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Les diagrammes suivants illustrent la mdthode. 1 2 3 1 2 3 4 5 6 MICROCOPY RESOLUTION TEST CHART (ANSI and ISO TEST CHART No. 2) 1.0 I.I 1.25 ilia |||i^ I'M M.O 1.4 2.5 [|| jj. iHii 2 2 2.0 1.8 1.6 ^ APPLIED IIVHGE inc 1 65J East Main Street Rochesier, New York 146C9 USA (716) 482 - 0300 - Plione (716) 288 ~ 5989 - Fax Xc^a^-^JLer^ A TEXT-BOOK OP EUCLID'S ELEMENTS. ,»- A TEX'l-BOOK OF EUCLID'S ELEMENTS BOOKS I. -VI. AM) XI. IJY H. 8. HALL, M.A. FOiniEHLY SCIIOLAU OF CHRISX's COLLEGE, CAMHllIDOE AND F. H. STLVENS, M.A. KOUMERLY SCHOLAK OK QUEEN'.S COLLEGE, 0.\F()l;]> MASTKKS OK T||K MH-ITAKY SIDK, CIJFTON COUM.K rrescrihed hij the Conncil of Public Instniction for u.c in the Schools of Nora Scotia. ILontion : MACMILLAN AND CO. AND NEW YORK. A. AND W. MACKINLAY. All JiifjhtH llescrved. ? ti P h VI ii ti SI n: E fa id St re ra w PKEB^ACE TO THE FII18T I^DITION. This volunus (3outjiiny the lirst Six Books of Euclid's EloTiieiits, toijfotlier with Apptnidicos givins,' tho most ij>-- ijoitant elementui-y dovolopmeuts of Euclide.ui Ooomctry. The text Jias been carefully revised, and special atten- tion given to those points wliich experience has shown to present ditiiculties to beginners. In tiie course of this revision the Enunciations have been altered as little as possible; and, except in Book V., very few departures have been made nom Euclid's proofs: in each case changes have been adopted only where the old text has been generally found a cause of diHlculty; and such changes are for the most part in favour of well-recog- nised alternatives. For example, the ambigu:ty has been removed from the Enunciations of Propositions 18 and 19 of Eook I.: the fact that Propositions 8 and 26 establish the complete identical equality of the two triangles considered has been strongly urged; and tlius the redundant step has been removed from Proposition 34. In Book II. Simson's ar- rangement of Proposition 13 has been abandoned for a well-known alternative proof. In Book III. Proposition 25 is not ,i,en at length, and its place is taken hy u VI I'UKKACi;. simpld JHiuivulciit. Proj^sitioiis .1.5 mul 'M) li.ivfj Ixkiu troiitfd ^'cnt'ially, niid it Ii.is not hfcti lliou;i,'ht lu'civssary to do iii()i(; thuii cfUl attiitition in u note to tlic* special cases. Finally, in Hook VI. w(* liuvo adopted an jUtenia- tivo pnjof of i'l-opositioii 7, a tlicoicni wliicji has been too much ncijlcctod, owinj,' to (hu cundu-ouH form in which it has lu'cii usually given. 'I'hcso are the ciiief dcniations from thi^ ordinary text as re,<,'ards nietliod .and airangenient of proof: they are points fanuliar as dilliculties to most teachers, and to name (hem indicates sutlicieutly, without further em-'iieration, the general principles which h;i\ e guided our revision. A few alternative proofs of ditllcult i)ropositions are given for tlu; convenience of those teachers who care to use them. With regard to Jiook V. we have established the princi- pjd propositions, both fi-om the algebraical and geometrical definitions of ratio .and projtortion, and we have endeavoured to bring out clearly the distiiiction between these two modes of treatment. In compiling the geometrical section of Book V. we liave followed the system iirst advocated by the late Pro- fessor l)e Morgan; and here we derived very material assistance from the exposition of the subject given in the text-book of the Association for the Improvement of Ceo- metrical Teaching. To this source we are indebted for the improved and more precise wording of dehnitions (as given on pages 2^G, 288 to 291), us well as for the order and substance of most of the propositions wliich .appear between pages 297 and oOG. But as we have not (except in the points above mentioned) adhered verbally to the text of the Association, wc .are anxious, while expressing in the fullest manner our obligation to their work, to exempt the I'KKKAIK. Vll Associdtiou from ull lesponailnlity for our ticiitmcut of tim subject. Olio purpos.! of tlie hook is to ^M-utluully fiiinili.iris.. iUr student with the use of le;riti„iate symbols .ind uhhrevia- tions; for a /,'eomotrieiil Jir;,'umeiit umy thus he thrown into a form which is not only more readily seized hy an a(lvanc«!d reader, hut is useful us a ^'uid(? to the way in which Euclid's pro[)osition8 may })e handled in written work. On the other hand, W(' think it very desirahh; to (h'fer the intro- duction of symbols until the be^dtuier lias learnt that they can only he properly used in Pure Geometry as abbrevia- tions for ^■erhal argument: and we hoix; thus to prevent the slovenly and inaccurate habits which are veiy apt to arise from their employment before this principle is fully recogniscid. Accordingly in Book I. we liave used no contractions or symbols of any