^. IMAGE EVALUATION TEST TARGET (MT-3) 1.0 1.1 1.25 2.0 EI4 U_ 11.6 V] 7 ^'./ as /A f/ w w Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 CIHM/ICMH Microfiche Series. CIHM/ICIVIH Collection de microfiches. Canadian Institute for Historical Microreproductions / Institut Canadian de microreproductions historiques Tachnical and Bibliographic Notas/Notas tachniquas at bibliographiquas TtMi toth Tha Institute has attamptad to obtain tha bast original copy availabia for filming. Faaturas of this copy which may ba bibliographically uniqua. which may altar any of tha imagas in tha reproduction, or which may significantly change tha usual method of filming, are checked below. □ Coloured covers/ Couverture de couleur □ Covers damaged/ Couverture end'ommagte □ Covers restored and/or laminated/ Couverture restaur^ et/ou pellicula D D D D D Covnr title missing/ Le titre de couverture manque I — I Coloured maps/ Cartaa g<^ographiquas en couleur Coloured ink (i.e. other than blue or black)/ Encre de couleur (i.e. autre que bleue ou noire) I — I Coloured plates and/or illustrations/ D Planches et/ou illustrations en couleur Bound with other material/ Relii avac d'autres documents Tight bindivig may cause shadows or distortion along imarior margin/ La re Mure serrie peut causer de I'ombre ou de la distorsion le long da la marge int^rieure Blank leaves added during restoration may appear within the text. Whenever possible, these have been omitted from filming/ II se peut que certaines pages blanches ajoutias lors d'une restauration apparaissent dans la taxte. mais, lor^qua cela Atait possible, ces pages n'ont pas At* filmias. Additional comments:/ Commentaires supplAmentairas: L'Institut a microfilmi la mailleur exemplaire qu'il lui a 6ti possible de se procurer. Las details de cet exemplaire qui sont peut-Atre uniques du point de vue bibliographiqua, qui peuvent modifier une image reproduite, ou qui peuvent exigar una modification dans la m^tlioda rormaia de filmage sont indiquAs ci-dassous. r~l Coloured pages/ D Pagea de couleur Pagaa damaged/ Pages endommagies Pages restored and/o( Pages restai;r4es et/ou pellicul^es Pages discoloured, stained or foxei Pages dicolories, tacheties ou piqudes Pages detached/ Pages ditach^es Showthroughy Transparence Quality of prir Quality intgale de i'impression Includes supplementary matarii Comprend du material supplimentaire Only edition available/ Seule Edition disponible I — I Pagaa damaged/ r~n Pages restored and/or laminated/ [ — I Pages discoloured, stained or foxed/ I I Pages detached/ rri Showthrough/ I I Quality of print varies/ I I Includes supplementary material/ r~n Only edition available/ The posa of til fllml Grig begi thai slon otha firat aion Drill The shal TINI whl< Map diffi antii begl righ reqi met Pages wholly or partially obscured by errata slips, tissues, etc., have been refilmed to ensure the best possible image/ Las pages totalemant ou partiailement obscurcies par un feuillet d'errata, une pelure, etc.. ont M fiimies A nouveau da facon di obtenir la meilleure image possible. This item is filmed at the reduction ratio checked below/ Ce document est filmi au taux da reduction indiquA ci-dassouc 1QX 14X 18X sx 26X 30X y 12X 16X aox 24X 28X 32X ails du >difier une nage The copy filmed h«r« has been reproduced thanks to the generosity of: Douglas Library Queen's University The Images appearing here are the best quality possible considering the condition and legibility of the original copy and In keeping with the filming contract specifications. L'exemplaire film* fut raproduit grice A la gin^rosit* de: Douglas Library Queen's University Las images suivantes ont 4t4 reprodultes avec ie plus grand soln, compte tenu de la condition at de la nettetA de I'exempieire fiimA, et en conformity avec las conditions du contrat de fllmage. Original copies in printed paper covers are filmed beginning with the front cover and ending on the last page with a printed or Illustrated Impres- sion, or the back cover when appropriate. All other original copies are filmed beginning on the first page with a printed or illustrated impres- sion, and ending on the last page with a printed or illustrated Impression. The last recorded frame on each microfiche shall contain the symbol — ^ (meaning "CON- TINUED"), or the symbol y (meaning "END"), whichever applies. IWIaps, plates, charts, etc., may be filmed at different reduction retios. Those too large to be entirely Included in one exposure are filmed beginning In the upper left hand corner, left to right and top to bottom, as many frames as required. The following diagrams illustrate the method: Lee exemplalres orlglnaux dont la couverture en papier est Imprimte sont f limAs en commenpant par Ie premier plat et en termlnant solt par la dernlAre page qui comporte une emprelnte d'impression ou d'illustration, solt par Ie second plat, salon Ie cas. Tdus las autres exemplalres orlglnaux sont fllmte on commenpant par la premlAre page qui comporte une emprelnte d'impression ou d'illustration et en termlnant par la dernlire pege qui comporte une telle emprelnte. Un des symboies suivants apparattra sur la dernlAre image de cheque microfiche, selon Ie cas: ie symbols — ► signlfle "A SUIVRE", Ie symbols V signlfle "FIN". Les cartes, planches, tableaux, etc., peuvent Atre flimte A des taux de reduction diff brents. Lorsque ie document est trop grand pour Atre reprodult en un seul cllchA, il est filmA i partir de Tangle sup^rieur gauche, de gauche H drolte, et de haut an bas, en prenant Ie nombre d'Images nAcessaire. Les diagrammes suivants illustrent la mAthode. rrata :o selure, 1 d D 32X 1 2 3 1 2 3 4 5 6 NEWCOMB'S Mathematical Course. — — -♦- — - I. SCHOOL COURSE. Algebra for Schools, I|120 Key to Algebra for Schools, . ... 1I20 Plane Oeometry and Trigonometry, with Tables, lAO The Essentials of Trigonometry, .... 1.25 II. COLLEGE COURSE. Algebra for OoUeges, *i g^ Key to Algebra for OoUeges, . . . ' . ' i'.6o Elements of Oeometry, j 50 Plane and Sphe -loal Trigonometry, with Tables! 2.00 Trigonometry (separate), i.gO Tables (A^orate), 140 Elements of Analjrtio Geometry . 1.50 Oaloolns (in preparation)^ Astronomy (Newoomb and Holden,) . . 2.BO The same, Briefer Course, .... .1.40 HEIifRY HOLT d CO.; Publishers, New York. NEWOOMB'8 MATHEMATICAL COURSE. ELEMENTS GEOMETRY BY SIMOlSr I^EWOOMB F^ofeasor of MathemaUcs, United StaUa Navy THIRD EDITION, REVISED. N"EW TOEK HENRY HOLT AND COMPANY 1884. / » Ck>FTRiaHTSD, BY HENRY HOLT & (DO. 1881. PEEFAOE. Is the present work is developed what is commonly known m the ancient or EucUdian Geometry, the ground covered being neariy the same as in the standard treatises of Euclid, Legendre, and Chauve- net. The question of the best form of development is one of such interest at the present time, among both teachers and thinkerw, as to justify a statement of the plan which has been adopted. It being still held in influential quarters that no real improvement upon Euclid has been made by the modems, a comparison with the ancient model will naturally be the first subject of consideration The author has followed this model in its one most distinctive fea- ture, that of founding the whole subject upon clearly enunciated defi- mtions and axioms, and stating the steps of each course of reasoning in their completeness. By the common consent of a large majority of educators the discipline of Euclid is the best for developing the powers of deductive reasoning. If the work had no other object than that of teaching geometry, a more rapid and cursory system might hftve been followed; but where the general training of the powers of thought and expression is, as it should be, the main object, it be- comes important to guard the pupil against those habits of loose thought and incomplete expression to which he is prone. This can be best done by teaching geometry on the time-honored plan. Notwithstanding this excellence of method, there are several points in which the system of Euclid fails to meet modem require- ments, and should therefore bo remodeled. The most decided failure 18 in the treatment of angular magnitude. We find neither in Euclid ""' " "'^ "'^ modem foiiowers any recognition of angles equal to 98847 iv PREFACE, or exceeding 180°, or any explicit definition of what Ib meant by the Bum of two or more angles. The additions to the old systeir of angular measurement are the following two : v Firstly. An explicit definition of the angle which is equal to the sum of two angles. Secondly. The recognition of the sum of two right angles as itself an angle. The term ''straight angle" has been adopted from the Syllalus of the English Association for the Improvement of Geotnetrieal Teaching. Although not unobjectionable, it seems to be as good a term as our language affords. The term geatrecUe Winkely used by the Germans, is more expressive. One of the most perplexing questions which the author has met in the preparation of the work is that of distinguishing the definitions of plane figures as lines and surfaces. In our recent text-books it is be- coming more and more common to define triangles, circles, etc., as por- tions of a plane surface. But, as soon as analytic geometry is reached, the circle is considered as a curve line and the triangle as three straight lines, while, even in elementary geometry, these terms, in a large ma- jority of cases, refer only to the bounding lines. It has seemed to the author that the confusion thus arising can best be avoided by defining plane figures neither as mere lines nor mere surfaces, but as things /orm^d by lines; to use the specific term area when extent of surface alone is referred to; and to use the words cireun\ference, perimeter, etc, , only in the sense in which they are used in higher geometry. Other leading features of the work, which may be briefly pointed out, are the following: I. The addition of an introductory book designed not only to pre- sent the usual fundamental axioms and definitions, but to practice the student in the aiialysis of geometric relations by means of the eye before instructing him in formal demonstrations. The exercises in sections 34 to 84 are first attempts in this direction, to which the teacher may add at pleasure until he finds that the pupil has thoroughly mastered the conceptions necessary for subsequent use. n. The application of the symmetric properties of figures in demonstrating the fundamental theorem of parallels. This system has been adopted from the Germans. III. After the second book, the analysis if the problems of con> PREFACE. y struction, whereby the pupil is led to discover the construction by reasoning. IV. The division of each demonstration into separate numbered steps, and the statement of each conclusion, where practicable, as a relation between magnitudes. It is believed that this system will make it much easier to carry the steps of the demonstration in mind. Each step is, when deemed necessary, accompanied by a reference to the previous proposition on which the conclusion is founded, not, however, to encourage the too frequent habit of requiring the pupil to memorize the numbers, but simply to enable him to refer to the proposition. He should always be ready, if required, to cite the proposition, but its number in the book is not of such importance that his memory need be burdened with it. A reference has not been considered necessary after a few repetitions. V. The theorems for exercise have been selected from native and foreign works with a view to present those best adapted, either by their elegance or their applications in the higher geometry, to inter- est the student. An attempt has been made to arrange those of each book in the order of their difficulty. VI. Some of the first principles of conic sections have been devel- oped for the purpose of enabling pupils who do not intend to study analytic geometry to have some knowledge of these curves. It is believed that a previous study of these principles will be a valuable preparation for the advanced treatment of conic sections. Vn. The most difficult subject to treat has been that of Propor- tion. The ancient treatment as found in Euclid is perfectly rigorous, but has the great disadvantages of intolerable prolixity, unfamiliar conceptions, and the non-use of numbers. The system common in our American works, of treating the subject merely as the algebra of fractions, has the advantage of ease and simplicity. But, assuming, as it does, that geometric magnitudes can be used as multipliers and divisors on a system which is not demonstrated, even for algebraic quantities, it is not only devoid of geometric rigor, but is not prop- erly geometry at all. The author has essayed a middle course between these extremes which he submits to the judgment of teachers with some reserve. On the annienf. nvflfnm mnrrni^ii'lpq <}.rp r>/%Tnv.««a^ -srSfH -^ ^^ vi PRBFAOa. their ratios by means of their multiples. For mstance, the magnitude A is considered to have to the magnitude B the ratio of 2 to 8 when 8A = 2B. This system has the undeniable advantage of admitting commensurable and incommensurable quantities to be treated on a uniform plan. But it has the disadvantage of not according with the natural and customary way of thini'.ing of the subject. When we say that the magnitude A is to B as 2 to 8, we mean that if A is repre- sented by the number 2, or is divided into 2 parts, B will be repre* sented by 8 of those parts. The author has considered it more im- portant to base the subject on natural and customary modes of thought than to adopt a system simple and rigorous, but not so based. The mode in which he has endeavored to avoid the difficulty, and to render the natural system as rigorous and nearly as simple as the other, will be s&m by an examination of the chapter on Pro- portion. Ym. Another difficult subject is the fundamental relations of lines and planes in space. In presenting it the author has been led to follow more closely the line of thought in Euclid than that in modem works. At the same time he is not fully satisfied with his treatment, and conceives that improvements are yet to be made. A collection of notes on the fundamental principles of geometry upon which the work has been based will be found in the Appendix. The author believes, from some trials, that the study of geometry as here presented can be advantageously commenced at the age of twelve or thirteen years. No especial knowledge of algebra is required for the first three books, but a previous familiarity with symbolic notation will facilitate the study of the second and follow- ing books, and may be found necessary to their advantageous use. From the fourth book onward a knowledge of simple equations is sometimes presupposed. TABLE OF OOJ^TEKTS. OHAPTiB ®^°^ ^' <^ENERAL NOTIONS. I. The Primary Concepts of Geometry '^J II. Comparison of Geometric Magnitudes a III. Of Symmetry *.*..".*.*.."*.*.'.*!*..'.'.* 15 IV. Logical Elements of Geometry !....!..!!!!.! 18 BOOK n. FUNDAMENTAL PROPERTIES OP RECTILINEAL FIGURES. I. Relation of Angles 2Jj n. Relations of Triangles .......... m IIL Parallels and Parallelograms ...... . . '. 53 IV. Miscellaneous Properties of Polygons. m V. Problems .f V.' .*.*.*."!!*'.' 76 VI. Exercises in demonstrating Theorems ............. '. . . . . . 35 BOOK IIL THE CIRCLE. L General Properties of the Circle 93 IL Inscribed and Circumscribed Figures ina IIL Propertiesof Two Circles *. ." jja IV. Problems relating to the (Jircle ii;* Theorems for Exercise .".................... 12« BOOK IV. OP AREAS. L Areas of Rectangles -og IL Areas of Plane Figures .li IIL Problems in Areas *.*.". !?? IV. The Computation of Areas 154 Theorems for Exercise " 1^ Numerical Exercises •..'."...!!..!!!!!!.. 163 BOOK V. THE PROPORTION OP MAGNITUDES. L Ratio and Proportion of Magnitudes in General ' 163 II. Linear Proportion * j«» IIL Proportion of Areas ..**' ^qa IV. Problems in Proportion OQS Theorems for Exerfiisfi «.^ ' 2iv viii CONTENTS, BOOK VI. REGULAR POLYGONS AND THE CIRCLE. OHATTBR PAO> I. Properties of Inscribed and Circumscribed Regular Polygons.. 214 II. Construction of Regular Polygons 210 III. Areas and Perimeters of Regular Polygons and of the Circle. . 224 Exercises 286 ly. Maximum and Minimum Figures 287 Theorems for Exercise 248 BOOK VII. OP LOCI AND CONIC SECTIONS. I. Lines and Circles as Loci 244 II. Limits of Certain Figures 240 m. TheEllipse 262 IV. The Hyperbola 268 V. The Parabola 266 VI. Representation of Varying Magnitudes by Curves 272 Exercises 274 GEOMETRY OF THREE DIMENSIONS. BOOK VIII. OP LINES AND PLANES. I. Relation of Lines to a Plane 277 II. Relations of two or more Planes 208 m. Of Polyhedral Angles 806 BOOK IX. OP POLYHEDRONS. I. Of Prisms and Pyramids 816 n. The Five Regular Solids 826 BOOK X. OP CURVED SURFACES. L The Sphere 886 n. The Spherical Triangles and Polygons 847 III. The Cylinder 866 rV. The Cone 868 BOOK XI. THE MEASUREMENT OF SOLIDS. L Superficial Measurement 864 IL Volumes of Solids 874 Problems of Computation 880 Thkobems fob Exbbcise in Gbometby of Thbeb Dimensions.. . 800 224 885 887 248 • 844 . 248 . 262 . 268 . 266 . 272 . 274 SYMBOLS AOT) ABBBEVIATIONS USED IN DEMONSTRATIONS. When a step of a demonBtration leads to a relatioii of two lines or other magnitudes, tha relation is expressed by symbols. = equals : states that two magnitudes are equal. il parallel to : states that two lines are parallel. ^, perpendicular : states that two lines are perpendiculai' to each other. = coincides with, or falls \ipon : states that two points, lines, surfaces, or figures coincide with each other. In recitation, the teacher may find it advantageous to have the student recit*? the reasoninp^ orally, but write th© conclusion of each step on the blackboard. In this case symbols or abbreviations of the more common words will shorteu the work. The following are recommended, though others are frequently used: Z , angh, /. line, • , point, Ar, area, R, or 90°, rigJit angle, 0, or 360°, circumferencG, S, or 180°, slraigkt angle. These abbreviations are not generally used in the printed booi^ the author believing that the full word, in its usual form, will make a stronger impression on the mind of the beginner than any symbolic representation of it. BOOK I. GENERAL NOTIONS. CHAPTER I. THE PRIMARY 'CONCEPTS OF GEOMETRY. !• BefmitiGn, Gteometry is the science which treats of magnitude, position, and fonn. Def, A geometiio magaitude is that which has extension in space, or what is familiarly called size, Geometiic magnitudes are of three orders : solids, surfaces, and lines. Besides these, there is a fourth concept — that of a point. /da Solids. Dtf, A solid is that which has length, breadth, and thickness. Iiength, breadth, and thioknesd are called the thr^e dimensions of the solid. All material bodies are solids because they have these three dimensions and no more. The solids of geometry are bodies supposed to have the size, form, and mobility of material solids, but no other properties. Surfaces. 5t. TlarP A mmmmmC^^^ •_ Xl- _ A I'll 1 .« ,^, ±.^j . j^ Biiiiav© ia mac whicn nas lengtn and breadth, but is not supposed to have thickness. Example 1. Let us conceive the solid AB to be divided 2 OENEBAL NOTIONS. I into two parts through CDEF without removing any part of it, so that the two parts touch each other. The diyision CDEF is then a surface. If the surface had any thickness it would be a part either of the one solid or of the other. But it is only the bound- ary between them, and therefore no part of either, and therefore has no thickness. Ex. 2. A sheet of paper is really a solid, because it must have some thickness. But the surface on which we write does not extend into the paper at all, and so has no thickness. it Lines. 4. Def. A line is that which has length, but is not supposed to have either breadth or thickness. If we suppose a surface cut into two parts, touching each other that which divides them is a line. It forms no part of either surface, and therefore can have no breadth. Def. A straight line is one which has the same direction throughout its whole length. Def. A curve line is one no part of which is straight. A straight line. A curve line. The Point. ^n.f« ■^'C" ^ ^"i"* ** ^^^^ '^'^"'^ ^ supposed to have position, but neither length, breadth, nor thickness. nthi; Ti! r^^T.* ?'°^ ™* '"'» *^» parts touching each part of either line, and therefore has no length. h^^ii, '"T!*.*" "«'««' ha^ ae three dimensions, length, breadth, and thickness. But we may make a dot a^d tWnk r„r "w ^°""'' , '■''^ ""' P°'"* ''""^d ^ tJ'e centre of the QEOMETBIO FTQURBS. 3 Generation of Magnitudes by Motion. 6. A line may be generated by the motion of a point, as when we press the point of a pencil on paper and move it. A surface may be generated by the motion of a line as a line is generated by the motion of a point. A solid may be considered as generated by the motion of a surface. The surface, as it moves, must be supposed to leave a mark in every point through which it passes. The Plane. 7. Def, A plane is a surface such that a straight line between any two of its points lies wholly on the surface. A plane is perfectly flat and even, like the surface of still water, or of a smooth floor. Geometric Figures. 8. Bef. A figure is any definite combination of points, lines, surfaces, or solids. Def, A plane figure is one which lies wholly in a plane. It is formed by points and lines. Plane geometry treats of plane figures. Parallel Lines. 9. Def. Parallel straight lines are such as lie in the same plane, and never meet, how far soever they may be ex- tended m both directions. ParaUelUnes. The Circle. 10. Def, A circle is a figure form- ed by a plane curve line, every point of which is equally distant from a point within it called the centre. The circumference of a circle is the line which forms it, Def. An arc of a circle is a part of the circumfer- ence. A circle. IT" Ml OBNBRAL NOTIONS. The Angle. 11. Dtf. An angle is a figure formed by two straight lines extending out from one point in different directions. Bef, The sides of an angle are the two lines which form it. c Dtf, The vertex of an angle An angle. is the point where the sides meet. Geometrical Symbols. 1». Any geometric concept, whether apoiTU, line, surface, solid, or angle, may be represented by one or more letters of the alphabet. A point is represented by a single letter near it. A line is repifesented by one letter, or by two or more letters showing its course. Other magnitudes or figures are represented by let- ters showing their outlines. Designation of angles. A particular angle in a figure is designated by three letters, as ABG, of which the middle one ^18 at the vertex, and one of the other two on each side. The angle is then read ABG. When there is only one angle formed at a yertex, it may be designated by a single letter at the yertex. , . » > CHAPTER II. COMPARISON OF GEOMETRIC MAGNITUDES. Mode of Comparison. ^\-^^£' Two magnitudes which can be so applied to each other that each shall coincide with the other throughout its whole extent are said to be IdenUoally equal. OOMPABIBOS OF AlfOLSa. f, ™r2 rtot^ *r° "'?«^*^de8 can be SO divided into parts that each part of the one is identically equal to a «e^te part of the other, they a^ said to be equa^ e^n^™ J^ "'^"^ ^^^"* * magnitude into two B The point B bisects the line A C. To trisect a magnitude means to divide it into three equal parts. a.."'^^'*u** ^^^^'- '''^^ ^°Sl^« ^^^ and DJEJIP axe said to be equal if the angle ^5(7 can be taken up Bqnal angles. t Wprf ^'^ p *? 1^ ?^^^^ ^^^ ^^ ^'^^^ "tanner that |a with the side £JIP, and the side 5^ with the side 15. Unequal Angles. If, on thus applying the angles to each other, the side BA r ills between the sides UJD and EF, a» in the dotted Hue, then the angle C5^ (which is tlie same as FJEJA) — ^.„ ^^j A^oB buiiii uie angle FED, and the angle FEB is said I I! i 6 OBNEBAL MOTIONS. 16. RE3fABK. The magDitude of an angle does not depend upon the length of its sides, but only upon their direction. ^ When we make the sides / / / ^^and ^C coincide, it is only ' ' ^ necessary that they shall coin- cide through the length of the ««. - dhorfpr fiirlp in n^,!^,. +« +« ^ xi! J"»ese four angles are an equal, not- snorter sme, m order to test the withstandinsr the difference in the equality or inequality of the ^®°8'^ <>' their sides. angles. Symbols of Comparison. Xt. The statement that any two magnitudes are equal is expressfed by writing the sign = between the letters or words which indicate them. The statement expressed by the sign =, that two magnitudes are equal, is called an equation. The statement that one magnitude is greater or less than another is expressed by writing the sign > or < between them, the opening of the angle being toward the greater magnitude. Examples. The expression A=z B means that the magnitude A is ectual to the magnitude B, The expression A> B means that the magnitude A is greater than the magnitude B, The expression A < B means that the magnitude A is lesis than the magnitude B, Sum and Difference of Magnitudes. 18. Def. The magnitude formed by joining two or more magnitudes together is called their sum. I BUM A2W DIFFERENCE. ^ The flum of two or more straight lines is the line obteined by putting them end to end in the same straight line. The sum of two angles ABC and PQR is the angle ABR Sum of anglea vZf n\Tl^n^ ^\''^^ ^^ *^ *^^ «id« ^^^ «o that the vertex (2 shall fall on the vertex B, and the sid^ QR on the opposite side of BC from BA, Bef, If the angles ABQ and OBR are equal, ea<5h me line 5C is said to bisect the angle ABU When from one magnitude a part equal to an- Notation of Sum and Difference. the!Si I^?;r w*""' T,^^i*^^^« i« expressed by writing me sign +, ;7^«*5, between them. ^ Examples. -^^^ ^- o Angle ^^C+ angle CBR = angle ^5i?. Line AB + line ^C = line A C. wrifTi?! 5^^'*^''''! ^®*'^^^'' ^"^^ magnitudes is expressed by Tmtmg their symbols with the sign -, minus, between the^^ the magnitude taken away being on the riffht.' ' l^XAMPLES. Angle ABR - angle ABC= angle CBR. Line AC- line AB = line ^C. o -D 8 OENERAX NOTIONS. ELlnds of Angles. 30. Dqf. When a straight line AB standing on another straight line CD makes the angles ABC and ABD equal, each of these angles is called a right angle, and the Une AB is said to be pexpendioiilar Right angles. to the line CD. 21. Def, When the two sides OA and OB of an angle go out in opposite direc- tions, so as to be in the same ^^ ^ straight line, the angle is called a <5 straight angle. straight angle. We may conceive a straight angle to have its vertex at any point of a straight line. A straight angle is by definition the sum of two right angles, because the sum of the two right angles ABC and ABD is (1,8) the angle CBDy which is a straight angle. 23. Def. An acute angle is one which is less than a right angle. Acute angle. Obtuse angle. 23. Def. An obtuse angle is one which is greater than a right angle. Example of forming: Angles by Addition. 24:c Let us take a surface with a circular boundary ABCDEFGH, and cut it into eight equal parts by four lines all passing through its centre 0. A circular disk of paper or pasteboard may be used to represent this surface. * ADDITION OF ANGLES. Then let us put the pieces to- gether again, one by one, beginning at A and going round in alpha- betical order, and let us study the angles thus formed. On adding the angle once we shall have a right angle AOC, On adding it again we shall have the obtuse angle A OD, greater than a right angle, but less than a straight angle. On adding it again we shall have a straight angle A OB, be- cause we cut the figure so that A OB should be in a straight line. On adding it again we shall ihave the convex angle A OF. This angle will be greater than a straight angle if we count it round in the direction we have added its parts, but it will be less if we count it in the shortest direction from A through II and G to F. The relation of these two ways of con- sidering an angle will be shown presently. By one more addition the angle formed will be the sum of a right angle and a straight angle, or of three right angles, when measured the one way, but equal to a right angle when measured the other way. If we add the angle twice more -o the whole space around O will be filled up, and the sum of the eight — £,xvij TTiii yu inQ anp*' ^ AOA counted all the wayrounu, »fhich is called a perigon. li ll I 10 GENERAL NOTIONS. Dtf. A perigon is equal to the sum of four right angles or of two straight angles. Angular Measure. The following summary includes a recapitulation of results from the preceding sections: 35. An angle is measured by how much one of the sides must be turned to make it coincide with the other side. Since one side can be brought into coincidence with the other by turning it in either direction, there are two measures to every angle. Example. In the figures the side OA can be brought into coincidence with OD by turning it either in the opposite direction to that in whtch the hands of a watch move, or in the same direction. The lesser measure of the angle AOD. The greater measure of the angle AOD. These two directions can be distinguished and the amount of motion be measured by describing an arc of a circle from one side to the other around the vertex of the angle as a centre. This arc must pass through the space over which the one arm must turn in order to coincide with the other. 36. In practice, angles are measured by degrees and sub- divisions of a degree, in the following way: Let a complete circle be drawn with its centre on the vertex of the angle. Let this circle be divided into 360 equal parts. Then each of these parts is called a degree. The sides of the angle will cut the circle in two points. Thft Tmrnhfir nf HAorrppa Tip+xirooTi +1^000 •K»i>i«+r. ;« +1,^ , of the angle, and the angle is said to be of that number of degrees. ANGULAR MEAiiUME, 11 luuuSuru The two sides of the angle divide the circle into two arcs corresponding to the two measures of the angle just de- scribed. Example. In the figure the side OA of the angle A OB cuts the circle at 20°, and the side OB at 360° Counting the degrees in both directions we see that the angle measures 240 m one direction, and 120° in the other. oaf^^""^^^'. P® i"^ ^* *^® *^« measures wiU always be obO , which IS therefore a perigon. 27. Def. The two measures of an angle are said to be conjugate to each other, or to represent conJu- gate angles. ^ .1, ^^l^o *^® ^'^njiigate measures will always be less than 180 and the other greater, except when each is equal to 180°. The greater measure is called a reflex angle. Hence, .iJ^t: ^ r^^"" ^""^^^ ^® """^^ ^^i^^ is greater than a straight angle, or greater than 180°. been L?" ^"^^"'"'''^ ''^^*'°^' ^^ ^^«^'« ^^^ ^^"^ ^l^^t has 1 perigon 1 straight angle • 2 straight angles 1 right angle 2 right angles 4 right angles : 90° 180° 360° pi 18 QBNERAL N0TI0N8. BXBRGISBS. Note. The following exercises are intended to familiarize the pupil with the idea of the magnitudes and measures of angles by causing him to make an eye-estimate of the magnitude of each angle, and, where applicable, a computation of their relations. It will be well for him to make a small paper protractor in order that he may check his estimates by some kind of measures, though rude. Whore he is asked to draw angles, it is Intended that he shall prac- tice the drawing without instruments, repeating hi8 first attempts until he obtains a drawing as accurate as he can make it>by the unaided eye. 1. What kind of an angle is each of the following, arsJ how many degrees do you judge it measures, the magnitude a -AO B- o -A "Oi of each angle heing measured from OA to OB in a direction the opposite of that of the motion of the hands of a watch ? 2. What is the magnitude of each of th^* ^ '' n^'ag angles, ^0(7and(70J?? A B- -A B- COMPARISON OF FIQUliSS. 18 3 Draw an acute angle ^OA Bisect it. Draw another, and trisect it. ' , ■— A trisect u'^*'' *"" ''^^'''^ ''''^'''" ^'^''''^ ''^' ^'^"^ *''^*'^«^ "^^ 6. Draw an angle of 176° and bisect it. trisect u''*'' ^ '^'^'^^^ ''''^^' ^"""^ ^''"'^ '^' ^'^^ ^"^"^^^ »»d 7. Draw a reflex angle and bisect it on the convex side. Then bisect the conjugate angle on the other side. Estimate the number of degrees in each of the angles thus formed. 8. Here are seven straight lines going out from the same point and making equal angles with each other. Now draw five other figures formed respectively of 6, 5, 4, 3, and 2 straight lines going out from the same point and making equal angles with each other. How many degrees m each angle thus formed ? 9. Draw, by the eye, angles of 60°, 90°, 120°, 160°, 210°, 240°, 270°, 300°, 330°. Comparison of Geometric Figures. 89. The only way in which we can decide whether two magnitudes are equal or unequal is by applying n I u ' ■ ( .( Ilillillili Mil 14 GENERAL NOTIONS. one to the other, or applying some third magnitude to both. We are to think of the geometric figures as mBdo of per- fectly stiff lines which can be picked up from the paper and moved about without bo:iding or undergoing any change of lorm or magnitude. If two straight lines are to be compared we ^ay one upon the other, and find whether the two c ds can be made to coincide. If so they are equal ; if not, unequal. "We may also take some measure (a scale of equal parts, for example) and apply it first to one line and then to the other. If two planes are to be compared, they may be applied without change to each other. If they are of different shapes, one may be cut to pieces and the parts laid upon the other. If the latter can thus be exactly covered, the two surfaces are equal ; if more than covered, the first is the larger ; if not covered, the second is the larger. Solids are compared by finding whether they will fill the same space, one or both of them being cut to pieces if necessary. But the geometer does not actually apply his figures to each other, but only imagines them so applied. He is thus able to learn things which are true of all figures of a certain kind, whereas by actual measurement one can only learn what is true of the particular figure which he measures. This will be better understood when it is seen how theorems are demon- strated. Trace of a Fig^ure. 30. When we imagine a figure moved away, we may also imagine that it leaves its outline fixed upon the paper. Such an outline is called a trace. The trace will occupy exactly the position which the figure itself occupied before being moved, will be equal to the figure in every respect, and will be repre- sented by the drawing of the figure. If another figure is found to coincide with the trace, it will be identically equal to the first figure. Since figures are supposed to be movable, the beginner may grasp the relation between a figure and its trace by imagining that he marks around the figure with a pencil. Then when the figure is taken away the marks will remain. aTMMETBF. 15 CHAPTER III. OF SYMMETRY. Symmetry with Respect to an Axis. 31. Let us take the figure in the murgin, turn it over, and put It back so that the line PQ shall fall in its original position, but the figure shall bo turned right side left. It will then fall into the position ^ represented by the dotted lines "J'o make this change of jiosition we may suppose the ends of the line PQ io be pivots, so that the figure can turn on this line as on an axis. In turning one side of the figure must be sup' An unsymraetricai figure. P^^^^^ ^^ sink below the paT)er and the ^Gremair.unmove.l.""""' '" "^ "'"''" '*' ^^"« '''^ "^ If we take this figure and turn it over on the axis PQ, the right side will tall on the trace of the left side, and "«« versa, so that the figure will oo cupy the same lines on the paper as be- fore It was moved. Such a flgare is said to be symmetrical with respect to nWon-" ^'"""' ""^ ^"^^""^'k defi- »«f ^' J^f- .^ *»"'* '» said to be Q tw ' ^^S turned over on ""^ *" «"<' "^'^ ^• I'lT,*^?"' every part of the fiimre i« ,> ti-g ..«-^«- symmetry "-evolution is called an axis of ! , 1, 1 \Vl f! n It J 11 I! ; it ' I I 16 GENERAL NOTIONS. BXBRCISBS. 1. Copy the following figures. Then imagine each one turned over so that the line ^i^ shall be changed end for end, and draw dotted lines showing where the rest of the figure would fall. E- — ^r E' — F E- -FE / E^ — ^' J E- -F E- -F E -F 2. Let each of the following figures be turned over on the line PQ us an axis. Then draw dotted lines showing where the figure will fall. 3. Draw the axis of symmetry of each of the following figures. If there is more than one such axis, draw them all. 4. How many axes of symmetry can be drawn to a circle ? aYMMETRT. 17 Symmetry with Respect to a Point. ,33. We may next suppose that the figure, instead of being turned over IS turned half way round on a fixed point without leaving the paper. For instance, suppose the an A flgupe unsymmetrloal with respect (o the potat 11. neied figure to have a pin stuck through it at the point M and to be umed half round on that jln. It will then tal« up the position shown by the dotted outline. ' „,i T!™n '''' ''«"'* •"*" ^"y "^""""J in *e same way, everv part of It wiU occupy the position ^' ^ which the opposite part occupied before the motion, and the position of the figure will be represented by the same drawing. Such a figure is said to be symmetrical with respect to the point M. Hence the following definition: 34. D^. A figure is said to .<~-^». «. ««> point « h?rn!?r u'**''^ "^^ ""P*"* *» » point when, being S is SL'°r'' "\^j^ p°^"*' «-ry P-' of th« A figure symmetrical with respect to the point M. to a circle ? I h iiiij 1 18 GENERAL NOTIONS. In the latter the figure does not leave the paper, but simply turns on it without turning oyer. Every part of the figure changes places with the part which is at an equal dis- tance on the other side of the pivot point. EXERCISES. Copy the following figures. Then suppose them turned half way round on the point M, and draw dotted lines show- ing where the figure will fall. M M M M ^/ZA > » > CHAPTER IV, LOGICAL ELEMENTS OF GEOMETRY. Definitions. 35- Def. A proposition is either a statement that something is true, or a requirement that somethiug shall be done. A proposition affirming something to be true may be either an axiom or a theorem. 36. Def. An axiom is a statement which we as- sume to he true without proof. For the axioms of geometry we try tc ^ake propositions which are self-evident and so need no proof. AXIOMS OF OEOMETRT. 19 37. Def. A theorem is a statement which requires to be proved. A proposition requii^ng something to be d(Jne may be a postulate or a problem. 38. Def. A postulate is something which we sup- pose capable of being done without showing how. 39. Bef, A problem is something which we must show how to do. 40. Def. A demonstration is the course of rea- soning by which we prove a theorem to be true. 41. Def. A corollary is a theorem which follows from some other theorem. 43. Def. A lemma is an auxiliary theorem, to be used in demonstrating some other theorem. 43. Def. A soholium consists of remarks upon the application of theorems. Axioms of Geometry. 44. Axioms of magnitude in general. Axiom 1. Magnitudes which are each equal to the same magnitude are equal to each other. Symbolic expression of this axiom. From Magnitude X = magnitude A, Magnitude Y = that same magnitude A, we conclude Magnitude X= magnitude Y. Ax. 2. If equals bemadded to equals, the sum will be equal. Symbolic expression. From we conclude and X= Y, A = B, T -^ V . r> jj ■XT r J- A-\-Y=B^X. j 1 ; i 1 i ^ 1 t ; ■ i i 1 i 1 : ! : i 1 iii }' ii i ' it l! I!! I 20 OBNBRAL NOTIONS. Ax* 3. If equals be subtracted from equals, the remainders will be equal. Symbolic expression. From ^ ^ we conclude A z= B, X-A==r-B. Ax. 4. Similar multiphjs of equals are equal to each other. Symbolic expression. If n be any number, then from X==A we conclude n times JT == /i times A, This may be regarded as a corollary from Axiom 2. Ax. 5. Similar fractions of equal magnitudes are equal. Ax. 6. If equals be added to unequals, that sum will be the greater which bias been obtained from the greater magnitude. Symbolic expression. From we conclude A > B, A-{-x> B^ r. Ax. 7. If equals be subtracted from unequals, that remainder will be the greater which is obtained from tne greater minuend. Symbolic expression. From we conclude A > B, X=Y, A- X> B- r. Ax. 8. The whole is greater than its part. DEMOIfSTBATION OF THEOREMS. gj 45. Axioms of geometric relation. Ax 9. A straight line is the shortest distance bfi tween any two of its points. ^isrance be- Ax.10. If two straight lines coincide in two or moy mts, they will coincide throughout their whole asiSS. ^^^^^^^^^----tersectinonly Ax. 11. Through a given point one straight lin« can be dmwn and only one, which shall be Mel to a given straight line. P^^ranei to The Demonstration of Theorems. [encJto a'C^" " ^^^'""^ *» demonstration by refer- the'hj^S*'on^hrot^ 'r". "f' "»' eo^espondto fnlfiUingthecondiS ' *™'^ P°^'"« ««"«> designate ttom, mZv a«, ™^^ ,^1"' ^°' '""'^ ""'y "^ «««« to (reader. • " ™y ««« supposed to be conceiyed in tlie mind of tlio thaf'ti./vil^^? *^° propositions are so relatfi,? 22 GENERAL HOTIONS. il m nilLhi .:! li lit ill liiiiiii "■ Theorem I.* 48, A straight line can he bisected in only a single point. Here the hypothesis supposes that we take any straight line whatever and bisect it. p To enunciate the hypothesis we call a o b one end of the line A and the other end B, and the point of bisection 0. Then the hypothesis means that the point is equally distant from A and B. The conclusion asserts that there is no other point than on the line which is equally distant from A and B. The proof is ^fEected by showing that to suppose any other point having this property is impossible. If there is such a point, call it Py and suppose it between A and (because we may call either end of the line A), Let us then suppose that PA is equal to PB. Because P is between A and 0, AP will be less than A 0, Because OB is by hypothesis equal to OA, PB, which is greater than OB, will be greater than OA. Therefore, if we suppose PA and PB equal, PA will be greater than OA and less than OA at the same time, which is absurd. Therefore there is no point on the line except which is equally distant from the ends of the line. Theorem II. 49. ^ straight line is symmetrical with respect to the perpendicular passing through its middle point Hypothesis. AB,a straight line; 0, its middle point; PQ, a perpendicular passing through 0. * These simple theorems are presented partly as exercises and explana- tions for the beginner, and partly as the- basis of subsequent theorems. The demonstrations are not necessarily to be recited in full as given, but the student should be encouraged and assisted in stating the sub- stance of the reasoning in his own language. DEMONSTRATION OF THEOREMS. 23 appose any -B Conclusion, The line AB is symmetrical with respect to the axis PQ. By reference to the definition of symmetry, § 83, the conclusion is found to mean that if the P line AB be turned over on the line P^ as an axis, it will fall on its own trace; that is, into its original position; being merely changed end for end. Demonstration. Sup- pose the line turned over on the axis PQ, By hy- pothesis and definition the angles FOB and POA are ^ equal. Therefore, after the nA -n * 11 • . X. ^*"® ^^ turned over, the side OAmW fall into the position OB, and vice versa. (8 14) Because the lengths OA and OB are equal (by hypothesis ] the point ^ will fall on 5, and vtce versa. So the conclusion is proved. Exercise for the pupil Prove in the same way that the line AB is symmetrical with respect to the point as a centre of symmetry (§ 34). Theorem III. 50. All straight angles arid all right angles are equal to each other, y ^/c To prove the first part of this proposition it is sufficient' to show that any two straight angles we choose A— Q g to take are equal. The hypothesis will ^ be that we have any two M ^ __j^ straight angles which we may call A OB and MQN. By the definition of a straight angle the hypothesis will mean that 04 and OB go out from in opposite directions nM /^?. '' * '*^^^^^* ^^^®' »^'=ang.^ Jf JV, therefore Line M'A'~ trace iVZ>; ) Line JV'B' = trace MaI\ (§ ^*) 3 Therefore the whole line A'B' will fall upon CD, and CD' upon AB (§45, Ax. 10). 4. Suppose, if possible, that the lines AB and CD, when produced, meet in the direction B .and D. Then, when the figure IS inverted the liies A'B' and C'D' will meet in the direction B' and D\ Because the new and old figures coin- cide when applied,^^ and i>Cmust also, when produced, meet m the direction ^ and Cas well as in the direction B and D 5. But the two straight lines AB and CD cannot meet each other m two points (§45, Ax. 10, Cor.). Therefore they do not meet on either side. Therefore they are parallel, by definition (§ 9). Q.E D ^oroUary 1. If any two corresponding angles, as C^Y and AMJV, are equal, then, because J/JVZ) is opposite to CNV it IS equal to It (Th. L), and the alternate angles MND and ^^jYare also equal. Hence— ...^*^' ^^^"^ i^'^^^^^^^rsal crossing two straight lines makes any two corresponding angles equal, those lines will be parallel 1:1 il 1 ( iu 'I iil ilii m m I i MSI -!• 30 BOOK IL RECTILINEAL FIGURES. 70. Corollary 2. Any two perpendiculars to the same straight line are parallel. For such straight hne is a transversal crossing the two perpendiculars, and making the angles all right angles. Theorem III. ^1, If a transversal cross two parallel straight lines, the four alternate and corresponding angles are equal to each other, and the other four are each equal to the common supplement of the first four. Hypothesis. XF, a transversal crossing the parallel lines AB and CD in the points and Q^ and forming with them the four alternate and corresponding angles a, a', a"f a'", and the other four alternate and corresponding angles h, V, b tt \nt .ttt \ttt Conclusions. 1. Angle a = angle a' = angle «" = angle a' II. Angle b = angle b' = angle A" = angle b' III. Any angle a -f- any angle b — straight angle. Proof. If the alternate angles a' and a" are not equal, draw through the line A'B', making the angle ^'0^ equal to its alternate angle OQD. Then — I. Because the alternate angles are equal. Line yl'^' II Hne CD. ' (§68) . But AB II CD, by hypothesis. (§67) Q.E.D. Q.E.D. (§51) RELATIONS OF ANGLES. 3^ Therefore we should have passing through two straight hues AB and A'B' each parallel to CD, which is impossible (§ 45, Ax. 11). Therefore Angle a' = angle a". Also, Angle a = angle a'. \ Angle a'" = angle a", f Therefore Angle a = angle a' = angle «" = angle a'". II. In the same way we may prove that Angle b = angle b' — angle b" = angle *'". III. Because ^ 0^ is a straight line. Angle a + angle b = straight angle. ( But all of the four angles a and b are equal. Therefore Any angle a + any angle b = straight angle. Q.E.D. 72. Corollary. If a line be perpendicular to one of two parallels, it will be perpendicular to the other also. Theoeem IV. 73. The sum of the three interior angles of a tri- angle is equal to a straight angle. Hypothesis. ABC, any triangle. Conclusion. Angle A + angle B -f angle C = straight angle. Proof. Through C draw a straight line MJV parallel to the opposite side AB. Then — 1. Because CA is a m _^ transversal between the /s: - - W parallels AB and MN, Angle J = alt. angle MCA (§ 72). 2. Because CB is a transversal between the same parallels, Angle B = alternate angle BCJV. 3. Angle C = angle A CB (identically). 4. Adding these three equations, Auffle^ 4- anorlo /? 4- nno-lp C— Mf^ a _]_ Arin \ T>n\r ~ angle MCN, = straight angle. 32 BOOK II, RECTILINEAL FIGURES. i I Therefore Angle A-\- angle B + angle C = straight angle. Q,E.D. 74, Corollary 1. If two angles of a triangle are given, the third angle mag be found by subtracting their sum from 180°. "75. Corollary 2. If two triangles have two angles of the one equal respectively to two angles of the other, the third angles will also be equal. EXERCISES. 1. If a triangle has two angles each equal to 60°, what will be the third angle? 2. If one angle of a triangle is a right angle, and one of the remaining angles is double the other, what will be the value of these two angles? 3. Prove that a triangle cannot have more than one right angle. Theorem V. 76. Each exterior angle of a triangle is equal to the sum of the two interior and opposite angles. Hypothesis. ABC, any triangle. /), any point on AB produced. Conclusion. Exterior angle CBD = angle A + angle C. Proof. 1. Because ABB are in one straight line, Angle B + exterior angle CBD = straight angle. 2. Angle B + angle A -f an- gle C = straight angle. (§ 73) 3. Cc':aparing (1) and (2), Angle A + angle B + angle C = angle B + ext. angle CBD. 4. Taking away the common angle B, Angle A + angle C = exterior angle CBD. Q.E.D. 77. Corollary. Any exterior angle of a triangle is greater than either of the interior and opposite angles. 78. T)ef. Two •nnrnllpl linpa. pa,r»h DTiinoc c\y\t ^rc\xn a point, are said to be similarly directed or oppositely lli RELATIONS OF ANGLES. 33 directed according as they go out in the same direc- tion or in opposite directions fioni their startine- points. ° Theoeem VI. 79. If the two sides 0/ one angle are respectwelv parallel to the two sides of another, and similarly directed, these angles are equal. A S «'^1 ^r? """"f'' ^^^ ^'^^ ^^^ h^^i^g the sides a J«n f ^ParaUel and similarly directed, and the sides ^P and^g also parallel and similarly directed. Condmion. Angle JVBQ = angle MAP. rroof. Produce the side JVB, if necessary, in either dn^ction nntil it shall intersect the side AP o the her angle, also produced if necessary. Produce NB past B to any point 8. Let C be the point of intersection of JSTB and AP. Then— crosslnftrm! ""^ " '"''''' '' ""^^ '"' ^ "^ " ^ ''''''''''''' Angle Jf^P = alternate angle ^C^. (§ 7i> Angle iV5^ = angle A OJS, 3. Comparing (1) and (2), Angle iV:^^ = angle Jf^P. Q.E.D. the other BiiMh^^ '' *^' °°' ^°^^" ^« ^^ ^^^ "<^t <^o^tamed within otner. But the same reasoning can be applied to both. n.f^' ^'?^^^^^^y 1- Jf t^e sides of ttoo anqhs are varalM and oppositely directed, the angles will l>e equT ^ '^ 84 BOOK II. RECTILINEAL FIOUHES. .1 ,(, B 81. Corollary 2. If the sides are parallel and the one pair are similarly directed but the other pair oppositely directed, the angles will he supplementary, ^ Theorem VII. S2. TJie Usectors of two adjacent angles on the same straight line are perp^ dicular to each other. \ Hypothesis. BOO smd CO A, adjacent angles on the straight line ^i?; A, OX, the bisectors of these ^ angles. ^ Conclusion. OL J. lO, that is, ZOK= right angle. Proof. 1. By hypothesis, ^ Angle COir= | angle BOO. Angle COL = ^ angle COA. 2. Therefore Angle A^OL = COR + COL, = i{BOC-i-COA), = i straight angle BOA, = right angle. Q E D Corollary. Since an angle can have but one bisector, we conclude: ' tn ifh ^/^'^V.^^^7^^, ^^' ^^^^^^ Of an angle, perpendicular to the bisector, bisects the adjacent angle. Theorem VIII. sa^fltrXMl^' '^'"'' <^i'-.V.««^fe. arc intke Hypothesis. AOL, BOC, two ^ opposite angles formed by the > / straight lines AB and CD; OJ, ^ the bisector of BOC; OK, the Msec- * tor of A OB. y^ B Conclusion. OJ and OK are in ^. the same straight line. Proof. 1. Because the angles RELATIONS OF ANGLES. 35 BO J and AOK mq halves of the equal opposite angles BOGy AODy they are equal (§67). 2. Angle AOJ = angle AOB - angle BO J. Angle KOJ = angle AOJ -\- angle yi Oif, = straight angle AOB - BOJ-^AOK, Or, 30J and ^OA" being equal, Angle KOJ = straight angle, and the lines O^and- OJ are in one straight line. Q.E.D. 86. Corollary. The four bisectors of the four angles formed by two interseating straight lines form a pair of straight lines perpendicular to each other. EXERCISES. 1. If, in the diagram of Th. II., angle BMX = ^b°, what will be the values of the other seven angles of the figure? 2. If, in the diagram of Th. VIII., angle BOJ = 20°, com- pute the values of the angles CO Ay AOD, and DOB. 3. Draw a triangle ABCy and suppose angle A = 60°, angle B = 40°. What angle will the bisectors of these angles A and B form with each other? Note. Apply Th. IV. to the triangle formed by the bisectors and the side AB. 4. What will be the values of the three exterior angles of the preceding triangle 4BC? 5. If, in the preceuing triangle ABO, the bisector of the angle A be continued until it cuts the side BC, what angles will it form with that side? 6. Compute the same angles algebraically in terms of the angles AyBy and (7, and show that the difference between the two adjacent angles formed by the bisector with the side BO is equal to the difference between the angles B and C. 7. If, in a triangle ABO, exterior angle A = 85° and ex- terior angle B = 150°, what will be the three interior angles? if II IP ■ i; mi llr I i III CHAPTER II REUTIONS OF TRIANGLES. Equilateral triangle. Isosceles triangle. Definitions. 86. Definition. An equilateral triangle is one in which the three sides are equal. Def. An isosceles triangle is one which has two equal sides. Def. An acute-angled tri- angle is one which has three acute angles. Def. A right-angled triangle is one which has a right angle. Def. An obtuse-angled triangle is one which has an obtuse angle. 87. Def In a right- angled triangle the side opposite the right angle is called the hypotlie- nuse. Blght-angled triangle. Obtuse-angled triangle. 88. Def When one side of a triangle has to be distinguished from the other two it is called a base, and the angle opposite the lase is called the vertex. Either side of a triangle may be taken as the base, but we commonly take as the base a side which has some distinctiTe property. In an isosceles triangle the base is generally the side which is not equal to another. In other triangles the base is the side on which it is sup- posed to rest. IS one m RELATIONS OF TRIANGLES. 37 89. i>(?/. An obUque line is one which is neither perpendicular nor paraUel to some other Hne. 90. JD^ Segments of a straight line are the parts into which it is divided. Theorem IX. 91. In an isosceles triangle the angles opposite the equal sides are equal to each other. Hypothesis. ABC a, triangle in which CA = CB. Conclusion. Angle A = angle B. Proof. Bisect the angle C by the line OD, meeting AB in B. Turn the tri- angle oyer on CD as an axis. Then— 1. Because, by construction, Angle BCD = A CD, Side CA = trace CB, and vice versa. 3. Because, by hypothesis, CA = CB, Point B = position A, ^^^ Point ^E position^. 3. Therefore line AB = traxie BA, being turned end for end. 4. Therefore angle CU^e trace CBA. 5. Therefore angle CAB = angle CBA (§ 14). Q.E.D. Corollary 1. Since, after being turned over, the triangle lalls upon its own trace, the triangle is symmetrical with respect to the bisecting line. Hence— 93. The bisector of the vertical angle of an isosceles tri- angle Is an axis of symmetry, and bisects the base at right angles. ^ 93. Cor. 2. Every equilateral triangle is also equiangular. Theorem X. 94. Conversely, if two angles of a trianale are mal,^ the sides opposite these angles are equaL and trie tnanale i.t iananplfis^ the triangle is isosceles. 38 BOOK 11. REOTILINEAL FIQUBE8. Hypothesis. ABC, a triangle in which angle A = angle -ff. Conclusion. Side CA = side CB. Proof. Through the middle point /> of the side AB pass a perpendicular, and turn the figure over upon this perpen- dicular as an axis. Then — 1. Because the axis bisects AB per- pendicularly, Point A = position B. ) ,„ .q. Point B = position A. ) ^» ^^ Therefore AB = trace BA. 2. Because angle A = angle B, Side AC- trace BC Side BC=tTace AC. 3. Therefore the point of intersection C7will fall into its original position. 4. Therefore ^C=jrC. Q.E.D. 95. Corollary 1. A line Usecting the hose of an isosceles triangle perpendicularly Passes through its vertex, Is an axis of symmetry, and Bisects the angle opposite the base. 96. Cor. 2. Bvery equiangular triangle is also equilateral Theorem XI. 97. If in any triangle one side be greater than another, the angle opposite that side will he greater than the angle opposite the other. Hypothesis. ABC, Vk triangle in which AB> BC> CA. Conclusion. Angle C > angle A > angle B. Proof. On a greater side, CB, take CD equal to a lesser side, ^^ CA, and join AD. Then — 1. Because the line AD falls within the triangle, Angle CAB > angle CAD. HELATIONS OF TRIAKGLES, 2. Because CA = CDy Angle CAD = angle CDA. 89 (§03) 3. Because CDA is an exterior angle of 'the triangle AlS A 1? ^^^^^/^ODA> angle ABD, (iii\ from Ir ^^ ^'^ "' '"' "^'^ '^^^ > ^^^' -S Angle GAB > angle ^^Z>. m tne same way may be shown. Angle C> angle ^. Q.E.D. Theorem XII. 98. CouTiersely, if one angU o^ a triavaJp h. Hypothesis. ABC, a triangle in which -^^gle O angle ^ > angle ^. ' ^^Proof. Prom(7draw CD, making angle ACD = angle (7^ A 1. Because angle ACD z= angle CAD, AD = CZ>. ' /£> q;.x or, from (1), AB^CD + DB, 3. Because CB is a straight J hne, ^ ^ 1^ -^B 4. Therefore, from (2), 0. in the same way may be shown BC>AC. Q.E.D. (Ax. 9) the order ofmagmtude of their opposite angles. the 40 BOOK 11. liEOTILINEjiL FIOUIUCS. That Is, if Angle A > angle B > angle C, then Side BC> side AG> side AB, and vice versa. Theorem XIII. 100. If from any point within a triangle lines he drawn to the ends of the hase^ the sum of these lines will he less titan the sum of the other two sides of the triangle, hut they will contain a greater angle. Hypothesis. ABC, any triangle; P, any point within it. Conclusions. I. AF -{■ FB < AC -^ CB. II. Angle AFB> angle ACB. Froof. Continue the line AF until it meets the side CB in Q. Then— (1)1. AQ angle A CB. 6. Because AFB is an exterior angle of the triangle FQB, Angle AFB > angle FQB, 7. Comparing (5) and (6), Angle AFB > angle A CB. Q.E.D. :iih::ii RELATWm OF TRIAmLES. 41 Theorem XIV. 101. HVom a point outside a straight line only one perpendicular can be drawn to such straiqht Line, and this perpendicular is the shortest distance from the point to the line. Hypothesis. AB, any line; P, any point without it; PO, u perpendicu- liir from P on AB] PQ, my other line from P to A B. Conclusions. I. PQ is not perpendicular to AB. II. PO < PQ. Proof. Turn tlie figure over on AB m [in axis, so that the jmint P shall fall into the position P'. Then— 1. Because AOP is a right angle, AOP' is also a right angle, and POP' lie in a straight line (§ 60, Cor.). 2. If PQA were also a right angle, it could be shown in the same way that PijP' is a straight line, and there would be two straight Imes POP' and PQP' having the points P and P common, which is impossible (§ 45, Ax. 10, Cor.). Iherefore PQA is not a right angle. Q.E.D 3. Because PO = OP', and PQ = QP% we have PO = iPP';PQ = ^^PQ.^Qp.^^ But pp' OQ. Conclusions. I. PR > PQ. II. Angle PRO < angle PQO. Proof. Turn the figure over on ^P as an axis. Let P' be the point on which P shall fall. Then— 1. Because P'Q = PQ, and P'P = PR, PQ = UPQ + QP'). PR = \{PR-\-RP'). 2. If Q is on the same side of with R, the point Q will be within the triangle PRP'. Therefore ppj-ppf'-^pnj-np' and, taking the halves of these unequal quantities. PR > PQ. Q.E.D. RELATIONS OF TRIANQLB8. 48 3. If ^ is on the opposite side of from R, take another point Q on the same side. The two lines PQ wiP then be equal to each other (§ 102). The second point Q will fall between and i2 (hypoth.). 4. Therefore we shall still have PR > PQ. Q.E.D. 5. Because PQO is an exterior angle of the triangle PQR, Angle PRQ < angle PQO (§ 77). Q.E.D. I 9 Theoeem XVIL 104. I. Mery point on the perpendicular bisector of a straight line is equally distant from the extremi- ties of the line. II. Every point not on the perpendicular bisector is nearer that extremity t(mard which it lies. Hypothesis. AB, a straight line; 0, its middle point; OP, a perpendicular from 0; P, any point on this perpendicular; Q, a point on the same side of the perpendicular with B. Conclusions. I. PA = PB. II. QB < QA. / Proof. 1. Because PO 1. AB, A^ and ^ = OBf we have PA = PB. (§ 102) 2. From Q drop a perpendicular QO' upon AB. Then QO' II PO. (§ 70) Therefore 0' falls on the side of toward B, and O'B < 0A\ Therefore QB < QA (§ 103). Q.E.D. 105, Corollary. Every point equally distant from the extren. :Hes of a line lies upon the perpendicular bisector of the line. For, if it did not lie on this hisenf.nr if. nnniA r./xf r^« equally distant from the extremities without violating conclu- sion II. of the theorem. 1 •.. O' \ \ :i B I' I m 1 11, 17 m liliii/i 44 BOOK II. RECTILINEAL FIGURES. Theorem XVIII. 106. EGery point in the bisector of an angle is equally distant from the sides of the angle; and every point within the angle, hut not on the Usector is nearer that side toward which it lies. ' Hypotlitsis, MON, any angle; and OQ, its bisector, so that angle MOQ = NOQ ; P, any point on OQ-, T, a point within the angle, but not on OQ-, PR, FS, per- pendiculars to OM and ON; TU, TV, perpendiculars from T to OM and OJV^. Conclusions. I. PR = PS. 11. TU> TV. Proof I. Turn the figure over on the axis OP. Then— 1. Because angle PON= POM, Line 0M= trace ON, and vice versa. 2. Because PR is a perpendicular dropped from P on R, PR = trace PS. (Th 101) 3. Therefore Point R = S,md. \ • J PR = PS. Q.E.D. Proof II. Let Q be the point in which TU crosses the bisector OQ. From Q drop the perpendicular QW nvon ON Join TW. Then— 4. Because TV is a perpendicular and TWm oblique line to ON, TW>TV. (§101) 5. Also, TQ^QW>TW. (Ax. 9) Or, because QU=QW (I.), ^ ^ TQ^QU=TU>TW. 6. Comparing with (4), TU> TV. Q.E.D. 107. Corollary. Every point equally distant from two non-parallel lines in the same plane lies on the Usector of the angle formed hy those lines. RELATIONS OF TBIAmLEa. 45 Theorem XIX. .7. i^^* ^ two triangles have two sides and the in- eluded angle of the one equal to two sides and the included angle of the other, they are identically equal. Hypothesis, -4.5 (7 and MNP, two triangles in which PM^ CA. PJ^= CB. Angle P = angle O. Conclusion. The two tri- angles are identically equal. H SZ'-^' .^^^^J *.^' ^""^^^^^^ ^^^ *« ^^O in such manner tha the vertex P shall fall on C, and PM on CA. Thr~ 1. Because PM = CA, Point if = points. 3. Because angle P = angle C, Side PN~CB. /g 1,^ 3. Because PN= CB, ^^ ' Point JV^= point 5. 4. Because M=A and N= B, Line MN~ line AB. (§ 45, Ax. 10, Cor. ) Therefore every part of the one triangle will coincide with the corresponding part of the other, and the two trT angles are identically equal by definition (§13). Q. E. D. Theorem: XX. .^/^^; ^ ^T ^l^'^^'Oles haw a side and the two adjacent angles of the one equal to a side and the tual ' ^^' ''^^'^' ^^''^ ""^^ ^^e7^^^ca% Hypothesis. ABCmd. MNP, two triangles in which AB = MJV. a »% /v I r\ A — — — 1 _ H .T" Conclusion. Angle B = angle N. The two triangles are identically equal. 46 BOOK II. RECTILINEAL FIQUBES. Proof. Apply the triangle ABO to the triangle MNP in such manner that A shall coin- cide with Mf and AB with MN, Then— 1. Because AB = MNj Point B = point iV^. 2. Because angle ^ = angle M, Side AC = aide MP. ^^ 3. Because angle B = angle JV", Side J?