120°, 70°, 80°, respectively: find the number of degrees in the angles AED and AFB. 8. In the circumscribed quadrilateral ABCD, the angles A, B, C are 110°, 95°, 80°, respectively, and the sides AB, BC, CD, DA touch the circumference at the points E, F, G, H respectively: find the number of degrees in each an- gle of the quadrilateral EFGH. / 9. If two opposite sides of an inscribed quadrilateral are equal, prove that the other two sides are parallel. Problems of Construction". 253. Hitherto, our investigations have been purely theoretical, and have been confined to the demonstration of certain properties of figures, assumed to exist, satisfying certain conditions ; and our figures have been assumed to be constructed under these conditions, although no methods of constructing them have been given. Indeed, the precise construction of the figures was not necessary, as they were required only as aids in following the demonstration of principles* BOOK IL-PROBLEMS OF CONSTRUCTION. 105 The constructions were only hypothetical. We now ap- ply these principles to determine methods by means of which such figures are to be approximately drawn; for, owing to the imperfection of our instruments, the ideal state contemplated in theoretical Geometry cannot be at- [tained by them. The determination of the method of constructing a given ^figure with given instruments is called di. proUem ; and the solution oi a problem requires us to shoio how the construc- tion can be affected by the use of the given instruments, and to prove that the construction is correct. The solution of a problem depends on the instruments that are used. The more restricted the choice of instruments, the more limited will be the problems which can be solved by their use ; and the more difficult will be the solution of many that are thus solved. It IS the recognized convention of Elementary Geometry that the only instruments to be employed are the ruler and compasses, with the use of which the student should be- come familiar. The ruler is used for drawing and produc- ing straight lines, and the compasses for describing circles and for the transference of distances; the straight line and the circumference being the only lines treated of in this subject. This convention is embodied in the three postu- lates given in (45). Problems, though important as applications of geometric truths, form no part of the chain of connected truths em- bodied in the theorems of Geometry, so that, though they may be studied advantageously in* connection with the theorems on which they directly depend, they are not a necessary part of the pure science of Geometry.* * Elements of Plane Geometry, Association for the improvement of Geometric Teaching, p. 69. 106 PLANE GEOMETRY, Proposition 2 1 . Problem. 254. To Used a given finite straight line. I -^ -f=^-i- 1 \ I / :d Given, the line AB. Required, to bisect it. Cons, With A and B as centres, and eqnal radii greater than one-half of AB, describe two arcs cutting each other at C and D. Join CD cutting AB at E. Then AB is bisected at E. Pro of^mQQ the two pts. C a nd D are equa llj distan t from K and B, the st. ImeUD Is jL to AB at its mid- ^HTe'pt:' ~^ (67) Q.E.F. EXERCISES. ^ 1. If two chords intersect at right angles within a circle, prove that the sum of tfae opposite intercepted arcs is equal to a semi-circumference. 2. If the angles A, B, C of a circumscribed quadrilateral ABCD are 120°, 80°, and 100°, and the sides AB, BC, CD, DA touch the circumference at the points E, F, O, H, find the number of degrees in each angle of the quadri- lateral E, F, G, IT. 3. Divide a line into four equal parts, BOOK Il.-PROBLEMS OF CONSTRUCTION. 107 Proposition 22. Problem. p 255. To bisect a given arc or a given angle. (1) Given, the arc AB. Required, to bisect it. Go7is. Join AB. With A and B as centres, and with equal radii, describe arcs intersecting at C and D. Draw CD cutting the arc AB at E. Then the arc AB is bisected at E. Proof, Since the two pts. C and D are equally distant _f rom A and B, the st. line CD is J_ to the chord AB at its mid. pt. (67), and therefore bisects the arc (202). (2) Give7i, the ZACB. Required, to bisect it. Cons, With C as a centre, and with any radius, describe an arc cutting CA and CB at D and E. With D and E as centres, and with a radius > | the st. line DE, describe two arcs intersecting at H. Join CH. Then CH bisects the Z ACB. Proof, Join DE. Since the two pts.^_and E^ are equally distant fromC and H, the st. line CH is _L to the chord DE at its mid. pt. (67), andlherefore bisects the arc DE and the Z'DCE (202). Q.e.f. EXERCISES. 1. Divide an arc into four equal parts. 2. Divide an angle into four equal parts. 108 PLANE GEOMETRY. Proposition 23. Problem. 256. At a given point in a given straight line to draw a perpendicular to that line, Oiven, the pt. C in the line AB. Required, to draw from C a i. to ">k AB. j First Method. i D I E Cons, With as a centre, and ^ ' C ' B with any radius, describe arcs of a O cutting AB at D and E. With D and E as centres, and with equal radii greater than DC, describe two arcs cutting each other at H. Join HC. Then HC is the required _L. Proof. Since the twojQts^ C and H are equally di stant from D an037the stT line CH is _L to DE at its mid, pt. C. (67). Second Method. Cons. With any pt. without AB as I o a centre, and with the distance OC as ^^AE a radius, describe a Oce cutting AB /' |\ at C and D. ^.-'^ ^ |/ Join DO, and produce it to meet A D \^ /' C B the Oce at E. Join EC. , " ' Then EC is the required _L. Proof, Since the Z ACE is inscribed in a semicircle, it is a rt. Z ; . •. EC is _L to AB at C (240). q.e.f., BOOK IL-PROBLEMS OF CONSTRUCTION. l09 Proposition 24. Problem. 257. From a given point without a given straight line to draw a perpendicular to that line. Given, the pt. and the line AB. Required, to draw a _L from to AB. Cons. With C as a centre, and with a radius sufficiently great, describe an arc cutting AB at D and E. With D and E as centres, and with a radius > ^ of DE, describe arcs cutting each other at G. Join CG, cutting AB at H. Then OH is the required _L. Proof. Since the two pts. C and G are equally distant from D and E, the st. line CG is J_ to DEaUts mid-pt. (67) Q.E.F. EXERCISES. 1. In an indefinite straight line AB find a point equally ( distant from two given points which are not lolh on AB. When does tliis problem not admit of solution? 2. In a given straight line MN" find a point P such that PA, PB, drawn from P to two given points A, B, on oppo- site sides of MN, may make equal angles with MN, no PLANE GEOMETRY. Proposition 25. Problem. 258. At a given point on a given straiglit line to con- struct an angle equal to a given angle, E N / A MB Given, the pt. A, the line AB, and the Z 0. Reqniredy to make at A an Z = Z C. Co7is, With centre C and any radius, describe an arc cutting CD and CE at G and H. Join GH. With centre A and the same radius CG, describe an in- definite arc MN. With centre M and radius = GH, describe an arc cutting the arc MN at F. Join AF. Then ZA= ZC. Proof. Since chord MN = chord GH, (Cons.) . • . arc MN = arc GH, and Z A = Z C. In equal Qs equal cJiords subtend egual arcs, and equal Zs at the centre (198). Q.E.F, EXERCISES. 1. Given an angle, to construct its supplement, 2. Construct an angle of 45°. 3. Construct an angle of 22J°. BOOK U.-FUOBLEMS OF CONSTRUCTION. Ill Proposition 26. Problem. 259. Two angles of a triangle being given to find the third. Given, Z A and / B of a A . Required, to find the third / of theA. ^ ^ H\ Cons, Draw the indefinite straight \ ^^F line CD. At any pt. E in this line, ^^v^-rJll makeZDEFn: ZB, C E D and zCEHrr zA. . (258) Then Z FEH is the Z required. Proof. Since Z CEH + Z HEF + Z FED = 2 rt. Z s, (53) and ZA + zB + requiredZ = 2 rt. Zs, (97) and Z A =: z CEH, and Z B = Z FED, (Cons.) . • . Z HEF is the Z required. q.e.f. Proposition 27. Problem. 260. Through a given point to draio a straight line parallel to a given straight line. Given, the pt. A and the line BC. /^ Required, to draw through A a /' line II to BC. / Cons. In BC take any pt. D, and B D C join DA. At pt. A make Z DAE = Z ADC. (258) Then '" EAF is || to BC. Proof Since Z DAE = Z ADC, (Cons. ) .-. EFislltoBC, being alt.-iiit. As. (75). q.E.F, 112 PLANE GEOMETRY. Proposition 28. Problem. 261. Oiven tivo sides and the i?icluded angle of a tri- angle, to construct the triangle. Given, two sides a, h, and the in- cluded Z A. Required, to construct the A . Cons, Draw the line AB = a. At A make the / BAD= / A. (258) On AD take AC = b. Join CB. Then ABC is the A required. Proof, To be supplied by the student. q.e.f. Proposition 29. Problem. 262. Given a side and the tivo adjacent angles of a triangle, to construct the triangle. Given, the side a and the adj. ZsA, B. Required, to construct the A . I a Cons. Draw the line AB = a. C/'^^^^ At A make the Z BAD = Z A. (258) '/ ^Jl^^--:. At B make the Z ABE= Z B. (258) ^ a B The lines AD and BE will intersect at some pt. C. Then ABC is the A required. Proof. To be supplied by the student. q.e.f. 263. ScH. 1. Since the third angle of a triangle can be found when two angles are given (259), therefore, if a side and any two angles of a triangle are given, the triangle may always be constructed (262). 264. ScH. 2. This problem is possible only when the two given angles are together less than two right angles. BOOK IL-PliOBLEMS OF CONSTRUCTION. U3 Proposition 30. Problem. 265. Given the three sides of a triangle to construct the triangle. / \ c/ \b / \ /- \ B a C Given, the three sides a, h, c. Required, to construct the A . Cons. Draw the line BC = «. With C as a centre, and a radius = h, describe an arc. With B as a centre, and a radius = c, describe an arc, cutting the former arc at A. Join BA and CA. Then ABC is the A required. Proof. To be supplied by the student. Q.E.F. A EXERCISES. 1. Show how the construction fails if one side is greater than the sum of the other two. 2. Construct an equilateral triangle on a given base. /, — A. ^ ^ A' '^, a--"E /^'^x b^ -> / 1 \ 1 \ a ^-"/ la 1 NT -V — .^=t^_ \ B^N. H ,• 114 PLANE GEOMETRY, Proposition 31 . Problem. 266. To construct a triangle having given two sides and the angle opposite one of them. Given, the sides a, h, and / A opp. a. Required, to construct the A . Case I. When a , =, or < CH. Therefore, there may be two As, one A; or none •it all. BOOK II.— PROBLEMS OF CONSTRUCTION. 115 Case IL When a ^h, and /_K is acute. In this case the arc described /^ with centre C and radius a, cuts ^^^ AD at the two pts. A and B; and b/ \a hence there is but one solution: the ^ / \ / isosceles A ABU. ^i^Z, ~~yB"D Case III. When a>h, "When Z A is acute the arc de- /^ scribed with centre C and radius a ^'^^ cuts AD in B and B', the latter pt. ^^/b ^^"^X at the left of A in DA produced. gV — -^ "/q"^ Then ABC is the A required, X._ ,,-•'' since it satisfies the given condi- tions, and there is only one solution, for the A AB'C does not contain the given / A. When Z A is ohtu.se, as / CAB', there is only one solu- tion: the A AB'C, since the A ABC does not contain the given obtuse Z . When Z A is right there are two C equal rt. As: ABC and AB'C. /|\ When Z A is right or obtuse, and V ^i V r; =: or < h, the problem is impos- __:x^ J_ ^ sible; for the side opposite the right B' •-...... A --"b or the obtuse Z is the greatest side of the A (117). Q.E.F. EXERCISES. - 1. Construct a right triangle having given the hypote- nuse and one side. 2. Construct au isosceles right triangle on a given straight line as hypotenuse. -^--D 116 FhANE GEOMETRY, Proposition 32. Problem. 267. Given two adjacent sides and the included angle of a parallelograniy to construct the parallelogram^. Given, the sides a, h, and the ZA. Required, to construct the C7. ^ Cons. Draw AB = «. At A make the Z B AE = Z A. On AE take AC = h /E^ With C as a centre, and a ra- c/^^ ^^ dius = a, describe an arc. b/ / With B as a centre, and a ra- / / dins = b, describe an arc cutting A ~a b the first at D. Join CD, BD. Then ABCD is the required OJ, Proof. Since AB = CD and AC = BD (Cons.), the figure is a /Z7 (132), and it is the one required; for two /Z7s are equal when they have two adj. sides and the included Z equal each to each (135). Q.E.F. EXERCISES. >- 1. Construct a square upon a given straight line. 2. Construct a parallelogram, having given two adjacent sides and one diagonal. 7\3. Construct a rhombus, having given the two diagonals. 4. On a given straight line as hypotenuse, construct a [N^ight triangle having one of its acute angles double the other. BOOK IL— PROBLEMS OF CONSTUUCTION, 117 Proposition 33. Problem. 268. To draw a tangent to a given circle from a given point either on or witlwut the circumfere7ice» Given, the O BCD, and the pt. A. Jieqtiired, to draw from A a tangent ^i to the O BCD. | Case I. When the pt. A is ifi the Qce. Cons. Draw the radius OA. At A draw AE _L to AO. (256) Then AE is the tangent required. (209) Case II. When the pt. A is without the O, Cons. Join OA; bisect it at E. Witli centre E and radius EO, describe a Oce, cutting the O BCD at B, D. Join AB, AD. Then either AB or AD is the tangent required. Proof. Join OB, OD. Z ABO = rt. Z , and Z ADO = rt. Z, being inscribed in a semicircle (240). .*. AB and AD are each tangent to the O at B, D, being i to the radius at its extremity (209). Q.E.F. ScH. When the pt. A is without the O, two tangents may always be drawn, and they wdl be equal; for rt.A AOB = rt.A AOD. (110) 118 PLANE Geometry, Proposition 34. Problem. 269. To inscribe a circle in a given triangle. Given, the A ABC. Required, to inscribe u in A ABC. Cons. Bisect the Z s A, B by the lines AO, BO meeting at 0. From draw OD i. to AB. With centre and radius OD, describe the O DEF. The O DEF is the required o. Proof, Since the pt. of intersection of the bisectors of the angles of a A is equally distant from the three sides of the A, (1G7) .-.the J_s OD, OE, OF are equal. .•. a O described with centre and radius OD will be tangent to the three sides of the A at the pts. D, E, F, and be inscribed in it. (191) Q.E.F. 270. ScH. The centre of the circle inscribed in a tri- angle is sometimes called its in-centre. Note.— The bisectors of the angles of a triangle are concurrent, the point of intersection being the centre of the circle inscribed in the triangle. 271. Dep. If the sides of a triangle are produced and the exterior angles are bisected, the intersections of the bisectors are the centres of three circles, each of which is tangent to one side of the triangle and the other two sides produced. These three circles are called escribed circles. BOOK II.— PROBLEMS OF CONSTRUCTION. 119 Proposition 35. Problem. ^272. To draw an escribed circle of a given triangle. / Given, the A ABC. Required, to describe a O touching AB, and C A, CB pro- duced. Co7is. Bisect the Z s BAE, ABD by the lines AO, BO, which intersect at 0. (255) From draw OG, OH, OK 1 to CE, AB, CD. With centre and radius ^ \ Oa, describe the O OHK. The O GHK is the required O. Proof, Since the pt. is on the bisector of Z BAE, it is equally distant from BA and AE. (160) .•.0G = 0H. Similarly, OH = OK. . • . the _Ls 00, OH, OK are equal. . • . a O described with centre and radius 00 will touch CE, AB, CD at the pts. G, H, K. . • . the O GHK is an escribed O of the A ABC. q.e.f. ScH, In the same manner the centres 0', 0" of the other two escribed Os may be found. Therefore there are in general four circles tangent to three intersecting straight lines. 120 PLANE GEOMETRY. Proposition 36. Problem. 273. To circumscribe a circle about a given triangle. Given, the A ABC. . ^ Required, to circumscribe aO about the /'^ /\ \ A ABC. / d/ \\ Cons. Draw DO, EO _L to AC, AB at \ />\ \j tlieir mid. pts. intersecting at (284). a\ e / ^ With as a centre, and a radius = OA, ^^ -^^ describe a O. This O is the one required passing through the vertices A, B, C. Proof. Since the pt. is in the _L bisectors OD, OE of AC, AB, it is equally distant from the pts. A, B, C. JSver^ pt. ill Hie ± bisector of a line is equidistant from tfie extremities of the line (66). . • . a O described with centre and radius OA must pass through the pts. A, B, C, and is .*. circumscribed about the A ABC. Q.E.F. 274. ScH. This construction is the same as that of de- scribing a circumference through any three given points not in the same straight line, or of finding the centre of a given circle, or of a given arc. Note.— Since the perpendicular bisector of a chord passes through the centre of the circle <202), therefore, The perpendicular bisectors of the sides of a triangle are concurrent, the point of intersection being the centre of the circle circumscribed about the triangle. The centre of the circle circumscribed about a triangle is sometimes called its circum-centre. EXERCISE. Prove (1) if the given triangle be acute-angled, the centi(^ of the circumscribe! circle falls within it; (2) if it be a right triangle, the centre falls on the hypotenuse; (3) if it be an obtuse-angled triangle, the centre falls without the triangle. BOOK II.-PROBLEMS OF CONSTRUCTION, 121 Proposition 37. Problem. 275. On a given straight line, to describe a segment which shall contain a given angle* Given, the st. line AB. Required, to describe on AB a seg- ^'^ '"^^^ ment containing / C. ^Z x\^ Co7is, Make / BAD = Z C. (2^8) / q/ ' From A draw AH J. to AD. (256) \ \y ! \V Bisect AB m E. (254) )^ ^ f^ From E draw EO _L to AB meeting \' AH at 0. (256) '"^^ With as a centre, and OA as a C^ radius, describe the Oce AHBF. The segment AHB is the segment required. Proof, Join OB and BH. Since is in the J_ bisector of AB, it is equally distant from the pts. A, B. (66) .•. a O described with centre and radius A must pass through B. Also, since AD is _L to AH at its extremity, .-. AD is tangent to the O at A. (209) Since AB is a chord through A, .\ Z BAD is measured by i arc AFB. (243) But f AHB is measured by J arc AFB. (238) .'. any Z in the segment AHB = zC. (Ax. 1) Q.E.F. Note.— In the particular case when the given angle C is a right angle, the seg- ment required will be the semicircle described on the given st. line AB. 129 PLANE GEOMETHT. Proposition 38. Problem. 1 S76. To draw a common tangent to two given circles. Given, the Os AR, BS, and let AR > BS. (1) Required, to draw an ex- terior common tangent to the two Os. Cons, With centre A and ra- dius = the difference of the radii of the two Os, describe a O. From B draw BC tangent to this O- (268) Join AC, and produce it to meet the Oce of the given O in D. Draw BE II to AD, and join DE. DE is a common tangent to the two given Os. Proof. But But Since AC = AD - BE, .-. CD = BE. CD is II to BE, .-. BCDE is a CJ, .-. DE is =: and II to CB. (Cons.) (Cons.) (133) (124) ZACB=:art. Z. A tang, to a q at any pt. is L to the radius at thai pi. (210). .*. BCDE is a rectangle. (77) .-. ZD = ZE = art. Z. .-. DE is a tangent to both Os. q.e.f. Since two tangents can be drawn from B to the O AR, therefore two common tangents may always be drawn to the given Os. These are called the direct common tangents. AVhen the given Os are external to each other and do not intersect, two more common tangents may be drawn, called the transverse co7mnon tangents. BOOK 1L—HXERCI8ES. TltEOliEMS. 123 (2) Required, to draw the transverse pair of common tangents. Cons, and proof. With centre A and radius = the sum of the radii of the two OS, describe a ©, and complete the con- struction, and proof, as in (1). exercises. Theobems. 1. If a straight line cut two concentric circles, the parts of it intercepted between the two circumferences are equal. 2. If one circle touch another internally at P, prove that the straight line joining the extremities of two parallel diameters of the circles, towards the same parts, passes through P. 3. In Ex. 2, if a chord AB of the larger circle touches the smaller one at C, prove that PC bisects the angle APB. 4. If two circles touch externally at P, prove that the straight line joining the extremities of two parallel diame- ters towards opposite parts, passes through P. 5. Two circles with centres A and B touch each other externally, and both of them touch another circle with centre internally : show that the perimeter of the tri- angle AOB is equal to the diameter of the third circle. 6. In two concentric circles any chord of the outer circle which touches the inner, is bisected at the point of conkict. 7. If three circles touch one another externally in P, Q, R, and the chords PQ, PK of two of the circles be produced to meet the third circle again in S, T, prove that ST is a diameter. 8. Points P, Q, R on a circle, whose centre is 0, are joined ; OM, ON are drawn perpendicular to PQ, PR re- spectively : join MN, and show that if the angle OMN 124 PLANE OEOMETRT. is greater than ONM, then the angle PRQ is greater than PQR. 9. A circle is described on the radius of another circle as diameter, and two chords of the larger circle are drawn, one through the centre of the less at right angles to the common diameter, and the other at right angles to the first through the point where it cuts the less circle. Show that these two chords have their greater segments equal to each other and their less segments equal to each other. 10. is the centre of a circle, P is any point in its cir- cumference, PN a perpendicular on a fixed diameter : show that the straight line which bisects the angle OPN always passes through one or the other of two fixed points on the circumference. 11. Two tangents are drawn to a circle at the opposite extremities of a diameter, and intercept from a third tan- gent a portion AB: if C be the centre of the circle show that ACB is a right angle. 12. A straight line touches a circle at A, and from any point P, in the tangent, PB is drawn meeting the circle at B so that PB is equal to PA: prove that PB touches the circle. 13. OC is dmwn from the centre of a circle perpendic- ular to a chord AB : prove that the tangents at A, B inter- sect in OC produced. 14. TA, TB are tangents to a circle, whose centre is ; from a point P on the circumference a tangent is drawn cutting TA, TB or those produced in 0, D: prove that the angje COD is half the angle AOB. 15. AB is the diameter and C the centre of a semicircle: show that the centre of any circle inscribed in the semi- circle is equidistant from C and from the tangent to the semicircle parallel to AB. 16. If from any point without a circle straight lines be drawn touching it, the angle contained by the tangents is double the angle contained by the straight line joining the I BOOK IL-EXEIWISES. TIIEOllEMS. 125 points of contact and the diameter drawn through one of them. 17. C is the centre of a given circle, CA a radius, B a point on a radius at right angles to CA; join AB and pro- duce it to meet the circle again at D, and let the tangent at D meet CB produced at E: show that BDE is an isosceles triangle. 18. Let the diameter BA of a circle be produced to P, so that AP equals the radius; through A draw the tangent AED, and from P di-aw PEC touching the circle at and meeting the former tangent at E; join BO and produce it to meet AED at D: then will the triangle DEC be equi- lateral. Let O be the centre of the given 0. Produce OC to a pt. F so that CF = CO: compare as PCO, PCF, etc. 19. APB is a fixed chord passing through P, a point of intersection of two circles AQP, PBR; and QPR is any other chord of the circles passing through P: show that AQ and RB when produced meet at a constant angle. 20. Two circles whose centres are A and B touch exter- nally at C; the common tangent at C meets another com- mon tangent DE at F: prove that (1) CF, DF, FE are equal; (2) each of the angles AFB, DCE is a right angle; (3) DE touches the circle described on AB as diameter. 21. The diagonals AC, BD of a quadrilateral ABCD in- scribed in a circle intersect at right angles at P: prove that the straight line drawn from P to the middle point of one of the sides of the quadrilateral is perpendicular to the op- posite sides. Bisect AB in E, produce EP to meet CD in F, etc. 22. If a side of a quadrilateral inscribed in a circle be produced, the Exterior angle is equal to the interior and opposite angle: and conversely, if the exterior angle of a quadrilateral made by any side and the adjacent side pro- duced be equal to the interior and opposite angle, a circle can be described about the quadrilateral, 126 PLANE GEOMETRY. ^ 23. ABOD is a quadrilateral inscribed in a circle with centre 0. If the angles BAD, BOD are together equal to two right angles, prove that the angle BCD is two-thirds of two right angles. 24. If two pairs of opposite sides of a hexagon inscribed in a circle are parallel, the third pair of opposite sides are parallel. See (342). 25. A circle is described passing through the ends of the base of a given triangle: prove that the straiglit line join- ing the points, in which it cuts the sides or sides produced, is parallel to a fixed straight line. 26. In the semicircle ABODE, the chord BD which is parallel to the diameter AE bisects the radius 00 at right angles : prove that the arc BO is double the arc AB. Join BO, CD; let OC cut BD in F; BF = FD, etc. 27. The straight lines which bisect the vertical angles of all triangles on the same base, on the same side of it, and with the same vertical angle, are concurrent. 28. AB, AO are chords of a circle: show that the straight line, which joins the middle points of the arcs AB, AO, cuts off equal portions of the chords. Let D, E be the mid pts. of arcs AB, AC; take D, E on opp. sides of diam. through A, etc. 29. BAO, BA'O are two angles in the same segment of a circle; AP, A'P' are drawn making the angles BAP, BA'P' equal to the angles BOA, BOA' respectively : prove that AP is parallel to A'P'. 30. OD is a chord of a circle at right angles to the diam- eter x\B ; E is any j^oint in the arc BO ; AE cuts OD in F : prove that the angles DFE, AOE are equal. 31. Prove that two of the straight lines which join tlie ends of two equal chords are parallel, and that the other two are equal. Let the equal chds. AB, CD be joined towards the same parts by BC, AD; towards opp. parts by AC, BD; ,•. chd. AB = chd. CD, etc. BOOK IL— EXERCISES. LOCI. 127 32. ABCD is a quadrilateral inscribed in a circle. Find the relation between the sides in order that AC may \)isect the angle BAD, and BD bisect the angle ABC. /DAC = /CAB, .-. arc DC = arc CB, .'. CB =CD, etc. 33. Two circles touch internally at 0, and a line is dra^n cutting one of the circles in P, P', the other in Q, Q' : show »iat PQ, P'Q' subtend equal angles at 0. Draw tang. OB, B being towards P, Q, etc. 34. Two circles touch externally at A; the tangent at B to one of them cuts the other in C, D : prove that BC and BD subtend supplementary angles at A. Draw common tang. AE meeting BC in E, Z EAB = Z EBA, etc. 35. If the opposite pairs of sides of a quadrilateral in a circle be produced to meet, and the angles so formed be bisected, the bisectors are at right angles to each other. Let ABCD be a cyclic quadl. ; let AD, BC meet in E, and AB, DC in F, etc. 36. If from any point on the circumference of the circle circumscribing a triangle, perpendiculars be drawn to the sides, show that the feet of these perpendiculars are col- liiiear (1()4). Let P be any pt. in the G ce of the O circumscribing the a ABC. From P draw PL ± to BC, PN ± to AB, etc. 37. In any triangle ABC, if be the orthocentre (171), and L, M, N the feet of the perpendiculars, the circle de- scribed through L, M, N will (1) bisect OA, OB, OC, and (2) will also pass through the middle points D, E, F of the sides of the triangle. Loci. 38. Find the locus of a point at a given radial distance from the circumference of a given circle. See (157). 39. On a given base as hypotenuse right triangles are de- scribed: find the locus of their vertices. See (240). ^ 128 PLANE GEOMETRY. 40. Find the locus of the centre of a circle which passes through two given points. See (159) and (202). 41. Find the locus of the centi-e of a circle which touches a given straight line at a given point, or which touches a given circle at a given point. 42. Find the locus of the centre of a circle which touches a given straight line, and has a given radius. 43. Find the locus of the centre of a circle which touches a given circle, and has a given radius. 44. Find the locus of the centre of a circle which touches two given straight lines. See (162). 45. Prove that the locus of the middle points of cqu^il chords in a given circle is another concentric circle. 4G. Prove that the locus of the middle points of all chords drawn through a given point in a given circum- ference, is a circle passing through the given point and having the radius of the given circle for a diameter. 47. Prove that the locus of the vertex of a triangle having a given base and a given vertical angle, is a circle. 48. A straight line moves so that its ends constantly touch two fixed lines at right angles to each other : prove that the locus of its middle point is a circle: find its centre and radius. 49. A and B are fixed points, and C is any point on a circle ; AC is produced to D so that CD is always equal to CB : find the locus of D. CD = CB, .-. ZCDB = ZCBD, etc. 50. Through one of the points of intersection of two fixed circles, centres A and B, a chord is' drawn meeting the first circle in P and the other in Q. Find the locus of the point of intersection of PA and QB. Let D be the intersection of 0s. (1) Let P, Q be each on the arc outside the other O ; let PA, QB meet in R, etc. . BOOK IL— EXERCISES. PROBLEMS. 129 IH 51. ABC is a triangle inscribed in a circle, P any point in the circumference of the circle, and Q is a point in PC such that the angle QBC is equal to the angle PBA. Prove that the locus of Q is a circle. If P is in arc AB, because ZQBC = ZPBA, and ZQCB = /PAB, .-. ZBQC = ^^^ BPA, etc. ^^m 52. Find the locus of the middle points of all chords of ^^Rcircle which pass through a given point. ^^KiBt P be the given iJt., O the cent, of the O, AB a chd. through P, etc. ^^ 53. From any point P on the circumference of a circle circumscribing a triangle ABC, perpendiculars PD, PE are let fall on the sides AB, AC : prove that the locus of the centre of the circle circumscribing the triangle PDE is a circle. Problems. If the construction of a problem does not readily appear to the student, he should present a careful analysis, that is, a course of reasoning by which the solution may be arrived at. In an analysis we generally begin by supposing the problems solved, and constructing the figure accordingly. We then reason out the conditions to be fulfilled, study the relations of the lines, angles, etc., in the figure, draw aux- iliary lines parallel or perpendicular, as the case may require, join given points or points assumed in the solution, describe circles when necessary, and endeavor to discover the de- pendence of the problem upon previously solved problems or theorems. The reverse process, or synthesis, then fur- nishes a construction of the problem. The analysis will furnish exercise for the student^s ingenuity in drawing useful auxiliary lines. The synthesis will furnish exercise for his invention in combining the different steps suggested by the analysis in the simplest form. 130 PLANE GEOMETRY. 54. In a given circle to inscribe a triangle equiangular to a given triangle. Given, the O ABO, and the aDEF. Required, to inscribe in the O a A equiangular to DEF. Analysis. Suppose the problem solved, and that ABC is the re- quired A . At A draw the tangent HAG. Then Z GAB = Z C, and Z HAC = Z B, (243) . • . the Z s of the A are the same as the Z s GAB, BAG, CAH. Hence, the following construction: Cons, At any pt. A on the ©ce, draw the tangent HAG. At A make Z GAB = Z F, aud Z HAC = Z E. (258) Join BC. Then ABC is the inscribed A required. (100) 55. Given two sides of a triangle and the straight line drawn from the extremity of one of them to the middle point of the other, to construct the triangle. Construct A ABC so that AB = 1st side, BD = ^ the 2nd, etc. 56. Given two sides of a triangle and the straight line drawn from their point of intersection to the middle point of the thh'd side, to construct the triangle. Construct the a ABE so that AB = 1st side, BE = 3nd side, etc. 57. Given the base, one of the angles at the base, and the sum of the sides of a triangle, to construct the trian- gle. 58. Given the base, one of the angles at the base, and the difference of the sides of a triangle, to construct the triangle. BOOK IIAeXERQISES. PROBLEMS. 131 59. Through a given point P between two given straight lines AB, AC^ draw a straight line which shall be termi- nated by AB, AO, and bisected in P. 60. If P be outside the lines AB, AC, draw PDE meeting AB, AC in D, E so that PD equals DE. 61. Find points D, E in the sides AB, AC respectively of the triangle ABC, so that DE may be parallel to BC and equal to BD. 62. Draw a straight line DE parallel to the base BC of the triangle ABC and meeting AB, AC in D and E, so that DE equals the sum of BD, CE. 63. From a given point without a given straight line draw a straight line making a given angle with this given line. 64. AB, AC are two straight lines, AE is an intermedi- ate straight line. Show how to draw the straight lines which are terminated by AB, AC, and bisected by AE. Take any pt. E in AE, draw EF 1| to AC meeting AB in F; from FB cut off FG = AF, etc. 65. Construct a triangle, having given a median and the two angles into which the angle is divided by that median. QQ, Construct a parallelogram, having given (1) two di- agonals and the angle between them, and (2) one side, one diagonal, and the angle between the diagonals. 67. From a given point draw three straight lines OA, OB, OC of given lengths so that A, B, C may be collinear, and AB equal to BC. Construct a a GAD having the sides OA, AD = the two outside lengths, and OD double the middle, etc. ^^, In a given straight line find a point whose distance from a given point in the line may be equal to its distance from another given straight line. Let P be the given pt. in the given line XY, AB the other given line. Draw PC J. to AB, bisect l XPC, etc. 69. Construct a triangle, having given the base, the ver- tical angle, and (1) the sum, or (2) the diiference of the sides. 132 PLANE GEOMETRY. Construct a right triangle having given : 70. The hypotenuse and the sum of the sides. 71. The hypotenuse and the difference of the sides. 72. The hypotenuse and the perpendicular from the right angle on it. 73. Tiie perimeter and an angle. 74. Construct an isosceles triangle, having the vertical angle equal to four times each of the base angles. 75. Construct an isosceles triangle, having one-third of each angle at the base equal to half the vertical angle. ■ Circles. 76. Through a given point inside a circle which is not the centre, draw a chord which is bisected at that point. 77. With a given point as centre describe a circle which shall intersect a given circle at the ends of a diameter. 78. Describe a circle which shall pass through a given point outside or inside a given circle and touch it at a given point. 79. Describe a circle with a given centre which shall touch a given circle. How many such circles can be drawn ? 80. Through a given point inside a given circle draw two equal chords which shall contain an angle equal to a given angle. 81. Describe a circle which shall touch a given circle at a given point, and also touch a given straight line. Let C be the cent, of ©, P the point. Draw the diam. A BCD i to given line XY, meeting the O in A; join PA, PB, and produce them to meet XY in E, F, etc. 82. Describe a circle passing through a given point and touching a given straight line at a given point. 83. With a given point outside a given circle as centre, describe a circle which shall cut the given circle orthogo- nally (228). 84. Given two points A, B on a circle, and a fixed straight line through A. Draw through B a straight liuQ BOOK IL— EXERCISES. CIRCLES. 133 cutting the circle in C, and the fixed line in D, so that AC shall be equal to CD. Let the fixed line meet the in F; bisect the arc AB in E, draw EC i to AF, etc. 85. Describe a circle which shall touch a given straight line at a given point, and pass through another given point not in the line. 86. Construct a triangle, having given the base, the ver- tical angle, and the median drawn from the vertical angle. h 87. ABC is a given straight line: find a point P such ^hat each of the angles APB, BPC may be equal to a given angle. 88. Find the point inside a given triangle at which the ides subtend equal angles. 89. About a given circle to describe a triangle equiangu- lar to a given triangle. On two sides describe segments of ©s containing an Z = § of 2 rt. Zs, etc. 90. If the escribed circles of the triangle ABC (272) touch BC, CA, AB externally in D, E, F respectively, prove that BD = EA, CD = AF, CE = BF. Book III. EATIO AND PEOPOETION. SIMILAE riGUEES. Definitions. 277. Four quantities are said to be in proportmi when tlie ratio of the first to the second is equal to the ratio of the third to the fourth. Thus, if g = 5» then, A, B, C, D are said to be in proportion. The propor- tion is written A : B = C : D, or A : B : : C : D, which is read *^ A is to B as C is to D." Note.— The first form is preferable, and the one most generally used in the higher mathematics; but the second form is the more usual one in elementary works. Let a and b denote the numerical measures of A and B (229), and c and d the numerical measures of C and D. Then, since the ratio of two quantities is the same as the ratio of their numerical measures (230), ,'. A :B = a :b, and C :J) = c : d. Hence, if four quantities A, B, C, D, are in proportion, their numerical measures a, Z>, 6', ^, are in proportion; that is, a : b =i c : d^ 134 BOOK IIL—BATIO AND PROPORTION. 135 Conversely, if tlie numerical measures of four quantities A,B,C,D, are in proportion, the quantities themselves are in proportion; that is, A : B = C : D, 2uhen A and B are quantities of one kind, and C and D are quantities of one hind, though the latter kind may be dif- ferent from the former. That is, all four quantities may he of the same kind, as, for instance, four straight lines, four surfaces, four angles, and so on; but the quantities in each pair must be of the same kind. The magnitudes we meet with in Geometry are more often incommensurahle (232) than commensurable. The preceding reasoning does not apply directly to the case in which two quantities are incommensurable, but it may be extended to this case. 278. To find the greatest common measure of tioo quan- tities. Let there be two quantities, as, j j ^ ^ ^ for instance, the two straight lines A E GB AB, CD. fe ^ Apply the smaller CD to the greater AB, as many times as possible, suppose twice, with a remainder EB. Apply the remainder EB to CD as many times as pos- sible, suppose once, with a remainder FD. Apply the second remainder FD to EB as many times as possible, suppose once, with a remainder GB. Apply the third remainder GB to FD as many times as possible. This process^ill terminate only if a remainder is found which is contained an exact number of times in the pre- ceding one ; and if it so terminates the two given lines are com mensurable, and the last remainder will be their great- est common measure. 136 PLANE GEOMETRY. Suppose, for example, that GrB is contained exactly twice ill FD. Then GB is contained exactly in OF, and therefore ex- actly in CD, and therefore exactly in AE, and therefore exactly in AB. Therefore GB is a common measure (231) of AB and CD. And, since every common measure of AB and CD must divide AE, it must divide EB or CF, and therefore FD, and therefore also GB. Hence, the common measure can- not be greater than GB. Therefore GB is the greatest common measure of AB and CD. By regarding GB as the measuring unit, the values of the preceding remainders are easily found, and finally, those of the given lines, from which their numerical ratio is ob- tained. Thus, FD = 2GB ; EB = FD + GB = 3GB ; CD= EBH-FD= 2FD+ GB = 5GB; AB = 2CD + EB = 10GB + 3GB = 13GB. Therefore the given lines AB and CD are numerically expressed in terms of the unit GB by the numbers 13 and 5 ; and their ratio is ^-^-. 279. When the quantities are incommensurable the above process never terminates. However far the opera- tion is continued, we never find a remainder which is con- tained an exact number of times in the preceding one. But as there is no limit to the number of parts into which AB and CD may be divided, we may obtain a remainder as small as we please, one that is less than any assignable quantity. Hence, although no fiyiite numerator and denominator however large can exactly express the ratio of two incom- mensurable quantities, yet by properly increasing both numerator and denominator we may obtain a ratio as nearly BOOK III.— RATIO AND FEOrORTION. 187 equal as we please to the required ratio, i.e., we way ohtain two numbers loliose ratio will expres>> the ratio of two in- commensurable quantities to any required degree of accu- racy. (232) Note.— We therefore conclude that ratio in Geometry may be treated in the same way as ratio in Arithmetic and Algebra. Indeed, the algebraic treatment is the easiest and the simplest. Euclid's treatment of ratio and proportion is now pi-actically disregarded. While his reasoning is exquisite and rigorous, it is remote from our practical notions on the subject. For these reasons, only simple algebraic proofs of the propositions in proportion will be given in this work. The student will perceive that the propositions do not introduce new ideas, but merely supply proofs, based on the geometric definition of propor- tion, of results already familiar in the study of Algebra. 280. Let us now consider the numerical proportion (277), which may be written in either of the three forms: a : b = c : d\ a '.b\\ c \ d\ — = —, b d The four terms of the two equal ratios (277) are called the ter^ns of the proportion. The first and fourth terms are called the extremes^ and the second and third, the means. Thus, in the above proportion, a and d are the extremes, and b and c the means. The first and third terms are called the antecedents, and the second and fourth the consequents. Thus, a and c are the antecedents, and b and d the consequents. The fourth term is called a fourth iiroportional to the other three. T^hus, in the above proportion, d is a fourth proportional to a, b, and c. In the proportion « : Z> = J : c, c is a third 'proportional to a and b, and Z^ is a mean jiroportional between a and c. 138 PLANE GEOMETRY. Proposition 1 . 281. If four quantities are in proportion^ the product of the extremes is equal to the product of the means. Hyp, Let a '.h — c\ d. To prove ad — he. Proof. By {m), | = ^. Multiplying by hd, ad = be, q.e.d. 282. Cor. If a:b = b:^c, ,', b' = ac, (281) .-. b = Vac, That is, the mean proportional between two quantities is equal to the square foot of their product. Proposition 2. 