i Ii A J,.b. THALE S _ '~-; ii -.... —J^'siissplN'^ag^^ C APLATO:^"^S^':: —'^S _:-iS -— ^:LaS -S~-JI. '"" P TT A 'nP O~~~~~~~~~U I ---— Sii I,,;;C~!~ — w i: -1 -i- Ii I7" PYTHAGORAS EUCLID FOUR DISCOVERERS AND FOUNDERS OF GEOMETRY (See pes 490G-498) BY FLETCHER DURELL, Pu.D. HEAD OF THE MATHEMATICAL DEPARTMENT, THE LAWRENCEVILLE SCHOOD NEW YORK CHARLES E. MAERRILL CO, 44-60 EAST TWENTY-THIIRD S-,TREET 1911 Durell's Maathematical Series Plane Geometry 341 pages, 12mo, half leather,. 75 cents Solid Geometry 213 pages, 12amo, half leather.. o 75 cents Plane and Solid Geometry 514 pages, 12mo, half leather $1.2 Plane Trigonometry 184 pages, 8vo, cloth... $1.00 Plane Trigonometry and Tables 298 pages, Svo, cloth. $1.25 Plane Trigonometry, with Surveying and Tables In preparation.... o. Plane and Spherical Trigonometry, with Tables 351 pages,, vo, cloth... $1.40 Plane and Spherical Trigonometry, with Surveying and Tables In preparation.... Logarithmic and Trigonometric Tables 114 pages, 8vo, cloth... o.. 75 cents Copyright, 1904, by Charles E. Merrill Co. [5] PREFAC1E ONE of the main purposes in writing this book has been to try to present the subject of Geometry so that the pupil shall understand it not merely as a series of correct deductions, but shall realize the value and meaning of its principles as well, This aspect of the subject has been directly presented in some places, and it is hoped that it pervades and shapes the presentation in all places. Again, teachers of Geometry generally agree that the most difficult part of their work lies in developing in pupils the power to work original exercises. The second main purpose of the book is to aid in the solution of this difficulty by arranging original exercises in groups, each of the earlier groups to be worked by a distinct method. The pupil is to be kept working at each of these groups till he masters the method involved in it. Later, groups of mixed exercises to be worked by various methods are given. In the current exercises at the bottom of the page, only such exercises are used as can readily be solved in connection with the daily work. All difficult originals are included in the groups of exercises as indicated above. Similarly, in the writer's opinion, many of the numeri(iii) I IV PREFACE c-al applications of greomietry call for special mnethods olf solution, and the thorough treatment of such exercises should be taken up separately anid systematically. [See pp. 304-331S, etc.] In the daily extempore work only such numerical problems are included as are needed to mak;,e cl-ear and definite the meaning, and value of the geometric lpriniciples considered. Every attempt has been made to create and cultivate the heuristic attitude on the part of the pupil. This has been done by the method of initiating the pupil into original work described above, by queries in the course of proofs, and also at the bottom of different pages-, anid also by occasional qlueries in the course of the text where definitions and discussilons are presented. In the writer s opinion, the timle has not y~'et come, for the purely heuristic study of Geometry in most schools, but it is all-important to use every means to arouse in the pupil the attitude and,energy of original investigation in the study of the subject. In other respects, the aim has been to depart as little as possible from the methods most generally used at lpreseut in teachingr geometry. The Practical Applications (Groups S-'S-91) heave beent drawii from mnany sources, biit tte, author wishes to cxIpress hiis especial indebtedniess to the Commilittee which As collectedi the _Real Appliedi Probl emsi" publishled fromt timel *to timl-e in,SK/Oool Sciem,w-, and ift, tcn ad of which Professor J. F. MAihlis of the Francis W. Parker School of Chicago is the chairman. Pagre 360 is duie almost entirely to Professor William Betz of the East High School of Rochester, N. Y. FLETC-HER DURELL, LAwREN.CEYILLE, N. J., Sept. 1, 1904. 'TO THE TEACHER 1. IN working original exercises, one of the chief dif~ ficulties of pupils lies in their inability to construct the figure required and to make the particular enunciation from it. Many pupils, who are quite unable to do this preliminary work, after it is done can readily discover a proof or a solution. In many exercises in this book the figure is drawn and tlhe particular enunciation made. It is left to the discretion of the teacher to determine for what other exercises it is best to do this for pupils. 2. It is frequently important to give partial aid to the pupil by eliciting the outline of a proof by questions such as the following: "On this figure (or, in these two triangles) what angles are equal, and why ' " What lines are equal, and why?" etc. 3. In many cases it is also helpful to mark in colored crayon pairs of equal lines, or of equal angles. Thus, in the figure on p. 37 lines AB and DE may be drawn with red crayon, AC and )F with blue, and the angles A and D marked by small arcs drawn with green crayon. If colored crayons are not at hand, the homologous equal parts may be denoted by like symbols placed on them, thus: c F C F C/L. __ R -x \ or thus: p B D R E A D In solving theorems concerning proportional lines, it is occasionally helpful to (denote tile lines in a pro portion v TO THE TEACHER Vi (either given or to be proved) by figures denoting the order in which the / lines are to be taken. Thus, if 0o: / OCG OD: OB, the relation may be indicated as on the figure. / 4/t 4. It is sometimes helpful to vary A the symbolism of the book. Thus, in dealing with inequalities a convenient symbol for "angle" is 29 as 4 A > 2; B. 5. Each pupil need be required to work only so many originals in each group as will give him a mastery of the particular method ilvolved. A large number of exercises is given in order that the teacher may have many to select from and may vary the work with successive classes. 6. It is important to insist that the solutions of exercises for the first few weeks be carefully written out; later, for many pupils, oral demonstration will be sufficient.and ground can be covered more rapidly by its use. 7. In leading pupils to appreciate the meaning of theorems, it is helpful at times to point out that not every theorem has for its olbect the demonstration of a new and unexpected truth (i. e., not all are "synthetic "), but that some theorems are analytic, it being their purpose to reduce an obvious truth to the certain few principles with which we start in Geometry. Their function is, therefore, to simplify and clarify the subject rather than to extend its content. TABLE OF CONTENTS PAGE DEFINITIONS AND FIRST PRINCIPLES....... o oa 9 BOOK I. RECTILINEAR FIGURES........o o o o 29 BooK II. THE CIRCLE................. 103 BOOK III. PROPORTION. SIMILAR POLY(GONS...... 177 BOOK IV. AREA OF POLYGONS.......o.. O 231 BOOK V. REGULAR POLYGONS. MEASUREMIENT OF THIE CIRCLE. 261 NUMERICAL APPLICATIONS OF PL ANI GEO.l:OI:lI'Y..... 304 BOOK VI. LINES, PLATNES AND ANGLES IN SPACE..... 319 Boo]i VII. POLYHEDRON-S................ 360 BOOI VIII. CYLINI)ERS AND C()NES........ o 403 BOOK IX. TtEr SPHERE............. 425 NJUMERICAL EXERCISES IN SOLIDI) GEOMETRY.. 469 MODERN (GEOMrETRIIC CONCTEPTS.............. 485 HISTORY OF G(EOMETRY............ 490 REVIEW EXEICISES IN I)PANE GEOIMETIRY.......... 499 REVIEW EXEICISES1 IN SOLID (IE OMIETRY......... 506 I'IPR ACTICAL APPLIC&TIONS OF G(EOMETY........... 510 FO)IIMULAS..,.................. 0538 (7) SYMB3OLS AND ABBREVIATIOMT A- plus, or increased by. - mninus, or dimtinishied by. X miultiplied by. divided by. equals; is (or are) egnal to. approaches (as a lintit). is (or are) equivalent to. > is (or are) greater than. < is (or are) less than. therefiore. J.perpendicular, perpentdicular to, or, is per-pendicular to. It perpendicuilars. 11parallel, or, is p~arallel to. IlJ parallels. 4 angle, angles A, triangle, trian~gles. i= parallelogram, parallelogramns. 0, 0) Circle, circles. Adj ajaccn t. Alt...alter,,atc. Art...article..Ax... axiom. Const~r..construiction. Car...corollary. De..,definition. Ex....exercse. Ext...exterior. Fig..fg,hgurc. Ilyp... hq)?IOthcsis. ident. M ien tity. uit...interior. Post...postulate. Prop...proposition. Pt... righIIt. ag. urpycstion. S up1. uppleinientary. St.... straiglt. Q. E. D. quad erat demonstrandum; that is, which was to be proved. Q. E. F. quad erat faciendum; that is, which was to be made. A few other abbreviations and symbols will- be introduced and their meaning indicated later on. (8) DEFINITIONS AND FIRST PRINCIPLES INTRODUCTORY ILLUSTRATIONS. DEFINITIONS 1. Computation of an area. Practical experience has taught men that certain ways of dealing with objects in the world about us are more advailt.aleous than others; thus, if it be desired to find the nunmber of square yards in tie area of a; floor, we do not mark off the floor { i into actual square yards and count I, I the number of square yards thus ' made, but pursue the much easier; course of measuring two lines, the length and breadth of the floor, obtaining 7 yards, say, as the length, and 5 yards as the width, and multiplying the length by the breadth. The area is thus found to be 35 square yards. Let the pupil determine the area of some convenient floor in each of these two ways, and compare the labor of the two processes. 2. Computation of a volume. Similarly, for example, in order to determine the number of cubic feet which a box contains, instead of filling the box with blocks of wood, each of the size of a cubic foot, and counting the number of blocks, we )pur- sue the much easier course of measurincg the number of linear feet in each inside edge of the box and multiplying together the three dimensions obtained; thus, if (9) 10 PLANTE GEOM'ETRY the inside dimnenlsions are 5, 4 and 3 feet, the volume is 5 X 4 X 3, or 60 cubic feet. Similarly, the direct method of measuring the number of bushels of wheat in a bin is to fill a bushel measure with wheat from the bin, time after time, till the bin is exhausted, and count the number of times the bushel measure is used. But a much less laborious method is to measure the three dimensions of the bin in inches and divide their product by the number of cubic inches in a bushel. Iet the pupil in like manner compare the labor of finding the number of feet of lumber in a given block of wood by actually sawing the block up into lumber feet, with the labor of measuring the dimensions of the block and comlputing the number of lumber feet by taking the product of the linear dimensions obtained. 3. Unknown line determined from known lines. The student is also probably familiar with the fact that, by computations based on the relations of certain lines whose lengths are known, tlhe lengths of other unknown lines may be determined without the labor of measuring these unknown lines. Thus, for instance, if a ladder 5 yards long lean against a wall and have its foot 3 yards from the -- wall, the height of the top of the ladder from the ground may be determined thus: (Height)2 = 2 - 32 = 25 - 9 = 16.. heiigt -- 4 ytards. 4. Economies in representing surfaces, lines, etc. Other principles of advantage of an even more general character occur in dealing with geometric objects. Thus, since only one straight line can be passed through two given points, the two points may be taken as a highly economized symbol or representative of the line, by the use of which much labor is saved in dealing with lines, and new results are made attainable. GEOMETRIC MAGNITUDES 11 Similarly, since only one flat, or plane surface can be passed through three given points (not in the same straight line), these three points may be taken as a symbol or representative of the flat surface. This gives the advantage not only of reducing an unlimited surface to three points, but also of giving for the plane a symbol made up of three parts. By varying one of thlese parts and not the others, the plane 1lmay be varied in one respect and not in others; also planes having cer'tai lI properties in (comm1on may be grouped together, and de.alt with in the groups formed. 5. Geometry as a science. Definition. The above illustrations serve to slhow howN advantaigeous it often is to deal with geometric magnitudes by certain methods rather than others. In the study of Geometry as a science we proceed to make a systematic examination of these methods. Geometry is the science which treats of the properties of continuous magnitudes and of space. GEOMETRIC MAGNITUDES 6. Solids. A physical solid is a portion of matter, as a block of wood, an iron weighlt, or a piece of marble. The portion of space occupied by a physical solid may be considered apart from the physieal solid itself, hence A geometric solid is tle portion of space occupied by a physical solid, or definitely determlied in any way. Hence, also, a geometric solid is a limited portion of s)pace. One advantage in using geometric solids lies in this. If we dealt with physical solids only, as blocks of wood, marble, iron, etc., we ehould need to determine the properties of each kind of physical solid, 1'2 PLANE GEOM ETRY separately; but, by determining the properties of a geometric solid, we determine once for all the properties of every physical solid, no matter what its material, that will exactly fill the space occupied by the given geometric solid. Hereafter in this book the term "solid" is understood to mean geometric solid, unless it be otherwise specified. 7. Other geometric magnitudes defined as boundaries. A surface is the bolundary of a solid. A line is the boundiary of a surface. A point is the boundary of a line. The solid, surface, line, and point are the fundamental geometric mlagnitudes. 8. Geometric magnitudes defined by their dimensions. A solid has three dimensions; viz., length, breadth and thickness. A surface has two dimensions, length and breadth. A line has one dimension, length. A point has no dimension. Hence a point is that which has position, but no magnitude. 9. Geometric magnitudes defined as generated by motion. A line is that which is generated by the motion of a point. A surface is that which is generated by the motion of a line (not moving along itself). A solid is that which is generated by the motion of a surface (not moving along itself). The three independent motions by which a solid is generated illustrate the fact that a solid has three dimensions. 10. Geometric magnitudes as intersections. The intersection of two surfaces is a line; of two lines is a point. It is sometimes more advantageous to regard geometric magnitudes from one of the above points of view, sometimes from another. LINES 13 11. A geometric figure is any combination of points, lines, surfaces, or solids. 12. The form or shape of.1 figure is determined by the relative position of its parts. 13. Similar geometric figures are those which have the,ame shape. Equivalent figures are those which have the same size. Equal or congruent figures are those which have the same sh(tpe and size, and can, therefore, be made to coincide. 14. A point is representefd to the eye by a dot and is named by a letter affixed to the dot, as the point A, o A. LINES 15. A straight line is a line such that, if any two points in it be fixed and the line rotated, every point in the line will retain its original position. A straight line is also sometimes described as a line which has the same direction throughout its whole extent; or, as the shortest line connecting two points. The word "line" may be used for " straight line," if no ambiguity results. 16. A curved line is a line no portion of which is straight. The word "fcurve7" is often used for "curved line." 17. A broken line is a line made up of different straight lines. 18. A rectilinear figure is" a figure composed only of straight lines; a curvilinear figure is a figure composed only of curved lines; a mixtilinear figure is a figure containing both straight and curved lines. 1 41 II ~~~PLA&NE GliOMETI4Y 19. Kinds of straight lir~e. A straig'ht, linle m-ay be definite or indefinite in leng, h. The line ot deffinit~e len -,th 1s sonetiines termeda seginen t 01' ScOf Othe(r kinds of straighit linie are defined in Arts, 26 and 41. 20. Naming a straight line. A straight line is named by naming two of its points, as the 1-3 ~~~~~A- -B line AB (a seet);or the line CD (indefinite in length). A segm nent or sect may also be de-noted by a single (small) letter, as the line, a. 21. The circumference of a eircle is a line every point of whieh is equally distant fromt a fixed point within, called the center. AN~GLES 22. An angle is the amount of opening between two straig~ht lines whieh meet at a point. The sides of a-n angle are the lines whos.e inter-sectio-n forms the cang'le; the vertex of an angle is the point in whieh the sides intersect. 23. Naming angles. (1) The most precise way of namingy an angle is to use three letters, one for a point on each side of the angle with the letter at the vertex between these two, as the angle ABC. Since the size of an angle is independent of the length of its sides, the points named on its sides may be taken at any 'place on its sides. Thus, the angles AOD)7 BODI) BOE, AOF a.re all the same angle. A B C C P A ANGLES E5, (2) Tn ease, tHere is hu)ft one ang-le at, a given vertex, the letter at the, vertex alone is sufficient to denote the, angle, as the angle 0 in the last figure. (3) Sometimes a letter or figure placed inside the- angle and near the vertex is a convenient symbol, as the angle a. 24. A straight angle is an angle whose sides lie in the same straigTht line, but wihextend in opposite directions from BA the vertex, as the angle BOA..I 25. A right angle is onie of two- equal angrles made b)y one strlai~lit line. 1ne(etiiv', anlother sr21t1in.Thus1" if the line ])Q meets line All so asto make angle PQ.jA equal to angle J)Q/}, ewoli of these anigles is a right angle. A1 (,ilit angle is Calso half of (A straight anigle. In11 ~ ~ ~ ~ ~ - 26. A perpendicular is a line that BA makes a right angile wNith a given line thus PQ In the last figrure is perpendicular to BA. The, foot of a perpendicular is the Point in which the perpendicular meets the line to which it is drawn, thus Q is the foot of the perpendicular PQ. 27. Ani acute angle is an angle C less than a righlt anglle, ais the angle A07 28. Anl obtuse angle is an angl01e greater than a (rigt angle bt lesds than a strig-hl-t. angle, as angle A Oj). 16 16 P~~YLANEI GEOMETR~Y 29. A reflex angle is an angile_____ greater than a, straight angle, but less0 th an two straigh t angcles, as angle AQEF. In this hook, angles larger than a straight Pf angle are not considered unless special mention is inado, ot' thiem. 30. An oblique angle is an angifle that is neit~her la righit nor a straight lang-le. Hence, oblique angiole is a g-eneral terat for acute, obtuse, and reflex angles. 31. Adjacent angles are an gles whi71ich have a common vertex and a comon side between themn, as angles A OB and 1300. 32. Vertical angles are angles whieh have a commnon vertex and the sides of one angle the prolongations of the sides of the other angle, as the angles AOC and BOD.D33. Complementary angles are two gether equal a right angle, as the anigles A OJ and JPOQ Hence, the complement of an angle is the difference between that ang~le and one righlt angl~e. 34. Supplementary angles are two angles which together equal two right angles (or a straight angle), as the angles A OP and P OR. Hence, the supplement of an angle is the difference between that angle angles. CY angles whlich to and two right ANGLLES 17 35. Angles as formed by a rotating straight lineo If the line OB start in the position OA and rotate to the position OB, it is said to generate the ZAOB. The size of an angle may, therefore, \ be considered as the amount of rotation \ of a line about a point, from the origi- /~ nal position of the line. If the rotation is continued far enough, a right angle ( ZAOLl ) is formed; afterward an obtuse angle ( Z AO)); then a straighlt angle ( Z AOE), and a reflex angle ( LA OF), etc. An advantage of this Inethod of forming or conceiving angles is that by continuing tile rotation of the moving line an angle of indefinite size may be formed. 36. Units of angle. A right angle is a unit of angle useful for many purposes. Sometimes a smaller unit of angle is needed. A degree is one-ninetieth of a right angle; a minute is one-sixtieth of a degree; a second is one-sixtieth of a minute. Besides these, otlher units of angle are used for certain purposes. SURFACES. DIVISIONS OF GEOMETRY. PARALLEL LINES 37, A plane is a surface such that, if any two points in the surface be joined by a straight line, the straight line lies wholly in the surface. Hence, a plane figure is a figure such that all its points lie in the same plane. 38. A curved surface is a surface no part of which is plane. B 18 PLANE GEOMETIRY 39. Plane Geometry is that branch of Geometry whihl treats of plane figures. 40. Solid Geometry is that branch of Geometry which treats of figures all points of which are not in the same plane. 41. Parallel lines are straight lines in the sam-le plane which do not meet, however far they be produced. EXERCISES. CROUP D Ex. 1. Draw a straight line. A curved line. A broken line. Ex. 2c Which of the capital letters of the alphabet are straight, which curved, which broken, which curved and straight lines combined? Ex. 3. Draw an acute angle. An obtuse angle. A reflex angle. Ex. 4. Draw two adjacent angles. Two vertical angles. Ex. 5. What is the complement of an angle of 43~? What is its supplement Ex. 6. What is the complement of 57~ 19'? of 62~ 23' 43"? What is the supplement of each of these? Ex. 7. At two o'clock, what is the angle made by the hour andl minute hands of a clock? at three o'clock? at five o'clock? Ex. 8. At 1:30 o'clock, what angle do the hands of a clock make? at 2:15? at 8:45? Ex. 9. What kind of an angle is the supplement of. an obtuse angle? of an acute angle? of a right angle? Ex. 10. What kind of a surface is the floor of a room? the surface of a baseball? the surface of an egg? the surface of a hemisphere? Ex. 11. If the surface ABCD move A. to the right, what solid is generated by - it? What surfaces are generated by its L I bounding lines? What lines by the ver. — tices of its, angles? f GEOMETRIC MAGNITUDES 19 Ex. 12. Draw two supplementary adjacent angles. Also two supplementary angles that are not adjacent. Also two adjacent angles that are not supplementary. Ex. 13. The sum of a right angle and an acute angle is what kind of an angle? Their difference is what kind of an angle? Ex. 14. The sum of an obtuse angle and a right angle is what kind of an angle? Their difference is what kind? Ex. 15. If three straight lines meet (but do not intersect) at a point in a plane, how many angles have this point as their common vertex? Draw a figure illustrating this and name the angles. Ex. 16. How many angles are formed if four lines meet (but do not intersect) at a point in a plane? Ex. 17. How many degrees are in an angle which equals twice its complement? Ex. 18. How many degrees in an angle which equals one-third its supplement? Ex. 19. What kind of an angle is greater than its supplement? What kind is less? PRIMARY RELATIONS OF GEOMETRIC MAGNITUDES 42. Certain primary relations of geometric objects have already been given in the definitions used for geometric objects. We now proceed to investigate the relations of geometric magnitudes more generally and systematically. 43. An axiom is a truth accepted as requiring no demonstration. 44. Two kinds of axioms are used in geometry: 1. General axioms, or axioms which apply to other kinds of quantity as well as to geometric magnitudes; for instance, to numbers, forces, masses, etc. PLNNJE GEOME~TRY 2.i. Geometric axioms, or axiomis whieh apply to geometrie niagnitades alonle. 45. The general axioms may be stated as follows: GENERAL AXIOMS 1. Thrings which, are, equal to the samew thing-, or to eqiwi things, (I(equ(ti to '(eah other. A2.If equals be addedl to equats, the sums are equtal. 3. If equals be subtraetedl from equals, the rmidr are equal. 4. Doubles of equals aire equal; or, in genieral, if equalls be muiltipl~ied by e(quals the prodlucts (tre equall. 5. IHalves of equals are equal; or, 'in gelwr(Ii, 'if equllals be (lividled by equals the quotients are equal. 6. The whole 'is equal to the sum~ of'its patrts. 7. The whole is greater than anty of its parts. 8. A quantity maiy be substitutedi for its equtal in any proce'ss. 9. If equials be. added to, or sutbtracted from), unequals, the( results are, unequal in the same order; 'if unewquals be adde1((l to 'unequals in the samew ordler, the 'results aire unequal int tat Orde'r. 10. Doubles, or halives, of unequals aire unequal in the same ordle'r. 11. If unequals be subtracted from equals, the remainders are unequal in the reverse order. 12. If, of three quantities, the first is greater than the, second, and the second is g'reater than the third, then the first is greater than the third. GEOMETRIC AXIOMS 21 46. The value of the general axiomso The axioms given above seem so obvious that the student at first is not likely to realize their value. This value may be illustrated as follows: If the distance from Washington to Philadelphia be known, and also the distance from Philadelphia to New York, the distance from Washington to New York may be obtained by adding together the two distances named; for, by axiom 6, the whole is equal to the sum of its parts. Thus the labor of actually measuring the distance from Washington to New York is saved. Again, if the heigiht of a schoolboy of a given age in Paris be measured, and the height of a like schoolboy in New York be measured, and the resullt of the measurement in each case is the same, we know that the boVs are of the same heilght, without the labor and cost of bringing the boys together and comparing their heights directly; for, by axiom 1, things which are equal to the same thing are equal to each other. Thus the general axioms are to be looked at not merely as fundamental equivalences, but also as fundamental economies. For many purposes the latter point of view is more important than the former. 47. The geometric axioms may be stated as follows: GEOMETRIC AXIOMS 1. Through two givent points only one straight line can be passed. 2. A g(eo)?,ctr ic figure may be freely moved in space uiithout any change in form or size. This axiom is equivalent to regarding space as u/iforwm, or oiomaloidal; that is, as having the same properties in all its parts. It has already been assumed in some of the definitions given. See Arts. 15 and 35. 3. Throutgh a given point one strctiglt line a(nd o ly one can be d ra:twn parallel to another gyienl straigt t linee. By Art.13, geometric figures 'zhich coincide are equal. 22 PLAJNE GEOMETRY 48. Utility or uses of the geometric axioms. By means of the first geometric axiom we are able to shrink or condense any straight line into two points. Later tlis advalntage gives rise to many other advantages. By the second geometric axiom, the knowledge which we have of one geometric object may be transferred to another like object, however widely separated in space. The utility of the third geometric axiom canh be made more evident when we come to use it in proving new geometric truths. 49. A postulate in geometry is a construction of a geometric figure admitted as possible. 50. The postulates of geometry may be stated as follows: 1. Through any two points a straight line imay be drawen. 2. A straight line may be extended 'indefinitely, or it vmay be limited at any point. 3. A circumference may be described about any given point as center, and with any givten radiuts. These postulates limit the pupil to the use of the straight-edged ruler and the compasses in constructing figures in geometry. One of the objects of the study of geometry is to discover what geometrie figures can be constructed by a combination of the elementary constructions allowed in the postulates; that is, by the use of the two simplest drawing instruments. 51. Logical postulates. Besides the postulates which are used in the actual construction of figures, there are certain other postulates which are used only in the processes of reasoning. Thus, for purposes of reasoning, a given angle may be regarded as divided into any convenient number of equal parts. Whether it is possible actually thus to divide this angle on paper by use of the ruler and compasses, is another question. 23 DEMONSTRATION OF PROPERTIES DEMONSTRATION OF GEOMETRIC RELATIONS 52. A geometric proof, or demonsttration, is a course of reasoning by which a relation between geometric objects is established. 53. A geometric theorem is a statement of a truth concerning geometric objects which requires demonstration. Ex. The sum of the angles of a triangle equals two right angles. 54. A geometric problem is a statement of the construction of a geometric figure which is required to be made. Ex. On a given line to construct a triangle containing three equal angles. 55. A proposition is a general term for either a theorem or a problem. Thus, propositions are subdivided into two classes: 1, Theorems; 2, Problems. 56. Immediate inference is of two kinds: 1. Changing the point of view in a given statement. Thus the statement, "two straight lines drawn through two given points must coincide," may be changed to "two straight lines cannot inclose a space." 2. Reasoning which involves but a single step. Ex. "All straight angles are equal;".. "All right angles are equal." (Ax. 5.) 57. A corollary is a truth obtained by immediate inference from another truth just stated or proved. 58. A scholium is a remark made upon somre particular feature of a proposition, or upon two or more propositions which are compared. 24 PLANE GEOMETRY 59. Hypothesis and conclusion. A proposition consists of two parts: 1. The hypothesis, or that which is known or granted. 2. The conclusion, or that which is to be proved or constructed. Thus, in the proposition, "if two straight lines are perpendicular to the same line, they are parallel," the lhypothesis is, that two given lines are perpendicular to another given line; the conclusion is, that the two given lines are parallel. 60. The converse of a proposition is another proposition formed by interchanging the hypothesis and the conclusion of the original proposition. Thus, theoreml, "every point in the perpendicular bisector of a line is equidistant from the extremities of the line;" Converse, "every point equidistant from the extremities of a line lies in the perpendicular bisector of the line." Or, in general, theorem, "if A is B, then X is Y:" converse, " if X is Y, then A. is 1.;'" Frequently a converse is formed by interchanging part only of an hypothesis with part or all of the conclusion, or vice versa. Thus, theorem, "if A is B and C is D, then X is Y.'" co0nersc, "if A is B and X is Y, then C is 1)." The converse of a theorem is not necessarily true. Thus, it is true that all right angles are equal, but it is not true that all equal angles are right angles. 61. The opposite of a theorem is a theorem formed by making both the hypothesis and the conclusion of the original theorem negative. Ex. theorem, "if A is B, then X is Y;' opposite, "if A is not B, then X is not Y." DEMONSTRATION OF PROPERTIES 25 If a direct theorem and its opposite are both true, the converse is known to be true without proof. Also, if a theorem and its converse are both true, the opposite is known to be true without proof. Thus certain economies arise in the demonstration of theorems. 62. Methods of geometric proof. Several principal methods of proving theorems are used in geometry. 1. Direct demonstration. 2. Proof by superposition, in which two fig'ures are proved equal by nmaking one of them coincide witll the other. 3. Indirect demonstration, which consists essentially in showing that a given statement is true by showing that its negative cannot be true. Other special methods of proof will be pointed out as they occur in the course of the work. 63. Forni of a proof. The statement of a theorem and its proof consist of certain distinct parts which it is important to keep clearly in mind. These parts are: 1. The general enunciation, which is the statement of the theorem in general terms. 2. The particular enunciation, or statement of the theorem as applied to a particular figure used to aid the mind in carrying forward the proof. 3. The construction of supplementary parts of the figure (not necessary in all proofs). 4. The proof. This must include a reason for every statement. 5. The conclusion. The letters Q. E. D., standing for "quod erat demonstrandumr" and meaning "which was to be proved," are usually annexed at the end of a completed demonstration. 26 PLANE GEOMETRY EXERC1ISES. GROUP 2 Ex. 1. In the figure measure AB; then measure BC. A --- — ---- Now find AC without measuring it. What axiom have you used? Ex. 2. If Z AOB=60~, Z BOC=,80~ and C Z COD= 130~; find without measuring them A AOC and BOD (reflex). What axiom have you used? Ex. 3. Prove by repeated use of the first part of Axiom 1 that magnitudes equal to equal magnitudes are equal to each other; (thus, D given A=., R3=y, and x=y. Prove A==B). Ex. 4. (Give a numerical illustration of Axiom 1. Ex. 5. Show by the axioms that a part is equal to a whole diminished by the remaining part. Ex. 6. Show that Axiom 1 is a special case of Axiom 8. Ex. 7. If a = x + y and x = y show by use of the axioms that x = ia(. Ex. 8. Draw a line and produce it so that the produced part shall equal another given line. Ex. 9. By use of the compasses, mark off on a given line a part twice as long as another given line. Ex. 10. On a given line mark off, by fewest uses of the compasses, a part four times as long as another given line. Ex. 11. Draw three straight lines and denote them by 1, n, and n. Then draw a line 1 + l —n, and also a line - 2m, - 3n. PROPERTIES OF LINES INFERRED IMMEDIATELY 64. If two straight lines have two points in common, the lines coincide throug.hout their whole extent (Art. 47, Geom. Ax. 1). Hence, two straight lines can intersect in but one point. 65. If two straight lines coincide in part, they coincide throughout, PROPERTIES OF ANGLES 27 66. Only one straight line can be drawn connecting two given points. 67. Two straight lines cannot enclose a surface. 68. A given straight line (sect) can be divided into two equal parts at but one point. For (by Ax. 5) halves of equals (or of the same thing) are equal. PROPERTIES OF ANGLES INFERRED IMMEDIATELY 69. All straight angles arc equal. 70. A straight acngle can be divided into two equal angles by but one line at a given point in the given straight line. For (Ax. 5) halves of the same magnitude are equal. 71. Hence, at a given point in a straight line but one perpendicular can be erected to the line. 72. All right angles are eqtual. For all straight angles are equal (Art. 69) and halves of equals are equal (Ax. 5). 73. The sum of the two adjacent angles formed by one straight line mteeting an-. other straight line equals two right angles. For the angles formed are supplementary adjacent angles (Arts. 31, 34). 74. If two adjacent angles are together eqtual to a straight angle (or two right (aglqes), their exterior sides form one and the samne straight line. For their exterior sides form a straight angle, and hence must lie in a straight line (Art. 24), 28 PLANE GEOMETRY 75. The compZeP2ents of two equal angles are equal (Art. 72 and Ax. 3!); the suprplemnents of twio equal 1agles are equal (Art. 69, Ax. 3). 76. The suimn of tll the angles about a point equals four right (agles. Thus, Za+Zb+Zc+Z +Ze= Fig. 2 4 rt. A. 77. The sum of all the angles about a point on the same side of a straight line passing through, the psioint equals _ two right angles. i.3 Thus, Zp -+Zq +Zr=2 rt.. iEXERCISES. GROUP 3 Ex. 1. How many different straight lines are determined by three points not in the same straight line? Ex. 2. How many straight lines are determined by four points in a plane, no three of them being in the same straight line? Ex. 3. If, in Fig. 2 above, A a, b,, d = 40~, 50~, 60~, 70~ respectively, find Z c. Ex. 4. If,' in Fig. 3 above, the lines forming the anglo q are perpendicular to each other and Z p = 47~, find the other angles of the figure. Ex. 5. Measure Za of Fig. 1 on preceding page. Find Lb without measuring it. Now measure Z b and compare the two results. A P Ex. 6. Given QB I AB, PB 1 BC, and Q ZABC=130~; find the other angles of the figure. B C Ex. 7. Arrange five points in a plane so that the fewest number of straight lines may pass through them, no line to pass through more than three points. PLANE GEOMETRY BOOK I RECTILINEAR FIGURES PROPOSITION I. THEOREM)2 78. If one straigt line intersects~ aniother straight line, the opposite or vertical angles are equal. A D Given the straight lines AB and CD intersecting at the point 0. To prove Z A OC=Z ~DOB and Z AOD = ZCOOB. Proof. Z AOC+ ZA OD =2 rt. A,1 Art. 73. the stmot of Iwo adjaceni~i anigles formed, by onec straight line meeting ((ntothe? str-aight line equals two righit angles). Also Z BOD+ ZA OD -2-rt. A, (same reason). ZA0C+ZA0D=Z-BD0+ZAOD, Ax., (thbigs equ~al to thte samte th inig are equal to each other). Subtracting ZAOD from the two equals, Z AOC =Z BODI Ax. 3. (if equals be subtracted frota equals, thte remaitiders are equal). In like manner it may be proved that Z A D =Z COB3.E.D Ex. If in the above figure Z~DOB"=7O0 find the other angles without measuring them. (29) no 20 ~1300K I. PLANE GEOMETRY PROPOSITION- II. THEOREM 79. If, from, (I 1oint in a perpendicular to a Oi-en line, tw(o obiiq'ue tles be dr1-awn cudttig #4 o'n thle givsen zlne eqalw segments fromn the foot of the perpendicutlar, the, obliquie lines are equal and make equal angles wvith the pe-rpendicular. D P Given a line AB w~ithi CD) I to it at the point, C, and ER and E'Q drawn from any point as P in (71), cutting off CR - QC on AR4. To prove PRi -- EQ and Z (P R -- Z EQ. Proof. Fold over the figure DCB about -DC as an axis till it comes in the plane DC'A. Geom. Ax. 21. Then Z DCB =Z DCA (all righit A are =). Art. 72, line CB will take the direction of CAl. But CR,: CQ. Hyp. point 11 will fall on point 9. Hence line PR wvill coincide with liie PQ, Art. 66. (only one straight line can be dratvn con~nething tivo girven points) And j CPR will coin-cidle with z (E'Q. ER =P9, and Z CP R /C(1), Art. 47. (geomnetric figures wvhich coincide are equal). Q. E. D Ex. 1. Point out the hypothesis and the conclusion in the general enunciation of Prop. I. Also point them out in the particular enunciation. Do the same for Prop. II. Ex. 2. If three straight lines intersect at a point, how many of the angles formaed is it necessary to measure, in order to determine all the anglesf LINES AND ANGLES 3 31 PROPOSITION III. THEOREM 80. From a given point without a straight line but one perpendicular can be drawn to the line. P AR '4, P Given the straight linie AR, -P any point without AB, PQ I AB, and PR any other line drawn fromi P to AR.To prove that P R is not I AR. Proof. Produce PQ to PI mak-ingo QP' = PQ. Draw RP'. Then -RQ IPP'. Hyp., P'Q =PQ. Conistr. RP =RP' and Z PRQ =Z P'RQ, Art. 79. (if, from a poin t in a I to a gircn line, two oblique lines be drawn cutn ojf' on the fliv' it line equal segmentts fromn thre foot of thre IL the o blique lines a-re equal and mnaiw equal X with the IL lbnt PRP' is not a straigyht line, Art. 66. (only one straighit line can be drawn connecting twro giren points). PRP' is niot a straighit Z. /PRQ), the half of Z PRP', is not a rigtZ Ax10. PR is not I ARB. only one perp~endienlar cani be dr-awn from P to AR. ___________ ~~~Q. E. A Ex. Three straight lines. intersect at a point. Two of the adjacent anglies formed at the point are 300 and 40". Find all the other angles at the point. 32 BOOK I. PLANE GEOMET RY TRIANGLES 81. A triangle is a portion of a plane bounded by three straight lines, as the triangle ABC. 82. The sides of a triangle are the lines which bound it; the perimeter of a triangle is the sum of the sides; the angles of a triangle are the angles formed by the sides, as the angles A, B and C; the vertices of a triangle angles of the triangle. B are the vertices of the 83. An exterior angle of a triangle is one side and by another side produced, as the angle BCD. With reference to the angle BCD, the angles A and B are termed the opposite inte- / rior angles. an angle formed by B. O 84. Classification of triangles according to relative length of the sides. A scalene triangle is a triangle in which no two sides are equal. An isosceles triangle is one in which two sides are equal. An equilateral triangle is one in which all three sides are equal. 6calcnme Isosceles TRIANGLES 33 85. Classification of triangles with reference to character of their angles, A right triangle is a triangle one of whose angles is a right angle. An obtuse, triangle is a triangle one of whose angles is an obtuse angle. An acute triangle is a triangle all of whose angles are acute angles. An equiangular triangle is one in which all the angles are equal. Kiiglit Obtutse Acute Equiangular 86. The base of a triangle is the side upon which the triangle is supposed to stand, as AB. The angle opposite the base is called the vertex angle, as angle ACB; the vertex of a triangle is the vertex of the vertex angle of the triangle. The altitude of a triangle is the perpendicular from the vertex to the A - \ base or base extended, as CD. L 87. In an isosceles triangle, the legs are the equal sides, and the base is the remaining side. 88. In a right triangle, the hypotenuse is the side opposite the right angle, and the legs are the sides adjacent to the right angle. 89. Altitudes, bisectors, medians. In any triangle, any side may be taken as the base; hence the altitudes of a triangle are the three perpendiculars drawn one from each vertex to the side opposite. C 34 BOOK I. PLANE GEOMETRY A bisector of an angle of a trianglle is a line wlic divi(es. tllis angle ilnto two equal parts. This I)isect:or i usllaltlly 1)pr,dceed to meet the side oplposite tlhe given an:gle. A median of a triangle is a line drawn fr>om a vertex of the triangle to the middle point of the opposite side. H1ow many medians has a triangle? 90. Two mutually equiangular triangles are triangles having their corresponding angles equal. 91. Homologous angles of two mutually equiangular triangles are corresponding angles in those triangles. Homologous sides of two mutually eqniangular triangles are sides opposite homologous angles in those triangles. We shall now proceed to determine first, the properties of a single triangle, as far as possible, then those of two triangles. 92. Property of a triangle immediately inferred. The sum of any two sides of a triangle is greater than3 the third side. For a straight line is the shortest line between two points (Art. 15.) Ex. 1. Point out the hypothesis and conclusion in the general enunciation of Prop. III; also point them out in the particular enunciation. (As each of the next fifteen Props. is studied, let the pupil do the same for it.) Ex. 2. Find the angle whose complement is 18~; whose supplement is 76~ Ex. 3. If the complement of an angle is known, what is the shortest way of finding the supplement of the angle? If the supplement is known, what is the shortest way of finding the complement? Ex. 4. In 25 minutes, how many degrees does the minute-hand of a clock travel? How many does the hour-hand? Ex. 5. Draw three straight lines so that they shall intersect in three points; in two points; in one point. TRIANGLES PROPOSITION IV. THEOREM 93. Any side of a triangle is greater than the difference between the other tiwo sidles. Given AB any side of the A AB C, and AC > B3 C. To prove AR >A C- J C. Proof. ABR +BC >AC, Art. 92, (the suiim of any two sules of a timangle is greater than the third side). Sulbtracting BC from each ineaiber of the inequality, AB >AC- BC, Ax. 9. if equals be subtracted fromt unequals, the remnainders are unequal 'in the samne order). Q. E. D. 94. COR. The Perpendicular is the shortest linLe that can be drawnen from a given, point to a gicen line. For, in the Fig. page 81, PP' <PR + R11', Art 92. Or, 2 PQ <2 PR,. Ax.8..1. PQ <PR. Ax. 10. Hence, DRFK. The distaence from a point to a line is the perpendicular drawn from the point to the line. Ex. 1. If one side of an equilateral triangle is 4 inches, what is its perimeter? Ex. 2. Is it possible to form a triangle whose sides are 6, 9 and 17 inches? Try to do this with the compasses and ruler. Ex. 3. Is it possible to form a triangle in which one side is 10 inches and the difference of the other two sides is 12 inches Ex. 4. On a given liiie as base, by exact use of ruler and cornpasos, construct an equilateral trianile. 3 6 86 ~~BOOK- I. PLANE GEOMETRYC) PROPOSITION V. YTHE~OREM 95. -If, from a point wvithin a triangtle, t -( "i in~?ri;V1 to the extrem11ities of onie side of thle trian'gle, /;.'u.of lthe othe(r two sides of the triangle is greater than thle sam11 of the two lines so drawvn. Given P any point within the triangle ABC, and( PA and PC lines drawn from P to the extremities of the, side A C. To prove AB +BC >AP +PC. Proof. Produce the line AP to meet BC at Q. Then AB + BQ > AP + PQ, A rt. 15. (a stcaightI line i.s the shortest lhie conectbing two Ipoints). IAlso 1)Q -h QC > PC, (samie reason). Adding~ these inequalities, AB + BQ + _PQ~ QC > AtP + P Q+ PC. A x..9. Substituting, BC for its equal BQ + QC, AB +BC +PQ >AP +PQ +PC. A x.8S. Subtracting PQ from each side, AB+ BC>AP+ PC. Ax.9. Q. E. D. Ex. Oni a given line as base, by exact use of ruler and compasses, construct au isosceles triangle each of whose legs is double the base. TRIANGLES 37 PROPOSITION VI. THEOREM 96. Two triangles are equal if two sides and the ineluded angle of one are equal, respectively, to tuwo sides and the included angle of the other. B E Given the triangles ABC and DEF in which AB=-DE, AC=DF, and ZA= ZD). To prove A ABC =A DEF. Proof. Place the A ABC upon the A DEF so that the line AC concides with its equal DF. Geom. Ax. 2. Then the line A B will take the direction of DE, (for Z A = Z D by lyp. ). Also the point B will fall on E, (for line AB = line DE by hiyp.). Hence the line BC will coincide with the line EF, Art. 66. (only one straight line can be drawn connecting twUo points)... ABC and DEF coincide... A ABC= A DEF, Art. 47. (geometric figures which coincide are equal). Ex. 1. What kind of proof is used in Prop. VI? (See Art. 62). Ex. 2. If A.4, B and C=600, 70~, 500, 4B=16, AC=19, BC=18: also Z D=60~, DE=16, DE'=19; find A E and F and side E' without measuring them. f-) -1 i3s 38 ~~BOOK I. PILANTE GEO)METRY Pi-otposirlo.N VII. Tiuinoniu-t 97. T71Wo tpiiji/l(5 are eqpual if tivo angles (end the, i'en, Chidedl Side of one are-" e(plai, res-pec~tiely, to tw-o an,1gles and the 'inc laded side of the othe~r. B F A CF Given the I ABC and -)-EF in which Z A ZD L C i ZF, and A C =UPF. To prove A ABC A DEF. Proof. Place the A ABC upon the A DEF so that AC shall coincide with its equal DF. Geom. Ax. 21. Then AB will tak-e the direction of DE, (for ZA = Z Dby hyp.), and the point B will fall somewhere on the line DR or DR produced. Also the line CB will take the direction of TyE,? (Ifor Z C = Z F by hy). ), and the point B will fall on rr or rr' produced.. point B falls on point FE, Art. 64. (twvo straight lines cane intersect in but one point). AABC and DEE coincide.. A ABC= ADEF, Art. 47. (geometric figures wvhich coin-cide- are equal). Q. E.D Ex. 1. What kind of proof is used in Prop. VII. ' Ex. 2. IfZAA) B, C=65, 550, 600, AB= 21, AC — 18, BC=- 27; also A DP F==650, 600, and DF= 18; find DE, EF, and Z E. Ex. 3. Construct by exact use of ruler and compasses a scalene triangle whose sides are 2, 3 and 4 times a given line. TRIANGLES 39 PROPOSITION VIII. THEOREM 98. Twvo right triangles are eqital if the hypotenuse and -an acute angle of one are equal to the hypotenuse and an acute angle of the other. B A C D F Given the right A ABC and DEE in whieh hypotenuse AB= hypotenuse DiE, and ZA = Z D. To prove A ABC= A DEF. Proof. Place the A ABC upon the A DEF so that thb side AB shall coincide with its equal, the side DE, the point A coinciding with the point D. Then the line AC will take the direction of DE, (for ZA =ZD by hyp. ) Also the side BC will coincide with the side FE, Art. 80. (from a given poinit, E, without a straight lneC, DE, but one I can be drawn to thre line). A ABC and DEE coincide..:. A ABC= A DEE, Art. 47. (geometric Jiqures whlticit coilichie are equal). Q. E. D. Ex. C"onstruct exac(tlv an equilateral triangle, each of whose sides shall be double a given line. 40 BOOK I. PLANE GEOMETRY PROPOSITION IX. THEOREM 99. In an isosceles triangle the angles opposite the equal sides are equal. B AL- * C Given the isosceles A ABC in which AB=BC. To prove Z A = Z C. Proof. Let BD be drawn so as to bisect Z ABC. Then, in the ~ ABD and DBC, AB=BC. Iyp. Also BD BD, Ident. And Z ABD -Z CBD. Constr... A ABD A CBD, Art. 96. (two & are equal if two sides and the included Z of one are equal, respectively, to two sides and the included Z of the other.).. ZA= ZC, (homologous A of equal ), Q. E. B Ex. 1. On a given line as base, construct exactly an equilateral triangle above the line and another below it. Ex. 2. On a given line as base, construct exactly an isosceles triangle whose leg shall be equal to a given line; make the same construction below the given line and join the vertices of the two isosceles triangles, TRIANGLES 41 PROPOSITION X. THEOREM (CONVERSE OF PROP. IX) 100. If two angles of a triangle are equal, the sides opposite are equal, and the trilangle is isosceles. B A a Given the triangle ABC in which ZA = ZBCA. To prove AB =BC. Proof. If the sides AB and BC are not equal, one of them must be longer than the other. Let AB be longer than BC. On AB mark off AD=BC Dravw DC. Then, in the & ABC and ADC, AD=BC, Constr. AC=AC, Ident. ZDA C= ZBCA. Hyp.. A )DAC=A BCA, Art. 96. (two A are equal if two sides and the included Z of one are equal, respectively, to two sides and the included Z of the other). Or a part is equal to the whole, which is impossible. Ax. 7. Hence AB cannot be greater than BC. In like manner it may be shown that AB is not less than BC. Hence AB=BC. ~ _________~__ Q. E. D, Ex. 1. What method of proof is used in Prop. X? Ex. 2. If in a triangle DEF, Z D)=36~, / E=3(~ and DF=12, find EF', Draw a figure and oI it mark the value of the parts named, 4 t2 1300K I. PL.ANE GEOMETRY PROPOSITION XI. THEOREM 101. Two trian~gles are requal -if thre thwre sPiebs ofon are eepual, respeectiet~cy, to thie three sid1C8 of the othier. B.A C D Given the AL ALC and DEE in which AIBWDE, -BC= FF, and AC=> P. To prove A AI))C= A DEF. Proof. Place the A ARIC so that its longest side AC,4 shall coineide with its equal DlF in the A\ J)EE anid the vertex B shall fall on the opposite side of DF? from -E. Geoin. Ax. 2. Draw the line EB. Then DE=DB (Hyp.):,. A DEB is 'isosceles. Def.. Z p =Z 1, Art. 99. (in an isoseeles A tile A opposite Nhe equal Sides are equal).. Iii like manner, in the A BEE, Zq Z s. Zp)+ Z q Zr + Zs, A x.. Or ZD1)EE Z DBYl Ax.6 A DEE =A -DEEF Art. 96. At.oz are equal if tvo si'des and the included Z of one are eyual, respectirely, to twvo sidles ant the 'inilched Z of th~e other-). AAB C=A DEE. Axi. Q. E.D. Ex. 1. Construct two equilateral triangles on thesame base, one above and the other below, and join the two vertices. Prove that the line joining the vertices bisects the vertex- anules, and also bisects the base at right angles. Ex. 2. Hence, at any point in a given straight line, construct exactly by use of ruler and compasses a perpendicular to that line. TRIANGLES 43 PROPOSITION XIIL THEOREM 102. Tio right triangles are equal if the hypotenuse ant a leg of one, are equal to the htypotenase and a leg of the other. B E A C A'D Given two right & ABC and DEE having the hypote'nuse AB ==hypotenuse DE, and BCZ=EE,. To prove A ABC= A DEFR Proof. Place the A ABC so that BC shall coincide with its equal, RF, and A fall on the opposite side of RE, f rom D, at Al. Geom. Ax. 2. Then A'Eanid ED will formia straighlt line, A'EJ) Art. 74. (if two adj. A are together equal to tw~o r-t. A, thcir ext. sides forms one and the some straight line). But A'E-=E1). Hyp. A A'E D is isosceles. Def.. Z A'=-ZD, Art. 99. (in an isosceles A the A opposrte the epucd sides are equtal.). A A'EE- A DEE, Art. 98. (two right & ar-e eqmal if the hypolenuse and an acute Z of one are equal to the hypotenuse and ant acute Z of the other. Or A ABC= A DEE. Ax.i, ______________ ~~Q. E. D. Ex. 1. By making the same constructionA as in Ex. 1, p). 42, bisect any given straight line.o Ez.Biseict any given angle AO01,-a 44 - BOOK I. PLANE GEOMIETRY PROPOSITION XIII. THEOREMA 103. An exterior anigle of a triangle is greater than either opposite interior an~gle. B E~- / A- D C Given Z BCD an exterior Z of the A ABC. To prove Z BCD greater than Z ABC or Z BA C. Proof. Let E be the middle point of the line BC. Draw AR and produce it to F, mnakinig FER- AR. Draw EC. Then, in the A AEB and FEC, AE - FE, and BE = O-E, Constr. Z~ BRA Z PERC(being vertical A). Art. 78. A AEB A FRC, Art. 96. (twvo A are equal it' two sides and the included Z /' one are equal, reCspectively, to twco sizes and the inlu(lced Z of the other). A. ABE, F = Z6FRE, (being homologous A of equal A). But /BCD is greater than Z FCE, Ax. 7. (the wvhole is greater than any of its parts). Substituting L ABE for its equal Z FCE, Ax. 8. Z BCD is greater than Z ABE, that is, than Z ABC. Similarly, by drawing a line from B through the midpoint of AC and by producing BC through C to a point H, it may be shown that A(ACH (= ZBCCD) is greater than ZBAC. Q. E. D. Ex. On a given line, as base, construct exactiy an isosceles triangle each of whose legs equals half a given line. TRIANGLES 45 PRoPosITIoN XIV. THE OREM 104. If two sides of a triangle are unequal, the angles opposite are unequal, and the greatr angle is opposite the greater side B Given the side BC > side A-B in the A ABO. To prove Z BA C greater than Z C. Proof. On the side BC take B)D equal to AD and draw A-D. Then, in the isoseeles AABD, Z r== Zs, Art. 99. (in an isosceles a the Zis opposite the equal Sides are equal). Z BA C is greater than Z r, Ax. 7. (the whtole is greater than any qf its parts). Z BA C is greater than Zs. Ax. 8. But Zs is an exterior Z of the A ADC. a s is gr'n-eater than Z C, Art. 103. (an eXt. Z of a A is greater than either opposite imit. Z Much more, then, is s ZiB C (which is greater thian Zs) greater than Z C, Ax. 12. (if, of three quantities, the first is greater than the secomd, and thes oacowlcl is greater thtan the third, then the first is greater than the th0(1). Q. E. D. 105. NoTE. The essential steps of the above proof may be arranged in a single statement, thus: LZtA(C> r =Zs> Z C.'. Z BA C is greater than Z C. Ex. 1. Which is the longest side of a right trianjae T of an obtuse trianole 1 Ex. 2. Construct exactly an equilateral triangle, each of whose Bides is half a given line. 46 BOOK I. PLANE GEOMETRY PROPOSITION XV. THEOREM (CONVERSE OF PROP. XIV) 106. If two angles of a triangle are lunequal, the sides opposite are luequal, tand the greater side is opposite the greater angle, B Given ZA greater than Z C in the A ABC. To prove BC > AB. Proof. BC either equals AB, or is less than AB, or is greater than AB. But BC cannot equal AB, for, if it did, Z A would equal Z C, Art. 99, (being opposite equal sides in an isosceles A). But this is contrary to the hypothesis. Also BC cannot be less than AB, for, if it were, Z A would be less than Z C, Art. 104. (if two sides of a A are unecqual, the X opposite are unequal, andl the greater Z is opposite the greater side). This is also contrary to the hypothesis... B >AB, (fr' it neither equals AB, nor is less than AB). Q. B. Bo Ex. 1. Draw a triangle the altitude of which falls on the base produced. What kind of a triangle is this? Ex. 2. Draw a triangle the altitude of which coincides with one side. What kind of a triangle is this? _ Ex. 3. By exact use of the ruler and compasses, draw a perpendicular to a given line from a given point without the line,. TRIANGLES 47 PROPOSITION XVI. THEOREMI 107. If two triangles have two sides of one equal, respectively, to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second. B E A, Ch' F F Given the A ABC and DEE in wiich AB1 =DE, BC=EF, and L ABC is greater than ZE. To prove A C > DF1 Proof. Place the A DEF so that the side DE coincides with its equal, the side AB, and F takes the position 7'. Geom. Ax. 2. Let the line BH bisect the Z F1 C and meet the line AC at H. Draw 1tH. Then, in the A F'BE and EHC, Y'B=BC, Hyp. BH= 3BH, Ident. Z F'BH= Z CBHT. Constr.. A FBH= A BHC. (Why?).. F11 = CI1, (ho1mlogous sides of eqlul A). But A H + F F > A F, Art. 92. (the sum of any two sides of a A is greater ltan the thi'rd sile). Substitutilg for HFJ' its equal HC, All + HC, or AC > AF'. Ax, 8... AC > DF. Ax. 8. Q. E. D. Ex. 1. Draw a figure for Prop, XVI in which the sides and angles are of such a size that If, falls within thle triangle AJBC. Ex. 2. Draw another figure in which "y falls on the Bide AC, 4S 43 VO1O0K t. iPLAVN1 GEOIFO~RY PROPOSlITON XVII. THEOREM (CONVERSE OF PROP. XVI) 108. If two sides of a triangfle are equial, respectively, to twol Sides oJ anothetr triangl~ye, bat the third side of the, first is greater than the thlrtd sidle of the seco~nd, the.n the angl~e op)posite the third side? of the first triangqle 'is greate) than the angle opposite the third side of the second. B A - E Given the A ABDC and D EE having EF, but A C > _DE. To prove Z B greater than Z E. Proof. The Z B either equals Z E, or or is greater than Z E. AB=D-E, BC= is less than Z, But L Bdoes not equal Z E, for, if it did, A ABC would= A DEF, Art. 90-. (twvo As are equal if twoe sidles and the incinded Z of one are equal, 'respeetively, to twvo sidles and the iueiluded Z of the other), and A(,C wouleV, eqmal 1)I (homologous sides of equal A which is contrary to the hypothesis. Also if Z B were less than Z E, side A C would be less than side DE, Art. 107. (if twvo A \ hare two sides of onie equal, 'respectioelt, to twoe sidles of the other, but the inchlued Z of the first greater that the inicluded Z of the second, them the third side of the fi rst is greater than the third side of the second). But this is also contrary to the hypothesis. Hence Z B is oreater than Z E, (for it neither equals Z E, nor is less than Z E). Q Ex. On a' given line (1) con- m struct a triangle whose other two sides are equal to two given lines (m and 4) LiNts 4 49 PROPERTIES OF LINES PROVED BY USE OF TRIANGLES PROPOSITION XVIII. THEOREM 109. Of lines drawvn from the same point in a perpendicultar and cutting off unequal segments from the foot of the perpendicular, the more remote is the greater. P -I? P Given P0 J- AB, PT anid, PQ oblique to AB, and OT > OQ. To prove PT'> PQ. Proof..Produce I1O to the point PI makingy OP'= OP. On AB take ORzz= 09. Draw PR, P'Q, P'R, P'T. Then PQ =P-R, Art. 79. (if frow a poinit in. a I,a qfiren line, twvo oblique lines be drawn, eattiug oftfon thre given linte equal segments frou the foot of the I, the oblique lines are equal). In A PTIP', PT + TP > P-R~ -RI-" Art. 95. (~'f, from, a point ni thiin a A, two hines be drown~ to the extrenitides of' a sitde /fthe A, the sumi of the other tw~o sidjes of the A is greater tan, the sian of thre twvo lines so dIrawvn) But OT is.1. PP' and PT and P-T cut off equal segmlents, P0 a(-nd P)!9, from the foot of the I- AO0. Hence PT =P'lT. In like manner PR PIRK Art. 79. ilenee, by substitution, 2 PT> 24 PR. Ax. 8. PT > P.Ax. 10. PT > P9Q. A x. 8. Q. E. D. D 50 BOOK I. PLANE GEOMETRY PROPOSITION XIX. THEOREM (CONVERSE OF PROP. II) 110. Equal oblique lines drawn from a point in a perpendicular cut off equal segments from the foot of the perpendicular. Given PC PR=PT. To prove ~ AB, PR and PT oblique to AB, and CR= CT. Proof. In the right A RPC and CPT, PC= PC. Ident. Hyp Also PR -PT..*. A RPC=/A CPT, Art. 1, (two right,A are equal if the hypotelnuse and a leg of one are equal t, t/h hypotenuse and a leg of the other)... RC= CT, (homologous sides of equal A). Q. E. D. 111. COR. Of two unequal lines drawn from a poi2n in a perpendicular, the greater line cuts off the greater segment from the foot of the perpendicular. Thus, if PT > PQ (Fig. of Prop. XVIII), OT cannot = OQ (Art. 110); nor is OT < OQ (Art. 109).'. OT > OQ. Hence, also, from a given point only two equal straight lines can be drawn to a given line. LINES 5 1 LIES5 PROPOSITION XX. THEOREM. 112. I. Etiery point in the perpendicular bisector of a line is equally distant from the extremtities of the line; and II. Every point not in the perpendicular bisector is unequally distant from the extremities of the line. C Given 0 the middle point of the line AP), 00 I A.B, P any point in 00, and Q any point not in 00. To prove AP=PB, but QA and QB unequal. Proof. I. AP= PB, Art. 79. (if, from. a point in. a I to a given, line, two obique lines be drown cutting oilr on the given line equal seglmen ts, etc. ). II. Since Q is not in the line 00, either A Q or QB must~ cut the line 00. Let AQ intersect 00 in the point Ra and join -RB. Thea AR= RB, (by first port of' thtis theorem). To each of these equals add RQ. Then AR + RQzzRBD-VRQ. A x.2. But RR-~+ RQ > QR. ~ W h y) by substitution, AR -F RQ, or AQ > QB. Ax. 8. Q. E. D. 113. CoR. Twco points each eqnitiistant froin the extremities of a line determnine the perpendicular bisector of the line. This corollary gives a useful method of determiniiing tme, perpendicular bisector of a given straight line,, by deterinining two points only of the perpendicular bisector, P- ) 0Z 52 130COOR 1. PIANE (IEO]MEARY LOCI 114. DEF. The locus of a point is the path of a point moving, accordingr to a given geomietric law. Thus, if a point move in a plane so as to be always two inches distaut fromn a given poinlt, its locus is the circurnferenc~e Of a Circle whose center is the given point, and whose radius is a line two inches in length. Thus, also, the, locus of a point; moving so as to be cqniidistant from two given p. rallel lines is a straight line lying midway between the two given lines. The locus of a p)oint may (consist of two or more separate lines or parts. Thus, the locus of a point moving so as to be always at a given distance fromn a given line is two lines, one on either side of the given line, at the given distance from it. 1 15. Demonstration of loci. In order to prove that a given line is the locus of a given point moving accord inp, to a gYiven geometric law, it is necessary: 1. rp prorne tat eieery )oint in the, gi~een. line sat isfies the givcen? hf law eor C-iti'i)l.. 26. To provec that eeevry 1.oint not in the gfiieen line (does not Satisfy the giive'n law? Or co)?dition'. Instead of 21, it may be proved that every point which satisfie~ the given condition lies in the given line. Hence, in Prop. XX it has been- proved that the per pendicular bisector of a line is the locus of all points eq'aidistant from the extremities of thie line. 116. Use of loci. Loci are useful in determining a point (or points) which shall satisfy two or move geomet-' LOCI 53 rical conditions. For, by finding the locus of all points which satisfy one oifthe given conditions, and also finding the locus of all points which satisfy a second condition, and then finding the intersection of these two loci, we obtain the point (or points) which satisfy both conditions at the same time Thus, if it be required to find the points which are two inches from one given point and three inches from another given point, the two given points being four inches apart, the required points are the intersections of the circumferences of two circles. Let the pupil make a construction and obtain the required points. Ex. 1. Draw the locus of a point moving at the distance of one inch from a given point. Ex. 2. Draw exactly the locus of a point moving at a distance of one inch from a given line. Ex. 3. Draw exactly the locus of a point moving so as to be equidistant from the extremities of a given line one inch lonlg. Ex 4. How many points in a plane are necessary to determine two parallel lines? Three parallel lines? (See Art. 47.) Ex. 5. Are two triangles equal if three angles of one equal the corresponding three angles of the other? Illustrate by drawing a figure. Ex. 6. Draw three isosceles triangles on the same base and connect their vertices. What truth is illustrated by this figure? Ex.7.7. Draw a straight line and locate a point 2 inches from it. By the use of loci, locate the points which are 11' inches from the given line and at the same distance from the given point. 54 54 ~~BOOK L. PLANE GEOM IETE PROPOS ITION XXI. THEOREM 117. I. Every point in the bisector of an angle is equi - distant fromn the sides of the antgle; and II. CONVERSELY, everY -point 07qI4id'sftlnt from~ the sides of and angle l-tes in the bisector of the angle. A Q B I. Given PB the bisector of the angle ABC, P any, Point in PB, PQ( I BA, anjd PR I BC. To prove PQ = PR. Proof. In the rt. A\ PBQ and PBR, PB PB. Ident. Also Z PB3Q Z PBR. Hp A ITQ~APBR, Art. 98. (two rt. A~ arc equal if the hypotenuse aud an acute Z of one, etc.). Hence PQ =PR, (homologous sides of equalA) II. Given Z A BO, PQ I AB, PR I BO, and PQ = PR. To prove that PB is the bisector of Z ABC. Proof. In the right A& PBQ and PBR, PB= PB. Ident. Also PQ= PR. Hyp. LOCI 55.'.A PBQi=APBR, Art. 102. (two rt. A are equal if the hiypotenu~se and a leg of one, etc.). A- Z ABP= Z CRPl (htomologous Z of equal A). Or Z ABC is bisected by BP. Q. Li. D. 118. Cop.. In. Prop. XX.T it has been proved that the bisector of an angle( is the lo(-as of all point's equidistant from the sides of 1/u (tnt/i, for it haes beeti I)voved that every point in the, given line satisfies the given law or condlition, and that every p)oint which satisfies the giyell condition lies in the given line (see Art. 11;5). 119. DEP. A transversal is a line that intersects two or more other lines. Thus, E F is a trans - versal of the Fines AB and CD. If two lines are cut by a transversal, it is convenient to give the eight angles of intersection special names. A C, F a, b, g, h, are called exterior angles. c, d,1 e, f, are called interior angles. c, f form a pair of alternate-interior angles. b, fform a pair of exterior-interior angles on the samne side of the transversal. Let the puipil name another pair of alternate-interior angles; also name canother ipair of exterior -interior anglers on the same side of the transversal; also namne a Vpair of interior angyles on the scame side of the transversal. BOOK I. PLANE GEOMETRY PARALLEL LINES 120. DEF. Parallel lines have already been defined (Art. 41) as straight lines which lie in the same plane and do not meet, however far they be produced. What is the fundamental axiom concerning parallel lines? (see Art. 47.) PROPOSITION XXII. THEOREM 121. Two straight lines in the same plane, perpendicular to the same straight line, are parallel, P Given the same plane. To prove lines AB and CD I line PQ and in the AB 11 CLD. Proof. If AB and CD are not parallel, they will meet if sufficiently produced. Art. 120. We shall then have two i from the same point to the line PQ. But this is impossible, Art. 80. (from a given point without a straight line but one I can be drawn to the line). Hence AB and CD never meet. ', AB and CD are parallel. Def. Q. E. D. PARALLEL LINES 5),7 1 22. COR. Tivo-staight lines parallel to a third straight line are parallel to each other; Lines parallel~ to parallel lines are parallel; Lines perpendicular to parallel lines are parallel; Lines perpendliealar to non-parallel lines are not parallel. PROPOSITION XXIII. TtiEOREM1 123. If a straifflit line is pJen' lt)' to W one ofCtO gitlen parallel lincs 'it 'is perpen~dicutlar to the othier also. A -~~~~B CD Given Al-, II ("D, and P9 I AB. To prove PQ I CD. Proof. Let CF be drawn 1- PQ at C. Then C-F II A 1, Art. 121. (two straight lines in. the samiepliine IL sanwe straight line are ) But CD 11 AB. Hp CF coincides with CD9 Geom.Ax.3. (through a given point one straight, lin)e, anad only one, can be drawn another gi~ren~ straight line). But Prq I CF. Constr. Hence P91 ~- CID, (for CD) coincides withi CF, to which, PQ is L). 58 BOOK I. PLANE GEOMET1 Y PROPOSITION XXIV. THEOREM 124. If ftwo parallel straight lics aret ct by a4 transversal, the alternate interior (anlgles atre equal. P o — _i___ __D___ Given te 1 ines A ld CF b the trve Given the | lines AP3 and CDi) cut by the transversal PQ at the points G and E respectively. To prove Z AGE = Z GED. Proof, Through R, the middle point of EG, let the line HF be drawn I AB. Then HF I CD, Art. 123. (if a straight line is 1- one of two I| litnes, it is. the other also). In the rightit GR0 and ERF, GR = ER, Constr. Z GRH = Z ERF. (Why?).. A GRH = A ERF, Art. 98. (two right & are equal if the hypotenuse and an acute Z of one the hypot. and an acute z of the other). Z. HGR = Z REF, or, Z A GE = Z GED, (homologous A of equal A). _____ I the aove Qgure let the pupl sw. E. D. ]x. In the above figure let the pupil show that Z BGE = Z GEC. PARALLEL LINES 59 PROP. XXV. THEOREMI (CONVERSE OF PROP. XXIV) 125. If two straight lines are cut by a transversal, making the alternate interior angles eqaiu, the two straight lines are parallel. A -B~~~~~~'Given the two lines AB and CD cut by the transversal PQ at the points F and G, making Z AFG = Z FGD. To prove AB II CD. Proof. Throngh F let the line KL be drawn 1I CD. Then Z KFG=Z FGJ), Artt. 124. (if two fl st. lines are cut by a transversal, the ciit. iut. A are equal). But Z AFG Z FGD. Hyp. Hence ZKFG6 ZAFG. Ax 1. ILL coincides with AB. But KlL C JD. Constr. Hence AB H CD, (for AB coincides with XL, which is 1I CD). _ _ _ _ _ _ _ _ _ _ _ _ _ _ ~~Q. B. D, Ex. 1. If Z BFG F FGC, prove that AB and CD are parallel. Ex. 2. By exact use of ruler and compasses, at a a giveu point (P) in B O p a given straight line (OA) construct an angle equal to a given Z (B). Ex. 3. By exact methods, through a given point draw a line parallel to a given line. 60 1300K1 1. PLANE G(1EOMETRY PROPOSITION XXVI. THEOREM 126. If two (para(llel lincs are cut by a t'ranstversal, the exterior interior a yles (re equall. ypD Given the 11 lines AB and C D cut by the transversal PQ at the points F and G respectively. To prove Z PIFB, Z FGD. Proof. Z I'FB - Z AFG. (Why?) Z FGD = Z A1 G, Art. 124. (being alt. Mnt. A of parallel lines)... Z PFBI = Z FGD. Ax. 1. In like manner it may be shown that Z 1PA = Z TFGC Q. E. D. PROP. XXVII. THEOREM (CONvERSE OF PROP. XXVI) 127. If t uo straigit lines are c(t y a! trtanstersal,,ma(ing the exterior interior atgles equal, the two straight lines are parallel. Given, on Fig. of Prop. XXV, Z PFB Z FGD. To prove AB 11 CD. Let the pupil supply the proof. Ex. If, in the Fig. to Prop. XXVI, Z PFB equals 67~, find the othtL seven angles in the figure without measuring them, PARALLEL LINES 61 PROPOSITION XXVIII. THEOREM 128. If two parallel lines are cut by a transversal, the sum of the interior angles on the same side of the transversal is equal to two right angles. A- B --- ------ Given the 11 lines A and CD cut by the transversal PQ at the points F and G respectively. To prove Z BFG + Z FGD 2 rt. X. Proof. Z FGD = Z PFB. Art. 126. To each of these equals add Z tBFG. Then BFG + Z FGD = Z PF i + Z B7FGA. Ax. 2. But Z PFB' + Z BFG =2 rt. z. Art. 73.. LBFG +- FGI)=2 rt.. Ax. 1. Q. E. D. PROPT. XXIX. THEOREM-tR (CONVERSE OF PROP. XXVIII) 129. If two straight lines are ceft by a transversal, mnaking the sum) of thfe interior tangles on. tlhe same side of the transversal eql q tol two t rig7h angles, the two lincs are varallel. Given, on Fig. of Prop. XXV, Z BFG + / FGD = 2 rt. X. To prove ARB | CD. Let the pupil supply the proof. Ex. If, on Fig. of Prop, XXVIII, Z F'lB -t- 'Z QIGD=180, are AB.nd CDl |? 62 B]OOK I. PLANE (1GEOETRY PROPOSITION XXX. THEORE-I 130. Two angles,whose sides tare pa(ratllel, each to each, are either equal or suq)pllementary. A / ^Q D H Given AB ii DH, and BC 11 GF. To prove A ABC, DEF and GEH equal, and A AB~ and DEG supplementary. Proof. Produce the lines BC and HD to intersect in P. Then Z ABC = Z QPR, and Z QPR = Z DEF, Art. 12;. (being ext. int. A of II lines). /. Z ABC = Z DEF. (Why?) Also Z DEF = Z GEH. (Whyv).. Z ABC = Z GEi. (Why?) Again X DEE and DEG are supplementary. Art. 73. -. ABC and DEG are supplementary. Ax. 8. Q. E. D. 131. NOTE. It is to be observed that in the above theorem the two angles are equal if, in the pairs of parallel sides, both pairs extend in the same direction from the vertices ( A B and DEF), or both pairs in opposite directions ( B and GEH); and that they are supplermentary if one pair extends in the same direction, and the other pair in opposite directions (AB and DEG). The directions of the sides are determined by connecting the vertices of the angles and observing whether the lines considered lie on the same side or on opposite sides of the line drawn. PARALLEL LINES PROPORITION XXXI. THEOREML 132. Twvo angles whose sides are perpendicular each to each are eith-er equal or supplementary. F D A - - - - - - B C Given BA I EF, and BC I _DI)' To prove Z ABC= Z BIN), anid X -ABC and FED' SUPplementarv. Proof. At B let line BK be drawn I BE and in the same direction with BA; also EG I DD' and in the same direction with BC. Then BA 11 BK, and BC fl EG, Art. 121. (two stra-ighit linies in the samie plante, 1. the samie straight linle, are J) Z ABC= ZKBVVG, Arts. 130, 131. (two A wh~ose sidies are fl, each to each, anil extendl in the samie direction101 fromi the vertices are. But Z KEG is complement of Z DEK, Art. 33. (frZ DEG is a rt. Z by conistr.) Also Z FED is- complement of Z DElL, Art. 33. (fior ZEFEK is a r-t. Z by contstr,). ZBBD=ZKEG Arti. 5 (compl)emienits of thre samie / a re-. Z ABC= ZFED. Axi. Bnt Z BE D' is supplement of Z FED. Art. 34. ZF DL' is supplemient of ZABC'. Ax. 8. Q. E.D 1 33. NOTE. In the above theorem the two angles are equal if the Sides, considered as rotating about tho vortices, are taken in the samie order (thns BC' is to the rikht of BA, and ED to the. right of 1B" Z ABC= ZEFED); -but the angles are supplementary if the corre,4 -Ponding sides are takeni in tho opposite order (thus, BC is to the rili~Ot of BA but ED' is to the left of BEF.'. Z ABCi~supplement of Z FEP'). 64 OOK I. PLANE GEOMETRY PROPOSITION XXXII. TH:EOREMI 134. The sum) of the angles of a t rityjle is equal to two right cangles. F, Given the A ABC. To prove ZA +ZB +ZB Z CA-= rt, X. Proof. Produce tlhe side AC to the point 7) 1and through C let the line CE be drawn Tl A. Then Z ECr() + Z BCE + Z -;CA= 2 rt. A, Art. 77. (th1e sum of all the A Z, abt a point on thie samet side of a straight line lpassing thlouyhl the point '2 rt. A ). But ZECD = ZA, Art. 126. (being ext. int. A of parallel lines). Also Z BCE = Z B, Art. 124. (7)cig alt. intf. A of parallel linles). SIubstituting for Z ECD its equal, Z A, and for Z 1,CE its equal, /Z I, Ax. 8. Z A + Z B + Z BCA - 2 rt.. 0. E. D. 135o COR. 1. An exterior angle of a triangle is equal to the sum of the tiwo opposite interior angles. 136o COR. 2. The sum of any two angles of a triangle is less than two right angles. 1370 COR. 3. In a right triangle the sum of the two acute angles equals one right angle. PARALLEL LINES 65 138. Corz. 4. A triangle can have but one right, or one obtuse angle. 139. CoR. I5. If two angles of one triangle equal two an~gles of another triangle, the third angle, of the first triangle equals the third angle of the second. 140. CoR. 6. If an acute an~gle of on~e right tria~ngle equals an acute angle of another reight triantgle, th~e retabbn inig acute an~gles of the triangles are equal. 141. CoR. 7. _Tao triangles are equtal if tao angles and a, side of one are equal to twvo angles and the homologous side of the second. 142. COR. 8. Taco right tr~ianigles are equal 'if a leg and an acute angle of one are equal to a leg and the homologous acute angle of the other. Ex. I. If two anglIes Of a triangle are 560 and 620', find the remaining angle. Ex. 2. if one acute -angle of a right triangle is 360 15', find the (ther acute angle. Ex. 3. low many dlegrees in eacht angle of an equilateral triangle? Ex. 4. How many de-rees in each acute an-de of,in i sosceles right triangle? Ex. 5. Is it possible to have a triangle whose anglIes are 450, 620c, 720? EX. 6. If one angle of a triangle, is 420, find the sum of the other two -angles. Ex. 7. If two angles of a triangle are 380 and 650, find all the eIXterior angles of the triangle. Ex. 8. IC the vertex, ang-le of anr isoscelles triangle is 380, finld each angle a t t'le l,)ase. Ex. 9. If an ang-le at the base of an Isosceles trian gle is 5iP, find the, vertex auri. EX. 10. in exterior anode at the base of an isoscelts triangle is 102", find( all th ean gles of the trianigle. E 66 BOOK300 I. LANE GEOMETRY QUADRILATERALS 143. A quadrilateral is a portion of a plane bounded by four straighlt liles. The sides of a quadrilateral are the bounding lines; the angles are the angles made by the bounding lines; the vertices are the vertices of the angles of the quadrilateral. The perimeter of a quadrilateral is tlh]e sum of the sides. 144. A diagonal is a straight line joining two vertices that are not adjacent. 145. A trapezium is a quadrilateral nio two of whose sides are paralllel. 146. A trapezoid is a quadrilateral which has two, and only two, ot its sides lparaillel. 147. A parallelogram is a quadrilateral whose opposite sides are parallel. Trapezium Trapezoid Paraillelogramn 148. A rhomboid is a parallelogram whose angles are oblique angles. 149. A rhombus is a rhomboid whose sides are equal. 150. A rectangle is a parallelogram whose angles are right angles. 151. A square is a rectangle whose sides are equal. Rhomboid Rhomlus Rectangle Square QUADRILATERALS 67 152. The base of a parallelo- c E gram is the side upon which it is I supposed to stand, as AB. The opposite side is called the upper A F base (CD). The altitude of a parallelogram is the perpendicular distance between the bases, as EF. 153. The bases of a trapezoid are its two parallel sides. The legs of a trapIzoid are the sides which are not parallel. The altitude of a trapezoid is the perpendicular distance between tile bases. T'le median of a trapezoid is the line joining the midpoints of the leg s. 154. An isosceles trapezoid is a trapezoid whose legs are equal. Ex. 1. Draw a quadrilateral with three acute angles and one obtuse angle. Ex. 2. Is every rhombus a rhomboid Is every rhomboid a rhombus? Ex. 3. What is the difference between a square and a rhombus? What properties do they have in common? Ex. 4. Find the perimeter of a square foot in inches. By aid of the following classification: Trapezium. Quadrilateral. Trapezoid..... Isosceles trapezoid. Parallelogram Rectangle... square. Rhomboid.. rhombus. Ex. 5. Determine what four names the rhombus is entitled to. Ex. 6. Determine what properties the rhombus, square and rectangle have in common. Ex. 7. A diagonal of a rhombus divides the rhombus into how many triangles? What kind of triangles are these? 68 BOOK I. PLANE GEOMETlY PROPOSITION XXXIII. THEOREM 155. The opposite sides of a parallelogram are equal, and its opposite Iangles a)re also equal. A D Given the parallelogram ABCD. To prove AD B, = BCA, =DC, Z= ZD, and Z LB ) — Z BCD. Proof, Draw the diagonal AC. Then, in the A BC and ADC, AC= AC, (Why?) Z BCA Z = Z CA.D, Art. 124. (being alt. int. A of parallel lines). ZBAC= ZACD, (sa)ne reason).. A ABC= AA CD, Art. 97. (two /A are equal f two A and the if inci(cld side of one are cqual rcss)cctirclf to two A anl the inclt(et(Id site of the ot her)..'. BA = 1C, tB= 7)C and Z B = Z L), (hom)ilo/uls pa'rts (of' cwual A). In like manner, 1b drawing the diagonal BD, it may be proved that Z BAD= Z BCD. Q D 156. COR. 1. A diagonal divides a parallelogram into two equal triangles. 157. COR. 2. Parallel lines comprehended between parallel lines are equal. 158. COR. 3. Two parallel lines are everywhere equidistant. __ Ex. 1. In the above figure, prove Z BAnD= L BCD by use of Ax. 2. Ex. 2. Prove the opposite angles of a parallelogram equal, by use Of Art. 130. QUADRILATERALS 69 PROPOSITION XXXIV. THEOREMS 159. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram. ------ I Given the quadrilateral ABCD in which AB= CD and BC=AD. To prove ABCD a Z7. Proof. Draw the diagonal AC. Then, in the A ABC and ADC, ACA=C. (Why?) BC= AD. (Why?) AB= CD. (Why?) 'o A ABC=A ADO. (W hy?).. Z BAC= Z A CD. (WNhy?).'. AB I| CD, Art. 125. (if two lines are cut by a transversal, 'makcig tihe alt. int. A equal, the lines are [i). Also Z B(A. - Z A. D. (Why?). BC AD. Art. 125..'. ABCD is a _7, Art. 147. (a '=7 is a qua(tdrilatera l 'iltose oppo) ite sides are j). Q. E. Do 70 BOO7K I. PLANE GEOMIETY PROPOSITION XXXV. T HEOREMl 160. If two sides of a (luladrilateral are equal and par" allel, tihe other two sides are equal (ad p(tarllel (11d tlhe figure is a parallelogram. Z — ---- -C D/ Given the quadrilateral ABCD in which BC = and! AD To prove ABCD a DZ7. Proof. Draw the diagonal AC. Then, in the A ABC and ADC, AC=AC. (Why?) BC==AD. (Why?) Z BCA- Z CAD, Art. 124. (being (ltt. int. A of p)arallel linlcS). A 13C A A 1)C. (Why?) /. Z L C= Z A CD. (Why?).A. IB 1| CD, Art. 125. (ij two lines are clut by a tra.ile'rsal, )Ia]kiii the alt. it. A c. llqua, the lines are 11)... ABCD is a /C7. Art. 14' Q. E. D. Ex. 1. Show that in a,53 each pair of adjacent angles is supplementary. Ex. 2. One angle of a parallelogram is 43~; find the other angles. Ex. 3. If, in the triangle ABC, ZA=G60~, ZB=70~, which is the longest side in the triangle? Which the shortest? QUADRILATERALS 71 PROPOSITION XXXVI. THEOREM 161. The diagonals of a parallelogram bisect each otier. B_ ___ ^C Given the diagonals A C and BD) of the ~-7ABCD, intersecting at F. To prove AF=FC, and BF=FD. Proof. Let the pupil supply the proof. [SUG. In the A BFC and AFD what sides are equal, and why? What A are equal, and why? etc.] Q. E. D. Ex. 1. How many pairs of equal triangles are there in the above figure? Ex. 2. If one angle of a parallelogram is three times another angle, find all the angles of the parallelogram. Ex. 3. If two angles of a triangle are j2~ anld ~,o find the third angle. Ex. 4. If two angles of a triangle are x~ and 90~ +X ~, find the third angle. Ex. 5. If one angle of a parallelogram is a~, find the other angles. Ex. 6. Construct exactly a aangle of 600. Ex. 7. How large may the double of an acute angle be? how small? Ex. 8. How large may the double of an obtuse angle be? how small? 72 BOOKR I. ]PLANE GEOMETYt PROPOSI'TION XXXVII. THE(O).EM 162. Tio parallelogrrlamils (aro (eqt l if tCo) adjacentt sidet: andt t te incllt(ed e(I ngle of )one are elt qal, re)spccti ely, to two adjacent sides and the i'ncludcd angle f tl e othlir. A 1 --- A i' Given the 7 A BCl D and A./B! C'D' in whliclh A- IZ =A I' AD-= 'D', and ZA = ZA'. To prove Z7 ABCD = Z7 Ar'B'CD',. Proof. Apply the Z 7 A'B'C'J) to the Z7 ABCD so that A'D' shall coincide with its equal AD. Then A'B' will take the direction of AB (for ZA'= ZA); and point B' will fall on B (for A'B'=,B). Then B'C' and BC will both be 11 AD and will both pass through the point B... B'C' will take the direction of BC, G(eom. Ax. 3. (through a given l)oint one straiglit linec, a(tl otily one, c-,n 7)c dr('(1ct1 i (Iao1It'Cr given straight lil0t<). In like manner, I)/C' Imust take the direction of )C..'