(7= side iVP. 4. Because the sides A and ^C fall upon MP and iVP respectively, the vertex C will fall upon the vertex P. Therefore the two triangles coincide in all their parts and are identically equal. . Q.E.D. Theorem XXI. 110. If two triangles have the three sides of the one respectively equal to the three sides of the other, they are identically equal, and have the angles op- posite the equal sides equal. Hypothesis. Two tri- angles, ABCm.dDEF, in which AB = DK BG = EF, CA = FD. Conclusion. The two triangles are identically equal. Proof. Take up the triangle ABC and apply the side ^i5 to the equal side DE oi the other triangle, letting the vertex C fall on the opposite side of DE from that on which the triangle DEF lies. Let C" be the point in which the vertex C falls. Join FC. Then— 1. Because FD == AC, wid DC — AC, it follows that !/ %' RELATIONS OF TRIANGLES. 47 FD = DC, and the triangle FBC is isosceles. Therefore Angle DFC = angle DC'F, (§ 91) 2. For the same reason the triangle FFG' is isosceles, and Angle BFC = angle FC'F. 3. Adding the equations (2) and (1), we find Angle BFC + angle HFC = angle I)C'F-\- angle BC'F. But Angle DFO' + angle BFO' = angle i9i^^. Angle DC'F-{- angle EC'F= angle />C"^. Therefore Angle DC'iS' = angle DFE. 4. Butangle^CJ5 = angle Z)C"^, by construction. There- fore Angle ACB = angle i)i?!£'. 5. The two giyen triangles, having the angle C= angle F and the sides which contain these angles equal, are identically equal (§108). Q.E.D. ^ Theoeem XXII. 111. If two triangles agree in the lengths of two sides and also in the angle opposite one of these sides, the angles opposite the other of the equal sides will be either equal or supplementary, and if they are equal the triangles are identically equal. BD E D E P Hypothesis, Two triangles ABC and DBF, in which CA = FD. CB = FK Angle A = angle D, Conclusion. Either angle E = angle B or Angle E = straight angle — angle B, and in thfl formfir ostap. t.liA twn frianorlfifl aro irlAyifi/Jollir AiNnol Proof. Apply the triangle DEF to the triangle ^^C in such manner that DF shall fall on the equal side A C. Then i|; 'h 48 BOOK II. nECTILINEAL FIGURES. Kl \ B D E D 1. Because i>i^= ^C, PointDE^; point i^E a 2. Because angle D = angle A, Base I) E= base A B. 3. Because F.E = CB, the point B will fall on a point of the base AB which is at a distance from C equal to CB, 4. There will be two such points equally distant from the foot P of the perpendicular from C on AB (§ 103). Because FF! = CB, one of these points will be B. Let B' be the other. 5. If iSf falls on 5, Triangle ABC= triangle DBF, identically. 6. If B falls on B', then, because CB' = CB, the triangle CB'B will be isosceles, and Angle CB'P = angle CBP, (§91) 7. Because AB'P is a straight line. Angle CB'A = supplement of angle CB'B, = supplement of angle CBP. (6) 8. But in this case angle CB'A = angle B. Therefore Angle B — supplement of angle CB'P, = supplement of angle B (7). Q.E.D. 113. Corollary. If the triangle ^5C should be right- angled at B, we shall have Angle B = straight angle — angle B, and the two possible angles B would have the same value. Hence the two triangles would then be identically equal. Scholium. 113. The nrecedinsr theorems of the idftutitv of triaticyl«g are also expressed by saying that when certain parts of a RELATIONS OF TRIANQLES. 49 A hinged triangle. triangle are given, the other parts are determined. T>he parts of a piano triangle are the three sides and the three angles. The three angles ai'e not all independent, because whenever two of them arc given the third ftiay be found by subtracting their sum from a straight angle (§ 74). Whenever three independent parts of a triangle are given the remaining parts may be found. In other words, when three independent parts are given there is only one triangle (or, in the case of § 111, two triangles) having those parts. When the three sides of a triangle are given, we may imagine ourselves to have three stiff thin rods which we can fasten end to end in the form of a tri- angle. When the angles are not given, we may suppose the rods to be fastened together by hinges at the angles. Theorem XXI. shows that al- though the hinges may be quite free the rods cannot turn upon them when linked together. If they could turn, we could make the rods into several triangles by turning the rods on the hinges, and these tri- angles would not be identically equal. When two sides and the included angle, as AC, BO, and the angle (7 are given, which is the case corresponding to rheoi;em XIX we must suppose the side AB removed and the hinge at C tightened, so that the two rods cannot turn, and we are required to find a third rod of such length as to fi into the space AB. Theorem XIX. shows that this rod must have a definite length. Suppose next, as in Theorem XXIL, that AC, BC, md the a^gle A are given, while the base AB and the angles B and C are not given. We may then suppose a long rod extending out from A. The side AC of given length must be fastened at A, and the hinge tight- ^ „ ened so that AG cannot turn, becaule the angle '^ is given. Bf'--^ -- f' ! 50 BOOK II. RECTILINEAL FIGURES. ,i ^i Wi m X / The rod CB of given length is hinged at C, and this hinge is left loose because the angle C is not given. We are then to swing the side CB around on until the end B touches the base, when it is to be fastened and the angle well fixed. There will be two points, B' and B', where the junction may be made, and only two. We may choose which point we will, and the triangle will then be fixed. The two angles at B will, by the last theorem, be supplemen- tary. If CB should be shorter than the perpendicular from C upon /- \ AB, there would be no triangle which could be formed from the given parts. A SuppoF;e, lastly, thaj; one side AB and the two adjacent angles A and B are given, the other two sides being of in- definite length. We must then turn the two sides on the hinges and tighten the latter at the required angles, when the sides will cross each other at a definite point, and will make a definite angle with each other. This corresponds to the case of Theorem XX. ^ B i'l Theoeem XXIII. 114. If two triangles agree in the length of two sides, that triangle in which these two sides include the greater angle will have the greater base. Hypothesis. ABC&nd. DBF, two triangles in which AB = DE, BC = BF, Angle B < angle B. Conclusion. Ba.8e DF> hase AC. Proof. Apply the side ^^ of the one triangle to the equal side DF of the other in such manner that B shall fall upon F, and A upon D. Let C be the position in which G falls. Bisect the angle C'FF, and let iVbe the point in which the bisector meets the base DF. Join C'iV. Then— RELATIONS OP TRIANGLES. 1. In the two triangles C'^JVand FEN, Angle NEC = angle NEF, by constrnction. Side EC = side EF, by hypothesis. Side E2i = side EN, identicaUy. 01 Therefore triangle ENC = triangle ENF, identically, (S 110) and NG' = NF. 2. Therefore DF= DN-\- NF= DN-\- NG\ 3. Because ^ 67 is a straight line, I)N+NO'>DO\ (Ax 9) Comparing (2) and (3), * ' DF>Da% ®r DF>AO. Q.E.D. 115. Corollary. Conversely, if two triangles have two sid^softhe one equal respectively to two sides of the other, hut the third sides unequal, the angle opposite the greater of the unequal sides will he the greater. For these angles could not be equal without violating IheoremXIX., nor could the angle opposite the lesser side De greater without violating Theorem XXIII. Theorem XXIV. IX^. If three or more lines, maJcing equal angles mm each other, he drawn from a point to a straight tine, tji'^tmir^ of lines will intercept the greater length which IS farther f Torn the perpendicular. Hypothesis. PC, a straight line; 0, any point outside of 52 BOOK II. RECTILINEAL FIOUBES. \ it; OAy OB, OGy three lines from to PC, making angle A OB = angle BOC\ OP, the perpen- dicular from upon P, Conclusion. The intercept BG, on the side of OB away from the perpen- dicular, will be greater than the inter- cept AB, on the side toward the per- pendicular. Proof, From B draw the line BSy making angle OBS — angle OBA, and meeting OG'in S. Then — 1. Because, in the triangles OBA and OBS, Angle BOA = angle BOS (hypothesis), Angle OBA = ajigle OBS (construction), Side OB = side OB identically, these triangles are identically equal (§ 109), and Angle OSB (opp. OB) = angle OAB (opp. OB). Side BA = side BS. 2. Therefore angle GSB (supplement of OSB) = angle OAP (supplement of OAB). 3. Because G is farther from P than A is. Angle SGB < angle OAP. Therefore Angle SGB < angle GSB, and side BG (opposite greater angle GSB) > side BS (oppo- site lesser angle SGB). (§ 98) 4. Comparing with (1), BG>AB. Q.E.D. (§ 103) PARALLELS AND PARALLKLOOIiAMS. 63 CHAPTER III. PARALLELS AND PARALLELOGRAMS. Definitions. 117. Def. A quadrilateral is a figure formed by four straight lines joined end to end. The sides of a quadrilateral are the lines which form iii. 118. Def. A parallelogram is a quadrilateral in which the opposite sides are parallel. Whenever two parallels cross two other parallels, the intercepted portions of the parallels form a parallelogram. 119. Def. ThediagO- a parallelogram. nals of a quadrilateral are two lines joining its oppo- site angles. Theorem XXV. 120. Straight lines which are parallel to the same straight line are parallel to each other. Hypothesis. The line i par- allel to the line a. The line c also parallel to the line a. a — Conclusion. The lines i and c are parallel to each other. 5 Proof. Draw any transver- / sal as ifiV^ across the three lines, " " " —70 intersecting them in the points / A, B, a N 1. Because h is parallel to a, Angle 5 = corresponding angle ^. (871) Angle O = corresponding angle A, M / / / IT 54 I^OOK IT. HEGTIUNEAL FlQUUm. 3. Comparing (1) and (2), Anglo B = corresponding angle C 4. Therefore line b || lino c (§ 69). Q.E.D. Theorem XXVI. 121. The opposite angles of a parallelogram are equal to each other. Hypothesis. A BCD, any parallelogram. Conclusion, Anglo A = opposite angle D. Angle B = opposite angle G. Proof. Continue CD to any point Jf, and BD to any point N. js Then— \ 1. Because DN is parallel to AC and similarly directed, and DM parallel toAB and similarly directed, Angle BAC= angle MDN. (§ 79) 2. Angle MDN = opp. angle BBC, (§ 67) 3. Comparing (1) and (2), Angle BDC= angle BA O. In the same way it may bo proved that Angle ^CZ> = angle ^^/). Q.E.D. -M Theorem XXVII. 123. Anp two adjoining angles of a parallelo- gram are supplementary. Proof. AiLj such pair of angles as A and B are interior angles between the parallels A C and BD, and are therefore supplementary (§ 71). 123. Corollary 1. All the angles of a parallelogram may he determined when one is given, the angle opposite to the given one being equal to it, and the other two angles each equal to its supplement. PARALLELS AND PARALLELOGUAm. 55 -M 124. Corollary %, 1/ ttvo parallelograms have one angle of the one equal }n one angle of the other, all the remaining angles of the one will be equal to the corresponding angles of the other. 125. Corollary Z. If one angle of a parallelogram is a right angle, all the other angles are right angles. Theorem XXVIII. 126. A pair of parallel straight lines intercept equal lengths of parallel transversals. Hypothesis. AB and CD, any pair of parallel straight lines ; MN, MS, parallel transverfjals crossing them at the points M, iV, E, and S. Conclusion. MN = R8. a \M ME = NS. Proof. Join the two oppo- site points E and JV by a third C— ^ \o j^ transversal EJV^, and compare the two triangles MJVE and ^ES. 1. MENS is a parallelogram, by definition. Therefore Angle EMN = angle ESJV^. (§ ng) 2. Because EJ^ ia a transversal crossing the parallels AB and CD, Angle MEN= alternate angle ENS. (8 71) Angle MNE = alternate angle NES. 3. Because the two triangles have the side EN common and the adjacent angles equal, they are identically equal (§109). Therefore Side ES (opp. angle N) = side MN (opp. equal angle E). Side ME = corresponding side NS. Q.E.D. 127. Corollary 1. The opposite sides of a parallelogram are equal. ^ 128. Corollary 2. The diagonal of a parallelogram divides it into two identically equal triangles. If the two transversals are perpendicular to the parallels ■ — -j-.'va xv,xxQt.ixo mil uiuiiiiuru me aisiance of the parallels. Hence 56 BOOK n. BETILINEAL FIOURES. i! / 1!S9. Corollary^. Two parallels are everywhere equaUy distant ! Theorem XXIX. 130. If three or more parallels intercept equal lengths upon a transversal crossing them, they are equidistant. Hypothesis. A transversal crossing the parallel lines a, h, c, d, etc., at the respective points AjB, 0, />, etc., in such wise that AB== BO = CD, etc. Conclusion. The distance between any two neighboring parallels, as «, b, is equal to the distance between any other two, as c, d. Proof. From two or more of the points of intersection A, B, 6', etc., drop perpendiculars upon the neighboring parallels and consider the triangles thus formed. 1. Because the angles A, B,C, __ etc., i.re corresponding angles be- tween parallels. Angle A — angle B = angle G, etc. (§ 71) 2. The angles X, Y, S, etc., are equal by construction, because they are all right angles. 3. AB = BG=^ CD, etc., by hypothesis. 4. Therefore any two triangles,.as ABX, CDS, have the angles and one side of the one equal to the angles and one corresponding side of another, and are identically equal. 5. Therefore AX=BY= CS, and the parallels are equi- distant. Q.E.D. Theorem XXX. 131. Conversely, Equidistant parallels intercept equal lengths upon any transversal. PARALLELS AND PARALLELOGRAMS. 67 Hypothesis. The lines a, by c, d, etc., parallel and equi- distant; MNy a transversal inter- secting them at the points^. By Cy Dy etc. Conclusion. Any one inter- cepted length on the transversal, as ABy equal to any other inter- cepted length, as CD. Proof. Take up the figure and lay it down again in such manner that the point A shall fall into the position C, and the line a coincide with the trace c. Tuen 1. Because angle A = angle C, Transversal M^= its own trace. 2. Because the parallels a, b are equidistant with the parallels c, d. Point B = position D. 4. Therefore AB = CD. Q.E.D. Scholium. This proposition may also be proved by drop- ping perpendiculars, as in Theorem XXIX. i •M Theorem XXXI. 133. If three or more parallels are crossed hy two transversals and intercept equal lengths on one, then — I. The parallels will intercept equal lengths on the other transversal. II, The intercepted part of each parallel will he longer or shorter than the neighboring intercept ^ hy the same amount. Hypothesis. GP, 0^, any e two transversals; A, By Cy equidistant points on OP; P " 'P 68 BOOK II. RECTILINEAL FIGUMES. a, b, c, d, Cj etc., parallels through these points, meeting OQ in the points A^, B', C", etc. Conclusions. I. A'B' = B'0'= CD' = D'E\ etc. II. BB' -AA'= CQ'~ BB'= DD'- CC\ etc. Proof I. 1. Because the parallels a, h, c, etc., intercept equal intervals on OP, they are equidistant (§ 130). 2. Because the parallels are equidistant they intercept equal intervals on the transversal OQ. Therefore the inter- cepted intervals A'B', B'C, etc., are equal (§ 131). II. Through A' draw a line parallel to AB, meeting BB' in K, and through B' draw another line parallel to BC, meet- ing C, C in L; then — 3. Because, by construction, ^^'J^iTand BB'GL are paral- lelograms, BK=AA',) CL = BB', S ^§ 134:) 4. Because 5'^ 'X and C"J5'Z are corresponding angles be- tween the parallels ^'JTand B'L^ they are equal. 6. Because A'B'K and B'C'L are corresponding angles between the parallels BB' and CC", they are equal. 6. Comparing with (3) the triangles A' KB' and B'LC, have one side and the adjacent angles of the one equal to a corresponding side and two adjacent angles of the other. Therefore they are identically equal, and B'K^C'L. 7. By construction, BB' -BK= B'K. ^ CC - CL = CL. Comparing with 3 and 6, BB'-AA' = B'X=CC-BB', etc. Q.E.D. 133. Corollary 1. The amount by which each length exceeds the preceding one may be found by laying off from the point of intersection 0, on the line OP, a length equal to AB, and drawing a parallel to the lines a, b, etc. The length of this parallel between the sides OP and OQ will be equal to the difference between the lengths of any two con- secutive parallels. PARALLELS AND PARALLELOGRAMS. m 134. Corollary 2. If the points A, B, (7, etc., are found by measuring off equal distances from tlie point 0, so that OA = AB = BC, etc., q^ we shall have BB'= %AA', CO' = 3 A A', DD' = 4:AA% etc. etc. 135. Corollary 3. If through the middle point of one side of a triangle a parallel be drawn to a second side, it will bisect the third side and will be half as long as the second side. For, let OCD be the triangle; A, the middle point of OC, and ABj'd, line parallel to CD, meeting OD in B. Let a third parallel pass through 0. Then 0, AB, and CD are three parallels intercepting equal lengths upon the transversal OC. Therefore they also intercept equal lengths on OD. Moreover, CD exceeds AB 2ib much as ^^ exceeds the intercepted length of 0. But this length is nothing. There- fore CD = "HAB. Corollary 4. Because through A only one parallel to CD can be drawn, it follows that if B be the middle point of 02>, AB will be that parallel. Therefore: 136. Tlie line joining the middle points of any two sides of a triangle is parallel to the third side. Theorem XXXII. 137. 7)^ the opposite sides of a quadrilateral are equal to each other ^ it is a parallelogram. Hypothesis. A BCD, a quadrilateral, in which AB = CD. AC = BD. Conclusion. The figure AB(^T) \a a. TiJi.rallolrtorpQTTi Proof. Draw the diagonal BC. Then — 1. The three sides of the triangle ABC are by construe- li 60 BOOK 11. RECTILINEAL FIGUEE8. tion and hypothesis respectively equal to those of the triangle DBC. 2. Because the angles BCD and GBA are opposite the equal sides CA and BD, Angle BCD = angle CBA. (§ HO) 3. But these angles are alternatb angles between the lines CD and ^^ on each side of the transversal CB. Therefore CD II AB, 4. In the same way may be shown AC \\ BD. Therefore ABCDisa. parallelogram, by definition. Q.E.D. .Theorem XXXIII. 138. If any two opposite sides of a quadrilateral are equal and parallel^ it is a parallelogram. Hypothesis. ABCDj a c_ ——.D quadrilateral in which CD = and || AB. Conclusion. AC= and / II BD, and therefore / ABCD is a parallelogram. -^ Prof. Draw the diagonal BC. Then— 1. Because AB and CD are parallels, and 5 C is a transversal, Angle ABC=: alternate angle BCD, 2. In the triangles ABC and BCD, AB = CD, by hypothesis; ^C is common; and (1) the angles between these equal sides are equal. There- fore these triangles are identically equal, and AC= BD. Q.E.D. Angle ACB= angle CBD. 3. T'i 30 angles ACB and CBD being alternate angles between j± C and BD, ACWBD. Q.E.D. Theorem XXXIV. 139. If two parallelograms agree in the lengths of their sides and in one angle^ they are identically equal. PARALLBLB AND PARALLELOGRAMS. Qi Hypothesis, ABCD and HKMN, paraUelograms in which A.I3 — UK. AC = HM, Angled = angled. Conclusion. ABCD is identically equal to HKMN, Proof, ApyljABCD to HKMNy turning it ^ over, if necessary, in such manner that the angle A shall coin- cide with the equal angle H, and the side AB with the equal side HK. Then — ^ 1. Because these sides are equal, Point ^ E point JT. 2. Because angle A = angle If, AC=HM. 3. Because ^C=^J!/; Point C= point Jf. 4. Because the parallelograms have the ande G = an^lfi M (§ 121), ^ ^ Side CD = MN. 5. In the same way it may be shown that every side and angle of the one will coincide with a corresponding side and angle of the other. Therefore The two parallelograms are identically equal. Q.E.D. Scholium 1. A more simple but less general demonstration may be given by drawing diagonals between any equal angles. 140. Scholium 2. We have shown that a triangle ip com- pletely determined when its three sides are given. From the preceding propositions it follows that a quadrilateral k not completely determined when its four sides are given, but chat the angles may change to any extent without changing the lengths of the sides. Thus the parallelogram in the margin may be made to assume the SnnP.PHHIVO -(nvrna ahrwirn Vjtt i-\\n Ar^i- ted lines. This is the geometrical expression of the well-known fact It iU/ g-ff'iii ji,[(n n "jinnBaj BOOK II. RECTILINEAL FIGURES. thafc a frame of four pieces is not rigid unless fastened by a diagonal brace. Theorem XXXV. 141. The diagonals of a parallelogram bisect each other. Hypothesis. ABGDy a parallelogram of which the diagonals intersect at 0. Conclusions. CO = OB. AO^OD. Proof. In ttf triangles A OB and CODy Angle AOB = opposite angle COD. 7^ "^ X Angle BAO = alternate angle ODC, ) Angle ABO = alternate angle DCO. ) (§67) (§ 71) (§124) Side ^5 = side CD. Therefore the two triangles are identically equal, and the sides opposite the equal angles equal; namely, BO = OC. AO=OD. Q.E.D. MiaOELLANEOm PROPEBTIES. 68 CHAPTER IV. MISCELUNEOUS PROPERTIES OF POLYGONS. Befinitions. .1. ?^*^; -P'S^- A polygon is a figure formed by a chain of straight hnes returning into itself and inclos- ing a part of the plane on which it Hes. The lines are called sides of the polygon Polygons of elementarj geometry. ..l"" !^' ^'tT j^^^°^^<^^y *^« «ide8 of a polygon may cross each other but elementary geometry treats only of polygons the sides of which dc not cross. ^^ 143. The angles of the polygon measured on the side containing the in- closed space are called inte- rior angles. Example. One interior angle of the polygon ABODE is the angle EAB, measured by turning the side AB through the interior of the polygon until it coincides with AE. In the first polygon the ontAaa — o 01«A txi.z xcso Polygon the sides of which cross. , u .wx.^ ^M^wm^^ - •"■J evu vuv otucB ML WIUCU CFOSS. Straight angles. But in the polygon ABODE the interior angle AED is greater than a straight angle 64 BOOK II. RECTILINEAL FIGURES. 144. An exterior angle of a polygon is an angle between one side and the continuation of the adjacent side. '' Eemark 1. To form all the different exterior angles of a polygon It 18 sufficient to produce each side, taken in regular order, in one direction. The number of exterior angles will then be the same as that of the interior angles. The angles BAL, GBO, etc., are exterior angles Also Angle ABC-{- angle CBQ = straight angle. ' The angle PQR is an exterior angle which falls inside the polygon because the interior angle PQS is greater than a straight angle. Remark 2. It is evident that a polygon has as many angles as sides. Remark 3. If the interior angle of a polygon is greater than a straight angle, the continuation of one of the sides will fall within the polygon. The corresponding exterior angle is then regarded alge- braically as having the negative sign. Remark 4. It is evident that the sum of any interior angle and of the corresponding exterior angle is a straight angle; that is. Int. angle + ext. angle = 180°. By transposing the second term of the first members, we have Ext. angle = 180° - int. angle. If the interior angle is greater than 180°, the members of this eo^jiiition will be negative. MISCELLANEOUa PBOPERTlEa. 55 Classification of Polygons. 146. Polygons are classified according to the num- ' ber of their sides. The least number of sides which a polygon can have is three, and it is then a triangle. 146. Def. A quadrilateral is a polygon of four sides. 147. Def. A pentagon is a polygon of five sides. 148. Def, A hexagon is a polygon of six sides. 149. Def An octagon is a polygon of eight sides. 150. Polygons of any number of sides may be designated hj the Greek numerals expressing the number of sides. ^ 151. Def A diagonal of a polygon is a Kne join- ing any two non-adjacent angles. Question for the student. How many diagonals can be drawn from one angle of a polygon having n sides ? 152. A regular polygon is one of which the sides and angles are all equal. Classification of Quadrilaterals. 16 . Def A trapezoid is a quadrilateral of which two opposite sides are parallel. 154. If both pairs of opposite sides are parallel, the quadrilateral is a parallelogram. 155. i>e/. A rectangle is a parallelogram in which the four angles are equal. XT. \^^' Py- ^ ^^ombus is a parallelogram of which the lour sides are equal. * -I sides are equal Def A square is a rectangle of which the 66 BOOK 11. RECTILINEAL FIQURES. — f» Theorem XXXVI. 158. JJ^ each side of a polygon he produced in one direction^ the sa.r o/ all the exterior angles is equal to a circuiiifereiuie. Hypothesis. ABCDE, a polygon. Conclusion. Sum of n ex- terior angles ABC, etc. = two straight angles. Proof. From any point draw n straight lines, each parallel to one of the sides of the polygon. There will then be n angles around the point 0. 1. Because OP is parallel to BK and similarly directed, and OQ io BC and similarly directed, Angle POQ = exterior angle KBO. (§ 79) 2. In the same way it may be proved that each of the exterior angles of the polygon is equal to one of the angles around 0. 3. Because the number of exterior angles and of angles around is equal, the sum of all the exterior angles will be equal to the sum of all the angles around 0; that is, to a circumference. Q. E. D. 159. Scholium. If any of the interior angles of the polygon should be reflex, so that exterior angles fall within the polygon, such exterior angles must be regarded as alge- braically negative in forming the sum (§ 144). Theorem XXXVIL 160. The sum of the interior angles of a polygon is equal to a number of straight angles two less than the number of sides of the polygon. Hypothesis. ABCDEF, a polygon having n sides and n angles. MISCELLANEOUS PROPERTIES. 67 -P K Conclusion, Sum of the n interior angles equal to n — 2 straight angles, or ABC + BCD -f CDE + etc. = (w-2)180°. Proof. 1. The sum of each in- terior angle and it's adjacent exterior angles, as ABO-\-CBOy is a straight angle (§ 61). 3. Because the polygon has n such pairs of angles, the sum of all the interior and exterior angles is n straight angles. Xi 3. But the sum of the exterior angles alone is a circnm- ference; that is, two straight angles (§ 155). 4. Taking these two straight angles from the sum (2), we have left the sum of the interior angles aloue^ equal to n — 2 straight angles. Q.E.D. 161. Corollary 1. The sum of all the interior angles of a quadrilateral is equal to two straight angles; that is, to four right angles. 163. Corollary 2. Since in a rectangle all the four angles are equal, and their sum is four right angles, each of the angles is a right angle. Theorem XXXVIII. 163. If through each angle of a triangle a line he drawn parallel to the opposite side., the three lines will form a triangle the sides of which will he hisected hy the ertices of the original triangle. Hypothesis. ABC, any tri- B^ angle; DEF, another triangle formed by drawing, through Cy J^Z)parallelto^5; through Ay -fi^i^parallel to CB\ through By FD parallel to A C, Conclusion, EC— CD. EA = AF-. FB = BD. im es BOOK IL RECTILINEAL FIGUUES. Proof. The quadrilaterals ABCD and ABEC are paral- lelograms, by construction. Therefore CD = ABi ) EC=:AB.\ (§127) Whence EC=CD. Q.E.D. In the same way the other conclusionfl may be preyed. Theorem XXXIX. 164. The bisectors of the three interior angles of a triangle meet in a point equally distant frmi the sides of the triangle. Hypothesis. ABC, any triangle; AO, BO, COy the bisectors of its in- terior angles, A, B, and C. Conclusions. I. These three bisect- ors meet in a single point 0. II. This point is equally distant from the three sides of the triangle. Proof. Let be the point in which the bisectors of ^ and 5 meet. Then— 1. Because is on the bisector of the angle A, O'ls equally distant from the sides AB and .4Cof the angle A (§ 106). 2. Because is on the bisector of the angle B, is equally distant from the sides BA and ^6' of the angle '5. 3. Therefore is equally distant from ^(7 and BC. 4. Therefore it is upon the bisector of the angle formed by ^Cand BC] namely, of the angle C (§ 107). 5. Therefore the point is equally distant from the three sides, and the bisectors all pass through it. Q.E.D. Theoeem XL. 165. The bisectors of any two exterior angles of a general triangle meet the bisector of the opposite interior angle in a point which is equally distant from the three sides of the triangle. MISOELLANEOUa PROPERTIES. Hypothesis. ABC, any triangle, of whicli the sides AB and AC are produced indefinitely in the directions P and Q'y BO, CO, the bisectors of the exterior angles CBP and BCQ; 0, the point of meeting of these bisectors. Conclusions. I. The bisector of the angle BAC passes through 0. II. The point is equally distant from the lines BC, BP, and CQ. Proof, 1. Because is on > the bisector of the angle CBP, '^ it is equally distant from the sides -5(7 and BP. 2. Because is on the bisector of the angle BCQ, it is equally distant from the sides ^Cand CQ. 3. Therefore is equally distant from the three lines AP, BC,&TidAQ. Q.E.D. 4. Because is equally distant from AP and AQ, it is on the bisector of the angle made by those lines (§ 107). Therefore this bisector passes through 0. Q.E.D. Theorem XLI. 166. The perpendicular bisectors of the three sides of a triangle meet in a point, which point is equally distant from the three vertices of the triangle. Hypothesis. ABC, any triangle; R, B, Q, the middle points of the respective sides; PO, BO, lines passing through P and R perpendicular to BC and AB respectively. Conclusions. I. The point is equally distant from A, B, and C. il. The perpendicular bisector ot AC A passes through 0. \m^\m 70 BOOK II. RECTILINEAL FIGURES. Proof. 1. Because is on the perpendicular bisector of the line BCy it is equally distant from the ends, 5 and C of this line (§ 104). 2. Because is on the perpendicular bisector of the line ABy it is equally distant from the points A and B. 3. Therefore it is equally distant from the three points A By and G. Q.E.D. ^ 4. Because it is equally distant from A and C, it lies on the perpendicular bisector of the line JC(§ 105). Therefore this bisector also passes through 0. Q.E.D. Theorem XLIL 16*7. The perpendiculars dropped from the three angles of a triangle upon the opposite sides pass through a point. Hypothesis. A BCy any triangle; AQ, BR, CP, perpen- diculars from A, By C, upon BCy CA, AB, respectively. Conclusion. These perpendiculars pass through a point. Proof. Through Ay B, and (7, respectively, draw parallels to the opposite sides of the trian- gle, forming the triangle LMJ^. \, Then— • ^ 1. Ay B, and (7 will be the mid- dle point of the sides of the trian- gle LMN (§ 163). 2. Because LM is parallel to ABy and CP perpendicular to AB, CP is also perpendicular to LM. (§72) 3. Therefore CP is the perpen- dicular bisector of LM. In the same way BR and A Q are perpen- dicular bisectors of LN and MN respectively. 4. Because the three lines A Qy BRy and CP are the perpendicular bisectors of the sides of the triangle LMJVy thoy pass through a point (§ 166). Q. E. D. MiaCELLANBOUS PB0PEBTIE8. 71 168. Def. The line drawn from any angle of a tri- angle to the middle point of the opposite side is called a medial line of the triangle. Corollary. Since a triangle has three angles it may have three medial lines. >' Theorem XLIII. 169. The three medial lines of a triangle meet in a point which is two thirds of the way from each angle to the middle of the opposite side. Hypothesis. ABC, sl triangle; F, Q, R, the middle points of its respective sides; BE, AQ, two medial lines of the triangle; 0, their point of intersec- tion. Conclusions. I. The third medial line CF also passes through the point 0. II. \ QO =. ^OA. ( RO = iOB. Froof. Bisect AO in M, and OB in JV. Join RQ, RM. QN,MN. Then— 1. Because MN is a line joining the middle points of the sides OA and OB of the triangle OAB, MN=^AB. (§135) MN II AB. (§ 136) 2. Because i?^ is a line joining the middle pomts of the sides CA and CB of the triangle ABC, RQ = {AB, RQ II AB, 3. Therefore RQh parallel and equal to MN, whence the quadrilateral RQMN is a parallelogram (§ 138). ^ ?MmmmM.mmmmi^jmmmm 72 BOOK II. RECTILINEAL FIGURES. RQMn''^''^^ ^^^^ ^^"^^ diagonals of the paraUelogram ; ' OQ=OM.l OH =OJV.\ (§ 1^1) But, by construction, M and TiT are the bisectors of OA and Olf, Therefore OM=iOA. Whence ^^=^^^- QO=:iOA, JiO = iOB. Q.E.D. ^1,,-i ^1 *^® same way it may be shown that th« point in which the medial line CP cuts the medial line ^^ is two The three medial lines pass through the point 0, Q.E.D. Bjl'tllr^ ''''' ^^ '^' same way as withi^Q and PO = iOa Q.E.D. Theorem XLIV. 170. '^^e line drawn from the middle of one of Z.^^^;^ ^ /.a?/ ^;^.^r sum, and bisects the opposite side theSlfrsidelTt^^^^^^^^^ '' ^^^^^ -^ '^ - middle point of A C\ EF, a parallel to AB and CD, meeting BD in F. • Conclusions. I. EF is equidistant from AB and Ci>. 11. BF=zFD. lll-EF=^AB-^CD). cJrth ?ZT A^. f^t -^ ^^ are parallels inter- they a^^disS^C^^r^^r^ ''' *""""^^ ^^ MiaCELLANEOUa PROPEHTIES. 73 2. Because they are equidistant, they intercept equal lengths upon the transversal BD (§ 131). Hence BF^FD. Q.E.D. 3. Therefore EF - CD = AB - EF {% 132, II.), and by transposition %EF=AB-ir CD, c» EF=i(AB+CD). Q.E.D. Def. The line ^i^'is called the middle parallel of the trapezoid. Theorem XLV. 1^:11. If the two non-parallel sides of a trapezoid are equal, the angles they make with the parallel sides are equal. Hypothesis. ABCD, a trapezoid in which AB and 67) are parallel, and CA = DB. Conclusion. Angle CAB = angle DBA. Angle ACD = angle BDC. Proof. From C draw CE parallel to DB, meeting AB in E. Then— i. Because CE is parallel to DB, and CD to EB, DBCE is a parallelogram. 2. Because D^C^ is a parallelogram, CE=DB. 3. Because by hypothesis DB = CA, Therefore CE = CA. 4. Therefore A CE is ar* isosceles triangle, and Angle CAE= angle CEA. 6. Because CE and /)j5 are parallel. Angle DBE = corresponding angle CEA, 6. Comparing with (4), Angle CAE = angle DBE. Q.E.D. In a similar way, by drawing through A a parallel to BD, is it H oh it Twi Angle ^ Ci> = angle i?i) C Q. E. D. ffl glWaB' ^l^Ra a waBii M 74 BOOK II. RECTILINEAL FIGURES. B Theorem XLVI. 17:3. Conversely, if the angles at the base of a trapezoid are equal, the non-parallel sides and the other angles are equal. Hypothesis. ABCD, a trapezoid in which the sides AB and CD are parallel and Angle CAB = angle DBA. Conclusion. I. CA = DB. II. Angle ^C7)=angle5Z)a Proof. Make the same construction as in the last theorem. Then — 1. Because BA is a transversal crossing the parallels CE and DB, Angl6 CEA = angle DBE. 2. By hypothesis, Angle DBE = angle CAE. Whence Angle CAE = angle CEA. 3. Therefore A CE, having the angles at the base equal, is an isosceles triangle, and CA=^CE=zDB. Q.E.D. Therefore angle ACD = angle BDG (§ 171). Q.E.D. Eemark. This form of trapezoid is sometimes called an antiparallelogram. Theorem XLVII. 173.^ quadrilateral of which two adjoining pairs of sides are equal is symmetrical with respect to the diagonal Join- ing the angles formed by the equal sides, and the diagonals cut each otJver at right angles. Hypothesis. A B CD, a quadrilateral in which AB = AD. CB = CD. Conclusion. ACis &n axis of sym- metrv. A (^ cuts BD at riffht ancrles. Proof, Because, in the triangles ABC m^ ADC, MISCELLANEOUS PEOPEMTJES. 76 ACi& common, these triangles are identically equal, and Angle BA G (opp. BC) = angle DA (opp. A C). Angle ^(7^ (opp. AB) = angle DGA (opp. AD), Therefore, if the figure be turned over on the line ^f— The lines AB and CB will fall upon ADwidiCD respectively. The point J5 « « the point i). And^i) « « its own trace. Therefore the figure is symmetrical and the line BD at right angles to A G. Q. E. D. Lemma RESPECTijjq^G Identical Figures. 174. In identically equal figures, corresponding lines are equal. Note. Corresponding lines are those which coincide when the figures are applied to each other. From this definition the conclusion follows without demonstration. Corollary, In identical figures, any lines so defined that there can be but one line in each figure answering to the definition are corresponding lines. For if such lines did not coincide when the figures were brought into coincidence, the two lines would equally cor- respond to the definition. 175. Special appUcaiions. In identically equal triangles— The perpendiculars from equal angles upon the opposite sides are equal. The Usec*ors are of equal length. In ^ c JA>.ally equal quadrilaterals the diagonals are equal. its t. 76 BOOK II RECTILINEAL FIQURE8. CHAPTER V. PROBLEMS. Postulates. 176. It is assumed — inflpfi* .Tf'i^*-'' ^.'?'*^ f*'""^^^* ^^^^ "^^y be produced indetL^ite]y m eitlier direction. III. That a circle may be drawn around any point a n?ipv ''^;^""^^^^*« f *^««^ P^st^l'^tes may be fulfilled with a ruler and a pair of compasses, which are the only instru- ments recognized m pure geometry. But it is not necessary to confine one»s self to these instru- ments m solying all problems. When it is once well under- stood how a given probJem is to be solved by them, other instruments may be used for the actual drawing, such as the protractor, the square, and the parallel ruler. Peoblem I. 177. Onagirm straight line to marJc off a lenath equal to a gimnflnite straight line. Given AB, an indefinite straight line; «, a given finite straight line. , * Required. To mark off ; on J^ a length equal to a. ^' ^] ~ B Construction. From any \ point as a centre, on AB « describe the arc of a circle with a radius equal to «. K p be the point m which the circle intersects a\ OP will be the required length. wm oe me Proof AH the radii of the circle around are b v construe tion equal to .. OP is one of these radii; therefore itlqual L ^^ PROBLEMS. 77 g Problem II. 178. To construct a triangle of wMcTi the sides shall he equal to three given straight lines. Given. The three lines a, d, c. Required. To draw a triangle with sides equal to these lines. Construction. On an in- definite line take a length CB equal to any one of the ' . three given lines (Problem I.), From B as a centre, with a radius equal to one of the remaining lines, c, describe the arc of a circle. From the point C as a centre, with a radius equal to the third given line describe another arc of a circle intersecting the former one in a point A. Join CA and BA. ABC will then be the triangle required. Proof. From the mode of construction, the three lines AB, AC, and BC will be equal to the three given lines. Kemark. The two circles may intersect on either side of the line AB. Therefore tv;o triangles may be drawn which shall fulfill the given conditions. But these triangles will, by § 110, be identically equal. Problem III. 179. To bisect a given finite straight line. Given. The line AB. Required. To bisect it. Construction. From the end ^ as a centre, with a radius greater than the half of . , -'' AB, draw the arc of a circle CND. From J5 as a centre, with ■r\\ r\ actmrt r\ -»«rt /l i ii o r\-nr%-%tT 4-T% ^ ^-^-h^ CMD, intersecting the first circle in the points Cand D. Join CD. /'^^^. mT 'N ^B /y / ^ 78 BOOK 11. BEOTILINEAL FIGUBB8. The point in which the line CD intersects AB will then bisect AB. Proof. Join AC, BC, AD, BD. ^ Because AC, AD, BC, BD, are radii of the same or equal circles, they are equal, and the figure ADBC is a paral- lelogram (§134). Therefore the diagonals AB and CD bisect each other. (§ 138) Therefore ^J5 is bisected at 0. Q.E.F. 180. Corollary. Because the parallelogrom A BCD has all its sides equal, it is a rhombus, and its diagonals intersect at right angles. Therefore the abovo construction also solves the problem: To draw the perpendicular bisector of a given line. Problem IY. 181. To bisect a given angle. Let A CB be the given angle. Construction. From C as a centre, with any radius CA describe the arc of a circle, cutting the sides of the angles in the points A and B. From ^ as a centre, with the radius AB draw an arc of a circle. From 5 as a centre, with an equal radius draw another arc intersecting the other in 0. Join CO. The line CO will bisect the given angle A CB, Proof. In the triangles C^Oand C^Owehave „ . ~ ^. „' !■ by construction. OA = OB, ) ^ c CO = CO, identically. Therefore these two trian- gles are identically equal, and the angle OCA, opposite the side AO, is equal to OCB, oppo- site the equal side in the other triangle. Therefore the line CO bisects the angle A CB. Q. E. F. PROBLEMS. Problem V. 79 k B' 182. Through a given point on a straight line to draw a perpendicular to this line. Let MN be the given line; Of the given point upon it. Construction. From as a centre, with any radius OA de- scribe arcs of a circle cutting the given line at A and B. From ^ as a centre, with the radius AB draw the arc of .. / a circle. SI4A. From 5 as a centre, with the equal radius BA describe another arc intersecting the former one at O. Join OG. OG will be the required perpendicular. Proof. In the triangles CUO and GBO the three sides of the one are, by construction, equal to the three sides of the other. Therefore the angle GAG ib equal to the angle GOB, and both these angles are right angles, by definition. Therefore the line OGi& perpendicular to MN. Q.E.F. Problem VI. 183. From a given point without a given line to drop a perpendicular upon the line. Let P be the given point; MNj the given line. Gonstruction. Take any point K on the opposite side of the line. From P as a centre, with the radius PK describe an arc cutting the given line at A and B. Bisect AB in the point 0. Join PO, Tl\e line PO will be the perpendicular required. ^ r vxrj E xxo student. in viiO xaa\j problem, und to be supplied by the II 80 BOOK II. RECTILINEAL FIOUliES. PltOBLEM VII. 184. At a given point in a straight line to make an angle equal to a given angle. Given. Anmg\Q,EFG'y a straight line, J ^; a point on that line. Required. At to make an angle equal to EFG. Construction. 1. Join any two points in the sides FE and FG, thus forming a triangle. 2. FromO take on OB a distance 0K=- FE. 3. On 0^ describe a triangle OJOf whose sides KM, OM shall be equal to the sides EG, FG. 4. The angle MOK will be the required angle. Proof. Because the triangles OiOf and ^jE'G^ have all the sides of the one equal to corresponding sides of the other, they are identically equal. Therefore the angle MOK, opposite MK, is equal to the angle EFG, opposite the equal side EG of the other triangle. Q.E.F. Problem VIII. 185. To construct a triangle, having given two ndes and the included angle. Given. Two sides, «, J; the angle ^. Required. To construct a triangle having the sides Q^ equal to a, b, and their included angle equal to g. Construction. Draw an indefinite line AB. At A make the angle BA C equal to g. On AB take a length Q-nH r\tr\ A n f oi-.- uii ^-xi> i/iiKu a length A Q equal to a. PROBLEMS, g| Join PQ. APQ will then be the required triangle. The result is evident, but should be shown by the pupil. Problem IX. 186. Two angles of a triangle being given, to ccm- strw ' the third angle. Given. Two angles, c and e, of a triangle. Required. To find the third angle of the triangle. Constructmi. 1. At any point in an indefinite line AB make the angle BOG equal to the given angle c. \ ^ ^C 2. At the same point make the angle COD = given an- gle e. -a. ^ B Then the angle DO A will be the angle required. Proof. From Theorem IV., to be supplied by the student. Problem X. 187. To construct a triangle, having given one side and the two adjacent angles. Given. A finite straight ^ line, AB; two angles, c ^ and e. Required. To construct a triangle having its base equal to ABf and the an- gles at its base equal to c and e respectively. Construction. 1. On an indefinite straight line mark off the length AB. 2. At A make the angle BAM = angle c, 3. At B make the angle ABN=z angle e. 4. Continue the lines AM and Bly until they meet, and let C be their point of meeting. li I i: ^, ^;^^< IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I 1.25 l^|2.8 |50 "^ |2.5 1^ |£ ?? U m ^ t 1^ u »i I. ■UUl. N^ l-'A 1.4 1 ni.6 6" V] .<^ vl / ^^^.^ M •■•> 7 Hiotographic Sciences Corporalion 23 WEST MAIN STREET WEBbTER.N.Y. 14580 (716) S72-4503 ,-\ kV \^^ €^ % o^ W4 f/. 82 BOOR n. REOTILINBAL FIQUREa. h- ABOmH then be the required triangle. The proof should be supplied by the student. OoroUary. Since the third angle of a triangle can always be found when two angles are giyen, this problem, combined with tne precedmg one, will suffice for the construction of the triangle when one side and any two angles are giyen. Pboblem XI. 188. Two sides of a triangle and the angle appo- site one of them being given, to construct the triangle. Given, The two sides, «— ■ «, h; the angle B opposite the side h. Required, To con- struct the triangle. Construction. 1. Ai! the point B on the indefi- nite line BM make an angle MBR equal to the given angle B. 2. From ^ on the line B U"^^^ — — '-jf BR cut off a length ^C equal to the giyen line «, which is not opposite the given angle B, 3. From C as a centre, with a radius equal to the line h describe a circle cutting BM B.t the points D and D\ Either of the triangles BCD or BCD' will then be the required triangle. The proof follows at once from the construction. 189. Scholium. The fact that there may be two tri- angles formed from the given data has been explained in the echolium § 113. Problem XII. 190. Through a given point to draw a straight line which shall he parallel to a given straight line. Oiven. A straight line, AB) 2k point, P. Required. To draw a straight line through P parallel to AB, PROBLEMS. 88 Construction. 1. Take any point D in AB, and join PD 2. At P in the line '^ p -B P2> . make angle DPM j^ equal to the angle PDB, 3. Produce JE'P in the direction F, EPF will then be the . _ required straight line pass- ing through P parallel to AB. Proof. By Theorem III., because the angles PED and PDB are alternate angles. Problem XIII. 191. To divide a finite straight line into any given numher of equal parts. Given. A finite straight line, AB'y a number, n. Required. To divide AB into n equal parts. Construction. 1. IVom one end of AB draw an indefinite straight line, making an angle with AB different from a straight angle. 2. Upon the indefinite line lay off any equal lengths, A 1, 1 2, 2 3, etc., until n lengths are laid off. 3. Join B to the end n of the last length. 4. Through each of the points 1, 2, 3, .... w, draw a parallel to nB intersecting AB. The line AB will then be divided into n equal parts by the points of intersection. Proof. The parallels intercept equal lengths on the trans- versal An, by construction. Therefore they also intercept equal lengths on AB, and the number of intercepted lengths is n (§ 132). Therefore AB is divided iuto n equal parts. Q.E.F. I 84 BOOK 11. RECTILINEAL FIGURES. Pkoblem XIV. 19J8. T^o adjacent sides of a parallelogra/nh and the angle which they contain being given, to describe the parallelogram. Given. Two lines, ^Cand Gil; an angle, 0, Required. To form a paral- lelogram having GH and A C for two adjacent sides, and for the angle between these sides. Construction. 1. At one end ^ 'B A of the line A C make an angle q. equal to 0. 2. On the side of this angle take AB = the giren line GH. < 3. Through B draw a line BD parallel Ui AC. 4. Through C draw CD parallel to AB, intersecting BD in D. ABCD will be the required parallelogram. Proof. May be supplied by the student. 193. Corollary. To construct a square upon a given straighi line. This problem is a special case of the preceding one. in which the given sides are equal and the given angle is a xi^ht angle. To solve it: DrswACrndBDlAB. Join CD. And the square will be complete. ind the DBMONBTRATION OF THEOREMS. 85 \ line BD wen . in CHAPTER VI. EXERCISES IN DEMONSTRATING THEOREMS. The following theorems should be demonstrated by the student in his own way, so far as he is able. Analysis of a given Theorem, The first* step in the process of finding a demonstration is to state the hypothesis, referring to a diagram, which it is generally best the pupil should draw for himself. The state- ment should include not simply what the theorem says, but what it implies. Beference must be made to definitions until the hypothesis is resolved into its first elements. Next, the conclusion must be analyzed in the same way, in order to see not only what it says, but also what it implies. By the analyses of these two statements they must as it were be brought together, in order to see in what way they are related. The process of discovering this relation is one which the student must find for himself in each case, and for which no rule can be given. " Frequently, however, it will be necessary to draw additional lines in the figure, and to call to mind the various theorems which apply to the figures thus formed. To facilitate this, references to previous theorems which come into play are added. The relation being found, the demonstration must next be constructed in the simplest manner, but without the omis- sion of any logical step. This, also, is a matter of practice in which no general rule can be given. It is recommended that the teacher require the pupil to express each step of the demonstration with entire completeness and fullness. Some of the first theorems are so simple that the only serious exercise is that of constructing an artistic demonstration. The work thus becomes a valu- able exercise in language and expression as well as in geometry. The most common fault is that of passing over steps in the demon- stration because the conclusion seems to be obvious. One of the great objects of practice in geometry is to cultivate the habit of examining the logical fouadations of those conclusions which are accepted without 66 BOOK II. RECTILINEAL FIQURE8. critical examination. The feeling of security that a conclusion is right before its foundation has been examined is a most fruitful source of erroneous opinions; and the person who neglects the habit of inquiring into what appears obvious is liable to pass over things which, had they been carefully examined, would have changed the conclusion.' Remabk. The theorems are arranged nearly in the order of their supposed difficulty. The references give the theorems on which the demonstration n>ay be founded, or of which the method of proof has some analogy. It is not to be expected that the beginner will prove more than the first fifteen or twenty. Theorem 1. If a line be divided into any two parts and each of these parts be bisected, the distance of the points of bisection will be one half the length of the original line. Theorem 2. If any angle be divided into two angles and each of these angles be bi- sected, the angle between the J) bisectors will be half the original angle. Hypothesis. BOD, the original angle divided into the two angles JBOCand COD by the line OC. OP, OQ, the bisectors of BOO and COD. Otmdusion. Angle POQ = i angle BOD. Theorem 3. The perpendicu- lars dropped from two opposite angles of a parallelogram upon the diagonal joining the other angles are equal (§§ 128, 175). Theorem 4. If perpendiculars be drawn from the angles at the base of an isosceles triangle to the opposite sides, the line from the vertex to their point of crossing bisects both the angle at the vertex and the angle between the perpen- diculars. Theorem 5. If from any point of the base of an isosceles triangle perpendiculars be dropped upon the two equal sides, they will make equal angles with the base. DEM0N8TBATI0N OF THEOREMS. 87 Thboeem 6. If, on the three sides of an equUateral tri- angle, points equally distant from the thi-ee angles be taken in regular order and joined bj straight lines, these lines will form another equilateral triangle (§ 108). Theobem 7. If the perpendicular from any angle of a triangle upon the oppo- site side bisects this side, the triangle is isosceles. Theobem 8. If the diagonals of any quadrilateral bisect each other, it is a parallelogram (§§67, 68, 108). Theorem 9. If, on each pair of opposite sides of a parallelogram, we take two points equally distant from the opposite angles and join them by straight lines, these lines ^ ^ ^1 form another parallels ^^. ^^^«^ ^^* DT=.RB, v^rifr'"''^'' ^?* ^ *^' ^*'™"*« ^"^'«« ^^^^^ by a trans, versa crossmg two paraUels be bisected, the bisectors will be parallel to each other (§§ 68, 71). • Theorem 11. If either bisector of an interior angle between two paral- lels be continued until it meets the opposite parallel, it forms the base of an isosceles triangle of which the equal sides are the transyersal and the intercepted part of that parallel. Corollary, The two bisectors of the angles which a trans- versal makes with one parallel cut ofE equal segments of the other parallel on the two sides of the transversal. Theorem 12. If the four interior angles formed by a trans- v^al crossing two paraUels be bisected, and the bisectors produced until they meet, what figure will be formed ? (8 82^ his theorem is to be enunciated by the student wmmm 88 BOOK II. REOTILINEAL FIOURES. Theoeem 13. If the bisectors of the four interior angles of a parallelogram be continued until each one meetstwo others, they will form a rectangle. Theorem 14. A line drawn from any point of the bisector of an angle parallel to one side, and meeting the other side, will form an isosceles tri- angle. Eypothem. Angle AOG = angle COB. CP II BO. Conclusion. PO = PC. Theorem 16. If any two interior angles of a triangle be bisected and a line parallel to the included side be drawn through the point of meeting of the bisectors, the length of this parallel between the sides will be equal to the sum of the segments which it cuts off from the sides. Eypothem. Angle BAG = angle BAB. Angle BBC = angle BBA. MBN II AB. Conclusion. MIf=: MA + NB. Theorem 16. In an antiparallelogram— I. The angles at the ends of the upper side are equal. II. The sum of each pair of opposite angles is equal to a straight angle. III. The diagonals are equal to each other (§ 172, Rem.). Theorem 17. That portion of the middle parallel of a trapezoid which is intercepted between the diagonals is equal to half the difference of the parallel sides (§ 170, Del). Theorem 18. If the diagonals of a trapezoid are equal, it is an antiparallelogram. Theorem 19. The sum of the diagonals of any quadri- lateral is less than i^,he sum of the four sides, but greater thaix tlio half Sum. DEMONSTRATION OF THEOREMB. 89 Theorem 20. If from any point in the base of an isosceles tnangle a parallel to each side be drawn until it meets the , other side, the sum of these parallels will be equal to either of j the equal sides (§ 121). Theorem 21. If the middle points of the sides of any quadrilateral be joined by straight lines, these lines will form a paral- lelogram (§ 136). If the given quadrilateral has its pairs of adjacent sides equal (cf. § 173), the parallelogram formed from it will be a rectangle; and if it is a rectangle, the parallelogram formed from it will be a rhomboid. Theorem 22. If one of the equal sides of an isosceles triangle be produced beyond the vertex, and the exterior angle thus formed be bisected, the bisector will be parallel to the base of the triangle. Theorem 23. If the middle points of any two opposite sides of a quadrilateral be joined to each of the middle points of the diagonals, the four joining lines will form a parallelogram (§ 136). Theorem 24. If one diagonal of a quadrilateral bisects both of the angles between which it is drawn, the other diagonal will cross it at right angles. Theorem 25. If from the right angle of a right-angled triangle a perpendicular be dropped upon the hypothenuse, the two triangles thus formed will be equiangular to the original one. Theorem 26. If one of the acute angles of a right-angled triangle be double the other, the hypothenuse will be double the shortest side. 90 BOOK n. BBOTILIITEAL FIQUBE8. Theorem 27. Each side of any triangle is less than half the sum of the throe sides. Theorem 28. If one side of an isosceles triangle be produced below the base to a cer- tain length, and an equal length be cut oft above the base from the other equal side and the two ends be joined together by a straight line, this line will be bisected by the base. Hypoth ma. AG = BC; AB = BF, Conclusion. EN= NF. Theorem 29. The sum of the three straight lines drawn from any point within a triangle to the three yertices is less than the sum of the sides, but greater than half their sum. Theorem 30. If from the yertex of any triangle two lines be drawn, one of which bisects the angle at the ver- tex, and the other is perpendicular to the base, the angle between these lines will be half the difference of the angles at the base of the triangle. Theorem 31. If from any point inside of an equilateral triangle perpendiculars be dropped upon the three sides, their sum will be equal to the perpendicular from the vertex upon the base. What corollary may be deduced from this theorem? Theorem 32. If from two opposite angles of a parallelo- gram lines be drawn to the middle points of two opposite sides, these lines will divide the diagonal joining the other angles into three equal parts (§137). Theorem 33. If from either angle at the base of an isosceles triangle a perpendicular be dropped upon the opposite side, the angle between this perpendicular and the base will be one half the angle at the vertex of the triangle. of Cl] ai 01 ea th m St of St ci , BOOK III. THE CIRCLE. CHAPTER 1. GENERAL PROPERTIES OF THE CIRCLE. Major conjugate are. Definitions. 1^4. Bef, The oiroumference Minor conjugate mo. of a circle is the total length of the curve-line which forms it. 195. Def, An aro of a circle is a part of the curve which forms it. 196. Def. When two arcs to- gether make an entire circle, they are called coi^iigate arcs, and the one is said to be the coAJiigate of the other. 197. Def. When two conjugate arcs are equal, each of them is called a semicirole. 198. Def. When two conjugate arcs are unequal, the lesser is called the minor arc, and the greater the major aro. 199. Def A chord is a straight line between two points — of a circle. 200. Def A secant is a straight line which intersects a circle. 02 BOOK m, THE CinOLB. Remark. A secant may be considered as a chord with one or both of its ends produced, and a chord as that part of a secant contained within the circle. circle is a chord 201. Def, The diameter of a which passes through its centre. 20%. Def. A segment of a circle is composed of a chord and either of the arcs between its ex- tremities. 303. Def. A sector is formed of two radii and the arc included between them. To a pair of radii may belong either of the two conjugate arcs into which their ends divide the circle. 304. Def. Concentric olroles are those which have the same centre. 305. Def A tangent to a circle is any straight line which touches the circle without intersecting it. 306. Special Axioms relating to the Circle. I. A circle has only one centre. II. Every point at a distance from the centre less than the radius is within the circle. III. Every point at a distance from the centre equal to the radius is on the circle. IV. Every point at a distance from the centre greater than the radius is without the circle. Theorem I. 307. Circles of equal radii are identically equal. Hypothesis. Two circles of which and P are centres, and. Eadius OQ = radius PR. GENERAL PROPERTIES. Conclusion, The circles are identically equal. Proof. Apply the one circle to the other in such manner that the centre shall coincide with P, and OQ with PR. Then— 1. Because 0^= Pi?, Point Q = point R. 2. Because each point of the one circle is at the distance OQ from the centre, it will fall on the other circle. (§ 206, Ax. III.) Therefore the circles are identically equal. Q.E.D. Theorem II. 208. Equal arcs of equal circles are identically equals subtend equal angles at centre^ and contain equal chords. Hypothesis. AB, CD, equal arcs around the centres and P. 0A = OB = PC=PD. Conclusion. The an- gles A OB and CPD and the chords AB and CD are equal. Proof. Apply the sector OAB to the sector PCD so that the centre shall fall on P, and the radius OA on the radius PC. Then— 1. Because OA = PC, Point A = point C. 2. Because the radii are all equal, every part of the arc AB will fall on some part of the circle to which the arc CD belongs (§ 206, III.). 3. Because arc AB = a.vc CD, the point B will fall on D, and chord AB = chord CD. Therefore Chord AB = chord CD. Angle AOB = angle CFD, Q.E.D. 94 BOOK lU, THE OIROLE. Theorem III. 209, Equal angles between radii irmlude equal arcs on the circle and eqvAil cJiords, Hypothesis. OA, OB, OP, OQ, four radii of a circle such that Angle AOB — angle FOQ. Conclusions. Arc AB = arc PQ. Chord AB - chord PQ. Proof. Apply the sector A OB to the sector PO^ in such manner that OA shall coincide with OP, Then — 1. Because OA = OP, Point A = point P. 2. Because angle A OB = angle POQ, OB = OQ. 3. Because OB = OQ, Point B = point Q, Therefor3 AB = PQ. Q.7il.D. 4. Because all the radii are equal, the arcs will coincide between P and Q. Therefore the arcs are identically equal. Q.E.D. 210, Corollary. Sectors of equal angles %n equal circles are identically equal, and every line in the one sector is iden- tically equal to the corresponding line in the other (§ 174). Lemma: 211, A sum of two arcs at the cmtre subtends an angle equal to the sum of the angles which each arc subtends separately. Proof. The arc ^P subtends tH f.agle A OB, the arc BC sub- tends the angle BOC, and the arc ABC subtends the angle A OC. But A DC IB by definition the sum of the angles, and ABO iu the sum of the arcs* ^omma. ^h •nrhi/* vuO Hi a -^ >>. . J f' aZ le I' U 1^ GJSNERAL PROPERTIES, d5 The Measurement ol Angles by means of Arcs. 312. From the preceding theorems it follows that to every arc of given length in a given circle corresponds a definite angle, and to every angle corresponds a definite length of arc. To express corresponding arcs and angles in the shortest way, we call the arc corresponding to the angle ^0-B the arc angle A OB, and we call the angle corresponding to an arc the angle arc. Thus, in the following figure. Angle AOB=z angle arc AB, Angle A00 = angle arc ABO, Arc ABO=: arc angle A 00, Atg AB =z aro angle A OB. Combining Theorem III. with the above lemma, it follows that arcs can be taken as the measure of the corresponding angles, and vice versa. In the figure the circle is divided into eight sectors, and since ^^° = 45°, each v^f these sectors subtends an angle of 45°. Therefore Angle AOB= 45°; arc AB arc ABO arc A BOD arc ABODE arc ABODEF arc ABODEFG Angle ^6>C=: 90° Angle AOD =z 135° Aj\glQAOE= 180° Angle ^Oi^= 225° Angle ^06'= 270° Angle J 0^= 315° Angle ^0-4 = 360° O = 46°. = 90°. = 135°. = 180°. = 226°. = 270°. arc ABCDEFOH = 315°. arc ABODEFQHA = 360°. 913. The following are the principles to which we are thus led: I. In the same circle or in equal circles, the greater arc measures the greater angle. II. A minor arc, or an arc less than a semicircle, measures an angle less than a straight angle. ■\.:mij i .m AB, Conclusion. Minor arc CD > minor arc AB. Major arc CBAD< major arc BCDA Proof. From the centre of the tT^ circle draw the radii OA OB, OC, OD. TheL 1. In the triangles A OB and COD we have 0A = OC; OB = OD' AB <: m Therefore angle AOB < angle COD. ,. Therefore minor axe AB < minor arc CD (§ 213, 1.). 