283. Conversely, if the product of two quantities be equal to the product of two others, two of them may be made the extremes, and the other two the means, of a proportion. Hyp. Let ad = be. To prove a : b = c : d. Proof, Dividing the given equation by bd, (I _c ^ b~l' that is, a : b = c : d, q.e.d. In a similar manner it may be shown that the proportions a '. c = b \ d, b : a = d : c, b : d = a : c, c \ d ^=^ a \b, etc., are all true provided that ad — be. BOOK UL-HATIO AND PUOPORTION. 139 284. Sen. By tlie product of the extremes or of the meiiiis of a proportion is meant the product of the numerical measures, of those quantities. Hence, the product of tioo lines will be often used for brevity, meaning the product of the numbers luliich represent those lines. Proposition 3. 285. If fonr quantities are in proportioUy they are in proportion hy inversion; that is, the second term is to the first as the fourth term is to the third. Hyp, Let a -.h — c \ d. To prove h \ a = d \ c. Proof | = |. (277) • -^ • b-^ ' d' that is. b_d a c ' ,'. b : a = d : c, q.e.d. Proposition 4. 286. If four quantities are in proportion, they are in proportion by alternation; that is, the first term is to the third as the second term is to the fourth. Hyp. Let a : b = c : d. To prove a : c = b : d. Proof Since a : b = c : d, , ' , ad = be. (281) .'. a:c:=b:d, (283) Q.E.D. 140 PLANE GEOMETRY. Proposition 5. 287. If four quantities arc in proportion, they are in pro^jortion by composition; that is, the sum of the first and second is to the second as the sum of the third and fourth is to the fourth. Hyp. Let a : b = c : d. To prove a-{-b :b = c -{- d : d. Proof. ^ = -^. (277) a 1~ c '7v ■1+- ^J+>> a-{-b c-^-d that is, _ . d .'. a + b : b = c -\- d : d. Q.E.D. Similarly, a + b : a = c ^ d : c. Proposition 6. 288. If four quantities are in proportion, they are in proportion by division; that is, the difference of the first and second is to the second as the difference of the third and fourth is to the fourth. Hyp. Let a -.b = c : d. To prove a — b : b = c — d : d, (277) Proof. a c b ~d' • • b ^-d ^' that is. a — b c ^rz^ b ~ d ' .' . a — b : b — c — d : d. Similarly, a — b \ a — c — d \ c. Q.E.D. BOOK III.— RATIO AND PROPORTION. 141 Proposition 7. 289. Jf four qiiantities are in proportion, they are in proportion by composition and division; that is, the sum of the first and second is to their difference as the sum of the third and fourth is to their difference. Hyp. Let a:b — c\d. To prove a -\- b: a — b =^ c -\- d: c — d. Proof. By (287), — and by (288) by division, d. Q.E.D. Proposition 8. 290. The products of the corresponding terms of two or more proportions are projjortional. Hyp. Let a: b = c : d, and e: f = g \h. To prove ae : bf = eg : dh. r) y a c ^ e g Proof -7 = -J J and -. = f . •^ h d' f h ae _ eg ■'■ ¥"//'■ .*. ae: of = eg: dh. q.e.d. 291. A greater quantity is said to be a inultiple of a less, when the greater contains the less an exact number of times. BquimuUiples of two quantities are quantities which contain them the same number of times. Thus, ma and mb are equimultiples of a and b. V^o<;, h - ' d ' a — b b ~ c — d, ■ d ' a-\-b a — b c^d ' c-d' .\a + b :a-b: = c + d: c 149 PLANE OEOMBTRY. Proposition 9. 292. Wlien four qvantities are in proportion, if the iird and second he multiplied^ or divided, hy any quantity, as also the third and fourth, the resulting qua^itities will he in proportion. Hyp, Let a:l) = c:d. To prove ma :mh= nc\ nd. Proof, a c Multiply both terms of the first fraction by m, and both terms of the second by n. Then Similarly, w,a _ nc mh ~ nd ma : mh = nc : nd, q.e.d. a h _ c d m ' m n' n 293. ScH. Either m or n may be unity. In a similar manner it maybe shown that if the first and third terms he multiplied, or divided, hy any quantity, and also the second and fourth, the resulting quantities will he in proportion. 294. Cor. Since ma _ a mh h a _a h~h .'. ma : mh = a: h. That is, equimtdtiples of two quantities are in the same ratio as the quantities tliemselves. BOOK ilL-PROPORTIONAL LINES. 143 Proposition lO. 295. If four quantities are in proportion, their like poivers, or roots, are in jjroportion. Hyp. Let To prove and a',h = c: d. «» : Z.» = c" : (?«, 11 11 a^ \ !)■»■ — c^\ d\ Proof a c b~d' Kaising to the Extracting the n^^ power, n*^ root, 1 1 J« = c"^ ! d\ 111 J» = 6*» : d"" Q.E.D. Proposition 11. 296. If any number of quantities are in proportion, any antecedent is to its consequent, as the sum of any num- ber of the antecedents is to the sum of the corresponding consequents. Hyp. Let a : b = c : d = e : f ^ etc. To jjrove a\b = c\d— etc. — a -\- c -\- e -\- etc. \b-{-d +/+ etc. Proof. __ ab = b a^ ad = be, and af= be, etc., etc. (281) Adding, a{b + d +/-f <^tc.) = b{a -\- c -{- e -\- etc.) ,\a:b = c :d = etG. = a + c + e-^etG. :b + d+f +etG. (283) Q.E.D. Proportional Lines. 297. Def. Two straight lines are said to be cut propor- tionally when the parts of one line are in the same i-atio as the corresponding parts of the other line. Thus, AB and CD are cut pro- P ^ portionally at P and Q if A AP : PB = CQ : QD. k ^ 144 PLANE GEOMETRY. Proposition 1 2. Theorem. . 298. A straight line parallel to one side of a triangle divides the other tivo sides proportionallg. Hy2J. Let DE be II to BC in the A ABC. A, To prove AD : DB = AE : EC. '^' Case I. Wlieu AD and DB are com- mensurable. D/ ^-^'^ Proof. Take AH, any common meas- ure of AD and DB, and suppose it to be contained 4 times in AD and 3 times inDB. ^ ^ Then ^5 = |. (1) Through the several pts. of division of AD, DB draw || s to BC. They will divide AE into 4 equal parts and EC into 3 equal parts. Jf \\ s iniei'cepi equal lengths on any transversal, tliey int€rccpt equal lengths on evei'y transtei'sal (152). • • EC - 3 • ^^^ Therefore, from (1) and (2), DB~ EC ^ ^ . • . AD : DB = AE : EC. Case II. When AD and DB are incom- Aa inensnrable. Proof. In this case we know (232) that we may always find a line AG as nearly equal as ive please to AD, and such that G/ \H AG and GB are commensurable. Draw GHllto BC ; then AG AH ,p Tx ^ * ^ BOOK IIL-PROPORTIONAL LINES. 145 (233) Q.E.D Now, these two ratios being always equal while the com- mon measure is indefinitely diminished, they will be equal when G moves up to and as nearly as we please coincides with D. ADAE •**DB~EO* 299. Cor. 1. By composition (287), we have AD + J3B : AD = AE + EC : AE, or AB : AD = AC : AE. Also, AB : DB = AC : EC, and AB:AC = DB:EC. 300. Cor. 2. If Uvo straight lines AB, CD are cut hy any mimber of 2)arallels, AC, EF, GH, BD, the cor- responding intercepts are proportional. For, let AB and CD meet at 0. Then, by (299), OE OF AE ~CF EG "FH* (288) (286) t" e/ \f J \„ 1 \ B Similarly, AE : CF = EG : FH. EG : FH = GB : HD. EXERCISES. 1. From a point E in the common base AB of two tri- angles ACB, ADB, straight lines are drawn parallel to AC, x\D, meeting BC, BD at F, G : show that EG is parallel to CD. 2. In a triangle ABC the straight line DEF meets the sides BC, CA, AB at the points D, E, F respectively, and it makes equal angles with AB and AC : prove that BD : CD = BF : CE. 146 PLANE GEOMETRY. Proposition 13. Theorem. 301. Conversely, if a straight line divides two sides of a triangle proportionally, it is parallel to the third side. Hyp, Let DE cut AB, AC in a the A ABC so that -p=- = -j-=, / \ AD AE / \ / To prove DE || to BC. 9 ^ " V Proof If DE is not || to BC, ^/_ \^ draw DE' || to BC. B Then, But T^ = Ti- (Hyp-) (Ax. 1) .•.AE' = AE, which is impossible unless DE^ coincides with DE. .-. DEis II toBC. Q.E.D. 302. Def. When a finite straight line, as AB, is cut at a point X between A and B, it is said to be divided internally at X, l 1 — ^— — jY and the two parts AX and BX are called segments. But if the straight line AB is produced, and cut at a point Y beyond AB, it is said to be divided externally at Y, and the parts AY and BY are called seg- ments. The given line is the sum of two internal seg- ments, or the difference of two external segments. AB AC AD" AE' AB AC AD ~AE- AC AC AE' "AE" BOOK III— PROPORTIONAL LINES. 147 Proposition 14. Theorem. 303. The hisector of an angle of a triangle divides the opposite side into segments proportional to the adjacent sides. Hyp. Let AD bisect the Z A of the A ABC. To prove DB : DC = AB : AC. Proof Draw CE !| to DA, meeting BA produced in E. Since DA is H to CE, (Cons.) .-. Z BAD = Z AEC, being ext. -int. Zsof\\ lines {11), and . Z CAD = Z ACE, being alt. -int. As of \\ lines (72). But Z BAD = Z CAD. (Hyp.) .-. Z AEC = I ACE. (Ax. 1) .-. AE = AC, being opp. equal As (114). Because DA and CE are 1| , .-. DB : DC = AB : AE. (298) But AE = AC. (Just proved) .-. DB : DC = AB : AC. q.e.d. Cor. Conversely, if AD divides BO into tioo segments that are proportional to the adjacent sides, it bisects the angle BA C. 148 PLANE GEOMETRY. Proposition 1 5. Theorem. 304. Tlie bisector of cm exterior angle of a triangle divides the opposite side externally into segments 2^ropo7'- tio7ial to the adjacent sides. Hyp, Let AD bisect the ext. Z CAH of the A ABC. To prove DB : DC = AB : AC. Proof, Draw CE || to DA meeting BA at E. Since DA is || to CE, (Cons.) (77] and But .-. Z AEC = Z DAH, c.^r^\J^ U Z ACE = Z DAC. oulA. L (72) Z DAH = Z DAC, K^cH,/ (Hyp.) .-. Z AEC = Z ACE. (Ax. 1) .-. AE == AC. (144) Because DA and CE are II , .-. DB : DC = AB : AE. (299) But AE = AC. (Just proved) .-. DB : DC = AB : AC. q.e.d. 305. Cor. If A F be the bisector of the interior angle BAC, we have, from (303) and (304), BE : EC = BD : CD. 306. Def. "When a straight line is divided internally and externally into segments having the same ratio, it is said to be divided Imrmonically, Therefore (305), The bisectors of an interior and exterior angle at the vertex of a triangle divide the opposite side harmonically. BOOK IIL-SIMILAU FIGURES. 149 EXERCISES. 1. li a ', h = e '. fi and c : d = e \ f, show that a : J = c : d, 2. \i a-.l — l', c, show that a\ c — o} '.h^, 3. A, B, C, D are four points on a straight line, E is any point outside the line. If the parallelograms AEBF, OEDG be completed, show that EG will be parallel to AD. Similar Figures. 307. Def. Similar figures are those which are mutually equiangular (144), and have their homologous sides pro- portional. The homologous sides are those which are adjacent to the equal angles. (145) Thus, the figures ABCD, EFGH are similar if ZA= ZE, ZB= ZF, ZC= ZG, etc.; and AB_ BC op DA EF ~ EG ~ GH ~ HE ' The sides AB and EF are homologous, since they are adjacent to the equal angles A and B, and E and F, respec- tively; also the sides BC and EG are homologous, etc. The diagonals AC, EG are homologous. 308. The constant ratio of any two homologous sides of similar figures is called the ratio of similitude of the figures. 150 PLANE GEOMETBT. Proposition 1 6. Theorem. 309. Triangles luhicli are r}iutuaUy equiangular are similar. Hy2h In the As ABC, A'B'C, A. let Z A = Z A', Z B = Z B', A zc= zc\ / \ ^ To prove A ABC similar to qA \g / \ aA'B'C / A / \ Proof. Apply the A A'B'C ^ ^ ^ )^'_ to the A ABC so that the pt. A' coincides with A, and A'B' falls on AB. Let B' fall at D. Then, since Z A' = Z A, (Hyp.) . • . A'C falls on AC. Let C fall at E; then B'C falls on DE. Since ZADE=zABC, (Hyp.) .-. DEislltbBC. (78) .-. AB: AD = AC: AE, / (299) or AB : A'B' = AC : A'C. In the same way, by applying the A A'B'C to the A ABC so that the Z s at B' and B coincide, we may prove that AB : A'B' = BC : B'C. Combining these two proportions, we have AB : A'B' = AC : A'C = BC : B'C. . • . the two A s are similar. (307) Q.E.D. 310. Cor. 1. Ttuo triangles are similar when two angles of the one are eqnal respectively to tivo angles of the other. ' (97) 311. Cor. 2. A triangle is similar to any triangle cut off hy a li7ie parallel to one of its sides, 312. ScH. In similar triangles the homologous sides lie opposite the equal angles, BOOK IIL— SIMILAR FIGURES, 151 Proposition 1 7. Theorem. 313. Triangles which have their homologous sides pro- portional are similar. Hyp. In the As ABO, A'B'C, AB _ AC _ BC let ^,-g, - ^,^, - ^,^, . To prove Z A = Z A', Z B = ZB', etc. C\ ' ' Proof. On AB take AD == , A'B', and draw DE || to BO. Then, because DE is I| to BC, . • . the AS ABC, ADE are similar. ABACBC •'• AD~AE~DE* But A'B' = AD. AC AB AD A'C' BO B'C (311) (307) (Cons.) (Hyp.) Since the first ratio in each of these proportions is the same, the second and third ratios in eacli are equal respec- tively ; that is. AC AE AC ^ BO ^^ and :f^. = :=^^, BC B'C A'O'^ — J) J. Since the numerators are the same, . • . AE = A'C, and DE = B'C. .-. A A'B'0'= A ADE, having the three aides equal each to each (108). But A ADE is similar to the A ABC. . • . A A'B'C' is similar to the A ABC. q.e.d. EXERCISE. If the sides of the triangle in Prop. 14 are AB = 8, BO = 12, and AC = 10, find the lengths of the segments BD and CD. 152 PLANE GEOMETRY, Proposition 1 8. Theorem. 314. Two triangles which have an angle of the one equal to an angle of the other, and the sides about these angles proportional, are similar. Hyp. Ill the A s ABC, A'B'C, let AB _ AC A'C Z A = Z A', and A'B' A ABC similar to A'B', (311) To prove A A'B'C. Proof. On AB take AD and draw DE || to BC. Because DE is |I to BC, .• . the AS ABC, ADE are similar. .-.AB: AD = AC: AE, homologous sides of similar as are jn'oportional (307). But A'B' = AD. (Cons.) .• . AB : AD = AC : A'C. (Hyp.) .• . A'C = AE. (Comparing the two proportions) .-. A A'B'C = A ADE, having two sides and ilie included /. equal, eacJi to each (104). . • . A A'B'C is similar to the A ABC. q.e.d. 315. ScH. From the definition (307), it is seen that two conditions are necessary that polygons may be similar : (1) they must be mutually equiangular, and (2) their homologous sides must be proportional. In the case of triangles we learn from Props. 16 and 17 that each of these conditions follows from the other; so that one condition is sufficient to establish the similarity of triangles. This, however, is not necessarily the case with polygons of more than three sides; for even with quadrilaterals, the angles can be changed without altering the sides, or the proportionality of the sides can be changed without alter- ing the angles. BOOK IIL—SIMILAB FIGUIiES. im Proposition 1 9. Theorem. ( 316. Tioo triangles tuhich have their sides parallel or perpendicular, each to each, are similar. Hyp, In the As ABC, A'B'C, ^ ^, let AB, AC, BC be || or J_, respec- tively, to A'B', A'C, B'C. To prove A ABC similar to A A'B'C. B Proof. Since the sides are I| or J_ each to each, the included Z s are equal or supplementary (80, 83). . • . three hypotheses may be made : . (1) A + A' = 2rt.Zs,B + B' = 2rt.Zs,C + C' = 2rt.Zs. (2) A = A', B + B' = 2 rt. Zs, C + C = 2rt.Zs. (3) B = B', C = C'. The first and second hypotheses are inadmissible, since the sum of the Z s of the two A s cannot exceed 4 rt. Z s (97). Therefore the third is the only admissible hypothesis. the A s ABC, A'B'C are similar, being mutually equiangular (309), Q.E.D. 317. ScH. The homologous sides of the two triangles are either the parallel or the perpendicular sides. EXERCISE. If the sides of the triangle in Prop. 15 are AB = 16, BC = 10, and AC — 8, find the lengths of the segments BD and CD. 154 PLANE GEOMETUY, Proposition 20. Theorem. 318. If in any triangle a 2J(irallel he draion to the base, all lines from the vertex will divide the base and its parallel proportio7ially. Hyp. Let ABO be a A, WQ' \\ to BC, and AD, AE two lines intersect- ing B'C at D'E'. BD DE EC To prove ^_ = j^_ =-g^. Proof Since B'C is || to BC, the AS AB'D', ABD are similar; also the AS AD'E' and ADE; and the As AE'C and AEC. (311) AD BD , AD DE __BD^ ^i^- •*• AD' "" B'D'' AD' ~ D'E' ' their Jwmologous sides being proportional (307). BD _ DE •*' B'D"'~"D'E'' In the same way it may be shown that DE EC (Ax. 1) D'E' BD E'C DE B'D' ~ D'E' EC E'C Q.E.D. 319. Cor. If BD = DE = EC, we shall have B'D' = D'E' = E'C. Therefore, if the lines from the vertex divide the base into equal parts, they will also divide the parallel into equal parts, EXERCISES. 1. If BD = 8, B'D' = 6, DE = 10, and EC = 3, find the lengths of D'E' and E'C. 2. If AD = 20, AD' = 16, AE = 15, BD = 7, and DE = 5, find B'D', D'E', and AE', BOOK III.— SIMILAR FIGURES. 155 Proposition 2 1 . Theorem. 320. If two 2yolygo7is are composed of the same numler of triangles similar each to each, and similarly placed, the jjolygons are similar. Hyp, Let the as ABC, ACD, ADE of the polygon ABODE be similar re- spectively to the AsA'B'C, A'C'D', A'D'E' of the polygon A'B'C'D'E', and similarly placed. To prove the polygons are similar. Proof. Since the homologous Zs of similar A s are equal, (307) .-. ZB=ZB'. Also, and Z AOB Z (307) A(Ji5 = Z A'U'B', ) ACD = Z A'C'D'. [ Adding, Z BCD = z B'C'D'. In like manner, Z CDE = z C'D'E' ; etc. .*. the polygons are mutually equiangular. Since similar as have their homologous sides propor- tional, (307) AB BC AC AC A'B' ~ B'C ~ A'C" ^^^ A'C CD "CD' AB BC A'B' B'C CD CD 7 = etc. (Ax. 1) . • . the homologous sides of the polygons are proportional. . • . the polygons are similar, heing mutually equiangular and hamng tJieir Iwmologous sviies proportional (307). Q.E.Q, 156 PLANE GEOMETRY. Proposition 22. Theorem. 321. Conversely, two similar polygons may he divided into the same number of triangles, similar each to each, and similarly placed. Hyp. Let ABCDE, A'B'C'D'E' be two similar polygons divided into As by the diagonals AC, AD, A'C, A'D' drawn from the homologous Z s A and A'. To prove As ABC, ACD, ADE sim ilar respectively to As A'B'C, A'C'D', A'D'E'. Proof. Since the polygons are similar, ^, ^ AB BC ... ZB=zB,and^^ = g^. .-.A ABC is similar to A A'B'C. (314) .-. Z ACB= Z A'C'B'. (307) But Z BCD = Z B'C'D'. (307) . • . remaining Z ACD = remaining Z A'C'D'. ., BC _ AC , BC _ CD ,_^, ^iso, ]V(7 ~ K'Q" ^ WW ~ cTy ' (^^ ' ) .-. AC : A'C = CD : CD'. (Ax. 1) . • . A ACD is similar to A A'C'D'. (314) In the same way it may be shown that A s ADE, A'D'E' are similar. q.e.d. Cor. The homologous diagonals of two similar j^olygons are j^ropor'tional to the ho^nologous sides. BOOK III.—SIMILAH FIGUUE8. 157 Proposition 23. Theorem. 322. Tlie perimeters of tioo similar polygons are to each other as any tivo homologous sides. Hyp, Let ABCDE, A'B'C'D'E' be two similar polygons; denote their perimeters by P and P'. To prove P : P' = AB : A'B'. Proof. Since the polygons are similar, AB BO CD . ^_,. AB 4- BC + CD + etc. _ AB • ' • A'B' + B'C + CD' + etc. ~ A'B'* ,-.P:P' = AB: A'B'. EXERCISES. 1. From the ends of a side' of a triangle any two straight lines are drawn to meet the other sides in P, Q ; also from the same ends two lines parallel to the former are drawn to meet the sides produced in P', Q': show that PQis parallel to P'Q'. Let ABC be the a ; BP, CQ the lines. AC : AP = AQ' : AB, etc. 2. ABC is a triangle, AD any line drawn from A to a point D in BO ; a line is drawn from B bisecting AD in E and cutting AC in F : prove that BF is to BE as 2 OF is to AC. Draw EG || to AC meeting BC in G. DE = EA, .*. AC = 3 EG, etc. 3. If in Prop. 20, AD = 24, AD' = 20, AE = IG, BD ^ 8, and DE = G, jfind B'D', D'E', and AE', ^ vo 158 PLANE GEOMETRY. Numerical Eelations between- the Diffeeent Parts of a Triangle. Proposition 24. Theorem. 323. In a right triafigle, if a perpendicwlar he drawn from the right angle to the hypotenuse : (1) The two triangles' on each side of it are similar to the tvhole triangle and 'to each other, (2) Tlie perpendicular is a mean proportional between the segments of the hypotenuse, (3) Each side ahoiitthe right angle is a mean proportional between the hypoten'nse and the adjacent segment. Hyp. Let ABO be a rt. A , and AD the _L from the rt. Z A to the hypote- nuse BO. (1) To prove i\iQ As DAB, DAO, and ABO similar. Proof In the rt. As DAB, AOB, the acute Z B is common. . • . the A s DAB, AOB are similar. (309) Also, in the rt. A s DAO, ABO, the acute Z is common. . • . the A s DAO, ABO are similar. (309) . • . the A s DAB, DAO are similar to each other, as they are both similar to ABO. q.e.d. (2) To prove BD : AD = AD : OD. Proof Since the A s DAB, DAO are similar, .-.BD: AD = AD:OD. (307) (3) To prove BO : AB = AB : BD. Proof. Since the A s BAO, BAD are similar, BOOK III— NUMERICAL MEASURES IN A TRIANGLE. 159 ».-. BC : AB = AB : BD. (307) Iso, since the A s CAB, CAD are similar, .-. BC : AC = AC : CD. q.e.d. 324. Cor. 1. If the lines of the figure are expressed in \ Qumbers by means of their numerical measures (277), we have from the above three proportions, by means of (281), / / AD' = BD X CD, n' AB' = BC X BD, V AC' = BC X CD. By division, the last two equations give AB'_ BD AC' ~ CD- J Hence, tlie squares of the sides about the right angle are vroportio7ial to the adjacent segments of the hyjjotenuse. 325. Cor. 2. Since an angle inscribed in a semicircle is a right angle (240), therefore: (323) (1) The perpendicular fro7n any point in the circumference of a circle to a diam- " ^ ter is a mea7i proportio7ial betiveen the segments of the diameter. (2) The chord from the point to either extremity of the diameter is a mea7i proportional between the diameter and the adjacent segment, 326. Def. Tlie projection of a point A upon an indefinite straight line XY is the foot C of the perpendicular let fall from the point to the line. X C D Y The projection of a ji)iite straight line AB upon the line XY is the part of XY between the per- pendiculars dropped from the ends of the line AB. Thus, CD is the projection of AB upon the line XY. 160 PLANE GEOMETRY. Proposition 25. Theorem. 327. 21ie square of the niimher which measures the hypote^iuse of a right triangle is equal to the sum of the squares of the numbers which measure the other two sides. Hy]). Let ABC be a rt. A , with rt. / at A. To prove BC' = AB' + AO'. Proof Draw the ± AD. Then AB' = BC X BD, ^ and (324) (324) Q.E.D. AC =: BC^ CD. : Adding, AB' + AC' = BC(BD + CD) = BC". 328. ScH. 1. By this theorem one of the sides of a right triangle can be found when the other sides are known. If we know the two sides b and c about the rt. Z , the hy- potenuse a is given by the formula a'' = b'' + c\ .'. a = Vb' + c\ Thus, if ^> = 3, c = 4, we have a = V9~+16 = V2b = 5. If the hypotenuse a and one of the sides b of the rt. / be known, the other side c is found by the formula c' = a''- b' Va' - W Thus, for rt = 5, and Z> = 3, we have c — V25 — 9 = Vl6 = 4. 329. ScH. 2. If AC is the diagonal of a square ABCD, we have (327) AC' = AB' + BC', or AC' = 2 AB'. AC" _ , AC ^^ ,\ -=ri — 2, and -t-r = ^^• AB AB Thus, the diago7ial and side of a square are two incom- mensurable lines, since their ratio, the square root of 2, is an incommensurable number. BOOK III.— NUMERICAL MEASUllES IN A TRIANOLE. 161 Note.— To simplify the enunciations of the following theorems, the term square of a line will be given to the second power of the number which ex- presses the measure of the line. The term product of two lines will be given to the product of two numbers which express the magnitude of those lines, measured with the same unit. Proposition 26. Theorem. 330. 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 di7ninished hg tivice the product of one of these sides by the projection of the other side upon it. Hyp. Let B be an acute Z of the A ABC, and BD the projection of AB upon BC. To prove AC' = AB' + BC' - 2BCx BD. Proof. Because ADC is a rt. A, .-. AC' = AD' + DC', (327) = AD' + (BC-BDr,(fig.l) or = AD'' + (BD - BC)'' (fig. 2) BCf — 2 BC X BD, by Algebra*, BC - 2BC X BD. (327) Q.E.D. * The square of the difference of two numbers is equal to the sum of the squares of the two numbers diminished by twice their product. 162 PLANE GEOMETRY. Proposition 27. Theorem. 331. In an obtuse-angled triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other tiuo sides increased by twice the prod- uct of one of these sides by the projection of the other side upon it. C B D Hyp. Let B be the obtuse Z of the A ABO, and BD the projection of AB upon CB produced. To prove !(? = AB" + BO' + 2BC X BD. Proof CD = OB + BD. . • . CD' = CB' + BD' + 2B0 X BD. (Algebra*) Add AD to both sides. AC' =r AB' + So' + 2BC X BD. (327) Q.E.D. 332. Cor. From the three p>receding theorems, it follows that the square of the side of a triangle is less than, equal to, or greater than, the sum of the squares of the other two sides, according as the angle opposite this side is acute, right, or obtuse. Note.— By means of these three theorems we may compute the altitudes of a triangle whea the three sides are known. EXERCISES. ^ 1. If the side BC of an equilateral triangle ABC be pro- /duced to D, prove that the square on AD is equal to the squares on DC and CA with the product of DC and CA. 2. If ABC be an isosceles triangle having AB = AC, and if D be taken in AC so that BD = BC, prove that the square on BC = AC X CD. * The square of the sum of two numbers is equal to the sum of the squares of the two numbers increased by twice their product, BOOK IIL-NUMEIUGAL MEASURES IN A TRIANGLE. 163 I Proposition 28. Theorem. 333. The sum of tlie squa7'es on two sides of a tricmgle is equal to twice the square on half the third side increased hy tioice the square on the median upon that side. } Hyp, Let ABC be a A, AE the median bisecting BC. To prove AB' + AC' = 2 BE' + 2AE'. Proof Draw AD _L to BO. Then one of the Z s AEB, AEG must be obtuse and the other acute : let Z AEO be acute. In the aAEB,Xb'=BE'+AE'+ 2BExED.(l) (331) In.the A AEC, AC'= CE'+ AE'- 20ExED. (2) (330) Adding, and observing that BE = CE, we get AB'+ AC'= 2 BE'+ 2 AE'. Q.E.D. 334. CoK. Subtracting (2) from (1) in (333), we have AB'-A0'=2BCxED. Hence, the difference of the squares on two sides of a tri- angle is equal to twice the product of the third side by the projection ofihe median upon that side. Let the student prove the case in which AD falls without the triangle ABC, or coincides with AE. Note.— By means of this theorem we may compute the medians of a triangle when the three sides are known. 164 PLANE GEOMETRY. Proposition 29. Theorem. 335. If two cliords cut each other m a circle, the product of the segments of the one is equal to the product of the seg- ments of the other, D Hyp, Let the chords AB, CD cut at P. To prove AP X BP = CP X DP. Proof Join AD and BC. In the As APD, CPB, Z ADC = Z ABC, and Z BAD = Z BCD, As in the same segment of a © are equal (239). . • . A APD is similar to A CPB. (309) .-. AP:CP = DP:BP. (312) .-. APx BP = CPX DP. (281) Q.E.D. 336. Def. When four quantities, such as the sides about two angles, are so related that a side of the first is to a side of the second as tlie remaining side of the second is to the remaining side of the first, the sides are said to be recipro- cally proportional. Therefore — 337. Cor. 1. If tioo chords cut each other in a circle, their segments are reciprocally proportional. 338. Cor. 2. //' through a fixed point within a circle any number of chords are draion, the products of their segments are all equal, EXERCISES. 1. If AP = 4, BP = 5, and CD = 12, find the lengths of CP and DP. 2. If AB = 20, and CD = 24, find tlie lengths of CP and DP when CD bisects AB. BOOK IIL— NUMERICAL MEASURES IN A TRIANGLE. 165 Proposition 30. Theorem. 339. If from a point without a circle a tangent and a secant be drawn, the tangent is a mean proportional be- tween the whole secant and the external segment. Hyp. Let PC and PB be a tangent and a secant drawn from the point P to the OABO. ^- To prove PB : PC = PC : PA. Proof Join CA and CB. IntheAsPAC, PBC, ZACP= ZABC, eacJi being measured by i arc A C (238), (243). Z P is common. .-.AS PAC and PBC are similar. (309) .-.PB: PC == PC: PA. (312) Q.E.D. 340. Cor. 1. PC" = PB x PA. Therefore, the square of the tangent is equal to the prod- uct of the 2vhole secant and the external segment, 341. Cor. 2. Since PB x PA = PC', (340) and PB X PD = PC', (340) .-.PBx PA^PExPD. (Ax. 1) a Therefore, if from a point without a circle two secants he drawn, the product of one secant and its external seg^nent is equal to the product of the other and its external segment. 342. Cor. 3. If from a point without a circle any num- ber of secants are drawn, the products of the lohole secants and their external segments are all equal. .^.-.•* 166 PLANE GEOMETRY. Proposition 3 1 . Theorem. 343. If any angle of a triangle is bisected hy a straight line wliicli cuts the base, the product of the tioo sides is equal to the product of the segments of the base plus the square of the bisector. Hyp. In the A ABC let AD bisect the Z BAG. To prove AB X AC = BD x DC + AD\ Proof Describe a O about the A ABC, and produce AD to meet tlie Oce in E. Join CE. E Then in the As BAD, EAC, ZBAD=ZEAC. (Hyp.) ZABD = ZAEC. Zs in ilie same segment of a Q are equal (239). .*. AS BAD and EAC are similar. (309) .-. AB : AE = AD : AC. (312) .-. AB X AC =: AE X AD (281) = (ED + AD) AD ' = ED X AD + AD'. But ED X AD = BD X DC. (335) .-. AB X AC = BD X DC + AD'. q.e.d. Note.— By means of this theorem we may compute the lengths of the bisect- ors of the angles of a triangle when the three sides are known. EXERCISE. If the vertical angle BAC be externally bisected by a straight line which meets the base in D, show that the prod- uct of AB, AC together with the square on AD is equal to the product of the segments of the base. BOOKlIL—NUMEUWALMEASUUESmA TIUAMULE. 167 Proposition 32. Tiieorem. 344. The "product of two sides of a triangle is equal to \e product of the diameter of the circumscribed circle by the perpendicular let fall upon the third side from the vertex of the opposite angle. Hyp, In the A ABC lefc AD be J. to BO, and AE the diameter of the circum- scribed O. To prove AB x AC = AE x AD. Proof. In the AS ABD, AEC, ZB=ZE, each being measured by i arc AC (238), and Z ADB = Z ACE = a rt. Z . (Cons.) and (240) .-. AS ABD, AEC are similar. (309) .-. AB: AE = AD: AC. (312) .-. AB X AC = AE X AD. (281) Q.E.I). Note.— By means of this theorem we may compute the radius of the circle circumscribed about a triangle when the three sides are known. EXERCISE. The product of the two diagonals of a quadrilateral in- scribed in a circle is equal to the sum of the products of its opposite sides. SuaoKSTioN.— Make /DAE = /BAG. Then in the similar a s DAE, CAB, AD ; AC = DE : BC; .-. AD . BC = AC DE. Also in similar as DAC, EAB, AB:AC = BE:CD; .-. AB . CD = AC. BE. Add. AD . BC + AB . CD = AC . DB. 168 PLANE GEOMETRY. Problems of Constructiq]^. Proposition 33. Problem. 345. To divide a given straigld line into parts projjor- tio7ial to given straigld lines. Given, the line AB, and the lines P N M M, N, P. Reqiiiredf to divide AB into parts proportional to M, N, P. Cons. From A draw the indefinite straight line AH making any / with AB. On AH take DE = P. AC = M, CD = N, Join EB, and through D, C draw DG, CF II to EB. Then AB is divided at G and F into parts proportional to M, N, P. Proof, Since CF, DG, EB are || , (Cons.) AF _FG_GB •'• AC"~CD~DE* If two St. lines are cut by any number of \\8, the corresponding segments are proportional (300). ButAC = M, CD = ]Sr, DE = P. (Cons.) AF FG GB 346. ScH. If it be required to divide a line AB into any number of equal parts, take the same number of equal parts on AH, so that AC = CD = DE, and complete the con- struction as before ; then AF = FG = GB. EXERCISES. 1. Divide a straight line into five equal parts. 2. Give the construction for cutting ofl: two-sevenths of a given straight line. BOOK IIL— PROBLEMS OF CONSTRUCTION. 169 Proposition 34. Problem. 347. To find a fourth proportional to three given straight lines. Given, the three lines M, N, P. Required, to find a fourth propor- tional to M, N, P. Cons. Draw the two indefinite straight lines AG, AH, making any / with each other. On AH take AB = M, BC = N ; and on AG take AD = P. Join DB, and through draw CE II to DB. Then DE is the fourth proportional required. Proof. Since BD and CE are II , .-. AB:BO = AD:DE. A St. L \\to a side of a A divides the other two sides proportionally (298). But AB = M, BC = ^, AD = P. (Cons.) .-. M:ISr = P:DE. q.e.f. 348. ScH. If the lines IST and P are equal, then BC and AD are both laid off equal to N, and DE is the third pro- portional to M and N (280). The proportion in (347) then becomes M : N = N : DE. EXERCISES. 1. Construct a fourth proportional to the lines 3, 7, 11. 2. If from D, one of the angles of a parallelogram ABOD, a straight line is drawn meeting AB at E and CB produced at F; show that OF is a fourth proportional to EA, AD, mad AB. 170 PLANE GEOMETRT. Proposition 35. Problem. 349. To find a mean proportional between tiuo given straight lines. Given, the lines M and N. M Reqicired, to find a mean propor- "^ tional between M and N. Cons. On an indefinite straight line take AD = M, and DB = N. I \ On AB describe a semicircle, and i "Twr — ~U~^\" draw DO J. to AB. Then DC is the mean proportional required. Proof, Since AD:DC = DC:DB, (325) and AD = M, DB = N, > (Cons.) .-. M:DC = DC:]Sr. q.e.f. 350. ScH. The mean proportional between two lines is often called the geometric mean, while their half sum is called the arithmetic mean. EXERCISE. From a given point P in a circle a perpendicular PM is drawn to a given chord AB ; from A, B perpendiculars AC, BD are drawn to the tangent at P: prove that PM is a mean proportional between AC and BD. 351. Def. a straight line is said to be divided in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less. Thus, the line AB is divided in extreme and mean ratio at C if <^ C B AB : AC = AC : CB. BOOK IIL-PROBLEMS OF CONSTRUCTION. 171 Proposition 36. Problem. 352. To divide a given straight line in extreme and 7nean ratio. Given, the st. line AB. /'"* *"*'v. Required, to divide it in extreme / ,,''''^f | and mean ratio. i ,. ' Cons. At B erect the J. BO =i AB. /i^<' With centre C and radius CB de- .--'''^ V. ^ A F B scribe a O. Join AC, cutting the O at J), and produce it to meet the O again at E. On AB lay off AF = AD. Then F is the pt. of division required. Proof. Since AB is a tangent and AE a secant to the O, . • . AE : AB = AB : AD. (1) The tangent is a mean pt'opoo^tional between the wJiole secant and the external segment (339). .-. by division, AE - AB : AB = AB - AD : AD. (288) But AB = DE and AD = AF. (Cons.) .-. AE - AB = AD = AF, and AB - AD = FB. Substituting these values in the above proportion, we have AF : AB = FB : AF. 4^^ .-. AB:AF = AF:FB. ^-y^s) . ' . AB is divided in extreme and mean ratio at F. q.e.f. 353. ScH. If BA be produced to the left of A to a point F' so that AF' = AE, then F' is another point of division having the same property as F. Thus we have from proportion (1) in (352), by composition, AE H- AB : AE = AB + AD : AB. (287) But AE + AB = BF', and AB -[- AD = AF'. (Cons.) .-. BF': AF' = AF': AB, or AB : AF' = AF' : F'B. . • . AB is divided in extreme and mean ratio at F'. AB is said to be divided at F internally, and at F' ex- terjially, in extreme and mean ratio. Note.— This division of the straight line was called by the ancient geometers the golden section. 172 PLANE GEOMETRY. Proposition 37. Problem. 354. On a given straiglit line to construct a 2)olygon similar to a given ^polygon. Given, the polygon ABODE and the straight line A'B'. Required, to construct on A'B' a polygon similar to ABODE. Cons. Let AB be the side of the given polygon to which A'B' is to be homologous ; join AO and AD. At A' make Z B'A'O' = / BAO, and at B' make Z A'B'O' = Z ABO. Also at A' makez O'A'D' = Z CAD, and at 0' make Z A'O'D' = z AOD. Also at A' make Z D'A'E' = Z DAE, and at D' make Z A'D'E' = z ADE. Then A'B'O'D'E' is the required polygon. Proof. Since the A s ABO, A'B'O' are mutually equi- angular, (Ooiis.) .*. they are similar. (309) Also AS AOD, A'O'D' are similar. (309) Likewise A s ADE, A'D'E' are similar, (309) . •. the polygons ABODE, A'B'O'D'E' are similar, being composed of ttie same number of as similar each to each, ami similarly placed ( 3^0) . q. e. f. BOOK IlI.-APPLIGATIONS. 173 Applications. 1. To find the altitudes of a triangle in terms of its sides.* See (332). Let ABC be a A ; denote by a, l, c, the lengths of the sides opposite the ^/^ Z s A, B, C, respectively, and by h the \cy^ I altitude AD. y^ A Of the two Z s B and C, at least one is Z^ / acute ; suppose it to be the Z C. Then : In the aADC, A^ = h' - CD'. (328) In the A ABC, 6'^ ^a' j^¥ - 2a X CD. (330) .-.CD 2a r.V = {a^-\-l^-oy ^.a'h'-ia'+h'-cy ■^ 4^' ~ 4a" {2ah^-a^W-c'){2al-ce-V^-c^) ,,^ ,, 4.a' ^^ J -^^^S' 1 / -{a-hy\ . {a+i-^c){a+h- c)(c-\-a-'b)(c-a-\-y) (1) 4a' Put a+b + c = 2s. Then a-\-b- c = 2{s — c). c -\- a — ■b = 2{s- I). c — a -\- h = 2(s - a). Substituting in (1) and extracting the square root, we , 2 /' = - Ms - a){s - b){s - c). 2. To find the medians of a triangle in terms of its sides. See (333). * See Trait6 de G6otn6trie, par Roiich6 e4; Comberousse, p. 139. t The difference of the squares of two numbers equals the product of the pum «ind difference of the two numbers, 174 PLANE GEOMETRY. Denote by m the median on the side BO. Then, ■!)'-{- c' =^ 'Zm' ^ ^(^^\ (333) r,m = ^ i/2(^' + c') - a\ 3. To find the bisectors of a triangle in terms of its sides. See (343.) Denote the bisector AD, figure of (343), by d, and the sides as in Exs. 1 and 2. By (343), ^c = BD X DC + d\ .•.^•^ = Z»c~BD X DC. (1) By (303), BD ^ DC ^ BD^C ^ a _ •^^^c b b -\-c b -\-c ^ ' ...BD = ^^,andDC:= ^* b + o' b-\-c Substituting these values in (1) and reducing, and de- noting by s the semi-perimeter of the A , we have 4. To find the radius of the circumscribed circle in terms of the sides of the triangle. See (344). Denote the radius by R, figure of (344), and the sides as before. By (344), bc = 2^x AD. Substituting for AD its value in Ex. 1, and solving for E, we have P _ abc ~ 4 Vs{s-a){s-l)){s - v) ' 5. Find the altitudes of a triangle whose sides are 13, 9, and 6. Ans. 3.G41, 5.259, 7.88G. 6. Find the medians of the above triangle. Ans. 4.031, 9.069, 10.770. 7. Find (1) the bisectors of the angles of the above tri- angle, and (2) the radius of the circumscribed circle. Ans, (1) 3.833, 7.778, I0.407j (2) 7.416. BOOK IIL-EXERCI8ES. THEOREMS. 175 exercises. Theokems. 1. If two circles touch each other, either internally or externally, any two straight lines drawn from the point of contact will be cut proportionally by the circumferences. 2. If two circles intersect in P, and the tangents at P to the two circles meet the circles again in Q and E; prove that PQ : PR in the same ratio as the radii of the circles. 3. ABO, DEF are two isosceles triangles, BO, EF being the bases. If AB : B0^= DE : EF, show that the triangles are similar. 4. P is a point in a diagonal of a parallelogram. EPF, GPH join points in the opposite sides of the parallelogram. EH, GF cut the diagonal in which P is. Show that EH is parallel to GF. 5. The parallel straight lines AB, OD are joined by AD, BO which intersect in ; in AB, OD points E, F are taken such that AE : EB = DF : FO : prove that E, 0, F are col- linear. (1G4). C. The side BO of a triangle ABO is bisected at D, and the angles ADB, ADO are bisected by the straight lines DE, DF, meeting AB, AO at E, F respectively : show that EF is parallel to BO. Apply (303). 7. AB is a diameter of a circle, OD is a chord at right angles to it, and E is any point in OD; AE and BE are drawn and produced to cut the circle at F and G; show that the quadrilateral OFDG has any two of its sides in the same ratio as. the remaining two. Apply (303). 8. In the circumference of the circle of which AB is the diameter, take any point P, and draw PO, PI) on opposite sides of AP, and equally inclined to it, meeting AB at and D: show that AO : BO = AD : BD. Apply (303) and (304). 9. From tlie same point A straight lines are drawn mak- ing the angles BAO, OAD, DAE each equal to half a* right angle, and they are cut by a straight line BODE, which 176 PLANE GEOMETliY. makes BAE an isosceles triangle : show that BC or DE is a mean proportional between BE and CD. Apply (303) and (304). 10. ABCD is a quadrilateral having two of its sides AB, CD parallel; AF, CG are drawn parallel to each other meeting BC, AD respectively in F, G. Prove that BG is parallel to DF. 11. If D be the middle point of the side BC of a trian- gle ABC, and if any straight line be drawn through C meet- ing AD in E and AB in F; show that the ratio of AE to ED will be twice the ratio of AF to FB. 12. If be the centre of the inscribed circle of the tri- angle ABC, and AO meet BC in D; prove that AO is to OD as the sum of AB and AC is to BC. Apply (303). 13. ABCD is a quadrilateral inscribed in a circle; AD, BC meet in P, and the angle P is bisected by a straight line cutting AB, CD in E, F. Show that AB : BE = CD : DF. Apply (309). 14. ABC is a straight line, and ABD, BCE triangles on the same side of it, such that the angles ABD, EBC are equal, and AB: BC = BE: BD: if AE, CD intersect in F, prove that AFC is an isosceles triangle. See (314). 15. Two circles touch in C, a point D is taken outside them such that the radii AC, BC subtend equal angles at D, and DE, DF are tangents to the circles : if EF cut DG in G, prove that DE : DF = EG : GF. 16. C is the centre of a circle, and A any point within it ; CA is produced through A to a point B such that the radius is a mean proportional between CA and CB : show that if P be any point on the circumference, the angles CPA and CPB are equal. See (314). 17. AB, AC are the equal sides of an isosceles triangle ; the straight line bisecting AB at right angles meets BC in D : prove that the square on AB = BC X BD. BOOK III. -EXERCISES. THEOREMS. 177 18. The square on the base of an isosceles triangle is equal to twice the product of either side by the part of that side intercepted between the perpendicular let fall on the side from the opposite angle and the end of the base. Apply (330.) 19. ABC is a triangle having the sides AB, AC equal; if AB is produced to D so that BD is equal to AB, show that the square on CD is equal to the square on AB", together with twice the square on BC. 20. The sum of the squares on the four sides of a paral- lelogram is equal to the sum of the squares on the diagonals. Apply (333). 21. The base of a triangle is given and is bisected by the centre of a given circle : if the vertex be at any point of the circumference, show that the sum of the squares on the two sides of the triangle is constant. 22. In any equilateral the sum of the squares on the diag- onals is equal to the sum of the squares on the straight lines joining the middle points of opposite sides. Apply (155) and then Ex. 20. 23. The sum of the squares on the four sides of a quad- rilateral is equal to the sum of the squares on the diagonals, increased by four times the square on the line joining the middle points of the diagonals. Apply (333). 24. AB is a diameter of a circle, the chords AC, BD intersect in a point E inside the circle : prove that the square on AB is equal to the sum of the products of AC, AE and BD, BE. Apply (335) and (331). 25. AD, BE, CF, the perpendiculars from the vertices on the opposite sides of a triangle ABC, intersect in : prove that the products of AO, OD and BO, OE and CO, OF are equal to each otlier. Apply (335). 