kind, though we have introduced verbal alterations into the text wherever it appeared that con- ciseness or clearness would he gained. In Book 11. abbreviated forms of constantly re(;urring words are used, and the phrases therefore and is equal to are replaced by the usual symbols. In the Third and following Books, and in additional matter throughout the whole, we have employed all such signs and abbreviations as we believe to add to the clear- ness of the reasoning, care being taken that the symbols chosen are compatible with a rigorous geometrical method, and are recognised hy the majority of teachers. It nmst be understood that our use of symbols, and the removal of unnecessary verbiage and I'-^petition, by no means implies a desire to secure brevity at all hazards. On the contrary, nothing appears to us more mischievous than an abridgement which is attained by omitting via PRKFACK. steps, or condensing two or more steps into one. Such uses spring from the pressure of examinations; but an examination is not, or ought not to bo, a mere race; and wliile we wish to indicate generally in the later books how a geometrical argument may be abbreviated for the pur- poses of written work, we have not thought well to reduce the propositions to the bare skeleton so often presented to an Examiner. Indeed it does not follow that the form most suitable for the page of a text-book is also best adapted to examination purposes; for the object to be attained in each case is entirely diflerent. The text-book should present the argument in the clearest possible manner to the mind of a reader to whom it is new: the written proposition need only cojivey to the Examiner the assurance that the proposition has been thoroughly grasped and remembered by the pupil. From first to last we have kept in mind the undoubted fact that a very small proportion of those who study Ele- mentaiy Geometry, and study it with profit, are destined to become mathematicians in any real sense; and that to a large majority of students, Euclid is intended to serve not so much as a first lesson in viathematical reasoning, as the first, and sometimes the only, model of formal and rigid argument presented in an elementary education. This consideration has determined not only the full treatment of the earlier Books, but the retention of the formal, if somewhat cumbrous, methods of Euclid in many places where proofs of greater brevity and mathematical elegance are available. We hope that the additional matter introduced into the book will provide sufiicient exercise for pupils whose study of Euclid is preliminary to a mathematical edu- cation. PREFACK. IX Tlie questions distributed through tlie text follow very easily from the propositions to which they are attached, and we think that teachers are likely to find in them all that is needed for an average pupil reading the subject for the tirst time. The Theorems and Examples at the end of each Book contain questions of a slightly more difficult type : they have been very carefully classitied and arranged, and brought into close connection with typical examples worked out either partially or in full ; and it is hoped that this section of the book, on which nmch thought has l)een expended, will do something towards removing that extreme want of freedom in solving deductions that is so commonly found even among students who have a good knowledge of the text of Euclid. In the course of our work we have made ourselves acquainted with most modern English books on Euclidean Geometry: among these wo have already expressed our special indebtedness . the text-book recently published by tlie Association for the Improvement of Geometrical Teach- ing; and we must also mention the Edition of Euclid's Ele- ments prepared by Dr. J. S. :Mackay, Avhose histoiical notes and frequent references to original authorities ha\e been of the utmost service to us. Our treatment of Maxima and Minima on ))atre 239 is based upon suggestions derived from a discussion of tlie subject which took place at the annual meeting of the Geometrical Association in January 1887. Of the lliders and Deductions some are original ; but the greater part have been drawn from that large store of floating material which has furnished Examination Papers for the last 30 years, and nmst necessarily form the basis of any elementary collection. Proofs which have been ^ PREFACE. found in two or inoi-u boc^ks without Jicknowledgeiuent have been regarded a.s eoiiinion property. As regards figures, in .'iccordance with a usage not uncommon in recent editions of Euclid, we liave made a distinction between given lines and lines of construction. Throughout the book we have italicised those deductions on wiiich we desired to lay special stress as being in them- selves important geometrical results : this arrangement