3. Because the major and minni. o^« *« j.i. Q*"^*-^' Theore.aV. Th'e^^mllt'bT^i "*'*'"' ^"'"^^^ Theorem VI. . , ^}j^'^^^ery diameter divides the circle Mn t..^ Identically equal semicircles. ^ ^^^ hypothesis. AMBN, a circle* -^ ~"^ AB.a, diameter; 0, the centre ' Conclusion. Arc AMB = arc AI^B. x\ Proof. Draw any two radii OM ^ ^^i ^^ making equal angles with -^/3. Turn the semicircle AMB over on ^^ as an axis. Then— 98 BOOK III THE CIROLB. 1. Because angle ^OJf= angle ^OJV, ^ Radius 0M= radius ok. 2. Because OJf = ON, o Q. .. Point Jf= point JVT. 3. Since M may be any point whatever on the circle every point of the arc AMB will fall on a point of ANB ' Therefore the two arcs coincide and arridentically equal. Q.E.D. Theorem VII. cJtrp\frf "^^^^^^^^ equally distant from the centre^ and of unequal chords the greater is nearer the centre. Hypothesis. AB, CD, and MN, chords of a circle such that AB =CD>MN; 0, the centre of the circle; OP, OQ, OR, perpen- diculars from on AB, CD, and MN, respectively. Conclusions. I. OP = OQ. II. OP < OR. Proof. I Draw the radii OA, OB, OC, OD. Then- 1. In the tnangles ^ 05 and COD we have ^dius Oc=^ OB; OD^ OA', CD^AB (hyp.). 2. Therefore these triangles are identically equal, and the II. Turn the figure ORMN, composed of the chord and its peipendjou »r, around the centre in such manner that oS '^r^fr'^t-^' '''^ '^ the point in Which S^ 3. Because the radii are equal, M~ A 4 Because MN < AB, it subtends a less minor arc f8215^ and the point J^T will fall within the minor arc AB ^^ ^' 6. Therefore JfJVr will fall inside the minor segment ^5 aad 00 < OR. ^ ' 6. But because OP is a perpendicular on AB, Ot^ < 00. f u c\ u h th( Di GENERAL PB0PBBTIB8. 99 Comparing 6 and 6, OP < OR. Q.E.D. ' Theorem VIII. Hypothesis. CD, a chord meet- ing a circle in the points C and /), and not passing through the centre. Conclusion. CD is less than a di- ameter. jDf Proof. Let be the centre of the circle. Join OC and OD, and continue />C? across the centre to B, Then— 1. Because CD is a straight line, CD CD. Q.E.D. Theorem IX. 231. T'^e perpendicular from the centre of a circle upon a chord bisects the chord and the arc which contains it. Hypothesis. CD, a chord of a circle; Oy its centre; OPQ, a perpendicular from on CD, cutting the circle in v\ Conclusion. PC= PD. Arc CQ = arc QD. Proof. Join OC and Oi>. Then— 1. Because OC and OD are radii, they are equal. 2. Because the triangles COP and i>OP have 0(7 = OD, OP common. 100 BOOK III. THE CIRCLE. and OPG and OPD right angles, they are identically equal. Therefore PO=PD. Q.E.D. And Angle COP - angle DOP. 3. Because the arcs CQ and QD are subtended by the equal angles GOQ and DOQ, ^ Arc CQ = arc QD, Q.E.D. Theoeem X, 232. Comersely, a line bisecting a chord at right angles passes through the centre. Hypothesis. CD, a chord of a circle; PM, its perpendicular bisector. Conclusion. The centre of the circle lies on the line PM. Proof. 1. Because C and B are each upon the circle, the centre is, by defini- tion, equally distant from C and D. 2. Because PM is the perpendicular bisector of CD, every point equally distant from C and D lies upon this line (§105). p J* TJ'S'^1^'''*® *^® ""^^"^^^ ""^ *^® ^'''^^® ^ies upon the line JrM. Q.E.D, Theorem XL 223. Parallel chords or secants intercept eaual arcs between them. i' ^^^ Hypothesis. AB, CD, parallel straight lines, of which the first meets the circle in the points A and B, and the second in the points CsLiidD. Conclusion. Arc AC =a.rc DB. Proof. Let be the centre of the circle. From drop the radius OF perpendicular to one of the parallels. Then— IM OBNEBAL PB0PEMTIB8. 101 1. Because OF is perpendicular to one of the paraUels, it will be perpendicular to the other also (§ 72). 2. Because OF is perpendicular to AB, AToAF=aTGBF. 3. Because O^is perpendicular to CD, Arc CF=a,rcDF, 4. Subtracting (2) from (3), AiG AC = arc DB, Q.E.D. am) Theoeem XII. 234. Of lines passing through the md of any radius the perpendicular is a tangent to the circle and every other line is a secant * Hypothesis. 0, the centre of a circle ; OP, a radius ; MN, a line through P perpendicular to OP; MS, any other line through P. Conclusion. I. JfiV is a tangent | to the circle at P. II. MS is a secant. Proof I. 1. Because OP is a perpendicular from upon JfiV, it is less than any other line from to JfJV (§ 101). 2. Therefore every other point of MJVia farther from the centre than P is. 3. Therefore every point of ifiV except Pis outside the circle, while P is on the circle (§206, Ax. III.). 4. Therefore M]^ is a tangent (§ 205, def .). Q. E. D. Proof II. From drop a perpendicular upon MS, and let Q be the point of meeting. Then— 6. Because the line PM is, by hypothesis, different from PM, and 0PM is a right angle, the angle 0PM cannot be a right angle. 6. Therefore OP is not a perpendicular upon MP. 7. Therefore OQ, the perpendicular, will be a different line from OP. 102 BOOK III THE CIRCLE, 8. Because OQ is a perpendicular, it will be less than OP an oblique line (§ 101). ' 9. Therefore the point Q is inside the circle (§206, Ax. II. ). 10. Therefore i^iS* is a secant of the circle. Q.E.D. 225. Corollary 1. The radius to the point of contact of any tangent is perpendicular to the tangent. 336. Corollary 2. The perpendicular from the point of tangency passes through the centre of the circle. Theorem XIII. 227. Thco tangents drawn to a circle from the same external point are equal, and make equal angles with the line joining that point to the centre. Hypothesis, P, a point out- side a circle; PM, PN, two tangents from P touching the circle at M and N-, 0, the cen- tre of the circle. Conclusion, PM = PN, Angle OPJf= angle OPN, Proof, 1. In the triangles OMP and ONP we have Side OJf=side ON, Side OP = side OP, Angle OMP = angle ONP = right angle. 2. Therefore these two triangles are identically equal; namely, the side PM is equal to its corresponding side PN, and the angle 0PM to OPN. Q.E.D. OP, [I.). ^ of tof the les al; INaOBIBBD AND CIRCUMSORIBED FIGURES. CHAPTER II. INSCRIBED AND CIRCUMSCRIBED FIGUR'£S 103 Befinitions. ^S^\'' ^^A ^ rectilineal figure is said to be in- acnbed in a circle when all its ver- tices lie on the circle. The circle is then said to be circumscribed about the figure. ^ 239. Def. A figure is said to be circumscribed about a circle when x \ / / aJU Its sides are tangents to the circle. ^<^^^^^^^ The circle is then said to be in- ^^Sf^'^^^i polygon and -,^_.^^ - . ,, ^ ^^ ^ "^ "* * circumscribed circle. scribed in the figure. 330. Def. An inscribed angle is one of which the sides are two chords going out from the same point on a circle. 231. Def. An inscribed angle is \x said to stand upon the arc included a circumscribed polygon Detween the ends of its sides. *"** *" iMcribed ciicie. If the sides of the inscribed angle are PA and PC, and the circle is divided into two segments by the third chord A, A G, the angle APC'ib said to be inscribed m the segment ACBPA and to stand upon the arc AMO, 232. Def. A line is said to sub- ^-.„.^^ tend a certain angle from a certain «« point when the lines dmwn from^.'S\nd"2^ a^lS! tne point to the ends of the line ^ ^ *^« segment form that angle. t^^Auc. '*~~ "^"^ *^^ The line ^C subtends the angle ^PCfrom the point P. 104 BOOK HI. TUB CIBOLB. Theorem XIV. 333. If from any point on a circle lines he drawn to the end of a diameter and to the centre, the angle at the end of the diameter will be half that at the centre. Hypothesis. AB, & diameter of a circle; 0, the centre; P, any point on the circle. Conclusion. Angle PBO = i angle POA. Proof. 1. Because OP and OB are radii, they are equal. Therefore the triangle POB is isosceles; whence Angle OPB = angle PBO. 2. POA is an exterior angle of the triangle POB. There- fore Angle OPB + angle PBO = angle POA. (§ 76) 3. Comparing (1) and (2), Angle P^O = i angle PO^. Q.E.D. Theorem XV. 334. JEach angle between a chord and the tangent at its end is measured by half the arc cut off by the chord on the corresponding side. Hypothesis. AB,& tangent touch- ing the circle at T; TO, a chord from TtoC. Conclusions. 1. Angle A TC — \ angle of arc TA'C ora the side A. 2. Angle BTC = ^ angle of arc TB'DC on the Bide B. ^ T Proof. Let be the centre of the circle. From T draw the diameter TOD, and join OC. Suppose the chord from Tto fall between TO and TA. Then— 1. Because TA is a tangent and TO a radius, ATD ia a right angle. Therefore Angle BTC— right angle -f- angle OTO. Angle ATC= right angle — angle OTG. on at re. .B re- re) nt he )m I a Or INSCRIBED AND CIR0UM8CRIBED FIGUUEQ, I05 2. Angle TOC = straight angle TOD - angle COD, = 2 right angles - angle COD. Reflex angle TOG= straight angle TOD -f- angle COD, g g =2 right angles + angle COD, Angle COD = 2 angle OTC, (8 233) 4. Comparing (2) and (3), Angle TOC = 2 right angles - 2 angle Ora Keflex angle TOC = 2 right angles + 2 angle C?ra 6. Comparing with (1), Angle TOC = 2 angle A TO. Reflex angle TOC =2 angle -5^(7. Angle ATC=zi angle TOC, = i angle arc TM'C. Angle BTC = i reflex angle TOC, = i angle arc TB'C. Q.E.D. Theoeem XVI. 335. An inscribed angle is one half the angle of the arc on which it stands. Hypothesis. TC, TD, two chords meeting at a point Ton a circle. Conclusion. Angle CTD = i angle of arc CD (that arc be- ing taken on which Tdoes not lie). ^~ r^ fi, ^T^^i'^t.?^^*^^ ''®''*^® ^^ *^® circle. From draw 1. Angle C7!5 = | angle of arc CDB'T; ) 2. Angle Z>rj5 = ^ angle of arc DB'T. ' ) __„ _^. , (§234) 3. Subtracting the second equation from the first, and remarking that Angle CTB - angle DTB = angle CTD, Arc CDB'T- arc i?5'r = arc C2> we have angle CTD = i angle of arc CD whioh rine^ «^t '- dude ^. Q.E.D. ' "*' 106 BOOK III. THE OmOLB, Corollary 1. The angle of the arc CD is, by definition, the angle COD between the radii OC and OD, Therefore the preceding theorem may be expressed thus: \ 236. 7/*, from two points on a circle^ lines be drawn to the centre and to any third point on the circumference, the angle at the centre will be double the angle at the circumference. But, in applying the theorem, the an- gle at the centre, COD, must be counted round in such a direction as not to include the radius 07* to the angle at the cir- cumference. This angle will therefore be greater than 180° whenever the are CTD is J'^S^"'^-^ a mmor arc. segment, are equal 237. Corollary'^, All angles in- scribed in the same segment are equal, be- cause they are all halves of the same angle at the centre, 238. Corollary Z. All angles inscribed in a semicircle are right angles. For they are all measured by half a semi- circle. 239. Corollary 4. If a triangle be in- scribed in a circle, its angles will divide the circle into three arcs. The angle of each of these arcs will be double the opposite angle of the triangle, 240. Corollary 5. Every pair of an- gles inscribed in conjugate segments are supplementary. For if ACB and AFB dio jwa such angles, each of them is mer ';iiirf''> br one ^ half the opposite arc, and th^j-cfore their sum is half a circumference, or a straight angle; whence they are supplementary, by definition (§ 60). '% I a a the the I to T'D, ame '« B —-•»-«=• ^ A"' INSORIBBD And CIRCUMSOnitiED FIGURES. 107 Theorem XVII. 4 ^^ J: 7!^^^^0h three given points not in the sarne stratyht line, one circle, and only one, may be dravm^ Hypothesis. Ay B, C, three given points. Conclusion. There is only one point, 0, so siti'iiicd that it may be the rontre oi: a circle passing through tliean points. Frnnf. The centre of the circle must be equally distant from A, B, and C. JoinAB and BC, and let the lines m and n be the per- pendicular bisectors of AB and BC. Then— 1. Every point which is equally distant from A and B lies on the Ime m (§ 105). 2. Every point which is equally distant from B and C lies on the line n. 3. Therefore every point which is equally distant from all three points, A, B, and C7, lies on both the lines m and ni that 18, on their point of intersection 0. 4. But there is only one point of intersection. Therefore there is one point, and only one, equally distant from A B and C; namely, the point 0. * ' 6. Because OA = OB = OC, if with the centre and the radius OA we describe a circle, it will pass through A B and a Q.E.D. e ^ » iJvliolium. If the three points A, B, and C are in a straight line, the perpendiculars to the lines AB and BG are parallel (§ 70). Therefore in this case no point can be found which shall be equally distant from A, B, and O. Theorem XVIII. 242. m-om any point within a circle every diam- eter subtends an angle greater than a right angle, and from any point without the circle it subtends an angle less than a right angle. tiii 108 BOOK III. THE CIRCLE. Hypothesis. AB, any diameter of a circle; P, any point within the circle; Q, any point without the circle. Conclusions. I. Angle APE > right angle. II. Angle AQB < Hght angle. -^f Proof. I. Continue either side of the angle APE, say AP, until it meets the circle. Let R be the point of meet- ing. Join BR. ^^ Then— T. Because the angle APE is an exterior angle of the tri- angle ERPj it is greater than the interior angle PRB (§ 77). 2. Because the angle PRB — ARE is inscribed in a semi- circle, it is a right angle (§ 238). 3. Therefore APE is greater than a right angle. Q.E.D. II. The proof of this case is so near like that of case I. that it is left as an exercise for the student. Theorem XIX. 243. If a quadrilateral he inscribed in a circle^ the sum of each pair of opposite angles is two right angles. Hypothesis. AECD, a quadrilateral of which the four angles lie on a circle. Conclusions. Angle A -}- opposite angle (7=2 right angles. Angle E + opposite angle D = ^ right angles. Proof. Draw the diagonal ED. This diagonal will be a chord dividing the circle into two segments. Then — 1. Angle BCD = i angle arc BAD. (§ 235) Angle EAD = l angle arc BCD. 2. Adding those two equations, j = 2 right angles. :>mt B tri- mi- .D. 9 1. He, e a 35) i imCRIBEl) AND CmaUMaVIUBED FIQUIiES. 109 In the same way, by drawing the diagonal A G, may be shown Angle B -\- angle C=% right angles. Q. E. D. Theorem XX. 244. Comer sely, if the sum of two opposite angles of a quadrilateral is equal to two right angles, the four angles lie on a circle. — -^ ' Jiypothesis. ABGD, a quadrilateral D/1 9AC in which Angle A -\- angle 0= two right angles. ' Conclusion. The points J, ^, C7and a' B lie on the same circle. Proof. Describe a circle through the three points B, A, B. ^ nr^nn'' T^^^^'f ^^^* pass through C, it must intersect BC, or i>C7 produced, at some other point than C. Let Q be this point. Join BQ. Then— 1. Because the quadrilateraU5^Z>i8 inscribed in a circle. Angle BAB -f- angle BQB = two right angles. 2. Angle 5^i> + angle BOB = two right angles (hyp.). Therefore Angle ^^Z) = angle I (7A Which IS impossible, because BQB is an exterior angle of the triangle BQC (§ 77). ^ 3. In the same way it may be shown that the circle cannot intersect BO produced at any point beyond 0. Therefore the circle must pass through (7, and ABCB lie on one circle. Q.E.D. 345. Corollary. Bach exterior angle of an inscribed quadrilateral is equal to the opposite interior angle. Theorem XXI. S46. When two chords of a circle intersect each other, each angle is measured by a — half the sum of the arcs interested by its sides and the sides of its ver- / X p ^d tically opposite angle. Hypothesis. AB, CB, two chords Ca /n intersecting at the point F, 110 BOOK III. THE CIRCLE. I ■! Conclusions. Angle DPB z= APC = ^ angle arc BD -\- ^ angle arc AO. Angle APD = CPE = \ angle arc DA -\- ^ angle arc CB. Proof, JomBC. Then— 1. Because APC is an exterior angle of the triangle BCP, Angle APC = angle PBC + angle PCB. (§ 76) 2. Because PBC is an inscribed angle standing on the arc A C, Angle PBC= i angle arc A C, (§ 235) 3. Because PCB is an inscribed angle standing on the arc BDy Angle PCB = ^ angle arc BD. 4. Comparing (2) and (3) with (1), Angle APC=^ angle arc AC + ^ angle arc BD. Q.E.D. 5. In the same way may be shown Angle APD = i angle arc DA + ^ angle arc CB. Q.E.D. Corollary. Since vertically opposite angles are equal, we conclude — 24:1i. The sum of each pair of vertically opposite angles is measured by the sum of the corresponding intercepted arcs on any circle which includes the vertex of the angle, Theoeem XXII. 248. If two secants he drawn from a point out- side a circle^ the angle between them is measured by half the difference of the intercepted arcs. Hypothesis. PAB, PCD, two secants emanating from the point P without a circle, and intersecting the latter at the respective points A, B and (7, D. Conclusion. Angle APC —\ angle arc BD — ^ angle arc CA. Proof. Through A draw a par- **! allel to PD, intersecting the circle ..^ in the points A. and P. Then— 1. Because BP is a transversal of the parallels FA and DP, Angle APC = corresponding angle BAF, c i n I B to The J- Because BA^ i. ^ ,^^ ^^,^ ^^^^.^^ ^^ ^^ Angle 5^/-= ^ angle arc Bi; 3. Because fti "nV^ln ? ^^ " * '"^'^ "«= ^-O- parallels ^^and GO, intercepted between the /•Co^parinrSr^alar"--- « -3, Angle AFC= l ansle aro nn i t angle arc £D ^ ^ angle arc CA, Q.E.D. Theorem XXIII "PP^^^e sides is eoualLf^! -^ of the other pair *^ '"^ ^- ^. ^, i?, >y. *^® P°^^*s s -4^ = A8, 3- B«t.eha;e,ai2i;^''+^«+^« + ^^. ' ' -fUt* AU, u. (§337) 112 BOOK III. THE CIRCLE. ! I CHAPTER III. PROPERTIES OF TWO CIRCLES. 250. Def, Two circles are said to touch each other, or to be tangent to each other, when they meet in a single point but do not intersect. Theorem XXIV. 251. Two circles cannot intersect in more than two points. Hypothesis. 0, P, the centres of two circles ; Jf, N, Q, three of their points of intersection. Conclusion. The hypothesis is impossible. Proof. If and P were the centres of two circles pass- ing through Mf iV, and Q, then the two points and P would be equally distant from all three points M, iV, and Q, which is impossible (§ 241). Axiom V. 252. If the distance between the centres of two circles is greater than the sum of their radii, they will not meet each other. Theorem XXV. 253. i)^ the distance of the centres of two circles is equal to the sum of their radii, they will he tangent to each other. Hypothesis. and P, the centres of two circles, the sum of whose radii is equal to the line OP. Conclusion. The two circles have one point in common, and no more. Proof 1. On OP take a point M, such that OM shall be equal to the radius of the circle whose centre is at 0. The point ilf will be on that circle (§ 306, Ax. III.). PROPBRTIES OF TWO CIRCLES. 113 tone; M?Z ^eiTol"' *^^ '"" "' '"^^ ™^". ^e d,V . both circles. i" "' ^ will be at the same time on have ~/+^/>1^7],^~S'^* ^-. - shall 5. But because PJf and PiJ are radii ;f the same circle PH = PM. ' 6. Taking (5) from (4), „ _ OJi> OM. 7. Therefore the point R lies without the circle around O BelSr^efintist?*! T.* « -^Po^- h^ul"^^. are tangent ^rhZetJIIsT St ""'^""*"^' '"^^ Theorem XXVI. t^eir raau, tUey .ill SZt IZZSt'""'' ""^ Hypothesis. 0, P, the centres of two circles; OA, a radius of the greater one, on the line OP- PB fine!^'"'' «^ *^e lesser, on the same OA-BP<:OP «' ""d t^e nil. circle 1- IS withm the circle (§ 306, Ax. II.). 114 BOOK III. THE ClIiCLK ;i! I Continue OP until it intersects the circle P in Q. Then — 3. Because BP and PQ are radii of the same circle, the condition OA - BP < OP gives OA -BP^BP< OP-^PQ, or, which is the same thing, OQ > OA. Therefore the point Q is without the circle 0. 4. Because the point B is within the circle and the point Q without it, if we move a point along the circle P from B to Q, this point must cross the circle 0. 5. But there are two ways in which we can go from B to Q; namely, around either semicircle. Therefore there must be at least two points of intersection of the circles. 6. There cannot be more than two such points, because two circles would ihen pass through the same three points, which is impossible (§ 341). Therefore there are two. Q.E.D. Theorem XXVII. ' 255, If the distance of centres of two circles is equal to the difference of their radiiy they will touch each other in a single point. Hypothesis. 0, P, the centres of two circles such that the line OP is equal to the difference of their radii. Conclusion. These circles touch in a single point, and no more. Proof. 1. Produce the line OPy and on it take a point M, such that PM shall be equal to the radius of the circle P. M will then be on that circle. 2. Because OP is equal to the difference of the radii, the point M will also be on the circle 0. 3. Therefore the point M will be common to both circles. Now, take any other point R on the greater circle, and join OR and PR. Then— 4. Because OR is a straight line, OP-^PR> OR. PROPSRTIES OF TWO GIMGLES. 115 6. Because OR and OM &ve radii of fTi« oorv, • , OM=OF-^ PM, ^^ ^^^^^®' a^d fi n . ' OR=.OP^PM. part 0p7""^ with (4) and taking away the common 7. Because PJf is a radius of the circle P an^ pp • greater than this radius, the pointi. falls iuhecird^^^ 8. Therefore eyery Doint nf i\.. • i ^^z?^^' ^^- ^^'^ Without the Cele TKllX^iLr^Lr ^^ Axiom VI. ^•■^•^* the distance of the centrett^sfb: 'elthr^' ^"^"^''^ "^ greater than the sum of the radii or equal to the sum of the radii ' " of ttetd^ '"" -"^ S™^*^^ '"- ">« ^iff-nce or equal to the difference, or less than the difference the^eltef '■' ""'^'^'"'=^ *^ ^"""^"g ^'oUaries from tern^ll/^HnLlX- ^nT'fi^'f ""^ *°"'"' each other ex- lies wholly ruSS» i? • '^^ "'"" <^ ^^^> ^^'^ °°« "''«'« whoUy iSe le tttr ' "^ '''' '"""'^ "^ <§«««) " " 116 BOOK III THE CIRCLE. 258. Corollary 2. When ttvo circles touch each other, the two centres and the point of contact are in the same sk^aight line. 259. Corollary 3. Two circles cannot touch each other in more than one point, unless they coincide so as to form but one circle. Axiom VII. 260. When two circles inter- sect, the straight line which joins the two points of intersection is a chord of each circle. I ' Theorem XXVIII. 261. When two circles intersect each other ^ the straight line Joining their centres bisects their common chord at right angles. Hypothesis. 0, P, the centres of two circles which intersect each other; M, N, their points of intersection. Conclusion, The line OP bi- sects the line ilfiV at right angles. Proof. Let R be the middle point of the chord MN. Through R draw a perpendicular to the chord. Then — 1. Because MN is a chord of the circle P, its perpen- dicular bisector will pass through the centre P (§ 222). 2. Because MN is a chord of the circle 0, its perpen- dicular bisector will pass through the centre 0. 3. Therefore the perpendicular bisector passes through the centres of both circles. 4. But there can be only one straight line between these centres. Therefore the straight line OP bisects JfiV perpen- dicularly in the point R. Q.E.D. 2G2. Corollary. Conversely, the perpendicular bisector of a common chord passes through the centres of both circles. PROBLEMS. 117 I CHAPTER IV. PROBLEMS REUTING TO THE CIRCLE. Hereafter we shall generally present with ea^h problem an analysis; that is, a course of reasoning by which the solution may be arnved at. In an analysis we generally begin by sup- posing the problem solved, and reasoning out the conditions which must thus be fulfilled. It is expected that the analysis will generally enable the student to supply the proof himself, since the latter will com- prise the same reasoning as the analysis, but generally in the reverse order. o j i^o Problem I. 263. To find the centre of a gwen circle Given. A circle, A BOD. Required. To find its centre. --^^^ ~^ Analysts. The perpendicu- lar bisector of every chord of the circle passes through the ^f centre (§232). Therefore if we draw two such chords and bisect each of them at right angles, the centre will lie on each bisector; that is, it will be their point of intersection. Hence the following ^J'^^^iructio^ Draw any two chords of the circle, as AB Bisect each of these chords at right angles (8 179, Cor ) the c'rdr' "" " "'"' *'^^ '''''''''' "^^ blVe centre of chor^d?*'Thf f '''!• V' ""^^ "'""''"^y ^^*^^"y *° ^raw the cnords. The construction mav be found as follnwo. i^.^^ thirnf ^ ""^ f' T^' "' ^ ''^*^^^ ^^^^ ^^y radius"describ^ the arc of a circle. From any other point B, with the same ^, f /\ 118 BOOK III. THE CIROLE, radius doscribe another arc intersecting the first at the points P and Q. ^ t The straight line PQ, produced if necessary, will pass through the centre of the circle. In the same way another line passing through the centre can be found, and the centre itself will then be their point of intersection. Corollary. Since we need not use any definite portion of the circle in this construction, we may in the same way find the centre when only an arc of the circle is given. Then from this centre we may describe the whole circle. Hence this construction also enables us to complete a circle of which an arc is given. Problem II. 265. From a given point without a circle to draw a tangent to the circle. Given. A circle, TCT'\ a point P outside of it. Required. To draw through P a tangent to the circle. Analysis. Any tangent to the circle is at right angles to the radius drawn to the point of tangency(§325). Therefore the centre of the given circle, the point P, and the point of tangency will be at the vertices of a right-angled triangle. But a right-angled triangle is that inscribed in a semi- circle (§ 238). Therefore the point of tangency will be on the circle of which OP is a diameter. Construction. From the centre of the given circle draw the line OP, and bisect it at the point C. From C as a centre, with the radius GO = GP describe a circle OTPT, intersecting the given circle in Tand T\ JoinPrandPJ". The lines PTand PT' will each be a tangent to the given circle and will pass through P as required. PROBLEMS. 119 Pkoblem III. Given. A circle and a point P ^ upon it. ii?e^i«>e^. Through P to draw ai tangent to the circle. ( ^w«/j^5w. The tangent will be at right angles to the radius from the V 7^1 centre to P (§ 224). Hence we have ^-^^ tTlgh p' *'^^ ^^^"^ -^ ^-- « perpendicular to it tiJrdTatiroS rtr^^' r ^-^^- ^- Join AB. ' """^ ""^ *^^ ^q"^^ ai'cs i'^, PP. Bisect AB at right angles by a line 00 fore^aShr^r^ '"^ Pe^endicular to OQ and there- ^^S^^^^^ i^ = fhLS t^ ^S ortht" i' 'I bisects the arc AB (§§ 221, 222) ® ^"^ ^ ^- .because the arcs Pj' and PP are eoual Pic fi, v.- . mg point of the arc AB. TlhevetaJnn' 1 *^^ ^''^^*- 3. Because PTi.rZ -^^f^^foi'e 0^ passes through P. gent required P^^P^^^^^-^^r to this line, it is the tan- Scholium. The radina nn ,•« ^ x conseruction,since^"i^for;Vr7S'° *'^ """^ Problem IV oil. \ t^llTc^ - « ^-- ^-•-.^. Required. To inscribe a cir- cle within it. Analysis. The bisectors of the ^three angles ^, P, and (7 mee. in a point equally distant from the three sides of the tri- A- ^ 1 i m a r Mil Hlf J'l t. 1 , " j ii i 120 BOOK III. TllK CIIWLR. angle (§ 164). Thoroforo this point is the centre of the required circle. Construction. 1. Bisect any two angles of the triangle as A and D by the lines AO and Z?0, and let bo their point of meeting. 3. From drop a perpendicular OD upon any side, as AB. 3. From as a centre, with the radius OD describe a circle. This circle will be the required inscribed circle. Proof. The perpendiculars from upon each of the three sides of the circle are equal (§ 1G4). Therefore each of these sides is a tangent to the circle. if 268. Scholium. In the general triangle (§ 58) there are four circles, each fulfilling the condition of touching the three sides of the triangle. One of these is within the triangle as just described, and the other three are without it, each of them touching one of the sides from without. The latter are called escribed circles. rrprt! xr 'iicrr °' ''- '''' '^'^ J^T^LI^S' •-' '"' *•« ^'^ -'- ot the .^cHbod ^ 0' bisector of exterior angle SAB ^^^' " " " " ^i9P. BCR. QOA. 7iO" CO" CO"' AO'" n it (( (( tt ti t( Problem V. 269. To describe a circle which ^hnii ^ through three given points, ^"^^ ^""'^ Given. Three points, A, B, C. Required. To describe a circle which shall pass through each of them. Analysis. The centre of the .. circle must be equally distant from This circle will be thaf. r^f„,i.«^ ^.__-. ., points ^, B, and a '-a«J— , i^uBbing througii the Proof. As in §§ 166, 241. 122 BOOK III THE CIRCLE. 370. Corollary. Since, in the construction of this prob- lem, we describe a triangle having its angles at the given points, this problem is the same as that of circumscribing a circle about a given triangle. Problem VI. 271. To bisect a given arc of a circle, . Given. An arc AB. p Required. To bisect it. Analysis. 1. The radius which is perpendicular to the chord of the arc bisects both the chord and the arc. 2. The line which bisects the ~ Q chord at right angles passes through the centre of the circle, so that that part of it which is contained between this centre and the circle itself is a radius. 3. Therefore this line will bisect the arc of the chord. Construction. 1. Draw the chord AB between the two ends of the given arc. 2. Bisect this chord at right angles by the line PQ. 3. Let D be the point in which this bisector intersects the given arc AB. The point D will bisect the arc. Problem VII. 272. Upon a gimn line as chord, to describe an arc of a circle of wMcTi the in- scribed angle shall be equal to a given angle. Given. A line, ^-B; an angle, X, ^^ Required. On ^5 as a chc rd, to draw an arc of a circle such that any angle inscribed in it shall be equal to X. Analysi'^. 1. Suppose the whole circle of { ch an arc is required to PJiOIiLEMlS. 123 oho'nr^uppi: *r ^"' ^^^' "^ "-^^ «' 0- end of the dicular bisector of the chord AB will both „.T, t^ ^"T"" centre of the circle (§§ 223, 226). Hence ^ "^"^ **>' Construction. 1. At one extremity as J of rt« r . „ make an angle BAD ennni tn ti><, • . ' '"e ''"« ^^ a. At yl erect a perpendicular ^0 to the line ia The ::tt oVn^^ 1 "^■'* '-S'- b/the Le W. ^..e^^^oSrr^i; 7:r^' -^-'-^ «' -^^^ «„ eide'Vith Tb, X: "li^fi r'^, r '■- ^ ^^ -» eoin. circle, and Oik Ztvt '''" ''' " ^'^'^'' "f the Pboblem VIII so as to form its three exterior angles ^ ^^^ 3. At the end of each radius draw i tnnffnT,* +« +1, a.d produce these tangents untilX S^ril'^'/ ^^^^^^' This triangle will be that required. ^ Peoblem X. ciS: ^"^ ^^"""^ "" ^"^"^^t^ngmt to two given Given. Two circles P^^^T^v — -—-^ o i'^ and Q8, of which "^ ^ ^ ~ ^ QS is the lesser. Required, To draw a straight line which shall be tangent to both circles. ^ 8. Let ^ and be the centres of the circles. c paidiieis /^cy and PQ, they are equal (8 126^ ^p':^X ^i^:r' *- '•'^ ^'«— '"'^H; radu 'Mn f 1 J ' I fl 126 BOOK III THE GUWLE. i I il .pnf * Jf>. ^ ' T*'' ^ ^^ *^^ «^^"^^ ^ir^^^e draw a tan- gent to the inner circle, and let B be the point of tangencr 3. Join AB, and produce the joining line to P 4. From the centre draw the line OQ parallel to AP and meeting the circle in ^. v i^ c* i^i to ^^ 6. Join Pg. The line PQ will then be one of the tan- gents required. ^" Proo/. To be supplied by the student from the analysis. EXERCISES. Theorem 1. If two chords be drawn in a circle intersect- ing each other at right angles, their ends will divide the circle into four pros. Show that the sum of each pair of opposite ares is a semicircle. x- ri- » ^^ How will the theorem be modified if the chords do not intersect within the circle? Theorem 2. If, from the two ends of a chord, chords per- pendicular to It be drawn, they will be equal in length. Theorem 3. The shortest line between two concentric circles IS part of the radius of the outer one. Theorem 4. If the angles at the base of a circumscribed trapezoid are equal, each non-parallel side is equal to half the sum of the parallel sides (§ 227). Theorem 5. If from the centre of a circle a perpendicular be dropped upon either side of an inscribed triangle, and a radius be drawn to one end of this side, the angle between the radius and perpendicular will be equal to the opposite angle of the triangle. Theorem 6. If two equnl chords intersect, the segments of the one are respectively equal to the segments of the otner. Theorem 7. The only parallelogram which can be in- scribed in a circle is a rectangle. Theorem 8. From any point outside a circle a chord sub- tends an angle less than half its arc; from any point inside the circle an angle greater than half its arc. THEOREMS FOR EXERCISE 127 which which ^ Explain the relation of the two c( the chord divides the circle to the sid the subtended angle lies. Theoeem 9. If an angle between a diagonal and one side of a quadri- C lateral IS equal to the angle between the other diagonal and the opposite side, the same will be true of the three other pairs of angles corre^' ^ spondmg to the same description, and ^^P' -^OB^adb. the four vertices of the quadrilateral ^''^' %ba-'c?jPa he on a circle (§237). DAbZb%. TTTTTrtPTj-xr in A • , , ^^CZ) on a Circle. Theorem 12. If a circle pass " through the centre of another circle Py and from the centre of P a di- ^ ameter to the circle be drawn every chord of P passing (when produced) through the other end Q of this diameter is bisected by the circle (§§ 221, 238). Theorem 13. If two circles be drawn each touching a pair of parallel mes and a transversal crossing them, the distance between the centres of the circles is equal to the length of the transversal intercepted between the parallels (§ 227.) Theorem 14. If any number of tri- ang es have the same base and equal angles at the vertices, the bisectors of thes_e angles pass through a noinf. (^ 9si, E^ etc. Conclusion. Area ^.P = area (.!.<% + ^.^) + j.^ + yl.,n. Proof On BP erect the rectangle BPNM, of which the sides BM, PN, shall each be equal to the lino A, and MJV ""ni YJ'^r.^: ^^ ^' ^^ ^' ^^^- «^^«^ *he perpendiculars GO , DD'y EE'y etc., meeting MNm C", />', W, etc. Then— 1. Because the angles at C, D, E, etc., are all right angles, each of the quadrilaterals MBCC, C'CDD\ etc. is a rectangle. ' 2. The sum of all these rectangles is equal to the rectande BM. BP, by construction. 3. Therefore area BM. BPz= sum of areas A.a^A .1, etc., or, because BM = A, and BP = P, Area A.P = area {A.a + A.h + A*c + ^.^). Q.E.D. Schohum. When we give the symbols A, a, h, c, d, which we have supposed to represent lines, their algebraic significa- tion, we have P = a-^l + c^d, which gives the well-known formula ^{fi^-h + c-\-d)=:Aa-{-Al-\-AG-{-Ad, and expresses the distributive law in multiplication. Theorem II. 288. If a straight line he made up of two parts, the square of the whole line is equal to the sum of the squares of the two parts plus twice the rectangle con- tained mt flip /M/y/r/o ^ n AREAH OF mCTAmiEti. jg^ ITypothesis. AB, a straight lino dividnd nf p • * .. parta A J* and PB, aiviaed at P into the Condmion. Square on AB oquuls sum of squares on AP G and Pi^ plus twice the rectan- gle ^P./'/?. Or, in symbols, ^B'=AP'-\.PB'-\.2AP,PB. W Proof. On AB erect the square A BCD. On AC take ^/" = ^7:>. Through P draw PP" parallel to ^C7, and through P' draw P'P'" parallel to AB, Then— A^ ^ 1. Taking away from the equallines^;? AP ^\.^ .. i parts ^P, AP\ we have ' ^ ^' *^® ^'^''^^ ^^ = ^'^. CQ. DQ, and ^^ are'pi^^rl^f ' *'' quadnlaterak ^ft ngM angle. Therefore they are all rectangles (8 125) 4. Because of the parallelism of the lines JcPP" .„^ BB, and also of the lines ^BP'J'^^ l^lcuXZ^ ' P"'D = QP" = ptQ 6. Therefore Aiea APQP' = j^p\ Area QP"'DP" = pp» Area P'QP^'C = AP '. pp. Area PBP"'Q = AP pp aJ: Thtrf °"' """^ ""^« "P ^''o' -«» of the sqnare on uence: — ~^ '^^-' • "^■'^» (1) ' ^ -f^v 136 BOOK IV. OF AREAS. 289* If frwu the square of the sum of two tinea we take away the sum of their squares, we shall have left twice their rectangle. 290. Scholium. By hypothesis we have AB = AP-i- PB. Substituting this in the conclusion, wo have (AP + PBY = AP^ + 'HAP . PB + PB\ a well-known algebraic expression. ^ The geometric construction serves to exhibit to the eye the different parts of which this algebraic expression is made up. iii ill B K B- n Theorem III. 291. T7i& square upon the difference of two lines is equal to the sum of the squares upon the lines^ diminished by twice the rectangle contained by them. Hypothesis. AB, AG, two lines of which AC ia the longer; BC, their difference. Conclust07i. BC = AB' + AG' - 2AB . AG. Proof. On AG erect the square AGOH. On ^C erect the square BGEF, D On AB erect the square ABKL. Produce FE till it meets AOm D. Then— A 1. The whole area AKLBGHG = AB' + AG\ 2. Because EB — BG, and BL := AB,yfQ have EL = AB -{- BG = AG. Therefore Area KLDE = area AB . AG. 3. Because CH = AG, and GF = BG, we have FH = AG -== BG =z AB. Hence Area DFQH = area -4-5 . -4 0. i« we take vice their the eye 'ession is jOo lines \e lineSf *y them. C is the iH AliBA^H OF liEOTANQLEa. 187 4. If from the whole area (1) we take away the areas '2) and (3), we have loft the square BVEF; that is. BO* Therefore > -w^^ . 292, Scholium. Since BO =z AO ^ AB, vro have (AO-^ ABy = A0'- 2AB.A0+AB\ the algebraic formula for the square of the difference of two numbers. TlIEOKEM IV. 293. T7ie difference of the squares described on two lines IS equal to the rectangle contained by the sum and difference of the lines. Hypothesis. -4 i?, vlC, two straight lines of which ulC is tiio greater, and each of which is to have a square described upon it. % ~ ' iD Conclusion. AC'-~AB^=:(AC-{.AB){AO-AB). p I'roof. On AO describe the square A ODE. On AE take AF= AB. From F draw FH parallel to AO, meeting OD in H, and from B draw ^(7 parallel to AE and meet- ing FH in Q. Then— iV^^' vl^\^Tr''^ ^' '^ *^' '^'^ *^«°^^^ i* °iay"be shown that EH and 6^(7 are rectangles, and ^ 6^ a square. 2 Because AE= AO {h^ construction), and AF = AB, to /a Thirllf ~ ^^^ ^"^ ^^ ^«' ^^ --^-^-^ equal Rectangle EH = AO {AO - AB). 3. Because /'^ and ^C are parallel, OH = AF = AB While 5C ,s, by construction, equal to AO^ AB. Therefore Rectangl GO = AB (AO- AB) 4. The sum of the rectangles AO(AO~^AB) and ^5 \ / A n yi/j). (^C — AB\ = reotana-lo ( AH ^ a p ^r i^ ^n^ ^^ffe^f f« between the squares on the lines i^ and AC IS made up of the sum of these rectangles. - I U m IH < II 138 BOOK IV. OF ABEAa. Therefore AC -- AB" = (AG -{• AB) (AG -- AB), Q.E.D. 294. Scholium. Expressing the areas of the squares and rectangles in algebraic language, this theorem gives a» - 5» = (a + b) (a - b). ■• ♦ > CHAPTER II. AREAS OF PUNE FIGURES. Theorem V. 395. The area of a parallelogram is equal to that of the rectangle contained by its base and its altitude. Hypothesis. ABGD, any parallelogram of which the side ABiB taken as the base; AE, the altitude of the parallelo- gram. Gonclusion. Area A BCD — rectangle AB . AE. Proof. From A and B draw perpendiculars to the base AB, meeting CD produced in E and F. Then — 1. Because A BCD and ABEF&re both parallelograms, EF = AB, and CD = AB. (§ 127) Therefore EF = CD. BF = AE. BD = AG. 2. If from the line ED we take away EF, FD remains; and if we take away CD, EC remains. Because the parts taken away are equal (1), FD = EC. ARBA8 OF PLANE MOUBBS. 139 3. Comparing with the last two equations of (1) it is seen that the triangles BFD and AEC have the three sides of the one equal to the three sides of the other. Therefore Triangle AEC = triangle BFD, 4 Prom the trapezoid ABED take away the triande AEC, and there is left the parallelogram ABCB, From the same trapezoid take away the equal triangle BED and there 18 left the rectangle ABEE, Because the triangles are equal Rectangle ABEE = parallelogram ABCD, * Therefore Area ABCD = rectangle AB . AE. Q.E.D. 296. Corollary 1. All parallelograms upon the same base and between the same parallels are equal in area, because thev are all equal to the same rectangle. 297. Cor. 2. Parallelograms having equal bases and equal altitudes are equal in area. 4T. ??^' 7 ^^^' ^' ^^ ^^^ parallelograms having equal bases, that has the greater area which has the greater altitude Of parallelograms having equal altitudes, that has the greater area which has the greater base. Theorem VI. 399. ^Ae area of a triangle is equal to half the area of the parallelogram formed from any two of its sides, having an angle equal 4o that betwem those Hypothesis. ABC, any triangle; PQRS, a parallelogram im m which J if 11 140 BOOK IV. OF AREAS. PQ^ R8 = AB, PR= QS=AG. Angle PQ8 = angle CAB. Conclusion. Area ABC = ^ area PQR8, Proof. Draw the diagonal P8. Then— 1. Because of the equations supposed in the hypothesis, the triangles PQ8 and ABC have two sides and the included angle of the one efqual to two sides and the included angle of he other. Therefore the triangles are identically equal, and Area ABC = area PQ8. (§ 108) 2. In the same way is shown Area ABC = area PR8. 3. The sum of the areas PQ8 and PR8 makes up the whole area of the parallelogram PQR8. Therefore, com- paring (1) and (2), Area ABC^^ area PQR8. Q.E.D. 300. Corollary. A diagonal of a parallelogram divides it into two triangles of equal area. Theorem VTI. 301. The area of a triangle is one half the area of the rectangle contained by its base and its altitude. Hypothesis. ABC, a triangle having the base AB and the altitude CD. Conclusion. Area ABC^^VQctAB.CD. Proof. Through B draw BG parallel to AC, and through G draw CG parallel to AB, meeting BGinG. Then- 1. ABCG is a parallelogram having the base AB and the altitude CD. Therefore Area ABCG = rect. AB . CD. (§ 295) 2. Because AB and CD are each sides of this parallelo= gram. Area ABC = ^ area ABGG. (§ 299) I AREAS OF PLANE FIG USES. 141 3. Comparing (1) and (2), Area ABC = i rect. AB . CD. Q.E.D. 302. Corollary 1. All triangles on the same base, having their vertices in the same straight line parallel to the base, are equal to each other in area. 303. Cor. 2. If several triangles have their vertices in the same point, and their bases equal segments of the same straight line, they are equal in area. 304. Cor. 3. If a triangle and a —^ ^ parallelogram stand upon the same base and between the same parallels, the area of the parallelogram will be double that of the triangle. Theorem VIII. 305. The area of a trapezoid is equal to that of the rectangle con- o tained by its altitude and half the sum of its parallel sides. Hypothesis. ABCD, a trapezoid of which the sides AB and CD are -* 1 parallel; CE, the altitude of the trapezoid. Conclusion. Area ABCD = ^{AB + CD) . CE. Proof. Draw either diagonal of the trapezoid, say BC. Then — 1. Because ABG is a triangle having AB as its base and (7^ as its altitude. Area ABC = ^AB . CE. (§ 301) 3. Because BCD is & triangle having CD as a base and an altitude equal to the distance of the vertex B from CZ>— that is (because AB and CD are parallel), to CE— Area BCD = i CD . CE. 3. The sum of these areas makes up the whole area of the i'i»i;cauiQ. inereiore Area of trapezoid = ^AB . CE -f ^CD . CE = UAB + CD) CE{% 187). Q.E.D. II' il ii 142 BOOK IV. OF ARBAH Theoeem IX. 306. If through any point on the diagonal of a parallelogram two lines be drawn parallel to the sides, the two parallelograms on each side of the diagonal will be equal. Hypothesis. ABGD, a paraUelogram; P, any point on the diagonal AD ; RS, MN, lines passing through P, parallel to AB and AG re- spectively, and meet- ing the four sides in the points P,/^, J/, JV. A' H T y-...^..,._^r....A X N Conclusion. Area RPMO = area NB8P. Proof. 1. Because the lines AD, AP, and PD are the diagonals of the respective parallelograms ABGD ANRP and PSMDy we have ' Area ACD — area ABD. Area ARP - area J iVP; ) ,„ Area PMD = area PSD. S ^» ^^"^ 3. From the area A GD take away the areas ARP and PMD, and we have left the area RPMG. 3. From the equal area ABD take away the equal areas ANP and PSD, and we have left the area NB8P. There- fore Area RPMG = area NBSP. Q.E.D. 307. Definition. In the foregoing constructions the parallelograms ANRP and PSMD are called paraUelo- grams about the diagonal AD. RPGM and NBP8 are called the complements of parallelograms about the diagonal AD, Theoeem X. 308. In a right-angled triangle the sauare of the hypothenuse is equal to the sum of the squares of the other two sides. AREAS OF PLANE FIGUBEB. 143 Hypothesis. ABO, a triangle, right-angled at A: BAGF, AGKH, EC ED, squares on its respective sides. Conclusion, Area BA GF -f area A CKH F = area BCED, Proof, Through A draw AL parallel to BD and CE, meeting DE in L. , Join FG and AD. The proof will now be an'anged as follows: We shall show (1) that the triangles FBG and ABD are identically equal; (2) that the area BAGF is double that of the triangle FBG; (3) that the area BL is double that of the equal triangle ^5Z). From this will follow area ABGF= area BL. It may be shown in the same way that the area of the square on AG is equal to that of the rectangle GL, from which the theorem will follow. 1. In the triangles ABD and FBG we have BA = BF,), . , , . BD = BO \ ^ Wothesis. Angle DBA = right angle DBG+ angle ABO, Angle FBG = right angle FBA + angle ABO. Therefore the two triangles haying two sides and the in- cluded angle of the one equal to two sides and the included angle of the other are identically equal, so that Area ABD = area FBG. 2. Because BA G and BAG are both right angles (hypoth- esis), GA and A are in the same straight line. Therefore the triangle FBG is on the same base FB, and between the same parallels FB and GO, as the square BA GF. Therefore 18 ^"*; - ^t '•■% ' Wm • ] m. SI ^ ^ ' "Sr i fii \r^ r »iti •! wi lifcaHi 3. Because ^Z- is (by construction) parallel to BD, the triangle ABD is upon the same base BD, and between the 144 BOOK IV, OF ABBA& same jarallels BD and ^ A as the rectangle BL, Therefore Area BL = % area ABB. \ 4. Comparing (?) and (3) with (1), I Area BA QF = area BL, ' 5. In the same way, by joining B£: and AK it may be shown that "^ Area A QKH = area CL, 6. Adding (4) and (5), Area {,BA OF-\-A OKH) = area (JBL + CL) = square BCED, Q.E.D. 309. Scholium. This proposition is called the Pythago- rean proposition, because it is said to have been discovered by Pythagoras, who sacrificed a hecatomb of oxen in gratitude for so great a discovery. It is one of the most important propositions in geometry, as upon it is founded a great part of the science of measurement. It also furnishes the basis of trigonometry. Corollary. An important special case of this problem occurs when the two sides of the triangle are equal, or when AB =z AC. Since the squares on AB and AC are then equal, we have BC = AB' + AC' = 2AB\ If we complete the square by drawing BD parallel to AC and CD parallel to AB, A BCD will be a square, and BC its diagonal. Hence: 310. The square on the diagonal of a square is double the square itself. Theorem XL 311. If from the right angle of a right-angled triangle a perpendicular he dropped upon the hy- pothenuse the square of this perpendicular will be e^r^^^v vu v,ov icvvwivyiG oj me iwo parts of the hv- pothenuse, ^ . \ A^BA8 OF PLANE FIGURES, 146 Hypothesis. ABC, a triangle, right^gled at A; AD, a perpendicular from A on BC, Conclusion. AD" = BD.DG. Proof. 1. Because BAD and CAD are both right-angled at i), AB^ - BD^ = AD\ X .g_._. AC^ - DC' = AD\\ ^»^^^^ 2. Adding these two equations, AB' -i-AC- BD' J DC = 2AD' 3. Because BA C is right-angled at A AB' -i- AC = BC\ ' 4. Comparing (2) and (3), BC - BD' - DC = 2AD' 6. Because the line BCis the sum of the lines BD and DC BC-BD^-DC^=2BD.Da (8 289) 6. Comparing (4) and (5), ^® ^ 2AD' = 2BD . DC, (§ 308) or AD' = BD.DC. Q.E.D. Theorem XII. 313. The square on a side opposite any acute angle of a triangle is less than the sum of the squares on the other two sides hy twicetJie rectangle contained by either of those sides and the projection of the other side upon it. Hypothesis. ABC, any triangle having the angle at A acute; CD, the perpendicular ^ dropped from C on D, and there- fore AD the projection oiACon AB. Conclusion. BC^ = AC + AB^-2AB.AD. Proof. 1. Because CDB is right-angled at D, j__ 2. Because A CD is right-angled at D, CD' = AC-AD\ "i" '11 i ■^ y-'M 1 146 BOOK IV. OF AREAS. 3. Putting this value of CD* in (1), BC :=z BD' - AD' -{. AG\ 4. BD' - AD' = (BD + AD) (BD - AD), = AB (BD - AD), = AB (BD + AD- 2AD), = AB {AB - 2AD), = AB'-^2AB,AD. 6. Substituting this last value in (3), BC = AO'-\-AB'- 2AB . AD, Theorem XIII. (§ 293) (§ 287) (§ 287) Q.E.D. 313. In an obtuse-angled triangle the square on the side opposite the obtuse angle is greater than the sum of the squares on the other two sides by twice the rectangle contained by either of those sides and the projection of the other side upon it. Hypothesis. ABC, a triangle, obtuse-angled at ^; CD, the perpendicular from C upon AB q produced, so that DA is the pro- jection of CA on AB. Conclusion. BC = AC -{- AB' + 2AB.AD. Proof. 1. Because CDB is right-anglp'l at D, BC = BD' 4- CD'. JD A ^B 2. Because CD A is right-angled at D, CD' = AC - AD'. 3. Putting this value of CD' in (1), BC = AC -\- BD' - AD'. 4. BD' - AD' = (BD - AD) {BD -f AD), (8 293) = AB {BD + AD), = AB{AB-\-2AD), = AB' + 2AB . AD. (§ 287) 5. Substituting this last value in (3), BC = AC' + AB'-i-2AB.AD. Q.E.D. 314. Scholium. The method of demoiistration is the same in the last two problems, except that in Th. XII. the AREAS GF PLANE FIGURES. 147 line AB is the stim of the lines AD and BD, and in Th XIII It is equal to their difference. But if we regard the projecl tion AD as algebraically negative when it falls outside the triangle, as in Th. .XIII., then Th. XII. will express both theorems, because the subtraction of the negative rectangle AB . AD would mean that it was to be added arithmetically. Theorem XIV. 315. The projections of a straigJU line upon parallel straight lines are equal. Proof. Let AB be the line projected, and MN, PQ, its projections upon two parallel ^b lines. *- ^^ 1. Because these lines are parallel, the perpendiculars AM At and AP, BQ and QN, form two I j straight lines. ~jL 1 2. Because the lines PJfand gjV^are perpendicular to the same straight line MN, MP II NQ. 3. Therefore MNPQ is a parallelogram, and MN=:PQ{^m), Q.E.D. Theorem XV. 316. The sum of the squares upon the two diag- onals of a parallelogram is equal to the sum of the squares upon the four sides. ^ Proof. Let ^5CZ) be the paral- /^ lelogram, having an acute angle at / "^^^ A, Then— ^ ^ 1. In the triangle ^^(7, A^ ^B BC^ = AB^ + AC^-2ABx proj. of ^ C on ^^. (8 312) AD'== AO^ + CD^J^^OD X proj. of AC on CD. (1 313 ^. Because AB and CD are parallel, Proj. of AC on CD = proj. of ^ O' on AB. (8 315) Also, becansfi 4 nCIT) {a p T.o,«,n^i^ — „_, ^° ■' AB = CD, Therefore the laat two terms of the equations (1) are equal. \ in r'i i ! l i .,.i l ji[ 148 BOOK IV. OF AREAS. 3. Adding the equations (1), tlie last terms of the equa- tions (1) cancel each other, and wo have BC -{■AD' = AB'-\-A C + A 6"-f CD\ = AB"" + BD" + CD' + A C"(because A C^BD), That is, the sum of the squares on the diagonals AD and BC is equal to the sum of the squares upon the four sides. Q.E.D. « » • CHAPTER 111. PROBLEMS IN AREAS Problem I. 317. To construct a triangle which shall he equal in area to a gimn polygon. Given. A polygon, ABCDEF. Required. To construct a triangle equal to it in area. Construction. 1. Join the ends of any pair of adjacent sides, say BC and Ci>, by the line DB. % Through the intermediate angle C draw a line parallel to BD, meeting AB produced in B'. F< 3. The polygon AB'DEF will have the number of its sides one less than ABCDEF, because the two sides BC, CD are replaced by the one side B'D, and it will be equal in area. 4. By performing the same operation upon AB'DEF, the number of sides will be still further diminished by one, and the operation may be repeated until the number of sides is reduced to three. Proof. 1. Because the triangles DCB and DB'B are on the same base, DB, and have their vertices on the line CB^ parallel to that base, they are equal in area (§ 302). PROBLEMS. 149 2. The original polygon is made up of the parte Area A BDEF -f area D CB, and the new one, AB'DEF, is made up of Area ABDEF+ area BB'D. 8. Because the area DCB = BB'Dy Area AB'DEF = area ABCDEF. 4. In the same way it may be shown that each transformed polygon IS equal to the one from which it is formed, so that the last one of aU, which is a triangle, is equal in area to the original polygon. Problem II. 318. To describe a parallelogram which ^hall he equal in area to a given triangle, and ham one of its angles equal to a given angle. Given. A triangle, ABC] an angle, X. Required. To construct a parallelogram having one of its angles equal to X, and its area equal to the area ABO. Construction. 1. Bisect the base AB ot the tri- angle at the point D. 2. Through C draw a line Ci^ parallel to AB. ^i 3. At D make the an- gle BDE = X, and continue the side until it meets OF in E 4. Through B draw BF parallel to DE. DBFE will then be the parallelogram required. The proof is left as an exercise for the student. 319. Corollary. If it be required to construct a rect- angle which shall be equal to the triangle, we have only to make the Ime DE perpendicular to AB. Problem III. 330. To describe a smm.rp. onMnJi 0*^77 7.., ^ 7 .•„ area to a given rectangle. ^-V' TS t \ - ^ ■ ■ 1 ' ^ |i ■ f 160 BOOK IV. OF AJiEAS. I ! ! I m 1 D Given. A rectangle, A BCD. Required. To construct a square of the same area. Analysis. Theorem X. teaches that in the right-angled triangle ABGy the square upon AD is equal to the rectangle BD . DC. Therefore, if we can construct a right-angled triangle in which the perpendicular from the right angle upon the hy- pothenuse shall divide the latter into two parts equal respect- ively to two adjacent sides of the given rectangle, this perpen- dicular will be the side of the square required. This result will be reached by describing a semicircle upon a diameter equal to the sum of two adjacent sides of the rect- angle, because all the angles in the semicircle are right angles. Construction. 1. Produce AB to the point P, making BP = BG. 2. On AP as a di- -^- JL ameter describe a semi- circle. 3. Produce BC until it cuts the semicircle in Q. The square on BQ will be the square ..'equired. Proof. 1. If we join AQ, PQ, the triangle AQP will be right-angled, because it is inscribed in a semicircle. 3. Because BQiaa. perpendicular from the right angle Q upon the base AP, we have BQ' = AB.BP = AB.BC. Q.E.D. 331. Corollary. By the three preceding constructions a square may be constructed equal in area to any given polygon. The polygon is first transformed into a triangle by Problem I. ; this triangle into a rectangle by Problem II., Cor.; this rect- angle into a square by Problem III. Peoblem IV. 322, To describe a rectangle wMcTi shall he equal in area to a gicen pctrallelogTam. Given. A parallelogram, A BCD. N \ \ / / / / / 1 1 L' \ P I l4-.~^i^;| PROBLEMS. 161 Required. To describe a rectangle having the same area. Construction. Produce the side CD to F, and at A ' and B erect perpendiculars to AjB, meeting £!F in B and F. ABFF will be the rect- angle required. Proof. The proof is given by Theorem V. 333. Corollary. If, instead of a rectangle, we wish to describe a parallelogram having a given angle, we have only to make the angle BAF equal to the given angle. Peoblem v. 324. On a gwen line as a base to describe a paraUelogram equal to a given parallelogram in area ana tn angles. Given. A parallelo- gram, ABCD; a line, BM. Required. To de- scribe on BM9, parallel- ogram having the same area as ABCB and equi- / angular to it. * b-.o^Tvf'!f' V^'* 5Jf be drawn in the same straight Ime with AB. Then— ^ duced ki^iT^^ ^^''^'^ ^-^^ parallel to BC, meeting Z^C'pro- 2. Draw the diagonal NB, and produce it until it meets BA produced in P. 3. Through P draw PQR parallel to AM, and meeting CB produced m Q and WM produced in R. pij- ., \ "^"'- xcvjuixcii piiiuhuiogram, having Oii; = i^il/ as Its base, and equal to ABCD in area and in its andes. Proof. From Theorem IX. 162 BOOK IV. OF AHEAS. !l i Problem VI. . 325. On the base of a given triangle to describe another triangle equal in area and hamng a gitm angce, q t\ Given. A triangle,^ J5C; an angle, 0. Required. On the base AB to describe a triangle equal to ABC m area, and having an angle equal to 0. Construction. a* 1. Through C draw CD parallel to AB. 3. At A make the angle BAD = 0, and produce the side until it meets CD in D. ' f 3. Join BD. ABD will be the triangle required. Proof. From Theorem VII., Cor. 1. Problem VII. 326. To form a triangle equal in area to a given triangle, and having its base on the same straight line and its vertex in a given point. Given. A triangle, ABC\ a points P' Required. To describe a triangle equal to ABC in area, having its base on AB and its vertex in P. Construction, 1. Join^P and PB. 3. Through C draw CD^ parallel to AB, meeting AP in D. 3. Draw DQ parallel to PB, meeting AB in 0. 4. Join Pq. 6 V AQr will be the required triangle on the base AB\ equal m area to ABC, and having its vertex in P, I H--^M PROBLEMS. 153 Proof. Join BD. Then— 1. Because AB and CD are parallel, ' Area ABC— area ABD. 3. Becaust) D(^ and P5 are parallel, Area BPB = area ^P^. 3. Area ABD + BPD - BPQ = area A QP, 4. Comparing with (2), Area ABD = area AQF. 5. Comparing this with (1), Area Jl^C7 = area ^^P. Q.E.D. The construction and demonstration of the following problems are left as exercises for the student. Problem VIII. 327. To construct a square which shall be equal to the sum of two given squares. Eemark. If we form a right-angled triangle of which the sides about the right angle are equal to the sides of the given squares, the square upon the hypothenuse will be that required (§ 308). Peoblem IX. 328. To construct a square which shall be equal to the difference of two given squares. Peoblem X. 329. To construct a square which shall be equal to one half ^ gi'aen square. Peoblem XI. 330. To divide a triangle into any given number of equal tri- annlfis i7 ■if I linn a o vertex to the base. WW :f!lfi p 1^ It \ ■r < i '1 ■ It . t M t 1 i 1 .,ilii 11 I i I ! i Nii 154 BOOK IV. OF ABEA8. CHAPTER IV. THE COMPUTATION OF AREAS. 331. Geometrical problems may be divided into two gen- eral classes, depending on the kind of solution which is to be obtained. ^ I. Problems of pure geometry. In problems of construe- tion the solution consists in drawing a figure which is to con- form to the conditions of the problem. The answer to such a problem is given, not in numbers or algebraic expressions but as a geometrical figure simply. The problems we have hitherto considered belong to this class, and they are the only kind recognized as belonging to pure geometry. II. Problems of numerical geometry. In problems of the second class the solution appears not merely as a line or figure drawn upon a plane, but as a calculated length or a calcula^ 3d extent of area. For example, the result may be expressed by saying that a line the length of which is re- quired is seven centimeters or other units in length, or that a surface contains a certain number of square units The number of units in either case may be expressed " either by algebraic symbols or by the numbers of arithmetic. Such problems may be considered as belonging to numerical or algebraic geometry. Relatims of the ttvo methods. In pure geometry the division of magnitude into definite units is not recognized. If an angle is given, it is supposed to be given by drawing it not by stating the number of degrees. The angle itself may not be drawn at all except in imagination. So, also, a given length is a length of a given line, and not a number of units of any kind. Pure geometry was almost the only kind cultivated by the ancients, because the methods of algebra were not known to COMPUTATION OF AREAS. 166 them. Hence it is sometimes called the ancient geometry. But it does not suffice for modern wants, where numbers of miles, feet, acres, etc., are required. One great advantage of the modern method arises from the application of algebraic signs to lines. In the ancient geometry, whenever the position of a point is changed to the opposite side of a line, we have to suppose a different theorem or a different case of the same proposition. But in the modern geometry the difference is expressed by changing the algebraic sign of the distance of the point from the line, and the general statement of the proposition remains the same. The general investigation of lines and areas by algebraic methods requires an application of trigonometry, and, in the cases of curve lines, of the integral calculus; but there is a general method applicable to the computation of areas both m the surveying of land and in the integral calculus, the principles of which can now be explained. Problem XII. 332. To find hy measurement and calculation the area of a given polygon. n Let ABODE be the polygon. C Draw any straight line MN. From each angle of the polygon drop a perpendicular upon the ^ line MN. Let A', B', C\ etc., B' be the points at which these perpendiculars meet the line. The area ^M^CC" includes E' the whole polygon plus an area AEDCC'A' between the poly- gon and the line. Therefore, if from the first of these areas we take away the second, the remainder will be the area of the polvffon. Each of these two areas is made up of several trapezoids namely: ^ ' il |£ tt li. \ 1 166 BOOK IV. OF AREAS. First area = trapezoid A' ABB' + trapezoid B'BCC, Second area = trapezoid A' ABE' + trapezoid E'EDD* + trapezoid D'DCC, The first area includes the quantities to be added, the second those to be subtracted. The area of each trapezoid is the rectangle of its altitude into half the sum of its parallel sides (Th. VIII.). In par- ticular. Area A' ABB' = J (A' A + B'B) A'B\ Area B'BCC = ^ (B'B + CO) B'C. Area O'CDD' = ^ (C'C + D'D) G*D\ etc. etc. etc. To express these areas in algebraic form let us put j3i, p%y pz, etc., for th6 lengths of the several perpendiculars; that is, p\ = A' A, P'i = B'B, pz= CO, p* = D'D. ps = E'E. Let us also take an arbitrary point on the line, to measure distances from, and put yi = 0A\ y^ = 0B\ yz = 0C\ ^4 = OD'. yr> - OW. Then A'B' = yi- yu B'C = y3- yu CD' = 2^3 - yu D'E' = yi- y5. E'A' = ys - yu The expressions for the areas vvIU then be: Area A' ABB' = | {r-,... f p^) {^^ _ y^y B'BCC =z}^{p,^p,)(y,-y,y OG'Djy = ^ (j04 + jt73) {yz - y,). n'JJEE' = i ( j05 +^4) («/4 - y,). E'EAA' =^{p,J^p,)(y,^^y{), ! COMPUTATION OF AREAS. 