26. With a point inside a circle ADC as centre a circle is described cutting the former circle in D, E ; a chord AOB is drawn to the first circle : if the product AO, OB is 178 • PLANE GEOMETRY, equal to the square on the radius of the second circle, prove that D, 0, E are collinear. Apply (335). 27. If two circles intersect, prove that tangents drawn to them from any point in their common chord produced are equal to each other. 28. AEB, PEDQ are respectively a diameter and a chord of a circle at right angles to each other, ADC, BC are otlier chords. Prove that the product of AB, AE is equal to the product of AC, AD. (341). 29. If AB be a diameter of a circle, and APQ a straight line cutting the circle again at P and a fixed straight line perpendicular to AB at Q; the product of AP, AQ is con- stant. (341). 30. A, B, D are three points in order on a circle ; C is the middle point of the arc AB ; if AC, AD, BD, CD be joined, prove that the square on CD is equal to the square on AC with the product of AD, BD. 31. A circle is described round an equilateral triangle, and from any point in the circumference straight lines are drawn to the angular points of the triangle : show that one of these straight lines is equal to the sum of the other two. 32. From the extremities B, C of the base of an isosceles triangle ABC, straight lines are drawn at right angles to AB, AC respectively, and intersecting at D : show that BC X AD = 2 AD X DB. Problems. 33. AD bisects the angle BAC. From any point in AB draw a straight line to AC which shall be divided by AD in a given ratio, i.e., into parts which have the same ratio which two given straight lines have to each other. 34. Through a given point between two given straight lines draw a straight line which shall be terminated by the given lines and divided by the point in a given ratio. 35. Upon a given portion AC of the diameter AB of a semicircle another semicircle is described. Draw a line BOOK IIL—EXBHGISES. PROBLEMS. 179 throiigli A so that the part intercepted between the semi- circles may be of given length. 36. ABC is a segment of a circle; the chord AC is divided into two parts at D. Divide the arc ABC into two parts so that the chords of the parts shall have the same ratio which AD, DC have to each other. Complete the ® ABCE ; bisect arc AC opp. B in E ; produce ED to B, etc. 37. Divide the straight line AB into two parts at C so that the square on AC may be equal to twice the square on BC. Make zBAD = J^ rt. z, draw BD ± to AB, make zADC = ZDAB, etc. 38. From a given point as centre describe a circle cutting a given straight line in two points, so that the product of their distances from a fixed point in the straight line may be equal to a given square. 39. ABC is a given obtuse-angled triangle : find a point D in the side BC opposite the obtuse angle, so that BD X DC = Ai5'. Let O be the cent, of the ® about a ABC. Join AO, on it describe | ® ADO, cutting BC in D ; join AD and produce it to E, etc. 40. Describe a circle which shall pass through two given points and touch a given straight line. Two solutions. (339). Let A, B be the pts., XY the line ; produce AB to meet XY in C, etc. 41. Describe a circle to pass through a given point and touch two given straight lines which are not parallel to each other. Two solutions. Let given pt. P be between the given lines AB, AC ; bisect Z BAC, etc. 42. From a given point on the circumference of a given circle draw two chords so as to be in a given ratio and to contain a given angle. Let P be the given pt.; draw diameter PA ; make Z APB = given z, etc. 43. From any point A on one of two intersecting straight lines OA, OB draw two straight lines AB, AC cutting OB ill B and C, such that the triangles AOB, OAC may be similar, and the sides AB, AC have a given ratio to each other. 44. To describe a circle which shall pass through two given points and touch a given circle. Book IV. AEEAS' OF POLYGONS. Measurement of Areas. 355. Def. The area of a surface is its numerical meas- ure (229), referred to the unit of surface. To form a unit of surface, we take any unit of length, as an inch, a foot, etc., and construct a square on it. This square is the corresponding unit of surface, or the superficial unit. Two surfaces are equivalent when they have equal areas and cannot coincide; they are equal when they can coin- cide. Proposition 1 . Theorem. 356. If two rectangles have equal altitudes, their areas are to each other as their bases. Hyp. Let ABCD, AEHD be two "^ rectangles, having equal altitudes. ABCD _ AB AEHD ~ AK To prove Case I. Wien AB and AE are com- mensurahle. Proof. Take AF, any common meas- ure of AB and AE, and suppose it to be contained 5 times in AB and 4 times in AE. AB 5 AE~4" j A f I - B 3 H j 1 1 i A F Then Draw J_s to AB and AE at the several points of division. The rectangle ABCD will be divided into 5 rectangles, and AEHD into 4 rectangles. 180 D KC 1 A QB D H BOOK IV.-MEASUREMENT OF AREAS. 181 These rectangles are all equal. (136) ABCD _ 5 •'• AEHD~4' ABCD _ AB •*• AEHD~ AE* (Ax. 1) Case II. When AB and AB are incommensurable. Proof, In this case we know (232) that we may always find a line AG as nearlg equal as ice please to KB, and such that AG and AE are com- mensurable. Draw GK II to BC. Then, since AG and AE are com- mensurable, AGKD AG ,n Tx A E ...^^jjj^ = ^-^.(CaseI) A Now, these two ratios being always equal while the com- mon measure is indefinitely diminished, they will be equal when G moves up to and as nearly as lue j^lcase coincides with B (233). ABCD _ AB •'• AEHD~AE' ^•^•''• 357. Cor. Since the altitudes of rectangles maybe con- sidered as their bases, and their bases as their altitudes, therefore : The areas of rectangles having equal bases are to each other as their altitudes. Note —In these propositions the words "rectangle," "triangle," etc., are often used to denote the "area of the rectangle," the "area of the triangle," etc. EXERCISES. The area of a rectangle is 1728 square feet, and its base is 2 yards : what is the area of a second rectangle whose base is 5 yards and whose altitude is the same as that of the first ? 182 PLANE GEOMETRY, Proposition 2. Theorem. 358. The areas of any tivo rectangles are to each other as the products of their bases by their altitudes. Hyp, Let R and R' be two rectangles, b and Z>' their bases, a and a' their altitudes, ^ U a X b To prove R' Xb'' c: 1 a ^ ^ J 9 Proof Construct the rect- angle S having the same base b as that of R, and the same altitude a' as that of R'. Then _^ D-- a' rectangles of equal bases are to each other as tJwir altitudes (357) ; -S b and ^ = y, rectangles of equal altitudes are to each oilier as tlieir bases (356). Multiplying these ratios, and canceling the common fac- tor, we have R^_ a X b R' ~ «' X y 359. ScH. By the product of two lines is meant the product of the numbers which represent those lines when they are measured by a common linear unit (277). EXERCISE. Find the ratio of two rectangles, the base of the first bemg 3 yards and its altitude 13 feet, the base of the sec- ond being 8 feet and its altitude 39 inches, Reduce to the same unit, and compare. AnS, 4^. Q.E.D. BOOK IV.- AREAS OF POLYGONS. 183 (358) Proposition 3. TFieorem. 360. The area of a rectangle is equal to the product of its base and altitude. Hyp, Let K be the rectangle, b the base, and a the altitude expressed in numbers of the same linear unit; and let S be the square whose side is the linear unit. To prove area of K = « X ^. Proof ^-=__^^X^. But since S is the unit of surface, . • . R -f- S == the area of R. (355) .'. area of R = a X b, q.e.d. 361. ScH. 1. The statement of this Proposition is an abbreviation of the following: The number of units of area in a rectangular fig^ire is equal to the product of the number of linear units in its base by the number of linear units in its altitude. When the base and altitude can be ^^ ^ expressed exactly in terms of some com- mon unit, this proposition is rendered evident by dividing the figure into squares, each equal to the unit of meas- ure. Thus, if AB contain 4 linear units and AD 3, and if through the ^ ^ points of division parallels are drawn, it is seen that the rectangle is divided into three rows, each having four squares, or into four coluvuis, each having three squares. Hence the whole rectangle contains 3 X 4, or 12 squares, each equal to the unit of measure. Similarly, if AB and AD contain m and n units of length respectively, the rectangle will contain mil units of area. 362. ScH. 2. The area of a square is equal to the second poumr of its side, being the product of two equal sides. In Geometry, the expression " rectangle of two lines ■" is frequently used for the " product of two lines," meaning the product of their numerical measures. ! 184 PLANE OEOMETBY. Proposition 4. Tiieorem. 363. Tlie area of a parallelogram is equal to the prod- uct of its base and altitude. Hyp, Let ABCD be a CJ, AB its F__D E ^ base, and BE its altitude. \ / \ To prove ZZ7 ABCD = AB x BE. j / I / Proof Draw AF J_ to AB meet- t U ing CD produced to F. ^ ^ Then ABEF is a rectangle hav- ing the same base and altitude as the dJ ABCD. Then AD = BC, and AF = BE, being opp. sides of a CO (129). .-. A ADr= A BEC, Tiaving the hypotenuse and a side respectively equal in each (110). Take the A ADF from the trapezoid ABCF, and there is left the z=7 ABCD. Take the A BEC from the trapezoid ABCF, and there is left the rect. ABEF. r, OJ ABCD = rect. ABEF. (Ax. 3) But area of rect. ABEF = AB x BE. (360) . • . area of OJ ABCD = AB X BE. (Ax. 1) Q.E.D. 364. CoE. 1. Parallelograms having equal bases and equal altitudes are equivalent, because they are all equiva- lent to the same rectangle. 365. Cor. 2. Any two parallelograms are to each oilier as the products of their bases by their altitudes; therefore parallelograms having equal bases are to each other as their altitudes, and parallelograms of equal altitudes are to each other as their bases. BOOK IV.— AREAS OF POLYGONS. 185 Proposition 5. Theorem. 366. Tlie area of a triangle is equal to half the product of its base and altitude. Hyp, Let ABC be a A, BO its \; base, and AD its altitude. \ . To prove A ABC = I BC X AD. \ / Proof From B draw BH || to \ / CA, and from A draw AH || to CB '^/ meeting BH in H. BCAH is a ZZ7 having the base BC and the altitude AD. (124) .-. A BAC = A BAH, a diagonal of a OO divides it into two equal As (130). But CJ BCAH = BC X AD, the area of aOJ is equal to the pi'oduct of its base and altitude (363). .-. A ABC = iC7BCAH=:4BC x AD. q.e.d. 367. Cor. 1. Triangles having equal bases and equal aUiludes are equivalent ; and therefore triangles on the same base, and having their vertices in the same straight line parallel to the base, are equivalent, 368. Cor. ^, If a triangle and a parallelogram have the same base and are between the same parallels, the area of the triangle is half that of the parallelogram. 369. Cor. 3. Any two triangles are to each other as the products of their bases by their altitudes ; therefore tri- angles having equal bases are to each other as their alti- tudes, and triangles of equal altitudes are to each other as their bases. 370. Cor. 4. Of tiuo parallelogra7ns or two triangles on equal bases, that is the greater which has the greater alti- tude ; and of two ])arallelograms or triangles of equal alti- tudes, that is the greater which has the greater base. 186 PLANE GEOMETRY. Proposition 6. Theorem. 371. The area of a trapezoid is equal to the product of the half sum of its parallel sides hy its altitude. Dr, .C AH B Hyp, Let ABCD be a trapezoid, AB and CD the || sides, and DH the altitude. To prove area ABCD = i ( AB + CD) DH. Proof Draw the diagonal BD, dividing the trapezoid into two AS ABD and DCB, having the common altitude DH. Then area A ABD = i AB x DH, and area A DCB = J CD X DH. The area of a A equals ^ the product of its base and alt. (366). .• . area ABCD = J ( AB + CD) DH, since the two as make up the area of the trapezoid. Q.E.D. 372. Cor. Since the median EF = i (AB + CD) (156), therefore the area of a trapezoid is equal to the product of the median joining the middle points of the non-parallel sides hy the altitude, ,'. area ABCD :=rE X DH. 373. ScH. The area of any polygon may be found by dividing it into triangles, and finding the areas of the sev- eral triangles. But in practice the method usually employed is to draw the longest diagonal AF of the polygon, and upon AF let fall the perpendiculars BM, CN, DP, EO, GQ, thus decomposing the polygon into right triangles and trapezoids. By measuring the lengths of the perpendicu- lars, and the distances between their feet upon AF, the areas of these figures are readily found, and their sum will be the area of the polygon. BOOK IV.—aOMFAlilSOJSr OF AREAS. 187 Comparison of Areas. Proposition 7. Theorem. 374. The square described on the hyiMenuse of arujlit triangle is equivalent to the sum of the squares described on the other two sides, Jlyp. Let ABC be a rt. A, rt. angled at A, and BE, AK, AF squares on BC, AC, AB. To 2jrove sq. on BC=sq. on AB-fsq. on AC. Proof Through A draw AL || to BD, and join AD, FC. Since Z BAG is a rt. Z , (Cons.) and Z BAC is a rt. Z , (Hyp.) .-. CAGisast. line. (52) . • . sq. BG is double the A FBC, having ilie same base and between the same \\ s (368), and rect. BL is double the aABD. Again, since FB = AB, and BC = BD, (Cons.) and ZFBC=ZABD, eoA^h being tlie sum of art. Z. and the common Z ABC, .-. A FBC=:aABD, having two sides and the included Z equal each to each (104). .-.sq. BG = rect. BL. (Ax. 6) In like manner, by joining BK, AE, it may be proved thatsq. HC = rect. CL. . • . whole sq. BDEC = sum of sqs. BG and HC. (Ax. 1) . • . sq. on BC = sq. on AB -f sq. on AC. q.e.d. Note.— This proposition is commonly called the VytUagorean Proposition, be- cause it is said to have been discovered by Pythagoras (born about 600 B.C.). Tlie above demonstration of it was given by Euclid, about 300 b.c. (Prop. 47, Book I. Euclid). 188 PLANE GEOMETRY. Proposition 8. Theorem. 375. The areas of two triangles having 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. Hyp. Let ABC, ADE be the two As having the common Z A. A ABC AB X AC To prove ^aDE^ADxAE' Proof Join BE. Since the As ABC, ABE have their bases AC, AE in the same straight line, and the common vertex B, they have the same altitude. A ABC AC aABE AE' AS of the same altitude are to each other as their bases (369). Also, the A s ABE, ADE have the same altitude, since their bases AB, AD, opp. the common vertex E, are in the same straight line. A ABE AB * • A ADE AD* Multiplying these equalities, we have A ABC AB X AC (369) aade-adxae- ^^^^^ Q.E.B. 376. Cor. If the products of the sides including the equal angles are equal, the triangles are equivalent. EXERCISE. If two triangles have an angle of the one supplementary to an angle of the other, their areas are to each other as the products of the sideB including these angles, BOOK IV.— COMPARISON OF AREAS. 189 Proposition 9. Theorem. 377. The areas of similar triangles are to each other as the squares of their homologous sides. B Hyp. Let ABC, A'B'C be similar As. A ABC BC' To prove XT^B^ ^ g^,^ ' Proof. Since Z B == / B', (Hyp.) A ABC _ BA X BC _ BA BC^ • • A A'B'C ~ B'A' X B'C B'A' ^ B'C* ^ ^^ BA BC But B'A' B'C'^ homologous sides of similar as are proportional (307). BA BC Substitute in the above equality for ^^^ its equal .^v^, A ABC BC BC W A A'B'C ~ B'C ^ B'C ~ JB^/^ * ^'^'^' 378. Cor. The areas of tivo similar triangles are to each other as the squares of any ttvo homologous lines. EXERCISES. 1. ABC is a triangle. AE and BF, intersecting in G, are drawn to bisect the sides BC, AC in E and F. Com- pare the areas of tlie triangles AGB, FGE. 2. The base and altitude of a triangle are 18 and 8 re- spectively; wliat is the altitude of a similar triangle whose base is 12 ? 3. A has a triangular piece of ground, the base of the triangle being 40 rods; what is the base of a similarly- shaped lot containing 4 times as much ? A71S. 80 rods. 190 PLANE GEOMETRY. Proposition 10. Theorem. 379. Tlie areas of similar polygo7is are to each other as the squares of their homologons sides, Hyj). Let S and S' denote the areas of the two similar polygons ABODE and A'B'C'D'E', in which AB and A'B' are homologous sides. To prove ^ S AB Proof. Join AC, AD, A'C, A'D', dividing the two polygons into the same number of similar As, and simi- larly placed. (321) A ABC Also, A A'B'C A ACD AB^ CD' A A'C'D' cq5> and A ADE DE A A'D'E' ~ 5^ similar As are to each otliei'' as tlie squares of ilieir Iwmologous sides (377) AB _ M)_ _ OT^ _ ^ B'C But A'B' A ABC CD' A ACD " D'E' ~ A ADE etc. A A'B'C ~ A A'C'D' ~ A A'D'E'* (307) (Ax.l) A ABC + A ACD + A ADE •''a A'B'C + A A'C'D' + A A'D'E' A ABC "aA'B'C AB" A'B' (296) A^' Q.E.D. 380. Cor. 1. TJie areas of similar polygons are to each other as the squares of any tioo homologous lines. 381. Cor. 2. The homologous sides of similar polygons are to each other as the square roofs of their areas. BOOK IV.—PBOBLEMS OF CONSTRUCTION. 191 Problems of CoNSTRucTioiq". Proposition 1 1 . Problem. 382. To construct a triangle equivalent to a given poly- gon. Given, the polygon ABODE. Required f to construct a A equal to it in area. Cons. Join BD. From C draw CF || to DB, meet- ing AB produced in F ; join DF. In like manner Join AD, draw EG II to DA, meeting BA produced in G, and join DG. Then area DGF = area ABODE. Proof. A BDF = A BDO, having the same base BD and their vertices G and F in tlie ") St. line GF \\ to the base (367). ^ Adding the polygon ABDE to each of these equalities, we have area AFDE = area ABODE. (Ax. 2) Again, A ADG = A ADE. Adding the A ADF to each, we have area GDF t area AFDE 4 area ABODE. Q.E.F. 383. ScH. In the same manner a triangle may be found equivalent to a polygon having any number of sides, since each operation of the above process diminishes the number of sides of the polygon by one. 192 PLANE geometry: Proposition 12. Problem. 384. To construct a square equivalent to a given paral- lelogram. Given, a^ABCD, CE its ^ V ^ q altitude. | | V jV Esquired, to construct a I \ j \ square of the same area. I J V eB Cons, Find a mean pro- portional between AB and CE (349), and represent it by EG. The square described upon the line EG will be equiva- lent to the CJ ABCD. Proof, Since AB : EG = EG : CE, (Cons.) .-. FG' = ABx CE. (281) .• . area EGHK = area ABCD. q.e.f. 385. Cor. 1. To construct a square equivalent to a given triangle, ice take for its side a mean j^roportional be- tween the lase of the triangle and one-half its altitude* o^-rx] 386. Cor. 2. A square can ie found equivalent to any given polygon, hy first constructing a tria^igle equivalent to the given polygon (382), and then constructing a square equivalent to the triangle (385). Proposition 13. Problem. 387. To construct a square equivalent to the sum ofttoo P Q give7i squares, q p Given, P and Q the sides of the squares. Required, to construct a ■ = square equivalent to their sum. ^v^ D E A B BOOK IV.— PROBLEMS OF CONSTRUCTION. 193 Cons, DrawAB = P. At A construct the rt. /_ A, draw AC = Q, and join BC. Construct the square DEFG, witli each of its sides equal toBC. Then DEFG is the square required. Proof, W = AB' + AC'. (374) .-. DE' = P^ + Q\ Q.E.F. 388. ScH. In the same wa}^ a square may be found equivalent to the sum of any number of squares; for the same construction which reduces two of them to one will reduce three of them to two, and these two to one, and so of others. Proposition 14. Problem. 389. To construct a square equivalent to the difference of ttvo given squares. Given, P and Q the sides of p _ the squares. G F Q Required, to construct a f" ~\ Cjv,^ square equivalent to their dif- I ference. ! Cons. Draw the indefinite st. D e A "^B~H line AH. At A construct the rt. /A, draw AC = Q, the shorter side of the given lines. With centre C, and a radius = P, describe an arc cutting AH in B. Construct thq square DEFG, with each of its sides equal to AB. Then DEFG is the square required. Proof, AB' = BC' - AC'. (374) ... DE':=P-Q\ q.P^.F. 194 PLANE GEOMETRY. Proposition 15. Problem. 390. To cQustrtict a rectangle equivalent io a given square, and having the sum of two adjacent sides equal to a give7i line. Given, the square S, and the c ^^^- ^ ^^^^--P line AB = the sum of the sides, j / * j\ Required, to construct a rect- [ \\ angle == S, having the sum of A E B its base and altitude = AB. Cons. Upon AB as a diameter, describe a semicircle. At A erect AC _L to AB and = a side of S. Draw CD 1| to AB, cutting the ©ce at D, and draw DE i, to AB. Then AE and EB are the base and altitude of the re- quired rectangle. Proof, AE X EB = DE', mice DEis a mean proportional hetioeen AE and EB{d25), and DE = AC = a side of S. (Cons.) .-. AE X EB = S. Q.E.R 391. ScH. When the side of the square S exceeds half the line AB the problem is impossible. EXERCISES. 1. Construct a square equivalent to the sum of two squares whose sides are 6 and 8 inches. / d 2. Construct a square equivalent to the difference of two squares whose sides are 15 and 25 feet. > 3. The perimeter of a rectangle is 144 feet, and the length is three times the altitude : find the area. 4. On a given straight line construct a triangle equal to a given triangle and having its vertex on a given straight Mine not parallel to the base. 5. Construct a parallelogram that shall be equal in area and perimeter to a given triangle. :^ <7 C P BOOK IV.— PROBLEMS OF GONSTRtfCTtON. 195 Proposition 1 6. Problem. 392. To construct a rectangle equivalent to a given sqiiare, and having the difference of tioo adjacent sides equal to a given line. Given, the square S, and Cq,^- -^^ S \ the line AB = the difference ;,*^n. of the sides . A \ ^^-s: \ b Required, to construct a \ rectangle = S, having tlie dif- ference of its base and alti- tude = AB. Cons. Upon AB as a diameter, describe a O. At A erect AC _L to AB and = a side of S. Through C and the centre of the O draw CH, cutting the Oce at D and H. Then CII and CD are the base and altitude of the re- quired rectangle. Proof. CH X CD = AC , a tangent is a mean proportional between tlie whole secant and ilie ext. seg. (339). .-.CHxCD^S, and the difference between CH and CD is DH = AB. .•. CH X CD is the required rectangle. Q.KF. EXERCISES. 1. The bases of a trapezoid are 14 and 16 feet ; the non- parallel sides ai-e each 6 feet : find the area of the trape- zoid. 2. Construct a rhombus equal to a given parallelogram and having one of the sides of the parallelogram for one side of the rhombus. C n .<: 1 ^\ I A V'" B - 196 PLANE GEOMETRY. Proposition 1 7. Problem. 398. Tioo similar polygons being given, to construct a similar polygon equal to their sum, p Given, two homologous sides P and "^ ^' Q of two similar polygons R and S. Required, to construct a similar^ polygon equivalent to their sum. Q Cons, Draw AB = P. At A construct the rt. Z A, draw AC = Q, and join BC. . On BC, homologous to P and Q, construct a polygon T similar to R and S, as in (354). Then T is the polygon required. Proof -5 = ^,and|^^. (379) .... R + S F + Q' Adding, —y^ = ^u > But BC* = P' + Q'. (Cons.) .•.T = R + S. Q.E.F. 394. ScH. To construct a polygon similar to two given similar polygons, and equivalent to their difference, we find the side of a square equivalent to the difference of the squares on P and Q (389), and on this side, homologous to P and Q, construct a polygon similar to the given polygons R and S (354). This will be the polygon required (379). BOOK IV.-ritOBLEMS OF CONSTRUCTION. 197 Proposition 1 8. Problem. 395. To find tivo straiglit lines proportional to two given polygons, Give7i, two polygons R and S. ^^ Required, to find two st. lines ^--^^ j\ proportional to R and S. /' | \ Cons. Find two squares equiva- .^^. 1. \ •lent to the given polygons R and S ° (386) ; let P and Q be the sides of these squares. Construct the rt. Z A, draw AB = V, and AC = Q. Join BC, and draw AD _L to BC. Then BD and DC are the lines required. Proof, Since AD is a JL from the rt. / A on the hypote- nuse BC, .-. AB": AC' = BD:DC, tJie sqs. of the sides about ifie rt. Z are proportional to the adj. segments of tJie hypotenuse (324). But AB = P, and AC ^ Q. (Cons.) .•.F:Q' = BD:DC, and . • . BD, DO are proportional to the areas of the given polygons. Q.E.F. EXERCISE. Bisect a quadrilateral by a straight line drawn fr^m one of its vertices. Let ABCD be the quad.; bisect BD in E, let E lie between AC and B; through E draw EF ll to AC to meet PC in F; join CE, EA, AF, tljen AFCP = ^APCD, p — r\ S c K D B "K 198 PLANE GEOMETRY. Proposition 19. Problem. 396. To construct a square loliicli shall have to a given square the ratio of tioo given li7ies. Given, the square S, and the ratio P : Q. Required, to construct a square which shall be to S as P is to Q. Cons, Draw the st. line AK; take AD = P and DB = Q. Upon AB as a diameter, describe a |^Oce. At D erect the _L DC, and join AC, BC. On CB, or CB produced, take CH = a side of S. Draw HEj^ t^BA. _ '"'""Then CE is the side of the required square. Proof. Since CD is a _L from the rt. / C to the hypote- nuse AB, .•.Cr:Cg' = AD:DB. (324 But CA : CB = CE : CH, a St. line \\io a side of a A cuUilie other two sides proportionally (298), and CA' : CB' = CE' : ClI'. (295) .-. CE' : CH' = AD : DB =P : Q. But CH' = S. .-.CE^-.S^PlQ. Q.E.F. 397. ScH. To construct a polygon similar to a given polygon S, and having to it the given ratio of P to Q, we find, as / \ in (396), a side x so that x^ shall / ^ / be to 6' (where s is a side of S) ~^~ s as P is to Q, and upon x as a side q homol(f^ous to s, construct the polygon R similar to S (354) j this will be the polygon required. (379) BOOK IV,— PROBLEMS OF CONSTRUCTION. 199 Proposition 20. Problem, ^ 398. To construct a polygon equivalent to a given poly- gon F, and similar to a given polygon Q, Given, two polygons P aud Q. Esquired, to construct a poly- gon equivalent to P and similar toQ. ^ Co7is, Find jt? and ^, the sides p ^ of squares equivalent respectively /^ toPandQ. (386) / \ Take any side of Q as AB, and \ Q ^ find a fourth proportional A'B' to p -^f ^,jt?, andAB. (347) Upon A'B', homologous to AB, construct Q' similar to Q. (354) Then Q' is the polygon required. Proof, Since Q' is similar to Q, (Cons..) .•.Q;Q' = AB' : A7F'\ (379) But q:p = AB: A'B'. (Cons.) .'.(i:q' = f:p\ (Ax. 1) But P = i/, and Q = q\ (Cou^.) .•.Q:Q' = Q:P; and.-.Q' = P. . • . Q' is equivalent to P and similar to Q, Q.E.F. 200 PLANE GEOMETBY. ApPLICATIOI!^S. 1. To find the area of an equilateral triangle in terms of its side a,* Let h denote the alt., and S the area, of the A . Then h = Va' - T - ^ ^3. But (3GG) S = ^^^ . 2. To find the area of a triangle in terms of its sides and the radius of the circumscribing circle. Let a, h, c denote the sides and h the alt. of the A , and R the radius of the circumscribing O. By Ex. 4 (354), be = 2R/^. .-. abc = 2Jla7i = 4ES. (3G6) ••^-4R- Therefore, tlie area of a triangle is equal to the prodicct of the three sides divided hy four times the radius of the circumscribing circle. 3. To find the area of a triangle in terms of its sides. By Ex. 1 (354), /i = - V s{s - a){s - b){s - c). a And since S = ^ah, . • . S = Vs{s-a){s-b){s-c), 5. Find the area of an equilateral triangle if a side = 1 foot. Ans. 0.433 sq. ft. 6. Find (1) the area of the triangle whose sides are 3, 4, and 5 feet, and (2) the radius of the circumscribing circle. Ans, (1) 6 sq. ft. ; (2) 2.5 ft. * Rouch^ et Comberousse, p. 286. BOOK IV.— EXERCISES. THEOREMS, 201 exercises. Theorems. ^ 1. A triangle is divided by each of its medians into two xparts of equal area. 2. A parallelogram is divided by its diagonals into four triangles of equal area. 3. ABC is a triangle, and its base BC is bisected at D; if H be any point in the median AD, show that the tri- angles ABH, ACH are equal in area. 4. In AC, a diagonal of the parallelogram ABOD, any point H is taken, and HB, HD are drawn: show that the triangle BAH is equal to the triangle DAH. 5. If two triangles have two sides of one respectively equal to two sides of the other, and the included angles supplementary, show that the triangles are equal in area. 6. ABCD is a parallelogram, and E, H are the middle points of the sides AD, BC ; if Z is any point in EH, or EH produced, show that the triangle AZB is one quarter of the parallelogram ABCD, 7. If ABCD is a parallelogram, and E, H any points in DC and AD respectively, show that the triangles AEB, BHC are equal in area. 8. ABCD is a parallelogram, and P is any point within it: show that the sum of the triangles PAB, PCD is equal to half the parallelogram. 9. Find the ratio of a rectangle 18 yards by 14^ yards to a square whose perimeter is 100 feet. 10. The medians AD, BE of the triangle ABC are pro- duced to meet the straight line, drawn through C parallel to AB, in F, & respectively: prove that the triangle AGC is equal in area to the triangle BCF. AD = DF (105), etc. 11. ABCD is a parallelogram ; through A any two straight lines AE, AF are drawn meeting BC in E and CD in F; 202 PLANE GEOMETRY. through D, DG is drawn parallel to AE meeting AF in G. Show that the parallelogram having AE, AG as adjacent sides is equal to ABCD. Draw EH || to AG meeting DG produced in H; produce BC to meet DH in K; CO AH = ^Z7 AK; why? etc. 12. If is the centroid (172) of the triangle ABC, prove that the triangles AOB, BOO, COA are equal in area. 13. Show that the line joining the middle points of the parallel sides of a trapezoid divides the area into two equal parts. 14. If any point in one side of a triangle be joined to the middle points of the other sides, the area of the quadri- lateral thus formed is one-half that of the triangle. 15. Prove that any straight line which bisects a parallel- ogram must pass through the intersection of its diagonals. Let EF bisect ^Z/ABCD meeting AB in E, DC in F, BD in O; join ED, FB, etc. 16. APB, ADQ are two straight lines such that the tri- angles PAQ, BAD are equal. If the parallelogram ABCD be completed, and BQ joined cutting CD in K, show that CR = AP. 17. If the straight line joining the middle points of two opposite sides of any quadrilateral divide the area into two equal parts, show that the two bisected sides are parallel. 18. Through the vertices of a quadrilateral straight lines are drawn parallel to the diagonals: prove that the figure thus formed is a parallelogram which is double the quadri- lateral. 19. Through D, E the middle points of the sides BC, CA of a triangle ABC, any two parallel straight lines are drawn meeting AB or AB produced in F and G: prove that the parallelogram DEGF is half the triangle ABC. 20. In a trapezoid the straight lines, drawn from the middle point of one of the non-parallel sides to the ends of the opposite side, form with that side a triangle equal to half the trapezoid. 21. Prove that the points F, A, K^ in the figure of Prop. 7, are colli near. BOOK IV,— EXERCISES. THEOREMS. 203 IV 22. In the figure of Prop. 7, if FG, KH be produced to meet in P, prove that PA produced cuts BC at right angles. 23. Points E, F, G, II are taken respectively in the sides AB, BC, CD, DA of a rectangle ABCD; if EF = GH, prove that AG' + CH' = AF' + CE'. 24. From D, the middle point of the side AC of a right triangle ABC, DE is drawn perpendicular to the hypote- nuse AB : prove that BE' = AE' + BC'. 25. ABCD is a parallelogram. If AC is bisected in and a straight line MON is drawn to meet AB, CD in M, N respectively, and OR parallel to AB meets AN in R; prove that the triangles ARM, CRN are equal. NC = AM (105), .-. AN is II to CM (133), etc. 26. If through the vertices of a triangle ABC there be drawn three parallel straight lines AD, BE, CF to meet the opposite side or sides produced in D, E, F: prove that the area of the triangle DEF is double that of ABC. Let D be in BC; E, F in CA, BA produced; a EFB = a ECB, etc. 27. ABC, DEF are triangles having the angles A and D equal, and AB equal to DE : show that the triangles are to each other as AC to DF. 28. The side BC of the triangle ABC is bisected in D: prove that any straight line through D divides the sides AB, AC into segments which are proportional. 29. On the sides AB, AC of a triangle ABC points D, E are taken such that AD is to DB as CE is to EA : if the lines CD, BE intersect in F, prove that the triangle BFC is equal to the quadrilateral ADFE. 30. ABCD is a square. A line drawn through A cuts the sides BC, CD, produced if necessary, in E, F. Prove that the triangle CEF is to the triangle ABC as the differ- ence of the lines CE and CF is to BC. 31. is a point inside a triangle ABC; AO, BO, CO produced meet the sides in D, E, F, respectively. If AO is to OD as BO to EO as CO to OF, prove that is the cen- troid (172) of the triangle ABC. 204 PLANE OEOMETRT. 32. ABO is a triangle, and points D, E are taken in BC, CA respectively, such that BD = 4 DC and AE = ^AC; AD, BE intersect in 0. Prove that A AOB = i A ABC. A BAD : A ADC = BD : DC = 1 : 2; sim. a OBD : a ODC =1:2, etc. 33. ABCD is a quadrilateral having two of its sides AB, CD parallel; AF, CG are drawn parallel to each other meeting BC, AD respectively in F, G. Prove that BG^ is parallel to DF. Produce CB, DA to meet in E, etc. 34. ABC is a triangle, G its centroid, D the middle point of BC, AE its perpendicular to BC. Show that if the rect- angle of which AE, EC are adjacent sides be completed, the fourth corner being F, then FG produced bisects BE. Produce FG to meet BC in H; HD : AF = 1 : 2, etc. 35. ABCD, AB'C'D' are two squares, BAB', DAD' being straight lines ; B'C meets AD in E, and CD meets AB' in F. Prove that AE ^AF. 36. A triangle ABC has the side AB = 2AC ; from C is drawn CD to a point D in AB such that/ ACD = Z ABC: show that ABCD == 3aACD. 37. If in an isosceles triangle ABC the two equal sides AB, CA be divided at F and E respectively in any given ratio ; the straight line FE will meet BC produced at a point D, such that CD : BD = AF' : FB', or = CE' : EA'- Draw AG II to CB meeting EF produced in G, etc. 38. ABC is an isosceles triangle having the sides AB, AC equal, and is such that if BD be drawn bisecting the angle ABC and cutting AC in D, then AD is equal to BC. Show that me angle C is double the angle A. 39. Perpendiculars are drawn from the vertices of a tri- angle on any straight line through the centroid of the triangle : prove that one of these perpendiculars is equal to the sum of the other two. 40. Perpendiculars are drawn from the vertices and the centroid of a triangle on a given straight line : prove that the perpendicular from the centroid is the arithmetic mean of the perpendiculars from the vertices. Let ABC be the a , O its centroid, D tiie mid. pt. of BC; draw AL, BM, CN, DR, OK ± to given line; bisect AO in E; draw EH j. to LM, etc. BOOK IV.— EXERCISES. THEOREMS. 205 41. If the vertical angle C of a triangle ABC be bisected by a line which meets the base in D, and is produced to a point E, so that the rectangle of CD and CE is equal to that of AC and CB; show that if the base and vertical angle be constant the position of E is fixed. Since CD x CE = ACxCB,.-.CD : AG = BC: C£,and Z ACD = zBCD, .'. ZCAB = ZCEB, etc. 42. ABC is a triangle inscribed in a circle; from A straight lines AD, AE are drawn parallel to the tangents at B, C respectively, meeting BC, produced if necessary, in D, E: prove that BD is to CE as the square on AB is to the square on AC. Let BF, CG be the tangs, : ZADB = alt. /DBF = ZBAC (238 and 243), and ZB is common to both as BAC, BAD, etc. 43. If P be a point on a diameter AB of a circle, and PT be the perpendicular on the tangent at a point Q : show that PT X AB = AP X PB + PQ'. Produce QP to meet the in R; draw diam. RS: Z TQP = ZQSR, etc. 44. Prove that the square constructed on the sum of two straight lines is equiva" lent to Ihe sum of the squares con- structed on the two lines, together with twice the rectangle of the lines. 45. Prove that the square constructed on the difference of two straight lines is equivalent to the sum of the squares constructed on the lines, diminished by twice the rectangle of the lines. 46. Prove that the rectangle of the sum and the difference of two straight lines is equivalent to the difference of the squares constructed on the lines, 47. ABC is an isosceles triangle, CA, CB being the equal sides; BO is drawn at right angles to BC to meet CA pro- duced in 0: show that the square on OB is equal to the square on OA with twice the rectangle OA, AC. . 48. If a straight line be divided into two equal and D G H D' E D 206 PLANE GEOMETRY. also into two unequal parts, the rectangle of the unequal parts with the square on the line between the points of section, is equal to the square on half the line. (Euclid, B. II, prop. 5.) 49. The square on the straight line, drawn from the ver- tex of an isosceles triangle to any point in the base, is less than the square on a side of the triangle by the rectangle of the segments of the base. 50. The sides AB, CD of a quadrilateral ABCD inscribed in a circle are produced to meet in P; and PE, PF are drawn perpendicular to AD, BO respectively: prove that AE is to ED as CF to FB. 51. The sides AB, AD, produced if necessary, of a par- allelogram ABCD, meet a line through C in E, F respect- ively; CB, CD, produced if necessary, meet a line through A in G, H respectively. Prove that GE is parallel to HF. Problems. 52. Construct a square equal to three given squares. 53. Construct a square which shall be five times" as great as a given square. 54. Construct a triangle equal in area to a given quadri- lateral. 55. Construct an isosceles triangle equal in area to a given triangle and having a given vertical angle. 56. Divide a straight line into two parts, so that the sum of the squares on the parts may be equal to a given square. 57. Trisect a triangle by straight lines drawn from a given point in one of its sides. 68. Find a point inside a triangle ABC such that the triangles OAB, OBC, OCA are equal. 59. Construct a square that shall be one-third of a given square. 60. Divide a straight line into two parts so that the rect- angle contained by the whole and one part may be equal to a given square. Let AB be the given line. Draw BD ± to A.B so that BD" = given square; join AD, draw DE ± to AD, meeting AB produged in E. BOOK IV.— EXERCISES.— PROBLEMS. 207 61. Produce AB to so that the rectangle of AB and AC may be equal to a given square. 62. Construct a parallelogram equal to a given triangle and having one of its angles equal to a given angle. 63. Given three similar triangles: construct another sim- ilar triangle and equivalent to their sum. (377.) 64. Construct a triangle similar to a given triangle ABC which shall be to ABC in the ratio of AB to BC. Find DE a mean proportional between AB, BC, etc. 65. Bisect a triangle by a straight line parallel to one of its sides. Bisect AC of the a ABC in D; from AC cut off A£ a mean proportional to AC, AD, etc. 66. A is a point on a given circle: draw through A a straight line PAQ meeting the circle in P and a given straight line in Q, so that the ratio of PA to AQ may be equal to a given ratio. Draw AB to any pt. B in the given line MN; produce BA to C so that BA : AC = given ratio, etc. 67. From the vertex of a triangle draw a line to the base, so that it may be a mean proportional between the segments of the base. About the given a describe a © , etc. 68. Show how to draw through a given point in a side of a triangle a straight line dividing the triangle in a given ratio. 69. Construct a triangle equal to a given triangle and having one of its angles equal to an angle of the triangle, and the sides containing this angle in a given ratio. 70. Draw through a given point a straight line, so that the part of it intercepted between a given straight line and a given circle may be divided at the given point in a given ratio. *' Let P be the given pt,, XY the given line, A the cent, of given : produce AP to meet XY in Q; in PA take B so that QP : PB == given ratio, etc. Book V. EEGULAR POLYGONS. THE CIRCLE. MAXIMA AND MINIMA. Regular Polygons. 399. Def. a regular polygon is a polygon which is both equilateral and equiangular; as, for example, the equilateral triangle and the square. Proposition 1 . Theorem. 400. If the circumference of a circle he divided into any numher of equal arcs, (1) the cliords of these arcs form a regular polygon inscribed in the circle, and (2) the tan- ge?its at the points of division form a regular polygon cir- cumscribed about the circle. Hyp. Let the Oce be divided into equal arcs at the pts. A, B, C, etc.; let AB, BO, CD, etc., be chords of these arcs, and GBH, HCK, etc., be tangents. (1) To prove that ABCD is a regular polygon. BC = arc CD = etc., (Hyp.) BC = chd.CD = etc., Proof Since arc AB = arc .-. chd. AB = chd in ilie same O equal arcs are subtended by equal cliords (194), and ZFAB=: ZABC= / BCD = etc., all As inscribed in equal segments are equal (239). . • . ABCDEF is a regular polygon. (399) (2) To prove that GHK ... is a regular polygon. Proof In the As ABG, BCH, etc., AB = BC = CD = etc., and zCABr=zGBA = zHBC=ZHCB=etc., being measured by halves of equal arcs (243). . • . the As ABG, BCH^ etc., are all equal and isosceles. 208 BOOK V.—REGVLAll POLYGONS. 209 .-. AG = BG = BE = CH = etc., /G = zH=ZK = etc.( li. i^S ] .-. GH = IIK = etc., --kr^tLSf^* ^) and .*. GHKL ... is a regular polygon. (399) Q.E.D. 401. Cor. 1. If a regular inscribed 'polygon is given, the tangents at the vertices of the given polygon form a regular civ ciimscrihcd polygon of the same numler of sides, 402. Cor. 2. //" a regular i7i- .r ^^ ^r scribed polygon ABCD , . . . is given, / v^ j ^^ the tangents at the middle poi7its M, /y \ I /^ \\^ N, P, etc., of the arcs AB, BG, CD, A/ M^"^^ ^^ . etc., form a regular circumscribed nK^^ /^\\9/ polygon ichose sides are parallel to V\ / \ /jp those of the i^iscribed polygon, and yC^ ^^X / ^ tvhose vertices A', B' , C", etc, lie on ^ the radii OAA', OBB', etc For, the sides AB, A'B' are || , being _L to OM, (204) and (210), and the same for the others ; also, since B'M = B'N (268), the pt. B' must lie on the bisector OB (160) of the Z MON. 403. Cor. 3. If the chords AM, MB, BN, etc., be draion, the chords form a regular inscribed polygon of double the number of sides of ABCD .... 404. Cor. 4. If through the poinds A, B, C, etc., tan- gents are draiun intersecting the tangents A' B' , B'C, etc, a regular circumscribed polygon is formed of double the munber of sides q/' A'B'C'D' 405. ScH. 'It IS clear that the area of any inscribed polygon is less than the area of the inscribed polygon of double the number of sides ; and the area of a circum- scribed polygon is greater than that of the circumscribed polygon of double the number of sides. 210 PLANE OEOMETBT, Proposition 2. Theorem. 406. A circle may he circumscribed about a regular poly g 071, and a circle may be inscribed in it. Hyp, Let ABCDEF be a regular polygon. (1) To prove that a O may be circumscribed about it. Proof, Bisect the / s A and B^ and let the bisectors meet at 0. Since zFAB=ZABa (Hyp.) .