. n' lust fall on C, Art. 64. (two straight liines can initerscet in bult ole point)..~. / ABCD = z7 A'B'C'D', Art. 47. (geometric figures wh ich coincide are equal). Q. E. D. 163. COR. Two rectangles woiich have equal bases and equal altitudes are equal. Ex. Construct exactly an angle of 30~, POLYGONS 73 POLYGONS 164. A polygon is a portion of a plane bounded by straight lines, as ABC)DE. The sides of a polygon are its bounding lines; the perimeter of a polygon is the sum of its sides; the angles of a polygon are the angles formed by its sides; the vertices of a polygon are the vertices of its angles. A diagonal of 1a polygon is a straiglht line joining two vertices which are not tdjacent, as ID in Fig. 1. B I A FN E K Fig. 1 Fig. 2 165. An equilateral polygon is a polygon all of whose sides are equal. 166. An equiangular polygon is a polygon all of whose angles are equal. What four-sided polygon is equilateral but not equiangular? Also, what four-sided polygon is both equilateral and equiangular? 167. A convex polygon is a polygon in which no side, if produced, will enter the polygon, as ABCDE (Fig. 1). Each angle of a convex polygon is less than two right angles and is called a salient angle. 168. A concave polygon is a polygon in which two or more sides, if produced, will enter the polygon, as FGHIJK (Fig. 9), 74 BOOK I. PLANE GEOMETRY Some angle of a concave polygon must be greater than two right angles, as angle GHI of Fig. 2. Such an angle is iermed a re-entrant angle. If the kind of polygon is not specified in this respect, a convex polygon is meant. 169. Two mutually equiangular polygons are polygons whose correspondincg angles are equal, as Figs. 3 and 4. ^ _____ aj /a' (['I p /(j ( lj Fig. 3 Fig. 4 Fig. 5 Fig. 6 170. Two mutually equilateral polygons are polygons whose corresponding sides are equal, as Figs. 5 and 6. From Figs. 3 and 4 it is seen that two polygons may be mutually equiangular without being mutually equilateral. What similar truth may be inferred from Figs. 5 and 6? 171. Names of particular polygons. Some polygons are used so frequently that special nalnes have been given to them. A polygon of three sides is called a triangle; one of four sides, a quadrilateral; one of five sides, a pentagon; of six sides, a hexagon; of seven sides, a heptagon; of eight sides, an octagon; of ten sides, a decagon; of twelve sides, a dodecagon; of fifteen sides, a pentedecagon; of n sides, an n-gon. Ex. 1. Let the pupil illustrate Arts. 169 and 170 by drawing two pentagons that are mutually equilateral without being mutually equiangular, and another pair of which the reverse is true. Ex. 2. Can two triangles be mutually equilateral without being mutually equiangular? What polygons can? Ex. 3. How does the number of vertices in a polygon compare with the number of sides? POLYGONS 75 PROPOSITION XXXVIII. THEOREM 172. The szum of the angles of any polygon is equal to two right angles taken as many times, less two, as the polygon has sides. L /2 5 silds 6 si[S 7 sides 3 triangles 4 trilngles 5 triangles Given a polygon of n sides (the above polygons of 5, 6, 7 sides being used merely as particular illustrations, to aid in carrying forward the proof). To prove the sum of its X =(n-2) 2 rt. X. Proof. By drawing diagonals from one of its vertices the polygon is divided into (n- 2) triangles. Then the sum of the X of each triangle==- rt. X. Art. 134. (tihe sult of the A of a A is equal to 2 rt. A). Hence the surm of the X of the (n-2) a = (n -2 ) 2 rt. X. Ax. 4. But the sum of the X of the polygon is equal to the sum of the A of the (n.-2). Ax. 8. Hence the sum of the A of the polygon (n - 2) 2 rt. X. Ax. 1. Q. E. D. 173. COR. 1. The sum, of the angles of a polygon equals (2n-4) rt. A. 174. COR. 2. In an, equiangular polygon of n sides each angle eals (n-2) rt. A 2 - r. n, n 1100K T.. PLANE' GE]"OMETR1" — 1Y PROPOSITIONXXI THERM 175. Thle 81,Im of the C'x/Crior ((fl(/l s (f (t polygon for-me-d four Hig/t ((/1(11(. Given a polygon of v. sides having its sides produced in succession. To prove the sum of the exterior A = 4 rt. A Proof. If the interior A of the polygfon be denoted by A, B, C, and the corresponding' exterior A by a, b, c ZA + Za=2rt. A, (Wy) Z B+ Zb= 2 rt. A, (Why) etc. Adding, Ijut. A + ext. A = it times 2~ it. A =2 P r't. A. Ax. 2. BLAu it. A =(n-2) 2 rt. A ~2 )a rt. A - 4rt. A. Art. 18 Ex t. A =4 rt. A. ______________ ~~Q. E. D. Ex. 1. -What does the sum of the interior angles of a hexa-on equal"? of a heptagon? of a decagon? Ex. 2. Each angle of an equiangular pentagon contains how many degrees? of an equiangular hexagon? octagon? decagon - Ex. 3. Would a quadrilateral constructed of rods hinged at the ends (i. e., at the vertices of the quad rilateral) he ri,-id? Would a triangle so constructed be rigid? Would a pentagon I MISCELLANEOUJS THEORIEM\S 77 MIVSCELLANEOUS THEOREMS PRiOPOSITION XL. THEOREM '176. If three, or more parallels intercept equal parts on one transversal, they 'intercep~t equal parts on every transversal. A F D \H ~~T Given AP, BQ, CR and _DT parallel lines intercepting eqnal parts AB, BC and CD on the transversal AD. To prove that they intercept eqnal parts PQ, QR and RT on the transversal PT. Proof. Throng'h A., B and C let AF, B>G and C~H be drawn parallel to P'T and meeting the lines BQ, CR and DT' in the points F, G and H- respectively. Then the lines AF, BG and CIT are, Art. 122. (twvo straight hues 11a third straight hune are 11 each, other). In the ABF, BCG and (!-DH, Z ABF -= Z BCGG CDHI (being ext. int. A of It lines).Ar.16 Also Z/BAF- Z CBGz ZDCHT, (same reason). Aiit AB= BC C-D. Hyp. LAAI}F= A CG =ALCDI[. Art. 97. IF=-BG — = CII. (Why) Bat AF-J)Q, IBG=OQR, CI=zRT, Art. 157. (parallel lines cuuyireuhendc betwecen paralel lines are equal) PQ Z=-QR =ZRI. A x., Q. D 78 8)0BOOK 1. PLANE G1EOMET]Y PROPOSITION XLI. THEOnREI 177. The line wchich1 joiins the mipoinits of two sides of a triangle is )(arallel to the third side, (cnd is equal to onehalf the third side. A B, Given D the midpoint of AIB, and E the midpoint of AC in the triangle ABC. To prove DE 11 BC and =ABC. Proof. Through B let BL be drawn 11 AC, and meeting DE produced at L. Then, in the A ADE and BDL, ADi=BD, (Why?) Z ADE = BDL. (Why?) Z DAE Z 7) IL. (Why?).A. A E17 I. (Why?).l. iE-DL, or D7)E- LE(, and A- E-S' L. (Why?) But AR=]E (C (Hyp.).. EC- BL.. A 1. Also EC 1 7BL. Constr..B. JLEC is a Z 7, Art. 160. (if o silcs of ta quorilatcral are eltqual auc parallcl el e figure is a )..'. DE 1 B C. Art. 147. Also LE= BC, (opp. sides of a /Z7 are =). Art. 155... ~ LE, or DE=~ BC. Ax. 5. Q. E. Do 178. COR. The line vwhich bisects one side of a triangle and is parallel to another side bisects the third side. Thus, given AD=DB and DE l BC, then will AE=EC. For, suppose a line FG drawn through A 11 BC, MISCELLANEOUS THEOREEMS 79 Then the three parallels FG, DE, BC will intercept equal parts on AB. Hyp. o they intercept equal parts on AC. Art, 176. ', AE=EC. PROPOSITION XLII. THEC2EEM 179. The line which joins the midpoints of the legs of a trapezoid is parallel to the bases ald ecqual to onei-half their sum. -f/ - DY \ Given the trapezoid ABCD, E the midpoint of the leg AB, and F the midpoint of the leg CD. To prove that a line joining E and Fis || AD and BC and A= - (AD + BC). Proof, Draw the diagonal BD and take G its midpoint. Draw EG and GF. Then, in'the A A PB, E, G |1 AD 1 and - A1 ), Art. 177. (the line which, joins the l minipoifts o' tao,.s' o /'s o 'a 2 is 1e the tird side and( =- one-hl(lf the third)l seFItc). Also, in the A EDC, GF ii BC and -= BP, (sat,, rcason)..CT F and Al) botlh 11 B(; oo, C II \ 1), Art. 122. (tiwo straight ulnes 1I a tirtd stral'/t t (atr I\ eatct other);.' EG and GF are both l1 AD. EG and G' form one.lnd the stme straight line EF, Geor. Ax. 3. (thrutgh a givet n point one line, and7 only one, caln, be drawn ji to another gi.cen oinc)..F. EF 1 A1l) and 1C. Also EG - A D, and GF =? BC. Adding, EG -+ GF, or BE - (AD + BC). Ax. 2. Q. E. D. 180. COR. A line drawntt bisecting one leg o(f a trapezoid and parallel to Ihe base bisects the other leg also, 80 3BOO0K I..PLNE GEO! MET'Y' PROPOSITION XLIII. THEOREM 1 81. Tih b5isectors 'of the anglcs qof a tr/, n(Ic inters('ct at a common Ioii (caCled tJhc in-center). B 13,4@ ~ C 4 C A C -- Fig. 1. Fig. 2. Given the A ABC( with the lines AJi, BQ, CRI (Fig. 1) bisecting the Z, BAC, ABC, ACIB, respectively. To prove that AP, BQ, CR intersect in a common point. Proof. Let AP, the bisector of Z IBA1C, and (CR, the bisector of Z BC'A, intersect at the point.0 Thllen, being in bisector AP, is equidistant from AiB and AC, Art. 117. (every point in the bisector of acn / is equidistant fromg the sides of the Z ). Also 0, being in bisector (R, is equidistant from At( and BiC, (SnCe r'etso'). lence () is equi(listlnt from1, the sides.l and I ('. A x, 1..() is in tlie lisector of / AC, Art. 117. (every poiln equidistnt fro the sides of (il Z lies in the ISrctor of' tt e / ). Hence B(Q, or IQ procuced, palsses through 0. Hence the bisectors, AP, BQ, (,lCR, of the three z of the A ABC' intersect at 0. Q..D. 182. COR. The point in which the three bisectors of the angles of a triangle intersect is equidistant from the three sides of the triangle. 1 83. DEF. Concurrent lines are lines which pass through the same point. Ex. Find other Z of above figure if Z BA C = 772~ and Z BCA = 44~. MISCELLANEOUS THEOREMS 81 PROPOSITION XLIV. THEOREM 184. The perpendicular bisectors of the sides of a triangle intersect at a common point (called the circumcenter)o B D \ELD F A F C - F c Fig. 1. Fig. 2. Given tle A A l(C with DP, EQ, lIFt (Fig. 1, thle _ bisectors of tle sides AB, BC, CA1, respectively. To prove tlat DP, EQ, FR intersect at a common point. Proof. Let DP and EQ intersect at tle point 0. Then 0, being in I_ DP, is equidistant fromn the points A and B, Art. 112. (every point in the pe'pendliculcar bisector of a line is equally distantf fromn the extremities of the line). Also 0, being iln _ EQ, is equidistant from thle points B and C, (same ireas8on). Hence () is equidistant from. and ('. Ax. 1... is in tlie I_ bisector of A C, Art. 115. (the _ bisector of a line is the locus of all points equidistanlt from the extremities of the line). Hence Fi'R, 0or FR prolduedl, )passes tlhrough 0. Hence the perpendicutlar bisectors, I)P, F1Q: IFJ,' of tbe three sides of the A ABC mIeet in tle point ). Q. E. D. 185. COlt. T71c l)poi, ins i tIlllie tie pesl),oiclulr tis<etors 0' the sides of a triu /lle meet is equidlistc tl from thle vertices of the trian gl(. F EOOK I, PLANTE G4EORIMET PROPSITIO.N XLV. THEoREM_) 1 86, The perpendiculars from? the, sertices of a triangle to thje opposite sidcs mneet in a pondi (callrd the ortho-center). B F /C /, Given AD, BE, and CE the perpendieulars from the vertices A, B, and C of the A ABC to the opposite sides. To prove that AD1, BE, and CE intersect in a cornmnon point. Proof. Throngyh 4, B, and C let the lines P1R, P0~ and Qbi be drawn ii BC, AC, and AB, respectively-, and forming the A PQR. Tlhen A )IPArt. 123, (ber AD I _BC, and a, line I one of two it bines is I the othier als'o). Also A PRO3 a nd ARO R are =. Constr. A,4 P = DC, a n.d A R =B C, Arit. 155. (thIie oppo s Ite s id(es of a = a re..AP =AB. Ax. 1. lie-nce, in the A PQR, AD is the perpendienlcar bisector of side -P.R. In like manner it may be shown that BE is the perpendienlar bisector of PQ, and that CE is the perpendienlar bisector of QR. Hence AD, BE, and (JR are the perpendienlar bisectors of the sides of the A PQR. AD, BE, and CE, meet in a common point, Art. 184.. Mohepependicular bisectors of the sides of a A~ are concurrcnt). Q. E. D. MISCELLANEOUS THEOREMS 83 PROPOSITION XLVI. THEOREM 187. The medians of a triangle intersect (or are concurrent) in a point (called the centroid) which cuts off twoTo prove that AD, BF and CE intersect at a point which cuts off two-thirds of each median from its vertex. Proof. Let the medians AD and CE intersect at O. Take R the midpoint of AO, and S the midpoint of 0C, and draw RE, ED, DS, and SR. Then ED 1i AC and = AC. Art. 177. Also, in the A A OC, Rs 1| AC and = AC. Art. 177. Hence ED 11 RS and == S. Art. 122 and Ax. 1.. REDS is a Z7. (Why?) Hence ES and RD bisect each other. (Why?).But AR=RO, and CS=SO. Constr... AR= O= OD, and CS=S=-OE. Ax. l. Hence CE crosses A)D at a point 0, such that A 0=- AD.i In like manner it may be shown that BF crosses AD at the point 0. Hence the medians AI, BE, and CE intersect at point 0, which cuts off two-thirds of each median from its vertex. Q. E. D. 84 BOOK I. PLANE GEOMETRY 188. Properties of rectilinear figures to be proved by the pupil. Proof of equality of triangles; Other properties of rectilinear figures will now be g'iven which are to be demlonstrated by the pupil. These theorems will be arranged ia groups, according to the method of proof to be used, followed by a group of general or nixed exercises. Let the pupil form a list of the conditions that make two triangles equal. (See Arts. 96, 97, 98, 101, 102, 141, 142.) EXERCISES. GrOUP 4 EQUALITY OF TRIANGLES Ex. 1. Given ABC any triangle, BO the bi- B sector of ZABC, and AD 1 B0; prove I ABO A BOD. [SUG. In the & A BO and DB)O what lines A4. are equal? What A are equal? etc.] Ex. 2. If, at any point in the bisector of an angle, a L be erected and produced to meet the sides of the angle, how many triangles are formed?. Are these triangles equal? Prove this. Ex. 3. If, through the midpoint of a given straight line, another line be drawn, and produced to meet the perpendiculars erected at the ends of the given line, the triangles so formed are equal. Ex. 4. If two straight lines bisect each other and their extremities be joined, how many pairs of equal triangles are formed f Prove this. Ex. 5. If equal segments from the base be laid off on the sides of an isosceles triangle, and lines be drawn from the extremities of the segments to the opposite vertices, prove that two pairs of equal triangles are formed. EXERCISES. EQUALITY OF TRIANGLES Ex. 6. If, upon the sides of an angle, equal segments be laid off from the vertex, and lines be drawn from the ends of these segments to any point in the bisector of the angle, prove that the triangles formed are equal. Ex. 7. If two sides of a triangle be produced, each its own length, through the vertex in which they meet, and the extremities of the produced parts be joined, prove that a new triangle is formed which equals the original triangle. B 'D Ex. 8. Given AB=DC, and BC=D.4 prove A BJC= A DAC. What other pair of equal / triangles is there in the figure? A - Ex. 9. Two right triangles are equal if their corresponding legs are equal. Ex. 10. The altitudes from the extremities of the base of an isosceles triangle upon the legs of the triangle divide the figure into how many pairs of equal triangles? Prove this. Ex. 11. In a given quadrilateral two adjacent sides are equal and a diagonal bisects the angle between these sides. Prove that the diagonal bisects the quadrilateral. Ex. 12. If, from the ends of the shorter base of an isosceles trapezoid, lines be drawn parallel to the legs and produced to meet the other base, prove that a pair of equal triangles is formed. 189. Proof of the equality of lineSo There are several methods of proving that two lines (segments) are equal. One of the principatl methods of proving that two lines are equal is by proling that two triangles, in iwhich the given lines form homologous parts, are equal. 86 BOOK I. PLANE GEOMETRY EXERCISES. CROUP b EQUATLITY OF LINES EX. 1. Given ABCi any triangle, BO the bi- B sectorof Z ABC, and A4I) 1 Oi); prove AOOD=. /T) Ex. 2. If, at any point in the bisector of an / angle, a perpendicular be erected to the bisector A - and produced to meet the sides of the angle, the perpendicular is divided into two equal parts at the given point. Ex. 3. If two sides of a triangle be produced, each its own length, through the common vertex,, the line joining the extremities of tie produced,/ parts equals the third side of the triangle (i e., /C DE == BA). B/ — A Ex. 4. If equal segments from the base be laid off on the legs of an isosceles triangle, liles drawnl from the ends of these segments to the opposite vertices are equal. Ex. 5. Given AR 11 QB, and At' =P —; A P prove RP = PQ. Ex. 6. The bisector of the vertical an- gle of an isosceles triangle bisects the base. Ex. 7. The altitudes of an isosceles triangle upon the legs are equal. Ex. 8. The diagonals of a rectangle are equal. Ex. 9. The medians of an isosceles triangle to the legs are equal. Ex. 10. If two altitudes of a triangle are equal, the triangle is isosceles. Ex. 11. The perpendiculars to a diagonal of a parallelogram from a pair of opposite vertices are equal. Ex. 12. If the equal sides of an isosceles triangle be produced through the vertex so that the produced parts are equal, the lines joining the extremities of the produced parts to the extremities of the base are equal. Ex. 13. If the base of an isosceles triangle be trisected, lines drawn from the vertex to the points of trisection are equal, EXERCISES. EQUALITY OF ANGLES 87 Lines may also be proved equal by showing that they are: opposite equal angles in a triangle; or opposite sides of a parallelogram; or parallel lines comprehended between parallel lines, etc. (See Arts. 100, 155, 157, etc.). Ex. 14. If the exterior angles at the base of a triangle are equal, the triangle is isosceles. Ex. 15. In the bisectors of the equal angles of an isosceles triangle, the segments next to A P the base are equal (AO = OP). A Ex 16. In A ABC, given '~, D AB = AC, DE 1| BC; prove D / K^\.AD = AE. -^^I4 LC Ex. 17. Given AB = DC, BL and BC= AD; prove AE= EC. 190. Proof of the equality of angles may be obtained in several different ways. One of the principal methods of proving that two angles are equal is by protvig that two triangles, in which the given angles form homologous parts, are equal. EXERCISES. CROUP 8 EQUALITY OF ANGLES Ax. 1. Given ABC any A, BO the bisector B of the ZABC, and AD L BO; prove ZRBA4 Z BDO. / Ex. 2. If, at any point in the bisector of an A 0 angle, a perpendicular be erected and producedl to meet the sides of the angle, the perpendicular makes equal angles with the sides B D of the angle. Ex. 3. Given A1 -- DC.; d AD.= B:;\ provet ZB -./ 1..- --- IS 8 88 ~~1300K I. PL.ANE GEOME'TIY Ex. 4. If eq-ual segments from the base be laid off on lie leg s of an isosoeles triangle, the lines dr-awn from tHe extlremilties of thie iiegments to thie opposite vertices make equal angles with the base. Ex. 5. The median to the base of an isosceles triangle is perpendienlar to the base. Ex. 6. The altitudes upon the legs of an isoseeles triangle make equal angles with the base. Ex. 7. The (liagonals of a rhomhu-s biseet its ang-les. Angles mnay also be, proved equ~al b~y proving that they: atre oppos~ite eqfua.tides in anr isoscele~s triaimgle; or1 are vertical angles; or are compleme'nts (or supptemcn Is) of etj ua angles; or by the use of the properties of parallel liner.; or that their siides are parallel, or ])erpevndicular. (See Arts. 75, 78, 99, 124, 126, 130, 132J. - - n -1 I I. - I - - -— 1 -. - - - I I tix. 8. In an isoscemes triangle Tinc exierior angles cado by producing the base are equal. D Ex. 9. Given AC == CB, and DE f D E ARB; prove Z CDE = Z CED. C ELx. I0. Given Cli AR, and Z D CE, 'A B =ZECA, pmove ~B LA. C Ex. 1 1. Conversely given Z A =Z B, and CE 1AR; 13 A.n —r af in f.t' hiQ aQ / + -T) 0 1- - - I-. -- - - - Ex. 12. Given BD the bisector of the angle ABC, and PR I1 CR; prove PBR, an isosceles A. Let the pupil state this theorem B. in general language. EXERCISES. PARALLEL LINES 89 A. Ex. 13. Given AB = AC, and BD= CE; prove ZBCD= Z CBE. How many pairs of equal A in the figure? of equal lines? of equal A? Ex. 14. A line drawn D H E through the vertex of an angle, perpendicular to the bisector of the angle, makes equal angles with the sides of the given angle. Ex. 15. If a straight line which bisects one of two vertical angles be produced, it bisects the other vertical angle also. 191. Proof that two lines are parallel mnlty be obtained b.y showing that: the lines are cut by a transversal, makin g the alternate interior aingles equal; or making the exterior anterior angles equal; or making the interior angles on the samec side of the transversal supplementary; or that the lines are opposite sides of a parallelogram; or that one of the lines joins the midCpoint.s of two sides of a triangle, c d the other linee is the third side of the triangle. (See Arts. 125, 127, 129, 147, 177, etc.) EXERCISES. GROUP 7 PARALLEL LINES Ex. 1. If two sides of a triangle be produced, each its own lengthi tarough the common vertex, the line joining their extremities is parallel to the third -_ side of the triangle. \^ / \\ ~Ex. 2. The bisectors of two alternate interior angles of par- allel lines are parallel. Ex. 3. Given ZA = B, and Z DCE= ZECA \ prove CE || "A. B. --- 90 90 11100BOK 11. PLANh (LOET1l Ex. 4. The b)isectors of thle opposite angles of a jparllleloggramn aro parallel. Ex. 5. Linles perpenl-dicullar to parallel li-nes Care l)arallel (or Ex. 6. Two straight lines are parallel if two points on one line are equidistant from the other line, 192. The proof of a numerical pr operty of the rectilinear figures (of Book I) usually depends on one of the f ol0Wlow The 'uon of the an~gles about, a given) point on? the saone side of a straight linie passinig throutgh the, point is 1800; or the sani of the angles about a point is 30600; or the sum~i of the anigles of a, triangle is 1800; or the sum~i of the 'inerior an~gles~ ofparialle'l l1ines on the samie side of a transversal is 1800; or the sumn of the interior anglles of ai polygont of n sides is (n-2) 1800; &r the sum~i of the exterior angles of a polygon is 3600, (See Arts. 76, 77, 128, 134, 17"2, 175,) EXERCISES. CROUP 8 NUMERICAL PROP~ERTIES Ex. 1. If an exterior angle of a triangle is 123' and an opposite interior angle is 180, find the other two angles of the triangle. Ex. 2. Find the angle formed by the bisectors of the two acute 'sngles of a right triangle. Ex. 3. If two angles of a triangle are 5O' and 600, find the angle fOrnied by their bisectors. ' Find the same if the two angles contain i~ and q0. Ex. 4. 11 tlie vertex angle of an isosceles triangle is 40' and a -perpenidicuilar'i-' drawn from an extremity of the base to the oppos'-Ae side, find tho al-gles of the figure. EXERCISES. NUMTERICAL PROPERTTES 9 9 I Ex. 5. If the ver-tex ang-le of an isosceles triangle is 400, finid the ati~gle included between the altitudes drawn from the extremities of the base to the opposite sides. Ex. 6. How many degrees in each angle of an equiangular dodecagon? of an equiangnla r n-gon? Ex. 7. How many diagonals are there in a pentagon? in a hexagon? in a decagon? in an n-gonY The methods of proving that a given angle is a right angle (or that a given line is perpendicular to another given line), or that one angle is the supplement of another, aire elosely related to the aibote wteihods of' obtainitbg th'e nte Cal vPU lOPS of g iren aangles. Ex. 8. Any pair of adjacent angles of' a parallelogramn is stipplemeontarv. Ex. 9. If one angle of a parallelogramn is a right angle the figure is a rectangle. Ex. 10. The bisectors of two suppleinentary adjacent angles form a right angle (are perpendicular). Ex. 11. The bisectors of two interior angles oii the same side of a transversal to two parallel lines form a right angle. D) Lx. 12. If one of the log s (AR) of an isosceles trianole he produced its own leng-th (RI)) and its cx-,,- B tremity (D)) be, joined to the other end of the base (C), the line last drawn (DC) is perpendicular to the base. A a 193. Algebraic method of proving theorems. The prootr cf certain properties of a geometric ffigure is oftteii f acilita-ted by tile use, olfai algebraic symbol for (ti unknown anyie or Wfl mik'noen line oj' thie figuire, and the as of ani equation1 or Othe'r algebraic net kod yf 8olation, !)2 B1;()OOKi Ii. PLANE GE:)OMETR:'Y EXERCISES. OGROUP 9 AGI,(:EBRAIC METHOD Ex. 1. Find the number of degrees in an angle which equals twice its complement? [SucG. Let x.=tie complement, etc.] Ex. 2. Find the number of degrees in an angle which equals its supplement? in one which equals one-third its supplement? Ex. 3. The angular space about a point is divided into four angles which are in the ratio 1, 2, 3, 4. Find the number of degrees in each angle. [Su(,. x 2 +3 x -x 4 x=360~, etc.]. Ex. 4. The angles of a triangle are in the ratio 1, 2, 3; find the angles. Ex. 5. Two angles are supplementary and the greater exceeds the less by 30~; find the angles. Ex. 6. Find all the angles of a parallelogram if one of them is double another angle. Ex. 7. One of the base angles of a triangle is double the other, and the exterior angle at the vertex is 105~. Find the angles of the triangle. Ex. 8. How many sides has a polygon the sum of whose angles is fourteen right angles? [SuG. 2 (n1-2)=14; find n.] Ex. 9. How many sides has a polygon the sum of whose angles is ten right angles? twenty right angles? 720~? Ex. 10. How many sides has an equiangular polygon one of whose angles is seven-fourths of a right angle? Ex. 11. How many sides has a polygon the sum of whose interior angles equals the sum of the exterior angles e Ex. 12. How many sides has a polygon the sum of whose interior angles equals three times the sum of the exterior angles? Ex.13. If the base of any triangle be produced in both directions, the sum of the exterior angles thus formed, diminished by the vertex angle, is equal to two right angles. SGue, 1800~-a + 1800- b-(180~ —a-b) - t, te.] EXERCISES. AUXILIARY LINES 93 Ex. 14. The bisectors of the base angles of an isosceles triangle include an angle which is equal to the exterior angle at the base. [SuG. To prove a-=b, denote one of the base A by 2 x, etc.] Ex.15. In an isosceles triangle the altitude upon one of the legs makes an angle with the base which equals one-half the vertex angle. [SuG. To prove a-=i b, show that a=9O90- x, b=1S10 -2X, etc.] x rb lx. 16. If the opposite angles of a quadrilateral are equal, the figure is a parallelogram. [SuG. 2x 1 + 2y=380, etC.] Li: 194. Use of auxiliary lines. Inequalities. The demionstration of a property of a greomietrical figuire is frequently facilitated by drawing one or more auxiliary lines on the figure. For examples of the use of such lines, see Props. III, V, IX, etc., of Book I. Some of the principal auxiliary lines used on rectilinear figures are: a line connecfting two given'i points; a line through a given point parallel to a gliven lizin; a line through a given po3inlt plez)rCi)en(lwu 7j2ta to at give line; a line making a given angle w(,.lith a given line; a line prodcwed its owivn length, etc. EXERCISES. GROUP 10 AUXILIARY LINES Ex. I. In the quadrilateral ABOI given AB:AD, and 13= CCD; prove ZBiI) = Z. I). [SL'G. Draw AC, etc.] Ex. 2. Prove that the angles at the base of an isosceles trapezoid are equal. B A 1 D) 94 - 94 L()( ~K I PLANE G1 EOM ET11Y Ex. 3. State andl prove the converse of Ex.. 21. Ex. 4. Given AB (C); prove b Z a + iZC. Ex. 5. Conversely, given Z h -D — a -V c;prove JBI D. Ex. 6. Thle median to the hypotenuse of a right triangle is one-half the hypotenuse. Ex. 7. If one acute angle of a right triangle is doubie the other, the hypotenuse is double the shorter leg. [Sue. Draw the median to the hypotenuse, etc.] a1 C D I? ~L. 111 1 Ex. 8. In an isosceles triangle, thle sum of the perpendiculars drawn from any point in the base to the legs is equal to the altitude uI)on one of the legs. in some cases At is useful to draw twlu:o or mnore auxiliary lbics. Ex. 9. Show that the median of a trapezoid equals one-half the sum of the two bases hy drawing a line through the midpoint of one log of thle trapezoi(I, parallel to the other leg and meeting one base and the other base produced. Ex. 10. Lines joininq the midpoints of the sides of a quadrilateral taken in order form a parallelogram. A [Sue. Draw the diagonals of the quadrilateral anml use Art. 177.] C x. 11. If the opposite I sides of a hexagon are equal aniid one pair of sides (AB and CD) are parallel, the opulosite angles of the hexagon are equal. Ex. 12. Given AB=BC, and AD=CJS; prove DF= BE. EXERCISES. INEQUALITIES 93 Let the student for m a list of the prbiciples proved in Book I concerning uwneqal lines and unequal angles. (See Arts. 92, 93, 950, etc.). EXERCISES. CROUP 110 IN EQUALITIES Ex. 1. In the triangle ABC let -Pbe any point 'in the side BC; prove A.B + BC > A41 -'-.PC. Ex. 2. In the quadrilateral AJBCI) let F be any poinit in the side 1W prv perimeter of APCD) > perimeter of AI)D. Ex 3. In the triangle ABC let 1) be any points in AI; and F anly point in B C. Prove A B + B C > Al ) - IDF+ IC. Ex. 4. The sum. of the four sides of a (luadrilater~il is greater t1han the sumn of the diagonals. Lx. 5. if, from any point within a triangle, lines be drawii to the 'vertices, the sum of the lines drawn is greater than one-half the sumn of the sides of the triangle. ESUG. Use Art. 92 three times, etc.] Ex. 6. If, from any point, within a triangle, lines be drawn to the vertices, thle sum, of the lines drawn is less than the sum of the sides of the triangle. ESUG, I-se Art. 95.] Lx. 7. The median to,any side of a triangle is /0 less than half the sum of the other two sides. [Suo,. Produce the median its own. len-th.] L x. 8. If B is the vertex of an isosceles triangle ABC,,and BC be produced to the point, I), Z IPA I) is greater than Z BI-)A~. L.9 in thie figure p). 71, -MC > zI I- prlovet /RFC graerta Z B FA. [Sun)v~. Use Art.. 108.] Lx. 10. In the samne figture, show that FC < B' Lx. 11, Tn thre, quadIrillate-ral A I-,l), Al1) is the, longest side anld B;Ctthe shortest; ~ri)Ve 4K`,1B( greater 0tian /AI(,and / PCI) greate I Ihian Z JBA- 1). Ex. 12 Linies are, drawn from A, B~, C, I), f tiIPou pInIts inI a srai -iigtit line, to the point, If outside of the, line Winchl anligles Onl titie figure are,, less than angle A(_'F Which angles are greater It 9 G 96 ~~~1300K T. PLIA'NE GEOMFIETRY 195. Indirect demonstrations. Loci. In Boo)k I three, methods of indireet proof have been used. 1. The reduction to an absurdity (rednetio ad absurdum), that is, the proof that the negatitee of a j~ratheoren~ leads to an ab-suiditI (see Prop. X). 2.The method of exclusion, that is, showingy thiat an)?y other statemnent than the giren theorentw can not be tire (see Props. XV, XVII). This method is a speeial ease of the pree~eding, the negative of a given theorem being' divided in it into two parts which are separately shown to be imnpossible. 8. The method of coincidence, that is, proof that- a1 gqilrn lin coinicides with aniothier line whichl jblutil t ('1rtain 1H'qutired condition~s (see Props. XVII, XXIII, XXV, ete.), EXERCISES. CROU.P 12 INDIRECT, OR 'NEGATIVE DEMONSTRATIONS Prove the following by an indirect method: Ex. 1. Every point within an angle and not in the bisector of Ihe angle is unequally distant from the sides of the angtle. [SuG. In the given angle take P any point not in the b)isector of the angle. Then, if P is not unequally distant from AG and OiB, it mast be equally distant from them, et c. ] Ex. 2. If two straight lines are cut by a transversal, makling t~he al~ernate interior angles unequal, the lines are not parallel. Ex. 3. The line joining the midpoints of two sides of a triangle, is parallel to the third side. [SUG. Through one of the midpoints draw a line 1I to the third side, show that it bisects the second side and that the line joining the, midpoints coincides with it.1 EXERCISES. LOCI 97 Ex. 4. If, from a point P in a line AB, lines PC and PD be drawn on opposite sides of AB making the angle APC equal to the angle BPD, PC and PD are in the same straight line. [SUG. From P draw PQ in the same straight line with PC and show that PD coincides with it.] Ex. 5. The bisectors of two vertical angles are in the same straight line. Ex. 6. In the triangle ABC, D is any point in the side AB, and E is any point in the side AC. Prove that BE and DC cannot bisect each other. EXERCISES. GROUP.33 LOCI Ex. 1. What is the locus of all points at a given distance, a, from a given line? Prove this. Ex. 2. W'hat is the locus of all points equidistant from two given parallel lines? Prove this. By use of known loci (see Arts. 112-118), prove the following: Ex. 3. The diagonals of a rhombus are perpendicular to each other. [Sru. See Art. 113.] Ex. 4. The median of an isosceles triangle is perpendicular to the base. Ex. 5. The line that joins the vertices of two isosceles triangles on the same base is perpendicular to the base. 196. General method of obtaining a demonstration of a theorem. Analysis. A clue to the solution of some of the more difficult theorems is often obtained by proceeding thus: Assume the proposed theorem as true; observe what other relation almong the parts of the figure muwst then be true; proceed backwalrd thus, step by step, till the required theorem 'is found to detped onl some (known'i trut'?; tlhen, startiig with this known truith, revcrse the s'te/s tlkent, (id thus buill up t direct lproo f of the required thleoem. This method is called solution by analysis. a 98 BOOK I. PLANE GEOMETRY The following is a simple example of the use of this method: Ex. Given AB and AC the legs of an isosceles triangle and D any point on AB; prove DC greater than DB. ANALYSIS. If DC > DB, Z B is greater than 'Z DCB Art. 104. Hence substituting for Z B its equal, Z ACB, we B C have Z AC is greater than ZDCB. Ax. 8. But we know that Z ACB > Z DCB. Ax. 7. Hence, DIRECT PROOF (or SYNTHESIS) Z ACB is greater than ZDCB, Ax. 7. '. Z B is greater than Z DCB. Ax. 8. '. DC > DB. Art. 106. Q. E. D. The first part (analysis) of the above process is to be purely mental work on the part of the pupil, in investigating a given theorem; the second part (the direct proof, or synthesis) is to be written out as the required solution. In working the following exercises, this method will be found to be necessary in the solution of only a few of the more difficult theorems. EXERCISES. CROUP ea THEOREMS PROVED BY VARIOUS METHODS Ex. 1. If two opposite angles of a quadrilateral are bisected by the diagonal connecting their vertices, the quadrilateral is bisected by this diagonal. Ex. 2. Perpendiculars drawn from the extremities of the base of a triangle to the median to the base, are equal. Ex. 3. If the perpendiculars from the extremities of the base of a triangle to the other two sides are equal, (1) these perpendiculars make equal angles with the base, (2) the triangle is isosceles. MISCELLANEOUS EXERCISES 99 Ex. 4. If the lines AB and CD intersect, then 3AB+CD > 2AC-DB. Ex. 5. The vertex angle of an isosceles triangle is 44~, and one of the base angles is bisected by a line produced to meet the opposite side. Find all the angles of the figure. Ex. 6. In the figure of Prop. V prove that LAPC is greater than L/ABC. Also prove the same in another way by means of an auxiliary line drawn through B and P. Ex. 7, Perpendiculars drawn from the midpoint of the base of a;a isosceles triangle to the legs are equal. Ex. 8 State and prove the converse of Ex, 7. Ex. 9. In an isosceles triangle an exterior angle at the base equals a right angle increased by one- half the vertical angle. Ex 10. In a re entrant quadrilateral the exterior angle at the re entrant vertex equals the sum of the three opposite interior angles ( Z d = Z a - Z b - Z c). [SUc Draw an auxiliary line.] Ex 11 In the equilateral triangle ABC, BC is produced to D; prove BD > AD > AB. Ex. 12. 'If from a point in the bisector of an oblique angle lines are drawn parallel to the sides of the angle, the quadrilateral formed is a rhombus. Ex. 13. If two angles of a quadrilateral are supplementary, the other two angles are supplementary. Ex 14. If the median of a triangle is perpendicular to the base, the triangle is isosceles. Ex. 15. Given AB-AC. Z BAC = 4 Z B, and DF J BC, prove A EFA equilateral. F A B<E - C D^ ---- [ANALYSIS. It A FA1 is equilateral, ZL'E-f=6;0~;.~. Z I)EB = 600; L. Z B=JOo. Hence. to get a direct proof, show that C 30~ by USing LAC = 4ZC.] 100 BOOK I. PLANE GEOMETiRY Ex. 16. If the diagonals of a quadrilateral bisect each other at right angles, what kind of a figure is the quadrilateral? Prove this. Ex. 17. From the point in which the altitudes drawn to the legs of an isosceles triangle intersect, a line is drawn to the vertex. Prove that this line bisects the angle at the vertex. Ex. 18. If from a point within an acute angle perpendiculars are drawn to the sides of the angle, the angle formed by these perpendiculars is the supplement of the given angle. Ex. 19. Lines joining the midpoints of the sides of a triangle divide the triangle into four equal triangles. Ex. 20. If, in the parallelogram JABCD, BP = DQ, then JQCP is a parallelogram. A P [ANALYSIS. If JQCP is a7, AP=and II QC..'. begin the direct proof by show- ing that AP = and is 11 QC.] Q Ex. 21. If the diagonals of a parallelogram are equal, what kind of a figure is the parallelogram? Prove this. Ex. 22. If the angle A of the triangle ABC is 50~ and the exterior angle BCD is 120~, which is the largest side in the triangle? Ex. 23. Two triangles are equal if two sides and the median to one of these sides in one triangle are equal, respectively, to two homologous sides and a median in the other. Ex. 24. Two isosceles triangles are equal if the base and an angle of one are equal to the base and the homologous angle of the other. Ex. 25. Two equilateral triangles are equal if an altitude of one is equal to an altitude of the other. Ex. 26. If two medians of a triangle are equal, the triangle is isosceles. [SUG. On Fig. to Prop. XLVI, taking AD=EC, prove A OC isosceles, AAZEC= AL)DC etc, How could this theorem be investigated.by analysis t ] MISCELLANEOUS EXERCISES 101 Ex. 27. Prove the sum of the angles of a triangle equal to two right angles by drawing a line through the vertex of the triangle parallel to the base. Ex. 28. The homologous medians of two equal triangles are equal. Ex. 29. The bisectors of an angle of a tri- B angle and of the two exterior angles at the other vertices are concurrent. Ex. 30. If ABC is an equilateral triangle and AP = BQ = CR, then P'(R is an equilateral A triangle. Ex. 31. If the two base angles of a triangle be bisected, and through the point of iutersec- B tion of the two bisectors a line be drawn parallel to the base, the part of this line intercepted between the two sides equals the sum of the P Q segments of the sides included between the parallel and the base (i. e., prove PQ=AP+ QC). A C Ex. 32. Two quadrilaterals are equal, if three sides and the two included angles of one are equal to three sides and the two included angles of the other, respectively. Ex. 33. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. Ex. 34. Lines joining the midpoints of the sides of a rectangle in order form a rhombus. Ex. 35. Lines joining the midpoints of the sides of a rhombus form a rectangle. Ex. 36. The bisectors of the angles of a parallelogram form a rectangle. Ex. 37. The bisectors of the angles of a rectangle form a square. Ex. 38. If lines be drawn through the vertices of a quadrilateral Parallel to the diagonals, a parallelogram is formed which is twice as large as the origina quadrilateral. 1]02 BOOK I. PLANE GEOM(ElRY Ex. 39. Lines drawn from two opposite vertices of a parallelogram to the midpoints of a pair of opposite sides trisect a diagonal of the parallelogram. Ex. 40. On the diagonal AC of a parallelogram AiCD equal parts, AP and CQQ, are marked off. Prove BI' PQ a parallelogram. HowX many pairs of equal triangles does tlie figure iontain? Ex. 41. The opposite angles of an isosceles trapezoid are supplementary. Ex. 42. In an isosceles trapezoid, the diagonals are equal. Ex. 43. If the upper base of an isosceles trapezoid equals the sum of the legs, and lines be drawn from the midpoint of the upper base to the extremities of the lower base, how many isosceles triangles are formed? Prove this. Ex. 44. The lines joining the midpoints of the sides of an isosceles trapezoid, taken in order, form a rhombus or a square. Ex. 45. The bisectors of the angles of a trapezoid form a quadrilateral whose opposite angles are supplementary, BOoK I THE CIRCLE 197. A circle is a portion of a plane bounded by a curved line, all points of whlich are equally distant from a point within called the center. The circumference of a circle is the curved line bounding the circle. The tetrm (ircle may also be used for the bounding line, if no ambiguiity results. A circle is named by naming its center, as the circle O; or by naming two or more points on its circumference, as the circle ACD. 198. A radius of a circle is a / straight line drawvn from the center to any point on the circumference, B / as AO. A diameter of a circle is a straight line drawn through the center and terminated by the circumference, as BC. 'ig. 1 199. An arc is any portion of a circumference, as AC. A semi- circumference is an arc equal to F one-half the circumference, as BAC. A quadrant is an arc equal P to one-fourth of a circumference, as BJD. 200. A chord is a straight line ____ joining the extremities of an are, T as BF. M 2 Vr'..* -' (103) 104 BOOK TII. PLANE GEOMETRY Every chord subtends two arcs, A minor arc is the smaller of two arcs subtended by a chord. A major arc is the larger of two arcs subtended by a chord. Thus, for the chord EF the minor arc is EPF, and the major are is ETF. Conjugate arcs is a general term for a pair of minor and major arcs. If the are subtended by a given chord is mentioned, unless it is otherwise specified, the minor arc is meant. 201. A tangent to a circle is a straight line which, if produced, has but one point in coImmon with the circle, as M3X. Hence, a tangent touches the circumference in one point only. A secant is a straight line which, if produced, intersects the circumference in two points,'as GI-r. C B o P R B A A Pi.3 Fig. ~D PT [ig.'3 Fig. I Fig. 5 202. A segment of a circle is a portion of the circle bounded by an arc and its chord, as ABC (Fig. 3). Into how many segments does each chord divide a circle? A semicircle is a segment bounded by a semicircumference and its diameter. 203. A sector of a circle is a portion of a circle bounded by two radii and the arc included by them, as POQ (Fig, 3), THE CIRCLE 105 204. A central angle is an angle whose vertex is at the center and whose sides are radii, as the angle POQ (Fig. 3). An inscribed angle is an angle whose vertex is in the circumference and whose sides are chords, as the angle ABC (Fig. 4). An angle inscribed in a segment is an angle whose vertex is in the are of the segmlent and whose sides are chords drawn from the vertex to the extremities of the are. Let the pupil draw a circle, a segment ii it, and an angle inscribed in the segment. 205. Two circles tangent to each other are circles which are tangent to the same straight line at the same point. They are tangent internally or externally according as one circle lies entirely within or entirely without the other. See the figures, page 122. Concentric circles are circles which have the same center. Let the' pupil draw a pair of concentric circles. 206. A polygon inscribed in a circle is a polygon all of whose vertices lie in the circumference of the circle, as ABCDE (Fig. 4). A circle circumscribed about a polygon is a circle whose circumference passes through every vertex of the polygon. 2070 A polygon circumscribed about a circle is a polygon all of whose sides are tangent to the circle, as PQRST (Fig. 5). A circle inscribed in a polygon is a circle to which all the sides of the polygon are tangent. Concyclic points are points lying on the same circurm ference, 106 3B00K IT, PLANE GEOMETRY PROPERTIES OF THE CIRCLE INFERRED IiMEDIATELY 208. Radii of the same circle, or of equal circles, are equal. 209. The diameter of a circle equals twice its radius. 210. Diameters of the same circle, or of equal circles, are equal. 211. If two circles are equal (i. e., may be made to coincide, Art. 13), their 'radii are equal, and conversely. 212. A diameter of a circle bisects the circle. For, by placing the two parts of the circle so that the diameters coincide and their arcs fall on the same side of the diameter, these arcs will coincide (Art. 197). 213. A straight line cannot intersect a circle in more than two points. For, if a straight line can intersect a circle in three (or more) points, three or more equal lines (radii) can be drawn from the same point (the center) to the straight line. But this is impossible (Art. 111). PROPOSITION I. THEOREM 214. A diamneter of a circle is longer than any other chord. D B Given AB a diameter, and CD any other chord in the circle 0. To prove AB > CD. THE CIRCLE10 107 Proof. Draw the radii 00 and OD. Then, in the A 90D, 0 + OD > CD. (Why?) Substituting for 90 its equal OA, and for OD its equal GB, A x. 8. OA -1 01B, or ARB> GD,, Q. E. D). PRoPosiT~io-N- TI. THEOREM 215. In the, samne circle, Or ini equal circles, equal Cenv tral aniylesbintercept equal aircs on the ci rciovference. A F? A B o 0 Given the equal circles 0 and 0', and the equal central A AO.1 and A'0'B'. To prove the are AR -are A'B'. Proof. Apply) the circle 0' to the circle 0 so that the center 0' coincides with the center 0, and the radius 0'A' with the radius OA. Then the radius 0'B' will fall on the radius GB, (~for by hiyp, Z AOB = ZAI'O'B'). And B' will f all on B, Art. '208. (for OB=O'B', being radii of equal 0). Hence are A'R' will coincide with AB, Art. 197. (for all poin ts of eacht arc arec ep idistant from tMe center).arc A4BR'= arc AR. Art. 47. Q. E. D. I-OS TLOOK IT. PLANE CIEOMEITY, PROPOSITION ITI. THEOREM (CONVERSE OF PROP Ii) 216. Int the same circle, Or in equaal circles, equial re sutbten)td equal angles at the center-. A B A B' 00 Given the equal circles 0 and 0', and the equal ares AB and A'B' subtending, the central A 0 and 0'. To prove /0 = ZO0.-i Proof. Apply the circle 0' to the equal circle 0 so that the center 0' coi ncides with the center 0 and the point A' with the point A. Then the point B' will fall on B, (for arc A'B'= are- AB by hiyp.). Hence the radius O'A' will coincide with OA, and radius 0'B' -with OB, Art. 66. (betweent two po~ls oaily one straighit liae can be driven). / 0 and / 0' coincide. Z 0 =Z 0'. Art. 47. Q. E.D 217. CoR. In the same circle,9 or in equal circles, of two unequal central angles the greater -angle intercepts the greater arc, and, conversely, of two unequal arcs, the greater arc subtends the greater angle at the center. Ex. Draw a circle and in it a segment which is less than the sector having the samue are. Also one that is greater. Also one that is equal. THE CIRCLE 109 ]PROPOSITION IV. THEOREMI 218. In the same circle, or in equal circles, equal chords subtend equal azrcs. Given the equal circles 0 and 0', and chord AIB'. the chord AB -: To prove are AB - aire A'B'. Proof. Draw the radii OA, OB, 0'A', O'B'. Then, in the A~ AOB and A'O'-B', AB =AIBI. (Why?) AO=zA'O', and BO-B'O'. (W hy?7) LA AOBI3,AA'0'B'. (Why?) A' / iO 0' O. ('Why?7) arc ABL are AIR'. Art. 215. (in the same 0, or -in equal 0, equal central A intercep~t cqual arcs onl the circumferencee). EX. 1. In the above figure, if chord AB=1 in., chord A'B'~1 in.? and arc- AB=1It in., find the length of arc A'iB'. EX. 2. Draw a circle and mark off a part of it that is both a segment and a sector. E-X. 3. If the distance fromi the center of a circle, to a line is greater than the radius, will the line intevrect the eircumf drence 110 lBOOK II. PLANE GEOMETRY PROPOSITION V. THEOREM 219. In thf sanml circle, or in equal circles, equal arcs are subtended by equal chiords. A A' 0 0 Given the equal circles 0 and 0', and arc AB-=are A'B'o To prove chord AB = chord A'I'. Proof. Draw the radii AO, BO, A'0', B'O'. Then, in the A AOB and AO'B', Z/0- Z0', Art. 216. (for arc AB =A'B', a(d, in the samCe 0, or inl equal 0, equal arcs subtlcnd ecqal A at tlhe center). Also OA = O'A', and OB = OB'. (Why?),.. AAOB= A AAO'B'. (Why?). AB = A'B'. (Why?) Q. Eo D. Ex. 1. In the above figure if arc AB==1t in., arc A'B'=i~ in., and chord AB= 1 in., find chord A'B' without measuring it. Ex. 2. Draw two circles so that the radius of one is the diameter of the other. Ex. 3. To which of the classes of figures mentioned in Art. 18 does a sector belong? a segment? THE CIRCLE ill~ PROPOSITION VI. THEOREM 220. In the same circle, or in equal circles, the greater of two (minor) arcs is subtended by the greater chord; and, CONVERSELY, the greater of two chqrds subtends the greater (minior) arc. A 0~~~~~ Given the equal circles 0 and 0', and arc AB > arc -Di. To prove chord AD > chord -DE. Proof. Drawx the radii OA, OB, 0'D), 0'F. Then, in the & AOB and DO'F, OA= O'D, and 0B 0'P. (Why?) Z 0 is greater than Z 0', Art. 217. (for a-rc AB > arc DE, and, in the same 0, or in equal 0, of tieo unequal arcs the greater arc subtends the greater, angle at the center).-. chord AD > chord DFE, Art. 107. (if two & heave two sides of one equat to two sides of thte other, but 'the included angle of, the first greater, etc.). CONVERSELY. Given the eqlal circles 0 and 0', and chord AD> chord DFE. To prove arc AD> arc DiE. Proof. In the & A OB and DO'F, OA = O'D, and OD0= O'IF (Why?7) AD > DFE. (Why?)y Z 0 is greater than Z 0' (Why arcc A > arch J), Art. 217. (in the same 0 o r in equal 0 of tw o iiequal central A the gireater angle intercepts the greater arc. le I Xj a M L JI'l 112 L~~OOK II. PLANXE GEOMETi,r izY PROPOSITION VII. THEOREM 22 1. A diaetmeer perpendlicular to a chord bisects the chord antd the arcs su'btended by the chord. Given the circle 0, and the diameter PQ I chord A-B. To prove that PQ bisects the chord AR and the arcs APR a nd A QB. Proof. Draw the radii OA and 013. Then, in1 the rt. &OAR and OBRE OA ==OB. (Why?) OR - OR. (Why?) AXOARE=A OBR. (Why?) ARPi= R, and Z A OE Z B OT. ( Why?) arc A4 P) - are RIP, Art. 215. (in the samef 0, o;r in. 0, =centra Xu ~'in tereept -areci on. the ciref.. LA-OQ00 ZROQ (Art. 7 5)...are A4Q are EQ. (Why?) Q.E. D. 222. Con. 1. -1 diameter -whicht bisect~s a chord (shorter than a diamieter) is~ J-erpemldicidlar to the chord. 223. COR. 2. The perpeudicular bisector of a chord passes through the center of the circle, and bisects the arcs subtended by the chord. 224. CoR. 8. A line fromt the center perpendicular to a chord bisects the chord. 225, CoR. 4. A line passing through the midpoints Of a chord and its arc passes through the center, and is a diame - ter perpendicular to the chord. THE CIRCLE 113 PROPOSITION VIII. THEOREM 226. In the same circle, or in equal circles, equal chords are equidistant from the center; and, CONVERSELY, chords which are equidistant from the center are equal. Given the circle 0 are equal. To prove that AB center. in which the chords AB and CD and CD are equidistant from the Proof. Let OE be drawn I AB, and OF I CD. Draw the radii OA and 0C. Then OE bisects AB, and OF bisects CD, Art. 224. (a line from th1e cnc tcr 1 a chord bisects the chord). Hence, in the rt. A OAE and OCF, OA = OC. (Why?) AE = CF. Ax. 5.. A OAJ 1 = A OC1t. (Why?) O' 0F- OF. (Why?) CONVERSELY. Given circle 0, and AB and CD equidistant from, the center. To prove AB- CD. Proof. Let the pupil supply the proof, 114 0 OOK II. PLANE GEOMETRY PROPOSITION IX. THEOREM 227. In the same circle, or in equal circles, if two chords are unequal, they are unequally distant from the center, and the less chord is at the greater distance from the center. D a Given in the circle 0 the chord CD < chord AB. To prove that chord CD is at a greater distance from the center than chord AB. Proof. Let OG be drawn 1 CD, and OF J AB. Then chord AB > chord CD. Hyp..'. arc AB > arc CD, Art. 220. (in the same 0, or in equal 0, the greater of two minor arcs is subtended by the greater chord, and conversely). Mark off on the arc AB the arc AE=arc CD, and draw the chord AE. Chord AE = chord CD, Art. 219. (in the same 0, or in equal 0, equal arcs are subtended by equal chords). Let OH be drawn ~ AE, and intersecting AB at L..'. OH = OG, Art. 226. (in the same 0, or in equal 0, equal chords are equidistant from the center). But OH > OL. Ax. 7. Also OL > OF. (Why?) Much more then OH, or its equal OG > OF. Ax. 12. 9. D. THE CIRCLE11 115 PROPOSITION X. THEOREM1 (CONVERSE OF PROP. IX) 228. ITn the same circle, or in equal circles, if two chords are unequtally distant fro~n the center, the more rei tote is the less. B A Given in the circle 0 the chord CDf farther from the center than the chord AB. To prove chord CD < chord -4B. Proof. Let OHl be drawn I GD, and 0(1 I AB. 011> 0(G. (Why?) On OH mark off Oh = OG. Throughl L let the chord ELF be drawn A OH,. Then chord EFW chord AR3, Art. 2296. (in the samne.0, or -in equal 0, chwr(Is?whch are equidistant from the Center are, equal). Bnt the are CD < arc EF. Ax. 7, chord CD K chord EFE, or its equal AR, Art. 2120. (in rthe same 0, or in equal CD, the geatce- of twovominor ares is subentelde by thte greater chtordl, and conversely). Q. E.D. 116 -BOOK 11. PL-ANE GEOMETRY PR1OPOsTC1TIN XI. TUEOREM 229. A1 straight inte perpendicular to a it'ius ait its extremity is tange0t to the circle. 0 B- A 1' r Given the circle 0, the radius OA, and the line BC I OA at its extremitv A1. To prove that BC is tangent to the 0. Proof. Take P, any point on the line BC except A, and draw OP. Then OP > OA. (Why?) Hence the p)Oint P lies without the circle. every point in the line BC(, except A, lies outside the 0. BC is tangent to the circle, Art. 2G1. (a tangent to a 0 is a straight line which, etc.). E. E. D. 230. COR. I. The radius drawvn to the point of contact is perpendicutlar to a tangent to a circle. 231. COR. 2. A perpendicular to a tangent at the point of contact passes thr-ough the center of the circle. 232. COR. 3. The perpendicular drawn from the center of a circle to a tang'ent passes through the point of contact, THE CIRCLE17 PROPOSITION XII. THEOREM 233, Two parallel lines intercept eqiual arcs on a cir cumference. A B A 13 C D-C ---Q E P Q D Fig 1 'Fig. Fi-, 3 CASE I. Given AB (Fig,. 1) tangent to the 0 PCD at P, CD a seclant 11 ALB and int-ersecting- the circunifei'enice in C and D. To prove arc PC = are PD. Proof. Drawi the diameter PQ. Then PQ I AB. Art. 230. PQ I GD. Art. 12-3. are, PC =are PD- Art. 221. (a diameter I chiordl bisects the chlord andI the arcs subtended by thle Chord). CASE II. Given AB and CL) (Fig-. 2) H1 secants intersecting the circunmferetice ini Az, B Iand (~ D respec tively. To prove are A4 C — arc -BD. Proof. Let a tangent EF be drawn I1 A.Bcand touching the circle at P. (h Then FE II CD. hy? Then are AP are BL]. Case 1. Also tare CL] are JDL]' (W'hy? Hence are ACG are -BD. Ax. 3. i18 BOOK H1. PLANE (EMINE, TRlY CASE 1II. Given AB and CD) (Fig,. ~3) I1 tang ents touch. ing the 0 at P and Q respectively. To prove are RE Q - aire PFQ. Proof. Let the pupil supply the proof. 234. COR. The strcaight line wehich joins the points of contact of two r~arallel tan)gents is a diami~ete~r, PROPOSITIONT XIII. TH-EOREM 235. Through threc pobits, not in thie samev straight line, one circumference, ctn at oly one, can be drawen. C A Given A B and C any three points not ill the same straight line. To prove that one cireninference, anid only one, can be,drawn through A, B and C. Proof. Draw the straight lines AB and BC, and let I be erected at the midpoints, 1) and F, of AB and BC1 respectively. These.I will intersect at some point 0, Art. 122. (lines I non-parallel lines are not But 0 is in the I bisector of AB. condt, THE CIRCLE 119. 0 is equidistant from the points A and B. Art. 112. In like manner, 0 is in the I bisector of BC, and is equidistant from the points B and C. Hence 0 is equidistant from the three points A, B and C. Ax. 1L Hence if a circumference be described with 0 as a center and OA as a radius, it will pass through A, B and C. Also DO and EO intersect in but one point. Art. 64. Hence there is but one center. Again, 0 is equally distant from the points A, B and C; hence there is but one radius. With only one center and only one radius, but one circumference can be described. Hence one circumference, and only one, can be drawn through the points A, B and C. Q. E. Do 236. NOTE. The theorem of Art. 235 enables us to shrink or economize a circle into three points; or to expand any three points into a circle. Ex. 1. How many circumferences can be passed through four given points in a plane, each circumference passing through three, and only three, of the given points? Ex. 2. Draw two circles so that they can have a common chord. Ex. 3. Can two circles which are tangent to each other have a common chord? Ex. 4. Can two circles which are tangent to each other have a common secant? Ex. 5. Draw two circles which can have neither a common chord nor a common tangent. Ex. 6. Is it possible to draw two circles which cannot have a colmmon secant I 120 BOOK II. PLANE GEOMETrLY PROPOSITION XIV. THEOREM 237. The two tangents drauwn to a circle fromn a point outside the circle are eqtal, tad make equal langles with a line drawnf from the point to the center. A Given PA and PB two tangents drawn from the point P to the circle 0. To prove PA-PB, and Z APO= Z BPO. Proof. Let the pupil supply the proof. 238. DEF. The line of centers of two circles is the line joining their centers. 239. DEF. Two circles which do not meet may have four common tangents. A common internal tangent of two circles is a tangent which cuts their line of centers. 240. DEF. A common external tangent of two circles is a tangent which does not cut their line of centers. Ex. 1. In the above figure, prove that the line drawn to the center from the point in which the two tangents meet makes equal angles with the radii to the points of contact. Ex. 2. Draw a circle with a radius of 3 in. and another with a radius of 2 in., with their centers 4 in. apart. Will these circles intersect t If their centers were 6 in. apart, would they intersect V THE CIRCLE 121 PROPOSITION XV. THEOREM 241. If two circles intersect, their line of centers is perpendicular to their common chord at its middle point. Given the circles 0 and 0', intersecting at the points A and B. To prove 00' _l AB at its middle point. Proof. Draw the radii OA, OB, O'A, O'B. Then OA = OB, an O'A = O'B. (Why?) Hence 0 and 0' are two points each equidistant from A and B... 00' is the J bisector of AB. Ex. 1. Draw two intersecting circles and show centers is less than the sum of the radii. Draw two circles in which the line of centers Ex. 2. Equals the sum of the radii. f I' Art. 113. Q. E. D. that the line of 3 A and B..'. 00' is the J bisector of AB. Art. 113. Q. E. D. Ex. 1. Draw two intersecting circles and show centers is less than the sum of the radii. Draw two circles in which the line of centers that the line of 122 B1OOK II. PLANE GEOMETRY PROPOSITION XVI. THEOREM[ 242, If two circles are tangent to each other, the line of centers passes through the point of contact. A 0 A O'B B Given the circles 0 and 0' tangent to each other at the point R. To prove that the line 00' passes through R. Proof. At the point R let PQ, a tangent to the given 0, be drawn. Also let AB be drawn I PQ at R. Then AB passes through 0 and also through 0', Art. 231. (a IJ to a tangent at the point of contact passes through the center). line AB coincides with the line 00'. Art. 64... 00' passes through the point R, (for it coincides 'ith AB which passes through R). Q. E. Q. How many common internal, and how many common external tangents have two circles Ex. 1. If they touch externally? Ex. 2. If they touch internally? Ex. 3. If they intersect? Ex. 4. If one circle lies wholly within the other? Ex. 5, If one circle lies wholly without the other V THE CIRCLE 123 EXERCISES. GROUP I6 Ex. 1. The line joining the center of a circle to the midpoint of a chord is perpendicular to the chord. Ex. 2. A, B, C and D are four points taken in succession on the circumference of a circle, and arc AB=arc CD. Prove that chord AC==chord BD. Ex. 3. Tangents drawn at the extremities of a diameter are parallel. Ex. 4. PA and PB are tangents to a circle drawn from the point P. 0 is the center of the circle. Prove that PO is the perpendicular bisector of the chord AB. Ex. 5. If the perpendiculars from the center upon two chords are equal, the arcs subtended by these chords are equal. Ex. 6. A, B, C and D are points taken in succession on a semi-circumference, and arc AC is greater than are BD; prove that chord AB> chord CD. Ex. 7. State the converse of the preceding A theorem and prove it. / Ex. 8. Given RA, BQ and QB tangents of the circle 0; prove RQ=RA + QB. ~ Ex. 9. If a quadrilateral be circumscribed about a circle, show that the sum of one pair of B Q opposite sides equals the sum of the other pair. Ex. 10. If a hexagon be circumscribed about a circle, show that the sum of three alternate sides equals the sum of the other three sidesEx. 11. If a polygon of 2n sides be circumscribed about a eircle, the sum of n alternate sides equals the sum of the other n sides. Ex. 12. A circumscribed parallelogram A is equilateral, B Ex. 13. If two circles are tangent C externally, the coruimon internal tangent bisects the common external tangent (i. e., Prove PA = PB). 1 4 300K II. PLANE CEOMETRY Ex. 14. If two circles are tangent (either externally or internally) tangents drawn to them from any point in the con- mon tangent are equal. B Ex. 15. Two circles whose centers are 0 and f A/ 0 \ 01 are tangent internally at P. The line PAB is P\ / )drawn intersecting the circumferences at A and B. Prove that 0A and O'B are parallel. MEASUREMENT. RATIO. 243. Measurement. For many purposes, the most advantageous way of dealing with a given magnitude is to take a certain definite part of the magnitude as a unit, and to determine the number of times this unit must be taken in order to make up the given magnitude. Ease and precision in dealing with magnitudes are thus obtained. Geometric magnitudes thus far have been treated as wholes, the object being simply to determine whether two given magnitudes are equal, or unequal, or to determine some similar general relation. Hereafter geometric magnitudes will frequently be treated as if composed of units. To measure a given magnitude is to find how many times the givenl magnitude contains another magnitude of the same kind taken as a unit. 244. The numerical measure of a magnitude is the number which expresses how many times the unit of measure is contained in the given magnitude. Thus, when a boy says that he is five feet tall, he means that, if a foot rule be applied to his height, the foot rule will be contained five times. A quantity is often measured to best advantage by measuring a related but more accessible quantity, which has the same numerical measure as the original quantity. This process is indirect measurement. Thus, the temperature of the air is measured indirectly by measuring the height of a column of mercury in a thermometer tube. So the number of times a unit of angle is contained in a given angle is often ascertained MEASUREMENT most readily by determining the number of times a unit of arc is contained in a given arc. 245. The ratio of two magnitudes of the same kind is their relative magnitude as determined by the number of times one is contained in the other. Hence, it is the quotient, or indicated quotient, of the two mnagnitudes. 36 in. 36 Thus, the ratio of 3 ft. to 1 ft. 7 in. is 36 in, or 36 i)1 in. 19 The use of ratio is illustrated by the fact that several indicated quotients when taken together may be sinimplified by cancellation before a final determination of their value is lmade. Two maynitldes of the same kind Ihavc the same ratio as their numlZerical mieas (lrs. 246. Measurement as a ratio. An important particular instance of ratio is that ratio in which one of the two magnitudes compared is a unit of measurement. Hence, the numerical measure of a magnitude is the ratio of the magnitude to the unit of measure. Thus, the numerical measure of the height of a boy is the ratio of, his height (5 ft.) to the unit of measure (1 ft.), or 5. 247. Commensurable magnitudes are magnitudes of the same kind which have a coninOll unit of measure. Thus, 12 ft. and 25 ft. have the common unit of measure, 1 ft., and hence are commensurable magnitudes; also 13t bus. and 7} bus. have a common unit, I peck, and are commensurable. 248. Incommensurable magnitudes are magnitudes of the same kind which have no common unit of measure. Ills. $5 and $/3; 7 yrs, and i/15 yrs.; so the side and the diagonal of a square may be proved to have no common unit of measure; likewise the diameter and the circumference of a circle. In general, a ra(tio which is expressed by a surd number, as\/2 or 3, is a ratio between incomnmensurable magnitudes (called an incommensurable ratio,) 126 B()OOK TI. PLANE GEOMETRY METHOD OF LIMITS 249. A variable quantity, or a variable, is a quantity whic11h -may have an indlefinite numb)ner of different values ul(dei thle c(onditions of a theorem or p'robileim. Thus, the distance a railroad train goes varies with the numbet of hours which the train travels. 250. A constant is a quantity which remains unchaniged in value in a,\-ivc problecii or discussioin. Thus, if a polygon l)e inscribedl in a circle and the nulmber of sides of the polygon Ibe doubled, quadrupled, etc., the perimeter of the polygon will be variable, but the circumference of the circle will remain unchanged, and hence be a constant. 251. The limit of a variablle quantity is a constant quantity which, un1der tile conditions of a givel discussion, tile given variable may approacll as learly as we Iplease in value, but iwhich tlie variable call never etqual. Thus, in tlhe illustration of Art. 250, the circumference of the circle is the limit of the perimeter of the inscribed polygon; also,-the area of the circle is the limit of the area of the inscribed polygon. From the definition of limit, it follows that the difference between a variable and its limit may be made as small as we please but can not become zero. As another illustration of a variable and its limit, we may take the case of a point P travelling along a given line AB, in such a way that in the first second it passes over AP1, one-half the line AB; in the second second over half the remaining part of the line, and arrives at P',; in the third second over onehalf the remainder of the line, and _ _ arrives at P3, etc. It is evident A P1 P B that the point P can never arrive at B; for, in order to do this, in some one second the point would need to pass over the whole of the remaining distance. In this illustration, AP, the distance traveled by the moving point, is a variable (depending on the number of seconds), and AB is its limit. If the distance AB be denoted by 2, the distance traveled will be denoted by 1 + i+ 1 + I.... and will vary according to the number of terms of the series taken. METHOD OF LIMITS 127 252. Use of variables and limits. Many of the properties of limits are the same as the properties of the variables approaching them. Hence a demonstration of a difficult theorem may often be obtained by first finding the properties of relatively simple variables and then transferring these properties to their more complex limits. 253. Properties of variables and limits. 1. The limit of the sumt of a number of varidr'les equals the sum of the limits of these variables. For, since the difference between each variable and its limit may be made as small as we please, the sumn of all these differences may be made as small as we please (since it is a finite number of differences, with each difference approaching zero). 2. The limit of a times a variable equals a times the limit of the variable, a being a constant. For, if the difference between a variable and its limit may be made as small as we please, a times this difference may be made as small as we please. 3. The limit of lth part of a variable is -th part of the a a limit of the variable, a being a constant. For, if the difference between a variable and its limit may be made as small as we please, -th part of this difference may be made as small as we please. 4. If a variable _ O, a times (a being finite) or - part of the variable _'0; and if a diminiish in the one, or increase in the other process, the limit is still zero. Ex. 1. Are 21 gal. and 3} qt. commensurable? If so, what is the common unit of measure? Ex. 2. If c denote a constant and v and v' varia'les, is the value of each of the following a variable or a constant: —,, -? Illustrate, 12S 128 ~1300K II. PEAN E GEOMETRY I-'ROPO'SITION XVII TEOE 254. If two etari(dlbis are always rq00i, and, each anO, (tche liin it, th(~iiF lijilits r (U# q alo. A Given AB the limit of the variable AP, AC the limit of the variable AQ, and AP=ZAQ al~i~ays. To prove ABz=Ac. Proof. If the limit AB does not equal the limit AC, one of these limits, as AC, mnust be larger than. the other. Then, on AC, tak-e AD equal to AD. Pi ut AQ miay hiave a value greater than-. AD. Art. 251. Hence AQ would he grreater than AD, (f'or AQ > AD wit-cht AB).. A Q > AP, Ax. 12. (for AB > A4P). -But this is contrary to the hypothesis that AQ and AP are always equal. AC cannot be gyreater than AB. In like manner it may be shown that AB is not greater than A C. AB=A.C. Ez. What method of proof is used in Prop. XVII T MEASUREMENT OF ANGLES j()9 PROPOSITION XVIII. THEOREM 255. It the same circle, or in equal circles, tmwo central angles have the same ratio as their intercepted arcs. CASE I. When the intercepted arcs are commensurable. 8`~~~~~~~~~~~~,i \\ Given the equal ~ 0' and 0, with the central A A B'0I' and AOB intercepting the commensurable arcs A'B' and AD. To prove Proof. Let are PQ (from a circle = 0 and 0') be a common measure of the arcs AB' and AB. PQ will be contained in are A'B' an exact nmnber of times, as 7 times, and in AD an exact number of times, as 5 times. Tare AB' 7 Then Art. 2-i, are AD 5A From 0' and 0 draw radii to the several points of division of the ares A'B' and AD. Then the ZAO'B' will be divided into 7, and the LA0_B into 5 small angles, all equal, Art. 216. (in the same o, or in =s equal arcs subtend equal a at the center). ZA'0'D'I 7 Art, 245. ZA0 O.B 54 W Hence -AB' Ax- R A13~H I 130 BOOK II. PLANE GEOMETRY CASE II. ir7ne the intercepted arcs a(re iwcomnen, surable. 0 1 0' Given t lie elual (i (' tand 0, with the central Z A'O'B' and AOB intercepting the incommensurable ares A'B' and AB. To prove Z A'OI'B' A' ZA OB A' Proof. Let the arc AB be divided into any number of equal parts, and let one of these parts be applied to the arc A'B'. It will be contained in A'B' a certain number of times,with an arc DB' as a remainder. Hence the arcs A'D and AB have a common unit of meas ure. Cotnsr Z t' jO'D) AI'D. * '. Oi = WAB Case l If now we let the unit of measure be indefinitely diminished, the arc DB', which is less tl:ha tlie unit of measure, will be indefinitely diminished. Hence arc A'D- arc A'B' as a limit, and Z A'O'D A'O'B' as a limit. Art 251. ZA'O'D A/'O' Hence Z A O' becomes a variable approaching Z --- A~ B ZA eLc Ze AA O as its limit; Art. 253, 3. AID AIBl Also becomes a variable approaching. as its limit. Art. 253,. MEASUREMENT OF ANGLES 131 Z Al O__ A'D But the variable the vaiable always. ZAOB AB Case I. ZA' O'3' AImi At A'.. the limit = OB the limit - Art. 254. ZAOB AB (if two variables are always equal, and each approaches a limit, their limits are equal). 256. DEF. A degree of arc is one three hundred arnd sixtieth part of the circumference of a circle. 257. COR. The number of degrees in a cenltral (tgcle equals the nlitber of degrees,in thf intercepted at(r; that is, a central angle is measured by its intercepted arc. Ex. 1. What is the ratio of a quadrant to a semi-circumference? Ex. 2. What is the ratio of an angle of an equilateral triangle to one of the acute angles of an isosceles right triangle? Ex. 3. Draw two circles so that the center of each circle is on the circumference of the other. Ex. 4. Draw three circles so that the center of each is on the circumference of the other two. [SUG. First draw an equilateral triangle,] Ex. 5. Draw three circles each of which shall be tangent to the other two. Ex. 6. Draw two concentric circles and a line which is a tangent to one of these circles and a chord of the other. Ex. 7. Draw two concentric circles and a line which is a secant of One and a chord of the other. 13 -01 132 ~1300K II. PLiNE CGEOMEL R PROPSITION XIX. THEOREM 258. An inscribed angle, is measuired by one -half it.s intercepted arc. A A A 0 'I (A DD Fig. 1 Pi,,. 2 Fig. 3 CASE I. Wheni the center of the circle lies in one side of the inscribed angle. Given Z BAIC (Fig. 1) inscribed in the eircle 0, and Afl passing through the center 0. To prove that Z BAG is measured by ~ are 130. Proof. Draw the radius 00. Then,9 in the A OAAC, OA 00. (Why?) ZA Z C. (Why?) But Z BOC ZA + ZC. Art. 135. Z BOCr2 ZA. Ax.S. But Z BOC is imo-sured hy arc BC, A rt. 257. (a central Z is fleasnredl by its bi tcrcqptcd (Le). NNo. of auguiar degrees in Z BOC=No. of,arc degrees in BC'. Hence Z A is measured by ~-arc LC. Ax. 5. CASE U1. When the center of' the circle lies wvithin the i)iscr'ibed angle. Given the inscribed Z BAGC (Fig. 2), with the center of the circle 0 lying within the angle. To prove that Z BA C is measured by ~ are B C. Proof. Draw the diameter AD. Then Z BAD is measured by ~ arc BD. Case I. Let the pupil complete the proof. MEASUREMENT OF ANGLES 133 CASE III.;When the center of the circle is without the inscribed angle. Given the inscribed Z BAC (Fig. 3) with the center 0 outside the angle. To prove that / BAC is measured by 3 arc BC. Proof. Let the pupil supply the proof. 259. NOTE. By use of the above theorem, if the number of degrees in the intercepted arc be known, the number of degrees in the inscribed angle can be determined immediately. Thus, if the arc BC (Fig. 3) contains 48~, the angle BAC contains 24~; also, if it be known that the angle BAC contains, say 27~, the arc must contain 54~. A A G A " D Eig. d Fig. 5 Fig. 6 260. COR. 1. All angles inscribed in the same segment, or in equal segments, are equal. Thus A A, A', A" (Fig. 4) are all equal, for each of them is measured by one-half the arc BDC. 261. COR. 2. An angle inscribed in a semicircle is a right angle, for it is measured by one-half a semicircumference. Thus FH (Fig. 5) is a diameter.. Z FGH is a rt. Z. 262. COR. 3. An angle inscribed in a segment greater than a semicircle is an acute angle; an angle inscribed in a segment less than a semicircle is ({. obtuse angle. Ex. If, im Fig. 1, p. 132, ZA contains 23~, how many degrees are there in are BC? in arc AC? L34 [34 00OK IL. PLANNE GEOMTETRZY PROPOSITIONN XX. THEOREM 263. An angle formned by htwo chiordls intersecting withijn a circum~)ferenice is mneasured by half thre sumi of the intercepted arcs. D A) C B Given the chords AB and CD) in the 0 A DBC, intersecting, within the circumference atV the point P. To prove that Z DPB is measured by.1 (are DB + are AC). Proof. Drawv the chord AD. Then, in A ADP, ZDP-B-Z A+ ZDi Art. 135. (an ext. Z of a A is equal 1to the sumb of'thie twto opp. imt. Z). But Z A is measured by ~: are DB, Art. 258. (an inscribed Z is m~easur'ed by onie-7talf its iutaerceepte( arc). Also Z D is measured by are AC. (Why'?) Hence Z DPB is measured by (are DB + are AC). Ax. 2. ______________ ~~~Q. E. D. Ex. 1. In the above figure, if arc DB contains 640 and arc AC conktains 380, how many degrees in Z APC?in Z APD? Ex. 2. If arc AD - 520 and Z APP = 1240, find arc GB. Ex. 3. If, in Fig. 1, p. 132, arc AC contains 1120', how many degrees are there in the ~,< A MEASUREMENT OF ANGLES 135 PROPOSITION XXI. THEOREM 264. An angle formed by a tangent and a chord drawn from the point of contact is measured by half the intero cepted arc. C v D A j B Given the 0 PCD, and Z APC formed by the tangent APB and the chord PC. To prove that Z APC is measured by ~ arc PRC. Proof. Let the chord CD be drawn 11 AB. Then Z APC = Z PCD. (Why?) But Z PCD is measured by ~ arc PD. (Why?).'. ZAPC is measured by ~ are PD. Ax. 8. But arc PRC = are DP, Art. 233. (two 11 lines intercept equal arcs on a circumference)... Z APC is measured by ~ arc PRC. Ax. 8. Also Z BPC, the supplement of Z APC, is measured by i arc PDC, Ax, 3. (Jobr arc 'PDC is the conjugate of arc PRC). Q E.. B. Ex. 1. In the above figure, if are PRC contains 124~, how many degrees are in the angle A'C? Ex. 2. If arc CD = 96)i, find the angles on the figure. 136 13BOO II. PT,.LANE GEOMETRY PROPOSIrION XXII. TXXI T OREIM 265. An angle formed by two secants, or by tc'o tagents, or by a secant (ad a( tangenlt vmeetin wit/hout t/ji circu('ference is mteasured by half th (e (df';'c of/ 17e i1trcp'('ted ar's. P P C D D C/ X\n n J \ -A ~ YGC - D E Fig. t Tig., ig. a I. Given the 0 ACDB (Fig. 1), and the ZAPB formed by the two secants PA and PB, meeting at the point P without the circumference. To prove that Z P is measured by A (are AB-arc CD). Proof. Draw the chord CB. Then ZACB= ZP+ Z. (Why?).Z. ZP= ZACB- ZB. x. 3. But Z ACB is measured by ~ are AB. Art. 258. Also Z B is measured by - arc CD. (Why?).. Z P is measured by A (arc AB -arc CD). Ax. 3. II. Given the 0 ADB (Fig. 2), and Z CPB formed by the tangent PC and the secant PB. To prove that Z P is measured by A (arc AEB —arc AD)o Proof. Let the pupil supply the proof. EXERCISES ON THE CIRCLE 137 III. Given Z APB (Fig. 3) formed by the tangents PC and PD. To prove that Z P is measured by; (arc A-EB — arc AFB). Proof. Let the pupil supply the proof. 266. NOTE. By means of Props. XVIII-XXII, angles formed by chords, secants, or tangents, or combinations of these, are all reduced to central angles and hence may readily be compared. Ex. 1. If, in Fig. 1, p. 136, are CI) = 34~ and are B = 108~, draw -AD and find all the angles of the figure. Ex. 2. If, in Fig. 3, p. 136, angle P=80~, find arcs AlFB and AEB. EXERCISES. GROUIP 6 Ex. 1. What new methods of proving two lines equal are afforded by Book II? Ex. 2. What new methods of proving two angles equal? of proving two angles supplementary? of proving an angle a right angle? Ex. 3. What methods of proving two arcs equal? Ex. 4. In a quadrilateral inscribed in a circle, each pair of opIposite angles is supplemlentary. Ex. 5. A chord forms equal angles with the tangents at its extremities. Ex. 6. If an isosceles triangle be inscribed in a circle, the tangent at its vertex makes equal angles with two of its sides and is parallel to the third side. (Is the converse of this theorem tiru'?) Ex. 7. If two chords in a circle intersect within the circle at right angles, the sum of a pair of alternate arcs equals a semicircumference. Ex. 8. Given 0 the center of a circle and 0 AC a tangent; prove Z BAIC=%'Z O. Ex. 9. If one side of an inscribed quadrilateral be produced, the exterior angle so A formed equals the opposite interior angle of the quadrilateral. 18 '11-00K 11. PIANE CE)I~I Ex. 10. If two tiangents to a circle include an angle of 600 at their point of intersection, the chord joining the points of contact forms with the tangents an equilateral triangle. Ex. 11. The chordl Al)' equals the chord CD) in a given circle,, and the chlords, if produced, intersect at the point P. Prove secant PA== secant P1C. Ex. '12. A circle is circumsceribed about the triangle ABC, and P ig the midpoint of the arc A4L. Prove theat the angle _All equals onehalf thie angle C. Ex. 13. If A,7 B, C, D and E be poin~ts taken in succession on the circumfterence of a circle, and the, arcs A4R, BC CL n B eeul prove that the angles ABC, BC]) an ld CDE are equal. Ex. 14. An inscribed ang~le formed by a diameter and a chord has its intercepted arc bisected by a radius -which is parallel to the chord. Ex. 15. Two secants, PAIl' and PCD, intersect the circle JllDC- Prove that the triangles PII)C and PAD) are mutually equianugular. Ex. 16. Ciiven -AC' a tangent and -A/Ill1 CE; prove L\ ACD) and ABE mutually equiangular. Ex. 17. Given -ABC an inscribed triangle, AE IL BC, and CD I.AB; prove arc BD~arc BE. A D Lx 18. If the diagonals of an inscribed quadrilateral are diameters, the quadrilateral is a rectang (le. Ex. 19. Tangents through the vertices of ain BU inscribed rectangle form a rhombus. Ex. 20. Two circles intersect at P and Q. P and P1B are diameters. Prove that QA1 and QB form a straight 04 line.AB Ex. 21. Two equal circles intersect0 at P and Q, through P a line is drawn terminated by the circumferences at A4 and B. Show that QA1 equals QB. [Scc. I A and B are measured by what arcs 1 EXERCISES. AUXILIARY LINES 139 267. Use of auxiliary lines. In demonstrating theorems relating to the circle, it is often helpful to draw one or mnore of thb following auxiliary lines: A radius, a diiaineter, a chord, a perpendicular front the center upon, a chord, an arc, a circumJerenee, etc. EXERCISES. GROUP BY AtUXILLXRY LINES Ex. 1. Given circle 0, arc. A=arc )C, A BY I 0A, and -B IL 0 1C); prove BMRBX. N. Ex. 2. If from any point in the circumference of a circle a chord and a tangent be drawn, the 0 N perpendiculars drawn to them from the midpoint of the arc subtended by the chord are equal. Ex. 3. Given 0 the center of a circle, and 01D B I chord AC; prove ZAOD= ZBP. Ex. 4. From the extremity of a diametcr chords are drawn making equal angles with the diameter. C Prove that these chords are equal. [SuG. Draw Is from the center to the chords, etc.] Is the converse of this theorem true Ex. 5. If through any point within a circle equal chords be drawn, show that the line drawni from the center to the point of intersection of the chords bisects thleir angle of intersection. Ex. 6. Tanirgenits PJ and P1), are drawn to a circle whose center is 0. Prove that angle ' equals twice angle 0AB. Ex. 7. If in a circle two equal chords intersect, the segments of one, chiord equal the seginerits of the other chord. Ex. 8. A parallelogram inscribed in a circle is a rectangle. [SUG. Draw the diagonals of the ~_ use Art. `18, and Ex,"1, lP 100.J 140 B3OOK II. PLANE EIOMETI'TRY Ex. 9. If a quadrilateral be circumscribed about a circle, the angles subtended at the center by a pair of opposite sides are supplementary. [SUG. Draw radii to the points of contact and show that there are four pairs of equal X at the center.] Ex. 10. If an equilateral triangle ABC be inscribed in a circle and any point P be taken in the arc AB, show that 'PC= PA —+PB. [SuG. On PC take PMl equal to PA, draw AM.1 and prove A PAs and MAIC equal.] Ex. 11. Two radii perpendicular to each other are produced to intersect a tangent, and from the points of intersection other tangents are drawn to the circle. Prove that the tangents last drawn are parallel. [SucG. Draw radii to the three points of con- 1 tact, etc.] Ex. 12. A straight line intersects two concentrie circles. Show that the segments of the line \ intercepted between the circumferences are equal (prove AB = CD). Ex. 13. A common tangent is drawn to two circles which are exterior to each other. Show that the chords drawn from the points of fangency to the points where the line of centers cuts the circumferences are parallel. Ex. 14. A circle is described on the radius of another circle as a diameter, and a chord of the larger circle is drawn from the point of contact of the two circles. Prove that this chord is bisected by'the circumference of the smaller circle. [SuG. If the chord is bisected, a 1 from the center of the larger to the chord will also bisect the chord, etc.] A Ex. 15. Two circles are tangent ex- D\ ternally at the point P. Through P any p two lines APB and CPD are drawn terminated by the circumferences. Show that the chords AC and BD are parallel. Co, - [SuG. Draw the common tangent at P, If AC and BD are |, what A must be equal I] EXERCISES MAXIMA AND MINIMA 14 1 Ex. 16. Two circles intersect at the points P and Q. Lines AP'R and CQD are drawn, terminated by the circumferences. Show that AC and BD are parallel. Ex. 17. If a square be described on the hypotenuse of a right triangle, a line drawn from the center of the square to the vertex of the right angle bisects the right angle. [SUG. Describe a circumference on the hypotenuse of the right triangle as a diameter.] Ex. 18. If two circles are tangent externally at P, and a common tangent touches them at A and B, respectively, the angle APB is a right angle. Ex. 19. If in the triangle ABC the two altitudes BD and AE are drawn, the angle ABD equals the angle AED. [SUG. Describe a semicircumference on AB as a diameter.] 