167 Ihe required area of the polygon we have found to be given by subtracting the last three areas from the first two Now this subtraction may be indicated by simply changing the algebraic signs of the quantities to be subtracted— a change which will be effected by changing the factor ^8 — jji into ^4 — yz. y^ — yi into yt, — yu yi — y\ into y\ — y^. The expression for the area will then be: Area ABODE = i {p^ + jp.) {y^ - y,) + Hp3-\-]h) (yi-yi) + i(^4+i»3) (y* -yz) -{-^{pi +j06) {yi ~y,). It will be seen that the formula is uniform with respect to the different values of p and y taken in order, each value of p being added to that next following in order, and each value of y subtracted from that next following in order. If we execute the multiplications indicated, one half the partial products will cancel each other, and the area will reduce to ilpiy^ — p^yx •j-piya — piyi -\- pzyi — p4yz -{-piys — p'\y4 -{-p^yx —pxys] In principle this method is that used by surveyors in computing the area of irregular pieces of land. It also in- volves the best system of measuring areas in more advanced mathematical investigations. The student may be supposed to find the values of pi, p^, etc., yx, yi, etc., by measurement on the actual figure. Their calculation, when all the sides and angles are given, is a prob- lem of trigonometry. The criterion whether a trapezoid is to be put into the additive or the subtractive column is this: If, in crossing over anvsiflpnf flio T^nlTTn•/^« ^»^/^r« +v,<^ ^„4-r,:Ar. to the inside, we pass to the inside of the trapezoid bounded by that side, the area of that trapezoid is additive, or aig( orai- cally positive. } r .1 ,1 1 i;.' I I :i i 168 BOOK IV. OF AREAS. If, in passing inside of the polygon, we pass out of the trapezoid, the area of that trapezoid is subtractive, or alge- braically negative. Examples. If we pass into the polygon over the side AB, We pass into the trapezoid A'ABB*. Therefore the area ol this trapezoid is additive. (See diagram on p. 155.) The same applies to B'BCC. If we pass into the polygon over the side CD, we pass out of the trapezoid C'CDD\ Therefore the area of this trape- zoid is negative. The same remark applies to the trapezoid bounded by DE and by EA. EXERCISES. Measure off and compute the area of each of the following polygons in square centimetres, inches, or other scale measure. It will be noticed that owing to the re-entrant angles of the second figure there is a double overlapping of some of the trapezoids. But this makes no change in the application of the formula, which always gives correct results when the algebraic signs of the quantities arc properly interpreted. 333. Algebraic expression for the Area of a Triangle. Because a triangle is completely determined when three ol its sides are given (§ 110), its area must admit of being ex- pressed algebraical^ in terms of its sides. The required ex- pression is found as follows: AREA OF A TRIANGLE. 159 Let ABC be the triangle, and CD the perpendicular from C upon D. Put ; a, the side BC; by " " AC', (t t( AB; JO, the perpendicular CD, Then Area ABC X 2 = cp. (§ 301) To find p we have, from the / , right-angled triangles CDA andjj- -J- CDB, » p^ = AC'- AD' = BC - BD' = BC' - (AB - ADY or \ / > f = h'- AD' = a'-{c- ADy = a'-c' + 2c.AD-AD\ We must use these equations to eliminate the quantity iD from the expression for^. Equating the second and fourth members of the last line of equations, we find *' = a' - c" + 2c.AD: whence 2c Substituting the square of this in the expression for p\ we 4c» —435 -. Squaring the above pxpression for the area, 4(Area^^c ' = cy. 16(Area ABCy = 4c>' = 43V» - (5» + c' - ««)» This expression, being the difference of two squares, may be transformed into the product » j "« (2bc -\-b'-i~c'- a') (2bc - b'-c'-{- an The first three terms in the first factor are a perfect square- namely, the square of b + c; and the first three of the second factor are the negative of the same square. Therefore each lactor can again be factored, making the product {b-^^c + a){b + c~a){a-i-b-c)(a~b + c), s = U^ -\- b -{- c) or 28 = a -\-b + e, i ; m i'! 1 1 ' :!! :i 160 wo have BOOK IV. OF AHEAa. b -\- c -\- a = 'Zs. b-^c — a = 2(v - a). a-\-i — c ~ ;^o — v), a — b -\-c = 2{s — b). Substituting these values and dividing by 16, wo have (Area ABOy = s{s- a) {s - b) (a - c) and Area ABC = Vs{s - a) (s - b) (s'^c), the required expression. TlIEORKMS FOR EXERCISE. Theorem 1. The dififerencc of the squares upon any two sides of a triangle is equal to the difference of the squares of the projections of these sides upon the third side. Theokem 2. The sum of the squares upon the diagonals of a quadrilateral is equal to twice the sum of the squares upon the four lines joming the middle points of its sides, taken consecutively. (See Th. 21, p. 89.) Theorem 3. If we join the middle points of two opposite sides of a quadrilateral to two opposite angles, the two triangles thus formed will have half the area of the quadrilateral. HypotJmis. FG =FD, AE= EB. Conclusion. Area AFB + EBG = area AECF = i area ABGB. Theorem 4. The four triangles into which a parallelo- gram is divided by its diagonals are of equal area. Theorem 5. If from any point on the diagonal of a parallelogram lines be drawn to the opposite angles, the parallelogram will be divided into two pairs of equal tri- angles. Area OAD = area OAB. Area OCB = area OGB. K-i^ B mEOUEMS FOR EXER0I8E. igj Theorem 6. Tin- parallelogram formed by joining the middle points of the consecutive sides of a quadrilateral has halt the area of the quadrilateral (§ 13G). ; Theorem 7. If through the middle point of one of the non-parallel sides of a trapezoid we draw a line parallel to the opposite side, and complete the parallelogram, tlie area of the parallelogram will be equal to that of the trapezoid. Theorem 8. If we join the middle of one of the non- parallel sides of a trapezoid to the ends of the opposite side the middle triangle will lave half the area of the trapezoid. ' Theorem 9. If two triangles have two sides of the one equal to two sides of the other re- q spectively, and the included an- gles supplemen- tary, they are equal in area. Hypothesis. GA = MK. GB = ML. Angle AGB-\- angle KML = 180°. Condumn. Area ABG = area KLM. Theorem 10. The sum of the squares upon the diagonals of a trapezoid are equal to the sum of the squares upon the non-parallel sides plus twice the rectangle of the parallel sides. Gonclumn. AG^ -\- BBP - AD"^ -f 5C« + 2AB. GD. Theorem 11. If from any point within a polygon perpen- diculars be dropped upon the sides, the sum of the squares of one set of alternate segments is equal to the sum of the squares of the other set. Aa? -I- Bb^ + Gc^ -f Bd^ + Ee^ = aB^ + hG^ + cL'^ + ~aE^i ^ ,^.. ''I 51 ■ i ;if x lif !l ::*••■ . 1 i 1 i , i ^B 1 ^H '^ra^l iili ill i 162 BOOK IV. OF AUEA8. Numerical Exercises. 1. In a right-unglod triangle the lengths of the sides oon- tttiniug the riglit angle are 9 and 12 feet. What is the length of the hyiwtheniiso? What is the area of the triangle? 2. If the length of the hypothenuse is 10 feet, and that of one side 8 feet, wliat is the length of the remaining side? What is the area of tlio triangle? 3. In a riglit-angled triangle the perpendicular from the right angle upon the hypothenuse divides the latter into segments wliich are respectively 9 and 10 feet. Find the lengths of the perpendicular and of the two sides, and the area of the triangle. 4. What three different expressions for the area of a triangle may we obtain from § 301 by taking different sides as the base? What theorem hence follows? 5. What is the area of the triangle of which the respective sides are 15, 41, and 52 metres? 6. If the diagonal of a rectangle is 13 feet, and one of the sides 12 feet, what is the area? 7. Show how the altitude and area of a trapezoid may be computed when its four sides are known. Refer to the computation of the altitude p of a triangle in § 833. 8. If each side of an equilateral triangle is unity, find its altitude. 9. Draw an equilateral triangle, ABO. Show that the bisectors of each interior angle will bisect the opposite side perpen- dicularly. Show that if the bisector of be produced beyond the point in which it meets the other bisectors and intersect the opposite side in D, and if wo take DF=DO and join AF, BF, then— I. 0^1 i^^ and O^jP will be equilateral triangles. II. The points A, C, B, F lie on a circle. III. The lines OA, OB, and 0(7 will all be equal. Also, supnosinff the lenerth of each side of the trianjylp. ABO io be unity, compute the lengths of OCand OD. BOOK V. THE PROPORTION OF MAGNITUDES, CHAPTER I. RATIO AND PROPORTION OF MAGNITUDES IN GENERAL 334. Definition. When a greater magnitude con- tains a lesser one an exact number of times, the greater one is said to be a multiple of the lesser, and the lesser is said to measure the greater, and to be an aliquot part of the greater. 335. Def. When a lesser magnitude can be found which is a measure of each of two greater ones, the latter are said to be commensurablei and the former is said to be a common measure of them. 336. Def. When two magnitudes have no com- mon measure they are said to be incommensurable. 337. Def. When one of two commensurable magnitudes contains the common measure m times, and the other contains it n times, they are said to be to each other as m to n. Example. If the magni- tude A contains the measure a 5 times, and B the same measure 3 times, then ^ is to ' ' ' ' ^ as 5 to 3, and « is a common measure of A and B. Exercises. Draw, by the eye, pairs of lines which shall be to each other as 3 to 4; as 2 to 5; as 4 to 7; as 5 to 6; as 7 to %. A a B a a "1 ) 1 1 i ' n / \ \ nejl^ '[■( s/r:MA I i 164 BOOK V. PROPORTION OF MAQNITUDES. 338, Corollary, If ^ is to -6 as m to n, then, by defini- tion (§ 337), the With part of A will be equal to the nth. part of B', or, in symbohc language, A B A 5 A 5 A 5 A 5 A 5 m n Note. The mth. part of a mag- nitude is indicated by a fraction of L which the symbol of the magnitude Whole magnitude A. is the numerator and m the denominator. Notation.. The statement that two magnitudes A and B are to eae>i other as the numbers m and n is written symbolically A : B :: m . n, or A : B = m : n. Note. The second form, or that of an equation, is preferable, and is most used by mathematicians; but the first form is more common in elementary books. 339. Def. If a pair of magnitudes A and B are to each other as two numbers m and n, and another pair P and Q are also to each other as m:n, then we say that A is to 5 as P to Q, and the four magnitudes A, B, P, and Q are said to be proportional or to form a proportion. Notation. The statement that the four magnitudes A, B, P, and Q are proportional is expressed in the symbolic form ' A : B :: F : Q, or A : B = P : Q; which is read: ^ is to ^ as P is to Q. 340. Def. The symbolic statement that four magnitudes are proportional is called a proportion. Example. If A contains a twice, and B contains it three times; if also P contains jo twice, p _ p p and Q contains it three times, ' ' ' then _ Q -—P-P p A : B :: P : Q. ' ' ' ' 341. I)ef. The four quantities which form a pro- portion are called terms of the proportion. B a a a a ^1%,^ AXIOMS. 166 343. Def. The first and fourth terms of a pro- portion are called the eictremes; the second and third, the means. Example. In the last proportion A and Q are the ex- tremes, B and P the means. 343. Def. The first and third terms, which pre- cede the symbol : , are called antecedents; the second and fourth, which follow the symbol : , are called consequents. Example. In the last proportion A and P arc the ante- cedents, B and Q the consequents. 344. Def. If the means are equal, each of them is said to be a mean proportional between the ex- tremes, and the three quantities are said to be in pro- portion. Axioms, 345. Ax. 1. If there be a greater and a lesser magnitude of the same kind, the greater may be divided into so many equal parts that each part shall be less than the lesser magnitude. Note. By magnitudes of the same kind are meant those which are both numbers, both lines, both surfaces, or both solids. Ax. 2. If a greater magnitude be a certain number of times a lesser one, then any multiple of that greater one will be the same number of times the correspond- ing multiple of the lesser. Symbolic expression of this axiom. If mag. G = ix mag.Z, nG=:i X nL. then Ax. 3. If a lesser magnitude be a certain aliquot part of a greater one, then any multiple of the lesser one will be the same aliquot part of the correspond- ing multiple of the greater. ic exnrp-sisi'.nn n-F ihi'o /»..m*/^*v, t* , r mag^. (? -. .— j_. ,„., ,,j vfi-vv ltAH///t. XX. UlUlHtX/ ^^^ T Symbol then 11 ( ( ^l . u I 'E W i t i um III 166 BOOK V. PROPOUTION OF MAGNITUDES. 346. Theorem. Equimultiples of commensurable magnitudes are 'proportional to the nri/agnitudes them- selves. Hypothesis. A and B, two a w commensurable magnitudes; P, a magnitude i times as ' ' great as A; Q, a magnitude i ^' '' times as great as B, i being Qj ■ any number whatever. Conclusion. P : Q :: A : B. Proof. 1. Let the Ty^r-gnitude A he to B asm to n. This will mean that if we divide A into m parts and B into n parts, these parts will be equal, or A_B m~ n* 2. Because P ~ iA, if we divide P into m parts, we shall have for each part P I'A A ' 'Z; = ~ = '^X-' (§345, Ax. 3) mm m \o ^ / 3. In the same way, if we divide Q into n parts, n n 4. Comparing these results with (1), m n Therefore P '. Q \\ m \ n. (§ 339) Comparing with iX), P '. Q w A \ B. 347. Corollary 1. In a similar way it may be shown that similar aliquot parts of magnitudes are to each other as the magnitudes themselves. That is, whatever be the whole number it, k ' h " ^ '^• 348. Cor. 2. Similar fractions of magnitudes are pro- portional to the magnitudes themselves. } "J y^xrt wvvx'f :/ ^> V iiiC T V -cP:i:Q::iP:iQ, RATIO OF TWO MAGNITUDES. 1^7 and, by the original theorem, iP : iQ :: P : Q, ; from which follows • • -]^P''-^Q'-''P'' Q- Ratio of Two Magnitudes. 349. Consider any two numbers, which we may call m and n. If we divide each of them into n parts, each part of m will be -, and each part of n will be - = 1. By Corollary 1 of the last theorem these parts will be to each other as the original numbers; that is, ,.f' m : n - -1 n Therefore, if two magnitudes A and B are to each other as m to n, they will also be to each other as- to 1, or n A : B m n 1. 350. Def. When two magnitudes are to each /rrt Other as m to n, the fraction ~ is called the ratio of lb the magnitude A to the magnitude B. Corollary. When wo say A : B :-. m : n, we mean that A contains m parts, and B contains n equal parts (§ 338). Hence: 351. The ratio of a magnitude A to another magnitude B is the quotient formed by dvnding the number of parts in A by the numher of equal parts in B. 352. Scholium. There are three ways of conceiving of the ratio of two magnitudes, which all lead to the same result. I. If we have two magni- tudes A and B, the ratio of A a to B is the numerical factor by which we must multiply ® the consequent, B, in order to produce the antecedent, A. There may then be three cases: A n\ in 5i1 .11 I 'I 'I L I '\ 111 H I 1^1 I' * 168 B005' V. PROPORTION OF MAGNITUDES. I. If ^ is a multiple of B, the ratio is a whole number. The multiplication is then effected by adding B to itself the proper number of times. I 2. If ^ and B are commensurable, the ratio is a vulgar fraction. If A contains the common measure m times, and ' B contains it n times, the ratio is — . We may conceive the multiplication to be effected by dividing B into n parts, and taking m of these parts to make A. 3. If ^ and B are incommensurable, the ratio will neither be a whole number nor a vulgar fraction. If we attempt to express it as a decimal, the figures will go on without end. II. The ratio ot A to B may also be conceived of as a number expressing the magnitude of A when we take B as the unit of measure. This amounts to the same thing as I., be- cause when we multiply unity by any factor we produce the factor itself. III. If A and B are numbers, instead of geometric magni- tudes, the ratio of ^ to 5 is the quotient -g. The consistency of these ways of conceiving a ratio is es- tablished by the following definition: 353. JDef. To multiply a magnitude 5 by a nu- merical factor r means to find a magnitude wMcli shall have the same rati o to B that r has to unity. Hence the expressions A : B :: r : 1 and A=rB are equivalent. The preceding definition of a ratio gives us another defini- tion of .« proportion, namely: 3 W. Four magnitudes are proportional when the ratio of the §rst to the second is equal to the ratio of the third to the fourth. 355. Bef. If the terms of a ratio are interchanged, the new ratio is called the inverse of the original one. |-i.~»t i\ RATIOS OF INCOMMENSURABLE MAGNITUDES. 169 Thus the ratio B : A\b the inverse oi A : B, It A : B :: m : 71, then A : B = ~ and B : A = -:the mn ^^ ^ product of these ratios is — = 1. Therefore: nm 356. Theorem. The product of Uoo inverse ratios is unity. Ratios of Incommensurable Magnitudes. 357. If two magnitudes are incommensurable (§ 336), they may still be considered as having a ratio, but this ratio cannot be exactly expressed by a fraction. Let us suppose that in dividing the magnitude B into n parts, A is found to contain m of these parts and a fraction of another part. Then the ratio of ^ to J5 will be greater than -, and less Wl -f- 1 W2 1 ^ than ; that is, less than — \- -. The number n may n n n *' here be as great as we please. 358. Theorem. If four incommensur able magni- tudes A, B, P, and Q are so related that, on dividing the antecedents A and P each into n equal parts, Q shall contain the same whole number of parts if P that B contains of A^ fractions being neglected, and this however great the number n,—then the ratio of Qto P is equal to the ratio of B to A^ and A, B^ P, and Qform a proportion. Hypothesis, ^ B e lees than a (§ 345, Ax. 1). Ill 11 mmi ' M ! \r4 m ■'ii I HJIiii II -ili !lll|! 170 BOOK V. PROPORTION OP MA0NITUDB8. If m be the whole number of times which A contains the »th part of B, A:B>- andA : B <- 4--. n n n By hypothesis P contains the wth part of Q this same number m of times, plus a fraction. Therefore P ' Q> - and P : Q < — h - . n n n Since both ratios are greater than ^- and less than -4--, n n ^ )i' their difference ra:ijt be lesb than - and therefore less than n a, because ~ 1000. - we take a divisor > 100000, 100000 etc. etc. etc. Transformation of Proportions. 361. Def. Inversion is when the terms of each ratio in a proportion are interchanged to form a new proportion. TRANSFORMATION OF PROPORTIONS. Ill ^1 B : Theorem of Inversion. From the proportion A : B .'. P :Q we may conclude by inversion B '.Av.Q'.P, Proof. From § 355. 363. Def. Alternation is when the means of a proportion are interchanged to form a new proportion. The new proportion is then said to be the alternate of the original proportion. Theorem op Alternation". In any proportion the antecedents Moe the same ratio to each other as the consequents. Hypothesis. If A : B :: P : Q— Conclusion. Then ^ j ^ ; A: P :: B: Q. q Proof. 1. Let the ' ' ' ratios A : B and P : Q each be -, so that mth. part of ^ = nth part of B, which call a. mth. part of P = ^ith part of Q, which callj?. 2. Then A = ma. B = na. P = mp. Q = npo 3. Hence A : P :: ma : mp :: a : p; ) B : Q :: na : np :: a :p. ) 4. Therefore A : P :: B : Q. Q.KD. 363. Qorollary. If the extremes he interchanged, the pro- portion will still be true. Fcr by alternation we have A '. P '.: B '.Q, and then by inversion, and putting the second ratio first, Q:B::P:A. (§346) fj; \l I ' ! I 364 B^f antecedents and consequents are compared with either antecedents or consequents to form a new proportion. ^_LJ L I ;| 1 ■i' m T mil I II i 172 BOOK V. PROPORTION OF MAGNITUDES. Theorem op Composition. If we have the proportion A : B :: F : Q, (1) we may conclude A : A + B :: P : P+Q;) A-{-B: B:: P^Q: Q.\ (2) Proof. Let the equal ratios in (1) be m : n. Using the same notation as in the preceding theorem, we find A=ma\ B = na. A-\- B = {m-{- n)a, P = mp', Q =np. p -f. Q = (m _|_ n)p, A : A-\- B = m '. m-\-n. P : P-\- Q = m : m-^-n A-\- B : B = m -\- n : n, P -{■ Q ' Q = m -{- n : n. Whence the conclusions (3) follow by comparing the equal ratios. 365. Bef. Division is when the difference of antecedents and consequents is compared with either antecedents or consequents to form a new proportion. Theorem of Division. If we have the proportion A:B::P:Q, (1) we may conclude A:A-B::P:P-Q;l Y9\ A-B :B ::P- Q :Q.\ ^'^> Proof. By the same process as in the last theorem. 366. Theorem. If we have the several proportions A : B :: P ; Q, A' : B :: P' : Q, etc. etc., we may conclude A + A'-i- etc. : 5 :: P + P' 4- etc. : Q. Proof. The proportions show that if A n m A* = —iB. etc. n' MULTIPLE PHOPOBTIONS. 173 then whence P' = %Q, etc., ^ + ^' + etc. = (^ + ^' + etc.)i?, P + P' + el.. = (^+-' + ctc.)c. The conclusion now follows from the equality of the co- efficients. Multiple Proportions. 367. When three or more ratios are equal, a proportion may be formed between any two of them. Thus, if A:B = M:N=P'.Q=.XiY, etc., (1) we may form the proportions A:B::M:W, A'.B'.'.XiY, M:N'.:X'.Y, etc. etc. The equality of such ratios is generally expressed by writ- ing all the antecedents with the sign : between them, fol- lowed by the consequents in the same order. Thus (1) would le expressed in the form A :M:P'.X=B:N:Q: Y,) ,., or A'.M:P:X',:B:NiQ:Y.) ^^ Here the first consequent {B) corresponds to the first antecedent {A)-, the second {N) to the second (if), etc. 368. Bef. A proportion of six or more terms expressed in tlie form (2) is called a multiple propor- tion. Simple proportions may be formed as follows from a multiple proportion. 369. Theoeem. Ill a multiple proportion any antecedent is to its consequent as any other ante- cedent to its consequent. mm i '■ i ' 1 174 BOOK V. PliOPOHTiijN OF MAGNITUDES Example. In the proportion (2), us the first antecedent, A, is to the first consequent, B, so is the third antecedent, P> to the third consequent, Q\ or, in symbolic language, A'.BwP'.Q, Proof, This theorem follows at once from the form of expression in (1) and (2). 370. Theorem. In a multiple proportion any two antecedents are to each other as the correspond- ing consequents. Example. In (2), as A is to P, so is B (the consequent of ^) to ^ (the consequent of P); or A'.P'.'.B'.Q. Proof. Any such proportion is the alternate of one of the original proportions expressed by the continued proportion. Thus one of the original proportions expressed in (1) is A : B :: P : Q, and the above proportion is its alternate. 371. Theorem. In any proportion the sum of any number of antecedents is to the sum of the corresponding consequents as any one antecedent is to its consequent. Proof. Let the proportion be that in (2), and let the ratio of each antecedent to its consequent be — , so that n A : B :: m : n, M '. N :: m :n, P : Q :: m :n, etc. etc. Then B A m m n N m n etc. etc. MULTIPLE PROPORTIONS. 17C By adding these equations wo have A^M-^r -\- etc. __ yy -f jyr 4- g 4- etc. m n > that is, the wth part of ^ + Jf + P + etc. is oqu.ii to the wth part oiB -\- N -\- Q -{- etc., so that A + Jf + P + etc. : /y + jv+ g + etc. :: m : n, which is by hypothesis t..u ^ume as the ratio of each ante- cedent to its consequent. Because this reasoning is correct how great soever wo sup- pose the numbers m and 7i, the theorem is true whether the magnitudes are commensurable or incommensurable (§ 359). 372. Theorem. In any proportion the difference of any two antecedents is to the difference of the cor- responding consequents as any antecedent is to its consequent. Proof. By taking the difference of the first two equations of §371, we have A - M _ B- N m ~ n ' which shows that A — M : B — ]^ :: 7n : n, the same ratio which each antecedent has, by hypothesis, to its consequent. In the same way it may be shown that any other difference of the corresponding magnitude has this ratio. 373. Theorem. If in a series of ratios the conse- quent of each is the antecedent of the next, the ratio of the first antecedent to the last consequent is equal to the product of the separate ratios. Hypothesis. We have the separate ratios A:B. B : C. GiD. Oonchtsion. The ratio of A to D is the product of the ratios A : B, B : C, C : D. J ij- i t I i : ' V^^'^0 ^J^^ .^M ^ ^ .^^ IMAGE EVALUATION TEST TARGET (MT-3) 1.0 1.25 LilZi 12.5 ■50 ■'^■" DIIIMB ^ 1^ iiiir <^ 116 ill u Mi 12.2 IM 2.0 18 U 11.6 I ew W 'C, m //% /A yj,M Hiotngraphic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. i4580 (716) 1172-4503 <^ \ #^^ ^c^ mm 176 BOOK V. PROPORTION OF MAGNITUDES. Proof, Let the values of the respective ratios A : B, B:G,0:Dhe-, i, A so that w J q A: B ::7n:n,0T ratio A : B = ^, n B : 0:: p:q,0T ratio B : C=^, Then G iB ::i : ;, or ratio C : i> = 4-. J A m B P i B ^ q' • • .7 (1) Divide the first equation by p and the second by n, mp~' np* np ^ nq' A__G_ mp~ nq' Divide this equation by i and the last of (1) by nq, Therefore G Therefore that is. mpt nqi A nqi D D nqj mpt nqi ' nqj n q ^ i 374. Def. When the ratio of the first antecedent to the last consequent is formed by multiplying a series of intermediate ratios, the ratio thus obtained is said to be compounded of these intermediate ratios. LINEAM PROPORTIONS, 111 CHAPTER II. LINEAR PROPORTIONS. Definitions. 375. Def. Similar figures are those of which the angles taken in the same order ^ are equal, and of which the sides between the equal angles are proportional. Example. The figure A BCD is similar to A'B'C'D' when Angle A = angle A'y Angle B = angle B% etc., and AB : BO : CD : DA :: A'B' : B'C : CD' : D'A\ 376. Def. In two or more similar figures any- side of the one is said to be homologous to the corre- sponding side of the other. Example. In the above figr re, the sides AB and A'B' are homologous, the sides BG and B'O' are homologous, etc. etc. etc. 377. Bef. When a finite straight Kne, as AB, is cut at a point P be- tween A and B it is ! f i ? said to be divided^ B intenially at P, and the two parts AB and BB are called segments. 378. Bef. If the straight line AB is produced and cut at a point Q outside of A and P, it is said to 06 divided eztemallv at a an^ f>»^ Hr. -o V are called segments. mU 178 BOOK V. PROPORTION OF MAGNITUDES. Corollary 1. A line cut internally is equal to the sum of its segments. Cor. 2. A line cut externally is equal to the difference of its segments. 379. Bef. Two straight lines are said to be similarly divided when the diiferent segments of the one have the same ratios as c n d the corresponding segments " ~ "• of the other. -J M b Example 1. If the line AB m divided at M and the line cd at N in such manner that AM: MB :: CN : ND, the lines A B and CD are similar- ly divided at M and N. C N Q ly Example 3. If the lines AB and CD are divided at M, P, N, A M Y 3 and Q in such wise that AM : MP : PB :: CN : NQ : QD, they are similarly divided. 380. Def. If three straight lines, a, 5, c, are so related that a \h '.',h : c, the line h is said to be a r^'^an pro- portional between a and c. 01- Theorem I. 381. If two straight lines are similarly divided^ each part of the first has the same ratio to the corre- sponding part of the second that the whole of the first has to the whole of the second. Hypothesis. Two straight ^ lines, AB and A'B\ divided '~ at P, Q, P', and Q', so A^ p' tf g that i. B I ■ ■ I AP : PQ'. QB :: A'P' : P'Q' : Q'B\ Conclusion. AP : A'P' :: AB : A'B', \ PQ: P'q :: AB : A'B'-A LINEAR PRQP0UTI0M8. 179 0, are so or, expressed as a multiple proportion, AF : FQ: QB : AB :: A'F' : F'Q' : Q'B' : A'B\ Froof. In the proportion of the hypothesis the sum of the antecedents is AB, and the sum of the consequents A'B' Therefore (§371) AB : A'B' :: AP : A'F' :: FQ : F'Q\ etc. Q.E.D. Theoeem II. 383. A line cannot be divided at two different points, both internal or both external, into segments having the same ratio to each other. Hypothesis I. A line, AB, divided internally at the points ^ P and Q, Conclusion. The ratio AF : FB will be different from the ratio A Q : QB. Fro^f. Let the ratio AF : FB be ^. Then JP will con- tain m parts, and FB n equal parts. Because AQi^ greater than AF, it will contain more than m parts; and because QB is less than FB, it will contain less than n parts. Therefore the numerator of the ratio AQ .QB will be greater than w, and its denominator less than n, whence it must be greater than - and cannot be equal to it. Therefore there is no other point of division than F for which the ratio of the segments will be the same as ^P : FB. Hypothesis II. A line, AB, divided externally at the pomts F and Q. Conclusion. The ratio ^^- ^ ^ ^ AF : BF will be different from the ratio AQ : BQ. Froof. Let m be the number of equal parts in AF: n, the number in BF-, and s, the number in FQ. Then m AF :BF = ^Q n n~{-8 180 BOOK V. PROPORTION OB' MAGNITUDES. If we reduce these fractions to a common denominator and take their difference, we find it to be {m — n) 8 n {n-\-s)* Because m and n are necessarily different, this fraction cannot be zero, and the ratio AP : BP is different from AQiBQ. Q.E.D. 383. Corollary 1. When the point of division P is nearer to B than to A, AP is greater than BP, and the ratio AP : BP is greater than unity. When it is nearer to A than to B, AP is less than BP, the ratio is less than unity. 384. Cor. 2. If we suppose the point P to move from A toward B, the ratio AP : PB will be equal to zero as P starts from Al will be unity when P is half wa} between A and B, and will increase without limit as P approaches B. 385. Cor. 3. A line cannot be divided externally into segments having the ratio unity. 386. Cor. 4. Two different points may be found, the one internal and the other external, which shall divide a line into segments having the same given ratio. EXERCISES. 1. Draw a line, AB, and cut it internally in several points so that the ratios of the segments shall be 1 : 6, 2 : 5, 3 : 4, 4 : 3, 5 : 2, 6 : 1. 2. Cut the same line externally in the ratios 2 : 9, 1 : 8, 8 : 1, 9 : 2, 11 : 4. 3. A line 7 inches long is to be divided into segments hav- ing the ratio 4 : 5. How lo'ng are the segments? 4. A line 6 inches long is to be divided externally into segments having the ratio 5 : 8. How far is the point of division from each end of the line. 5. If the line ^5 (§ 377) is 3 centimetres in length, and the points P and Q divide it both internally and externally into segments having the same ratio 1 : 2 (§ 386), find the lengths AP, AB, and AQ. LINEAR PROPORTIONS. 131 Theorem III. 387. fftwo straight lines are cut by three or more parallel straight lines, any two intercepts on the one are as the corresponding intercepts on the other. Hypothesis, a, by c, three parallels intersecting the line p in the points A, B, G, and the « line q in A\ B', G\ y V Conclusion. «■ "^^ AB : BG :: A'B' : B'C\ Proof. Diyide AB into any ^ b/" — 1l ^ number m of equal parts, and through the points of separation draw lines parallel to AA^ and *: /n /r' BB\ / i Cut off from BC parts equal to those of AB. Let the number of parts in BC be n plus a fraction. Through the pomts of separation draw lines parallel to BB\ Then— Because the lines;? and q are cut by parallels intercepting equal lengths on p, the intercepts on q are also equal (§132, 1. ). Therefore A'B' is divided into m equal parts, and B'C is divided into n equal parts plus a fraction. Because this is true however great the numbers m and n, we conclude AB : BC :: A'B' : B'C (§ 358). Q.E.D. Corollary. If the points A and A' coincide so that » and q cross each other at A, the figure 4 AGC will form a triangle; and the conclusion will, by the same demonstration, be b/ X^b > AB : BG :: AB' : B'G*. ^^_ >^q, But55' is parallel totheside CC'ot the triangle. Therefore: 388. A straight line parallel to one side of a triangle divides the other two sides similarly. Theoeem IV. ^- ^^^' I( ^^^ **^^* ^-^ ^ i'Tiangle he similarly aimaed,the line joining the points of division will uoparauec to the rmaining sids of the triangle. i Hi 189 BOOK V. PROPORTIOJSr OF MAQNITUDEa Hypothesis. ABCy a triangle of which the sides CA and CB are cut (internally or externally) in the points D and E in such manner that CD : AD :: CE : BE. Conclusion. DE\\AB, Proof. If DE is not parallel to AB, draw DE' ^ parallel to ^^ and meeting CB in E". Then CD: AD:: CE' : BE', (§ 388) and the line CB is divided at two points, E and E'l into parts haying the same ratio CD : AD, which is impossible (§ 382). Therefore the points E and E' coincide, and the line DE is the same as the parallel i>ii^'. Q.E.D. Theokem V. 390. Equiangular triangles are similar, and the sides between the equal angles are homologous to each other. Hypothesis. ABC and DEF, equiangular triangles such that Angle C = angle F. Angle A — angle D. Angle B = angle E. Conclusion. AB : AB : AB: BC: or BC :: DE : EF, AC:: DE: DF, CA :: DE: EF : FD. Proof. Apply the triangle EDF to ABC so that -P shall coincide with C and FD shall fall on CA, Let D' be the point of CA on which D falls. Then — 1. Because angle F = angle C, UneFE=CB. ijui; Jii ue ine poiux, on wniun ^ laus. LINEAR PROPOJEiTIONa. jgg 4. Therefore '''^"^^- (8 69) CZ)' : AD' :: C^' ; BE'; and, by composition (§ 364), GD' : CA :: GE' : GB; or, because GD' = FD and Cj^' = FE, T .1, I>F '. AG ',x EF '/bG, Q.E.D. m the same way it may be shown that the other nronnr tions of the conclusion are true. ^ ^'' 391. CW/ary. ijT/rom owe ^n«w^/e anotJier he cuf nff by a hue parallel to one of its sides, thetna^ttUuslut S will be similar to the original triangle. ^ Theoeem VI. 392. If the sides of one triangU have to each other the same ratios as the sides of another these triangles are equiangular and similar '^''''^' ^^''' Hypothesis. Two triangles, ABG and DEF, such that AB : BG : GA :: BE : EF : FD. Conclusions, Angle 6^ (opposite AB) = angle i^ (opposite DE), Angle A { « BG) = angle D( <^ fF) Angle B( « <7^) = angle ^( « fD) Proof. On the sides FD and FEot the triangle />;?;?» t^ke the points M and N such that ^ ^^ FM=:GA, EJ^ = GB, and join MJV. Then— 1. Because FM = GA and JYF = BG, and because SGiGA y.EF.FD (hyp.), we have - ly'F : FM II EF : FD. ^L Hi ll lill !l!i 184 BOOK V. rUOPORTION OF MAQNITUDE8, %. Therefore the sides FD and FE are divided similarly at M and N, whence MN II DE, ^§389) 3. Because these lines are parallel, the triangles FDE and FMN are similar (§ 391). Therefore MN : NF'. FM :: DE : EF : FD, (§ 390) 4. Because NF= ^Cand FM - GA, MN :BC :CA :: DE : EF : FD, 6. Comparing with the hypothesis, MN = AB. Therefore the triangle FMN is identically equal to CAB^ and, comparing the angles opposite the equal sides, Angle G = angle F. Ang\e A = angle FMN = angle D; ) /„v 7. ) ^ ^ Q.B.D. Angle B = angle FMN = angle ^. Theorem VII. 393. Two triangles having one angle of the one equal to one angle of the other^ and the sides containing these angles propor- tional^ are similar. Hypothesis. ABG and A'B'G'y two tri- angles in which Angle G — angle C". aA' : G'B' :: GA : GB. Conclusion. The triangles are similar; that is. Angle A — angle A\ Angle B = angle B', AB'.BG: GA :: A'B' : B'G' : C'A', Proof. In GA take GP = G'A'^ and draw PQ parallel to AB, Then— LINEAR PROPORTIONS. 185 I. The triangle CAB is similar to CPQ. (9, 391 \ 11. CPiOQy.CA :GB. (ggsg) 2. By hypothesis, CM' : O'B' :: CA : CB, and CP = G'A'y by construction. Therefore GP : G'B' :: GA : GB. 3. Comparing with (i), II., G'B' = 6'(>. Therefore the triangles G'A'B' and CPg have two sides and the included angle equal, and are identically equal. 4. Comparing«with (1), I., Triangle G'A'B' similar to triangle GAB. Q.E.D. Theorem VIII. 394. Mectilineal figures similar to the same figure are similar to each other. Hypothesis. Two ^. figures, P and Q, each r^\ similar to the figure A.\ \ Gonclusion. P and i ^ / Q are similar to each \^ / other. \ / Proof. 1. Let any ^ side of P, a' for in- stance, be to the ho- mologous side aotA asmm, and the side a be to the homologous side a' of ^ as jt> : ^, so that a' m « = -, n a: a'' = ^. Then, because the antecedent of one of these ratios is the consequent of the next, «-:a" = ^. nq (§ 373) In the same way it may be shown that every side of P has to tiie nomoloorniia airlQ nf /y +>.« ««,», j.z- ... o ^''^ ^^ a "iv o»iuc luiiu mp : no. J u \\ H' ■' 186 SOOK V. PROPORTION OF MAQNITUDES. 2. Because the angles of P and Q are severally equal to the corresponding angles of A, they are equal to each other. 3. The figures P and Q having their homologous sides in the same ratio mp : nqj and their angles equal, are similar by definition (§ 376). Q.E.D. Theorem IX. 395. Similar polygons may he divided into the same nv/mber of similar triangles. Hypothesis. Two simi- lar polygons, ABCDE and A'B'C'D'E', divided into triangles by lines drawn from the corresponding angles G and \C to all the non-adjacent angles. Conclusion. Triangle CDE similar to CD'E', Triangle CEA similar to G'E'A', etc. etc. Proof. 1. In the triangles CDE and CD'E', angle D — D* (hyp. and def.), and CD : C'Z)' :: DE : D'E\ Therefore these triangles are similar and equiangular, and CE : G'E' :: CD: G'D' :: DE : D'E*. (§ 393) 2. From the equal angles DEA and D'E'A' take the equal angles DEC and D'E'G', and we have left Angle CEA = angle G'E'A'. Also, from (1) and the hypothesis of similarity of the two figures, CE : CE' :: EA : E'A\ Therefore the triangles CAE smd G*A'E* are also similar. 3. In the same way it may be shown that all the other triangles into which the polygons are divided are similar. Q.E.D. 396. Def. Similar figures are said to be similar- ly placed when so placed that each side of the one shall be T>arallel to the homologous side of the other. LINEAR PROPORTIONS, 187 Theorem X. t 307. fftwo similar figures are similarly placed, then — I. All straight lines joining a vertex of one to the corresponding vertex of the other meet in a point when produced, II. The point of meeting divides the lines exter- nally into segments having the same ratio as the ho7nologous sides of the figures. ■••^> Hypothesis. ABCD and A'B'C'D\ two eimilar figures in which AB '. BC : CD : DA :: A'B' : B'C : CD' : D'A% and of which each side is parallel to its homologous side in the other. Conclusion. I. The lines A A', BB', etc. (which we shall call junction lines), being produced, meet in a point P. II. AP : A'P :: BP : B'P, etc. :: AB : A'B\ Proof. 1. If the junction lines AA' and BB' are not parallel, they must meet in some one point. Let P be this point. 2. Because AB and A'B' are parallel, the triangles PAB and PA'B' are similar, so that AP : A'P :: BP : B'P :: AB : A'B\ (§391) 3. It may be shown in the same way that if we call the U' 188 BOOK V. PROPORTION OF MAGNITUDEQ. m 1 1 ii A'B' (hyp.), AB : A'B\ point in which the junction lines BB' and CO' meet, we shall have BQx B'Q'.i BG: B'C'x ^ or, because BG : B^G' :: AB BQ : B'Q : 4. Comparing with (2), BQ : B'Q :: BP : B'P, C\ Therefore the line BB' is cut externally at P and at Q into segments having the same ratio; namely, the ratio of the homologous sides of the figures. But a line can be cut only at a single point into segments having a given ratio (§ 382). Therefore the points P and Q are the same; that is, the lines A A' and C(7' cross BB' at the same point. 6. In itie same way it is shown that all the other junction lines intersect dt the point P, and that the segments termi- nating at P have the same ratio as the homologous sides of the figures. Q.E.D. 398. Corollary. If the similar figures are equal, the junction lines will all be parallel and the point of meeting will not exist. 399^ Def. The point in wliich the lines joining the equal angles of two similar and similarly placed figures meet each other is called the centre of simili- tude of the two figures. Theoeem XI. 400. A perpeniiczUar from the right angle to the hypothenuse of a right-angled triangle divides it into two triangles, each similar to the whole triangle. Hypothesis. ABG. a triangle, right-angled at G; GD, a perpen- dicular from (7 on AB. KyUfiVtaCSlUfi. J-iiC l/X iUiIlgilJB JlJJX^, -
    X/, UUU \JJJU iiiO aii similar to each other, so that GB'.GA'. AB ::. DC : DA : AG :: DB : DG : BG. LINEAR PB0P0BTJ0N8. 189 Proof 1. In the triangles ABC and A CD the angle A is common, and the angle G = angle D because each of them is a right angle. Therefore the third angles are also equal, and the tri- angles are equiangular. Comparing the sides opposite equal angles, CBiCAiAB ::DC:DA:Aa Q.E.D. 3. In the same way is shown CBiCA: AB :: DB -. DC : BG. Q.E.D. Corollary 1. Comparing the equiangular triangles ADC and GDB, we have AD : DC :: DC : DB. Therefore DC is a mean proportional between AD and DB, or: 401. The perpendicular from the right angle upon the hypothenuse is a mean proportional between the segments into tohich it divides the hypothenuse. Corollary 2. It has been shown that lines from any point of a circle to the ends of a diameter form a right angle with each other. Therefore: 408. If from any point of a circle a perpendicular he dropped upon a diameter, it will be a mean proportional between the segments of the diameter. Theorem XII. 403. If between two sides of a triangle a parallel to the hase he drawn, any line from the 'oertex will divide the base and its parallel similarly. Hypothesis. ABC, & triangle; DE, a parallel to AB, intersecting ^ C in D nnd BG in E-, GN, a line from G, inter- bocting DE in M and A B in N. Conclusion. DM : ME .: AN i NB. Proof. 1. Because in the triangles GDM I n I 190 BOOK V. PROPORTION OF MAGNITUDES. sides i>if and AN are parallel, these two triangles are equi- angular and similar (§ 391). 2. Comparing the homologous sides opposite the equal angles, DM: AN:: CM: ON. (§390) 3. In the same way it may be shown that the triangles CEM and CBN are similar, so that ME : NB :: CM : GN. 4. Comparing with (2), DM : AN :: ME : NBy or, by alternation, DM : ME :: AN : NB (§ 362). Q.E.D. Corollary. ltAN= NB, the ratio will be one of cqutJity and we shall hajire DM— ME. Therefore: 404. The line drawn from any vertex of a triangle so as to bisect the opposite side will also bisect any line in the tri- angle parallel to that opposite side. Theoeem XIII. 406. The bisector of an interior angle of a tri- angle divides the opposite side into segments having the same ratio as the two adjacent sides. Hypothesis. ABC, any triangle; CDf the bisector of the angle C, yt meeting the side AB in D, so that Angle AGD = angle DGB. Conclusion. AD :DB ::AG: CB. Proof. Through B draw BG parallel to DC, meeting AG pro- duced in G. Then — 1. Because DG and. BG are ■narallelj Angle CGB — corresponding angle ACD. Angle CBG = alternate angle DGB. .lO LtNEAR PROPORTIONS. m 3. Comparing with the hypothesis, Angle OQB .- angle CBG. Therefore the triangle BQCis isosceles, and CB = CO, 3. Because DC and BG are parallel, AD'.DB V.AC'. CG, :; AC : CB (hy 2). Q.E.D. M Theobem XIV. 406. The bisector qf an exterior angle of a tri- angle dimdes the opposite side exterimlly into seg- ments having the same ratio as the two adjacmt sides. Hypothesis. ABC, 9, triangle; CM, the continuation of AG-y CD, the bisector of the exterior angle at C, meeting AB produced in D, so that angle BCD = angle MCD. Conclusion. AD.BD ::AC:BC. A- Prooof. Through B draw BG parallel to DC, meeting ^C in G. The proof will then be so much like that of Theorem V. that it is left as an exer» else for the student. Corollary. Since the bisectors of the interior and ex- terior angles of a triangle ^ each divide the opposite side into segments having the ratio of the other two sides, the ratios of the two divisions are the same. That is. A- ^' N P B AP :BP iiAQ'.BQ. Q 4.A7 VV horn Q lino la rlliri<1/irl lT>+rt-t»*»on-rr n-nA '» ii'-^-J* X.V j-XXiT_- i!j ^J-i T X-vA^^XJ. J.XXI/T7X J^CtiJ_L jr CXiXXVL externally into segments having the same ratio, it is said to be divided harmonically. 192 BOOK V. PBOPOBTION OF MAONITUDM. The preceding corollary may therefore be expressed thus: 408. The bisectors of an interior and exterior angle at the vertex of a triangle divide the base harmonically » Theorem XV. 409. If a line AB he dimded harmonically at the points P and Q, the line PQ will he divided harmom- cally at the points A and B, x P b Q t 1 — f H Hypothesis. A line AB divided internally at P and ex- ternally at Q, so that AP : BP :: AQ : BQ, Conclusion. PB : QB :: PA : QA, Proof. From the ratio of the hypothesis we have, by in- version, 4 BP : AP :: BQ : AQ. Then, by alternation, BP: BQ :: AP : AQ. Q.E.D. HarnK^nic Points. 410. Def. The four points A, B and P, ft of which each pair divide harmonically the line termina- ted by the other pair, are called four harmonic points. Scholium. The relation of four harmonic points may be made clear to the beginner by supposing the line terminating in one pair to partly „ overlap the line termi- r- — ^ i i nating in the other; ^ ^ thus, when the points are harmonically aranged, the line AB is divided harmonically at the points P and Qy and the line P^ at the points A and B. Theorem XVI. 411. The hypothenuse of a right-angled triangle in difiidp.d harmn.nnio.nll'H h'ii n/n/ii nnn.ir nf l/ivifis tTivnoinh the right angle, making equal angles with one of the sides. LINEAR PROPORTIONS. 193 Hypothesis. ABC, a triangle, right-angled at Ci CM, CN, two lines from G, mak- ing angle MCA = ACN, ^d and meeting the base in M .,-''' and JV. Conclusion. The base AB is divided harmoni- cally at M and N. Proof. Produce MC to **' A ^ any point B. Then— 1. Because angle MCA = ACN, CA is the bisector of MCN. Therefore CB, perpendicular to CA by construction, IS the bisector of the exterior angle NCD (§ 83). 3. Because CA and CB are the bisectors of an interior and exterior angle of the triangle MCN, the base MN of this triangle is divided harmonically at the points A and B (§408), 3. Therefore the line AB \^ divided harmonically at the points M and JST (§ 409). Q. E. D. Theorem XVII. 413. The diameter of a circle is dimded har- monically hy any tangent and a perpendicular passing through the point oftangency. Hypothesis. AB, a diameter of a circle; CT, a tangent at T, cutting the diameter (pro- duced) in C; 77), a perpendicu- lar from Tupon AB. Conclusion. The diameter AB \^ cut harmonically at C^* andZ). Proof. Join TA and TB, and produce TD until it cuts the circle in C7". T5 -> — 1. Because AB is a diameter perpendicular to the chord TU, it bisects the arc TU (§231). Therefore Arc TA=z\ arc TU= arc^ £/: 3. Also, Angle ClA = ^ arc TA. (§334) Angle ^ 77) =:i archer. (§235) Therefore Angle CTA = angle ^77). ill I r 194 BOOK V. PROPORTION OF MAGNITUDES. 3. Because the angle ATB is inscribed in a semicircle, it is a right angle. And because the angles CTA and A TD on each side of TA are equal, the line AB is divided harmoni- cally at C and i) (§ 411). Q.E.D. »»i CHAPTER ML PROPORTION OF AREAS. B Theorem XVIII. 413. If two rectangles have equal altitudes, their areas are to each other as their bases. Hypothesis. ABGD and PQB8, two rectangles in which the altitude PR is equal to the c D altitude A G. Conclusion, Area P8 : area ADi.PQ: AB, Proof. 1. Let PQ and AB be to each other asm : n. Then, if P^ be divided into m parts and AB into n parts, the mth part ot PQ will be equal to the nth. part of AB. Through the points of division m parte, draw lines parallel to the sides of each rectangle. Then the area PS- will be divided into m equal parts and AD into n parts, each equal to the parts of PS, Therefore Area PS : area AD :: ^Q : AB, 2. Because this proportion is true how great soever may be the numbers m and n, it remains true when the sides AB and PQ are incommensurable. Therefore the proportion Area PS : area AD :: PQ : AB is true in all cases (§ 359). Q.E.D. Corollary 1. Because the area of a triangle is one half the area of a rectangle having the same base and altitude (§ 301), • i n parts. B s rl I T t ! L 1 Q AREAS, 195 and because aliquot parts of magnitudes have the same ratio as the magnitudes themselves (§ 347), therefore: 414. The areas of triangles having equal altitudes are to each other as their bases. 415. Cor. 2. The areas of all triangles having their ver- tices in the same point and their bases in the same straight line are to each other as their bases. For all such triangles have the same altitude. Theorem XIX. 416. The area of a rectangle is expressed hy the product of its base and altitude. Hypothesis. ABGD, a rectangle; X, the unit of area. D, ^ P Q BiT N Conclusion. When AB and CD are exprtssed in numbers of which the side of X is the unit, then the area ABCD is expressed in units by the product AB X AD. Proof. Let us put a, the number of units in AB; bf the number of units in AD^ that is (§ 352, II.), a = ratio AB : side X, b = ratio AD : side X. The numbers a and b may be either entire, fractional, or in- commensurable. Construct a rectangle MNPQ, having MN = side X and MP = b. Consider MP as a base of this rectangle. Then— 1. Because MN = side X, Area MNPQ : X .. MP -. side X; that is. Area MNPQ : X = b. linrw TLtD A r» iUOC JSLJ. -_i uXL/f ABCD : area MJSTPQ :: AB : MN :: AB : side X; , Area ABCD : area MJSTPQ = a. 9. R 196 BOOK V. PROPORTION OF MAQNITUDES. Compounding the ratios (2) and (1), Area ABCD \ X— ah. (§ 373) That is, the area ABCD is expressed by ab when X is taken as the unit (§ 352). Q.E.D. Corollary 1. Because a triangle has one half the area of a rectangle, with the same base and altitude, we conclude: 417. Th6 area of a triangle is represented by one half the product of its base and altitude. Cor. 2. ItAB = AD, then a = b and ab = a\ Hence: 418. The area of the square on a line is expressed alge- braically by the square of the number of units in the line. We thus prove, for all cases, the theorems proved for whole num- bers only in Book IV., §§ 284, 285, 419. Cor. d. The areas of rectangles or triangles having equal bases are th each other as their altitudes. Theorem XX. 420. If four straigM lines are proportional^ the rectangle contained hy the extremes is- equal to the rectangle contained hy the means. Hypothesis. Four straight lines o, b, c, d, such that a '. b '.: c : d. Conclusion. Rect. a.d = rect. b.c. Proof. Form the rectangles ad and be, , ^ , ^q and place them so that the sides d and c shall be in one straight line and the sides a and b in an- other straight line, crossing the first at P. Complete rect- angle P^. Then— 1. Because the rectangles PQ and a.d have the same alti- tude a, and the bases c and d. a • e I A ■! d P % If—' ill—. A w Kect. PQ : rect. a.d :: c : d. (§413) ABEA3. 107 2. Bacause, taking a and b as bases, the rectangles PQ and b.c have the same altitude c, and the bases a and b, Eect. FQ : rect. b.c ::a :b; ' or, because a : J :: c : eeat,use of this proportion, B' t. AP.PB = rect. CP.PD (§ 420). Q.E.D. Scholium. This theorem and the corollary of the following atxj iwciinvjai micii ttu v>OiiDi«.ci. liiic v;iiuiui3 us Cutting c»uii other externally when they do not meet within the circle. AJUSAS. 1 201 Theorem XXV. 429. I/from a point witJiout a circle a secant and tangent be drawn, the rectangle of the whole secant and tJie part outside the circle is equal to the square of the tangent. Hypothesis, P, a point with- out a circle; PT, a tangent touch- ing the circle at T; PB, a secant cutting the circle at A and B. Conclusion. PA.PB = PT\ Proof. Join TA and TB. Then— 1. Because the angle ^^T stands upon the arc TA, Angle ABT=i angle arc A T. (§ 235) 2. Because PT'i^ a tangent, and ^T a chord. Angle PTA = \ angle arc A T. (§ 234) 3. Therefore angle ABT^ angle PTA, and the triangles P^Tand PTB have the angles at P identical and two other angles equal. Therefore the third angles PTB and PA T are also equal, and the triangles are similar (§ 390). 4. Comparing homologous sides, we have PA : PT:: PT : PB, 5. Therefore Rect. PA.PB = PT' (§ 420). Q.E.D. 430. Corollary. Because the rectangles formed by all secants from P are equal to the square of the same tangent PT, they are all equal to each other. Theorem XXVI. 431, When the bisector of an angle of a triangle meets the base, the rectangle of the two sides is equal 10 the rectangle of the segments of the base plus the square of the bisector. 302 BOOK V. mOPOHTIOI/ OF MAGmTl.DE8. ' Hypothesis. ABC, any triangle; CD, the bisector of the onglo at (7, cutting the base at i>. Conclusion, \ Reot. CA, CB = rect. AD,DB + CD\ -''""x^X Proof, Circumscribe a circle /^ y^'^x \N A OBE «.roui?d the given triangie, and ;[ y^ \ \ continue the bisector till it meets ^\ fe yl* the circle in E, Join BE. Then— \ 1. In the triangles CaD and GEB \ we have '"-... _ Angle AGD" angle BCE (hyp. ). "^ Angle CAD = angle BEC (on same ore BC), Therefore these triangles are equiangular and similar. 2, Comparing the sides opposite equal angles, , CA : CD i: CE : CB. Whence Kect. CA.CB = rect. CE.CD, = Teot{CD-\-DE)CD, = CD' + rect CD.DE, (§ 287) = CD* + rect. AD.DB (§ 428). Q.E.D. Theorem XXVII. 432. M a riglit-angled triangle the area of any polygon upon the hypothenme is equal to the sum of the areas of the similar and similarly described polygons upon the two other sides. Hypothesis. ABC, a triangle, right-angled at A. We also put a, a', a", the areas of any three similar and similai-ly described poly- gons, upon the sides BC, CA, and ABy respectively. a = a -j- a i^OflCiUSlOil. Proof. From A drop AD ± BC. Then- PBOBLEMB IN PROPORTION. 203 1. Because ABC is right-angled at A, and AD is a per- pendicular upon the hypothenuse, DC : CA :: CA : BC. ' DB : BA :: BA '. BG, 2. Because a\ a", and a are the areas of similar poly- gons described upon CA, BA, and 46", a' :a:: DO : BC. a":a::BD:BC. (§427) 3. Therefore, taking the sum of the ratios (§ 366), a' + a" : a :: BD-{ DC : BC. 4. Because BD-\-DC=BC, ibhe second ratio is unity; therefore the first also is unity, and «' + «" = «. Q.E.D. Scholium. This result includes the Pythagorean proposi- tion (§308), as a special case in which the polygons are squares. » » > CHAPTER IV. PROBLEMS IN PROPORTION. A— Problem I. 433. To divide a straight line similarly to a given divided straight line. Given. Aline, AB-, another line,C7/>, divided at the points M und N. Required. To divide AB t similarly to CD. Analysis. By §388 two sides of a triangle are similarly divided by any lines parallel to the base. Therefore, if we put together the lines AB and CD m such a way as to form two siflofl /-if o +«;«».^i^ «n i: , parallel to the third side will divide these two sides similarly. r M illi 204 BOOK V. PROPORTION OF MAQNITUDEa. Construction. 1. Form a triangle ABD, such that AB = given line AB, AD = given line CD, \ BD = any convenient length. 2. Through the points M and N draw MM' and NN' parallel to BDy meeting AB in M' and N'. Then AM' '.M'N' :N'B',:AM:MN',ND. (§388) Therefore the line ^5 is divided at the points M' m^N' similarly to CD, Q.B.F. Problem II. 434. To divide a straight line internally into segments which shall be to each other as two given straight lines. p Given. Two straight lines, y p and q ; a third straight line, AB, Required. To divide AB into segments having the A'^ ^ ^B same ratio 2ii& p to q. Construction. 1. From one end of the line AB draw an indefinite straight line AD. 2. From this line cut off ^ C = ^ and CD = q. 3. 3om DB. 4. From C draw a line parallel to DB, and let E be the point in which it cuts AB. The line AB will be cut internally at E into the segments AE and EB^ having to each other the same ratio as the lines p and q. Proof. As in Problem I. Problem III. 435. To divide a straight line externally so that the segments shall he to each other as two given straight lines. /3V^./.*. A a4-T.oir»lif lirjo A Ti' fwo nfliP.r linfis. «, «, Required. To cut AB externally so that the segments shall have to each other the ratio p : q. PROBLEMS IN PROPORTION. 206 Construction. 1. From either end of the line AB as A draw an indefinite straight line ^C. ' ' 2. On this line cut oft AC = ; the greater line p, and from C, ^ — ~~ toward A, cut off CD = the * ^<^ lesser line q. ^<^^'^^^^^^^ 3. Join DB. ^^^^^ \ 4. From C draw a line par- . ^^^^ \ _^\ allel to DB, and let B be the B " E point in which it cuts AB produced. The line AB will be divided externally at B, so that AH : BU ::p : q. Proof. As in Problem I. 436. Corollary 1. If ^ = g, the point D would fall upon A, and the line DB would coincide with ^^. The line drawn through C parallel to DB would then be parallel to AB, so there would be no point of intersection B. Therefore there would .be no external point for which the ratio of the see- ments would be unity. ^ 43 Y. Cor. 2. If we combine Problems 11. and III on the same straight line, using the same ratio, we shall divide the line harmonically, and the ends of the line, together with the points of division, will form four harmonic points. Problem IV. 438. To find a fourth proportional to three given straight lines. Given. Three straight lines, a, h, c. Required. To find a fourth line, X, such that a : b :: c : X. Analysis. To solve the prob- lem it is only necessary to form two similar triangles, one of which / ^\ shall have a and n for f.wn r.f i+« ^ " — ^^ sides, while the other shall have b as the side homologous to a, The side bomologovis to c win then be x. 3 \ !f n , 4 ^H j^ give AB '.AD ::AG:AE, which is the required proportion. Problem V. 439. To find a rnean proportional between two given straight lines. Given. Two straight lines, AD and DB, Required. To find a third line which shall be a mean proportional between them. a D Analysis, The perpendicular from any point of a circle upon the diam- eter is a mean proportional between the segments into which it divides the diameter (§ 402). Hence Construction. 1. On an indefinite line take the segments AD and DB^ equal to the given lines. 2. On ^ 5 as a diameter describe a semicircle. 3*. At D erect the perpendicular DC, meeting this circle in C. DC will then be the required mean proportional between ADmdDB. ^ , . ^. ceonQ Proof, This may be supplied by the student from §§^0S and 401. 440. Def A straight line is said to be divided in extreme and mean ratio when it is divided into two +« aiinh fhnt thft crreater sesrment is a mean proportional between the lesser one and the whole line. -B -.0 - — r^x-^ a \ s \ D b B PROBLEMS IN PROPORTION, Problem VI 207 441. To divide a straight line in extreme and mean ratio. Construction. Let AB be the given straight line. Then— 1. At one end B of the given straight line erect a perpendicular BOy and take BO = ^AB, 2. Around Cas a centre, with a radius CB describe a circle. 3. Join AC, produce the line A Cy and let ^ and i> be the points a- in which it intersects the circle. 4. From ^ as a centre draw an arc cutting off from AB the length ^^ = ^^. ^ "um ^zj The Une AB will then be cut at i^in such manner that FB : AF :: AF '. AB; that IS, it wiU be cut at i^in extreme and mean ratio Proof, 1. Bemuse AB is perpendicular to the radius CB at Its extremity, B, it is a tangent to the circle. Therefore 2.Bydivision,^^^^^--^^^^^^- (§^^9^ 4) AE:AB-AEi:AB:AD-^AB, (§365) An r^ ^"^ ^"^ ^^^ diameter ED, so that AB - AB = An^ED= AE = AF. Making these substitutions in (2) . AF:FB::AB:AF; ^ ^' or, by inversion, ' FB :AF::AF:AB. 4. Therefore the segment AF is a mean proportional between the segment FB and the whole line AB ^ "^""^ 443. Scholium. This division of the straight line has take i^(7 = FB. the linfi /4 P win v>« iT o A d ^ cut m extreme and mean ratio FB : AF :: ^j^ ; J^ 6^. For, by hypothesis. Il III 1 <\ 208 BOOK V. PROPORTION OF MAQNITUDEB, whence, by inversion and diyision, AF- FB : FB :: AB -AF: AF, Because OF = FB, this proportion is the same as AG : GF:: GF : AF. In the same way, by taking on GF a line Gh equal to -4 G^ we shall divide GF in extreme and mean ratio at h, and so on indefinitely. 443. Corollary. The two segments formed ly the ** golden section" are incommensurable with each other. For suppose ^i^were composed of m parts, and FB of n equal parts. Then when we cut off FG, we should have AG = m — n parts. Cutting this off from GF, we should have in hF a still smaller number of parts. But no part would ever be divided by cut- ting, because when we subtract one whole number from an- other, the remainder is a whole number. Therefore, because every time we cut off we reduce the number of parts by one or more, our last section would take away all the line that was left, and so would not cut it in extreme and mean ratio. | '■ } '■ ( > | i ■■■ | lUmtraUon. If AF had eight ^ O A F B parts, and FB five parts, then by successively cutting off we should have left AQ = Z parts = Qh. hF = 2 parts. Cutting off these two parts from Oh, we should have one part left, and after two more cuttings nothing would be left. The Phyllotaxis. 444. Let us bend the line AB around into a circle, the two ends, A and B, coming together. Let us then take the distance BF in our dividers, and, starting from the point AB, keep measuring off equal steps round and round the circle. The end of each step is numbered from through 1, 2, 3, etc. Then— 1. At every step we shall find the circle divided into arcs of two or three different lengths. PROBLEMS IN PROPORTION. 209 2. At every step the dividers will fall upon one of the longer arcs, and will divide it in extreme and mean ratio. 3. The division-marks will be scattered around the circle more evenly than by any other system of division. It has been considered by botanists that the leaves of plants are arranged around the stem on such a plan as this. This arrangement is then called the phyllotaxis. Problem VII. 445. Upon a gimn straight line to construct a polygon similar to a gimn polygon. Given. A polygon, ABODE) a straight line, PQ. Required. To con- struct upon PQ a poly- gon similar to ABODE. Oonstruction. Let AB be the side of the given polygon to which PQ is to be homologous. Divide the given polygon into triangles by diaff- onals from the point A. UponP^ describe the triangle PQR equiangular to ^ J? a Upon PR describe the triangle PRS equiangular to A OD, and continue the process through all the triangles into which ^5Ci)^ is divided. The polygon PQRSTmW be the one required. Proof. By §395. Problem VIII. 446. To describe a polygon wMch shall he equal to one and similar to the other of two given polygons. Given. Two polygons, P and Q. Required. To construct a third polygon, equal in area to P and similar to Q. Analysis. Because the required figure is similar to Q, the square of any one of its sides will have the same ratio to thfi square of the homologous side of Q that its area has to the area of Q. ii 210 BOOK V. PROPORTION OF MAGNITUDES. Because the area is that of P, this ratio of the squares of homologous sides is the ratio of the area of P to the area of Q. Therefore if we con- / p struct two squares, the one equal in area to P and the other to Q, the sides of these squares will have the same ratio which each side of the required figure has to the homologous side of Q. Construction. Construct the side rf of a square equal in area to P, and the side g of another square d equal in area to Q (§ 321). Take any side MN of Q and find a fourth proportional h to g, d, and MN (§ 438). h Upon h describe a polygon X similar to ^, and having l as the side homologous to MN. This polygon X will be equal in area to P, as well as similar to Q. Proof. Because h and MN are homologous sides of the similar polygons X and Q, Area X : area Q :: h^ : MN*; or, because by construction h : MN :: d : g, Area X : area Q :: d* : g\ But, by construction, d" = area P and g* = area Q, Therefore Area X : area Q :: area P : area Q. Whence Area X= area P. Q.E.F. Theorems for Exercise. Theobem 1. If the ends of two intersecting chords be joined by straight lines, the two triangles thus formed will J)e similar to each other. Theorem 2. If two chords of circles subtend arcs which together make up a semicircle, the sum of their squares is equal to the square of the diameter. Theorem 3. If from any point within a parallelogram lines be drawn to the four vertices, the sum of the areas of THEOREMS FOR EXERCISE. 