-. Z0AB= ZOBA, (Ax. 7) and . • . OA = OB. (jyMj^ /\] Join 00, OD. In the AsOAB and OBO, AB = BO (399), OB is common, and Z OBA = Z OBO. (Cons.) .-. aOBA= a OBO. (104) .-. Z0AB= zOCB = i ZBOD, and OA = 00. In the same way it may be shown that each Z of the BOOK v.— REGULAR POLYGONS. 211 polygon is bisected by the line joining it with the pt. 0, and that OA = OB = OC = OD = etc. . • . the O described with centre and radius OA will pass through each of the angular points. (2) To prove that a O may be inscribed in ABCDEF. Proof, Since the sides AB, BC, etc., are equal chords of the circumscribed circle, they are equally distant from the centre. (206) Therefore, if a circle be described with the centre and the perpendicular OH as a radius, this circle will be in- scribed in the polygon. q.e.d. 407. Defs. The point 0, which is the comiWn centre of the inscribed and circumscribed circles, is callM^the centre of the regular polygon. The radius of a regular polygon is the radius OA of the circumscribed circle. The apothem of a regular polygon is the radius OH of the inscribed circle. The aiigle at the centre of a regular polygon is the angle between tw^o radii drawn to the extremities of any side, as AOB. 408. CoK. 1. The inscribed and circumscribed circles of a regular polygon are concentric, / 409. Cor. 2. The 7)erpendicular bisectors of the sides of (I regnhir pohjgoii (ill jxiss iJiroinjli it^i ccjitre, ^ 410. Cor. 3. Tlie radiu,^ draiun to any vertex of a regular polygon bisects the angle at the vertex, 411. Cor. 4. If liyies be draiun from the ceiitre of a regular polygon to each of its vertices, the polygon will be divided into as m-any equal isosceles triangles as it has sides. 412. Cor. ^. Each angle at the centre of a regular polygon is equal to four right angles divided by the member of sides of the polygon. 413. Cor. 6. The interior angle of a regular polygon is tlw supplement of the angle at the centre. 212 PLANE GEOMETRY. Proposition 3. Theorem. 414. Regular ])oly (J ons of the same 7iumber of sides are similar. Hyp. Let P and P' be two regular polygons of the same number of sides. p^ To prove that P and P' are similar. Proof Since the polygons are regular, .-. AB = BC = CD =etc., and A'B' = B'C = CD' = etc. AB _ B^ _ ^ _ •'•A'B'"" B'C'~C'D'~ Also, if the polygons have each n sides, the sum of all the int. Zs of each polygon is (2;^ — 4) rt. Zs. (148) Since the polygons are equiangular (399), and have each the same numbel- of sides, (Hyp.) 2^j 4 . • . each Z of each polygon = — rt. Z s. (150) .-. ZA= ZA', ZB::^ zB', ZC= ZC, etc. Hence the polygons are mutually equiangular and have their homologous* sides proportional. . • . the polygons are similar. (307) Q.E.D. 415. Cor. 1. Tlie perimeters of regular polygons of the same number of sides are to each other as any two homolo- (jous* sides. (322) 416. Cor. 2. The ai^eas of regular polygons of the same nmnher of sides are to each other as the squares of any two homologous"^ sides. (379) * Since the polygons are regular, any side of one may be taken as homologous to any side of the other. BOOK V.-REGULAR POLYGONS. ^13 Proposition 4. Theorem, 417. Tlie perimeters of any two regular polygons of the same number of sides are to each other as the radii of their circumscribed circles, or as the radii of their inscribed circles. Hyp. Let P and P' be a H B a' h' B ^ the perimeters of two regn- iilar polygons, and 0' their centres, OA and O'A' the radii of their circum- scribed OS, OH and O'H' the radii of their inscibed ©s. To prove P OA OH P' ~ O'A' ~ O'H' Proof P AB P'- A'B'- In the isosceles As OAB, O'A'B', ZAOB=zA'0'B'. (415) (412) / . the isosceles A s OAB, O'A'B' are similar, (309) and .-. the rt. As OAH, O'A'H' are similar, (309) AB _ OA , OA _ OH * • A'B' - O'A' ' ^''"^ O'A' " O'H' • ^^^ ^^ . P _J0A __0H • • P' ~ O'A' "" O'H' • ^ ^ Q.E.D. 418. Cor. Tlie areas of regular polygons of the same number of sides are to each other as the squares of the radii of their circu^nscribed, or of their inscribed circles, (380) 214 PLANE GEOMETRY. Proposition 5. Theorem. 419. The area of a regular polygon is equal to half the product of its perimeter and apothem. Hyp. Let P denote the perimeter and R the apothem OH, of the regular polygon ABCDEF. To prove area ABCDEF = J P X R. Proof Join OA, OB, OC, etc. The polygon is divided into equal as whose bases are the equal sides of the polygon and whose common altitude is the apothem. Then, area OAB = i AB X OH, the area ofaA=i the product of its base and altitude (366). Similarly, area OBC = ^ BC X OH, and so on for all the A s of the polygon. . • . area OAB + area OBC + etc. = 4(AB + BC + etc.) OH. . • . area of the polygon ABCDEF = | P x R. Q. e. d. 420. CoK. The area of any polygon that circumscrihes a circle is equal to half the product of its peri7neter and the radius of the circle. BOOK v.— THE MEASURE OF THE CIRCLE. 215 The Measure of the Circle. 421. Def. a variable quantity, or simply a variable, is a quantity which may have an indefinite number of differ- ent successive values. A constaiit is a quantity whose value does not change in the same discussion. 422. Limit. When the successive values of a variable, under the conditions imposed upon it, approach more and more nearly to the value of some constant quantity, which it can never equal, yet from which it may be made to differ by as small as we please, the constant is called the limit of the variable ; the variable is said to approach indefinilely to its limit. 423. For example, suppose C D E a point to start at A and move a B along AB towards B under the condition that, during successive seconds of time, the point moves first half the distance from A to B, that is to C ; then half the remaining distance, or to I), then half the remaining distance, or to E, and so on indefinitely. Then the distance from A to the moving point is a variable whose limit is the distance AB. For, however long the point may move, under these conditions, there will always remain some distance between it and the point B, so that the distance from A to the moving point can never equal AB ; but as the moving point can be brought as near as we please to B, its distance from A can be made to differ from the distance AB by as little as we please. If we call the distance AB 2, then the distance the point moves the first second will be 1, the distance moved the next second will be J, the distance moved the third second will be J , and so on. Therefore the whole distance from A 216 PLANE OEOMETRY. - to the moving point tit the end of n seconds is the sum of n terms of the series 1 + i + i+i + etc. Now it is evident that, however many terms of this series are taken, the sum can never equal 2; but by taking a great number of terms the sum can be made to diifer from 2 by as little as we please. Hence we say that the limit of the sum of the series as the number of terms is indefinitly in- creased is 2. The limit of the fraction 1, as x is indefinitely increased, is zero ; for, by increasing x at pleasure, ? may be made to approach as near as we please to the value zero, but can never be made exactly equal to zero. 424. Principle of Limits. Theorem. If two variables are always equal and each approaches a limit, the tioo limits inust be equal. For, two variables that are always equal have always the same common value, that is, they are really but a single variable; and it is clear that a single variable cannot at the same time approach indefinitely to two unequal limits. 425. Cor. If two variables, while approaching their resjjective limits, are always in the same ratio, their limits are m the same ratio. For, if X and y are two variables in the constant ratio m, so that X = my, then the two variables x and my are always equal to each other, and their limits are equal (424); there- fore, if a and b are the limits of x and y respectively, we have a = mb; that is, a and b are in the same constant ratio m, EXERCISE. The apothem of a regular pentagon is 3 and a side is 2: find the perimeter and area of a regular pentagon whose apothem is G. BOOK v.— TUB MEASURE OF TEE CIRCLE, 217 Proposition 6. Theorem. 426. Every coiivex curve is less than any enveloping line whatever that has the same extremities* Hyp. Let ABO be a convex curve, and ADIIC any line enveloping it and terminating at A and 0. To prove that ABC < ADHC. Proof. Of all the lines enveloping the area ABC, there must be at least one line shorter than any other. Now ADHC cannot be this line. For, draw the tangent AH to the curve ABC. Then, the line AHC < the line ADIIC, dnce the st. line AH is < the curve ADH (Ax. 10). . • . ADHC is not the shortest line. In the same way it may be shown that no other envelop- ing line can be the shortest. . • . the curve ABC is less than any enveloping line. Q e.d. 427. Cor. 1. The circumference of a circle is less titan the perimeter of any circumscribed polygon, and greater than the perimeter of any inscribed polygon, 428. Cor. 2. The perimeter of a regular inscribed polygo7i is less 'than the perimeter of a regular inscribed polygon of double the number of sides, 429. Cor. 3. The perimeter of a regular circum,scribed polygon is greater than the perimeter of a regular circum- scribed polygon of do\ible the number of sides. ^18 PLANE GEOMETRT. Proposition 7. Theorem. 430. If a regular jiolyg on he inscribed in a given circle, and another regular polygon he circumscribed about the same circle, and if the number of sides of the polygons be in- creased indefinitely, then the perimeter of each of the poly- gons approaches the circumference of the given circle as its limit, and the area of each approaches the area of the circle as its limit. Hyp, Let be the centre of the given O, and AB, CD homologous sides of the regular inscribed and circum- scribed polygons of the same number of sides ; let ;; and P denote the pe- rimeters, s and S the areas of the in- scribed and circumscribed polygons respectively. To prove that the limit of p and P is the Oce of the O, and the limit of s and S is the area of the O. Proof Join OF and 00. Then, since the polygons are similar, ,\V'.p = OQ'.0¥, (417) and S : 5 = OC : OF (418) I^ow let the number of sides be increased indefinitely, j always remaining the same in each polygon; then the angle j FOC will diminish indefinitely, and the pt. C will approach j the pt. F, as near as we please. J . • . DC will approach OF as its limit, and OC" will approach OF'^ as its limit. BOOK v.— THE MEASURE OF THE CIRCLE. 219 .*. ultimately P andju, and also S and s, will differ from each other by less than any assignable quantity. .'. the Oce of the O, whjcliJ.s._always iirtjQrm ediate be;: tween P and p (427), is the limit of either of the perime- ters ; and the area of the O, which is clearly always interme- diate between S and s, is the limit of either of the areas. Q.E.D. 431. Cor. When the number of sides of the polygons is increased indefinitely, the radius of the circle is the limit of the radius of the circumscribed polygon and the apothem of the inscribed polygon, 432. Def. Similar arcs, sectors, and segments are those which correspond to equal angles at the centre, in circles of different radii. EXERCISES. 1. The apothem of a regular pentagon is G and a side is 4: find the perimeter and area of a regular pentagon whose apothem is 8. 2. If a point be taken within a regular polygon, prove that the sum of its distances from the sides of the pol3^gon is equal to the radius of the inscribed circle multiplied by the number of sides of the polygon. 3. ABODE is a regular pentagon; AC and BE are joined so as to intersect in F: prove that CDEF is a parallelo- gram. 4. The radius of a circle is 8 : find the apothem, perimeter, and area of the inscribed equilateral triangle. 5. The radius of a circle is 10: find the perimeter and area of the regular inscribed octagon. 6. The radius of a circle is 4: find the area of the in- scribed square. 1 220 PLANE GEOMETRY. Proposition 8. Theorem. 433. The circumferences of two circles are to each other as their radii, and the areas of tivo circles are to each other as the squares of their radii. Hyp. Let C and C be the Oces, K and R' the radii, and S and S' the areas of the two OS. To prove C : C = R : K', and S:S'=R^R". Proof Inscribe in the Os two regular polygons of the same number of sides. Let P and P' be their perimeters, and A and A' their areas. Then, because the polygons are regular, with the same number of sides, P : P' = R : R', (417) and A: A'^R'iR' (418) Now let the number of sides of each polygon be increased indefinitely, the number remaining always equal in each. Then P and P' will approach C and C as their limits, (431) and A and A' will approach S and S' as their limits. (431) Since the above ratios remain the same whatever be the number of sides in the polygons, so long as the number is the same in each (417), their limits are in the same ratios. (425) C : C = R : R', and S : S' = R^ R'^ 434. CoK. 1. ^, = and S_ S' 4R^ 4R"' Q.E.D. (294) C _2R C' ~ 2R'^ Therefore, the circumferences of circles are to uich other as their diameters, and their areas are to each other as the squares of their diameters. BOOK v.- VALUE OF 7t. 221 435. Cor. 2. Since similar arcs and sectors are like parts of the respective Oces and Os to which they belong (432), therefore : Similar arcs are to each other as their radii, and similar sectors are to each other as the squares of their radii, 436. Cor. 3. By alternation, we have, from (434), C : 2 R = C : 2 R'. That is, the ratio of one Oce to its diameter is the same as the ratio of any other Oce to its diameter. Therefore : The ratio of the circumference of a circle to its diameter is constant. This constant ratio is usually represented by the Greek letter n, so that — = 7r; .-. C = 2;rR. 437. ScH. The ratio 7r_is incommensurable, and can V/ therefore be expressed only approximately in numbers, /> though the letter n is used to represent its exact value. We give here the value of tt, its reciprocal, and its logarithm: It - 3.14159 26535 89793 23846 'I - = 0.31830 98861 83790 67153 n log 7t = 0.49714 98726 94133 85435 222 PLANE GEOMETliY. Proposition 9. Theorem. 438. TJie area of a circle is equal to half the product of A- ^7 radius, U its radius and circumference, Hyp, Let S be the area, K the and the Oce of the O. To prove S = J R X C. Proof Circumscribe about the O auy regular polygon; let P be its perimeter, and A its area. Then A = J P X 11. (419) Now let the number of sides of the polygon be increased indefinitely. Then the perimeter of the polygon will approach the Oce of the O as its limit, (431) and the area of the polygon will approach the ai-ea of the O as its limit. (431) But the above equality remains true whatever be the number of sides of the polygon. • • . in the limit we have S=:iRxC. 439. Cor. 1. By (436) C = 2;rR. Substituting above, we obtain (425) Q.E.D. Therefore, the area of a circle is equal to the square of the radius multiplied by the constant ratio it. 440. Cor. 2. Let s denote the area of a sector, and c the arc. BOOK V.-THE MEASURE OF TUE CIRCLE. 223 Then, since the sector is the same part of the O that the arc is of the Oce, r.s'.^ = c'.Q, (294) or . _ g X S _ g X jR X C s = -JR X g. (438) Therefore, the area of a sector is equal to half the product of its radius and arc. Proposition lO. Theorem. 441. The areas of tivo similar segments are to each other as the squares of their radii. Hyp, Let S and S' be the areas of the two similar sectors AOB, A'O'B', T and T' the areas of the two similar As AOB, A'O'B', R and R' the radii. To prove S'-T'~R'*' Proof, Since Z0= ZO', (432) .•.S:S' = R^R'% (435) and T:T' = R»:R'^ . (377) ST R' •*'S'~ T'~R"* • S-T R' ^ ^. . . , . g/ _ ry^, = j^ (by division). Q.E.D. EXERCISES. 1. What is the area of a circle whose radius is 40 feet ?tnf^ 2. What is the diameter of a circle whose circumference is 57 yards? 224 PLANE GEOMETRY. Pkoblems of Construction. Proposition 1 1 . Problem. 442. To inscribe a square in a given circle. Given, the O ABCD, with centre 0. Required, to inscribe a square in it. Cons. Draw the diameters AC, BD J_ to each other. Join AB, BC, CD, DA. Then ABCD is the square required. Proof, Since the Zs at are all rt/s, (Cons.) .*. the sides AB, BC, etc., are equal, in the same equal Zs at the centre intercept equal chords (196), and the Z s BAD, ADC, etc., are rt. Z s, evei'y £ inscribed in a \Q is art. Z (240). .*. the figure ABCD is a square. (120) Q.E.F. 443. Cor. 1. If tangents he draiun to the circle at the points A, B, C, D, the figure so formed will he a circum- scrihed square. 444. Cor. 2. To inscribe, and circumscrihe, regular polygons of 8 sides, bisect the arcs AB, BC, CD, DA, and proceed as before. By repeating this process, regular inscribed and circum- scribed 2Jolygons of 16, 32, .... , and, in general, of 2n sides, may he drawn, _ EXERCISES. 7 ^ 1. What is the area of a square inscribed in a circle whose area is 48 feet ? 2. What is the area of a regular hexagon inscribed in a circle whose area is 560 square feet ? BOOK v.— PROBLEMS OF CONSTRUCTION. 22o Proposition 12. Problem. 445. To inscribe a regular hexagon in a given circle. Given, the O ABC, with centre 0. Required, to inscribe a regular hexa- gon in it. Cons. With any pt. A on the ©ce as pi a centre, and AO as a radius, describe an arc cutting the ©ce in B. Join AB, BO, OA. ^ Then AB is a side of the hexagon required. Proof. Since the A OAB is equilateral, (Cons.) .-. it is equiangular. (113) .-. Z AOB is 4 of 2 rt. Z s, or | of 4 rt. I s. (103) .*. the arc AB is \ of the Oce. .*. the chord AB is a side of the regular inscribed hexa- gon. (400) .-, the figure ABCDEF, completed by drawing the chords BC, CD, DE, EF, FA, each equal to the radius OA, is the regular inscribed hexagon required. Q.E.F. 446. CoK. 1. If the alternate vertices of the regular hex- agon he joined, we obtain the inscribed equilateral triangle ACE. 447. Cor. 2. If tangents be drawn to the circle at the poi7its A, B, C, D, E, F, the figure so formed will be a reg- ular circumscribed hexagon. 448. Cor. 3. To inscribe, and circumscribe^ regular polygons of 12 sides, bisect the arcs AB, BC, CD, etc., and proceed as before. By repeating this process, regular inscribed and circum^ scribed polygons of 24, 48, etc., sides may be drawn. m PLANE GEOMETRY. Proposition 13. Problem. 449. To inscribe a regular decagon in a given circle. Given, the O ABCE. Reqtiired, to inscribe in it a regular dec- agon. Cons. Draw any radius OA, and divide it in extreme and mean ratio at D, so that OA : OD = OD : AD. (352) With centre A and OD as a radius, describe an arc cut- ting the Oce at B, and join AB. Then AB is a side of the required inscribed decagon. Proof. Join BO, BD. Since and OA (Cons.) (Cons.) OD = OD : AD, OD = BA, . • . OA : BA = BA : AD. . • . the A s OAB and BAD are similar, having Z A common and the including sides proportional (314). .-. ZABD= ZAOB. Because OA = OB, and BA = OD, . • . OB : BA =r OD : AD, .-. ZABD= zDBO. .-. ZAB0 = 2 ZABD = 2 zAOB. .-. also ZBAO = ZAB0 = 2 ZAOB, being opp. the equal sides of an isosceles A (111). Z ABO + Z BAO H- Z AOB = 5 Z AOB = 2 rt. Z s. Z AOB = I of 2 rt. Z s, or ^o of 4 rt. Z s. the arc AB is j\ of the Oce. the chord AB is a side of the regular inscribed deca- (400) the figure ABCE . .. . , formed by applying AB ten times to the Oce, is the regular inscribed decagon required. Q.E.F. (307) (Radii) and (Cons.) (303) gon BOOK v.— PROBLEMS OF CONSTRUCTION. 227 450. Cor. 1. If the alternate vertices of the required decagon be joi7ied, a regular pentagon is inscribed. 451. Cor. 2. If tangents be drawn at the points atiuhich the circumference is divided, the figure so formed ivill be a regular circumscribed decagon. 452. Cor. 3. To inscribe, and circumscribe, regular polygons of 20 sides, bisect the arcs AB, BC, etc., and pro- ceed as before. By repeating this process, regular i?iscribed and circum- scribed polygons 0/40, 80, etc., sides may be draion. EXERCISES. 1. The side of an inscribed square is equal to the radius of the circle multiplied by \^2. 2. The side of an inscribed equilateral triangle is equal to the radius multiplied by V^. 3. The apothem of an inscribed square is equal to half the radius multiplied by V^, 4. The apothem of an inscribed equilateral triangle is equal to half the radius. 5. The apothem of a regular inscribed hexagon is equal to half the radius multiplied by V^. 6. The area of a circumscribed square is double the area of the inscribed square. 7. Kequired the area of an equilateral triangle inscribed in a circle whose radius is 4. 8. Kequired the area of a square inscribed in a circle whose radius is 5. 9. Kequired the area of a regular hexagon inscribed in a circle whose radius is 8. 10. Kequired the area of a circle whose circumference is 100. 11. Kequired the area of a circle inscribed in a square whose area is 36, 228 PLANE GEOMETRY. Proposition. 1 4. Problem. 453. To inscrihe a regular pentadecagon in a given cir- cle. Given, the O ABC. /- ->^ Required, to inscribe in it a regular / ^ polygon of 15 sides. / Ae Cons. Lay off HB equal to a side I L of a regular inscribed hexagon. (445) V / Lay off HA equal to a side of a regular \^ -3^^^ inscribed decagon. (449) *^^""^^^^A**^ Join AB. Then AB is a side of the required inscribed pentadeca- gon. Proof, Arc AB = arc HB — arc HA = -J- of Oce — ^V of Oce = ^^ of Oce. . • . the chord AB is a side of the regular inscribed penta- decagon. (400) .'.the figure ABCD . . . , formed by applying AB 15 times to the Oce, is the regular inscribed pentadecagon required. q.e.f. 454. CoK. 1. If tangents be dratun at the points at which the circumference is divided, a regular circum- scrihed pentadecagon is obtained. 455. Cor. 2. To inscribe, and circumscribe, regular polygons of 30 sides, bisect the arcs AB, BC, etc., and proceed as before. By repeating this process, regular inscribed and circum- scribed polygons q/*60, 120, etc., sides may be draiun. 456. ScH. These are the only polygons that could be constructed by the ancient geometers, by the use of the rule and compass. About the beginning of the present century, Gauss, the great German mathematician, proved that whenever 2" + 1 is a prime number, and n an integer, a regular polygon of that number of sides could be in- scribed in the circle, by the rule and compass. Therefore, it is possible to inscribe regular polygons of. BOOK v.— PROBLEMS OF COMrVTATION. 229 17 sides, and of 257 sides ; but the construction of the lat- ter polygon is so lengthy, it is not likely that it has ever been attempted. Proposition 1 5. Problem. 45 7. Give7i the radius and tlie side of a regular inscribed polygon, to compute the side of a similar circumscribed poly g 071. Given, AB a side of the regular in- ^^ scribed polygon, and OF = K, the radius of the O ABH. Required, to compute CD, a side of the similar circumscribed polygon. Cons. Join 00, OD. They will cut the Oce in A and B. The similar as OOF, OAE give OF AE OF OE* 4=^=:^. .-.CD^ RX A B OE • In the rt. A OAE, OE = Voa' — AE' ' . OE = y^ /AB CD = 2Rx AB 458. SCH. (307) (328) Q.E.F. V 4R^ - AB^ When R = 1, this becomes If we know the sides of the regular polygons of n, 2n, 4:n, .... sides, inscribed in the circle of radius 1, we have, by this formula, the sides, and therefore the perimeters, of the regular circumscribed similar polygons. 230 PLANE GEOMETRY. Proposition 16. Problem. 459. Given the radius and the side of a regular in- scribed polygon, to compute the side of the regular in- scribed poly go7i of double the number of sides. Given, AB, a side of the regular in- ^ scribed polygon, and 00 = R, the radius A/ of the O ABD. Required, to compute AO, a side of the regular inscribed polygon of double the number of sides. Cons, is the mid. pt. of the arc AB. Produce 00 to D, and join OA. OD is _L to AB at its mid. pt. AO is a mean proportional between OD and OE .-. AO"^ = OD X OE = R (2R - 20E). But OE = i \/4R» -^^. .-. AO = |/r (2R - V'iw - A&) 460. ScH. When R = 1, this gives (403) (204) (325) (457) Q.E.F. AQ = \/'2- V'4-AB' By repeated applications of this formula, we may com- pute successively the sides, and therefore the perimeters, of the regular inscribed polygons of 2n, 4:n, 871, 16n, .... sides, n being the number of sides of the first polygon. EXERCISES. 1. If the radius of a circle is 4, find its circumference and area. 2. If the circumference of a circle is 28, find its diameter and area. BOOK V— METHOD OF rEUIMETERS. 231 Proposition 1 7. Problem. 461. To conqmte the ratio of the circumference of a cir- cle to its diameter. Given, the Oce C, iind the radius K. Required, to find the number tt. Cons, C = 2;rk (436) Let R — 1, then n = |C. That is, the number rr = semi Oce of radius 1. Therefore, the semi-perimeter of each polygon inscribed in this O ce is an approximate vakie of tt : and as the num- ber of sides of the polygons increases indefinitely, the lengths of the perimeters approach to that of the Oce as the limit. (430) Hence, by the process of (4G0), we may obtain a succes- sion of nearer and nearer approximations to the length of the semi Oce. It is convenient to begin the computation with the in- scribed hexagon. . •. making AB = 1, we have from (460), the following: Number of Sides. Semi-perimeters. Number of sides. Semi-perimeter 6 3.00000000 96 3.14103198 12 3.10582854 192 3.14145255 24 3.13262861 384 3.14155772 48 3.13935026 768 3. 1415847 L The last two results show that the first four decimals do not change as the number of sides is increased. Hence the approximate value of tt is 3.1415, correct to the fourth decimal place. q.e.f. In practice we generally take tt = 3.14159. 232 PLANE GEOMETRY. 462. ScH. The above is called the method of perimeters, For the meiliod of isoperimeters seeRouche et €omberousse. Edition of 1883,' p. 194. Note.— The number n is of such fundamental importance in geometry that mathematicians have sought for its value in a great variety of ways, all of which agree in the conclusion that it cannot be expressed^ exactly in decimals, but only approximately. Aichimedes (born 287 B.C.) was the first to assign an approximate value of ir. He proved tliat it is included between the numbei*s Z\ and m, or, in decimals, between 3.1428 and 3.1408; he therefore assigned its value correctly within a unit of the third decimal place. The number 3f, or V is often used in rough computations. Adrian Metius, a Dutch geometer of the 16th century, has given us the much more accurate value of f f §, which is correct to within a half-millionth, and which is remarkable for the manner in which it is formed by the first three odd numbers 1, 3, 5, each written twice. More recently, the value has been found to a great number of decimals by the aid of series. Dase and Clausen, German computers, carried the calculation to 200 decimal places, independently of each other, and their results agreed to the last figure. The first 20 figures of their result are as follows : 3.14159 26535 89793 23846. (437) This result is far beyond all the wants of mathematics. Ten decimals are sufficient to give the circumference of tlie earth to the fraction of an inch, and thirty decimals would give the circumference of the whole visible universe to a quantity imperceptible with the most powerful microscope.* EXERCISES. 1. The area of the regular inscribed hexagon is equal to twice the area of the inscribed equilateral triangle. 2. The square of a side of the inscribed equilateral trian- gle is three times the square of a side of the regular inscribed hexagon. 3. The area of a regular inscribed hexagon is half the area of the circumscribed equilateral triangle. 4. If tlie diameter of a circle be produced to C until the produced part is equal to the radius, the two tangents from C and their chord of contact form an equilateral triangle. ♦ Newcomb's Geom., p. 235. BOOK V.-EXERCISE8. 233 5. Divide an angle of an equilateral triangle into five equal parts. Describe a O about tbe a , then use (453). 6. The square inscribed in a semicircle is equal to two- fifths the square inscribed in the whole circle. 7. The area of a given circle is 314.16; if this circle be circumscribed by a square, find the area of the pai*t between the circumference and the perimeter of the square. 8. The area of a circle is 40 feet ; find the side of the inscribed square. 9. Find tlie angle subtended at the centre of a circl^ by an arc 6 feet long, if the radius is 8 feet long. 10. Find the length of the arc subtended by one side of a regular octagon inscribed in a circle whose radius is 20 feet. 11. Find the area of a circular sector whose arc contains 18", the radius of the circle being 4 feet. 12. Find the area of a circular sector, the chord of half the arc being 20 inches, and the radius 45 inches. 13. The radius of a circle is 5 feet : find the area of a circle 7 times as large. 14. The radius of a circle is 7 feet : find the radius of a circle 16 times as large. 15. Find the height of an arc, the chord of half the arc being 10 feet, and the radius 16 feet. 16. Find the area of a segment whose height is 4 inches, and chord 30 inches. 17. Find the area of a segment whose height is 16 inches, the radius of the circle being 20 inches. 18. Find the area of a segment whose arc is 100°, the ra- dius being 12 feet. 19. Find the area of a sector, the radius of the circle being 24 feet, and the angle at the centre 30°. 20. A circular park 500 feet in diameter has a carriage- way around it 25 feet wide : find the area of the carriage- way. 234 PLANE GEOMETRY. Maxima and Minima. 46 3 . Def. a maxim iwi quantity is the greatest quantity of the same kind ; and a minimum quantity is the least quantity of the same kind. Thus, the diameter of a circle is a maximum among all inscribed straight lines ; and a perpendicular is a minimum among all the straight lines drawn from a given point to a given straight line. 464. Isojjerimetric figures are those which have equal perimeters. We give here a few simple but important propositions bearing on this part of Geometry. Proposition 1 8. Theorem. 465. Of all triangles formed with two given sides, that in which these sides are perpendicular to each other is the maximum. Hyp. Let ABC, A'BC be two As having the sides AB, BC, respectively equal to A'B, BO; and let /ABC be a rt. Z. To prove A ABC > A A'BC. Proof Draw A'D _L to BC. B D c Since the _L is the shortest distance from a pt. to a line, (58) . • . A'B > A'D. But AB = A'B. (Hyp.) .-. AB> A'D. But AB and A'D are the altitudes respectively of the A s ABC, A'BC. . • . A ABC > A A'BC. (369) Q.E.D. c^"' ^,x'^ ! iG V sj._ /f 1 BOOK v.— MAXIMA AND MINIMA 235 Proposition 1 9. Theorem. 466. Of all triangles liaving equal perimeters and the same base, the isosceles triangle is the maximum. Hyp. Let the As ABC, ABD have equal perimeters and the same base AB, and let the A ABC be isosceles. To 2)rove A ABC > A ABD. Proof, Produce AC to H, so that CH = CB = CA, and join HB. Then ZABHisart. /, being inscribed in ilie |0 wliose cent, is Cand radius is CB (240). Produce HB, take DL = DB, join AL, and draw CG, DM II , and CE, DP J_, to AB. Then AD + DL = AD + DB = AC + CB = AH. But AD + DL > AL, .-. AH> AL. .•.BH>BL, (65) and .•.|BH>iBL. But ^ BH = BG = CE, and |BL = BM = DP. (61) .-. CE>DF, and .•.aABC> aABD. (369) Q.E.D. 467. CoK. Of all the triatigles of the same 2^crimeter, that which is equilateral is the maximum, Por, the maximum triangle having a given perimeter must be isosceles whichever side is taken as the base. 236 PLANE GEOMETRY. Proposition 20. Theorem. 468. Of all isoperinietric polygons having the same number of sides ^ the maximiun 2)olygon is equilateral. Hyp. Let ABODE be the maximum polygon of given perimeter and given ^^<^^^^^'^^"- number of sides. A^^^^™ -T^i^C To prove it is equilateral. \ / Proof Join AC. \ / If possible, let the sides AB, BC of A Z the A ABC be unequal, and let AB'C be an isosceles A having the same base AC, and AB'-f B'C = AB + BC. Then, aAB'C>aABC. (466) But the area ACDE remains unchanged. . • . area AB'CDE > area ABCDE. (Ax. 4) But area ABCDE is the maximum. (Hyp.) . • . area AB'CDE < area ABCDE, and .*. AB and BC cannot be unequal. .-. AB = BC. In the same way it may be proved that BC = CD = DE = etc. Q.E.D. EXERCISE. If the diagonals of a quadrilateral are given in magnitude, the area of the quadrilateral is a maximum when the diago- nals are at right angles to each other. BOOK v.— MAXIMA AND MINIMA. 237 Proposition 2 1 . Theorem. 469. Of all 2)olygons formed of sides all given hut one, the maximum can be inscribed in a semicircle with the un- determined side for its diameter.* Hyp. Let ABCDEF bo the maximum polygon formed of the given sides AB, BC, CD, DE, EF. To j^trove ABCDEF can be in- scribed in a semicircle. « Proof Join any vertex, as D, with the extremities A and F of the side not given. Then, the A ADF must be the max. of all A s formed witli the given sides AD and FD; for, if it is not, by in- creasing or decreasing the / ADF, keepmg tlie sides AD and FD unchanged in length, the A ADF may be increased, wliile the rest of the polygon, ABCD, DEF, remains un- changed; so that the whole polygon ABCDEF is increased. But this is contrary to the hypothesis that the given polygon is a maximum. . • . the A ADF must be the max. of As formed with the given sides AD and FD. . • . the Z ADF is a rt. Z . (4G5) . • . D IS on the semiOce whose diameter is AF. (240) .*. any vertex is on the semiOce whose diameter is AF. Q.E.D. * Peirce's Geometry, p. 92 238 PLANE GEOMETRY. Proposition 22. Theorem. 470. Of all polygons formed loith the same given sides, that which can be inscribed in a circle is the maximum. Hyp, Let ABODE be a polygon inscribed in a O, and A'B'C'D'E' any other polygon with the same sides as the first, but which can- not be inscribed in a O. To prove ABODE >A'B'C'D'E'. Proof Draw the diameter DH. Join AH, HB. Upon A'B' (= AB) cons. aA'B'H' = aABH, and join D'ir. Then, since the polygon HAED is inscribed in \0 with diameter HD, and Adding, Subtracting, .-. HAED > H'A'E'D', HBCD > H'B'C'D'. AHBCDE > A'H'B'O'D'E'. AABH=r aA'B'H'. •. ABODE > A'B'O'D'E'. (469) (469) (Ax. 5) Q.E.D. 471. Cor. The maximum of all isoperimetric polygons of the same number of sides is regular. For, it is equilateral, (468) and it can be inscribed in a O, (470) . • . the polygon is regular. (400) EXERCISE. Find a point in a given straight line such that the tan- gents drawn from it to a given circle contain the greatest angle possible. BOOK v.— MAXIMA AND MINIMA. 239 Proposition 23. Theorem. 472. Of regular polygons ivith a given perimeter, that tvhicJi has the greatest number of sides has the greatest area. Hyp. Let P be a regular poly- gon of three sides, and Q a regular polygon of four sides, with the same given perimeter. To 2)rove Q > P. Q Proof. In any side AB of P take any pt. C. The polygon P may be regarded as an irregular polygon of four sides, in which the sides AC, CB make with each other a st. Z . Then, since the polygons P and Q are isoperimetric, and have the same number of sides, (HyP-) . • . the irregular polygon P < the regular polygon Q. (471) In the same way it may be shown thut Q < the i-egular isoperimetric polygon of five sides, and so on. q.e.d. 473. Cor. Tlie circle has a greater area than any poly- gon of the saine p)&rimeter, EXERCISES. 1. Of all triangles of given base and /'' area, the isosceles is that which has the / / greatest vertical' angle. 1 // 2. The shortest chord which can be \ — drawn through a given point within a ^^^ --'' circle is the ' perpendicular to the diameter which passes through tliat point. 240 PLANE GEOMETRY. Proposition 24. Theorem. 474. Of regular polygons having ihe same given area, the greater the number of sides the less ivill he theiierimeivw Hyp. Let P and Q be regular polygons having the same area, and let Q have the greater number of sides. To prove the perimeter of P > that of Q. Proof. Let R be a regular poly- gon having the same perimeter as Q and the same number of sides as P. Then, since Q has the same perimeter as R and a greater number of sides, (Cons.) .•.Q>R. (4;'.>) But Q=P, (Hyp.) .-. P> R. .*. the perimeter of P > the perimeter of R. (370) . •. the perimeter of P > the perimeter of Q. q.e.d. 475. Cor. The circumference of a circle is less than the perimeter of any polygon of the same area. Proposition 25. TFieorem. 476. Given two intersecting straight lines AB, AC, and a point P between them ; then of all straight lines which pass through P and are terminated hy AB, AC, that ivhich is bisected at P cuts off the triangle of minimum area. Hyp. Let EF be the st. line, termi- nated by AB, AC, which is bisected f> at P. To prove A AEF is a minimum. Proof Let HK be any other st. a" line passing through P. » BOOK v.— MAXIMA AND MINIMA. 241 Through E draw ED || to AC. Then aHPF = aDPE. (105) But aDPE < aKPE. .-. aHPF < aKPE. Add area AEPII to each. /. aAEF < aAKH. (Ax. 4) In the same way it may be shown that A AEF < any otlier A formed by a st. line through P. .*. A AEF is a minimum. q.e.d. Proposition 26. Problem. 477. To find at what point in a given straight line the angle subtended hy the line joining two given points, lohich are on the same side of the given straight line, is a maximum. Given, the st. line CD, and the pts. A, B, on the same side of CD. Required, to find at what pt. in CD the Z subtended by the st. line AB is a maximum. Cons. Describe a O to pass through " '' A, B, and to touch the st. line CD. (339) and Ex. 40 in (354) Let P be the pt. of contact. Then the Z APB is the required max. Z . Proof. Take any other pt. in CD as Q, and join AQ, BQ. Then, Z AQB < Z APB. (246) . • . Z APB > any other Z subtended by AB at a pt. in CD. Q.E.F. 242 PLANE GEOMETRY. Proposition 27. Problem. 478. In a straight line of indefinite length find a point such that the sum of its distances from two given point f<, on the same side of the given line, shall be a mininmm. Given, the st. line CD, and tlie pts. A, B, on the same side f"''^; of CD. Required, to find a pt. P in C- CD, so that the sum of AP, PB is a minimum. Cons. Draw AF J. to CD; P ,- and produce AF to E, making FE = AF. Join EB, cutting CD at P. Join AP, PB. Then, P is the required pt., and of all lines drawn from A and B to a pt. in CD, the sum of AP, PB is a minimum. Proof. Let Q be any other pt. in CD. Join AQ, BQ, EQ. Then, rt. A AFP = rt. A EFP. (104) .-. AP = EP. Similarly, AQ = EQ. But EB < EQ + QB. (96) .-. AP + PB < AQ + QB. . • . the sum of AP and PB is a minimum. q.e.f. 479. CoE. TJte sum of AP and PB is a minimum, when these lines are equally inclined to CD; for Z APC = Z EPC = Z BPD. Note.— In order that a ray of light from A may be reflected to a point B, it must fall upon a mirror CD at a point P where ZAPC = /BPD ; i.e., by (478) the ray pursues the shortest path between A and B and touching CD. BOOK v.— EXERCISES. THEOREMS. 243 exercises. Theorems. 1. If AB be a side of an equilateral triangle inscribed in a circle, and AD a side of the inscribed square, prove that three times the square on AD is equal to twice the square on AB. 2. Show that the sum of the perpendiculars from any point inside a regular hexagon to the six sides is equal to tliree times the diameter of the inscribed circle. 3. The area of the regular inscribed hexagon is twice the area of the inscribed equilateral triangle. 4. The area of the regular inscribed hexagon is three- fourths of that of the regular circumscribed hexagon. 5. The area of the regular inscribed hexagon is a mean proportioual between the areas of the inscribed and circum- scribed equilateral triangles. G. If the perpendicular from A to the side BC of the equilateral triangle ABC meet BC in D, and the inscribed circle in G; prove that GD = 2 AG. See figure of (269). 7. If three circles touch each other externally, and a tri- angle be formed by joining their centres, and another tri- angle by joining their points of contact, the inscribed circle of the former triangle will be the circumscribed circle of the latter. 8. If ABCD be a square described about a given circle, and P any point on the circumference of the circle; prove that the sum of the squares on PA, PB, PC, PD, is three times the square on the diameter of the circle. Use (3:33). 9. ABC is a triangle having each of the angles B, C double the angle A; the bisectors of the angles B, C meet AC and the circle circumscribing the triangle ABC respec- tively in D, E : prove that ADBE is a rhombus. 10. ABCDE is a regular pentagon inscribed in a circle, P is the middle point of the arc AB: show that the dif- ference of the straight lines AP, CP is equal to the radius of the circle. 244 PLANE GEOMETRY. 11. In the last exercise prove that the sum of the squares on PC and BC is equal to four times the square on the radius. 12. ABCDE is a regular pentagon, BE is drawn cutting AC^ AD in F, G respectively : show tliat the sum of AB and AE is equal to the sum of BE and FG. Describe a ® about the pentagon, etc. 13. In a circle a regular pentagon and a regular decagon are inscribed; the middle points of the adjacent sides of the pentagon are joined: prove that the sides of the penta- gon so formed are equal to the radius of the circle inscribed in the decagon. 14. Every equilateral polygon inscribed in a circle is equiangular. 15. Every equilateral polygon circumscribed about a cir- cle is equiangular, if the number of sides be odd. 16. Every equiangular polygon inscribed in a circle is equilateral, if the number of sides be odd. 17. Every equiangular polygon circumscribed about a circle is equilateral. 18. The figure formed by the five diagonals of a regular pentagon is another regular pentagon. 10. If the alternate sides of a regular pentagon be pro- duced to meet, the five points of meeting form another reg- ular pentagon. Let a denote a side of a regular polygon inscribed in a circle whose radius is R; then; 20. In a regular decagon, a = f(4/5-l). Jse (449\ 31. In a regular pentagon, T? / ' a^|VlO-2V5. (459 reciprocally) 22. In a regular octagon. a = R V2 - V2. (459) BOOK V. NUMElilCAL EXERCISES. 245 23. In a regular dodecagon, a = 11/2-4/3. (459) 24. In a regular pentad ecagon, « = ?( /lO + 2 V5 + V^- VWj. (453) 25. The side of the regular inscribed pentagon is equal to the hypoten^ise of a right triangle whose sides are the radius and the side of the regular inscribed decagon. Numerical Exercises. 26. The diameter of a circle is 5 feet; find the side of the inscribed square. Aris. 3.535 ft. 27. The apothem of a regular hexagon is 2; find the area of the circumscribing circle. Ans, b\ 7t, 28. There are two gardens; one is a square, and the other is a circle; and they each contain an acre: how much further is it around one than the other? Ans. 17.268 yards. 29. There are two circles; the diameter of the first is 18 inches, and the area of the second is 2-J times tliat of the first: what is the diameter of the second? Ans. 30 inches. 