268. DEF. A maximum is the greatest of a class of magnitudes satisfying certain given conditions, and a minimum is the least (see Arts. 470, 471). For instance, of the chords in a given circle, which is the maximum? EXERCISES. GROUP l8 MAXIMA AND MINIMA Ex. 1. Of the chords drawn through a given point within a circle, determine which is the greatest, and also which is the least. Ex. 2. Find the shortest line, and also the longest line, that can be drawn / -- from a given external point to the cir- p --- — -- -' cumference of a circle. A [SUG. 0 being the center, prove in A PCO, PA < PC; by A PDO, PB > PD.] Ex. 3. Find the shortest line, and also the longest line, that can be drawn to the circumference of a circle from a point within the circle. [SUG. O being the center, prove, by use of z OPC, PB < P'C, etc.] D 142 BOOK IL. PLANE GEOMETRY Ex. 4. If two circles intersect, show C ar that. of lines drawn through a point of. A H int jrsection and terminated by the cir- ( I cumferen3es, that line is a maximumln 0 ' D which is parallel to tlle -- A line of centers. /,P n[SUG. Prove ItS < 00'.'. CtPD < APB.] (-/0 a* Ex. 5. Given AB I OB in circle O; prove \ OAB the maximum of all A having their vertices on the circumference and their sides passing through O and B respectively. [SUG. Draw a circle on OA as a diameter.] EXERCISES, QROUP Ns DEMONSTRATIONS BY INDIRECT MIETHODS Prove the following by an indirect method (see Art. 195): Ex. 1. A segment of a circle which contains a right angle is a semicircle. Ex. 2. If a rectangle be inscribed in a circle, its diagonals are diameters. Ex. 3. Prove the second part of Prop. VI by an indirect method. Ex. 4. Prove Prop. X by an indirect method. Ex. 5. A straight line connecting the midpoint of a chord and the midpoint of the are subtended by the chord is perpendicular to the chord (use the method of coincidence). Ex. 6. A line joining the midpoints of two parallel chords passes through the center. [SuG. Draw a 1. to each chord from the center and show that these Is are in the same line, etc.] Ex. 7. If the opposite angles of a quadrilateral are supplementary, a circle can be circumscribed about the quadrilateral. [SUG. Pass a circumference through three vertices of the quadrilateral; if it does not pass through the remaining vertex, etc.] Ex. 8. Prove the converse of Prop. XII. EXERCISES. CONSTANTS ANID LOCI I4I 269. A geometrical constant is a geometrical managnitude which varies iii some respect, as in position, buit remains constant in. size. Thus the angles inscrib~ed in a given semicircle vary in position but are all of the same size; viz., a right angle (see Art. 261). EXERCISES. GROUP 20 DETER-MINATION' OF CON'STAN'TS' AN-D LOCI Ex. 1. AR anidAC aLre tangents to acirc(le. Pis any point on the circumference outside the triangle ARC. As P moves, prove that the sumn of the ZA. and / BPC is constant. Ex. 2. Tn Ex. 8, p. 123, if AR and RBQ be produced to meet at T, show that the perimeter of the triangle TRQ equals the sum, of TAI and TB, and~ hence that the perimeter of triangle TRQ is constant, no matter how f) may vary in position between A and B. Ex. 3. Show on the same figure that, if 0 is the center, L~ROQ is constant -as P varies in position. Ex. 4. Two circles intersect in the points c A and B. From any point P oa one circum-A ference lines PAC and PBD are drawn, terminated hy the other circumference. Show that the chord CD is constant. [SuG. Draw _BC and prove Z CR1) aB constant. ]D Ex. 5. Given OA I OB, and CD) a line ofB riven length m-oving- so that 1) is alway's in OA1C and C ine ON) and ]'tlire midpoinit of CD) pr~ove that OP is constant ini lengr i (see EX. 6, p. 14). The determination of/ l oci is often facilitated b?/ showcing that seinet g1Vi tn_____ magnituade, is a constant. Ex. 6. Find the locus of a point muevin' so that it is at a given distance a from agiven circumfe~rence whose radius is r. 144 1300K II. PLANE GEOMETRtY Ex. 7. Find the locus of the midpoints of the radii of a given circle. Ex. 8. Find the locus of the midpoints of all chords of a given length drawn in a given circle. Ex. 9. Find the locus of the vertices of all right triangles having a given hypotenuse. Ex. 10. Find the locus of the midpoints of all the chords drawn from a given point on a given circumference. [SG. IDraw a line from the center to the given point and perpendiculars from the center upon the chords.] /Q Ex. 11. In Ex. 5, p. 143, find A_ the locus of P. Ex. 12. QP is a line of given length and moves so that Q is always in a given circumference, and QP is always parallel to a fixed line. Find the locus of P. EXERCISES. 10ROUP 2l THEOREMS PROVED BY VARIOUS MIETTHODS Ex. 1. The line which bisects the angle formed by a tangent and a chord bisects the intercepted arc also. Ex. 2. An inscribed trapezoid is isosceles. S Ex. 3. Given TA and TB1 tangents, arc A41= T A C 80~, arc BD= 95~, and arc D)(=150~; find all the angles of the figure. Ex. 4. If two tangents to a circle are parallel, the line joining their points of contact is a diameter. [SUG. Draw radii to the points of contact and D use an indirect method of proof.] Ex. 5. A rectangle circumscribed about a circle is a square, [SUG. Use the preceding theorem.] MISCELLANEOUS EXERCISES 145 Ex. 6. Given O the center of a circle, A and BP= the radius; prove Z AOC=3ZP. /B Ex. 7. Find the angle formed by the C P side of an inscribed square and the tangent through the vertex of the square. Ex. 8. From an external point a secant is drawn through the center of a circle, and also two other secants making equal angles with the first secant. Show that the secants last drawn are equal. Ex. 9. ABC is a triangle and on the side AB the point P is taken and on BC the point Q, so that angle BPQ equals angle C. Show that a circle may be circumscribed about the quadrilateral APQC. Ex. 10. Two circles are tangent to each other externally, and a line is drawn through the point of contact terminated by the circumferences. Show that the radii from the extremities of this line are parallel. Ex. 11. If a circumference be described on the leg of an isosceles triangle as diameter, the circumference will bisect the base of the triangle. Ex. 12. The chord of an arc is parallel to the tangent at the midpoint of the arc. Ex. 13. If a triangle be inscribed in a circle, the sum of the angles inscribed in the segments exterior to the triangle is four right angles. Ex. 14. Find the corresponding theorem for an inscribed quadrilateral. Ex. 15. Find the locus of the centers of all circles passing through two given points. Ex. 16. If two unequal chords intersect in a circle, the greater Chord makes the less angle with the diameter through the point of intersection of the chords. State also the converse of this theorem. Is the converse true? Ex. 17. The sum of the legs of a right triangle equals the sum of the hypotenuse and the diameter of tihe inscribed circle. J 146 BOOK II. PLANE GEOMETTRY Ex. 18. The sides.4B, BC and AC of a triangle tolch the inscribed circle at the points P, Q and R. Show that angle PQR and one-half angle A are complementary. [SuG. Draw radii from the center 0 to P and R. Then ZIPQR= Z POA, etc.] Ex. 19. From the point in which the bisector of an inscribed angle meets the circumference, a chord is drawn parallel to one side of the angle. Show that this chord equals the other side of the angle. Ex. 20. Given AB a diameter, AP= D the radius, AD and PC tangents; prove C A CED equilateral. [SUG. Draw OC and CA; then in rt. B A OCP, CA=radius, ZP=30~, etc.] \ 0 Ex. 21. Perpendiculars are drawn from the extremities of a diameter upon a tangent. Show that the points in which the perpendiculars intersect the tangent are equidistant from the center. CONSTRUCTION PROBLEMS 270. Postulates. As stated in Art. 49, a postulate in geometry is a construction of a geometric figure admitted as possible. The postulates used in geometry, are as follows (see Art. 50): 1. Through any two points a straight line may be drawn. 2. A straight line may be extended indefinitely, or it may be limited at any given point. 3. A circumference may be described about any given point as a center and with any given radius. CONSTRUCTION PROBLEMS 147 271. The meaning of the postulates is that only two drawing instruments are to be used in making geometrical constructions; viz., the straight-edge ruler and the compasses. With these two simplest drawing instruments it is desirable to be able to construct as many geometrical figures as possible. 272. Form of solution of a problem. The statement of a problem and of its solution consists of certain distinct parts which it is important to keep in mind. These are 1. The general enunciation. 2. The particular enunciation. (1) Given, etc. (2) To construct, etc. (or some other construction phrase, as "to draw," "to bisect," etc.). 3. The construction. 4. The assertion. 5. The proof of the assertion. 6. The conclusion (indicated by Q. E. F., quod erat faciendunm, "which was to be done"). 7. The discussion of special or limiting cases, if such cases occur in the given problem. In the figures drawn in connection withl proble1ms, the gilen lines are drawn as heavy lines, the lines required as light lines, and the auxiliary lines as dotted lines, 148 BOOKK TI. PLANE GEOfMETRY PROPOSIrrIONx XXIII. PROBLEMI 273. From a given point iwithout a given line to draw a perpendicular to the line. P A C ______ d B Given the lille AB a1nd the point P outside AR. To construct a perpendicular from the point P to the line AB. Construction. With P as a center and with any convenient radius, describe an a rc intersecting lB in two points as at C and D. Post. 3. From C and D as centers and with convenient equal radii,greater than 1 CD, describe arcs intersecting at Q. Post. 3. Draw the line PQ and produce it to meet AB at R. Posts. 1, 2. [Assertion]. Then PR is the I required. Proof. P is equidistant from the points C and D. Constr. Also Q is equidistant from the points C and D. Constr..'..PR I CD. Art. 113. (two points, each equidistant from the extremities of a line, determine the I bisector of the line). Q. E. F CON STRUCTION PROBLEMS 149 PROPOSITION XXIV. PROBLEM 274. At a given point in a given line to erect a perpendicular to that line. R-_^~ R — """N 0, 7',// \' A cD ~/A c P A C I P- D B C'- ~'....P Fig. 1 Fig..2 Given the point P in the line AB To construct a perpendicular to the line A B at the point P. METHOD I. Construction. From P (Fig. 1) as a center with a convenient radius, describe an arc cutting off the equal segments PC and PD on the line AB. Post. 3. From C and D as centers and with equal radii, greater than PD, describe arcs intersecting at R. Post. 3. Draw PR. Post. 1. [Assertion]. Then PR is the I required. Proof. Let the pupil supply the proof. METHOD II. Construction. Take any point 0 (Fig. 2) without the line AB, and with OP as a radius, describe a circumference intersecting the line AB at C. Post. 3. Draw CO, and produce CO to meet the circumference at R. Draw RP. Posts. 1 and 2. [Assertion]. Then RP is the I required. Proof. Let the pupil supply the proof. Discussion. When is Method II preferable? 150 BOOK II. PLANE GEOlETr PROPOSITION XXV. PROBLEM 275. To bisect a gicen line. /Ic I F I 21__ _ __B — Given the line AB. To bisect line AB. Construction. With A and B as centers and with equal radii, greater than ~ AB, describe arcs intersecting in C and D. Post. 3. Draw the line CD intersecting AB in F. Post. 1. Then AB is bisected at the point F. Proof. Let the pupil supply the proof. Q. E. Fo PROPOSITION XXVI. PROBLEM 276. To bisect a given arc. IF Given the arc AB. To bisect arc AB. CONSTRUCTION PROBLEMS 151 Construction. With A and B as centers, and with convenient equal radii, describe arcs intersecting at C and D. Post. 3. Draw the line CD intersecting the arc AB at F. Post. 1. Then arc AB is bisected at the point F. Proof. Draw the chord AB. Then CD I chord AB at its middle point. (Whyt).. CD bisects the chord AB, Art. 223. (the J bisector of a chord passcs through the center and bisects the arcs subtended by the chord). Q. E. T PROPOSITION XXVII. PROBLEM1 277. To bisect a given tangle. A C " I — - B Given angle A OB. To bisect angle A OB. Construction. With 0 as a center and with any conve~ nient radius, describe an arc intersecting OA at C and OB at D. Post. 3. With C and D as centers and with convenient equal radii, describe arcs intersecting at F. Post. 3. Draw OF. Post. 1o Then Z AOB is bisected by line OF. Proof. Let the pupil supply the proof,;x. Construct an Z of 45~, 152 BOOK II. PLANE GEOMETRY PROPOSITION XXVIII. PROBLEM 278. At a given point in a given straig/f line to coan struct an angle equlal to a given angle. '{X _ D B P F. D) B P F Given the Z A, and the point P in the line BC. To construct at the point P an angle equal to ZA, and having PC for one of its sides. Construction. With A as a center and with any convenient radius, describe an arc meeting the sides of Z at D and E. Post. 3. Draw the chord DE. Post. 1. With P as a center and with a radius equal to A), describe an arc cutting PC at F. Post. 3. From F as a center and with a radius equal to the chord DE, describe an arc intersecting the arc HF at G. Post. 3. Draw PG. Post. 1. Then Z GPF is the angle required. Proof. Let the pupil draw the chord FG and complete the proof. Ex. 1. On a given line construct a square. Ex. 2. Construct an angle of 30~. Ex, 3. Hence construct an angle of 15~. CONSTRUCTION PROBLEMS 153 PROPOSITION XXIX. PROBLEM 279. Through a given point without a given straight line to draw a line parallel to a given line. D 7t /, \ E ~ ~______F C v Given any point P without the line A B. To construct a line through P II AB. Construction. Through P draw any convenient line CD meeting AB in C. Post. 1. At P in the line CD construct Z DPF equal to Z PCB. Art. 278. Then EF is the line required. Proof. Let the pupil supply the proof. Ex. 1. Through a given point draw a line parallel to a given line by constructing a parallelogram of which the given point is one vertex. Ex. 2. On a given line as a diameter, construct a circle. Ex. 3. Construct an arc of 60~ having a radius of 1 in. Ex. 4. On the figure, p. 148, if the point Q be constructed below the line AB, will the perpendicular required be likely to be more accurately, or less accurately constructed? Ex. 5. From a given point on a given circumference, how many equal chords can be drawn? Ex. 6. Through a given point within a given circumference, how many equal chords can be drawn? Ex. 7. Fromn a given point external to a given circle, how many equal secants can be drawn I 154 - 1OOK300 II. PLAkNE GEOMETRY nOPosIvoN SXXX- I)ROI3LEM 280. To ditiride a gih-(U Straight lite imb ui a requtired numbe-r of c((lf ol poros. Given the line AD. To divide AB into a rcqnirecl number (as three) equal parts. Construction. From A draw the line AP nmaking a convenient angle with AB. Post. 1. Take AC any line of convenient length and apply it to AP a number of times eqnal to the nnmber of parts into which AB is to he divided. Post. 2. From E, the end of the measure when last applied to AP, draw LB. Post. 1. Through the other points of division on AP, viz., C and D, draw lines 11 LB and meeting AB at P and G. Art. 279. Then AB is divided into the required nnmber of parts at F and G. Proof. A C =CD = DE. Constr. AF = PU = GB, Art. 176. (if three or more parallels intercept equal partsy on one transversal, they intercep~t equal parts on every transversal). Ex. i. Construct a right triangle whose legs are 1 in. and 1P in. E. 9. Construct aixectangle whose base is 0 in. and altitude 1 in. CONSTRUCTION PROBLEMS 155 PROPOSITION XXXI. PROBLEMI 281. To construct a triangle, given two sides and the included angle. --— A --- —---- l / C Given m and n two sides of a triangle and P the angle included by them. To construct the triangle. Construction. At the point A in the line AB construct A equal to the given Z P. Art. 278. On the line A B lay off A C equal to m. Post. 2. On the line AF lay off AD equal to n. Post. 2. Draw DC. Post. 1. The A ADC is the triangle required. Q. E. F. Ex. 1. Construct a triangle in which two of the sides are 1 in. and l[in., and the included angle is 135~. Ex. 2. Construct an isosceles triangle, in which the base shall be 1i in. and the altitude 2 in. Ex. 3. Construct the complement of a given acute angle. Ex. 4. Construct the supplement of a given angle. Ex. 5. I-low is the figure on p. 154 constructed with the fewest adjustrmenrts of thle compasses? Ex. 6. Draw a line (segment) and mark off three-fifths of it 156 BOOK IIf. PLANE GEOMIETRY PROPOSITION XXXII. PROBLEM 282. To construct a triangle, given two angles and thk included side. c P B ----- Given the X P and Q and the included side n. To construct the triangle. Construction. Take any line AD and on it mark AB equal to m. Post. 2. At A construct an angle equal to Z P. Art. 278. At B construct an angle equal to Z Q. Art. 278. Produce the sides of the A A and B to meet at C. Post. 2. Then A ACB is the triangle required. Q. E. P. Discussion. Is it possible to construct the triangle if the sum of the given angles is two right angles? Why? Is it possible if this sum is greater than two right angles? Why? Ex. I. Construct a triangle in which two of the angles are 30~ and 45~, and the included side is 1t in. Ex. 2. Construct the complement of half a given angle. Ex. 3. Construct an angle of 120~; of 150~; of 135~, Ex. 4. Trisect a given right angle. CONSTRUCTION PROBLEMS 157 PROPOSITION XXXIII. PROBLEM 283. To construct a triangle, given the three sides. fI _ / \p A mn BGiven m, n and p, the three sides of a triangle. To construct the triangle. Construction, Take the line AB equal to m. Post. 2. With A as a center and with a radius equal to n describe an are, and with B as a center and with a radius equal to p describe another arc. Post. 3. Let the two arcs intersect at the point C. Draw CA and CB. Post. 1. Then A ABC is the triangle required. Discussion. Is it possible to construct the triangle if one of the sides is greater than the sum of the other two sides? Why? What kind of a figure is obtained if one side equals the sum of the other two sides 1 Ex. 1. Construct an angle of 22i~. Ex. 2. Divide a given circumference into four quadrants. Ex. 3. Construct the figure on page 82, using the concurrence of the three altitudes as a test of the accuracy of the work. 158 BOOK II. PLANE GEOMETRY PROPOSITION XXXIV. PROBLEM 284. To constrluct at triangle, given two sides and an angle opposite one of tie)m. Given m and n two sides of a A and Z /P opposite n. To construct the triangle. Construction. Several cases occur, according to the relative size of the given sides and the size of the given angle, CASE I. When n > sn (and Z P is acute). B/., ~ -- p, \,'c,.______ — C ^ — - -A c —wAt the point A construct Z EAD equal to Z P. Art. 278. On AE take AB equal to m. Post. 2. With B as a center and with a radius equal to n, describe an arc intersecting AD at C and C'. Post. 3. Draw BC and BC'. Post. 1. Two A, ABC and ABC', are obtained, containing the sides m and n; but only one of theni, A ABC, contains the ZP. ~. A BC is the triangle required. CASE II. When n = m (and Z P is acute). Make the same construction as in the preceding case. The arc drawn intersects the line AD in the points A and C. Hence the isosceles A ABC is the triangle required, CONSTRUCTION PROBLEMS 159 CASE III. When n < m (and Z P is acute). B,' m/ m/\ I........ P g -- A c C D Make the construction in the same way as in Case I. Two A, ABC and ABC', are obtained. each of which contains the sides m and ni and an angle equal to Z P opposite the side n..'. A ABC and ABC' are the triangles required. Discussion. In Case I, if ZP is a right Z, let the pupil construct the figure and show that there are two A answering the given conditions. If ZP is an obtuse angle, let him construct the figure and show that there is but one answer. In Case II, if ZP is right, or obtuse, what results are obtained? In Case III, if Z P is acute and n = the L from B to AD, how many answers are there? also, if n < this i, how many? If /P is right, or obtuse, what result is obtained? Ex. 1. Construct a triangle in which two of the sides are 1 in. and 1i in., and the angle opposite the latter side is 45~. Ex. 2. Construct a triangle in which two of the sides are 1; in. and i in., and the angle opposite the latter side is 45~. Ex. 3. Construct a triangle in which two of the sides are 1i in. and t in., and the angle opposite the latter side is 30~. Ex. 4. Construct the figure of page 80, using the concurrence of the three bisectors as a test of the accuracy of the work, 160 100K II. PLANE (EOMETlY PROPOSITION XXXV. PROBLEM 285. To construct a parallelograCm, giteni two sides adi the itncltued ( ngle. n 'vn ~ 3II4~z~~33kLL~~ j~ L P'p / I/ — n B m B -7h^ A Given m and nt two sides, and P the included Z of a /7, To construct the parallelogram. Construction. Take line AB equal to m. Post. 2. At the point A construct Z CAB equal to Z P. Art. 278. On the side AC lay off AD equal to n. From D as a center and with a radius equal to m, anl( from B as a center with a radius equal to n, describe arcs intersecting in E. Post. 3. Draw ED and EB. Post. 1L Then ABED is the parallelogram required. Proof. Let the pupil supply the proof. PROPOSITION XXXVI. PROBLEMI 286. To circumscribe a circle about a given triangle. Given the A ABC. To construct a circumscribed ( about ABC. CON-STRUCTION PROBLEMS11 161.Construction. Erect Ii -D and EG at the midpoints of the sides AC and AB, respectively. Art. '2174. From 0, the points of intersection of these A, with a radius equal to OA, descriibe the 0 ABC. Post. 3, Art. 197. Then 0 ABC is the circle reqnired. Proof. Let the pupil supply the proof. PROP~OSITION XXXV II. IPROBLE.M 287. To ins~cribe a circle int a given tria~ngle. 0 Given the, A A BC. To construct an inscribed circle ini the triangle. Construction. Draw the line Al) bisectingc Z-BAC, and G-E bisecting Z BCA. Art. 277. From 0, the intcrscctioii of AD and CE, draw the line OP I. AC. Art. 273. From 0 as a center with a radius OP, describe a 0. Post, 3. Then circle 0 is the circle reqoitred. Proof. Let the pnpjil supp~ly the p~.roof. Q E Ex. 1. Find theo (enter, of a Driven ciretumference. 'Ex. 2. Construct. the fi rure on pi-e SI, using thie conecurrenoe of the three perpondicular bisectors as a test of thie accuracy of the Work. 162 BOOK II. PLANE GEOMETRY PROPOSITION XXXVIII. PROBLEM 288. Throuygh (t gicen point on the circumference to draw a tangent to a circle. P j,/' Given any point P on the circumference of the circle 0. To construct a tangent to the circle at the point P. Construction. Draw the radius OP. At the point P construct a line AB I OP. Art. 274. Then AB is the tangent required. Proof. Let the pupil supply the proof. 289. An escribed circle is a circle tangent to one side of a triangle ad A to the other two sides produced. / Thus the circle 0 is an escribed circle of the A ABC. A center of an B c escribed circle, as 0, is called an excenter of the triangle. Ex. 1. Draw a triangle and all of its escribed circles. Ex. 2. Construct the figure on page 83, using the concurrence of the three medians as a test of the accuracy of the work. CON-STRUCTION PROBLEMS16 163 PROPOSITION XXXIX. PROBLEM 290. From a given point 'withiout a circle to draw a tangent to the circle. J0 Given P any point withont the circle 0. To construct a line throuorh P tang-ent to the cirele 0. Construction. Draw, the line PO. Post. 1. Bisect the line P0 at JV. Art. 275. From -31 as a center with MDP as a radius, describe a circumnference intersecting the given circumiference at A and B. Post. 3. Draw P-A and Pill. Post. 1. Then P-A and PB are the tangents required. Proof. ZPPAO is iniscribed ia emiircle. Const~r. ZLP-AO is arighlt an o1e. (Why?) PA1 is tangrent t~o tlie (circle 0. (W~hy?) In like aniaier RB is tangent to the circle 0. 164 1300K II. PLAN-E GEOMETh''TY PROPOS-,ITION XL. P nROBM 291. Ujpon a given strcijh I ii)w to describ~e cl sr' mer, w~hih shall contain a, g~iven inscribed angle. K- P A Given the straight line, A B an~d the Z P. To construct on the line AL1 a seo-ment of a circle such that any angle inscribed in the se~Ynient shall equal ZJP. Construction. At the point B in the line AB construct Z ABC equal to Z P. Art. 2978. Construct -DR the I bisector of line AB. Art. 274. At BconsructBO IBCand intersectingr DE at 0O. Arts. 274, 122. From 0 as a cen-ter with OB as a radius (describe the eircle AMB, Post, 3. Then AiIIB is the required seoment. Proof. Let AQB be any Z inscribed in the segment A QB. Then Z A QB is measured by 11 are-c AFB. Art. 258. But BC is tangent to the circle AhI1B. (Why?) ZABC is measured by rcAB (Why?) Z Q =LABC, or ZLB. (Why?):.any Z inscribed in segment A]JIB Z P. Q. E.. CONSTRUICTION PROB3LEMS' 165 PROPOSITION- XLT. THEOREM. 292. To find the common unit o~fnmeasure of two commneub surable straight lines, and hence the ratio of the lines. K Ai Given two lines A Il and CD, of which CD is the shorter. To construct at coinmoni unit of measure of J I] anid (D,A and hence, obtain the ratio of A-B and CiD. Construction. Apply CD to AB as many times as possible; Post. 2. Say twice, wiith a remiainder KB. Then apply K-B to C'D as manyi times as possible; Post. 2. Say, three times, -with a remainder Li). Apply LI) to KB as mnany tinmes as possible; Post. 2. Say three times, -with a remnainder PB. Apply -PB to Li) as many times as possible; Post. 2. Say it is contained in Li) exactly twice. Then PB> is the conmmon unit of rmeasu-re, or I P and CL1). Proof. LD =2PB. KB> K-P + PB)'=3LD +PB = 7PB. Axs. 6,8S. CD~ C-!L-V L - )KI3KB + LD-)0"3-PB. AIY = AI K+ fB0 Ci) + KB 1i3 PB. 4J I- 53 PBL, 53 Hence CD 28W 3 Q. E.'v 1 66 166 BOOK-101' 11. PL1.ANK'E CTE 0AMIET R EXERCISES IN CONSTRUCTION PROB3LEMS 293. Analysis of problems. The nethodl of analysis (see Art. 196) is of especial v-alue in the solution of eonstruetion problemis. In gieneral, to investigate the solution of a problemi by this niethod: -Draw ai figure ifl w(hich, the ireqwired conisiruction is clssumi~ed ais malde Draw auxiliairy lines, if necessar-y Observe the relations betwecen the ptsof this figure, in orde~r to dliscover a An own relaition o)1 whlich, the 'required construction depend's; -Having discovcered the requir~ed r~elattion, co'nstruct (anothler figure by the direct use of this relation?. Ex. Through a given point within a circl-,e dra-w a chord -which shall be bisected by the given point. ANALYSIS. Let A be the given point within the0 given circle 0, and let PQ be a chord bisected atQ the point A. A bisected chord suggests a line QAA joining the point of bisection with the center 0, P and that (Art. '224) OA I PQ. SYINTHESiS, or D~iIRECT SOLUTION. Taking,another figure containing the data of the problem, connect the point A with the center of the circle by the line OA. Through A4 draw a line I QA (Art. 274), anid meeting the circumference at the points P and Q'. FQ is thetchord reqluired. EXERCISES. CROUP 22 CONSTRUCTION OF STRAIGHT LINES. Ex. 1. Draw a line parallel to a given line, and tangent to a given circle. [Suo. Suppose the required line drawn; then the radius to the point of tangency, if produced, is I given line, etc.] Ex. 2. Draw a line perpendicular to a given line, and tangent to a given circle, EXERCISES. PROBLEMS 167 Ex. 3. From two points in the circumference of a circle, draw two equal and parallel chords. Ex. 4. Through a given point draw a line which shall make a given angle with a given line. Ex. 5. Through a given point draw a line which shall make equal angles with the sides of a given angle. [SUG. The bisector of the given Z will be L the required line, etc.] Ex. 6. Through a given point between two given parallel lines, draw a line of given length with its extremities in the two parallel lines. Ex. 7. Through a given point A within a circle, draw a chord equal to a given line. Ex. 8. From a given point in the circumference of a circle, draw a chord at a given distance from the center. Ex. 9. Through a given point on the circumference of a circle, draw a chord which shall be bisected by another given chord. [SUG. Draw the radius to the given point and on it as a diameter describe a circle, etc. When is the solution impossible?] 294. Construction of points and of loci. In constructing a point to meet certain given conditions, it is often helpful to constr.uct the locus of a point answlering one of the given conditions and observe in nwhat point or points it meets a given line, or meets an.other locus ansuerri)g another givenl condition. EXERCISES. CROUP 23 CONSTRUCTION OF POINTS AND LOCI Ex. 1. Find a point P in a given line AB D equidistant from two given points C and D). ' [SUG. Construct the locus of all points C equidistant from C and D and observe where A= —'..B it intersects the given line AB.] Ex. 2. Find a point P in a given circumference, which is equidistant from two given points, (C and 1). 68 WBOOK II. PTANE T (EOETiY Ex. 3. Find a point P in a given line, whichl is equidistant from two given intersecting lines. Ex. 4. Find a point in a given line, which is at a given distance, d, from a given point. Ex. 5. Find a point which is at a given distance, a, from a given point A, and at another distance, b, from another given point B. Discuss the limitations of this problem. Ex. 6. Find a point equidistant from two given points, and at a given distance from a given straight line. [SUG. Draw the locus of all points equidistant from the two given points, and also the locus of all points at the given distance from the given straight line, etc.] Ex. 7. Find a point equidistant from two given points, and at a given distance from another given point. Ex. 8. Find a point equidistant from two given points, and also from two given intersecting lines. On the other hand, the determination of certain loci is equivalent to thle covnstrhction of all points which satisfy one or wore given conditions. Ex. 9. Find the locus of the center of a circle, which touches a given line at a given point. [Sut(. Construct a number of circles touching the given line at the given point and observe the relation of their centers.] Ex. 10. Find the locus of the center of a circumference with a given radius, r, which passes through a given fixed point. Ex. 11. Find the locus of the center of a circle, touching two given intersecting lines. Ex. 12. Find the locus of the center of a circle, touching two given parallel lines. Ex. 13. Find the locus of the center of a circle of given radius, r, which toucles a given straight line. Ex. 14. Find the locus of the center of a circle of given radius r, which touches a given circle, EXERCISES. PROBLEMS 169 EXERCISES. CROUP 2a CONSTRUCTION OF RECTILINEAR FIGURES Construct Ex. 1. An equilateral triangle, given the perimeter. Ex. 2. An equilateral triangle, given the altitude. Ex.- 3. An isosceles triangle, given the base and altitude. LX. 4. An isosceles triang-le, giveu the base and an angple at the base. Ex. 5. An isoseeles tr-ian gb, giveni thie veirteK- an112rl and~ the altitude. Ex. 6. A right triangle, given a leg and the acute, angIle a~djacent. Ex. 7. A righit triangle, given a leg an,.d the acute angle opposite. Ex. 8. A right triangle, given the hypotenuse and an acute angle. Lx. 9. A right triangle, given the hypotenuse and a leg. Lx. 10. A triang-le, given the altitude and the sides including the 'vertical angle. [SuG. Throughi the feet of the altitude draw a line ~L altitude and of indefinite length, etc.] Ex. I1. A triang-le,gie two sides and the altitude upon cue of them. Ex. 12. A sqluare, given the diagonal. Lx. 13. A rhombus, civen the two diagonals. ELx. '14. A rhomblus, given one angle andi one diagyonal. Lx. 15. A parall''1elogram11, given two adjacent sides and an altitude. Ex. 16. A paralle101,~logra, given a side, the altitude upon that side and an anglIe. Ex. 17. A parallelog'ram, given the diagonals and an angle included b)y them. Lx. 18. A quadrilateral, given thie sides,and one angle. 170 BO3OK II. PLANE GEOMETRY 295, Use of auxiliary lines in constructing rectilinear figures. In constructing polygons, auxiliary lines are frequently of service. Thus it is often of especial value to costruct, fist, either the inscribed or the circumscribed circle, and afterward the required triangle or quadrilateral. Ex. Construct an isosceles triangle, given the base, b, and the radius, r, of the inscribed circle. CONSTRUCTION. Draw a circle 0 with radius F equal to r. Draw a tangent at any point A. On this tangent mark off AB and AC each A equal - b. From B and C draw tangents BF / and CF to the circle. Then BCF is the a required triangle. EXERCISES. GROUP 25 CONSTRUCTIONS. AUXILIARY LINES Construct Ex. 1. An isosceles triangle, given the base and the radius of the circumscribed circle. Ex. 2. A right triangle, given the radius of the circmscribed circle and one leg. Ex. 3. A right triangle, given the radius of the circumscribed circle and an acute angle. Ex. 4. A right triangle, given the radius of the inscribed circle and an acute angle. [SUG. Draw the inscribed circle and at its center construct an angle equal to the supplement of the given angle.] Ex. 5. A triangle, given the base, the altitude and the vertex angle. [SUG. On the given base construct a segment which shall contain the given vertex angle. See Art. 291.] Ex. 6. A triangle, given the base, the median to the base, and the vertex angle. Ex. 7. A triangle, given one side, an adjacent angle, and the radius of the inscribed circle. (See Ex. 4.) EXERCISES. PROBLEMS 171 Ex. 8. A triangle, given one side, an adjacent angle, and the radius of the circumscribed circle. The use of auxiliary straight lines may be illustrated as follows: Ex. 9. Construct a triangle, given the perimeter and two angles. ANALYSIS. Suppose the required triangle ABC already constructed and let A ABC and A CB be the given angles. Produce BC to D and E, A making DB = AB and CE = AC(. - Then DE= given perimeter. Also ZD = LDAB.'. ZABC = 2I). '.- Similarly ZACB=2 ZE. Hence D B C CONSTRUCTION. Take DE the given perimeter; at D construct an angle = / of one given angle; at E construct an angle = 1 of the other given angle. Produce the sides of these angles to meet at A. Construct ZDAB =ZD and ZCAE = LZ. Then A ABC is the required triangle, etc. Construct Ex. 10. An isosceles triangle, given the perimeter and the altitude. [SuG. Bisect the perimeter and construct the altitude I to it at its midpoint.] Ex. 11. An isosceles triangle, given the perimeter and the vertex angle. [SUG. If the vertex Z is known, the base A may be obtained.] Ex. 12. A right triangle, give an aacute C angle and the sum of the legs. [SUG. Given AB the sum of the legs, construct LZ=45~, etc.] ) DB Ex. 13. A right triangle, given an acute angle and the difference of the legs. Ex. 14. A right triangle, given the hypotenuse and the sum of the legs. Ex. 15. A riliht triangle, given au acute angle and the sum of the hypotenuse and onu leg, 172 BOOK II. PLANE GEOMETRY Ex. 16. A triangle, given an anIgle, a side and the sum of tl other two sides. Ex. 17. A triangle, given an angle A, the sum of the sides _ and BC and the altitude upon AB. 296. Reduction of problems. In many cases a protlei may be solved b)y reducing the problem to a problem alread solved. (This is a special kind of analysis.) Ex. Construct a parallelogram, given the diagonals and one side. ANALYSIS: Suppose the D7 -ABCF to be the required z- alread constructed. Let AF be the given side. If the diagonals a, given, half each diagonal is given (Art. 161). Hence in the A AOF the three B - sides are given. Hence the required / problem reduces to the problem of con- / structing a triangle whose three sides.are given (Art. 283). Hence CONSTRUCTION. Let the pupil supply the direct construction. EXERCISES. CROUP 26 REDUCTION OF CONSTRUCTION PROBLEMS Construct Ex. 1. A right triangle, given the altitude upon the hypotenuse and the median upon the same. Ex. 2. A rectangle, given the perimeter and a diagonal (see Ex. 14, p. 171). Ex. 3. A rectangle, given the perimeter and an angle made by the diagonals. Ex. 4. A triangle, given the three angles and the radius of thl circumscribed circle. [SUG. The sides of the A are the chords of the segments of the 0 containing the given A.] Ex. 5. A triangle, given two sides and the median / / to the third side. / EXERCISES. PROBLEMS 173 Ex. 6. A triangle, given the three medians. [SUG. Reduce this to the preceding Ex. See Art. 187.1 Ex. 7. An isosceles trapezoid, given the bases and an angle. Ex. 8. An isosceles trapezoid, given the bases and a diagonal. Ex. 9. A trapezoid, given the four sides. Ex. 10. A trapezoid, given the bases and the two diagonals. [Sue. Reduce to Art. 283 by producing. the lower base a distance equal to the upper base, etc.] 297. Construction of circles. The construction of a required circle is frequently a good illustration of the preceding method of reducing one construction problem to another, For the construction of a circle frequently redluces to the problem, of finding a point (the center of the circle) which answlers given conditions. (See Art. 294.) Ex. Construct a circle which shall touch two given intersecting lines and have its center in another given line. This problem is equivalent to the problem of finding a point which shall be in a given line and be equidistant from two other given lines. (See Ex. 3, p. 168.) In some cases, however, the construction of a required circle must be made by an independent method. EXERCISES. CROUP 27 CONSTRUCTION OF CIRCLES Construct a circle with given radius, r, Ex. 1. Which passes through a given point and touches a given line. Ex. 2. Which has its center in a given line and touches another given line. Ex. 3. Which passes through two given points. Construct a circle Ex. 4. Which touches two given parallel lines and passes through 4 given point, 174 BOOK) I. PLANE GEOME3II;TRY Ex. 5. \Vlich passes through two given points and ]ias its center on a given line. Ex. 6. Which touches three given lines, two of which are parallel. /A Ex. 7. Which passes through a given point A and touches a given line BC at a ___ given point B. [Sue. Draw 1AB and t.B construct a I to BC.] Ex. 8. Which touches a given line and alsotouches a given circle at a given point A. - L;,' Ex. 9. Which touches a given line AB A, -— B/ at a given point A and touches a given A e/,circle. EXERCISES. GROUP 28 PROBLEMS SOLVED BY VARIOUS MIETHODS Ex. 1. Through a given point draw a line which shall cut two given intersecting lines so as to form an isosceles triangle. Ex. 2. Construct an isosceles triangle, given the altitude and one leg. Ex. 3. In a given circumference find a point equidistant from two given intersecting lines. Ex. 4. Draw a circle which shall touch two given intersecting lines, one of them at a given point. Ex. 5. Draw a line which shall be terminated by the sides of a given angle, shall equal a given line, and be parallel to another given line. Ex. 6. Construct a triangle, given one side, an adjacent angle, and the difference of the other two sides. Ex. 7. Find a point in a given circumference at a given distance from a given point. MISCELLANEOUS EXERCISESo PROBLEMS 175 Ex. 8. Construct a parallelogram, given a side, an angle, and a diagonal, Ex. 9. Through a given point within an angled draw a straight line terminated by the sides of the angle and bisected by the given point. [SUG. Draw a line from the vertex of the angle to the given point and produce it its own length through the point.] Ex. 10. Construct a triangle, given the vertex angle and the segments of the base made by the altitude. [SUG. Use Art. 291.] Ex. 11. Construct an isosceles triangle, given the angle at the vertex and the base. Ex. 12. Draw a circle with given radius which shall touch a given circle at a given point. Ex- 13. Construct a right triangle, given the hypotenuse and the altitude upon the hypotenuse. Ex. 14. Construct a triangle, given the base and the altitudes upon the other two sides. [SUG. Construct a semicircle on the given base as a diameter.] Ex. 15. Find a point in one side of a triangle equidistant from the other two sides. Ex. 16. Construct a triangle, given the altitude and the angles a9 the extremities of the base. Ex. 17. Construct a rhombus, given an angle and a diagonal. Ex. 18. Draw a circle which shall pass through two given points and have its center equidistant from two given parallel lines. Ex. 19. Construct a triangle, given one side, an adjacent angle and the radius of the circumscribed circle. Ex. 20. In a given circle draw a chord equal to a given line and parallel to another given line. [SUG. Find the distance of the given chord from the center, by constructing a right triangle of which the hypotenuse and one leg are given.] 1 7 B OOK Ill. PLVNE GEO'METRY Ex. 21. onstruc a ragle, given an an-le, thle bisector of that angle, and the altitude from another vertex. Ex. 22. Find thie locus of the points of contact of tangents (Irawnl from a given point to a series of circles hav-ing a given center. [SUG. Use Arts. 229 and 261.] Ex. 2.3. Given a line Ill anrd tw o pointi. 'C C and D, on the same side of JJ Find a, D point P in A4B such that Z -JJIC C i-DPi). [SUG. Draw a IL from C to if)' and pro- j ~ duce it its own length, etc. Ex. 24. Given a line - II a two points C and( 1) en thie same side of AB; find a point P in l1B such that CP? + P'D shall he a minimuma. Ex. 25. Draw a common external tangent to two given circles. ---- Ex. 26. Draw a common internal < tangent to two given circles. B0oK II PROPORTION. SIMILAR POLYGONS THEORY OF PROPORTION 298. Ratio has been defined, and its use briefly indicated in Arts. 245, 246. 299. A proportion is an expression of the equality of two or more equal ratios. As, b d' This reads, "the ratio of a to b equals the ratio of c to d," or, "a is to b as c is to d." 300. The terms of a proportion are the four quantities used in the proportion. In a proportion the antecedents are the first and third terms; the consequents are the seco nd and fou(rth terms; the extremes are the first and last terms; the means are the second and third terms. A fourth proportional is the last term of a proportion (provided the means are not equal). Thus, in a: b=c: d, d is a fourth proportional. 301. A continued proportion is a proportion in which each consequent and the next antecedent are the same. Thus, a: =b:c=c: d=d: e is a continued proportion. A mean proportional is the middle term in a continued proportion containing but two ratios. L (177) 178 BOOK III. PLANE GEOMETRY A third proportional is the last term in a continued pro. portion containing but two ratios. Thus, in a: b=: c, b is a mean proportional, and c is a third proportional. PROPOSITION I. THEOREM 302. In any proportions, the 1)roduct of the extremes is equal to the p)rod(uct of the cmeants. Given the proportion a: b=): d. To prove a(d = bc. a c Proof. = Hyp Multiply each nmember by bd. ad = bc. Ax. 4. Q. E. D. PROPOSITION II. THEOREM 303. The mean proportional between two quantities is equal to the squcare root of their product. Given the proportion a: b=b: c. To prove b = /ae. Proof. a: b =: c. Hyp... b=ac, Art. 302. (in any proportion, the product of the extremes equals the product of the meaans )..b. b= ac. Q. E. Do. Ex. 1. Find the fourth proportional to 2, 3 and 6; also to 3, 4, 3o Ex. 2. Find the mean proportional between 3 and 6. Ex. 3, Find the third proportional to 3 and 5. THEORY OF PROPORTION17 179 PROPOSITION III. THEOREM 304. If the product of twvo quantities is equal to the product of two other quatmiities, one pair may be made the extremes and the other pair the means of a proportion. Given ad= be. To prove a, b = c:d. Proof. (1(1 be. Hlyp. Divide each mneimber b)y bd. Then Or a c bvd Ax. 5. a:b =c:d. Q. E.D 305. COR. 1. ITf the, anPteceden1ts equal, the consequtents are equal. of a proportion are Thus, if a: x =a: Y, thenr x = y. Let the pupil supply the proof. 306. Con. 2. If three termts of onte proportionj are equa(l to the corresp~onding three tervis of anothier proportion,1. the fourth termns of the two prop~ortions atre equal. Thus, if a: b =c x, and a:b =c:y, then x =y. Let the pupil supply the proof. Ex. 1. Write ab=pq as a proportion in as many different ways as possible. Ex 2. Write x (x-1) = 6 as a proportion; aItso x';i 5 180 BOOK~ III. PL'ANE GE'OMETRY PROPOSITION IV. THEOREMT 307. ITf four quantities are in proportion, the?! are in) proportion by alternation; that is, the first term is to the third as thre second is to the fourth. Given the proportion a b - c d. To prove a c c=h (d. Proof. a b = c (1. Hyp.. ad~bc, Art. 302. (in any proportion, thre product of the extremes cqnals the product of thle w('ans) Writing a and d as the extremes, and e and b as the m-eans of a proportion, a: c — ) d I Art. 304. (if the product of tro quantities 'is equal, to the product of two other quantities, one pair 'may be made th~e extrenics andI the othecr pa'ir the means of a proportion). Q.D PROPOSITION V. THEOREM 308. If four quantities are in pro]portion, they are in proportion bay inversion;that, is, the Second term is to the first as the, fourth is to the third. Given the proportion a:b c d. To prove b a' d c. Proof. a:b=ec d. Ilyp, ad = be. (Why?) Writing b and c as the extremes, and a and d as the means, b:a==d: c. (Why Y) ______________ ~~ ~~Q. B, E ~.D Ex. Trausform x-a==b:eso that xshall be the last terim THEORBY OF PROPORTION 181 PROPOSITION VI. THEOREM 309. If folr qutaltities are in proportion, they are in proportion by composition; that is, the sum of the first two terms is to the secoed terml (.s the snum of the last two terms is to the last. term. Given the proportion a: b=c: d. To prove a b,: b=c+d: d. Proof. ayp. b - Add 1 to each member of the equality. Ax, 2. a c + 1=- + b d +_ + b c + d Or b d That is a + b: b= c+d: d. Let the pupil showi also that a T- b: a-c +- d: c. Q. E. D. PROPOSITION VII. THEOREhM 310. If folr qan.tities are in proportion,, they are in proportion by division; that is, the cliference of the first two is to the second as the (lifference of the last two is to the last. Given the proportion a b =c: d. To prove a-b: b=c-d: d. Proof. b d eHyp 182 iBOOK iii. PLANE GEOMETIZY Subtract I. fromi eah embler of the equiality. Or b (I That is a.-b: bzc-(:d. (1.. D Let the pupil show also that (a - b:a. e -c PROPOSITION VIII. THEOREM 311. If foqjr quantlities are in, proportionj, they care in proportion by composition and division; that is, the sum~p of the first two is to their differ-ence (is the setnt of the last twlo is to their difference. Given the proportion a: b e: d. To prove a + b: a - b = d: c-d Proof. a: bzz=: e1. Hyp. By comnposition a+1 (I Art. 3009. Also by division a IArt. 31 0. b (I Dividingy the corresponding imembers of the twvo equalities, a+b e) + d A.5 a -b c -( dx5 That is a-i- b: a - b~c + (:c cd. Q E.D~ Ex. 1. What does the proportion 12:3~=8:2 become by eompositiont also by division? Ez. 2, What does 2x - 5:5= 3x - 7: 7 become by compositionV THEORY OF PROPORTION 183 PROPOSITION IX. THEOREM 312. In a series of equal ratios, the sum of all thie antecedents is to the sum of all the consequents as any one antecedent'is to its consequent. Given a:b=c: d= e:f=g: h. To prove a+ c e+-e g: b + d +f+ h=a: b. Proof. Denote each of the equal ratios by r. Then -r- *.. a=br. Similarly c-=dr, e=fr, g=7ir. Ax. 4. Hence a + c + e- g=(b- d+ f + h) r. Ax. 2. Dividing by b - d + - + h, + + =r=-. Ax. 5. Ddny d +in+hh That is a + c+e + g: b +d+f+ -h=a: b. Q. E. D. PROPOSITION X. THEOREM 313. The products of the correspondigfl termos of two or more proportions are in proportion. Given a: b=c: d, e: f= y: h, and j: k=: m. To prove aej: bfk=cgl: dhm. Proof. -=- e - - Hyp. b d' f h' k m Multiplying together the correspondiing terms of these equalities, lAx. 4. bfk d7 hm That is aej:bfk=cgl: dhm., Q. EB 184 8BOOK III. PLANE GEOMETRY PROPOSITION XI. THEOREM 314. Lik'e powers or like roots of the term)s of a propor. tion are in proportion. Given the proportion,a: b=c: d. To prove an: l = c": d, and a.: a l =_ c-:d-. a C Proof. *yp b d Raising both imembers to the nt1 power, (!n O C7 That is a"l:7 1 c n:d( In like manner aH b-:: dn. Q. E. D. PROPOSITION XII. THEOREM 315. Equi multiples of two quantities have the same ratio as the quantities themselves. Given the two quantities a and b. To prove Ma: m =a: b. Proof. (= _(' Ident. b) b Multiply each term of the first fraction by vm. ma a a )mb b That is ma: mb=a: ~b. Q. E. D. Ex. 1. Transform p: q=x y in all possible ways by the use of the properties of a proportion. Ex. 2. Transform 7:3=28:12 by composition and division, Ex. 3..Also 2x + 5: 2x-5 - 2- 1: XI2-1L THEORY OF PROPORTION 185 316. NOTE. In the theory of proportion as just presented, the quantities used are assumed to be commensurable, but the same theorems may also be proved for proportions whose terms are incommensurable by use of the method of limits. For each incommensurable ratio may be made the limit of a corresponding commensurable ratio; then, by showing that the variable commensurable ratios are equal, it may be proved that the limiting incommensurable ratios are equal. It is also to be noted, that, in the above theorems, the terms of a ratio must be of the same kind of quantity; that is, both be lines, or both be surfaces, etc. Hence, in order that a proportion be treated by alternation, for instance, all four of the terms must be of the same kind. Ex. 1. Find the fourth proportional to a, 2 a, 3 x. Ex. 2. Find the third proportional to a - b and a-b. Ex. 3. Find the value of x in the proportion, 4: 5=x: 15. Ex. 4. Find a short method of determining whether a given proportion is true or not. Use this method to determine whether the following proportions are true: (1) 4:6=3:9. (2) 5 a: 2 a=15:6. Ex. 5. Transform the following proportion by composition and division, and afterward find the value of x; x +- 1: x-1=7: 5. Ex. 6. Construct exactly the figure of page 77 with the fewest pos. sible adjustments of the compasses. Ex. 7. Draw an obtuse triangle and construct its three altitudes, using the concurrence of the altitudes as a test of the accuracy of the work. Ex. 8. Arrange nine points in a plane in such a way that the greatest number of straight lines may pass through them, each line to pass through three points and only three, 186 BOOK iII. PlLANE (GEOMETRY PRnoosrr ION XI1I. THEOREMI 317. A line parallel to one side of a triangle and meet. ing the other two sides, divides these sides proportionally. A A K \ B C B C Fig'. 1 Fig'. 2 Given the triangle ABC and the line DE 1} base BC and intersecting the sides AB and AC in the points D) and E, respectively. To prove DB: AD=EC: AE. CASE I. When DB and AD (Fig. 1) are commensurableo Proof. Take any coillonl unit of mleasure of DB and AD, as AK, antd let it be contained 'i /I); a certaill num lber of times, as 'n times, tlandI ill,). i imin. Then - - Art. 245. Through the points of division of DB and AD draw lines I| BC. These lines will divide EC into )?, and AE into m parts, all equal. Art. 176. EC n -E m (Why?) AD* J)BAE(Why?) AD AE PROPORTIONAL LINES 187 CASE II. Whenz DB and AD (Fig. 2) are incolnmmensuralttle. Let the line AD be divided into any number of equal parts and let one of those parts be applied to DB. It will be contained in DB a certain number of times, with a line PB, less than the unit of measure, as a remainder. Draw PQ || BC. Then DP and AD are commensurable. Constr. DP. EPQ_. Al) AI~E C'ase I. If now the unit of measure be indefinitely diminished, the line PB, which is less than the unit of measure, will be indefinitely diminished. Hence DPl DB, and EQ = EC as a limit. Art. 251. DP DB tHnce - becomes a variable with -as its limit. AD AD Art. 253, 3. Also E becomes a variable with EE as its limit. AE AE Art. 253, 3. DP EQ But the variable — =the variable -A always. Case I. AD) AE DB EC. the limit — =the limit At. 254. AD A254 Q. E. D. 318. COR. 1. 3By composition, Art. 309. DB + AD: AD=EC + AE: AE. Or AB: AD=AC: AE. In like manner AB: DB=AC: EC, or, in general language, if a line parallel to the base caut the sides of a trianlgle, a side is to a segmentt of that side as the other side is to the corresponding segment of the second side. 319. COR. 2. Using Fig. 2, PB AP DP AP PB 7DP — ~-;also -__ QC AQ; s Ea Q AQ "Q* C E Q Hence, if two lines are ct by a number of parallels, the corresponding segments are proportional. 188 1100OK7 III. PLANE GEOMEF~TIRY PROPOSITION XIV. THErormM (Co-NVERISE OF PROP. XIII) 320. If a stv-aiqhtd lifle divide,, two sid(s of a triangle proportionally, it is Jparailel to the third sidle. A DF B Given the L ABC and the line DE intersecting, AB and AC so that AB: A-D=AC:- AF. To prove _DE 11 BC. Proof. Throghi -D draw the line _DK 11 BC and meeting, the side AC in K. Then AR: AD -_AC: AK', Art. 318. (if a line ji base cut the sides of a A, a sidleis~ to ac segment of that side as the other side is to the correspondliog segment of the spcond Side). But AB:AD=i'C: AF. Hyp. AF= AK, Art. 306. (,if thtree terms of one Proportion are eqoawl to the corresponidingi three terms of another proportion, thie purtit terms of the two j)roportions are equal). Hence the point K falls on F, and the line iDA coincides with the line DF. Art. 66. But the line DK 11 BC. Constr.. DF II B C, (for DF coincides with DR, which is 11BC). _____________ ~~~Q. E. D. Ex. 1. In the:figure of Prop. XIII, if AD=6, DB=4, and AE=9, find EC. Ex. 2. Also, if AD= 121, DB=8, and AC= 15, find AL an EC. EXERCISES 189 EXERCISE8 GROUP 29 REVIEW EXERCISES Make a list of the properties of Ex. 1. One straight line (in connection with such points as may be related to the line). Ex. 2. Two straight lines that meet, or intersect (in connection with such angles as may be formed by the given lines). Ex. 3. Two or more parallel lines (in connection with transversals and angles formed). Ex. 4. Right angles. Ex. 5. Complementary angles. Ex. 6. Supplementary angles. Ex. 7. A single triangle (in connection with lines within the triangle, as altitudes, medians, etc.). Ex. 8. A right triangle. Ex. 13. A trapezoid. Ex. 9. An isosceles triangle. Ex. 14. An isosceles trapezoid. Ex. 10. Two triangles. Ex. 15. A parallelogram. Ex. 11. Two right triangles. Ex. 16. A rhombus. Ex. 12. A quadrilateral. Ex. 17. A polygon. Ex. 18. A single circle, or circumference (in connection with related points). Ex. 19. Arcs of a circle (in connection with related lines and angles). Ex. 20. Chords in a circle (in connection with related lines or arcs). Ex. 21. Central angles in a circle. Ex. 22. Tangents to a circle. Ex. 23. Secants to a circle. Ex. 24. Two circles. 190 BOOK III. PLANE GEOMETrRY SIMILAR POLYGOlNS 321. DEF. Similar polygons are poltygons having their homologous angles equal and their homologous sides pro. portional. B B' A >C A C' Thus, if the fig(ures AiBCODE and A'B'C'I)'E' are similar the angles A, B, C, etc., must equal the anglles A', B', C', etc., respectively; also AB: A'B' = BC: B' C= CD: CDt, etc. Hence it is constantly to be borne in mind that similarity in shape or form of rectilinear figures involves two distinct properties: 1. The h7omologoos angles are eqcual. 2. The homologous sides are proportional. It should also be clearly realized that one of these properties may be true of two figures, and not the other. Thus, in the rectangle A and the rhomboid B, the cor~ responding sides are proportional but the corresponding angles are not equal. A B 7 -C D Also, in the rectangle C and the square 1), the corresponding angles are equal but the corresponding sides are not proportional. However, it will be found that, in the case of triangles, if one of the two properties is true, the other must be true also. 322. DEF. The ratio of similitude in two similar figures is the ratio of any two homologous sides in those figures, SIMILAR POLYGONS 191 PROPOSITION XV. THEOREM 323. If two triangles are mutually equiangular, they are similar. A A'.B C B' C' Given the A ABC and A'B'C' with LA = LA', LB = ZB', and ZC== C'. To prove the A ABC and A'B'C' similar. Proof. The given A are mutually equiangular. Hyp. Hence it only remains to prove that their homologous sides are proportional. Art. 321. Place the A A'B'GC upon the A ABC, so that ZA' shall coincide with its equal, the LA, and B'C' take the position FH. Then Z AFf = Z B. Hyp..*. FHI BC. (Why?).. AB: AF=AC: AH. Art. 317. Or AB: A'B'-A C: A1/C. Ax. 8. In like manner, by placing the A A'B'C' upon the A ABC so that the L B' shall coincide with its equal, the Z B, it may be proved that AB: AB'=BC: BC'. Hence the A ABC and A1B'C' are similar. Art. 321. 324. COR. 1. If two triangles 7have two angles of one equal to two angles of the other, the triangles are similar; also, If two right triangles havet an acute angle of one equal to an acute angle of the other, the triangles are similar. 325. COR. 2. If two triangles are each similar to the same triangle, they are similar to each other. 192 BOOK III. PLANE GEOM.ETRY PROPOSITION XVI. THEOREMA 326. If two triangles have their homologous sides pro8 portional, they are siilar. A A' B C' B' C' Given the A ABC and Al'BIC', in which AB:A-E-= AC: ACI= BC: B'C'. To prove the A ABC and A'B'C' similar. Proof. The given A have their homologous sides proportional. Hyp. Hence it only remains to prove that the A are mutually equiangular. Art. 321. 1. On AB take AF equal to A'BR, and on AC take AH equal to A'C'. Draw FH. Then AB: AF=AC:AIL. Hyp. ~f H II BC, Art. 320. (if a straight lilne divides two sides of a A proportionally, it is i the third side). Z AFH = Z B, and Z AFP = Z C. (w.hy?).. AFHf and ABC are similar, Art. 323. (if two are mutually equiangular, they are sirmilar). 2..'. AB: AF=BC: FH. Art. 321. Or AB: A'BI=BC: FH. Ax. 8. But AB: A'B'=BC: BIGC. yp. SIMILAR POLYGONS19 193 PH r ' Art. 306. Hence the A AFII and A'B'C' are equal. Art. 101.. A A'B'C' is similar to A ABC, Ax. 3. (for- its equaa A AFIH is similar to A ABC). Q. E. D. PROPOSITION XVII. THEOREM 327. If two triangles have an. angle of one equal to anr angle of the others, and the includling sides proportional, the trian)gles are shinilar. A F -- -- -- - B Given the As ABC and AID'(C', in which Z A Z A'I and AB: A'B' =AC: A'C' To prove the, A ABC and A'B'C' similar-. Proof. Place the A A'B'C' upon the A ABC so that Z Al shall' coincide with its equal, the Z A, and BIC' take the position PH. Then AB: AF=AC:All. Hyp. Hence, F11 11 BC. (Why?) Z APJII Z B, and Z AHP= ZiC. (Why?) AABC and AFH ar ia. Art. 323. Or A ABC and A'B'C' arc similar. Ax. 8. ______________ ~~Q. E.D. Ex. In the A -ABC, AB=6, AC=S, BC=10; inthesimilarA A'B'C'D V,'B'=9. Find dA'C'and B''. 1!)94 1PO()(K 1III. ' LAN J.V-K; '1' (-LOI.E 1'T1 PROPOSITION XVIII. THEOREM 328. If two trianigles have their sides p.arallel, or per pendicitlar, each to each, the triangles are similar. A A A' B' B C B' C 2 Fig. 1 Fi.: C' Fi:,.: Given tle A A'B'C' (Fig. 2) with its sides II, and tt1. A AH''C' (Fig. 3) with its sides I, to corresponding sides of the A ABC. To prove & ABC and A'B'C' similar. Proof. The X A and A/ are either equal or supplemenltary, (for the sides forming them, are I or J ). Arts. 130,132. Similarly, the A B and B', and C and C' are either equal or supplementary. Hence one of the three following statements must be true concerning the angles of the A: either 1. The & contain three pairs of supplementary X and ZA + ZA'=- rt. A, ZB + Z- B'-= rt. A,, Z C + C'= 2 it. A; or 2. The A contain two pairs of supplementary A, as ZA= ZA', Z B + Z B-0 rt., Z C + Z C= t. A; or 3. The & contain three pairs of equal A and ZA= ZA', Z B= ZB', Z C Z C', Art. 139. (if two A of a A = two A of another A, the third A are eqall). The first two of these statements are impossible, for the sum of the A of two A cannot exceed four rt. X. Art. 134. Hence the third statement is true, and the A ABC and A1'' C"' are Iiutnally equiangular, and therefore similar. Art:323 Q. E. D. SIMILAR I'OLYGONS 195 PROPOSITION XIX. THEOREM 329. If two polygons are similar, they may be separated into thle same number of triangles, similar, each to each, and similarly placed. B B' A C A O' CC E D E' D' Given tlhe simil-ar polygN-onis AB.DEI)> a(nd A''C'DWE', divided into triangles )y the diagonals AC, AD, aind A'C', A'YD drawn from the corresponding vertices A and A'. To prove that A ABC, ACD, ADE n re similar to the A A'JW'C', A'C'D', A'D' E', respectively. Proof. 1. Z l. = Z 1>'. Art. 321. Also A AB: AI'=: 'C'. Art. 3221..A. AA C anid A'B'C' are similar, Art. 327 (if tlwo L h(e( n Z of onre - (1n, Z of thie other and the incidtlin si(tds prop)ortional, Ithe A ('re simrilai). 2. Again Z BCD = ZB'C'D', Art. 321. (homologous X of similtr polygotns). Also Z BCA =- Z CA', Art. 321. (homologolus A of the slnilar & ABC and Acw'BIC'). Subtracting Z A CD = Z A' C''. A 3. B3ut BC: B'C'= CD ( CI'), Art. 321. (homologous sides of similar polygqons are proportional). And BC: B' C'- AC A' C, Art. 321. (being homologous sidls of the similar &/ ABC and A'B'C'). Hence AC: ~A' CD: (Cj. Ax. 1..'. the A ACI(, and A'C'I)' lare similar. Art. 327. 3. In like manner it can be shown that the A I)I E and A')'E' are similar. Q. E. D. 196 BOOK III. PLANE (rEOMET):' PROPOSITION XX. THEOREM (CONxEuRSE OF PR(-Oc,. XIX) 330. If two polygons are colmposed of the se nsae umber of triangles, similar, each to each, and similarly placed, the polygons are similar. B.'AI. --- —— ` A K Of E D E' D Given the two polygons ABODE and A'B'C'D'L', in which the Q ABC, ACI), A-DE are similar, respectively, to the A A'B'G', A'C'D', A'D'E', and are similarly placed. To prove the polygons ABODE and A'B'C'D'E' similar. Proof. Z B = Z B', Art. 321. (being homologous A of similar A). Also Z BCA = ZBC'A'. (Why?) And Z A CD = Z A' ' D'. (Why;i Adding equals, Z BC1) = Z B C'/D. (Why?, In like manner it may be shown that Z CDE= Z (C'IE', Z BAE = B'A'E', etc. Hence the polygons are mutually equiangular. AB BC Also A'B' (homologous sides of sinilar ). Art. 321. BC AC CD AC BC CCD Again =C, and. - B' C AlC' CG'D' A'/Ci' B' C')' CD _DE AE ( Why In like manner -... A'E C/J' D 'E' A'E' Hence the homologous sides of the given polygons ar( proportional... the polygons ABCDE and AL'B'C'D'E are similar. Art. 321 Q. E. Do PROPORTIONAL LINES 197 331. NOTE. It is often convenient to write a series of equal ratios, like those used in the above proof, as follows: AB = BC _ /C =\ CD _ ( D\ DE _ AE A'B'B'C' 4 'C' C'W A'D'} \A E'' E'I a ratio which is used merely to show the equality of two other ratios being inclosed in parenthesis. PROPORTIONAL LINES AND NUMERICAL PROPERTIES OF LINES PROPOSITION XXI. THEOREM 332. In any triangle, the bisector of ran angle (divides the opposite side in to segments!which are proportional to the other two sides. F\ '.A 0 D B Given the A A BC, with the line A D bisecting the Z BA C, and meeting BC at I). To prove I C: I B - C (: B. Proof. Draw the line CF A| AD, and meeting' A L produced at F. Then DC: 1)B=AF: AB. Art. 317. But Zr= /p/. Art. 124. And Zs- Zp. Art. 126. Also Zp'= Zp. Hyp..Z. / s. (Why?).. A CF is isosceles, and A C= F. (Why?) Substituting A C for its eclual,AFF, in the above proportion, DC: DB=AC: AB. Ax. S. __________ _Q. E. D. Ex. If, in the above figure, AB=16, AC( — 12, and 14, find DC and DtB. [Svu. Let DC=-x, l)-B=14 —x, etc.] 1398 13O0K Ii. PLANEJ GE1OMETRY 333. A line divided internally is a line (livided into two parts by a point taken between its extremities. A r B Thus, the line AB is divided internally at t-he point P into the segmnents PA and PB-. To divide a give~n line internally in a r(iven ratio (as 2: 7), di vole it into a nLnilte1r Of equal parts, equal to the 8um of the ternms of the ra~io (as 2 -+ 7, or 9 equal parts). 334. A line divided externally is a line whose parts ate con1sidered to be the segmients included betweeni a point on the litie produeed and the extremities of tite given line. F' A ~~~B Thus, the line AB is divided externally at the point P', into the two segmnents P'A and PtmB. To divide a given line externally in a given ratio (as 2:7,divide, it into a number of eq ual p)arts, eq11( nil) the dQ'ec c f tile 0115(fth ratio (as 7 -2, or 5 parts) anid p' i-mce it till the produced part, equals the smialler terni of the ratio times,- the unit ii no foumid. 335. A~ line divided harmonically is a line, dividedl both interntdllv and externally1- in the satne ratio. Thus, if the line AB is divided internally, so that PA PB=-1: 2 and externally, so that P'A: P'B= 1:2i, then PA:PB==PA:P'B, and the line is divided hartnonically Ex. 1. Divide a given line harmonically in the ratio 1 3. Fx. 2. Divide a given line harmonicall y in the ratio 2 5 PROPORTIONAL LINES 199 PROPOSITION XXII. THEOREM 336. In a.y triangle, the bisector of an exterior angle d'ivdes externally the opposite side produceld into segments which are proportional to the other two sides. Given the A1 ABC with its exterior Z /BAG bisected by the line AD meeting the opposite side produced at D. To prove )DB: DC=AB: AC. Proof. Draw the line BF 1 AD and meeting AC at F. Then DB: DC=AF: AC. (Why?) (Why?) But And Zr = Z1/. Zs = Z1p. Z/p= Zp..'. Zr= Zs. Also (Why?) ( W0 hly?) (Why?) (Why?).. L A BF is isosceles, and AB= AF. Substituting A1B for its e(ual, AF, in the above pror)ortion, DB: DC=AB: AC. Ax. 8. Q. E. D, 337. Con. The lines bisecting the interior atd exterior anygles of (a triangle at a given vertex, diidle tihe olposite Wside harm ot ic lly, 20) BOOK 111. PLANE (GEMETRY PROPOSITION XXIII. THEOREM 338. The homologous altitudes of two siwilar triangles have the same ratio as any two homologous sides. B B' A I C A' D' C Given the similar & ABC, A'B'C', and BD, BrDI any two homologous altitudes in these A. B 7) B A B C A l 0 To prove To prove B' - B/A I B' C A' C Proof. In the right & ABI) and Al/B/D', ZA = Z A', Art. 321. (icing homoloqlus X of the similar &A A BC and A'B'C')... A ARIBD and A'B'lD' are similar, Art.:,. (if two rt. Q have an acute Z of one = an acute Z of the other, the & are similar). B D B A B. BDf~B'A/ Art. 321. B'J)' B'A But, in the similar A ABC and A'B'C', BA BC AC B'A~' B'OC'A'C'6' Art. 321. BDHe BA BC AC Ax Hence g/bgtBD Ax. 1. Q. E. D. Ex. If, in the above figure, AC= 18, A'C' 12, and BD= 15, find B'D'. PROPORTIONaAL LINES20 C 01 PROPOSITION XXIV. THEOREM 339. If three or miore lines pass through the same point and intersect two parallel lines, they intercept proportional segments on, the parallel lines. D' C' B' A' A' D' 0 AR C D A B C D Given the transversals OA., OR, OC, Of) intersecting the parallel lines AD and A'D' in the points A, B, C, D and A', B', C', D', respectively. To prove A'B' B'C C'D AIB BC'1 CIDn Proof. A'D' II Al). Hp. Therefore the base A of the & A''O, BR'C'O, etc., are equal to corresponding base A of the & ABO, BOO, etc. Art. 12; or0 Art. 124. Hence the & A'B'O, B'C'O, etc., are similar to the & ABO, BCO, etc., respectively-. Art. 324. Arte. 321. A'B' BO B 'C' (IO'O CID' Q. E. ~D. Ex. if the sides of a polygon are 2, 3, 4, 5, 6 feet, tnod, in a similar polygon, the side homologous to 6 is 9, find all the other sides of the second polygon, I 220 2 202 ~BOOK Ill. PLANE GEOMLETR71Y PROP. XXV. THEOREAI (CONVERSE OF PR~oP. XXIV) 340. If three or more non-parallel straight lines 'intercep~t proportio~wl segmcnts on two paralil~s they ])ItSs througha the sante pobint (that is, are, concurrent). A ii Given the transversals AA', BB', CC' intersectingr the' parallel lines A C and Ai'I' so thatAL= C To prove that the lines AA', BB', CC' are concurrent. Proof. Produce the lines -AA' and BR' to meet (at some point 0. Draw the line OC and let it intersect the line A' C' at somie point P). Then Art.:889. 13 11.ence B'11 zB'C'. Art point P coincides with point C'. line CC' coincides with line CP. A the line CC', if prodneed, passes through 0, (for it coincides with the liuc CP, which passes throug 0).. AA', BB', CC' mneet in 0. "3-p. 8. 06. 4. 64 q ItE. DI. PROPORTIONAL LINES 203 PROPOSITION XXVI. THEOREM 341. The perimieters of two similar polygons have the same ratio as any two homologous sides. B Ia' Given the two similar polygons ABC L )Eand A'B'C'D'E", with their perimeters denoted by P and P' and with AB and A1'I any two homologous sides. To prove that P: P'AB: AB'. Proof. AB: A'B'=BC: B'C'=CD: ( CD'etc. Art. 321. Henc e Ai+ B->C+ Cl)+, etc. A: 'B/'+ C B C'+ C()', etc., =- A1 At!, Art. 312. (ai ( scrics of ('(/ qu l ratios, /hC si 0 of alt /1 l thr (( rcede ts iS, s ) t1,', t (of alt thc consequentts as anfy o)r ant(cc(tc(( t is to its cons(UC.e)et). Or IP:~ I ' = —: 1. A B'. Q. E. D. 341 (I) (Ci. nL, t11o,i.sl.lar /p//o..s', atil tavo /ololo(/lolt liit (,ls ar t(o CeIch oth.r (s ta l/ ollcr two ihoaologouo s lilros; a nd the pcritoitlcrs arc to c('ct otlh r a(s a( /l tt(o /)lomolotogts lit,( s. Ex. If the perilnetcr of a oiven fiel(l is 210 rodls and (l side of this field is t(o i li'coloe-o(ls sie of a simlilar field (s 3: 2, tind the peri1eter of the seco(nd field. 4 ) &.J04 204 1300K III. PLANE, OGEMETRY PROPOSITON XXVII. THEOREM 342. In a right triangle, IL The altitude to the hypotenuse is a mean pro~portional between the segments of the hypotenuse; II Each leg is a iiean proportional betwceen the hypote nsie and the segmient of the hypotenutse adIjacent4 to the gi~ven leg. B A F Given the right A ARC and RE the altitude upon the hypotenuse A. C. To prove I. AIF BF=z RE: FC. -i C: A,4 3=zAR.: A F A AC BC~ R C: FC. Proof. rphe ZA is common to the rt. AL ARE and ARC. A ABF and ABC are similar. Art. 324. Similtarly Z C is common to the rt. ABEC and ABC, and A -BFC and ABC are similar. ABAR and BEC are similar, Art. 325. (if two A are sim-ilar to the same A, they are simtilar to each other). Hence, I. In the & ABE and BEC, AE: RE= BE: FC, Art. 321. (homologous sides of'similar A~ are proportional), PROPORTIONAL LINES 205 II. In the similar ~ ABC and ABF, AC: AB=AB: AF. Art. 321. Also, in the similar A ABC and BFC, AC: BC= BC: FC. Art. 321. Q. E. B. 343. COR. The perpendicular to the diameter from any point in the circleemference of a circle is a mean proportional between the segments of the diameter; and the chord joining the point to an extremity of the diacm.eter is a mean proportional betweeen the ldiameter andz - the segment of the diameter adjacent to the chord. 344. DEF. The projection of a point upon a line is the foot of the perpendicular drawn from the point to the line. A - B C L A' B' r P C" Q Thus, if AA' be perpendicular to L.M, A' is the projection of the point A on the line LM. 345. The projection of a line upon another given line is that part of the second line which is included between perpendiculars drawn from the extremities of the first line upon the second line. Thus, the projection of AB on LM is A'B'; of CD on PIQ, is C'D. Ex. 1. If the segments of the hypotenuse of a right triangle are 3 and 12, find the altitude on the hypotenuse; also find the legs of the triangle. Ex. 2. If, in the figure of Art. 343, the diameter is 20 and the longer chord 16, find the segment of the diameter adjacent to the chord, 206 BOOK 1TI1. PLANE G(EOI,)ETrLY PROPOSITION XXVIII. THEOREIM 346. In a right triangle, the asquare of the hypotenuse is equal to the sum of the squtares of the legs. B A 1;' Given the right triangle IA l C witih A C the hypotenuse. To prove A C =.-. 1) t+ BC1. Proof. Draw the line BF I AC. Then A C: A B- AB: Al F I.. AC X AF?= A~B2. Arts. 342, 302. Also AC: BCBRC: FC. iAC X FC= BC2. (WNhy?) Adding equals, AC (AF + FC) =-Al - 1B2. Ax. 2. Or AIC2= - B + BC'. Axs. 6, 8. Q. E. D. 347. COR. 1. nl, a right triangle, the square of either leg is equal to the square of the hypoteluse minuls the squllare ol the other leg. 348. COR. 2. In the square ABCD, the diagonal divides the square into two righlt triangles. Hence AC=AB2 + tBCh92 A/B... AC-=ABV1/0, or - 1/ AC -,AB 2 AR 1 / Hence the diagonal cald the side of a A D square are in commensu rab le. Ex. 1. If the legs of a rt. A are 11 in. and 2 in., find the hypotenuse. Ex. 2. In the figure on p. 204, show that AB2: BC2= AF: FCG NUMERMICAL PROPERTIES27 2 0 7 PROPOSITION XXIX. THEORE-M 849. lin any oblique triangle, the square of a side opposite an acute angle is equal to the. sumi of the squares of the. othier two sides, diminished by twice the product of one of those sides by the projection of the other side upon it. BB A D C D A 4 Fig. I ~~~~~~~~~Fig.29 Given acute Z CGin A ABC, and DC the projection of the side BC on. the side A C. To prove:f25 L+72 2 AC X DC. Proof. If D falls onl A C (Fig. I1), J)D= A C - -DC. If DT falls on AC produced (Fig. 2), AD-= DC-A4 C. In either case, AD-=AO C2 4- AC X DC. Ax. 4. Adding, BD9- to eachi of these equals, ADJ2+ D~ C D ~I -2 AC X ) C. Ax. 2 But, in the rt. A ABD, ADAB,7-h Art. 3406. and, in the rt. A DBC, 77C+ B B7c~ (Whvy?) Substituting these values in the above equality. AB2-C+C —2AC >K DC. Ax. 8. _________ _ _ ~~~Q. E..O, Ex. 1. A line 10 in. long makes an an~le of 450 with a second line; find the projection of thle first line on the second. Ex. 2. Find the samne, if thle angle is 600. Ex. 3. If the side of an eqnilateral triangle is a., find its projec~tion On thle base. Ex. 4. if, in Fig. 1, BC= 1O, AC- 12, and Z C=-600, find AB. 208 BOOK II(. PLANE (GEO)MEIJtY PROPOSITION XXX. TiOllE:~o)mr 350. In any obtuse triangle, the square of the side opposite an obtuse angle is equal to the sunt of the squares of the other two sides, increased by twice the product of one of the sides by the projection of the other side ulpon that side. B A C D Given the obtuse /ACB in the A ABC, and CD the projection of BC on AC produced. To prove AB2= A -~_C2+B + 2 AC X CD. Proof. AD=-A(7+ CD. Ax. 6.A. AD2-C+ ~CD+2 A C X CD. Ax. 4. Adding BD2 to each of these equals, AD2I+ ~BDJ (A + C2+ 9D+ ABD2+ 2 AC X CD. Ax. 2. But, in the rt. A ABD, AD2+ ~ D2 = 7B2 (Why?) And, in the rt. A CBD, 6CD2+ B-D2= BC-. (Why ) Substituting these values in the above equality, A2_B= -Co+ 2 C+D 2D. Ax. 8. Q. E. D. 351. COR. If the square on one side of a triangle equals the sum of the squares on the other two sides, the angle opposite the first side is a r;ight angle; for it cannot be acute (Art. 349), or obtuse (Art. 350). Ex. 1. If, in the above figure, BC=10, AC=2, and ZBCA=120~, find AB. Ex. 2. If AB=20, BC=14, and AC=12, find CD. NUMERICAL PROPERTIES 209 PROPOSITION XXXI. THEOREM 352. If, in any triangle, a median be drawn to one side, I. The sum of the squares of the other two sides is equal to twice the square of half the given side, increased by twice the square of the median upon that side; and II. The difference of the squares of the other two sides is equal to twice the product of the given side by the projection of the vmedial iupon that side. A B I3 F C Given the A ABC, AB>AC, AM the median upon BC, and MF the projection of AM on BC. To prove I. '-+ iA-2 2 B= 2 - + 2 All. II. ~B2- AC2=2 BC X MF. Proof. In the A B3MA and AMC, BM1z=MC, AM=AM, and AB>AC. Hypo.. Z AfMB is greater than Z AMC. Art. 108... /LAMB is obtuse (for it is greater than half a straight Z ). In obtuse A ABM, ABB-2 l2~AT 2 2 BM X MF. Art. 350. In acute A ACCM, AC2 C A-2+A M2 — 2 2 MC X EF. Art. 349. Adding, TB2+ 22 BM2 + 2 2 2, Ax. 2 (for MC= BM). Subtracting, AB2 -- C2i BC X MF, Axs. 2, 6. (for BM+MC= BC). Q. E. D. N 12d 1 0 210 1B00K HI,. PLANNE GEOMETRY 353. Formula for median of a triangle in terms of its sides. In the Fig., p. '209, denoting, AB by c, AC by b, BC by a, and AM by in, by Art. 35.2, b2 + (2 - 29M + (),whence iit n (b c)a2 Plioiposll'1ioN XXXII. THTEoRE-s 354. Itf two chords in a circle intersect,7 the productt of the seginents of one chtord is equal to the producet of/ the seg-,Ments of the other chord. D Given the 0D ADBC with the chords AB and CD interseeting at the point F. To prove A F X F B Cl,' X 121. Proof. Draw AD and CB. Then, in the & AED~] and CFB, Z Ai =Z C, Art. 2158, (each be~igng cant red by -i arc i)B). Also ZD= ZB. (Why) Hence the 1 AED and C YB are similar. (Why).'..AF:CF=F~D:FB. (Why?7) And AF XFB=CF X PD. (Why? Q. E. D. 355. CoR. 1. If through a fixed point within a circle a chord be drawn, the product of the segments of the chord is constant, in whatever direction the chord be drawn. PROPORTIONAL LINES 211 356. DEF. Four directly proportional quantities are four quantities in proportion in such a way that both the antecedents belong to,one figure and both the consequents to another figure. Four reciprocally proportional quantities are four quantities in-proportion in such a way that the means belong to one figure and the extremes to another figure. 357. COR. 2. The segments of two chords intersecting in a circle are reciprocally proportional (tile segments being considered as parts of tle chords, not of the A). PROPOSITION XXXIII. THEOREM 358. If, fron a given point, a secant and a tangent be drawn to a circle, the tangent is the mean proportional between the whole secant and its external segment. A Given AB, a tangent, and AC, a secant, to the circle BCF, and AF the external segment of the secant. To prove AC: AB=AB: AF. Proof. Draw the chords BC and BF. Then, in the A ABC and ABF, ZA = ZA. Ident. Z C = Z ABF. Arts. 258, 264..the A ABC and ABF are similar. (Why?).A. 1C: ABz=AB: AF. (Why?) Q. E. 911) 1300K III. PLANE GEOMETRY 3'59. COP,. 1. If, froin a given 'point, a tangent and a secant be drawn to a circle, the product of the whole secant cMd its external segment is eqlaul to the squtare of the tangent. 360. COR. -. if, from, a given p1oinlt witihout (a circle, a secant be drawn, the proitact of the secant and its external segment is constant, in whatever direction the secant be draewn. For the product of each secant and its external segment equals the square of the tangent, which is constant, 361, COR. 3. If two secants be drawnv from an external point to a circle, the whole seClants and their external segments are reciprocally proportional. PROPOSITION XXXIV. THEOREM 362. The square of the bisector of an angle of a triangle is equal to the prodact of the sides forniiug the angle, diminished by the produact of the segments of the third side formed by the bisector. / /_ A~~I Given the A ABC, CF (or t) the bisector of ZA CB, and m aud n the segments of AD formed by UP. To prove t2-ab - mn. Circumscribe a 0 about the A ABC. NUMERICAL PROPERTIES21 213 P'roof. Pr'oduce CF to -meet the circumf erence at if,. and draw the chord B13ff Then, in the & A CE and GlIB, Z A = Z11. (Why?7) And ZACF= ZilCB. (Why?7) Hence the & A CE dnd ('fIB are similar. (Why?7) b -tx-It:a. (Why) t (x-f-t)=zab. (Why?7) Or tX~1 2 ab. Subtracting tx, t2 ab -tx. Ax. 3. But tX vmn. Art. 35-4. Snbstitutingornin for tx, t2=zab-rnn. Ax. 8. Q. E. D. 363. Formula for bisector of an angle of a triangle. m n n= b: a (Art. 3'32).:. mn n it:m b + a b (Art. 309), or c m =a +b:b (Ax. 8).'. be- (Art. 302, Ax. 5). In like mnanner i= - Substitutingr for fin and n, t2= al ci - 2 ab(au + b) + c)(a + b) - c) 12~b-(Y + b) (a +J I) 2 Then t 1as( ) Ex. 1. On the figure, p. 1210, let AF~=10, F-B=4, and FC=S, Find E-D. Ex. 2. On the figure, p. 211, let AC=16 and AB= 12. Find AF. Ex. 3. On the figure, p. 212, let AC=16, CB==12, and AB=14. Find CE. 214 214 ~BOOK III. PLANE GEOME TRY PROPOSITION XXXV. THEOREM 364. In any triawyle, the product of any two sides i equal to the prod uct of the diameter of the circumwscribed cjircle by the altitude upon the third side. B 4 OFC D Given the A ABC, ABOCD a circumscribed 0, BD the diameter of this 0, and BE the altitude upon AC. To prove AB X BC=BD X BE. Proof. Draw the chord DC. ThenI in the A AB'E) and DB C, Z A / D. Zi)CB isart. Z. A' AB and DRC are similar. AB21) BJI)=BE BC. ulAB X BC= BD)X BE. (Why?) (W hy?9) (Wh y?') (Why?) (Why?7) Q. E. D. 365. Cop. The diameter of a circle circumscribed about a triangle equals the product of two sides of the triangle divided by the altitude upon the third side. Ex. In the above figure, if AB = 8, BC = 6, and BF = 4-6, find the radius of the circumscribed circle. CONSTRUCTION PROBLEMS 215 CONSTRUCTION PROBLEMS PROPOSITION XXXVI. PROBLEM 366. To construct a fourth proportional to three given lines. - -R,-< D A ta B n 0 P Given the lines m, n, p. To construct a fourth proportional to m, n, and p. Construction. Take any two lines, A P and AQ, making any convenient Z A. On AP take AB = m, and BC= n. On AQ take AD=p. Draw BD.* Through the point C draw CRG 1 DB, and meeting AQ at IR. Art. 279. Then DR is the fourth proportional required. Proof. AB: BC=AD: )R. Art. 317. Or q,: n=p: DR. Ax. 8. Q. E. F. PROPOSITION XXXVII. PROBLEM 367. To construct a third proportional to twio given lines. Let the pupil supply the construction and proof. [SUG. Use the method of Art. 366, making p=n.] Ex. 1. Construct a fourth proportional to three lines, t, 1 and 1 in. long. Ex. 2. Construct a third proportional to two lines, 2 and 1 in. long. * From now on, references to the postulates will be omitted where these are not necessary to complete solmeu: other reference. 