211 each pair of opposite triangles is half the area of the paral- lelogram. ^ " Theorem 4. If two equal tri- angles on the same base be cut by . a/..\b...Bk a line parallel to the base, equal areas will be cut off from them. Area ABC = DEF. Theorem 6. Lines drawn through the point of contact of two circles, and terminated by the circles, form four chords which are proportional, and the \ lines through the ends of which ^| are parallel. CoTidusions, I. AP : PC :: BP : PD. II. ACWDB. Theorem 6. If a circle be described touching two paral- lels, and from the points of tan- ^ gency secants be drawn intersect- ing the circle in the same point, and terminated by the opposite parallel, the diameter of the circle will be a mean proportional be- tween the segments of the paral- lels. HypotTism. The lines PB and QA intersect at B, a point of the circle Condusion. BQ : PQ :: PQ : AP. Corollary. The same thing being supposed, prove Kect. RA.RB = rect. RP.RQ, Theorem 7. If through any ver-Gr- ^ tex of a parallelogram a line be \ drawn meeting the two opposite sides produced without the paral- lelogram, the rectangle of the pro- duced portion of such sides is equal to the rectangle of the sides of the \ 'ixl tliiciugi aiii. \ ^' Eypothem. OD. BF= AB. BC. ' I 212 BOOK V. PROPORTION OF MAGNITUDES. Theorem 8. If two triangles have one angle of the one equal to one angle of the other, and the perpendiculars from the other two angles upon the oppo- site sides proportional, they are similar. Jlypothem. Angle = angle C AP.BQr.A'P :BQ'. Condumn. The triangles are similar. Theorem 9. If the four sides of an inscribed quadrilateral taken consecutively form a pro- portion, the diagonal having the means on one side and the extremes on the other side divides it into two triangles of equal area. (Book IV., Ex. Th. 9.) Theorem 10. The rectangle of two sides of a triangle is equal to the rectangle of its altitude above the third side and the diam- eter of the circumscribed circle. Theorem 11. The area of a triangle is equal to the product of the three sides divided by twioo the diameter of the circumscribed circle. Theorem 12. If two parallel tangents to a circle are inter- cepted by a third tangent, the rectangle of the segments of the latter is equal to the square of the radius of the circle. Theorem 13. If a chord be drawn parallel to the tangent at the vertex of an inscribed triangle, the portion of the tri- angle cut off by the chord is similar to the original triangle. Theorem 14. If from the middle point A of the arc sub- tended by a fixed chord a second chord be drawn inter- secting the fixed one, the rectangle contained by the whole of that second chord and the part of it intercepted between the fixed chord and the point it is a constant whatever be the direction of the second chord. Show to what square or area the constant area is equal. Theorem 15. If in two triangles any angle of the one is equal to some angle of the other, their areas are to each other as the rectangles of the sides which contain the equal angles. NUMERICAL EXEBCISEa. 213 Theorem 16. If two chords of a circle intersect each other at right angles, the sum of the squares of the four segments is equal to the square of the diameter. Theorem 17. If on each of the sides of an angle having its vertex at two points A and B on the one side and P and Q on the other be taken such that OP : OA :: OB : OQ, the four points A, B, P, and Q will lie on a circle (§§ 244, 392). Theorem 18. If at any point outside of two circles a point be chosen from which the tangents to the two circles shall be equal in length, and from this point secants to each circle be drawn, the four points of intersection will lie on a third circle. Apply § 429, 4, to the case of each circle. Theorem 19. Conversely, if two circles be intersected by a third, and secants be drawn through each pair of points of intersection, the tangents to the circles from the point of intersection of the secants will be equal in length. Theorem 20. If the common secant of two intersecting circles be drawn, the tangents to the two circles from each point of this secant will be equal in length. NuMEEicAL Exercises. 1. If one of two similar triangles has its sides 50 per cent longer than the homologous sides of the other, what is the ratio of their areas ? 2. The owner of a rectangular farm containing 10,000 square yards finds that it measures 5 inches X 20 inches on a map. What are its length and breadth ? II 1 1 ! ' '? 1 1 i h ''i 1 ! '* ^ ik 1 1 i 1 1 1 ' ' ^^S ^H •/.I l^aKl ^^1 mm ■ 1' m I BOOK VI. REGULAR POLYGONS AND THE CIRCLE. CHAPTER I, PROPERTIES OF INSCRIBED AND CIRCUMSCRIBED REGUUR POLYGONS. Theorem I, 447. If a circle he dimded into any numher of equal arcs^ and a chord he drawn in each arc, these chords will form a regular polygon. Hypothesis. A, B, 0, D, E, equidistant points around a circle, separating it into equal arcs; AB, ^ BGf etc., the chords of those arcs. Conclusion. The polygon -4 -BC/)^ is regular (§ 152). Proof. I. Because the sides are by ^\ hypothesis all chords of equal arcs, they are all equal to each other. II. Take any two angles of the poly- D gon, say ^^Cand CDE. Join ^Cand GE. Then— 2. Because the arcs A C and GE are equal, being sums of equal arcs, Chord AG^ chord GE. (§ 208) 3. Hence in the triangles ABG and GDE we have AB = GD, BG = DE, AG^GE. Therefore these triangles are identically equal, and Angle ABG= angle GDE. UfaCBIBED AND CIIWUMaVRIBED POLYGONS. 215 4. In tho sumo way it may bo shown that any other two angles of tho inscribed polygon wo choose to take are equal. 6. Comparing (1) and (4), tho polygon is shown to be regular (§ 152). Q.E.D. Scholium. Tho equality of the angles of the polygon may be proved with yet greater elegance by showing that they are all inscribed in equal segments. Theorem II. 448. If a circumscribed polygon touch a circle at equidistant points around it, it is regular. Hypothesis. A circumscribed polygon whose sides touch tho circle at tho equi- distant points ABODE. Conclusion. This polygon has all its sides and angles equal. Proof. Let be the centre of the circle. Join OA, OB, etc. Then — 1. Because tho intercepted arcs AB, BO, etc., are equal, we shall have Angle AOB= B00= ODD, etc. Turn the figure around on the point until the radius OA coincides with the trace OB. Then — 2. Because of the equality of the angles A OB, BOO, etc., OB will fall upon the trace 00; 00= trace OD, etc. 3. Because the radii are equal, the point A will fall on B, B on (7, etc. 4. Because each radius is perpendicular to the tangent at its extremity, each side will fall upon the trace of the side next following. 5. Therefore each point of intersection will fall on the trace of the point next following. 6. Therefore each side and angle is equal to the side and angle next following, and the polygon has all its sides and angles equal. Q.E.I). Remark. Another demonstration may be found by draw- in or liTiPa frnm f) +.n f.Vio ancrloo t\f +lia nrtWrrATi onrl rkT»rk-«rJr»rf flio equality of all the triangles thus formed. Ill' lit' I ■i\ Il J I I 216 BOOK VI. BEQULAR POLYGOM AND THE OIMCLE. Theoeem III. 449. A circle may be inscribed in any regular polygon^ or circumscribed about it Proof. I. Let ABODE be the regular polygon. Bisect any two adjacent angles of the polygon, as A and B, and let be the point of meeting of the bisecting lines. Then — 1. In the triangle AOB the angles OAB and OB A are by construction the halves of the equal angles EAB and ^^C(hyp.). Therefore ihe angles are equal and the triangle is isosceles. 2. Join 00. In the triangles AOB and BOO we haye BO = AB {hjg.). OB = OB (common sidie). Angle OAB = angle OBC (halves of equal angles). Therefore these two triangles are identically equal, and 0(7r= OB. Angle 00 B = angle OB A = I angle B = ^ angle 0. 3. In the same way it may be shown that if we join to the other angles of the polygon, the triangles thus formed will all be identically equal. Therefore OA = 0B=00= 0D=: OB. Hence if a circle be drawn around as a centre with a radius equal to either of these lines, it will pass through all the points A, B, O, D, B, and will be circumscribed around the polygon. Q.E.D. II. Let Oa be the perpendicular from upon AB. 4. Because the triangles OAB, OBG, etc., are identically equal (2), the perpendiculars from upon AB, BO, etc., are equal (§ 175). Therefore if a circle be drawn around as a centre, with a radius Oa, this circle will also pass through the feet of the perpendiculars dropped from upon BO, upon CD, etc. 6. Because each of the sides AB, BO, etc., is perpen* INSCRIBED AND CIRCXIMaCBIBBD POLYGONS. 217 dicular to a radius at its extremity, it is a tangent to the cipcie* Therefore the circle is inscribed in the polygon. Q.E.D. 450. Corollary 1. The inscribed and circumscribed circles of a regular polygon are concentric. 4:51. Cor, 2. The bisectors of the angles of a regular polygon all meet in a point, which point is the centre of both the inscribed and circumscribed circles, and is equally distant from all the angles and all the sides of the polygon, 4:52. Def. The common centre of the inscribed and circumscribed circles is caUed the centre of the regular polygon. 463. Cor. 3. The perpendicular bisectors of the sides of a regular polygon all pass through its centre. 454. Cbr. 4. If lines be dra^vn from the centre of a regu- lar polygon to each of its vertices, the polygon will be divided into as many identically equal isosceles triangles as it has sides. Theorem IV. 455. All regular polygons having the same num- oer of sides are similar to each other. Proof 1. If the polygons have each n sides, the sum of all the n angles of each is equal ton -2 straight angles (§ 160). 2. Because the angles of each polygon are equal, each angle of each polygon is the nth part otn-2 straight angles; that IS, the angles of one polygon are each equal to the angles of the other. ^ 3. Because the sides of each polygon are equal to each other (hypothesis), the ratio of any side of the one to any side of the other is the same whatever side be chosen. Therefore the polygons are similar (§ 375). Q.E.D. Theorem V. 456. Regular inscribed and circumscribed poly- gons of the same number of sides may be so placed that their sides shall be parallel, and each vertex of the one on the same radius with a vertex of the other. Ill I ! ? I III J ii in iiiiiii ^i ■ r A II ; II ^ 11 i m 11 vl 1 'i ffi 2il 1 '\ 218 BOOK VI. REGULAR POLYGONS AND THE OIROLE. Hypothesis. ABCDE, an inscribed regular polygon; A'B'C'D'E'f a circumscribed poly- gon of the same number of sides, hav- ing the side^'-B' tangent to the circle at jT, the middle point of the arc AB- Conclusions. I. A'B' \\ AB ; B'C II BG, etc. II. The vertex ^' is on the radius OA produced, and all the other ver- tices, B'y C", etc., are on the radii OBy 00 f etc., produced. Proof 1. 1. Because A'B' is tangent at the middle point of the 2ixc ABi it is parallel to the chord of that arc. 2. Because the two polygons are equiangular (Th. IV.) and have a pair of homologous sides parallel, all the other homologous sides are parallel. Q.E.D. Proof II. Because the line B'O is drawn from the inter- section of the two tangents, B'U and B'T, to the centre of the circle, it bisects the arc TU. Therefore it passes through the middle point B of this arc, and OBB' are in the same straight line from the centre of the circle. 457. Oorollary. The polygons may also be so placed that the circumscribed polygon shall touch the circle at the angles of the inscribed polygon. Theorem VI. 458. The greater the number of sides of a regular circumscribed polygon theless will «^ . . f,,^, be its perimeter. Hypothesis. BB', one side of a reg- ular circumscribed polygon having n sides; 00', one side of another such polygon having n-\-l sides; OA, a per- ■nendicular from tlie centre UT»on BB\ Oonclusion. The perimeter of the polygon having n sides is greater than that having w -{- 1 sides. C0N8TRU0TI0N8. 219 Proof, 1. Because the one polygon has n equal sides, and the other w -f 1 equal sides, we have Anglo 0" 0(7 = -^. Angle B'OB= ^^^\ Therefore Angle O'OG : angle B'OB ::n:n-{-l, (§337) And because AOO andAOB are respectively the halves of C'OCand B^OB, Angle AOO : angle A OB ::n :n-{-l. That is, if we divide the angle A 00 into n equal parts, A OB will be % + 1 of these parts. Hence GOB will be one'of the parts. 2. Divide the angle AOC into n parts by straight lines meeting AB in the points «, b, etc. Because OA is a perpendicular upon BB', each of the seg- ments J «, ah, 10, etc., will be longer than the segment next preceding (§116). Therefore n times segment OB will be greater than the sum of the n segments which make up AG, 3. The perimeter of the polygon of n sides is nAB = nAG-{-nGB, and that of the polygon of w + 1 sides is {n-\-l)AG = nAG-\-AG, 4. Because n6'^ is greater than ^C, nAB>{n + l)AO; that is, the perimeter of the polygon of n sides is the greater. Q.E.D. • ♦ > CHAPTER II. CONSTRUCTION OF REGUUR POLYGONS. ^^^' J^]^^^^^^ i- of this book shows that a regular poly- gon of anj, uumber of sides may be inscribed in a circle by dividing the latter into as many equal parts as the polygon has sides, and joining the points of division by chords. Hence 220 BOOK VI. REQULAB POLYGONS AND THE GIROLE. the problem of constructing such polygons is reduced to that of dividing a circle into any required number of equal parts. The following theorem may be seen almost without demon- stration. • If we can divide a circle into any number of equal parts, we can also divide it into twice that number. For if we divide it into n parts, we have only to bisect each of these parts, which we do by § 271, and it will be divided into 2n parts. 460. It is also easily seen that the problem of dividing an arc into any number of parts may be reduced to that of dividing the angle cor- responding to the arc into the same number of parts. For let ABhk the arc, and its centre. If we divide the arc into any number of equal parts, and join to the points of division, the angle A OB will be divided into that same number of equal parts. Hence, by dividing the arc we divide the angle, 461. Conversely, if, having an angle A OB, we draw the arc AB of a circle around the vertex . as a centre, and divide the angle ^,-\'' "?"^^^ A OB into any number of equal parts by the lines Oa and Ob, meeting the Av, arc in a and b, the points a and b will divide the arc into that same number of equal parts. Hence, by dividing the angle we divide the arc. The problem of bisecting the angle (or arc) is so simple that it offered no difficulty to the ancient geometers. By bisecting the halves of each arc it might be divided into fourths, and so on ; therefore there was no difficulty in dividmg any angle or arc into 3, 4, 8, 16, etc., equal parts. This being so easy, it was naturally sought to trisect the angle, or divide it into three equal parts. This problem of the tri section of the angle was long celebrated. But geometers never succeeded in sol" "ng it, and it is now considered impossible by the constructions of elementary geometry. COmTRUCTIONS. 221 Problem I. 462. To dimde a given circle into 2, 4, 8, 16, etc., equal parts. Any diameter, as AB, will divide the circle into two equal parts at the points A and B. Drawing through the centre an- other diameter perpendicular to this, the circle will be divided into four equal ^ I parts by four radii, at right angles to each other. By joining the ends of these radii a square will be described in the circle. Bisecting each of the right angles at the centre by radii like ON, the circle will be divided into eight equal parts. By joining the points of division, a regular octagon will be inscribed in the circle. The process of bisection may be continued indefinitely, so as to divide the circle into 3"» equal parts, where m may be any positive integer. null id Problem II. 463. To divide the circle into 3, 6, 12, 24, etc., equal parts. Analysis. 1. Suppose the division into six parts effected, and the points of division to be A, B, C, 2>, E, F. 3. Draw the radii OA, OB, etc., and join AB, BG, CD, etc., forming Q,Ji regular hexagon. 3. Because each of the three equal angles AOB, BOO, GOD, is one third the straight angle A OD, they are each angles of 60°. And because the sides OA, OB, are equal, the nnrrloa f) J Ti n^j/l /ITf 4 <,r v----t-i-- i.cii\j. \^SJJ1. tlic £ «1 1 4. But the sum of the three angles is 180°. Therefore the three angles are each equal to 00°; that is, the triangle OAB EiU 322 BOOK VI. REGULAR POLYQOm AND THE CIRCLE. is equiangular and therefore equilateral, and the other five triangles, being identically equal to it, are also equilateral. 6. Therefore each of the sides, AB, BG, CD, DE, EF, Fa, is equal to the radius OA of the circle. Hence Construction. 1. Starting from any point A on the circle, cut off the distances AB, BG, GD, etc., each equal to the radius. 2. Six equal measures will T to the point A, and the circle will be divided into six ei^i parts M. the points A, B, (7, etc. 3. The alternate points A, G, E, or B, D, F, divide the circle into three equal parts. 4. By bisecting the arcs AB, BG, etc., the circle will be divided into 12 parts; by bisecting these arcs, the number of parts will be 24, etc. Corollary. The perimeter AB -\- BG-\- GD-\- BE -\- EF + FA of the hexagon ABGDEF is six times the radius of the circle, and therefore three times its diameter. Because each of the six sides is a straight line, it is less than the corresponding arc; that is, side AB < arc AB, etc. Therefore The sum of the six arcs, or the whole circumference of the circle, is more than three tirnes the diameter. Problem III. 464. To divide a circle into 5, 10, 20, etc., equal parts. Analysis. Let be the centre of the circle, and the arc AB one tenth part of the circumference, or 36°. Join OA, OB, and AB. Then— 1. Because the sum of the three angles of the isosceles triangle A OB is ^ 180°, the sum of the angles A and B 7 is 180° - 36° = 144°, and each of these { angles is 72°. Therefore each of the \ angles OAB and OB A is double the angle at 0. OL' — • CONSTRUOTIONS. 223 2. If we bisect the angle OB A by the line BPy meeting OA in P, the angles OBP and PBA will each be 3G°, and we shall have Angle APB = 180° - angle PAB - angle PBA (8 74) = 180° - 72° - 36° = 72°. 3. Therefore the triangle BAP is isosceles and equiangu- lar, and therefore similar, to the triangle OAB. Also, because angle POB = angle PBO, the triangle OPB is isosceles. Hence PO = PB = AB. 4. Because of the similarity of the triangles BAP and OAB, AP : AB :: AB : AO, or ' AP : PO :: PO : AO, 5. Therefore the radius OA is divided in extreme and mean ratio at the point P (§ 440). Hence we conclude: 465. If the radius of a circle be divided in extreme and mean ratio, the greater segment will be the chord of one tenth of the circle. Construction. Divide the radius OA of the circle in ex- treme and mean ratio at the point P (§ 441). Around ^ as a centre, with a radius equal to the greater segment, OP, describe a circle, and let B and Che the points in which it intersects the first circle. The arc AB will be one tenth of the circle and i^C will be one fifth of the circle. By measuring BCoE five times around the circle, the latter will be divided into five equal parts. By successive bisection the circle will be divided into 10, 20, 40, etc., equal parts. Q.E.F. !!l l!ii 1 III tin III! U Problem IV. 466. To divide a circle into fifteen equal parts. Construction. 1. From any point A cut oft AB equal to one third the circle (^ 463V 2. From the same point A cut off AD, equal to one fifth the circle (§ 466). 224 BOOK YL REGULAR POLYGONS AND THE CIRCLE. 3. Arc BD will then be (^ — \) of the circle; that is, ^ of it. 4. Therefore, if we bisect the arc BD, we shall have an arc equal to ^ of the circle. By measuring this arc off 16 times, the circle will be divided into 15 equal parts. Q.E.F. Scholium. The foregoing divi- sions of the circle are all that were known to the ancient geometers. But, a about the beginning of the present century. Gauss, the great mathematician of Germany, showed that whenever any power of 3, increased by 1, made a prime number, the circle could be divided into that number of parts by the rule and com- pass. Thus: 2^'-|- 1 = 3, a prime number. 2» + 1 = 5 2* + l= 17 2" + 1 = 257 Therefore, besides the old solutions, the circle can be divided into 17 or into 257 equal parts. The division into 17 parts by construction is, however, too complicated for « the present work, and that into 257 parts is so long that no one has ever attempted to really execute the construction. it + etc. = iOP X perimeter. Q.E.D. 469. Corollary 1. Because the perimeter of each cir- cumscribed regular polygon is less the greater the number of its sides (§458), it follows that the area of the circumscribed regular polygon is 1p«?s the greater the number of its sides. Cor. 2. It is easily shown that the area of the circumscribed square is equal to the square upon the diameter of the circle. Therefore: 470, The area of any circumscribed regular polygon of more than four sides is less than the square upon the diameter of the circle. 4*71. Scholium. If a regular polygon be inscribed in a circle, and another regular polygon of the same number of sides be circipiscribed about it, the area of the outer poly- gon will be greater than that of the inner one by the surface contained between the perimeters of the two polygons. This surface will be called the included area." ° When the two polygons are so pla<5ed that their respectivo j il III I li;f 226 BOOK VI. BEQULAB POLYGONS AND THE CIRCLE. sides are parallel (§ 466), the included area will be formed of as many identically equal trapezoids, AA'B'By BB'G*Cy etc., as the polygons each have sides; and each apothegm, as OT, will again divide each of these trapezoids into two iden- tically equal trapezoids. When rhe polygons are so placed that the sides of the circumscribed polygon shall touch the circle at the vertices of the inscribed polygon (§457), the included area is made up of as many identically equal triangles as the polygons have sides. Theorem VIII. 473. WTieT^ the inscribed and circumscribed regu- lar polygons ham more than four sides, the included area is less than the square upon one side of the inscribed polygon. Hypothesis. 0, the centre of the circle; BG, a side of the inscribed polygon; FG, a side of xhe circumscribed polygon, placed paral- lel to BC. Conclusion. If the polygons be completed, the included area will be ^i less than the square upon CB. Proof. Join OBF. Drop the perpendicular OQ from upon FG; draw the diameter COE and join BR. Let n be the number of sides of each polygon. Then — 1. Because tl 3 area of the inscribed polygon is made up of 2n triangles identically equal to OBF, and the circum- scribed polygon of 2n triangles equal to OFQ, the inscribed polygon will be to the circumscribed one as the area OBP is to the area OFQ. That is, if we put A, the area of the circumscribed polygon, a, the area of the inscribed polygon, we shall have A :a :: area OFQ : arcE OBP, AREAS. 227 Hence, by (§365) 2. Because the linos BO and FO are parallel, the tri- angles OJJP and OFQ are similar. Because of this similarity. Area OFQ : area OBF :: OQ' : 0P\ (§423) 3. Comparing with (1), and because OQ = OB, both being radii of the same circle, Aia I'.OB^ : 0P\ 4. The included area is equal to A — a. division, A-a:A:'.0B^- OP^ : OB*, 6. Because OFB is a right-angled triangle, OB' - OP' = PB\ Making this substitution in (4), and putting A the diameter of the circle (whence OB = ^D), s, the length of the side CB of the inscribed polygon (whenca PB = ^8), we shall have A-a:A::is':iD'::s':D\ (346) Therefore 8* X A _ A^ A — a = B' ~ Z) 6. But A < D\ or ^ -. i)» < 1 (§ 470). Therefore A-a<8\ Q.E.D. Corollary, % sufficiently increasing the number of sides of the polygons, we can make each side as short as we please and therefore its square as small as we please. Hence: ' 473. If the number of sides of the inscribed and cir- cumscribed polygons be indefinitely increased, the included area will become less than any assignable quantity. Problem V. 474. From the areas of the inscribed and circum- scribed polygons of n sides to find the areas of those hamng 2n sides. Given. 0, the centre of the circle; AB. one of t.h« «i^os Of the inscribed polygon of n sides; OD, one of the sides of tHe circumscribed polygon of n sides, placed paraUel to AB Hill fl Jill t\ 228 ^OOK VI. UEQULAli rOLYUOm AI/JJ X^^ CUWLE. and tangent to the circle at F\ OEF, the perpendicular from the centre upon the tungeut; AF^ FJJ, two sides of the in- scribed polygon of 2n sides; A 0, BH, tangents to the circle at A and B; wherefore OH is one side of the circumscribed polygon of 2n sides (though not parallel to any side of the inscribed polygon of 2n sides). Also, the areas of the triangles OAB and OCB are supposed given, and from them those of the inscribed and circumscribed polygons are found by multiplying by n. Required. I. To find the area OAF (from which the area of the second inscribed polygon is obtained by multiply- ing by 3n). ' II. To find the area OOH (from which the area of the second circumscribed polygon is found by multiplying by 2w). Solution. Let us put ty the area of the triangle OEA, which is one half that of the given area OAB; T, the area of the triangle OFC, which is one hdf that of the given area OCB; t\ the required area OAF; T', the required area OGH. Then — 1. 1. Because the triangles OAF skud O^i^have the same vertex A, and their bases on the same straight line OF, their areas are as OF to OF, or t:t' :: OF: OF. (§416) 2. Because the triangles OAF and OF have the common vertex F, and their bases on the same straight line OC, t' : T '.: OA : OG. 3. Because ^^and GF are parallel, OF : OF :: OA : OG. 4. Comparing with (1) and (2), t:t' ::f : T; that is, the area t' is a mean proportion between t and T. or V= VIT, AJiBAJS. 229 GF II. 6. Because the triangloa FOO and A GOhcLYi OA = OF, and 00 coinnion, thoy arc identically o, the diameter, or 2r; A^ the area of the circle. By § 480 we have A^nr" = \7tD\ But we have just proved that A=^rC=\DG, Therefore \DG = InD^; or 0=ytD, D that is, the number it = 3.14159 the circumference of the circle to its diameter. This number ic is of such fundamental importance in geometry that mathematicians have devoted great attention to its calculation. The preceding method, by which we have found it to six decimals, is the easiest afforded by elementary geometry, but more rapid methods are afforded by the higher mathematics. Dasb, a German omputer, car- ried the calculation to 200 places of decimals. The followmg are the first 36 figures of his result:* 3.141 692 653 689 793 238 462 643 383 279 602 884. The result is here carried far beyond all the wants of mathematics. Ten decimals are sufficient to give the circumference of the earth to the fraction of an inch, and thirty decimals would give the circumference of the whole visible imiverse to a quantity imperceptible with the most powerful microscope. EXERCISES. 1. Assuming the radius of a circle to be 5 metres, com- pute, by the process of § 475, the area of the inscribed and circumscribed regular hexagon, dodecagon, and polygon of 24 sides. . is itself the ratio of Crelle's Journal, Vol. 27, p. 198. 236 JiOOK VI. BEQULAR P0LYO0N8 AND THE OIROLE. Note. In computations like this, the student should not be satisfied by working blindly with the formula), but should reason the results out by the same process employed to reason out the formula}. In the present case the computation of the area of the hexagon is easy; and that of the figures of 12 and 24 sides can then be executed as in § 476. 2. If an equilateral triangle bo inscribed in a circle, show that the perpendicular from any vertex upon the opposite side is three fourths the diameter of the circle. 3. Using the preceding theorem, compute the length of sides and the area of the equilateral triangle inscribed in the circle whose radius is unity. 4. Show that tiie altitude of a circumscribed equilateral triangle is three times the radius of the circle. 5. Without using any of the preceding theorems, show that the radius pf the circle circumscribed about an equilateral triangle is double the radius of the inscribed circle. 6. What conclusion thence follows respecting the relation of the areas of the two circles? (§ 481.) 7. If the radius of a circle is r, what is the length of each side of the circumscribed equilateral triangle ? 8. In a circle of radius r, find the sides of the inscribed and circumscribed squares and their areas. 9. From the results of the preceding example find the areas of the inscribed and circumscribed regular polygons of 8, 16, 33, and 64 sides, and thence the area of the circle, as in §§ 475, 479. 10. A bought a piece of pasturage 30 yards X 40 yards in B's field, and then tied his cow in the centre with a rope just long enough to reach to the corners of his piece. Over how much of B's part of the field could A's cow feed? 11. Four equal circles of radius a have their centres on the corners of a square, and touch each other. What is the radius of the circle in the centre touching each of them? 12. What must be the diameter of a circle in order that its area may be 100 square feet? (Apply § 480. ) 13. In a regular polygon of n sides, what angle (in degrees) floes a xine froni any vertex to the ccntro make with the sides meeting at that vertex? (§ 160.) MAXIMUM FIOUUES. 237 14. If from one vertex of a regular polygon of n sides lines be drawn to all the other vertices, what angles will they form with each other? (Apply § 235.) 16. What is the area of a circle circumscribed about a square whose side is a? 16. If the apothegm of a regular hexagon is h, what is the area of the ring included between its inscribed and circum- scribed circles? ■• » » CHAPTER IV. MAXIMUM AND MINIMUM FIGURES. 485. Def. A maximum figure of a given class. figure is the greatest ^ 486. Def. A minimum figure is the least of a given class. Eemark. If a figure is entirely unrestricted, there can be no such thing as a maximum or a minimum, because a figure, if not restricted, can be made as great or as small as we please! Hence a maximum or minimum figure means one subject to certain conditions; for example, required to have a certain perimeter, or to be included between certain limits, or to have some relations among its parts which prevent it from becom- ing indefinitely great or indefinitely small. Having defined the conditions of the figure, we may imagine ourselves to construct all possible figures fulfilling these conditions. This collection of possible figures will con- stitute a class. The greatest among them will be the maxi- mum; the least, the minimum. * 48T. Def. Isoperimetrical figures are those which have the same perimeter. !„ 238 BOOK VI. BEQULAR POLYGONS AND THE CIRCLE. Theobem X. 488. If two sides of a triangle he given, its area will he a maximum when these sides are at right angles. Proof. Let AB and AP be the two given sides of the triangle. At whatever angle we fix these |t^ p» sides the area will be equal to y AB X altitude, and so will be the rv;--,, greatest when the altitude is j ^'\ / greatest (§ 301). -^ -'-J- If JP is perpendicular to ^i5, AP will itself be the altitude. In any other position, as AP' or AP", thp altitude A'P' or A"P" will be less than ^P(§101). Therefore the triangle of greatest area is BAP, in which AP L AB. Q.E.D. 489. Problem VIII. Having given A straight line M, Two points E and F on the same side of the line: It is required to find the point P on the line M for which the sum of the distances PE -^ PF shall le a minimum. Solution. From one of the given points, as F, drop a perpendicular upon the line M, and produce it to the point F' at an equal distance on the other side. Because the line M is the per- pendicular bisector of FF', every point upon it will be equally dis- tant from F and F* (§ 104). Therefore, if P' be any point upon the line, we shall have EP' -f P'P = EP' + P'i^. The distance EP' + P'P' will be a minimum when P' is in the straight line from E to F'. Therefore the required point P is the point in which the straight line from Eio F' intersects the given line. Draw PF. " MAXIMUM FI0URE8. 239 Because the line M is the perpendicular bisector of the line FF*, we haye Angle MPF' = angle MPF; also. Angle MPF' = opp. angle £PP; whence Anglo BPP' = angle MPF. The solution of the problem is therefore expressed in the following lemma: 490. Lemma. The sum of the distances from a movable point on a straight line to two fixed points on the same side of the line is a minimum when those distances make equal angles with the straight line. Scholium. If the line JIf is a section of a mirror, the lines BP + PF are those which would be followed by a ray of light emanating from a candle at F and reflected to F, because it is a law of optics that the angles of incidence and reflection, namely FPP' and FPM, are equal. Hence: The course taken ly a ray of light emanating from one point and reflected hy a plane surface to another point is the shortest path from the one point to the reflector, and thence to the other point. Theorem XL 491. If the hase of a triangle and the sum of the other two sides he given, the area will he a maximum when these sides are equal. Hypothesis. APB, an isosceles tri- angle on the base AB; AP'B, another triangle on the same base AB, in which AP'-\-P'B = AP + PB. Conclusion. Area. AP'B < area^P^. Proof. Through P draw PJ^IMJ?. ^ Because the areas of the triangles are ^ proportional to their altitudes, it is sufficient to show that the vertex P' must fall below the parallel PF, 1. Because angle PAB =^angle PBA (§ 91), and PF || AB, the sides AP and PB make equal angles with PF. i A ^40 BOOK VI. UKQULAIi POLYGONS AND THE CIRCLE. P' cannot lie on FF, because then wo 2. The vertex should have AF' H- F'B >AF-\. FB. (§ 490) 3. If possible, suppose the vertex F' to fall at any point R above FE. The sides RA and RB will then include a segment of the line FP between them. From any point Q of this segment draw QA and QB. Then AR+RB>AQ + BQ. (§100) AQ -{- QB > AF -{. FB, (§490) Therefore AR+RB> AF-\-FB. Because R may be any point above ^ FF, the vertex F' cannot fall above the line FF. 4. Since it can fall neither above nor upon this lino, it must fall below it, and we must have Alt. of P' < alt. of F; Area AF'B < area AFB. Q.E.D. whence Theorem XII. 492. Among all isoperimetrical polygons of a given number of sides, that of maximum area has all its sides equal. Froof, If possible, let ABODEFhe the maximum polygon of given perimeter and number of sides in which some two ^ adjacent sides, as AB and BC, are un- equal. Join AOy and describe on ^ Can isosceles triangle AB'Oy such that *. AB'-\.B'0=AB + Ba Then— ^ 1. Because ABO is isosceles and AB''i-B'C=AB-{-BO, Area AB'C > area ABC. 2. Because the area ACBBFremamB unchanged. Area AB'CDEF> area ABCDEF. 3. But the polygon AB'GDEF has the same perimeter and number of sides as the polygon ABCDEF. Because the xoi-iner has a gTeater area, ^q latter cannot be the polygon of maximum area. (§491) II MAXIMUM FIOUHIBS. 241 4. Therefore no polygon having two adjacent sides unequal can be a polygon of maximum area; and because any polygon with unequal sides must have some two adjacent sides imequal, rid^S'TEtE^'''"" "' "'"'"""" ^'^^ "--^ ^»- ^ '^ Theorem XIII. •493. 7)r a line of gimn length, which may he mrved at pleasure, is required to 7tave its extremities upon an indefinite straight line, it will inclose a maximum area when hmt into a semicircle. Hypothesis. MN, an indefinite straight line: ADB a curve line wliieh may bo bent at ' pleasure and have its extremities, A and B, rest upon MN, Conclusion. The inclosed area ADBA cannot be a maximum unless ^i)5 is a semicircle. -M" ^ — g Proof. If the line is not a semicircle, there must be some point £ upon It such that the angle ADB shall not be a right angle (§§ 238, 241). Join DA, DB. ^ The ^reaADBA between the curve and the straight line IS then divided into three parts, which we may call AD DB and the triangle ADB. ' ' Bend the curve at the point D, leaving the two branches AD and BD unchanged, and sliding the ends A and B along I^ the line MN, so that the curve shall take up the form AD'B, in which AD'B is a right angle. y^__jc .^ Because the triangles ADB ^ B and AD'B have their sides AD and DB = AD' and D'B and AD'B is a right angle, ' Area AD'B > area ADB, (§ 488) while the other two areas, AD and DB, remain unchanged. Therefore the inclosed area ADBA can be increased with- out ciianging the length of the curve, whence this area ia not a maximum. m ff.M 242 BOOK VI HMO ULAR POLYGOm AND rilE CIRCLE. Honco tho curve cannot inclose a maximum area unless AB subtends a right angle from every one of its points, and it is then a semicircle. Q.E.D. Theorem XIV. 494. Of all areas inclosed by equal perirrieters^ the circle is a maximum. Hypothesis. ABCD, a closed line of given length. Conclusion. ABCD cannot inclose the maximum urea unless it is a circle. Proof. Take any two points, A and G, on the curve so as to divide it into two equal parts. Join A G. Now if ABQ and ADC are not a] both semicircles, suppose the curve- line 7l7?C bent into a semicircle with- out changing its length, the foot A remaining unchanged in position, and let this semicircle be AB'C\ We shall then have Area AB'C'A > area ABGA, (§ 493) If ADC is not a semicircle, we may bend it into the semi- circle AD'C such that Area AD'C'A > area ADCA. Because the two semicircles are equal in length, they are halves of equal circles and the diameters are equal, so that the two points C coincide. Adding the two areas, we shall have Area of circle AB'G'D'A > area ABGDA. • Therefore the area ABCD wiU not be a maximum when it is not a circle. Q.E.D. Theorem XV. 495. A polygon of wMcTi all the sides are given incloses a maximum area when it can he inscribed in a circle. TUKOUEMa FOli EXMIWWE. 243 Hypothesis, if, a polygon inscribed in acirclo; N anv other polygon, having its > » J sides equal in length, num- ber, and arrangement with those of the polygon if. E Conclusion. f Area if > area JV. \ Proof. 1. Upon the sides of N describe arcs of circles equal to the arcs upon the sides of M. Then Area M= area of circle - area of segments. Area N = urea of distorted circle -^ area of segments. 2. Jiecausc each segment around a side of A^ is identically equal to the segment around tlie corresponding side of M the areas of the two sets of segments are equal. ' 3. Because the circumferences of the true circle around M and the distorted circle around A^are equal. Area of circle > area of distorted circle. (8 494^ 4. Compai-ing with (1), ^ Area if > area a: Q.E.D. Corollary. It has been shown that the maximum polygon of given perimeter and number of sides has equal sides. if a polygon with all its sides equal be inscribed in a circle. It must have its angles equal and bo regular (§ 447). Hence: 496. A polygon of wMcli the perimeter and number of sides are given incloses the maximum area when it is regular. Theorems for Exercise. 1. The inclosed area between two concentric circles is equal to the area of a circle whose diameter is that chord of the outer circle which is tangent to the inner one. r^nlv; J^' T""' ^^ \'^''^' '' *^ *^^* ^* ^^y Circumscribed polygon as its circumference to the perimeter of the polygon. 3. The area of the regular inscribed hexagon is a mean proportiona between the areas of the inscribed and circum- scribed equilateral triangles. .f ft' "^r' ^ffP"^?^ 1'^'^"'^^^ octagon is equal to the rectangle of the sides of the inscribed and circumscribed squares. 11 BOOK VII. OF LOCI AND CONIC SECTIONS, CHAPTER I. LINES AND CIRCLES AS LOCI. 497. To fix the position of a point on a plane, two inde- pendent conditions are required. Example. If a point is subject to the condition that it must be two inches below one line and one inch to the right of another perpendicular line, its position is completely fixed. But if the only condition is that it must be two inches below a given horizontal line, its position is not fixed, but it may move along a line two inches below the given one. This last line is then called the locus of the point. 498. Def, The locus of a point is a line or group of lines to which the point must be confined when subject to some one condition. Problem I.* 499. To find the locus of a point which must he at a giten distance from a given straight line. Let AB be the given straight line, and call a the *'*"r ~ ^ given distance. :* Draw MN and PQ parallel ^*""i — ; ^ to ABj at the distance a on ;* ^ either side of it. p—*.-- Q * It is recommended to the student that, before beginning to draw the iocus in these problems, he mark a number of points each fulfilling the required condition, and continue marking until he sees what the locus will be. "> LINES AND CIRGLBS AS LOCI. 246 Every point on either of the lines MN and PQ is at the given distance a from the given line AB (§ 129). It is evident that no other point in the plane can be at that distance a. Therefore the two lines MN and PQ form the required locus of the point at the distance a from AB. 600. Corollary. If the condition were th^t the point must be at the distance a above the line AB, the locus would be the line JfiV^ alone. If the point must be at the distance a Mow th'^ line AB the locus would be P^ alone. ' Problem II. 501. To find the locus of the point which is equi- distant from two given straight lines. Let AB and CD be the two lines, and their point of intersection. Let each of the four angles at ^ 0— namely, BOD, DOA, AOG, COB— be bisected by the respective M lines OJV, OQ, OM, OP. Every point on the bisecting ^ lines will be equally distant from the two given lines (§ 106), and every other point will be un- equally distant. Therefore these bisectors form the required locus. By § 85 they form a pair of straight lines at right angles to each other. Therefore the locus of the point which is equidistant from two given straight lines is a pair of lines at right angles to each other, bisecting the angles formed by the given lines. Scholium. There are two ways of thinking of the rela- tion of a point to its locus which both amount to the same thing. 1. That a row of points, as numerous and close as we choose, lie on the locus. Every one of these points will then fulfill the given condition. i. rfi ''i bl' •B 246 BOOK VII. OF LOCI AND OONIG 8E0TI0m. 2. That the point slides along the locus. The point wiU then fulfill the condition so long as it does not leave the locus. Examples. 1. If we make any number of points on the lines MN and PQ, every one of these points will be equally distant from the lines AB and CD. 2. If we slide a point along the lines MN and PQy it will always be as far from the line AB as from the line CD, Pjjoblem III. 502. To find the locus of the point subject to the condition that it shall he equally distant from two given points. tp Let A and B be the given j points. Join them by a straight j lino, and bisect this line at OA, jO by another line PO^ at right j angles to it. \ Then every point on PQ • ! will be equally distant from the two points A and B, and every other point will be unequally distant (§ 104). Therefore P^ is the locus of the point which is equally distant from A and B. Therefore the locus of a point equally distant from two fixed points is the perpendicular bisector of the straight line joining the points. Peoblem IV 503. To find the locus of the point which is at a given distance from a given point. Let be the given point, and a the given distance. Around as a centre describe a circle -^^^'"---^ with the radius a. / v Every point on this circle will be at the / distance a from 0, and every point either ( inside or outside the circle will be at a less ^ or greater distance from (§ 206). Therefore every point on the circle ful- / LINES AND 0IR0LE8 AS LOCI. 247 fills the required condition of being at the distance a from and no other point does. ' Therefore the locus of the point at the distance a from the point IS a circle of which is the centre and a the radius. Problem V. 504. To find the locus of the point from which a given line suhtmds a right angle. Let ^ J5 be the given line. On ^i? as a diameter describe the circle APB. If P be any point on this circle, the angle APB will be a right angle (§ 238). Prom any point inside or outside of the circle the angle will be greater or less than a right angle (§2*2). Therefore every point on the circle ^ fulfills the required condition, and no \ / other point does. \^ / Therefore the locus of the point '*'* *''' from which the line AB subtends a right angle is the circle described around ^i? as a diameter. We may also say : If the point P slides around the circle APBy the angle APB will always be a right angle. Corollary 1. This result teaches us a curious method by which a circle maybe described. Drive two pins A and B into the surface rep- resenting the plane. Take a common square, and fasten a pencil-pomt into its interior angle P. Then slide the square around on the two pins, and the pencil-point will describe a circle. The pins will be at the extremities of a diameter of the circle. 505. Cor. 2. It may be shown in the same way ^that the locus of all the points from which a given line subtends a given angle different from a right angle is formed of two arcs of circles. (Compare §230.) 248 BOOK Vll. OF LOCIANU CONIC SECTIONS. Problem VI. 506. To find the locus of the point svbject to the condition that its distances from two given points shall have a given ratio to each other. Let A and B be the two given points, and let the given ratio be that of m : w. Let P be any position of p the required point. Join AB, PA, and PB. .^ , , ^ Bisect the angle ^P^ in- Q B ^ ternally by the line PQ, cutting AB internally at Q, and bisect the adjacent exterior angle by PR, cutting AB ex- ternally in R. Then, by the given condition, PA : PB ::m :n. Therefore AQ :BQ ::m : n. (§405) AR : BR '.'.mm. (§ 406) Because of the equality of these ratios, the line ABi& cut hai monically in the points Q and R (§ 407). Because the condition requires that the lines PA and PB constantly have this same ratio m : n, it follows that the bisectors in question constantly pass through the same points Q and R, wherever the point P may move. But these bisectors are at right angles to each other (§ 82). Therefore the angle QPR is a right angle, and the locus of P is the same as the locus of the point from which the line QR subtends a right angle. Therefore the required locus is the circle described around QR as a diameter, the points Q and R being fixed by the c nditions AQ : BQ ::m : n. AR : BR :: m : n. Problem VII. 507. To find the locus of the point from which two adjacent segments of the same straight line sub- tend equal angles. LIMITS OF FIQUBES. 249 Let AB and BO be the two adjacent segments, and P any position of the point. By the condition we must have Angle APB = angle BFG. Therefore PB is the bisector of the angle A PC, and in con- sequence the lines PA and PC fulfill the condition BA:PC::AB:Ba Because the points A, B, and C are fixed, the ratio ABiBC is a constant. Therefore the ratio PA : PC is also a con- stant, and the locus is that of the point whose distances from A and O have a given ratio, AB : BO, to each other. This locus is a circle ( § 50G). Note. The locus may be found independently of Prob. VI. by drawing PD at right angles to PB, and then reasoning as in Prob. VI. • ♦ » CHAPTER II. LIMITS OF CERTAIN FIGURES. Theorem I. 508. If the vertex of an isosceles triangle be carried away from the base indefinitely, each angle at the base will approach a right angle as its limit. Hypothesis, ABO, an isosceles triangle in which CA = OB; OD, the perpendicular from the vertex O upon the base, bisecting the latter. f Oonclusion. If the vertex O be carried out indefinitely along the line DO, produced past /, each of the angles DBO and DAO will approach a right angle as their limit. Proof, Through B draw a line BI parallel to DO, and therefore perpendicular 4.^ 13 71 tU JJU. 1. If BI is not the limit of BO, let j^ some other line, BI', be that limit. .«« 260 BOOK VII. OF LOGI AND CONIC SECTIONS. 2. Because BI' is not parallel to DG, it will meet it if suf- ficiently produced (§ 45, Ax. 11). Call the point of meeting y. 3. By carrying the vertex G beyond y,the angle DBG will become greater than DBI\ 4. Because this is true howeyer small the angle I'BI^ the angle DBG has no limit less than the right angle DBL 6. DBG can never become equal to a right angle, because then the triangle GDB would have the angles D and B both right angles. 6. Therefore the right angle DBI is the limit of the angle DBGy as the vertex Cgoes out indefinitely along the lineZ>»/. In the same way it may be shown that the limit DAG is a right angle. Q.E.D. 509. Gorollaryl. As the vertex C goes out indefinitely, each of the sidep BG and -4 C' will approach the position of par- allelism to the perpendicular GD as their limit, and will there- fore approach indefinitely near to parallelism with each other. Gor, 2. The same thing being supposed, because the angle DGB and DGA are each supplements of GAD and GBDy and these angles approach indefinitely near to right angles, we conclude: 510. The angle at the vertex approaches zero as its limit. Theorem II. 511. IftJie vertex of a right-angled triangle he lengthened out indefinitely, the adjacent side will approach the length of the hypothenuse as its limit. Hypothesis. ABG, Vi right-angled triangle of . which the base AB is fixed, but of which the (] vertex G may be carried out indefinitely along the line A G produced. Gonclusion. However great the distance AB, the vertex G m&j be carried out to such a dii^tance that the difference GB - GA shall be le^s than any length we can assign. Remabk. We mav fixnrpsH thn pnni-»liiqi/ip tr. fv,?« form: How many miles soever may be the base A7i, we can cany the vertex C so far out that the exco:. J CB LIMITS OF FIGURES. 251 over OA shall be less than a foot, less than an inch, less than the hundredth of an inch, and so on indefinitely. Proof, From A draw AD ± BC. Then— 1. Because CD A and CAB are right angles, CA is greater than CD but less than CB. 2. The triangles BDA and ^^Care similar (§ 400). 3. As the vertex C moves out indefinitely, the ratio BA: BC will approach zero as its limit. Therefore the ratio BD : BA will also approach zero as its limit; that is, AB being constant, the point D will approach B as its limit. 4. Then CA, being between CD and CB, will approach CB as its limit. Q.E.D. [ - Theorem III. 512, If the radius of a circle increase indefinite- ly., an arc of the circle of gjiisen lengtJi will approach a straight line as its limit. Proof, 1. Let RT be the length of the given arc. 2. At R erect the line RO perpendicular to RT, and join x OT. 3. From as a centre describe an arc equal to 72 r and passing through R. 4. Let the centre move out indefinitely along the line RO produced, the circle still passing through R, so that OR is its radius. 5. If K be the point in which the circle intersects the hue OT, we sliall have OK = OR. Therefore, as moves out mdefinitely, the point K m\\ approach T^as its limit (§ 511), and the arc of the circle will approach the straight line RT as its limit. Q.E.D. Theorem IV. £^13. Tf P.n.n7}. n-f hnn 1/inne>o nnJinAh /JA-4P^^ J.-. « slant quantUy, are extended indefinitely, their ratio will approach unity as its limit. m if f III ¥ \ m\ 252 BOOK VII. OF LOCI AND CONIC SECTIONS. Proof. Call A the lesser line; />, the constant difference of the two lines; A -\-D, the greater line; r, the ratio of -4 + /> to A ; a, the quantity by which r exceeds unity. Then A + D-rA - (I -^ a)A = A-^ aA, Therefore D = aA, or D : A :: a : 1, Now, we can increase A so that the ratio D : A shall be less than any quantity we may assign. Therefore a may be made less than any assignable quantity, whence r may differ from unity by less than any such quantity; that is, the limit ofrisl. Q.E.D. > > « CHAPTER III. THE €LUPSE. 614. Def. An ellipse is the locus of the point, the sum of whose distances from two fixed points is a constant. Each of the two fixed points is called a focus of the ellipse. 515, To describe an ellipse. Let B and Fhe the foci. Take a thread of which the length shall be equal to the sum of the distances of each point ">f the curve from the foci, which sum is supposed to be given, and fasten one end in each focus. Stretch the thread tight by pressing a pencil-point against it, and move the latter round, keeping it pressed against the thread. The pencil-point will describe an ellipse. Proof. Let P be any point of the curve described by the pencil-point. The sum of the distances of this point from 6UU ioui IB x-jd -j- s-jr, Dui x'ji, -f- I'^f makes up the whole length of the thread which remains constant. Therefore the THE ELLIP8EL 263 1b sum of the distances of P from the foci is equal to this constant, whence P is by definition a point of the ellipse. 616. Axes of the ellipse. In drawing the ellipse there will be two points, A and /?, where the two parts of the thread will overlap each other. The line AB is called the major azifl of the ellipse. Let us put I = the length of the thread. Then AB-{-AF=l;l /^ £!B-\-FB = l\ ^^^ Adding these equations, and noting thai AF = AB 4- EF and EB-EF-\- FB, we have %AE-\-'iEF-^%FB = 'U, Dividing this equation by 3, AE-^EF'\-FB=il, ^"^ AB = l Hence: 61 7. The major axis of the ellipse is equal to the sum of the distances of each point of the ellipse from the foci. The same equations (a) also give AE-\-AF=EB + FB, or 2AE-{-EF=:EF-i-.2FB; whence AE=FB, Hence the foci are equally distant from the ends of the major axis. If be the middle point of the major axis, the distance OA is called the semi-major axis, and is represented by the letter a. Then 2a is the major axis; whence 2a = I, the length of the string. 618. Minor axis. In drawing the ellipse there will be two points, G and J9, equidistant from the two foci. Join these points by the line CD, intersecting the major axis in 0. Af Because, in the quadrilateral ECFD, CE = ED = DF=: FC, this quadrilateral is a rhombus, and the diagonals J^and CD bisect each other at right angles (§173). Hence CD is the perpendicular bisector of the major axis AB, CD is called the minor axis of the ellipse. 254 DOOK VII. OF LOCI AND CONIO SECTIONS, 519. Dtf. The minor axis of the ellipse is that segment of the perpendicular bisector of the major axis which is terminated by the curve. , v Also: The point is called the centre of ^he ellipse. The major and minor axes aro callea the principal axes of the ellipse. The distance of the centra ^ r^om each of the foci is called the linear eccentricity of the ellipse. The ratio of the linear eccentricity to the semi- major axis is called the eccentricity of the ellipse. That is, Eccentricity = OE : OA = — . 520, It is common to use the notation: by the semi-minor axis of the ellipse = 00. c, its linear eccentricity. e, the eccentricity of the elhpse. A The relation between the two eccentricities is then expressed by the equations e = — (because c = OJS). c = ae. , 621. To find the length of the minor axis of an ellipse. Because BOO is a right-angled triangle, and BO = a, b' = a'~ c' =a' (1 - e'). Whence, by extracting the square root, b=a Vr^^, which enables us to determine the length of the minor axis when the major axis and the eccentricity are known. Note. In the older geometry the length OB, which we have called the linear eccentricity, was called the "eccentricity" simply. But the i^-r^iCi ii3 use lliC 'orvi eeeestiieitj tu uumguuic mc ratio oi tiuo lengta OE to tlie major axis, according to the above definition. THE ELLIPSE. B m 255 an ellipse is any straight Tangent 522. D^, A chord of line terminated by two points of the ellipse. 523. Def, A diameter of an ellipse is any chord passing through the cen- tre. R9>± A *«_ . i Diameter. Chord. Theoeem v. .^^^'9^^^!/ point without the ellipse the svrn ellipse. Conclusions. I. PE -\- PF > 2a. 11. QE+QF< 2a. Proof. I. Let T be the point in which the line BP intersects the ellipse. Join TF. Then— ^P + PF= FT 4- TP 4- PF TP-^PF>TF EP -\-PF> FT-^TF. ET+TF^U. (Def. of ellipse.) 1. ^erefore 2. Therefore TT I, . ^P + PP>2a. Q.E.D. J^72 T^ ' ^^ ^^*^^ '^ ^«^*« *^e ellipse in P. Join ^•ff. Then we prove, as in (I.), ^ *^°^^ EQ + QFoiut bout can- Pi^. the P + and in- line vlth (§ 104) (§ 102) (§67) Theorem VII. 528. Tlie line from any focus to the opposite point of the other focus, with respect to a tangent passes through the point of tangency, ' Hypothesis. VT, a tangent to an ellipse having E and F as foci ; F", the opposite point of F with respect to the tangent; ^- P, the point in which the lino iS'iF* intersects VT. Conclusion. P is the point of tangency. Proof, 1. Because Fr is the perpendicular bisector of FF'y PF = PF. Angle TPF = angle TPF' « t> . = angle ^Pr. ,^ „., 2. Because the angles FPT and FPVaro equal, P is the point at which the tangent touches the ellipse (§526). Q.E.D. Corollary. Because P is a point of the ellipse, we have ^P + PP=2a, and because PP' = PF, we have also FF' = 2a. Therefore the opposite point of any one focus is at the dis- tance 2a from the other focus, and we conclude: 529. The locus of the opposite point of one focus, with respect to a moving tangent, is a circle around the other focus with the radius 2a. In other words, if a tangent roll round on an eHipse, the opposite point of either focus will describe a circle round the other focus as a centre with the radius 2a. 530. This theorem and corollary afford an elegant method of drawing any number of tangents to an ellipse with- out drawing the ellipse itself. We need only to know the positions of the foci and the length of the major axis. Construction. Let F and F be the given foci. Around either focus, as F, with a radius equal to the 11 iffl 1 ^IH ■*l£^H UHH T' 1 ^^^1 ■ : il ' WM 1 i SBKBU 1 ^^^H ? * . ! j||^B WM /d^J^ 268 BOOK VII. OF LOOI AND OONIO SEOTIONS. major axis we describe a circle. This circle will then be the locus of all the opposite points of F with respect to the tangents. We draw any line from F to the circle, and bisect this line at right angles by another line. The bisecting line will be a tan- gent to the ellipse. By drawing a number of such lines any number of tangents may be drawn. Theorems for Exercise. I. Each principal axis of an ellipse is an axis of symmetry. II. The ellipse is symmetrical with respect to its centre as a centre of symmetry. III. Every diameter is bisected at the centre. rV. The tangents at the two ends of a diameter are parallel. • ♦ CHAPTER IV. THE HYPERBOLA. 631. Def, An hyperbola is the locus of the point the difference of whose distances from two fbced points is a constant. Each of the two fixed points is called a focus of the hyperbola. 63S. Any number of points of an hyperbola may be found by the intersection of two circular arcs, thus: Let 2a be the constant dif- ference between the distances of a point of the curye from the foci. From either focus as ^ a centre, with an arbitrary radius r describe an arc of a circle. THE HYPERBOLA. 259 the ry. as iel. nt ed he le. From the other focus, with the radius r + 2a, describe an arc intersecting the other arc. The point of intersection will be at the distance r from one focus and r + 2a from the other; the difference of those distances is 2a, whence the point lies on the hyperbola. 633. Corollary. Since there is no limit to the radius r, the hyperbola extends out to infinity. 534. Major axis of the hyperbola. If the constant 2a were greater than the distance between the foci F and F', there would be no point the difference of whose distances from i^and F' could be as great as 2a, and so there would be no hyperbola. Therefore 2a must be less than FF'. Again, if we pass along the line FF' from F to F\ the difference of the distances will be FF' when we start, it Will diminish to zero at the middle point of the line, and will then increase to FF' at the end F'. Hence there must be two points on the line for which this difference is 2rt; that is, two points of the hyperbola. Let A and B be these points. We must then have, by the conditions of the locus, BF- BF' = 2a; that is, FA + AB - BF' = 2a. AF' -AF= 2a; that is, ^ FA + AB -\^ BF' = 2a. The sum of these equations divided by 2 gives m, . AB = 2a. Their difference gives FA = BF'. From these two equations we readily see that the curve cuts the line FF' at the distance a on each side of the middle point of that hue. 535. Def. The distance between the points at which the hyperbola cuts the line joining its foci is called the major axis of the hyperbola. From what has been said we see that the major axis is equal to the common difference of the di&lances from each point of the curve to the foci. Since a point of the hyperbola may be either nearer to F than to F' by 2a, or nearer to F' than to F, the hyperbola consists of two branches. Also, if we draw the perpendicular bisector of FF' through S tl Um BOOK VII. OF LOCI AND GONIG 8EGTI0N8. 0, every point of this bisector being equally distant from F and F'y no point of it can be a point of the hyperbola: Hcljo 536. The two branches of the hyperlola are completely separated. Remark 1. Most of the properties of the ellipse and hyperbola correspond to each other in that where one has sums of lines, the other has differences; where an angle is formed in one, the adjacent angle will be formed in the other, etc. The student should compare the corresponding theorems. Remark 2. Since each branch of the hyperbola extends out tc infinity, it may be considered as dividing the plane into three distinct parts, one within each branch and one between the branches. The two first portions may be considered as belonging to one class, and as lying within the hyperbola— i.e,y within one of its branches— and the last as lying without the hyperbola. Theorem VIII. olH. From every point without the hyperbola the difference of the distances from the foci is less than the major axis. From every point within the hyperbola that differ- ence is g, eater than the major axis. Hypothesis. E, F, foci of an hyperbola; P, any point without the hyperbola; Q, any point within the hyperbola. Conclusions. I. PF-PF<2a. II. QE-QF> 2a. Proof. I. Let N be the point in which the line PF intersects tlie curve, and call ^ the amount by which PF exceeds PF Then A = PE-PF=PE- PN- NF, Because N is on the curve, 9,/» — • JP.W — ATJST Because PE is a straight line, EN 4- PN > PE', whence PE - PN < EN. ' THE HYPERBOLA. 