30. The circumference of a circle is 78.54 inches: find (1) its diameter, and (2) its area. Ans. (1) 25 ins.; (2) 490.875 sq. ins. 31. A circle and a square have each a perimeter of 120 feet : which contains the greater area, and how much ? Ans. the circle, 245.95 sq. ft. 32. What is the width of a ring between two concentric circumferences whose lengths are 160 feet and 80 feet? Ans. 12.732 feet. 33. Find the side of a square equivalent to a circle whose radius is 40 feet. 34. The radius of a circle is 15 feet; find the radius of a circle just three times as large. 346 PLANE GEOMETHY. 35. The area of a square is 225 square feet: find the area of the inscribed circle. 36. The length of an arc of 180" is ttE, where E is the radius of the circle (436) : find the leugth of the arc of 25"" 45' in the circle whose radius is 9 inches. Ans, 4 ins. 37. Find the number of degrees in the arc whose length is equal to the radius of the circle. Ans. 57° 17' 44".8. 38. Find the number of degrees in the arc whose length is'18 inches, the radius being 5 feet. Ans. 17° 11' 19". 39. Find the length of the arc of 75° in the circle whose radius is 5 feet. Ans, 6.545 feet. 40. Find a side of the circumscribed equilateral triangle, the radius of the circle being K. Ans. 2R V3. 41. Find a side of the circumscribed regular hexagon, 2T? the radius of the circle being R. Ans. . VI 43. Find the length of the arc subtended by one side of a regular dodecagon in a circle whose radius is 12.5 feet. Ans. 6.545 feet. 43. Find the area of a sector of 60° in the circle whose radius is 10 inches. Ans. 52.3599 sq. ins. 44. The area of a given sector is equal to the square con- structed on the radius: find the number of degrees in the arc of the sector. Ans. 114° 35' 29".6. 45. Find the area of the segment of 60° in the circle whose radius is 2 feet. Ans. 0.362344 sq. ft. 46. Find the radius of the circle in which the sector of 45° is .125 square inches. Ans. .564 inches. 47. Two tangents make with each other an angle of 60°: required the lengths of the arcs into which their points of contact divide the circle, the radius being 21 inches. Ans. 44 inches, 88 inches. 48. Venice is due south of Leipsic 5° 55': how many- miles apart are they, the radius of the earth being 4000 miles? Ans. 413 miles. 49. The three sides of a triangle are 9, 10, 17 inches BOOK v.— EXERCISES IN MAXIMA AND MINIMA. 247 respectively; find (1) its area, and (2) the area of the in- scribed circle. Aiis. (1) 36 sq. ins.; (2) 4;r, See (398), Ex. 3. 50. Mount Washington is visible from a point at sea 87 miles off: required the height of the mountain. Ans. 6270 feet. Problems. 51. To circumscribe a square about a given circle. 52. To inscribe a regular octagon in a given square. 53. Given the base of a triangle, the difference of its sides, and the radius of the inscribed circle: construct the triangle. From the given base AB, cut off AD = given difference; bisect DB in E, draw EO ± to AB and = the given rad., etc. 54. Describe a circle which shall touch a given circle and two given straight lines which themselves touch the given circle. 55. In a given circle inscribe a triangle whose angles are as the numbers 2, 5, 8. See (453). 56. To inscribe a regular hexagon in a given equilateral triangle. 57. To construct a circle equivalent to the sum of two given circles. 58. In a given equilateral triangle, construct three equal circles tangent to each other and to the sides of the tri- angle, and find the radius of these circles in terms of the side of the triangle. 59. Construct an isosceles triangle having each angle at the base double the third angle. 60. Given the vertical angle of a triangle and the radius of the circumscribed circle: find the locus of the centre of the inscribed circle. Describe a O ABC with the given radius, draw AB cutting off segment ACB containing the given Z ; bisect arc AB opp. to C in D, draw any chord DC; bisect ZCAB by AO meeting CD in O, O is the centre of tlie inscribed O: ZDOA = ZOCA + zOAC = ZBAD + ZOAB =ZDAO; .-. DO = DA; ..etc. 61. Given a vertex of a triangle, the circumscribed cir- 248 PLANE GEOMETRY. cle and the centre of the inscribed circle: construct the triangle. Let A be tlie vertex, BACD the O, O the centre of the inscribed o ; join AO, .produce it to D; with centre D and radius DO describe the o BOC; .•. BD = DC .-.arc BD = arc DC; ZDBO = zDOB = ZOAB + zOBA = ZDBC4- ZOBA; .-. zOBC = ZOBA,and ZOAB = ZOAC; .-.etc. ExEECisEs IN Maxima akd Mii^ima. 62. Two sides of a triangle are given in length: how must they be placed that the area of the triangle may be a maximum ? (465). 63. Given the base and vertical angle of a triangle; con- struct it so that its area may be a maximum. 64. Of all triangles of given base and area, the isosceles is that which has the least perimeter. (479). 65. Divide a given straight line so that the rectangle con- tained by the two segments may be a maximum. 66. A straight rod slips between two straight rulers at riglit angles to each other: in what position is the rod when the triangle formed by the rulers and the rod is a maxi- mum ? 67. Show that the greatest rectangle which can be in- scribed in a circle is a square. QS. Of all triangles of given vertical angle and altitude, the isosceles is that which has the least area. 69. Of all rectangles of given area, the square has the least perimeter. 70. Of all polygons having the same number of sides and equal areas, the perimeter of an equilateral polygon is a minimum. 71. A and B are two fixed points without a circle: find a point P on the circumference such that the sum of the squares on AP, BP may be a minimum. (333). BOOK V.-EXERC1SE8 IN MAXIMA AND MINIMA. 249 72. A bridge consists of three arches, whose spans are 49 ft., 32 ft., and 49 ft. respectively: show that the point on either bank of the river at which the middle arch sub- tends the greatest angle is 63 feet distant from the bridge. See (477). 73. If the sum of the squares of two lines be given, their sum is a maximum when the lines are equal. 74. Of all triangles having the same base and vertical angle, the isosceles triangle has the sum of the sides a max- imum. 75. Of all triangles inscribed in a circle, the equilateral triangle has the maximum perimeter. SOLID GEOMETRY. Book VI. PLANES AND SOLID ANGLES. Definitions. 480. K plane has been defined (11) as a surface in which the straight line joining any two of its points lies wholly in the surface. A plane is indefinite in extent, so that, however far the straight line is produced, all its points lie in the plane. But to represent a plane in a diagram, it is necessary to take a limited portion of it, and it is usually represented by a parallelogram supposed to lie in the plane. 481. A plane is said to be determined by certain lines or points, when it is the only plane which contains these lines or points. 482. Any number of planes may he passed through any given straight line. For, if a plane is passed through any given straight line AB, the plane may be turned about AB as an axis, and made to occupy an infinite num- /f| ber of positions, each of which will be l^ a different plane passing through AB. Hence a given straight line does not determine the posi- tion of a plane. 483. A plane is determined hy a straight Ime and a point without that line. For, if the plane containing the / " .q straight line AB, turn about this line as / an axis until it contains the given point I ^ C, the plane is evidently determined. 250 ^ 1/ B 7 BOOK VI.— DEFINITIONS. 251 If it is then turned, in either direction, about AB, it will no longer contain the point C. 484. A plane is determined hy three points not in the same straight line. For, if we join any two of the points by a straight line, this line and the third point determine a plane. (483) 485. A plane is determined by tivo intersecting straight lines. For, a plane passing through AB and any point in AC, in addition to the point of intersection A, contains the two straight lines AB, AC (480), and is determined. (483) 486. A plane is determined by two parallel straight lines. For, two parallel straight lines lie in the same plane (68); and since this plane contains either of these parallels and any point in the other, it is deter- mined. (483) 487. A straight line \^ perpendicular to a plane when it I is perpendicular to every straight line which, it meets in ' that plane. Conversely, the plane in this case is said to be perpendic- ular to the line. The point in which a line meets a plane is called the foot of the line, 488. A straight line is parallel to a jolane when it never meets the plane, however far both may be produced. Conversely, the plane in this case is said to be parallel to the line. 489. Two planes are parallel when they do not meet, however far they may be produced. 490. The projection of a point on a plane is the foot of the perpendicular let fall from the point to the plane. 252 SOLID GEOMETRY. 491. The projection of a line on a plane is the locus of the projections of all the points of this line. 492. The a7igle which a straight line makes with a plane is defined to be the acute angle between the straight line and its projection on the plane, and is called the inclina- tion of the line to the plane. 493. By the dista^ice of a point from a plane is meant the shortest distance from the poitit to the plane. 494. The line that determines the projection of a point on a plane is called tha 2)roJectin(/ line of that point. The jilane including all the projecting lines of a straight line is called the projecting plane of the line. Lines and Planes. Proposition 1 . Tiieorem. 495. If two planes cut each other, their common inter- section is a straight line. Hyp, Let AB, CD be two planes which cut each other. To prove their common inter- section is a straight line. Proof. Let E and H be any two pts. common to both planes. Join E and H by the st. line EH. Because E and H are in both planes, the st. line EH lies in both planes. (480) Because a st. line and a pt. out of it cannot lie in two planes, {'^^^) . • . EH contains all the pts. common to both planes. . • . EH is the common intersection of the two planes. Q.E.D. I BOOK VL-LINE8 AND PLANES. 253 Proposition 2. Tiieorem. 496. If oUique lines are draiofi from a i)oint to a plane: ■ (1) Two oUiqiie lines meeting the plane at equal distances from the foot of the perpendicular are equah (2) Of two oblique lines meeting the plane at unequal dis- tances from the foot of the perpendicular, the 7nore remote is the longer. Hyp. Let OP be _L to the plane MN, and PA = PB, and PC > PA. To prove (1) OA = OB, and (2) 00 > OA. (1) Proof Since OP is com- mon, and PA = PB, (Hyp.) .-. rt. A OPA = rt. A OPB. (104) .•.OA=rOB. ^ (2) Proof On PC take PB = PA, and join OB. * Then, OC > OB. (63) But OB = OA. (Just proved) .-. 0C> OA. • Q.E.D 497. Cor. 1. The perpendicular is the shortest distance from a p)oint to a plane; therefore, hy the distance of a point from a plane is 7neafit the peipendicular distance from the point to the plane. (493) 498. Cor. 2. Equal oblique lines from a point to a plane meet the plane at equal distances from the foot of the per- pendicular ; . and of two unequal oblique lines, the greater meets the plane at the greater distance from the foot of the perpendicular. 499. Cor. 3. The locus of the point in a 2^la7ie at a given distance from a fixed 2Joint loithout the plane, is a circle whose centre is the foot of the perpendicular. 254 SOLID GEOMETRY. o "rv \^ ^N \\\ M, / \\^^l / / P 5«rn:^rr_47''' /' / / 'X'V^'' / / /A/ /' / / 1 / / / / i /// ^ (//'' K Nf Proposition 3. TFieorem. ^ 500. 7/*^ straight line is jJ^^pendicula?' to each of tioo straight lines at their point of intersection, it is perpenr dicidar to the plane of those lines. Hyp. Let OP be J. to PA, PB at the pt. P. To prove OP is _L to the plane MN of these lines. Proof. Join AB, and through P draw in MN any other st. line PC cutting AB in C. Produce OP to 0' making PO' =: PO, and join 0, 0' to each of the pts. A, B, C. Since PA, PB are I. to 00' at its mid. pt., (Hyp.) (Cons.) .• . OA = O'A, and OB = O'B. (66) .• . A OAB = A O'AB, and .• . z OAC = Z O'AC. (108) Also, A OAC = A O'AC, having two sides and ilie included Z equal each to each (104). , . • . OC = O'C, and . • . PC is ± to 00' at its mid. pt. P. (67) .* . OP is JL to any st. line in MN passing through its foot P. . • . OP is J. to the plane MN. (487) Q.E.D. ] 501. Cor. \. At a given point in a plane, only one perpendicular to the plane can he erected. For, if there coukl be two J_s at the same pt. P, pass a plane through them whoso intersection with the plane MN is AP; then these two J_s would be both J_ to the line AP at the same pt. P, which is impossible. (51) 502. Cor. 2. From a given point without a plane only one perpendicular can he draica to the plane. For, if OP, OA be two sue]] _Ls, the A OPA contains two rt. Z s, which is impossible. (1^1) y BOOK VI. -PERPENDICULARS TO LINES. 255 Proposition 4. Theorem. 503. Cot^veri>6ly, all the i^erpendicidars to a straight line at the same point lie in a plane perpendicular to the line. Hyp, Let PA, PB be two st. lines J_ to OP at P, and PC any other line _L to OP at P. To prove PC is in the same plane with PA, PB. Proof. Let the plane passing through PO and PC cut the plane APB in the line PC. Then OP is _L to PC. (487) But in the plane OPC only one _L can be drawn to OP atP, (51) .-.PC and PC coincide, and PC lies in the plane APB. Q.E.D. Note.— Hence a plane is determined by one point and the normal to the plane at that point. 504. Cor. 1. At a given point in a straight line one plane, and only one, can he drawn j)erpendicular to the li7ie. 505. Cor. 2. If a right angle he turtied roimd one of its arms as an axis, the other arm ivill generate a plane. 506. Cor. 3. Tlirough a given point luithont a straight line one plane, and only one, can he drawn perpe7idicular to the line. For, in the plane of OP and the pt. C, the J^ CP can be drawn to OP; then the piano genei-ated by turning PC round OP will be J_ to OP; and it is clear that there is only one such J_ plane. 256 SOLID GEOMETRY. Proposition 5. Theorem. 507. If from the foot of a perpendicular to a plane a straight line is drawn at right angles to anfline in the plane, and its intersection with that line is joined to any point of the perpendicular, this last line will be per2)en- diciilar to the line in the plane. Hyp. Let OP be a _L to the plane MN, PA a J_ from P to any line BC in MN, and OA a line joining A with any pt. in OP. To prom that OA is ± to BC. Proof. Take AB = AC, and join PB, PC, OB, OC. Since PA is _L to BC at its mid. pt., and PB = PC, OB = OC. (496) Then since and A are each equally distant from B and C, .-. OA is_LtoBC. (67) Q.E.D. 508. Cor. 1. Tlie line BC is perpendictdar to the plane of the triangle OP A . For, it is i. to the st. lines AP, AO at pt. A. (500) Cor. 2. The line PA measures the shortest distance hetiveen OP and BC. EXERCISES. 1. Find the locus of points equally distant from two given points. 2. Given a straight line and any two points: find a point in the straight line equally distant from the two points. BOOK VL—PERPENDICULAR AND PARALLEL LINES. 257 Proposition 6. Theorem. n/ 509. Tivo straight lines perpendicular to the same plane are parallel. Hyp. Let AB; CD, be _L to the plane MN at the pts. B, D. j^ To prove AB || to CD. Proof. Join DB, DA, and ^w DE J_ to BD in t(ie plane MN. Then CD is JL to ED, and BD is J. to ED, and AD is _L to ED. . •. CD, AD, BD, are all in the same plane, all the Is to a si. line at the samepL lie in the same plane (503). . • . AB and CD lie in the same plane. they are both JL to BD. .-. AB is II to CD. and thev are both J to BD. (487) (70) Q.E.D. 510. Cor. 1. If one of two parallel lines is perpen- dicular to a plane, the other is also perpendicular to that plane. A c For, if AB is jj to CD, and J_ to the j^ plane MN, then a J. to MN at D will be jj / to AB (509), and will coincide with CD / (501). . • . CD is _L to MN. ^N 511. Cor. 2. Two straight lines that are parallel te a third straight line are fjarallel to each APE other. For, each of the lines AB, CD, is J_ lo a plane MN that is _L to EF (510); y . • . AB and CD are I| . (509) N^ 7 U/ M 258 SOLID OEOMETRT. Parallel Planes. Proposition 7. Theorem. 512. If two straiglit lines a7'e parallel, each of them is parallel to every plane passing through the other and not containing loth lines. Hyp, Let AB, CD be two || st. lines, and MN any plane passing through CD. To prove AB || to MN". Proof The || s AB, CD lie wholly in the plane ABCD. . • . if AB meets the plane MN, it mnst meet it in some pt. of the intersection CD, of the two planes. But AB is II to CD, and so cannot meet it. (HyP-) . • . AB cannot meet the plane MN. .-. AB is II to MN. Q.E.D. 613. Cor. 1. A line parallel to the intersection of tivo planes is parallel to each of those planes . 514. Cor. 2. Through any given straight line, a plane can be passed j^arallel to any other given straight line. For, through any pt. C of CE draw CD II toAB; then the plane of DCE is II to AB. (512) 515. Cor. 3. Tlirough a given point a plane can he passed parallel to any two given straight lines in space. For, draw through the given pt. 0, in the plane of the given line AB and O, the line A'B' i| to AB, and in the plane of the given line CD and 0, the line CD' || to CD; then the plane of the lines A'B', CD' is || to each of the lines AB and CD, BOOK VL— PARALLEL PLANES. 259 Proposition 8. Theorem. 516. Two planes perpendicular to the same straight line are pitrallel. ^c Hyp. Let the planes MN and PQ be AB. prove MN II to PQ. Proof. If the planes are not jj they will meet if sufficiently produced. (489) There will then be, through a pt. of their intersection, two planes J_ to the same st. line AB. /J p N / = ../ But this is impossible. .-.MN is II to PQ. (506) Q.E.D. 517. Cor. 1. If a straight line is parallel to a plane, the intersection of the plane witJi any plane passed through the line is parallel to the line. Let the student show this. 518. Cor. 2. If a straight line and a plane are parallel, a parallel to the line drawn through any point of the plane lies in the plane, (517) Hem. a polygon in space may be formed by joining end to end any number of straight lines, as defined in (137). ,But in Plane Geometry the lines are all confined to one plane, while there is no such restriction upon polygons in space. 260 SOLID GEOMETRY, Proposition 9. Tiieorem. 519. The intersections of two parallel planes hy plane are parallel lines. Hyp. Let the li planes MN, PQ be cut by the plane AD into the lines AB and CD. To prove AB || to CD. Proof, The lines AB, CD are in the same plane AD. Because AB lies in the plane MN, and CD in the I| plane PQ, and because these planes cannot meet, .• . AB and CD cannot meet. a third AB is II to CD. Q.E.D. 520. Cor. Parallel straight lines included between pai*- allel planes are equal. For, the plane of the || lines AC, BD intersects the I| planes MN, PQ in the || lines AB, CD. .• . ABCD is a O, and .• . AC = BD. (129) EXERCISES. 1. Show that all the propositions in Plane Geometry which relate to triangles are true of triangles in space, how- ever situated. See (485). ^c^u , 2. Show that those propositions are not true of polygons of more than three sides situated in any way in space. D BOOK VI.— PARALLEL PLANES. 261 Proposition lO. Theorem. 521. ^ straight line perpendicular to one of two parallel planes is j)erpendicular to the other. HifiJ. Let MN and PQ be || planes, "^y 7 and "let AB be _L to PQ. / A. C / To prove AB ± to MK I- '- Proof. Through A draw any st. line p AC in the plane MN", and pass a plane / through AB, AC, intersecting PQ in ^ BD. Then AC is || to BD. (519) But AB is J_ to BD. (487) .-. AB is Jl to AC. (71) And since AC is any line drawn through A in the plane MN, /. AB is i. to the plane MK (487) Q.E.D. 522. CoK. 1. Through a given point one plane can he passed parallel to a given 2^lane, and only one. For, if AB is J_ to PQ, a plane passing through the pt. A, J. to AB, is II to PQ. (51G) Also, since every plane || to PQ is _L to AB (521), and since only one plane can be passed through the pt. A J. to AB (504), therefore only one plane can be passed through a given pt. parallel to a given plane. 523. CoK. 2. Two parallel j^lctnes are everywhere equally distant. For, all st. lines J_ to the plane PQ are also J_ to the |i plane MN (521) ; and being J_ to the same plane, they are II (509) ; and being included between || planes, they are equal (520). Hence the planes are everywhere equally distant. * (497) 524. Cor. 3. If tioo intersecting straight lines are each parallel to a given plane, the plane of these lines is parallel to the given plane. See (518), (516). 262 SOLID OEOMETRY. Ur Proposition 1 1 . Theorem. 525. If two angles not in the same j^lane have their sides respectively parallel and lying in the same direction, they are equal and their planes are parallel. Hyp, Let Zs A and A' lie in the planes MN, PQ respectively, and let AB be II to A'B' and AC be || to A'C. (1) To prove ZA= /A'. Proof, Take AB = A'B', and AC = A'C, and join AA', BB', CC, BC, B'C. ; Then, since AB is = and || to A'B', .-. AA' is = and || to BB'. In like manner A A' is = and || to CC. Because BB' and CC are each = and || to AA', .-. BB' is = and || to CC. (511) .-. BC is = and || to B'C. (133) .-. A ABC = aA'B'C, having tlie three sides equal each to each (108). .-. ZA=ZA'. (2) Tojyrove MN |I PQ. Proof, Since the lines AB, AC are each || to the plane PQ, ' (512) .-. their plane MN is || to PQ. ' (524) Q.E.D. EXERCISE. Find the locus of points equally distant from three given points. BOOK VL— PARALLEL PLANES. 263 Proposition 12. Theorem. 526. If two straight lines are cut by three parallel planes, the corresponding segmeyits are proportional. Hyp. Let AB, CD be cut by the || U^ _^ planes MN, PQ, RS, in the pts. A, E, B, /Kr'-^f / and C, F, D. ^" To prove EB CP FD- isn M Proof. Join AD cutting the plane / PQ in H. ^ Join EH and FH. Then because the || planes PQ, RS are cut by the plane ABD, in the lines EH, BD, .-. EH is II to BD. .-. AE:EB = AH:HD. In like manner HF is || to AC .-. AH:HD = CF:FD. .-. AE:EB = CF:FD. (519) (298) (519) Q.E.D. EXERCISES. 1. To draw a perpendicular to a given plane from a given point without it. 2. To erect a perpendicular to a given plane from a giv- en point in the plane. 3. Prove that through a given line of a given plane, only one plane perpendicular to the given plane can be passed. 264 SOLID GEOMETRY. DiEDRAL Angles. DEFINITIONS. 527. When two planes intersect they are said to form with each otlier a diedral angle. The two planes are called the faces, and their line of in- tersection, the edf/ey of the diedral angle. Thus, the two planes AC, AE are the faces, and the intersection AB is the edge, of the diedral angle formed by these planes. A diedral angle is read by the two let- ters on the edge and one in each face, the two on the edge being read between the other two; or, simply by the two letters on the edge. Thus, the angle in the figure is read either DABF or AB. 528. If a point is taken in the edge of a diedral angle, and two straight lines are drawn through" this point, one in each face, and each perpendicular to the edge, the angle between these lines is called the plaiie angle of the diedral angle. This plane angle is the same at whatever point of the edge it is constructed. '' Thus, if at any point G we draw GH and GK in the two faces AC and AE respectively, and both perpendicular to AB, the angle HGK is equal to the angle DAF, since the sides of these angles are respectively parallel. (525) The plane of the plane angle HGK is perpendicular to the edge AB (500); and conversely, a plane perpendicular to the edge of a diedral angle at any point cuts the faces in lines perpendicular to the edge. (487) BOOK VI.—DIEDRAL ANGLES. 265 529. A diedral angle may be conceived to be generated by revolving a plane about a line of the plane. Thus, suppose a plane, at first in coincidence with a fixed plane AC, to turn about the edge AB as an axis until it comes into the position AE; then the magnitude of the diedral angle DABF varies continuously with the amount of turning of tliis plane about AB. Tlie straight line DA, perpendicular to AB, generates the plane angle DAF. 530. Two diedral angles are equal when one of them can be applied to the other so that the edges coincide, and the two plane faces of the one coincide respectively with the two plane faces of the other. The magnitude of a diedral angle depends only upon the relative position of its faces, and is independent of their extent. A__ ^c 531. When two diedral angles have a com- mon edge, and the intermediate face common to both, they are said to be adjacent. Thus, the angles CABD, DABE are adja- cent angles. Two diedral angles CABD, DABE, are added together by placing them adjacent to each other, giving as the sum the diedral angle CABE. 532. When the adjacent diedral angles which a plane forms with another plane on opposite sides are equal, each of these angles is called a right diedral angle ; and the first plane is said to be jjerpendicular to the other. Thus, if the adjacent diedral angles ABCM, ABClSr are equal, each of these is a right diedral Jicgle, and the planes AC and M"N are perpendicular to each other. Through a given line in a plane only one plane can be passed perpendicular to the given plane. E/1 A M 7" 266 SOLID GEOMETMT, Proposition 13. Theorem. 533. Two diedral a7igles are equal if their plane angles are equal. Hyp. Let the plane / s CAD, A C'A'D' of the diedral /s AB, A'B' be equal. To prove diedral Z AB = diedral Z A'B'. B C.A" B' C Proof, Apply the diedral Z AB to the diedral ZA'B', so that the plane ZCAD coincides with the equal plane Z C'A'D'. Because the pt. A coincides with the pt. A', and the plane CAD with the plane C'A'D', (485) .*. AB, the _L to the plane CAD, coincides with A'B', the J. to the plane C'A'D'. (501) Because AB coincides with A'B', and AC with A'C, .'. the plane BC coincides with the plane B'C. (485) Similarly, the plane BD coincides with the plane B'D'. .-. diedral Z AB = diedral Z A'B'. (530) Q.E.D. 534. ScH. A diedral angle is called acute, right, or ob- tuse according as the corresponding plane angle is acute, right, or obtuse. EXERCISE. Prove that through a line parallel to a given plane; only one plane perpendicular to the given plane can be passed. BOOK VL—DIEDRAL ANGLES. 267 Proposition 14. Theorem. 535. The ratio of any two diedral aiigles is equal to the ratio of their i^lane angles. Htjp. Let CABD, C'A'B'D' be two diedral Zs, and let CAD, C'A'D' be their plane Z s. AB ZCAD Toprove _^,^__,__. Proof. Take any common measure of the Zs CAD, C'A'D', and suppose it to be contained three times in Z CAD and 4 times in Z C'A'D'. Then Z CAD : Z C'A'D' = 3:4. Pass planes through the edges AB, A'B' and the several lines of division of the plane Z s CAD, C'A'D' made by the common measure. The^e planes divide CABD into 3, and C'A'B'D' into 4 equal parts. (533) .-. CABD : C'A'B'D' = 3 : 4. .-. CABD : C'A'B'D' = zCAD :Z C'A'D'. This proof may be extended to the case where the plane angles are incommensurable, by the method employed in (234), (298), and (356). q.e.d. 536. ScH. Since the diedral angle and its plane angle increase and decrease in the same ratio, the plane angle is taken as the measure of the diedral angle. See (236). 268 SOLID GEOMETRY. Proposition 1 5. Tl-ieorem. 537. If iioo planes are iderpendicular to each other, a straight line drawn in one of them 2)erpendicular to their intersection is perpendicular to the other. Hyp. Let the planes PQ and MN be _L to each other, and let AB be drawn in PQ J_ to their intersection QC. To prove AB _L to MN. Proof In the plane MN draw BD X to QC at B. Because BA and BD are each _L to QC, (Hyp.) and (Cons.) . • . Z ABD is the plane Z of the rt. diedral Z PCQN. (528) Because PQ is _L to MN, (Hyp.) .-.ZABDisart. Z. (53G) . • . AB is _L to QC and BD at their pt. of intersection. . AB is _L to the plane MN. (500) Q.E.D. 538. Cor. 1. If tioo ^jlanes are perpendicular to each other, a straight line through any j^oint of their intersec- tion perpendicular to one of the planes %uill lie in the other. See (501). 539. Cor. 2. If tivo planes are perj)e7idicular to each other, a straight line from any poi7it of one plane peipefi- dicular to the other will lie in the first plane. See (502). BOOK Vl.—DIEDHAL ANGLES. 269 Proposition 1 6. Theorem. 540. If a straight line is perpendicular to a plane, every plane passed through the line is perpendicular to that plane. Hyp. Let AB be _L to the plane MN, and PQ any plane passed through AB, intersecting MN in QO. To prove plane PQ J_to plane MN. Proof. In the plane MN draw BD i. to QC at B. Because AB is _L to MN, (Hyp.) Q 'n .• . AB is J_ to QC and BD. (487) . • . Z ABD is the plane Z of the diedral Z PCQN. (528) But Z ABD is a rt. Z . (Proved above) .-. PQis_LtoMN. Q.E.D. 541. Cor. A plane perpendicular to the edge of a diedral angle is perpendicular to its faces. 542. ScH. When three straight lines, AB, CB, DB, are perpendicular to one another at the same point, each line is perpendicular to the plane of the other two, and the three planes are perpendicular to one another. EXERCISE. If two planes which intersect contain two lines parallel to each other, the intersection of the planes will be parallel to the lines. 270 SOLID GEOMETRY. Proposition 1 7. Theorem. 543. TJirough a given straight line oblique to a plane, a plane can he passed perpendicular to the given plane, and hut one. Hyp. Let AB be the given st. line oblique to the plane MIS'. To prove that one plane can be passed /"[ j 7 through AB _L to MN, and but one. / ^^ 'D / Proof. Draw AD J_ to MN from any / / pt. AofAB. ^ Through AB and AD pass a plane AC, Because the plane AC passes through the _L AD^ (Cons.) .'. the plane AC is _L to the plane MN. (540) Also, because any plane passed through AB _L to MN" must contain the _]_ AD, (539) . • . the plane AC is the only plane JL to MN that can be passed through AB. (485) Q.E.D. 544. CoR. 1. Tlie projection of a straight line on a plane is a straight line. For, the J_s from all pts. of AB to [MN lie in the plane AC J_ to MN (539), and therefore these J_s all meet MN in the intersection CD of the two planes, which is a straight line. (495) 545. Cor. 2. If a lifie intersects a p)ln'ne, its p)rojection passes through the point of intersection. BOOK Vl.'-rERPENDICULAR PLANES, 271 Proposition 18. Theorem. 546. If Uvo intersecting planes are each perperidicular to a third plane, their intersection is perpendicular to the third plane. Hyp. Let the planes PQ, RS, in- tersecting in AB, beJ_to the third plane MN. To prove AB _L to MN. Proof, At the pt. B erect a_Lto the plane MN. This _L lies in each of the planes PQ and ES. Because this _L lies in each plane PQ and RS, it is their line of intersection AB. . AB is JL to the plane MN. Q.E.D. 547. Cor. 1. A plane perpendicular to each of two in- tersecting planes is perpendicular to their intersection, 548. Cor. 2. If the planes PQ and RS include a right diedral angle, the three platies PQ, ES, MN, are perjjen- dicular to one another ; the intersection of any two of tliese planes is perpendicular to the third plane ; and the three intersections are perpendicular to one another. Compare (54S). EXERCISE. Show that three planes in general intersect in one point: what are the exceptions ? 272 SOLID GEOMETliY. Proposition 19. Theorem. 549. Every i^oint in the 2:)lane bisecting a diedral angle is equally distant from the faces of the angle, Hyi). Let AM be the plane which bisects the diedral Z CABD, and let PE, PH be_Ls from any pt. P in the plane AM to the planes CB and DB. To prove PE = PH. Proof Let be the pt. where the plane EPH cuts the line AB. Join EO, OP, OH. Because PE is_Lto CB and PH is _L to DB, (Cons.) . • . the plane EPH is _L to each of the planes CB and DB. (540) . • . the plane EPH is _L to their intersection AB. (547) . • . the Z s POE, POH are the plane Z s of the diedral Z s CABM and DABM. (528) Because the diedral Z s CABM, DABM are equal, (Hyp.) .-. ZP0E= ZPOH. (536) .-. rt. aPOE = rt. A POH, hamng the hypotenuse and an acute Z = each to ea^h (106). .•.PE = PH. Q.E.D. EXERCISES. 1. If three planes have a common line of intersection, the normals drawn to these planes from any point of that line are in one plane. 2. If from any point perpendiculars be drawn to the faces of a diedral angle, the angle between these perpen- diculars is the supplement of the diedral angle between the planes in which the point is situated. BOOK VI.—LMAST ANGLE WITH A PLANE. 2T6 Proposition 20. Theorem.* 550. TliG acute angle between a straight line and its projection on a plane is the least angle which the line makes luith any line of the plane. Hyp. Let BO be the projection of ^/ GK (497), or its equal A'A. BOOK VL-EXERG1SE8. 275 EXERCISES 1. Parallel lines intersecting the same plane make equal angles with it. (492) 2. If a plane bisects a line perpendicularly, every point of the plane is equally distant from the extremities of the line. 3. If three lines in space are parallel, in how many planes may they lie when taken two at a time ? 4. If four lines in space are parallel, in how many planes may they lie when taken two at a time ? 5. How many different planes may the sides of a quad- rilateral in space contain when taken two and two ? 6. If two lines not in the same plane are intersected by the same line, how many planes may be determined by the three lines taken two and two ? 7. A straight line makes equal angles with parallel planes. 8. The sum of two adjacent diedral angles, formed by one plane meeting another, is equal to two right diedral angles. 9. If two planes intersect each other, the opposite or vertical diedral angles are equal. 10. When a plane intersects two parallel planes, the alternate-interior diedral angles are equal, and the exterior- interior diedral angles are equal. 11. Show that two observations with a spirit-level are sufi&cient to determine if a plane is horizontal: and prove that for this purpose the two positions of the level must not be parallel. 12. To draw a straight line perpendicular to a given plane from a given point outside of it. 13. To draw a straight line perpendicular to a given plane from a given point in the plane. ^76 SOLID GEOMETRY, POLYEDRAL ANGLES. DEFINITIONS. 553. When three or more planes meet in a common point, they are said to form a polyedral angle at that point. The common point in which the planes meet is the vertex of the angle, the intersections of the planes are the edges, the portions of the planes between the edges are t\iQ faces, and the plane angles formed by the edges are i\iQ face-angles. Thus, the point S is the vertex, the straight lines SA, SB, etc., are the edges, the planes SAB, SBC, etc., are the faces, and the angles ASB, BSO, etc., are the face-angles of the polyedral angle S - ABCD. 554. The edges of a polyedral angle may be produced indefinitely ; but to represent the angle clearly, the edges and faces are supposed to be cut off by a plane, as in the figure above. The intersection of the faces with this plane forms a polygon, as ABCD, which is called the base of the polyedral angle. 555. In a polyedral angle, each pair of adjacent faces forms a diedral angle, and each pair of adjacent edges forms a face-angle. There are as many edges as faces, and therefore as many diedral angles as faces. 556. The magnitude of a polyedral angle depends only upon the 7'elatlve position of its faces, and is independent of their extent. Thus, by the face SAB is not meant the triangle SAB, but the indefinite plane between the edges SA, SB produced indefinitely. 557. Two polyedral angles are equal, when the face and BOOK Vl.—POLYEDRAL ANGLES. 277 diedral angles of one are respectively equal to the face and diedral angles of the other, taken in the same order, 558. A polyedral angle of three faces is called a triedral angle; one of four faces is called a tetraedral angle; etc. 559. A polyedral angle is convex when its base is a con- vex polygon. (141) 560. A triedral angle is called isosceles when it has two of its face-angles equal ; when it has all three of its face- angles equal it is called equilateral. 561. A triedral angle is called rectangular, hi-rect- angular, or tri-rect angular, according as it has one, two, or three, right diedral angles. The corner of a cube is a tri-rectangular triedral angle. 562. Two polyedral angles are symmetrical, when the face and diedral angles of one are equal to the face and diedral angles of the other, each to each, but arranged in reverse order. Thus, the triedral angles S - ABC, S' - A'B'C are sym- metrical when the face-angles ASB, BSC, CSA are equal re- spectively to the face-angles A'S'B', B'S'C, C'S'A', and the diedral angles SA, SB, SC to the diedral angles S'A', S'B', S'C. When two polyedral angles are symmetrical, it is, in general, impossible to bring them into coincidence. The two hands are an illustration. The right hand is symmetrical to the left hand, but cannot be made to coin- cide with it. The right glove will not fit the left hand, but is symmetrical to it. 563. Opposite or vertical polyedral angles are those in which the edges of one are the prolongations of the edges of the other. 278 SOLID GEOMETRY, Proposition 22. Theorem. 564. Two opposite 2)olyedral angles are symmetrical. B Hyp. Let S - ABC, S - A'B'C be two opp. triedral Z s. To prove they are symmetrical. Proof. Because the face Z s ASB and A'SB' are vertical Zs, .-. ZASB= ZA'SB' (53) Similarly, Z BSC = Z B'SC, etc. Also, because the diedral Z between two planes is the same at every pt., (528) . * . the diedral Z s whose edges are SA, SB, etc., = re- spectively the diedral Z s whose edges are SA', SB', etc. But the edges of S - A'B'C are arranged in the reverse order from the edges of S - ABO. .• . S - ABC is symmetrical to S - A'B'C. q.e.d. EXERCISE. Pass two parallel planes, one through each of two straight lines which do not meet and are not parallel. Let AB, CD be the lines : draw AE || to CD, CF || to AB. .*. plane AEB is li tg plane CFD. BOOK VL—TBIEDRAL ANGLES. 279 ;^ Proposition 23. Theorem. 565. The sum of any two face-angles of a triedral angle is greater than the third, S ,V'^C Hyp. Let S-ABO be a triedral Z in which /ASC is the greatest face / . To prove ' Z ASB + Z BSC > Z ASC. Proof, In the plane ASC draw SD, making Z ASD=:ZASB. Draw AC cutting SD in D, take SB = SD, and join AB, BC. Because the Z s ASB, ASD have SA common, SB = SD, ZASB =ZASD, (Cons.) .-. aASB=aASD,.-. AB=AD. (104) In A ABC, AB + BC> AC. (96) But AB = AD. (Proved above) Subtracting, BC > DC. (Ax. 5) Because in the A s BSC, DSC, SC i^ common, SB = SD, and BC > DC, .-. ZBSC> ZDSC. (120) But ZASB=ZASD. (Cons.) Adding, Z ASB + Z BSC > Z ASC. Q.e. d. 280 SOLID GEOMETRY. Proposition 24. Theorem. 566. The sum of the face-cinglcs of any convex poly edral angle is less tha^ifour right angles. Hyp. Let tlie convex »polycdral / S be cut by a plane making tlie section ABODE a convex polygon. To 2^rove Z ASB + I BSC, etc., < 4 rt. Z s. Proof From any pt. within the polygon ABODE draw OA, OB, 00, OD, OE. There are thus formed two sets of A s, one with their common vertex at S, and the other with their common vertex at 0, and an equal number of each. . • . the sum of the Z s of these two sets of A s is equal. (97) Because the sum of any two face Z s of a triedral Z > the third, (565) and .-. ZSAE+ ZSAB>ZEAB, Z SBA + Z SBO > Z ABO, etc. Taking the sum of these inequalities, we find that the sum of the Z s at the bases of the A s whose common ver- tex is S > the sum of the Z s at the bases of the A s whose common vertex is 0. .•. the sum of the Zs at S < tlie sum of the Zs at 0. the sum of the Z s at S < 4 rt. Z s. (5G) Q.E.D. BOOK VL—TRIEDUAL ANGLES. 281 Proposition 25. Theorem. 567. Two triedral angles y which have the three face- angles of the one equal respectively to the three face-a7igles of the other, are either equal or symmetrical. B •■ B' B' Hyp, In the triedral Zs, S and S', let the face Zs ASB, BSO, OSA = the face Z s A'S'B', B'S'C, C'S'A'. To prove S - ABC = or symmetrical to S' - A'B'C. Proof, In the edges SA, S'A', etc., take the six equal distances SA, SB, SO, S'A', S'B', S'C; and join AB, BO, CA, A'B', B'O', O'A'. Then, AS SAB, SBO, SO A = As S'A'B', S'BT/, S'O'A'. .-. A ABO = aA'B'O'. (108) In the edges SA, S'A', take SD = S'D'. In the faces ASB, ASO, and A'S'B', A'S'C, draw DH, DK, and D'H', D'K' _L to AS and A'S', meeting AB, AC and A'B', A'C, in H, Kand H', K'. Join HKand H'K'. Because AD = A'D', and Z DAH = Z D'A'H', . • . rt. A ADH = rt. A A'D'H'. (107) .• . AH = A'H', and DH = D'H'. In the same way, AK ==: A'K', and DK = D'K'. . • . A AHK = A A'H'K', and HK = H'K'. (104) . • . A HDK = A H'D'K', and Z HDK = Z H'D'K'. (108) . • . diedral Z SA = diedral Z S'A'. (536) Similarly, diedral Zs SB, SO = diedral ZsS'B', S'C. Now if the ^ual Z s are arranged in the same order, as in the first two figures, the two triedral Z s are equal. (557) P>ut if the equal Z s are in reverse order, as in the first and third figures, the triedral Z s are symmetrical. (5G2) Q.E.D, 282 SOLID GEOMETRY. EXERCISES. Theorems. 1. If a straight line is parallel to a plane, any plane per- pendicular to the line is perpendicular to the plane. See (503) and (540). 2. If a line is parallel to each of two intersecting planes, it is parallel to their intersection. 3. If a line is perpendicular to a plane, every plane paral- lel to that line is also perpendicular to the plane. 4. If a line is parallel to each of two planes, the intersec- tions which any plane passing through it makes with the planes are parallel. 5. If a straight line and a plane are perpendicular to the same straight line, they are parallel. 6. If two straight lines are parallel, they are parallel to the common intersection of any two planes passing through them. 7. If the intersections of several planes are parallel, the normals drawn to them from any point are in one plane. 8. If a plane is passed through one of the diagonals of a parallelogram, the perpendiculars to it from the extremities of the other diagonal are equal. 9. Prove that the sides of ah isosceles triangle are equally inclined to any plane through the base. 10. From a point P, PA is drawn perpendicular to a given plane, and from A, AB is drawn perpendicular to a line in that plane: prove that PB is also perpendicular to that line. 11. If the projections of any line upon two intersecting planes are each of them straight lines, prove that the line itself is a straight line. See (495). 12. If a plane is passed through the middle point of the common perpendicular to two straight lines in space, and parallel to both of these lines, prove that the plane bisects BOOK VI.— EXERCISES. THEOREMS. 283 every straight line which joins a point of one of these lines to a point of the other. See (526). 13. The projection of a straight line on a plane is a straight line which is either parallel to the first straight line or meets it in the point where it cuts the plane, ac- cording as the first line is parallel to the plane or not. 14. From a point P, PA and PB are drawn perpendic- ular to two planes which intersect in CD, meeting them in A and B; from A, AE is drawn perpendicular to CD: prove that BE is also perpendicular to CD. 15. Two planes which are not parallel are cut by two parallel planes: prove that the intersections of the first two with the last two contain equal angles. See (519), (525). 