216 21OK300K III, PLANE GEOMETRY PRIoPOSITION XXXVIII. PROBLEM1 368. To construct a mnean proportional between two given, lines. n Given the lines m and n. To construct a mean -proportional between m and n. Construction. On the line AP take AB=r m, and BC==n. On AC as a diameter construct a semi-circumference. Art. 275, Post. 3. At B erect a I to AC meeting the semi-circumference at R. Art. 274. Then BR is the mean proportional required. Proof. AB: BR=BR: BC, Art. 343. (the I. to the diameter from any point ini the circumlference of a circle is a mean proportioniai betwveen thte segments of the dliameter). Substituting for AB and BC their values mn and in, m: BR==BR:n. Ax. 8. Q. E. F. Ex. 1. Construct the third proportional to two lines, 1 and 1i in. long. Ex. 2. Construct a mean proportional between two lines, 1 and 2 in. long. Ex. 3. Taking any line as 1, construct V&/2. CONSTRUCTION PROBLEMS 217 PROPOSITION XXXIX. PROBLEM 369. To divide a given straight line into parts propor. tional to a number of given lines. ---- ^\ \ \ An \ m --- -- * ~-= D\ \\ n Given ad P Given the straight lines AB,, Z n and p. To divide AB into parts proportional to m, n and p. Construction. Draw the line AP, making any convenient angle with AB. On APmark off AC=w, CD=n, and DF=p. Draw BP. Through the points C and D draw lines 11 BF, and meeting AB at R and S. Art. 279. Then AR, RS and SB are the segments required. AR RS SB Proof. - FArt. 319. A C CD DF (if two lines are cut by a nmnber of parallels, the corresponding segments are proportional). For A C, CD and DF substitute their equals mi, n and p. AR, RS SB Then = - Ax. 8. m n p Q. E. F. 370. DEF. A straight line divided in extreme and mean ratio is a straight line divided into two segments such that one of the segments is a mean proportional between the whole line and the other segment. Ex. Divide a given line into parts proportional to 2, 3 and 4. 218 2BOOK III. PLANE GEOMETRY PROPOSITION XL. PROBLEM 371. To diivide a given straight t ine in extremie and mnean ratio. A BP) Given the line A-B. To divide AB in extreme and mean ratio. Construction. At one end of the given line, as B, coil struet a I OB equal to - 4B. Arts. 274, 275, From 0 as a center and with OB as a radins, describe a circumference. Draw AO meeting the circumference at C, and produce it to meet the circunmference again at F. On AB mark off AP eqnal to AC; on BA produced take AQ=AF. phen A1BAR is divided in extreme and mean ratio internally at P, and externally at Q. Proof. 1. AF: AB=AB3: AC. Art. 358.. AF-AB:AB==AB-AC: AC. (Why) But OFz=20B=AB, and AC=AP. Hence AF-A B=AP, and AB-AC= PB. AP: AB= 6PB- AP, or AR: AP = AP: PB. Ax. 8. Art, 308. 2. AF:AB=AB:AC. (Why 7) AF~+AB: AF= AB+ AC: AB. (Why 7) But AF + AB = QB, and AB + A C= AF= QA.,', QB: QA= AR: AB. Ax. 8, Q, E. F0 CONSTRUCTION PROBLEMS 219 PROPOSITION XLI. PROBLEM 372. Upon a given straight line, to construct a polygon similar to a given polygon, and similarly placed. C7. \\ Given the polygon ABCDE and the line A'B' To construct on A'B' a polygon similar to ABCDE and similarly placed. Construction. In the given polygon, draw the diagonals AC and AD, dividing the polygon into triangles. At B',on the line A'B', construct Z A'B'C' equal to / B; and at A' construct Z B'AC'' equal to Z BAG. Art. 278. Produce the lines B'C' and A'C' to meet at C'. In like manner, on A'C' construct A A'C'I)', equiangular with A ACD and similarly placed; ani on A'D' construct the A A'D)E', equiangular with A ADE and similarly placed. Then A'B'CID'E' is the polygon required. Proof. The A A'B'C', A'C'D', etc., are similar to the A ABC, ACD, etc., respectively. Art. 324. Hence the polygons A'B'C')'E' and ABCDE are similar. Art. 330. Q. E. F. 220 BOOK III. PLANE GEOMETRY EXERCISES. GROUP 30 SIILAR TRIANGLES Let the pupil make a list of all the conditions that make two triangles similar (see Arts. 323, 324, 325, etc.). Ex. 1. Given AD I BC, and BFL AC; prove A ADC and BFC similar. A Ex. 2. In the same figure, prove the A AOF and BFC similar. What other triangle on this / figure is similar to A BFC? F Ex. 3. Two isosceles triangles are similar if C D B their vertex angles are equal. Ex. 4. Two isosceles triangles are similar if a base angle of one equals a base angle of the other. Ex. 5. Given arc AC=arc BC; prove A APC and AFC similar. Ex. 6. Prove that the diagonals and bases of a trapezoid together form a pair of similar triangles. Ex. 7. AB is the diameter of a circle, BD is a tangent, and AD intersects the circumference at E. Prove the triangles ABE and ADB similar. Ex. 8. BC is a chord in a circle, AQ is the diameter perpendicular to BC and meeting it at X; A4P is any chord intersecting BC in M. Prove the A AMN and APQ similar. Ex. 9. The triangle ABC is inscribed in a circle; the bisector of the angle A meets BC in I) and the circumference in P. Prove the triangles BAD and APC similar. Ex. 10. A pair of homologous medians divide two similar triangles into triangles which are similar each to each. Ex. 11. Two rectangles are similar if two adjacent sides of one are proportional to the homologous sides of the other. Ex. 12. Two circles intersect in the points A and B. AC and AD are each a tangent in one circle and a chord in the other. Prove the A ABC and ABD similar. [SuG. Prove LBAD= LACB, ete.] EXERCISES. PROPORTIONAL LINES 22'L CC I 373. Proof that lines are proportional. In order to prove that certain lines are proportional, or have proportional relations, it is usually best to showz that the given lines are homologous sides of similar triangles. Sometimes, however, other methods of proof are used (as the theorems of Arts. 354 and 358); but these, if investigated, are usually found to be the method of similar triangles in disguise. EXERCISES. GROUP 30 PROPORTIONAL LINES Ex. 1. On the figure of Ex. 1, p. 220, prove AD>X BC=BFX AC, and BCX OD=BOX FC. Ex. 2. On the figure of Ex. 5, p. 220, prove CP: CA=CA: CF. (Hence as P moves the product of what two lines is constant? ) Ex. 3. The diagonals of a trapezoid divide each other into proportional segments. Ex. 4. In the isosceles triangle ABC, AB-=AC, on the side AB, the point P is taken so that PC equals the base. Prove AB X PB= 137-2. Ex. 5. In a triangle the median to the base bisects all lines parallel to the base and terminated by the sides. Ex. 6. If PQ is any line through F, the midpoint of the line AB, and AP and BQ are perpendicular to PQ, show that the ratio PF: FQ is constant. Ex. 7. The triangle ABC is inscribed in a circle. F is the midpoint of the arc AC, and BF intersects the line AC in E. Prove AB: BC = AE: EC. [SUG. Use Art. 332.] Ex. 8. If two circles intersect, the common chord, if produced, bisects the common tangent. [Sue. Use Art. 358.] Ex. 9. If two circles intersect, tangents drawn to the two circles from any point in the common chord produced are equal. 222 BOOK III. PLANE GEOMETRY Ex.. 10. Given AD, T and BC I; A D prove PQ - RT. PQ PT r' T [SUG. Show that I(- X' by show- T ing them equal to a common ratio.] B C Ex. f1. Lines are drawn from a point 0 within the triangle, to the vertices of the triangle IBC. From B' aly point in OB, B'A' is drawn parallel to BA andl meeting OA in A', and B'C' is drawn parallel to BC and meeting 0C in C'. Prove A'B':.AB=B'C: BC, and the triangles ABC and Al'B'C' similar. Ex. 12. Given ABCD a, and P any point in BC produced; prove AR2= RQX X PP. [SUG. Compare the similar zA p ABR and RQD; also the similar A ARD and RBP.] EXERCISE8. GROUP 32 NUMERICAL PROPERTIES OF LINES Ex. 1. If AD is the altitude of the triangle ABC, AB2-C2=_ -BD DC2. Ex. 2. If the diagonals of a quadrilateral are perpendicular to each other, the sum of the squares of one pair of opposite sides equals the sum of the squares of the other pair of sides. Ex. 3. The square of the altitude of an equilateral triangle is three-fourths the square of one side. Ex. 4. If AB is the hypotenuse of a right triangle, and the leg BC is bisected at K, A1-2 - A~-2 = 3 C2. Ex. 5. PQ is a line parallel to the hypotenuse AB of a right triangle ABC, and meeting AC in P and BC in Q. ProveQ 2 + BP2= jB-2+PQ2. Ex. 6. In the right triangle ABC, BE and CFbisect the legs AG and AB in the points E and F. Prove 4BE2 + 4CF2-=5C2. EXERCISES. AUXILIARY LINES 223 EXERCISES. CROUP 33 AUXILIARY LINES Ex. 1. Given ABCD a rectangle; prove p2 + p-c2= i 2 i2 2- PD.' Ex. 2. The common tangent of two circles divides the line of centers into segments which have the same ratio as the diameters A D of the circles. [SUG. Draw radii to the points of contact.] Ex. 3. AB and AC are the legs of an isosceles triangle and BF is an altitude. Prove 2AC X PC=BC2. C Ex. 4. Given the chords AB and CD perpendicularto each other and intersecting at 0; prove A, A2+ O B2 + OC-2+ ~OD2=(diameter)2. [SUG. Draw the diameter BE and the chords \AC, BD, DE. Prove AC=ED, etc.] E Ex. 5. ABC is an inscribed isosceles triangle of which AB and AC are the legs. AD is a chord meeting BC in E. Prove AB2 AD X AE. Ex. 6. If C is the vertex of an isosceles triangle ABC, and D is a point in the base produced, then CD2= CB2 + AD X BD. Ex. 7. Two circles touch at the point T. PTPi' and QTQ' are lines drawn meeting the circumferences in P, Q and P', Q/ respectively. Prove the triangles PTQ and P'TQ' similar. [SUG. Draw the common tangent at 2.] Ex. 8. Two circles touch at the point T, through T three lines are drawn meeting the circumferences in F, Q, R and Pf Q/, i', respectively. Prove the triangles PQR and P'Q'R' similar. Ex. 9. If A is the midpoint of CD, an arc of a circle, and 4P is any chord intersecting the chord CD in Q, prove that JPX AQ is a constant. 224 BOOK III. PLANE GEOMETRY Ex. 10. In an inscribed quadrilateral, the product of the diagonals is equal to the sum of B the products of the opposite sides. [SUG. Draw BF so that Z CBF= ZABD and use similar triangles.] A- D Ex. 11. The sum of the squares of the sides of any quadrilateral is equal to the sum / Ctr of the squares of the diagonals, plus four times the square of the line joining the midpoints of the diagonals. - D EXERCISES. GROUP 34 INDIRECT DEMONSTRATIONS Ex. I. If the sum of the squares on two sides of a triangle is greater than the square on the third side, the angle included by the two given sides is an acute angle. Ex. 2. If D is a point in the side AC of the triangle ABC, and AD: DC=AB: BC, then DB bisects angle ABC. Ex. 3. A given straight line can be divided in a given ratio at but one point. Ex. 4. If the sides of two triangles are parallel, each to each, and a straight line be passed through each pair of homologous vertices, these lines, if produced, will meet in a common point. Ex. 5. If each of three circles intersects the other two, the three common chords intersect in one point. EXERCISES. GROUP 35 THEOREMS PROVED BY VARIOUS METHODS Ex. 1. In the figure on p. 204, show that ABX BF=BCX AF. Ex. 2. In the same figure, if FC=3AF, show that ABL2: jC2-1 3. Ex. 3. AB is the diameter of a circle and PB is a tangent. If AP meets the circumference in the point Q, prove that AP X A Q-AB,'2 MISCELLANEOJUS EXERCISES. THEOREMS 225' Ex. 4. If the line bisecting the parallel sides of a trapezoid be produced, it meets the legs produced in a common point. [SUG. See Art. 340.] Ex. 5. In similar triangles, homologous medians have the same ratio as homologous sides. Ex. 6. A diameter AB is produced to the point C; CP is perpendicular to AC; PB produced meets the circumference at Q. Prove the triangles A QB and JPCY1 similar. Ex. 7. If PA and PB are chords in a circle, and CI is a line parallel to the tangent at P and meeting PA and PB at C and D, the triangles PAB and PCI) are D similar. Ex. 8. Given AB a diameter and AD and BC C tangents, AC and DB intersecting at any point F on the circumference; prove AB a mean proportional between the sides AD and BC. Ex. 9. In any isosceles triangle, the square A B of one of the legs equals the square on a line drawn from the vertex to any point of the base plus the product of the segments of the base. Ex. 10. A line drawn through the intersection of the diagonals of a trapezoid parallel to the bases and terminated by the legs is bisected by the diagonals. [SUG. See Ex. 10, p. 222.] Ex. 11. If a chord is bisected by another chord, each segment of the first chord is a mean proportional between the segments of the second chord. Ex. 12. In a parallelogram the sum of the squares of the sides equals the sum of the squares of the diagonals. Ex. 13. If two circles are tangent externally, and a line is drawn through the point of contact terminated by the circumferences, the chords intercepted in the two circles are to each other as the radii. Ex. 14. Three times the sum of the squares of the sides of a triangle equals four times the sum of the squares of the medians. 0 d-d -1 G 1B00K III, IJANNE GE0ME1fTli1 Ex. 15. Find the locus of the midpoinits of lines in a triangle parallel to the base and terminated by the sides. Ex. 16. Given AB the (lialneter, AP, PYQl, BRi, tangents; prove E-Q X Q91? a constant (:z radius squared). Ex. 1 7. 0 is the center of a circle and A is any point within the circle; G.A is produced to B, so that GA X GB equals the radius squared. If P is any point in the circumference, the angles GPA and GBP are equal. [SuG. Use Art. 3217.] Ex. 18. Given AF==FB, and. CIIfl Al? prove -HP P P= 11K: EKA. [ SUe. HP:F P = CHI: B, etc.] FI, AB K Ex. 19. If from any point P within the triangle ABC the perpendiculars P9, PR, PT are drawn to the sides AB, AC, BC, respectively9 then ]=ARi~7~ C2J~ -+BQ2 +17i. EXERCISES. GROUP 36 PROBLEMS Given three lines a,, b, c, Lx. 1. Construct x=(1-; alIs o xv=ab Lx. 2. ConIstruct X= i (2 -b2, i. e., -1/(a- b) (a. —b). Lx. 3. Construct x 1/3 ab, i. e., 1(3 a) 1). ELx. 4. Construct x = i'a2 - bc, i. e., V"2 _(Vbc)'2 Ex. -5. Given a line denoted by 1, construct V/3; also i5 LEx. 6. Divide a line into three parts proportional to 2, ~ Ex. 7. Divide a line harmaonically in the ratio 3:5. EXERCISES. PROBLEMS 227 Ex. 8. Divide one side of a triangle into segments proportional to the other two sides. Ex. 9. Divide a line into segments in the ratio 1: /2. Ex. 10. Given a point P in the side AB of a triangle ABC, draw a line from P to AC produced so that the line drawn may be bisected by BC. [Sue. Suppose the required line, PQR, drawn meeting BC in Q and AC in R. From P draw PL || AC. Compare the APLQ and QRC.] Ex. 11. Through a given point P in the arc subtended by the chord AB draw a chord which shall be bisected by AB. o [SUG. Suppose the required chord drawn, viz., / PQR. Jointhe center Owith P and (. What kind A /TQB of an angle is OQP, etc.?] Ex. 12. In an obtuse triangle draw a line from the vertex of the obtuse angle to the opposite side which shall be a mean proportional between the segments of the opposite side. [SUG. Circumscribe a circle about the triangle and reduce the problem to the preceding Ex.] Ex. 13. Find a point P in the arc subtended by the chord AB such that chord PA: chord PB-2: 3. [SuG. Suppose the required construction made, and also the chord AB divided in the ratio 2:3 at the point Q. How do the angles APQ and QPB compare?] Ex. 14. Given the perimeter, construct a triangle similar to a given triangle. Ex. 15. Given the altitude of a triangle, construct a triangle similar to a given triangle. Ex. 16. In a given circle inscribe a triangle similar to a given triangle. Ex. 17. About a given circle circumscribe a triangle similar to a given triangle, Ex. 18. By drawing a line parallel to one of the sides of a given rectangle, divide the rectangle into two similar rectangles. 228 BOOK III. PLANE GEOMETRY Ex. 19. Inscribe a square in a given A --- —triangle. / [SuG. If ABC is the given triangle, sup- /~- ' pose D)GFE the required inscribed square..Join BE and produce it to meet All 1 BC. Prove AH=A1K, etc.] B CG F C 374. The method of similars in solving geometrical problems is best shown by the aid of an example. Ex. In the side BC of a triangle _41AC( find a point D such that the perpendiculars from it to the other sides shall be in the ratio 3: 1. 1 B CONSTRUCTION. At any point I in AC Q erect a 1 PQ of any convenient length. <D In a direction I AB draw RQ = Q'P. Join RP. From S draw S'T 1 RQ. Produce / AT to D1. Then D is tle required point. A- C Let the pupil supply the proof. EXERCISESo QROUP 37 PROBLEMS SOLVED BY MIETHOD OF SIMILARS Ex. 1. In one side of a triangle find a point such that the perpendiculars from it to the other two sides shall be in the ratio m: n. Ex. 2. Find a point the perpendiculars from which to the three sides of a given triangle shall be in a given ratio. / [SUG. Use Ex. 1 twice.] Ex. 3. Construct a circle / which shall touch two given / lines and pass through a ' R given point. [SUG. Let OA and OB be O- B the given lines and P the " given point. Draw any 01R touching the two lines (OB at Y) and intersecting OP produced at X. Draw the chord XY, etc.] Ex. 4 Inscribe a square in a given semi-circle. [SUG. Circumscribe a semi-circle about any given square, by taking the midpoint of the base of the square as a center, and the line from this midpoint to a non-adjacent vertex as a radius, etc.] Ex. 5. Solve Ex, l1', p. 228, by the method of sirilars. EXERCISES. PROBLEMS 229 375. Algebraic analysis of problemso The conditions of a problem may often be stated as an algebraic equation; by solving the equation, the length of a desired line in terms of knouwn lines may then be obtained, and the problem solved by constructing the algebraic expression thus obtained. Ex. Find a point P in the line AB such that AP 2=- -- -- -3p2. A P B P ANALYSIS AND CONSTRUCTION. Denote AB by a, AP by x, and PB by a- x. Then x=3(a-x)2. 3a o 2a 1/3. 2.. 2-()ax= -3(t, and x = -- 3 2 Construct a 1/3, whence construct 3a- 1; lay off the line obtained, as AP, on AB; this gives the point P of internal division. Similarly, the construction of 3a- - gives P', the point of external division. EXERCISES. CROUP 38 PROBILEMS SOLVED BY ALGEBRAIC ANALYSIS Ex. 1. Find a point P in a given line A BI such that JA 2= 2BP2. Ex. 2. Construct a right triangle, given one leg a and the projection, b, of the other leg on the / hypotenuse. [SUG. Denote the projection of a on the hypotenuse by x. Then aZ= x (x -+ b), etc.] b x Ex. 3. Inscribe a square in a given semicircle. Ex. 4. From a given line cut off a part which shall be a mean proportional between the remainder of the line and another given line. Ex. 5. Given AC and CB arcs of 90~, and a a C given line. Draw the chord CQ intersecting AB in ' so that PQ=a. [SUm. x' --- _ r2. (;x=(r +y) (r-y), etc.] A B Ex. 6. Given the greater segment of a line di- Vie' 'i extreme and mealn ratio, construct the line, 230 BOOK III. PLANE GEOMETRY EXERCISES. GROUP 39 PROBLE3MS SOLVED BY VARIOUS METHODS Ex. 1. Construct two lines, given their sum (a line AB) and their ratio (m: n). Ex. 2. Construct two lines, given their difference and their ratio. Ex. 3. Divide a trapezoid into two similar trapezoids by drawing a line parallel to the bases. [SUG. Conceive the figure drawn, and compare the ratio of the bases in the two trapezoids formed.] Ex. 4. Construct a mean proportional between two given lines by use of Art. 358. Ex. 5. Construct a circle which shall pass through two given points and touch a given line. Ex. 6. From a given point draw a secant to a circle so that the external segment shall equal half the secant. [SUG. Draw a tangent to the 0 and use the algebraic method.] Ex. 7. From a given external point P, draw a secant meeting a circle in A and B so that PA: AB=m: n. [SUG. Draw a tangent to the circle from the point P and denote its length by t. Denote PA by mx and AB by nx. Then m (m + n) x2 mtl =t2, or t: nmx=vmx, etc.] 'Lt " 4- nH, Ex. 8. Through a given point P draw a straight line so that the parts of it, included between that point and perpendiculars drawn to the line from two other given points, shall be in a given ratio. [SUG. Join the last two points, and divide the line between them in the given ratio.] Ex. 9. Construct a straight line so that the perpendiculars on it from three given points shall be in a given ratio. [SUG. Let P, Q, R, be the given points and im: i: p the given ratio. Divide PQ in the ratio m: n and QR in the ratio n: p, etc.] Ex. 10. Upon a given line as hypotenuse construct a right triangle one leg of which shall be a mean proportional betwe - the other leg and the hypotenuse. BOOe IV AREAS OF POLYGONS 376. A unit of surface is a square whose side is a unit of length, as a square inch, a square yard, or a square centimeter. 377. The area of a surface is the number of units of surface which the given surface contains. It is important for the student to grasp firmly the fact that area means not mere vague largeness of surface, but that it is a number. Being a number, it can be resolved into factors, it may be determined as a product of simpler numbers, and handled with ease and precision in various ways. 378. Equivalent plane figures are plane figures having equal areas. Thus two triangles may have equal areas (be equivalent) and yet not be of the same shape, that is, not be equal (congruent). 379. Abbreviations. Instead of "area of a rectangle," for example, it is often convenient to say simply "rectangle." So instead of "the number of linear units in the base," we use simply "the base." In like manner, for "product of the number of linear units in the base by the number of linear units in the altitude," a common abbreviation is "product of the base by the altitude." (231) 232-) c BOOK IV. PLANE GEOMETRY COMPARISON OF RECTANGLES PROPOSITION I. THEOREM 380. If two rectangles Ih(ve the same altitude, they ire to each other as their bases. B C F G It~~~ ti~~ ~ t~I AK D E II Given the rectangles EFGHI and ABCD, having their altitudes EF and AB equal. To prove EFGH: ABCD= EH: AD. CASE I. When the bases are commensurable. Proof. Take some common measure of EHl and AD, as AK, and let it be contained in ElH n times and in AD m times. Hence EH: AD=n: m. (Why?) Through the points of division of the bases of the two rectangles draw lines perpendicular to the bases. These lines will divide EG into n, and AC into m small rectangles, all equal. Art. 163. Hence EFGH: ABCD=n: m. (Why?).'. EFGH: ABCD = EH: ADo (Why 7) COMPARISON OF RECTANGLES 233 CASE II. Wlhen the bases are incomnzensurable. B C F PG.A D E QI Proof. Divide the base AD into any number of equal parts, and apply one of these parts to Elf. It will be contained in Eli a certain number of times with a remainder QIL, less than the unit of measure. Draw QP _ ElH, meeting FG at P. Then EQ and AD are commensurable. Constr. EFPQ EQ EFPQ= Q. Case I. "ABCD AD If now the unit of measure be indefinitely diminished, the line QH, which is less than the unit of measure, will be indefinitely diminished. Q- ElI.as a limit; Eli'PQ ' EFGHas a limit. Art. 251. Hence FQ becomes a variable with EFGT as its ABCD ABCD limit; also 1-Q becomes a variable with AD as its limit. AD A D Art. 253, 3, E`PQ _ EQ But the variable -BCD the variable always. ~~~ABO~D AD ~ Case I. the lint the limit D (why?) Q. E. D. 381. Con. If twlo rectacvlles have equal bases, they are to each other (ia their (altitiudles, 234 B3OOK IV. PLANE GEOMtETRY5 PR OPOSTr ION JI1 T 1 EO M 382. Thee areas of any two recttngles are to cach other as the prodscts of their bases by their altitudes, a --------- ------- ^^ a CR S b b' b — --- Given the rectangles R and R', having the bases b and b', and the altitudes a and a', respectively. To prove R b X a' -7= b-X ca Proof. Construct a rectangle, S, having its base equal to that of R, and its altitude equal to that of f'. Then — =-. Art. 381. S b Also Art. 380. Taking the product of the corresponding members of the two equalities, ' b X a. Ax. 4. R/ b X a' Q. E. D. Ex. 1. Find the ratio of the area of a rectangle whose dimensions are 12 X 8 in. to that of one whose dimensions are 9 X 2 in. Ex. 2. How many bricks, each 8 X 5 in., will it take to cover a pavement 60 X 9 ft.? AREAS OF POLYGONS AREAS OF POLYGONS PROPOSITION III. THEOREM 383. The area of a rectangle is equal to the product of its base by its altitude. h R _ 1 Given the rectangle R, with a base containing b, and an altitude containing h7 units of linear measure. To prove area of R= b X h. Proof. Let U be a square each side of which contains 1 unit of linear measure. Then U is the unit of surface. Art. 376. R bXh __ - -=bX h. Art. 382. "U 1X1 But -I is the area of R, Art. 377. (by! definition of area).. rea of = b Xh. Q. E.. 384L NOTE. By use of this theorem, the problem of finding the area of a rectangle is reduced to the simpler problem of measuring the two linear dimensions of the rectangle and taking their product. (See Art. 1.) Ex. 1. Find, in square feet, the area of a rectangle 8 yds. long and 5 ft. wide. Ex. 2. The area of a rectangle is 60 sq. ft. and its altitude is 5 ft. Find the base, 2 36 BOOK IV. PLANE GEOMETIY PROPOSITION IV. THEOREM 385. The area of a parallelogram is equal to the product of its base by its altitude. K B F C 1/ h A b D Given the Z7 ABCD with the base AD (denoted by b) and the altitude DF (denoted by I). To prove area of ABCD-=b X h. Proof. From A draw AK 11 DF, and meeting CB produced, at K. Then AK IL CK. Art. 123,.'. AKFD is a rectangle with base b and altitude h. (Why?) In the rt. A AKB and DFC, AB = DC. (Why?) AK= DF. (Why?).A. A AKB== DFC. (Why?) To each of these equals add the figure ABFD; Then rectangle AKtFD = -1 A 7 CD. A2. But area of the rectangle AKFD=b X h. Art. 383..'. area 1 ABCD = b X h. (Whv?) Q. E. D. 386. COR. 1. Parallelograms which have equal bases and equal altitudes are equivalent. 387. COR. 2. Parallelograms which. have equal bases are to each other as their altitudes; Parallelograms which have equal altitudes are to each other as their bases. 388. COR. 3. Any two parallelograms are to each other as the products of their bases and altitudes, AREAS OF POLYGONS 237 PROPOSITION V. THEOREM 389. The area of a triangle is equal to one-half the product of its base by its altitude. B. D h \ / A b C Given the A ABC with the base AC (denoted by b), and the altitude FB (denoted by h). To prove area of A ABC=. b X 7h. Proof. Draw BD |I AC, and CD I| AB, forming the D7 ABDC. Then BC is a diagonal of E7 ABDC... A BC= - 7 ABDC. Art. 156. But area 7 ABDC= b X h. (Why?).. area A ABC=a- b X h. Ax. 5. Q. E. D. 390. COR. 1. Triangles which have equal bases and equal altitudes (or which have equal bases and their vertices in a line parallel to the base) are equivalent. 391. COR. 2. Triangles which have eq ual bases are to each other as their altitudes; Triangles 'which have equal altitudes are to each other as their bases. 392. COR. 3. Any two triangles are to each other as the products of their bases and altitudes. Ex. 1. Find the area of a parallelogram whose base is 9 ft. 8 in. -nd whose altitude is 2 ft. 3 in. Ex. 2. Find the altitude of a triangle whose area is 180 sq. in. and whose base is 1 ft. 3 in, B130K IV. PLANE GEOMETRY PROPOSITION VI. THEORnEM 393. If a, b, c denote the sides of a triatngle opposite the agles A, B, C, respectively, and s=- (a + b - c), the area of the triangle =/s (s-a) (- b)(s - c)o B A Given the A ABC with the sides opposite X A, B and C, denoted by a, b and c, respectively, ( (a + b +-c) denoted by s, and A an acute angle. Toprove area AABC=l/s (s-a-l) (s- b) (s —c). Proof. Draw the altitude BD and denote BD by h. Then a= b2 + c2- 2 b X AD. Art. 349..2 b X AD=b2^ + c-a2-. Axs. 2, 3.?2+ c2 — W.' {AD= --- --- Ax. 4. b2b But h2=c2-AD2)=(c+ -l)) (c-AD) Art. 347. ( +b2- C-2- b2 + - a2) Ax. S. 2b 2b 2 c+b2+c2-b a 2bc-b2-2+a "Y 2b J b 2b J [(b + c)2 a2] [a2 -(b- c)2] 4 b2 (b c + a) (b + c-a) (a+ b-c) (a-b+c) 4 b2 Now a-+ b c=2 s.' a+ — b -c=2s-2c, etc. Hyp., Axs, 4, 3. AREAS OF POLYGON'S 2s (2s- 2a)(0'2ds-2e) (2s-2b) 16s(s-a) (s-b) (s-c) h2= ~4 b2 4 b2, 'Is(s -a) (s -b) (s-c) b But area A ABC-= 4 b X hi. Art. 390. area LABC= Vs (s -a) (s -b) (s-c). Ax. 8. Q. B. D. PROPOSITION VII. THEOREIM 394. The area of a trapezoid is eqatal to oite-half the SUM of its bases, multiplied by its altitude. B I)c A bD Given the trapezoid ABC]) with the bases AD and BC (denoted by b'and b'), and the altitude PB (denoted by h). To prove area of ABCD = A. (b + b') X h. Proof. Draw the diagonal B.D. Then area of A ABD~ = b X h. (Wily) Anid area of A BUID-, b' X h;. (Why) Adding, area of ABCD- (b' ~ b) h,, (W hy) Q. E. D. 395. COR. The area of a trapezoid equtals the produtct of the median of the trape zoid by the attitude. For the mediani of a trapezoid equals ouc-haif the sum of the, bases (Art. 179). 240 13OOK IV. PLANE GEOMETRY 396. SCHOLIUIM. The area of a polygon of four or more sides can usually be found 'in one of several ways; as, by dividing the polygon into triangles sand taking the sum of the areas of te tr iangles; or, by drawing the longest diagonal of the )polygon and drawing perpendlicutlars to this diago1al from. the vertices 7whicih it does not meet, and obtaining -- -- - the sum of the areas of the triangles (and trapezoids t1hus formed. Ex. 1. Find the area of a parallelogram whose base is 1 yd., and whose altitude is 1 ft. Ex. 2. Find the area of a triangle whose sides are 5, 6, and 7 in. Ex. 3. Find the area of a trapezoid whose bases are 18 and 10 in., and whose altitude is 6 in. Ex. 4. If the area of a trapezoid is 135, and its bases are 12 and 18, find its altitude. Ex. 5. The measurement of the area of a parallelogram reduces to the measurement of what two straight lines? Ex. 6. The measurement of the area of a triangle reduces to the measurement of what lines? Ex. 7. The measurement of the area of a trapezoid reduces to the measurement of what lines? Ex. 8. Prove the theorem of Art. 385 by drawing perpendiculars from B and C, instead of from A and D. COMPARISON OF POLYGONS 241 COMPARISON OF POLYGONS PROPOSITION VIII. THEOREM 397. If two triangles have an angle of one equal to an angle of the other, their areas are to each other as the products of the sides including the equal angles. A Given the A ABC and ADF having Z A incommonf To prove A ABC ABXAC AD ADF AD XAF Proof. Draw the line DC. Then the A ABC and ADC may be regarded as having their bases in the line AD, and as having the common vertex C. A ABC AB A ADC AD A ADC AC In like manner AADE-A Multiplying the corresponding members ABC AB X AC tes, A ADF AD X AF Art. 391. (Why?) of these equaliAx. 4. Q. E. D. Ex. In the above figure, if AB=12, AC=18, AD)=30, and AF=32, find the ratio of the areas of the A ABC and ADF. p 94;2 BOOK IV. PLANE GEOMET:IY PROPOSITION IX. THEOREM 398. The areas of two similar triancgles are to each other as the squares of any two homologous sides. C,A F B A' F', B' Given the similar A C ACand A'B'C' with AB and A'B' homologous sides. A ABC A_ I To prove A A Proof. Draw the homologous altitudes CF and CIFI. ABC AB XCF AB, CF Then = B ' X, Art. 392. LA I'B'C' A 'BX C'F' AL'l C'F' (anyl trw' A arc to each other as the ))prot(cts of their bases and altitudes). But Crt.= - Art. 338. C'FJ ' A'W Substituting A- for its equal —,' Ax. 8. A ABC AB AB _ "A A AIBC A A R'B' AB' Q. E. D. Ex. 1. If a pair of homologous sides of two similar triangles are 4 ft. and 5 ft., find the ratio of the areas of the triangles. Ex. 2. Prove Prop. IX by use of Prop. VIII. COMPARISON OF POLYGONS 243 PROPOSITION X. THEOREM 399. The areas of two similar polygons are to each other as the squares of any two homologous sides. B BA A ------- C E D D Given the similar polygons ABCDE and A'B'CtD'Et, with their areas denoted by S and S', respectively, and with AB and A'B' any pair of homologous sides. To prove 8: S'=AB: A'B'2. Proof. Draw the diagonals AC, AD and A'C', A'D' from the homologous vertices A and A'. These diagonals will divide the polygons into similar A. Art. 329. A ABC AB2 A A-B-C — A= Art. 398. A A'BIC AB;2 t ABC _ A( (2 A A C A _ AD)E LAAB'C' \Aw'2) AaC'D'y ~ \jCiy2]) A A'D'E' (Why?) A AB _ AC A CD A ADE * A A'BX' AA\XC' AAD'E (Why?) ' A A'BCI = Al CID' = A AID'EE ABC ACD + A AE A ABC Ar. 31 -- ~ Art. 312. LA A'B'C'-+ A A.C'ID' A A'D'E' A A 'B'C S A ABC. * -== - ------ * Ax. 6. "S A AB'C S ABd ' - A- Ax. 1. S'1 AB2 Q. E. D. Ex. If a pair of homologous sides of two similar polygons are 1 and 2 ft., find the ratio of th r of the r f t polygons, 04-4. 244 RBOOK IV. P~LANE GEOMETRKY PROPOSITION XI. THEORIEM 400. in' a right triangle, the sqiare On the hypotelluse is eqptivalent to the sum of the squtares on the twvo legs. F F LD Given A-D the square on AC the hypotenuse of the rt. A\ A4BC, and BE and BK tile squares on tile legs Bi.and -BC respectively. To prove AD =-czBF+ BK. Proof. Through B draw BL Ii AE, and meeting ED in L. Draw BE and FC. Then A ABC and ABG are rt. A. (Why?..GBC is a straight line. (Why? In the ~ BAE and.FAC, AB=AF, and A-E=AC. (Why?) Also Z BAE =Z PA C, Ax. 2. (for each=-ZJBAC +a rt. Z). A BA E -AFAC. But rectangle AL2 BAE. (for AL has the saine base, AE, and the same altitude, EL, as Also square BE~cz2 A FAC. rectangle AL-sqaare BE. In like manner rectangle CL~czsquare BK. Adding, AL +CL, or AD=-BF -BK. (Why?) A BAE). (Why?) Ax, 1. Ax. 2. ~. E. D. 401. COR. The square on either leg of a right triangle is equivalent to the square on the hypotenuse diminished bV the square on the other leg. .CONSTRUCTION PROBLEMS 245 CONSTRUCTION PROBLEMS PROPOSITION XII. PROBLEM 402. To construct a square equivalent to the sume of two given squares. ~4S R *__ ^, --—....... --- —---. Given two squares S and1 S. To construct a square equivalent to S +- 5'. Construction. Construct a right angle BAC. Art. 274. On one side of this angle take AB equal to a side of 8, and on the other side take AC equal to a side of S9. Draw BC. On a line equal to BC construct the square R. Then B is the square required. Proof. RB=C-2 AB2 -+ Aa Art. 400.. R B S - 8' Ax. 8, Q. E. F. 403. CoR. To construct a square equivalent to the sum of three or more given squares. At C in the above figure erect a line CD I BC (Art. 274), and equal to a side of the third given square. Draw DB. 1)B will be a side of a square equivalent to the sum of three given squares, etc. Ex. 1. Construct a square equivalent to the sum of two squares whose sides are i in. and 1 in., respectively. Ex. 2. By use of Art. 403, taking a given line as unity, construct V/2; also i/3. For example, construct a line -/2 inches long; also one -3 inches long. 246 BOOK IV. PLANE GEOMIETRY PROPOSIiION XIII. P'ROBLEM 404. To construct a squatre eqpuivatlcnt to the difference of two given squtares. Given the squares S and S'. To construct a square equivalent to the difference of 8 and S'. Construction. Construct a right angle BA AE. Art. 274. On one side B take AB equal to a side of the smaller given square S'. From B as a center with a radius BC, equal to a side of the larger square, describe an arc intersecting AK in C. On a line equal to AC construct the square R. Then R is the square required. Proof E C -BC-, At. 41. Proof. B AC = A C B C-AB, Art. 401. (the square on either leg of a right triangle is equivalent to the square on the hypotenuse diminished by the square on the other leg). Hence R-S-S'. Ax. 8. Q. E. F. Ex. Construct a square equivalent to the difference of two squares whose sides are 1 in. and i in., respectively. CONSTRUCTION PROBLEMS 247 PROPOSITION XIV. PROBLEM 405. To constract a square equhivalent -to a given paral7elogram. -~C \r S /1 Given the =7 A-BCI)D with base b and altitude h. To construct a square equivalent to ABl GD. Construction. On the line J~G take EF equal to h and FG equal to b. On EG as a diamieter construct a semicircle. Art. 275, Post. 3. At F erect a I. meeting the semicircunmference at K. A rt. '2'7 4 On a line equal to EKC construct the square &. Then S is the required square, Proof. S8K= But ~ ~~~ -2 But ~~~KE = X hi Art. 343. (a -1 from any poinit in. a, circnn?,fercnee to a diameter is a mean proportional between the segmients of the di'ameter). But area = ALWCD -b X hi. (W.hy) Saf-area ~17 ABUD. Q. F 406. COR. To construtct a square equuialevt to a given triangle, construct the mean proI,)ortioncal between the base and half the altitude of the triangle and construct a square oni, hs mnean proportional. 248 BOOK IV. PLANE GEOMETRY PROPOSITION XV. PROBLEM 407, To construct a triangle equivalent to a given polygon. BF F A E G Given the polygon ABCDE. To construct a triangle equivalent to ABCDE. Construction. Let A, B, C be any three consecutive vertices in the given polygon. Draw the diagonal AC. Draw BF 11 AC (Art. 279), and meeting AE produced at F. Draw FC. In the polygon FCDE take the three consecutive vertices C, D, E, and draw the diagonal CE. Draw DG 11 CE, and meeting AE produced at G. Draw CG. Then A FCG is the triangle required. Proof. A ABC LAFC, Art. 390. (having the same base AC, and their vertices in, a line BF I] the base). Also A ACE = A ACE. Ident. And A ECD = A ECG, Art. 390. (having the same base CE and their vertgces in line DG |1 base). Adding, AABC + A ACE + A ECD =c A AFC + A ACE- + A ECG. Ax. 2. Or polygon ABCDE AL FCG. Q. E. F. 408. COR. To construct a square equivalent to a gijen polygon, use Arts. 407 and 406. Ex. Construct a triangle equivalent to a given hexagon. CONSTRUCTION PROBLEMS 249 PROPOSITION XVI. PROBLEM 409. To construct a rectangle equivalent to a given square, and having the sum of its base and altitude equal to a given line. Ic sXI 71 ---- A E B Given the square S and the line AB. To construct a rectangle equivalent to S, and having the sum of its base and altitude equal to AB. Construction. On AB as a diameter describe the semicircumference ADB. Art. 275, Post. 3. At the point A erect a _, AC, equal to a side of S. Art. 274. Through C draw a line 11 AB (Art. 279), and meeting the circumference at D. Draw DE I AB. Art. 273. Construct the rectangle R with a base equal to EB and an altitude equal to AE. Then R is the rectangle required. Proof. DE2== AE X EB. (Why?) But DE=CA. (Why?).A. =AE X EB. (Why?) Or SIR. Q. E. F. 410. COR. The above problem is equivalent to the problem: Given the sum and product of two lines, to construct the lines. 250 BOOK IV. PLANE GIE;OMETRY PROPOSITION XVII. PROJLEM 411. To construct a rectangle equivalent to a given square, and having the difference of its base and altitude equal to a,.. given line. e \ e ad te lin Given the square S and the line A B. To construct a rectangle equivalent to S, and having the difference of its base and altitude equal to AB. Construction. On AB as a diameter describe the circumference ADBF. Art. 275, Post. 3. At A erect the L AC equal to a side of S. Art. 274. Draw CF through the center 0, and meeting the circumference at the points D and F. Construct a rectangle R with base equal to CF and altitiude equal to CD. Then R is the rectangle required. Proof. CF: CA = CA: CD. Art. 358..'. GA= CF X CD. (Why?) -. S-R. (Why?) Also the difference of the base and altitude of R= CF- CD=DF=AB. Q. E. F. 412. COR. The above problem is equivalent to the problem: Given the difference and the product of two lines. to construct the lineo, CONSTRUCTION PROBLEMS 251 PROPOSITION XVIII. PROBLEM 413. To construct a polygon similar to two given similar polygons, and equivalent to their sum. M Pt/ A B,4' ' K ~ ' B" Given the similar polygons P and P. To construct a polygon similar to P and P', and equivalent to their sum. Construction. Take any two homologous sides, AB and A'B', of P and F. Draw 1MK 1 KL (Art. 274), making MKA = AB, and KL Draw ML. On A"B", equal to ML, as a side homologous to AB construct the polygon P" similar to P. Art. 372. Then P" is the polygon required, P AB 2 P AIR2 Proof. p; 2 also -— A 2 Art. 399. P" A"B"2 P" A"B"2 (the areas of two similar polygons are to each other as the squares of their homologous sides). p + P Al-2 + AIB2 Adding, * --- -- ' Ax. 2. But l~lX2 +_ T'_2 = 1i12 But 2 MK2 + ~KL%== Ml2, Art. 346. Or ABA2 + AI'2= AB- 2. Ax. 8. AhJ2 + '- 7721 Ax 8 2'A^ Ax. AB AP+ P'1 - o. =l 1 (Ax. 1.).P' P +P' /P". Ax. 4. p-' Q. E. o 252 BOOK IVo PLANE GEOMETRY PROPOSITION XIX. PROBLEM 414. To construct a square which shall have a given ratio to a given square. -,,,~ B, /.C, / C' a hD '-.; x Given the square S and the lines )m and n. To construct a square which shall be to S in the ratio: nm. Construction. Take AB equal to a side of S and draw AF, making a convenient angle with AB. On AF take AD equal to m, and DF equal to n. Draw DB. Draw FC 11 DB, meeting AB produced in C. Art. 279. On AC, as a diameter, construct a semicircumference AKC. Art. 275, Post. 3. At B erect a I BK meeting the semicircumference at K. Art. 274. Construct a square 8' having a side equal to BK, or x. Then S' is the square required. Proof. x2=a X b. (Why?) Also a: b=m: n. (Why?) S a2 a2 a m Hence -- -= Axs. 8, 5. 2 ab b n QA. 1. CONSTRUCTION PROBLEMS 253 PROPOSITION XX. PROBLEM 415. To construct a polygon similar to a given polygon, and having a given ratio to it. Given the polygon P and the lines m and n. To construct a polygon P' which shall be similar to P, and be to P in the ratio n: m. Construction. Construct a square which shall be to the square on AB as n: m. Art. 414. Let A'B' be a side of this square. Upon A'B' as a side homologous to AB construct a polygon P' similar to P. Art. 372. Then P' is the polygon required. P AB Proof, A- (Why?) P' A'B'2 2 m But AB m Constr. AB2 n Hence P- (Why?). - n Q,. 254 BOOK IV. PLANE GEOMETRY PROPOSITION XXI. PROBLEM 41 6. To construct a polygon similar to one given polygonz and equlicalent to another given polygon. m 9A B ' J > JA B A' B' Given the polygons P and Q. To construct a polygon similar to P, and equivalent to Q. Construction. Construct a square equivalent to P, and let m be one of its sides. Art. 408. Construct a square equivalent to Q, and let n be one of its sides. Art. 408. Construct A'B', the fourth proportional to m, n, and AB. Art. 366. On A'B', as a side homologous to AB, construct a polygon P' similar to P. Art. 372. Then P' is the polygon required. 2 2 P m AB- P Proof _o Q- _ A- B _'. Constr., Arts. 314,399. Q n A'B' P' P P -Q p Ax. l.'. P,'Qo Art. 305. Q.;. F. EXERCISES, THEOREMS EXERCISES. CROUP 40 THEOREMS CONCERNING AREAS Ex. 1. The diagonals of a parallelogram divide the parallelogram' into four equivalent triangles. Ex. 2. Any straight line drawn through the point of intersection of the diagonals of a parallelogram divides the parallelogram into two equivalent parts. Ex. 3. If, in the triangle ABC, D and F are the midpoints of the sides AB and AC, respectively, the area of ADF equals one-fourth the area of ABC. [SUG. Use Art. 397.] Ex. 4. If the midpoints of two adjacent sides of a parallelogram be joined, the area of the triangle so formed e(luals one-eighth the area of the parallelogram. Ex. 5. If, in the triangle ABC, D and F are the midpoints of the sides AB and AC, respectively, the triangles ADC and AFB are equivalent. Ex. 6. In a right triangle show, by obtaining expressions for the area of the figure, that the product of the legs equals the product of the hypotenuse by the altitude upon the hypotenuse. Ex. 7. If two triangles are equivalent, and the altitude of one is three times the altitude of the other, find the ratio of their bases. Ex 8. If two isosceles triangles have their legs equal, and half of the base of one equivalent to the altitude of the other, the triangles are equivalent. Ex. 9 If two triangles have an angle of one the supplement of an angle of the other, their areas are to each other as the products of the sides including these angles. a b Ex. 10. Prove geometrically that (a+b)2='- 2+b2 +2ab. Ex. 11. Similarly prove (a-b)2 = a2-b2 ---2ab. Ex 12. Similarly prove (aA-1))(a-b)=a2-b2. Ex. 13. The line joining the midpoints of the parallel sides of a trapezoid divides the trapezoid into two equivalent parts. 256 BOOK IV. PLJANE (GEOMETRY Ex. 14. The lines joining the midpoint of c quadrilateral to the vertices not joined by the diagonal divide the quadrilateral into B two equivalent parts. Ex. 15. Given QR and TS passing through P, any point on the diagonal AC A of a /Z7, QR 11 AD, and TS I AiB; prove QBTP-. PRDS. Ex. 16. Given OB=ODn; prove AxBC- AAADC. Let the pupil also state this as a theorem in general language. EXERCISES. CROUP 42 USE OF AUXILIARY LINES )ne diagonal of a T 0 P B S D B Ex, 1. Given ABCD a /7 and P any point inside ABDC; prove APAD -- APBC A PPAB+- A PCD. P LID ~g~66 I —~ Dn Ex. 2. The area of a triangle is equal to one half the product of its perimeter by the radius of the inscribed circle. Ex. 3 If the extremities of one leg of a trapezoid be joined to the midpoint of the other leg, the middle one of the three triangles thus formed is equivalent to half the trapezoid. Ex. 4. The area.of a trapezoid is equal to the product of one leg by the perpendicular on that leg from the midpoint of the other leg. Ex 5. If the midpoints of the sides of a quadrilateral be joined in order, the parallelogram thus formed is equivalent to one-half the quadrilateral. I., "I Ex. 6. A quadrilateral is equivalent to a triangle two of whose sides are the diagonals of the quadrilateral, the angle included by these sides being equal to one of the angles formed by the intersection of the diagonals. EXERCISES. THEOREMS 257 EXERCISES. CROUP 42 THEOREMS PROVED BY VARIOUS METHODS Ex. 1. If through the midpoint of one leg of a trapezoid a line be drawn parallel to the other leg to meet one base and the other base produced, the parallelogram so formed is equivalent to the trapezoid. Ex. 2. If the midpoints of two sides of a triangle be joined to any point in the base, the quadrilateral so formed is equivalent to half the triangle. Ex. 3. If P is any point on AC the diagonal of a parallelogram ABCD, the triangles IAPB and APD are equivalent. Ex. 4. If the side of an equilateral triangle be denoted by a, the area of the triangle equals - Ex. 5. Find the ratio of the areas of two equilateral triangles, if the altitude of one equals the side of the other. Ex. 6. If perpendiculars be drawn from any point within an equilateral triangle to the three sides, their sum is equal to the altitude of the triangle. Ex. 7. If E is the intersection of the diagonals AC and BD of a quadrilateral, and the triangle 4 DE is equivalent to the A BEC, then the lines AB and CD are parallel. Ex. 8. If, in the quadrilateral ABCD, the triangles ABC and ADC are equivalent, the diagonal AC bisects the diagonal BD. Ex. 9. If two triangles have two sides of one equal to two sides of the other, and the included angles supplementary, the triangles are equivalent. Az~l Ex. 10. If, in the parallelogram ABCD, F is the midpoint of the side BC, and AF intersects BD in K, the triangle BKFT= — the parallelogram ABCl). Ex. 11. Given PQ 11 AC, and PR I1 AB; prove AQAR a mean proportional between GBQP and APRC. 'o, _ ^ I Ex. 12. i' is any point in the side IBC of the parallelogram A4 PCD and DIP pro(duced meets.11) produced in Q. Show that the triangles 1JB.A and CPQ are equivalent. A D Gu__ e 258 BOOK IV. PLANE GEOMETRY EXERCISES. CROUP 43 PROBLEMS IN CONSTRUCTING AREAS Ex. 1. Construct a square having twice the area of a given square. Ex. 2. Construct a square having three times the area of a given square. Ex. 3. Construct a square equivalent to the sum of three given squares. Ex. 4. Transform a given triangle into an equivalent isosceles triangle having the same base. Ex. 5. Transform a given triangle into an equivalent triangle having the same base, but having a given angle adjacent to the base. Ex. 6. Transform a triangle into an equivalent triangle with the same base, but naving another given side. Ex. 7. Transform a parallelogram into an equivalent parallelogram having the same base, but containing a given angle. Ex. 8. Construct a triangle similar to a given triangle and containing twice the area. To construct a similar triangle containing five times the area; how is the construction changed? Ex 9. Bisect a given triangle by a line parallel to the base. Ex. 10. Construct a polygon similar to two given similar polygons, and equivalent to their difference. Ex. 11. Draw a line parallel to one side of a given rectangle, and cutting off five-sevenths of the area. Ex. 12. Bisect a parallelogramby a line perpendicular to the base. Ex. 13. Through any given point draw a line bisecting a given parallelogram. F Ex. 14. Construct a triangle equivalent to R a given triangle, ABC, having a given base AD, but the ZBAC adjacent to the base unchanged. [Sua, Draw CF II DB) etc.] D C EXERCISES. PROBLEMS 259 Ex. 15. Transform a parallelogram into an equivalent parallelogram having a given base, but the angle adjacent to the base unchanged. Ex. 16. Transform a given triangle into an equivalent right triangle having a given leg. [SuG. Use Ex. 14, then Ex. 5.] Ex. 17. Transform a given triangle into an equivalent right triangle having a given hypotenuse. [SUG. Find the altitude upon the hypotenuse of the new triangle by finding the fourth proportional to what three lines? Ex. 18. Transform a re-entrant pentagon into an equivalent triangle. Ex. 19. Transform a given triangle into an equivalent equilateral triangle. [SUG. See Art. 416.] Ex. 20. Bisect a triangle by a line perpendicular to one of its sides. [SUG. See Art. 416.] Ex. 21. Construct a square equivalent to two-thirds of a given square. EXERCISES. CROUP 44 PROBLEM. S SOLVED BY ALGEBRAIC ANALYSIS Ex. 1. Transform a given rectangle into an equivalent rectangle with a given base. Ex. 2. Transform a given square into a right triangle having a given leg. Ex. 3. Transform a given triangle into an equivalent isosceles right triangle. Ex. 4. Draw a line cutting off from a given triangle an isosceles triangle equivalent to one-half the given triangle. [SUG. Use Art. 397.] Ex. 5. Through a given point in one side of a given triangle draw a line bisecting the area of the triangle. Ex. 6. Transform a given square into a rectangle which shall have three times the perimeter of the given square. 260 BOOK IV. PLANE GEOMETRY EXERCISES. GROUP 46 PROBLEMS SOLVED BY VARIOUS METTIODS Ex. 1. Bisect a given parallelogram by a line parallel to the base. Ex. 2. Transform a parallelogram into an equivalent parallelogram having the same base and a given side adjacent to the base. Ex. 3. Construct a square which shall contain four-sevenths of the area of a given square. Ex. 4. In two different ways construct a square having three times the area of a given square. Ex. 5. Trisect a given triangle by lines parallel to the base. Ex. 6. Find a point within a triangle such that lines drawn from it to the vertices trisect the area. Ex. 7. Find a point within a triangle such that lines drawn from it to the three vertices divide the area into parts which shall have the ratio 2: 3: 4. [SuG. If one of the small A contains i the area of original A, a line through its vertex cuts off 9 the altitude, etc.] Ex. 8. Divide a triangle into three equivalent parts by lines through a given vertex. Ex. 9. Divide a triangle into three equivalent parts by lines drawn through a given point 1' in one of the sides. [Suc. Use Art. 39T.] Ex. 10. Divide a given quadrilateral into three equivalent parts by lines drawn through a given vertex. Ex. 11. Through a given point in the G base of a trapezoid draw a line bisecting the /\ area of the trapezoid. Ex. 12. Bisect the area of a trapezoid A 'B by a line drawn parallel to the bases. F - [SUG. Construct A GEF similar to A \ ABG and equivalent to i sum of what two AU y BOO0K V IREGULAR POLYGONS. MEASUREMENT OF THE CIRCLE 41 7. DEE. A regular polygon is a polygon that is both equilateral and equiangular. PROPOSITION 1. THEOREM 418, Ant equiilateral polygon that is hiscribed hi a circle is also eqa ian gala r and regular. Given ABC.. K an inscribed polygfon, with its sides AB, BC, CD, etc., equal. To prove the polygon ABC... K equjangular and regular. Proof. Arc AB~arc BC~ arc C'D, etc. Art. 218. arc ABC= arc (JICD-arc CIDE, etc. Ax. 2. Z ABC= Z.BCUL) Z CDE, etc., Art. 260. (all S inscribcd in the same segment, or in equal segm~ents, are equal). the polygron AiBC... K is equiangular. the polygoni ABC... K is regfular. Art. 417. Q. E.D. (261) 262 262 1300K V. PLANE (GEOMET 7~ 419. COR. 1. If the ar1cs SUbtended, by the sides of a regular inscribod p)olygon be bisected, and each, point of bisection' be joinewl to the nearest vertices of the polygon, a regular inscr-ibed polygoni of doable the n amiber of sidles is3 formed. 420. COR. 2. The perimieter of an intscribed polygon is less than the por'uneter, of an, inscribed polygont of doatble the number of sides. PROPOSITION II, THEOREM 421. A circle wtay be circumscribed about, and a circle may be inscribed in, any regular p)olygon. 'P P~~~~~~~~~~~C scribed in, ABCDE. Proof. I. Through A, B and C (Fig. 1), any three successive vertices of the polygon ABODE, pass a circumference. Art. 235. Let 0 be the center of this circumference. Draw the radii OA, OB, 00. Also draw the line OD. Then, inuL OBC, OB=O00. (Why?) 9 LOBCL=ZOCB. (Why?) REGULAR POLYGONS 263 But Z ABC = Z BCD, Art. 417. (being A of a regular polygon) Subtracting, Z OBA = Z OCCD. (Why?) Ience, in the A OAB and OCD, OB= OC. (Why?) AB= CD. (Why?) Z OBA = Z OCD, (just proved). A. A B0 A 0A C0. (Why?).. 01)-0. (Why?) Hence the circumference which passes through the vertices A, B and C, will also pass through the vertex D. In like manner, it may be proved that this circumference will pass through the vertex E. Hence a circle described with 0 as a center, and OA as a radius, will be circumscribed about the given polygon. II. The sides of the polygon ABCDE (Fig. 2) are equal chords in the circle 0. Hence they are equidistant from the center. Art. 226..'. a circle described with 0 as a center, and the distance from 0 to one of the sides of the polygon as a radius, will be inscribed in the given polygon. Q. E. D. 422. DEF. The center of a regular polygon is the comll mon center of the inscribed and circumscribed circles, as the point 0 in the above figure. 423. DEF. The radius of a regular polygon is the radius of the circumscribed circle, as OA in the above figure. 424, DEF. The apothem of a regular polygon is the radius of the inscribed circle. "If.J64 264 ~1300K V. PLANE GEOMAETRY 425. DEF. The angle at the center of a regular polygon is the angle, between two radii drawn to the extremities of any side, as the angle A OB. 426. Cor,. The angle ait the ceniter of a regular polygon, is equal to foutr right (angles (Idiviledi by the nam ber of sides. Hence, if 'n denote the number of sides in the polygon, the angle at the Henter, of a reguilar polygon, eqaals4rt also the angle betwveen ani ajiot hem and the nearest radius, ina regat tar polygon of n sides, eqaals PROPOSITrION III. THEOREM 427. If the circumnference of a circle be divided into any num)ber of equal arces, I. The chords of these arcs formb a regular inscribed polygon; II. Tangents to the circumference at the points of division form a regular ci rNunmscr'ibed polygon. T A P E11 SG Given the circumference ABC, divided into the equal arcs AB, BC, C~D, etc., the chords AB, BC, etc., and PQ, QR, etc., lines tangent to the circle at B, C, etc. To prove ABGDE a regular inscribed polygon, and PQR$ST a regular circumscribed polygon. RMG1YLAR POLYGONS 2Q65" Proof. I. The chords AB, BC,7 CD, etc., are equal. (Why) polygon ABODE is equilateral and regular. Art. 418. II. In the APB), BQC, GRD, etc., AB=BC= CD, etc. (Why?) Also Z~PAT3 Z PBA =Z QJC =Z QCB =Z RCD, etc. Art. 264. (each being vieasiored lof half of owe of' the eqaal arcs AI-,, BC,! CID, etc.)., AAPB, JiQC, CRI), etc., arc ctitial, istee tinls (Why?) jJ)= ZQ ZR, etc. (Why) And AP=JPB=BQ=QC, etc. (Why?.PQ =QR =RS, etc. Ax. 4. PQRST is a regfular p~olygon. Art. 417. Q. E. D. 428. Co-R. 1. If the arcs AB, BC, CD, etc., be bisected, andl a tangent be drawn, at each. phiut of bisection, a circ um - scri bed regular polygon of double the number of sides of PQRST will be formed. 429. COR. 'I2. The perimeter of a circaumscribed regular Polygon is greater than, that of a circumscribedl regular polygon of doutble the number of sides. Ex. 1. Find the number of~ degrees in the central angle of a 'regular pentagon. Of a regular hexagon. Uf a square. Ex. 2. What is the short name for an inscribed equilateral quadrii~atuaj? 2_166 266 ~BOOK V. PLANT (4EOMETIRY PROPOSITIo-N IV. THEOREM 430. Tangent-s to a circle -at the midpoints of the arcs subtended by the sides of a regular 'inseribed polygon ftorml a regn bar cire~tcwscribed po lygon 'a'hose sides are parc lid to thre correspo i ding sides of Ike, 'inscribed polygon. A TA 1 ' B' SQ D C Given the regular polygon ABCDE inscribed in the 0D ACD; P, Q, Pi, etc., the midpoints of the arcs AB, BC, CD, etc.; and A'B', BIC', C'D', etc., tangents to the circle at P, Q, Pi, etc. To prove A'BW'ID'E' a regular polygon with its sides I corresponding sidcs of the polygon ABCDE. Proof. The arcs AB, BC, CD, etc., are equal. Art. 218. the arcs AP, PB, BQ, QC, etc., are equial. Ax. 5. the arcs PQ, QJI, -RS, etc., are eqnal. A.4 A'B'C'D'E' is a regrular polyg-on, Art. 427. (if the circum~ference of a 0 be divided, etc.). Side AB I OP. Art. 113. A'B' I Z)P. (Why ~?)..AB II A'B'. (Why?) In like manner, each pair of homologous sides in the two polygons is parallel. Q.D 431. CoR. Jiornologous radii of an inscribed and a circumscribed regular, poly~jon, wh~ise sides are parallel, coincide in direction.. Thus, in the above figure A POA and POA' each 2t:.OA and OA' coincide in direction. lb ~~~~~~~~~Art. 420 - REGULAR POLYGONS 267 PROPOSITION V. THEOREM 432. Two regular polygons of the same number of sides are similar. C" \ K DD B~ - ' Given K and K' two regular polygons, each of n sides. To prove Kland K' similar. Proof. Each Z of K= - ) 2 t _, Art. 174. il (in an equiangular polygon of n sides, each - 2) 2 =rt. ) Similarly each Z of KA' ( —2) 2 rt A Hence K and K' are mutually equiangular. Ax. i, Art. 169. Also AB=BC. 1 Art. 417, Ax. 5. And A'B'- '. ' B'.- (Why?) A A'B' AB BC ** BC 'C' A ' B'C. (Why?) BC CD 1E In like manner have thei- ho e pt tional. Hence K and K' are similar. Art. 321. Q. D. D. 433. COR. The areas of two regular polygons of the same number of sides are to each other as the squares of any two homologous sides, 1100K V. PLANtGE 0OMETRtY PROPOSITION VI. THEOREM 434. I. The perimeters of twco regular polygons of the same number of sides are to each other ais the radii of their circumiscribed circles, or as the radlii of their iniscribed circles; IT. Their areas are to each other as the squares of these radii. D D C ~~ F ~C' E Given AC and AI'C' two regular polygons, each of n sides, with centers 0 and 0', and wvith perimeters dle-noted by P and PI, radii by Ru and RII, and apothems by r and ri, respectively. To prove. I. P: PI = P: RI'-r r. IL. Area AGC area ACi'G=-R2 R12 =r2:ri2. Proof. I. The polygons AG and A'G' are similar. Art. 432. Hence 1>: P' =AB:AWB. Art. 341. But, in the & OAB and 0YAWB Z A OB =/ A'0B', (for each Z = ) r. i Art. 426. Also OA 0R, =0'A' 0 'B', (for each A is isosceles). ~ OAR and 0'A'B' are similar. Art. 327. AB: A'B'= OA: 0'A'. Art. 321. And AB: A'B'- OL: OIL'. Art. 338. P:P'= OA:O'A'= OL:OIL/. Ax. 1, Or P: P'=R: R'=r: r'. II. Area AGC: area AlG'= ARB A: Art. 399. But AB'~ A'BI2R = R '2=r2 r'. Art. 314. area A; C area Al'G'= R I '2. A.1 Q. E. D., REGULAR POLYGONS 269 PROPOSITION VII. THEOREM 435. If the number of sides of a regular inscribed polygon be indefinitely increased, the apothent of the polygon approaches the radius as a limit. D 0 G - B Given the regular inscribed polygon A...D of n sides, with radius OA and apothem OL. To prove that, as n is indefinitely increased, OL apyroaches OA as a limit. Proof. In the A OAL, AL > OA - OL, Art. 93. (any side of a A is greater than the dcifference betwieen the other two sides.) But, as n *, AB-O.'. AL 0. Art. 253, 3. Hence OA OL O.. OL OA, or r -R. Or the limit of thp apothem OL is the radius OA. Q. E. D. 436. COR. As nt, R2-r2'0. For, R2- 2= (R +r) (R-r). But, as n t o, R+rR -+R or 2 R, and R-rO-. Ax. 8..R. R —2 X or 0. A.8. Ex. A pair of homologous sides of two regular pentagons are 2 and 3 ft. Find the ratio of the areas of the polygons. q P K1300K V. PLANE GEOMETRY PROPOSsTION VIII. THEOREM 437. The length of any line inclos~ing the ciracumference o!f it circle, andl iot?pasSing wOillin the circimnference, is greater than the length of the circuminference. D B K F Given the circumference AKEF, and ABCDEF any line which does not pass within AKEF. To prove circumference AKEF < perimeter ABCDEF. Proof. Let K be any point on the circumference AKL'F not touched by the line ACDEF. At K draw a tangent to the given circle, meeting AGEE at B and D. Then the straight line BKD < line BCD. Art, 15. 'To each of these unequals add the line DEFAB. Then line ABKDEF < line ACDEF. Ax. 9, Hence every, line enveloping the circular area AKEF, except the circumference AKEF, may be shortened. Hence the circumference AKEF is shorter than any line envelopiui it. A w 190 &,A- My. REGULAR POLYGONS 271 438. COR. 1. The circumference of a circle is less than the perimeter of any polygon circumscribed about the circle. 439. COR. 2. The circumference of a circle is greater than the perimeter of any polygon inscribed in the circle, for each side of an inscribed polygon is less than the arc subtended by it. 440. COR. 3. The difference between the perimeters of an inscribed and a circcumscribed polygon is greater than the difference between either perimeter and the circumference of the circle. PROPOSITION IX. THEOREM 441. If the number of sides of a regular inscribed, or of a regular circumnscribed polygon be indefinitely increased, I. The perimeter of each polygon approaches the circumference as a limit; II. The area of each polygon approaches the area of the circle as a limit. Given a circle of circumference C and area A, with regu~ lar inscribed and circumscribed polygons, each of n sides, with their perimeters denoted by P and P', and their areas by K and K', respectively. To prove that as n is indefinitely increased, P and P' 273 B3OOK V. PLANE GEOM'ET R each approaches C as a linit, and K and K' each approaches A as a limit. Proof I. Denote the apothems of the two polygons by R and r. Then Pi R. Art. 434. P r Hence - Art. 310, Let nz=o; then R-r-O 0.Art. 435. R r. -.'. ---=0, Art. 253, 4. (for the de(nominator, r, r, increase s s increaseS). pi- P Hence p 0 (Ax. 8).. '. P' - P 0, (Art. 253, 4). But P- C < P' — (Art. 440).~. P'- C 0, or P'-O. Also C- P < P- P (Art. 440):. C -P-), or P-C. 17 p/2 K-/ If /?2 __ r2 II. — = — (Art. 434) '. 1= ---' Art. 310. K r K r Let n -_ *x; tlhen l2 - 2 0. Art. 436. 112 _ 7.2 o ___ ' 0, Art. 253, 4, (for the (rdominator, 'r, increases). Hence -- 0 (Ax. 8). -. ' - 0, (Art. 253, 4). But Ki —A < K'-K (Ax. 7).. K'-A --, or K'-IA Also A - < K- K(Ax. 7).. A —K 0, or K-A. ~~_____~_____.~Q. E. oD Ex, Find the perimeter and trea of a square field, one of whose sides is 10 rods. Find the same in a square field, one of whose sides is 20 rods. Is it more economical, therefore, to fence land in large or small fields t REGULAR POLYGONS 273 PROPOSITION X. THEOREM 442. Twvo circumferences ha~ve the samne ratio as their radii, or as their diameters. 0 R Given the circles 0 and 0', with circumferences denoted by C and C', radii by Pi and Pi', diameters by D and DR, respecti velv. To prove C: C' - R: R' -= D: D'. Proof. In the given 0 let regular poly7gons of the same number of sides be inscribed. Denote the perimeters of the inscribed polygons by P and PI, respectively. Then Art. 434. P1 P Hence, by alternation) Art. 307. If, now, the number of sides of the similar inscribed polygons be increased iniefinitely, P becomes a variable approaching C as a limit. Art. 441. pi- C Hence -R becomes a variable approaching as a limit. Art. 253, 3, Pi C' Similarly, the variable p approaches as a limit. P Pi But the variable - the variable always. C C' Art. 234. C C' R P'R! Art, 307. Rt Pi C 2R I) Also (Art. 315.) Ax. 8, A' 9R1 C' 'JR' d' 27 4 274 ~BOOK V. PLANE GEO0METRZY PROPOSITION, XI. THEORM~x 443. la eiery circle, the ratio of the (irrumiference toth diamieter (that is, the Numzber 01' tbimes the diamietcr is contai~ned in the circai)'ference) is the samne, antd may be, denoted by ant approp~riate symbol 7) C~~~~~~~~C Given two 0D with circumferencees denoted by C and C', and diameters by D and DI, respectively. C C/ To prove - = =-t C)DI Proof. C =D Art. 442. C G' By alternation Art. 307. Hence, in any griven cirle L ~1 Chas as valne equal to that of in some standard circle. D ~~~C the value of - is the same in all circles. DC De-noting this constant by 7t, in every circle-C t D Q. E. D. 444, Formula for the circumference in terms of the radius.C We have C=7tD=:7t(2 R). C=:2 tR. Ex, Find the circumference of a circle whose radius is 10 inches. REGULAR POLYGONS 275 445. Formula for length of arc of a circle. arc central Z centrall / are cetra Z-(Art. 255.).'. are enra X 27tR. circumference 360 ( 3600 central Z Aarce 1 - X o PROPOSITION XII. Ti-rpoiNint 446. -The area, of a regular ])oIygolo is equal to one-half t1w product oj its perinieiter by its aIpot henm. Et '0, —~ Given the regular polygon ABCDE with area denoted by K, perimeter by 1', and apotbem. by r. To prove K-k Xr. Proof. Draw the radii 0A, 0B, 00, etc., dividing the polygon into as many & as the polygon has sides. All the & thus formed have the same altitude, r. Art.205. the area of each A= ~- product of its base by r. Art. 389. Hence the sum of the areas of the = product of the surm of the bases of the & by r, or —k P X r. But the sum of the areas of the & equals the area of the polygon. Ax. 6. Henei K== ); P X r. Ax. 8. Q. E.D. 447. DEF. Similar sectors are sectors in different circles which have equal angles at the center. 448. DEE. Similar segments are segments in different Circles whose arcs subtend equal angles at the center-. 276 276LOK V. PLANE GfEOMETRY PROPOSITION X-III. THEOREM 449. The area OJf a circle is equal to one-haif thle product of its circilinjerenee by its radius. Given a 0) with circumiferenCe denoted by C, radius by k, and area hr K. To prove K= ~ C X r. Proof. Circumscribe a regular polygon about the given circle, and denote its perimeter by P and its area by If. In this case the apothemn of the regular polygon is R. Hence K i T P X P. Art. 446. Let the number of sides of the circumscribed polygon be increased indefinitely; then K' becomes a variable approaching Y Kas its limit; Art. 441. Pbecomes a variable approaching C as its limit; Art. 441. Anid variable 4 PX 1'Xlapprojoaches I)- CX Has a limit. Art. 253, 2. But K7' I _ X R alwayvs. Art 4G6 Ilenco X Art. 254. Q. E. D. 450. Formula for the area of a circle in terms of the radius R. IK=3 C X R; but C=2. 7tR. Art. 444, i2 tPR) R. Or K= =7R2. Ax. 8. XD2~ A gain K==D:. K - Ax, 451. COR. 1. The (area of a circle is equal to the square of the radius, multiplied by 7r; or to one-fourth the square oJ the diameter, multiplied by 7t, MEASUREMENT OF THE CIRCLE 277 452. COR. 2. The areas of two circles are to each other as the squares of their radii, or as the squares of their diamr eters. K 7tR2 pR2 iK i nD D2 For h- = R2 -:2i also - - d2' 453. The area of a sector is equal to one-half the product of its radius by its arc. Hence area of sector cerl X R. Art. 445. central Z Or area of sector = 3 -X nR-. 454. COR. 3. Similar sectors are to each other as the squares of their radii. Ex. 1. The measurement of the area of a circle can be reduced to the measurement of the length of what single straight line t? Can it be reduced to the measurement of any other single straight line? to the measurement of a single curved line? Find the area of a circle, Ex. 2. Whose radius is 10 ft. Ex. 6. Whose radius is 2b ft. Ex. 3. Whose diameter is 10 ft. Ex. 7. VWhose radius is R; iR; 2fR. Ex. 4. Whose radius is b ft. Ex. 8. Whose radius is RV3. Ex. 5. Whose radius is Ab ft. Ex. 9. Whose diameter is -RA/2. Ex. 10. If the radius of one circle is 10 times as great as the radius of another circle, how do their areas compare? Also how do their circumferences compare? Ex. 11. A wheel with 6 cogs is geared to a wheel with 48 cogs. How many revolutions will the smaller wheel make while the larger wheel revolves once? Ex. 12. A 2 in. pipe will discharge how much more water in a given time than a 1 in, pipe Y 278 dL I 8 BOOK V. PLANE GEOMETRY PROPOSITION XIV. THEOREM 455. The areas of two similar segments are to each other as the squares of their radii. 0 A 2 \: a: \ A-7, O0 A., % I1 \ AR/ 7 \: \ A' C'i z Given the similar segments ABC and A'B'C' in cirole, whose radii are R and R'. To prove segment ABC: segment AB''C'=R2: B'2. Proof. Draw the radii OA, OB', OA ', 0'B'. Then the sectors OAB and O'A'B' are similar. Arts. 448, 447. And A OAB and O'A'B' are similar. Art. 327. sector OAB -P 1' A OA R2 sector OA/1B R/2 an A OAB B / Art. 454, 398. sector OA-R A OAB (hy.'........... —..........~ (Whly ~) sector O'A'J' A L 'A'J' sector OA1B sector O'A'B' A OAB A O'A'B' sector OAB -A OAB sector O'A'B'/- O'A'B' A OAB A 0AI'B' (WI segment ABC segment A'B'C' A OAB A O'A'B' segment ABC _ / A OAB O R2 segment A'B'C'- A O -iA'' ~K' Arts. 307, Q. E. I y?) ly?) 398. )o CONSTRUCTION PROBLEMS a CONSTRUCTION PROBLEMS PROPOSITION XV. PROBLEM 456. To inscribe a square in a given circleo 279 Let the pupil supply a solution. 457. COR. By bisecting the arcs AC, CB, BD, etc., and drawing chords, a regular octagon may be inscribed in the circle; by repeating the process, regular polygons of 16, 32, 64,... and 2'" sides mlay be inscribed, where n is a positive integer greater than 1. How can a regular polygon of 21L sides be circumscribed about a given circle? PROPOSITION XVI. PROBLEM 458. To inscribe a regular hexagon in a given circle. A C~\\r J:\- V, Let the pupil supply a solution. 280 BOO3 V. PLANE GEOMETRY 459. COR. 1. The side of a regular iinscribed hexagon equals the radius of the circle. 460. COR. 2. By joining the alternate vertices of a regular inscribed hexagon, an equilateral triangle can be in~ scribed in a given circle. 461. COR. 3. By bisecting the arcs AB, BC, CD, etc., and drawing chords, a regular polygon of 12 sides can be inscribed in a given circle; by repeating the process, regular polygons of 24, 48,.... 3 X 2 sides can be inscribed. Similarly, how can a regular polygon of 3 X 211 sides be circumscribed about a given circle? PROPOSITION XVII. PROBLEM 462. To inscribe a regular decagon in a given circle. Given the circle 0. To inscribe a regular decagon in the given 0. Construction. Draw any radius OA, and divide OA in extreme and mean ratio at K, OK being the greater segment. Art. 371. With A as a center and OK as a radius, describe an arc cutting the given circumference at B. Then the chord AB is a side of the decagon required, CONSTRUCTION PROBLEMS 281 Proof. Draw OB and KB. Then OA: OK= OK: KA. Constr. But AB=OK,.. OA: AB=AB: KA. Ax. 8. Also Z OAB = Z KAB. Ident. ~., OAB and KAB are similar. Art. 327..'. Z 0 -- ABK (homolog. X of similar A). And A AKB is isosceles, Art. 321. (being similar to A AOB, which is isosceles). AB=KB=KO. Ax. 1. Hence Z 0= /Z BO. (Why?).. ZABO =2 ZO. A. A 2... ZOAB=2 Z O. Art. 99. ZO==Z0. Adding, Z ABO + Z OAB + Z O==5 Z O. But ZABO + Z OAB+ Z 0=2 rt. A. Art. 134..'. 5 Z0=2 rt. X. Ax. 1. Z0= O of2rt. A, or - of4 t. A..'. arc AB is -iQ- of the circurrerence. Hence, if the chord AB be applied ten times in succession to the circumference, a regular decagon will be inscribed in the given circle. Art. 418. Q. E. F. 463. CoR. 1. By joining the alternate vertices of a regullar inscribed decagon, a regular pentagon can be inscribed in a given circle. 464. COR. 2. By bisecting the arcs AB, BC.. and drawling chords, a regular polygon of 20 sides can be inscribed in a givenl circle; by repeating the process a regular polygon of 40, 80,... 5 X 2 sides can be inscribed. How can a regular polygon of 5 X 2'1 sides be circumscribed about a given circle I 282 BOOK V. PLANE GEO.METRY PROPOSITION XVIII. PROBLEM 465. To inscribe a regular polygon of fifteen sides (pentedecagon) in a given circle. Given the 0 AK. To inscribe a regular pentedecagon in the given 0. Construction. Draw the chord AC equal to the radius of the given 0, and the chord AB equal to a side of a regular decagon inscribed in the circle. Art. 462. Draw the chord BC. Then BC is a side of the required pentedecagon. Proof. AC is a side of a regular inscribed hexagon. Art. 459.. arc AC=-6 of the circumference. In like manner arc AB= -11 of the circumference. Art. 462..'. arc BC= --, or -i- of the circumference. Ax. 3. Hence, if chord BC be applied fifteen times in succession to the circumference, a regular pentedecagon will be inscribed in the given circle. Art. 418. Q. E. F. 466. COR. By bisecting the arcs BC, CD,.. etc., drawing chords, and repeating the process, regular polygons of 30, 60,.. 15 X 2n sides can be inscribed in a given circle. How can a regular polygon of 15 X 2z sides be circumo scribed about a given circle? COMPUTATION PROBLEMS 283 COMPUTATION PROBLEMS PROPOSITION XIX. PROBLEM 467. Given the side and radius of a regular inscribed polygon, to find the side of a regular inscribedpolygon of double the number of sides, in terms of the given quantities, A 0 Given the circle 0 with radius R, AB a side of a regular inscribed polygon, and AC a side of the regular inscribed polygon of double the number of sides. To determine AC in terms of AB and R. Solution. Draw the radii OA and OC. Then OC is the I bisector of AB. (Why?) But arc ACB < semicircumference. (Why?).'. AC < a quadrant. Ax. 10... ZAOC is an acute Z. Art. 257. Hence, in A OAC, Ai2=OA2+~O-2-2 OCX OD. Art. 349. Or AC2=2R2- R X OD But, in the rt. A OAD, OD= OA2-AD2, Art. 347. Or D=R2- (AB)2. Ax. 8.. OD= /R2 —i AB2= 4 R -AB. Substituting for OD its value thus obtained, AC=2R2-R 4 R2 -AB2. Ax. 8. Or AC-= R (2 R — /4 R2 - AB2), 284 BOOK V. PLANE GEOMETRY 468. Special Formulas. If the radius PR, be taken as 1, AC = 1/2- 4 - AB2. If a side of the regular inscribed polygon of n sides be denoted by S,, and R=1, then S2n -/ V-4- S"PROPOSITION XX. PROBLEM 469. To compute approximately the numerical value of 7. Given a ( whose radius is 1, and whose circumferenee is denoted by C. To compute C, i. e., 2 t, and hence find the numerical value of 7t approximately. Computation. 1. Inscribe a regular hexagon in the given circle, and denote its side by S6. Then So =. Art. 459..'. perimeter of the inscribed hexagon = 6. 2. Inscribe a regular polygon of double the number (12) sides. Then, by the second formula of Art. 468, S&2= V2- v/4- = 0.51763809 +. Denoting the perimeter of a regular inscribed polygon of n sides by Pn, P12= 12 (0,51763809 ) = 6.21165708. MAXIMA AND MINIMA28 285 3. In the second formula of Art. 468, let n= 129. S2=V12 -V4 -(0.51763S09 + )2- 9.61082N38 +, etc. Henc'e P4 =6. 263 245 7 72. Computing 848, P48, etc., in like manner, the following results are obtained: 812 =V2 - V4'~-1 =.51763809...P12 =6.211165708. 8 24 =VI/2 -v/4a.5 1 706-8 09x - 2 =.261052-38...P24 =6.2065257022 8481 ==V"2-1V4-(.26105238)2_ -.103080(W.. P48 =6.27870041. 89 ==V2 —V4 -(.l30oso626_)2 =065438 17... P(6 =6.298206396. ASI 2 1'-4 -(.0&43817) =03212,346.:.PID2=6.28290510. 8384 4- (.0372346==016362-28..P,384=6.28311544. s768=V2 - 14 4-(.01636228) I.0182... P768=6.2S316941. By continuing the computation it is found that the first six decimal figures in the value of the perimeter of the inscribed polygon remain unchanged. C, o r 217t =6. p88169 approximately.7t= 3.14159 approximately. MAXIMA AND MINIM~A 470. IDE. A maximum (see Art. 268) is the greatest of a group of magnitudes, all of which satisfy certain given conditions. Thus, the diameter is the maximum chord which can be drawn in a circle. 47 1. DEF. A minimum is the smallest of a group of magituesall of which satisfv certain ie odtos manitds,2ngiecodtns Thus, of all lines which can be drawn f romn a given point to a given le the perpendicular is the minirmum. Certain maxiima and minimna have already been studied, and we -now proceed to investigate more particularly those relating to regular polygonls and the circle, BOOK V. PLANE GEOMETRY PROPOSITION XXI. THEOREM 472. Of (all triangles which have two sides equal, that triatygle in whbich these sides incllude a right angle is the maximum. C A~4 D B.' if Given the & ABC and A'B' C in which AB=A'B', CA C'A', and LA' is a rt. Z. To prove A AB'C' > A ABC. Proof. In the A ACB, draw CD 4L AB. Then CD < CA. (Why?) G CD < C'A'. Ax. 8. But the & ABC and A'B'C' have equal bases. Hyp..these A are to each other as their altitudes, CD and ClIA. Art. 391..A A'B'G' > A ABC. (for C'A' > CD). Qb E. D. 473. Isoperimetric figures are figures having equal perimeters. Ex. 1. Find the minimum line that can be drawn between two given parallel lines. Ex. 2. What is the largest stick that can be placed on a rectangular table 12 x 5 ft., and not have an end projecting over a side of the table? MAXIMA AND MINIMA 287 PROPOSITION XXII. THEOREM 474. Of all isoperimetric tr-iangles which have the same base, the isosceles triangle is the maximium. HI Given the A ABC and ADD~ having the sasme Ibase AB~ and equal perimeters, AC= CD, and AD and DbB uaequal. To prove A ACB > A ADB. Proof. Produce AC to I, makingy CF= AC. Draw FB, and from D as a center, with a radius equal to DB, describe an arc cutting EB produced in G. Draw DG and AG. Draw CHI and DK I AB; also CL and DP I PG. Then Z ABF is a right Z, Art. 261. (for it may be inscribed in a sewicircle whose center is C and whose diameter is ACE). Also ADG is not a straight line, (for-, if it were, the A DAB and DBA wourld be complements of the = A DGB and IBG, respectirvely, and hence wvoldd be equal, aind.'. A DAB wonid be isosceles, which is contrary to the hypothesis). AF - A C + CB = AD ~ DBZ =ZA D + DG. Constr. Hyp. But AD + DG > AG (Art. 92)..-. A F> AG. Ax. 8. BF > B. Art. 111. BF > B DG, or CII > KD. Ax. 10. A ACB > A ADB. Art. 391l. Q. L. ) 288 BOOK V. PLANE GEOMETRY PROPOSITION XXIII. THEOREM 475. Of isoperimetric polygons h7aving the samze number of sides, the maximum is equilateral. B B' A --------------- Given ABCDE the maximum of all polygons having a given perimeter, and a given number of sides. To prove ABCDE equilateral. Froof. If ABCDE is not equilateral, at least two of its sides, as AB and BC, must be unequal, If this is possible, on the diagonal AC as a base, con~ st'ruet a triangle having the same perimeter as ABC, and having side AB'B = C. Then A AB'C > A ABC. Art. 474. To each of these unequals add the polygon A CD)E. AB'CDE > ABCDE. Ax. 9. But this is contrary to the hypothesis that ABCDE is the maximum of the class of polygons considered. Hence AB=BC, and ABCDE is equilateral. Ex. Of all circles which are described on a given line as chord, which is the minimum? MAXIMTA AND MINIMIIIA 289 PROPOSITION XXIV. THEOREM 476. Of all polygons having all sides given buat one, the mnaximtnum can be inscribed in a semicircle having the undetermined side as a diameter. D B -4) Given the polygon ARODEF, the maximum of all polygons having the sides BC, CD, DE, HF, PA, in aommon9 and AD undetermined. To prove that AD is the diameter of a semicircle in which AIBCDEF can be inscribed. Proof, Draw lines from any vertex, E, to A and B. The A BEA must be the maximum of all & haviag the sides BE and. E..A, (for, if it is not, bq increasing or decreasing the angle BEA, the A (l3EA can be chanlged till it is a maximumn, the rest of figure, BCDE and EFA, meantimne remtaining unchanged; thus the arca of the poly, gon ABCDEE 'would be increased, which is contrary to the hypothesis that ABCDEF is a maximumn)..'. ZBEA is a right Z. Art 472. E is on the semicircumference of which AD is the diameter. In like manner the other vertices, C, D and F, must h(e on the semicircumference which has AD for a diameter. Q. F. D,~ S 290 BOOK V. PLANE GEOMETRY PROPOSITION XXV. THEOREMT 477. Of all polygonrs formed writh the same given sides, that Wkich can be inscribed in a circle is the maximnum. B B ~?cx K;~? / Given ABODE a polygon wb ich can be inscribed in a 0, and A'B'CI'D'E a polygon which has the same sides as ABO-DE, but which cannot be inscribed in a 0. To prove ABODE > A'B'C'D'E'. Proof. From any vertex, A, of ABCODE draw the diameter AK, and join K to the adjacent vertices 0 and D. Upon O'D' constrnct the triangle O'K'D' equal to A ODK, and draw A'IK. Then area ABORT> area A'B'OK'K. Art. 416. Also area AEDK > area A'E'D'K. (Why) Adding, ABOKDE > A IB'O'KDI'E'. (Why?) But A OKD=- CAO'KD'. Constr. Subtracting, ABODE > A''CIOD'E'. (Why Q. E. D. 478. NOTE. It might happen that one of the parts of the second figure formed by the diameter A'K', as A'B'C'K', could be inscrihed in a semicircle, and.", = ABCK. How, then, would the above proof be modified?7 479. COR. Of all isoperimetric polygons of a given number of sides the maximum polygon is regular. For it is equilateral (Art. 475), and can be inscribed in a tircle (Art, 477), and is, therefore, =~u1ar (Art, 417). MAXIMA AND MINIMA 291 PROPOSITION XXVI. THEOREM 480. Of two isoperimetric regular polygons, that which has the greater number of sides has the greater area. F I / A'^' 'I AL —.- --- ^\: ---D Given K a regular polygon of any number of sides, as three, and K' a regular polygon of one more, or four sides, and let K and K' have equal perimeters. To prove K' > K. Proof. From any vertex, C, of K, draw a line CD to any point D of the side AB, which meets one of the sides of z/C. Construct the A DCF, having CF=DA, and DF= CA... A DCF= A CDA. Art. 101. Adding A CBD, DFCB 1 K. Ax. 2. Hence the polygon DFCB has the same perimeter as K', and the same area as K. But DFCB is an irregular, while K' is a regular polygon of four sides..'. K > DFCB. Art. 479, K'. ' > K. Ax.. In like manner it Inay be shown that a regular polygon of one more, or five sides, > K, and so on. Q.;. ~. 481. COR. Of isoperimetric planoe figures, the circie is the maximum1. That is, tlhe area of a circle is greater than the area of any polygon with equal perimeter, 292 BOOK V. PLANE GEOMETRY PROPOSITION XXVII. THEOREM 482. Of two equivalent regular polygons that which has the less number of sides has the greater perimeter. --— P —m} K K| H Given K and K' two regular polygons having the same area, and having their perimeters denoted by P and P', but K' having the less number of sides. To prove P' > P. Proof. Let H be a regular polygon having the same number of sides as K', and the same perimeter as K. Then K > H. Art. 480..K. K > H. Ax. 8... P' > P. Art. 399. Q. E. D. Ex. 1. Find the area of a triangle in which two of the sides are 6 and 12 in., and the included angle is 90~. Find the area of another triangle having two sides of 6 and 12 in., and the included angle 60~. Ex. 2. How long is the fence about a garden 60x40 ft.? How many square feet in the area of the garden? Find also the length of fence and area of a garden 50 ft. square. Ex. 3. Find the area of an equilateral triangle, a square, a hexagon, and a circle, in each of which the perimeter is 1 ft. Ex. 4. Find the perimeter of an equilateral triangle, a square, and a circle, in each of which the area is 24 sq. in. Ex. 5. What principle of maxima and minima is illustrated in each of the four preceding Exs. SYMMETRY 293 SYMMETRY 483. Symmetry of polygons. Many of the properties of regular figures can be obtained in a simple and expeditious way by the use of the ideas of symmetry. 484. An axis of symmetry is a line such that, if part of a figure be folded over upon it as an axis, the part folded over will coincide with the remaining part of the figure. EXERCISES. GROUP 46 Ex. 1. How many axes of symmetry has an isosceles triangle? an equilateral triangle? Ex. 2. How many has a square t a regular pentagon Ex. 3. How many has a regular hexagon? a regular heptagon? Ex. 4. How many has a regular octagon? a regular polygon of n sides? Ex. 5. How many has a circle? 