961 ' 1^ Therefore PE-^PN-NF< MN - NF, or J < "Ja. Q.E.D. II. Let M be the point in which QE first intersects the curve, and call A the excess of QE over QF Then A = QE~QF, Because if is on the curve, U=zME-MF =:QE- {MQ + MF), Because ^i^is a straight line, MQ-\-MF>QF. Therefore to form A we take from QE a less line than we do to form 2a, whence A > 2a. Q.E.D. 538. Corollary. Since every point on the plane must be either within the hyperbola, without it, or upon it, we con- clude that, conversely: Every point the difference of wliose distances from the foci is less than %a lies without the hyperbola. Every point the difference of whose distances from the foci is greater than 2a lies within the hyperbola, 539. Problem. Having a straight o^ne passing between the foci, it is required to find that point upon it at which the differ- ence of the distances from the foci shall be the greatest. Solution. Let E and F be the foci; MN, the line; F, the focus nearest the line. Let F' be the opposite point of F relatively to the line, and let EF' produced intersect the line MN on P. Let Q be any point at pleasure on tne line, difference of the distances. Then- Call A the whence QF= QF\ A= QE- QF=QE- QF' i I 1 lit 1 1 ! 262 BOOK VII OF LOOI AND CONIC SECTIONS. 2. So long as g is at any point of the line except P, QEF will form a triangle, and we shall have A^qE^qF' ^t P both an ellipse and an hyperbola be passed, having L and /"as the foci, the tangents to the two curves at P will be per- pendicular to each other (§ 82)." 541. Def. The ellipse and hyperbola are called the conio sections, or simply oonics. 543. Def. Confooal conies are those which have the same foci. 543. Def. A family of confocal conies means an indefinite number of conies having the same foci. 544. Scholium. Two curves which intersect are said to cut each other at an angle equal to the angle between their tangents at the point of intersection. The reason of this appellation is that a curve at any point is considered to have the same direction as its tangent at that point. ; If I ' Theoeem X. 545. In a family of confocal ellipses and hyper- holas, all the ellipses cut all the hyperbolas at right angles. Proof. Let P be any point of intersection of an ellipse with a confocal hyperbola, and PK PF the lines from P to tiic foci. Because P is a point of the ellipise, the tangent to the ellipse at P bisects the exterior angle at P. Because P is a point of the ie laiijgpnii co ine hyprrbola at P bisects the interior angle EPF, 1 1 264 ^OOK Vtl. Otr LOCI AND CONIG 8ECTI0N8. Therefore these tangents are perpendicular to each other (§83), and the curves inter- sect at right angles (§ 644). 546. Asymptotes of the hyperlola. Because the tan- gent to the hyperbola at the point P bisects the angle FPE, it divides the line EF between the foci at the point Q into two such seg- ments that Now suppose the point P to move out upon the hyperbola to infinity, j'he ratio EP : FP will then approach unity as its limit, because the difference between its terms is the finite quantity 2«, while each term may increase to infinity (§ 613). Therefore the point of intersection Q approaches the centre as its limit; and, using the convenient language of infinity, we may say: 547. A tangent to the hyperbola at infinity passes through the centre of the hyperbola. 548. Be/. The tangents at infinity are called asymptotes of the hyperbola. 549. 2h construct the asymptotes. From E and F let the lines ER and FS, parallel to the asymptotes, be drawn. As the point P moves along the hyperbola to infinity, EP and FP will approach EP and FS as their limits (§ 508). On the lines EE and EP take segments EM and EM^ each equal to 2a; then, since PE-PF- 2a, we have PM'= PF, so that the triangle M'PF'is, isosceles and angle M' = angle F. Therefore, as P recedes to infinity, the point M' approaches iJf as its limit, the angles M' and i^both approach right angles as their limit (§ 508). The triangle EMF is therefore right-angled at M. Hence the direction of the asymptote is found thus: On the line EF as a base erect a right-angled triangle EMF, of which the side ^Jf shall be equal to 2a. THE HTPEBBOLA. 265 lii The asymptote will be the line through the centre of the hyperbola parallel to the side EM. To consfruct the required triangle we notice that, since EMFis a right angle, the vertex M lies on the circle of which -fi^^is the diameter. Hence the construe-, , », tion: ^ ^^ On EF as a diameter describe a circle. . From either ^ or ^as a centre, with a g^ radius 2a describe arcs of a circle cutting the first circle in the points if and M\ 3om EMfm^ EM\ ^ "IT^M' \ Through the middle point of the circle draw lines paral- lel to ^Jf and J^if'. ^ These lines will be asymptotes of the hyperbola of which E and i^are the foci, and 2a the major axis. 550. It is easy to prove that if we draw the arcs of circles around the centre F, the chords then found will be parallel to EM and EM', and will therefore lead to the same pair of asymptotes. We therefore conclude that the asymptotes of the i%y;pp,rhola consist of a pair of straight lines, intersecting each other in the centre of the figure* 266 BOOK VIL OF LOOI AND OONIO 8EUTI0Jf& CHAPTER V. THE PARABOLA. 661. Definition. A parabola is the locus of a point equidistant from a fixed point and a straight line. r 553. Def. The fixed point is called the focus of the parabola. 553. Def. The straight line ^ is called the directrix of the parabola. 554. Def. A straight line through the focus, and perpen- ,^, distances PE and PF ar. dicular to the directrix, is called ®^"*^ ^^ whatever point of the thp a■«^i9 of fho -noTKiKnlQ curve P may be placed. AF iu tiie axis 01 me paraOOla. the axis of the parabola. Remark. Since there is no limit to the possible distance of a point from both the focus and directrix, every parabola extends out to infinity. Theoeem XI. 555. I. Mery point without the parabola is nearer to the directrix than to the focus. 11. Every point within the para- hola is nearer to the focus than to the directrix. Proof I. Let P be a point without the parabola. The line from this point to the focus must then intersect the curve. Let S' Q be the point of intersection. Drop the perpendiculai's QS upon the directrix and ^2^ upon Pi?. Then— R S THE PARABOLA. Because Q is on the parabola, Adding PQ to these equal lines 267 also. Because QTP ==m^PQ; PR=:TRJ^TP, is right-angled at T, PQ > TP, Therefore pp > pj^ Q E D Proof II. Let P' be a point within'the parabola. From P drop a perpendicular P'S' upon the directrix. Let oZ the point in which it intersects the parabola Jofn FO' Then we prove, as in the case of the ellipse and hypeZl J^ ' 8'Q' = Q'F, S'P'=:Q'P^Q'P\ Q'P+Q'P^>P'Fr whence s'P' -> p'p Q E D Corollary Since every point in the piane must be either c^4tr ' ''' " "^*'^^* ^' ^^ concludVtl'a:: Theorem XII. focus and the perpendicular \ upon the directrix is a tangent p ^0 the parabola, ^ hypothesis. P, any point of a^ anT^K^l^^^.^^^^^^^^ focus and HW the directrix; PP thA perpendicular upon th^ dirttri. ^P;i=t.S!""2l\/'-'^"^-Sle the Oonclusi parabola i-s-v -I.- J. 'on. PViaa at P. tangent to j; ! f\ m 268 £00K VIL OF LOCI AND OONIO SEOTIONS. Proof, Let V be any point of the line MP V. Join VF, VR, and from Fdrop the perpendicular VW on the direc- trix. Then — 1. In the triangles RP V and FP F, Angle RPV= angle FP V (hyp.). PR = PF {P being on the curve). P V common. Therefore the triangles are identically equal, and VR = VF. 2. Because FW is a perpendicular upon the directrix, VW < VR; whence VW < VF, and the point F is therefore without the parabola (§ 556, II.). 3. Because F may be any point of the line MP V except P, every point of this line except P is without the parabola, and the line touches the parabola at P without intersecting it. Q.E.D. 558. Scholium. The property of the ellipse and parab- ola expressed in this theorem and in Theorem VI., relating to the ellipse, ^^^ leads to the use of these curves in reflec- tors designed to bring rays of light to a focus. Since the curve and the tan- t':^''" — gent have the same direction at the \'?\f?t point of tangency, rays of light are re- Vf^X""''" — fleeted by the curve as they would be by \:V\ the tangent at the point of reflection. ^\„.^\ Because the angles of incidence and ^v?v reflection are equal, it follows that if parallel rays of light, perpendicular to the directrix and therefore parallel to the axis, fall upon a parabolic reflector, they will all be reflected toward the focus. Conversely, if a light be placed in the focus of a parabolic reflector, all the rays from the focus _ will be parallel after reflection. yf^^^^Z-^'' '^ In the case of the ellipse, thecorre- / /^'-/J'-— '"^""--/.M spending property leads to all the rays ri%rvy-"-~^J»V*rJ55'/« which emanate from one focus being ^« "^ reflected to the other focus. I i a f a Si s: u t] THE PARABOLA. 269 oin VF, le direc- ve). brix, »56, II.). V except >arabola, irsecting 1 parab- rix and eflector, •arabolic "SI \ Relations of the Ellipse, Parabola, and Hyperbola. Theorem XIII. 559. The parabola may be regarded as an ellipse of which the major axis is infinite, ' Proof, Let E and F be the foci of an ellipse, and A one eud of its major axis. On the line FA produced take «, AM=EA. -PJf will then be equal to the major axis (§§ 616, 617). From the farther focus -Pas a centre, with a radius FM describe an arc of a circle. Let P be any point of the ellipse. Join EP, FP, and pro- duce FP until it meets the circle in R. Then- 1. Because FP -}- EP = major axis = FJi, we have PE = PR. Therefore each point of the ellipse is equally distant from the focus E and from the arc MR, 2. Now let the focus F recede to infinity along thA line ^^ produced. 3. If MT be the perpendicular to MA at M, the arc MR will approach MTaa its limit (§ 613); and PR will approach parallelism to MF, and therefore perpendicularity to MT as its limit (§ 509). HencQ each point P of the " ellipse will approach a position in which it is equally distant from the focus E and the line MT That is, it will approach a parab- ola having E as its focus and MT as its directrix. Eence: 560. If one focus of an ellipse recedes to infinity, the ellipse will become a parabola about the other focus, Q.E.D. 561. Passage from the ellipse to the hyperbola. Starting as m the last section, let us place the second focus, F, on the opposite F"-- side of the point Jf from E; and let us, as before, draw an arc around the centre i^with a radius FM, Let FR be any radius of this arc; M A -.B Bi I : Hi >. IMAGE EVALUATION TEST TARGET (MT-3) M 4^ A f/j % 1.0 I.I lllll^ M mil 2.0 1.8 • 1.25 1.4 \b ^ 6" - — ^ ► V] VW" /: ^"^ oT!fe .V :»' ''^^J'^ 'I M Oj;^ Photographic Sciences Corporation ^ ■^ ^ NS fV ^v^ ^\ *. ^ ''<:^ '% 23 W£ST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4Sd3 270 BOOK YII. OF LOOI AND CONIO SECTIONS. and produce FR to a point P, such that PR = PE, We shall then haye PF — PE = FR. Therefore P will lie upon an hyperbola of which E and F are the foci, and FM= FR, the major axis (§ 635). Then, supposing F to recede to infinity in the direction MF, we show, as before, that P will approach a parabola of which E is the focus, and a perpendicular to FM through M the directrix. Scholium. The ellipse, parabola, and hyperbola therefore all belong to one class of curves. It is shown in solid geometry that they may all be formed by the intersection of a cone with a plane, from which property is derived the term conic section. Tangents as Limits of Secants. 562. Becafise one straight line, and no more, may be drawn between two points, two points determine a straight line passing through them. If the two points lie on a curve, the straight line passing through them is a secant of the curve. 663. Def. The tangent to a cnrve is the line which the secant approaches, as its limit, when the two points which determine the secant come indefi- nitely near together. This is a more general definition of a tangent than that heretofore given for the circle, and applied to the ellipse. By means of it the fundamental properties of tangents to the circle and conic sections may be established, as follows: 564. The Circle. Let be the centre of a circle, and OA, OB, two b| of its radii. Through A and B draw a secant. a1 Then in the isosceles triangle OAB we have Angle A -f angle B -\- angle = ISO"*, or 2 angle A -\- angle = 180°; whence Angle A = angle P =: 90° ~ ^ angle 0. Now let the T>oint B a'o'nrQaGh indefiniti^l'*'' ji^**"** ^/^ -4- -rx' --v ««V'lVfl. VT^ .£XB THE PARABOLA. 271 P Q The angle will then approach zero as its limit. Therefore the angle A will approach 90° as its limit, and, the tangent being the limit toward which the line ^5 ap'- proaches, must be perpendiculai- to OA. We thus arrive at the property of the tangent demonstrated in § 226. 565. llie Ellipse. Let E and F be the foci of an ellipse and P and ^ two points .upon it. ' Through P and Q pass a secant. Join PE, PF, QE, QF. Now let the point Q approach indefinitely near to P, and, as the E approach becomes nearer, let us look at P and Q through a microscope of which the magnify- ing power may increase indefinitely. Then at the limit, because the angles at E and F become zero, EP and EQ, as also FP and FQ, will seem parallel in the microscope. From P drop the perpen- dicular PR upon EQ, and from Q the perpendicular QS upon FP, Then— 1. EQ-EP=zRQ. FP-FQ = 8P, From the fundamental condition of the ellipse, EQ-\-FQ^EP-\-FP, we haye EQ-EP=zFP-^ FQ, whence RQ = SP. 2. Because the right-angled triangles PRQ and P8Q have Hypothenuse PQ common. Side QR = side P8, they are identically equal, and Angle PQR = angle QPS, But PQR is the angle which the tangent makes with the line EQ to the focus at E, and QP8\s the angle with PP. Be- cause at the limit EP \\ EQ and PP|| QF, the tangent PQ iuakes equal angles with the lines to the foci. (§511) !=»,!=" 272 BOOK VII OF LOCI AND CONIO SECTIONS. 566. The Hyperbola. The same reasoning will apply to the hyperbola, except that the foci E and F will be on opposite sides of the tangent, and in con- sequence the tangent will bisect the angles to the foci. In the case of the parabola the same reasoning will always apply, the lines UP and BQ being re- placed by perpendiculars to the directrix. ♦ • CHAPTER VI. represe'ntation of varying magnitudes by CURVES. 567. The changes of a varying magnitude may be repre- sented to the eye by a curve on a system which we shall illustrate by showing the changes of the National Debt of the United States between 1860 and 1880. In the following figure the horizontal line WJT is divided up to represent the different years. On tl Tiiddle of each 5/i / / wi=^ r--'r 00 s year we erect a perpendicular proportional to the magnitude of the debt at that time, as given in millions of dollars on each perpendicular. Then by noting the length of these per- CURVES. 273 >ply to / BY repre- I shall of the ivided I each nitade ars on le per- pendiculars we have the different magnitudes of the debt presented to the eye. These perpendiculars are called ordinals. f The ordinates only show what the debt was on July 1st of each year, and we may wish to know what it was at other times, supposing it to have varied in a regular way. This we do by drawing a curve through the tops of all the ordinates. Then the height of the curve above the base WJT will show the length of the ordinate, and therefore the amount of the debt. Having drawn the curve, we may erase the ordinates entirely, and get the amount of the debt at any time by measuring the height of the curve above the base line at the point corresponding to that time. 668. Def. A curve showing the magnitude of a varying quantity at any time is caUed a graphic representation of that quantity. 669. The preceding method may be used to show to the eye the relation between two varying quantities connected by an algebraic equation. The following is an example of the process. Let us have the equation y = X + ^• We suppose x to hcve in succession a number of different values, say -4,-3,-2,-1, 0, 1, 2, 3, and 4, and for each of these values we calculate the corresponding value of y. We arrange these values together, thus: l^^^^^ofx -4,-3,-2,-1, 0,1, 2,3, 4. CoiTesp.valuesof2^.. 6, 3i, 2, li, 1, Ji, 2, 3i, 5. We next draw a hori- \ zontal base WX, and lay \ the values of x off on it, the positive values being laid off toward the right, the negative ones toward the left. Then from each point on the line we meas- ure upwards a length equal to the corresponding value of y, and there make a point. TO", N . - ., .. " -4 -J -I -I 274 BOOK YU. OF LOOI AND OONIO SEOTIONS. Let the points be a, J, c, etc. The heights of these points show the several values of y. Should any values of y be negative, they extend below the line. Then we draw a curve through all the points. The height of this curve above the base-line will then represent the value of y corresponding to any value of x between the limits, -f 4 and — 4, so that the eye can see at a glance how y varies. 670. Dtf. AbsoisBas are the values of a?, laid off upon the base-line. 571. Def. The base-line is called the azlf of abfloissas, or the axis of X 672. Drf. The equation which gives rise to a particular curve is called the equation of that curve. Exercises. Plot in the above way the curves correspond- ing to the following equations between the limits a; = — 4 and a; = 4. y = a; -f- 4. 2 y = ~ _ 3a; - 1. 3. y = — - 3a: 2 these IS of y [raw a above inding that id off If of to a rve. ipond- = -4 GEOMETBY OF THEEE DIMENSIONS. BOOK VIII. OF LINES AND PLANES. 573. Def Geometry of three dimensloM treats HT^^ ^ ^^^^ ^""^ ''''* ''''''^''^^ to a single .^f''''"'*7''f.i^''' dimensions is also called the ^.om^/ry of space and solid geotnetry. ^ ^"Hn^ry ^7^ ^/- ^a>^el Plaaeg are those which never meet, how far soever they may be produced. 6 T5. Def. A straight line is said to be paraUel to a plane when it never meets the plane, how far soever It may be produced. 676. The different parts of a figure are said to lie in one plane when a plane can be passed through them alL Remark. Whenever a plane can be passed through a system of lines or points in space, the theorems of plane geometry apply to the figure formed by such lines and points. Otherwise they do not so apply, 677. Axiom I. If two or more points of a straight line lie in a plane, the whole Ime lies m that same plane. Ax. II. Any number of planes mav be passed through the same straight line, and a plane may be turned round on any hne lying Ax. III. Only one plane can pass through a,™, a line and a point without that line paS^^w^^l??^ samestiilghtli,;!^ 276 BOOK VIII. OF LINES AND PLANES. EXERCISES. The following exercises arc inserted here in order that the student may. by working upon them, acquire a clea\^«"««P\^°^ «'J?^"^f ^''■ ^nJe between the relations of lines in space and in a plane before pro- ceeding to the study of the theorems of the geometry of space If four stTfl rods be Jointed together, or merely held together so as to form a auadrUateral it may assist in Exercises 8 and 4. A beginner in the subS^^ it useful to construct the diagrams in space with wires, strines. and rods attached to a flat board. Tpolygon in space may be formed by joimng end to end any number of finite straight lines, as defined in Book II., 8142 The only change is that in Book II. the hnes are all supposed to be confined to one plane, whereas there is no such restriction upon polygons in space. Now: 1 Explain that all the propositions of Book I. whic) reto to triangles are true of triangles in spaxie, however ''^'^Thif may be done by Theorem I., which follows, by showing that the three sides must be all in one plane. j. . t «^i„„«„« a. Explain that these proportions are not true of polygons of four or more sides situated anyhow in space. Show that a polygon of four or more sides (which may be formed of stifl^traTgM may be so joined that its sides shall not all be m one ^^^""t How many different planes may the sides of a quadri- lateral in space contain when taken two and two ? 4. Show that the two diagonals of a quadrilateral in space do not necessarily intersect. ^ , a «^„«^ «« thA Show that each pair of adjacent sides may be turned round on the vertices joining tnem to the opposite pair. 5 But if we draw three lines in a triangle, one from each angle to any point of the opposite side, each of these lines will intersect the two others. 6 Any pair of parallel lines must lie in the same plane. But there may be three lines each parallel to the other ^o, and yet they may not all three lie in any one plane. How many planes will they lie in when taken two and two? 7. If four lines are parallel, how many planes may they lie in when taken two at a time? . 8. A transversal intersecting a pair of parallel lines lies in the same plane with them. student 3 dlfifer- >rc pro- If four form a r in the li wires, to end lok II., are all ao such which lowever ; that the >olygon8 formed of be in one I quadri- in space nd on the rom each lese lines ae plane, ther two, le. How ►? may they aes lies in RELATION OF LINES TO A PLANE. 377 th«til ^r ^'"""^ ,"°.* ''' *^' '^"^^ P^*^« «^« intersected by tl Z! ?^"Tr^' ^^ °^«"y planes may be determined by me three hnes taken two and two? ^ -» ♦ • CHAPTER I. REUTION OF LINES TO A PUNE Theorem L 578. prot^^^ two intersecting straight lines me pcane, and one only, may pass. in ^F'ointo ^""'^ '^'*'^^^ ^'''®'' "^^ ^^ ^^* ^'n Meeting Conclusion. One plane, and A. only one, can pass through them. Proof. 1. Let any plane pass through the line AB. ^ 2. Turn this plane around on AB as an a-ria «*,+,•! •* any point C of the line CD (§ 577 II ) * '"^*' conditions (§ 577 ifl) Q E D ^ ^ "'' *""^*' *''* 580. Cor. a. TAroM^fA anv three, nninu «/.' .V «. , *««y« «„. onefhm, andonlione, n^pasT' "" """^ 278 BOOK Yin. OF L1NB8 AND PLANES. Theorem II. 681. Tkoo planes intersect each other in a straight line. Hypothesis. MN, PQ, two planes intersecting each other along the line AB. Conclusion. The line , . „ AB 18 a, straight line. Proof. Let A and B be any two points common to both planes. Draw a straight line M from A to B. Then — 1. Because the points A and B are both in the plane MN, the straight line AB lies wholly in the plane MN (§ 677, I.). 2. Because the points A and B are both in the plane PQ, the straight line AB lies wholly in the plane PQ. 3. Hence this straight line lies in bo+h planes, and there- fore forms their line of intersection. 4. The planes cannot intersect in any point not in this line, because then two planes would each contain a line and a point without it, which is impossible (§ 577, III.). Therefore ^^ is the only line of intersection. Q.E.D. 683. Corollary. The line of intersection of two planes is a line lying wholly in both planes. Theorem III. 583. If a straight line he perpendicular to two straight lines in a plane, it will be perpendicular to every other straight line lying in the plane and pass- ing through its foot. Hypothesis. MN, any plane; AB, CD, two lines in this plane; OP, a common perpendicular to these lines at their point of intersection 0; BS, any other line lying in the plane, __J wwi XT 1- •^ a w th Tl wl BBLATtON OP LINES TO A PLAm. M 279 Conclusion, The linA np ;« „i Proof. In the lines ' Perpendicular to m ^B and CZ> take the points ^, j5, C and D, 80 that AO^BO, CO =D0, Join ^i) and A C, and let i? and S be the points m which the joining lines intersect the third line R8, (These lines must interserf J?Si i.«« L^. -me plane with it and notScf to it T"'* ''^^ *" "' "'« Jom PC, PR, pj, p^ p -^ Oc'^^nn '^r^'"' <'^^-<' ^S?. tH.et.an«leat?laer,^:^^-f'-^^: BD = Aa Angle (>^i? = angle 0^^. f^gJe OC^ = angle OZ)^. toiLpSi^Tor;^''''' *^« equality Of ^O.i 3. Because OP is perp^n,Uc„,ar to ^^.^ and OA = OB. and, in the same way, PC= pn 4. Beoanse of these equalities, and otBD=AO 5. Because, m the triangles PAR and PBS PA = P^, ^i? = B8, f hooo + • . ^^^® P^^ = angle P^^ • these tnangles are identically equalf and ' pT> -pa Therefore ^gle POsZ^x^gh pqs- whence POifa^dPO^are right ai:,e?^', OP . ^c "• "^"■'"'" ^'^' ""•? •« »-y '^e-whateveriyin^gf-the (3) (2) W 280 BOOK VIIL OF LINES AND PLANES. plane and passing through 0, the line OP is perpendicular to every such line. Q.E.D. 584. Def. A line meeting a plane so as to be per- pendicular to every line lying in the plane and pass- ing through the point of intersection is said to be perpendicular to the plane. 586. Corollary, From a given point in a plane only one perpendicular to the plane can be erected, and from a point without a plane only one perpendicular can be dropped upon the plane. Theorem IV. 586. Conversely, all lines perpendicular to an- other line at the same point lie in the same plane. Hypothesis. OP, any straight line ; 0-4, OC, two lines perpendicular to OP at 0; OB, any third line per- pendicular to OP at 0. Conclusion. OB lies in the same plane with OA and OC. Proof. 1. If OB is not in the plane A 00, pass a plane through PO and OB, and let OB' be the line in which it intersects the plane AOC. 2. Because OP is perpendicular both to OA and OC, it is also perpendicular to OB', which lies in this plane (§ 583). 3. Because OB' is in the plane POB, we have in this plane two straight lines, OB and OB', both perpendicular to OP, which is impossible. 4. Therefore OB and OB' are the same straight line, and OB lies in the plane A OC. Q.E.D. 587. Corollary. If a right angle be turned round one of its sides as an axis, the other side will describe a plane. Theorem V. > 688. ^ a plane bisect a line perpendicularly, every point of ihe plane is equally distant from the ends of the line. BBLATlOif OF Lims TO A PIANB. 281 intone tt afl" "" '""'^'""'"'" '^ ""> »"- ^^. middle point of the |P line ; i?, any point in the plane. Conclusion, R ig equally distant from P and Q. Proof. Join OR, Then— 1. Because P^ is perpendicular to the '0 plane, it is perpendicular to OR in that plane (§ 684). 2 Because OR is perpendicular to PQ at its middle point 0, It 18 equally distent from P and Q (§ 104). ^i D 589. Cbro//«ry. Conversely, mr^^ ;>om^ ^A,-^;, •, ,'„ distant from two fixed points is in the plane bisecting TriZ angles the line joining the points, ^ ^ Theorem VI. ir/^rtUrP^^' *'" I«!^"^-'- to a plane and Q. Conclusion. These perpendiculars are par- allel. Proof. In order to proye the parallelism « of the lines, we must show — I. That they can never meet; T Tf+i, ?■ T?"** *^®y ^^e in the same plane. - ^-;- ■, '.1 !■ 282 BOOK VIIL OF LINES AND PLANES. II. From the point P draw in the plane JlfiV the line PAy perpendicular both to PQ and to PR. In the line Q8 take QB ^ PA. Join BP, BA, QA. Then— 1. Because, in the right-angled triangles QPA anl PQB, AP = BQy PQ common, these triangles are identically equal, and BP = AQ. 2. Because, in the triangles BQA and BPA^ AQ = BP, BQ = AP, AB common, these triangles are idexitically equal. But BQA is a right angle (§ 584). Therefore tue corresponding angle APB is also a right angle. 3. Therefore from the point P of the line AP there pro- ceed tLiree straight lines PQ, PB, and PR, all at right angles to AP. Hence these tlnee lines are in one plane; that is, PR is in the' plane fixed by the two lines PQ, PB, 4. But QS is also In this plane, because it intersects these lines (§ 579). Therefore QS and PR are in the same plane. 5. Kence the lines PR and QS are in the same plane and never meet, and are therefore parallel. Q.E.D. Theorem VII. 591. Conversely, if one of several parallels is porpendicv^ar to a plane, each of the others is also perpendicular to tftat plane. Hypothesis. A plaiie, MN', two parallel lines, PR and QS, intersecting the plane at P and Q in such manner that QS is perpendicular to the plane. Conclusion. Pi2 is also per- pendicular to the plane. Jf^ Proof. 1. n PR is not perpendicular to the plane, let PR' be perpendicular to it. /' 2. Then PR' is parallel to ^ QS (§ 590). 3. Therefore' through the point P we hare two straight lines, PR and PR', both parallel to QS, which is impossible. BT. B 7" h I RELATION OF MNES TO A PLANS. ggs 4. Therefore the perpendicular to the plane at Pis the line Pi?. Q.E.D Corollary. It two or more lines are each parallel to the same straight line, and a plane be drawn perpendicular to this straight line, it will also be perpendicular to the other lines, and they wiU be parallel to each other. Hence: 593, Lines parallel to the same straight line are paral- lel to each other. V Theoeem VIII. 693. Frora any point above a plane lines Tneeting the plane at equal distances from the foot of the per- pendicular are equal, and the line meeting the plane at the greater distance from this foot is the greater. Hypothesis. MN, a plane; P, any point outside of it; Oy the foot of the per- pendicular from P; A OG, any straight line in the plane through O'j Ay B, two points in the plane equally dis- tant from 0; G, a point more distant from than A is. Conclusions. I. PB = PA. II. PC > PA. Proof. 1. In the triangles POA and POB, PO is common. OA = OB (hyp.). Angle POA = angle POB (both right angles). Therefore hypothenuse PB = hypothenuse PA. Q.E.D. 2 Because is the foot of the perpendicular from P on AC, and 0C> OA, BC>PA{^10B). Q.E.D. „„^. .-^,^,,^,y ^. ^j i,nrougn ihe centre of a circle a line he passed perpendicular to its plane, each point of this Aue IS equally distant from all points of the circle. acz= 284 BOOK VIIL OF LIITES AND PLANES. ' 595. Cor. 2. Equal lines meet the plane at equal dis- tances from the foot of the perpendicular, and greater lines at greater distances. ^ This corollary may be expressed as follows: 596. The locus of the point in a plane at a given dis- tance from a fixed point without the plane is a circle drawn around the foot of the perpendicular as a centre. 597. Def. The pr<\jection of a point upon a plane is the foot of the perpendicular dropped from the point upon the plane. Example. If MN be a plane, and P a point outside of it, and if the perpendicular from P upon the plane meet the latter in P', then P' is the projection of P upon the plane Mm 598. Def. Tlie projection of a line upon a plane is the locus of the feet of the perpendiculars dropped from every point of the line upon the plane. Example. The line P'R' is the projection of the line PR upon the plane ifiV". Theorem IX. 599. The projection of a straight line upon a plane is itself a straight line, and the straight line and its projection are in one plane. Hypothesis. ifJV,aplane; P^i?, a straight line; P'Q'R', the projection of PQR upon JOT. Conclusion. P'Q'R* is a straight line, and lies in one plane with PQ. Proof. 1. Because the lines PP' and RR' are perpendiculars to the ^N plane MN^ they are parallel to each other, and therefore in one plane (§ 690). BBLATION OF LINES TO A PLANE. 335 2. Pass a plane through these lines. Because this plane contams the points P and R, it will contain the point which lies on the Kne PR. ^ ^' 3. Because the line QQ' is perpendicular to MN, it is also parallel to PP^(%m); and because it contains 'the pott Q, the plane PP'QQ' is the same as the plane PP'Rr' Therefore the foot Q' lies in this same plane. ' ^^ ^^ ' T ^;fi^o?wl *^® intersection of two planes is a straight me (§ 681), the foot Q', which lies on the intersection of the two planes, is m a straight line with P' and R' 5. Because ^ may be any point on PR, the* projection of every point of PR is in the straight line P'R\ Q.E.D /,w^^^' ^,V^\y ^' Va line intersect a plane, its projec^ tion passes through the point of intersection. ^^ 601. Cor. 2. If aline le perpendicular to a plane its pro^ectnn upon the plane is a point; namely, thepTnt n which %t intersects the plane. Theoeem X. ar,nJ^^.nJ{^, ^m^ mjJ^^^ciJ a plane, it makes a less angle with its projection than with any other line in the plane passing through the point ofintersecZi vi.S'aroAS;^^^^^^^ "^' ^ ^^^^ ^^^--^^^^ *^^« jection of OD upon the plane; 0^ any other line m the plane passing through 0. Conclusion. The angle ^OA is less than BOB. Proof. Take OB = OA, and join AB and DB. Then— of ij, ^'""^'^ ^^^ '' ^ ^^'' '^^gl« (^ being the projection DB > DA. (§593) d8d BOOK VITL OF LINES AND PLANES. 2. Because the triangles DO A and DOB have the side DO common, and OB = OA, while the third side DB is greater than the third side DA, Angle DOB > DOA (§ 116). Q.E.D. 603. Def. The angle between a line and its pro- jection on a plane is called the inollnation of the line to the plane. l: 1! Theorem XI. 604. ^ a line intersect a plane — I. The angle which it makes with a line in the plane passing through its point of intersection is greater, the greater the angle this last line makes with its pfojection. II. The line makes equal angles with lines at equal angles on both sides of its projection. Hypothesis. MN, a plane; OD, a line intersecting it in O'y OA, the projec- tion of OD on the plane; OB, OB', two lines from making equal angles with OA; 00, a line making a still greater angle with OA. Conclusions. I. Angle DOC > angle DOB. II. Angle DOB = angle DOB\ Proof I. Take OB, OB', and 0C7aU equal to OA. Join DB', DA, DB, DC, AB, AB', BC. Then— 1. Because the points B, A, B', C are all in the same plane and equally distant from 0, they lie on a circle having as its centre. 3. Because angle AOC> A OB, the distance ACia greater XI it- - -i J J r»- iiuan lui) uuoiu ^x>; XT *— ~ DC>DB. (§ 693) RELATION OF LINES TO A PLANE. 287 3. In the ti-ianglos ODB and ODC we have 0/? common; 0C= OBy DC> DB, Therefore ? Angle DOG (opp. DC) > angle DOB (opp. DB), (§ 115) Q.E.D. Proof II. 4. In the triangles A OB and u4 OB' we have 0^ common; OB = 0-ff' (construction); Angle AOB = angle ^ 05' (hyp. ). Therefore these triangles are identically equal, and AB = AB\ 6. Because DAB and DAB' are both right-angled at A and because /?^ is common and AB = ^i5', these trianries DAB and jD^jB' are identically equal, and DB = DB'. 6. Therefore the triangles DOB and J905', having their sides equal, are also identically equal, and Angle Z>05 = angle Z?(?5'. Q.E.D. Join Theorem XII. 606. At the point of intersection, a line in the plane perpendicular to the projection of a line is perpendicular to the line itself. Hypothesis. OD, a line intersecting the plane in 0- OA the projection ot OD or , ^, upon the plane; POQy a line in the plane perpen- dicular to OA. Conclusion. Line POQ ± OD. Proof. Take the points P and Q at equal distanc as from 0, and join AP, AQ, 1. Because the points P and Q are at equal distances from biiv xv^uu {y ui mc purpuudicujar ^ c/ on PQ, we have !288 BOOK VIIL OF LINES AND PLANES. 2. In the trianglea DAP and DAQ, DA is common; AP=^AQ', s (1) Angle DAP = angle DAQ (hyp.). Therefore DP = PQ. 3. In the triangles /)0i^ and i>Og, i>0 is common; DP = DQ; (2) OP = 0^ (construction). Therefore Angle DOP = angle DOQ; and because PO^ is a straight line, both these angles are right angles. Q.E.D. Corollary. If the line OD is not perpendicular to the plane MN, there can be only one line in this plane perpen- dicular to OD. For if there were two such lines, OD would be perpendicular to the plane (§ 583). Hence, because POQ is the only perpendicular to OD in the plane : 606. Conversely, a line in a plane perpendicular to an intersecting line is perpendicular to the projection of the inter- secting line. Sdwlium. An astronomical illustration of these theorems is afforded by conceiving one's self to be looking at the sun in the south. The plane is that of the horizon, in which the observer must suppose himself situated at the point 0. Let the line OD be that toward the sun. (It is not necessary to suppose it cut off at D or any other point, because our theorems do not refer to its length.) Then the horizontal line OA from the observer to that point of the horizon under the sun will be the projection of the line to the sun. By Theorem X. the angular distance of the sun from this point will be less than from any other point of the horizon. This angle is called the sun's altitude. If we suppose a horizontal east and west line, Theorem XII. shows that this line will always be at right angles both to the direction of the sun and to the south line which passes directly below the sun. If we take a series of points along the horizon, starting from the point directly below the sun, Theorem XI. shows that the angular dis- tance of these points from the sun will increase up to the opposite point of the horizon from that below the sun. BBLATION OF LINES TO A PLAXB. 289 (1) (8) are the Theorem XIII. 607. When two straight lines are parallel, each of them is parallel to e&erp plane passing through the other and not containing both lines. Hypothesis, AB, CD, two paraUel straight lines; MI^, any plane passing through CD, Conclusion. AB ia a.- parallel to the plane JfiV. Proof. Let us call P* the common plane of the parallels AB and CD. Then— ~B 1. Because AB lies wholly in the plane P, if AB meets the plane Mli at any point, that point will be common to the plane P and the plane MN. 2. But the only points common to these two planes are on their line of intersection; namely, the line CD (§ 682) Therefore if AB eyer meets the plane JfJV, it must meet this line CD. 3. But, by hypothesis, it is parallel to CD, and so cannot meet it. 4. Therefore it cannot meet the plane JO^, and therefore IS parallel to it (§ 575). Q.E.D. 5. But should the plane JfiV^ coincide with the plane P, the line AB will then lie in JfiV^as it does in P. Theorem XIV. 608. Angles of which the sides are parallel and similarly directed are equal. Hypothesis. BOC and P'O'C", two angles in which OPilO'P'and 0C\\ O'C, Conclus ion. Angle BOC— angle B* 0' C\ * The letter Pis here employed not as a mark on the diagram, but as a short and convenient appellation of the plane referred to. Such a use of letters is of constant occurrence in the higher geometrv and should be well understood. ^ ^' '^90 BOOK VIIL OP LINES AND PLANES. Proof. On the sides B and B' take OB = 0'B\ and on the sides C and C take OC = 0'C\ Join BB', CC\ 00', Then— 1. Because 05 and 0'^' are equal and parallel, the q^ figure OO'BB' is a parallel- ogram (§ 138), and BB' = and || 00\ 2. In the same way, CC: = and || 00\ 3. Therefore BB'CQ' is a parallelogram, and BC 4. Therefore the triangles i^OC'and B'O'C, having the three sides of the one equal to the sides of the other, are iden- tically equal, and Angle BOC = angle B'0'C\ Q.E.D. 609. Corollary. It may be shown, as in Book II., Th. VI., that if the sides of the angles are parallel and oppositely directed, the angles will still be equal, and that if one pair of sides is similarly directed and the other oppositely directed, the angles will be supplementary. Theorem XV. 610. Parallel lines intersecting the same plane make equal angles with it. Hypothesis. OA, PB, two parallel lines intersecting the plane MN in and P; OA', PB'y the projections of certain portions of these lines on the plane. Conclusion. Angle A 0^'=:angle5P5'. Proof. At the points and P erect the per- pendiculars OR and P8. Then— 1. Because OR and PS are perpendicular to the same plane, they are parallel (§ 590), while OA \\ PB^ by hypothesis. Hence Angle AOR = angle BPS. (§ 608) and BBLA TION OF UNES TO A PLANS, ggj an/'^^T'" ^^i' * traneversal crossing the parallels A' A and OR It IS m the same plane with them. Also becau^ OR ± plane if^; OR 1 OA' m that plane (§584^ The^m^ ehmgs are true of Pi?', PB, and PS. 3. Because A' OR and B'PS are right angles. ^ 0^' is thfi complement of the angle A OR, and BPB^ it t^e compleme^ of the equal angle BPS. Comparing with (1), '''''^^^^'^'"'^ Angle A OA' = angle BPB\ Q.E.D. Theoeem XVI. n^^^\' ^^^^^^ ^^^ ^^'^^^ ^ot in the same plane me and only one, comTnon perpendicular can hed^raZ Hypothesis. AB, CD, two lines not in the same plane and therefore passing each other without intersecting to bi"^rn/^r '' ^" ^"^^'"^ ^^ -°^^' p-p-^-^- Ipf ,> r"-^' ^' T^'"''''^^ '^''' ^^^^' «^y ^A pass a plane, and let t turn round on OD until it is parallel to AB. Let ^^ be this plane Let A'B' be the projection of AB on the 1. Every point of A 'B' is fixed by dropping a pernendion ar from some point of AB. Let p\, the^ofn' ofThicr^ IS the projection. Then PO 1 plane MN (§ 597) lar t; bofbT f ^ ^^P;^r^li«"I^r to JfiV, it is perpendicu- lar to both the lines A'B' and CD, which lie in MN. o. Hecanae PO is nArnGnri]V.«]o^ f^ .«/n/ .^ • pendicular to AB, which is parall 3 ^'5' (8 72 Therefore OP is perpendicular to both t OD. Q.E.D. lines AB and 292 BOOKVIIL OF LINES AND PLANEa. II. If there is any other common perpendicular, let it be P'Q, Through ^ draw, in the plane JfiV,^i?lMi?. Then— 4. Because P'Q is perpendicular to AB, it is also perpen- dicular to QR, which is parallel to AB. 6. Because P'Q is perpendicular to both the lines QE and CD, it is perpendicular to their plane JfiVi 6. Put, because A'B' is the projection of AB, the foot of the perpendicular from P' on the plane must fall on some point of A'B'. Let 0' be this point. Therefore from the point P' are dropped two perpendiculars P'O' and P'Q upon plane ifiV", which is impossible (§585). Therefore P'Q is not a common perpendicular to the lines AB and CD, and PO is the only common perpendicular. Q.E.D. Theorem XYII. 613. The least distance between two indefinite lines which do not intersect each other is their corn- mon perpendicular. Hypothesis, a, h, two lines in space, the one being sup- posed to lie behind the other, so that they do not intersect. Conclusion. No line which is not perpendicular to both lines can be the shortest line between them. "-^ P b Proof. If possible, suppose that some line PQ which does not make a right angle with a is the shortest line. From P drop a perpendicular PR upon a. Then PR < PQ, Therefore PQ is not the shortest line. JtXLATIOJfS OF PLANBa. 293 Shortest """^ *' """^ ^° ""^ ^^'^^^ '»'"« Ji"" ■»"»' be the Mn^XZy^ "°' " "'"' ""*'"« " "«•" '»«'« »'"» both i>>» ® CHAPTER II. REUTIONS OF TWO OR MORE PUNES. Theorem XVIII. inauL w*^ ^"'''* <^re parallel ^any two intersect- tng lines on the one are both parallel to the other plane parallel to the plane PQ. Conclusion, The planes i/"i\randP^ are parallel. J'roof, 1. If the planes are not parallel, they must intersect in a straight line lying in both planes, and therefore in the plane MM Let us call this line X * ^ ^Aan'rwtr 'c:t?he 5:: 7o' "^^ "^ f ' r. ^"^ '>"'- are paraUel. Q.E.D '^ ^" ^''««*°"^ *hese planes intVstt ImliV '"^ '''^'^^ ^^-o P<^ranel pUnes traersect a third plane are parallel. 294 BOOK VIIL OF LINES AND PLANES. Let ns call A and B the parallel planes,* and Xthe third or intersecting plane. Then — 1. The lines of intersection are in one plane, because they both lie in the plane X. 2. Because one of the lines of intersection lies in the plane A, and the other in the plane By parallel to it, and because these planes never meet, the lines can never meet. Therefore the lines are in one plane and can never meet, and so are parallel, by definition. Q.E.D. , Theorem XX. 616. Parallel planes intercept equal segments of parallel lines. ' • Hypothesis. MN, PQ, two parallel planes; AB, CD, two parallel lines intersecting the planes in the points A, B, C,D. Conclusion. AB = CD. Proof. 1. Join ^ C and BD. Consider the plane containing the parallels AB and CD. Because the four points A, B, C, and D all lie in this plane, the joining lines A C and BD lie in it. 2. But because the lines AC and BD also lie in the respective planes MNand PQ, they are the lines of intersec- tion of these planes with the plane A BCD. 3. Because the planes ifiVand PQ sre parallel (hyp.). ILo lines of intersection A C and BD are parallel (§ 614). 4. Because AB || CD (hyp.) and AC || BD, as just shown, A BCD is t* parallelogram. Therefore Ali- (7i>(§127). Q.E.D. * This theorem iti so iiimple that the student can imagine the figure which is to embody the hjrpothesis and conclusion better than it can be j'ork'ppaonto/i in a dlaTam. We therefore ffive- the demonstration with- out a diagram. 2 i 1 RELATIONS OF PLANES, 905 Theorem XXI, 616. Planes perpendicular to the same straight line are parallel or coincident. ^ Hypothesis. Two planes, MN and PQi OH, a line ner- pendicular to each of these planes. > ^« per Conclusion. The planes are parallel Proof. If they are not parallel, tKey must intersect. If they in- tersect, call Xany point on the line of intersection and join OX, RX, Then- 1. Because OX is in the plane S.'' " P^^P^"^^^"!^^ *« OR, a perpendicular line to the to 0k^'°'"'' ^^'' '"^ *^' ^^'"' ^^' '^ '' ^^«« perpendicular 3. Therefore from the same point, X. we have two r.«r rrs^/' ""' ^''' '" ''^ ^' «t-4hTiSrwS 4. Therefore the planes never meet, and so are parallel. hJi^J'i ^'"■""''7- Conversely, a straight line perpendictl Ur to aplam ts also perpendioular to every paralhl plane. Theokem XXII. *«*; f ^*™^S^ ^««« ■m« student angles. ^ Propositions respecting ordinaryplane TT /7T1 T , -^ ^niersemon as a commnft firing ---. ^^^v,o«.- ,«,-«am ««^fes are equal. " I ■If if- mtmi^m 298 BOOR Vlll. OF LINES AND PLANES. Corollary. When two planes intersect, we may take either of the two adjacent dihedral angles as measuring their incli- nation. Y.Ifa plane intersect two parallel planes, the alternate and corresponding dihedral angles are equal. Note. When speaking of the angle between two planes, the adjective dihedral may be omitted when no ambiguity will arise from the omission. Theoeem XXIV. \ 635. Iffrmt anypoint perpendiculars he dropped upon two intersecting planes, the angle between these perpendiculars will he equal to the dihedral angle hetween the planes, adjacent to the angle in which the point is situated. p. Hypothesis. MN, RS, two planes (of which the parts in the diagram are supposed to be rectangu- ^ lar) intersecting along the "^ line AB; P, any point in ^ the obtuse dihedral angle TBS; PO, PQ, perpen- diculars upon the planes from P. Conclusion. Angle OPQ = dihedral angle SBJV. Proof. From P drop the perpendicular PC upon AB. Join Cd, CQ. Then— 1. Because PO is perpendicular to the one plane and PQ . to the other, CO is the projection of CP on the plane RS, and CQis the projection of CP on the plane ifJV(§§ 598, 600). Therefore, J 5 being perpendicular to CP, the angles -4 CO and ACQ are right angles (§ 606). Therefore CP, CO, and CQ are in one plane, (§ 586) which plane must also contain PO and PQ. 2. Let D be the point in which PQ and OC intersect. Because, in the triangle POD, is a right angle, — ^1^ nr>n — nun — «'^T«^i«»»^'ir»+ />-P /IDD = complement of CDQ, = DCQ. RELATIONS OF PLANES. ►S >'l«l 299 Therefore Angle OPQ = angle OCQ, 4. Because CO and CQ are each perpendicular to A£ TZZe TnToP^^'^hl r'^ b^etweentlVplanel: anereiore Angle OPQ = dihedral angle SBJV, Q E D situated the point from which they are Xppfd we J^^fi '.' hem to be supplementary. Thi/follow Zm CcSr^' .on^that the angle. SB^ ( = OOQ) and -.^r Ts^X fart^: TAu, rt&' ^^'^'^ '"^ •'"'"* ^ ^^-'^^ perpendicular upon the plane MJV shall fall to the left of AB, and therefore not inter- sect the plane RS. Then, if we imagine ourselves to look M directly along the line AB so as to see both planes edgewise, the figure will present this appearance. whilh^olTteSr t^ a.-'^^^ralin angle. OPO and O^fwilTf IwoS'tr.le? two angles will be supplementary. K. th ' i« ^ dihedral angle between the planes. ' °''*™* to 0^^."^'° '"^"'^ "'"^° *'"'" *« ""gl* ^-^^ is still equal 63'?. Coroffiary 1." If from any point C of the K«, „f xnters^tun of two plane, two perpmdiculars: 00 CO L erected, one tneach plane, andfror^O and Q perpe'ndiXt ;liii 300 BOOK VIU. OF LINES AND PLANES, Theoeem XXV. 639. ^fa line he perpendicular to aplane^ etery plane containing this line is jlIso perpendicular to the plane. Hypothesis, MN, a plane ; PO, any line perpendicular to it, intersecting it in 0; ^(7, any plane con- taining the line PO. Conclusion. The plane AG i& perpen- dicular to the plane MN. Proof. From draw OD in the plane JfiV^ perpendicular to ^j5. Then — 1. Because PO L plane MN, it is perpendicular to AB and OD in that plane, and Angle POD= right angle. (§ 584) 2. Because OP and OD are both perpendicular to AB, their angle POD measures the dihedral angle between the planes which intersect along AB (§ 623). 3. Therefore, from (1), this dihedral angle is a right angle, and the planes are therefore perpendicular. Q.E.D. Theorem XXVI. 630. If two planes he perpendicular to each other ^ every line in the one, perpendicular to their common inter section^ is perpendicular to the other. Hypothesis. AG, MN, two plant's inter- secting at a right angle along the line AB; OD, a line in the plane MN perpendicular to AB, Conclusion. OD is CI pCllU. lU UXOii V\J cipciiu.iC plane AG, .Oil v\j v1l\ Proof. In the plane ^(7 draw OP 1 AB. Tbon- BBLATIOm OF PLANm. g^j rtp/i^^""* ^^ " perpendicular to ^5 (hyp) ^a *„ Theoeem XXVII. 631. If two planes be perpendicular to each othpr the line .4 J9; OP, a line perpendicular to the plane ABCD. Conclusion. OP II plane M'JT, Proof. From anypoint R of the plane MN drop a perpendicular i?0 upon AB. Then— It (§607). Q.ED ' ^^' ^^ contains fioo „ "^* °* '"*^'"««''«<>° -4 A Hence: ott.r. '''''"''*">'' PorpenatcnUr to tU one mil lU in the »„- „ Theoeem xxvni. 302 BOOK nil. OF LINES AND PLANES. Hypothesis. PQ, m two planes, each perpendicxaar to plane MN; AB, their ° -^ line of intersection. Conclusion. AB L pla^io ^^' Proof. 1. Because the plane PQ is per- pendicular to MN, il ^rt^^^^^o. a perpendicular to MN, it will lie in the P^^ne P(2 (§ 632). ^^^ I- srorffh^i^i^^^^ two'plLes, or AB-. whence^ ^ is the perpendicular to the nXrB-'- 5-%!;Sndicular to two planes is n^Sicular to their line of intersection, and because ^1 plLTiendicular to the same line are parallel or com- cident (§ 616), we conclude: 634. All planes perpendicular to the same two planes are either parallel or coincident. Theorem XXIX. 636. Iftwoplanes are respectively perpendicular to two intersecting Ivnes, their Une of inter sect^on ^s perpendiealar to the plane of the lines. Hypothesis. OH, Oh two lines intersecting at 0; Plane UN 1 line OJI\ Plane KL 1 line 0/; TJV, the line of inter- section of these planes. Conclusion. VY L plane EOI. Proof. 1. Because the plane MN is perpendicular to OH, it is perpendicular to every plane passing through OH. That is, Plane MN RELATIONS OF PLANES. 303 2. In the same way^ Plane KL ± plane HOI, 3. Because ea<5h of the planes is perpendicular to HOL and UViB their line of intersection, ?7r X plane ^0/ (§ 633). Q.E.D. Belatlons of Three or more Planes. 636. Remark. When three planes, which we may call ^, r, and Z, mutuaUy intersect, there will be three lines of intersection: One line formed by the planes X and F; One line formed by Fand Z; One line formed by Z and X. Theoeem XXX. 637. The three lines of intersection of three planes are either parallel or meet in a point. also'^Ilf' ^* ""^ ^^ *^^ ^^""^^ ^^^""^^ '^^ ^' ^""^ ^* ^^'^ ^® a, the line of intersection of Xand Y' h, the line of intersection of Fand zl c, the line of intersection of Z and X Then— 1. Because the lines a and 5 both lie in the plane T they are either parallel or intersect each other. The same may be shown for i and c, and for c and a. 3. Suppose a and b to intersect. Because a lies in both the planes Xand F, and h lies in both Fand Z, the point where they intersect must lie in all three planes X, f; and Z Therefore it must lie on both the planes Xand Z, and therefore on their line of intersection c. The three lines a, b, and c will then all meet at this point. 3. If a and b are parallel, c cannot meet either of them, because, by (2), where it meets the one it must meet the otuer aiBo. xnerefore, in this case, none of the lines will V^ZI^T^ ^x? ""^ *^® ''^^®^'' *^^ ^6ca«se each pair lies in the same plane they must be all parallel. i " Ih!* 304 BOOKVIIL OF LINE8 AND PLANES, «Qft Cnrollar'U 1. // the three lines of intersection of J^ne'^:^^^^^^ ^/- ^^ree planes all pass through thatpmnt, ^^^ ^^^^^ ^^^ ^^^^^^^^ ^^ ^^^^ ^^^^ of mm is also pa/allel to the intersection of any two planes, one of which passes through each of the lines. Theorem XXXI. v 640. If the Ivnes of intersection of three planes are parallel, any fourth plane Verpendi^^^^^ of the three planes is also perpendicular to the mra ^ HypothesL The parallel lines a, h o are the lines of intersection of three planes, « which wo ckW the planes a&, Ic, and cff; MN, a plane perpendicular to the two planes al and Ic. Conclusion. ifiV is also perpendicular to the plane ac. Proof 1. Because the . plane uk is perpendicular to both the planes a& and &., it is T^prnendicular to their line of intersection I (§ 633). ^ TBecau^ MNi. perpendicular to 5, it is perpendicular to the lines a and c, parallel to 5 (§ 591). 3. Therefore it is perpendicular to the plane ac, wnicn passes through those lines (§ 629). Q.E.D. Scholium. The most remarkable position of three planes is that in which each plane is per- pendicular to the other two. By the preceding theorems each line of :^4^,v^n/>o+irkTi will he per- pendicular to the other two lines of intersec- tion and to the third plane. i! . ^,„,,„... POLYHEDRAL ANQLB& 806 n of mgh each ineSf anes > two ird. les of c, it is dicular , which CHAPTER III. OF POLYHEDRAL ANGLES. A polyhedral angle. O, the vertex; OA, OB. 00, etc., the edges ; AOB, BOO, etc., the faces. 641. Def, When three or more planes pass through the same point, they are said to form a poly- hedral angle at that point. A polyhedral angle is also called a solid angle. Each plane which forms a poly- hedral angle is supposed to be cut off along its lines of intersection with the planes adjoining it on each side. "| / \ "'>C 642. Def, Edges of a poly- hedral angle are the straight Jines along which the planes intersect. 643. Def. Faces of a poly- hedral angle are the planes which form it. 644. Def. The vertex of a polyhedral angle is the point where the faces and edges all meet. 645. The edges of a polyhedral angle may be produced indefinitely. But to make the study of the angle easy, the faces and edges may be supposed cut off by a plane. The intersection of the faces with this plane will then form a poly- gon, as ^^(72)^. This polygon is the base of the polyhedral angle. 646. Each pair of faces which meet an edge form a dihedral angle along that edge. There are as many edges as faces, and therefore as many dihedral angles as faces. Hence two classes of angles enter into any polyhedral angle, namely: I. The plane angles AOB, BOO, COD, etc., called also face angles, which the edges form with each other. The planes on which these angles are measured are the faces. 806 BOOK VUL OP LINSa AND PLANES. i Example. The plane of the angle A OB is the face bounded by OA and OB. „ a , II. The dihedral angles between the faces, called awo edge angUs. By § 623 each of these angles is measured by the plane angle between two lines, one in each face perpen- dicular to the edge of the dihedral angle. If the cutting plane ABODE vfQXQ perpendicular to one of the edges, say OB, then the dihedral angle along OB would be measured by the plane angle ABG. But this plane cannot be perpendicular to more than one edge, so that to measure the dihedral angles in this way we must have as many cutting planes as edges. 647. Def. Two polyhedral angles are identioally equal when they can be so applied to each other that al the faces and edges of the one shaU coincide with the corresponding faces and edges of the other. In order that such coincidence may be possible, the face and dihedral angles of the one must all be equal to the face and dihedral angles of the other, taken in the same order, each to each. Positive and Negative Rotations. 648. When a person looking down upon a point sees a. motion around that point in a direction the opposite of that of the hands of a watch, the motion is said to be positive relative to his standpoint. If the motion is in the other direction, I it is said to be negative. A motion which is positive when seen from one side will appear negative when the observer views it from the other side Positive rotation, of the plane, or when the figure is turned over so as to be seen from the other side. For illustration, imagine one's self seeing the hands of a watch by looking through it from behind. To avoid ambiguity one side of the plane of motion may be taken as positive and the other side as negative. Then a POLYHEDRAL AKQLSa. d07 positive rotation is that which appears positive when seen from the positive side, or negative as seen from the negative side. Astronomical illustration. If one could look down upon the earth from above the north pole, the earth would appear rotating in the positive direction. If he should look down upon it from above the south pole, it would appear rotating in the negative direction. Remark. The habit of regarding the motion opposite that of the hands of a watch as positive arose from the direction of the north pole being taken as positive, because astronomy was developed among the people of the northern hemisphere; these people regarded as positive the direction in which tlie earth would appear to rotate when seen from the north. 649. As there are positive and negative rotations, so letters, angles, and lines may succeed each other in a positive or negative di-^ rection. 650. NoTATioia". A polyhedral angle is desig- nated by a letter at its ver- letters succeed each tex followed by a hyphen S."" "^^ "^"'^ and the letters at the vertices of its base. Letters succeed each other in the negative ordor. 661. Def. Symmetrical polyhedral angles are those which have their plane and di- "^ hedral angles equal, each to each but arranged in reverse order, the one posi- tive and the other negative when seen from the vertex. B BT- Example. The tri- ^' hedral angles 0-ABG and O'-A'B'C* are symmetrical when \ 't I 308 BOOK nil OF LINES AND PLANES. Angle ^Oi? = angle A'0'B\ " BOC=z " B'0'C\ " 00 A = " CO' A', Dihedral angle along edge OA = dihedral angle along edge O'A ', " " , ** ** 0B= " " " *< O'B' " " '< " 00= " " " " O'c' The two symmetrical polyhedral angles may be so cut that the base ABO shall be identically equal to the ^^ base ^'i?'C". But, in order to bring these bases into coincidence, one of the figures must be turned over and the bases brought together with the vertices in opposite directions, as in the figure. Hence two symmetrical polyhedral angles cannot in gen- eral bo brought into coincidence. 653. Be/. Opposite polyhedral an- gles are those each of which is formed by the continuation of the edges and faces of the other beyond the common vertex. Example. If the lines AO, BO, 00, and DO are produced through to the respective points A', B', C, D', then the polyhedral ^< angle 0-A'B'C'D' is the opposite of the angle O-ABOD, Theorem XXXII. 663. Opposite polyhedral angles are symmet- rical. Hypothesis. 0-^ 5 Ci), any polyhedral angle: O-A'B'O'D' POLYHEDHAL ANGLES. 809 Conclusion. Angle 0-A'B'C'D' is symmetrical to O-ABCD, Proof. 1. Because ^0^1' and BOB' are in the same straight line, Pace angle A' OB' = opp. angle A OB. In the same way it may bo shown that every other face angle of the one is equal to the opposite face angle of the other. ^ 2. Because the linos ^0^' and BOB' ^kbb through the same point, they are in the same plane. Therefore the face A OB' IS in the same plane with the face AOB. In the same way, every other pair of corresponding faces are in the same plane. 3. Because the dihedral angle between two planes is every- where the same (§ 619), each of the edge angles OA', OB', etc., IS equal to the corresponding one of the edge angles OA, OB, etc., of the other angle. 4. If one should look down upon the figure from above, the letters ABCD and A'B'O'D' would each succeed each other in the positive order. Hence if the opposite angle is turned over into the position 0-A"B"C"D", the order of the letters wiU appear negative, and therefore the opposite of those in the original polyhedral angle. 5. Therefore the two polyhedral angles, being equal in all their face and edge angles, but having them arranged in reverse order, are symmetrical. Q.E.D. 654. Def. A trihedral angle is a polyhedral angle which has three edges, and therefore three faces. Remark. In a trihedral angle each ^^^v^^,^. tsxie has an opposite edge, and each ed^e K^^^gj! °PP!t*~^^- an opposite face. ^ ^'ace obc is opp. ^gi oZ A trihedral angle. 310 BOOK VIII. OF LIMES AND PLANES, ^--C Theorem XXXIII. 655. jj/ two face angles of a trihedral angle are equal, the edge angles opposite them are also equal. Hypothesis. 0-ABO, a trihedral angle in which face angle BOA =1 face angle BOG. Conclusion. Edge angle OA — edge angle OC. Proof. From any point P of OB drop the perpen- diculars PM _L OAi PN L 00; PD 1 plane ^Oa Join DM, DN. Then— 1. Because PM intersects the plane A OC in Jf, and PD is perpendicular to this plane, MD is the projection of PM upon this plane (§ 600). Therefore, because OA 1 PM, OA 1 MD. (§606) 2. Because MD and MP are pependicular to OA in the planes forming the dihedral angle OA, Dihedral angle OA = plane angle PMD. (§ 623) 3. In the same way, we show Dihedral angle OC = plane angle PJVD. 4. In the right-angled triangles POM and, PON, The side OP is common, Angle PON = angle POM (hyp.); therefore these triangles are identically equal, and PN= PM. 6. In the right-angled triangles PDN&nd PDM, Side PD is common, PN = PM; (4) therefore these trianglen are identically equal, and Angle PND = angle PMD. 6.. Comparing this result with (2) and (3), Dihedral angle OA = dihedral angle OC. Q.E.D. POLTHBDBAL ANGLES. 311 (4) Theorem XXXIV. 656. In a trihedral angle the greater face angle and the greater edge angle are opposite each other. Hypothesis. 0-ABC, a trihedral angle in which foce angle BOA > face angle BOO. Conclusion. Edge angle OC > edge angle OA. Proof. Make the same constructions as in the last theorem. Then — 1. In the same way as in the last theorem fol- lows: Edge angle OA = plane angle PMD. Edge angle 00 = plane angle PND. 2. In the right-angled triangles POJfandPOiV the line OP IS the common hypotheuuse; and because angle POM is greater than angle PON, Line PM > PN. 3. In the right-angled triangles PDNm^ PBM, because the side PB is common and the hypothenuse PM greater than the hypothenuse PJV, Angle PJ^B > angle PMB. 4. Comparing this result with (1), Edge angle 00 > edge angle OA. Q.E.D. Corollary 1. From these two theorems it follows, as in Bk. II., §115: 657. If one edge angle is greater than another, the face angle opposite it is greater than that opposite the other. For the face angles could not be equal without violating Theorem XXXIIL; nor could that opposite the lesser edge angle be the greater without violating Theorem XXXIV. 658. Cor. 2. If the edge angles be arranged in the order of magnitude, the face angles opposite them will be in the same order of magnitude, so that the smallest, mean, and greatest angle of the one class will be opposite the smallest, mean, and greatest angle of the other, respectively. im i 312 BOOK VIU. OF LINES AND PLANBB. Theorem XXXV. 659. In a trihedral angle in which each of the face angles is less than a straight angle, the sum of any two face angles is greater than the third. Hypothesis. 0-ABG, a trihedral angle in which AOG la the greatest face angle. Conclusion. The sum of the face angles A OB and BOO is greater than AOC. Proof. Through draw in the plane A 00 a, line OD, making angle AOD = angle A OB. Let the base ABC he so cut off that we shall have OB = OD^ Then— 1. Because the triangles OAB and OAD have Side OA common, Side OD = OB, ) ^^, „i.