16. If two parallel planes be cut by three other planes which have a point, but no line common to all three, and no two of which are parallel, the triangles formed by the intersections of the parallel planes with the other three planes are similar to each other. See (519), (525). 17. The projections of parallel straight lines on any plane are themselves parallel. See (509), (525), (519). 18. Two straight lines which intersect are inclined to each other at an angle equal to two-thirds of a riglit angle, and to a given plane, each, at an angle equal to half a right angle. Prove that their projections on this plane are at right angles to each other. 19. In any triedral angle, the planes bisecting the three diedral angles, all intersect in the same straight line. See (549). 20. In any triedral angle, the planes which bisect the three face-angles, and are perpendicular to these faces respect- ively, all intersect in the same straight line. From the vertex S, take equal distances SA, SB, SC, on the three edges; the intersections of the three ± bisecting planes with the plane of ABC are ± bisect- ing lines of the sides of ABC, ^nd h^ve a common intersection (168); .• . etc. 284 SOLID GEOMETRY. 21. In any triedral angle, the three planes which pass through the edges, perpendicular to the opposite faces respectively, all intersect in the same straight line. See (546). 23. ABCD is a face and AE a diagonal of a cube, BG is drawn perpendicular to AE, and DG is joined : prove that DG is perpendicular to AE. 23. If two face-angles of a triedral angle are equal, the diedral angles opposite them are also equal. 24. An isosceles triedral angle and its symmetrical tri- edral angle are equal. 25. If BAG, CAD, DAB be the three face-angles of a triedral angle, prove that the angle between AD and the straight line bisecting the angle BAG is less than half the sum of the angles BAD, CAD. See (565). 26. A triedral angle is contained by the three face-angles BOG, CO A, AOB; if BOG, COA are together equal to two right angles, prove that CO is perpendicular to the line which bisects the angle AOB. Loci. 27. Find the locus of points which are equally distant from three given points not in the same straight line. 28. Find the locus of points which are equally distant from two given intersecting straight lines. 29. Find the locus of points which are equally distant from two given parallel planes; or whose distances from the parallel planes are in a given ratio. 30. Find the locus of points which are equally distant from three given planes. See (549). 31. Find the locus of points which are equally distant from the three edges of a triedral angle. 32. Find the locus of points which are equally distant from the three faces of a triedral angle. 33. Find the locus of pointG whicli are equally distant I BOOK Vl—EXERCISES PROBLEMS. "^St^ from two given planes, and at the same time equally distant from two given points. 34» Find the locus of a point such that the sum of its distances from two given planes is equal to a given straight line. 35. Find the locus of a point such that the sum of its distances from three given planes is equal to a given straight line. Problems. 3G. Pass a plane perpendicular to a given straight line through a given point not in that line. 37. Pass a plane through a given straight line, perpen- dicular to a given plane. See (543). 38. Pass a plane through a given point parallel to a given plane. See (522). 39. Find the point in a given straight line which is equally distant from two given points not in the same plane with the given line. 40. Draw a straight line through a given point in space, so that it shall cut two given straight lines not in the same plane. 41. Draw a straight line through a given point in a given plane, so that it shall be perpendicular to a given line in space. 42. Two given straight lines do not intersect and are not parallel : find a plane on which their projections will be parallel. 43. Divide a straight line similarly to a given divided straight line lying in a different plane. Let ACDB be the divided line, NMLF the other line ; draw EHKG II to AB, and CH, DK, BG II to AE ; also KL, HM. EN II to GF: .• . (525), (526). 44. Given three straight lines meeting at a point: draw through the given point a straight line equally inclined to the three. Book VII. POLYEDKONS. Definitioi^s. 568. K polyedron is a solid bounded by planes. The portions of the bounding planes, limited by their mutual intersections, are the faces of the polyedron; tlie intersec- tions of the faces are the edges, and the intersections of the edges are the vert ices, of the polyedron. A diagonal of a polyedron is a straight line joining two vertices not in the same face. 569. A polyedron of four faces is called a tetraedron; one of six faces, a hexaedron; one of eight faces, an octa- edron; one of twelve, a dodecaedron; one of twenty, an tcosaedron. 5 70. A polyedron is convex when the section made by any plane intersecting it is a convex polygon. Only convex polyedrons are treated of in this work. 571. The volume of a solid is its numerical measure (229), referred to some other solid called the unit of vol- ume. 572. Two solids are equivalent when they have equal volumes. 573. Two polygons are said to be parallel when their sides are respectively parallel. 286 BOOK VII.—P0LYEDR0N8. DEFINITIONS, 287 Prisms and Parallelopipeds. 5 74. A prism is a polyedron two of whose faces are equal and parallel polygons, and the other faces are parallelograms. The equal and parallel polygons are called the bases of the prism; the par- allelograms are the late7'al faces; the lat- eral faces taken together form the lateral or convex surface; and the intersections of the lateral faces are the lateral edges. The lateral edges are parallel and equal (511) and (520). The area of the lateral surface is called the lateral area. 575. The a//iY?<^e of a prism is the perpendicular dis- tance between its bases. 576. A prism is said to be triangular, quadrangular, hexagonal, etc., according as its bases are triangles, quadri- laterals, hexagons, etc. 577. K rigid prism is one whose lateral edges are perpendicular to its bases. A regular prism is a right prism whose bases are regular polygons : therefore its lat- eral faces are equal rectangles. An oblique prism is one whose lateral edges are not perpendicular to its bases. 578. A right section of a prism is a sec- tion by a plane perpendicular to its lateral edges. 579. A truncated prism is the part of a prism included between the base and an ob- lique section made by a plane cutting all the lateral edges. 580. A parallelopi2)ed is a prism whose bases are parallelograms : hence all the faces are parallelograms. 581. K right ptarallelopiped \s one whose lateral edges are perpendicular to the bases: hence the lateral faces are rectangles. "^^ij 288 SOLID GEOMETRY. 582. A rectangular parallehpvpcd is one whose faces are all rectangles. The three edges of a parallelepiped which meet at a vertex are its dimensions. o 583. A ciihe is a rectangular parallelepiped whose six faces are all squares. 584. The cube whose edge is the unit of length is taken as the unit of volume. (^'^1) Eem. In a right parallelepiped the adjacent lateral faces may make any angle with each other, while in a rectangular parallelepiped the faces are perpendicular to each other. Proposition 1 . Theorem. 585. The sections of the lateral faces of a prism made hy parallel pkmes are equal polygons. ^^^ Hyp. Let the prism MN be cut by the II planes AD, A'D'. To prove ABODE = A'B'C'D'E'. Proof Because the intersections of two II planes by a third plane are par- allel, (519) .*. AB, BC, CD, etc., are || respec- tively to A'B', B'C, CD', etc. .-. ZABC:=ZA'B'C', ZBCD = ZB'C'D'. (525) Because || s included between || s are equal, (1^1) . • . AB = A'B', BC = B'C, etc. .• . the polygons ABCDE, A'B'C'D'E' are mutually equi- lateral and mutually equiangular. .• . ABCDE = A'B'C'D'E'. (14G) Q.E.D. 586. Cor. Any section of a prism hy a plane parallel to the base is equal to the base; and all right sections are equal BOOK VIL— EQUAL PRISMS. 289 (Hyp.) (567) Proposition 2. Theorem. 587. If three faces including a triedral angle of a prism are equal respectively to three faces including a tried^rd angle of a second pris?n, and siniilaidy placed, the two prisms are equal. Hyp. Let the faces AD, AG, BH, of the triedral Z B, be = respectively to thefacesA'D',A'a',B'H' of the triedral Z B'. To prove prism AI = prism A'l'. Proof, Since the face ZsABO,ABG,CBG = the face ZsA'B'C, A'B'G', C'B'G', respectively, . • . triedral Z B = triedral Z B'. . • . the prism AI may be applied to the prism A'l' so tliat the base AD shall coincide with A'D', the face AG with A'G', and the face BH with B'H', the pts. D, E fall- ing on D', E'. Since the pts. F, G, H coincide with F', G', H', . • . the planes of the upper bases will coincide. (484) Since the lateral edges of the prisms are equal and par- allel, (574) .-. the edges DI, EK will coincide with D'l', E'K', re- spectively, and the pts. I, K with the pts. V, K'. .*. the prisms coincide throughout, and .*. are equal. Q.E.D. 588. CoK. 1. Two right prisms which have equal bases and equal altitifdes are equal, 589. CoK. 2. Two truncated prisms are eqiial if three faces including a triedral angle of the one are equal re- spectively to three faces including a triedral angle of the other y and similarly placed. 290 SOLID GEOMETRY. Proposition 3. Theorem. 590. An oUique prism is equivalent to a right prism havi7igfor its lase a right section of the oblique prism and for its altitude u lateral edge of the oblique prism. B C Hyp, Let ABCDE-I be an oblique prism, and A'D' art. section of it. Produce AF to F', making A'F' = AF, and through F' pass the plane F'l' i. to AF', cutting all the edges BG, CH, etc., produced, in the pts. G', H', etc., and forming the rt. section F'l' || to A'D'. (51G) To prove prism AI = prism A'l'. Proof In the truncated prisms AD' and FI', ABODE = FGHIK. (574) Because AF = A'F', and BG = B'G', .• . AA' = FF', and BB' = GG'. Because AG and A'G' are CJs, (574) .• . AB is = and || to FG, and A'B' is = and || to F'G'. .-. quadl. AB' = quadl. FG', being mutually equilatei'al and equiangular (146). Similarly, BC = GH'. . •. the truncated prisms AD' and FI' are equal. (589) Taking AD' from the whole solid, the right prism A'l' remains, and taking FI' from the same solid, the oblique prism AI remains. .*. prism AI = prism A'l'. q.e.d. BOOK VIL—LATEBAL AREA OF A PRISM. 291 Proposition 4. Theorem. 591. The lateral area of a ^jrism is equal to the product of the ])erimeter of a right section by a lateral edge. B Hyp, Let FGHIK be a rt. section, and AA' a lateral edge of the prism AD'. To /jroz^e lateral area of AD' = AA' (FG + GH + etc.). Proof Since the rt. section FI is J. to the lateral edges, (578) .• . FG, GH, etc., are _L to AA', BB', etc., (487) and .* . FG, GH, etc., are the altitudes of the /Z7s AB', BC, etc., which form the lateral area of the prism. .-. area AA'B'B = A A' X FG, (363) and area BB'C'C = BB' X GH, etc. Because AA' = BB' = etc., (574), and FG + GH + etc., = the perimeter of the rt. section, .'. adding the above areas, we have lateral ai-ea AD' = AA' X FG + AA' X GH + etc. = AA'(FG+GH+HI+etc.). Q.E.D. 592. Cor. The lateral area of a right prism is equal to the product of the perimeter of its base by its altitude. 292 SOLID GEOMETHY. ' Proposition 5. Theorem. 693. Tlie ojjjjosite faces of a paraUelopiped are equal and parallel. Eic aH B C Hyp. Let ABCD-G be a paraUelopiped. To prove face AF = and || to DG. Proof Since AC is a z=7, (580) .-. AB = and||toDC. Since BG is a OJ, (580) .-. BF = and II to CG. Since BA and BF are || respectively to CD and CG, .• . Z ABF = Z DCG, and plane AF is || to plane DG.(525) .-. ^ AF =^DG, having two sides and the included Z equal, each to each (135). .'. face AF = and || to face DG. It may be proved in the same way that face AH = and || to face BG. Also, face AC = and || to face EG, by definition. (574) Q.E.D. 594. ScH. Since the opposite faces of a parallelepiped are equal and parallel parallelograms, any face may be taken for the base. 695. Cor. Hie twelve edges of a paraUelopiped may he divided into three sets, each set having four equal and par- allel lines. BOOK VIl.—PARALLELOPIPEDa, 293 Proposition 6. Theorem. 596. Tlie four diago7ials of a parallelopiped bisect one another. Hyp. Let ABCD-G be a paral- lelopiped. To prove that the four diagonals r AG, BH, CE, DF bisect one an- other. A< Proof. Through the opp. and || edges AE, CG pass a plane. Join AC, EG. Because AE and CG are = and H, (574) . • . the figure ACGE is a E7. .*. its diagonals AG, EC bisect each other in the pt. 0. (134) In the same way it may be shown that AG, BH, and AG, DF, bisect each other at 0. . • . the four diagonals bisect one another at the pt. 0. Q.E.D. 597. Cor. 1. The diagonals of a rectangular parallelo- piped are equal. 598. Cor. 2. The sq^iare of a diagonal of a rectangular parallelopiped is equal to the sum of the squares of the three edges meeting at any vertex. For, if AG is a rectangular parallelopiped, the rt. A s ACG, ABC give AG' = AC" + CG' = AB' + BC' + BF'. 599. ScH. The point through which all four diagonals pass is called the centre of the parallelopiped. 294 80LID QEOMETBT. Proposition. 7. Theorem. 600. The plane passed through kvo diagonally opposite edges of a parallelopiped divides it into two equivalent tri- angular pris7ns. Hyp, Let the parallelopiped ABCD-G be divided by the plane AEGO, passing through the opp. edges AE and CG. To prove that the triangular prisms ABC-G and ADC-G are equivalent. Proof, Take a rt. section KLMO of the parallelopiped, in- tersecting the plane AEGO in the line KM. Because the faces AF and AH are || respectively to DG and BG, (593) 1 LJ ^1 AT"' A / \ i '^ - -4/ /m KL is II to OM, and KO is || to LM. (519) KLMO is a O. (124) .'. the intersection KM of the plane AEGO with the rt. section is the diagonal of the LJ KLMO, and divides it into two equal As KLM, KOM. (130) The oblique prism ABO-E is equivalent to a rt. prism whose base is KLM and altitude AE, and the oblique prism ADO-E is equivalent to a rt. prism whose base is KOM and alt. AE. (590) But since these two rt. prisms have equal bases KLM and KOM, and the same altitude, they are equal. (588) prism ABC-G is equivalent to prism ADC-G. q..e.d. BOOK VII.—PARALLEL0PIPED8, 295 Proposition 8. Theorem. 601. Two rectangular parallelopipeds having equal bases are to each other as their altitudes. Hyp. Let AB and CD be the altitudes of two rectangular par- allelopipeds, P and Q, having B equal bases. rojt?roveP:Q = AB:CD. Proof, Suppose the altitudes AB and CD have a common measure, which is contained m times in AB and n times in CD. \ V V Then AB : CD = m : n. Divide AB and CD by this common measure. Through the several pts. of division of AB and CD pass planes J_ to these lines. The parallelopiped P will be divided into m, and Q into n, parallelopipeds, all equal to one another. (588) and '. P : Q = m : 7^; . P:Q = AB:OD. Q.E.D. I The case in which the altitudes AB and CD are incom- mensurable, is included in this demonstration by the rea- soning in (232) ; or the proof may be extended as in (234), (208), and (356). 602. ScH. This theorem may also be expressed as fol- lows : Tivo rectangular parallelopipeds tuhich have two dimen- sions in common are to each other as their third dimen- sions. 296 SOLID GEOMETRY. Proposition 9. Theorem. 603. Two rectangular imrallelo'pijjeds having equal ah titades are to each other as their bases. Hgp. Let the rectangles ab and p q a^y be the bases of two rectangular parallelopipeds, P and Q, having the common altitude c. m F ab To prove ^ = -^-^,. Proof, Construct a third paral- lelepiped R whose dimensions are a, y, and c. Then, because P and R liave the two dimensions a and c in com- mon, (Cons.) P_^ '''n~b'' i\ v^ E (602) (602) Q.E.D. And because R and Q have the two dimensions ¥ and c in common, (Cons.) R_^ Multiplying these equalities, we have P_ ab^ Q "" a'b' ' 604. ScH. This theorem may also be expressed as fol- lows: Two rectangular parallelopipeds having one dimension in common are to each other as the products of the other two dimensions, EXERCISE. Eind the length of the diagonal of a rectangular paral- lelepiped whose edges are 1, 4, and 8. BOOK VIL^PABALLEL0PIPED8. 297 Proposition lO. Theorem. 605. Any two rectmigular paraUelojnpeds are to each other as the products of their three dimensions. Hyp, Let P and Q be two rect- angular parallelopipeds whose dimensions are a, h, c, and a\ ¥y c', ^ respectively. ^ P ale Proof. Make a third rectangu- lar parallelepiped E whose dimen- sions are «', h\ and c. Then, because P and K have the dimension c in common^ L \ Of R ~ a'h'' (604) And because R and Q have the two dimensions «', h' in common, R_ c (602) Multiplying these equalities, we have P _ ahc Q ~ a/b'c' Q.E.D. EXERCISE. Find the ratio of two rectangular parallelopipeds whose dimensions are 4, 7, 9, and 6, 14, 15, respectively. 298 SOLID GEOMETRY. Proposition 1 1 . Theorem. 606. The volume of a rectangular parallelopiped is equal to the product of its three dime^isions. Hyp. Let P be the rectangular parallelopiped, a, h, and c its dimen- sions, and let Q be the cube whose edge is the linear unit. Q then is the lit of volume. (584) To prove V=ahc, Proof P ax b X c Q~1X IX 1 But since Qis the unit of volume, A Q Q = abc. = the volume of P. (605) (571) Q.E.D. . • . volume of P = abc. 607. ScH. The statement of this theorem is an abbre- viation of the following : TJie number of units of volume in a rectangular parallel- opiped is equal to the product of the numbers lohich measure the linear units in its three dimensions. Compare (361). When the three dimensions of a rectangular parallelopiped are each exactly divisible by the linear unit, the truth of the theorem may be shown by dividing the solid into cubes, each equal to the unit of volume. Thus, if AB contain the linear unit 3 times, AC, 4 times, and AD, 5 times, these edges may be divided respectively into 3, 4, and 5 equal parts, and then planes passed through the several points ^ ^ of division at right angles to these edges will divide the solid into cubes each equal to the unit \D BOOK VIL—PARALLELOPIPEDS. 299 of volume. Hence the whole solid contains 3 X 4 X 5, or 60 cubes, each equal to the unit of volume. 608. Cor. 1. Since a x bi^ the area of the base, and c is the altitude, of the parallelepiped P ; therefore the above result may be expressed in the form : The volume of a rectangular" pai^allelopiped is equal to the product of its hase and altitude, 609. Cor. 2. The volume of a cube is the tliird power of its edge, being the product of three equal factors ; if the edge is 1, the volume is 1 X 1 X 1 = 1; if the edge isrt,the volume \^ a X a X a =^ a^. Hence it is that in arithmetic and algebra the ^"^cube" of a number is the name given to the " third power '^ of a number. EXERCISES. 1. Find the surface, and also the volume, of a rectangular parallelepiped whose edges are 4, 7, and 9 feet. 2. Find the surface of a rectangular parallelepiped whose base is 8 by 12 feet and whose volume is 384 cubic feet. 3. Find the volume of a rectangular parallelepiped whose surface is 208 and whose base is 4 by 6. 4. Find the length of the diagonal of a rectangular parallelepiped whose edges are 3, 4, and 5. 5. Find the ratio of two rectangular parallelepipeds whose dimensions are 1, 4, 8, and 3, 4, 5, respectively. 6. Find the surface of a rectangular parallelepiped whose base is 7 by 9 feet and whose volume is 315 cubic feet. 7. Find the volume of a rectangular parallelepiped whose surface is 416 and whose base is 4 by 12. 8. A man wishes to make a cubical cistern whose con- tents are 186624 cubic inches: how many feet of inch boards will line it ? 9. Find the side of a cube which contains as much as a rectangular parallelepiped 20 feet long, 10 feet wide^ and 6 feet high. 300 SOLID GEOMETRY. Proposition 1 2. Theorem. 610. Tlie volume of any parallelopiped is equal to the product of its base and altitude. Hyp. Let B'O be the d',^ ?:..__ h;_ q' altitude of the oblique parallelopiped AC. To prove that vol.AC'=ABODxB'0. \ H' :j=' Proof. Produce the edges AB, DC, A'B', D'C, and take EF = AB. C M The rt. parallelopiped EG', formed by the sections EE'H'H, FF'G'G J_ to the produced edge EF is equivalent to AC. (590) Now produce the edges HE, GF, G'F', H'E', and take KL = HE. The parallelopiped KNN'K'-L formed by the sections LMM'L', KNN'K' J_ to the produced edge KL is equiva- lent to EG', and . • . to AC. (590) .*. given parallelopiped AC and the last, KM', are equivalent. But, since the rt. sections LM', KN', are rectangles, . • . KM' is a rectangular parallelopiped. (582) Also, since area ABCD = area EFGH = area KLMN, (363) (360), and the three solids have the same altitude B'O, (523) .-.the volume KM' = KLMN X B'O. (608) .-. the volume AC = ABCD X B'O. Q.e.d. BOOK VII.- VOLUME OF A PEISM. 301 Proposition 13. Theorem. 611. Tlie volume of a tria7igular prism is equal to the product of its base and altitude, , _ ^, Hyp, Let H denote the altitude of the triangular prism ABC-B'. Toprove vol. ABC-B' = ABC X H. Proof Complete the parallelopiped ABCD-D', having its edges = and |1 to AB, BC, BB'. vol. ABC-B' = i vol. ABCD-B', area ABC = ^ area ABCD. vol. ABCD-D' = ABCD X H. .-. vol. ABC-B' = J ABCD X H, r= ABC X H. Then, vol. ABC-B' = i vol. ABCD-B', (600) and area ABC = 1 area ABCD. (130) But vol. ABCD-D' = ABCD X H. (610) Q.E.D. 612. Cor. 1. The volume of any prism is equal to the product of its base and altitude. For, any prism may be divided into triangular prisms by passing planes through a lateral edge AA' and the cor- responding diagonals of the base. Then the volume of the given prism is the sum of the volumes of the triangular prisms, that is, the sum of their bases, or the base of the given prism, multiplied by the common altitude. 613. Cor. 2. Two prisms are to each other as the prod- ucts of their ^ bases and altitudes; two jt?n^*?m.^^^ into the three pyramids E-ABC, b E-ACD, and E-CDF. The first pyramid E-ABC has the base ABC and the vertex E. The second pyramid E-ACD and the pyramid B-ACD, have the same base ACD, and the same altitude, since their vertices E and B are in the line EB || to the base ACD. . • . E-ACD is equivalent to B-ACD. (633) But the pyramid B-ACD is the same as D-ABO. . • . E-ACD is equivalent to D-ABO. The third pyramid E-CDF and the pyramid B-ACP, have equivalent bases CDF and ACF (367), and the same altitude, since their vertices E and B are in the line EB || to CF, or to the plane ACFD. . • . E-CDF is equivalent to B-ACF. (633) BOOK VII.—POLYEDRONS. 315 |b But tli^ pyramid B-AOF is the same as F-ABC. .-. E-ODF is equivalent to F-ABC. . • . ABC-DEF is equivalent to the sum of the three pyra- mids E-ABC, D-ABC, F-ABC. q.e.d. 639. Cor. 1. The volume of a truncated right triangular prism is equal to the product of its base hy one-third the sum * of its lateral edges. For, the lateral edges AD, BE, CF, being _L to the base, are the altitudes of the three pyramids whose sum is equiva- lent to the truncated prism. (638) .-. volume ABC-DEF = lABC X AD + lABC X BE + lABC X CF, (632) = ABC X i(AD + BE + CF). 640. Cor. 2. The volume of any truncated triangular prism is equal to the product of its right section hy one- third the sum of its lateral edges. For, the rt. section GHK divides the truncated triangular prism ABC-DEF D- into two truncated rt. prisms whose vol- umes are GHKx4(AG + BH + CK), (039) A< and GHKxKC^D-f HE + KF). B . • . their sum is GHK X J(AD + BE + CF). * Called the arithmetic mean of its lateral edges, also its mean height. 316 SOLID GEOMETRY, Similar Polyedroks. 641. Similar poly edrons are those whose corresponding polyedral angles are equal, and which have the same num- ber of faces, similar each to each, and similarly placed. Homologous faces, edges, aiigles, etc., in similar poly- edrons are faces, edges, angles, etc., which are similarly placed. 642. Cor. 1. The liomologotis edges of two similar poly- edrons are proportional to each other, (317) 643. Cor. 2. The homologous faces of two similar poly- edrons are proportional to the squares of any two homolo- gous edges, (379) 644. Cor. 3. The entire surfaces of two similar poly- edrons are proportional to the squares of any two homolo- gous edges, (296) Proposition 23. Theorem.* 645. Tico similar ijoly edrons may he decomposed into the same number of tetraedrons, similar each to each, and similarly placed. Hyp, Let ABCD:E-H and abcde-h be two similar poly- edrons, H and h being homolo- gous vertices. To prove that they may be decomposed into the same B C number of tetraedrons, similar each to each, and similaily placed. Proof, Decompose the polyedron AG into tetraedrons, by dividing all the faces not adjacent to H, into As, and drawing st. lines from H to their vertices. ' BOOK VII.-SIMILAR P0LYEDR0N8. 317 lu the same manner, decompose the polyedron ag into tetraedrons having their common vertex at li homologous toH. The two polyedrons are then decomposed into the same number of tetraedrons, similarly placed. In the two homologous tetraedrons H-ABC and h-ahc, since the faces AD, HA, HC are similar respectively to the faces ad, lia, lie, (641) .*. the As ABC, HAB, HBC are similar respectively to the As dbc, hob, hlc. (321) Q- AC! AB ,AH AB ,^^^, Since = — y-, and —7- = —:-, (307) ac ab ah ah ^ ' , • . AC : ac — AH : ah, X A • AC BC ,H0 BO ,^^^, And since = -. — , and -,— = -^ — , (307) ac be he be ^ ' .-. AC : «c;=HC:7ic. . • . the faces HAO and hac are similar. (313) Also, the corresponding triedral Z s of these tetraedrons are equal. (567) .•. the tetraedrons H-ABC, h-abc are similar. (641) In the same way it may be proved that any two homol- ogous tetraedrons are similar. .*. the two similar polyedrons may be decomposed into the same number of tetraedrons, similar each to each, and similarly placed. q.e.d. 646. Cor. Any two homologous lines i7i two similar polyedrons are proportional to any tivo hovwlogous edges. 318 SOLID GEOMETRY. Proposition 24. Tlieorem. 647. Two similar tetraedrons are to each other as the cubes of their homologous edges. Hyp, Let SABC, S'A'B'C be | two similar tetraedrons ; and let /j\ ? V and V denote their volumes. / I \ /l\ rr V AB To prove v> = ^TgP A^ Proof, Let the vertices S and S' be homologous. b" ^ Then, since the triedral Zs S and S' are equal, (641) V^ _ SA X SB X SC _ _SA SB_ SC^ •'• V ~ S'A' X S'B' X S'C ~ S'A' ^ S'B' ^ S'C" ^ '^ ^_^__SC__AB ■tiui S'A' "" S'B' ~ S'C ~ A'B' * ^ ^ V _ AB AB ^ _ AB' ' * • V"' ~ A'B' ^ A'B' ^ A'B' ~ A^'^ * '^'^'^' 648. Cor. 1. 7^/^o similar polyedrons are to each other as the cubes of their homologous edges. For, two similar polyedrons may be decomposed into the same number of tetraedrons, similar each to each, and similarly placed (645); and any two homologous tetra- edrons are to each other as the cubes of their homologous edges (647), or as the cubes of any two homologous edges of the polyedrons (646). Therefore the polyedrons them- selves are to each other as the cubes of their homologous edges. (296) 649. Cor. 2. Similar prisms, or pyramids, are to each other as the cubes of their altitudes, 650. Cor. 3. Similar polyedrons are to each other as the cubes of any two ho^nologous lines. BOOK VII.— REGULAR FOLYEDRONS. 319 Regular Polyedrons. 651. A rcfjnlar polyedron is one whose faces are all equal regular polygons, and whose polyedral angles are all equal. Proposition 25. Theorem.* 652. There can he only five regular convex poly edrons. Proof. At least three faces are necessary to form a poly- edral angle, and the sum of its face-angles must be less than 360°. (566) 1. Because the angle of an equilateral triangle is 60^, each convex polyedral angle may have 3, 4, or 5 equilateral triangles. It cannot have 6 faces, because the sum of 6 such angles is 360°, reaching the limit. Therefore no more than three regular convex polyedrons can be formed with equilateral triangles ; the tetraedron, octaedron, and icosa- edron. 2. Because the angle of a square is 90°, each convex poly- edral angle may have 3 squares. It cannot have 4 squares because the sum of 4 such angles is 360°. Therefore only one regular convex polyedron can be formed with squares ; the hexaedron, or cube, 3. Because the angle of a regular pentagon is 108°, each convex polyedral angle may have 3 regular pentagons. It cannot have 4 faces because the sum of 4 such angles is 432°. Therefore only one regular convex polyedron can be formed of regular pentagons; the dodecaedron. Because the angle of a regular hexagon is 120°, and the angle of every regular polygon of more than 6 sides is yet greater than 120°, therefore there can be no regular convex polyedron formed of regular hexagons or of any regular polygons of more than 6 sides. Therefore there can be only five regular convex poly edrons. 320 SOLID GEOMETRY. Proposition 26. Problem.* 653. To construct the regular polyedrons, having given one of the edges. Given, the edge AB. Required, to construct the regular polyedrons. 1. The regular tetraedron. Cons, Upon AB construct the equi- lateral A ABC. At its centre erect OD _L to the plane ABC so that AD = AB. Join DA, DB, DC. Then ABCD is a regular tetraedron. Proof, Since the faces are each = the A ABC, (496) . • . the triedral Z s A, B, C, D are all equal. (567) . • . ABCD is a regular tetraedron. 2. The regular hexaedron, or cube. Cons, Upon AB construct the square ABCD. Upon the sides of this square construct the four equal squares AF, BC, CH, DE, J. to the plane ABCD. Then AG is a regular hexaedron. ^ ° Proof, Since the faces are equal squares, (Cons.) . • . the triedral Z s A, B, C, D, E, F, G, H are all equal. (567) . • , ABCI)-E }s a regular hexaedron, or cube. BOOK VIl.-REGULAR POLYEDRONS, 321 3. The regular octaedrofu Cons, Upon AB construct the E square ABCD. /i^\ At its centre erect EOF ± to the / h-K^K plane ABCD, so that OE = OP = OA. a^'---^^^^^ Join the pts. E and F to all the ^^\^^p^1^/^ vertices of the square. Wl-'y^ Then E-ABCD-F is a regular octa- ^^ edron. Proof. The lines from E and F to A, B, C, D are all equal. (496) And since the rt. A s AOE, AOB are equal, (1^4) . • . AE = AB. . • . the twelve edges of the octaedron are all equal, and the faces are eight equal equilateral A s. Since the diagonals BD and EF are equal and bisect each other at rt. Z s, (Cons.) . • . BEDF is a square == ABCD. And since AO is i. to BD and EF, . • . AO is _L to the plane of BEDF. (500) . •. pyramid A-BEDF = pyramid E-ABCD. (633) . • . polyedral Z A = polyedral Z E. Similarly, it can be shown that any other two polyedral Z s are equal. , • . E-ABCD-F is a regular octaedron. 32^ SOLID GEOMETRY, 4. The regular dodecaedron. Cons. Upon AB construct the regular pentagon ABODE, and join to its sides the sides of five other equal pentagons, so inclined to the plane of ABODE as to form five triedral Z s at A, B, 0, D, E. I i There is then formed a convex surface FGHIK, etc., composed of six equal regular pentagons. Oonstruct a second convex surface fyhik, etc., equal to the first. Then the two equal convex surfaces may be combined so as to form a single convex surface, which is the regular dodecaedron. Proof. Because the face Z s of the triedral Z s in FK are equal respectively to the face Zs of the triedral Zs in fkf (Cons. ) . • . the triedral Z s of FK and fk are equal, each to each. {b^'^i) Now suppose the convexity of FK to be up, aod the con- vexity oifk to be down. Put the two surfaces together so that the pt. and the side OP shall coincide with the pt. n and the side no, re- spectively. Then two consecutive face Z s of one surface will unite with a single face Z of the other. Thus the three faces 1, 5, 7, will enclose a triedral Z :, since the diedral Z contained by 1 and 5 is the diedral Z of the equal triedral Z formed at E. The vertex N will coincide with m, and a like triedral Z will be formed at that pt., and so of all the others. . • . the triedral Z s are all equal, and the solid is a regular dodecaedron. BOOK VIL— REGULAR POLYEDRONS. 323 5. TJie regular icosaedron. Cons, Upon AB construct a regular pentagon ABODE. At its centre erect a _L to its plane, and in this _L take a pt. S so that SA — AB. Join SA, SB, SO, SD, SE. Then S- ABODE is a regular pentagonal pyramid (496), and each of its faces is an equilateral A . Oomplete the polyedral Z s at A and B by adding to each three equilateral A s each equal to SAB, and making the diedral Z s around A and B equal. There is then formed a convex surface ODEF, etc., com- posed of ten equal equilateral A s. Oonstruct a second convex surface cdef, etc., equal to the first. Then the two equal convex surfaces may be combined so as to form a single convex surface, which is the regular icosaedron. Proof. Suppose the convexity of DG to be up, and the convexity of dg to be down. Put the two surfaces together so that the pt. D, where two faces meet, falls upon the pt. 6', where three faces meet. Then two consecutive face Z*s of one surface will unite with three consecutive face Zsof the other. Thus, the two faces 1 and 9 will unite with the three faces 10, 11, 12, forming a polyedral Z of five faces equal to S, without in any way changing the form of either surface, since the 324 SOLID GEOMETRY. three diedral Zs contained by 1 and 9, 10 and 11, 11 and 12 are those which belong to such a polyedral Z . The vertex C will fall at h, and a like polyedral / will be formed at that pt., and so of all the others. .'. the polyedral /s are all equal, and the solid is a regular icosaedron. • q.e.f. 654. ScH. Models of the regular polyedrons may be easily constructed as follows: Draw the following diagrams on cardboard, and cut them out. Then cut half way through the board in the dividing lines, and bring the edges together so as to form the re- spective polyedrons. TETRAEDRON OCTAEDRON DODECAEDRON ICOSAEDROhT BOOK VIL—PBOPERTIES OF P0LTEDR0N8. 325 General Properties of Polyedrons. euler's theorem. f Proposition. 27. Theorem.* 655. hi any polyedro7i the numher of edges increased by 2 is equal to the whole number of vertices andfaces,l Hyp, Let E, F, and V denote the number of edges, faces, and vertices respectively of the polyedron S-ABCD. > To prove E + 2 = V + F. / Proof. Beginning with one face / ABCD, we have E = V. ^"' If we annex to this a second face SAB, by applying one of its edges, as AB, to the corresponding edge of the first face, we form a surface having one edge AB, and two vertices A and B co7}wion to both faces ; therefore, the whole number of edges is now one more than the whole number of vertices. . •. for 2 faces ABCD and SAB, E = V + 1. If ,we annex a third face SBC, adjacent to both ABCD and SAB, we form a new surface having two edges SB and BC, and three vertices S, B, C in common with the preced- ing surface ; therefore the increase in the number of edges is again one more than the increase in the number of vertices. .-. for 3 faces, E = V + 2. In like manner, for 4 faces, E = V -|- 3. And so on until every face but one has been annexed. . •. for (F - 1) faces, E = V + F - 2. But annexing the last face adds no edges nor vertices. .-. for F faces, E = V + F - 2, or E + 2 = V + F. Q.E.D. t The proof of this theorem was first published by Euler (1752). % This proof is due to Cauchy. 326 SOLID GEOMETRY. EXERCISES, 1. How many edges has the regular tetraedron ? 2. How many edges has the regular hexaedron ? 3. The frustum of a regular four-sided pyramid is 8 feet high, and the sides of its bases are 3 feet and 5 feet : find the height of an equivalent regular pyramid whose base is 10 feet square. Proposition 28. Theorem.* 656. The sum of the face-angles of any jwlyedron is equal to four right angles taken as ma?iy tifues as the poly- edral has vertices less two. Hyp. Let E, F, and N denote the number of edges, faces, and vertices respectively, and S the sum of the face Z s of any polyedron. To prove S = 4 rt. Z s ( V - 2). Proof Since each edge is common to two faces, . • . the whole number of sides of the faces considered as independent polygons is 2E. If we form an exterior Z at each vertex of every polygon, the sum of the interior and exterior Z s at each vertex is 2 rt. Z s. And since the whole number of vertices in the faces is 2E, . • . the sum of all the interior and exterior Z s of the faces is 2 rt. Z s X 2E, or 4 rt. Z s X E. Since the sum of the exterior Z s of each face is 4 rt. Zs, (151) .' . the sum of the ext. Z s of the E faces is 4 rt. Z s X F. .-. S + 4rt. Zs X E =:4rt. Zs X E. .•.S=r4rt.Zs(E-F). But E-F = V-2. (655) .-. S===4rt.Zs (V- 2). Q.E.D, BOOK VIL— EXERCISES, THEOREMS, i327 exercises. Theokems. 1. Show that a lateral edge of a right prism is equal to the altitude. 2. Show that the lateral faces of right prisms are rect- angles. 3. Show that every section of a prism made by a plane parallel to the lateral edges is a parallelogram. 4. The lateral areas of right prisms of equal altitudes are as the perimeters of their bases. 5. The opposite faces of a parallelepiped are equal and parallel. 6. If the four diagonals of a quadrangular prism pass through a common point, the prism is a parallelepiped. T. If any two non-parallel diagonal planes of a prism are perpendicular to the base, the prism is a right prism. See (546). 8. Any straight line drawn through the centre of a parallelepiped, terminating in a pair of faces, is bisected at that point. 9. The volume of a triangular prism is equal to the product of the area of a lateral face by one-half the per- pendicular distance of that face from the opposite edge. 10. In a cube the square of a diagonal is three times the square of an edge. 11. In any parallelepiped, the sum of the squares of the twelve edges is equal to the sum of the squares of its four diagonals. 12. In any quadrangular prism, the sum of the squares of the twelve edges is equal to the sum of the squares of its four diagonals plus eight times the square of the line joining the common middle points of the diagonals taken two and two. I yjl3. A plane passing through a triangular pyramid, I 328 SOLID GEOMETRY. parallel to one side of the base and to the opposite lateral edge, intersects its faces in a parallelogram. 14. The lateral surface of a pyramid is greater than tlie base. Project the vertex on the base, etc. 15. The four middle points of two pairs of opposite edges of a tetraedron are in one plane, and at the vertices of a parallelogram. (617). 16. The three lines joining the middle points of the three pairs of opposite edges of a tetraedron intersect in a point which bisects them all. 17. In a tetraedron, the planes passed through the three lateral edges and the middle points of the edges of the base intersect in a straight line. The intersections of the planes with the base are medials of the base; .* . etc. 18. The lines joining each vertex of a tetraedron with the point of intersection of the medial lines of the opposite face all meet in a point, which divides each line in the ratio 1 : 4. Note,— This point is the centre of gravity of the tetraedron. 19. The plane which bisects a diedral angle of a tetra- edron divides the opposite edge into segments which are proportional to the areas of the adjacent faces. See (303), (369). /20. The volume of a truncated triangular prism is equal to the product of the area of its lower base by the perpen- dicular upon the lower base let fall from the intersection of the medial lines of the upper base. 21. The volume of a truncated parallelopiped is equal to the product of the area of its lower base by the perpendicu- lar from the centre of the upper base upon the lower base. 22. The volume of a truncated parallelopiped is equal to the product of a right section by one-fourth the sum of its four lateral edges. See (640). 23. Any plane passed through the centre of a parallelo- piped divides it into two equivalent solids (640). BOOK VIL— NUMERICAL EXERCISES. 329 24. The portion of a tetraedron cut off by a plane parallel to one of its faces is a tetraedron similar to the whole tetraedron. 25. When two tetraedrons have a diedral angle of the one equal to a diedral angle of the other, and the faces in- cluding these angles similar each to each, and similarly placed, the tetraedrons are similar. 26. Two polyedrons composed of the same number of tetraedrons, similar each to each, and similarly placed, are similar. Numerical Exekcises. 27. Find the lateral area of a right prism whose altitude is 14 inches and perimeter of the base 16 inches. \ Ans, 224 square inches. 28. Find the volume of* a prism the area of whose base is 24 square inches and altitude 7 feet. Ans, VQ8 cubic feet. 29. Find the surface of a cube whose sides are each 11 inches. | A7is, 726 square inches. 30. Find the surfac£^f a cubical cistern whose contents are 373,248 cubic inches. ' Ans\ 180 square feet. 31. Find the lateral surface of a right prism whose alti- tude is 2 feet, and whose base is a regular hexagon of which each side is 10 inches long. 32. Find the depth of a cubical cistern which shall hold 1600 gallons, each gallon being 231 cubic inches. A71S, 5.98 feet. 33. If the dimensions of a rectangular parallelepiped are 20.5 feet, 18 1 feet, and 6.75 feet, what is the edge of an equivalent cube ? Ans. 11.4 feet. 34. Find the depth of a cubical box which shall contain 100 bushels of grain, each bushel holding 2150.42 cubic inches. Ans. 4.9 feet. 35. If the dimensions of a rectangular parallclopiped are 9,16, and 25, what is the edge of an equivalent cube? 36. Find tlie dimensions of the base of a rectangular 330 SOLID GEOMETRY. parallelepiped whose volume is 60, surface 94, and alti- tude 3. 37. Find the ratio of two rectangular parallelopipeds, if their altitudes are each 4 feet, and their bases 6 feet by 3 feet, and 12 feet by 8 feet, respectively. 38. Find the ratio of two rectangular parallelepipeds if their dimensions are 5, 6, 8, and 10, 12, 16, respectively. 39. Find the volume of a right triangular prism, if the height is 8 inches, and the sides of the base are 6, 5, and 5 inches. 40. Find the lateral area of a right pyramid whose slant height is 4 feet, and whose base is a regular octagon of which each side is 3 feet long. 41. Find the volume of a right quadrangular pyramid whose altitude is 12, and whose base is 4 feet square. 42. Find the volume of a pyramid whose altitude is 20 feet, and whose base is a rectangle 8 feet by 7. 43. Find the lateral area of a right pentagonal pyramid whose slant height is 18 inches, and each side of the base 6 inches. Ans. 270 square inches. 44. Find the volume of a pyramid whose altitude is 20 inches, and whose base is a regular hexagon, each side be- ing 6 inches. Ans, 623.5386 cubic inches. 45. Find the lateral area and volume of a right triangular pyramid, the sides of whose bases are 3, 5, and 6, and whose altitude is 6. See (398), Ex. 2. 