485. A center of symmetry for a polygon is a point such that any line drawn through the point and terminated by the perimeter is bisected by the point. EXERCISES. GROUP 47 Ex. 1. Has an equilateral triangle a center of symmetry? Has a square? Ex. 2. Has a regular pentagon a center of symmetry? Has a regular hexagon? Ex. 3. In general, which regular polygons have a center of symmetry, and which do not? Ex. 4. Has a circle a center of symmetry? Ex. 5. Which is the most symmetrical plane igure studied thus far V 294 BOOK V. PLANE GEOM.YET'RY SYMMETRY WITH RESPECT TO A LINE OR AXIS 486. Two points symmetrical with respect to a line or axis are points such that the straight line joining them is bisected by the given line at right angles. Thus, if PP' is bisected by AB, and PP' is I AB, the points P and P/ are symmetrical with respect to the axis AB. 487. A figure symmetrical with respect to an axis is a figure such that each point in the one part of the figure has a point in the other part symmetrical to the given point, with respect to an axis. B 488. Two figures symmetrical with respect Af4 C to an axis are two figures such that each point in the one figure has a point in the other ---- figure symmetrical to the given point, with respect to an axis. A C-; B' SYMMETRY WITH RESPECT TO A POINT OR CENTER 489. Two points symmetrical with respect to a point or center are points such that the straight line joining them is bisected by the, point or center. Thus, if PP' is bisected by the point C, C is a center of symmetry, with respect to P and P'. /' 490. A figure symmetrical with respect,7 to a point or center is a figure such that each point in the figure has another point in the figure symmetrical to the given point with respect to the center., 491. Two figures symmetrical with respect <\. to a center are figures such that each point in Cl ' one figure has a point in the other figure 77 symmetrical to it with respect to the center B SYMMETRY 295 PROPOSITION XXVIII. THEOREM 492. If a figure is symmetrical with respect to two axes which are pperpendicular to each other, it is symmetrical with respect to their point of intersection as a center. Y P P P' 7-;- D Given the figure ABC.. H symmetrical with respect to the two axes XX' and YY'; and XXA' IL lYY and intersecting it at 0. To prove ABC... IH symmetrical with respect to O. Proof. Take any point P in the perimeter of the figure, and determine the points P' and P", symmetrical with respect to P, by drawing PKP' L XX', and PLP" L YYI'. Draw KL, PO', OP". Then PP' I Yr', and PP" II XX/. Art. 121. But PK= KP. Art. 486. Also PK= and II LO. Art. 157. ~. KP/=and I LO. Ax. 1.. KLOP' is a Z7, and KL=and 1| P'O. Art. 160. In like manner it may be shown that KL= and II OP"..'. PO = and [I OP'". Ax. 1, Art. 122.. P'OP" is a straight line bisected by point 0. Geom. Ax. 3. Hence any straight line drawn through 0, and terminated by the perimeter, is bisected at 0, (for P is any point on the perimeter). 0A O is a center of symmetry for the given figure. Art. 490, Q. E. D. 2.96 BOOK V. PLANE GEOMETRY EXERCISES. CROUP 48 THEOREMS CONCERNING REGULAR POLYGONS AND THE CIRCLE Ex. 1. The diagonals of a regular pentagon are equal. Ex. 2. If ABCDE is a regular pentagon, the B triangles ABK and ABC are similar. Ex. 3. In the same figure BC=KC. Ex. 4. Also KCDE is a parallelogram. Ex. 5. Also AC=AB+ BK. Ex. 6. The diagonals of a regular pentagon divide each other in extreme and mean ratio. Ex. 7. The apothem of a regular inscribed triangle equals one-half the radius (r==R). Ex. 8. The altitude of an equilateral triangle equals one and a half times the radius of the cir- \, cumscribed circle. Ex. 9. The altitude of an equilateral triangle is to the diameter of the circumscribed circle as 3: 4. Ex. 10. The side of an equilateral triangle equals ~VB/3, the radius of the circumscribed circle being denoted by i. Ex. 11. The apotheni of a regular inscribed hexagon equals onehalf the side of a regular inscribed triangle (= —/3). Ex. 12. The side of a regular circumscribed triangle is double the side of the regular inscribed triangle. Ex. 13. Find the side of an inscribed square in terms of R?; also the side of a circumscribed square. Ex. 14. Find the apothem of an inscribed square, and also of a circumscribed square, in terms of R. Ex. 15. Find the areas of the inscribed and circumscribed squares in terms of 1?, and show that one of these is double the other. lx. 16. Show that the area of, regular inscribed triangle is3- /3Rq EXERCISES. THEOREMS 297 Ex. 17. Shew that the area of a regular inscribed hexagon is 313R2 What, then, is the ratio of the area of a regular inscribed triangle to that of a regular inscribed hexagon T Ex. 18. The area of a regular inscribed hexagon is three-fourths the area of the regular circumscribed hexagon. Ex. 19. The area of a regular inscribed hexagon is a mean proportional between the areas of a regular inscribed and a regular circumscribed triangle. [SUG. On a figure similar to that of Prop. IV, p. 266, let OP intersect AB in E, and compare the A OiKA, OAP, and OPA'.] Ex. 20. The area of a regular inscribed polygon of 2n sides is a mean proportional between the areas of regular inscribed and circumscribed polygons of n sides. Ex. 21. The area of a regular inscribed octagon equals the area of the rectangle whose base and altitude are the sides of the circum scribed and inscribed squares respectively. Ex. 22. The area of an inscribed regular dodecagon equals three times the square of the radius. Ex. 23. An angle of a regular polygon is the supplement of the angle at the center. Ex. 24. Diagonals drawn from a vertex of a regular polygon of n sides divide the angle at that vertex into n- 2 equal parts. Ex. 25. The diagonals formed by joining the alternate vertices of a regular hexagon form another regular hexagon. Find also the ratio of the areas of the two hexagons. Ex. 26. If squares be erected on the sides of a regular hexagon, the lines joining their exterior vertices form a regular dodecagon. Find also the area of this dodecagon in terms of b, a side of the hexagon. Ex. 27. The square of a side of an inscribed equilateral triangle equals the square of a side of an inscribed square added to the square of a side of an inscribed regular hexagon. Ex. 28. The area of a circle equals four times the area of a circle described on its radius as a diameter, 298 BOOK V. PLANE GEOMETI'RY Ex. 29. The area of a circular ring equals the area of a circle whose diameter is the chord of the outer circle tangent to the inner circle. Ex. 30. Given AaBbC, AcB, BdC, semi- d circles; prove that the sum of the two ores- c cents AcBa and BdCb equals the area of the I / right triangle ABC. A C Ex. 31. An equiangular polygon inscribed in a circle is regular if the number of its sides be odd. [SUG. In the figure to Prop. III, p. 264, arc AEDC=arc, EDCB., are A/E=arc BC,.'. side AE=side BC, etc.] EXERCISES. GROUP 49 MAXIMA AND MINIMA Ex. 1. Of isoperimetric parallelograms with the same base the rectangle is the maximum. Ex. 2. Of equivalent parallelograms with the same base, which has the minimum perimeter? Ex. 3. Of isoperimetric rectangles which is the maximum? Ex. 4. Divide a given line into two parts such that their product is a maximum. Ex. 5. Find a point in the hypotenuse of a right triangle such that the sum of the squares of the perpendiculars drawn from the point to the legs shall be a minimum. Ex. 6. How shall a mile of wire fence be stretched so as to contain the maximum area? Ex. 7. Find the area in acres included by a mile of wire fence if it be stretched as a square, a regular hexagon, and a circle respectively. Ex. 8. Of all triangles with the same base and equal altitudes, the isosceles triangle has the least perimeter. Ex. 9. Of all polygons of a given number of sides inscribed in a given circle, the maximum is regular. [SUG. Prove the maximum polygon (1) equilateral, (2) equiangular, EXERCISES. SYMMETRY 299 Ex. 10. Find the maximum rectangle inscribed in a circle. Ex. 11. Find the maximum rectangle that can be inscribed in a semicircle. [SUG. Inscribe a square in the circle.] Ex. 12. Of trapezoids inscribed in a semicircle (having the diameter as one base), find the maximum. Ex. 13. Divide a given straight line into two parts such that the sum of the squares of these parts shall be a minimum. EXERCISES. CROUP B0 SYTMMETRY Ex. 1. A rhombus has how many axes of symmetry? Has it a center of symmetry? Ex. 2. What axis of symmetry has a quadrilateral which has two pairs of equal adjacent sides? Has such a figure a center of symmetry? Ex. 3. A parallelogram is symmetrical with respect to the point of intersection of its diagonals. Ex. 4. A segment of a circle is symmetrical with respect to what axis? Ex. 5. Has a trapezium a center of symmetry? An axis of symmetry? Ex. 6. How many axes of symmetry have two equal circles taken as one figure? Have they a center of symmetry? Ex. 7. W'hat axis of symmetry have any two circles? Ex. 8. How must two equilateral triangles be placed so as to have a center of symmetry? So as to have an axis of symmetry? Ex. 9. If two polygons are symmetrical with reference to a center, any two homologous sides are equal and parallel and drawn in opposite directions. 300 B)OOK V. PL'ANE GEOMETtYl EXERCISE8. GROUP 8u VARIOUS THEOREMS Ex. 1. The square on the diameter of a circle equals twice the square inscribed in the circle. Ex. 2. The altitude of an inscribed equilateral triangle is to the radius of the circle as 3: 2. Ex. 3. The diagonal joining any two opposite vertices of a regular hexagon passes through the center. Ex. 4. The radius of an inscribed regular polygon is a mean proportional between its apothem and the radius of the circumscribed regular polygon of the same number of sides. Ex. 5. If the sides of a regular hexagon be produced, their points of intersection are the vertices of another regular hexagon. Also find the ratio of the areas of the two hexagons. Ex. 6. Of all lines drawn through a given point within an angle, and terminated by the sides of the angle, the line which is bisected at the given point cuts off the minimum area. Ex. 7. Each angle of a regular polygon of n + 2 sides contains 2 right angles. Ex. 8. The diagonals from a vertex of a regular polygon of - +- 2 sides divide the angle at that vertex into n equal parts. Ex. 9. The sum of the perpendiculars drawn to the sides of a regular polygon of n sides from any point within the polygon equals n times the apothem. Ex. 10. An equiangular polygon circumscribed about a circle is regular. [Suo. Draw radii from the points of contact, and lines from the vertices of the polygon to the center.] Ex. 11. An equilateral polygon circumscribed about a polygon is regular if the number of its sides is odd. [SUG. See Figure of Prop. III, p. 264. Prove AT=BQ. Draw radii and prove Z T= Z Q, etc.] EXERCISES. MISCELLANEOUS THEOREMS 301 Ex. 12. How must two equal isosceles triangles be placed so as to have a center of symmetry? How must they be placed so as to have an axis of symmetry? If, in a regular inscribed polygon of n sides, Sn denotes a side and rn denotes the apothem, show that Ex. 13. For the triangle, S3=R-V3, r3=iR. Ex. 14. For the square, S4=RV2, r4=RV/2. Ex. 15. For the hexagon, Se=R, r6=4R-/3. R(5 5- ) RV/1O +_2/5 Ex. 16. For the decagon, Sio=, r,1 = 2 4 JR/10 —21/5, _R(V/5+-l). Ex. 17. For the pentagon, S5=- 2 -V r5- ( + 2 4 Ex. 18. For the octagon, S8=R1/2 —/2. Ex. 19. For the dodecagon, S12 = R1/2 —/3o Ex. 20. Prove that S52=R2+Sio2. Ex. 21. Prove that S52=S2 + S102. Ex. 22. If ADB, AaC, CbB are semicir- cles and DC JL AB, prove that the area bounded by the three semicircumferences equals the area described on DC as a diameter. C B Ex. 23. If p,, denotes the perimeter of an inscribed polygon of n sides and P, the perimeter of a circumscribed polygon of n sides, 2pt P, prove that Pn,, -P ~-PR e \-+ '^s 302 BOOK V. PLANE GEOMETRY EXERCISES. QRQOUP S2 PROBLEMS Circumscribe about a given circle Ex. 1. An equilateral triangle. Ex. 2. A square. Ex. 3. A regular pentagon. Ex. 4. A regular hexagon. Ex. 5. A regular octagon. Ex. 6. A regular decagon. Ex. 7. Construct a regular pentagram, or 5-pointed star. Ex. 8. Construct a hexagram, or 6-pointed star. Ex. 9. Construct an 8-pointed star. Ex. 10. Construct an angle of 36~. Ex. 11. Construct angles of 18~, 9~, 72~. Ex. 12. Construct angles of 24~, 12~, 60, 48~. Ex. 13. Divide a given circumference into two parts which shall be in the ratio of 3: 7. Ex. 14. Construct a regular pentagon which shall have twice the perimeter of a given regular pentagon. Ex. 15. Construct a regular pentagon whose perimeter shall equal the sum of the perimeters of two given regular pentagons. Ex. 16. Construct a regular pentagon whose area shall be twice the area of a given regular pentagon, Ex. 17. Construct a regular pentagon whose area shall be equal to the sum of the areas of two given regular pentagons. Ex. 18. Construct a circumference which shall be twice the cirOumference of a given circle. EXERCISES. PROBLEMS 303 Ex. 19. Construct a circle whose area shall be three times the area of a given circle. Ex. 20. Construct a circumference equivalent to the sum, and another equivalent to the difference, of two given circumferences. Ex. 21. Construct a circle equivalent to the sum, and another equivalent to the difference, of two given circles. Ex. 22. Construct a circle whose area shall be two-thirds the area of a given circle. Ex. 23. Bisect the area of a given circle by a concentric circumference. Ex. 24. Divide the area of a given circle into five equal parts by drawing concentric circumferences. On a given line construct Ex. 25. A regular pentagon. Ex. 26. A regular hexagon. Ex. 27. A regular dodecagon. Ex. 28. A circle equivalent to a given semicircle. Ex. 29. Inscribe a regular octagon in a given square. Ex. 30. Inscribe a circle in a given sector. Ex. 31. Inscribe a square in a given segment. Ex. 32. In a given equilateral triangle inscribe three equal circles, each of which touches the other two circles and a side of the triangle. Ex. 33. In a given circle inscribe three equal circles which shall touch each other and the given circumference. NUMERICAL APPLICATIONS OF PLANE GEOMETRY METHODS OF NUMERICAL COMPUTATIONS 493. Cancellation, In numerical work in geometry, as elsewhere, the labor of computations may frequently be economized. Those methods of abbreviating work, which are particularly applicable in the ordinary numerical applications of geometry, may be briefly indicated, as follows: To simplify numerical work by cancellation, group together as a twhole all the numerical processes of a given problem, and make all possible cancellations before proceed~ iag to a final numerical reduction. Ex. Find the ratio of the area of a rectangle, whose base and alti. tude are 42 and 24 inches, to the area of a trapezoid, whose bases are 21 and 35 and altitude 12. By Arts. 383, 394 3 area of rectangle_ 42X24 _<X 3 R X~ 3, Ratio. area of trapezoid 6(21 + 35) -~X5 a 494. Use of radicals and of 7t. Where radicals enter in the course of the solution of a numerical problem, it frequently saves labor not to extract the root of the radical till the final answer is to be obtained. Ex. 1. Find the area of a circle circumscribed about a square whose side is 8. The diagonal of the square must be 81/2 (Art. 346)..'. the radius of 0 =4-/2.,. by Art. 449, area of 0 = -r(41/2)2=327r=100.6, Area. (304) NUMERICAL COMPUTATIONS 305 Similarly in the use of 7, it frequently saves labor not to substitute its numerical value for 7 till late in the process of solution. Ex. 2. Find the radius of a circle whose area is equal to the sum of the areas of two circles whose radii are 6 and 8 inches, respectively. Denote the radius of the required circle by x. Then, by Art. 449, r x' = 36 rr + 64 7r..'. O = 100r. '. = 100, and x=10, Radius. 495. Use of x, y, etc., as symbols for unknown quantities. In some cases a numerical computation, is greatly facilitated by the use of a specific symbol for aJ n unknowln quantity. Ex. In a triangle whose sides are 12, 18, and 25, find the segments of the side 25 made by the bisector of the angle opposite. Denote the required segments of side 25 by x and 25-x. Then 12: 18=x: 25- x (Art. 332).. 18x=12 (25-x) (Art. 302) / 2 x=10.) And 25 -x=15. Segnments. 2 496. Limitations of numerical computations. Owing to the limitations of human eyesight and of the instruments used in making measurements, no measurement can be accurate beyond the fifth or sixth figure; and in ordinary work, such as is done by a carpenter, measurements are not accurate beyond the third figure. As all numerical applications of geometry are based on practical measurements, it is.not necessary to carry arithmetical work beyond the fifth or sixth significant digit. Other methods of facilitating numerical computations, as by the use of logarithms, are beyond the scope of this book. T 306 PLANE GEOCMETRY NUMERICAL PROPERTIES OF LINES EXERCISES. CROUP i3 THE RIGHT TRIANGLE Ex. 1. Find the hypotenuse of a right triangle whose legs are 12 and 35. Ex. 2. The hypotenuse of a right triangle is 29, and one leg is 20. Find the other leg. Ex. 3. If a window is 15 ft. from the ground and the foot of a ladder is to be 8 feet from the house, how long a ladder is necessary to reach the window? Ex. 4. Find the diagonals of a rectangle whose sides are 5 and 12. Ex. 5. If the base of an isosceles triangle is 8 and a leg is 5, find the altitude. Ex. 6. Find the diagonal of a square whose side is 1 ft. 6 in. Ex. 7. The diagonals of a rhombus are 24 and 10. Find a side. Ex. 8. One side of a rhombus is 17 and one diagonal is 30. Find the other diagonal. Ex. 9. In a circle whose radius is 5, find the length of the longest and shortest chords through a point at a distance 3 from the center. Ex. 10. In a circle whose radius is 25 in., find the distance from tile center to a chord 48 in. long. Ex. 11. If a chord 12 in. long is 5 in. from the center of a circle, how far from the center of the same circle is a chord 10 in. long? Ex. 12. A ladder 40 ft. long reaches a window 20 ft. high on one side of a street and, if turned on its foot, reaches a window 30 ft. high on the other side. How wide is the street? Ex. 13. If one leg of a right triangle is 10 and the hypotenuse is twice the other leg, find the hypotenuse. Ex. 14. Find the altitude of an equilateral triangle whose side is 6. Ex. 15. Find the side of an equilateral triangle whose altitude is 8, NUMERICAL EXERCISES. LINES 307 Ex. 16. Find the side of a square whose diagonal is 15. Ex. 17. One leg of a right triangle is 3, and the sum of the hypotenuse and the other leg is 9. Find the sides. Ex. 18. A tree 90 ft. high is broken off 40 ft. from the ground. low far from the foot of the tree will the top strike? Ex. 19. The radii of two circles are 1 and 6 in., and their centers are 13 in. apart. Find the length of the common external tangent. Ex. 20. The sides of a triangle are 10, 11, 12. Find the length of the projection of the side whose length is 10, on the side 12. EXERCISES. GROUP b4 TRIANGLES IN GENERAL Ex. 1. The sides of a triangle are 12, 18 and 20. Find the segments of the side 20, made by the bisector of the angle opposite. Ex. 2. In the same triangle, find the segments of the side 20, made by the bisector of the exterior angle opposite. Ex. 3. If the legs of a right triangle are 6 and 8, find the hypotenuse, the altitude on the hypotenuse, and the projections of the legs on the hypotenuse. Ex. 4. Is a triangle acute, obtuse, or right, if the three sides are 5, 12, 14; 5, 11, 12; 5, 12, 13; 4, 5, 6? Ex. 5. If the sides of a triangle are 6, 7 and 8, compute the length of the altitude on 8. Ex. 6. Also the length of the median on the same side. Ex. 7. Also the length of the bisector of the angle opposite the side 8. Ex 8. If two sides and a diagonal of a parallelogram are 8, 12 ad 10, find the other diagonal. [SUG. Use Art. 352.] 308 PLANE GEOMETRY Ex. 9. Two sides of a triangle are 10 and 18, and the median to the third side is 12. Find the third side. Ex. 10. Two sides of a triangle are 17 and 16, and the altitude on the third side is 15. Find the third side. Ex. 11. The hypotenuse of a right triangle is 10, and the altitude on the hypotenuse is 4. Find the segments of the hypotenuse and the legs. Ex. 12. Find the three medians, the three bisectors, and the three altitudes of a triangle whose sides are 13, 14, 15. EXERCISES. CROUP 51 CIRCUMFERENCES AND ARCS Using r 2-2 7 Ex. 1. Find the circumference of a circle whose radius is 1 ft. 9 in. Ex 2. Find the radius of a circle whose circumference is 121 ft. Ex. 3. A bicycle wheel 28 in. in diameter makes, in an afternoon, 3,000 revolutions. How far does the bicycle travel? Ex. 4. What is the diameter of a wheel which makes 1,400 revolutions in going 8,800 yds.? Ex. 5. If the diameter of a circle is 20, find the length of an arc of 60~; also of 83~. Ex. 6. If the length of an are is 14 and the radius is 6, find the number of degrees in the arc. Ex. 7. If the are of a quadrant is 1 ft. in length, find the diameter. Ex. 8. Two concentric circumferences are 88 and 132 in. in length, respectively. Find the width of the circular ring between them. Ex. 9. If the year be taken as 365i da., and the earth's orbit a circle whose radius is 93,250,000 miles, find the velocity of the earth in its orbit per second, NUMERICAL EXERCISESo LINES 309 Find the radius and circumference of a circle circumsdribed about Ex. 10. A square whose side is 5. Ex. 11. An equilateral triangle whose side is 4. Ex. 12. A rectangle whose sides are 12 and 5. Ex. 13. Find the central angle of a sector whose perimeter equals one-half the circumference. Ex. 14. Find the diameter of a circle circumscribed about a triangle whose sides are 7, 15, and 20. Ex. 15. Find the radius of a circle whose circumference equals the perimeter of a square whose diagonal is 10. EXERCISES. CROUP b@ CHORDS, TANGENTS, AND SECANTS Ex. 1. Two intersecting chords of a circle are 11 and 14 in., and the segments of the first chord are 8 and 3 in. Find the segments of the second chord. [SUG. Denote the required segments by x and 14- x.] Ex. 2. In a circle whose radius is 12 in., a chord 16 in. long is passed through a point 9 in. from the center. Find the segments of the chord. Ex. 3. Two secants drawn from a point to a circle are 24 and 27 in. long. If the external segment of the first is 6 in., find the external segment of the second. Ex. 4. From a given point a secant whose external and internal segments are 9 and 16 is drawn to a circle. Find the length of the tangent drawn from the same point to the circle. Ex. 5. From a given point a tangent 24 in. long is drawn to a circle whose radius is 18 in. Find the distance of the point from the center. Ex. 6. If a diameter 60 in. long is divided into 5 equal parts by Chords perpendicular to it, find the length ot he chrtis, 310 PLANE GEOMETRY Ex. 7. If a mountain 3 miles high is visible 150 miles at sea, what is the diameter of the earth? Ex. 8. If the earth is a sphere of radius 4,000 miles, how far will the light of a lighthouse 100 ft. high be visible at sea? EXERCISES. CROUP B7 LINES IN SIMILAR FIGURES Ex. 1. If the sides of a triangle are 6, 7 and 8, and the shortest side of a similar triangle is 18, find the other sides of the second triangle. Ex. 2. If a post 5 ft. high casts a shadow 3 ft. long, find the height of a steeple which casts a shadow 90 ft. long. Ex. 3. In a triangle whose base is 14 and altitude 12, a line is drawn parallel to the base and at a distance 2 fromi the base. Find the length of the line thus drawn. Ex. 4. The upper and lower bases of a trapezoid are 12 and 20 and the altitude is 8. If the legs are produced till they meet, find the altitude of each of the two triangles thus formed. Ex. 5. If the upper and lower bases of a trapezoid are bl and b) and the altitude is h, find the altitude of each of the triangles formed by producing the legs. Ex. 6. If the perimeters of two similar polygons are 300 and 400 and a side of the first is 27, find the homologous side of the second. Ex. 7. If the perimeter of a regular polygon is three times the perimeter of a regular polygon of the same number of sides, what is the ratio of their apothems? Ex. 8. If the circumferences of two circles are 600 and 400 ft., what is the ratio of their diameters I Ex. 9. In the preceding example, if a chord of the first circle is 30, what is the length of a chord in the second circle, subtending the same number of degrees of arc? NUMERICAL EXERCISES. AREAS 311 COMPUTATION OF AREAS EXERCISE8. GROUP c8 AREAS OF TRIANGLES Ex. 1. Find the area in acres of a triangular field whose base is 300 ft. and altitude 200 ft. Ex. 2. Find the area of a triangle whose sides are 10, 17, and 21. Ex. 3. Find the area in acres of a field whose sides are 60, 70, and 80 chains. Ex. 4. Find the area in acres of a triangular field each of wlb'se sides is 10 chains. Find the area of Ex. 5. An isosceles triangle whose base is 16, and each of whose legs is 34. Ex. 6. An equilateral triangle whose altitude is 8. Ex. 7. A right triangle in which the segments of the hypotenuse made by the altitude upon it are 12 and 3; also, in one in which the segments are a and b. Ex. 8. An isosceles right triangle whose hypotenuse is 12. Ex. 9. A right triangle in which the hypotenuse is 41 and one leg is 9. Ex. 10. Find in two ways the area of a triangle whose sides are 6, 5, 5. Ex. 11. A side of a given equilateral triangle is 4 ft. longer than the altitude. Find the area of the triangle. Ex. 12. The area of an isosceles triangle is 144 and a leg is 24. Find the base. Ex. 13. The area of an equilateral triangle is 4/3. Find a side. Ex. 14. The area of a triangle is 1125, and a c:: c=-'::34, Find a, b, c. Ex. 15. The area of a triangle is 6 sq. in., and two of its sides are 3 and 5 in. Find the remaining side. 312 PLANE GEOMETRY EXERCIGE$. CROUP a9 AREAS OF OTHER RECTILINEAR FIGURES Find the area of Ex. 1. A parallelogram whose base is 24 ft. 6 in. and whose altttude is 12 ft. 9 in. Ex. 2. A trapezoid whose bases are 12 and 20 in. and whose altitude is 1i ft. Ex. 3. A rhombus whose diagonals are 9 ft. and 2 yds. Ex. 4. A quadrilateral in which the sides AB, BC, CD, DA are 12, 13, 14, 15 and the diagonal AC is 17. Ex. 5. A quadrilateral in which the sides are 27, 36, 30, 25 and the angle included between the first two sides is a right angle. Ex. 6. A square whose diagonal is 12 in. Ex. 7. Find the number of boards, each 4 yds. long and 6 in. wide, which are necessary to cover a floor 48 X 24 ft. Ex. 8. How many persons can stand in a room 15 X 9 ft., if each person requires 27 X 18 in.? Ex. 9. The baseball diamond is a square each side of which is 90 ft. What fraction of an acre is its area? Ex. 10. A rectangular garden contains 4,524 sq. yds. and is 20 yds. longer than wide. Find its dimensions. Ex. 11. Each side of a rhombus is 24 ft. and each of the larger angles is double a smaller one. Find the area. Ex. 12. Find the area of a rhombus one of whose sides is 17, and one of whose diagonals is 30. Ex. 13. The area of a trapezoid is 4 acres, one base is 120 yds., and the altitude is 100 yds. Find the other base. Ex. 14. The bases of an isosceles trapezoid are 20 and 36 and the legs are 17. Find the area. NUMERICAL EXERCISES, AREAS 313 Ex. 15. The base of a triangle is 20 and the altitude 18. Find the length of a line parallel to the base which cuts off a trapezoid whose area is 80 sq. ft. [Sue. Denote the altitude of trapezoid by 18 —x and find its "pper base by similar triangles.] Ex. 16. The perimeter of a polygon, circumscribed about a circle Whose radius is 20, is 340. Find the area of the polygon. Ex. 17. The area of a rectangle is 144 and the base is three times the altitude. Find the dimensions. Ex. 18. Find the area of a regular hexagon one of whose sides is 10. Ex. 19. Find the area of a regular decagon inscribed in a circle whose radius is 20. Ex. 20. Find a side of a regular hexagon whose area is 200 sq. in. EXERCISES. GROUP 60 AREAS OF CIRCULAR FIGURES Ex. 1. Find the area in acres of a circle whose radius is 100 yds. Ex. 2. Find the radius in inches of a circle whose area is 1 sq. yd. Ex. 3. Find the area of a circle whose circumference is p. Ex. 4. Find the radius of a circle whose area equals the sum of the areas of two circles whose radii are 9 and 40 in. Ex. 5. Find the radius of a circle whose area equals the sum of the areas of three circles whose radii are 20, 28, 29. Ex. 6. In a circle of radius 50 find the area of a sector of 80~. Ex. 7. Also of a segment of 60~; of a segment of 300~; of a segment of 240~. Ex. 8. In a circle whose radius is 7, the area of a sector is 45 sq. ft. Find the number of degrees in its angle. Ex. 9. In a circle whose radius is 10, find the sum of the segments formed by an inscribed square, 314 PLANE GEOiMETET Ex. 10. A circular mill-pond, 3 mile in diameter, contains a circu. lar island, 10() yds. in diameter. Find the water surface of the pond in acres. Ex. 11. The same pond is surrounded by a driveway 30 ft. wide. Find the area of the driveway. Ex. 12. Two tangents to a circle, whose radius is 15, include an angle of 60~. Find the area included between the tangents and tile radii to the points of contact. Ex. 13. Find the length of the tether by which a cow must be tied, in order to graze over exactly one acre. Ex. 14. Three equal circles touch each other externally. Show '7r that the area included between them is R2(- 3 -,-)' Ex. 15. If the area included between three equal circles which touch each other externally is a square foot, find the radius of each circle in inches. EXERCISES. GROUP Bn AREAS OF SIMILAR FIGURES Ex. 1. The homologous sides of two similar triangles are 3 and 5. Find the ratio of their areas. Ex. 2. The homologous sides of two similar polygons are 4 and 7, and the area of the first polygon is 112. Find the area of the second polygon. Ex. 3. The radius of a circle is 6. Find the radius of a circle having three times the area of the given circle. Ex. 4. The areas of two circles are as 16 to 9, and the radius of the first is 8. Find the radius of the second. Ex. 5. The sides of a triangle are 5, 6, 7. Find the sides of a similar triangle containing 9 times the area of the given triangle. Ex. 6. If, in finding the area of a circle, a student uses D=50 as R=50, how will the area as computed differ from the correct area? MISCELLANEOUS NUMERICAL EXERCISES 315 Ex. 7. In a triangle whose base is 24 in. and altitude is 18 in., the altitude is bisected by a line parallel to the base. Find the area of the triangle cut off. Ex. 8. In the triangle of Ex. 7, what part of the altitude must be cut off in order that the area of the triangle be bisected? Ex. 9. In a circle whose diameter is 30 in., what are the diameters of concentric circumferences which divide the area into three equivalent parts? Ex. 10. If a circle be constructed on the radius of a given circle, and segments, one in each circle, be formed by a line drawn from the point of contact, find the ratio of the segments. EXERCISES. CROUP 62 GENERAL NUMERICAL EXERCISES IN PLANE GEOMETRY Ex. 1. The leg of an isosceles triangle is 10 and the base is 16. Find the altitude and the area. Ex. 2. Find the area of a triangle whose sides are 25. 39, 40. Also find the radius of a circle equivalent to this triangle. Ex. 3. Find the area of a regular hexagon inscribed in a circle whose radius is 2. Ex. 4. The sides of a triangle are 7, 8 and 9 inches. Find the sides of a triangle of four times the area. Also, of twice the area. Ex. 5. The temple of Herod is said to have accommodated 210,000 people at one time. If each person required 27X IS in., and one-third the space inside the temple be allowed for walls, sanctuaries, etc., what were the dimensions of the temple, if it was a square? Ex. 6. If the sides of a triangle are 12, 16 and 21, what are the segments of the side 21 made by the bisector of the angle opposite? Ex. 7. The sides of a quadrilateral in order are 5, 5, 4, 3, and the first two of these sides contain an angle of 60~. Find the area. Ex. 8. Find the diameter of a wheel which, in a mile, makes 480 revolutions. 316 PLANE GEOMETRY Ex. 9. The area of a trapezoid is 112 and the two bases are 12 and I6. Find the altitude. Find the radius of a circle equivalent to Ex. 10. A square whose side is 10. Ex. 11. An equilateral triangle whose side is 12. Ex. 12. A trapezoid whose bases are 16 and 18 and altitude 9. Ex. 13. A semicircle whose radius is 15. Ex. 14. Find the diameter of a circle whose area shall be equivalent to the sum of two circles whose diameters are 144 and 17. Ex. 15. A circle, a square, and an equilateral triangle each have a perimeter of 12 yds. Find the area of each figure. Ex. 16. In a circle whose area is 400, the area of a sector is 125. Find the angle of the sector. Ex. 17. How many acres are included within a half-mile running track, if the track is in the shape of a rectangle twice as long as it is wide? Ex. 18. In a square whose side is 6 in., find the area of the inscribed and of the circumscribed circles. Ex. 19. Find the area of the circle circumscribed about a rectangle whose sides are 40 and 9. Ex. 20. One leg of a right triangle is 12, and the difference between the hypotenuse and the other leg is 8. Find the area. Ex. 21. Find the area of an isosceles right triangle whose hypotenuse is 20 ft. Ex. 22. In a triangle whose sides are 16, 18, 20, find the length of the altitude, median, and bisector of the angle opposite the longest side. Ex. 23. A line 16 inches long is divided internally in the ratio of 3:5; find the segments. Also find the segments when the line is divided externally in the same ratio. Ex. 24. A line 16 inches long is divided in extreme and mean ratio, Find the segments. MISCELLANEOUS NUMERICAL EXERCISES 317 Ex. 25. If a line is divided in extreme and mean ratio and the smaller segment is 4, find the whole line. Ex. 26. A triangle whose altitude is 20 is bisected by a line parallel to the base. Into what segments is the altitude divided t Ex. 27. Find the area of a square inscribed in a circle whose radius is 5. Ex. 28.. In a circle whose diameter is 20, a chord is passed through a point at a distance 6 from the center, perpendicular to the diameter through that point. Find the length of this chord, and of the chords drawn from its extremities to the ends of the diameter. Ex. 29. Each leg of an isosceles trapezoid is 10, and one base exceeds the other by 16. Find the altitude. Ex. 30. If three arcs, each of 60~ and having 10 for a radius, are each concave to the other two, find the area included by them. Ex. 31. Find the area of a trapezoid whose legs are 4 and 5, and whose bases are 8 and 11. Ex. 32. A square piece of land and a circular piece each contain 1 acre. How many more feet of fence does one require than the other? Ex. 33. If the base of a triangle is doubled and the altitude remains unchanged, how is the area affected? If the altitude is doubled and the base remains unchanged? If both the base and the altitude are doubled? Ex. 34. The side of an equilateral triangle is 12; find the area of the inscribed, and of the circumscribed circle. Ex. 35. In a given circle a chord of 60~. is 16. Find the chord of 1200. Also of 30~. Ex. 36. In a given triangle, equivalent to a rectangle whose sides are 40 and 20, the base is 32. Find the altitude. Ex. 37. Find the side of an equilateral triangle equivalent to a circle whose diameter is 10. Ex. 38. The area of a rhombus is 156 sq. in., and one side is 1 ft. l in. Fiud the diaganalB. 318 PLANE GEOMETRY Ex. 39. The sides of a triangle are 8, 10, 12. Find the areas of the triangles made by the bisector of the angle opposite the side 12. Ex. 40. In a circle of area 275 sq. ft., a rectangle of area 150 sq. ft. is inscribed. Show how to find the sides of the rectangle. EXERCISES. CROUP 63 EXERCISES INVOLVING THE METRIC SYSTEM Ex. 1. Find the area of a triangle of which the-base is 16 dm. and the altitude 80 cm. Ex. 2. Find the area of a triangle whose sides are 6 m., 70 dm., 800 cm. Ex. 3. Find the area in square meters of a circle whose radius is 14 dmn. Ex. 4. If the hypotenuse of a right triangle is 17 dm. and one leg is 150 cm., find the other leg and the area. Ex. 5. If the circumference of a circle is 1 m., find the area of the circle in square decimeters. Ex. 6. Find the area in hectares, and also in acres, of a circle whose radius is 100 m. Ex. 7. If the diagonal of a rectangle is 35 dm. and one side is 800 mm., find the area in square meters, and also in square inches. Ex. 8. Find the area of a trapezoid whose bases are 600 cm. and 2 m., and whose altitude is 80 dm. Ex. 9. If a rectangular field is 700 dm. long and 200 m. wide, find its area in hectares and in acres. Ex. 10. In a given circle two chords, whose lengths are 15 di. and 13 dm., intersect. If the segments of the first chord are 12 dm. and 3 dm., find the segments of the second chord. Ex. 1. Find in decimeters the radius of a circle equivalent to a square whose side is 1 ft. 6 in. Ex. 12. Find in feet the diameter of a wheel which, in going 10 kilometers, makes 5,000 revolutions. SOLID GEOMETRY BOOK VI LINES, PLANES AND ANGLES IN SPACE DEFINITIONS AND FIRST PRINCIPLES 497. Solid Geometry treats of the properties of space of three dimensions. Many of the properties of space of three dimensions are determined by use of the plane and of the properties of plane figures already obtained in Plane Geometry. 498. A plane is a surface such that, if any two points in it be joined by a straight line, the line lies wholly in the surface. 499. A-plane is determined by given points or lines, if no other plane can pass through the given points or lines without coinciding with the given plane. 500. Fundamental property of a plane in space. A plane is determined by any three points not in a straight line. For, if through a line con-.c necting two given points, A A and B, a plane be passed, the plane, if rotated, can pass through a third given point, C7, in but one position. B The importance of the above principle is seen from the fact that it reduces an unlimited surface to three points, thus making a vast economy to the attention. It also enables us to connect different planes, and treat of their properties systematically. (319) 320 BOOK VI. SOLID GEOMETRY 501. Other modes of determining a plane. A plane may also be determined by any equivalent of three points not in a straight line, as by a straight line and a point outside the line; or by two intersecting straight lines; or by two parallel straight lines. It is often more convenient to use one of these latter methods of determining a plane than to reduce the data to three points and use Art. 500. 502. Representation of a plane in geometric figures. In reasoning concerning the plane, it is often an advantage to have the plane represented in all directions. Hence, in drawing a geometric figure, a plane is usually represented to the eye by a small parallelogram. This is virtually a double use of two intersecting lines, or of two parallel lines, to determine a plane (Art. 501). 503. Postulate of Solid Geometry. The principle of Art. 499 may also be statel as a postulate, thus: Through any three points not in a straight line (or their equivalent) a plane may be passed. 504. The foot of a line is the point in which the line intersects a given plane. 505. A straight line perpendicular to a plane is a line perpendicular to every line in the plane drawn through its foot. A straight line perpendicular to a plane is sometimes called a normal to the plane. LINES AND PLANES 321 506. A parallel straight line and plane are a line and plane which cannot meet, however far they be produced. 507. Parallel planes are planes which cannot meet, however far they be produced. 508. Properties of planes inferred immediately. 1. A straight line, not in a gi/ren plane, can intersect the given plane in but one poinet. For, if the line intersect the given plane in two or more points, by definition of a plane, the line must lie in the plane. Art. 498. 2. The intersection of two planes is a straight line. For, if two points common to the two planes be joined by a straight line, this line lies in each plane (Art. 498); and no other point can be common to the two planes, for, through a straight line and a point outside of it only one plane can be passed. Art. 501. Ex. 1. Give an example of a plane surface; of a curved surface; of a surface, part plane and part curved; of a surface composed of different plane surfaces. Ex. 2. Four points, not all in the same plane determine how many different planes? how many different straight lines? Ex. 3. Three parallel straight lines, not in the same plane, determine how many different planes? Ex. 4. Four parallel straight lines can determine how many different planes? Ex. 5. Two intersecting straight lines and a point, not in their plane, determine how many different planes? 32 BOOK VI. SOLID GEOMETRY PROPOSITION I. THEOREMS 509. If a straight line is perpendicular to each of two other straight lines at their point of intersection, it is perpendicular to the plane of those lines. A t!it F Proof. Through B draw BG, any other line in the plane MN. Draw any convenient line CD intersecting BC, BG and BD in the points C, G and D, respectively. Produce the line AB to F, making BF= AB. Connect the points C, G, D with A, and also with F. Then, in the A A CD and FCD, CD= CD. Ident. A C= F, and AD =DF. Art. 112... A ACD=A FCD. (Why?) /. Z ACD=Z FCD. (Why?) Then, in the A ACG and FCG, OG= CG, (Why t) AC= CF, and Z AGG = Z FCG. (Why?). A G=- A FGG. (Why t) LINES AND PLANES 323 AG= GF. (Why?).: B and G are each equidistant from the points A and F. B.. BG is L AF; that is, AB L BG. Art. 113..'. AB 1 plane M7IY, Art. 505. (for it is L any line, BG, in the plane MV, through its foot). 'Q. E. D. PROPOSITION 1I. THEOREM 510. All the perpendiculars that can be drawn to a given line at a given point in the line lie in a plane perpendicular to the line at the given point. rI B -\ P Given the plane M~ and the line BC both I line AB at the point B. To prove that BC lies in the plane M~N. Proof. Pass a plane AF through the intersecting lines AB and BC. Art. 503. This plane will intersect the plane MN in a straight line BF. Art. 508, 2. But AB J plane Ml (Hyp.).-. AB 1 BF. Art. 505. Also AB 1 BC. Hyp.. in the plane AF, BC and BF 1 AB at B. oB. B and BF coincide. Art. 71. But BF is in the plane MN.:..BC must be in the plane MTN, (for BC coilcides witih B 't whchc lies in the plane MX). Q, B; o0 0,)4 0_, BOO34 0SOI. 20LD GE31 F0FA7I 5 1. Con. 1. At it guiti point B t thie sh Stylih t tine A-B, to construct a plane perpevdicular to thre line A4 B3. Pass a plane AF through AB in any convenient di' ieion, and in thje plane AF at the point B constrnut _BF I ABIj' (Art. 274). Pass another plane through AB, and in it construct BP I AB. Through the lines BE and BP pass the plane A J15 (Art. 503). MY is the required plane (Art. 509). 512. CoR. 2. Thraough a given external point, P, to pass a plane perpendicular to a, given line, AB. Pass a plane through AB and P (Art. 503), and in this plane draw PB I AB (Art. 273). Pass another plane thronuh A4T B, as AT, and in AF draw BF I AB at B (Art. 274). Pass a plane through BP and BE (Art. 503). This will be the plane required (Art. 509). 513. COR. 3. Through lta given point but one plane can be passed perpendicular to a given line. Ex. 1. Five points, no four of which are in the same plane, determine how many different planes? how many different straight lines? Ex, 2. A straight line and two points, not all of which are in the same plane, determine how many different planes? Ex. 3. In the figure, p. 322, prove that the triangles GAD and GDF are equal. Ex. 4. In the same figure, if AB=;8 and BC=6, find FfC. LINES AND PLANES 325 PROPOSITION III. PROBLEM 514. At a given point in a plane, to erect a perpendicular to the plane. ---------------------- I K' - TtiT Iv. - t Given the point A in plane MIN. To construct a line perpendicular to MN at the point A. Construction. Through the point A draw any line CD in the plane I1N. Also through the point A pass the plane PQ L CD (Art. 511), intersecting the plane MN in the lineRS. Art. 508, 2. In the plane PQ draw AK I line RS at A. Art. 274. Then AK is the 1- required. Proof. CD I plane PQ. Constr.,. CD) A K. Art. 505. Hence A K I C7D. But AK I- -S. Constr..'. AK I - plane MY. Art. 509. Qo. F. 515. COR. At a Yic(,n point 'in a plane but one perpendicular to the plane cain be drtawn. For, if two s could be drawn at the given point, a plane could be passed through them intersecting the given plane. Then the two I would be in the new plane and - to the same line (the line of intersection of the two planes, Art. 505); which is im~ possible (Airt. 71). 32G B0OO0I VI. SOLID GEOMETRY PROPOSITION IV. PROBLEM 516. From a given point without a plane, to draw a line perpendicular to the plane. A ~\ ^ ~': \ Given the plane MY and the point A external to it. To construct from A a line I plane IN. Construction. In the plane 1MN draw any convenient line BC. Pass a plane through BC and A (Art. 503), and in this plane draw AD 1 BC. Art. 273. In the plane MN draw LD I BC. Art. 274. Pass a plane through AD and LD (Art. 503), and in that plane draw AL I LD. Art. 273. Then AL is the I required. Proof. Take any point C in BC except D, and draw LC and A C. Then I ADC, ADL and LDC are right Q. Constr..-. A (c=:AD + DC-. Art. 400. AC. C A = 'AL + 2 + DC. Art. 400, Ax. 8. -2 — 2 + L —,2 2 =-A7L2 - L. Art. 400, Ax. 8.. Z ALC is a right Z. Art. 351. But AL J LD. Constr..' AL 1 MN. Art. 509. Q. F. P. 517, COR. But one perpendicular can be drawn from a given external point to a given plane, LINES AND PLANES 327 PROPOSITION V. THEOREM 518. I. Oblique lines drazwn from a point to a plane, meeting the plane at equal distances from the foot of the perpendicular, are equal; II. Of two oblique lines drawn from a point to a plane, but meeting the plane at unequal distances from the foot of the perpendicular, the more remote is the greater. \Ap Mai: I B, \I Given AB 1_ plane MN, BD=BC, and BH > BC. To prove AD=-AC, and AH > AC. Proof. I. In the right A ABI) and ABC, AB = AB, and BD = BC. (Why?). A. A ABD=A ABC, (Why?).'. AD =A C. (Why?) II. On Bi take BF= BC, and draw AF. Then AF= AC (by part of theoremi just proved). But All > AF. (Why?).'.AH> AC. Ax. 8. Q. E. D. 519. COR. 1. CONVERSELY: Equtal oblique lines drawn from a point to a plaine meet the pl(lne (t equal distances from the foot of the perpendicular drawn' fr'om the same point to the plane; and, of two uneequal lines so drawn, the greater line meets the ple lae t the greater distance fromE the foot of the perpeendicu lar. 32) S 328 POOR1 0 VI. SOLID GFOMEO~TRY 520, Con 2),9. The locuts of a. point, Inl space eiju.idistantt from all the poinits in thre circantiiferencwe of a circle is at straight line passingy throuigh, the center of 1 le circle and perpendicular to its plane. 521. Con. 3. The perpendicular is~ the shortest linie that can, be draiwn from, a gicen point to a given p~lane. 522. DEF~. The distance from a point to a plane is the perpendicular dr-awn from the point to the plane. PROPOSITION VI. THEORE-M 523. If from the foot of a perpen~dicuilar to a plane a line be drawen at rights amglcs to any line in the plane, the line dIraw'en from, the point of intersect ion, so formed to any poin t 'in the perpendicular, is p)erpendicular to the line of the plane. A D Given AB I plane MTY, and BF I (J), any line in MAX. To prove AF I CD. Proof. On CD) take EP and EQ equal segments. Draw AP, BP, AQ, BQ. Then BP=-=B Q. Art. 112 Hence AP= AQ. Art. 518.. in the line AF, the point A is equidistant f rom P and Q, and F is equidistant from P and Q. AJF I CD. (Why 7) Q. E. D. Ex. In the above figure, if AB=61 AF=5, and AQ= 10, find QF, BF and BQ, LINES A.ND PLANES32 3249 PROPOSITION VII. THE~OREM 524. Twvo straight lines perpenidicatlar to the same plane are parallel. tA B /, 11 Given the lines A B and CD I plane JIMY To prove AR I IGD. Proof. Draw BD, and througrh D, in the plane MY, draw PHf 1 RD. Draw AD. Then RD I PH. Constr. AD I PHFI. Art. 523. CD I P11. Art. 505. RD, AD and CD are all I PHf at the point D. RD7 AD and CD all lie in the same plane. Art. 510. AR and CD) are in the samne plane~. (-W hy?) But AR1 and CD are I RI.Art. 505.. AB and CD are H.Art. 121. Q. E. D. 525. CoR,. 1. If one of two pa(rallel lines iS per'PendicUlar to a planie, the othier is gperp#~licidar to the plane also. For, if A and CD) be 1K and AR I Plane PQ, a line drawn from C I PQ A C Dmnst be 11 AR:. Art. 524. But CD imust coineide with the line B D -so drawni (Art. -47, 'I);.-. C)IPQ 9 ct) ) O'o0 218OOK VI. SOLTI) GEOMETRY 526. COR. 2. If itco straight lines are each parallel to a third straiglht tlion, they ctre parallel to each other. For, if a plane be drawin L to tihe third line, each of the two other lines must be -L to it (Art. 525), and therefore be 1 to each other (Art. 524). PROPOSITION VTII. THEOREM 527. If a straight line external to a given plane is parallel to a line in the pylane, thea l hefirsl line is parallel to the given plane. A B C 1)ID Given the straight line AP 1i line CD in the plane MN. To prove AB) ]1 planle MIX. Proof. Pass a plane through the 11 lines AB and CD. If AB meets 3MN it must meet it in the line CD. But AB and CD cannot meet, for they are II. Art. 120... AB and MN cannot meet and are parallel. Art. 506. Q. E. Do 528. COR. 1. If a straight line is parallel to a plane, the intersection of the plane with any plane passing through the given line is parallel to the given line, LINES AND PLANES 331 529. COR. 2. Through a given line A (CD) to pass a plane parallel to another given line (AB). --. —.-R Through P, any point in CD, draw QR I| AB (Art. 279). Through CD and QR pass a plane (Art. 503). This will be the plane required (Art. 527). If AB and CD are not parallel, but one plane can be drawn through CD II AB. PROPOSITION IX. THEOREM 530. Two planes perpendicular to the same straight line are parallel. Given the planes MN and PQ L line AB. To prove MIN II PQ. Proof. If MN and PQ are not parallel, on being produced they will meet. We shall then have two planes drawn from a point per~ pendicular to a given line, which is impossible. Art. 513..'. MN and PQ are parallel. Art. 507. Q. E. D 3 332 00OOK YI. SOLID GEOMETRY PROPOSITION X. THE 531. If two parallel planes are cut by a third plane, the intersections are parallel lines. Given MN and PQ two 1I planes intersected by the plane RS in the lines AB and CD. To prove AB II CD. Proof. AB and CD lie in the same plane RS. Also AB and CD- cannot meet; for if they did meet the planes M1YI and 'PQ would meet, which is impossible. Art. 507. AB and CI are parallel. Art. 41. Q. E. D. 532. COR. 1. Parallel lines inscluded between parallel planes are equal. For, if AC and BD are two parallel lines, a plane may be passed through them (Art. 503), intersect~ ing MN and PQ in the 11 lines AB and CD. Art. 531..'. ABDC is a parallelogram... AC=BD. Art. 147. Art. 155, LINES AND PLANES 333 533. COR. 2. Two parallel planes are everywhere equi. distant. For lines I to one of them are I1 (Art. 524). Hence the segments of these lines included between the 11 planes are equal (Art. 532). PROPOSITION XI. THEOREM 534. If two intersecting lines are each parallel to a given plane, the plane of these lines is parallel to the given planee A D \.-'"- \ K L k i< _______, ___ Given the lines AB and CD, each || plane PQ, and intersecting in the point F; and IMN a plane through AB and CD. To prove MNX PQ. Proof, From the point F draw FH. PQ. Pass a plane through FC and FH, intersecting PQ in BfK; also pass a plane through FB and FPi, intersecting PQ in HL. Then HK II FC, and HL II FB. Art. 528. But FH 1 IK and HL. Art. 505..~ FH I FC and FB. Art. 123. T 1 X1Y. Art. 509. M' f 11~ PQ, Art. 530. Q. p. 3 334 BOOK VI. SOLID GEOMETRY PROPOSITION XII. THEORE3i 585. A straight line epeendicular to one of two parallel planes is perpendicular to the other also. Given the plane MN 1l plane PQ, and AB I PQ. To prove AB I MN. Proof. Through AB pass a plane intersecting PQ and MN in the lines BC and AF, respectively; also through AB pass another plane intersecting PQ and MN in BD and AH, respectively. Then BC 11 AF, and BD 11 AH. Art. 531. But AB I BC and BD... AB 1 AF and AH..'. AB I plane MN. Art. 505. Art. 123. Art. 509. 0. B. D. 586. COR. 1. Through a given point parallel to a given plane. Let the pupil supply the construction. to pass a plane 537. COR. 2. Through a given point but be passed parallel to a given plane, one plane ca LINES AND PLANES 335 PROPOSITION XIII. THEOREM 538. If two angles not i tMhe same plane have their corresponding sides parallel and extending in the same direco tion, the angles are equal and their planes are parallel. '\ A(.. C -Q Given the Z BAC in the plane 3fN, and the ZB'A'C' in the plane PQ; AB and A'B' I1 and extending in the same direction; and AC and A'C II and extending in the same direction. To prove Z BA C = Z B'A'C, and plane MN plane PQ. Proof. Take AB=A'B', and A C A C'. Draw AA', BBs, CC', BC, B'C'. Then ABB'A' is a ~Z7, Art. 160. (for AB and A'Bt are= and I{ )..'. BB' and AA' are==and I1. Art. 155. In like manner CC' and AA' are=and |..'. BB' and CC' are=and I1. (Why?).. BCC'B' is a Z, and BC= B'C. (-hsy?). A BC= A 'B'C'. (Why?). ZA = ZA'. (Why?) Also AB II A'B',.'. AB | plane PQ. Art. 527. Similarly AC II plane PQ., plane MATl 1 plane PQ. Art. 534~ -f 4 I~S^ 336 BOOK VI. SOLID GEOMETEY PROPOSITION XIV. THEOREM 539. If two straight lines are intersected by thJrre ptrallel planes, the correspondiing segments of thcese lines are proportional. Given the straight lines AB and CD 11 planes JMN, PQ and RS in the points H, D, respectively. intersected by the A, F, B, and C To prove AF_ CH FB )HD Proof. Draw the line AD intersecting the plane PQ in G. Draw FG, BD, GH, AC. Then FG I1 BD, and GH 11 AC. AF. G FB GD Art. 531. Art. 317. And CH AG HD GD AF CH BH FB HD. _________..Q. Ex. 1. In above figure, if AF=2, FB=5, and CH=3, fin Ezx 2. If CH6-3, D)=4,, and AB=10, find AF and BF. (Why?) (Why?) E. D. d CD. DIHEDRAL ANGLES 337 DIHEDRAL ANGLES 540. A dihedral angle is the opening between two intersecting planes. From certain points of view, a dihedral angle may be regarded as a wedge or slice of space cut out by the planes forming the dihedral angle. 541. The faces of a dihedral angle are the planes forming the dihedral angle. The edge of a dihedral angle is the straight line in which the faces intersect. p 542. Naming dihedral angles. A dihedral angle may be named, or denoted, by naming its edge, as the dihedral angle AB; or by naming four points, two on the edge and one on each face, those on the edge coming between the points on B the faces, as P-AB-Q. The latter method is necessary in naming two or more dihedral angles which have a common edge. 543. Equal dihedral angles are dihedral angles which can be made to coincide. 544. Adjacent dihedral angles are dihedral angles having the same edge and a face between them in common. 545. Vertical dihedral angles are two dihedral angles having the same edge, and the faces of one the prolongations of the faces in the other. 546. A right dihedral angle is one of two equal adjacent dihedral Canles formed by two planes. v 338 - OO(K VI. SOLID GEOMETRY 547. A plane per- | pendicular to a given |ii i plane is a plane forming a right dihedral i! — angle with the given \ plane. Many of the properties of dihedral angles are obtained most conveniently by using a plane angle to represent the dihedral angle. 548. The plane angle of a dihedral an- gle is the angle formed by two lines drawnl P --- —--- one in each face, perpendicular to the edge at the same point. Thus, in the dihedral angle C-AB-F, - _ D if PQ is a line in the face AD perpendicular to the edge AB at P, and PR is a line F in face AF perpendicular to the edge AB at P, the angle Q(PR is the plane angle of the dihedral angle C-AB-F. 549. Property of plane angles of a dihedral angle. The mag.itr.de of th7 plane angle of a dihedral angle is the same at every point of the edge. For let EAC be the plane / of the dihedral Z E-AB-D at the point A. Then PR 11 AE, and PQ 11 AC (Art. 121.).. ZIRPQ = ZEAC (Art. 538). 550. The projection of a point upon a plane is the foot of a perpendicular drawn from the point to the plane. A 551. The projection of a line t B upon a plane is the locus of the pro- f i jections of all the points of the line \ 1,il. \ on the plane. Thus A'B' is the \ B \ projection of AB on the plane MN. - - DIHEDRAL ANGLES 339 PROPOSITION XV. THIEOREM 552. Tiwo dihedral angles a e eqcul if their plane anglei are equal. -A A 1F 13 F' D D D Given Z DBF the plane Z of the dihedral Z C-AB-F, Z D'B'F' the plane Z of the dihedral Z C'-A'BD-F', and Z DBF = D'B'F'. To prove Z C-AB-F = Z C'-A'B'-FY. Proof. Apply the dihedral Z C'-A'B'-F' to Z C-AB-F so that / D'B'F' coincides with its equal, DBF. Geom. Ax. 2. Then line A'B' must coincide with AB, Art. 515. (for A'BI and AB are both I plane DBF at the point B). Hence the plane A'D'D! will coincide with plane ABD, Art. 501. (through twco intcrsecting lines only on e plane can be pas'sed). Also the plane AIYB'1'v will coincide with the plane ABE, (sa1me reason)... Z Ca-A'' —F coincides with Z C-Ai -F and is equal to it. Art. 47. Q, E. D. 553. COR. The vertical dihedral (tngljes for)med tI! tro intersectin.g plranes are equarl. In like manner, many other l)roperties of plane angltes are true of dihedral angles. 340 -)0B1OK VI. SOLID GEOMETRY PROPOSITION XVI. THEOREM 554. Two dihedral angles lave the samel ratio as their i I (I 1 D' D --- — ''^ <iiniti;! r-^^pl^^ c illl I Fi. 1i F I ~i iiFi..2 Given the dihedral XA C-AI-1 and C'-A1''-)' having the plane X CAD and C'A'DI, respectively. To prove Z C'-A'B'- Z: (-.A B-1) Z CA'D': Z CAD. CASE I. When the plane C'A' T1an' d CAD (Figs. 2 and 1), are commesurable. Proof. Find a common measure of tfle X C'A'D' and CAD, as Z CAK, and let it be contained in Z C'A'D' n times, and in Z CAD mi times. Then Z C','])' Z CAt= ~ 4. Through A'B' and the lines of division of Z C'A'D' pass planes, and through AB and the lines of division of Z CAD pass planes. These planes will divide the dihedral Z C'-A'B'I-D into n, and Z C-AB-D into t parts, all equal. Art. 552... Z C-A'B'-D': / C-A B-D=n: i. Hence Z C'-A'B'-D': Z C-AB-D - Z C'A'D': CAD. (Why?) DI1HEDl)RAL ANGLES 341 CASE II. Whien the, plane angles C'AID' and CAD (Figs. 3 and 1) are ivncommnensurable. Proof. Divide the -Z CAD into ainy number of equal parts, and a'pply, one of these parts to the Z C'A!D'. It will be contained a (certain unmber of times with a remainder, as Z LA'D', less than the nnit of measure. Hence the X C'A'L and CAD are commensurable. C'- AWB-L:ZC- AB -PD C'A'L: ZAD. CaselI If now wie let the unit of measuire b~e indefinitely dimintished, the Z LAI',which. is less thani the unit of measure, will be indefinitely diminished. Z C'A'J~ Z C'A —'D' as a limit, and / C'-A'4J}'-.L CI / C'-'B' — D' a s a, limi t. Art. 251L Hence Z C'-A'B'-L becomes a variable, with zC-AJB-D Z CG-A1B1 — as its limit; Art. 2~53, 3. Also ZCALbecomes a variable with asIAD its Z CAD) ZCAI) limit. Art. 253, 3, But the variable Z '-AR-L- the variableZCAI Z C- A B-D Z CA!) always. Case 1. Z C'-A'IB'-D' Z C'A'D' the limit -y P =the limit D Art. 254. Z C-A. 13'-D/ CA - _____________ ~~Q. E.D. Ex. 1. How many straight, lines are necessary to indicate a dihedral aingle (as ZE-AB-I), p. 33S)? How inany straight lines are 'necessary to indicate the plane angle of a dihedral angle I Hence, what is the advantage of using a plane angle of a dihedral angle instead of the dihedral angle itself Ex. 2. (live thr~ee additional properties of dihedral angles analogous to p~roperties of plane ang-les given in Book 1. r) I cl (-,) 4'-' 842 BO11)OR YLJo. l GEOME1TRY PROPOS-ITION XVIIL THEOREM -N 555. If a stri~ight line is perpe~ndicitlar to a plane, every Wlane drawn throu~gh, that line is perpendicular to the plane. Given the line AR I plane MNL, and the plane PQ passing througli AB and intersecting, MN in BQ. To prove PQ J-II MN. Proof. In the plane MX draw BC I RQ at B. But AB IRQ. Art. 505.. Z ARC is tile plane Z of the dihedral Z P- BQ - 11. Art. 548. Bunt Z ARC is a rigyht Z/ Art. 505. (f1or ARB I AMN by h yp. ).. PQ ~ ALf. Art. 547. Q. E. D. 556. CoR,. A plane perpendicular to) the edge of a dihedral angle is perpenidicular to each, oJfthe two faces formni~ng the, dihedral angle. DIHEDRANL ANGLES 343 PROPOSITION XVIII. THEOREMI 557. If twvo planes are perpendicular to each other, a straight line drawn in one of them perpendicular to their line of intersection is perpendicular to the other plane. P A V~~~~~~~~~I Given the plane PQ I plane Mi and intersecting it in the line RQ; and AB a line in PQ I RQ. To prove AB I plane MNLAT. Proof. In the plane MN draw BC I RQ. Z ABC is the plane Z of thc dihedral FP- RQ-M. Art. 548. Z ABC is a rt. Z, Art. 554. fir- Q - QM is an rit di(hedral Z ). AB I BC and RQ at their intersection. A B I p I a lane il-. (Why?) Q. F. D. 558. COR. 1. If two planes are perpendica tar to each other, a perpendicular to one of' them, at any point of their intersection will lie in thre other plane. For, in the above fignure, a I erected at the point B in the plane MN must coincide with ABI? lying in the plane PQ and I MIN, for at a given point in a plane ouly one I can be drawn to that plane (Art. 515). 559. CCR. 2. If twvo planes are perpedcdic~dar to each other, a perpendicular to one piac)Ie, Jroi, a pobit ini the other plane, wtill lie in the other phtiue. 34~4 BOOK VI. ISOLII) GEO-METRY PROPOSITION XI1X. THEOREM 560. If twtlo intersecting planes are each, perpenid'icular to a third plant~e, their line of 'intersection is perpenidica tar to the third plane. Given the planes PQ and RS I plane MN, and ihtersecting in the line AB. To prove AB I plane MY Proof. At the point B in which the three planes meet erect a I to the plane -,If-V. This nimust lie in thc plane ]'Q. and also in the plane RS. Art. 558. Ihence this I mnust coincide, with AB, the intersection, of PQ and 1R$. Art. 508,2 AB I plane M-N. Q. E.D. 561. CoR. if twco planes, including a right dihedral' a~ngle, are each perpendicular) to a thi-rd planie, the inter-setion of any two of the planes is perpendicular to the third plane, an~d each of the three lines of 'intersection is per-pen. dicular to the other twvo. Ex. 1. Name all the dihedral angles on the above figure. Ex. 2. If L CBQ=3O0, find the ratio of each pair of dihedral A I)IIIEDBRAL ANGLES34 345 PROPOSITION XX. THEOREM 562. Every point in the plane which bisects a given dihedral angle is equidistant fr om the, faces of the dihedral angle. D -A Given plane CB? bisectingc the dihedral LZA-BR-D, P ainy point in plane BC, -P9 anid PT I- faces BA and BD, respectively-. To prove P9 = PT. Proof. Thronghi P9 and PT pass a plane intersectinDg AR in Q9R,, BD in RT,and BC in P-R. Then plane PQT I planes AB and RD. Art. 555.. plane PQT I line RB, the intersection of the planes AB and RD. Art. 560. R1B I RQ, RB and RT. Art. 505. A 91?]? and PIT arcil the plane XA of the dihedral A A-B-R-P and PBRJ.Art. 548. Bunt these dihedral A taie equhial. Hyp. z Qw[, A PRT. Art. 554. rt.A I)~kz.A IPRT (Why) PT 11. (Why?7) Q. E. 0. 563. The locus of all points cq~itlistaiit from, the faces of a dihedral anfylr is the plante bisectinig the dihedirat angle. 3G46 BOOK VT. SOI) CGEOMEITIRY PROPOSITION XXI. PROBLEM 564. Through any straiyght line not perpendicular to given plane, to pass a plane perpendicular to the given plane A \Givn te li AcB nt p e M. Given the lineg through AB plae A1 N. To construct a plane passing, through AB and L Mr. Construction. From a point A in the line AB draw a _ AC to the plane lMN. Art. 516. Through the intersecting lines AB and AC pass the plane AD. Art. 503. Then AD is the plane required. Proof. The plane AD passes through AB. Constr. Also plane AD L plane MiN, Art. 555. (for it containss AC, which is I- 3MN). 565. CoR. 1. Thro(gh a straight line not perpendicular to a given plane only owe plane can be passed perpendicular to that plane. For, if two planes could be passed through AB. plane MNV, this intersection AB would be I MN (Art. 560), which is contrary to the hypothesis. 566. CoR. 2. The projection upon a plane of a straight line not perpendicular to-that plane is a straight line. For, if a plane be passed through the given line I to the given plane, the foot of a I from any point in the line to the given plane will be in the intersection of the two planes (Art. 559), DIHE)DRI L ANGLES 347 PROPOSITION XXII. THEOREM 567. The acute angle which a line makes with its projection on a plane is the least angle which it mcakes wvith any line of the plane through its foot. Given line AB meeting the plane lMN in the point B, BC the projection of AB on MN, and PB any other line in the plane MN through B. To prove that Z ABC is less than Z ABP. Proof. Lay off PB equal to CB, and draw AC and AP. Then, in the Q ABC and ABP, AB==AB. (Why?) BC= BP. (Why?) But AC < AP. Art. 521.. ABC is less than Z ABP Art. 108. Q. E. D. 568. DEF. The inclination of a lhie to a plane is the ceute angle which the given line makes with its projection ipon the given plane. Ex. 1. A plane has an inclination of 47~ to each of the faces of a ihedral angle and is parallel to the edge of the dihedral angle; how aany degrees are in the plane angle of the dihedral angle? Ex. 2. In the figure on page 345, if PT.'-QT, how large is the iledral _ A -ll-I-)? it I' oIRT, how large is it? 34S BOOK VI. SOLID GEOMETEY PROPOSITION XXIII. PROBLEM 569. To dr(7', a co) mon perped/icular to any two lib, not in the same plane. \ 1 Gven the lies A the sae Given the lines AB and CD not in the same plane.' To construct a line perpendicular to both AB and CD. Construction. Thlrougllh A 7' lp:Ls- a plane MLN'l line CD. Art. 52' Through CD pass a plane (' i_ plane JIX (Art. 564), aI, intersecting plane iMN in the line EF. Then EF II CD (Art. 528),.'. EF must intersect A i (which is not I{ CD by hyp.) in some point K. At K in the plane CF draw LK l2 EF. Art. 274 Then LK is the perpendicular required. Proof. LK L EF. Constr.L. LK I C). Art. 12:. Also L K I plane MXY. Art. 557.. LK 1 line AR. (Why?.'. LK I both CD and AB. ' Q.. P. 570. Only one perpencicwlar can be drawvn between two lines not in the same plane. For, if possible, in the above figure let another line BD be drawn I AB and CD. Then, if a line be drawn through B II CD, BD I this line (Art. 123), and.'. L plane MY (Art. 509). Draw DE I line EF; then DF I. plane MN (Art. 557). Hence from the point D two _, DB and DE, are drawn to the plane MY1, which is impossible (Art. 517), POLYHEDRAL ANGLEIS34 349 POLYHEDRAL ANGLES' v 571. A polyhedral angle is the amount of opening between three or more planes meeting at a point. Such an angle may be regarded as a portion ofA Space cut out by the planes forming the angle. B 572. The vertex of a polyhedral angle is the point in wvhich the planes forming the angle meet; the edges are the lines in which the, palales intersect; the faces are the portions of the planes formingr the, polyhedral angle 'Which are included between the edges; the face angles are the angfles formied by the edg(es. Each two adjacent faces of a, polyhedt'al angle form a d-ihedral augle. The parts of a polyhedral angle are its- face angles and dihedral angles taken togethier. 57-3. Naming a polyhedral angle. A polyhedral angle is 'named either by namiing the -vertex, as, FV or by naming, the vertex and a point onl each edlge, as V-ARC. In case two or more polyhediral angles have the samie 'vertex, the latter method i s niece s sa ry. In the above polyhedral angle, the vertex is, V; the edges are VA, IYB, VC; the face angrles are AV7B, BVC, AV'C. V 574. Aconvex polyhedral angle is,- aE/ Poly., hedral ang~le in which a sectlion Made by a plane cutting, all the edg~esLs a convex polygoni, as V -ABCDE.B a 575. A trihedral angle is a polyheh1al angrle having, thee, faces; a tetrahedral angle is one laying four faces,-, etc. 350 BOO()K VI, SOLID GEOMETLY 576. A trihedral angle is rectangular, birectangular, or trirectangular, according as it contains one, tivo, or three right dihedral angles. 577. An isosceles trihedral angle is a trihedral angle two of whose face angles are equal. 578. Vertical polyhedral angles are polyhedral angles having the same vertex and the faces of one the faces of the other produced. v V' 579. Two equal polyhedral angles are polyhedral angles having their correspond- / c ing parts equal and arranged in the same order, as V-ABC B' and V-AIB'C'. Two equal polyhedral angles may be made to coincide. v' 580. Two symmetrical poly- hedral angles are polyhedral an- -- gles having their correspondingl B \ parts equal but arranged in reverse order. If the faces of a trihedral angle, V-ABC, be produced, they will form a vertical trihedral angle, "' V-A'B'C', which is symmetrical to V-ABC. For, \ -- ':A if V-A'B'Cf be rotated forward about a horizontal C\ axis through V, the two trihedral angles are seen to have their corresponding parts equal but ar- / ranged in reverse order. A - Similarly, any two vertical polyhedral angles are symmetrical. POLYiHElDREAL ANGLES 351. 581. Equivalence of symmetrical polyhedral angleso It has been shown in Plane Geometry (Art. 488) that two triangles (or polygons) symmetrical with respect to an axis have their corresponding parts equal and arranged in reverse order. By sliding two such figures about in a plane b e.fb they cannot be made to coincide, but by lifting one of them up from the plane in which it lies and turning it over it may be made to coincide with the other figure. Symmetrical polyhedral angles, hol:wcver, cannot be made to coincide in any way; hence somic indirect method of showing their equivalence is necessary. See Ex. 29, p. 358, and Arts. 789-792. Ex. 1. Name the trihedral angles on the figure to Prop. XX. If ZPRQ=90~, what kind of trihedral angles are those on the figure? If Z PRQ=30~, what kind are they? Ex. 2. Are two trirectangular trihedral angles necessarily equal? Prove this. Ex. 3. Are two lines which are perpendicular to the same plane necessarily parallel? Are two planes which are perpendicular to the same plane necessarily parallel? Are two planes which are perpendicular to the same line necessarily parallel? Ex. 4. Let the pupil cut out three pieces of pasteboard of the form ndicated in the accompanying figures; cut them half through where Ihe lines are dotted; fold them and fasten the edges so as to form hree trihedral angles, two of which (Figs. 1 and 2) shall be equal mnd two (Figs. 1 and 3) symmetrical. By experiment, let the pupil ind which pair may be made to coincide, and which not. Fig. 1 igo Fig, 3 35)2 1 OOI(K VI. SOLID GEOMETRY PROPOSITION XXIV. THEOREM 582. The sum,, of tanr thco fait cc ales of a trihed angle is greater than the third ftace angle. S Given the trihedral angle S-ABC, with angle ASC its greatest face angle. To prove Z ASB + Z I;'SC greatter thanl Z ASC. Proof. In the face ASC dra(w l.1), makilg Z ASD= Z ASB. Take S7 = SIB. In the face ASC draw the line ADC in any convenient direction, and draw AB and BC. Then, in the A ASP and ASD, SAS; =. (Why?) SB = SI, and Z ASPB = Z,ASD. (Why?).. A SB=/A ASi). (Why?) A.. iAB=AD. (Why?) Also AB + BC > AC. (Why?) Hence, subtracting the equals AB and AD. BC > DC. (Why?) Hence, in the A BSC and DSC, SC= SC, SB SD, and BC > DC. (Why?).. Z BSC is greater than Z DSC. Art. 108. To each of these unequals add the equals Z ASB and Z ASD... LASB -+ Z BSC is greater than ASC. (Why?) Q. E. D. Ex. In the above figure, if Z.ASC equals one of the other face angles at S, as AdSB, how is the theorem proved? POLYHEDRAL ANGLES 353 PROPOSITION XXV. THEOREM 583. The snim of the face angles of any convex polyhedral angle is less than foar right angles. A D Given the polyhedral angle AS-ARUD-E. To prove the sumi of the face A. at AS less than 4 rt. / Proof. Pass a plane cuitting the edges of the given polyhedral angle in the poiats A, B, C, ii, E. From any point. 0 in the polygon ABCDE draw OAl, OB, O C, 01), OF. Deniote the & having the common vertex AS as the AS& and those having, the common vertex 0 as the 0z. Then the sumn of the A of the AS & = the sum of A of the 0 &. Art, 134. But Z ASBA + Z ASBC is greater than Z ABC, Ar 52 ZASCB'+ Z SC )is greater thaiZ ABCD, etc.jAt52 the sumi of the base A of the AS & > the sum of the base A of the 0 ~.Ax. 9. the sumr of the vertex A of the AS & < the sum of tlxe vertex A or the, () Ax. 11. (if unequnts be sub/ractedf from epw Is, the r-emainders are unequal DIn rererse Or(d(r) But the sumn of the A at 0=4 rt. AX. (Why?) the sum of face A at AS < 4 rt. A. x. S. Q.E.D W 354 BOOK VI. SOLID GEOAMETRY PROPOSITION XXVI. THEOREO I 584. If tw'o trihedral -angles hae.' hie ithrCe face an lr, of one equal to th1e three face Jangyls of /t1e otl)er, the r,hedral angles hate their corresponidin dih odrt l an gles equal and are either equal or symmeltrical, accordintg (as thei,. correspotnding face angles are arr'aged in tie tsame or ij reverse order. S S S D DI D A A___ A C AD B B' BE Given the trihedral AX S —BC and S'-AaBdC', havinc the face A ASB, ASC and BiSC equal to the face A AS'BI', A'S'C' and B'S'C', respectively. To prove that the corresponding dihedral X of S-ABCe1 and S'-A'B'C' are equal, and that A S-BC and S'A'B'C' are either equal or symmetrical. Proof. On the edoges of the trihedral A take SIA, SB, SC, S'A', 8'B', S'C' all equal. Draw AB, AC, BC, A'B', A'C', B,'C' Then, 1. In the & ASB and A'S'B', SA=S/A", SB= S'B', and ZASB= ZA'S'B'. (Why?). / AASB= A A'S'B'. (Why?). AB=A'B'. (Why?) 2. In like manner AC-A'C', and BC= B'C'. '. A ABC-A A'B'C'. (Why?) EXERCISES ON THE LINE AND PLANE 35 3 "a" 5 3. Take D a convenient point in SA, and draw DE in the face ASB, and DF in the face ASC, each I SA. DE and DF meet AB and AC in points, E and F, respectively-, (for A SJB and S4C (trcacute). Similarly, take S'il)' SD and constrnet A D'E'F', Then, in the rt. A ADE and A'D'E', AD-A'D', and L DA E= D'A'E'. (Why?) A ADEF:A _zt'.IY'E. (Why'?) AJE=A'J]', and D)E-D'_E'. (WAhy?9) 4. In like mnanner it inay he shown that AFPA'P', and DF- D'F'. A A EI~=A AE'F'. (W h y?) And EFF- E'F' (Why.?) 5. Hence, in the A DEE and'D'EF'Y, DE=D'E', DF -=D'F' and EF-E'F'. (Why) A DE F =A D'E' F'. (Why) ZRJ)E-DF ZR'lD'F. (Wh y? But these Xare the plIane Z of thie dihiedral A -whose edges are SA and S'A'. dihedral Z B-A 5-4C dfihedral Z Bh-A'-1S'-C' Art. 552. In like manner it may hie shown that the dihedral A at SB and SIB' are equal; and that those at SC and SIC' aie equal.. the trihedral Xs S and 5' are either equal or symmetrieal. Arts. 579, 580. Q. E. D. EXERCISES. GROUP 64 TH'EOREMS CONCE RNING THE LINE ANI) PLANNE IN S-,-PACE Ex. I. A seg-ment of a line not parallel to a. plane is longer than its projectionI in the plane. Ex. 2. Eqjual straight lines drawn. frot a. point to a plane are equally inc~linjed to the plane, 356 BOOK VI. SOLID G(EOMETIRY Ex. 3. A line and plane perpendicular to the same plane are parallel. Ex. 4. If three planes intersecting in three straight lines are perpendicular to a plane, their lines of intersection are parallel. Ex. 5. If a plane bisects any line at right angles, any point in the plane is equidistant from the ends of the line. Ex. 6. Given IB J_ plane TMN, and 4AC -L plane RS; prove BC JL SR. Ex. 7. Given PQ L plane MNV, PIR L plane BL, and RS JL plane MN; prove QS j- AB. i Ex. 8. If a line is perpendicular to one of two intersecting planes, its projection on the other plane is perpendicular to the line of intersection of the two planes. Ex. 9. Given CE 1 DE, AE -L DE, and Z C-AD-E a rt. dihedral Z; prove CA L plane DAIE. p L B N Ex. 10. The projections of two parallel lines on a plane are parallel. (Is the converse of this theorem also true?) Ex. 1. If two parallel planes are cut by two non-parallel planes, the two lines of intersection in each of the parallel planes will make equal angles. Ex. 12. If a line is perpendicular to a plane, any plane parallel to the line is perpendicularto the plane. (Is the converse true?) Ex. 13. In the figure to Prop. VI, given A.B I MN and AF 1L DC; prove BF I DC. Ex. 14. Two planes parallel to a third plane are parallel to each other. [$vo. Draw a line - third plane.] EXERCISES ON THE LINE AND PLANE Ex. 15. The projections upon a plane of two equal and parallel straight lines are equal and parallel. Ex. 16. A line parallel to two planes is parallel to their intersection. Ex. 17. In the figure to Prop. XXII, if angle CBP is obtuse, prove the angle ABP obtuse. B Ex. 18. In a quadrilateral in space (i. e., a quadrilateral whose vertices are not all in the C same plane), show that the lines joining the midpoints of the sides form a parallelograim. Ex. 19. The lines joining the midpoints of the opposite sides of a quadrilateral in space bisect each other. Ex. 20. The planes bisecting tie dihedral angles of a trihedral angle meet in a line every point of which is equidistant from the three faces. [SUG. See Art. 562.] Ex. 21. Given OQ bisecting ZROS, R PQ l plane ROS, QR L OR, 0 and QS I OS; prove PR = PS, P1R _, 1R, and PS 1. OS. Ex. 22. In a plane bisecting a given plane angle, and perpendicular to its plane, every point is equidistant from the sides of the angle. [SUG. See Ex. 21; or through P any point in the bisecting plane pass planes I to the sides of the Z, etc.] Ex. 23. In a trihedral angle, the three planes bisecting the three face angles at right angles to their respective planes, intersect in a line every point of which is equidistant from the three edges of the trihedral angle. Ex. 24. If two face angles of a trihedral angle are equal, the dihedral angles opposite them are equal. Ex. 25. In the figure to Prop. XXIV, prove that ZASC+- BSC is greater than Z LAS) +- Z SD. Ex. 26. The common perpendicular to two lines in space is the shortest line between t hem. 35o8 1COK VI. SOLID GEOMETRY Ex. 27. iven M S, and 7' = 1>b; V,1 d_ prove Z J/I;C=, Z\ bc, \ and A AIC - tbc. \ Ex. 28. Two isosceles symmetrical c trihedral angles are equal. Ex. 29. Any two symmetrical trihedral angles are equivalent. [SUG. Take SA,, S C,, S'A', S S'i', S'C, all equal. Pass planes, SC, A'B'C'. Draw SO andi S'O' 1 these planes. Then the trihedral Z aroe divided into three pairs of isosceles B/ \ A' - B' symmetrical trihedral A, etc.] C c EXERCISES. GROUP 65 LOCI IN SPACE Find the locus of a point equidistant from Ex. 1. Two parallel planes. Ex. 3. Three given points. Ex. 2. Two given points. Ex. 4. Two intersecting lines. Ex. 5. The three faces of a trihedral angle. Ex. 6. The three edges of a trihedral angle. Find the locus Ex. 7. Of all lines passing through a given point and parallel to a given plane. Ex. 8. Of all lines perpendicular to a given line at a given point in the line. Ex. 9. Of all points in a given plane equidistant from a given point outside the plane. Ex. 10. Of all points equidistant from two given points and from two parallel planes. Ex. 11. Of all points equidistant from two given points and from two intersecting planes. Ex. 12. Of all points at a given distance from a given plane and equidistant from two intersecting lines. EXERCISES ON THE LINE AND PLANE 359 EXERCISES CGROUP 66 PROBLEMS CONCERNING THE POINT, LINE AND PLANE IN SPACE Ex. 1. Through a given point pass a plane parallel to a given plane. Ex. 2. Through a given point pass a plane perpendicular to a given plane. Ex. 3. Through a given point to construct a plane parallel to two given lines which are not. in the same plane. Prove that only one plane cal be constructed fulfilling the given conditions. Ex. 4. Bisect a given dihedral angle. Ex. 5. Draw a plane equally inclined to three lines which meet at a point. Ex. 6. Through a given point draw a line parallel to two given intersecting planes. Ex. 7. Find a point in a plane such that lines drawn to it from two given points without the plane make equal angles with the plane. [SUG. See Ex. 23, p. 176.] Ex. 8. Find a point in a given line equidistant from two given points. Ex. 9. Find a point in a plane equidistant from three given points. Ex. 10. Find a point equidistant from four given points not in a plane. Ex. 11. Through a given point draw a line which shall intersect two given lines. [SUG. Pass a plane through the given point and one of the given lines, and pass another plane through the given point and the other given line, etc.] Ex. 12. Through a given point pass a plane cutting the edges of a tetrahedral angle so that the section shall be a parallelogram. [SUG. Produce each pair of opposite faces to intersect in a straight line, etc.] BPOOK VII POLYHEDRONS 585. A polyhedron is a solid bounded by planes. 586. The faces of a polyhiedron are its lboui-iding pfantes; thle edges of a polyhedron ar-e the lines of intersection of its faces. A diagonal of a polyhedron is a straight line joining two of its 1~1I" vertices which are niot in the sarie face. The vertices of ai poly- Polyhedion bedron are the points iii which its edges meet or intersect. 587. A convex polyhedron is a polyliedIon in nhii h a section mnade by any phane is a Convex polygoll. Only convex polyhedrons are to be considered in this book. 588. Classification of polyhedrons. Polyhedrons are sometimes classified according to the numnber of their faces. Thns, a tetrahedron is a polyhedron of four faces; a hexahedron is a polyhedron of six faces; a-n octahedron is one of eigrht, a dodecahedron one of twelve, and an icosahedron one of twenty faces. Tetrahedron Cube Octahedron Dodecahedron icosahedron (360) POLYHEDRONS The polyhedrons most important in practical life are those deter~ mined by their stability, the facility with which they can be made out of common materials, as wood and iron, the readiness with which they can be packed together, etc. Thus, prism means "something sawed off." PRISMS AND PARALLELOPIPEDS 589. A prism is a polyhedron bounded by two parallel planes and a group of / planes whose lines of intersection are / parallel. / 590. The bases of a prism are the faces formed by the two parallel planes; Pri the lateral faces are the faces formed by the group of planes whose lines of intersection are parallel. The altitude of a prism is the perpendicular distance between the planes of its bases. The lateral area of a prism is the sum of the areas of the lateral faces. 591o Properties of a prism inferred immediately. 1. The lateral edges of' a prismn are e(qual, for they are parallel lines included between parallel planes (Art. 589) and are therefore equal (Art. 532). 2. The lateral faces of a prsism are patrallelogravms (Art. 160), for their sides formed by the lateral edges are equal and parallel. 3. The bases of a prismz are equal polygons, for their homologous sides are equal and parallel, each to each, (being opposite sides of a parallelogralm), and their homolf ogous angles are equal (Art. 538). 592. A right section of a prism is a section made by a Plane perpendicular to the lateral edges. C) ".) 0.) ( ) - 362 L~OOiK VII. SOLID GVOEWThTY 593. A triangular prism is a prismn whiose base is a, triangle a quadrangular pris m is (dic whiose base ijs a qjuadrilaterail, (t I. OhliQ1uo Prisms RI-"lit Prismn Regulari ilIsm 594. An oblique prism is a prism wiu ose 1 Attral ed'(1es aic obliqlue to the basess. 595. A right prism is, a prism whose lateral edges are perpendicular to the bases. 596. A regular prism is a right prism whiose bases are regular polygoiis. 