^,«+:«„ Angle ^ 05 = ^OAr"'^'*^"*'^^' they are identically equal, and AB = AD. 2. Because ^5C is a plane triangle. Sides AB + BC> third side A G. 3. Taking away from this inequality the equal lengths AB and AD, BC>DC. 4. Because, in the two triangles OCB and OCD, Side OCia common, OD = OB (const.), and CB > OD, we haye Angle BOC > angle OOD. 5. Adding the equal angif^s A OB and A OD, (§115) Angle AOB + angle £00 > angle A 00. Q.E.D. POLTHEDhAL ANGLB8. 313 Theorem XXXVI. 660. Two trihedral angles are either equal or symmetrical when the three face angles of the one are respecUvely equal to the face angles of the other. Hypothesis. 0-ABC,0''A'B'G', 0"-A"B"C'', three tri- hedral angles in which Face angle AOB = angle A'O'B' = angle A"0"B"- " " BOG= " B'Q'Q'- « B''0"G"'' " " COA = « G'O'A' = " C"0"^"- orders of angles ABG and A'B'G' positive when viewed from the vertex, and of A"B"G'' negative. Jf elusion The edge angles of the trihedral angles are also equal, and the trihedral angles O-ABC and 0' A'B'C' are equal, and 0".A"B-G- is symmetrical with them. fakP r-^* ^^*^.' f '' ^^^ ^'^'' ^""^ ^"^"^ respectively, take the equal distances OP, O'P' and r)"P" o«^ 1. In the triangles OPQ, 0'P'Q\ 0"P''Q" A 1 „^^ = ^'^' = ^"-^" (const.).' Angle P(9^ = P'0'^' = P"0"0"(hvp) OPQ = O'P'^' = 0"i>"r (all being rigKngles). Therefore these triangles are identicaUy equal, so that OQ = O'Q' = 0''Q", PQ=P'Q'=.P''Q'\ siiy 314 BOOK Vin. OF LINES AND PLANES. 2. In the same way, OPR, O'P'R', and 0"P"R" being right angles by construction, OR = O'R' = 0"R'\ PR = P'R' = P"R'\ V Imagine QR^ Q'R't and Q"R" to be joined, then — 3. In the triangles OQR, 0'Q'R\ 0"Q"R", etc.. Face angle QOR= Q'O'P' = Q"0"R" (hyp.), and the sides which include these angles are equal by (1) and (2). Therefore these triangles are identically equal, and QR = Q'R' = q"R' vt 4. Comparing with (1) and (2), the three triangles PQR, P'Q'R'j P'*Q*'R*' have their corresponding sides equal, and are therefore identically equal to each other. Hence Angle qPR = angle Q'P'R' = angle Q"P**R'', 5. Because QP, QR, etc., are each perpendicular to their edges, and lie in their respective faces, their angles measure the dihedral angle between those faces. Therefore • Edge angle OA = edge angle O'A^ = edge angle O'M". 6. In the same way may be shown Edge angle OB = edge angle O'B^ = edge angle 0"J5". Edge angle 00 = edge angle O'O' = edge angle 0"(7". Therefore the three trihedral angles have their edge angles all respectively equal. 7. Because in the first two trihedral angles the arrange- ment of the equal angles is the same, while in the third this arrangement is reversed, the first two angles are equal, and the third symmetrical with them. Q.E.D. 1 POLYHEDRAL ANGLES. being 316 Theorem XXXVII. fnl^^' 1^ ? ^^'^'^^^ P^^^y^edral angle the sum of the /ace angles ts less than a perigon (360°). thef^lf tfi. O.^^C/>^, a polyhedral angle of which aU the angles of the base are convex. Conclusion. Angles AOB -\- BOG + etc. + EOA < 360°. Proof. Let n be the number of faces A^nr. f"^"^ polyhedral angle. The base ABLDEmW then be a polygon. of n sides, l^t us also put 2, the sum of the face A angles A OB, BOO, etc. Then— 1. Because ABODE is a convex poly- gon of n sides, ^ Angle ABO + angle BCD + etc. = („ - 3) straight angles. anie« ^fT"" t""* ^T *°™ " *™°SH the sum of alMhe angles of these triangles is n straight angles. That is, 2 + angles {OAB + 05^ + OBG->t 00B+ etc.) = « straight angles. ande a^Tof wr tTf'^^' f ^^' ^^^form a trihedral angle at B, of which the face angles are OBA OBO ABC Angle OBA + OBO > J^a Angle OC^ + OCD > ^CZ). (§ 659) etc. etc. etc. 4. Taking the sum of these inequalities, we find Sum of the %n base angles of triangles OAB, OBO etc ^eater than the sum of the angles of the polygon ABODE- that 18, greater than n-^ straight angles. resi"(.7an'dVi:' *^ ^"" "' '^'^ "^ -^les, the -^ + -^ = w straight angles. ^ > (/i — 2) straight angles. The difference of these shows that or 2 <1 2 Strftiorlif, anncl Sxv,o, ^<360° Q.E.D. I'* ' . 1 4 I Jf n ;Fi«i Fiffi BOOK IX. OF POLYHEDRONS. CHAPTER I. . , OF PRISMS AND PYRAMIDS. :ll'IIlll{ 662. Definition. A solid is that which has length, breadth, and thickness. A solid is bounded by a surface. Eemark. The form, magnitude, and position of a solid are completely determined by the form, magnitude, and position of its bounding surface. Hence we may consider the surface as defining the solid. « 663. Def. A polyhedron is a solid bounded by planes. 664. Def. The faces of a polyhedron are its bounding A< planes. 665. Def. The edges of a polyhedron are the lines in which its faces meet. ^ A polyhedron. 666. Def. The vertices of The planes hab, hbc, akb, t ■, t ,-, . . . 6tc., are the faces. The lines a polyhedron are the points m^^.^^, etc., bounding the faces, i~. T_ . , , , -^ are the edges. The points A, B, which its edges meet. C, D, H, K&re the vertices. 667. Def. Diagonals of a polyhedron are straight lines joining any two vertices not in the same face. 668. Def A plane section of a polyhedron is the polygon in which its faces cut a plane passing through it PRISMS. 317 669. I)ef. Two polygons are said to be paraUel to each other when each side of the one is paraUel to a corresponding side of the other. Prisms. 670. Def. A prism is a polyhedron of which the end faces are equal and parallel polygons, and the side faces parallelograms. ^o > em. De/: The bases of a prism are its end faces. 673. Def. The lateral faces are aU except its pases. 673. Be/, The lateral edges of a prism are the intersections of its lateral faces. 674. Def. A right section of a prism is a section by a plane perpendicular to its lateral edges. 675. Def. A prism is said to be triangular. Quadrangular h«» . „f° hexagonal prism. »5»ufu,«iuaurailgUiar,neZ- ABCDEF and A'B'C'iyE'S^ agonal, etc. , according as its bases *^ ***« '^^^ ^^'^^ are triangles, quadrilaterals, hexagons, etc. ;,. ^J^^'P^f' The altitude of a prism is the perpen- dicular distance between its faces. 1 ^^Vl -^^^' ^ ^^"^ P^^"^ i» one in which the lateral faces are perpendicular to its bases. 1 ^^"^h ^^^' ^^ oblique prism is one in which the lateral faces are not perpendicular to the bases. 679. Def. A regular prism is a right prism whose bases are regular polygons. ;i.i-;;ii M mk. jk.«Bl 318 BOOK IX. OF POLYHEDJiONS. ! Ill TlIEORSM I. 680. The lateral edges of a prism are equal an4 parallel, and make equal angles with the bases. Hypothesis. ABC, A'B'C, two edges of the bases of a prism; ABA'B', BCB'C, the lateral faces joining those edges. Conclusion. AA', BB', CC are equal and parallel and make equal angles with the bases. Proof. 1. Because ABA'B' is a parallelogram (§ 670), Line BB' - and || AA', 2. Because BB'CC is a parallelo- gram, ! Line BB' = and || CC\ 3. Because AA' and CC are equal and parallel to the same line, they are equal and parallel to each other (§ 592). 4. In the same way it may be shown that all the other lateral edges are equal and parallel. Q.E.D. 5. Because the lateral edges are parallel lines, they make equal angles with either base (§ 610). Q.E.D. 6. Because the bases are parallel planes, the edges make equal angles with the two bases (§ 618). Q.E.D. Theoeem II. 681. The sections of the lateral faces of a prism by parallel planes are equal and parallel polygons. Hypothesis. ABCD-A'B'C'D\ a prism; EFOH and E'F'G'H, sections of the lateral faces by parallel planes. Conclusion. EFGH = and || WFG'W. Proof. 1. Because A A' and BB' are parallel lines inter- secting parallel planes, EE' = and \\_HF', _ _ _ (§ 615) Therefore the four lines EF, FF% F'E% and E'E form a parallelogram and Line ^ii^= and \\E'F'. PIiI8M% tal an(i es. a^es of a ^c b1 to the § 592). the other bey make ges make a prism ygons. 'OH and lanes. nes inter- (§615) 'E form a 3. In the ame way we find 819 Line FG = and || ro'. Line OH— and || 0'H\ etc. etc. A ^1%^}^ '^^®' ^^ *^® respective angles are paraUel Angle BFO = angle E'F'O', parauei. Angle FOH = angle F'0'h\ , (§ 608) ^^ etc. etc. ^ Therefore the polygons EFOH and ^'^'(?'^', having their sides t\ and angles, taken in order, all equal are identically equal. Q.E.D. ' 683. Corollary, Any section of */ a prism hy a plane parallel to the d bases is identically equal to the oases. Theorem III 1.7 T^T.' ff'^^/'f^'^ of a prism hy a plane paral- lel to the lateral edges is a parallelogram. Hypothesis. ABCA'B'G^ any prism: PORS a sec *Z,'^ *^^' P"sn^ by a plane parallel to ^ ^ ' '" AA'y BB'. *X zpiTi Conclusion. PQRS is a parallelo- gram. Proof. 1. Because the line AA' is parallel to the intersecting plane PQRS, it cannot meet either of the lines PR or Q8 which lie m that plane. Therefore . the lines A A', PR, and QS are parallel. ^ (§ 637) 3. Because the opposite sides AP jl ^''^^f^^:. ^^' ^""^ ^^ ^^ *be quadrilateral APA'R are parallel, the quadrilateral is a parallelogram, and „ , ,, ^^ = and II AA\ o. m the same way we may prove .. rpu . ^^""^ QS = B.n^\\ A A'. PO^^t ii^'^"^"^ " ^^' ^^^^ *be quadrilateral i^^//^6 IS a parallel o^rnni. Q.E.D. "f 320 BOOK IX. OF POLYHEDRONS. Parallelepipeds. 684. Def. A parallelopiped is a solid contained by three pairs of parallel planes. A parallelopiped is therefore a prism of which the lateral faces are two pairs of parallel planes. 685. Def. A rectangular parallelopiped is one whose faces intersect at right angles. 686. D^. A cube is a par- allelopiped whose faces are all squares. ^ paraUeloplped. Theorem IV. 687.^ The opposite faces of a parallelopiped are identically equal parallelograms. Hypothesis. ABCD-BFGH, any parallelopiped. Conclusion. The opposite faces ABCD and BFOH are identically equal parallelo- grams. Proof. 1. Because 5(7 and AD are the lines in which the parallel planes BCFG and ADEH intersect the third plane ADBO, BG II AD. (§ 614) 2. In the same way it may be shown that the lines AB and DC are parallel. Therefore ABGDiBVi, parallelogram. Q.E.D. 3. It may be shown in the same way that all the other faces are parallelograms. Therefore, by comparing opposite sides of successive parallelograms, AB = and || EF, and 5C = and II i^(?. 4. Because the sides BA and BG ot the angle ABG bxq parallel to the sides FE and FG of the angle EFGy Angle ABG = angle EFG. (§ 608) PAUALLEL0P1PED8. 821 Qtalned ich. the led. oed are therefore .he other opposite iBG are (§ 608) 5. Therefore the parallelograms A BCD and EFQII having their respective sides and one angle equal, are identi- cally equal. In the same way it may be shown that every other pair of opposite faces are equal. Q.E.D. Corollary 1. The edges of a parallelopiped are twelve in number y and may be divided into three eets, each set compris- ing/our equal and parallel lines. Cor. 2. The vertices of a parallelopiped are eight in number. Cor. 3. The diagonals of a parallelopiped are four in num- ber, and may be drawn from any angle of each face to the opposite angle of the opposite face. Theorem V. 688. The four diagonals of a parallelopiped all intersect in a point which bisects them all. Hypothesis. ABCD-EFGH, any parallelopiped. Conclusion. The four diagonals ^^ AGy BH, CE, and i>i?^ all inter- ^[ sect in a point (9, and are bisected by this point. Proof. Through the opposite parallel edges AB and HG pass a plane. Join AH, BG. Then — 1. Because the sides AB and -^ HG of the quadrilateral ABHG are equal and parallel, ABHG is a par- allelogram. Therefore AG and BH, the diagonals of this parallelogram, intersect and bisect each other. Let be the point of intersection. 2. In the same way it may be shown that BH and CE bisect each other. Therefore CE passes through the point of bisection 0, and bisects CE. 3. In the same way it may be shown that DF passes through and is bisected by 0. Therefore all the diagonals pass through and aie bisected by that point. Q.E.D. v and BO, EH and FO. Then— 1. AH and DE are diagonals of the parallelogram AD HE (§ 688, 1). Therefore AH' -^ DE' = 2AD' -{- 2AE\ (§316) 2. In the same way, BG' + CF' = 2BC* + 2BF* = 2BC' -f 2AE\ 3. Adding (1) and (2), AH' + BO' + OF' + DE' = 2(AD' + BC^) + 4^JSr«. 4. Because AD and ^C are dis'.gonals of the parallelo- gram ABGD, AD' + BC = 2AB' + 2A0\ (§316) 6. Substituting this result in (3), we have Sum of squares of diagonals = 4AB' -j-4:AC* -\- 4:AE*. 6. Since there are four edges equal to AB, four equal to AC, and four equal to AE, this sum is equal to the sum of the squares of all the edges, and Sum of squares of diagonals = sum of squares of edges. Q.E.D. Theorem VII. 691. The four diagonals of a rectangular paral- lelopiped are equal to each other. Hypothesis. ABCD-EFGH, a rectangular parallelopiped. Conclusion. The diagonals AH, BG, CF, and DE Skre all equal. PjiRALLELOPIPEDS. 823 \e four -"-^ B iADHE (§316) \AE\ >arallelo- (§316) AE\ equal to I sum of dges. Q.E.D. paral- lopiped. E are all Proof. 1. Because the faces AFvLnd AG are each at right angles to the fade AD (§ G85), their lino of intersection AE 18 also perpendicular to that „ face (§ 633), and to the line ^Z> in that face (§ 584). 2. Therefore ADEH is a rect- angle, and its diagonals AH and DE are equal. 3. It may be shown in the same way that any other two diagonals are equal. Therefore these diago- nals are all equal to each other. Q.E.D. Theorem VIII. 693. The square of each diagonal of a reetangu^ lar parallelopiped is equal to the sum of the squares of the three edges which meet at any vertex. Hypothesis. Same as in Theorem VII. Conclusion. AH'' = AB^ -{■ AC^ •{- AE\ Proof. 1. Because AEH is a right angle (Th. VII., 1), AH' = AE' + EH* = AE' + AI)\ 2. Because ABD is a right angle, AD' = AB' + BD' = AB'-\-AC\ 3. Comparing with (1), AH' = AE' ■^AB'-\-A G\ Q.E.D. 693. Scholium. This theorem might have been regarded as a corollary from the two preceding ones. But we have preferred an indejjendent demonstration, owing to its impor- tance. It may be considered as an extension of the Pytha- gorean proposition from a plane to space. Pyramids. 694. Def. A Dvramid is a Dolvhedron of whicli all the faces except one meet in a point. The point of meeting is called the vertex. ■ * ( » I ^^^^Wp 324 BOOK IX. OF P0LTHEDR0N8. Remark, The face which does not pass through the vertex is taken as the base. 696. Def. The faces and edges which meet at the vertex are called lateral faces and edges. 696. JDef. The altitude of a pyramid is the perpendicular dis- tance from its vertex to the plane of its base. 6911. Def. A pyramid is said to a pyramid, be triangular, quadrangular, pentagonal, etc., ac- cording as its base is a triangle, a quadrilateral, a pentagon, etc. 698. 'Def. If the vertex of a pyramid is cut off by a plane parallel to the base, that part v^hich remains is called a frustum of a pyramid. Theorem IX. 699. If a pyramid be cut hy a plane parallel to the base, then — I. The edges and the altitude are similarly di- vided. II . The section is similar to the base. Hypothesis. 0-ABCDE, a pyramid; OP, its altitude; dbcde, a section of the pyramid by a plane parallel to the base ABODE, cutting the altitude line at g. Conclusions. I. OF :Og::OB: Ob ::OC:Oc, etc. II. The polygon abcde is similar to the polygon ABODE. ^ Proof. 1. Because AB and ah are the intersections of parallel planes [/ with the plane OAB, ah II AB. (§614) PYRAMIDS. 825 >c 2. Therefore the triangles OAB and Odb ar^ simUar, and OA : Oa :: OB : Ob bB. whence Oa -. aA :-. Ob 3. In the same way it may be shown that 00, OP, etc are all divided similarly at c, g, etc. Q.E.D. *' 4. It may also be shown, as in (1), that ea^h side of the polygon abcde is parallel to the corresponding side of ABODE Therefore the angles of the two polygons are respectively equal (§ 608), and the polygons are equiangular to each other. 6. Because the bases ab and AB of the triangles Oab and OAB axe parallel, _ AB :ab = OA :0a. In the same way, BO: be:: OB: Ob :: OA : Oa, CD:cd:: 00 : Oc :: OA : Oa, etc. etc. 6. Therefore the polygons ABCDE and abcde, having their angles equal and the sides containing the equal angles proportional, are similar to each other. Q.E.D. Corollary 1. Because abcde and ABCDE ^ve similar poly- gons, their areas are as the squares of ab and AB: that is, as Oa' : 0A\ or Og' : OP' (§435). Hence: 700. ITie areas of parallel sections of a pyramid are pro- portional to the squares of the distances of the cutting planes from the vertex. Cor. 2. The base of a pyramid may be regarded as one of the plane sections, so that if two pyramids have equal bases and altitudes, the plane sections made by the bases are equal, and the distances of these sections from the vertices are also equal. Hence: 701. In pyramids of equal bases and altitudes, parallel plane sections at equal distances from the vertices are equal in area, ^ m\\ 'i I lit '4 326 BOOK IX, OF P0LTHEDB0N8. CHAPTER II. THE FIVE REGULAR SOLIDS. 703. Def. A regular polyhedron is one of which all the faces are identically equal regular polygons and all the polyhedral angles are identically equal. Remaek. a regular polyhedron is familiarly called a regular solid. 703. Problem. Tofind> Jiow many regular solids are possible. 1. Let us consider any vertex of a regular polyhedron. Since at least three faces must meet to form the polyhedral angle at each vertex, and since the sum of all the plane angles which make up the polyhedral angle must be less than 360° (§ 661), we conclude: Each angle of the faces of a regular solid must be less than 120°. 2. Since the angles of a regular hexagon are each 120°, and the angles of every polygon of more than six sides yet greater, we conclude: Bach face of a regular solid must have less than six sides. Such faces must therefore he either triangles, squares, or pentagons. 3. If we choose equilateral triangles, each polyhedral angle may have either 3, 4, or 5 faces, because 3 X 60°, 4 x 60°, and 5 X 60° are all less than 360°. It cannot have 6 faces, because each angle of the triangle being 60°, six angles would make 360°, reaching the limit. Therefore no more than three regular solids may be constructed with triangles. 4. Because each angle of a square is Therefore only one regular 6. Because each angle 90°, three is the only ■nnlvViAflral an or] ft. have square faces. alar pentagon is 108°, a which ilygons ual. called a solids ^hedron. lyhedral e angles lan 360° f le less 3h 120°, jides yet ix sides. ',ares, or ral angle 4 X 60°, ) 6 faces, es would 3re than IS. the only rl ftTijylfi. ~ — o — !S. 3 108°, a BEOULAR 80LID8. ' ^^l polyhedral angle cannot be formed of more than +}ir«n >. . ..J' "^^ *^«^?;f?^« conclude that not more than five re^kr e JSt^: ''^^^TTsJ'z "^ ^-^^ *^"^ «^"*'^'- CO^, and join them at the common vertex, 0, ABO win be an identi- cally equal equilateral triangle. Therefore a polyhedron will be formed having as faces four identi- cal equilateral triangles. This solid is a regular tetra- hedron. lelop'Jedof'l;^^^^^^^^^^^ is clear from §§ 685-693 ^ '''* ^*' construction T06. The Octahedron. Let -45CZ> be a square; 0, its centre; ^{f, a Ime passing through perpen- dicular to the plane of the square. On this line take the points P and A«=££---'-/-Or'-±i::^-^D Q, such that PA, PB, PO, and Pi> also QA, QB, QO, and QB, are each equal to a side of the square. The ^f^^f'^BOD^Q will be a regular PrlTT 1'' ^'' '^^'^ ''^^^ ^^"''J^teral triangles Pm/. 1. Because all the lines from Por to 7* P n a D are equal, all the eight triangles which form fh.^' ' equal and equilateral. °^ *^® ^*^^s are 3. Let a diagonal be drawn from A to D Because is the centre of the square jpnn +t.- ^• onal will y>oc,« fi, 1 ^ , . «4"'*re ^//6x', this diair- ^, ^, ^, and D are in one plane. 328 BOOK IX. OF POL THEDR0N8. 3. In the triangles APDy ABD, AQD we have Side AD common, All the other sides equal. Therefore the triangles are identically equal; and because ABD is a right angle, APD and AQD are also right angles, and the quadrilateral APDQ is a square equal to the square ABCD. . ^ ^ 4. Because the polyhedral angle at B has its four faces and its base APDQ equal to the four faces and the base ABCD of the polyhedral angle at P, Polyhedral angle B = polyhedral angle P. 5. In the same way it may be shown that any other two polyhedral angles are equal. Therefore the figure P-ABGD-Q is a regular solid having eight equilateral triangles for its faces. This solid is called the regular octahedron. 707. The Dodecahedron. Taking the regular pentagon ABODE as a base, join to its sides those of five other equal pentagons, so as to form five trihedral angles at A, B, 0, D, and E, respectively. Because the face angles of these trihedral angles are equal, the angles themselves are identically equal. (§ 660) Therefore the dihedral angles formed along the edges AK, BL, CM, etc., are equal to the dihedral angles AB, BO, etc. Therefore the face angles EAK, LBC, etc., are identi- cally equal to the angles of the original regular pentagons. Pass planes through JT^Pand OKF, etc., and let FP be their line of intersection. Then, continuing the same course of reasoning, it may be shown that the face angles GKF, FLJ, etc., are all angles of 108°, or those of a regular penta- gon. Completing this second series of five pentagons, we shall have left a pentagonal opening, which being closed, the surface Ol LU6 UUi Y UUtliAJli will J7C tJUiiipicluu. tto DiiUTTii 111 tuo Uxaijji tiixii. The solid thus formed has \% pentagons for its sides, and is called a regular dodecahedron. BEQULAB 80LID8. 829 because angles, square ir faces he base her two BCD'Q is faces. entagou stc. ) identi- yons. \.FP be le course js GKF, ir penta- we shall e surface XiOigX CtXi.i« deS; and o J^^\ ^^^ IcosaTiedron. Let five equilateral tnauffiee form a polyhedral angle at P, such that ^laugies form the dihedral angles along FA, FB etc., shall all bo equal. The base ABCDB will then form a regular pentagon. Complete the polyhedral angles at A, B, O, D, and B by adding to each three other equilateral triangles, and making the dihedral angles around^, ^,C; etc., all equal. It can be then shown, as in the case of the dodecahedron, that the lines F G H T 1 ^m form a second regular pentagon. ' ' ' ^' "^ ^^^ This solid is called the icosahedron. 709. The five regular solids are therefore- The tetrahedron, formed of 3 triangles. Ihe cube, or hexahedron, formed of 6 squares. The octahedron, formed of 8 triangles. The dodecahedron, formed of 13 pentagons. The icosahedron, formed of 20 triangles. Theoeem X TIO. The perpendiculars through the centres of the faces of a regular solid meet in a point which lyj^^^^fstantfr^ all the faces, f%o:i aUhe edges, and from all the vertices. lar fj.fr'- ^f^^f. ^^? ^^^^^^ *^« ^^««« 0^ a regu- lar polyhedron, intersecting along the edge ^5- thP centres of these faces; OR. OR. ^.rr..^^^^^.t,: .yl 1\ *^' Conclusion, These perpendiculars meet all the perpen- diculars through the centres of the other faces in a point R equally distant from all the faces, edges, and vertices J i 1 330 BOOK IX. OF POL THEDR0N8, Proof, From and Q drop perpendiculars upon the edge AB, Then— 1. Because these perpendic- ulars are dropped from the cen- tres of regular polygons, they will fall upon the middle point P of the common side AB, 2. Because PO and PQ are perpendicular io AB oi the same point Py and OR and QR arc perpendiculars to the intersecting planes, they will meet in a point (§ 6?.7). Let R be their point of meeting. Join PR. Then — 3. In the right-angled triangles POR and PQR we have Side PR common, pd = PQ (being apothegms of equal polygons). Therefore these triangles are identically equal, and OR = QR. Angle PRO = angle PRQ. 4. If S be the centre of any other face adjacent to ABODE, it can be shown in the same way that the perpen- dicular from S will meet QR in a point. 5. Because the angles between the faces Q and 8 are the same as between and Q, it may be shown that the perpen- dicular from 8 will meet QR in the point R. 6. Continuing the reasoning, it will appear that all the perpendiculars from the centres of the faces meet in the same point R. 7. If from R perpendiculars be dropped upon all the edges and all the vortices, these perpendiculars, together with those upon the corresponding faces and the lines like ^P and QB from the centres of the faces to the edges and vertices, will form identically equal triangles. Hence will follow the con- clusion to be demonstrated. Note. We have given but a brief outline of the demonstration, which the student may complete as an exercise. The conclusions may also be considered as following immediately from the symmetry of the polyhedron. the edge len — we haye s)., id acent to e perpen- 6^ are the e perpen- it all the the same the edges i^ith those ^ and QB tices, will ' the con- lonstration, asions may letry of the BBGULAB SOLIDS. 331 Theorem XI. 711. ^ regular solid is symmetrical with respect to all its/aces, edges, and vertices. Proof, Let ABO be a face of any regular polyhedron- A, B, and G will then be yer- ' tices. Let AD, BE, BF, etc.,be the edges going out from these yertices. Now moye the polyhedron so as to bring any other face into the position ABC. This can be done, because the faces are all identically equal. Because the polyhedral angles are all identically equal, whatever polyhedral angles take the positions A, B, 0, their faces and edges will coincide with the positions of the faces and edges already marked in the figure. Because the edges are all of equal length, the yertices at the ends of D, E, F, will fall into the same positions where the former yertices were. Continuing the reasoning, the whole polyhedron will be found to occupy the same space as before, eyery face, edge, and yertex falling where some other face, edge, or yertex was at first. Because this is true in whatever way the positions of the faces may be interchanged, the polyhedron is symmetrical. Q.E.I). 713. Corollary. Conyersely, if a polyhedron le such that, when any one face is brought into coincidence with the position of any other, every other face shall coincide with the former position of some face, the polyhedron is regular. Theorem XII. If a plane oe passed tlirougTi each vertex of - solid, at right angles to the radius, these planes will he the faces of another regular solid a regul ar n\ 5-!| 332 BOOK IX. OF POLTHEDRONa. Proof, 1. Let A, B, and (7 be any vertices of a regular solid, and its centre. Imagine planes passed through 4> B, and G, perpendicular to OA, OB, and 00 respectively, and cut off along their A.-^ lines of intersection, so as to form \, / y:^B the faces of another solid. We call this the new solid, and the original '^^"' ^ one the inner solid, 2. Because the inner solid is regular, if we bring any other of its faces into the position ABC, the whole solid will occupy the same position a? before (§ 711). 3. Because each face of the new solid is at right angles to the end of a radius to some vertex of the inner solid, f.nd these radii all coincide with the former positions, the plane of each face of the new solid will, when the change of position is made, take the position of the plane of some other face. 4. Therefore the edges in which these planes intersect will take the positions of other edges. 5. Therefore the vertices where these edges meet will take the positions of other vertices. 6. Therefore the new solid occupies the same space as before the change, and is consequently symmetrical with respect to all its faces, edges, and vertices. Therefore it is a regular solid (§ 712). Q.E.D. 714. Def. A pair of polyhedrons such that each face of the one corresponds to a vertex of the other are called sympolar polyhedrons. Theobem XIII. 716. Every regular solid has as many faces as its sympolar has wrtices, and as many edges as its sympolar has edges. Proof, 1. Continuing the reasoning of the last theorem, it is evident that the centres of the faces of the new solid coincide with the vertices of its sympolar. RBOULAR SOLIDS. 838 'egular y other occupy angles p solid, us, the ange of le other atersect 2. Because each edge of a regular solid is equally distant from the centres of two adjoining faces, each edge of the new solid will be equally distant from two adjoining vertices of the sympolar. 3. Since every two such vertices are connected by an edge of the sympolar, the new soHd will have as many edges as the sympolar, each edge of the one being at right angles and above the edge of the sympolar. Q.E.D. Let Ay B, and C be three vertices ^- ^ "^ -'* of the sympolar, and therefore the centres of three faces of the new solid. 4. By what has just been shown, P«, Ph, and Pc will be three edges of the new solid, meeting in a vertex at P, Therefore — 5. The new solid will have a vertex over the centre of each side of the sympolar, and so will have as many vertices as the sympolar has faces. Q.E.D. I rill take pace as al with at each e other aces as s as its theorem, lew solid Theorem XIV. 716. The sympolar of a polyhedron whose faces have 8 sides will have S-hedral angles. Proof. The vertex of one polyhedron being at P over the centre of the face of its sympolar, the edges meeting at this vertex are each perpendicular to an edge of the face of the sympolar. Therefore if the face has S sides, the polyhedral angle above it will have S edges, and therefore 8 faces. Q,E.D. •717. Corollary. Conversely, the sympolar of a poly- hedron, whose vertices are S-hedral will have S-sided faces. 718. Corollary. What pairs of regular solids are sympolar to each other can be readily determined from the preceding theorems. The tetrahedron has four vertices. Therefore its sym- polar has four faces, and is therefore another tetrahedron. 334 BOOK IX. OF POLTHEDROm. The cube has 8 trihedral vertices and 6 four-sided faces. Therefore its polar has 8 triangular faces and 6 four-hedral vortices. This is the octahedron. Conversely, the sympolar of the octahedron is the cube. They each have 8 edges. The dodecahedron has 12 pentagonal faces and 20 tri- hedral vertices. Therefore its sympolar has 12 five-hedral vertices and 20 triangular faces. This is the icosahedron. Each of these polyhedrons has 30 edges. These results are shown in the following table, where the headings at the top of each column refer to the solid on the left, and those at the bottom to its sympolar on the right. SoUd. i Number of sides to each face. Number of faces. Number of edges. Number of vertices. Number of edges at each vertex Tetrahedron. Cube. Octahedron. Dodecahedron. Icosahedron. 8 4 8 6 8 4 6 8 n 20 6 13 n 80 80 4 8 6 ao 12 8 8 4 8 6 Tetrahedron. Octahedron. Cube. Icosahedron. Dodecahedron. Number of edges at each vertex. Number of vertices. Number of edges. Number of faces. Number of sides to each face. Sjrmpolar Note that each column applies to two solids. For instance, the left- hand column shows the number of sides to each face of the solid named at the leit, and the number of edges at each vertex of the £olid named at the right. i faces, -hedral e cube. 20 tri- )-hedral ron. lere the on the ght. ihedron. hedron. 5. ihedroo. )cahedron. mpolar solid. , the left- id named named at BOOK X. OF CURVED SURFACES. CHAPTER I. THE SPHERE. Definitions. 719. Def. A curved surface is a surface no part of which is plane. ■720. Def. A spherical surface is a surface which is everywhere equally distant from a point within it called the centre. •731. Bef. A sphere is a solid bounded by a spherical surface. Note 1. A spherical surface may also be described as the locus of the point at a given distance from a fixed point called the centre. Note 2. In the higher geometry a spherical surface is called a sphere. We shall use this appellation when no con- fusion will thus arise. 732. Def. The radius of a sphere is the distance of each point of the surface from the centre. 733. Def. A diameter of a sphere is a straight line passing through its centre, and terminated at both ends by the surface. Corollary. Every diameter is twice the radius; therefore all diameters of the same sphere are equal. •^QA. V -a- -'-■ ■-.- w-rw m wf K_5.i.«iiiff^ I;t t n. >-l ; v% ; — 1. iSjut? lA^ a, D^^iici c its u, pia-ue which has one point, and one only, in common with the sphere. 1 IB 336 BOOK X OF CURVED SURFACEB. 726. Def. A line is tangent to a sphere when it touches the spherical surface at a single point. i 726. Def. Two spheres are tangent to each other when they have a single point in common. 727. Def. A section of a sphere is the curve line in which any other surface intersects the spherical surface. 728. Def Opposite points of a sphere are points at the ends of a diameter. Theorem I. 729. Etiery section of a sphere hy a plane is a circle of which the centre is the foot of the perpen- dicular f rem the centre of the sphere upon the plane. Hypothesis. AB, any sphere; 0, its centre; QRS, the curve in which a plane intersects the spherical surface; OC, the perpen- dicular from upon the cutting plane. Conclusion. QRS is a cirde^l having G as its centre. Proof. 1. Because the lines OQ, OR, OS&ve radii of the sphere, they are equal. 2. Because they are equal, they meet the plane QRS at equal distances from the foot G of the perpendicular (§ 595). Therefore the curve QRS is a circle around (7 as a centre. Q.E.D. 730. Corollary. The line through the centre of a circle of the sphere, perpendicular to its plane, passes through the centre of the sphere. 731. Def. The circular section of a sphere by a plane is called a circle of the sphere. 732. Def. If the cutting plane passes through the centre of the sphere, the circle of intersection is called THE BPUERB. 837 tien it other curve lerical points e ts a erpen- plane. W, the QRS a.t § 595). i centre. .E.D. ■ a circle mgh the :e by a ign rne 3 called ^?'l''l^*?''!\® ""^x.**"* "P^*"' ^"^ *^« *^o parts into which It divides the sphere are ciilled hemiipherei. 733. Corollary. All great circles of the sphere are equal to each other. ^ x^ i^ «•!*» xt, "^^l* ^y- "^^ ^"^^ P^^'^*^ ^^ w^ich a perpendicular through the centre of a circle of the sphere intersects the surface of the sphere are called poles of the circle. Cor. 2. Because the perpendicular passes through the centre of the sphere: 736. The two poles of every circle of the sphere are at the extremities of a diameter, and so are opposite points (§ 728). Theorem II. 736. Every great circle divides the sphere into two taentically equal hemispheres. Hypothesis. AB, a circle of the sphere, having the centre of the sphere in its plane and dividing the sphere into the parts M and iV; Conclusion. The parts Jfand JV are identically equal. Proof. Turn the part M, on 0>| as a pivot so that the plane ^^ shall return to its own position but be inverted. Then— 1. Because the centre remains fixed, the great circle AB will fall upon its own trace. 2. Because the surfaces if and JV are everywhere equally distant from the centre, they will coincide throughout. Therefore the parts are equal. Q.E.D. Theorem III. 737. Any two great circles Used each other. Proof. Let AB and CD be the two great circles. Because the planes of these circles both pass through the centre of "-ii r I 338 BOOK X OF OUBVBD SUBFACES. the sphere, their line of intersection is a diameter of the sphere, and therefore a diameter of each circle. Hence it divides each circle into two equal parts. Q.E.D. 738. Corollary. If any number of great circles pass through a point, they will also pass through the opposite point. Theorem IV. 739. Through three points on a sphere one circle, and only one, can be passed. Proof. 1. Three points determine the position of a plane passing through them (§ 580). 2. This plane cuts the sphere in a circle of the sphere. (§ 'J'29) 3. Because the three points lie both upon the sphere and in this plane, they lie in thi^ circle. 4. Only one circle can pass through these points (§ 241). 6. Therefore this circle, and no other, passes through the three points. Q.E.D. Theorem V. 740. 77: rough two points upon a sphere, not at the extremities of a diameter, one great circle, and only one, can pass. Proof. 1. Because the plane of the great circle must pass through the centre of the sphere, the centre and the two points on the surface determine its position (§ 580). 2. But if the points are at the extremities of a diameter, the centre is in the same straight line with them, and an infinite number of planes may pass through them (§ 577, II.). 741. Def. The arc between two points on a sphere means the arc of the great circle passing through these points. 743. Equal arcs upon the same sphere subtend equal angles at the centre. TBB SPHERE. 839 Proof, Because all great circles are equal, their arcs are arcs of equal circles (§ 733). Because their centres are in the centre of the spheres, and because equal arcs subtend equal angles at the centre (§ 308) their equal arcs subtend equal angles at the centre of the sphere. Q.E.D. 743. Corollary, Equal chords in the sphere suMend equal angles at the centre. 744. Corollary. The angular distance letween two points on the sphere may le measured either by the great circle join- ing them, or by the angle between the radii drawn to them. Note. By the distance of two such points is c< imonly meant their angular distance Theorem VII. 745. All points of a circle of the sphere are equally distant from a pole. Hypothesis. QR, a circle of the sphere AB-, P, P', the poles of the circle. Conclusion. Every point of the circle QR is equally distant from P and equally distant from P'. q^ Proof. 1. Because PP' is a line through the centre of the a I circle perpendicular to its plane, every point of this line is equally distant from all points of the circle (§ 594). 3. Therefore P and P', being on ^ the line, are each equally distant from all points of the circle. Q.E.D. 746. Corollary. Every point of a circle of the sphere is at an equal angular distance from the pole. 747. Bef. The polar distance of a circle of the sphere is the common angular distance of all its points from either pole. 'sU ^ffii 'W\ 340 BOOKX. OF OUBVED SURFACES. Cor, 2. The angular distance of two poles is a semicircle, because they are at the extremities of a diameter. Hence: 748. The sum of the distances of a circle from its two poles is a semicircle. 749. Cor. 3. Every point of a great circle of the sphere is half a semicircle or a right angle distant from each pole. 750. Def. A quadrant is an arc of one fourth a great circle, or half a semicircle. Corollary. A quadrant subtends a right angle from the centre of the sphere. Illustration. If AB is a great circle of the sphere, and P and P' its poles, then ArcP^P' = arc PBP' = semicircle. Arc PB =' arc PP' = quadrant. ^ Angle POB = angle BOP' = right angle. 751. Corollary. On a sphere the locus of a moving point one quad- rant distant from a fixed point is a great circle having the fixed point as its pole. Theorem VIII. 753. The poles of any two great circles lie on a third great circle, having their points of intersection as its poles. Hypothesis. AB, CD, two great circles intersecting in the points B and P'; P, P', the poles of AB; Q, Q', the poles of CD. Conclusion. The poles P, Q, P', Q' A lie on the great circle having E and P' as its poles. Proof. 1. Because P and P' are ■no^^+fl '^^ ^lio orroof. nirnlo A Ti of which P and P' are the poles. Arc PR = arc PB'= arc P'B = arc P'P'= quadrant (§ 749). THE SPHERE. 341 2. Because R and R' are points on the circle CD of which Q and Q' are the poles, ' Arc QR = arc QR' = arc Q'R = Q'R' = quadrant. 3. Because P, ^, P', and Q' are points each one quadrant distant from R and R', they lie on the great circle having R and R* for its poles (§ 761). Q.E.D. B B Theorem IX. •763. Conversely, if the poles of two great circles lie on a third great circle, the two great circles will intersect in the poles of that third circle. Proof, 1. Because all points one quadrant distant from P lie on the great circle AB, the poles of every great circle through P must lie somewhere on the circle AB, 2. In the same way, the poles of eyery great circle through Q lie on the great circle CD. ^j 3. Therefore the poles of the great circle through P and Q lie in both of the great circles AB and CD) that is, in the points R and R', in which AB and CD intersect. Q.E.D. •754. Def A great circle having a point B as its pole IS caUed the polar circle of the point R. The great circle containing the poles of other great circles IS called the polar circle of these other circles. ^ ^5^, Corollary 1. If any number of great circles have tfieir poles upon another great circle, they will all intersect tn the pole of that other circle. 756. Cor. 2. If any numher of great circles pass through a common point, their poles will all lie on the polar circle of that -w^'****' '^ 842 BOOK X. OF CURVED SURFACES. Theorem X. 757. The angular distance between two poles of circles is equal to the dihedral angle between the planes of the circles. Proof. 1. If P and Q are two poles of circles, then the perpendiculars from P and Q upon p the planes of the circles pass through the centre of the sphere (§§ 730, 734). c. 2. Because these perpendiculars pass through the centre of the sphere, a[ the arc P^ is equal to the plane angle between them (§ 744). 3. Because they are perpendicular to the planes of the circles, the angle they form is equal to the dihedral angle between those planes (§ 625). 4. Comparing (2) and (3), the arc P^ is shown to be equal to the dihedral angle between the planes. Q.E.D. Theoeem XI. 758. 7/" one great circle passes through the pole of another, their planes are perpendicular to each other. Proof. 1. Because one pole lies on the polar great circle of the other pole, the angular dis- tance of the poles is a quadrant. 2. Therefore the dihedral angle between the planes of the circles is a right angle (§ 757). Q.E.D. 759. Corollary. If any number of great circles vass through a com- mon point, the., planes will all be perpendicular to the plane of their " -r-k 1 -1 rt •!• /o iv(-j\ i^i • If three great circles Intersect For, by definition (§ 764), their at Q, their planes intersect polar circle passes through all their ^^ng oq, and their poles, p, ~ , -I "i" 1 • •i~ » -^> -^"'"5 lis on another great poles, and its plane is tneref'^re per- circle of which q is the poie, pendicular to all of their planes by *'^<* ^f which the plane opf'' *", . X ./ jg perpendicular to each of thf the theorem. given planes. Illtistration of $$ 758, 769. yoles of 3en the ■h&n. the 3n those be equal pole of h other. Bat circle sles Intersect *8 intersect lir poles, P, other great is the pole, plane OFF" each of thf THE BPHERE. Theorem XII. 343 •760. Two spherical surfaces intersect each other in a circle whose plane is perpendimlar to the line joining the centres of the spheres, and whose centre is in that line. Hypothesis. 0, 0', the centres of two spheres; ADB, the curve line in which their surfaces intersect. Conclusion. ADB is a circle having its centre on the line 00' and its plane perpendicular to that line. Proof. Let A and D be any two points on the curve of intersection. Join OA, O'A, OD, and O'D. From A and D drop perpendiculars upon 00\ Then— 1. In the triangles OAO' and OBO' the side 00' is common; OA = OB and O'A = O'B, because these lines are radii of the same sphere. Therefore Triangle OA 0' = triangle OBO' identically. a. Because these triangles are identically equal, perpen- diculars from A and B upon the base 00' are equal (8 175) and the feet of these perpendiculars meet 00' at equal distances from 0; that is, at the same point. Let C7be this point. + L?^T'^ *^^ .^'"^' ^^ ^^^ ^^ ^^« ^«*^ perpendicular /Lo.x .1 ""^^ '"^ ''''^ P^^"® perpendicular to this line (^586); and because they are equal, their ends are in a circle having its centre at C. Q.E.D. Theorem XIII. 761. A plane perpendicular to a radius of the svhere at ita ficrf.rfm'iUi -/o ^ ^ tyi^fui/c/oo vu c/ic dpfiere. Proof. Let OP be the radius to which the plane is per- pendicular. Then— I '" f- 844 BOOK X. OF CURVED SURFACES. 1. Because OP is a radius, the point P is common to the sphere and the plane. 2. Because OP is perpen- dicular to the plane, it is the shortest line from to the plane. Therefore every other point of the plane is without the sphere. 3. Therefore the plane is a tangent to the sphere at the point P. Q.E.D, 763. Corollary 1. Every line perpendicular to a radius at its extremity is tangent to the sphere (§ 725). 763. Cor. 2. Conversely , every plane or line tangent to the sphere is perpendicular to the radius drawn to the point of contact. \ Theorem XIV. 764. All lines tangent to a sphere from the same external point are equal, and touch the sphere in a circle of the sphere. Proof. Let be the centre of the sphere; P, the point from which the tangents are drawn; P8, PT, any two tangents touching the sphere at 8 and T. Then — 1. In the triangles PSO and P20 the side PO is common ; OS = OT (because they are radii), and angle OTP = angle OSP (both being right angles) (§ 763). Therefore these triangles are iden- tically equal, whence PS = PT. Q.E.D. 2. Because the triangles PSO and PTO are identically equ2, tlif perpendiculars from 8 xind T upon PO are equal, and meet FO at the same dist8t{»30 from P\ that is, at the same point Q. Since 8 and T may bo any two points in which tangents through P touch the sphere, all these points are in one plane, and in a circle having its coiifj-e at Q. Q.E.D / THE SPHEBB. Theorem XV. .846 765. Through four points not in the same plane one spherical surface, and no more, may pass. Proof, Let ^^(7Z) be the four points. Join ABy BC, CD, and bisect each of these hnes by a plane perpendicular to them. Let us call these respective planes the planes (ah), {he), {cd). Then— 1. If the sphere passing through A, B, C, and D exist, then, because its centre IS equally distant from A and B, it lies in the plane {ah) bisecting AB perpendicularly (§ 589). 2. In the same way, it lies in the other two bisecting planes, and therefore in their point of intersection if they nave one. •' 3. If they have no point of intersection, their three lines of intersection are parallel (§ 637). 4. Suppose these lines to be parallel. Because the plane ^jBC contains the line AB 1. plane {ah), Plane ABC L plane {ah), (8 639) For the same reason. Plane ABC ± plane {he). Because plane ABC 1 to both the planes (ah) and (he), and (cd) is a third plane having parallel lines of intersection with (ah) and (he). Plane ABC 1 plane (cd). (§ 640) Because plane ABCL plane (cd), and line CDJL plane (cd), by construction, therefore CD lies in the plane ABC( § 631), and A, B, C, D lie in one plane, which is contrary to the hy- pothesis; whence the lines of intersection are not parallel. 5. Therefore the three planes intersect in a point (§ 637), which point is equally distant from A, B, C, and D, and which may therefore be the centre of a sphere passing through A,B, (7, andi>. Q.E.D. ^ i' 6 s Corollarv. The fournoints Ann ot,/i n^ — v« a^v^- as the vertices of a tetrahedron. The edges will then be the six Unes formed by joining every pair of vertices, and we may 'N: I iij 346 BOOK X. OF CURVED SUBFACES. take any three of those edges to determine the positions ol the planes whose point of intersection is the centre of the sphere. Hence: 766. The six planes which bisect at right angles the six edges of a tetrahedron all pass through a point* SIX Theorem XVI. 767. ^ spJiere may be tangent to any four planes wMcTi do not intersect in a point, and of which tJie lines of intersection are not all parallel. Proof. Let AB, AC, AD, BG, BD, and CD be the lines of intersection of the four planes, taken two and two. Bisect any three of the dihedral angles which lie in one plane, as ABy BC, and AC, by other planes. Let be the point in which these planes meet. Then — 1. Because is on the bisector of the dihedral angle BA, it is equally distant from the faces ABC and ABD (§ 638). 2. In the same way is equally distant from the faces BCA and BCD, and from CAB and CAD. 3. Therefore the point is equally distant from the four planes; and if a sphere be described having its centre at and its radius equal to the common distant:, it will be tan- gent to ail four planes. Q.E.D. Corollary. Since we may take any three dihedral angles not meeting in a point to determine the centre of the sphere, we conclude: 768. The six planes which bisect the six edge angles of a tetrahedron intersect in a point. Scholium. It may be shown, as in the case of the tri- angle, that besides the sphere inscribed in the tetrahedron there are four escribed spheres, each touching one face externally and the other three faces produced. ions of of the the six planes \ch the the six le faces ;he four re at be tan- l angles sphere, '?es of a the tri- ahedron ne face SPHERICAL TRIANGLES AND POLYGONS 347 CHAPTER II. OF SPHERICAL TRIANGLES AND POLYGONS. 769. Def. If two great circles of the sphere inter- sect, they are said to make an angle with each other equal to the angle between their tangents at the point of intersection. Illustration. If the great circles a and I intersect at 0, and if H and K are their respective tan- gents at Oy then their angle is measured by the angle HOK between the tangents. 770. JDef. A spherical triangle is the figure fonned by joining any three points on the sphere by arcs of great circles. 771. Def. The points which are joined are called vertices of the tri- angle. 772. Def. The arcs of which a spherical triangle is formed are called sides of the tri- angle. 773. Def The angles which the sides make with each other are called angles of the triangle. a spherical tnangie. Remark 1. Between any two points two arcs of a gi-eat circle may be drawn, the one greater and the other less than a semicircle. In forming a spherical triangle the arcs less than a semicircle are supposed to be taken, unless otherwise expressed. Remark 3. In a spherical triangle, as in a plane tri- angle, each side has an opposite angle and two adianfiTi+. angles, and each angle has an opposite side and two adjacent Bides. '' y." aa}\ 348 BOOK X. OF CURVED SURFACES. Theobem XVII. 774. The angle between two great circles is equal to the dihedral angle betwa i.ovoi plants. Hypothesis. AB, CD, iwo g cat circles of the sphere; PQ, the line of intersection of their planes; QB, QF, their respective tangents at Q. Conclusion. The angle between the circles is equal to the dihedralAl angles along PQ. Proof. 1. By definition the angle between thQ circles is meas- ured by the angle EQF (§ 769). 2. Because QF and QE are tangents to the circles AB and CD, they are perpendicular to the diameter PQ; and because they lie in the planes AB and DC, their angle measures the dihedral angle between those planes (§ 623). 3. Therefore the angle between the great circles AB and CD is equal to the dihedral angle along PQ. Q.E.D. Theorem XVIII. 775. J^ on two great circles points he taken a quadrant distant from their points of intersection, the arc between the points measures the angle be- tween the circles. Hypothesis. AB, CD, two great circles of the sphere intersecting along the line PQ', B, D, two points each a quadrant distant from P and Q. Conclusion. The arc BD is equal to the angle BQD between the circles. A( ' >i^---l---^^ Proof. From the centre erect in each plane a perpendicular to PQ. Then— 1. Because these perpendiculars are erected from the centre, they will meet the great circle in the points B and D. J ^5 and L because jures the AB and ). taJcen a ^sectio% ngle he- le sphew III; Circle ii SPUmaOAL TBI ANGLES AND POLYGONS. 349 2. Because OB and OD are radii of the sphere, Arc BD = angle BOD. 3. Because OB and OD are perpendicular to PQ^ Angle BOD = dihedral angle of planes. 4. Because angle BQD = dihedral angle of planes (§623), Angle BOD = angle BQD = angular distance BD. Q.E.d! TnEOREM XIX. •776. ^^e angular distance between the poles of two great circles is equal to the angle between the circles. Proof. Let P aiid ^ bo the poles of the great circles AB and OD, and the centre. Join OP, OQ. Then— 1. Because P and Q are poles, by A definition. OP 1 plane ^P; ) (§ 734) Oq 1 plane CD. Therefore Angle PO^ = dihedral angle between planes, = angle between circles. Q.E.D. Corollary 1. From this and the preceding theorem, with Theorem XI., follows: 7*77. If through the poles P and Q of two great circles a third circle be passed intersecting the other circles at A, 0, P, and D, we shall have — I. Angle PQ = angle AC = angle BD, = angle between circles AB and CD, = dihedral angle BOD. II. The third ci. jle will intersect both the other circles at right angles. 778. Cor. 2. If two sides of a spherical triangle are quadrants, the angles opposite these sides will be right angles. Belatioii of a Spherical Triangle to a Trilie- dral Angrle at the Centre of the Sphere. Because the three sides of a spherical triangle are arcs of 51^0.0 unCiuo, Luuir pianea ail intersect at tne eeniiu of the sphere, where they form a solid trihedral angle. wm 850 BOOK X. OF CURVED SURFACES, By § 744 tho throo sidoa of tho triangle are measured by the three plane angles of tho solid angle, and by § 774 the three angles of the triangle are measured by the three dihe- dral angles of tho solid angle. Hence: 779. To every spherical triangle corresponds a trihedral angle at the centre of the sphere, having its six parts equal to the parts of the spherical triangle. Conversely, if 0-AliC be any trihedral angle, we may imagine a sphere with its centre at and an arbitrary radius OP, Then — The surface of tho sphere will in- intersect the edges OJ, OB, 00 at the points P, Q, and E. The same surface will intersect the planes OAB, OIW, OCA in tho arcs of great circles PQ, QU, HP, which arcs will form a spherical tri- ^^ angle. Hence: 780. Bvery trihedral angle may be represented by a triangle on a sphere having its ce?Ure at the vertex. From this it follows that the relations proved in §§ 655-660 between the faces and angles of a trihedral angle are true of spherical triangles, when for the face angles of the trihedral angle we substitute the sides of the spherical triangle, and for the dihedral angles tho angles of the triangle. We thus con- clude: '7 SI, If two sides of a spherical triangle are equal, the opposite angles are equal (§ 655). 782. In a spherical triangle the greater side and the greater angle are opposite each other (§ 656). 783. The sum of any two sides of a spherical triangle is greater than the third side (§ 659). 784. Two spherical triangles are equal when their sides are equal and similarly arranged (§ 660). WOR WT^ .^./v^. «/• 47>a iTiifoo et'rfpa r\fn snlrp.r'Ir.nl, f.rin.nnln is less than a circumference (§ 661). uurcd by g 774 tho ireo diho- trihedral 8 equal to we may |§ 655-660 ,re true of I trihedral e, and for thus con- equdly the e and the riangle is their sides rtl trianalo SPHBRICAL TRIAmiKti AND POLYUOm. 351 Symmetrical Spherical Triangles. 786. Def. Two spherical triangles are said to be oppoaite when the vertices of the one are at the ends of the diameters from the vertices of the other. Corollary 1 Since tho radii AO, BO, CO from* the ver- tices of a spherical triangle are tho edges of the trihedral angle correspond- ing to it, we conclude: 787. To two oppoaite spherical triangles correspond two opposite and symmetrical trihedral angles (§ 653). Corollary 2. Since tho lines AA\ BB\ and CC all intersect in the same point 0, each pair of them is in one plane, passing the centre of the sphere ^"^^^ ^^'' ^^'' ^^' »»« and containing the corresponding arcs Xtl^Lti'a^ S„Z"' AB and A'B', 5(7 and B'C, CA and CA\ Hence: • 788. The corresponding sides of two opposite triangles are formed of arcs of the same great circle. ^ 789. Symmetrical spherical triangles are those in which the sides and angles of the one are respectively equal to those of the other, but arranged in the reverse order. Theorem XX. 790. Opposite spherical triangles are symmetric cat. Proof. Let ABC and A'B'C be the triangles, and the centre of the sphere. Then— 1. Because the angles AOB and •*■ A' OB' are in one plane, we have Angle ^07? = opp. angle A' OB'-, and for the same reason, Angle BOC = opp. angle B'OC; Anglo GO A = opp. angle C"OA'. 3. For th« same reason. 852 BOOK X. OF CURVED SURFACES, Dihedral angle 0^ = dihedral angle 0A\ Dihedral angle OB = dihedral angle OB', Dihedral angle 0(7 = dihedral angle 0C\ Whence the corresponding angles of the two triangles are respectively equal (§ 774). 3. To an eye looking from the yertices A, B, C and the sides AB, BC, CA succeed each other in the negative order (§ 648); while A', B', C and the fides A'B', B'C, CA' succeed each other in the positive order. Therefore the triangles are symmetrical. Q.E.D. Note. This theorem should be compared with § 653, which is the equivalent L'leorem in the case of' a trihedral angle. 791. Corollary. Two symmetrical triangles cannot in general be made to coincide identically. For if we slide the triangle A'B'G' over to the other side of the sphere so that A' ~ A and C = C, the vertices B' and B will fall on opposite sides of A C. If we turn one triangle round so that B and B' shall fall on the same side of AC, the vertex A' = C and C = A. Theorem XXI. 793. If two symmetrical spherical triangles are isosceles^ tJiey are identically equal. Proof. In the preceding case suppose Side BA = side BO; we shall then have Angle A = angle 0. (§ 781) Therefore, in the symmetrical triangle, Side B'A' = side B'C = side BO = side BA. Angle A' = angle 0' = angle = angle A. Then if we slide A'B'O' over so that ^'e Cand O'^A, we shall have Side ^M'= side ^a ^'ideB'G' = sideBA. Vertox /?' = vertex B. Therefore the two triangles are identically equal. Q.E.D. Polar Triangles. 793. Bef. If the sides of one spherical triangle have their poles at the vertices of a second triangle, r\ea axe , G and legative bierefore ch is the nnot in iher side 5 B' and shaU f aU A, lies are (§781) I. E^, we Q.E.D. triangle riangle, 8PHERI0J L miANQLEa Am) POLYGONS. 353 the first triangle is called the polar triangle of the second. Theokem XXII. 794. ThepoUr triangle of a polar triangle is the original triangle. Hypothesis. ABC, a spherical triangle; A'B'C\ its polar triangle. , Conclusion. The polar tri- yK angle of A'B'O' is the original /' ^'^^ triangle ABC. Proof. 1. Because the great circles A'B' and A'C have their poles at G and B (hyp. and def.), , , , . their point of intersection A' is / a' ^ \ the pole of the circle BC (§ 753). •^'' — ---"'" ^ 2. In the same way it is shown that C" is the pole of the circle AB, and B' ot AC. 3. Because the three great circles AB, BC, and CA have their poles at C, A', and i?', ABCk, by definition, the polar triangle of ^'^'(7'. Q.E.D. ^ 795. Corollary. The relation between two polar tri- angles may he expressed hy saying thai the vertices of each triangle are the poles of the sides of the other. Theorem XXIII. 796. In two polar triangles each side of the one is the supplement of the opposite angle of the other. Hypothesis. ABC, A'B'C, a pair of polar triangles in which A'B' is the pole of the c' vertex C, etc. /^>kN Conclusions, Arc AB -\- angle C" = 180°. ^ X \ Xfi* Arc BC -f angle /I' = 180°. Arc A'B' -j- angle O = 180°. Arc B'C -f angle A = 180°. / etc. etc. ttc. jS[^> -A^ Proof. Produce the sides « AB and A C until they meet B' C in M and N. Then— 11 1. 'it| 1 11' !»}«=: 364 BOOK X. OF CURVED SURFACES. 1. Because A is the pole of MN, AM and AN are quad- rants and , ^ ._ ^^^. Arc MN = angle A. (§ 775) 2. Because ^' is the pole of ACN and C" is the pole of ABM, Arc B'N= arc C*M— quadrant, or 90''. 3. Arc B'C - arcs i?'JV'4- G'M ~ JfiV (identically); or comparing with (2), Arc B'C*- 180° - arc MN\ and comparing with (1), Arc B'O' - 180° - angle A, 4. Therefore arc B'G' + angle A = 180°. Q.E.D. In the same way all the other relations may be proved. ? Theorem XXIV. 797. The sum of the three angles of a spherical triangle is greater than a straight angle and less than three straight angles. Proof. Let A, B, and G be the three angles of the spherical triangle, and a', b', and c^ the opposite sides of the polar triangle. Then — I. A-\-a' B-\-h' G + c' = 130 = 180 = 180 (§ 796) Sum Whence A^B^G+a' + l'^-c'= 3.180° A + B^G= 3.180° - (a' + J' + c'). But because a', J', and c' are sides of a spherical triangle, «' _|_ J' 4- c' < 2.180°. (§ 785) Therefore A + B ■\- G > 180°. Q.E.D. II. Because each angle of the triangle is less than a straight angle, the sum of the three angles is less than three straight angles. Q.E.D. 798. Def. The spherical excess of a spherical tri- angle is the excess of tlie sum of its three angles over a straight angle. Gorollary. The spherical excess may de equal to any positive angle less than a circumference. f Tm OTLINDER 355 e quad- (§ 'S'75) pole of CHAPTER ill. THE CYLINDER. ically); ved. Tierical %d less of the 38 of the (§ ^96) briangle, (§ ^85) than a an three ical tri- les over to any 799. Def. A cylindrical surface is the surface which is generated by the motion of a straight line constantly touching a given curve, and remaining parallel to its original position. Illustration. If the straight line AB move around the curve AM, remaining parallel to the posi- tion AB during the motion, it will generate ^ a cylindrical surface. 800. Def. The generatrix is the line which generates the surface. 801. Def. The directrix is the curve which the generatrix touches. ^ 802. Def. A cylinder is a solid bounded by a cylindrical surface and two parallel planes, 803. Def. Elements of the cylinder are the differ- ent positions of the generatrix. Remaek 1. In elementary geometry the directrix is a circle whose plane is perpendicular to the generatrix. Kemark 3. Since the generatrix may extend out indefi- nitely in two directions, a cylindrical surface may extend out indefinitely in two directions. 804 . Def. The axis of a cylinder is a line through the centre of the directrix parallel to the elements. 805. Def. A tangent plane to a cjdinder is one which touches the cylindrical surface without inter- secting it. 806. Def. xV right section of a cylinder is the sec- tion by a plane perpendicular to the elements. 807. Def. A sphere is said to be inscribed in a cylinder when all the elements of the cylinder are tangents to the sphere. ■I' 366 BOOKX. OF CURVED SURFACES. Theoeem XXV. 808. A plane tangent to a cylinder is parallel to all the positions of the generatrix, and touches the cylindrical surface along an element. Hypothesis. KLMN^ a cylindrical surface; P,0', two points in which a plane may touch the surface without intersecting it. Conclusion. The line P^' is an ele- ment and lies in the plane, and all other elements are parallel to the plane. Proof. 1. Let KM be any element. If KM were not parallel to the trngent plane, it could be produced so far as to Mt^ intersect the plane, and then the cylindri- cal surface would also intersect the plane, which is contrary to the definition of a tangent plane. Therefore KM II tangent plane. Q.E.D. 2. Let P^ bo the element which passes through P. Be- cause P is in the tangent plane, and PQ cannot intersect the plane (1), Q is in the plane. But Q' is also in the plane, by hypothesis. Then the plane would intersect the curve MN at the points Q and Q', and therefore would intersect the surface also, which is contrary to the definition of a tangent plane. 