46. Find the volume of the frustum of a square pyramid, the sides of whose bases are 8 and 6 feet, and whose alti- tude is 12 feet. Ans. 592 square feet. 47. If a plane be passed parallel to the base of the pyra- mid in Ex. 44, midway between the vertex and base, find the lateral area and volume of the frustum. 48. The slant height of the frustum of a right pyramid is 6 feet, and the perimeters of the two bases are 18 feet and 12 feet: what is the lateral area of the frustum? BOOK VlI.—NUMElilCAL EXERCISES. 331 49. Find tlio slant height of the pyramid whoso frustum is given in Ex. 48. 50. Find the volume of the frustum of a regular triangu- lar pyramid, the sides of whose bases are 8 and G, and whose lateral edge is 5. 51. A pyramid 18 feet high has a base containing 1G9 square feet. How far from the vertex must a plane be passed parallel to the base so that the section may contain 81 square feet ? 52. The base of a pyramid contains 169 square feet; a plane parallel to the base and 5 feet from the vertex cuts a section containing 81 square feet: find the height of the pyramid. 53. A pyramid 16 feet high has a square base 10 feet on a side. Find the area of a section made by a plane parallel to the base and 6 feet from the vertex. 54. The base of a regular pyramid is a hexagon of which the side is 4 feet. Find the height of the pyramid if the lateral area is eight times the area of the base. 55. F'ind the total surface of a regular pyramid, (1) when each side of its square base is 12 feet, and jfche slant height is 24 feet; (2) when each side of its square base is 30 feet, and the perpendicular height is 94 feet; and (3) when each side of its triangular base is 8 feet, and the slant height is 24 feet. 56. Find the volume of a regular pyramid when each side of its square base is 80 feet, and the lateral edge is 202 feet. 57. Find the volume of a regular pyramid whose base is an equilateral triangle inscribed in a circle of radius 40 feet, and whose slant height is 48 feet. 58. Find the lateral edge, lateral area, and volume, (1) of a regular hexagonal pyramid, each side of whose base is 2, and whose altitude is 12; (2) of a frustum of a regular triangular pyramid, the sides of whose bases are 5 1^3 and t^3, and whose altitude is 3 ; (3) of a frustum of a quadrangu- lar pyramid, the sides of whose bases are II and 1, and whose 332 SOLID GEOMETRY. altitude is 12; and (4) of a frustum of a regular liexagonal pyramid, the sides of whose bases are 12 and 4, and whose altitude is 12. 59. Find the volume of the frustum of a regular square pyramid, the sides of whose bases are 40 and 16 feet, and whose slant height is 20 feet. GO. The volume of a frustum of a regular hexagonal pyramid is 12 cubic feet, the sides of the bases are 2 and 1 feet : find the height of the frustum. 61. Find the difference between the volume of the frus- tum of a regular quadrangular pyramid, the sides of whose bases are 4 and 3 feet, and the volume of a prism of the same altitude, whose base is a section of the frustum paral- lel to its bases and midway between them. 62. Find the dimensions of a cube whose surface shall be numerically equal to its contents. 63. A regular pyramid 8 feet high is transformed into a regular prism with an equivalent base: find the height of the prism. //64. A cube whose edge is 3 feet is transformed into a right prism whose base is a rectangle 3 feet by 2 feet: find the height of the prism. 65. The height of the frustum of a regular quadrangular pyramid is 12 feet, and the sides of its bases are 6 and 10 feet: find the height of an equivalent regular pyramid whose base is 24 feet square. 66. A mound of earth is raised with plane sloping sides and rectangular bases; the dimensions at the bottom are 80 yards by 10, at the top 70 yards by 1, and it is 5 yards high: find its cubical contents. 67. A bath 6 feet deep is excavated; the area of the sur- face at the top is 100 square yards, at the- bottom 81 square yards: find the number of gallons of water it will hold. 68. A railway embankment across a valley has the fol- lowing measvires: width at top 20 feet, at base 45 feet. BOOK VIL— NUMERICAL EXERCISES. 333 height 11 feet, length at top 1020 yards, at base 960 yards : find its volume. G9. The altitude of a prism is 9 feet and the perimeter of the base 6 feet: find the altitude and perimeter of the base of a similar prism one-third as great. 70. A pyramid is cut by a plane parallel to the base and midway between the vertex and base: find the ratio of the volume cut off to the whole volume. 71. The length of one of the lateral edges of a pyramid is 16 inches. How far from the vertex will this edge be cut by a plane parallel to tlie base, which divides the pyramid into two equivalant parts ? 73. The height of the frustum of a pyramid is 16 inches, and two homologous edges of its bases are 8 and 6 inches: find the ratio of the volume of the frustum to that of the entire pyramid. 73. Show that the volume of a regular tetraedron is equal to the cube of its edge multiplied by -f^ V2. 74. Show that the volume of a regular octaedron is equal to the cube of its edge multiplied by ^ V2. 75. Find the volume of a regular icosaedron whose edges are each 20 feet. 76. Verify Euler's Theorem in the case of all the regular polyeclrons. 77. Find the number of edges in a pyramid, and in a prism, on a polygon of n sides as base. Problems. 78. Given three indefinite straight lines in space which do not intersect, to construct a parallelo piped which shall have three of its edges on these lines. 79. To cut a solid tetraedral angle by a plane, so that the section shall be a parallelogram. Book VIII.* THE SPHEEE.t Circles of the Sphere and Tangent Planes. definitions. 657. K sphere is a solid bounded by a curved surface, every point of which is equally distant from a point within it, called the centre. A sphere may be generated by the revolution of a semicircle ACB about its diameter AB as an axis. 658. A radius of a sphere is a straight line drawn from tlie centre to any point of the surface. A diameter of a sphere is a straight line passing through its centre, and terminating at both ends in tlie surface. All the radii of a sphere are equal; and all the diameters are equal, since each is the sum of two radii. 659. A line or plane is tavgent to a sphere when it has but one point in common with the surface of the sphere. 660. Two spheres are tangent to each other when their surfaces have but one point in common. 661. The common point is called the j^oint of co7iiact, or pf>int of tangency. 662. Two spheres are concentric when they have the same centre. 663. Two spheres are equal when they have equal radii. * This book treats of the properties and relations of those parts of the surface of a sphere whieli are bounded by arcs of great circles. t In teaching tliis subject, as well as in teaching Spherical Trigonometry, the class-room should be furnished with a spherical black-board, on which the student should draw the diagrams of spherical surfaces. 334 BOOK VIIL—jDmCLES OF THE SPHERE. 335 Proposition 1. Theorem. 664. Every section of a sphere made hy a plane is a cir- cle. Hyp. Let AOB be a plane section of the sphere whose centre is 0. To prove that AOB is a O. Proof. Draw 00' J_ to the plane AOB meeting it in 0', and draw OA, 00 to any two pts. in the section. Since OA and 00 are radii of the sphere, .•.OA = OC. (658) And since OA and 00 are equal oblique lines from to the plane AOB, .•.0'A = 0'C. (498) But A and are any two pts. in the section AOB. .'. the curve AOB is a O whose centre is 0'. q.e.d. 665. Def. The circular section of a sphere by a plane is called a circle of the sphere. If the plane passes through the centre of the sphere, the section is a circle whose radius is equal to the radius of the sphere. This circle is called a great circle of the sphere. A section made by a plane which does not pass through the centre of the sphere is called a small circle. 666. Def. The two points in which a diameter of the sphere, perpendicular to the plane of a circle, meets the surface of the sphere, are called the j^oles of the circle, and this diameter is called the axis of the circle; thus, P, P' are the poles of the circle AOB, and PP' is its axis.* P' is said to be antipodal to P. 336 SOLID GEOMETRY. 661. Cor. 1. The line through the centre of a circle of the sphere, perpendicidar to its plane, passes through the centre of the sphere. {i) Therefore, the axis of a circle passes through its centre {Q^^), and all parallel circles have the same axis and the same poles. 668. Cor. 2. All great circles of a sphere are equal. CoQ^) 669. Cor. 3. All sinall circles at equal distances from the centre of the sphere are equal ; and of two circles un- equally distant from the centre, the nearer is the larger, and conversely. 670. Cor. 4. Every great circle bisects the sjjhere and its surface. For, if the two parts are separated, and then placed on the common base, with their convexities turned the same way, their surfaces will coincide, since all the points of either are equally distant from the centre. (GST) 671. Cor. 5. Any two great circles lisect each other. For, the common intersection of their planes passes through the centre of the sphere, and is a diameter of each circle. 672. Cor. G. An arc of a great circle may he draw7i through any two given points on the surface of a sphere. For, the two given pts. with the centre of the sphere de- termine the plane of a great O passing through the two given pts. (484) If the two pts. are the extremities of a diameter, the position of the circle is not determined: for the two given pts. and the centre being in the same st. line, an indefinite number of planes can be drawn through them. (482) 673. Cor. 7. An arc of a circle may he draimi through any three given points on the surface of a sphere. For, the three pts. determine a plane. (484) Note.— By tlie distance between two points on a sphere is meant the arc of a great circle joining them. BOOK VIIL— CIRCLES OF THE SPHERE. 337 Proposition 2. Theorem. 674. All points in the circumference of a circle of a sphere are equally distant from each of its poles. Hyp. Let P, P' be the poles of the O ACB, where A, C, B are any pts. on its Oce. To prove arcs PA, PC, PB equal. Proof. Since PP' is a line through the centre of the O ACB J_ to its plane, .-. the St. lines PA, PC, PB are equal. (496) .-. the arcs PA, PC, PB are equal. (198) Similarly, the arcs P'A, P'C, P'B are equal. q.e.d. 675. Def. The distance from any point in the circum- ference of a circle to its nearest pole is called the polar dis- tance of the circle. 676. Cor. 1. The polar distance of a great circle is a quadrant. Thus, PE, PG are quadrants, for they are the measures of the rt. Z s POE, POG, whose vertices are at the centre of the sphere. (237) 677. Cor. 2. If a point P on the surface of a sjjhere is at a quadrant^s distance from the two points E, G, of an arc of a great circle, it is the pole of that arc. For, the /s POE, POG are rt. Zs; .*. the radius OP is _L to the plane of the arc EG (500) ; and .' . P is the pole of the arc EG. (6G6) 678. ScH. In Spherical Geometry, the term quadrant usually means a quadrant of a great circle. 338 SOLID GEOMETEY. Proposition 3. Theorem. 679. Tlie shortest distajice on tlie surface of a sphere, 'between any two points on the surface^ is the arc of a great circle, not greater than a semi-circumference, joining the two ptoints. Hyp. Let AB be an arc of a great O, not greater than a semiOce, join- ing any two pts..A and B on the sphere; and let ACEB be any other line in the surface joining A and B. To prove arc AB < ACEB. Proof Take any pt. D in ACEB, pass arcs of great Os through A, D and B, D; and join 0, the centre of the sphere, with A, D, B. Then, since the Z s AOB, AOD, DOB are tlie face Z s of the triedral Z whose vertex is at 0, .• . Z AOD + Z DOB > Z AOB. (565) But the Z s AOD, DOB, AOB are measured by tlie arcs AD, DB, AB, respectively. (236) .• . arc AD -|- arc DB > arc AB. Similarly, joining any pt. in ACD with A and D by arcs of great Os, the sum of the arcs is greater than arc AD; and joining any pt. in DEB with D and B by arcs of great Os, the sum is greater than arc DB. If this process is indefinitely repeated, the line from A to B, on the arcs of the great Os, will continually increase and approach the line ACEB; that is, the sum of the arcs of the great Os will approach ACEB as the limit, and will always be greater than AB. / . arc AB < ACEB. q.e.d. BOOK VIIL-TANQENT PLANES. 339 Proposition 4. Theorem. 680. A plane perpendicular to a radius of a sphere at its extremity is tangent to the sphere. Hyp. Let the plane MN be JL to the radius OP at its extremity P. To prove MN tangent to the sphere. Proof. Take any other pt. H in the plane, and join OH. Becanse the _L is the shortest dis- tance from a point to a plane, (497) .-.OP < OH. . • . the pt. H is without the sphere. But H is any pt. of MN except P. . * . every pt. of MN except P is without the sphere. . • . plane MN is tangent to the sphere at the pt. P. (659) Q.E.D. 681. Cor. 1. Every straight li?ie perpendimtlar to a radivs at its extremity is tangent to the sphere. (659) 682. Cor. 2. Every plane or line tangent to a sphere is 2)erpe)idicular to the radius draiun to the 2Joint of contact. 683. Cor. 3. A straight line tangent to any circle of a sphere lies in the plane tangent to the sphere at the point of contact. 684. ScH. 1. Any straight line drawn in a tangent plane through the point of contact is tangent to the sphere at that point. 685. ScH. 2. Any two straight lines, tangent to the sphere at the same point, determine the tangent plane at that point. 340 SOLID GEOMETRY. Proposition 5. Theorem.* 686. Tlirougli any four points not in the same plane, one spherical surface, and only one, may pass. Hyp. Let A, B, 0, D be the four pts. not in the same plane. To prove that one spherical surface, and no more, may pass through A, B, C, D. Proof, Let E, H be the centres of the Os circumscribed about the As BCD, ACD, respectively. Draw EK _L to plane BCD, and HL _L to plane ACD. Every pt. in EK is equally distant from the pts. B, C, D, and every pt. in HL is equally distant from the pts. A, C, D. (496) Join E and H to F, the middle pt. of CD. EF and HF are each J_ to CD. (203) . • . the plane through EF and IIF is J_ to CD, (500) and . • . this plane is J_ to both planes BCD, ACD. (540) Since HL is _L to the plane ACD at H, (Cons. ) . • . HL lies in the plane EFH. (538) Similarly, it may be shown that EK lies in this plane, . • . the _Ls EK, HL lie in the same plane ; and, being J_ to planes which are not || , cannot be || , and . •. must meet at some pt. 0. Since is in the J_s EK and HL, it is equally distant from B, C, D, and from A, C, D. (49G) . • . is equally distant from A, B, C, D ; and the sphere described with as a centre and OA as a radius, will pass through the pts. A, B, C, D. BOOK VIIL—SrUEEE AND TEDRAEDRON, 341 Also, since the centre of any sphere passing through the four pts. A, B, C, D, must be in the _Ls EK, HL, (498) . • . the intersection is the centre of the only sphere that can pass through the four given pts. q.e.d. 687. Cor. 1. A sphere may he circumscribed about any tetraedron. 688. CoE. 2. The four perpendiculars to the faces of a tetraedron through their centres meet at the same point, 689. Cor. 3. Tlie six planes ivhich bisect at right angles the six edges of a tetraedron all intersect in the same point* Proposition 6. Theorem.* 690. A sphere may be inscribed in a given tetraedron. Hyp. Let ABCD be the given tetra- edron. To prove that a sphere may be in- scribed in ABCD. Proof. Bisect any three of the die- dral Z s which have one face common, as BC, CD, BD, by the planes OBC, OCD, OBD, respectively. Since is in the bisector of the diedral Z equally distant from the faces ABC and BCD. In the same way is equally distant from the faces ACD and BCD, and from BAD and BCD. . • . the pt. is equally distant from the four faces of the tetraedron. . • . a sphere described with as a centre, and with a radius equal to the common distance of from any face, will be tangeut to each face, and will be inscribed in the tetraedron. (659) Q.E.D. 691. Cor. The six planes lohich bisect the six diedral angles of a tetraedron intersect in a point. 342 SOLID GEOMETRY. Spherical Triaxgles and Polygok^s. definitions. 692. The angle between two intersecting curves is the angle included between their tangents at the point of intersection. When the two curves are arcs of great circles the angle is called a spherical angle, 693. A spherical 2)olyg on is a portion of the surface of a sphere bounded by three or more arcs of great circles. The bounding arcs are the sides of the polygon; the points of intersection of the sides are the vertices of the polygon ; and the angles which the sides make with each other are the angles of the polygon. A diagonal of a spherical polygon is an arc of a great circle joining any two vertices which are not consecutive. 694. A spherical triangle is a spherical polygon of three sides. A spherical triangle is right or oblique, scalene, isosceles, or equilateral, in the same cases as a plane triangle. Note. — Between any two points two arcs of great circles may be drawn, the one less, and the other greater, than a senii-circunifereuce. In the present treatise the arcs less than a semi-circumference will be taken, unless otherwise stated. 695. A spherical pyramid is a portion of the sphere bounded by a spherical polygon and the jilanes of tlie sides of the polygon. The centre of the sphere is the vertex of the pyramid, and the spherical polygon is its hase. 696. Two spherical polygons are equal if they can be applied one to the other, so as to coincide. 697. Since the sides of a spherical polygon are arcs, they are usually expressed in degrees, minutes, and seconds^ BOOK VIIL— SPHERICAL ANGLES. 343 Note.— Because the apparent position of the heavenly bodies is referred to an imaginary spherical surface whose centre we occupy, the geometry of the sur- face of the sphere early attracted attention. It cannot be studied to any great extent without a knowledge of Trigonometry, but a few important proposi- tions may be given, which illustrate this branch of geometry. Proposition 7. Theorem. / 698. A spherical angle is measured hy the arc of a great I circle described IV ith its vertex as a pole and inclnded be- tiveen its sides, produced if necessary. Hyp. Let ABO, AB'C be two arcs of great Os intersecting at A; let y^ Al\ AT' be the tangents to these / arcs at A; and let OBB' be a plane / through the centre _L to AC, in- I tersecting the sphere in the great O \ BB'. \^__ To prove that the spherical Z BAB' C is measured by the arc BB'. Proof Since TA and T'A are respectively in the planes of the arcs BA and B'A, and are JL to their intersection AC, (210) .-. Z TAT' = the diedral Z BACB'. Also, Z BOB' = the diedral Z BACB', (528) and it is measured by the arc BB'. (236) But the spherical Z BAB' is measured by Z TAT'. (692) . • . it is measured by the arc BB'. q.e.d. 699. CoR. 1. A spherical angle is equal to the diedral angle between the pla^ies of the tivo circles. 700. Coft. 2. If two arcs of great circles cut each other, their vertical angles are equal. 701. Cor. 3. The angles of a spherical triangle are equal to the diedral angles betioeen the p^lanes of the sides of the triangle. 344 80LID GEOMETRY. Relation of a Spheeical Polygon to a Polyedkal Angle. 702. Because the planes of all great circles pass through the centre of the sphere, therefore the planes of the sides of a spherical polygon form a polyedral angle at the centre whose face-angles AOB, BOO, etc, are measured by the sides AB, BO, etc., of the polygon (236), and whose diedral angles OA, OB, etc., are equal to the angles A, B, etc., of the spherical polygon ABO, etc. (699) We may therefore speak of all the parts of a spherical polygon as angles, meaning ther-eby the face-angles, and the diedral angles between tlie faces, of the polyedral angle whose vertex is the centre of the sphere, and base the spherical polygon. It follows, therefore, that the properties of a spherical polygon and a^ polyedral angle are mutually convertible. Hence : 703. From any relation proved between the face, and diedral, angles of a polyedral angle, we may infer the same relation hetween the sides and angles of a spherical polygon. And conversely: From any relation proved hetween the sides and angles of a spherical polygon, we may infer the same relation hetiveen the face, and diedral, angles of a polyedral angle. Therefore: 704. Each side of a spherical triangle is less than the sum of the other ttvo sides. (565) 705. The sum of the sides of a spherical polygon is less than a circumference. (566) 706. Tiuo mutually equilateral triangles on the same, or on equal, spheres, are mutually equiangular, and are either equal or symmetrical, (567) BOOK VIIL—SPUERICAL TRIANGLES, 345 Symmeteical Spherical Triangles. 707. Symmetrical Spherical Tria7igles are those in which the sides and angles of the one are equal respectively to the sides and angles of the other, but arranged in the reverse order. Thus, the spherical triangles ABO and A'B'C are symmetrical when the vertices of the one are at the ends of the diameters from the vertices of the other. * The corresponding triedral angles 0-ABO and O-A'B'C are also sym- metrical. (567) In the same way, we may form two symmetrical polygons of any number of sides. Two symmetrical triangles are mutually equilateral and equiangular; yet in general they cannot be made to coin- cide. Thus, if in the symmetrical triangles ABO, A'B'O', AB is made to coincide with A'B' to bring the vertex upon the corresponding vertex 0', the two convex surfaces would have to be brought together. The triangles are in fact right-handed and left-handed, and, though corre- sponding and equal in every detail, can no more be con- ceived as superposed on one another so as to occupy the same space, ^han the form of the right hand on that of the left hand. • Antipodal. 346 SOLID GEOMETRY, Proposition 8. Theorem. 708. Tioo symmetrical spherical triangles are equivalent. Hyp, Let ABC, A'B'C be two sym- metrical spherical As with their ho- mologous vertices diametrically oppo- site each other. (707) To prove area ABC = area A'B'C. Case I. When the triangles are isos- celes. Proof. Let BA = BC, and B'A' = B'C. If Z B be placed on the equal Z B' (702), the convexities of the sphere being on the same side, the side BA will fall on B'C, and BC on B'A'. And since BA = B'C, and BC = B'A', . • . A will fall on C, and C on A'. . • . the two A s coincide throughout and are identically equal. Case II. When the triangles are not isosceles. Proof. Let P and P' be the poles of the small Os pass- ing through the pts. A, B, C, and A', B', C, respectively.* Draw the great arcs PA, PB, PC, and P'A', P'B', P'C. Then PA = PB = PC. (674) Similarly, P'A' = P'B' = P'C. Since two symmetric As are mutually equilateral, (707) .-. P'A' =z PA, P'B' = PB, P'C = PC. .-.the AS PAC and P'A'C, PCB and P'C'B', PBA and P'B'A' are respectively isosceles symmetric A s. .• . they are identically equal. (Case I) Because sum of As PAC, PCB, PBA = area ABC; and sum of As P'A'C, P'C'B', P'B'A' = area A'B'C, . • . area ABC = area A'B'C, q.e.d. * The circle which passes through the three points A, B, C can only be a small circle of tlie sphere; for if it were a great circle, the three sides AB, BC, CA, would lie in one plaue, and the triangle ABC would be reduced to one of its sides. BOOK VIIL- SPBEHICAL TRIANGLES. 347 Proposition 9. Theorem.* 709. Two triangles on the smne, or on equal spheres, having two sides and the included angle of one equal re- spectively to tioo sides and the included angle of the other, are either equal or equivalent. Hyp. Let ABO, DEF be two A s having the side AB = DE, the side AC = DF, and the ZA= ZD. Case I. When the given parts of the two AS are arranged in the same order. B To prove A ABC = A DEF. Proof. The A ABC may be placed on the A DEF, as in the corresponding case of plane A s, and will coincide with it. (104) Case II. When the given parts are arranged in inverse order, as in A s ABC, DEF'. To prove as ABC and DEF' equivalent. Proof. Let the A DEF be symmetrical with the A DEF', having its sides and Z s equal respectively to those of DEF'. Then in the A s ABC, DEF, we have AB =^ DE, AC = DF, z A = zD, and the parts arranged in the same order. . • . A ABC = A DEF. (Case I) But A DEF is equivalent to A DEF'. (708) . • . A ABC is equivalent to A DEF', Q.E.D. 348 SOLID GEOMETRY. Proposition lO. Theorem.* 710. Two triangles on the same, or on equal spheres, having a side and the two adjacent angles of one equal re- spectively to a side and the two adjacent angles of the other, are either equal or equivalent. Proof, One of the A s, or the A symmetric with it, may be applied to the other, as in the corresponding case of plane As. (105) Q.E.D. Proposition 1 1 . Theorem.* 711. Tivo mittually equilateral triangles, on the same, or on equal spheres, are either equal or equivalent. Proof, They are mutually ^equiangular, and equal or symmetrical. (706) . • . they are either equal or equivalent. (708) Q.E.D. 712. Cor. 1. In an isosceles spherical triangle, the angles opposite the equal sides are eqiiah A For, in the A ABO, let AB = AC ; pass the arc AD of a great O through A and / the mid. pt. of BC ; then the A s ABD / and ACD are mutually equilateral, and / . • . mutually equiangular. (706) b^ — 713. CoR. 2. Tlie arc of a great circle drawn from the vertex of an isosceles spherical triangle to the middle of the base is perptendicular to the base, and bisects the vertical angle. BOOK VIIL— POLAR TRIANGLES. 349 Polar Triangles. - . 714. One spherical triangle is called the polar triangla. of a second spherical triangle when the sides of the first triangle have their poles at the vertices of the second. Thus, if A, B, C are the poles of the arcs of the great circles B'C, C'A', A'B', respectively, then A'B'C is the polar triangle of ABC. The great circles, of which B'C, '^^^^^C^^^C C'A', A'B' are arcs, will form three 0^ other spherical triangles on the hemisphere. But the one which is the polar of ABC is that whose vertex A', homolo- gous to A, lies on the same side of BC as the vertex A ; and in the same way with the other vertices. Proposition 12. Theorem. 715. If the first of ttvo spherical triangles is the polar triangle of the seco7id, then the second is the polar triangle of the first » A' Hyp. Let A'B'C be the polar A of //^ ABC. To prove that ABC is the polar A of b'^^>^ A'B'C. Proof. Because B is the pole of A'C, . • . BA' is a quadrant. Because C is the pole of A'B', . • . CA' is a quadrant. . • . A' is the pole of BC. (677) In like manner, B' is the pole of AC, and C the pole of AB. Also, A and A' are on the same side of B'C, and so of the other vertices. . • . ABC is the polar A of A'B'C. q.e.d. 350 SOLID GEOMETRY. Proposition 13. Theorem. 716. In tico polar triangles, each angle of one is the sv})- plement of the side opposite to it in the other. Hyp, Let ABC, A'B'C be a pair of polar As in which A, B, 0, and h! , B', C denote the Z s, and a, l, c, and a', l', c' denote the sides. To prove A = 180° -a', B = 180° - h', C = 180° - c', A' = 180° - a, B' = 180° - h, C = 180° - c. Proof. Produce the sides AB, AC, until they meet B'C at D and H. Because A is the pole of B'C, AD and AH are quad- rants. (676) .-. ZAniarcDH. (698) Because B' is the pole of AH, arc B'H = 90° Because C is the pole of AD, arc CD = 90°. . • . arc B'H + arc CD = 180°. But arc B'H + arc CD = arc B'C + arc DH. . • . arc B'C + arc DH = 180°. But arc B'C = a', (Hyp.) and arc DH = Z A. (Proved above) .•.A+«' = 180°, or A = 180° - a\ In the same way all the other relations may be proved. Q.E.D. 717. ScH. Two polar triangles are also called stipple- mental triangles. BOOK VIIL-SPIIERICAL TRIANGLES. 351 Proposition 14. Theorem. 718. The sum of the angles of a spherical triangle is greater than two, and less than six, right angles. Hyp, Let A, B, C denote the three Z s of the spherical A ABC. . To prove A + B + > 180° and < 540°. Proof. Let a\ h' . c' denote the opposite sides respectively of the _ polar A A'B'C Then "^ 'G' A = 180° - a', B = 180° -V,Q> = 180° - c\ (716) Adding, A + B + = 540° - (a' + &' + o% Because a* , V , c' are sides of a spherical A , .•.rt' + 6' + ^' <360°. (705) .-. A+ B-h C > 180°. Also, because each Z of the A is < 2 rt. Z s, .• . A + B + C < 6 rt. Z s or < 540° q.e.d. 719. Cor. A spherical triangle may have tioo, or even three, right angles ; also two, or even three, obtuse angles. 720. If a spherical triangle has two right angles, it is called a M-rectangular triangle ; and if a spherical triangle has three right angles, it is called a tri-rectangular tri- angle. * EXERCISE. In a bi-rectangular triangle the sides opposite the right angles arc quadrants. 352 SOLID GEOMETRY. Proposition 15. Theorem.* 721. Ttuo 7nutvally equiangular triangles, on the same, or on equal spheres, are mutually equilateral, and are either equal or equivalent. Hyp. Let the spherical A s P and Q be mutually equiangular. To prove that A s P and Q are mutually equilateral, and either equal or equivalent. Proof, Let P' be the polar A of P, and Q' the polar A of Q. Since P and Q are mutually equiangular, (IlyP-) .• . their polar A s P' and Q' are mutually equilat- eral. (716) And since P' and Q' are mutually equilateral, .• . they are mutually equiangular. (706) But since P' and Q' are mutually equiangular, .* . As P and Q are mutually equilateral. (716) .'. As P and Q are either equal or equivalent. (711) Q.E.D. Note.— Mutually equiangular spherical triangles are mutually equilateral, only when the triangles are on the same, or on equal, spheres. When the spheres are unequal, the homologous sides of the triangles are no longer equal, but are pro- portional to the radii of their spheres (433) ; the triangles are then similar, as in the case of plane triangles. 722. CoK. 1. If tivo angles of a spherical triangle are equal, the triangle is isosceles. BOOK VTIL— SPHERICAL TRIANGLES. 353 For, since the polar A is isosceles (716), the Zs opp. the two equal sides are equal (712); .-. the given A is isosceles. (716) 723. Cor. 2. If three planes are passed through the centre of a sphere, each perpendicular to the other two, they divide the surface of the sphere into eight equal tri-rect angular triangles. (702) and (721) Proposition 1 6. Theorem. 724. hi a spherical tria?igle, the greater side is opposite the greater angle, and conversely, (1) Hyp, Let ABC be a A having ZACB>ZB. To prove AB > AC. Proof, Draw CD, an arc of a great O, making Z BCD = Z B. Then DB = DC. (722) Add AD to each. Then AD + DB = AD + DC. But AD + DC > AC. . •. AD + DB > AC, or AB > AC. (2) Hyp. Suppose AB > AC. To prove ZACB > ZB. Proof. If Z ACB =: Z B, then AB = AC, (722) which is contrary to the hypothesis. And if Z ACB < Z B, then AB < AC, (Proved above) which is also contrary to the hypothesis. .-. ZACB > ZB. Q.E.D. (704) S64: SOLID GEOMETRY. Relative Areas of Spherical Figures, definitions. 725. A lune is a portion of the sur- face of a sphere included between two semi-circumferences of great circles ; as ACBD. The a7igl6 of a lune is the angle be- tween the semi-circumferences which form its sides; as the angle CAD, or the angle COD. (699) 726. On the same, or on equal, spheres, lunes of equal angles are equal, as they are evidently superposable. 727. A spherical tvedge, or ting u la, is the part of a sphere bounded by a lune and the planes of its sides; as AOBCD. The diameter AB is called the edge of the ungula, and the lune ACBD is called its base, 728. The spJierical excess of a spherical triangle is the excess of the sum of its three angles over a straight angle. (718) Thus, if the angles of a spherical triangle are denoted by A, B, C, and its spherical excess by E, we have E = A + B + C- 1st. Z. The spherical excess of a spherical polygon is the excess of the sum of its angles over as many straight angles as it has sides, less two. EXERCISES. 1. Each side of a tri-rectangular spherical triangle is a quadrant. (720) 2. If the three angles of a spherical triangle are 100°, 120°, 95°, find its spherical excess. BOOK VIII.— AREAS OF SPHERIGAL FIGURES. 355 3. Given a spherical triangle whose sides are 80°, 100°, 115°; find the angles of its polar triangle. 4. Given a spherical triangle whose angles are 65°, 87°, 115°; find the sides of its polar triangle. 5. Show that the sum of the angles of a spherical pen- tagon is greater than six, and less than ten, right angles. Proposition 1 7. Theorem. 729. If t too arcs of great circles intersect on the surface of a hemisphere, the sum of the ttco opposite triangles thus formed is equivalent to a lune whose angle is equal to the angle letiveen the given arcs. Hyp. Let the arcs ACA', BOB' >^^^^>>g intersect on the surface of the hemi- /^ /^ / jX sphere CABA'B'. / ,_^-/:L/b' \ To prove A ABO + aA'B'O ti^^^^^ equivalent to the lune CAC'B. V BT" / / j Proof. Because the As A'B'C n1/' / y and ABC are symmetrical, (707) C'^'^^^ .• . area A'B'C = area ABC. (708) Add area ABC to each. .• . area ABC + area A'B'C = area ABC + area ABC = area of lune CAC'B. Q.E.D. 730. ScH. It is evident that the two spherical pyra- mids, which have the triangles ABC, A'B'C for bases, are together equivalent to the spherical wedge whose base is the lune CAC'B. EXERCISE. If the sides of a triangle are 75°, 110°, and 130°, show that the angles of its polar triangle arc respectively 105°, 70°, and 50°. 356 80LID GEOMETRY. Proposition 1 8. Theorem. 731. TJie area of a lune is to the surface of the sphere as the angle of the lune is to four right angles. Hyp. Let ACBD be a lune, and ECDH the great O whose poles are A and B; let L denote the area of the lune, S the surface of the sphere, and A the Z of the lune in degrees. To prove ^ = ^o' . Proof Since the Z of the lune is measured by the arc CD, (698) .-. A : 360° = arc CD : ©ce ECDH. (236) (1) Wlien the arc CD and the Qce ECDH are commen- surablc. Apply the common measure to the Oce ECDH, and let it be contained m times iii CD and n times in ECDH. Then, arc CD : Oce ECDH = m : n. Through the pts. of division of ECDH and the axis AB pass arcs of great Os; they will divide the whole surface of the sphere into n equal lunes (726), of which the lune ACBD will contain fn. . • . L : S = wi ; 7^. .-. L : S == arc CD : ©ce ECDH = A : 360°. BOOK VIII.- AREAS OF SPHERICAL FIGURES. 357 (2) Wien the arc CD and the Qce BCD II are income meyisurahle. The proof in this case may be extended as in (234), (298), and (356). Q.e.d. 732. Cor. 1. Two Junes on the same, or on equal, spheres, are to each other as their angles, 733. Cor. 2. If T denote the area of the tri-rectan- gular A (720), 8T will express the surface of the whole sphere. (723) Then, if the rt. Z. he taken as the unit, we have L : 8T = A : 4. (731) . • . L = 2A X T. Hence, if the right angle he taken for the unit of angles, the area of a lune is equal to tioice its angle mutiylied hy the area of the tri-rect angular triangle. 734. Cor. 3. The volume of a spherical wedge is to the volume of the sphere as the angle of the lune is to four right angles. For, the lunes being equal, the spherical wedges are also equal (727) ; hence two spherical wedges are to each other as the angles included between their planes. 735. Cor. 4. If the right angle he taken as the unit of angles, the tri-rect angular triangle as the unit of surfaces, and the tri-rectangular p>ym't^id as the unit of voluines, then the area of a lune, and the volume of an singula, are each expressed hy tiuice its angle. (733) and (734) EXERCISES. 1. What part of the surface of a sphere is a lune whose angle is 60° ? 90° ? 120° ? 2. What part of the volume of a sphere is an ungula whose angle is 50° ? 90° ? 100° ? 358 80L1JJ GEOMETRY. Proposition 1 9. Tiieorem. 736. The area of a spherical triangle is equal to its spherical excess. Hyp. Let A, B, denote the numerical measures of the / s of the .. spherical A ABC, the rt. Z being / the unit of / s, and the tri-rectangu- i lar A the unit of areas. ^'V — -. 7(9;j7-oyeareaABC=A4-B+C— 2. Proof. Continue any one side as AB so as to complete the great O ABA'B'. Continue the other two sides AC and BC till they meet this O in A' and B'. Then, area ABC+area A'BC^lune ABA'C=2A, and area ABC+area AB'C=lune AB'CB=2B. (735) Also, since the A s ABC and A'B'C are together equiv- alent to a lune whose angle is C, (729) .-. area ABC + area A'B'C = 2C. But the sum of the areas of the As ABC, A'BC, AB'C, A'B'C is equal to the area of the hemisphere, or 4. . • . adding these three equations, we have 2 area ABC + 4 = 2A + 2B + 2C. .• . area ABC = A + B + C-2. q.e.d. 737. Cor. 1. By a method similar to that in (736), in connection with (730) and (735), it may be p|-oved that The volume of a triangular spherical pyramid is equal to the sp)herical excess of its base {the volume of the tri- rectangular i^yramid being the unit of volume). BOOK VIII.-AREA OF SPHERICAL POLYGON. 359 738. Cor. 2. If the three vertices of a triangle are on a great circle, its three sides must coincide with that circle, and each angle must equal 180°. The area of the tri- angle is then a hemisphere, and its spherical excess 4 rt. / s 01 360°. Therefore the area of the surface of the whole sphere is 720°. Hence The area of a spherical triangle is to that of the surface of the sphere as its spherical excess, in degrees, is to 720°. i""^ Proposition 20. Theorem. 739. The area of a si)herical polygon is equal to its spherical excess. Hyp. Let K denote the area, n the . 3- — ^.^^ number of sides, and S the sum of the y/^"'"""^^^ / s of the spherical polygon ABODE, a-^^T ^^-^ ] the rt. Z and the tri-rectangular A ^^\^ Jc being the units of Zs and areas re- ^^^/ spectively. ^ To 2)rove K = S-2(n — 2). Proof From any vertex, as A, draw diagonals, dividing the polygon into {n — 2) A s. Then, since the area of each A = the sum of its Z s minus 2 rt. Z s(736) ; and since the sum of the Z s of the {71 — 2) As = the sum of the Z s of the polygon, or S, .-. K = S-2(7i-2). Q.E.D. Note.— In the last three propositions, only the ratios of the areas are ex- pressed. If the absolute area is required, the area of the surface of the sphere must be known. 360 SOLID GEOMETRY. exercises. Theorems. 1. The intersection of the surfaces of two spheres is a circle whose plane is at right angles to the line joining the centres of the spheres, and whose centre is in that line. 2. If lines be drawn from any point of the surface of a sphere to the ends of a diameter, they will form with each other a right angle. 3. If any number of lines in space pass through a point, the feet of the perpendiculars upon these lines from another point lie upon the surface of a sphere, i u t^nvvo-^^M 4. If two straight lines are tangent to a sphere at the same point, the plane of these lines is tangent to the sphere. 5. On spheres of different radii, mutually equiangular triangles are similar. See (721), note. 6. The sum of the two arcs of great circles drawn from the extremities of one side of a spherical triangle to a point within it, is less than the sum of the other two sides. 7. If from any point on the surface of a sphere two arcs of great circles are drawn perpendicular to a circumference, the shorter of the two arcs is the shortest arc that can be drawn from the given point to the circumference. 8. Any lune is to a tri- rectangular triangle as its angle is to half a right angle. 9. Spherical polygons are to each other as their spherical excesses. 10. Two oblique arcs drawn from the same point to points of the circumference equally distant from the foot of the perpendicular are equal. 11. Of two oblique arcs, the one which meets the cir- cumference at the greater distance from the foot of the perpendicular is the longer. BOOK VIII.—NUMEKICAL EXEUC18ES. 361 12. The arc. of a great circle, tangent to a small circle, is perpendicular to the radius of the sphere at the point of contact. 13. If three spheres intersect one another, their planes of intersection intersect in a right line pei-pendicular to the plane containing the centres of the spheres. 14. Prove that this line is the locus of points from which tangent lines to the throe splieres are equal. 15. Through a fixed point, within or without a sphere, three lines mutually at right angles intersect the sphere ; I prove that the sum of the squares of the three chords is con- stant, depending only on the radius of the sphere and the distance of the point from the centre. IG. Prove also that the sum of the squares of the six seg- ments is constant. 17. Prove that the area of a spherical triangle, each of whose angles is | of a right angle, is equal to the surface of a great circle. 18. If through a point any secant OPP' is drawn to cut a sphere in P, P', prove that OP. OP' is constant. 19. Find the radii of the spheres inscribed and circum- scribed to a regular tetraedron. Numerical Exercises. 20. If the sides of a spherical triangle are respectively 65°, 112°, and 85°, how many degrees are there in each angle of its polar triangle ? 21. If the angles of a spherical triangle are respectively 90°, 115°, and 70°, how many degrees are there in each side of its polar triangle ? 22. Given the spherical triangle whose sides are respec- tively 80°, 90°, and 140°, to find the angles of its polar triangle. 23. What part of the surface of a sphere is a lune whose angle is 45°? 54°? 80°? . S6:^ SOLID OEOMETRT. 24. What part of the volume of a sphere is an uiigula whose angle is 72°? 36"? 25. If the angle of a lune is 50°, find its area on a sphere whose surface is 72 square inches. Ans. 10 square inches. 26. Find the area of a spherical triangle whose angles are respectively 75°, 100°, and 115°, on a sphere whose sur- face is 72 square inches. Ans, 11 square inches. 27. Fmd the area of a spherical triangle each of whose angles is 70°, on a sphere whose surface is 144 square inches, Ans. 6 square inches. 28. Find the area of a spherical triangle whose angles are G0°, 90°, and 120°, on a sphere whose surface is 64 square inches. Aus. 8 square inches. 29. Find the area of a splierical pol\^gon of six sides each of whose angles is 150°, on a sphere whose surface is 100 square inches. Ans. 25 square inches. 30. Find the area of a bi-rectangular triangle whose vertical angle is 108°, on a sphere wliose surface is 100 square inches. A7is. 15 square inches. 31. Find the area of a spherical pentagon whose angles are respectively 138°, 112°, 131°, 168°, and 153°, on a sphere whose surface is 40 square feet. Ans. 9 square feet. 32. Find the area of a spherical triangle whose angles are 61°, 109°, and 127°, on a sphere whose surface is 10 square inches. Ans. 1.625 square inches. 33. Find the area of a spherical quadrangle whose angles are 170°, 139°, 126°, and 141°, on a sphere whose surface is 400 square inches. 