597. A truncated prism is tbat part of a prisim included betwceu a base anid a section made by a plane oblique to the b~ase and cutting~ all the Lateral edges. Trunciatedl Pi-ism 598. A parallelopiped is a prism whose bases aie paiallelogrvaiis. Heiele, al U Ihe fices of at paralielop',iped are parallelograms, Oblique Rxight Rectangular Cube parallelopiped P ir llelopiped Paraillelopiped 599. A right parallelopiped is a parallelopiped whose lateial edo'es are perpendicular to the bases, PRISMS AND PARALLELOPIPEDS 363 600. A rectangular parallelopiped is a right parallelopiped whose bases are rectangles. Hence, all the fatces of a rectangular parallelopiped are rectangles. 601. A cube is a rectangular parallelopiped whose edges are all equal. Hence, all the faces of a clbe are squares. 602. The unit of volume is a cube whose edge is equal to some linear unit, as a cubic inch, a, cubi foot, etc. 603. The volume of a solid is the number of units of volume which the solid contains. Being a number, a volume may often be determined from other numbers in certain expeditious ways, which it is one of the objects of geometry to determine. 604. Equivalent solids are solids whose volumes are equal. Ex. 1. What is the least number of faces which a polyhedron can have? Ex. 2. A square right prism is what kind of a parallelopiped? Ex. 3. Are there more right parallelopipeds or rectangular paralelopipeds? That is, which of these includes the other as a special case? Ex. 4. Prove that if a given straight line is perpendicular to a ' Pen plane, and another straight line is perpendicular to another i'nle, and the two planes are parallel, then the two given lines are a'i-allel. 364 BOOK VII. SOLID GEOMgETRY PROPOSITION I. THEOREIj 605. Sections of a prisim made by parallel planes cuttitng all the lateral edges are equal polyg)ons. Given the prism PQ cut by 1i planes forming the sections AD and A'D'. To prove section AD= section AlDI'. Proof. AB, BC, CD, etc., are 1 AI'B, B'C', C'D', etc., respectively. Art. 531,.A. AB, BC, CD, etc., are equal to A'B', B'GC, CD'F etc., respectively. Art. 157. Also A ABC, BCD, etc., are equal to A A'B'C', B''DI', etc., respectively. Art. 538, '. ABCDE-A'B'C'D'E', Art. 47. (for the polygons have all their parts equal, each to each, and.'. can be, made to coincide). Q. E. Bo 606. COR. 1. Every section of a prism made by a plane parallel to the base is equal to the base. 607. COR. 2. All right sections of a prism are equal. PRISMS 365 PROPOSITION II. THEOREMI 608. The lateral area of a prism is equal to the product of the perimeter of a right section by a lateral edge. Given the prism RQ, with its lateral area denoted by 8 and lateral edge by E; and AD a right section of the given prism with its perimeter denoted by Po To prove S= P X E. Proof. In the prism RQ, each lateral edge=E. Art. 591, 1. Also AB (L GH, BC J IJ, etc. Art. 505. Hence area C RHI= R AB X G = AB X E, area ^7 GJ= BCX E, Art. 385. area D IQ - CD X E, etc. j But S, the lateral area of the prism, equals the sum of the areas of the z forming the lateral surface..'. adding, S= (AB - BC + CD + etc.) X E. Ax. 2. Or S=PXE. Q. E. D. 609. COR. The lateral area of a right prism equals the product of the perimeter of the basf by the altitude. Ex. Find the lateral area of a right prism whose altitude is 12 in., and whose base is an equilateral triangle with a side of 6 in, Also fnd the total area of this figure, #I -. 3 BOOK VIL, SOITD GEOMETRY PROPOSITION ITT. THEOREM 610. if tw(,o prisnlS hbaire the three faces including a tri3herail (ngie of once eqial, respectitel!, to the three faces incluldiig a trilhedral angle of the other, and similarly placedl, the priSmns are equal. F (ULj F('JJJ -i i A1 D D B C B C Given the prisms AJ and A'J', having the faces AK, Al), AG equal to the faces A'K', A'D', A'G', respectively, and similarly placed. To p-rove AJ= A'J'. Proof. The face A EAR, EAB and BAF are equal, respectively, to the face A E'A'E', E'A'B' and B'A'FI. Hyp. trihedral Z A =trihedral Z Al Art. 584. Apply the prism A'J' to the prism AJ, making each of the faces of the trihedral Z A' coincide with corresponding equal face of the tri-hedral LA. Geom. Ax. 2..,. the plane F'J' will coincide in position with the plane E), Art. 500, (for the poinis G', f1, I' coincide with G, F, K, respectively). Also the point C' will coincide with the point C. C'fl' will take the direction of CH. Geom. Ax. 3. H'. R will coincide with H. Art. 508, 1. In like manner J' will coincide with J. Henie the prisms AJ and A'J' coincide in all points. A,. AJ=A'JA. Art. 47. 611. COR. I. Twco truncated prismns are equal if the three faces inlcl'udlinlg a trihedral angle of one are equal to the three jqvos including a trihedral angle of the other. PRISIMS36 367 612. CoR. 20. Twvo right pr-isms are equal if they harve equal bases a~nd equal altitudes. PROPOSITION IV. THE~OREM~~ 613. An oblique prismi is epdiralent to a, right prisin whose base is a right section of the oblique prism and wchose altitude is equal to a lateral edge of the obliqute prism. F K A B C Given the 'ob~lique prismn AD, wi~th the right section FJ; also the riglit prism FJ' whose later d edges ar-e each equal to a lateral edg'e of AD'. To prove Al) — FYJ'. Proof. AA1' FF'. Iyp. Subtractingy FPA' f rom each of these, AF= A'F'. (Wh~y) Siimilarly B G =B' G'. Also A B =A'B, and FG = F' G'. Art. 155. And X of f ace A G =homnolouous X of f ace A' G'. Art. 130. face AG==face _A'G', Art. 47. (for they hare all their parts eqpal, each to each, and.'. can be made to coincide). In like manner face AK-face A'K'. B3Ut face AD- fice 'AI'D. Art, 591,~ 3.. trunicated prismi AJ~ trvmcatedA prism A'14J'. Art, 611. To each of these, equals add the solid F-)'. AD'c~FJ'. (Why?7) Q. A. D. 368 BO00K VII. SOLID GEOMETRY PROPOITIOsIo V. THEOREM). 614. The oppo-)ste l(ateral face's of a jparalleloPiped are equtal awd parallel. 11 A 13 Given the parallelopiped AH wvith the base AC. To prove AG~and II ~ll, and A-J= and ii BfH. Proof. The base AC is a =. Art, 598...AB==and 11 EC. (Why.?) Also the lateral f(ace 11 is a z=. Art. 591, 2.., A4F a iid I I Ed'. (Why?7) Z~ P3AFz Z CEJ'. Art. 538. And =7 A G Z/17TH. Art. 162. Also plane AZG I1 planie H.Art. 538. Ini like miannier it inayr be shown that Ad' and BRH are eqnal and parallel. Q 615. CoR.. Any tao opposite faces of a Paallel11opiped way be taken as the bases. Ex. 1. How many edges has a parallelopiped? How many faces?7 flow many dihedral angles? How many trihedral angles Ex. 2. Find the lateral area of a prism whose lateral edge is 10 and whose right section is a triang le whose sides are 6, 7, 8 in. Ex. 3. Find the lateral area of a right prism whose lateral edge is 16 and whose hase is a rhombus with diagonals of 6 and 8 in. PRISMS36 369 PROPOSITION VI. THEOREM 616. A plane passed thi ough two diagonally opposite edges of' a para~lielopi'pedl divides the paralielopiped into two equivalent triangular priszs. F K P B C Given the parallelopiped AlT with a plane passed through the diagyonally opposite edges AF and CIL, forming the triangular prisms AJW-G an-d AD C-K. To prove ABC-G~c:ADC-K. Proof. Construct a plane 1L to one of the edges of the Prism formaing the, right section 1'Q118, having the diagonal Pit formed by the intersection of the plane FIHC1A. Theni P9 11 S, and 911 0, Pt, S. (-Why) IIP911 is a (. (hy) A PQR ==A P181. (Why) But the triangcular prism ABC-G =: a prism whose base is the righlt section P'QR anid whose caltitude is AF. Art. 613. Also the triangncrlar lprisln ADO-K o a pr-ism whose base is the right, section P811 and whose altitude is AF. (Why?) But the prismns having the equtil basess, P911 and PS11, an1d the same altitnde, AF, are equail. Art. 612. AIB(>G =_ADC-Kf. Ax 1. IX 370 BOOK VII. SOLID GE'OMETRY PROP1OSIT ION V1II THEOREM 6 17. ITf two ia tangigalar par a tic lop ipe (I ha, te equal b~ases they are to each other as thuer altihtaelL P AA Given the rectanguflar parallelopipe~ls PI anid P havin, equal bases an-d the altitues A'II' and AB. To prove PI: Pz-zA'ID':1A2. C AsE I. When the altitudes A'D' arni ABl are co?)j mie nsura ble. Proof. Find 'a common m-easure of At'Bl and AB, as AK, and let it be contained in A'B' n times and in ADB tunes. Then A'IB':A B~n Through the points of division of AIR' and A-[)p,, planes parallel to the bases. These planes will divide PI into a1, and P into ma small rectangutlar parallelopipeds, all equial. Art. 612. P P AIRD' AB. (Why) CASE II. When the altitudes A'B' and AD are incomtLet the pupil supply the proof, using the method of limits. (See Art. 554). 618. DEr. The dimensions of a rectangular parallelopiped are the three edges which meet at one vertex. PARALLELOPIPEDS 371 619. CoR. If 'two rectan~gular parallelopi0peds have two dimensg~ons in coinmon, they are to each other as their third dimensions. PROPOSITION VIU1. TH1EOREM 620. Two rectlangular parallhlopipcds having equal altitudes are to each other as theur bases. P C) (1~~~~~~~~~~~~~~~~~~~~~~~~~7 b b Given the rectangular parallelopipeds P and PI having the common altitude a,, and the dimensions of their bases bc and b', c', respectively. -P b)X c To prove 'b Proof. Construct the rectangular parallelopiped, Qwhose altitude is a and the dimensions of whose base are b and el. Then P C Art. 619. Q c' Also PI bl Wy1 Multiplying the corresponding members of these equalities, P b~e P'b'Xc' ~~~~~Ax. 4. Q. E. D. 621. CoR. Two rectangulilar pa mu 1tOieP eds having1 one 'limnension in conunon are to eatch other ais the 1)r~odllcts of the, Qther tivo (linmemis ions. 37' BOOK VII. SOLID (GEOMETLY PROPOSITION IX. THEORE3I 622. Auy trwo rectangular paralelop)ipeds are to each other as the products of their three dimensions. f) ' I I:,;',::ii b b' Given the rectangular parallelopipeds P and P' having the dimensions a, b, c and a', b', c', respectively. P a Xhb Xc To prove P X b X -;=at a IX b' Xc/' Proof. Construct the rectangular parallelopiped Q having the dimensions a, b, c'. Then -= Art. 619, Q c' Also Q a X b Art. 621. PI a' X b' Multiplying the corresponding members of these equalities, P _XbXc = — Ax. 4. P' a X b X c' Q. E. D. Ex. 1. Find the ratio of the volumes of two rectangular parallelopipeds whose edges are 5, 6, 7 in. and 7, 8, 9 in. Ex. 2. Which will hold more, a bin 10x2x7 ft., or one 8x 4x5 ft.? Ex. 3. How many bricks 8 x 4 x 2 in. are necessary to build a wall 80x6 ft. x8 in. l PARALLELOPIPEDS 373 PROPOSITION X. THEOREM 623. The volume of a rectangular parallelopiped is equal to the product of its three dimensions. p u b 1 Given the rectangular parallelopiped P having the three dimensions a, b, c. To prove volume of P= a X b X c. Proof. Take as the unit of volume the cube U, whose edge is a linear unit. Then P X Art. 622. U IX1XI The volume of P is the number of times P contains the P unit of volume U, or U Art. 603. volume of P=a X b X c. Q. E. D. For significance of this result, see Art. 2. 624. CoR. 1. The zolzme of a cube is the cube of its edge. 625. COR. 2. The volume of a rectangular parallelopiped is equal to the product of its base by its altitude. Ex. 1. Find the number of cubic inches in the volume of a cube whose edge is 1 ft. 3 in. How many bushels does this box contain, if 1 bushel=2150.42 cu. in.? Ex. 2. The measurement of the volume (if a cube reduces to the measurement of the length of what single straight line? 374 BOOK VII. SOLID GEOMETRY PROPOSITION XI. THlEOREM 626. The voltsme of Lany parallelopiped is equal to t/i prodact of its base by its altitcde. P 9 i II i in B \j Given the oblique parallelopiped P, with its base denoted by B, and its altitude by E. To prove volume of P=B X H. Proof. In P produce the edge CD and all the edges parallel to CD. On CD produced take FG= CD. Pass planes through F and G 1 the produced edges, forming the parallelopiped Q, with the rectangular base denoted by B'. Similarly produce the edge GI and all the edges 1l GI. Take IK= GI, and pass planes through I and iK I the edges last produced, forminig the rectangular parallelopiped BR, with its base denoted by B". Then P- Q R. Art. 613. Also B:B'=B". Art. 386. But volume of R-B" X H. Art. 625..volume of P=B" X Hi. Ax. 1. Or volume of P=B X I1. Ax. 8. Qe E. o. [Outline Proof. P Q- = -B" X = B X H.] PRISMS 375 PROPOSITION XII TE=OREM 627. The volaie of a triang-aWn pr ism is equal to the product of its base by its altitude. 'K LIT Given the triangular prism PQR-M1, with its volume denoted by V, area of base by B3, and altitude by H. To prove Y= B X -IT. Proof. Upon the edges P9, QR, 911, construct the parallelopiped QK. Hence QK= twice PQIR-Mf. Art. 616. But volume of QIW area -PQRt-7 X IT. Art. 626, 2, X H. Ax. 8.. twice volume PQR —I — 2B X TH. Ax. 1. volume PQR- X1f= 1XH. Ax. 5. Q. B. D. 'Ex. 1. If the altitude of a triangular prism is 18 in., and the base is a right trian gle whose legs are 6 and 8 in., find the volume. Ex. 2. Find the volume of a triangular prism whose altitude is 24, and the edges of whose base are 7, S, 9. Also find the total surfac. 376 3BOOK VII. SOLID GEOMETRY PROPOSITION XIII. TIHEOREM 628. The voliume of (ny prism, is equal to the product of its base by its altitude. T,1 El L A B Given the prism AK, with its volume denoted by V, area of base by B, and altitude by IH. To prove V= B X I. Proof. Through any lateral edge, as AF, and the diagonals of the base, AC and AD, drawn from its foot, pass planes. These planes will divide the prism into triangular prisms. Then V, the volume of the prism AK, equals the sum of the volumes of the triangular prismis. Ax. 6. But the volume of each triangular prisnm its base X 1L. Hence the sum of the volumes of the L prisms - the sum of the bases of the A prisms X H. -1 X R. Ax. 8. V= B X H. Ax. 1. Q. E. D. 629. CoR. 1. Two prisms are to each other as the products of their bases by their altitudes; prisms having equivalent bases and equal altitudes are equivalent. 630. COR. 2. Prisms having equivalent bases are to each other as their altitudes; prismis having equal altitudes are to each other as their bases. 0 I PYRAMIIDS PYRAMIDS 631. A pyramid is a pol!hedron bounded by a group of planes passing through a common point, and by another plane cutting all the planes of the group. 632. The base of a pyramid is the face formed by the cutting plane; the: — lateral faces are the faces formed by the Pyrarmidl group of planes passing through a conimon point; the vertex is the common point thrlough whlich the group of planes passes; the lateral edges are the intersections of the lateral faces. The altitude of a pyramid is the perpendicular from the vertex to the plane of the base. The lateral area is the sunm of the areas of the lateral faces. 633. Properties of pyramids inferred immediately. 1. The latferatl fatccs of a lpyr7ti id (ti1' tr'iatgles (Art. 508, 2). 2. The base of ac pyraminid is ca polygonl (Art. 508, 2). 634. A triangular pyramid is a pyramid whose base is a triangle; a quadrangular pyramid is a pyramid whose base is a quadrilateral, etc. A triangular pyramidl is also called a tetrahedron, for it has four faces. All these faces are triangles, and anLy one of them may bt taken as the base. 635. A regular pyramid / is a pyramid whose base is a I regular polygon, and thle foot / ' of whose (altitude coincides. - with the center of the base. 378 BOOK VII, SOLID GEOMETRY 636. Properties of a regular pyramid inferred immedi ately. 1. The lateral edges of a regula(r pyrmid tre eqtual, fop they are oblique lines drawn fromn a point to a plane cutting off equal distances from the foot of the per. pendicular froin the point to the plane (Art. 518). 2. The lateral fices of a regular py/ramlid are equal isosceles triangles. 637. The slant height of a regular pyramid is the altitude of any one of its lateral faces. The axis of a regular pyramid is.its altitude. 638. A truncated pyramid is thle portion of a pyramid included between the base and a section cutting all the lateral edges. 639. A frustum of a pyramid is the part of a pyramid included between the base and a plane parallel to the base. The altitude of a frustum of a pyramid is the perpendicular distance between the planes of its bases. 640. Properties of a frustum of a pyramid inferred immediately. 1. The labtral faces of a frustum of a pyramid are trapezoids. 2. The lateral faces of a frtstumt of a regular pyramid are equal isosceles trapezoids. 641. The slant height of the frustum of a regular pyramid is the altitude of one of its lateral faces. Ex. 1. Show that the foot of the altitude of a regular pyramid coincides with the center of the circle circumscribed about the base. Ex. 2. The perimeter of the midsection of the frustum of a pyramid equals one-half the sum of the perimeters of the bases, PYRAMIDS 3 379 PROPOSITION XIV. THEOREM 642. The lateral ar ea of a regular pIrainid, is equtal to half the product of the slant height by the perimieter of the base. 0 A4 D Given O-ABCDJ? a regular pyraumid with its lateral area denoted by S, slant height by L, and perimeter of its base by P. To prove S= EL X P. Proof. The lateral faces OAB, 0)C, etc., are equal isosceles A. Art. 686)G, 2. Hence each lateral face has the same slant height, L..8 the area of each lateral facee= -- L X its base. A the sum of all the lateral faces - L X sum of bases. ZZ L, X P. /SI'L X P. Ax. S. Q F D. 643. Coa. Thie lateral area of f a thefrutstum of a regular )';/ra1(1 I's equal to one-half the sum of the perimieters of its bases multiplied A D by its slant height. 1 Ex. Find the lateral airea of a regular square pyramidI wuosOe slant, heiglit is 32, and an edge of whose base is 16. Find the total area also, t 38 0 380 1BOOK VIt. SOLTID GEOMETRY PROPOSITION XV. THEOREM 644. If a p-yramid 'is cutt by a plane parallel to the base, I. The lateral edges ai,(d the altituwle are divided p-o~. po~rtionally; II. The section is a polygon si~milar to the base. Cjiven the pyramid S-ABODE, with the altitude SO cut by a plane MY, which is parallel to the base and intersects the lateral edges in a, b, c, d, f and the altitude in o. Sa Sb) Sc So To prove I. = IL. The section abcdf similar to the base ABODEF. Proof. I. Pass a plane through the vertex 5 11 limy. Then SA, SB, SOC.. SO, are lines intersected by three Iplanes. Sa Sb Sc SoAr.59 SA B SO.. -. r.59 IL. ab II AB. (Why?7) A~Sab and SAB are similar. Art. 328. In like manner the & Sbc, Sed, etc., are similar to the i~SBO, SOD, etc., respectively. abl 1b _) be _ (Si c' ed -~~- — = e~~~tc PYRAMIDS 381 That is, the homologous sides of abcdf and ABCDF are proportional. Also Zabc = Z ABC, Zbcrd = Z BCD, etc. Art. 538.. section abcdf is similar to the base ABCDF. Art. 321. Q. E. D. 645. COR. 1. A section of a pyranmid parallel to the base is to the base as the square of its distance from the vertex is to the square of the altit(de of the pyramid. abcdf _ at)2 For = (Why?) ABCDF A-B2 ~But ab Sa So l b" So" But —. —=.. = — (Why?) AB S- S0 ABI B O2 abcdf _ So '. __f So2. __ ~(Why?) ABCDF So2 AP3@i 5so9 646. COR. 2. If two pyramids having equal altitudes are cut by a plane parallel to their bases at equal distances from the vertices, the sections have the same ratio as the bases. Let S-ABCDF and V-PQR be two pyramids cut as described. The abcdf 70 _ (W y T'heln- adf — o (Art. 645); also pqr (hy?) ABCDF SO PQR VT But VT=SO, and t= So. (Why?) abcdf pqr rabd(( f ABCDF AB OD - PQR por (Why?) ABCDF -' ' pqr = PQR647. COR. 3. If two pyr amids hiave equal altittudes (nd!equivalent bases seions wrde by planes parallet to the bases at equal distances from the vertices are equlialent. 0? 3) BOOK VTI. SOLID GEOMETRY PRctIOt ITrION XVI. THEOREM 648. The volume)C of a t)ria)lngilar pyramid is the limit of the sut)i of the tolltes of fa ss of sei of inscribed, or of a series of circtise'rib(d Ipris s of cqual altitude, if the nlumber of prismis be indefin itely in/clease(d. 0 0 /1 Given the triangular prism O-ABC with a series of inscribed, and also a series of circumscribed prisms, formed by passing planes which divide the altitude into equal parts, and by making the sections so formed first upper bases, then lower bases, of prisms limited by the next parallel plane. To prove 0-ABC the limit of the sum of each series, if the number of prisms in each be indefinitely increased. Proof. Each inscribed prism equals the circumscribed prism immediately above it. Art. 629... (sum of circumscribed prisms)-(sum of inscribed prisms) = lowest circumscribed prism, or ABC-K. If the number of prisms be indefinitely increased, the altitude of each approaches zero as a limit. Hence volume ABC-K'-O, Art. 253, 2. (for its base, ABC, is constant while its altitude -0)..'. (sum of circumscribed prisms)-(sum of inscribed prisms) -0.. volume 0-ABC-(either series of prisms) - 0. (for this difference < difference between the two series, which last difference 0)... O-ABC is the limit of the sum of the volumes oi either series of prisms. Q. E. D. PYRAMIDS 383 PROPOSITION XVII. THEOREM 649. If two triangular pyramids have equal altitudes and equivalent bases, they arc ectquialcent. 0 0 B' Given the triangular pyramids O-ABC and 0'-ABf'C' having equivtent bases ABC and A'B'C', and equal altitudes. To prove 0-ABCG O'-A'B'C'. Proof. Place the pyramids so that they have the common altitude HT, and divide H into any convenient number of equal parts. Through the points of division and parallel to the plane of the bases of the pyramids, pass planes cutting the pyramids. Using the sections so formed as upper bases, inscribe a series of prisms in each pyramid, and denote the volumes of the two series of prisms by ' and PT. The sections formed by each plane, as KLM and K'L'MI', are equivalent. Art. 647,.. eaeh prism in 0-AB =c= corresponding prism in O'-AB'!C (Art. 629).. '. = ". Ax. 2. Let the number of parts into which the altitude is divided be increased indefinitely. Then V and 1' become variables with O-ABC and O'-A'B'C' as their respective limits. Art. 648. But 1T T1' always. (Why?) 'oo -A4 -B 7C O'-A'B' C'. (W\hy?). E. D. 384 384 ~BOOK VII. SOLID GEOMIETRY PROPOSSITION XVIII. THEOREM. 650. Thr Vmlaiffc of a triangular pyjramvid is eqjual to oie -fliirl tlir pr-oduct oJ;its base l)y iHS a It i Iae Given the triangular pyramid 0-A BC, havingr its volume ceuoted by V, the area of its base by B, and its altitude by HI. To prove T7= t B X Hf. Proof. On ABC as a base, with OB as a lateral edge, coiistruct the prism ABCU-DOE. Then this prism will be composed of the original pyra~ mid 0-ABC and the quadrangyular pyramid 0-ADEG. Througrh the edgres Of) and OC pass a plane intersecting the face ADEC in the liucDC, and dividhingthe quadrangular pyr~amid into the triaingullar pyramids O-ADC and 0-DEC. Thien 0-A- I)C~cO -DFC. Art. 649. (for they~ hare the commi~on icertex 0, awd the eqiual bases ADC anld DEC). B-'ut 0-DEC mjay be regfarded as having C as its vertex. and DOE as its base. Art. 634. 0' -DECu:~- 0-AB C. Art.- 649. the prism is made np of three equivalent pyramids. 0-ABCz-L the prism. Ax. 5. But volume of prism =B X H. Art. 627. 0-A BC, or V- IB X H. Ax. 5. _ _ _ _ _ _ _ _ _ _ _ _ _ _ Q.~~ ~ E. D. Ex. Finid the volume of a triangular pyramid whose altitude is 12 ft., and whose base is an equilateral triangle with a side of 15 ft. PYRAMIDS38 085 PROPOSIT1ON XIX. THEOREM 651. The volumne of any pyramid is equal to onte-third ~the product of its base by its (altitude. 0 F B C Given the pyramid ()-ABCDF, having, its volume denoted by -V, the area of its base by B, and its altitude by H. To prove V= % B X H. Proof. Through any lateral edge, as OD, and the diagonfals of the base drawn from its foot, as AD and BD, pass P'lanes, dividing the pyramid into triangular pyramids. Then Y, the volume of the. pyramid 0 -ABCDE, will equal the sum. of the volumes of the triangular pyramids. B~ut the volume of each A pyramid =- its base X H. Art. 650. Hence the sum of the volumes of A~ pyramids 4I sum of their bases X H. Ax. 2. =&B XH. Ax.8S. IF= k B XI. Axi. Q. E. D. 652, CoR,. 1. The volumtes of twvo pyramids are to each Other as the products of their batses an~d altihtudes; pyram,,ids having equivalentt bases antd equal altitudes areeqiaet 653. CoR,. 2. -Pyiramids having equivalent ba~ses are to lach other as their altitudes; pyramids having equal altitudes are to each. other as their, bases. 654. SCHOLIUM. The v~olume of (any polyhedron may be found by dividing thc polyhedroni in o pyramids, fin~ding the ~'OlUhe of each ptqrawnid scparatcly, and taking their sumi, 386 BOOK VII. SOLID GEOMETRY PROPOSITION XX. THEOREM 655. The frustum of a triangular pyramid is equivalent to the sum of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the lower base, 'the upper base, and a mean proportional between the two bases of the frustulm. D F B Given ABC-DEF the frustum of a triangular pyramid, having the area of its lower base denoted by B, the area of its upper base by b, and its altitude by H. To prove ABC-DEE =-0 three pyramids whose bases are B, b and 1/Bb, and whose common altitude is H. Proof. Through E and AC, E and DC, pass planes dividing the frustum into three triangular pyramids. Then 1. E-ABC has the base B and the altitude H. 2. E-DFC, that is, C-DEF, has the base b and the altitude H. 3. It remains to show that E-ADC is equivalent to a pyramid having an altitude H, and a base that is a mean proportional between B and b. Denoting the three pyramids by I, II, III, I L ABE AB A C A ADC II II L ADAD ~DE D LF DFC II (Arts. 653, 391, 644, 321. Let the pupil supply the reason for each step in detail). I II *- = I-(Ax. 1.) or II=1/IXIII. Art. 303.. E-ADC= I/(. HXB) ( H X b) = H /BX b. Hence, ABC-DEFV sum of three pyramids, as described. Q.. D. PYRAMIDS 387 656. Formula for volume of frustum of a triangular pyramid. v= I- (B + b + RBb). PROPOSITION XXI. THEOREM 657. The volume of the frushtu. of any pyramid is equivalent to the sumi of the volumes of three pyramt(ids, whose common altitude is the altitude of the frbstumin, and Zwhose bases are the lower base. the upper base, and a mean proportional between the two bases (:' the fr stem. T A D, base by b, and its altitude by H. ___ To prove -= f (B -+ b + 1/ Bb). Proof. Produce the lateral faces of Ad to meet in K. Also construct a triangular, py-ramid with base PQR,qnivalent to ABCDF, and ini thle samne plane n with it, and with an altitude eq'ual to tlie altitude of K-1ABCD)F. Prolice the plane of ad to cut. the secoid pyramid in pqr. Then pqr 7 abedf. - Art. 647,.'. pyramid K-ABCD F =c ppyramid T-PQR. Art. 652. Also pyramid K-,abcdf - py-rataiid- Tpqr. (Why ) Subtracting, frustum A-d =c f rulstuml Pr. Ax. 3. But volume P - - H ( B -+ b + T i ). volume AI-d II (lb -i iWk. (Why?) ** * * -* **- * * * ""Q.-K B. -~~~~~~~~~ 98 BOOK VII. SOLID GEOMETRY PROPOSITION- XXII THEOREM 658. A truncated triangatlar prismw is (pmui alent to thK, sumwi of threne pyramitls, of which the base of the,, prisnl is the comm)non base, and whose vertices are the thuee vertices of the inclined section. B B B Fig. 1 Fig. 2 Given the truncated triangular prism ABC- PQB. To prove ABC-PQB =c= the sum of the three pyramids P-ABC, Q —ABC and -RABC. Proof. Pass planes through Q and A C, Q and PC, dividing the given figure into the, three pyramids Q-ABC, Q-APC and Q-PBC. 1. Q-ABC has the required base and the required vertex Q. '2. Q-APC -- B-APC, Art. 652. (,Jbr they have the same base, APC, and, the same altitude, their vertices being in a line fl base APC). But B -APC may be regarded as having P for its vertex, and ABC for its base, as desired. Art. 634. 3. Q-PRC B-ARC (see Fig. 2). Art. 652. (for the base ARC z base PRC (Art. 390); and the altitudes of tie two pyramids are equa?, the vertices Q and B being in line 11 plane PA CR, in which the bases lie). But B-ABC may be regarded as having B for its vertel, and ABC for its base, as desired, Art, 634 - PRISMATOIDS 0OIS9.. ABC-PQR =o: sum of three pyramids whose common base is ABC, and whose vertices are P, Q, R. i. g. D. Fig, 13 Fi 659. COR. 1. The volumne of a truncated 'right triangiular prism)j (Fig. 8) is ealqv to the product of its base by one-third the sitamn of its lateral edges. 660. COR. 2. The volume of any truncated triangular prism (Fig. 4) is equal to the product of the area of its right section by one-t third the suni of its lateral edges. PRISMATOIDS 661. A prismatoid is a polyhedron bounded by two polygons in parallel planes, called bases, and by lateral faces which are either trimn gles, trapezoids or parallelograms. 662. A prismoid is a prismatoid in which the bases have the same number of Sides and have their corresponding sides parallel. Ex. The volume of' a truncated right parallelopiped equals the area of the lower base multiplied by one-fourth the sum of the lateral edges (or by a p)erpendteular from the center of the upper base to thre lower base). 390 1300K VII. SOLID GEOMETRY PROPOSITION XXIII. TnIOIM) 663. Th1e volumt of a pr isatfoid is (eqacl to one-ssixti the product of its altitude by tlie sium of its bases and of four times the area of its mi(tsectioln. F K Fig' 1 Fig. 2 Given the prismatoid ABCD-FGK, with bases B and b, midsection M, volume Vf, and altitude 11. To prove T=;I (B + b + 4 3). Proof. Take any point 0 in the midsection, and through it and each edge of the prismatoid let planes be passed. These planes will divide the figure into parts as follows: 1. A pyramid with vertex 0, base ABCD and altitude ~ H, and whose volume.~. = i-; X B. Art. 651, 2. A pyramid with vertex ), lbase FGK, and altitude A H, and whose volume.~. = 1 X b. (Why?) 3. Tetrahedrons like 0- 1Al(; whose volume may be determined as follows (see Fig. 2): AB-=2 PQ. (Why?).. A Ai G B= 4 A PGQ. Art. 398... 0-AGB=4 O-PGQ. Art. 653. But O-PGQ (or G-I'QO)= PQO X i- H= H X PQO. A. - BG - H X 4 A PQO. (Why?).. the sum of all tetrahedrons like -A GB3 H X 4 M.. H. V-I XB+ -I HIX b-x + 1 X 4 3I. Or v= 1 ( B+ b + 4 I ). Q... Do REGULAR POLYHEDRONS 391 REGULAR POLYHEDRONS 664. DEF. A regular polyhedron is a polyhedron all of whose faces are equal regular polygons, and all of whose polyhedral angles are equal. Thus, the cube is a regular polyhedron. PROPOSITION XXIV. THEOREM 665. But five regular polyhedrons are possible. Given regular polygons of 3, 4, 5, etc., sides. To prove that regular polygons of the same number of sides can be joined to form polyhedral X of a regular polyhedron in but five different ways, and that, consequently, but five regular polyhedrons are possible. Proof. The sum of the face A of any polyhedral angle < 360~. Art. 583. 1. Each Z of an equilateral triangle is 60~. Art. 134. 3 X 60~, 4 X 60~ and 5 X 60~ are each less than 360~; but any larger multiple of 60~=or > 360~... but three regular polyhedrons can be formed with equilateral A as faces. 2. Each Z of a squtare contains 90~. Art. 151. 3 X 90~ is less than 360~, but any larger multiple of 90= or > 360~... but one regular polyhedron can be formed with squares as faces. 3. Each Z of a regular pentagon is 108~. Art. 174. 3 X 108~ is less than 3600, but any larger multiple of 108~ > 3600~ 39 2J 8BOOK VIIT. SOLI'T]) GEOMETRY.'but one re-gular p(lyhliedirn can be formed with regu. lat pentagons as faces. 4. Each Z of a regular lhexagon is 120~, and 3 X 120~ _360'.'. no regular polyhedron can be formed with hexagons or with polygons with a greater number of sides as faces... but five regular polyhedrons are possible. Q. E. D. 666. The construction of the regular polyhedrons, by the use of cardboard, may be effected as follows: I)raw on a piece of cardboard the diagrams given below. Cut the cardboard half through at the dotted lines and entirely through at the full lines. Bring the free edges together and keep them in their respective positions by some means, such as paltingl strips of paper over them. Tltraltf(lron.'' I I Octahedron Hexahedlron DodecalledrQ, I% I I,, I / ' I N I I O 1 I I,, I - / -4/ ---XheIcosahedlrQn POLYIHEDRONS 393 POLYHEDRONS IN GENERAL PROPOSITION' XXV. TiEOREM 667. In atny polyhedron, the number of edges increased by heo equals the numnber of rertices i'ncreased by the numiber of fac OS. P T B ~~~~B Given the polyhedron AT, with the number of its vertices, edges and faces denoted by V1, F and F, respectively To prove E +2=Vr+ F. Proof. Taking~ the siiignle face AB'CD, the number of edgyes ecquals the number of vertices, or, F- V. If another face, CR TD, he annexed (Fig,. 2), three new et~ges, (JRt~ RT, TI), are added and hto new v-ertices, II and T. the number of edgyes gyains onle on the number of vertices, or F=I+l1. If still another face, R,>Q-RC, be annexed, two new edges, RBQ and QIt, are added, and one new vertex, (2. A- E- V-F20 VWith each new face that is annexed, the number of edges gains one on tile number of vertices, 'till but one face is lacking. ITile last face increases neither tile luailber of edges nor of vertices. H1ence numbr of edg'es gains one on number of vertices, for, *#very ithee except two, the first and the last, or gains P —2 in all. tor tiht,entire figuire, E=~ V — F- 2. That iS -E~+2 -V~7 F. A x. 2. Q. EI. 0`9 4e BO41300K VII. SOLID GEOMETLY PROPOSITION -XXVN 1. THEOREM 668. The sum of the fiJcc angles of any polyhedron, equals four right angles taben, as mtany tuinzes, less Jwo, nt the polyhedron has -etY b-tees. Given any polyhedron, with the sum of its f ace angles denoted by 8, and tie number of its vertices, edges and faces denoted by V, E, F, respectively. To prove S= (07 2) 4 rt. A. Proof. Each edge of the polyhedron is the intersection of two faces,.'. the number of sides of the faces=2 E. the sum of the interior and exterior A of the faces= 2 E X 2 rt. A, orFE X 4 rt. As. Art.a73. But the snm of the exterior A of each face=4 rt. A. Art. 175., the sum of exterior A of tie F faces = F X 4 rt. Z AxBL. 4. Subtractinlg the sum of tie exterior A from the sum of all the Z, the sum of the interior A of the F faces= (E X 4 rt. A )-(F X 4 At. Z ). Or S=(E-F)4rt. A. But E+2=V+F. Art. 66 Hence E-F= V-2. Ax. 3, Substituting for E-F, S= ( V-2) 4 rt. A. Ax. 8. Q. ~. ~. Ex. Verify the last two theorems in the case of the Cube. COMPARISON OF POLYHEDRON~S 9 395 COMPARISON OF POLYHEDRONS. SIMILAR POLYHEDRONS PROPOSITION XXVII1. THEOREM 669. If tuwo tet rahedirons hav~e a trihedral angle of one equal to a trihediral angle of the other, the?, are to each otheras the products of the edges including the equal trihedral angles. C C 0 0 ~ ~~~~~~~~~~~~~~~~~ B Given the tetrahedrons 0-ARC and 0'-A'B'C', with their volumes denoted by Vand 1V', respectively-, and having the trihedral A 0 and 0' equal. Toprve V= OAXOBXOC To proveI 01A1 X 0'R' X IC Proof. Apply the tetrahedron O'-A'R'C' to 0-ABC so that t~he trihedral Z 0' shall coincide with its equal, the trihedral Z 0. Draw CP and 0'P' I plane OAB, and draw OP the projection of 00 in the plane OAB. Takingo OAR.I and OA'R' as the bases, and CP and C'P' as the altituides of the pyramids C-OAR3 and C'-OA'R', respectivelyV A OAB XOP A OAR XCj-~ Art. 652. But ALOARB OAX OR Art. 39 7 ~L OA'B' OA' X OB' In the similar rt. & OCP and OC'P', — =- (Why) V7 OA X OBXO OC A,4X OB XOC VI OA' X OR' X 00' 0'A/ X O'R' X 01 Ax. 5 Q.E.D 396 1-B0OK Vii. SOLID GEOMETRY 670. Dui,: Similar polyhedrons are polyhedrons havicu the same number of faces, similar, each to each, and sra"I-i larly placed, and having their corresponding polyheda angles equal. PRoPO'SITION XXVIII. THEOREM 671. Aaqi two sitnilar polyhedrons miay be decomp~aed into the samne vamber of tetrahtedrons, sibniiar, each to each, and similarly placeti. Given P and PI, twvo similar polyhedrons. To prove that P and PI may be decomposed into the same number of tetiahedrons, similar, each to each. Proof. Take ITi and HI any two homologrous vertices of Pand PR. Draw homologous diagonals in all the faces of P and PI except those faces which meet at H and H', separating the faces into corresponding similar triangles. Througfh fi and each face diagonal thus formed in P, and throughl RII and each face diagonal in P', pass planes. Each corresponding pair of tetrahedrons thus formed may be proved similar-. Thns, in H-ABC anid Ii'-A'13C', the & HIBA and H'B'A' are similar. Art. 329. In like manner & HBC and H'B'C' are similar; and ~ ABC and A'B'C' are similar. SIMILAR POLYGONS39 397 HA (HB\_ HC _B1C_ AC Also - \fB',B','~jj* Art. 321. & AHC and A'Hf'C' are similar. Ait. 326. Hence the corresponding f aces of H-ABC and fl'-A'B'C' are similar. Also their homologous trihedral!~ are equal. Art. 584. tetrahedron H-ABC is simiflar to H'-A'B'C'. Art. 670. After removing HT-ABC from P, and ff-A'B'C' from PI, the remaining polyhiedronis arie similar, for their faces are similar, and the remaininig polyhedral X are equal. Ax. 3. By continuing this process, P aind P' may be decomposed into the same number of tetrahedrons, similar, each to each, and simnilarly placed. Q. E. D. 672. COR. 1. The hon)iologous edges of similar polyhedrons are proportioital; Any two hom,?ologous lie es~ 'in two si m iitarl polgh edrons heave the same ratio as any other two hoiiologous libies. 673. COR. 21. Aniy two hom-oloyoits laces of two similar pot jhedrons are to each, other as thec squares of anty two homologous edges or lines; The total areas of any two shinalar polyhedirons are to -each other as the squares of ctng to hoimologouis edges. EX. 1. In the figure, P. 395, if the edges meeting at O are 8, 9, 121 in., and those meeting at 0' are 4, 6, 5 in., find the ratio of the volumes Of the tetrahedrons. 1Rx. 2. If the linear dimensions of one room are twice as great as the corresponding dimensions of another room, how will their surf aces (and.'. cost of papering) compare? How -will thour volumes compare Ex. 3. How many 2 in. cubes can be cnt from a 10 in. cube? Mx. 4. If the, bases of a prismoid are rectangles whose dimensions 4r a and b all and altitude, is H, find the formula for the volumes 398 BOOK VII. SOLID GEO.METIIY PROPOSITION XXIX. THEOREM 674. The volues of twvo similar) tetra~hedronis are to each oth~er as the cubes of any pair of homologous edges. 0 B Given the similar tetrahedrons 0-ABC and O'-A'B'C'. v OAi To prove - Proof. V OA X OBYXOC (Why?7) VI O'A' X O'B' X 0'C' OA xOR O O'A' 0-B' 0'C' OA OD 00 But -.Art. -672. O'A' O'BI OW'I VO0 04 OA OA~ F (fjjf~~~/~4/ O'A' ~Ax. 8. Ex. 1. In the above figures, if AB=2 A'1B', find the ratio of, V to 77'. Find the same, if AB= 14 AIB'. Ex. 2 The measurement of the volume of a regular triangular prism reduces to the measurement of the lengths of how many straight lines? of a f rustum of a regular -square pyramid? Ex. 3. Show how to'constrUct o~ut of pasteboard' a vegular pft~mn, s -pe~tiUelopiped, and a, truncated square prism. SIMILAR POLYGONS 399 PROPOSITION XXX. THEOREM 675. The volumes of any tlo similar polyhedrons are to each other as the cubes of any two homologous edges, or of any other two homologous lines. Given the polyhedrons AK and A'K' having their volUmes denoted by V and V', and HB and H'B' any pair of homologous edges. To prove _ HB3 TV Krt/B'3 Proof. Let the polyhedrons be decomposed into tetrahedrons, similar, each to each, and similarly placed. Art. 671. Denote the volumes of the tetrahedrons in P by vi, v2, '3. and of those in PI by v'i, 'v 2, '3... Then H3. Also 1'...3 Art. 674, A. I. Vtl '1 t2 Vt3 HB~ (for each of these ratios — ) H'B'3 vl- + V2 + -'3 -- V,'1 T1 ~'+ —,- ~,-=+-==,; that is, -= — Art. 312, 7v1 - v% q- '3 q --,') ' F' v I V __ H]:.'. t -,-.__ 'Ax. 1. 400 BOOK VII. SOLID GEO5METRY EXERCISES. CROUP 67 THEOREMS CONCERNIN( POLYHEDRONS Ex. 1. The lateral faces of a right prism are rectangles. Ex. 2. A diagonal plane of aI prism is parallel to every lateral edge of the prism not contained in the plane. Ex. 3. The diagonals of a parallelopiped bisect each other. Ex. 4. The square of a diagonal of a rectangular parallelopiped equals the sum of the squares of the three edges meeting at a vertex. Ex. 5. Each lateral face of a prism is parallel to every lateral edge not contained in the face. Ex. 6. Every section of a prism made by a plane parallel to a lateral edge is a parallelogram. Ex. 7. If any two diagonal planes of a prism which are not parallel to each other are perpendicular to the base of the prism, the prism is a right prism. Ex. 8. What part of the volume of a cube is the pyramid whose base is a face of the cube and whose vertex is the center of the cube? Ex. 9. Any section of a regular square pyramid made by a plane through the axis is an isosceles triangle. Ex. 10. In any regular tetrahedron, an altitude eqluals three times the perpendicular from its foot to any face. Ex. 11. In any regular tetrahedron, an altitude equals the sum of the perpendiculars to the faces from any point within the tetrahedron. Ex. 12. Find the simplest formula for the lateral area of a truncated regular prism of n sides. Ex. 13. The sum of the squares of the four diagonals of a paralo lelopiped is equal to the sum of the squares of the twelve edges. [SUG. Use Art. 352.] Bx. 14. A parallelopiped is symmetrical with respect to what point? Ex. 15. A rectanglar parallelopiped is symmetrical with respect to how many planes? (Let the pupil make a definition of a figure symmetrical with respect to a plane. See Arts. 486, 487.) EXERCISES ON POLYHEDRONS 4.01 Ex. 16. The volume of a pyramid whose lateral edges are the three edges of the parallelopiped meeting at a point is what part of the volume of the parallelopiped? Ex. 17. If a plane be passed through a vertex of a cube and the diagonal of a face not adjacent to the vertex, what part of the volume of the cube is contained by the pyramid so formed? Ex. 18. If the angles at the vertex of a triangular pyramid are right angles and each lateral edge equals a, show that the volume of the pyramid is -' Ex. 19. IHow large is a dihedral angle at the base of a regular pyramid, if the apothem of the base equals the altitude of the pyramid? Ex. 20. The area of the base of a pyramid is less than the area of the lateral surface. Ex. 21. The section of a triangular pyramid by a plane parallel to two opposite edges is a parallelogram. If the pyramid is regular, what kind of a parallelogram does the section become? Ex. 22. The altitude of a regular tetrahedron divides an altitude of the base into segments which are as 2 1. Ex. 23. If the edge of a regular tetrahedron is a, show that the ave3 a V'6 slant height is -; and hence that the altitude is —, and the vole 2 3 ume is /2 12 Ex. 24. If the midpoints of all the edges of a tetrahedron except two opposite edges be joined, a parallelogram is formed. Ex. 25. Straight lines joining the mid(points of the opposite edges of a tetrahedron meet in a point and bisect each other. Ex. 26. The midpoints of the edges of a regular tetrahedron are thbe ertices of a regular octahedron, Z 1B00K VII. SOLID GEOMETRY EXERCISES. GROUP 68 PRlOBLEM3IS CONCERNING POLYHEDRONS Ex. 1. Bisect the volume of a given prism by a plane parallel to the base. Ex. 2. Bisect the lateral surface of a given pyramid by a plane parallel to the base. Ex. 3. Through a given point pass a plane which shall bisect the volume of a given parallelopiped. Ex. 4. Given an edge, construct a regular tetrahedron. Ex. 5. Given an edge, construct a regular octahedron. Ex. 6. Pass a plane through the axis of a regular tetrahedron so that the section shall be an isosceles triangle. Ex. 7. Pass a plane through a cube so that the section shall be a regular hexagon. Ex. 8. Through three given lines no two of which are parallel pass planes which shall form a parallelopiped. Ex. 9. From cardboard construct a regular square pyramid each of whose faces is an equilateral triangle. EXERCISES. CROUP 69 REVIEW EXERCISES Make a list of the properties of Ex. i. Straight lines in space. Ex. 9. Right prisms. Ex. 2. One line and one plane. Ex. 10. Parallelopipeds in geno eral. Ex. 3. Two or more lines and one plane. Ex. 11. Rectangular paralleloEx. 4. Two planes and one line. pipeds. Ex. 5. Two planes and two lines. Ex. 12 Pyramids in general. Ex. 6. Polyhedrons in general. Ex. 13. Regular pyramids. Ex. 7. Similar polyhedrons. Ex. 14. Frusta of pyramids. Ex. 8. Prisms in general. Ex. 15. Truncated prisms. BOol(, VIIT CYLINDERS AND CONES CYLINDERS 676. A cylindrical surface is a curved surface generated by a straight line which moves so as / constantly to touch a oiven fixed curve and constantly be parall 1l to a given fixed straight line. /,, Thus, every shadow cast by a point B of light at a great distance, as by a star or the sun, approximates the F cylindrical form, that is, is bounded Cylindrical surface by a cylindrical surface of light. Hence, in all radiations (as of light, heat, magnetism. etc.) from a point at a great distance, we are concerned with cylindrical surfaces and solids. 677. The generatrix of a cylindrical surface is the moving straight line; the directrix is the given curve, as CDE; an element of the cylindrical surface is the moving straight line in any one of its positions, as DF. 678. A cylinder is a solid bounded y a cylindrical surface and by two parallel j;il/ Planes. // The bases of a cylinder are its parallel plane faces; the lateral surface is the i cylindrical surface ineluded between the Parallel planes forminig its base::s; tlhe alti- ('lilder tude of a cylinder is the distance between the bases. The elements of a eyliiltdr are tle elements of the cylindrical surface bounding-' it. (-ttj) 404 1BOOK VIII. SOLID GEO3METIRY 679. Property of a cylinder inferred immediately. _4ti the ele,7zents of a cylindelr are eqlual, for they are parallel lines included between parallel planes (Arts. 532, 676). The cylinders most important in practical life are those determined by their stability, the ease with which they can be mlade from common materials, etc. 680. A right cylinder is a cylinder whose elements are perpendicular to the bases. 681. An oblique cylinder is one whose elements are oblique to the bases. l 682. A circular cylinder is a cylinderlt whose bases are circles. Oblique circular 683. A cylinder of revolution is a cylin- cylinder der generated by the revolution of a rectangle about one of its sides as an axis. Hence, a cylinder of revolution is:a right circular cylinder. Some of the properties of this solid are derived ' || most readily by considering it as' generated by a revolving rectangle; and others, by regarding it as a particular kind of cylinder derived from the general Clinder of definition. revolution 684. Similar cylinders of revolution are cylinders generated by similar rectangles revolving about homologous sides. 685. A tangent plane to a cylinder is a plane which contains one element of the cylinder, and which does not cut the cylinder on being produced. Ex. 1. A plane passing through a tangent to the base of a circular cylinder and the element drawn through the point of contact is tangent to the cylinder. (For if it is not, etc.) Ex. 2. If a plane is tangent to a circular cylinder, its intersection with t.le slane of the base is tangent to the base, CYLINDERS 405 686. A prism inscribed in a cylinder is a prism whose lateral edges are elements of the cylinder, and whose bases are polygons inscribed in the bases of the cylinder. I- i '41,~llll l 'ii Inscribed prism Circumscribed prism 687. A prism circumscribed about a cylinder is a prism whose lateral faces are tangent to the cylinder, and whose bases are polygons circumscribed about the bases of the cylinder. 688. A section of a cylinder is the figure formed by the intersection of the cylinder by a plane. A right section of a cylinder is a section formed by a plane perpendicular to the elements of the cylinder. 689, Properties of circular cylinders. By Art. 441 the area of a circle is the limit of the area of an inscribed or circumscribed polygon, and the circumference is the limit of the perimeters of these polygons; hence 1. The volume of a circular cylin)der' is the limit of the volume of an inscribed or circulscribed prism. 2. The lateral area of a circutlar cylinder is the limit of the lateral area of an inscribed or circiumscribed prism. Also, 3. ~By methods too advancedfor this book, it may be proved that the perimeter of t righrt jction. ti the limit of the perimeeter of a righlt section of an insr'ibed or circl!i scr'ied prism. 40f BOO1()K V\1 I.4OLi, (, EOM{ ITH'i PRCzOPOSITION i. r l-IEORE.ML 690. Ecery section of a ciylinlder made by a plane passing througli an element! is a p(tarllelogram. Given the cylinder AQ cut by a plane passing through the element AB and forming the section ABQP. To prove A BQ P a Z 7. Proof. AP i BQ. Art. 531. It remains to prove that PQ is a straight line 1l AB. Through P draw a line in the cutting plane |I AB. This line will also lie in the cylindrical surface. Art. 676..'. this line muist coincide with PQ, (for the line drawn lies in. both tl e cutting p1la)lc alnd the cylindrical suifacc, tienlcc, it,n.st 7e their intersection).. PQ is a straight line ]| AB... AIQP is a /. (Why ) Q.. D. 691. CoR. Every section of ca 'riqlt cylinder made by a plane passing through an element is a rectangle. Ex. 1. A door swinging on its hinges generates what kind of a solid? Ex. 2. Every section of a paiallelopiped made by a plane intero secting all its lateral edges is a parallelogram, CYLINDERS 407 PROPOSITION IT. THEOREM 692. The bases of a cylinder are equal. Given the cylinder AQ with the bases APR and CQD. To prove base APB=base CQD. Proof. Let AC and BD be any two fixed elements in the surface of the cylinder AQ. Take P, any point except A and B in the perimeter of the base, and through it draw the element PQ. Draw AB, AP, PB, CD, CQ, QD. Then AC and BD are = and I1. (Why?).'. A) is a Z7. (Why?) Similarly AQ and BQ are Z7..' AB= CD, AP= CQ, and BP= DQ. (Why?).. A APB = Z CQD. (Why?) Apply the base APB to the base CQD so that AB coincides with CD. Then P will coincide with Q, (for A PB= CQI)). But P is any point in the perimeter of the base APB..'. every point in the perimeter of the lower base will coincide with a corresponding point of the perimeter of the upper base... the bases will coincide and are equal. Art. 47. Q. E. D. 408 408 BoOR~~1101 VTII. SOLID41 (C T'OMETLYk 693. CoR. 1. The sections of a cylinder 7mado by tj(wo parallel planes cattting all thte elements are eqial. For the sections thus formed are the bases of the cylinder included between the eutting planes. 694. CoR. 12. Anty slection of a cylinder parallel to the1 base is equal to the base. PROPOSITION' III. THEOREMT 695. The lateral area of a circular;), cyliinler is equal7 to the product of the per-im~eter of a1 right section of the cylin. der by a-i elemtent. A Given the eireular cel-limier AJ, havinga its lateral area denoted by 8, an element hy EP and the perimeter of a right section by P. To prove S =PX E Proof. Let a prism with a regular polygon for its base be inscribed in the cylinder, Denote the lateral area of the inscribed prism by 8', and the perimeter of its right section by P'. Then the lateral edge of the inscribed prismi is an elemnent of the cylinder. Constr CYLINDERSIS 4(1 409 S'P= 'X E. Art. 608. If the number of lateral faces of the inscribed prism be indefinitely increased, S' will approach S as a limit. Art. 689, 2. P' will approach P as a limit. Art. 689, 3. And PI X E will approach PX E as a limlit. Art. 253, 2. But S' PI'XE always. (Why 9) S=zPXE. (Whyp ) Q. E. D. 696. COR. 1. The lateral area ofa caeyituder of re1vol lion is eqypwl to the, product of the ciir- wife-renlee of its base by its altitude. 697. Formulas for lateral area and total area of a cylinder of revolution. Denioting the lateral area of a cylinder of revolution by 5, the total area byr T. the radius by R, and the altitude by fi. T = 2Ifl+ 7t2... T=2 R (H+ R). Exc 1. If, in a cylinder of revolution. H=lO in. and R=7 in., find S and T. Ex. 2. If the altitude of a cylinder of revolution equals the radius of the base (RI=-i), whai:lt dlo the formullas for S and TI become in terms of J-1?? also, in telinus of H1? Ex. 3. What do they become, if the altitude equals the diameter of the base? Ex. 4. In a cylinder of revolution, what is the ratio of the lateral areBa to the area of the base? to the total area 7 41o BOOK VITT. SOLT GEOMETRY PROPOSITION IV. THEOREM 698. The volume of a circular cylinder is equal to the product of its base by its altitude. t /; I / ',/i Gen the circula cyline J having is volu Given the V, its base by ine, han its avolume by denoted by V, its base by B, and its altitude by Z. To prove V= B X i. Proof. Let a prism having a regular polygon for its base be inscribed in the cylinder, and denote the volume of the inscribed prism by TFU, and its base by B'. The prism will have the same altitude, H, as the cylinder. /. V' X H. (Why?) If the number of lateral faces of the inscribed prism be indefinitely increased, TV will approach V as a limit. Art. 689, 1. B' will approach B as a limit. (Why?) And B' X H will approach B X H as a limit (Why?) But V'I=B' XH always, (Why?).. F=B X RW. (Why ) Q E., B, 699. Formula for the volume of a circular cylinder, By use of Art. 450, CYLINDERS 4 411 PROPOSITION V. TTEOREMi 700. The lateral areas, or the total areas, of two simitar cylinders of revolution are to each other as the squares of their radii, or as the squares of their altitudes; and their volumes are to each other as the cubes of their radii, or as the cubes of their altitudes. L L! --- — ---- Given two similar cylinders of revolution having their lateral areas denoted by S and S', their total areas by T and T/, their volumes by V and V', their radii by R and R', and their altitudes by H and H', respectively. To prove S T: T R2: T'-R: R'=H2: jH/2; and V ': =R3:/=fl: H3 H RI? H+ R/ Proof. R=- 7- Arts. 321, 309. S 92 tRR'H RX H R H R: H2 S 2 R' RX/ ~ = r — = IF' (w y?) T 2 7R -R ) R H RI R It IF T 27FR (Hf-+RJ) h HH+R H2 Hf Tl ' 2 '7R' (-H' + ') R -I' + H -H'- 112 (Why?) V T tHH R2 1P," H R: I V X jFHR~ V tl t' _a t: Also (Why?) T -;, = '~ ' - '-' H —.... Q. E. D. Ex. If a cylindrical cistern is 12 ft. deep, how mluch more cement is required to line it than to line a similar cistern 6 ft. deep? How tuch more water will the former cistern hold? 412 BIOOK VITT. 'O(IAD GEOMETrY'II CONES 701. A conical surface is a sur- face generated by a straight line.t which moves so as constantly to touch a given fixed curve, and constantly v pass through a given fixed point. Thus every shadow cast by a near point of light is conical in formi, that is, is bounded 0 by a conical surface of light. Ience, the study of conical surfaces and solids is im- portant from the fact that it concerns all cases of forces radiating from a near point. 702. The generatrix of a conical c0 A surface is the moving straight line, as B' AA'; the directrix is the given fixed Conical surface curve, as ABC; the vertex is thle fixed point, as 0; an element is the generatrix in any one of its positions, as BB'. 703. The upper and lower nappes of a conical surface are the portions above and below the vertex, respectively, as O-ABC (lnd lO-AI'C'(Y. Usually it is convenient to limit a conical surface to a single nappe. 704. A cone is a solid bounded by a conical surface and a plane cutting all the elements. 705. The base of a cone is the face formed by the cutting plane; the lateral / surface is the bounding conical surface; the vertex of the cone is the vertex of the conical surface; the elements of the cone Oblique circula~ are the elements of the conical surface; cone the altitude of a cone is the perpendicular distance from- the vertex to tlhe plalne of the base. CONES 41 ') 706. A circular cone is a cone whose base is a circle. The axis of a circular cone is th, line drawn from the vertex to the center of the base. 707. A right circular cone is a circular cone whose axis is perpendicular to the plane of the base. An oblique circular cone is a circular cone whose axis is oblique to the base. 708. A cone of revolution is a cone generated by the revolution of a ri'ht triangle about one of its legs as an axis. Hence a cone of revolution and a right circular cone are the same solid. Cone of revolution 709. Properties of a cone of revolution inferred immediately. 1. The altitade of a cone of reoluttion is the axis of the cone. 2. All the eleiments of a cone of rel'oltion (t'e equnal. 710. The slant height of a cone of revolution is any one of its elements. 711. Similar cones of revolution a;re cones generated by similar right triangles revolvilng L)lout honologous sides. 712. A plane tangent to a cone is a plane iwhich contains one element of tihe cone, but whicli does not cut the conical surface on being produced. Ex. 1. A plane passing through a tangent to the base of a circular cone and the element drawn through the point of contact is tangent, to the cone, Ex. 2. If a plane is tangent to a circular cone, its intersection with the plane of the batse is tangent to the cone. 414 414 l~~I()( )I VITT. NOLTA) E( 'ETME1Y 713. A pyramid inscribed in 7_. a cone is a pyramnid whose lateral K edlges are elements of the cone vialty arid whose base is a polygon in.I li seribed in the base of the cone. H 714. A pyramid c ir cu mscribed about a cone is a py-ramid whose lateral faces are tan guent to the cone and whose base is a polygon circumscribed about the base of the cone. 715. Properties of circular cones. By Art. 441 the area of a circle is, the limilt of the area of an inscribed, or of a circumscribed polygon, and the eircnmference is the lim-it of the perimeters of these p)olyg~on~s;- hence 1. The vollu~ne o f a circu(lar COne6 is the limit of the, Vol. I(Ine of an inscrib~ed or eirciutnsc'i bed pyramid. 2. The lateral arca of a circular cone is the Ulit of the, late~ral are~a of (0? in~scribed 0o' circa wec,~ribed pyramnid. 716. A frustum of a cone is the por - tion of the cone included between the base of the cone and a plane parallel to thle/ bcase. The lower base of the frustum is t ihe ba-se of the cone, and the upper baseO i the frustum is the section made by the Frustum -of a cone. plane parallel to the base of the cone. What must be the altitude and the lateral surface of a frustum of a cone; also the slant height of the frustum of a cone of revolution V CONES 415 PROPOSITION VI. THEOREM 717. Every section of a cone adce by a plane passing through its vertex is a triangle. S PR Given the cone S-A.APBQ with a plane passing through the vertex S, and nmaking the section SPQ. To prove SPQ a triangle. Proof. PQ, the intersection of the base and the cutting plane, is a straight line. (Why?) Draw the straight lines SP and SQ. Then SP and SQ must be in the cutting plane; Art. 498. And be elements of the conical surface. Art. 701..'. the straight lines SP and SQ are the intersections of the conical surface and the cutting plane. o. the section SPQ is a triangle, Art. 81. (for it is bounded by threet straight lines). Q. E. D. Ex. What kind of triangle is a section of a right circular cone made by a plane through the vertex? 416 BOOK VIII. SOLID GEOMETRY PROPOSITION VII. THEOREM 718. Etery section of a circulla) coine made by a plane parallel to the base is a circle. c a c- AB Given the circular cone SAB with apb a section made by a plane parallel to the base. To prove apb a circle. Proof. Denote the center of the base by 0, and draw the axis, SO, piercing the plane of the section in o. Through SO and any element, SP, of the conical surface, pass a plane cutting the plane of the base in the radius OP, and the plane of the section in op. In like manner, pass a plane througlh SO and SB forming the intersections OB and ob... OP I op, and OB I ob. (Why?).. A SPO and SBO are similar to A Spo and Sbo, respectively. Art. 328. op /So ob OPp (SOb ) OB (Why T) But OP=OB. (Why?).. op =ob. (Why?).. opb is a circle. (Why ) Q. E. D. 719. COR. The axis of a circular cone passes through the center of every section parallel to the base. CONES 417 PROPOSITION VIII. THEOREM 720. The lateral area of a cone of revolution is equal to half the product of the slant height by the circumference of the base. I'\ Given a cone of revolution having its lateral area denoted by S, its slant height by L, and the circumference of its base by C. To prove S=4 C X L. Proof. Let a regular pyramid be circumscribed about the cone. Denote the lateral area of the pyramid by 8', and the perimeter of its base by P. Then S'= P P X L. Art. 642. If the number of lateral faces of the circumscribed pyramid be indefinitely increased, 8' will approach S as a limit. Art. 715, 2. P will approach C as a limit. Art. 441. And ~ P XL will approach i CX L as a limit. Art. 253, 2. But S' = PX L always. (Why?).' S=- CX L. (Why?) Q. E. D. 721. Formulas for lateral area and total area of a cone of revolution. Denoting the radius of the base by R, s5= (2 n RX L).:. S-=iRL. Also T = InRL -- RI'.. T =-R (L + R) 418 BOOK VIII. SOLID (GEOMETRY PROPOSITION IX. THEOREM 722. The volume of a circular cone is equal to one-third of the product of its base by its altitude. P~ JGiven a circular cone having its volume denoted by V its base by B, and its altitude by H. To prove V=- B X H. Proof. Let a pyramid with a regular polygon for its base be inscribed in the given cone. Denote the volume of the inscribed pyramid by VI, and its base by B'. Hence V'=* B' X H. Art. 65i. If the number of lateral faces of the inscribed pyramid be indefinitely increased, VI will approach V as a limit. (Why?) B1 will approach B as a limit. (Why t) And ~ B' X H will approach * B X H as a limit. (Why?) But VI=-& B' X H always. (Why?).'. V= B X H. (Why?) Q. E. D. 723. Formula for the volume of a circular cone. V-= tR2H. Ex. 1. If, in a cone of revolution, H=3 and R=4, find S, T and F. Ex. 2. If the altitude of a cone of revolution equals the radius of the base, what do the formulas for S, T and V become? CONES 419 PROPOSITION X. THEOREM 724. The lateral areas, or the total areas, of two similar cones of revolution are to each other as the squares of their radii, or as the squares of their altitudes, or as the squares of their slant heights; and their columes are to each other as the cubes of these lines. Given two similar cones of revolution having their lateral areas denoted by S and S', their total areas by T and T', their volumes by V and V', their radii by R and R', their altitudes by H and H', and their slant heights by L and L', respectively. To prove S: S=T: T= R2: R'2= H2: H'2= L2: Lv2; and V: V'=R3: R3=13: H33B3: L3 L:3. I H?r L L+ R Proof. H RiL (Why ) RL _R L R2 2 H2 '-T RIL R- R X L - = - L-,' ( Why?) T,tR (L+R) R L R+R R2 L2 T' tR' (L' + R,) R' L' + R R L1'2 H (Why?) ~it2H fft/R2 H2 Raf H_ a j LJ (wJ V, =tR'H' H"X LI" Q. E. D. 725. DEF. An equilateral cone is a cone of revolution such that a section through the axis is an equilateral triangle. 420 BOOK VIII. SOLID GEOMETRY PROPOSITION XI. THEOREM 726. The lateral area of a frustum of a cone of revolution is equal to one-half the sum of the circumiferences of its bases multiplied by its slant height. I \ Given a frustum of a cone of revolution having its lateral area denoted by S, its slant height by L, the radii of its bases by R and r, and the circumferences of its bases by C and c. To prove S-= ( + c) >( L. Proof. Let the frustum of a regular pyramid be circumscribed about the given frustum. Denote the lateral area of the circumscribed frustum by S', the perimeter of the lower base by P, and the perimeter of the upper base by p. The slant height of the circumscribed frustum is L. Hence S'=r (P+p) X L. Art. 643. Let the pupil complete the proof. 727. Formula for the lateral area of a frustum of a cone of revolution. S=~ (2 7tR + 2 nr) L.. Sn (R + r) L. 728. COR. The lateral area of a frustum of a cone of revolution is equal to the product of the circumference of its.idsection by its slant height. CONES 421 PROPOSITION XII. THEOREM 729. The volume of the frustum of a circular cone is equivalent to the volume of three cones, whose common altio tude is the altitude of the frustum, and whose bases are the lower base, the upper base, and a mean proportional between the two bases. Given a frustum of a circular cone having its volume denoted by V, its altitude by H, the area of its lower base by B, and that of its upper base by b. To prove V=- H (B+ b + B X b). Proof. Let the frustum of a pyramid with regular polyo gons for its bases be inscribed in the given frustum. Denote the volume of the inscribed frustum by VI, and the areas of its bases by B' and b'.:. F=. H (B' + b'V d VB X b'). (Why i) If the number of lateral faces of the inscribed frustum be indefinitely increased, V' will approach V, B' and b' approach B and b respectively, and B' X b' approach B X b, as limits. Art. 715. Hence, also, B' + b'+V/B' X bl will approach B + b + 1VB X b as a limit. Art. 253. But F=- H (B' + b' +/B' X b') always (Why?). 7=Vi H (B + b + VB X b). (Why?) Q. B. De 4,~22 4 -d - BOOK VIII. SOLID GEOMETRY 730. Formula for the volume of the frustum of a circu, lar cone. V= i H (RJ? + tr2 + VnR2 X Ant)... V= x7t (R2 +.r2' + t R). Ex. 1. The measurement of the volume of a frustum of a cone of revolution reduces to the measurement of the lengths of what straigiht lines? Ex. 2. If a conical oil-can is 12 in. high, how much more tin is required to make it than to make a similar oil-can 6 in. high? How much more oil will it hold? Ex. 3. The linear dimensions of a conical funnel are three times those of a similar funnel. How much more tin is required to make the first? How much more liquid will it hold? Ex. 4. Make a similar comparison of cylindrical oil-tanks. Of conical canvas tents. EXERCISE$S CROUPl 70 THEOREMS CONCERNING CYLINDERS AND CONES Ex. 1. Any section of a cylinder of revolution through its axis is a rectangle. Ex. 2. On a cylindrical surface only one straight line can be drawn through a given point. [SuG. For if two straight lines could be drawn, etc.] Ex. 3. The intersection of two planes tangent to a cone is a straight line through the vertex. Ex. 4. If two planes are tangent to a cylinder, their line of intersection is parallel to an element of the cylinder. [SuG. Pass a plane J to the elements of the cylinder.] Ex. 5. If tangent planes be passed through two diametrically opposite elements of a circular cone, these planes intersect in a straight line through the vertex and parallel to the plane of the base. Ex. 6. In a cylinder of revolution the diameter of whose base equals the altitude, the volume equals one-third the product of the total surface by the radius of the base. EXERCISES ON THE CYLINDER AND CONE 423 Ex. 7. A cylinder and a cone of revolution have the same base and the same altitude. Find the ratio of their lateral surfaoes, and also of their volumes. Ex. 8. If an equilateral triangle whose side is a be revolved about one of its sides as an axis, find the area generated in terms of a. Ex. 9. If a rectangle whose sides are a and b be revolved first about the side a as an axis, and then about the side b, find the ratio of the lateral areas generated, and also of the volumes. Ex. 10. The bases of a cylinder and of a cone of revolution are concentric. The two solids have the same altitude, and the diameter of the base of the cone is twice the diameter of the base of the cylinder. What kind of line is the intersection of their lateral surfaces, and how far is it from the base? Ex. 11. Determine the same when the radius of the cone is three times the radius of the cylinder. Also when r times. Ex. 12. Obtain a formula in terms of r for the volume of the frustum of an equilateral cone, in which the radius of the upper base is r and that of the lower base is 3r. Ex. 13. A regular hexagon whose side is a revolves about a diagonal through the center as axis. Find, in terms of a, the surface and volume generated. Ex. 14. Find the locus of a point at a given distance from a given straight line. Ex. 15. Find the locus of a point whose distance from a given line is in a given ratio to its distance from a fixed plane perpendicular to the line. Ex. 16. Find the locus of all straight lines which make a given angle with a given line at a given point. Ex. 17. Find the locus of all straight lines which make a given angle with a given plane at a given point. Ex. 18. Find the locus of all points at a given dis- tance from the surface of a given cylinder of revolution. Ex. 19. Find the locus of all points at a given dis --- tance from the surface of a given cone of revolution. 424 BOOR VIII. SOLID GEOMETRY EXERCISES. QROUP 7T PROBLEMS CONCERNING THE CYLINDER AND CONE Ex. 1. Through a given element of a circular cylinder, pass a plane tangent to the cylinder. Ex. 2. Through a given element of a circular cone, pass a plane tange t,: to the cone. Ex. 3. About a given circular cylinder circumscribe a prism, wiih a regular polygon for its base. Ex. 4. Through a given point outside a circular cylinder, pass a plane tangent to the cylinder. Ex. 5. Through a given point outside a given circular cone, pass a plane tangent to the cone. [Su(o. Through the vertex of the cone and the given point pass a line, and produce it to meet the plane of the base.] Ex. 6. Into what segments must the altitude of a cone of revolution be divided by a plane parallel to the base, in order that the volumn e of the cone be bisected? Ex. 7. Divide the lateral surface of a given cone of revolution into two equivalent parts by a plane parallel to the base. Ex. 8. If the lateral surface of a cylinder of revolution be cut along one element and unrolled, what sort of a plane figure is formed? Hence, out of cardboard construct a cylinder of revolution with given altitude and given circumference. Ex. 9. If the lateral surface of a cone of revolution be cut along one element and unrolled, what sort of a plane figure is formed? Hence, out of cardboard construct a cone of revolution of given slant height. Ex. 10. Construct an equilateral cone out of pasteboard. Ex. 11. Construct a frustum of a cone of revolution out of paste. board. BooK IX THE SPHERE 731. A sphere is a solid bounded by a surface all points of which are equally distant from a point within called the center, 732. A sphere may also be defined as a solid generated by the revolution of a semicircle about its diameter as an axis. Some of the properties of a sphere may be obtained more readily from one of the two definitions given, and some from the other. A sphere is named by naming the point at its center or by naming three or more points on its surface. 733. A radius of a sphere is a line drawn from theil tenter to any point on the surface. A diameter of a sphere is a line drawn through the center and terminated at each end by the surface of the sphere. 734. A line tangent to a sphere is a line having but one point in common with the surface of the sphere, however tar the line be produced. (425) 4;2G BOOK IX. SOLID GEOMETRY 735. A plane tangent to a sphere is a plane having bu one point in common with the surface of the sphere, how ever far the plane be produced. 736. Two spheres tangent to each other are sphere. whose surfaces have one point, and only one, in common 737. Properties of a sphere inferred immediately. 1. All radii of a sphere, or of equal spheres, are equal. 2. All diameters of a sphere, or of equal spheres, are equal. 3. Two spheres are equal if their radii or their diame. ters are equal. PROPOSITION I. THEOREM 738. A section of a sphere made by a plane is a circle. Given the sphere 0, and PCD a section made by a plane cutting the sphere. To prove that PCD is a circle. Proof. From the center 0, draw OA.L the plane of the section. THE SPHERE 427 Let C be a fixed point on the perimeter of the section, and P any other point on this perimeter. Draw AC, AP, OC, OP. Then the A OAP and OAC are rt. A, Art. 505. OP= OC. (Why?) OA OA. (Why?) A OAP=A OA.C (Why?).'. AP=AC. (Why?) But P is any point on the perimeter of the section PCD..'. every point on this perimeter is at the distance AC from A... PCD is a circle with center A. Art. 197. Q. E. D. 739. COR. 1. Circles which are sections of a sphere made by planes equidistant from the center are equal; and conversely. 740. COR. 2. Of two circles on a sphere, the one made by a plane more remote from the center is smaller; and conversely. 741. DEF. A great circle of a sphere is a circle whose plane passes through the center of the sphere. 742. DEF. A small circle of a sphere is a circle whose plane does not pass through the center of the sphere. 743. DEF. The axis of a circle of a sphere is the diameter of a sphere which is perpendicular to the plane of the circle. Thus, on figure p. 426, BB' is the axis of PCD. 744. DEF. The poles of a circle of a sphere are the Extremities of the axis of the circle. Thus, B and B', of agure p. 426, are poles of the circle PCD. 428 BOOK IX. SOLID) GEOMETRY 745. Properties of circles of a sphere inferred im-m, diately. 1. The axis of a circle of a sphere passes through ti, center of the circle; and conversely. 2. Parallel circles have the same axis and the same poliS 3. All great circles of a sphere are equal. 4. Every great circle on a sphere bisects the sphere li its surface. 5. Two great circles on a sphere bisect each other. For the line of intersection of the two planes of ti circles passes through the center, and hence is a diamet, of each circle. 6. Through two points (not the extremities of a diametr on the surface of a sphere, one, alnd only one, great circle c. be passed. For the plane of the great circle must also pass throng the center of the sphere (Art. 741), and through thre points not in a straight line only one plane can be passe (Art. 500). 7. Through any three points on the surface of a sphere not in the samte plane with the center, one small circle, an only one, can be passed. 746. DEF. The distance between two points on the sui face of a sphere is the length of the minor are of a grea circle joining the points. Ex. 1. If the radius of a sphere is 13 in., find the radius of circle on the sphere made by a plane at a distance of I ft. from th center. Ex. 2. What geographical circles on the earth's surface are great and what small circles? Ex. 3. What is the largest number of points in which two circle on the surface of a sphere can intersect T Why? THE SPHERE 429 PROPOSITION II. THEOREM 747. All points in the circumference of a circle of a sphee re are equally distant from each pole of the circle. IPj yi',j~j A - I' 3 0!;,!::!:: ':::,:,-.;:::p::: i,! Given ABC a circle of a sphere, and P and PI its poles. To prove the arcs PA, PB, PC equal, and arcs PA, P'B, PIC equal. Proof. Draw the chords PA, PB, PC. The chords PA, PB and PC are equal. Art. 518..'. arcs PA, PB and PC are equal. Art. 218. In like manner, the arcs P'A, P'B and P'C may be proved equal. Q. e. D. 748. DEF. The polar distance of a small circle on a sphere is the distance of any point on the circumference of the circle from the nearer pole. The polar distance of a great circle on a sphere is the distance of any point on the circumference of the great circle from either pole. 749. Cor. The polar distance of a great circle is the quadrant of a great circle. 430 BOOK IX. SOLID GEO' TTRY PROPOSITION III. THEOREM 750. If a point on the surface of a sphere is at a qualr rln/t's distance from two other points on the surface, it i the pole of the great circle through those points. P B Given PB and PC quadrants on the surface of the sphere 0, and ABC a great circle through B and C. To prove that P is the pole of ABC. Proof. From the center 0 draw the radii OB, OC, OP. The arcs PB and PC are quadrants. (Why?).* A POB and POC are rt. A. (Why ) '. PO 1 plane ABC. (Why ).'. P is the pole of the great circle ABC. (Why?) Q. E. D. 751. COR. Through two given points on the surface of a sphere to describe a great circle. Let A and B be the given points. From A and B as centers, with a quadrant as radius, describe arcs \ oB on the surface of the sphere intersecting at P. With P as a center and a quadrant as a radius, describe a great circle. THE SPHERE 431 PROPOSITION IV. THEOREM 752. A plane perpendicular to a radius at its extremity is tangent to the sphere. c Given the sphere 0, and the plane MN I the radius OA of the sphere at its extremity A. To prove MN tangent to the sphere. Proof. Take P any point in plane MN except A. Draw OP. Then OP > OA. (Why?) *. the point P is outside the surface of the sphere. But P is any point in the plane MN except A.. plane MN is tangent to the sphere at the point A, (for every point in the plane, except A, is outside the surface of the sphe're). Art. 73.. Q. D. 753. COR. 1. A plane, or aU line, which is tangent to a sphere, is perpendicular to the radius drawn to the point of contact. Also, if a plane is tangent to a sphere, a perpendicular to the plane at its point of contact passes through the center of the sphere. 754. COR. 2. A straight line perpendicular to a radius of a sphere at its extremity is tangent to the sphere. 755. COR. 3. A straight line tangent to a circle of a Sphere lies in the plane tangent to the sphere at the point of contact. 4#-,))t2 42BOOK IX. SOLID GEOMEYTR 756. COR. 4. A straight line drawen in a taBigent PI(I,, and th-roitgh the point of contact is tangent to the sph11e)ere that point. 757. COR. 5. Twvo straight lines tanizgent to a sphierc a given point determinie the tangent plane at that point. 758. DEP. A sphere circumscribed about a polyhedro is a sphere in whose surface lie all the vertices of tioolyhedron. 759. IEF. A sphere inscribed in a polyhedron is sphere to which all the faces of the polyhedron are taugeli, PROPOSITION V. PROBLE-M 760. To circuimscribe a sphere abolut a given tetrat hedron. Given the tetrahedron ABCD. To circumscribe a sphere about ABCD. Construction aud Proof, Coustruct E and of circles circumscribed about the & ABC spectively. Draw E1 I. plane ABC and FK I- plane F the centers and BCD, reArt. 286. BC-D. Art. 514. Draw EG aiid PG to G the midpoint of BC. THE SPHERE 433 Then EG and FG are I BC. Art. 113... plane EGF 1 BC. Art. 509... plane EGiF 1 plane ABC. Art. 555. E. EH lies in the plane FGE. Art. 558. In like manner FK lies in the plane FGE. The lines EG and FG are not I|, (for they meet in the point G)... the lines EH and FK are not!1. Art. 122. Hence EH must meet FK in some point 0. But EHf is the locus of all points equidistant from A, B and C; and FK is the locus of all points equidistant from B, C and D. Art. 520... 0, which is in both EH and FK, is equidistant from A, B, C and D. (Why?) Hence a spherical surface constructed with 0 as a center and OA as a radius will pass through A, B, C and D, and form the sphere required. Q. E. F. 761. COR. 1. Four points not in the same plane edeter mine a sphere. 762. COR. 2. The four perpendiculars erected at the centers of the faces of a tetrahedron meet in a point. 763. COR. 3. The six planes perpendicular to the edges of a tetrahedron at their midpoints intersect in a point. 764. DEF. An angle formed by two curves is the angle formed by a tangent to each curve at the point of intersection. 765. DEF. A spherical angle is an angle formed by two intersecting arcs of great circles on a sphere, and hence by tangents to these arcs at the point of intersectionu BB 434 BOOK IX, SOLID GEOMETRY PROPOSITION VI. PROBLEM 766. To inscribe a sphere in a given tetr(thdlron. i!"": j[:. I D. __________R_____:_:: W, __1 Given the tetrahedron ABCD. To inscribe a sphere in ABCD. Construction and Proof. Bisect the dihedral angle DAB-C by the plane OAB; similarly bisect the dihedral A whose edges are BC and AC by the planes OBC and OA C, respectively. Denote the point in which the three bisecting planes intersect by 0. Every point in the plane OAB is equidistant from the faces DAB and CAB. Art. 562. Similarly, every point in OBC is equidistant from the two faces intersecting in BC, and every point in OAC is equidistant from the two faces intersecting in AC... O is equidistant from all four faces of the tetrahedron. Ax. 1. Hence, from 0 as a center, with the L from 0 to any one face as a radius, describe a sphere. This sphere will be tangent to the four faces of the tetrahedron and.. inscribed in the tetrahedron. Art. 759. 767. COR. The planes bisecting the six dihedral angles of a tetrahedron meet in one point, THE SPHERE 435 PROPOSITION VII. PROBLEM 768. To find the radius of a given material sphere. P.. 1'~ V~,, xP / D-, p -o / Fig. 1 Fig. 2 lg,. I Given the material sphere O. To construct the radius of the sphere. Construction. With any point P (Fig. 1) of the surface of the sphere as a pole, describe anyfconvenient circumference on the surface. On this circumference take any three points A, B and C. Construct the A ABC (Fig. 2) having as sides the three chords AB, BC, AC, obtained from Fig. 1, by use of the compasses. Art. 283. Circumscribe a circle about the A ABC. Art. 286. Let KB be the radius of this circle. Construct (Fig. 3) the right A kpb, having for hypotenuse the chord pb (Fig. 1) and the base kb. Art. 284. Draw bp' I bp and meeting pk produced at p'. Bisect pp' at 0. Then op is the radius of the given sphere. Proof. Let the pupil supply the proof. Yu I.. F. 436 BOOK IX. SOLID GEOMETlRY PROPOSITION VIII. THEORE5M 769. The intersection of two spherical surtfaces is the circumference of a circle whose plane is perpendicular to the line joining the centers of the spheres, anid whose center is i1n that line. o --- —-- ---— o' Given two intersecting (d 0 and 0' which, by rotation about the line 00' as an axis, generate two intersecting spherical surfaces. To prove that the intersection of the spherical surfaces is a 0, whose plane ~ 00', and whose center lies in 00'. Proof. Let the two circles intersect in the points P and Q, and draw the common chord PQ. Then, as the two given D rotate about 00' as an axis, the point P will generate the line of intersection of the two spherical surfaces that are formed. But PR is constantly J1 00'. Art. 241..'. PR generates a plane 1 00' Art. 510. Also PR remains constant in length..P describes a circumference in that plane. Art. 197. Hence the intersection of two spherical surfaces is a 0, whose plane L the line of centers, and whose center is in the line of centers. Q. E. D. The above demonstration is an illustration of the use of the second definition of a sphere (Art. 732). THE SPHERE 437 PROPOSITION IX. THEOREM 770. A spherical angle is measured by the arc of a great circle described from the vertex of the angle as a pole, and included between its sides, produced, if necessary..A. -— f A' Given / BAC a spherical angle formed by the intersection of the arcs of the great circles BA and CA, and BG an arc of a great circle whose pole is A. To prove / BA C measured by arc BG. Proof. Draw AD tangent to AB, and AF tangent to AC. Also draw the radii OB and 0C. Then AD I AO. Art. 230. Also OB IL AO (for AB is a quadrant)..-. OB I|AD. (Why ) Similarly OC I AF..'. BOC= Z DAF Art. 538. But Z BOC is measured by arc BC. Art. 257... ZDAF, that is, ZBAC, is measured by arc BC. (Why?) Q. E. D. 771. COR. A spherical angle is equal to the plane angle of the dihedral angle forimed by the planes of its sides, 438 BOOK IX. SOLID GEOMETRY SPHERICAL TRIANGLES AND POLYGONS 772. A spherical polygon is a portion of tie surface of a sphere bounded by three or more arcs of great circles, as ABCD. The sides of the spherical polygon are the bounding arcs; the vertices - are the points in which the sides in- tersect; the angles are the spherical angles formed by the sides. The sides of a spherical polygon are usually limited to arcs less than a semicircumference. 773. A spherical triangle is a spherical polygon of three sides. Spherical triangles are classified in the same way as plane triangles; viz., as isosceles, equilateral, scalene, right, obtuse and acute. 