3. Therefore the points Q and Q' are the same, and PQ' is an element. 4. Because P and Q are both in the plane, the line P^ is in this plane, and is the line of contact between the cylinder and plane. Q.E.D. Theoeem XXVI. 809. If a sphere he inscribed in a cylinder, the ^irfaces will touch on a great circle the plane of which forms a right section of the cylinder. Proof. Let be the centre of ttne sphere, and A3 any element of the cylinder. Through pass a plane perpen- f t THE CYLINDER, 857 frallel to cTies the I P. Be- ersect the plane, by irve MN irsect the \ tangent and PQ' ne PQ is ) cylinder ider^ tlie ')lane of I AB any s perp€n- diculM- to the elements of the cylinder, which call the plane (/. Then — ^ 1. Because the sphere is inscribed in the cylinder, the line AB is tangent to the sphere (§ 807). ^ 2. Let P be the point of tanffencv. Join OP, Then ^ OP L AB. (§ 763) Therefore AB being X plane 0, OP lies in the plane 0; whence P also lies in this plane, and is the only point in which AB meets the surface of the sphere. +n \ .^""^T ^^ T^ ^' ^""^ ^^'^^^*' a" tlie elements touch the sphere m the plane perpendicular to them 4. Because the plane passes through the centre of the sphere and the points of tangency are all on its intersection with the surface, these points are on a great circle of the spnere. Q.lli.D. o.V^7fZ^' It *? 'P^''"' ^'' ms^nh^^di in the same cylmder, then the planes of contact, being perpendicular to spheres are m the axis perpendicular to the planes, their dis- tance IS also equal to the distance between those centres Hence: lenffl^n'f fr'^^^'f '""''f'^ *^ i^^^ ^^rne cylinder intercept lengths of the elements equal to the distance letioecn the centres of the spheres. ^'"i^/vd Theorem XXVII. i. n^l)'^'"''^ ^^""''^ ''^^'"^ ^^^ <^y^^'dTical surface i^J!^^'^^'''''' ^^^'' '"'^ ^^^''' '''*^^'' ""^ * cylindrical sur- Conclusion. APB is an ellipse. Proof. In the cylinder inscribe two spheres of which ^ and Q are the centres, in such a position that each of them snail touch the piano A .« Let ^-and F be the points of contact with the plane. LBt P oe any point oi. the section APB, HI the element 'M 358 BOOK X. OF CURVED 8USFACE8. passing through P, and H and I the points at which this element touches the spheres and Q. Join PF and PB. Then— 1. Because the plane ^^ is tangent to the sphere at B, and P is a point in this plane, PB is a line tangent to the sphere at^. 2. Because P/f is another tangent from P to the same sphere, PB r=. PH. (§ 764) 3. We find in the same way, for the A sphere Q, PF-PL Whence PE-\-PFr:z PII^ PI= HI, or PF + PF=OQ, (§810) 4. Since P may be any point of the section, the sum of the distances of every point of the section from F and F is equal to the same constant length OQ. Therefore the sec- tion is, by definition, an ellipse around ^and Pas foci (§ 614). ^ Q.E.D. 813. Corollary 1. The distance of the centres of the in- scribed tangent spheres is equal to the major axis of the ellipse. 813. Cor. 2. The tangent spheres touch the plane of the ellipse at its respective foci. 814. Cor. 3. Parallel plane sections of a cylindrical surface are identically equal. » » CHAPTER IV. THE CONE. 815. Bef. A conical surface is the surface gener- ated by the motion of a straight line which constantly passes through a fixed point and touches a curve. TEE CONE. 359 ich this h B Hi e snm of ind F is the sec- i (§ 614). i.E.D. f the in- le ellipse. me of the lindrical se gener- nstautiy rve. 816. A cone is the solid formed by cutting off a portion of a conical surface by a plane. A cone is completely bounded by the conical surface and the plane. 817. The base of a cone is its plane surface. Remark. In the higher geometry a conical surface is called a cone, and we shall use this abbreviation when con- venient. 818. Def. The generatrix is the generating line of a cone. 819. JDef. The directrix of a cone is the curve along which the generatrix moves. 830. The vertex of a cone is the fixed point through which the genera- trix passes. 831. Bef. Elements of the cone a cone, are the straight lines occupying the different positions of the generatrix. Remark 1. In elementary geometry the directrix is sup- posed to be a circle. Rema iK 2. Since the generating line may extend on both sides of tne vertex, a complete conical surface consists of two surfaces meeting in a point at the vertex and extending out indefinitely in both directions. 833. Def. The two parts of a complete conical Suixace are called nappes of the cone. 833. Def. A tangent plane to a cone is a plane touching the conical surface without intersecting it. 834. Def A sphere is said to be inscribed in a cone when ail the elements of the cone are tangents to the sphere* iLviii m ^'m 'iVi 360 BOOK X. OF CURVED aUBFACES. if I i 11 825. D^. The axis of a cone is the straight line from the vertex through the centre of the directrix. 836. Def. A right cone is one in which the vertex is in the line passing through the centre of the directrix perpendicular to its plane. Note. In the following propositions the word cone means a right cone, though eome of the theorems are true of other cones. Theorem XXVIII. 827. A plane tangent to a cone touches it along an element, and passes through the vertex. Hypothesis. 0-AB, a conical surface; M, a point at which a plane touches the surface. Conclusions. I. The plane passes through 0, II. OMj and no otlier element, lies in it. Proof. I. If the plane did not pass through 0, then, because it does pass through M, OM would intersect the plane at My and the plane could not be a tangent. Therefore the plane passes through 0. Q.E.D. II. Because the points and M both lie in the plane, the element OM lies wholly in it. If any other element than OM may lie in the plane, let OA be that element. The plane would meet the directrix AB vX two points A and K, and therefore would intersect it and would not be a tangent plane. Therefore the plane touches the cone only along the element OK, Q.E.D. Scholium, In II. of this demonstration it is assumed that no part of the directrix is a straight line. If such were the case, a portion of the conical surface would be a plane, and the tangent plane might coincide with this plane surffice. ' lit line t along kt which ano, the lane, let iirectrix ersect it ong the aed that wrere the a.ne, and fice. THE GONE. Theorem XXIX. 361 828. If a sphere he inscribed in a cone, the sur^ faces touch on a circle all points of which are equally distant from tJie vertex. ^ Proof. When a sphere is inscribed in a cone, each element IS, by definition, a tangent to the sphere. Hence all the elements are tangents to the sphere from the vertex, and are equal by Theorem XV. (§ 764). From the same theorem it follows +hat the points of contact lie upon a circle. Q.E.D. 839. Corollary. If hoo spheres be inscribed in the same cone, the segments of the elements intercepted between the points of tangency are all equal. Theorem XXX. 830. Bmry complete plane section of one of a cone is an ellipse. Hypothesis. APB, a plane section of one nappe of passing through all the elements. Conclusion. APB is an ellipse. Proof Let and Q be the centres of two spheres mseribed in the cone, and tangents to the cutting plane at the respective points E and F. Let P be any point of the section, and a and H the points in which the element VP touches the respective spheres. Join PE, PF. Then-. 1. Because PE and PG are tan- gents to the aume sphere, PE = PG, nappe 86S BOOK X. OF VUBYED SUBFACES, 2. In the same way, whencfc PF\-PF=OH. 3. Because and H are the points in which an element touches two inscribed spheres, the line OH has the same length for all the elements (§ 829). Therefore the sum of the distances PB -f PF is the same for every point of the section, and this section is an ellipse, ])y definition. Q.E.D. 831. Corollary. The points in which the inscribed spheres touch the cutting plane are the foci of the ellipse. Theokem XXXI. 832. E:i}ery section of hotli nappes of a cone by a plane is an hyperbola. Hifpothesis. AB, a section of two nappes of a cone by the same plane; F,the vertex of the cone. Conclusion. AB is an hyper- bola. Proof. Let and Q be the cen- tres of two spheres inscribed in the cone, and tangents to the cutting plane at the points F and F. Let P be any point of the sec- tion; PVG, the element through P; and G and H, the points in which this element touches the spheres. Join PF, PF Then— 1. Because the plane AB is tan- gent to both spheres at the points F and F, and P is in this plane, PF and PF are tangents to the respective spheres. 2. Because PF and PG are tangents to the sphere from the same point P, PF = PG. rt in and r> TT -« i-^. »j%»»-4-rt 4-^^ ■i'T% «^ nif\ V% ^\*ty\ n o. Decause 1''jc' Hud. I'ti arc langenis lo me spnero \^y PF = PH. THE CONE. 803 4. Subtracting this equation from (2), PE--PF=^ PG-PH= 0H= 0V-{- VH, 6. But the lengths VQ and VH are each constant for all the elements of the cone. Therefore since every point of the section must bo on some element of the cone, the difference of the distances of all such points from i^and Fib the same. fj. Therefore the section is an hyperbola Laving its foci at ^andi^(§531). ^ E.D. 833. Corollary. TJie major axis of the hyperMa is equal to the common length of the segments of the elements contained between the points in which they touch the inscribed spheres. Scholium. Let MNR8 be a portion of a conical surface, of which AB'iq the axis, extending out indefinitely; andM< let an indefinite plane pass through a point X Then— 1. If the plane makes with the axis an angle greater than A VH, it will in- tersect one nappe of the cone com- pletely, and will not touch the other nappe. The section is then an ellipse. 2, If the plane makes with the axis an angle less than that of the cone, it will partly intersect both nappes but will cut through neither of them. The section is then an hyperbola. ?• ?i^® ?^T ""^^^^ ^'^^ ^^® ^^^s an angle equal to the angle AVM of the cone, it will cut into one nappe of the cone indefinitely, but will not cut the other. If we imagine the plane ^Xto turn upon X until it assumes the position PX, the lower of the two tangent inscribed spheres will be moved off indefinitely. Hence the focus of the elliptic section of the cone will move off indefinitely, and the ellipse will become a parabola (§ 560). Hence: 834. Corollary. The section of a cone bv a vlane mralM io an element is a parabola, " " m IMAGE EVALUATION TEST TARGET (MT-3) /. 1.0 I.I li£ 12.0 6" 1.8 IL25 IIIU 111.6 V] vl ^^^1^ y^ V Hiotographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 ^v '"^ 4^ \ \ ""O q\ «^ BOOK XL THE MEASUREMENT OF SOLIDS. CHAPTER I. SUPERFICIAL MEASUREMENT. 835. Def. A right parallelepiped is a parallelo- piped in which the lateral faces are at right angles to the base. \ Remark 1. Since any face may be taken as the base, any parallelopiped in which one face is at right angles to the four adjoining faces is a right parallelopiped. Kemark 2. A right parallelopiped differs from a rect- angular one (§ 685) in that its base need not be a rectangle. 836. Bef. The lateral area of a prism or pyramid is the combined area of its lateral faces (§§ 672, 695). 837. Def, A prism is said to be inscribed in a cylinder when its bases coincide with the bases of the cylinder, and its lateral edges are elements of the cylinder. 838. Def. A pyramid is said to be inscribed in a cone when its base is a polygon inscribed in the base of the cone, and its lateral edges are elements of the cone. 839. Def. A regular pyramid is a pyramid of which the base is a regular polygon, the perpendicu- lar through whose centre passes through the vertex of the Dvramid. SUPERFICIAL MEA8UBEMENT. 366 Theorem I. 841. The lateral area of a prism t. Theoeem III. 844. The lateral area of a regular pyramid is equal to half its slant height into the perimeter of its base. Proof. Let V-ABCDE be a regular pyramid, and VN its slant height. Then— 1. Considering ^J5 as the base of the triangle VAB, FJVwill be its alti- tude. Therefore Area VAB =^ ^VK , AB. 2. Because the slant heights are all equal to FJV", Area VBC = iFJV^. BC, Area VCD = iFiV^. CD, etc. etc. SUPERFICIAL MEASURBMEMT. 367 3. Taking the sum of all these areas, Lateral area = i VN{AB + ^C+ CD + etc.), - iFiV. perimeter ABODE. Q.E.D. Theoeem IV. 846. The lateral area of a frustvm ni' n ^^....i pyramid is equal to its slant heigkt ^^ ^''^'' trUo half the sum of the perimeters F^-i^i of its bases. Proof, Let ABCDH-FOmj he the frustum. Then— 1. Because the lateral faces are trape- ^ zoids, having FG || AB, OH \\ BO, etc., b C 8. Taking the sum of'all these 1^^, ^^ '**^> Lateral area = * (AB + BC+ etc. ^ FG + GH+ etc.) X slant height. ' :^« t f 21 1 1"- = ?«"'»«'«■• of lower base; Therefore "' "" perimeter of upper base. Lateral area = i sum of perimeters x slant height. Q.B.D. Theobem V. * n- „ ^ -^ .,, . ^J^^ifJ^J^ be the right cone. Inscribfi in It a pyramid having a regular base of any number of sides. Then— 1. Because the edges of the pyramid coincide with the elements of the cone they are all equal and the pyramid is regular. Therefore Lateral area of pyramid = ^ slant height ^ X perimeter of base. , -^et the number of sides be indefinitely increased. Then— ^ 868 BOOK XL MEA8 UREMENT OF SOLIDS, 2. The perimeter of the base of the pyramid will approach the circumference of the base of the cone as its limit. 3. Therefore th') slant height of the pyramid will approach the slant height of the cone as its limit. 4. The surface of the pyramid will approach the surface of the cone au its limit. Therefore Lateral area of cone = i slant height X circumf. of base. Q.E.D. 847. Corollary 1. The lateral area of the frustum of a right cone is equal to the product of its slant height into half the sum of the circumference of its bases. Cor. 2. If the frustrum of the pyramid be cut midway by a plane parallel to the base, the section of each face will be half the sum of the bottom and top edges (§ 170). Hence the perimeter of the section will be half the sum of the perime- ter of its bases'. The same will be true of the cone. Hence: 848. The lateral area of a frustum of a cone is equal to its slant height into the circumference of its mid-section. Spherical Areas. Theorem VI. 849. Two symmetrical spherical triangles are equal in area. * Proof. Let ABC and A'B'C be the two symmetrical triangles placed opposite each other (§786); P, the pole of the circle of the sphere passing through A, B, and C. From P draw the diameter POP'. Then— 1. Because P is the pole of the circle through A, B, and G, Arc PA - arc PB = arc PC. (§ ^45) 2. Because A', P', C\ and P' are opposite points to^, B,C, and P, respectively, P'^' JPA, P'B' = PB,^ P*C' = PC', whence P'A' = P'B' = r'C\ SUPERFICIAL MEASUREMENT. 3. Therefore ^PCand A'P'C' BPA and B'P'A ' 369 /nr Ta -a ?\T ^^^P^c^ively isosceles symmetrical triangles and identically equal (§§ 784, 792). ^ 4. Because the sum of the three triangles APO etc makes up^^C; and the sum of the three equal trianglj^ ^ P'5', etc., makes up ^'^'C, therefore Area A'B'C = area ABC. Q.E.D. 850. Def A lune is a portion of the surface of a sphere bounded by two great semi- circles. ^ 861. Be/. The angle of the lune IS the angle between the great semi- circles which bound it. Corollary. The angle of a lune is equal to the dihedral angles between the planes of its bounding semicircles (§ 774). Ahae, Theorem VII. 852. On the same sphere, or on equal spheres lunes of equal angles are identically equal. Proof. Let the two spheres be applied so that their centres Turn one sphere ronnd on its centre so that the vertex and one semicircle ol itelune shall coincide with the lo^ spending vertex and semicircle of the other Because the angles are identically equal, the planes of the two semicirc es will then coincide, and therefore the bound! mg semicircles will also coincide. The lunes 4 therefore identicaUy equal, by definition. Q.E.D. werelore Theorem VIII. n^^^' '^^J^^'f^ O^^t '^^eiee mutually intersect the areas of the two triangles on opposite sides of rniy vertex are together eq,ml to the area of a lune •i..,..^vy u,v^ ^ivyvvjvi iiita ai me vertex. ■SH 370 BOOK XL MEASUUBMENT OF SOLIDS. Hypothesis. AB, Gil, MN, any throo great circles; P, any vortex where two of them crosa. Conclusion. Areas FGM+FIIN= ImioFHQNP. Proof. 1. Because PMG and QNN are opposite triangles formed between gI the same great circles, they are sym- metrical triangles (§ 790). Therefore Area PMG = area QNH. 2. Therefore Area Pmi + area PMG = area PNH-\- HNQ, = area of lune Pi/(?iVP. Q.E.D. Theorem IX. 854. The area of a lune is to the whole surface of the sphere as the angle of the lune to a circumfer- ence. Proof. 1. Let PAQB be a lune of which P and Q are the vertices. From the vertex P to ^ pass n semi- circles making equal angles with each other. Th'^ whole surface of the sphere will then be divided into n equal lunes. (§ 853) 2. Let m be the number of those equal lunes contained in the given lune PAQB. The ratio of this lune to the surface of the sphere will then be w : ». 3. The angle APB is made up of m angles, of which n are required to make up the whole circumference around P. Therefore the ratio of the angle APB to a circumference is m : n. 4. Because these ratios are equal however great the num- bers m and w, they remain equal when the angle of the lune and the circumference are incommensurable. Therefore Area of lune : surface of sphere :: angle of lune : circumf. Q.E.D. aUPERFIOIAL MEASUREMENT. 37^ 855. Corollary. I. U there ai-e lunes of andcs A B etc., we have ' ' Area (lune A + luuo J5-f etc.) = area lune (^ + i? + etc.). of \^\ ^''''' ^' ■^'^'^ ''"''* ""^ ** liomiBphcre is that of a lune Theorem X. /• ^^yV P^ ^r^ ^-^ "" spherical triajigle is prcmor- tional to its spherical excess. Proof. Let .li?(7 be the triangle. Con tmue any one side, as Bty so as to form the great circle BGEF. Continue the other two sides b till they meet this circle in E and F /( Then— 1. Area ABC -}■ area ACF = lune ' . BABC = lune of angle B. \/ Area ^^C+ area^J5i?^= luneC^i^^ fV"'"^'^*^ = lune of angle C. ^^"-., Area ^^C7+area AFB= lune ACA^B = Inne It Inglo A. 2. BeomseBCBFiB a great circle, the sum of thi^ffur areas ABC, A CE, ABF, and ^^^ is a hemisphere. S fore, if we add the three equations (1), we have 2 area ^5C + hemisphere = lune angle (^+^+(7). (8855) /QoL ""'^ *^® hemisphere is the same as a lune of 180° (§ 856), we may write the last equation 3 area ABC -\- lune 180° = lune angle (^ + J5 4- (7V whence, by transposition, ^ ■ t" /> 2 area ABC= lune {A-\-B-\-C- 180°) and Area ABC = \ lune (^ + 5 + c - 180°) • nv ^' ^*t^.+ ^r ^^^° ^'' ^y definition, the spherical excess of the triangle ABC Because the area of the lune is proportional to its angle, the area ABCi% proportional to the ' same angle or to ^ + j5 + C - 180°. Q.E.D Corollary. If the three vertices of a triangle are on one great circle its three sides will coincide with that circle and each of its angles wiU be 180°. Its area will then be a nemispnere, and its spherical excess 3.180° — 180° = 36"° 372 BOOK XI. MEASUREMENT OF aOLWa. I Doubling these quantities, wo have the area of the whole sphere corresponding to a spherical excess of 720°. Henco 868. The area of a spherical triangle is to that of the whole sphere as its spherical excess is to 720°. Scholium, Every spherical triangle divides the surface of the sphere into two parts, of which one may be considered within the triangle, and the other without it. We may con- sider either of these parts as the area of the triangle, if, in applying the preceding proposition, we measure the angles through that part of the spherical surface whose areas we are considering. If the inner angles are A, B, and C, the^anglcs measured on the outer surface will be 360° — A, 360° — B, and 360° - G (cf. §§ 25-27). Subtracting 180° from the sum of these three angles gives 900° - (A -^ B -\- C) as the outer spherical excess. If the inner area becomes indefinitely small, AJ^B-\-C will approach to 180°, and the outer spherical excess will approach to 900° - 180° = 720°, which is there- fore the spherical excess for the outer angles of the triangle whose outer area is that of the whole sphere. 869. Def. A zone is that portion of the surface of a sphere contained between two parallel circles of the sphere. 860. Def. The altitude of a zone is the perpen- dicular distance between the planes of its bounding circles. Theorem XL 861. The area of a zone is equal to the product of Us altitude hy the circumference of a great circle. Proof. Let HKRSLI be a zone of a sphere whose centre is at 0, and let the plane of the paper be a section of the sphere through the centre, perpendicular to the base HI of the zone. The zone is then generated by the motion of the arc HKR around the axis OT, perpendicular to its base. In the arc HKR inscribe the chords HK, KR. Let M be the middle point of HK. Join OM, and from M drop the perpendicular MD upon OT, Then— BVPEUmOIAL MEASUliEMENT, le whole tlenco at of the arface of msiderod may con- rle, if, in le angles ^ wo are [le angles 50° - By L the sum ;he outer jly small, spherical is there- ) triangle irface of 3 of the perpen- ounding 873 product it circle. )se centre bhe sphere : the zone, arc HKR Let Jf be drop the 1. By the revolution around the axis OT, the chord UK will describe the lateral surface of '.he frustrum of a cone of "^^^l Sn *'"®? ""'" ^^ ^^ X circumference of a circle of which MD IS the radius (§ 848). That is, Area of frustrum = ^nMD.HK. (8 484) 2. Because OMK is a right angle, also ^^^^ ^^^ ^ ^°°^P* ^^^ = ^°°^P- ^^^y Angle Q = angle D = right angle. Therefore the triangles OMD and HKO are similar, and HE : KQ :: MO : MDi whence MD.KH=MO.EG, 3. Hence, from (1), Area of frustrum = 2;r. OM.KO, 4. In the same way, Area of frustrum KR8L = ^nOP x alt. of frustrum. Inscribe in the arc IfKM an indefinite number of equal chords and consider the frustra they describe. Then— 6. The perpendiculars OM, OF, etc., will approach the radius of the sphere as their limit. 6. The sum of the lateral surfaces of all the frustra will approach the surface of the zone as their limit. 7. Because the area of each frustrum approaches the limit 27r X radius of sphere X alt. of frustrum, the sum of all the areas will approach the limit 27e radius of sphere x sum of altitudes of frustra = 27r radius of sphere x alt. of zone, which limit is the area of the zone. 374 BOOK XL MKA8UREMENT OF 80LID8, II II 8. But ^n radius of sphere = circumf. of great circle. Hence: Area of zone = alt. of zone X circ. of groat circle. Q.E.D. Corollary 1. Let AB and CD bo two parallel tangent planes to a sphere. Since the preceding a — ;p^ — ^;^: b theorem is true of zones of all altitudes, it will remain true how near soever we suppose the bases of the zones to the tangent planes. If, then, we suppose the bases of the zones to approach the tangent planes as their limit, the alti- ^ tude of the zone will approach to the diameter of the sphere, and its surface to the surface of the sphere. Hence: 862. The entire surface of a sphere is equal to the pro- duct of its diameter into its circumference. Cor, 2. If we put r for the radius of the sphere, we have Diameter = 2r, Circumference = 2;rr; whence . ^ . Surface = 47rr". Now we have found the area of a circle of radius r to be Trr* (§480). Hence: 863. The area of the surface of a sphere is equal to the area of four great circles. The area of a hemisphere is equal to twice that of a great circle. • » . CHAPTER II. VOLUMES OF SOLIDS. 864. Bef. The volume of a solid is tlie measure of its magnitude. 865. Def, The base of a solid is that one of its faces which we select for distinction. 866. Def, The altitude of a solid is the perpen- VOLUMEQ, 875 ont'bJ^^^ '""^ *^ '^^^^^* P--^ «P- the plaae of S ^ "^^^ ParaUelopiped is a pamllelopiped any angle with each other, whereas the rectangur.7pard lelopiped has all its faces perpendicular to each other ««if ^?* ?.^* '^^^ '^^^^ ®^ volume is the volume of a cube of which each edge is the unit of length. Tolumes of Polyhedrons. Theorem XII. 4H^^\?^^^^ ?^***^* ^^^^'^^ ^Q.'^l altitudes and identically equal bases are identically equal IJ yiyfo right prisms in which ^ J^ J^ i^ U and ^"^ ^^^^^ = ^*«« A'B'G'D'E' (identically) Alt. AF = alt. A'F\ Conclusion, The two j prisms are identically ^/^ _. ^^.^ eqnal. v< •' >r ..-''7^"^^^!' Proof, Apply the bases to each other so that they shall wholly , coincide. Let A' ~ A, ^ B' = B, etc. Then— - ^ jy — "c^ 2. Because ^^= ^'jp' q T +1, Point /^= point i?*. with-a^rJX„"rn::rtr2 r^^**?."* '^ °- -U coincae £- — — Q T.^xtv-i vx uiiv utiier. 'i" , /I/ 376 BOOK XL MEASUREMENT OF SOLIDS. 4. Therefore every edge of the one will coincide with a corresponding edge of the other. 5. Therefore every face of the one will coincide with the jone8pon«ling face of the other, whence the figures will be identically ec^ual. Q.E.D. Theoeem XIII. 870. Bight prisms having equal bases and alti- tudes are equal in volume. Hypothesis, JfiVand PQ, two equal bases of right prisms having equal altitudes. A B C / /a in Conclusion. These prisms are equal in volume. Proof. By the definition of equal magnitudes (§13) the hypothesis implies that the bases can be divided into parts such that each part of the one base shall be identically equal to a corresponding part of the other base. Let A, B, and C be the parts of MN, and A*, B\ and 0' the corresponding identically equal parts of PQ. Let each prism be divided into smaller prisms by planes perpendicular to the base, and intersecting it along the bounding lines which divide the bases into the parts A, B, C, A', etc. Then, because the bases A. and A' are identically equal and have equal altitudes, Prism on base A = prism on base A^ identically. ( §869) In the same way, each part of the one prism is identically equal to a corresponding part of the other. Therefore the two prisms, being made up of these equal part(3, are identically equal. Q.E.D. 871. The volumes of right prisms having equal bases are proportional to their altitudes. VOLUMES. 377 e^'S'- if\ir''r "'''' ^™- - -^-^ f^e G £ 8 Q M SI .>- ,^-- /- a -1 — w H ,^1 .^'•^■ D base ^5CZ> is equal to MNOF, and the altitude AJS is to the altitude MQ as m : n. Conclusion. Vol. Jjy : vol. MT ::m : n. Proof, Divide the prism ^iTinto m parts of equal alti- tude by planes parallel to the base. Let prism MT be di- vided into n parts in the same way. Then— Tall Stis^,;" '^''^ ^^ ^^ -*^ ^ P-^«^ these parts l^laL^mn^r^^''' P"'"^^ ^"™^ identically equal Dases and altitudes, they are equal in volume (8 869) 3. Because the volume Jfr is composed of « mrts of which m parts make up the volume ^^, therefore ' Volume Aff : volume MT :: m : n. Q.E D 4. Because this is true how great soever the numbers m and n, it remains true for all cases (§ 359). Theorem XV. ^./f^^' ^^ /^^^\ P""^^^^ of equal altitudes the mlumes are to each other as the areas of their bases +. +f '*'''''^' u^®* *^® ^^^^^ ^® *^ ^^^1^ «ther as the number m to the number ^. This means that if the one base be divTded into m equal par s, n of these parts will make up the other base tuae to the given prisms. These prisms will all be equal (§ 870). One given prism will be made up of m of these eaual prisms, and the other of n of them. ^ Therefore the volumes will be to fiar^h othov as m to -z- that is, tne volumes will be as the areas of the baseJ. Q.E d' ^orollary. Because a parallelepiped is a prism, we con^ 378 BOOK XI. MEASUREMENT OF SOLIDS. 873. If a right prism and a right parallelopiped have equal altitudes, their volumes are as the areas of their bases. Theorem XVI. 874. The volumes of rectangular parallelopipeds are proporticmal to the products of their three dimen- sions. Proof Let ^ be a par- allelopiped of which the dimensions are a, b, and h; W, one of which the dimen- sions are c, d, and k. * Cut off from K a paral- lelepiped L of altitude k. Then— 1. Because L and W have equal altitudes, Vol. L : vol. W :: area a.b : area c,d, :: ab : cd. 2. Because K and L have equal bases, Vol. K : vol. L :: &\t h : alt. Is. 3. Multiplying these ratios. Vol. K : vol. W :: abh : cdk. Q.E.D. Corollary 1. If the dimensions c, d, and h of the volume W are each unity, the product cdJe will be unity, and W will be the unit of volume (§ 868). The above conclusion will then become Vol. K : 1 :: abh : 1, which gives Vol. K = abh; that is: 875. The volume of a rectangular parallelopiped is measured by the continued product of its three dimensions. Scholium. If the dimensions rt, J, and g are all whole numbers, this result may be shown in the following simple way: Being g units in height, it may be divided up into g layers each a unit in height. (§872) (§416) (§871) VOLUMES. 379 ptd have 'bases. opipeds \dimenr w U (§878) (§416) (§871) h of the nity, and onclasion ^piped is b. Bemg h units in breadth, each of these g layers may be divided into b rows, each containing a units of Yolume. Thus the total numb.er of units of volume will be abg. Cor, 2. Because the base of a rectangular parallelepiped IS a rectangle, its area is equal to the product of its two dimensions. The third dimension is then the altitude of the parallelepiped. The preceding result may therefore be ex- pressed in the form: 876. The volume of a rectangular parallelopiped is ex- pressed by the product of its base into its altitude. Cor. 3. Since a rectangular parallelopiped is a kind of right prism, every right prism is, by § 870, equal in volume to a rectangular parallelopiped having an equal base and alti- tude. Therefore we conclude : 877. The volume of every right prism is expressed by the product of its base and altitude. Theorem XVII. 878. All parallelopipeds hamng the same base and equal altitudes are equal in xolume. Case I. Hypothesis. ABCD-EFQH ssl^ ABCD-MNOP, two parallelopipeds hav- ing the same base ABCD P q_ h o and equal altitudes. In this case the edges ^ FE and NM, also GH and OP, are supposed to lie in straight lines. Proof. 1. Because EF and MN are each equal and parallel to AB^ we have MN = EF and ME = NF. 2. Considering the two triangular ^xismsAEM-DHP and BFN-COO, it is proved, from the equality and parallelism of all their parts, that they are identically equal. 380 BOOK XI. MEASUnSMENT OF SOLIDB. 3. From the solid ABFM-DCQP take away the solid AEM-DHP, and we have left the parallelopiped ABOD- FFQH, 4. From the same solid ABFM-DGOP take away the equal solid BFN-CQOy and we have left the parallelopiped ABGD-MNOP, 5. Therefore Volume ABCD-EFGH= volume ABCD-MNOP. Q.E.D. Case II. Let the upper base be in any position, as IJKL, Produce the parallel edges FE and OH to the points M and P, and pro- duce KJ and LI to N and Mf forming the parallelo- gram MNOP. Join AM, BN, CO, DPi forming the parallelopiped ABCD- MNOP. Then— 6. Because the parallelopiped ABCD-IJKL has the same base and altitude as ABCD-MNOP, and has the bounding edges JK and IL of its upper face in the same straight line with the edges PJf and ON, we have, by Case I., Vol. ABCD-MNOP = vol. ABCD-IJKL, 7. We have, for the same reason, Vol. ABCD-EFOH = vol. ABCD-MNOP, 8. Comparing with (6), Vol. ABCD-EFOH = vol. ABCD-IJKL. Q.E.D. Corollary. Whatever be the oblique parallelopiped ABCD- IJKL, we may construct upon the same base a right paral- lelopiped ABCD-EFOH to which the above demonstration will apply. Therefore, from (§877): 879. The volume of any parallelopiped is equal to the product of its base and altitude. Theorem XVIII. 880. A diagonal plane divides any parallelopiped into two triangular prisons of equal volume. * tie solid ABCD- way the alopiped Q.E.D. I IJKL, VOLUMES. 881 ihe same ounding ight line Q.E.D. ABGD- it paral- istration il to the lopiped ^ ^^T b ^^P^^^''^^- ABCD-EFOH, a right parallelovi- ped, of which B DBF is a diagonal I'^rmmopi plane. Conclusion. Vol. ABD-EFH=z vol. DBG-HFQ. Proof. 1. Because ABCD is a parallelogram, the diagonal BD di- vides its area into two equal parts. 2. Therefore the two right - ^riiMtuS^-"' "^^-^^^ ■"- ^^--^ "- -^ t-'^ 3. Therefore these prisms are equal (§ 870). Q E D of wwT ^I'raF'^^'l''' ^"^^^^-^FGH, any parallelepiped of which A CQE is a diagonal plane. Conclusion. Vol. .4 CD-EHG = vol. J CB-EGF. Pro(/.^ Through the vertices ^ and ^ pass th; planes AUK and ^ZifA^ perpendicular to the parallel edges AE, BF, CG, and Let 7, ^, JT and Z, if, iV^ be the points in which the cutting planes meet these edges, produced when necessarv. Then— ^ 1. Because the cutting planes are perpendicular to the same edges AE^ ^k etc., they are parallel. Therefore the ^^"^ solid AUK-ELMN is a right paral- lelepiped, and Vol. AJK EMN = vol. AJI-EML. 3. Because the edges of both parallelepipeds paraUel to AE&ie also equal to it (§ 687, cor. 1), HN= ED; GM = CJ; FL = BI-, also, by comparing the sides of the parallelograms, EH = AD', EN = AE; EG = AC, etc.; and because EMG and AJO are both right angles, by con- struction, "^ Angle EMG = angle AJC, (Case I.) 382 BOOK XL MEASUREMENT OF SOLIDS. 3. Therefore if the solid E-MQHNhe applied to the solid A'JCDK&o that the line Jg^if shall fall on AJ, then Triangle EMN = AJK, MO = JG, EO=AG, NH=KD, QH=CD, Therefore these two solids are identically equal. 4. If from the prism ACD-EOH\fQ take away the solid E-MGHN and add the equal solid A-JGDK, we shall have the right prism AJK-EMN. Therefore Vol. ACD-EGH = vol. AJK-EMN. 6. In the same way may be shown, Vol. ABG'EFQ = vol. AIJ-ELM. 6. Oomparing with Case I., Yol AGD'EGH =i yol, ABG'EFG. Q.E.D. Theokem XIX. 881, The volume of any prism is eqzcaZ to the prodtcct of its base by its altitude. Case I. A triangular prism. Proof. Let ABG-DEF be any tri- angular prism. Draw BP II AG; GP II AB; EQ II DF; FQ \\ DE Then— 1. Because ABPG and DEQF are, by construction, equal parallelograms with the sides of the one parallel to the coiTesponding sides of the other, the solid ABPG'DEQF ia a parallelopiped. Therefore Vol. ABPG'DEFQ = base ABPG X altitude. Area ABG = i base ABPG, VOLUMES. 383 8. Because BCFE is a diagonal plane of the parallelepiped. Vol. ABC'DEF = i vol. ABPC-DEQFy (§ 880) = i base ABPG X altitude, (1) = area ABC X altitude. (g) Case II. Any prism. Q.^.^. Let ABCDE-FQHIJ be any prism. Divide the prism into tri- angular prisms by passing planes through ACHF, ADIF, etc. These planes will divide the bases into triangles. Then — 1. Because ABC-FGH is a tri- angular prism. Vol. ABG-FGH = base ABG X alt. of prism. 2. In the same way. Vol. AGD-FHI = base AGD x alt. of prism. Vol. ADE-FIJ = base ABE x alt. of prism. etc. etc. 3. Adding these volumes, we have Sum of volumes = sum of bases x alt. of prism. 4. The sum of these volumes makes up the whole volume of the prism, and the sum of the triangular bases makes up the whole base of , the prism. Therefore volume of prism = base x altitude. Q.E.D. Theorem XX. 882. All pyramids having equal bases and equal altitudes g,re equal in liolume. Hypothesis. 0-ABCD and P-TUVWZ, two pyra- mids in which area ABCD = area r?7FPrZ, and alti- tude of = altitude of P. Gonclusiofi. The vol- umes of the two pyramids are equal. 884 BOOK XT. MEASUREMENT OF SOLIDS. Proof. Divide each pyramid into the same number n of slices by equidistant planes parallel to the base. Let us put s, the thickness of each slice; «, the common altitude of each pyramid; b, the area of the base of each pyramid. Then— 1. Because the altitude is divided into n parts, the thick- ness of each slice will be n 2. Because the number of slices is the same in each pyra- mid, the distances of corresponding slices from the vertex will be the same in the two pyramids. If we put I, the distance of any section from the vertex, r, the area of the section, we shall have in each pyramid r : h :: r : a\ (§ 700) Also, the areas of corresponding sections are the same in the two pyramids (§ 701). 3. Let and P be two corresponding slices from the same pyramid. Put r', the area of each upper base; r, the area of each lower base. Because the lower base of each is greater than the upper base, each slice is greater than the prism of altitude s and base r', but less than the prism of altitude s and base r. Hence s X r' < volume of each slice < s X r. . 4. Hence the difference between the volumes of corre- sponding slices of the two pyramids must be less than sr — «r'; that is, less than s{r — r'). 5. Let us call the areas of the several sections from the vertex to the base n, rg, rs, etc. We then have Difference of top slices < sri. Difference of second slices < s(r2 — rj. Difference of third slices < .?(r. — r-). etc. etc. Difference of bottom slices < s(r„ — r»_i). Adding up all these differences, and noticing that r» = J, we find Difference of volumes of pyramids < sb. YOLUMES. 385 That is: The difference of the yolumes of the pyramids is less than the Yolume of a prism of equal base, and having for its alti- tude the thickness of a slice. 6. But we may take the slices so thin that this volume sh shall be less than any assignable quantity. Therefore the volumes of the pyramids differ by less than any assignable quantity; that is, they do not differ at all. Q.E.D. Theorem XXI. 883. The wlwme of a pyramid is one third the wlume of a prism hamng the same base and altitude. Case I. Let P-ABC be a triangular pyramid. Through AC pass a plane AGFD ^ parallel to the opposite edge BP, ^^ '^' Complete the triangular prism ABC-DP F by drawing the edges PD, PF, DF, AD and Ci^paraUel to BA, BC.ACyBP. Divide the quadrangular pyramid P-A GFD into two triangular prisms by the plane PAF, Then— 1. Because AGFD is a parallelo- gram, the areas ADF and A GF are equal. Therefore Vol. P-ADF= vol. P-AGF{% 882). 2. In the same way, considering PBG and PFGbs the equal bases of two triangular pyramids having their vertices at A Vol. A-PBG = vol. A-PFG. ' 3. Comparing (1) and (2), and noting P-^C^and A-PFG 9,TQ the same pyra- mid, we see that the prism ABG-DEF is divided into three equal pyramids, of which one is the original pyramid. Hence Vol. P'ABG = \ vol. ABG-DEF, Case II. P-ABGDE, any pyramid. Through P pass the planes PAG, 386 BOOK XI. MEASUREMENT OF SOLIDS. FAD, etc., dividing the pyramid into the triangular pyramida F'ABO, P-ACD, etc. Let a be the altitude of the pyramid. Then, by Oase L, Vol. P-ABG = i prism ^^C-alt. a. Vol. P'ACD = i prism ACD-a\t a. Vol. P-ADE = i prism ^i>^-alt. o. The sum of these pyramids makes up the given pyramid, and the sum of the prisms is a prism having the base ABCDB a,nd the altitude a. Therefore, adding, Vol. P-ABGDE = i prism ABCDE-iXi, a. Q.E.D. Corollary. Because the volume of a prism is equal to the product of its base by its altitude, we conclude: 884. The volume of a pyramid is one third the product of its base by its altitude. Volumes of Bound Bodies. Theorem XXII. 886. The voltime of a cone is eqiml to one third the product of its base hy its altitude. Proof. In the base of the cone inscribe a regular poly- gon of any number of sides, and upon it erect a pyramid of which the vertex shall be in the vertex of the cone. Then — 1. Because the angles of the base of the pyramid are on the surface of the cone, and its vertex in the vertex of the cone, the lateral edges of the pyramid will lie on the conical surface, and its altitude will be equal to the altitude of the cone. Let us call a the common altitude of cone and ppamid. Then— 2. Volume of pyramid = ^ « X base of pyramid. 3. Let the number of sides of the base of the pyramid be indefinitely increased. Then the base of the pyramid will approach the base of the cone as its limit, and its volume will approach the volume of the cone as its limit. Therefore Volume of cone = ^ a x base of cone. Q.E.D. VOLUMES. 887 Theorem XXIII. '^ 886. The wlvme of a cylinder is eqvM to the product qf its base by its altitude. Proof, Inscribe in the cylinder a prism of which the number of sides may be increased without limit. Then the base of the prism will approach the base of the cylinder as its limit, and the volume of the prism will approach the volume of the cylinder as its limit. Because the volume of the prism is continually equal to the product of its base by its altitude, the volume of the cylinder must also be equal to the product of its base bv its altitude. Q.E.D. Theorem XXIV. 887. The volume of a sphere is equal to one third tts radius into the area of its surface. Proof. Make a great number of points on the surface of the sphere, and join them by arcs of great circles so as to divide the whole surface into spherical triangles. The planes of these arcs will form the lateral faces of triangular pyramids hav- ing their vertices in the centre of the sphere, and the angles of their bases resting upon the surface. Because the volume of each pyramid is ^ base x altitude, the combined volume of all is i sum of bases x altitude. Let the number of spherical triangles be indefinitely in- creased. Then the sum of the bases of all the pyramids will approach the surface of the sphere as its limit, and the alti- tudes will all approach the radius of the sphere as their limit. Therefore Volume of sphere = i radius X surface of sphere. Q.E.D. Corollary. We have found (§ 862) for a sphere of radius r, Surface of sphere = 4n'r". 888 BOOK XI. MEASUREMENT OF SOLIDS. Multiplying this by ^r, we have Volume of sphere = f^rr*. This result admits of being memorized in the following way. Suppose a cube circumscribed about the sphere. Be- cause each of its edges is 2r, its volume will be 8r'. Oom- paring with the expression for the volume of the sphere, we find Vol. sphere : vol. cube .: ^w : 2. Now if TT were exactly 3, this rutio would bo 1 : 2; that is, the volume of the sphere \M>ald be one half that of the cube. And in reality the sphere is greater than half the cube in the same ratio that n is greater than 3, which is nearly ^ part (§ 484). Therefore if we fit a sphere into a cubical box, it will occupy a little more than half the volume of the box. PEOBLEMS OF OOMPUTATIOK 1. The altitude of a right cone is 4 metres, and the diam- eter of its base 6 metres. Compute its slant height, lateral surface, area of base and volume. Ans. Slant height, 6 m. ; lateral surface, — -; area of base, 9;r; volume, IStt. 2. The lateral aroa ^ of a right cone being given, what relation must subsist between its altitude a and the diameter D of its base? Ans. irrD V{n' j- iZ)») = A. 3. The lateral surface of a right cone is double the area of its base. What is the ratio of its altitude to (he radius of its base? What must be the ratio in order that the lateral sur- face may be n times the area of the base? Ans. V'd and VnH^, 4. Find the ratio of the volume of a sphere to that of its right circumscribed cylinder. Ans. Vol. of sphere = f vol. of cylinder. Note. The circumscribed cylinder is that whose bases and ele- ments are all tangents to the sphere. 6. The slant height of a right cone = diameter of its base = 2a. Express its altitude, lateral area, and volume, and the radius, surface, and volume of its inscribed and cir- cumscribed spheres. Ans. Alt. of cone, i^3a; lateral area, 2;ra': volume, — : Rad. of insc. sphere, ~; surface, —-; volume, — '; Rad. of circ. sphere, ~; surface, i^; volume, ??^ V 3 ^ 9 4^ * 6. The radius of a sphere is bisected at right angles by a plane. What is the ratio of the two parts into which the plane divides the spherical surface? Ans. 3 : 1. 7. If a plane cut a cylinder at an angle of 45" with the elements, what will be the ratio of the axes of the ellipse of intersection? Ans. i^ : 1. THEOEEMS FOE EXEEOISE IN GEOMETRY OF THREE DIMENSIONS. BOOK vm. 1. A line parallel to each of two intersecting planes is parallel to their line of intersection. 2. Two lines, one perpendicular to one plane and one to another plane, form equal angles with the planes to which they are not perpendicular. 3. If a straight line be perpendicular to a plane, every line perpendicular to that line is parallel to the plane. 4. The supplement of any face angle of a trihedral angle is less than the sum, but greater than the difference of the supplements of the two other face angles. 5. If, on a line intersecting a plane perpendicularly, two points, A and B, equally distant from the plane be taken, and these points be joined to three or more points of the plane, the joining lines will form the edges of two symmetric poly- hedral angles having their vertices at A and 3, 6. If a plane be passed through one of the diagonals of a parallelogram, the perpendiculars upon it from the extremi- ties of the other diagonal are equal. 7. If the intersections of several planes are parallel, all perpendiculars upon these planes from the same point in space lie in one plane. 8. If any number of planes are respectively perpendicular to as many lines, and these lines all lie in one plane, or in parallel planes, the lines of intersection of the planes are all parallel to each other. 9. All points whose projections upon a plane lie in a straight line are themselves in one plane. How is this plane defined? 10. If two straight lines are on opposite sides of a plane, parallel to it, and equally distant from it (but not parallel to THEOREMS IN GEOMETRY OF THREE DlMENSIOm 391 each other), the plane will bisect every line from any point of the one line to any point of the other. 11. If any two straight lines ^ and 5 are parallel to a plane P, aU lines joining a point of ^ to a point of B are cut by the plane P, internally or externally, into segments having the same ratio. 12. Corollary, If through the ends of a harmonically divided line two planes be passed perpendicular to the line, and through the harmonic points of division two lines A and B be drawn, each parallel to the planes, but not in one plane, then every line joining a point of ^ to a point of B is cut harmonically by the two planes. 13. A plane parallel to two sides of a quadrilateral in space divides the other two sides similarly. BOOK IX. 1. If any two non-parallel diagonal planes of a prism are perpendicular to the base, the prism is a right prism. 2. If the four diagonals of a quadrangular prism pass through a point, the prism is a parallelopiped. 3. A plane passing through a triangular pyramid, paral- lel to one side of the base and to the opposite lateral edge, mtersects its faces in a parallelogram. 4. The four middle points of two pairs of opposite edges of a tnangular pyramid are in one plane, and at the vertices of a parallelogram. Note. The six edges of a triangular pyramid may be divided into three pairs, sach that the two edges of a pair do not meet each other. Since each edge meets two other edges at one vertex, and two yet other edges at the adjoining vertex, there is but one edge left to pair with it. Ihe pair is called a pair of opposite edges. 5. The three lines joining the middle points of the three pairs of opposite edges of a triangular pyramid intersect in a point which bisects them all. 6. The four lines joining the vertices of a triangular pyra- mid to the centres of the opposite faces intersect in a point n....^.. -i.Tiviva ^teuix ux iiiiciii Hi Liie raiio i : 3. Note The centre of a triangle is the point of intersection of its three medial lines (§§ 168, 169). 392 THEOBEMa FOB EXERCISE 7. The middle points of the edges of a regular tetrahedron are at the vertices of a regular octahedron. 8. The eight vertices of a cube are cut ofE by eight planes, each passing through the middle points of the three edges which diverge from each vertex. Explain the structure of the polyhedron thus formed, giving the number, form, and relation of its faces, edges, and vertices. Describe its sympolar polyhedron, showing that each face is a rhombus, and explain the number and form of its edges and vertices. Note. The sympolar of any polyhedron may be formed by draw- ing an edge across each edge of the given polyhedron, and uniting all the edges crossing the sides of each face into a single vertex. BOOK X. 1. If lines be drawn from any point of a spherical surface to the ends of a diameter, they will form a right angle. 2. Conversely, the locus of the point from which a finite straight line subtends a right angle is a spherical surface hav- ing the line for a diameter. 3. If any number of lines in space pass through a point, the feet of the perpendiculars from another point upon these lines lie upon a spherical surface. 4. If any number of lines in a plane pass through a point, the feet of the perpendiculars upon these lines from any point not in the plane lie on a circle. 6. If the axis of an oblique circular cone is equal to the radius of the base, every plane passing through the axis of the cone intersects the conical surface in lines forming a right angle at the vertex. When the axis of the cone is less than the radius of the base, all the angles thus formed are obtuse, and when greater they are acute. 6. All parallel lines tangent to the same sphere intersect any plane in an ellipse. t JL. XJ"' BOOK XI. ?he surface of a sphere is equal to the lateral surface of its circumscribed cylinder. Note. See Problem 4, p. 889. IN OEOMETRT OF THREE DIMENSIOm. " 393 3. If the slant height of a right cone is equal to the diame- ter of its base, its lateral area is double the area of its base 3. The lateral area of a pyramid is greater than the area of its base. 4. The volume of a triangular prism is equal to the area of any lateral face into half the perpendicular from the opno- site edge upon that face. ^^ 5. Any plane passing through the middle points of a pair of opposite edges of a triangular pyramid bisects its volume. 6. If the three face angles of a triangular pyramid around the vertex are all right angles, the square of the area of the ba^e IS equal to the sum of the squares of the areas of the tnree lateral faces. 7. The bisecting plane of any edge angle of ft triangular pyramid divides the opposite edge into segments propor- tional to the areas of the adjacent faces. 8. Equidistant parallel planes intercept equal areas of a spherical surface. LOCI. 1. Find the locn of the point in space whose distances from two fixed points are in a given ratio. 3. Find the locus of the point equally distant from two parallel lines. 3. Find the locus of the point equally distant from two intersecting straight lines. 4. Find the locus of the point equally distant from three given points. 6. Find the locus of the point equally distant from the sides of a triangle. 6. Find the locus of the point equally distant from the three edges of a trihedral angle. 7. Two given lines being on opposite sides of a plane parallel to it, and equidistant from it, find the locus of the point in the plane which is equally distant from the two lines. 8. Find the locus of the point from whinh fwn q/iiQ/»o«+ segments of the same straight line subtend equal angles. N ti( re in to rei th CO th fO] of po an im all sai ref obi del CO] to sol as imi ext bil: anc eqi as4 ext be APPEE^DIX. NOTES ON THE FUNDAMENTAL CONCEPTS OP GEOMETRY. The true basis and fonii of the fundamental axioms and defini- tions of Geometry have been the subject of extended discussion in recent times, especially among German mathematicians. The follow- ing summary of conclusions is given partly to show the direction towards which these discussions tend, and partly to explain the reasons for the particular forms of definitions and axioms adopted in the present work. Although the writer conceives that these views concur with the general conclusions of those who have investigated the subject, no one but himself is to be considered responsible for the form in which they are stated. I. Geometry has its foundation in observation. Clear conceptions of lines, as straight or curved; and, in general, the idea of relative positions in space, could never be acquired except through the eye and touch. The ancient axioms of Geometry proper, such as the impossibility of two straight lines inclosing a space, the equality of all right angles, and the necessity of two non-parallel lines in the same plane ultimately meeting if sufficiently produced, are not to be regarded, as they once were, as necessary conclusions apart from all observation, but only as necessary results of certain conceptions derived from observation. It has in fact been shown that a perfectly consistent Geometry can be constructed in which the axioms relatinff to straight lines are not fulfilled. n. The general concepts of Geometry— points, lines, surfaces, and sohds— are to be regarded as attaching to material bodies rather than as formed of mere space. A geometric solid, for instance, is an imaginary material body from which all qualities except those of extension and mobiUty are abstracted. The quality of impenetra- bility being abstracted, any two bodies may occupy the same space and may be brought into absolute coincidence if they are identically equal in their outlines. Surfaces, again, should rather be considered as extensions from which the idea of thickness is abstracted than as extensions absolutely without thickness. Similarly, a line need not be regarded as having no thickness, but may simply be considered as 396 APPENDIX. having the idea of thickness abstracted. A point is an object the magnitude of which we take no account of. This slight change of conception may perhaps be regarded as having little more than metaphysical interest. But it has a certain amount of practical value in releasing the young mind from a seem- ing necessity of conceiving portions of pure space as bodies and magnitudes with only one or two dimensions. In fact, it may be doubted whether any definitions of lines, points, and surfaces, in general, are of value to a young beginner. He naturally falls into the habit of applying the terms the right way. III. The following considerations have led to certain of the primary definitions adopted in the present work. 1. It may be doubted whether a straight line admits of any definition in the proper sense of the term. A student who does not know what a straight line is before it is defined will net know in consequence of the definition. The author therefore lays no stress upon the definition he has adopted, which is perhaps objectionable, but which has been chosen because most readily understood by a beginner. 2. A great majority of our writers upon Elementary Geometry make the mistake of trying to include the mode in which the angle is measured in the definition of it. The system of enunciating separate definitions of the angle and the method of measuring it has been adopted from Chauvenet, and its advantages are so obvious that they need not be pointed out. 3. That identically equal magnitudes are those which coincide is properly not an axiom, as used in the older geometry, but a defini- tion of the word " equal " and its derivatives. This will be obvious upon reflecting that the word must have some definition, and that all we can mean by it is that the two objects to which the term is applied coincide when brought together, or are made up of coin- cident parts. Had all bodies been immovable we should never have had the idea of equality. 4. A statement of what shall be meant by the sum of two magni- tudes, and especially of two angles, is absolutely necessary. The want of such a statement is one of the most serious defects in the Geometry of Euclid. Had Euclid enunciated a general definition of the sum of two angles, and adhered to it, his thirteenth proposition, that the angles which one straight line makes with another upon one side of it are together equal to two right angles, would have been unnecessary. 5. That a straight line is the shortest distance between i of its points is here considered au axiom rather than a definition. two APPENDIX. 397 in The r^on of placing it in thia category is simply that the idea of a straight line may be derived independently of any compariaoa of general measures of distance between the same two points. 6. Plane figures are defined after the modern instead of the ancient conceptions. As this will at first sight strike the teacher of the Euclidean Geometry as one of the most radical changes in the work, a comparison of the ideas on which the two systems of defini- tions are founded may be of interest. The ancient geometry was primarily a science of mere magnitude. Solids were bodies, and plane figures were pieces of a phine Of course other conceptions had to be brought in, but they were regarded as subsidiary. In modern geometry form and position are of equal importance with magnitude, and in order that all the conceptions associated with a figure may come in on terms of equality as it were it is necessary to confine the definition of a figure to what is 'really necessary to its formation. A flexibility and generaUty is thus given to the definitions which they cannot have under the older form It 18 not, indeed, claimed that, for the purpose of elementary instnic tion, one of these systems of definitions has any great advantage over the other. But it is important that the definitions should accord with the conceptions naturally formed ; with the language of every- day life, and with that of the higher modern geometry ; and theL considerations all point to the new system of definition. Let us iil^ circles and polygons as examples. In the older geometry a circle is a round piece of a plane or what, in ordinary language, is called a circular disk. In our laTlangu^^ a circle IS a curved line which the pupil can draw with a pS Hne ZT ^T^'^y^r^'^ '' "«^"« '""'^ '^'^^ -«"« this curv d lZ^LZ77{r'^ "' '"' '^:''' "^^'^^^"^^ *^« ^^'^ circumference 18 applied to far wider uses. But this nomenclature is changed as where he finds the circle and ellipse treated as curves When the eccentric ty vanishes the ellipse is said to become'Tt^hrc'rcum! ference of a circle, but the circle itself. The equation of the bound- ing curve of the circular disk is also called the equation of the circle. In a word, the old definition entirely vanishes, Z a new conception is attached to the word. A polygon, again, is completely determined by its bounding lines IndTed t thfvT '^"'/^^ '"^^^^^ *°y*^-^ ^-t these iLes Indeed m the higher modern geometry a Dolvcron i. nop«^.,«^ XrVn/r"'''"^'' ""^^ ^'"^"^"^ ^'^ ceWn^elations to'eth other, and the more strongly a definition inconsistent with this ia 398 / PPENDIX. impressed on the mind of the beginner, the more difficult he findB It to make the necessary change in the conceptions attached to the word. It may seem that the ancient form has some advantages, especially in conaldcriug areas, and it will also be remarked that the word circum- ference is used in the present work in its ordinary acceptation. A glauce at the true relations between geometric figures and words will make the state of the case quite clear, and will show a perfect analogy between the system of notation here adopted and the lan- guage of ordinary life. A geometric figure is to be regarded as combining a great number of associated conceptions, of which a certain number necessarily involve the others, and may therefore be regarded as essential. The essential conceptions are those which suffice to determine the figure. Since the figure may be determined in various ways, the only rule that can be followed is to choose for a definition the most simple and easily understood set of conceptions. Let us consider a polygon, for example. We have in the polygon a collection of associated con- ceptions: a certain number of sides, a certain number of interior angles, an equal number of exterior angles, a form, an area, and a perimeter — the latter being the sum of all the sides. No one of these has any special claim over the others to be considered as the measure of the figure. Two different polygons equal in area have no more right to be considered equal than two other polygons of different areas but equal perimeters. Nor have the concepts by which the figure is defined, however they may be chosen, any right to be con- sidered as the whole of the polygon because the associated concepts equally belong to it. Hence the proper course is to take the simplest defining conception; namely, the lines which the pupil draws when he constructs the figure, as being, not necessarily the polygon itself, but the things which determine or form it. Then its area, its angles, its perimeter, its centre (if it has one), its form, and any other concepts associated with it may be separately considered at pleasure. Again, with the circle we associate a circumference, an area, a centre, any number of radii, and any number of tangents we choose to draw. The word ''circle" is properly applied only to the whole assemblage of concepts; but since the circumscribing line is the fundamental determining thing, the word can be more properly applied to it than to any other of the associated concepts. When, however, the length of this line comes into consideration, or when the line is to be considered in antithesis to some other conception, the centre for instance, then the word circumference is used. The accordance of this mode of language with that of ordinary life will be seen by comparing it with the ideas which we attach to the APPENDIX. 399 word ''house." We may equally define a house as a space sur- rounded by walls or as walls enclosing a space. We habitually use the word in both senses without any ambiguity or confusion. %^ speak of building a house when we really mean building the walls and of living in the house when we mean living in the inirior ' V. The greatest improvement in the modern over the ancient geometry IS made in the extension of the idea of angular magS^ In Euclid ouJy angles less than ISQo are considerc^as S anv actual existence. Angular measures equal to or exceed nrthisHm^t are considered merely as sums of angles to which no viable gTo metric meaning is attached, and which are in fact treated as L?!w symbolic entities, like the imaginary quantities of moTe^n "atCm^^ tics. Some moderns have followed in his footsteps so slavishly L to acually apprise the pupil that an angle of 180° i not an angle Me„t the pupi might be led into the mistake of considering the sum o1 two right angles as having some conceivable meaning?^ We have already mentioned the failure of Euclid to give any defi- nition ofthe sum of two angles. Without such a definition we do not ^rZ^ V Tu ^ ^^"^ '*°^^'' "• ^'^^ «"^'^ * definition the sum eLln^,-^ / *"^.r ^''^°''' ^^^ ^"^'^ ^^'•'"^d by t^« straight lines extending from the same vertex in opposite directions. In modern geometry angular measure is unlimited, and a civen angle may have any number of measures diflFering from each other bv any entire number of circumferences. It is not, however, advisable to burden the beginner by attempting to impress this idea upon his mind but he should be led up to it gradually. Hence in commenc- mg to write the present work, the author started out by confining angular measures to the limit of 180°. He soon found, however that confusion would result from attempting to keep within thi^ limit, especially m considering the relation of angles inscribed in a circle He therefore adopted the plan of extending angular meas- ures to one circumference, and explaining in the beginning the two measures of the angle. He finds by expc rience that there is no diffi- culty in making this double measure clear to a very young beginner