34. Find the area of a spherical pentagon whose angles are 122°, 128°, 131°, 160°, and 161°, on a sphere whose surface is 150 square feet. 35. Find the angles of an equilateral spherical triangle whose area is equal to that of a great circle. Ans. 120°. 36. Find the angles of an equilateral spherical triangle whose area is equal to that of an equilateral spherical hex- agon, each of whose angles is 150°. Ans. 80°. BOOK VIII. -NUMERICAL EXERCISES, 363 37. Find the volume of a triangular spherical pyramid, the angles of whose bases are 80"", 90°, 130°; the volume of the sphere being 30 cubic inches. Ans. 5 cubic inches. 38. Find the volume of a quadrangular spherical pyra- mid, the angles of whose bases are 170°, 139°, 126°, 141°; the volume of the sphere being 10 cubic inches. Ans. 3 cubic inches. 39. Find the ratio of the areas of two spherical triangles on the same sphere, the angles being 60°, 84°, 129°, and 83°, 107°, 114° respectively. 40. Find the circumference of a small circle of a sphere whose diameter is 10 inches, the plane of the circle being 3 inches from the centre of the sphere. Ans, 25.133 inches. 41. Find the circumference of a small circle of a sphere whose diameter is 20 inches, the plane of the circle being 4 inches from the centre of the sphere. 42. Find the area of the circle of intersection of two spheres, their radii being 4 and 6 inches and the distance between their centres 5 inches. 43. Find the area of a small circle of a sphere whose di- ameter is 5 inches, the plane of the circle being 1 inch from the centre of the sphere. 44. The radii of two concentric spheres are 4 and 6 inches, a plane is drawn tangent to the interior sphere: find the area of the section made in the outer sphere. 45. Given two mutually equiangular triangles on spheres whose radii are 10 and 16 inches: find the ratio of two homologous sides of these triangles. See (721), note. ** Problems. 46. To pass a plane tangent to a sphere at a given point on the surface of the sphere. B64 SOLID GEOMETRY, 47. To pass a plane tangent to a sphere through a given straight line without the sphere. 48. To construct on the spherical blackhoard a spherical angle of 45°; 60°; 90°; 100°; 200°. From P, the pt. where the veilex is to be placed, with a quadrant describe an arc, which will represent one side of the angle required. From P as a pole, with a quadrant describe an arc from the side before drawn, to measure the re- quired angle. Lay off on this last arc from the first arc the measure of the re- quired angle; and through the extremity of this arc and P pass a great circle. For describing the arcs, the student can use a tape equal in length to half a great circle of the sphere, marked off into 180 equal parts. 49. To construct on the spherical blackboard a spherical triangle, having two sides 100° and 80°, and the included angle 58°. 50. To construct, as above, a spherical triangle, having aside 75°, and the adjacent angles 110° and 87°. 51. To construct, as above, a spherical triangle, having its sides 150°, 100°, 80°; also having its sides 50°, 85°, 160°. 52. To construct, as above, a spherical triangle, having two sides 120° and 88°, and the included angle 59°; then construct its polar triangle. 53. Given two points on the surface of a sphere, to de- scribe the great circle passing through them. 54. To bisect a given arc, or a given angle, on a sphere. 55. To draw an arc of a great circle perpendicular to a spherical arc, from a given point without it. 56. To erect a perpendicular to a given arc of a great circle from a given point in the arc. 57. Given three points on a sphere, to describe a small circle to pass through them. 58. To cut a given sphere by a plane passing through a given straight line so that the section shall have a given radius. 59. At a given point in a great circle, to draw an arc of a great circle making a given angle with the first. 60. Through a given point on a sphere, to draw a great circle tangent to a given small circle. BOOK VIIL-EXERCISES. PROBLEMS. 365 61. Through a given point on a sphere, to draw a great circle tangent to two given small circles. 62. To inscribe a circle in a given spherical triangle, and to circumscribe a circle about the triangle. 63. To construct a right spherical triangle, having (1) a side about the right angle and the hypotenuse; and (2) an angle and the opposite side. 64. To construct a spherical triangle, having given (1) the three sides; and (2) two sides and the included angle. 65. To describe a sphere to cut orthogonally two given spheres. Book IX.* THE THEEE BOUND BODIES. 740. The only solids bounded by curved surfaces, that are treated of in Elementary Geometry, are the cylinder, the cone, and the sphere, which are called the three round bodies. The Cylii^der. definitions. 741. A cylindrical surface is a surface generated by the motion of a straight line AB, called the generatrix, which constantly touches a given curve ACDE, called the directrix, and remains paralle. to its original position. The different positions of the generatrix are called eJe- ments of the surface. 742. A cylinder is a solid bounded by a cylindrical surface and two parallel planes. The cylindrical surface is called the lateral surface, and the plane sur- faces are called the bases. The altitude of a cylinder is the perpendicular distance between its bases. 743. A right section of a cylinder is the section by a plane perpendicular to its elements. 744. A circular cylinder is a cylinder whose base is a circle. The axis of a circular cylinder is the straight line joining the centres of its bases. 745. A right cylinder is one whose elements are perpen- dicular to its bases. * This book treats of the properties and relations of the cylinder, tlie cone, and the sphere, and shows how to find the convex surface and volume of each of these bodies. 360 BOOK IX.^TEE CYLINDEB. 367 ^' 14:6. A right circular cylinder, called also a cylinder of revolution, is generated by re- volving a rectangle about one of its sides. 747. Similar cylinders of revolution are those generated by similar rectangles revolv- ing round homologous sides. 748. A.fa7igciit plane to a cylinder is a plane which con- tains an element of the cylinder without cutting the sur- face. The element which the plane contains is called the element of contact. Any straight line in a tangent plane, which cuts the ele- ment of contact, is a tangent line to the cylinder. 749. A prism is inscribed in a cylinder, when its bases are inscribed in the bases of the cylinder and its lat- eral edges are elements of the cylinder. 750. A prism is circumscribed about a cylinder, when its bases are circumscribed about the bases of the cylinder. Proposition 1 . Theorem. 751. Every section of a cylinder made by a plane pasS" ing through an element is a parallelogram,. Hyp, Let the plane ABCD pass through the element AB of cylinder EH. To prove the section ABCD a OJ. Proof A plane passing through the element AB cuts the Oce of the base in a second pt. D. Through D draw DC || to AB. Then DC is in the plane BAD. . • . DC is an element of the cylinder. . • . DC, being common to the plane and the lateral sur- face of the cylinder, is their intersection. Also, AD is II to BC. (519) . •. ABCD is a co, (124) Q.E.D. 752. Coil. Every sectio?i of a right cylinder made by a plane passing through an element is a rectangle. (68) (741) 368 SOLID GEOMETRY. Proposition 2. Theorem, 753. The lateral area of a cylinder is equal to the j^e- rivieter of a right section of a cylinder omiltiplied hy an element. Hyp. Let S denote the lateral area, P the perimeter of a rt. section, and E an element of the cylinder AC. To prove S = P X E. Proof. Inscribe in the cylinder a prism ABCD-C, of any number of faces ; and let s denote its lateral area and p the perimeter of its rt. section. Then, since each lateral edge is an element of the cylinder. s = p X^. (749) (591) Kow let the number of lateral faces of the inscribed prism be indefinitely increased. The perimeter of the right section of the prism will ap- proach the perimeter of the right section of the cylinder as its limit. ^ (430) . • . the lateral area of the prism will approach the lateral area of the cylinder as its limit. Because, however great the number of the lateral faces, s=pXl^, and because p approaches P as its limit and s approaches S as its limit, ... S = P XE. Q.E.D, 754. Cor. 1. The lateral area of a cylinder of revolution is equal to the circumference of its hase multiplied hy its altitude. (43G) (439) BOOK IX.— LATERAL AREA OF A CYLINDER. 369 755. Cor. 2. If H denote the altitude of a cylinder of revolution, R the radius of the base, S the lateral area, and T the total area, we have S = 2;rR X H, T = 2;rRH + 2;rR' = 27rR(R + II). 756. Co.R. 3. Let S, S' denote the lateral areas ; T, T' the total areas; R,R' the radii of the bases; and H, H' the altitudes of two similar cylinders of revolution. Then, since the generating rect- angles are similar, (74?) ^ H •'• R' R + H H'~R' + H'' 2;rRH 27rR'H' R' ^H' and -7?^ = R(R + H) R'(R' + H') R' R + H R' + H' R"' TT8 xyi JjM "P'a- ('^^) Therefore, the lateral areas, or the total areas, of two similar cylinders of revolution are to each other as the squares of their altitudes, or as the squares of the radii of their bases. EXERCISES. 1. Required the lateral area, and also the total area, of a cylinder of revolution whose altitude is 25 inches and the diameter of whose base is 20 inches. 2. Required the convex surface of a right circular cylinder whose altitude is 16 inches, and diameter of the base 8 inches. 3. Requii'ed the altitude and radius of the base of a right circular cylinder whose lateral area is J as great as a similar cylinder of which the altitude is 20 feet and diameter of the base 8 feet. 370 SOLID GEOMETRY. (612) inscribed Proposition 3. Theorem. 757. The volume of a cylinder is equal to the product of its base hy its altitude. Hyp. Let V, B, H denote the volume of the cylinder, the area of its base, and its altitude, respectively. To prove V = B X H. 'Proof Inscribe in the cylinder a prisni!5 and let V and B' denote its volume and the area of its base. Then, since the altitude of the prism is H, .-. V' = B' X H. Now let the number of lateral faces of the prism be indefinitely increased. The base of the prism B' will approach the base of the cylinder B as its limit, and the volume of the prism V will approach the volume of the cylinder V as its limit. Because, however great the number of the lateral faces, we always have V = B' X H. . • . V = B X H. Q.E.D. 758. Cor. 1. If R denotes the radius of the base of a cylinder of revolution, then B =■ nW, (439) .•.V = ;rR'.H. (757) 759. Cor. 2. Let V, V denote the volumes, R, R' the radii of the bases, and H, H' the altitudes of two similar cylinders of revolution. Then, since the generating rectangles are similar, (747) R _ JI •'• R'~ H'' ^ _ R'H _ R^ H _ IT ^ R" . •*• V ~ R'^H' ~ R''' ^ W ~ W ~ W ^^ ^ Therefore, the volumes of siynilar cylinders of revohition are to each other as the cubes of their altitudes , or as the cubes of their radii. BOOK IX.— TUB GONE. 371 The Cone. DEFINITIONS. 760. A conical surface is a surface generated by the h motion of a straight line SA, called the generatrix, which constantly touches a given curve ABCD, called the directrix^ and always passes through a fixed point S, called the vertex. The different positions of the generatrix are called elements of the surface. If the generatrix extends on both sides of the vertex, the whole surface consists of two portions, S-ABCD and S-ahcd, lying on opposite sides of the vertex, which are called the loiver and tipjjer napjMs, respectively. 761. A cone is a solid bounded by a conical surface and a plane cutting all its elements. The conical surface is called the lateral surface. The base of a cone is its plane surface. The altitude of a cone is the perpendicular distance from the vertex to the base. 76^. A circular cone is a cone whose base is a circle. The axis of a circular cone is the straight line joining the vertex and the centre of its base. 763. A right cone is one whose axis is perpendicular to its base. The cones priucipally treated of in elementary geometry are right cones. 764. A right circular cone is a circular cone whose axis is perpendicular to its base. It is also called a cone of revolution, because it may be generated by the revokition of a right tri- angle SO A, about one of its perpendicular sides SO, as an axis: the hypotenuse SA generates the lateral surface, and OA generates the base. All the elements SA, 372 SOLID GEOMETRY. SB, etc., are equal; and any one is called the slant height of the cone. 765. Similar cones of revolution are those generated by similar right triangles revolving round homologous per- pendicular sides. 766. A tangent i^lane to a cone is a plane which contains an element of the cone without cutting the surface. The element which the plane contains is called the element of contact. Any straight line in a tangent plane, which cuts the element of contact, is a tangent line to the cone. 767. A frustum of a cone is the portion of a cone in- cluded between its base and a plane parallel to the base and cutting all the elements. If the cutting plane is not parallel to the base, the portion included between the base and plane is called a truncated cone. 768. The altitude of a frustum of a cone is the perpendicular distance between its bases. The slant height of the frustum of a cone of revolution is the portion of an element included between the parallel bases of the frustum. 769. A pyramid is inscribed m a cone when its base is inscribed in the base of the cone, and its lateral edges are elements of the cone. Any plane which cuts the cone, determines a truncated pyramid inscribed in the truncated cone. 770. A pyramid is circumscribed about a cone when its, base is circumscribed about the base of the cone, and its vertex coincides with the vertex of the cone. Any plane which cuts the cone, determines a truncated pyramid circumscribed about the truncated cone. BOOK IX.— THE CONE. 373 Proposition 4. Theorem. 771. Every section of a circular- cone made hy a plane parallel to the base is a circle. Hyp. Let the section ahc of the circu- lar cone S-ABC be || to the base. To prove that ahc is a O. Proof, Let SO be the axis of the cone, cutting the plane of the 1| section in the pt. 0, Through SO and any elements SxA., SB, pass planes cutting the base in the radii OA, OB, and the || section in the st. lines oa, oh. Then, since the planes ahc and ABC are"[| , to OA, and oh is || to OB. So , oh So oa IS (Hyp.) (519) oa OA ~ SO ' and OB ~ SO • But oa __ oh OA~OB* OA = OB. oa = oh. . (Eadii) Hence, all the st. lines drawn from the pt. o to the pe- rimeter of the section ahc are equal. .*. the section is a O. q.e.d. 772. TJie axis of a circular cone passes through the cen- tres of all the sections parallel to the base, EXERCISES. 1. Prove that every section of a cone made by a plane passing tlirough its vertex is a 'triangle. 2. Eequired the volume of a circular cylinder whose alti- tude is 25 inches and diameter of the base 20 inches. 374 SOLID GEOMETRY. Proposition 5. Theorem. 773. The lateral area of a cone of revolution is equal to the product of the circumference of its base hy half its slant height. Hyp. Let S, 0, L denote the lateral area of the cone, the Oce of its base, and its slant height, respectively. To prove S = X |L. Proof Inscribe in the cone a pyramid S-ABOD, having a regular base of any number of sides ; and let S' denote its lateral area, 0' the perimeter of its base, and L' its slant height. Then, since the edges of the pyramid are elQ«ftents of the cone (769), the pyramid is regular. .-. S' = C' X iL'. (628) Now let the number of lateral faces of the inscribed pyramid be indefinitely increased. C will approach C as its limit. - (430) . • . L' and S' will appi-oach L and S respectively, as their limits. Because, however great the number of lateral faces. S' = C X iL', . S = C X iL. Q.E.D. 774. Cor. \. If R is the radius of the base of a cone of revolution, and T is the total area, we have S = 27rR X IL = ttRL. (436) T r:= ;rRL + nW = nlX {h -\- R). BOOK IX.-FUWTUM OF A CONE. 375 775. CoK. 2. By the process that was employed in (756) we may show that the lateral areas, or the total areas, of two similar cones of revolution are to each other as the squares of their radii, or of their slant heights, or of their altitudes. Proposition 6. Tiieorem. 776. The lateral area of a frustum of a cone of revolu- tion is equal to half the sum of the circumferences of its bases multiplied by its slant height. Hyp. Let S, C, c, L denote the lateral area of the frustum, the Oces of its bases, and its slant height, respectively. To prove S = ^(C + c)L. Proof. Inscribe in the frustum of the cone the frustum of a regular pyramid ; and let S' denote its lateral area, C and c' the perimeters of its lower and upper bases, respectively, and L' its slant height. Then, S' = i(C' + c')^. (629) Now let the number of lateral faces of the inscribed frus- tum be indefinitely increased. C and c' will approach and c respectively, as their limits. (430) . • . L' and S' will approach L and S respectively, as their limits, .-. S=:i(C + c)L. Q.E.D. 777. Cor. TJie lateral area of a frustum of a cone of revolution is'^qual to the circumference of a section eqiddis- tant from its bases* 7nultiplied by its slafit height. * Called the mid-section. 376 SOLID GEOMETRY. Proposition 7. Theorem. 778. The volume of any cm e is eqital to one-third the 2)rodiict of its hase and altitude. Hyp. Let V, B, H denote the volume of the cone, the area of its base, and its alti- tude, respectively. To prove V = JB X H. Proof. Inscribe in the cone a pyramid, and let V and B' denote its volume and the area of its base. Then, V = JB' X H. (632) Now let the number of lateral faces of the pyramid be in- definitely increased. B' will approach B as its limit. (430) . • . V will approach V as its limit. .-.V^JBxH. Q.E.D. 779. Cor. 1. For the volume of a cone of revolution^ whose altitude is H and radius of the hase is R, we have •V-4;rR='H. (439) 780. Cor. 2. The volumes of similar cones of revolution are to each other as the cubes of their altitudes, or as the C2ibes of the radii of their bases. ('^59) EXERCISES. 1. Required the lateral area and volume of a right circu- lar cone whose altitude is 24 inches and radius of the base 10 inches. A?is. 81G.82 sq. ins.; 2513.3 cu. iiisT 2. Required the entire surface of a right circular cone whose altitude is IG inches and radius of the base 12 inches. Ans, 1206.3? sq. ins. BOOK IX.— VOLUME OF A FRUSTUM. 377 Proposition 8. Theorem. 781. The volume of a frustum of any cone is equal to the simi of the volumes of three cones, lohose common altitude is the altitude of the frustum, and luhose bases are the lower base, the upper base, and a mean proportional between the bases of the frustum, Hijp. Let V, B, b, H denote the volume i^^^^^T^ of the frustum, its bases, and its altitude, /li^ respectively. / // To prove V = ^(B + b + VB^). /cr-T'^X-. Proof Inscribe in the frustum of the ^^^ ^^^^'L ^ cone the frustum of a pyramid, and let V, B', b' denote its volume and the areas of its bases. Then, V = -?(B' + ^' + \^Wb'). (635) o Now let the number of lateral faces of the inscribed frustum be indefinitely increased. B' and V will approach B and b respectively, as their limits. (430) . • . V will approach V as its limit. .•.V=-^(B + &+ I^B^). Q.E.D. t) 782. Cor. 1. If the frustum is that of a right circ7ilar cone, and the radii of its bases are R and r, ice have B = 7tW, b = 7cr\ .-. V = i;rH(R' + r= + Rr). 783. Cor. 2. This formula may be put into the form Note.— The volume of a cask may be found approximately by this formula, in which H = total height of cask, R = radius of mid-section, and r = radius of end. The nearest approximation in the case of most casks Is given by the fornmla V = j7rn[2R2 + ra - i(R« - r")].* * S^ Rouch6 et Comberousse, p. 155. 378 SOLID GEOMETRY. The Sphere. definitions. 784. A zone is a portion of the surface of a sphere in- chided between two parallel planes. The altitude of the zone is the perpendicular distance between the parallel planes. The bases of the zone are the circumferences of the circles which bound the zone. If one of the parallel planes touches the sphere, the zone is called a zone of one base. 785. A spherical segment is a portion of the volume of a sphere included between two parallel planes. The altitude of the segment is the perpendicular distance between the parallel planes. The bases of the segment are the sections of the sphere made by the parallel planes. A segment of one base is a segment one of whose bounding planes touches the sphere. 786. A spherical sector is a portion of the volume of a sphere generated by the revolution of a circular sector about a diameter of the circle. 787. Let the sphere be generated by the revolution of the semicircle ACDEFB about its ^ diameter AB as an axis; and let CG ^y^ and DH be drawn perpendicular to the / axis. The arc CD generates a zone / whose altitude is GH, and the figure I ^^' CDHG generates a spherical segment eV'~7 whose altitude is GH. The circum- p^ ferences generated by the points C and ° D are the bases of the zone, and the circles generated by CG and DH are the bases of the segment. BOOK IX.-THE SPHERE. 379 The arc AC generates a zone of one base, and the figure ACG a spherical segment of one base. The circular sector OEF generates a spherical sector whose base is the zone generated by the arc EF; the other bounding surfaces are the conical surfaces generated by the radii OE and OF. If OF coincides with OB, the spherical sector is bounded by a conical surface and a zone of one base. If OE is perpendicular to OB, the spherical sector is bounded by a plane surface, a conical surface, and a zone. Proposition 9. Tiieorem. 788. Tlie area generated hy a straight line revolving about an axis in its plane, is equal to the product of the projection of the line on the axis by the circtwiference whose radius is the perpendicular erected at the middle 2)oint of the line and terminated by the axis. Hyp, Let AB be the revolving line, G CD its projection on the axis GO, and EO the X at the mid. pt. of AB and termi nating in the axis. To p>rove area AB = CD X 27rEO. Proof. Draw EF_L, and AH |i to GO. The area generated by AB is the lateral area of a frustum of a cone of revolution, whose slant height is AB and axis CD. . • • area AB = AB X 27rEF. (777) The A s ABH and EOF are similar. (316) . •. AB X EF = AH X EO = CD X EO. .-. area AB = CD X 27rEO. If AB meets the axis, or is || to it, thus generating a conical, or a cylindrical surface, the result is the same from (773) and (753). Q.E.i>, C 380 SOLID GEOMETRY, Proposition lO. Theorem. 789. The area of the surface of a sphere is equal to the product of its diameter hy the circumference of a great circle. Hyp. Let the sphere be generated by the A revolution of the semicircle ABDF about ^^^^ the diameter AF, let be the centre^ R the // radius, and denote the surface of the sphere ^/- by s. I To prove S = AF x 2;rR. \ Proof. Inscribe in the semicircle a regular \ semi-polygon ABCDEF, of any number of t'^^ sides. Draw m, Qc, etc., J_ to AF and OH 1 to AB. OH bisects AB. (201) Then, area AB = Ahx 27rOH. (788) Similarly, area BC = be X 2;rOH, and so on. In equal 0« equal cJiorda a/re equally distant from, the centre (206). Now the sum of the projections Kb , be, etc., of all the sides of the semi-polygon make up the diameter AF. . * . the entire surface generated by the revolving semi- polygon = AF X 2;rOH. Now let the number of sides of the inscribed semi-poly- gon be indefinitely increased. The semi-perimeter will approach the semi-circumference as its limit, and OH will approach the radius E as its limit. (430) . • . the surface of revolution will approach the surface of the sphere as its limit. .-. S = AF X 2;rR. ^ q.e.d, BOOK IX.— THE SPHERE. 381 790. CoK. 1. Since AF = 2K, .-. S = 2Rx 27rR = 4;rR'. (789) Therefore, the area of the surface of a sphere is equal to the area of four great circles, 791. Cor. 2. The areas of the surf.cts of two spheres are to each other as the squares of their radiif or as the squares of their diameters. 792. Cor. 3. The area of a zone is equal to the product of its altitude by the circumfereuce of a great circle. For, the area of tlie zone generated by the revolution of the arc BD . = bd X 27rn. 793. Cor. 4. Zones 07i the same sphere, or on equal spheres, are to each other as their altitudes. 794. Cor. 5. Since the arc AB generates a zone of one base whose area is AbX27rn= nkb X AF = ;?AB', (325) therefore, the area of a zone of one base is equal to the area of the circle whose radius is the chord of the zone. 795. Cor. 6. If a cylinder is circumscribed about a sphere, the total area of the cylinder = 6;rR". (755) Therefore, the area of the surface of a sphere is equal to tivO'thirds the total area of the circumscribing cylinder, EXERCISES 1. Find the area of the surface of a sphere whose radius is 4 inches. 2, Prove that two zones on different spheres are to each otlier as the products of their altitudes by the radii of the spheres. 882 80LID GEOMETRY. Proposition 1 1 . Theorem. • 796. Tlie volume of a sphere is equal to the area of its surface multiplied hy one-third of its radius. hyp. Let V denote the volume of a sphere, S tlie area of its sur- face, and R its radius. To prove V = S X JR. Proof. Conceive the whole sur- face of the sphere to be divided into a great number of equal spherical polygons. Form pyra- mids by joining the vertices of these polygons together successively and to the centre of the sphere. It is evident that these pyramids will have a common altitude. The volume of each pyramid = base X \ altitude. (631) .'. the sum of all the pyramids = sum of bases X \ al- titude. Now let the number of spherical polygons be indefinitely increased. The sum of the bases of all the pyramids will approach the surface of the sphere as its limit; and the altitude of each pyramid will approach the radius of the sphere as its limit. .*. the sum of the volumes of all the pyramids will ap- proach the volume of the sphere as its limit. ...V = SxiR. Q.E.D. Note.— This result might be obtained by regarding the sphere as the Hmit of a circumscribed polyedron, the number of whose faces was indefinitely in- creased. For, if pyramids ai-e formed having the faces of the polyedron as their bases, and the centre of the sphere as their common vertex, these pyra- mids will have a common altitude equal to the radius of the sphere. Then each pyramid = face X 3 altitude. .-. sum of pyramids = sum of faces X \ altitude. But in the limit sum of faces of polyedron = surface of sphere: and sum of volumes of pyramids = volume of sphere. .-. V = SX§R. BOOK IX.— THE SPHERE. 388 797. CoK. 1. The volume of a spherical pyrafnid is equal to the area of its base multij^lied by one-thii'd the ra- dius of the sphere. (79G) 798. Cor. 2. Since S = 47rR% (790) .•.V = |;rR^; (79G) or, if D denotes the diameter, V = \n^\ 799. Cor. 3. Tlie volumes of tivo s2Jheres are to each other as the cubes of their radii. 800. Cor. 4. I'iie volume of a spherical sector is equal to the area of the zone tvhich forms its base multiplied by one-third the radius of the sphere. For, a spherical sector, like the entire sphere, may he conceived as consisting of an indefinitely great number of pyramids wliose bases make up its surface, and whose com- mon altitude is the radius of the sphere. 801. Cor. 5. The volume of the cylinder circumscribed about a sphere = 27rR\ Therefore, the volume of a sphei^e is equal to tico-thirds the volume of the circumscribing cylinder. Note.— It was Archimedes who discovered that the surface and volume of the sphere are each two-thirds of that of the circumscribing cylinder. EXERCISES. L Show that the volumes of spherical sectors of the same sphere, or equal spheres, are to each other as the altitudes of the zones which form their bases. 3. Find the volume of a spherical sector, if the altitude of the zone which forms its base is 3 feet, and the radius of the sphere is G feet. 384 SOLID GEOMETUY. Proposition 12. Tiieorem. 802. Tlie volume of a splierical segiuent is equal to half the sum of its bases m,ultipUed hy its altitude, j^liis the volume of a sphere of which this altitude is the diameter. Hyp. Let AB be the arc of a O, and CD the projection of the chord AB on the diameter OM. Let AC = r', BD = r, CD =//, OA=R, and denote the volume of the segment generated by revolving the circular seg- ment AHBDC about OM by V. To prove V ^ \7th{r^ + r") + ^7th\ Proof. Draw the radii OA, OB. The volume generated by AHBDC is the sum of the spherical sector generated by OAHB and the cone gener- ated by OAC, diminished by the cone generated by OBD. Vol. sph. sect. OAHB = %7tWh. (800) Vol. of cone OAC = jTrr^OC. (778) Vol. of cone OBD = i;rr''OD. .-. V = fTTRVi + \7tr''00 - \7rr'' . OD. But 7^ = OC - OD, R' - r" = OC', and R'' - 7-' = OD .-. V = TtWh - j7r(0C' - OD') = iTthl^U' - (OC' + OC . OD + OD')] Since OC - OD = h, ... 00' - 20C . OD + OD' = h% and .• . OC'' + OC . OD + OD' = f (OC' + OD') - |-' V=|;r7*[i(r'+r")+|'] = i7rh(y' + r") + Jrf. Q.E.D. BOOK IX.— THE SPHERICAL SEGMENT. 385 803. Cor. 1. If the segment has but one base, as the volume generated by MABD, the radius r' = 0, and we have Therefore, the volume of a spherical segment of one base is equal to half the cylinder having the same base and the same altitude, plus the sphere of which this altitude is the diameter, 804. Cor. 2. When the segment has but one base, r' = (2R - h)h, . • . V = l7rh\2^ - h) + ^7t¥ (803) = 7rh\^ - ^h), 805. Cor. 3. Let V denote the volume of a frustum of a cone generated by the trapezoid ABDO about OM, and v the volume generated by the circular segment AHB. Then V = \7rh(r'' + r'' + rr'), (782) and V =i7rh(r' + r'') + }7rh\ (802) Subtracting, v = ^7r7i(r' + r" + h' — 2rr') = i7rAB\ h. Therefore, the volume generated by a circular segment revolving about a diameter exterior to its surface, is equal to one^sixth of the cylinder whose radius is the chord of the scg?nent a7id lohose altitude is. the projection of this chord on the axis. 886 SOLID GEOMETRY, exercises. Theorems. 1. The lateral area of a cylinder of revolution is equal to the area of a circle whose radius is a mean proportional be- tween the altitude of the cylinder and the diameter of its 2. If the slant height of a cone of revolution is equal to the diameter of its base, its lateral area is double the area of its base. 3. The volume of a cylinder of revolution is equal to the product of its lateral area by half its radius. 4. A plane through two elements of a cylinder of revo- lution cuts the base in a chord which subtends at its centre 7t an angle of - : compare the lateral areas of the two parts o of the cylinder. 5. A rectangle revolves successively about two adjacent sides whose lengths are a and h : compare the volumes of the two cylinders that are generated. 6. The two legs of a right triangle are a and h : find the area of the surface generated when the triangle revolves about its hypotenuse. 7. Prove that a sphere may be inscribed in a cylinder of revolution, and that it will touch it along the circumfer- ence of a great circle. 8. The lateral area of a given cone of revolution is double the area of its base : find the ratio of its altitude to the radius of its base. 9. On each base of a frustum of a cone of revolution, a cone stands having its vertex in the centre of the other base : find the radius of the circle of intersection of the two cones, the radii of the bases being^^ and r^. BOOK IX.-EXEIWISES. TIIEOUEMS. 387 10. If the altitude of a cylinder of revolution be equal to the diameter of its base,* the volume is equal to the product of its total area by one-third of its radius. 11. If the slant height of a cone of revolution be equal to the diameter of its base,t its total area is to the area of the inscribed sphere as 9 : 4. 12. In a frustum of a cone of revolution the inclination of the slant height to one base is 45°: find the lateral area, the radii of the bases being r, and i\ . 13. If the radius of a sphere is bisected at right angles by a plane, the two zones into which the surface of the sphere is divided are to each other as 3 : 1. 14. If a cylinder and cone, each equilateral, be inscribed in a sphere, the total area of the cylinder is a mean pro- portional between the total area of the cone and the area of the sphere. The same is true of the volumes of these bodies. 15. If a cylinder and cone, each equilateral, be circum- scribed about a sphere, the total area of the cylinder is a mean proportional between the total area of the cone and the area of the sphere. The same is true of the volumes. 16. A cone of revolution whose vertical angle is 60°, is circumscribed about a sphere: compare the area of the sphere and the lateral area of the cone. Compare their volumes. 17. The base of a cone is equal to a great circle of a sphere, and the altitude of the cone is equal to a diameter of the sphere : compare the volumes of the cone and sphere. 18. The^volume of a cone of revolution is equal to the area of its generating rectangle multiplied by the circum- ference generated by the point of intersection of the diago- nals of the rectangle. 19. The Volume of a sphere is to the volume of the cir- cumscribed cube as ;r : 6. * Called an equilateral cylinder. t An equilateral cone. 388 SOLID GEOMETRY. 20. The volume of a sphere is to the volume of the in- scribed cube as 7r : 2. 21. If d is the distance of a point P from the centre of a sphere whose radius is E, the sum of the squares of the six segments of three chords at right angles to each other passing through P is 6R* — 2iV, 22. If h is the height of an aeronaut, and R is the radius of the earth, the extent of surface visible = ^D~r~7~ • \i -\- fi Numerical Exercises. 23. Find the lateral area, total area, and volume, of a cylinder of revolution, the radius of the base being 4 and the altitude 10. 24. Find the lateral area, total area, and volume, of a cone of revolution, the radius of the base being 4 and the altitude 10. 25. Find the lateral area, total area, and volume, of a frustum of a cone of revolution, the radii of the bases being 7 and 2 and the altitude 3. 26. Find the lateral area of a frustum of a cone of revo- lution, the radii of the bases being 21 and 6 inches and the altitude 36 inches. Ans. 3308.1 cu. ins. 27. Find the volume of a frustum of a cone of revolu- tion, the radii of the bases being 4 and 2 feet and the alti- tude 9 feet. Ans, 65.97 cu. ft. 28. The slant height of a cone of revolution is 4 feet : how far from the vertex must the slant height be cut by a plane parallel to the base that the lateral area may be divided into two equivalent parts? 29. The altitude of a cone of revolution is equal to the diameter of its base : find the ratio of the area^ of the base to the lateral area. 30. Find the volume of an equilateral cylinder in terms of its total area, See Ex. (10). BOOK IX.—NVMEIUCAL EXERCISES. 889 31. The lateral area of a cylinder of revolution is 116f sq. ft., and the altitude is 14 feet: find the diameter of its base. Ans. 2.65 ft. 32. The volume of a cylinder is 15.7 cu. ft., and the diameter of its base is 2 feet: find its altitude. A7is, 5 feet. 33. The lateral area of a cylinder of revolution is TOtt and its volume is 175;r: find its height and the radius of its base. 34. The lateral area of a cone of revolution is 60 ;r and its slant height is 12 : find its volume. 35. Find the total area of a frustum of a cone of revolu- tion, the radii of its bases being 9 and 4 feet and its height 12 feet. Ans. ^Q^Qn. 36. The volume of a frustum of a cone of revolution is 920 cu. ft., and its height is 12 feet : find the radii of its bases if their sum is 8 feet. 37. Find the number of cubic feet in a log 12 feet long and 6f feet in diameter. Ans, 418.88 cu. ft. 38. Find the number of cubic feet in the trunk of a tree, 70 feet long, the diameters of its ends being 10 and 7 feet. 39. How many square inches of sheet-iron does it take to make a joint of 6-inch stovepipe 2^ feet long, allowing an inch and a half for the seam ? • 40. The heights of two cylinders of revolution of equal volumes are as 9 : 16; the diameter of one of them is 6 feet: find the diameter of the other. 41. A cylinder of revolution whose base is 5 feet in di- ameter and a cone of revolution whose base is 6 feet in di- ameter, have equal volumes: the height of the cone is 10 feet, find the height of the cylinder. 42. The height of a frustum of a cone of revolution is 6 feet, and the diameters of its bases are 3 and 2 feet : find the height of a cylinder of revolution of the same volume as the frustum, and whose base is equal to the mid-section of the frustum. See (777). 390 SOLIB GEOMETRY. 43. The volumes of two equilateral cylinders are to each other as 3 : 4 : find the ratio of their heights. 44. The volumes of two similar cones are 27 cubic feet and 216 cubic feet, and the height of the first is 9 feet: find the height of the other. 45. The volumes of two similar cones of revolution are to each other as 512 : 729 : find the ratio of their lateral areas. 46. The slant heights of two similar cones of revolution are to each other as 3:5: find the ratio of their lateral areas, and of their volumes. 47. The height of a frustum of a cone is f the height of the complete cone : find the ratio of the volume of the frus- tum to that of the cone. 48. The total areas of two similar cylinders of revolu- tion are to each other as 25 : 49 : find the ratio of their volumes. 49. The altitude of a cone of revolution is 10, and its slant height is 14: find the total area of the inscribed cylinder whose altitude is 6. 50. The volume of a cone of revolution is 392, and its slant height is to the diameter of its base as 100 : 56: find its altitude and the diameter of its base. 51. Find the surface and volume of a sphere whose di- ameter is (1) 16 inches; and (2) 17 inches. Ans. (1) 804}; 2144f : (2) 908; 2572.45. 52. Find the diameter of a sphere if the surface is (1) 1809 square inches; (2) 616 square inches; and (3) 9856 square inches. 53. The volume of a sphere is 113: find its diameter and its surface. 54. The volume of a sphere is 776 ;r: find its diameter and its surface. 55. The surface of a sphere is 7847r: find its radius and its volume. BOOK IX.— NUMERICAL EXERCISES. 391 56. The volume of a sphere is 2G8.08 cubic inches: find the altitude of the circumscribing cylinder. Ans, 8 inches. 57. The surface of a given sphere has the same numer- ical value as the circumference of a great circle: find the numerical value of its radms. 58. Find the surface of a zone of one base, its altitude being 10 feet, and the diameter of the sphere 100 feet. 59. Find the surface of a zone, its altitude being 6 feet, and the diameter of the sphere 20 feet. 60. Find the height of a zone whose area is equal to that of a great circle in a sphere of radius r. 61. Assuming the earth to be a sphere with a radius of 4000 miles, and the altitudes of the torrid and the temper- ate zones to be 3200 and 2052 miles respectively, find the areas of these two zones. 62. The surface and volume of a given sphere are ex- pressed by the same number: find its diameter. 63. Find the weight of an iron shell 4 inches in diameter, the iron being 1 inch thick, and weighing ^ of a pound to the cubic inch. Aiis. 7i lbs. 64. A sphere of radius r is cut by a plane so that the area of the greater zone is a mean proportional between the area of the smaller zone and the area of the sphere. 65. Find the surface of a sphere inscribed in a cube whose surface is 216. 66. Find the volume of a sphere circumscribed about a cube whose volume is 64, 67. If an iron ball 8 inches in diameter weighs 72 pounds, find the weight of an iron shell 10 inches in diameter, the iron being 2 inches thick. 68. A cone of revolution, the radius of whose base is 12, is inscribed in a sphere of radius 20: find the volume of the cone. 69. How high above the earth must a person be raised in order that he may see one-fifth of its surface? 392 SOLID GEOMETRY. 70. How much of tlie earth's surface would a man see if he were raised to the height of the diameter above it? 71. Find the vohime of a spherical sector, if the altitude of the zone which forms its base is 3 feet, and the radius of the sphere is 5 feet. 72. Find the volume of a spherical sector, if the area of the zone which forms its base is 5 square feet, and the radius of the sphere is 2 feet. 73. Find the volume of the spherical sector generated by revolving a circular sector about an axis in its plane per- pendicular to one of its limiting radii, the radius of the circuhir sector being 6 and its central angle 30°. 74. Find the volume of a triangular spherical pyramid, if the angles of the spherical triangle which forms its base are each 120°, and the radius of the spliere is 10 feet. 75. Find the volume of a quadrangular spherical pyramid, if the planes of the four faces of the pyramid make with each other angles of 80°, 100°, 120°, 150°, and a lateral edge of the pyramid is 4 feet. 76. Find the volume of a spherical segment, the radii of whose bases are 3 and 5, and whose altitude is 4. 77. Find the volume of a spherical segment, the diam- eters of whose bases are 24 and 20 inches, and whose alti- tude is 4 inches. 78. Find the volume of a spherical segment of one base whose altitude is 4 feet, the radius of the sphere being 10 feet. 79. Find the volume of a spherical segment of one base whose altitude is 6 inches and the diameter of its base 16 inches. 80. A sphere, 2 feet in diameter, is cut by two parallel planes, one at 3 and the other at 9 inches from the centre : find the volume of the segment included between them. 81. Find the volume of a spherical segment of one base whose altitude is 20 feet and the diameter of its base 60 feet. BOOK IX.-NUMERICAL EXERCISES. 393 82. The radius of a sphere is 6 inches: find the area of a spherical triangle whose angles are 95°, 100°, 120°. 83. The radius of a sphere is 7 feet : find (1) the area of a lune whose angle is 30°, and (2) the volume of a wedge whose angle is 3G°. 84. A sphere 6 inches in diameter has a hole bored through its centre with a 3-inch auger : find the remain- ing volume. 85. A boiler is a cylinder with hemispherical ends ; its total length is 20 feet and circumference 11 feet: find its surface and the quantity of water required to fill it half full. 86. A cone is circumscribed about a sphere, and its height is double the diameter of the sphere : show that the total surface and the volume of the cone are respectively double those of the sphere. 87. A circle, radius r, revolves about a line in its plane and at a distance d from its centre : find the volume of the ring which the circle generates. ^'M'<^'>'mm^/m^^ ^^: THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL<*I+^CREASE.T0^50 CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. t> 13 1933 SEP 13 193) r^^^kio ■"?: \-'\ ±\ v^-j. ^Sftw^SftSt^^fcK "^^i Wh ^^ UNIVERSITY OF CALIFORNIA LIBRARY