774. Relation of spherical polygons to polyhedral angles. If radii be drawn from the center of a sphere to the vertices of a spherical polygon on its surface (as OA, OB, etc., in the above figure), a polyhedral angle is formed at 0, which has an important relation to the spherical polygon ABCD Each face angle of the polyhedral angle equals (in number of degrees contained) the corresponding side of the spherical polygon; Each dihedral angle of the polyhedral angle equals the corresponding angle of the spherical polygon. Hence, corresponding to each property of a polyhedral angle, there exists a property of a spherical polygon, and conversely, THE SPHERE 439 Hence, also, a trihedral angle and its parts correspond to a spherical triangle and its parts. Of the common properties of a polyhedral angle and a spherical polygon, some are discovered more readily from the one figure and some from the other. In general, the spherical polygon is simpler to deal with than a polyhedral angle. For instance, if a trihedral angle were drawn with the plane angles of its dihedral angles, nine lines would be used, forming a complicated figure in solid space; whereas, the same magnitudes are represented in a spherical triangle by three lines in an approximately plane figure. On the other hand, the spherical polygon, because of its lack of detailed parts, is often not so suggestive of properties as the polyhedral angle. PROPOSITION X. THEOREM 775. The sum of tvwo sides greater than the third side. of a spherical triangle is Given the spherical triangle ABC, of which no side is larger than AB. To prove AC+BC > AB. Proof. From the center of the sphere, 0, draw the radii OA, OB, OC. Then, in the trihedral angle 0-ABC, Z A OC BOC > A B. A rt. 582.. A C - BC > AIB. Art. 774. Q. E. D. 440 3BOOK IX. SOLID GEOMETRY 776 o. CO. 1. Any side of a spherical triangle is great es than the difference between the other two sides. 777. COR. 2. The shortest path between two points on. the surface of a sphere is the arc of the great circle joinilng those points. For any other path between the two points may be made the limit of a series of ares of great circles connecting successive points on the path, and the sum of this series of arcs of great circles connecting the two points is greater than the single are of a great circle connecting them. PROPOSITION XI. THEOREM 778. The sum of the sides of a spherical polygon is less than 360~. B Given the spherical polygon ABCD. To prove the sum of the sides of ABCD < 360~o Proofo From 0, the center of the sphere, draw the radii OA, OB, OC, OD. Then ZAOB+ ZBOC+ Z COD+- DOA < 360~. (Why T) AB + BC + CD+ DA < 360~. Art. 774. Q. E. eD SPHERICAL TRIANGLES 441 779. DEF. The polar D ' triangle of a given triangle, is the triangle formed by taking the vertices of the given triangle as poles, and describing arcs of great cir- C cles. (Hence, if each pole be regarded as a center, the radius used in describing each arc is a quadrant.) Thus A'B'C ' is the polar triangle of ABC; also D'E'F' is the polar triangle of DEF. PROPOSITION XII. THEOREM 780. If one spherical triangle is the polar of another, then the second triangle is the polar of the first. A' Given A'B'C' the polar triangle of ABC. To prove ABC the polar triangle of AtB'C'. Proof. B is the pole of the arc A'C'. Art. 779.., arc A'B is a quadrant. (Why?) Also C is the pole of the arc A'B'. (Why I).. arc A'C is a quadrant. (Why?).A. is at a quadrant's distance from both B and C..' A' is the pole of the arc BC. Art. 750. In like manner it may be shown that B' is the pole of AC, and C' the pole of AB. Q. E. Do 44 -04 442 1,OOK IX. SOLID GEOMM"IY PROPOSITION XIII. THEoEKM 781. In1 ii s])her'ical triangle and its polar, each angle of one trianigle is the suipplemient of the side opposite in til other triangle. A' C~1 b 13 BC Given the polar zL ABC and A'B'C' with the sides of ABC denoted by a, b, c, and the sides of A'B'C' denoted by a', b', c', respectively. To prove A + a'-1800, B + b'= 1800, C + c'=1800, A'+a =1800, Bf + b z1800,Cf+C e =z1800. Proof. Produce the sides AB and AC till they meet B'C' in the points D and F, respectively. Then B' is the pole of AF:. arc B'F=900. Art. 78o. Also C' is the pole of AD.'. arc C'D 900. (Whys) Adding, B'F + C'ID= 1800. (Why.) Or B' F +FC1 '+DFz= ls0. Ax.6. Or BIC' + DF= 1800. But B'C' = a', and DF is the measure of the L A. Art. 77Q..,. A +-a= 1800. In like manner the other supplemental relations may be proved as specified. 782. DEF. Supplemental triangles are two spherical triangles each of which is the polar triangle of the other. This new name for two polar triangles is due to the property proved in Art. 781. SPHERICAL TRIANGLES 443 PROPOSITION XIV. THEOREM 783. The sum of the angles of a spherical triangle is greater than 1800, and less than 540'. A' C/ Given the spherical triangle ABC. To prove A + B + C > 1800 and <65400, Proof. Draw A'B'C', the polar triangle of ABO, and denote its sides by a', 6', c' Then A + a' 1800 + b = 1800 ' Art. 781. (' c'e - 1800) A + B + C+a'a b' + Y c' = 5400. A(1) x. But nla' + b1 + c' < 3600 Art. 778. 6(" + b' + e' > 0 -Subtracting each of these in turn from (1), A + B + CU> 1800 and < 540'. Ax. 11. Q. E. D. 784. CoR. A spherical triangle mnay have one, two or three right angles; or it mlay hazre one, two or, three obtuse angles. 785. DEE. A birectangular spherical triangle is a spherical triangle containing two right angles. 786. DE-F. A trirectangular spherical triangle is a spherical triangle containing three right angles. 444: B0OOK IX. SOLID GEOMETRY 787. CoR. The surface of a sphere may be divided into eight trirecttangtl(ar spherical triangles. For let three planes I to each other be passed through the center of a sphere, etc. 788. DEF. The spherical excess of a spherical triangle is the excess, of the sum of its angles over 180~. 789. DEF. Symmetrical spherical triangles are triangles which have their parts equal, but arranged in reverse order. B'B^Fig. 2 Fg' 3 Fig. 2 Fig. 3 Fig. 1 Three planes passing through the center of a sphere form a pair of symmetrical spherical triangles on opposite sides of the sphere (see Art. 580), as & ABC and A'B'C' of Fig. 1. 790. Equivalence of symmetrical spherical triangles. Two plane triangles which have their parts equal, but arranged inreverse order, may be C C made to coincide by lifting up one B A A B triangle, turning it over in space, and placing it upon the other triangle. But two symmetrical spherical triangles cannot be made to coincide in this way, because of the curvature of a spherical surface. Hence the equivalence of two symmetrical spherical triangles must be demonstrated in some indirect way. SPHERICAL TRIANGLES 445 791. Property of symmetrical spherical triangles. Two isosceles symmetrical spherical triangles are equal, for they can be made to coincide. PROPOSITION XV. THEOREM 792. Two symmetrical spherical triangles are equivalent. C...,A B' /.i... te,0 ^iI AA-. '. Given the symmetrical spherical A0 ABC and A'B'C', formed by planes passing through 0, the center of a sphere. (See Art. 789.) To prove A ABC A A 'B'C'. Proof Let P be the pole of a small circle passing through the points A, B, C. Draw the diameter POP'. Also draw PA, PB, PC, P'A', P'B', P'C', all arcs of great ). Then PA = PB- PC Art. 747. Also P'A' PA, P'B' PB, P'C' PC. Arts. 78, 215. P'A - P 'B' P'C'. Ax. 1. Hence PAB and P'A'/B are symmetrical isosceles A..'. A PArB A PA'B'.j Similarly A PAC= A PA'C'. Art. 791. And A PBC=A P'B'C'.J Adding A PAB+ A PA C+ A PBC A P'A'Bf+A P+A'C'+ A PABICG. Ax. 2. Or A ABC A'B'C'. Ax. 6. In case the poles P and P/ fall outside the A ABC atld A'B'C', let the puoil surply the demonstration, Q.. D. 416 BOOK IX SOLID G(EOMETRY Rl()POSITION XVI. ToVI.EM 793, On the same sphere, or on equal spheres, teo tri. antgles are equal, I. If two sides and the included angle of one are equal to fwo sides and the incltded angle of the other; or II. If two angles and the included side of one are equal to two angles anid the included side of the other, the correspondiqg equcal parts being arranged in the sae order in each case. A D C B F I. Given the spherical & ABC and DEF, in which A C= DF, CB=FE, and ZC = ZF. To prove A A BC= A DEF. Proof. Let the pupil supply the proof (see Book I, Prop. VI). II. Given the spherical & ABC and DEF, in which ZC ZF, ZB= ZE, and CB=FE. To prove A ABC= A DEF. Proof. Let the pupil supply the proof (see Book I, Prop. VII). Ex. 1. If the line of centers of two spheres is 10 in., and the radii are 12 in. and 3 in., how are the spheres situated with reference to each other? Ex. 2. The tank on a motor car is a cylinder 35 inches long and 15 inches in diameter. How many gallons of gasolene will it hold? Ex. 3. In an equilateral cone, find the ratio of the lateral area to the area of the base. SPHERICAL TRIANGLES44 447 PROPOSITION XVII. THEOREM 794. On the samne sphere, or on equtal sp~heres, twvo t~riangles are symtmetrical and equivalent, I. If two sides and the included angle of one are, equal too twvo sides anid the included angle of the other,; or II. If two angles and the included side of one are equal to twvo angles and the included, side of the other-, the correspondit (lnequal parts~ being arrangled int rel'erse order. A D E F~~~~~~~ I. Given the spherical z& ABC and DEF, in which AB DE, A C =DE, and Z A =Z D, the correspondingo parts being arranged in reverse order. To prove A ABC symmetrical with A -DEF. Proof. Constrnct the AJYE'.F'symmetr~ical with ADEF~. Then A ABC may be made to coincide with A D'E'E', Art. 793. (having two sides and, the 'included Z equal andi arranged in the same order). But A D'E'F' is symmetrical with the A -DEE. A' ABC, whtich coincides with A~ D'E'E', is syinmetrieal with A -I)EF. II. The second part of the theorem is proved in the 6v ne way. 448 BOOK IX. SOLID GEOMETRY PROPOSITION XVIII. THEOREM 795. If two triangles on the same sphere, or equal spheres, are mutually equilateral, they are also mutually equiangular, and therefore equal or symmetrical. A A' At -aO -— 0 O. B B' B' Given two mutually equilateral spherical A ABC and A'B'C' on the same or on equal spheres. To prove A ABC and A'B'C' equal or symmetrical. Proof. From 0 and 0', the centers of the spheres to vhich the given triangles belong, draw the radii OA, OB, OC, O'A', 0'B', O'C'. Then the face A at 0=corresponding face A at 0'. Art. 774. Hence dihedral A at 0= corresponding dihedral A at 0'. Art. 584. A'., of spherical A ABC=homologous A of spherical ZA A'B' C. Art. 774..'. the A ABC and A'B'C' are equal or symmetrical, according as their homologous parts are arranged in the same or in reverse order. Art. 789. 796. NOTE. The conditions in Props. XVI and XVIII which make two spherical triangles equal are the same as those which make two plane triangles equal. Hence many other propositions occur in spherical geometry which are identical with corresponding propositions in plane geometry. Thus, many of the construction problems of spherical geometry are solved in the same way as the corresponding construction problems in plane geometry; as, to bisect a given angle, etc, SPHERICAL TRIANGLES 449 PROPOSITION XIX. THEOREM 797. If two triangles on the same sphere are mutually equiangular, they are also mutually equilateral, and therefore equal or synmmetrical. / "", - ' Q___ Given the mutually equiangular spherical A Q and Q' on the same sphere or on equal spheres. To prove that Q and Q' are mutually equilateral, and therefore equal or symmetrical. Proof. Ccnstruct P and P' the polar A of Q and Q', respectively. Then A P and P/ are mutually equilateral. Art. 781..'. A P and P' are mutually equiangular. Art. 795. But Q is the polar A of P, and Q' of P'. Art. 780..'. A Q and Q' are mutually equilateral. Art. 781. Hence Q and Q' are equal or symmetrical, according as their homologous parts are arranged in the same or in::everse order. Art. 789. Q. E. Do 798. CoR. If tw'o mutually equiangular trianyles are o, al.eq(ual sph'res, their correspondiyg sides have the same rttio as the radii of their respectice spheres. cc .-/" 40U BOOK IX. SOLID GEOMETIRY PROPOSITION XX. THEIOREMI 799. In an1 isosceles spherical triangle the angles oppo site the equatl sides are equal. B U C Given the spherical A ABC in which AB= AC. To prove Z B Z C. Proof. Draw an are from the vertex A to D, the midpoint of the base. Let the pupil supply the remainder of the proof. PROPOSITION XXI. THEOREM (CONV. OF PROP. XX) 800. If two angles of a spherical triangle are equal, the sides opposite these angles are equal, and the triangle is isosceles. A' /^ /, I — Giv t s ic Given the spherical A ABC in which ZL=- Z C. To prove AB=A C. Proof. Construct A A'B'C' the polar A of ABC. Then A'C'=A'B'. Art. 781..Z /'-= Z B7. Art. 799. 4 AB=AC. Art. 781, Q. E. D. SPHERICA L TRIANGLES 4j 4,51 PROPOSITION XXII. THEOREM 801. In any spherical triangle, if two angles are unequal, the sides opposite these angles are unequal, and the greater side is opposite the greater angle, and CONVERSELY. B D A' C Given the spherical A A BC in whrleh Z BAGC is greater than GC. To prove PbC > BA. Proof. Draw the are A-D making, Z DAC equal to Z C. Then DA =-DC. Art. 500. To each of these equals add the are BI). BD. *.-D D A = RD + D C, or _BC. (Why?) But in A BDA,.t BD + DPA > B3'A. (Why?) BC. B C > BA. Ax. S. Let tile pupil prove the conversc by the indirect imethod (see Art. 106). _Q. E. D. Ex. 1. Bisect a given spherical angle. Ex. 2. Bisect a given are of a great circle on a sphere. Ex. 3. At a given point in an arc on a sphere, construct an angle equiial to a given spherical angle on the same sphere., Ex. 4. Find the locus of the centers of the circles of a sphere arm~ed by planes perpendicular to a given diameter of the given Tphere, 452 BOOK IX. SOLID GEOMETRY SPHERICAL AREAS 802. Units of spherical surface. A spherical surface may be measured in terms of, either 1. The cutstomary units of area, as a square inch, a square foot, etc., or 2. Spherical degrees, or spheridso 803. A spherical degree, or spherid, is one-ninetieth part of one of the eight trirectangular triangles into which the surface of a sphere may be divided (Art. 787), or o-7. part of the surface of the entire sphere. A solid degree is one-ninetieth part of a trirectangular angle (see Art. 774). 804. A lune is a portion of the surface of a sphere bounded by two semicircumferences of great circles, as PBP'C of Fig. 1. The angle of a lune is the angle formed by the semicircumferences which bound it, as the angle BPC. A D 805. A zone is the portion of the surface of the sphere bounded by?' two parallel planes. i. 1. A zone may also be defined as the surface generated by an are of a revolving Q semicircumferenee. Thus, if QFQ' (Fig. 2) Ii - generates a sphere by rotating about QQG, its diameter, any are of QFQ', as EF, - generates a zone. 806. A zone of one base is a zone L ' --- —-- one of whose bounding planes is tangent to the sphere, as the zone generated by the arc QE of Fig. 2. Fig. 2. 807. The altitude of a zone is the perpendicular dis tance between the bounding planes of the zone. The bases of a zone are the circumferences of the circles of the sphere formed by the bounding planes of the zone. SPHERICAL AREAS 453 PROPOSITION XXIII. THEOREM 808. The area generated by a straight line revolving about an axis in its plane is equal to the projection of the/ line upon the axis, mulltiplied by the ci)rcumferelce of a circle zwhose radius is the perpendicular erected at the midpoint of the line and termbiated by the axis. A A-e; ~A -- AC Pi F A P1 —QI VI/, y Y Y Fig. 1 Fig. 2 Fi2g. 3 Given AB and XY in the same plane, CD the projection of AB on XY, PQ the I bisector of AB; and a surface generated by the revolution of AB about XY, denoted as "area AB." To prove area AB= CD X 2 -,PQ. Proof. 1. In general, the surface generated by AB is the surface of a frustum of a cone (Fig. 1). ee area AB=AB X 2 YPR. Art. 728. Draw AF I' BD, then A ABF and PQR are similar. Art. 328... AB: AF=PQ: PR. (Why?). AB X PR AF X PQ, or CD X PQ. (Why?) Substituting, area AB= CD X 2 xtPQ. Ax. 8. 2. If AB 1 XY (Fig. 2), the surface generated by AB is the lateral surface of a cylinder... area ABG= CD X 2 tPQ. Art. 697. 3. If the point A lies in the axis XY (Fig. 3), let the pupil show that the same result is obtained. 0. E.D. 454 454 ~BOOK IX. SOLID GEOMETRY PRO1POSI1TON XX\. rTHEO 1R~E M~ 809. The area of the sitrfate of a sph err is equal to the produc(t' of the diarneter of the sphere by the circitmferencC of a reant circle. CL Given a sphere generated by the revolution of the senicircle ACE about the diameter AE, with the surface of the sphere denoted by S, and its radius by P. To prove S =A E X 21 Pt. Proof. Inscribe in the given semicircle the half of a regyular polygon of an even number of sides, as ABODE. DrAaw the apothemn to each side of the semipolygon, and denote it by a. From the vertices B, C, D draw Is to AFR Then urtea AB3=All X 21 7a. are~a BC= FO X 2-~ (ta.Ar.8 area CD= OK X 29 7ra. area DE= KE X 2 -,. Ja Adding, area ABCDE=AE X.2 7ta. If, now, the number of sides of the polygon be indefinitely increased, area ABO-DE approaches S as a limit. Art. 441. And a approaches Pt as a limit. (Why?). AE X 2 7ta approaches AE X 2 7t-R as a limit. (Why?~) But area ABODE=AE X 27a always. AS=A-EX 27tR. (WhyT) 9, B. D. SPHERICAL AREAS 455 810. Formulas for area of surface of a sphere. Substituting for AE its equal 2, S=4 nR2. Also denoting the diameter of the sphere by D, R=j D. ' =4 7(', or S= D. 811. COR. 1. The surface of a sphere is equivalent to four times the area of a great circle of the sphere. 812. CoR. 2. The areas of the surfaces of two spheres are to each other as the squares of their radii, or of their diameters. For, if S and S' denote the surfaces, R and R' the radii, and D and D' the diameters of two spheres, S 4 tiR2 R2 S _D2 D2 S- '-4 nR'2IR'-9; also ' D'= D'~ 4 B7tR/2 _R/2 S/In y2 D'-;, 813. Property of the sphere. The following property of the sphere is used in the proof of Art. 809: If, in the generating arc of any zonc, a broken line be inscribed, whose vertices divide the arc into equal parts, then, as the number of these parts is increased indefinitely, the area generated by the broken line approaches the area of the zone as a limit. HIence COR. 3. The area of a zone is equal to the circumference of a great circle multiplied by the altitude of the zone. Thus the area generated by the ar BC = FO X 2 7.R. 814. Colt. 4. On the same sphere, or on equal spheres, the areas of two zouil's are to eachl other as the altitudes of the zones. 2O'lC~S. Ex. 1. Find the area of a sphere whose diameter is 10 in. Ex. 2. Find the area of a zone of altitude 3 in., on a sphere whose rafdius is 10 in. 4 I"')' 6 43005 OK IX. SOLID GEOMETRY PROPOSITION XXV. THEOREM 815. The area, of a lane is to the arfeat of the, snirface of the sphere as the anYgle of the lzne is tofourii, rig/Ut angles. A F ii Given a sphere having its area denoted by S, aInd on the sphere the lune ABOD of / A with its area denoted by L. To prove L: 8=A0: 3600. Proof. Draw F73ff, the great 0 whose pole is A, intersecting the bounding ares of the lune in B and -D. CAsE I. When the are BD and the, circimiference FBH are comimensurable. Find a common measure of BD and F1311, and let it be contained in the are BD m times, and in the circumference FBHnit times. Then arc BD: circumfference T PBH= mit: n. Through the diameter AG, and the points of division of the circumference EBH pass planes of great 0. The arcs of these great 0 will divide the surface of the sphere in n small equal lunes, m of them being contained in the lune ABCD. L:S=mn: n. L L: S= are BD: circumference EBH. (Why?) Or L S=A0':3600. Art. 257. CASE H1. When the arc BD and the circumiference FB.U are inzcomm )ienstra ble. Let the pupil supply the proof. Q. E. D. SPHERICAL AREAS 457 816. Formula for the area of a lune in spherical degrees, or spheridso The surface of a sphere contains 720 spherids (Art. 803). Hence, by Art. 815, L L spherids A L A or - or -=A /' ^^A or = 2Asporeids; S 720 spherids 360' 2 L=2 that is, the area of a ltue in) spherical degrees is equal to twice the number of angular degrees in the angle of the lune. 817. Formula for area of a lune in square units of areao A - TJIA S=4 7r2R (Art. 810). 4 R —( or L3 90 818. COR. 1. On the same sphere, or on equal spheres, two latnes are to each other as their zangles. 819. COR. 2. Two lunes with equawl angles, but, on unequal spheres. are to each other as the squares of the radii of their spheres. T 7tR2A 7R'2A For L L' -: or L: L'=R-': 2. 90 90 o ' Ex. 1. Find the area in spherical degrees of a lune of 27~. Ex. 2. Find the number of square inches in the area of a lune ot 27~, on a sphere whose radius is 10 in. A solid symmetrical with respect to a plane is a solid in which a line drawn from any point in its surface I- the given plane and produced its own length ends in a point on the surface; hence Ex. 3. How many planes of symmetry has a circular cylinder? A cylinder of revolution? Ex. 4. Has either of these solids a center of symmetry? Ex. 5. Answer the same questions for a circular cone. Ex. 6. For a cone of revolution. For a sphere. Ex. 7. For a regular square pyramid. For a regular pentagonal pyramid. 458 BOOK IX. SOLID GEOMETRY PROPOSITION XXVI. THEOREM 820. If two great circles intersect on a hemisphere, the sum of two vertical triangles thus formed is equivalent to a lune whose angle is that angle in the triangles which is formed by the intersection of the two great circles. At B, B Given the hemisphere ADBF, and on it the great circles AFB and DFC, intersecting at F. To prove A AFC + A BFD _ lune whose Z is BFD. Proof. Complete the sphere and produce the given arcs of the great circles to intersect at F' on the other hemisphere. Then, in the & AFC and BF'D, ar AF= arc BF, (each being the supplement of the arc BF). In like manner arc CF= arc DF'. And arc AC= arc DB.. A FC- A BF'D. Art. 795. Add the A BFD to each of these equals;. A. AAFC - A BFD-/A BF'D + A BFD. Ax. 3. Or.. AFC +A BFD \lune FBF'D. Ax. 6, Q. E. D. SPHERICA.L AREAS49 459 PROPOSITION XXVII. THEOREMI 821. T'he winimber of spherical d(grees, or spherids, in the area of a spherical triangle is equal to the num~ber of angular degrees in the spherical excess of the triangle. A Given the spherical A ABC whose Z~ are denoted by A, B, C, and whose spherical excess is denoted by E. To prove area of A ABC FE spherids. Proof. Produce the sides AC and BC to meet AB produced in the points D and F, respectively. A ABC -4- CDB =Imic A BD C 2 A spbcrids.) Art. 816. ZA ABC -H~sA ACF= Itnc B CFIIU B spheri(Is. A ABC-HA CFD =luue of jB CA4_ 2 C- sp)hCieids. *rt820 -Adding, and observing that A ABC-HA CDB +t-AACF -I CED =hemisphere ABRDFC, 2 A ABC+I hemisphere =2 (A -F B, C) sphIerids. Or, 2 A ABC+I 360 spherids=21 (A +I P C) splieridIs. Ax. S. A. ABC -1 180 spherids = (A I-h + I- C) splierids. Ax. 5. A. ABC= (A -I B + C —i8O) spherids. Ax. 3. Or area A ABC-F sperid. A rt. 788S. Q. E. D. 460 BOOK IX. SOLID GEOMETRY 822. Formula for area of a spherical triangle in square units of area. Comparing the area of a spherical A with the area of the entire sphere area A: 4 -IR=E spherids: 720 spherids. 4 t, X E 1 R2E. area A = — 7, or area A-. 72O 180 823. The spherical excess of a spherical polygon is the sum of the angles of the polygon diminished by (n-2) 180~; that is, it is the sum of the spherical excesses of the triangles into which the polygon may be divided. PROPOSITION XXVIII. THEOREM 824. The area of a spherical polygon, in spherical degrees or spherids, is equal to the spherical excess of the polygon. A I/ Given a spherical polygon ABCDF of n sides, with its spherical excess denoted by E. To prove area of ABCDF= E spherical degrees. Proof. Draw diagonals from A, any vertex of the polygon, and thus divide the polygon into n —2 spherical A. The area of each A = (sum of its A -180) spherids. Art. 821... sum of the areas of the A = [sum of X of the A(n-2) 180] spherids. Ax. 2... area of polygon = E spherids, Art. 823. (for the sum of f of the A-(n-2) 180~=E~), SPHERICAL VOLUMES 461 SPHERICAL VOLUMES 825. A spherical pyramid is a portion of a sphere bounded by a spheri- / B cal polygon and the planes of the great circles forming the sides of the \-. polygon. The base of a spherical pyramid is the spherical polygon bounding it, and the vertex of the spherical pyramid is the center of the sphere. Thus, in the spherical pyramid O-ABCD, the base is ABGD and the vertex is 0. 826. A spherical wedge (or ungula) is the portion of a sphere bounded by a lune and the planes of the sides of the lune. 827. A spherical sector is the portion of a sphere generated by a sector of that semicircle whose rotation generates the given sphere. 828. The base of a spherical sector is the zone generated by the revolution of the arc of the plane sector which generates the spherical sector. Let the pupil draw a spherical sector in which the base is a zone of one base. 462O B)OKl IX. SOLID GEOMETliY 829. A spherical segment is a portion of a sphere included between two parallel planes. The bases of a spherical segment are the sections of the sphere made by the parallel planes which bound the given segment; the altitude is the perpendicular distance between the bases. 830. A spherical segment of one base is a spherical segment one of whose bounding planes is tangent to the sphere. PROPOSITION XXIX. THEOREM 831. The volume of a sphere is equal to one-third the product of the area of its surface by its radius. I f. I, t I X Given a sphere having its volume denoted by V, snrface by S, and radius by R. To prove V=i S X R. Proof. Let any polyhedron be circumscribed about the sphere. Pass a plane through each edge of the polyhedron and the center of the sphere. These planes will divide the polyhedron into as many pyramids as the polyhedron has faces, each pyramid having a face of the polyhedron for its base, the center of the SPHERICAL VOLTUMES 463 sphere for its vertex, and the radius of the sphere for its altitude..*. volume of each pyramid = base X R. (Why?).. volume of polyhedron =X (surface of polyhedron) X R. If the number of faces of the polyhedron be increased indefinitely, the volume of the polyhedron approaches the volume of the sphere as a limit, and the surface of the polyhedron approaches the surface of the sphere as a limit. Hence the volume of the polyhedron and - (surface of the polyhedron) X R, are two variables always equal. Hence their limits are equal, Or V= X R. (Why ) Q. E. D. 832. Formulas for volume of a sphere. Substituting S=4 tR2, or S=7tD2 in the result of Art. 831, 4 tRI3 tD3 V= --; also V=- - 3 6 833. COR. 1. The volumes of two spheres are to each other as the cubes of their radii, or as the cubes of their diamleters. V 47~R3 R" V DT '3~ DI For I - In also = D3 834. COR. 2. 'The rolunme of a spherical pyramid is rqcal to one-third the proldct of its base by the radius of the sphere. 835. CoR. 3. Thle volume of spherical sector is equal to Ofi- third the prodluct (f its base (the bountding zone) by the 'rat(lit of t1h spthe re. 464 BOOK IX. SOLID GEOMETRY 836. Formula for the volume of a spherical sector. De. noting the altitude of the sector by H and the volume by V, T= - (area of zone)X R, =- (2 'tRH) R. Art. 813..-. ==- n= rRH. PROPOSITION XXX. THEOREM 837. The voltme of a spherical segment is equal to onehalf the product of its altitude by the sum of the areas of its bases, plus the volulme of a sphere whose diameter is the altitude of the segment. Given the semicircle ABCA' which generates a sphere by its rotation about the diameter AA'; BD and CF semichords 1 AA', and denoted by r and r'; and DF denoted by H. To prove volume of spherical segment generated by BCFD, or V=(7tr2 +7r2) H+ - 73. Proof. Draw the radii OB and OC. Denote OF by h, and OD by k. Then V=vol. OBC + vol. OCF-vol. OBD..'. V=- R2H -+K. nrrt h — 7tr2k. Arts. 836, 723. But H= h-k, h2 =R2-r2, and k2=.R-r2, (Why?) SPHERICAL VOLUMES 46 465 V- 7t Pi, '2 (Ii-k) (R2-712) h,-(I?2-k k]. Ax. 8. 2At [2 R (h-k) R2 (h-1) —(h3-k3)1. 7t H[3 R12-(h12+ h- + k2)]. But h -2f hk + k2 112. Ax. 4. Subtract each member from 3 h12 -1 3 k2 and divide by 2. Axs. 3, 5. 12 Then h2 + hk + V=-R- (h2" + k)2 =3 R2-I (p.2 + '.1'2) f2 Q. E. D. 838. Formula for volume of a spherical segment of one base, In a spherical segment of one base r O, and?2= (2- R-H) H (Art. 343). Substituting for r and r' these values in the result of Art. 837, V~= 7t~ f 2 _(~.H 839. Advantage of measurement formulas. The student should observe carefully that, by the results obtained in Look IX, the measurement of the areas of 9ertain curved surfaces is reduced to the far simpler work of the measurement of the lengths of one or more straight lines; in like manner the measurement of certain volumes bounded byv a curved surface is reduced to the simpler work of lineatir measurements. A similar rermark applies to the results of Book VIII. Ex. 1. Find the volume of a sphere w hose radius is 7 in. Ex. 2. Find the volume of a sphere whose diameter is 7 in. Ex. 3. In a sphere whose radius is 8 in., find the volume of a spherical segment of one base whose altitude is 3. DD 466 BOOK IX. SOLID GEOMETRY EXERCISES. CROUP 72 THEOREMS CONCERNING THE SPHERE Ex. 1. Of circles of a sphere whose planes pass through a given point within a sphere, the smallest is that circle whose plante is perpendicular to the diameter through the given point. Ex. 2. If a point on the surface of a given sphere is equidistant from three points on a given small circle of the sphere, it is the pole of the small circle. Ex. 3. If two sides of a spherical triangle are quadrants, the third side measures the angle opposite that side in the triangle. Ex. 4. If a spherical triangle has one right angle, the sum of its other two angles is greater than one right angle. Ex. 5. If a spherical triangle is isosceles, its polar triangle is isosceles. Ex. 6. The polar triangle of a birectangular triangle is birectangular. Ex. 7. The polar triangle of a trirectangular triangle is identical with the original triangle. Ex. 8. Prove that the sum of the angles of a spherical quadrilateral is greater than 4 right angles, and less than 8 right angles. What, also, are the limits of the sum of the angles of a spherical hexagon? Of the sum of the angles of a spherical n-gon? Ex. 9. On the same sphere, or on equal spheres, two birectangular triangles are equal if their oblique angles are equal. Ex. 10. Two zones on the same sphere, or on equal spheres, are to each other as their altitudes. Ex. II. If one of the legs of a right spherical triangle is greater than a quadrant, another side is also greater than a quadrant. [SUG. Of the leg which is greater than a quadrant, take the end remote from the right angle as a pole, and describe an arc.] Ex. 12. If ABC and A'B'C' are polar triangles, the radius OA is perpendicular to the plane OB'C'. EXERCISES ON THE SPHERE 467 Ex. 13. On the same sphere, or on equal spheres, spherical triangles whose polar triangles have equal perimeters are equivalent. Ex. 14. Given OAO', OBO', and AB arcs of great o circles, intersecting so that Z OAB= Z O'BA; prove that A OAB= A O'A. A/ [Sue. Show that Z OBA= Z O'AB.] Ex. 15. Find the ratio of the volume of a sphere to the volume of a circumscribed cube. Ex. 16. Find the ratio of the surface of a sphere to the lateral surface of a circumscribed cylinder of revolution; also find the ratio of their volumes. Ex. 17. If the edge of a regular tetrahedron is denoted by a, find the ratio of the volumes of the inscribed and circumscribed sphe-as. Ex. 18. Find the ratio of the two segments into which a hemisphere is divided by a plane parallel to the base of the hemisphere and at the distance JR from the base. EXERCI8SE.o ROUP 73 SPHERICAL LOCI Ex. 1. Find the locus of a point at a given distance a from the surface of a given sphere. Ex. 2. Find the locus of a point on the surface of a sphere that is equidistant from two given points on the surface. Ex. 3. If, through a given point outside a given sphere, tangent planes to the sphere are passed, find the locus of the points of tangency. Ex. 4. If straight lines be passed through a given fixed point in space, and through another given point other straight lines be passed perpendicular to the first set, find the locus of the feet of the perpendiculars. 468 BOOK IX. SOLID GEOMETRY EXERCISES. GROUP 74 PROBLEIMS CONCERNSING THIE SPHERE Ex. 1. At a given point on a sphere, construct a plane tangent to the sphere. Ex. 2. Through a given point on the surface of a sphere, draw an arc of a great circle perpendicular to a given arc. Ex. 3. Inscribe a circle in a given spherical triangle. Ex. 4. Construct a spherical triangle, given its polar triangle. Given the radius, r, construct a spherical surface which shall pass through Ex. 5. Three given points. Ex. 6. Two given points and be tangent to a given plane. Ex. 7. Two given points and be tangent to a given sphere Ex. 8. One given point and be tangent to two given planes. Ex. 9. One given point and be tangent to two given spheres. Given the radius, r, construct a spherical surface which shall be tangent to Ex. 10. Three given planes. Ex. 11. Two given planes and one given sphere. Ex. 12. Construct a spherical surface which shall pass through three given points and be tangent to a given plane. Ex. 13. Through a given straight line pass a plane tangent to a given sphere. [Sue. Through the center of the sphere pass a plane J- given line, etc.] When is the solution impossible? Ex. 14. Through a given point on a sphere, construct an are of a great circle tangent to a given small circle of the sphere. [SUG. Draw a straight line from the center of the sphere to the given point, and produce it to intersect the plane of the small circle, etc.] NUMERICAL EXERCISES IN SOLID GEOMETRY For methods of facilitating numerical computations, see Arts. 493-6. EXEROISES CROUP 7T LINES AND SURFACES OF POLYHEDRONS Find the lateral area and total area of a right prism whose Ex. 1. Base is an equilateral triangle of edge 4 in., and whose altitude is 15 in. Ex. 2. Base is a triangle of sides 17, 12, 25, and whose altitude is 20. Ex. 3. Base is an isosceles trapezoid, the parallel bases being 10 and 15 and leg 8, and whose altitude is 24. Ex. 4. Base is a rhombus whose diagonals are 12 and 16, and whose altitude is 12. Ex. 5. Base is a regular hexagon with side 8 ft., and whose altitude is 20 ft. Ex. 6. Find the entire surface of a rectangular parallelopiped 8X12X 16 in.; of one pX qX rft. Ex. 7. Of a cube whose edge is 1 ft. 3 in. Ex. 8. The lateral area of a regular hexagonal prism is 120 sq. ft. and an edge of the base is 10 ft. Find the altitude. Ex. 9. How many square feet of til are necessary to line a box 20X6X4 in.? Ex. 10. If the surface of a cube is 1 sq. yd., find an edge in inches Ex. 11. Find the diagonal of a cube whose edge is 5 in. Ex. 12. If the diagonal of a cube is 12 ft., find the surface. (469) 470 SOLID GEOMETITRY Ex. 13. If the surface of a rectangular parallelepiped is 208 sq. in. and the edges are as 2:3 4, find the edges. In a regular square pyramid Ex. 14. If an edge of the base is 16 and slant height is 17, find the altitude. Ex. 15. If the altitude is 15 and a lateral edge is 17, find an edge of the base. Ex. 16. If a lateral edge is 25 and an edge of the base is 14, find the altitude. In a regular triangular pyramid Ex. 17. If an edge of the base is 8 and the altitude is 10, find the slant height. Ex. 18. Find the altitude of a regular tetrahedron whose edge is 6. Find the lateral surface and the total surface of Ex. 19. A regular square pyramid an edge of whose base is 16, and whose altitude is 15. Ex. 20. A regular triangular pyramid an edge of whose base is 10, and whose altitude is 12. Ex. 21. A regular hexagonal pyramid an edge of whose base is 4, and whose altitude is 21. Ex. 22. A regular square pyramid whose slant height is 24, and whose lateral edge is 25. Ex. 23. A regular tetrahedron whose edge is 4. Ex. 24. A regular tetrahedron whose altitude is 9o Ex. 25. A regular hexagonal pyramid each edge of whose base is a, and whose altitude is b. Ex. 26. In the frustum of a regular square pyramid the edges of the bases are 6 and 18, and the altitude is 8. Find the slant height. Hence find the lateral area. Ex. 27. In the frustum of a regular triangular pyramid the edges of the bases are 4 and 6, and the altitude is 5, Find the slant height. Hence find the lateral area. NUMERICAL EXERCISES IN SOLID GEOMETRY 471 Ex. 28. In the frustum of a regular tetrahedron, if the edge of the lower base is bi, the edge of the upper base is b2, and the altitude is a, show that L= -/S ( b - b2)2 +-4a2. Ex. 29. In the frustum of a regular square pyramid the edges of the bases are 20 and 60, and a lateral edge is 101. Find the lateral surface. EXERCISES. CROUP T7 LINES AND SURFACES OF CONES AND CYLINDERS Ex. 1. How many square feet of lateral surface has a tunnel 100 yds. long and 7 ft. in diameter Ex. 2. The lateral area of a cylinder of revolution is 1 sq. yd., and the altitude is 1 ft. Find the radius of the base. Ex. 3. The entire surface of a cylinder of revolution is 900 sq. ft. and the radius of the base is 10 ft. Find the altitudeo In a cylinder of revolution Ex. 4. Find R in terms of S and H. Ex. 5. Find H in terms of R and T. Ex. 6. Find T in terms of S and H. Ex. 7. How many sq. yds, of canvas are required to make a conical tent 20 ft. in diameter and 12 ft. high? Ex. 8. A man has 400 sq. yds. of canvas and wants to make a conical tent 20 yds. in diameter. What will be its altitude? Ex. 9. The altitude of a cone of revolution is 10 ft. and the lateral area is 11 times the area of the base. Find the radius of the base. In a cone of revolution Ex. 10. Find T in terms of S and L. Ex. 11. Find R in terms of T and L. Ex. 12. How many square feet of tin are necessary to make a funnel the diameters of whose ends are 2 in. and 8 in., and whose altitude is 7 in.? Ex. 13. If the slant height of a frustum of a cone of revolution makes an angle of 45~ with the base, show that the lateral area of the frustum is (ri2-r-12) 7r v/2. 472 SOLID GEOMETRY EXERCI8ES. CROUP 77 SPHERICAL LINES AND SURFACES Ex. 1. Find in square feet the area of the surface of a sphere whose radius is 1 ft. 2 in. Ex. 2. How many square inches of leather will it take to cover a baseball whose diameter is 34 in.? Ex. 3. How many sq. ft. of tin are required to cover a dome in the shape of a hemisphere 6 yds. in diameter? Ex. 4. What is the radius of a sphere whose surface is 616 sq. in.? Ex. 5. Find the diameter of a globe whose surface is 1 sq. yd. Ex. 6. If the circumference of a great circle on a sphere is 1 ft., find the area of the surface of the sphere. Ex. 7. If a hemispherical dome is to contain 100 sq. yds. of surface, what must its diameter be? Ex. 8. Find the radius of a sphere in which the area of the surface equals the number of linear units in the circumference of a great circle. Find the area of a lune in which Ex. 9. The angle of the lune is 36~, and the radius of the sphere is 14 in. Ex. 10. The angle of the lune is 18~ 20', and the diameter of the sphere is 20 in. Ex. 11. The angle of the lune is 24~, and the surface of the sphere is 4 sq. ft. Find the area of a spherical triangle in which Ex. 12. The angles are 80~, 90~, 1200, and the diameter of the sphere is 14 ft. Ex. 13. The angles are 74~ 24', 83~ 16', 92~ 20', and the radius of the sphere is 10. Ex. 14. The angles are 850, 95~, 135~, and the surface of the sphere is 10 sq. ft. NUMERICAL EXERCISES IN SOLID GEOMETRY 473 Ex. 15. If the sides of a spherical triangle are 100~, 110~, 120" and the radius of the sphere is 16, find the area of the polar triangleo Ex. 16. If the angles of a spherical triangle are 90~, 100~, 120' and its area is 3900, find the radius of the sphere. Ex. 17. If the area of an equilateral spherical triangle is onethird the surface of the sphere, find an angle of the triangle. Ex. 18. In a trihedral angle the plane angles of the dihedral angles are 80~, 90~, 100~; find the number of solid degrees in the trihedral angle. Ex. 19. Find the area of a spherical hexagon each of whose angles is 150~, on a sphere whose radius is 20 in. Ex. 20. If each dihedral angle of a given pentahedral angle is 120~, how many solid degrees does the pentahedral angle contain? Ex. 21. In a sphere whose radius is 14 in., find the area of a zone 3 in. high. Ex. 22. What is the area of the north temperate zone, if the earth is taken to be a sphere with a radius of 4,000 miles, and the distance between the plane of the arctic circle and that of the tropic of Cancer is 1,800 miles? Ex. 23. If Cairo, Egypt, is in latitude 30~, show that its parallel of latitude bisects the surface of the northern hemisphere. Ex. 24. How high must a person be above the earth's surface to see one-third of the surface? Ex. 25. How much of the earth's surface will a man see who is 2,000 miles above the surface, if the diameter is taken as 8,000 miles? Ex. 26. If the area of a zone equals the area of a great circle, find the altitude of the zone in terms of the radius of the sphere. Ex. 27. If sounds from the Krakatoa explosion were heard at a distance of 3,000 miles (taken as a chord) on the surface of the earth, over what fraction of the earth's surface were they heard? Ex. 28. The radii of two spheres are 5 and 12 in. and their centers are 13 in. apart. Find the area of the circle of intersection and also of that part of the surface of each sphere not included by the other sphere. 474 SOLID GEOMETRY EXERGCIES. GROUP 78 VOLUMES OF POLYHEDRONS Find the volume of a prism Ex. 1. Whose base is an equilateral triangle with side 5 in., and whose altitude is 16 in. Ex. 2. Whose base is a triangle with sides 12, 13, 15, and whose altitude is 20. Ex. 3. Whose base is an isosceles right triangle with a leg equal to 2 yds., and whose altitude is 25 ft. Ex. 4. Whose base is a regular hexagonswith a side of 8 ft., and whose altitude is 10 yds. Ex. 5. Whose base is a rhombus one of whose sides is 25, and one of whose diagonals is 14, and whose altitude is 11. Ex. 6. Whose base contains 84 sc. yds., and whose lateral faces are three rectangles with areas of 100, 170, 210 sq. yds., respectively. Ex. 7. How many bushels of wheat are held by a bin 30 x 10 x 6 ft., if a blhel is taken as 11 cu. ft.? Ex. 8. How many cart-loads of earth are in a cellar 30 x 20 x 6 ft., if a cart-load is a cubic yard? Ex. 9. If a cubical block of marble costs $2, what is the cost of a cube whose edge is a diagonal of the first block? Ex. 10. Find the edge of a cube whose volume equals the sum of the volumes of two cubes whose edges are 3 and 5 ft. Ex. 11. Find the edge of a cube whose volume equals the area of its surface. Ex. 12. If the top of a cistern is a rectangle 12 x 8 ft., how deep must the cistern be to hold 10,000 gallons? Ex. 13. Find the inner edge of a peck measure which is in the shape of a cube. Ex. 14. A peck measure is to be a rectangular parallelopiped with square base and altitude equal to twice the edge of the base. Find its dimensions, NUMERICAL EXERCISES IN SOLID GEOMETRY 475 Ex. 15. Find the volume of a cube whose diagonal is a. Find the volume of a pyramid Ex. 16. Whose base is an equilateral triangle with side 8 in., and whose altitude is 12 in. Ex. 17. Whose base is a right triangle with hypotenuse 29 and one leg 21, and whose altitude is 20. EA. 18. Whose base is a square with side 6, and each of whose lateral edges is 5. Ex. 19. Whose base is a square with side 10, and each of whose lateral faces makes an angle of 45~ with the base. Ex. 20. If the pyramid of Memphis has an altitude of 146 yds. and a square base of side 232 yds., how many cubic yards of stone does it contain? What is this worth at $1 a cu. yd. Ex. 21. A church spire 150 ft. high is hexagonal in shape and each side of the base is 10 ft. The spire has a hollow hexagonal interior, each side of whose base is 6 ft., and whose altitude is 45 ft. How many cubic yards of stone does the spire contain? Ex. 22. If a pyramid contains 4 cu. yds. and its base is a square with one side 2 ft., find the altitude. Ex. 23. A heap of candy in the shape of a frustum of a regular square pyramid has the edges of its bases 25 and 9 in. and its altitude 12 in. Find the number of pounds in the heap if a pound is a rectangular parallelopiped 4 x 3 x2 in. in size. Ex. 24. Find the volume of a frustum of a regular triangular pyramid, the edges of the bases being 2 and 8, and the slant height 12. Ex. 25. The edges of the bases of the frustum of a regular square pyramid are 24 and 6, and each lateral edge is 15; find the volume. Ex. 26. If a stick of timber is in the shape of a frustum of a regular square pyramid with the edges of its ends 9 and 15 in., and with a length of 14 ft., find the number of feet of lumber in the stick. What is the difference between this volume and that of a stick of the same length having the shape of a prism with a base equal to the area of a midsection of the first stick? 476 SOLID GxtOMETtY Ex. 27. Find the volume of a prismoid whose bases are rectangles 5 x 2 ft. and 7 x 4 ft., and whose altitude is 12 ft. Ex. 28. How many cart-loads of earth are there in a railroad cut 12 ft. deep, whose base is a rectangle 100 x 8 ft., and whose top is a rectangle 30 x 50 ft.? Ex. 29. Find the volume of a prismatoid whose base is an equilateral triangle with side 12 ft., and whose top is a line 12 ft. long parallel to one side of the base, and whose altitude is 15 ft. Ex. 30. If the base of a prismatoid is a rectangle with dimensions a and b, the top is a line c parallel to the side b of the base, and the altitude is h, find the volume. EXERCISES. GROUP 79 VOLUMES OF CONES AND CYLINDERS Ex. 1. How many barrels of oil are contained in a cylindrical tank 20 ft. long and 6 ft. in diameter, if a barrel contains 4 cu. ft.? Ex. 2. How many cu. yds. of earth must be removed in making a tunnel 450 ft. long, if a cross-section of the tunnel is a sermicircle of 15 ft. radius? Ex. 3. A cylindrical glass 3 in. in diameter holds half a pint. Find its height in inches. Ex. 4. If a cubic foot of brass be drawn out into wire A inch in diameter, how long will the wire be? Ex. 5. A gallon measure is a cylinder whose altitude equals the diameter of the base. Find the altitude. Ex. 6. Show that the volumes of two cylinders, having the altitude of each equal to the radius of the other, are to each other as B': Rt. Ex. 7. In a cylinder, find R in terms of V and H; also V in terms of S and R. Ex. 8. A conical heap of potatoes is 44 ft. in circumference and 6 ft. high. How many bushels does it contain, if a bushel is 41 cu. ft. Ex. 9. WNat fraction of a pint will a conical wine-glass hold, if its altitude is 3 in, and the diameter of the top is 2 i.n? NUMERICAL EXERCISES IN SOLID GEOMETRY 477 Ex. 10. Find the ratio of the volumes of the two cones inscribed in, and circumscribed about, a regular tetrahedron. Ex. 11. If an equilateral cone contains 1 quart, find its dimensions in inches. Ex. 12. In a cone of revolution find V in terms of R and L; also find V in terms of P and S. Ex. 13. Find the volume of a frustum of a cone of revolution, whose radii are 14 and 7 ft., and whose altitude is 2 yds. Ex. 14. What is the cost, at 50 cts. a cu. ft., of a piece of marble in the shape of a frustum of a cone of revolution, whose radii are 6 and 9 ft., and whose slant height is 5 ft.? Ex. 15. In a frustum of a cone of revolution, the volume is 88 cu. ft., the altitude is 9 ft., and I-=2r. Find r. EXERCI8E8. GROUP 80 SPHERICAL VOLUMES Ex. 1. Find the volume of a sphere whose radius is 1 ft. 9 in. Ex. 2. If the earth is a sphere 7,920 miles in diameter, find its volume in cubic miles. Ex. 3. Find the diameter of a sphere whose volume is 1 cu. ft. Ex. 4. What is the volume of a sphere whose surface is 616 sq. in.? Ex. 5. Find the radius of a sphere equivalent to the sum of two spheres, whose radii are 2 and 4 in. Ex. 6. Find the radius of a sphere whose volume equals the area of its surface. Ex. 7. Find the volume of a sphere circumscribed about a cube whose edge is 6. Ex. 8. Find the volume of a spherical shell whose inner and outer diameters are 14 and 21 in. Ex. 9. Find the volume of a spherical shell whose inner and outer surfaces are 20 wr and 12 w. 478 SOLID GEOMIETRY Find the volume of Ex. 10. A spherical wedge whose angle is 24~, the radius of the sphere being 10 in. Ex. 11. A spherical sector whose base is a zone 2 in. high, the radius of the sphere being 10 in. Ex. 12. A spherical segment of two bases whose radii are 4 and 7 and altitude 5 in. Ex. 13. A wash-basin in the shape of a segment of a sphere is 6 in. deep and 24 in. in diameter. How many quarts of water will the basin hold? Ex. 14. A plane parallel to the base of a hemisphere and bisecting the altitude divides its volume in what ratio I Ex. 15. A spherical segment 4 in. high contains 200 cu. in.; find the radius of the sphere. Ex. 16. If a heavy sphere whose diameter is 4 in. be placed in a conical wine-glass full of water, whose diameter is 5 in. and altitude 6 in., find how much water will run over. EXERCISES. QROUP 8S EQUIVALENT SOLIDS Ex. 1. If a cubical block of putty, each edge of which is 8 inches, be molded into a cylinder of revolution whose radius is 3 inches, find the altitude of the cylinder. Ex. 2. Find the radius of a sphere equivalent to a cube whose edge is 10 in. Ex. 3. Find the radius of a sphere equivalent to a cone of revolution, whose radius is 3 in. and altitude 6 in. Ex. 4. Find the edge of a cube equivalent to a frustum of a cone of revolution, whose radii are 4 and 9 ft. and altitude 2 yds. Ex. 5. Find the altitude of a rectangular parallelepiped, whose base is 3 x 5 in. and whose volume is equivalent to a sphere of radius 7 in. NUMERICAL EXERCISES IN SOLID GEOMETRY 479 Ex. 6. Find the base of a square rectangular parallelopiped, whose altitude is 8 in. and whose volume equals the volume of a cone of revolution with a radius of 6 and an altitude of 12 in. Ex. 7. Find the radius of a cone of revolution, whose altitude is 15 and whose volume is equal to that of a cylinder of revolution with radius 6 and altitude 20. Ex. 8. Find the altitude of a cone of revolution, whose radius is 15 and whose volume equals the volume of a cone of revolution with radius 9 and altitude 24. Ex. 9. On a sphere whose diameter is 14 the altitude of a zone of one base is 2. Find the altitude of a cylinder of revolution, whose base equals the base of the zone and whose lateral surface equals the surface of the zone. EXERCISES GROUP 32 SIMILAR SOLIDS Ex. 1. If on two similar solids L, L' and 1, 1 are pairs of homologous lines; A, A' and a, a' pairs of homologous areas, V, VF and v, v' pairs of homologous volumes, fL L = I: L l: I'-= A /: A /Al - 'v {'v. 2 2 show that A A-:L2: L '=a a' = F: rF. F: V'=L:LL'"=A A'=v: v. Ex. 2. If the edge of a cube is 10 in., find the edge of a cube having 5 times the surface. Ex. 3. If the radius of a sphere is 10 in., find the radius of a sphere having 5 times the surface. Ex. 4. If the altitude of a cone of revolution is 10 in., find the altitude of a similar cone of revolution having 5 times the surface. Ex. 5. In the last three exercises, find the required dimension if the volume is to be 5 times the volume of the original solid. Ex. 6. The linear dimensions of one trunk are twice as great as those of another trunk. How much greater is the volume? The surface? 480 SOLID GEOMETRY Ex. 7. How far from the vertex is the cross-section which bisects the volume of a cone of revolution? Which bisects the lateral surface? Ex. 8. If the altitude of a pyramid is bisected by a plane parallel to the base, how does the area of the cross-section compare with the area of the base? Ilow does the volume cut off compare with the volume of the entire pyramid? Ex. 9. Planes parallel to the base of a cone divide the altitude into three equal parts; compare the lateral surfaces cut off. Also the volumes; Ex. 10. A sphere 10 in. in diameter is divided into three equivalent parts by concentric spherical surfaces. Find the diameters of these surfaces. Ex. 11. If the strength of a muscle is as the area of its crosssection, and Goliath of Gath was three times as large in each linear dimension as Tom Thumb, how much greater was his strength? His weight? How, then, does the activity of the one man compare with that of the other? Ex. 12. If the rate at which heat radiates from a body is in proportion to the amount of surface, and the planet Jupiter has a diameter 11 times that of the earth, how many times longer will Jupiter be in cooling off? [SuG. How many times greater is the volume, and therefore the original amount of heat in Jupiter? How many times greater is its surface? What will be the combined effect of these factors?] CROUP 83 MISCELLANEOUS XNUMIERICAL EXERCISES IN SOLID GEOMETRY Find S, T and V of Ex. 1. A right triangular prism whose altitude is 1 ft., and the sides of whose base are 26, 28, 30 in. Ex. 2. A cone of revolution the radius of whose base is 1 ft. 2 in., and whose altitude is 35 in. Ex. 3. A frustum of a square pyramid the areas of whose bases are 1 sq. ft. and 36 sq. in., and whose altitude is 9 in. NUMERICAL EXERCISES IN SOLID GEOMETRY 481 Ex. 4. A pyramid whose slant height is 10 in., and whose base is an equilateral triangle whose side is 8 in. Ex. 5. A cube whose diagonal is 1 yd. Ex. 6. A frustum of a cone of revolution whose radii are 6 and 11 in. and slant height 13 in. Ex. 7. A rectangular parallelopiped whose diagonal is 2V/29, and whose dimensions are in the ratio 2: 3: 4. Ex. 8. Find the volume of a sphere inscribed in a cube whose edge is 6; also find the area of a triangle on that sphere whose angles are 80~, 90~, 150~. Ex. 9. Find the volume of the spherical pyramid whose base is the above triangle. Ex. 10. Find the angle of a lune on the same sphere, equivalent to that triangle. Ex. 1. On a cube whose edge is 4, planes through the midpoints of the edges cut off the corners. Find the volume of the solid remaining. Ex. 12. How is V changed if H of a cone of revolution is doubled and R remains unchanged? If R is doubled and H remains unchanged? If both H and B are doubled? Ex. 13. In an equilateral cone, find S and V in terms of R. Ex. 14. A piece of lead 20 x 8 x 2 in. will make how many spherical bullets, each e in. in diameter? Ex. 15. How many bricks are necessary to make a chimney in the shape of a frustum of a cone, whose altitude is 90 ft., whose outer diameters are 3 and 8 ft., and whose inner diameters are 2 and 4 ft., counting 12 bricks to the cubic ft.? Ex. 16. If the area of a zone is 300 and its altitude 6, find the radius of the sphere. Ex. 17. If the section of a cylinder of revolution through its axis is a square, find S, T, V in terms of R. Ex. 18. If every edge of a square pyramid is find b in terms of T. DD 482 SOLID GEOTMEITRI Ex. 19. A reuiilar square pyramid has a for its altitude andalso,r each side of the base. Find the area of a section made by a plane parallel to the base and bisecting the altitude. Find also the volumes of the two parts into which the pyramid is divided. Ex. 20. If the earth is a sphere of 8,000 miles diameter and its atmosphere extends 50 miles from the earth, find the volume of the atmosphere. Ex. 21. On a sphere, find the ratio of the area of an equilateral spherical triangle, each of whose angles is 95C, to the area of a lune whose angle is 80~. Ex. 22. A square right prism has an altitude 6a and an edge of the base 2o. Find the volume of the largest cylinder, sphere, pyramid and cone which can be cut from it. Ex. 23. Obtain a formula for the area of that part of a sphere illuminated by a point of light at a distance a from the sphere whose radius is R. Ex. 24. On a sphere whose radius is 6 in., find an angle of an equilateral triangle whose area is 12 sq. in. Ex. 25. Find the volume of a prismatoid, whose altitude is 24 and whose bases are equilateral triangles, each side 10, so placed that the mid-section of the prismatoid is a regular hexagon. Ex. 26. On a sphere whose radius is 16, the bases of a zone are equal and are together equal to the area of the zone. Find the altitude of the zone. Ex. 27. Find the volume of a square pyramid, the edge of whose base is 10 and each of whose lateral edges is inclined 60~ to the base. Ex. 28. An irregular piece of ore, if placed in a cylinder partly filled with water, causes the water to rise 6 in. If the radius of the cylinder is 8 in., what is the volume of the ore? Ex. 29. Find the volume of a truncated right triangular prism, if the edges of the base are 8, 9, 11, and the lateral edges are 12, 13, 14. Ex. 30. In a sphere whose radius is 5, a section is taken at the distance 3 from the center. On this section as a base a cone is formed whose lateral elements are tangent to the sphere. Find the lateral surface and volume of the cone. NUMERICAL EXERCISES IN SOLID GEOMETRY 483 Ex. 31. The volume of a sphere is 1,437i cu. in. Find the surface. Ex. 32. A square whose side is 6 is revolved about a diagonal as an axis; find the surface and volume generated. Ex. 33. Find the edge of a cubical cistern that will hold 10 tons of water, if 1 cu. ft. of water weighs 62.28 lbs. Ex. 34. A water trough has equilateral triangles, each side 3 ft., for ends, and is 18 ft. long. How many buckets of water will it hold, if a bucket is a cylinder 1 ft. in diameter and 11 ft. high? Ex. 35. The lateral area of a cylinder of revolution is 440 sq. in., and the volume is 1,540 cu. in. Find the radius and altitude. Ex. 36. The angles of a spherical quadrilateral are 800, 100), 120~, 120~. Find the angle of an equivalent equilateral triangle. Ex. 37. A cone and a cylinder have equal lateral surfaces, and their axis sections are equilateral. Find the ratio of their volumes. Ex. 38. A water-pipe - in. in diameter rises 13 ft f from the ground. How many quarts of water must be drawn from it before the water from under the ground comes out? If a quart runs out in 5 seconds, how long must the water run? Ex. 39. A cube immersed in a cylinder partly filled with water causes the water to rise 4 in. If the radius of the cylinder is 6 in., what is an edge of the cube? Ex. 40. An auger hole whose diameter is 3 in. is bored through the center of a sphere whose diameter is 8 inches. Find the volume remaining. Ex. 41. Show that the volumes of a cone, hemisphere, and cylinder of the same base and altitude are as 1: 3. Ex. 42. The volumes of two similar cylinders of revolution are as 8:125; find the ratio of their radii. If the radius of the smaller is 10 in., what is the radius of the larger? Ex. 43. An iron shell is 2 in. thick and the diameter of its outer surface is 28 in. Find its volume. Ex. 44. The legs of' an isosceles spherical triangle each make an angle of 75~ with the base. The legs produced form a lune whose aRea is four times the area of the triangle. Find the angle of the lune, 484 SOLID GEOMJ:TTRY cROUP M8 EXERCISES INVOLVING THE 3METRIC SYSTEM Find 8, T, V of Ex. 1. A right prism the edges of whose base are 6 m., 70 dm., 900 cm., and whose altitude is 90 dm. Ex. 2. A regular square pyramid an edge of whose base is 30 dmi., and whose altitude is 1.7 m. Ex. 3. A sphere whose radius is 0.02 m. Ex. 4. A frustum of a cone of revolution whose radii are 10 dm and 6 dm., and whose slant height is 50 cm. Ex. 5. A cube whose diagonal is 12 cm. Ex. 6. A cylinder of revolution whose radius equals 2 dm., ani whose altitude equals the diameter of the base. Ex. 7. Find the area of a spherical triangle on a sphere whose radius is 0.02 m., if its angles are 110~, 120~, 130~. Ex. 8. Find the number of square meters in the surface of a sphere, a great circle of which is 50 dm. long. Ex. 9. How many liters will a cylindrical vessel hold that is 10 dm. in diameter and 0.25 m. high? How many liquid quarts 7 Ex. 10. A liter measure is a cylinder whose diameter is half the altitude. Find its dimensions in centimeters. Ex. 11. Find the surface of a sphere whose volume is 1 cu. m, APPENDIX I. MODERN GEOMETRIC CONCEPTS 840. Modern Geometry. In recent times many new geometric ideas have been invented, and some of them developed into important new branches of geometryThus, the idea of symmetry (see Art. 484, etc.) is a modern geometric concept. A few other of these modern concepts and methods will be briefly mentioned, but their thorough consideration lies beyond the scope of this book. 841. Projective Geometry. The idea of projections (see Art. 345) has been developed in comparatively recent times into an important branch of mathematics with many practical applications, as in engineering, architecture, construction of maps, etc. 842. Principle of Continuity. By this principle two or more theorems are made special cases of a single more general theorem. An important aid in obtaining continuity among geometric principles is the application of the concept of negative quantity to geometric magnitudes. Thus, a negative line is a line opposite in direction to a given line taken as positive. - 0 + For example, if OA is +, OB is -. (485) 486 GEOMETRY. AIPPENDIX * Similarly, a negative angle is B an angle formed by rotating a line in a plane in a direction opposite _ from a direction of rotation taken - as positive. Thus, if the line OA rotating from the position OA forms the positive angle A OB, the same line rotating in the opposite direction forms the negative angle AOB'. Similarly, positive and negative arcs are formed. p In like manner, if P and P' are on opposite sides of the line AB and the area PAB is taken as positive, the area P'AB will be negative. p As an illustration of the law of P' continuity, we may take the theorem that the sum of the triangles formed by drawing lines from a point to the vertices of a polygon equals the area of the polygon. B Applying this to the quadrilateral ABCD, if the point P falls within the quadrilateral, A RAB + APBC + APCD +APAD= ABCD (Ax. 6). Also, if the point falls without the quadrilateral at P', A P'AB + A P'BC + A P'CD) + A P'AD = ABCD, since AP'AD is a negative area, and hence is to be subtracted from the sum of the other three triangles. 843. The Principle of Reciprocity, or Duality, is a principle of relation between two tleorems by which each theorem is convertible into tlhe otler by causing tle words for the same two geometric objects in each tleoren to exchange places. MODERN GEOMETRIC CONCEPTS 487 Thus, of theorems VI and VII, Book I, either may be converted into the other by replacing the word "sides" by "angles," and "angles" by "sides." Hence these are termed reciprocal theorems. The following are other instances of reciprocal geometric properties: 1. Two points determine a 1. Two lines (letermine a point. straight line. 2. Three points not in the same 2. Three planes not through the straight line tetermine a plane. same straight line determine a 1. iO t. 3. A straigqit line ad a point a 3. Ji straight line a(nd a plane determine a plane, determine a ploint. The reciprocal of a theorem is not necessarily true. Thus, two parallel straight lines determine a plane, but two parallel planes do not determine a line. However, by the use of the principle of reciprocity, geometrical properties, not otherwise obvious, are frequently suggested. 844. Principle of Homology. Just as the law of reciprocity indicates relations between one set of geometric concepts (as lines) and anoLher set of geometric concepts (as points), so the law of homology indicates relations between a set of geometric concepts and a set of concepts outside of geometry: as a set of algebraic concepts, for instance. Thus, if a and b are numbers, by algebra (a + b ) (a-b) a2 -2 Also, if a and b are segments of a line, the rectangle (a+b) X (a-b) is equivalent to the difference between the squares a' and b1. By3 means of this principle, truths which would be overlooked or difficult to prove in one department of thought 4SR GEOMIETRY. A'PPENl)IX are made obvious by observing the corresponding truth in another department of thought. Thus, if a and b are line segments, the theorem (a + b)2 +-(a - b)2=20(- +-b2) is not immediately obvious in geometlry, but becomes so by; observiing the like relation between the algebraic numbers a land b. 845. Non-Euclidean Geometry. Hyperspace. By vary - ing the properties of space, as these are ordinarily stated, different kinds of space may be conceived of, each having its own geometric laws and properties. Thus, space, as we ordinarily conceive it, has three dimensions, but it is possible to conceive of space as having four or more dimensions. To mention a single property of four dimensional space, in such a space it would be possible, by simple pressure, to turn a sphere, as an orange, inside out without breaking its surface. As an aid toward conceiving how this is possible, consider a plane in which one circle lies inside another. No matter how these circles are moved about in the plane, it is impossible to shift the inner circle so as to place it outside the other without breaking the circumference of the outer circle. But, if we are allowed to use the thirdl dimension of space, it is a simple matter to lift the inner circle up out of the plane and set it down outside the larger cir(le. Similarly if, in space of three dimensions, we have one spherical slhell inside a larger shell, it is impossible to place the smaller shell outside the larger without breaking the larger. But if the use of a fiourth dimension be allowed,-that is, the use of another dimension of freedom of motion,-it is possible to place the inner shell outside the larger without breaking the latter. 846. Curved Spaces. By varying the geometric, axioms of space (see Art. 47), different kinds of space may be conceived of. Thus, we may conceive of space such that through a given point one line may be drawn parallel to a given liue (that is ordinary, or Euclidean space); or such MODERN GEOMETRIC CONCEPTS 489 that through a given point no line can be drawn parallel to a given line (spherical space); or such that through a given point.more than one line can be drawn parallel to a given line (pseudo-spherical space). These different kinds of space differ in many of their properties. For example, in the first of them the sum of the angles of a triangle equals two right angles; in the second, it is greater; in the third, it is less. These different kinds of space, however, have many properties in common. Thus, in all of them every point in the perpendicular bisector of a line is equidistant from the extremities of the line. EXERCISES. GROUP 8S Ex. 1. Show by the use of zero and negative arcs that the principles of Arts. 257, 263, 258, 264, 265, are particular cases of the general theorem that the angle included between two lines which cut or touch a circle is measured )vy one-half the sum of the intercepted arcs. Ex. 2. Show that the principles of Arts. 354 and 358 are particular cases of the theorem that, if two lines are drawn from or through a point to meet a circumference, the product of the segments of one line equals the product of the segments of the other line. Ex. 3. Show by the use of negative angles A that theorem XXXVIII, Book I, is true for a quadrilateral of the form ABCD. [BCD is a negative angle; the angle at the vertex D is B the reflex angle ADC..] Ex. 4. What is the reciprocal of the statement that two intersecting straight lines deter- mine a plane? Ex. 5. What is the reciprocal of the statement that three planes perpendicular to each other determine three straight lines perpendicular to each other? IIT. HISTORY OF GEOM'.ETRY 847. Origin of Geometry as a Science. The beginnings of geometry as a science are found in Egypt, dating back at least three thousand years before Christ. Herodotus says that geometry, as known in Egypt, grew out of the need of remeasuring pieces of land parts of which had been washed away by the Nile floods, in order to make an equitable readjustment of the taxes on the samle. The substance of the Egyptian geometry is found in an old papyrus roll, now in the British museum. This roll is, in effect, a mathematical treatise written by a scribe named Ahmes at least 1700 B.C., and is, the writer states, a copy of a more ancient work, dating, say, 3000 B. C. 848. Epochs in the Development of Geometry. From Egypt a knowledge of geometry was transferred to Greece, whence it spread to other countries. Hence we have the following principal epochs in the development of geometry: 1. Egyptian: 3000 B. C.-1500 B. C. 2. Greek: 600 B. C.-100 B. C. 3. Hindoo: 500 A. D.-1100 A. D. 4. Arab: 800 A. D.-1200 A. D. 5. European: 1200 A. D. In the year 1120 A. D., Athelard, an English monk, visited Cordova, in Spain, in the disguise of a Mohammedan student, and procured a copy of Euclid in the Arabic language. This book lie brought back to central Europe, where it was translated into Latin and became the basis of all geometric study in Europe till the year 1533, when, (490) HISTORY OF GEOMETRY 491 owing to the capture of Constantinople by the Turks, copies of the works of the Greek mathematicians in the original Greek were scattered through Europe. HISTORY OF GEOMETRICAL METHODS 849. Rhetorical Methods. By rhetorical methods in the presentation of geometric truths, is meant the use of definitions, axioms, theorems, geometric figures, tie representation of geometric magnitudes by the use of letters, the arrangement of material in Books, etc. The Egyptians had none of these, their geometric knowledge being recorded only in the shape of the solutions of certain numerical examples, from which the rules used must be inferred. Thales (Greece 600 B.C.) first made an enunciation of an abstract property of a geometric figure. He had a rude idea of the geometric theorem. Pythagoras (Italy 525, B. C.) introduced formal definitions into geometry, though some of those used by him were not very accurate. For instance, his definition of a point is "unity having position." Pythagoras also arranged the leading propositions known to him in something like logical order. Hippocrates (Athens, 420 B. C.) was the first systematically to denote a point by a capital letter, and a segment of a line by two capital letters, as the line ABl, as is done at present. IHe also wrote the first text-book on geometry. Plato (Athens, 380 B. C.) made definitions, axioms and postulates the beginning and basis of geometry. To Euclid (Alexandria, 280 B. C.) is due the division of geometry into Books, the formal enunciation of tlleoremis, the particular enunciation. the formal (onistr'l(tioll 492 GEOMETRY. APPENDIX proof, and conclusion, in presenting a proposition. He also introduced the use of the corollary and scholium. Using these methods of presenting geometric truths, Euclid wrote a text-book of geometry in thirteen books, which was the standard text-book on this subject for nearly two thousand years. The use of the symbols A, Z, l, etc., in geometric proofs originated in the United States in recent years. 850. Logical Methods. The Egyptians used no formal methods of proof. They probably obtained their few crude geometric processes as the result of experiment. The Hindoos also used no formal proof. One of their writers on geometry merely states a theorem, draws a figure, and says "Behold! The use of logical methods of geometric proof is due to the Greeks. The early Greek geometricians used experimental methods at times, in order to obtain geometric truths. For instance, they determined that the angles at the base of an isosceles triangle are equal, by folding half of the triangle over on the altitude as an axis and observing that the angles mentioned coincided as a fact, but without showing that they must coincide. Pythagoras (525 B. C.) was the first to establish geometric truths by systematic deduction, but his methods were sometimes faulty. For instance, he believed that the converse of a proposition is necessarily true. Hippocrates (420 B. C.) used correct and rigorous deduction in geometric proofs. He also introduced specific varieties of such deduction, such as the method of reducing one proposition to another (Art. 296), and the reductio ad absurdunm. HISTORY OF GEOMETRY 493 The methods of deduction used by the Greeks, however, were defective in their lack of generality. For instance, it was often thought necessary to have a separate proof of a theorem for each different kind of figure to which the theorem applied. Thus, the theorem that the sum of the an-...", gles of a triangle equals two right angles was proved, (1) for the equilateral triangle by use of the regular hexagon; (2) for the right triangle by the use of a rectangle; (3) for a scalene triangle by dividing the scalene triangle into two right triangles. The Greeks appeared to fear that a general proof might be vitiated if it were applied to a figure in any way special or peculiar. In other words, they had no conception of the principle of continuity (Art. 842). Plato (380 B. C.) introduced thie method of proof by analysis, that is, by taking a proposition as true and working from it back to known truths (see Art. 196). To Eudoxus (360-B. C.) is virtually due proof by the method of limits; though his method, known as the method of exhaustions, is crude and cumbersome. Apollonius (Alexandria, 225 B. C.) used projections, transversals, etc., which, in modern times, have developed into the subject of projective geometry. 851. Mechanical Methods. The Greeks, in demonstrating a geometrical theorem, usually drew the figure employed in a bed of sand. This method had certain advantages, but was not adapted to the use of a large audience. At the time when geometry was being developed in Greece, the interest in the subject was very general. There was scarcely a town but had its lectures on the subject. The news of the discovery of a 494 GEOMETRY. APPENDIX new theorem spread from town to town, and the theorem was redemon. strated in the sand of each market place. The Greek treatises, however, were written on vellum or papyrus by the use of the reed, or calamus, and ink. In Roman times, and in the middle ages, geometrical filures were drawn in wax simeared on woodenl boards, called tablets. They were drawn by the use of the stylus, a metal stick, pointed at, one end for making marks, and broad at the other for erasing marks. These wax tablets were still in use in Shakespeare's time (see Hamnlet Act I, Sc. 5, 1. 107). The blackboard and crayon are modern inventions, their use having developed within the last one hundred years. The Greeks invented many kinds of drawing instruments for tracing various curves. It was due to the influence of Plato (380 B. C.) that, in constructing geometric figures, the use of only the ruler and compasses is permitted. HISTORY OF GEOMETRIC TRUTHS. PLANE GEOMETRY_ 852. Rectilinear Figures. The Egyp- tians measured the area of any four- b sided field by multiplying half the sum - of one pair of opposite sides by half the sum of the other pair; which was equivalent to using the (1a+c b+d formula, area = — X - 2 2 This, of course, gives a correct result for the rectangle and square, but gives too great a result for other quadrilaterals, as the trapezoid, etc. Hence Joseph, of the Book of Genesis, in buying the fields of the Egyptians for Pharoah in time of famine by the use of this formula, in many cases paid for a larger field than le obtained& The Egyptians had a special fondness for geometrical constructions, probably growing out of their work as temple HISTORY OF GEOMETRY 495 builders. A class of workers existed among them called " rope-stretchers," whose business was the marking out of the foundations of buildings. These men knew how to bisect an angle and also to construct a right angle. The latter was probably done by a method essentially thle same as forming a right triangle whose sides are three, four and five units of length. Ahmes, in his treatise, has various constructions of the isosceles trapezoid from different data. Thales (600 B. C.) enunciated the following thleorems: If two straight lines intersect, the opposite or vertical angles are equal; The angles at the base of an isosceles triangle are equal; Two triangles are equal if two sides and the included angle of one are equal to two sides and the included angle of the other; The sum of the angles of a triangle equals two right angles; Two mutually equiangular triangles are similar. Thales used the last of these theorems to measure the height of the great pyramid by measuring the length of the shadow cast by the pyramid and also measuring the length of the shadow of a post of known height at the same time and making a proportion between these quantities. Pythagoras (525 B. C.) and his followers discovered correct formulas for the areas of the principal rectilinear figures, and also discovered the theorems that the areas of similar polygons are as the squares of their homologous sides, and that the square on the hypotenuse of a right triangle equals the sum of the squares on the other two sides. The latter is called the Pythagoorean theorem. They also discovered how to constrlct a square equivalent to a given parallelogIram, and to divide a given line in mean and extreme ratio, 496 GEOMETRY. APPEENDIX To Eudoxus (380 B. C.) we owe the general theory of proportion in geometry, and the treatment of ineommensurable quantities by the method of Exhaustions. By the use of these he obtained such theorems as that the areas of two circles are to each other as the squares of their radii, or of their diameters. In the writings of Hero (Alexandria, 125 B. C.) we first find the formula for the area of a triangle in terms of its sides, K=l/s(s-a) (s-b) (s-c). H ero also was the first to place land-surveying on a scientific basis. It is a curious fact that Hero at the same time gives an incorrect formula for the area of a triangle, viz., K=-a(&)-c), this formula being apparently derived from Egyptian sources. Xenodorus (150 B. C.) investigated isoperemetrical figures. The Romans, though they excelled in engineering, apparently did not appreciate the value of the Greek geomr etry. Even after they became acquainted with it, they continued to use antiquated and inaccurate formulas for areas, some of them of obscure origin. Thus, they used the Egyptian formula for the area of a quadrilateral, ac-Ib cq-d K= —X X. They determined the area of an equilat2 2 eral triangle whose side is a, by different formulas, all 13a~2 incorrect, as K=-, K3 = (aS+a), and EK= a2. 853. The Circle. Thales enunciated the theorem that every diameter bisects a circle, and proved the theorem that an angle inscribed in a semicircle is a right angle. To Hippocrates (420 B. C.) is due the discovery of nearly all the other principal properties of the circle given in this book. HISTORY OF GEOMETRY 497 The Egyptians regarded the area of the circle as equivalent to. 4 of the diameter squared, which would make 7r =3.1604. The Jews and Babylonians treated t as equal to 3. Archimedes, by the use of inscribed and circumscribed regular polygons, showed that the true value of n lies between 31 and 3-?-; that is, between 3.14285 and 3.1408. The Hindoo writers assign various values to 7t, as 3, 38, 1/10, and Aryabhatta (530 A. D.) gives the correct approximation, 3.1416. The Hindoos used the formula 1Z2-_^/4_ B2 (See Art. 468) in computing the numerical value of 7t. Within recent times, the value of 7t has been computed to 707 decimal places. The use of the symbol t for the ratio of the circumference of a circle to the diameter was established in mathematics by Euler (Germany, 1750). HISTORY OF GEOMETRIC TRUTHS. SOLID GEOMETRY 854. Polyhedrons. The Egyptians computed the volumes of solid figures from the linear dimensions of such figures. Thus, Ahmes computes the contents of an Egyptian barn by methods which are equivalent to the use 3c of the formula V=aXbX 2. As the shape of these barns is not known, it is not possible to say whether this formula is correct or not. Pythagoras discovered, or knew, all the regular polyhedrons except the dodecahedron. These polyhedrons were supposed to have various magical or mystical properties. Hence the study of them was made very prominent. 49S GEOMETtRY. APPENDIX Hippasus (470 B. C.) discovered the dodecahedron, biut he was drowned by the other Pythagoreans for boasting of the discovery. Eudoxus (380 B. C.) showed that the volume of a pyra, mid is equivalent to one-third the product of its base by its altitude. E. F. August (Germany, 1849) introduced the prismatoid formula into geometry and showed its importance. 855. The Three Round Bodies. Eudoxus showed that the volume of a cone is equivalent to one-third the area of its base by its altitude. Archimedes discovered the formulas for the surface and volume of the sphere. Menelaus (100 A. D.) treated of the pIroperties of spherical triangles. Gerard (Holland, 1620) invented polar triangles and found the formulas for the area of a spherical triangle and of a spherical polygon. 856. Non-Euclidean Geometry. The idea that a space might exist having different properties from those which we regard as belonging to the space in which we live, has occurred to different thinkers at different times, but Lobatchewsky (Russia, 1793-1856) was the first to make systematic use of this principle. He found that if, instead of taking Geom. Ax. 2 as true, we suppose that through a given point in a plane several straight lines may be drawn parallel to a given line, the result is not a series of absurdities or a general reductio ad absurdum; but, on the contrary, a consistent series of theorems is obtained giving the properties of a space. III. REVIEW EXERCISES EXERCSES. CGROUP 86 REVIEW EXERCISES IN PLANE GEOMETRY Ex. 1. If the bisectors of two adjacent angles are perpendicular to each other, the angles are supplementary. Ex. 2. If a diagonal of a quadrilateral bisects two of its angles, the diagonal bisects the quadrilateral. Ex. 3. Through a given point draw a secant at a given distance from the center of a given circle. Ex. 4. The bisector of one angle of a triangle and of an exterior angle at another vertex form an angle which is equal to one-half the third angle of the triangle. Ex. 5. The side of a square is 18 in. Find the circumference of the inscribed and circumscribed circles. Ex. 6. The quadrilateral ADBC is inscribed in a circle. The diagonals AB and )C intersect in the point F. Arc AD _= 112~, arc A C= 108~, LAFC = 74~. Find all the other angles of the figure. Ex. 7. Find the locus of the center of a circle which touches two given equal circles. Ex. 8. Find the area of a triangle whose sides are 1 m., 17 dm.0 210 crm. Ex. 9. The line joining the midpoints of two radii is perpendicular to the line bisecting their angle. Ex. 10. If a quadrilateral be inscribed in a circle and its diagonals drawn, how many pairs of similar triangles are formed? Ex. 11. Prove that the sum of the exterior angles of a polygon (Art. 172) equals four right angles, by the use of a figure formed by drawing lines from a point within a polygon to the vertices of the polygon. (499) 500 GEOMETRY. APPENDIX Ex. 12. In a circle whose radius is 123 cm., find the length of the tangent drawn from a point at a distance 240 mm. from the center. Ex. 13. If two sides of a regular pentagon he produced, find the angle of their intersection. Ex. 14. In the parallelogram ABCD, points are taken on the diagonals such that AP=BQ=CR=DS. Show that PQRS is a parall elogram. Ex. 15. A chord 6 in. long is at the distance 4 in. from the center of a circle. Find the distance from the center of a chord 8 in. long. Ex. 16. If B is a point in the circumference of a circle whose center is 0, PA a tangent at any point P, meeting OB produced at 4, and PD perpendicular to OB, then PB bisects the angle APD. Ex. 17. Construct a parallelogram, given a side, an angle, and a diagonal. Ex. 18. Find in inches the sides of an isosceles right triangle whose area is 1 sq. yd. Ex. 19. Given the line a, construct a(2 —) Ex. 20. If two lines intersect so that the product of the segments of one line equals the product of the segments of the other, a circumference may be passed through the extremities of the two lines. Ex. 21. Find the locus of the vertices of all triangles on a given base and having a given area. Ex. 22. On the figure p. 206, prove that BC2+AF2=AB2+fFC2 Ex. 23. The area of a rectangle is 108 and the base is three times the altitude. Find the dimensions. Ex. 24. If, on the sides AC and BC of the triangle ABC, the squares, AD and BF, are constructed, AF and DB are equal. Ex. 25. If the angle included between a tangent and a secant is half a right angle, and the tangent equals the radius, the secant passes through the center of the circle. REVIEW EXERCISES IN PLANE GEOMETRY 501 Ex. 26. The sum of the areas of two circles is 20 sq. yds., and the difference of their areas is 15 sq. yds. Find their radii. Ex. 27. Construct an isosceles trapezoid, given the bases and a leg. Ex. 28. Show that, if the alternate sides of a regular pentagon be produced to meet, the points of intersection formed are the vertices of another regular pentagon. Ex. 29. If a post 2 ft. 6 in. high casts a shadow 1 ft. 9 in. long, how tall is a tree which, at the same time, casts a shadow 66 ft. long I Ex. 30. If two intersecting chords make equal angles with the diameter through their point of intersection, the chords are equal. Ex. 31. From a given point draw a secant to a circle so that the external segment is half the secant. Ex. 32. Find the locus of the center of a circle which touches a given circle at a given point. Ex. 33. If one diagonal of a quadrilateral bisects the other diagonal, the first diagonal divides the quadrilateral into two equivalent triangles. Ex. 34. In a given square inscribe a square having a given side. Ex. 35. A field in the shape of an equilateral triangle contains one acre. How many feet does one side contain? Ex. 36. If perpendiculars are drawn to a given line from the vertices of a parallelogram, the sum of the perpendiculars from two opposite vertices equals the sum of the other two perpendiculars. Ex. 37. Any two altitudes of a triangle are reciprocally proportional to the bases on which they stand. Ex. 38. Construct a triangle equivalent to a given triangle and having two given sides. Ex. 39. The apothem of a regular hexagon is 20. Find the area of the inscribed and circumscribed circles. Ex. 40. M is the midpoint of the hypotenuse AB of a right triangle ABC. Prove 8 M2G_ 2 AB - 2+ - C2 _50~2 mGEOMETRY. APPENDIX Ex. 41. Transform a given triangle into an equivalent right triangle containing a given acute angle. Ex. 42. The area of a square inscribed in a semicircle is to the area of the square inscribed in the circle as 2:5. Ex. 43. If, on a diameter of the circle 0, 0A ---=OB and AC is parallel to BD, the chord CD is perpendicular to AC. Ex. 44. Find the radius of a circle whose area is equal to onethird the area of the circle whose radius is 7 in. Ex. 45. State and prove the converse of Prop. XXI, Book III. Ex. 46. If, in a given trapezoid, one base is three times the othier base, the segments of each diagonal are as 1: 3. Ex. 47. If two sides of a triangle are 6 and 12 and the angle included by them is G60, find the length of the other side. Also tild this when the included angle is 45~; also, when 120~. Ex. 48. How many sides has a polygon in which the sum of the interior angles exceeds the sum of the exterior angles by 540~? Ex. 49. If the four sides of a quadrilateral are the diameter of a circle, the two tangents at its extremities, and a tangent at any other point, the area of the quadrilateral equals one-half the product of the diameter by the side opposite it in the quadrilateral. Ex. 50. An equilateral triangle and a regular hexagon have the same perimeter; find the ratio of their areas. Ex. 51. To a circle whose radius is 30 cm. a tangent is drawn from a point 21 dm. from the center. Find the length of the tangent. Ex. 52. If two opposite sides of a quadrilateral are equal, and the angles which they make with a third side are equal, the quadrilateral is a trapezoid. Ex. 53. If two circles are tangent externally and two parallel diameters are drawn, one in each circle, a pair of opposite extremities of the two diameters and the point of contact are collinear. Ex. 54. If, in the triangle ABC, the line AD is perpendicular to BD, the bisector of the angle B, a line through D parallel to BC bisects A C. REVIEW EXERCISES IN PLANE GEOMETRY 5 03 Ex. 55. Bisect a given triangle by a line parallel to a given line. Ex. 56. If two parallelograms have an angle of one equal to the supplement of an angle of the other, their areas are to each other as the products of the sides including the angles. Ex. 57. The sum of the medians of a triangle is less than the perimeter, and greater than half the perimeter. Ex. 58. If PARB is a secant to a circle through the center 0, PT a tangent, and TR perpendicular to PB, then PA: PI-PO: PB. Ex. 59. Two concentric circles have radii of 17 and 15. Find the length of the chord of the larger which is tangent to the smaller. Ex. 60. On the figure, p. 244, (a) Find two pairs of similar triangles; (b) Find two dotted lines which are perpendicular to each other; (c) Discover a theorem concerning points, not connected by lines on the figure, which are collinear; (c) Discover a theorem concerning squares on given lines. Ex. 61. One of the legs, AC, of an isosceles triangle is produced through the vertex, C, to the point F, and F is joined with D, the midpoint of the base AB. DF intersects BC in '. Prove that CF is greater than CE. Ex. 62. The line of centers of two circles intersects their common external tangent at P. PABCD is a secant intersecting one of the two circles at A and B and the other at C and D. Prove PA X PD= PBX PC. Ex. 63. Trisect a given parallelogram by lines drawn through a given vertex. Ex. 64. Find the area of a triangle the sides of which are the chord of an are of 120~ in a circle whose radius is 1; the chord of an arc of 90~ in a circle whose radius is 2; and the chord of an are of 60~ in a circle whose radius is 3. Ex. 65. Construct a triangle, given the median to one side and a median and altitude on the other side. Ex. 66. Two circles intersect at P and Q. The chord CQ is tangent to the circle QPIB at Q. APB is any chord through P, Prove that AC is parallel to BQ )504 G M(EOMETRY. APPENDIX Ex. 67. In the triangle ABC, from i), the midpoint of iB', j)j and DF are drawn, bisecting the angles ADB and ADC, and meeting AB at E and AC at F. Prove EFiT BC. Ex. 68. Produce the side BC of the triangle ABC to a point P, so that PB X PC = PA2. Ex. 69. In a given circle inscribe a rectangle similar to a given rectangle. Ex. 70. In a given semicircle inscribe a rectangle similar to a given rectangle.' Ex. 71. The area of an isosceles trapezoid is 140 sq. ft., one base is 26 ft., and the legs make an angle of 45~ with the other base. Find the other base. Ex. 72. Cut off one-third the area of a given triangle by a line perpendicular to one side. Ex. 73. Find the sides of a triangle whose area is 1 sq. ft., if the sides are in the ratio 2: 3: 4. Ex. 74. Divide a given line into two parts such that the sum of the squares of the two parts shall be a minimum. Ex. 75. If, from any point in the base of a triangle, lines are drawn parallel to the sides, find the locus of the center of the parallelogram so formed. Ex. 76. Three sides of a quadrilateral are 845, 613, 810, and the fourth side is perpendicular to the sides 845 and 810. Find the area. Ex. 77. If BP bisects the angle ABC, and DP D bisects the angle CDA, prove that angle P=~ sum B - /\ of angles A and C. Ex. 78. Two circles intersect at P and Q. A p Through a point A in one circumference lines APC and AQD are drawn, meeting the other in C and D. Prove the tangent at A parallel to CD. Ex. 79. In a given triangle, draw a line parallel to the base and terminated by the sides so that it shall be a mean proportional between the segments of one side. REVIEW EXERCISES IN PLANE GEOMETRY 505 Ex, 80. Find the angle inscribed in a semicircle the sum of whose bides is a maximum. Ex. 81. The bases of a trapezoid are 160 and 120, and the altitude 140. Find the dimensions of two equivalent trapezoids into which the given trapezoid is divided by a line parallel to the base. Ex. 82. If the diameter of a given circle be divided into any two segments, and a semicircumference be described on the two segments on opposite sides of the diameter, the area of tie circle will be divided by the semicircumfercnces thus drawn into two parts having the same ratio as the segments of the diameter. Ex. 83. On a given straight line, AB, two segments of circles are drawn, APB and A(QE. The angles QAP and QBP are bisected by lines meeting in R. Prove that the angle R is a constant, wherever P and Q may be on their arcs. Ex. 84. On the side AB of the triangle ABC, as diameter, a circle is described. EF is a diameter parallel to BC. Show that EB bisects the angle ABC. Ex. 85. Construct a trapezoid, given the bases, one diagonal, and an angle included by the diagonals. Ex. 86. If, through any point in the common chord of two intersecting circles, two chords be drawn, one in each circle, through the four extremities of the two chords a circumference may be passed. Ex. 87. From a given point as center describe a circle cutting a given straight line in two points, so that the product of the distances of the points from a given point in the line may equal the square of a given line segment. Ex. 88. AB is any chord in a given circle, P any point on the circumference, PAE is perpendicular to AB and is produced to meet the circle at Q; AN is drawn perpendicular to the tangent at P. Prove the triangles IYAM and PAQ similar. Ex. 89. If two circles ABCD and EBCF intersect in B and C and have common exterior tangents AE and DFP cut by BC produced at G aud 11, then GH2'=BC'2 —~AE2, 506 GEOMIETRY. iAPPEND X EXERCISES. GROUP 87 REVIEW EXERCISES IN SOLID G(EOMIETRY Ex. 1. A segment of a straight line oblique to a plane is greater than its projection on the plane. Ex. 2. Two tetrahedrons are similar if a dihedral angle of one equals a dihedral angle of the other, and the faces forming these dihedral angles are similar each to each. Ex. 3. A plane and a straight line, both of which are parallel to the same line, are parallel to each other. Ex. 4. If the diagonal of one face of a cube is 10 inches, find the volume of the cube. Ex. 5. Construct a spherical triangle on a given sphere, given the poles of the sides of the triangle. Ex. 6. Given AB 1 AIN, AE and BF -L MR; P prove EF iJ PJ. Ex. 7. The diagonals of a rectangular paral- M lelopiped are equal. F Ex. 8. What portion of the surface of a sphere is a triangle each of whose angles is 140~? Ex. 9. Through a given point pass a plane parallel to two given straight lines. Ex. 10. Show that the lateral area of a cylinder of revolution is equivalent to a circle whose radius is a mean proportional between the altitude of the cylinder and the diameter of its base. Ex. 11. The volumes of polyhedrons circumscribed about equal spheres are to each other as the surfaces of the polyhedrons. Ex. 12. Find S and T of a regular square pyramid an edge of whose base is 14 dm., and whose lateral edge is 250 cm. Ex. 13. If two lines are parallel and a plane be passed through each line, the intersection of these planes is parallel to the given lines. REVIEW EXERCISES IN SOLID GEOMETRY 507 Ex. 14. Given PII I p lane AD, P Z I'11 = Z PFH; prove Z I'PEF= Z PI'E. Ex. 15. If a plane be passed F through the midpoints of three C D edges of a parallelopiped which meet at a vertex, the pyramid thus formed is what part of the parallelopiped? Ex. 16. Find a point in a plane such that the sum of its distances from two given points on the same side of the plane is a minimum. Ex. 17. Given the points A, B, C, 1) in a plane and P a point outside the plane, AB perpendicular to the plane PBI, and A C per'pendicular to the plane ICI); prove that I'D) is perpendicular to the plane ICI). Ex. 18. In a sphere whoso radius is 5, find the area of a zone thei radii of whose upper and lower bases are 3 and 4. Ex. 19. Two cylinders of revolution have equal lateral areas. Show that their volumes are as R: R'. Ex. 20. The midpoints of two opposite sides of a quadrilateral in space, and the midpoints of its diagonals, are the vertices of a parallelogram. Ex. 21. How many feet of two-inch plank are necessary to construct a box twice as wide as deep and twice as long as wide (on the inside), and to contain 216 cu. ft.? Ex. 22. If two spheres with radii IR and r are concentric, find tile area of the section of the larger sphere made by a plane tangent to tile smialler sphere. Ex. 23. In the frustum of a regular square pyramid, the ed/lf(,s of the bases are denoted by bt and b1) and the altitude by H; provu that L-='/ (bl-b)z - ' 4. Ex. 24. If the opposite sides of a spherical quadrilateral are equ(al tnl oppo:lite angles are equal. 508 GEOMETRY. APPENDIX Ex. 25. Obtain the simplest formula for the lateral surface of a truncated triangular right prism, each edge of whose base is a, and whose lateral edges are p, q, and r. Ex. 26. The area of a zone of one base is a mean proportional between the remaining surface of the sphere and its entire surface. Find the altitude of the zone. Ex. 27. The lateral edges of two similar frusta are as 1: a. How do their areas compare? Their volumes? Ex. 28. Construct a spherical surface with a given radius, r, which shall be tangent to a given plane, and to a given sphere, and also pass through a given point. Ex. 29. The volume of a right circular cylinder equals the area of the generating rectangle multiplied by the circumference generated by the point of intersection of its diagonals. Ex. 30. On a sphere whose radius is 8~ inches, find the area of a zone generated by a pair of compasses whose points are 5 inches apart. Ex. 31. The perpendicular to a given plane from the point where the altitudes of a regular tetrahedron intersect equals one-fourth the sum of the perpendiculars from the vertices of the tetrahedron to the same plane. Ex. 32. Two trihedral angles are equal or symmetrical if their corresponding dihedral angles are equal. Ex. 33. On a sphere whose radius is a, a zone has equal bases and the sum of the bases equals the area of the zone. Finld the altitude of the zone. Ex. 34. A plane which bisects two opposite edges of a tetrahedron bisects the volume of the tetrahedron. Ex. 35. Find the locus of all points in space which have their distances from two given parallel lines in a given ratio. Ex. 36. If a, b, c are the sides of a spherical triangle, a', b', cl the sides of its polar triangle, and a > b> c, then a' < b < c. Ex. 37. A cone of revolution has a lateral area of 4 sq. yd. and an altitude of 2 ft. How much of the altitude must be cut off by a plane parallel to the base, in order to leave a frustum whose fateral area is 2 sq. ft. T REVIEW EXERCISES IN SOLID GEOMETRY 509 Ex. 38. The total area of an equilateral cone is to the area of the inscribed sphere as 9: 4. Ex. 39. Construct a sphere of given radius, r, whose surface shall be tangent to three given spheres. Ex. 40. The volume of the frustum of an equilateral cone is 300 cu. in. and its altitude is 20 in. Show how to find the radii of the bases. Ex. 41. On each base of a cylinder of revolution a cone is placed, with its vertex at the center of the opposite base. Find the radius of the circle of intersection of the two conical surfaces. Ex. 42. The volume of a frustum of a cone of revolution equals the sum of a cylinder and a cone of the same altitude as the frustum, and with radii which are respectively the half sum and the half difference of the radii of the frustum. Ex. 43. A square whose side is a revolves about a line through one of its vertices and parallel to a diagonal, as axis; find the surface and volume generated. Ex. 44. If a cone of revolution roll on another fixed cone of revolution so that their vertices coincide, find the kind of surface generated by the axis of the rolling cone. Ex. 45. An equilateral triangle whose side is a revolves about an altitude as an axis; find the surface and volume generated by the inscribed circle, and also by the circumscribed circle. Ex. 46. Find the locus of the center of a sphere which is tangent to three given planes. Ex. 47. If an equilateral triangle whose side is a be rotated about a line through one vertex and parallel to the opposite side, as an axis, find the surface and volume generated. Ex. 48. What other formulas of solid geometry may be regarded as special cases of the formula for the volume of a prismatoid? Ex. 49. Through a given point pass a plane which shall bisect the volume of a given tetrahedron. Ex. 50. In an equilateral cone and a cone whose opposite elements are perpendicular at the vertex, show that the ratio of tho vertical solid angles is as 2 —/3:2 —/2 PRACTICAL APPLICATIONS OF PLANE GOMOEITRY EXERCISES. GROUP 88 (BOOK I) 1. Take a piece of paper baring a straight edg n od h ae so as to formn a right angle. W~hat geometrical p~rinciple or d1efinition have you uised? 2. By use of a hoard with a straight edrc, test the, ac~curlacy of the aouitside of a carpent er's squiare b y a mnethoti indficatedI In the dar Iu. ow, then, would INl test the accuiracy of the in1side anigle of the square? NVIhat geonlictri cp~illiciple hav yo used in each cIase?. 3. In Ex. 2 prove that the error in the outsidle angle of the. carpenter's square, if there be any, equals one-half thc agic~l, x, between the outside lines of the squtare as shown in the diagram (dcnote the error hy c and show that c + c =x). This principle is important, because it is essentially the mietboll tised in correctingr the, axis of a telescope, and hence in correct ing hist niments of which the telescop~e is a part, as various surveying and astronomical instruments. 4. By use of a carpenter's square and a given straight edge, lay off a series of parallel lines. What property of parallel lines ha-ve you used? 7 5. Tell how to construct a carpenter's miter box. 6. The diagia(m1- Shows a dirawing instrument~ called T' the parallel rutlers. The (lotted out lines shows the rest 7 of thle instrllment in another posit ion, the part R8 remaining tixed. The distance PQ Ji? S, Eft) = QS~. Hence show that PQ is parallel to fRS. In like inarner show that P'Q' is parallel to RfS. Hienee show E'Q is parallel to) P'Q'. If a line be (Irawn perp~emdicullar to PQ and another perpendicular to P)/QI, the perpendicula Irs thuis drawn will be p~arallel to each other (Art. 122, lines perpendieuilai to parallel linies are parallel). 510 COPYRIGHT. 1911 IlY CIA.RLEs E. Mnmniu, ComPANy PRACTICAL APPLICATIONS51 f5- I t The above are cases of linked motion, a kind of mnechanism of wide importance. Observ~e for instance the system of links which eonncct the driving wheels of a locomotiv~e with the Ipiston in the cylinder, and also the jointed rods connecting the walking beamn of a steamboat with the engine. Look up) also, in the Century Dictionary, for instance, the words linkage, cell, andl parallel niot ion. 7. The distance betweeni two accessible places separatedl by an11 im1 -passable barrier (as between two houses separat el by apond) may be founid by the following inet hod whN\ A B no instrument, for mneasurinn angles is at, hand. L7 Let A andI B b~e the two lplaces. lake a (onveniieitt station C. and measureC AC anid BC. Produce AC to F, makingm( CFl A (. Produce B( to 1), miaking01- D DC -;BC. Measure 1)F. Prive _- lB - DF. If AC = CF':- 220 ft.; BC = CD ==190 ft.; and JDF 2210 ft., low,1 long is AR? 8. In the trusses of steel bridges, why are the beamis and rods arranged so as to formi a network of triangles as far as possible, and not of quadrilaterals, or pentagons, for instance? (See Ex. 3, p. 76; also.Art. 101.) How is this principle also mnade use of in forming the f rame of a wooden house, or box car, or to strengtnen a weak framte or fence of anya kinds 9. Draw a map for the following survey notes to the scale of 400 ft. to the inch. In laying -Staitions, Beari nus, Dist. inces off the angles draw a dotted north andl south line, throug-h N 4)0 p '10 ft., each station and then use a I) N.)9 ft protractor. S. 1'0 1 )0 ft. Keep your ilrawing for a I) S o W0 \\ Mt f t. la~ter use. 10. Obtain 01' niake up a set of surx cy notes slimilar to iho's lin Ex. 9 and iriake a drawinig for themn. 11. 1 t D( andl FC (p 512) be two wvalls perpendicullar to the planle Of thle Jap vp. LVt, smilall mirrors be ait taliedl to the(-se Wa'lls at A anid B. Let j C = 4)0 Let a ray of lidht- pass t hroiigll Q to A4, be retlect ed to B, ulid t cice m to I'. Prv tlit i t 11 B is a righ t angle. [Su,, Th law x of reflected light is thait the, angle of incidence equalzs the angle of reflection, or, in the figure, x = x and y == y. Then in thre tri angl A B C, x + y 1- 450 180', or x + y p 13"50. About tie points Atand b 2 x -P al2Iy + b =360'' a +b =900, etc.] 512 GEOMETtRY The preceding is the principle of an instrument called the "optical square"' used by foresters in constructing right angles. For a ray of c light coming from R through a small hole at B, above the mirror will make a right angle with the / \ ray comning from Q through P to A. x_ X - B 12. The velocity of light is determined by the /ic^ \ 'use of a rotating mirror in the following manner: F/ z \ Let AtiB be a mirror _ plane of the paper an(d D) "'.Q F rotating about 0 as a, pivot; let OP1, be _L A-,. Let LO be a ray of light striking the mirror at 0 and reflected through 11 to a small stationary mirror some miles distant, whence it is reflected back through ll to 0. On the return of the ray to 0, A4i/B will have rotated through a E small angle, a, to the position A2B2; 41 /a 0_ ---- -. hence the ray will be reflected in the j;__ /. - ] direction OR. Let the pupil show that a Z LOR. Since / LOR may readily be measured, Z a is known, and if the rate at which the mirror is rotat- L, R ing is known, the time occupied by the ray in traveling from 0 to the stationary mirror and back is determined. [SU. ZP1OP2 = Z A1OA2 (Art. 132) /P1OL = Z P1OM = x ( Z incidence = Zreflection).. Z P2OL =x - a Z. LOR = x + a (x - a) 2a If Z LOR = 2~ 19', M0 3 = 3 mi., and the mirror AB rotates 100 times per second, determine the velocity of light per second. 13. To prove that the image of a point in a plane mirror is on a perpendicular from the point to the mirror and as far behind the M mirror as the object is in front of it, let iMMI' be the mirror and P the point, and -p _____0 __ p PAE and PA'E' two reflected rays. Show -- < ^^ that A... /,t PAAM = Z EAM' = / P'AM. '.'. / PAMI' = Z P'AM'. HE Hence A PAA' = A P'AA'. l' E'.'. PA = P'A, etc. 14. From the above show the direction in which a billiard ball must be sent in order to strike a certain point on the table after striking one side of the table. PRACTICAL APPLICATIONS 513 15. If two mirrors OMA and 0M1' be perpendicular to each other make a construction to show 3f where the point P will appear to be to an eye at E, after the p-ta 70-Pa /E light from P has been reflected '\ y// / from both mirrors. bl & / [Sun. Prove BP" == BA + AP' x! = B A + AP.I [ 0 --- —- x[ —@7'x7 16. What would be the appli-,bl}~ 1,'/ cation of Ex. 15 to a ball struck, / I on a billiard table? p"/,t1,~ i '17. The path of a ray of light I I before and after entering a glass prism is given by the lines ARB and CD. The entire angle by which a,p ray of light is deflected on passing through the prism is denoted by x. Prove that ATci Kx = i + - P. (/s PBy and PCyare rt. /s.) C /D [SUG. By use of a quadrilateral BPCy, D/ / \ p = 1800 - P. Then use the quadrilateral A'^ 11' ~ all of whose sides are dotted lines.] EXERCISES. GROUP 89 (BooK II) 1. Given a fragment of broken wheel show how to find the radius of the wheel. 2. By use of the carpenter's square find the center of a given circle. 3. Show how a pattern maker by the use of a carpenter's square can determine whether the cavity made in the edge of a board or piece of metal is a semicircle. 4. Bisect a given angle by the use of a carpenter's square. 5. By use of squared paper divide a line 1- inches long into five equal parts. Into 7 equal parts. Into 3 equal parts. Can you make this division on papelr ruled in only one direction? 6. MIake up and work a similar example for yourself, 514 GEOMETRY 7. Draw a line AB 1 in. long and let 0 be its midpoint. Mark off OC -= in. and divide AC' into four equal parts at D, F, H. With 0 as a center draw circumferences through C, D, F, H, A. From 0 draw radii (one of which is OA) dividing the A_~ ---_ ^ l angular space about 0 into eight equal /./-..f... parts. Beginning at 0 draw the i/<XX,, X/ /d smooth curve C(123456 through the / //\1 / l!.6..\ \ points where the radii drawn inter/ A/ X, '< \ sect the circumferences as indicated 2L/ _'__L_ _ _5_ \on the diagram. The result will be /V\ > \ d 1 the outline of a special form of wheel c/ alled a cam, much used in machines, \^ | —,\ as in the sewing and harvesting -\-\\- j; / A machines, printing presses, automobiles, weaving looms, and machinery A^"^''B for making shoes. A cam converts a circular motion into a reciprocating or back and forth motion. The difference of length between OB and OC on the diagram is termed the "throw" of the cam. 8. Make a drawing of a can in which the throw shall be 1 in. and the longest radius 12 in. 9. Stretch a 100 ft. tape or part of it so as to construct an angle of 60~ as accurately as possible. 10. To extend a straight line AB beyond an obstacle, as a building, we may proceed as follows: At B measure off an angle C ABC = 60~. Produce CB to D, / // etc. Let the pupil complete the A- /.. K\ construction and prove that his method is correct. 11. Let the pupil solve the problem of Ex. 10 by the construction of right angles instead of angles of 60~. 12. To find the distance between two points A and B, one of which A p(c',lf| (B) is inaccessible, we may proceed as folC i-B lows: Extend BA to C. Measure a con_D -, y venient line AD and extend AD to E,, making DE = AD. From E run a line ]l E AC and meeting BD extended at F. Measure EF,. Prove that EF. AB. PRACTICAL APPLICATIONS 13. Show how to solve the problem of Ex. 12 by constructing a line at right angles to another line instead of one parallel to another. 14. If two streets meet, as in the diagram, show how a curve of given radius r may be 0s/ made to take the place of the angle A and be.A tangent with the curb of the two streets. 15. The curve in a railroad track is usually an arc of a circle tangent to each straight track T which it joins. If two straight tracks AB and P/ CD are joined by a circular curve tangent to -,/ ' both of them, and P is the point where.AB and A/ \ CD would meet if extended, prove Z TPD = 2 / T\ C. '/B D \ 16. One way of laying off a railroad curve tangent to a given track is as follows: Let ABP be the given straight track. At B construct a small angle PBC /RO (whose size depends on the degree of curvature which the curve is to have). On it mark off BC = 100 ft. B Construct Z CBD = / PBC. Take C as a center and 100 ft. as a radius, and describe an arc cutting BD at D. Construct E from D in like manner. Prove that, B, C, D, E all lie on / the arc of a circumference tangent to AB rp,/ at B. / \. [SUG. Pass a circumference through B, /_ '_ \ C, D. Prove that E lies on this circumfer-,/ C ence and that ABP is tangent to it.] D /, \ 17. Let AB and CD be two straight railr o'0 D road tracks connected by an arc AF of a AB/ '- circle whose center is 0, and an arc FC 0 whose center is 0', the arcs having a common tangent at F. Prove that the angle of intersection (x) of the two straight tracks, if produced, equals the sum of the central angles of the two tracks. [SUG. x Y + Z. Use Ex. 15.I A curve like AFC composed of two or more arcs of different radii is called a compound curve. Can you suggest why a compound curve should be used in connecting railroad tracks? 18. If the two arcs which compose a compound curve lie on opposite sides of.their common tangeut, the compound curve is called a rewrsa 516 GEOMETRY curve. Thus on the A 'B H I —1 4 _^ L__^ __..,. \ diagram, BCD is a reverse curve connecting the straight roads AB and DE. - - Discover the relation between./.Z.the angles x, y, z. [SUG. Use trianD;/ ~ gle HFE.] Reverse curves are much used in architecture and ornamen. tal work. I 19.. What is a railroad frog? If a curved track crosses a straight track, show that the angle of the frog (x) equals the central angle of the curved track (o). /i 20. If two tracks which curve in the same direction cross each other, find the relation between the angle of the frog and the central angles of the two curved tracks. 21. Find the same when the two curved tracks curve in opposite directions. 22. Show how to locate a gas-generating plant (P) along a given n straight road AB so that the length of pipe /, connecting it with two towns (C and D) C.. i/ shall be a minimum. " (Use Ex. 24, p. 176.) ~A P B a~23. Discover and state an application of Ex. 24, p. 176I, similar to one given in Ex. 22. 24. If a treasure has been buried 100 ft. from a certain tree, and equidistant from a given straight road and a path parallel to the road, show how the treasure may be found. 25. Make up and work a similar example for yourself using the principle of Ex. 2, p. 167. Also using that of one of Exs. 3-8, p. 168. 26. Prove that the latitude of a place #pn the P earth's surface equals the elevation of the pole 1 (that is, on the diagram, prove Z QEA = Z PAO). p Z 27. Given the sun's declina~td /~ ",p tion (i.e. distance north or south \ i /S \ /' X of the celestial equator), show how to determine the latitude H E ( of a place by measuring the zenith distance of the sun, Also hy measuring the altitude of the sun above the hori.Qn. PRACTICAL APPLICATIONS 517 28. How was Peary aided by the principles of Exs. 26 and 27 in determining whether he had arrived at the North Pole? 29. If on April 6 (the day of the year on which Peary was at the North Pole) the sun was 6~ 7' north of the celestial equator, how high above the horizon should the sun have been as observed by Peary? At what hour of the day was this? 30. The sextant is used almost daily by every navigator in determining the altitude of the sun above the horizon, and hence the latitude of the ship. The construction of the sextant is based on the principle that if a ray of light be S reflected from two mirrors'in succession, the change in the direction of the ray of light iy equals twice the angle made by the mirrors. Prove this law. /,i [SUG. Let M and N be the mirrors and the/ / reflected ray be SMNO. It is required to prove that Z y = 2 Z x. It is to be noted. / Y that ZFNM, being an exterior angle of triangle i,A,_ 0 NS'M, equals x + i. Hence angle MNO equals 180~ - 2 (x + i). Hence in triangle MNO, \x/! y + 180~ - 2 (x + i) + 2 i = 180~, etc.] '\ In the sextant, angle y is the elevation of the sun above the horizon, and x can be read on the rim of the instruB ment in a simple manner, since the mirror N is fixed in position while / rotates. It may be observed that the principle of this example is the same as S / / that proved in Ex. 12, p. 512. 31. The diagram shows a some(yO< \ \ \ uwhat simple substitute for the sex1Tk; is z tant called the angle-meter. O is the center of the arc BC and MM' is a fixed mirror. The instrument is held so that a ray SO from the sun ' —/ ' when reflected from the mirror passes C into the eye (A) in a horizontal direction. Show that the elevation of the sun above the horizon = 2 Z x. The rim BC is graduated so that the angle x can be read on it. 518 GEOMETRY The angle-meter may be used to measure horizontal as well as vertical angles. What is the advantage in using a mirror on an instrument of this sort? That is, why is not the position of the sun observed directly across a graduated arc? 32. Draw an easement cornice (AB) tangent to the rake cornice BC, \ ea and passing through a required point A. (The, - 4// zsame construction is used in laying out the ease\ //// ments of stair rails, etc.)? /s / ((Use Ex. 7, p. 174.) A / 33. A segmental arch is a compound curve composed of the arcs of three circles. The method of constructing a segmental arch is as follows: Let AB be the span and CD the altitude of the required arch. Complete the rectangle GADC. Draw the diagonal AC. Bisect the angles:i-A -- GAC and GCA. Let the bisectors meet at E. If N n Draw EH perpendicular to AC, meeting AB at V 1D _ N and CD produced at H. Make DK equal to A DN. Then show that N is the center and NA t, the radius, H the center and HE the radius, H and K the center and KB the radius for the arcs composing the arch. (See Hanstein's Constructive Drawing.) [SUG. At E draw a line perpendicular to EH. Then / NEA Z NAE. (Complements of equal angles are equal) etc.] 34. What is called a Persian arch may be constructed as follows: Let AB be the span and CD the altitude of the required arch. Draw E D F the isosceles triangles ADB. Divide AD into three —. ----,-~~~-..equal parts at H and G. Construct HK I AG at H, and meeting AB produced at K. Produce KG to meet EF which has been drawn through D I /H \ AB. With K as a center and KA as a radius, and -. E as a center and EG as a radius, describe arcs A. C R'-"4 meeting at G. Prove that these arcs have a common tangent at G, and therefore form a compound curve. (See Hanstein's Constructive Drawing.) 36. Construct a Persian arch in which the arc DG = arc NG. [SUG. Bisect line AD instead of trisecting it.] 36. Construct Persian arches in which the chords NG and DG have PRACTICAL APPLICATIONS 519 various ratios, and decide which of these arches you think is the most beautiful. 37. Construct segmental arches of various shapes and decide which of these you think is the most beautiful. 38. Construct a trefoil, given the radius of one of its circles. [SUG. Construct an equilateral triangle, one of whose sides equals twice the given radius. Take each vertex of the triangle as a center, and half of one of the sides as a radius and describe arcs.] 39. Inscribe a trefoil in a given equilateral triangle. [SuG. Bisect the angles of the triangle, etc.] 40. Discover the method of construction of each of the figures or diagrams on the next page, and then reproduce each of thetm in a drawing. Squared paper may be used to advantage as an aid in making some of these constructions. See Fig. 10, p. 520. 41. Construct a diagram similar to Fig. 10, but making a rectangle instead of a square the basis of the drawing. 42. Also one making the rhomboid the basis. 43. Construct a trefoil and develop into an ornamental design by placing small circles and arcs within its parts and larger circles outside. 44. By use of squared paper invent designs sinilar to Fig. 10 on p. 520, but formed by two intertwining lines. 45. Collect a number of pictures of ornamental designs, tracery, scroll work, etc., such as are used in architecture, wall paper, and similar patterns, whose construction depends on the principles of Book II. Show how these designs are constructed, and reproduce them in drawings. EXERCISES. GROUP 90 (BOOK III) B 1. To find the distance between two accessible objects At and B separated by a barrier, take a convenient point C, measure AC and BC. Produce, j/ by BC( to I) and AC to I' so that DC = some frac-.1 \1 t tion, as, of CB, and 'C = the same fraction of \.'C, and measure FD. H ow, then, is At (corn1 puted'? Prove tlis. 520 GEOMETRY a 2a a4 4 2 5 I lx= 30"1 I 6 10 1/ Ei0 At~ PRACTICAL APPLICATIONS 521 Let BC = 300 yd., DC = 60 yd., AC = 240 yd., FC = 48 yd., FD = 52 yd., find AB. Why is this method of finding AB often more convenient than that of Ex. 7, p. 511. A 2. What similar method does the adjoining dia- / gram suggest for finding AB? 3. In the diagram of Ex. 2, in case a river ran C F B between A and B, and also between B and F, what, measurements would be necessary in order to determine the length o:f AB? Of BF? 4. Work again Ex. 2, p. 310. 5. In case the sun is not shining and shadows cannot be used, the height of an object like a.:!~ ttree or steeple can often be /:*'~ (tl^cdetermined by a method in-; ~- ' dicated in the drawing. - What distance must be meas>/.~ ~~ ^^^ ured and why to determine -: _ /,.. ' the height of the tree? *..,... " -/- -... 6. Show how to find the _~ - -_1 height of a tree by placing -; / '.4.~;,'::::-: -,. i'i-:-L- ': 1a mirror in a horizontal posi^^-^J -- * -^-' tion on the ground and standing so as to see the reflection of the top of the tree in the mirror. If the observer's eye is 5- ft. from the ground, tile observer stands 6 ft. from the mirror, and the mirror is 120 ft. from the tree, how high is the tree? 7. Foresters often determine the / heighlt of a tree by an instrument!\ called "Faustman's Height Meas-. \ urer." The principle on which this instrument is constructed is shown in the diagram. /___ _ If the distance from.A to the foot of the tree is 150 ft., B(' = 6 inches, \ iand CF = 4. inches, find the height of the tree. 8. Thle distance froml A to B in Ex. 7,-p. 511, might have been determined by a graphical method as follows: Measure AC, CB, andl angle 522 GEOCMETRY ACB. On paper make a'dralwing of the triangle A CB to a convenient scale. On this drawing lmea:sure t lie i which represents AB and hence determine the length of 1AB. By use of this method w-e are s:ved the labor of marking out and measuring the lines CD, CF, and DF. Apply this nmethod to the measurement of two objects in your neighborhood which are separated by an impassable barrier. In like manner show how the distance from a given point to another inaccessible place may be determined. 9. Also the distance between two places both of which are inaccessible. 10. WThat is the measuring instrument called the diagonal scale'? How is the principle of similar triangles used in it? Show how the diagonal seale aids in the actclrate measurement of lines and hence in the accurate determination of a distance such as AB in Ex. 7, p. 511. 11. In the triangle OAB, A is a right angle, and OA B is 1. By a method which is beyond the scope of this.839 book, the length of AB is 40 /40 computed and found to be O i- A1 P r:oi t. 8.839 +. Using this fact find RQ in the second triangle. 12. By use of the accompanying table, find RQ if angle P is 10~. 20~. 70~. 80~. Tables giving the other sides of all possible right triangles when one side is unity have been computed and when used as in Exs. 11 and 12, form the basis of the subject of Trigonometry. By iuse of this science, after measuring the length of a single line a -lng le AB OB few miles long on the earth's surface, we -- can-determine the distances and relative 10.176 1.015 positions of other places thousands of 200.364 1.064 miles away, without measuring any in- 30.577 1.155 tervening lines. By use of these results 400.839 1.305 as a basis, the distance of the moon is 5 ~ 1.192 1.556 determined as approximately 240,000 6;00 1.732 2.000 miles; of the sun as 92,800,000 miles; 70~ 2.747 2.924 and of the nearest fixed star as 20,000,- 80~ 5.671 5.759 000,000,000 miles. _ PRACTICAL APPLICATIONS 523 A knowledge of these distances has led to important improvements in methods of navigation and have thus facilitated travel and commerce and increased their benefits for C us all. 13. On the diagram given AB is 6,000 ft., and the angles as indicated, compute the length of CD, by use of the table given in Ex. 12. 14. To the diagram in Ex. 13 annex another triangle CFD giving it angles which are multiples of 10~, and compute the / / length of CF. \ 15. A pantograph is an instrument for ' drawing a plane figure similar to a given h plane figure. It is used for enlarging or A 000 't B reducing maps and drawings. It consists of four bars, parallel in pairs and jointed at c, b, C, and E, as shown in the diagram. cbEC is a parallelogram. The rods may be joined so that any required C Ac CE ><k,^ ratio A turns upon a fixed pivot, and pencils are carried 1A^.s^-. ^ b at b and F. Show that A, b, and B are always Ab Ac in a straight line, and that -_ always equals the given ratio - AB AC [SUG. Draw a line from A to b, and a line from A to B. Prove the triangle Acb and ACCB similar (Art 193); hence show that Ab and AB coincide, etc.] 16. Using the fact that a triangle whose sides are 3, 4, and 5 units of length is a right triangle, show how, by stretching a 100-ft. tape, to construct a right angle as accurately as possible. (Among the ancient Egyptians a class of workmen existed called rope stretchers, whose business was to construct right angles in this general way.) 17. The strongest beam which can be cut from a given round log is found as follows: Take AB, a diameter of the log, an(t trisect it at C and D. Draw CAE and DFI _LAB;lIit meeting the circlmnfelren('e at. E and F' respectively. Draw A F, F, BE, B and; -I. \I. \ ' Prove A FBE:a mr ethan)le; also FB; A1 = 1' v:, \'-, or app)roximately as ): a. 524 GEOMETRY [SUG. FB is a mean proportional between DB and AB. Therefore FB2 -= B1 (Art. 343).1 In like manner FtA = A.R B2, etc.] 18. A sphere S weighing 100 lb. rests on the inclined plane AB. AC contains 8 units of length, BC 6 units. A m) force which prevents S from rolling down,/ I the plane would be equivalent to a lifting force of how many pounds exerted on S and parallel with AB' BEXj^ \ [SUG. Resolve the weight of S (represented, by SP) into two forces, one perpendicular to AB and the other parallel to AB. Prove the P triangles ACB and SPR similar, and obtain AB: BC =SP:SR, or 10:6 = 100 b.: SR.] The principle involved in this example is of great practical importance. Thus in many machines useful results are often obtained by representing a force by the diagonal of a rectangle (or parallelogram) and separating this force into two component forces represented by the sides of the rectangle (or parallelogram), only one of these components being effective. This principle makes possible the action of the propeller of an aeroplane or steamboat, of the best water wheels and wind mills, and indeed of all turbine wheels. It also determines the lifting power of the planes of an aeroplane. 19. A wagon weighing 1800 lb. stands on the side of a hill which has a rise of 18 ft. for every 100 ft. taken horizontally. Hence what force must a horse exert to keep such a wagon from running down hill, friction being neglected? 20. Make up and work a similar example for yourself. 21. Show how the diameter of the earth may be determined by the following method: Drive three stakes in a level piece of ground (or in a shallow piece of water) in line, each two successive stakes being a mile apart, and let each stake project the same distance above the ground (or water). By use of a leveling instrument, determine the amount by which the middle stake projects above a horizontal line connecting the tops of the end stakes. This distance will be found to be 8 in. [SUG. Use Art. 343. An are a mile long on the earth's surface may be taken as equal to its chord. Then from the diagram of Art. 343 we obtain the following proportion, the diameter of the earth: 1 mi. = 1 mi.: 8 in.] 22. Also show that the distance that the middle stake projects above PRACTICAL APPLICATIONS 525 the horizontal line connecting the top of the two end stakes varies as the square of the distance between the end stakes. Thus if the two end stakes were placed three times as far apart as in Ex. 21 (that is, 6 miles apart instead of 2 miles) the bulge of the earth between them would be 32 or 9 times what it was originally. Hence determine the projection of the middle stake (or bulge of the earth) when the end stakes are 4 mi. apart. Also 8 mi. 16 mi. 32 mi. 23. At the seashore an observer whose eye was 10 ft above sea level observed a distant steamboat whose hull was hidden for a height of 12 ft. above water level by the bulge of the earth. About how far off was the steamboat? 24. A seaman in a lookout 42 feet above water level with al glass could barely see the topsail of a distant ship, and estimated this topsail to be 45 ft. above sea- level. Estimate the distance of the observed vessel from the seaman. 25. Work again Exs. 7-8, p. 310, Group 56. 26. The moon's distance from the earth's center approximately equals 60 times the earth's radius. A body falling at the earth's surface goes 193 in. in 1 sec. Hence, if the law of A C gravitation is true, the distance the moon falls B 193 in. in 1 sec. toward the earth will be --- - In the 602 diagram, let 0 be the earth's center, ABE the 0 moon's orbit, and CB or AD the distance the moon falls toward the earth in 1 sec. Taking the month as 27 da. 7 hr. 43 min. 11 sec., 193 in. show that AD = - approximately. This is the calculation used 602 by Sir Isaac Newton in testing the truth of the law of gravitation. 27. Any rectangular object, as a book, door, or photograph, is considered to be of the most artistic shape when its length and breadth have the same ratio as the segments of a line divided in mean and extreme ratio (Art. 370). In accordance with this rule, if a window is 6 feet high, how wide should it be? The division of a line in extreme and mean ratio has been termed the Golden Section. By many the Golden Section is regarded as a fundamental principle of esthetics, having applications in determining the proportions of the ideal human form, in explanation of musical harmonies, etc, $8. Make up and work an example similar to Ex. 27. 526 GEOMETRY EXERCISES. GROUP 91 (BooK IV) 1. The supporting power of a wooden beam (of rectangular cross section and of given length) varies as the area of the I, lcross section multiplied by the height of the beam. ~8 B / 4 j If the cross section of a given beam is 4" X 8", compare the supporting power of the beam when it rests on the narrow edge (4") with its supporting power when it rests on its wide edge (8"). 2. Two beams of the same length and material have cross sections which are 2" X 4" and 3" X 8" respectively. Find the ratio of the supporting power of the two beams. 3. In Ex. 17, p. 523, compare the supporting power of a beam cut from a log in the manner indicated, with that of a square beam cut from the same log. 4. Also with that of a beam whose width equals l of the diameter. 5. When an irregular area like ABCD is A D calculated by means of equidistant offsets, the following rule is used: To the half sum of the initial and final offsets add the sum of all the B C intermediate offsets, and multiply the sum by the common distance between the offsets. Prove this rule. 6. Show how the rule of the preceding problem could be used to calculate an area whose entire boundary is an irregular curved line. 7. Surveyors often determine N the area of a piece of land, as of ABCD, by taking ---- an auxiliary line as NS, measuring the perpendicular distances from A, B, C, D, E to NS, and the inter- G.C cepts on NS between these perpendiculars, and -H combining the areas of the various trapezoids (or K — triangles) formed. Supply probable numbers for the lengths of lines on the diagram, and compute the L E area of ABCDE. aB ~d a, aa, T 8, Frequently (as when the center of Lu. a, a. a.,..3 T..,, the curve cannot be seen from the curve)' 1 a railroad curve is laid out by construc" ba'C ting a series of equidistant offsets perpeudicular to the tangent of the curve. PRACTICAL APPLICATIONS 527 Thus, if ABT is a straight track and BC is a curve to be laid out tangent to AB at B, mark off Bai, ala2, a2Cta, all = d, and construct the JI offsets albl, a2b2, a3b3 by using the formula atnb = r - Vr2 - n2d2 where r is the radius of the curve. Prove that this formula is correct. 9. The diagram represents two straight parallel railroad tracks connected by a compound curve (in this. case called a cross-over track) composed of pr- -- 0 -two tracks with common tangents at D and E with centers O2 and 01, and having equal / radii. Taking the magnitudes as indicated.-1 F on the figure, show that KPO'2D is a rect- /; 'i angle (use Arts. 122, 160). lJR'; ~ 10. Find a formula for r in terms of I, _ d and w. (Use the right triangle 01 POz =' and Art. 400.) Also for I in ternms of r, d, and w. 11. If w = 4' 8-", d = 10 ft., r = 150 ft., / find 1. 12. If a steamboat is traveling at, the rate of 12 miles an hour, and a boy walks across her deck at right angles to her line of motion at the rate of 3 miles an hour, draw a diagram to show the direction of his resultant motion. From this determine the speed at which he is going. 13. Make and work a similar example for yourself concerning a mail bag thrown from a train. 14. Also concerning a breeze blowing into a window of a moving trolley car. 15. Two forces, one of 300 lb., the other of 400 lb., act at right angles on the same body. Find their resultant. 16. Make up and work a similar example for yourself. 17. A river is flowing at a rate of 4.25 mi. per hour, and a man is rowing at right angles with the current at a rate of 3.75 mi. per hour.,What is the resultant velocity of the man? 18. If a star has a velocity of 15 mi. a second toward the earth and a velocity of 20 mi. a second at right angles with a line drawn from the star to the earth, find the velocity of the star in its own path. 19. A body is movingin the straight lineA D past the point 0 (p. 528). The distance passed over in a unit of time = A.B = BC = CD. When the body reaches B it is acted on by a force which impels it toward 0 - 528 GE3OMETRY a distance BF in a unit of time. Prove that the area swept over in tih unit of time )by a line dtrawn fromo the body lo 0) will hbe unchang-e: A by the action of the new force. (That is, on tih: dtiagiram prove that: AO('BB AOPB.) \ D This principle accounts for the fact. that the earth (or another planet, or a comet) moves faster in in >.-. '-C orblit the nearer it is to the sun. 20. By tracing the oullline of a inap on squalrel paper, show how to find the area of an irregular figure as of some country or part of a country. Bvuse of this method find the area of some part of the state in which you live. EXERCISES. GROUP 92 (BOOK V) 1. A cooper in fitting a head to a barrel takes a pair of compasses and then adjusts them till, when applied six times in succession in the chine, they will exactly complete the circumference. He then takes the distance between the points of the compasses as the radius of the head of the barrel. Why is this? 2. Draw a square and convert it into a regular octagon by cutting off the corners. [SUG. Draw the diagonals of the square, bisect their angles of intersection, etc.] 3. In heating a house by a hot-air furnace the area of the cross section of the cold-air box should equal the sum of the areas of the pipes conducting hot air from the furnace. If a given furnace has three hotair pipes, each 6" in diameter, and one pipe 8" in diameter, and the width of the cold-air box is 15", how deep should the box be? 4. What is the most convenient way of determining the diameter of a hot-air pipe if you have no callipers and the ends of the pipe are not accessible? 6. A half-mile running track is to have equal semicircular ends and parallel straight sides. The extreme length of the rectangle together with the semicircular ends is to be 1000 ft. Find the width of the rectangle. [SUG. Denote the length of the radius of the semicircular ends by x and that of one of the parallel side straight tracks by y, and obtain a pair of simultaneous equations.] PRACTICAL APPLICATIONS 529 6. A belt runs over two wheels one of which has a diameter of 3 ft. and the other of 6 in. If the first wheel is making 120 revolutions per minute, how many is the second wheel making. How many revolutions must the first wheel make in order that the second may make 300 revolutions per minute? 7. If in laying a track a rail 10 ft. long is.bent through an angle of 5~ 10', what is the radius of the curve? Assuming (what is not strictly true, owing to friction against the sides of the pipe, etc.) that the rate of flow through a cylindrical pipe is proportional to its area of cross section: 8. Work again Ex. 12, p. 277. 9. If a 1 l-in. pipe is replaced by a 1-in. pipe, how much is the flow of water increased? 10. A 3-in. pipe is to be replaced by one which will deliver twice as much water per minute. Find, to the nearest quarter of an inch, the diameter of the new pipe. 11. A 2-in. steam pipe conveying steam from the boiler to the radiators in a school building is found to supply only two thirds the needed amount of steam. What is the smallest even size of pipe that will convey the needed amount? 12. A city of 40,000 people is barely supplied with water by 12-in. mains from the reservoirs. If these mains are torn out, and S1-in. mains substituted, what future population of the city is allowed for'? 13. A steel bar 1 in. in diameter will hold up 50,000 lb. What load would a bar i in. in diameter carry? What would be the diameter of a round (cylindrical) bar to carry 150,000 Ib.? 14. If the center of symmetry of a flat, homogeneous object is the center of mass, find the center of mass of a square; of a rectangle; regular hexagon; circle. 15. It is evident that if the medium of a triangle (BM) be placed on a knife edge the triangle will balance (for if PP' B be 1I AC, the pull on P is balanced by the pull on P'). Hence find the center of mass for any triangle. For a' regular pentagon. / It is useful to be,ablc to determine the center of mass of an object by geometry or by any other At -i C IniItcai, since a k1nowledge of the center of nmass of a body often enlt)lus tius to treat the body in a simple way, for example, aSs 11 the boly were concentrated at a single point, 530 (G EOM ETRY 16. If a box has a triangular end, subject to the same pressure at all points, at what. single point on the end must a. supporting pressure be applied? 17. The cross section of a cylinder is a circle. The weight-supporting strength of a horizontal cylindrical beam of given material and length varies as the area of the cross section times its radius. Compare the weight-supporting power of two solid horizontal iron cylindrical beams of the same length and quality of iron, the radii being 3 in. and 6 in. respectively. (Point out and use the short way of getting the desired result.) 18. The cross section of a hollow cylinder (i.e. of a tube) is a circular ring. Denote the outside radius of the tube by R and the inside radius by r. Then it may be shown that the weight-supporting power of a hollow cylindrical tube of given length and material varies as the area of the cross section (i.e. of the ring) times R + R If R = 4 in. and r 3 in., compare the weight-supporting power of the tube, with that of a solid cylindrical beam of the same length and same cross sectional area. 19. Make up and work a similar example. In general a cylindrical tube is stronger than a solid cylindrical beam of the same length and containing the same amount of material. Hence in a framework, as in that of an airship where the maximum strength must be obtained from a given amount of material, the metallic rods and posts are all tubular. In like manner, bamboo rods, since thev are hollow, tre used in an aeroplane instead of solid wooden rods wherever possible. For the same reason, the bones of flying birds, and many bones in men and animals are hollow and not solid. D 20. To construct a square which shall be approximately equivalent to a given circle 0, divide the radius OA into four equal parts and produce /^A X< each end of two perpendicular diameters a disC\ 0o trance equal to one fourth of the radius, and connect the extremities of the lines thus formed. Show that taking the square thus formed as equivalent to the circle is the same as taking X- = 3-}. Also find the per cent. of error in taking the square as equivalent to the circle. 21. A short way to construct a regular inscribed pentagon and also a five-pointed star (or pentagram) is as follows: Draw a circle 0 and PRACTICAL APPLICATIONS 531 two diameters AB and CD at right angles. Bisect the radius OB at F, and with F as a center and FC as a radius describe an arc cutting AO at H. Then CH is the length of a side of C the regular inscribed pentagon, by joining " whose alternate vertices the pentagram may be formed. As a proof of this, by use of right tri- angles, prove O F CH = R 25 (See Ex. 17, p. 301.) 22. Carpenters sometimes use the following D method of approximately determining the length of a circumference whose radius is known: Let 0 be the given circle. Draw OA and OB, radii A d ght It right angles. Draw AB and the radius OEIIAB at D. Measure DE. Then take circumference E = 6 AO +DE. Find the per cent. of error in this method, taking ir = 3.14159-. 23. The following designs are important in architecture or in ornamental work (thus the first is a detail in a stained glass window in an early French cathedral; the second is the plan of the base of a column in an English cathedral). Discover a method of constructing each of the following designs, and reproduce each of them in a drawing: 532 GEO METRY 24. Construct a square. Take each vertex of the square as a center and one half a side of the square as a radius and describe arcs which meet. Erase the square and you have a quatrefoil. By drawing other circles and arcs of circles, elaborate the quatrefoil into an ornamental design (see design 11, p. 520). 26. In like manner construct a cinquefoil by use of a regular pentagon, and develop it into an ornamental design. 26. Treat a regular hexagon il the same way. 27. A pavement or mosaic may be formed out of regular polygons in the following ways. Show how to make each diagram in the simplest way. X i X_ "n i i i / '^i/ '^n? ^iy \ i I APPLICATIONS OF SOLID GEOMETRY TO MECHANICS AND ENGINEERING EXERCISES. GROUP 93 (BooKs VI AND VII) 1. A carpenter tests the flatness of a surface by applying a s traight edge to the surface in various directions. How does a plasterer test the flatness of a wall surface? What geometric principle is used by these mechanics? 2. Explain why an object with three legs, as a stool or tripod, always rests firmly on the floor while an object with four legs, as a table, does not always rest so. Why do we ever use four-legged pieces.of 30' furniture? \ D / 3. How can a carpenter get a corner post of a house in a vertical position by use of a carpenter's square? What geometrical principle does he 20 /B use? '~~/ Is~\> 4. The diagram is the plan 1/ ' \ of a hip roof. The slope of each face of the roof is 30~. '__________ ____ _ \Find the length of a hip rafter 2A 2 C (Ias AB. [Suc. Draw a triangle DBC representing a section of the roof at 20 DBC on the plan. Hence it may be shown tlhat BC( = /- v3. In like manner by taking a section through B BF, it is found that AC = 10. Hence in the triangle ABC, AB may be found.] 5. Find the area of the entire roof represented in Ex. 4. jD 6. Make drawings showing at. what 1(o angle the two ends of a rafter like BC in Ex. 4 must be cut. 533 5.34 GEOMETRY 7. Make drawings showing at what angle a jack rafter like 12 in Ex. 4 must be cut. [SvG. To determine how the end 1 of the jack rafter must be cut use the principle that two intersecting straight lines determine a plane (Art. 501). The cutting plane at 1 must make an angle at the side of the jack rafter equal to angle CBH, and on the top of the jack rafter equal to angle ABC.] 8. What is a gambrel roof? Make up a set of examples concerning a gambrel roof similar to Exs. 4-7. 9. By use of Art. 645, show that a page of this book held at twice the distance of another page from the same lamp receives one fourth the light the first page receives. 10. The supporting power of a wooden beam varies directly as the area of the cross section times the height of the beam and inversely as the length of the beam. Compare the supporting power of a beam 12 ft. long, 3 in. wide, and 6 in. high with that of a beam 18 ft. long, 4 in. wide, and 10 in. high. Also compare the volumes of the two beams. EXERCISES. GROUP 94 (BOOKS VIII AND IX) 1. A hollow cylinder whose inside diameter is 6 in. is partly filled with water. An irregularly shaped piece of ore when placed in the water causes the top surface of the water to rise 3.4 in. in the cylinder. Find the volume of the ore. 2. What is a tubular boiler? What is the advantage in using a tubular boiler as compared with a plain cylindrical boiler? If a tubular boiler is 18 ft. long and contains 32 tubes each 3 in. in diameter, how much more heating surface has it than a plain cylindrical boiler of the same length and 36 in. in diameter? (Indicate both the long method and the short method of making this computation and use the short method.) 3. If a bridge is to have its linear dimensions 1000 times as great as those of a given model, the bridge will be how many times as heavy as the model? Why, then, may a bridge be planned so that in the model it will support relatively heavy weights, yet when constructed according to the model, falls to pieces of its own weight? Show that this principle applies to other constructions, such as buildings, machines, etc., as well as to bridges. PRACTICAL APPLICATIONS 535 4. Work again Exs. 23-25, 27, p. 473. 6. Make and work for yourself an example similar to Ex. 24, p. 473. 6. Sound spreads from a center in the form of the surface of an expanding sphere. At the distance of 10 yd. from the source, how will the surface of this sphere compare with its surface as it was at 1 yd.? How, then, does the intensity of sound at 10 yd. from the source compare with its intensity at a distance of 1 yd.? Does this law apply to all forces which radiate or act from a center as to light, heat, magnetism, and gravitation? Why is it called the law of inverse squares? 7. If a body be placed within a spherical shell, the attractive forces exerted upon the body by different parts of the shell will balance or cancel each other. Hence a body inside the earth, as at the foot of a mine, is attracted effectively only by the sphere of matter whose radius is the distance from the center of the earth to the given body. Hence, prove that the weight of a body below the surface of the earth varies as the distance of the body from the center of the earth. [SUG. If TV denote the weight of the body at the surface, and w its weight when below, R the radius of the earth, and r the distance of the body from the center when below R -— 'T-1 Athe surface, show that r3 13 IV =w = -- -; - 3- R' r R2 ' r2 8. The light of the sun falling on a smaller sphere, as on the earth or the moon, causes that body to cast a conical shadow. Denoting the radius of the sun by R, the radius of the smaller sphere by r, the distance between the two spheres dr by d, and the length of the shadow by 1, show that I - R Find 1 when d = 92,800,000 mi., R = 433,000 mi., r = 4000 mi. 9. If in a lunar eclipse the moon's center should pass through the axis of the conical shadow, and the moon is traveling at the rate of 2100 mi. an hour, how long would the total eclipse of the moon last? How long if the moon's center passed through the earth's conical shadow at a distance of 1000 mi. from the axis of the cone? o)~536,) (.I GEOMETRY 10. If the moon's diameter is 2160 mi., find the length of the moon's shadow as caused by sunlight. 11. If the distance of the moon from the earth's center varies from 221,600 mi. to 252,970, show how this explains why some eclipses of the sun are total and others annular. Why, also, at a given point on the earth's surface is an eclipse of the sun a so much rarer sight than an eclipse of the moon? Why, also, is its duration so much briefer? PORMUL-A8 OF PLANE (4EOMNETRY SYMBEOLS a, b, c=sides of Wangie ABC. s=i (a b +c). Ie= altitude on side c. qnc=median on side c. tcrisector of angle )j1)Osite side c. I and m-line segmy~ents,,. P== perimeter. Sit=side of a regular polygon of a sides. P=:radius of a circle. D) —diamaeter of a circle. ('=circumference of a circle. i-~radius of an inscribed circle. 71; approx. (or 1.1416-). A= area. /)=base of a triangle. 1i =altitude of a triangle. bl, and 7b2-bascs of a trapezoid. LENGTHS OF LINES 1. In a right triangle, C being the right angle, C2 =a2+ b2. Art. 346. 2. In a rig-ht triangle, I and m being the projections of a and b on c, and lh, the altitude onc J21l m. XArt. 342. W=l> I, 7r= >cc 3. In an oblique triangle, ~nbeing the projection of b on c, if a is opposite an aeute Z, 0 b + 42 - 2 cX m. Art. 349. if a is opposite an obtuse Z, a' = b2 ~' + 2 c >K mi. Art. 350. 4. h = C2 I/ s (s-a) (s-b) (s-c). Art. 393, 5. mic- V12 (a(2 -1 b) -C2. Art. 353. 2 6. te = ab/'a b s (s -c). Art. 363. 7. If I and ait are the scgments of c made by the bisector of the angle opposite, a 5b= I in. Arts, 332, 336. ~538)' FORMULAS OF PLANE GEOMETRY 3 8. If I and m are the segments of a line, a, divided in extreme and mean ratio, and 1 > m, a:: l i. Art. 370.. - P -sP1:.a. Art. 341. 9. In similar polygons, { P '= l Art. 3 1. { a( =b b'. Art. 321. 10. In circles, C: C'=R: 1'; also C: C'=D: D. Art. 442. 11. C=2 7rR, or C = r ). Art. 444. central angle 12. An arc = c-et - R. Art. 445. 180o 13. In inscribed regular polygons, ^. / It ('2 R1 - 1/,i-4 2. Art. 467. AREAS OF PLANE FIGURES 1. In a triangle, K'=- blh. Art. 389. 2. In a triangle, K=V/s (s —a) (s -b) (s- c). Art. 393. 3. In an equilateral triangle, K=- Ex. 4, p. 257, 4. In a parallelogram, K=)b X h. Art. 385. 5. In a trapezoid, K==- h (b 4-+ b2). Art. 394. 6. In a regular polygon, K=- r X P. Art. 446. 7. 'In a circle, KI —rrR' or K= i rrD2. Art. 449. 8. In a sector of a circle, K=- RX are, Art. 453. central Z or K=- 36-o R2. Art. 453. 9. In a segment of a circle, K=sector -- A formed by the chord and radii of the segment. 10. In a circular ring, E=7-r ( — R'2). Art. 449. 11. In any two similar plane figures, K: K'-f12: (aI; Art. 399. also a: a'= //K: 1/7K' Art. 314. 12. In two circles, K: K'='12 R'=) 1D2 I'2(C:1 C/2; Art. 45,2 also R: ~R'=D D'= C: C ='-,/ K':f. Art. 314. FORMULAS OF SOLID (ILEOMETRY SYMBOLS B, b=areas of the lower and upper bases of a frustum. E=lateral edge (or element); or = spherical excess. 11= altitude. 1, b, h=length, breadth, height. L=slant height. M=area of midsection. P=perimeter of right section. P, p_= perimeters of lower and upper bases of a frustum. 'r, 'r'=radii of bases. 8=area of lateral surface; or = area of surface of sphere, etc. T=area of total surface. F= volume. FORMULAS FOR AREAS 1. In a prism, S=EX P. 2. In a regular pyramid, S= L XP. s. In a frustum of a regular pyramid, S= (P- +p) L. 4. In a cylinder of revolution, S=-2 WrH 1= 2)7P R(Pi,+H). 5. In a cone of revolution, S=7rRL. T=7rR (L Rt ). 6. In a frustum of a cone of revolution, S=7rL (B + r). 7. In a sphere, S=4 rpt2, or S=7TD2. 8. In a zone, S=2 7rjeH. 9. In a lune, 8= 90 10. In a spherical triangle, 8 180 7rR2E I1. In a spherical polygon, S= (ISO0 Art. 608. Art. 641, Art. 643. Art. 697. Art. 697. Art. 721, Art. 721. Art. 727. Art. 810. Art. 813, Art. 817, Art. 822, Art. 824, FORMULAS OF SOLID GEOMETRY 541 FORMULAS FOR VOLUMES 1. In a prism, V=BX H. Art. 628. 2. In a parallelopiped, == X )X h. Art. 626. 3. In a pyramid, V= B- X H. Art. 651. 4. In a frustum of a pyramid, TV= H (B b + - /Bb). Art. 656. 5. In a prismatoid, V= — H ( B -t- b + 4 1). Art. 663. 6. In a cylinder, V=BX H. Art. 698. 7. In a cylinder of revolution, F=7R2RH. Art. 699. 8. In a cone, V=- BX fH. Art. 722. 9. In a circular cone, V= - 7rR2I. Art. 723. 10. In a frustum of a cone, V=:: H(B+ b + l/Bb). Art. 729. 11. In a frustum of a cone of revolution, fV 7rH (~ -r-2 +1- ). Art. 730. 12. In a sphere, V= - 7rr3, or J'-1= Tp) Art. 832. 13. In a spherical sector, r=V RJIIt. Art. 836. 14. In a spherical segment of two bases, r=-~ (7'2 q+ rr./2) H -- -H rH3. Art. 837o 15. In a spherical segment of one base, V=w7r2 (R- i, H). Art. 838. CONSTANTS 1 acre=43,560 sq. ft. '2=1.2599 + 1 bushel-=2150.42 cu. in. p'3= 1.4422 + 1 gallon=231 cu. in. -.3183 + i/2=1.4142 -L V '-1.7725 - /3= 1.7321- =0.5642+ GG( 542 542 GEOMETRY. APPEND)IX SUMMiY1iARY OF METRIC SYSTEM TABLE FCR LENGTH 10 millimeters (umm.) 10 cI. 10 din. 10 M. 10 DDT. 10 rnm. 10 Km. 1 centimeter (cm.) =1 decimeter (di.) = Imeter (i.). =1 Dekameter (Din.) =1 Hektometer (Hm.) =1 Kilometer (Km.) =1 -11yriameter (Mlm.) Similar tables are used for the unit of weight, the gram; for the unit of capacity, the liter; for the unit of land measure, the are; and for the unit of wood measure, the stere. TABLE FOR SQUARE MEASURE 100 sq. mm.=1 sq. cm. 100 sq. cm. =1 sq. din., etc. TABLE FOR CUBIC MEASURE 1000 cu. mm.=1 cu. cm. 1000 on. cm. =1 cu. din., etc. A titer =-1 en. din. A qra?)i==weig-ht of 1 An ore = 100 sq. m. A s&tee =1 cu. m. cu. cm. of water at '39.21 Fahrenheit. EQUIVALENaTS 1 meter =39.37 inches. I liter =1.057 liquid quarts, or.9581 dry quarts. 1 kilogram=2.2046 lbs. av. 1 hektare =2.471 acres. 1 sq. in. =1550- sq. in. INDEX OF DEFINITIONS AND FORM9ULAS Abbreviations Algebraic analysis Algebraic method Alternation. Altitude of cone of cylinder. of frustum of cone of frustum of pyramid of parallelogram of prism of pyramid of spherical segmlent of trapezoid of triangle. of zone. Analysis, solution by Angle acute PAGE 8, 231 A. 229 91.180. 412. 403. 414.378. 67. 361. 377. 462 * 67 33 A. 452. 97. 14 15 ngle of lune t.,-1 V I1,ll.1 1 eO [U1" 11 ~U I i:l 1 re-entrant reflex.. right salient sides of. spherical straight tetrahedral trihedral vertex. vertex of ngles, adjacent. alternate-inter io: complementar y. exterior exterior-inte rior homologous. interior of polygo. of quadrilateral of spherical polygon of triangle opposite-interior supplementary vertical PAGE.. 452.. 349o o. 74 o. 16 o 15. 73.. 14. 433 15. 349. 349. 33.. 14. 16 55. 16 55a 55 3). ). 34.. 73. 66 ~. 438.. 32. 77. 263 493. 490 at center of regular polygon.. 64 central... 105 dihedral.... 337 exterior in triangle formed by a rotatinl straight line 17 formed by two cuvels. 33 inscribed in circle 1 05 inscribed in sergent of circle. 05 oblique... 1 Antece(dents Apotl henm Appololius Arabs.. - I obtuse 15 543 ,44 IN-DEX OF DEFINITIONS AND FOBSIULA~S PAGE Arc 103 formiula for. 25 majjor. 104 minor.... 104 Archimedes 497, 498 Arcs, conjugate. 104 Area of surface. 231 Aryabhatta.4 97 August, E. F....498 Auxiliary lines 93. 170 Axion.19 Axioms, general. 19, 20 geometric 29 21 Axis of circle of sphere. 427 of circular cone. 413 of regular pyra mid. 3 78 of sphere. 42 7 of symmetry PAGE Center of synnactry.. 293 Centroid. S3 Chord.103 Circle 103, 490 arc of. 103 center of. 103 circumference of. 103 circuinference of, formula11"t for. 274 circumscribed 105 (lianieter of. 103 formula for area of 270 great.427 inscribed..05 radius of... 103 sector of.104 small.427 Circles, concentric. 105 escribed. 102 tangent.105 Circum-center of tii- 1a<e. 81 Circumference. 14, 103 Classification of polvlicwlions 360 of triangles 32, 33 Commensurable 125 Complement 16 Composition..... 11 and division.... 2 Babylonians Base of cone of isosceles triangle of pyramid of spherical pyramid of spherical sector of triangle Bases of cylinder of frustum of cone of frustum of p~yralmin d of parallelogram of prism of spherical segment of trapezoid of zone Bisector of triangle Bodies, the Three Round Center of circle of regular polygon of sphere 49 7 412 33 377 401 33 403 414 378 67 301 402 * 67 * 4 52 34 498 103 * 263 * 422 5 Conclusion Concurrent lines Concylic points Conel altitude of itXi5 of. base of * circular. 24 105 4123 412. 413 * 412 413 circular, formula for volume of. 418 circular, fornmlas for lateral and total area.. 417 INDEX OF DEFINITIONS AND FORMULAS PAGE Cone, elements of.. 412 lateral surface of...412 oblique circular.. 413 of revolution... 13 right circular... 13 vertex of.... 412 Cones, similar... 413 Congruent figures...13 Conical surface....412 directrix of....412 element of..... 412 generatrix of.. 412 nappes of... 412 vertex of. 412 Conjugate arcs... 104 Consequents... 177 Constant... 126 geometrical.. 143 Constants.. 541 Continued proportion..177 Continuity, principle of.. 485 Converse of a theorem. '24 Corollary... 23 Cube.......363 Curved spaces... 488 Curved surface.. 17 Cylinder... 403 altitude of..... 403 bases of... 403 circular..... 404 circular properties of.. 405 elements of.... 403 lateral surface of.. 403 oblique... 404 of revolution... 404 of revolution, formulas for lateral and total areas of...... 409 right..... 404 right circular....404 PAGE Cylinder, right section of. 405 section of.. 405 Cylinders of revolution, similar... 404 Cylindrical surface.. 403 directrix of.... 403 element of. generatrix. n....403. 403 Decagon.. 74 I i Degree of arc.. 131 angular... 17 spherical.. 452 Demonstration, indirect. 25, 96 Determined plane 319, 320 Diagonal of polygon.. 73 of polyhedron... 360 of quadrilateral.. 66 Diameter of circle.. 103 of sphere... 425 Dihedral angle....337 edge of... 337 faces of.... 337 plane angle of.... 338 right..... 337 Dihedral angles, adjacent. 337 equal..... 337 vertical...337 Dimensions.. 12 Directrix of conical surface. 412 of cylindrical surface. 403 Discussion... 147 Distance from point to line. 35 Distance from point to plane 328 on surface of sphere..428 polar, of great circle.. 429 polar, of small circle. 429 Division... 181 Dodecagon.. 74 Dodecahedron. 360 54 4 I NDE R X OF DEFINITIIONS AsN D FO RM- IULLAS PAGE Duality, Principle of. 486 Edge of dihiedral angle.. 837 Edges of polyhedral angle 349 of polyllyedron.. 0 Egyptians 490, 491, 492, 494, 497 Elements of conical surface 412 of cylindrical surface.. 403 Enunciation, general 25 particular 25 Equal figures.. 13 Equivalent figures. 13, 231 solids.363 Euclid.. 491 Eudoxus. 493, 496, 498 P Fourth proportionlal Frustum of conuL altitude of bases of formula for lateral area of formula, for volnnme of lateral surtace o1 slant height of Frustum of pyramid altitude of bases of slant heigyht of 177 411 414 414 420 4 22 414 414 378 378 378 378 Euler European Extreme and mean ratio Extremes Faces of dihedral ang-le of polyhedral angile of polyhedron Figure, curvilinear geometrical mixtilincar 497 490.217 177 337 349 360 13 13 13 GeneratriX of conical surface.412 of cylindrical surface 4 403 Geometsric solid 11 Geomletry.1 epochs in development of 490 historv of...490-498 modern.485 Non-Euclidean. 488, 498 plane.17 rectilinear 13 Figures, rectilinear.. 494 Formulas of Plane Geometry, for lengths of lines. 528, 539 of Plane Geometry, for areas of plane figures,. 539 of Solid Geometry, for areas. 540 of Solid Geometry, for volumes... 541 Foot of perpendicular.. 15 of line. 320 I origin of lplane projective solid. Gerard. 408 Greeks. 490, 492, 493, 494 Harmonic division. *,, 19 98 IHeptagon..... 74 Iero.. 496 490 i18 4S5 18, 319 Hexagon H-lexahedron Hindoos 74.360 490, 492, 497 Hippasius.. 498 Hhippocratus.. 491, 492, 490 History of geometry. 490-498 Homologous angles. 34 sides...... 34 INDEX OF DEFINITIONS AND FORMULAS PAGE Hlomology, Principle of.487 Hlyperspace. 488 Hypotenuse 3 Hypothesis.2-4 Icos'ahedron. 360 In-center of triangle.83 Inclination of line to plane 347 Incommensurable..12 5 ratio 12 Inference, immediate.23 Inversion ISO Isoperimetric figures 2,. 26 Jews..49 7 Line divided internally of centres. parallel to plane perlpendicular to plane straight. PAGE.198.120.321.320 1 '3 Lines, auxiliary concurrent Lobatchewsky Locus. Log0,ical methods Lune angyle of formiula for area of, spherical degrees formiula for area of, square units of arca 93, 170 * 80 * 498 5 52 * 492 * 452 * 452 inl * 467 i n * 457 Lateral arc pyrami of prism of pyram Lateral edo of pyran Lateral fac, of pyrani Lateral su of Cyliuc of f rustij Legys of iso of right of trape2 ULiit Limits, mc Line broken curved. divided divided divided meau!a of f rustum of d, f ormula f or.379.~~~361 id.. 3 77,es of prism 6 I6 alid 3 377 as of prism. 361 aid.. 37 7 rface of cone.412 ter....403 man of com-. 414 sc(el('5; triangle.339 triangle...33 coid... 67 Ithod of..126 12 13 13 externially. 19 harmionically 198 in extremle, aul1( Major arc of circle.. 104 Magnitudes, commensurahle 125 incommensur~ble 1 4125 M1a~ximumr.n 141, 285 Mean proportional.. 177 Means... 177 Mechanical mnethods.493 Mledian of trapezoid.67 of triangle. 34 Menelaus. 498 Mlethod, algebraic.91 of limits...126 Methods, logical 4 49-21 mechanaical. 493 of num11e~ieal Icomp< a"'1 -tions. 3(4-5 rhetorical. 491 Metric system, summnary of 5412 Mfinor arc of circle.. 104 Minul'te.17 'Modern emtr 485 rat,14io 217 5,4S 8 INDEX OF DEFINITIONS AND FOEIMULAS PAGE Nappes of cone. 412 Negative quantities. 485, 480 N-gon..74 Non-Euclidean geomletry 48S, 498 Numerical measure.. 124 computation, methods of Octagon Octahedron Opposite of a theorem.Origin of geometry Ortho-center 304-3305 7 4. 360 *. 24 * 490 * 82 Parallel lines Parallelogram 06 altitude of.....07 bases of.. 7 Parallelopiped.. 362 right.....2 rectangular. 363 Parallel planes. 321 Pentagon..74 Pentedecagon..74 Perimeter of p)olygon 73 of quadrilateral 60 of triangle.. 32 Perpendicular lines. 15 planes.338 Pi (7r).274 Plane. 17, 319 figure... 17 determination of 319, 320 Planes, parallel.. 321 Plato.491,493 Point.. 12 Points, concylic.. 105 Polar distance of circle.. 429. trjangle,., 441 PAhf rL Pole.427 P'olygon...... 73 angles of ' 73 circumnscribed.. 105 concave.. 73 convex... 73 diagonal of 73 equiangular 73 equilateral... 73 inscrihed.. 105 perimeter of 73 regular... 261 reg'ular, angle at center of 204 regyular, apothemn of 263 regular, center of. 2036 regular, radius of.263 sides of.. 73 spherical.438 vertices of.. 73 Polygons, mlutually equiangular. 74 mutually equilateral. 74 similar. 190 symmetry of. 293 Polyhedral angle.. 349 convex. 349 edges of. 349 face angles of.. 349 faces of.349 vertex of. 349 Polyhedral angles, equal 350 symmetrical. 350 vertical.. 350 Polyhedron. 360 convex.. 360 diagonal of 360 edges of.. 300 faces of.. 360 regular...391 section of 360 INDEX OF DEFINITIONS AND FORMULAS 549 Polyhedron, vertices of PAGE. 360 PAGE. 177 Proportional, fourth. Polyhedrons.. 497 classification of.. 360 similar.... 396 Postulate.. 22 of solid geometry... 320 Postulates.. 22, 146 logical...... 22 Prism.....361 altitude of..... 361 bases of.. 361 circumscribed about cyl inder inscribed in cylinder lateral area of lateral edges of lateral faces of oblique.. quadrangular regular.. right right section of triangular truncated Prismatoid bases of. formula for volume of Prismoid Problem. 405 405.361.361.361.362. 362. 362. 362 361. 362. 362. 389.389. 390. 389 23 mean... 177 third..... 178 Proposition.. 23 Pyramid.. 377 altitude of.....377 axis of..... 378 base of.. 377 circumscriled about cone 414 frustum of.... 378 inscribed in cone... 414 lateral area of.... 377 lateral edges of... 377 lateral faces of.. 377 quadrangular.. 377 regular.. 377 regular, slant height of. 378 spherical. 461 triangular..... 377 truncated.... 378 vertex of.... 377 Pythagoras. 491, 492, 495, 497 Quadrant of circle.. 103 Quadrilateral.. 66 angles of.... 66 perimeter of.66 sides of... 66 vertices of... 66 Radius of circle.. 103 of regular polygon. 263 of sphere.. 425 Ratio... 125 of similitude.. 190 Reciprocally proportional. 211 Reciprocity, principle of. 486 Rectangle... 66 Rectilinear figures.. 494 Rhetorical methods.. 491 Rholmbhoid.... 66 Projection of line on line. 205 of line on plane.. 33 of point on line.. 205 of point on plane. Proof... by superposition forms of methods of. Proportion continued terms of... 338 23.. 23 25 25.25. 177 177 ~ 77 i ) INDEX OF DEFINITIONS AND FORMULAS PAGE Rhombus...... 66 Right section of cylinder. 405 of prism.. 361 Romans.44, 496 I t Solids, physical Spaces, curved Sphere axis of. center of circumscribed PAGE *... 11. 488 *.. 425.. 427.. 425 Round Bodies, The Three C ni -i. Z IUkJIIUI 1 1.11.I.. Secant line.... Second..... Sect Section of polyhe(dron Sector of circle of circle, forlnula for of sphere Sectors, similar. Segment of circle of line of sphere Segments, similar Semicircle Semicircumference Sides of angle of polygon. of quadrilateral of triangle.. Similar cones of revolution cylinders of revolution figures polygons polyhedrons sectors of circles segments of circles Slant height of cone. of frustum of cone of frustum of pyramid of regular pyramid Solid 1 geometry... 18 Solids, equivalent geometric 198 23 104 17 14 360 104 277 461 275 194 14 462 275 104 103 14 73 66 32 413 404 63 190 396 275 275 413 414 378 377 1,12,319 363 11 about poly hedron... diameter of formula for area of sur 432 425 face of. 455 formula for volunim of. 463 great circle of. 427 inscribed in polyhedron. 432 poles of.... 427 radius of...425 small circle of.. 427 Spherical angle.... 433 degrees.... 452 excess.. 444,460 Spherical polygon.. 438 angles of..438 sides of.... 438 vertices of... 438 Spherical pyramid...461 base of.. 461 vertex of.... 461 Spherical sector... 461 base of.. 461 formula for volume of. 464 Spherical segment.. 462 altitude of..... 462 bases of... 462 formulas for volume of. 465 of one base... 462 Spherical triangle... 438 bi-rectangular... 443 formula for area of, in square units of area. 460 tri-rectangular...443 INDEX OF DEFINITIONS ANID FORMULAS 551 p Spherical triangles, supplemental.. symmetrical Spherical wedge Spherid.. Square Straight line. Superposition Supplement Surface area of... conical curved. cylindrical.. unit of... Symbols Symmetry, axis of. 293 center of.. 293, 'AGE 442 444 461 452 66 13 25 16 12 231 413 17 403 234 8 294,294 Trapezoid, isosceles. median of. Triangle acute. altitudes of PAGE I. 67.. 67 32, 74 33 33 center of, for a polygon. 293 Tangent circles.. 105 Tangent, common external. 120 common internal... 120 line to sphere... 425 plane to cone. plane to cylinder. plane to sphere. 413. 404. 426 I base of. 33 bisectors of... 34 bisectors of, formula for 213 classifications of. 32, 33 equiangular.. 33 equilateral..... 32 isosceles. 32 medians of. - 34 obtuse.. 33 polar. 441 right.. 33 scalenre.. 32 spherical... 438 vertex angle of o 33 vertex of. 33 Trihedral angle. 349 bi-rectangular.... 350 isosceles... 350 rectangular... 350 tri-rectangular. 350 spheres... 426 Terms of a proportion.. 177 Tetrahedral angle.. 349 Tetrahedron.. 360 Thales.. 491, 495,496 Theorem..... Theory of limits. Third proportional. Transversal Trapezium Trapezoid. altitude of.. bases of.... 23. 126. 178 55 ~ 66. 66. 67. 67 Ungula...461 Unit of measure.. 124 of surface. 231 of volume ~ 363 Units of angle 17 of spherical surface, ~ 452 Variable.... 126 Vertex angle. 33 Vertex of angle. 14 of cone.. 412 of polyhedral angle. 349 of pyramid.. 377 of spherical pyramid. 461 .~ 0'2 I ND EX: O F D ETINITIONS. AND FO-RMUA Vertex of triangle Vertices of polygon of polyhedron of quadrilateral of spherical polygon of triangle Volume of solid PAGE 33 * 73 * 360 * * 66 * 438 * 32 363 Wedg~e, spherical Xenodorus Zone. altitude of bases of of o e base PAG I * 461 * 496 * 452 * 452 * 452