we think will be useful to teachers who have little time to devote to riders, or who wish to sketch out a suitable course for revision. We have in conclusion to tender our thanks to many of our friends for the valuable criticism and advice which we received from them as the book was passing through the press, and especially to the Rev. H. C. Watson, of Clifton College, who added to these services much kind assistance in the revision of proof-sheets. H. S. HALL, F. H. STEVENS. July, 1888. PREFACE TO THE SECOND EDITION. In the Second Edition the text of Books I— VI. has been revised ; and at the request of many teachers we have added the first twenty-one Propositions of Book XL together with a collection of Theorems and Examples illustrating the elements of Solid Geometry. September, 188'J. Igeiiient CONTENTS. BOOK I. PAGE Definitions, Postulates, Axioms ••....! Section I. Propositions 1—26 22 Section II. Parallels and Parallelogkams. Propositions 27—34 . , ka Section III. The Areas of Parallelograms and Triangles. Propositions 35 — 48 (;(^ Theorems and Examples on Book I. Analysis, Synthesis I. On the Identical Equality op Triangles . II. On Inequalities III. On Parallels IV. On Parallelograms V. Miscellaneous Theorems and Examplls VI. On the Concurrence of Straight Lines in a Tri ANGLE VII. On the Construction of Triangles with given Parts VITI. On Areas . . IX. On Loci X. On the Intersection of Loci 87 90 93 95 96 100 102 107 109 114 117 xn CONTENTS, BOOK II. Definitions, Ac Propositions 1 — 11 .... Theorems anu Examples on Book II. PAGE 120 122 144 BOOK III. Definition.s, itc 149 Propositions 1 — 37 153 Note on tiik Method of Limits as Applied to Tangencv . 213 I. II. III. IV. V. Tliroirnifi mid K.ramjjlrs an Book III. On the Centre and Chords of a Circle . . . 215 On the Tangent and the Contact of Ciacij-.s. The Common Tangent to Two Circles, Problems on Taugency, Orthogonal Circles 217 On Angles in Segments, and Angles at the Centres and Circumferences of Circles. The Orthocentre of a Triangle, anrl properties of the Pedal Triangle, Loci, Simson's Line . , . 222 On the Circle in Connection with Rectangles, Further Problems on Tangency 283 On Maxima and Minima ....... 239 Harder Miscellaneous Examples . . . . 246 .BOOK IV. Definitions, itc. Propositions 1 — Ki . Note on Hegular Polygon.s 250 25] 274 Theorems and Examples on Booh IV I. On the Triangle and its Circles. Circumscribed, Inscribed, and Escr'bed Circles, Tlie Nine-points Circle 277 II. Miscellaneous Examples ....... 283 CONTENTS. Xlll PA«E 120 122 144 . 149 . 153 . 213 215 s on TUE.S 217 ' the 222 . 233 . 239 . 246 . 250 . 251 . 274 BOOK Y. PAOE Introductory 285 Dkfimtions 286 Summary, with Algebraical Proofs, of xhk Principal Theorems or Book V. 292 Proofs of the Procositions derivk]) from the Geometrical Definition of Proportion ....... 297 BOOK VI. Definitions Propositions 1 — T>. Theorems and Examples on Book VI. I. On HAR:\roNic Section II. On Centres of Similaiutv and Similitude . III. On Pole and Polar IV. On the Eadical Axis of Two or More Circles V. On Transversals . VI. Miscellaneous Examples on Book VI. . BOOK XI. Definitions Propositions 1 — 21 .... Exercises on Book XI. . Theorems and Examples on Book XI. 307 308 359 363 365 371 374 377 383 393 418 420 The . 277 . 283 nit line its but bei] tha hetl its/< «>.vcr; pcrft 'J EUCLID'S ELEMENTS. BOOK I. Definitions. 1. A point is that wliidi has position, but no ma"- nitude, ° 2. A line is that Avliicli has length without breadth. The extremities of a line are points, and the intersection of two lines IS a ijomt. 3. A Straight line is tliat which lies evenly between its extreme points. Any portion cut off from a straight Hne is called a segment of it. 1. A surface is that wliidi lias length and breadth, but no thickness. The boundaries of a surface are lines. 5. A plane surface is one in whicii any two points being taken, the straight line between them lies .vholly in that surface. A plane surface is frequently referred to simply as a plane. ^^'^^; Euclid regards a point merely as a mark of position, and lie therefore attaches to it no idea of size and shape. Similarly he considers that the properties of a line arise only from Its length and position, without reference to that minute breadth whicli !?n^P' .^^ f "^* ^"^'"^^^y ^^""'^ 'f ^'''^""^^i/ d>-^w>h even though tlie most perfect instruments are used. The ilufinition of a surface is to be understood in a similar way. 2 euclid'h elkments. G. A plane angle is tlio inclination of two straight ]inc\s to one another, -whifli nujct togutlicr, but arc not iw the same straiu'lit line. The point at which the straight lines meet ia called the vertex of the angle, and the straight lines themselves the arms of the angle. When r^cveral angk'S jiro at one point O, any one of them ia expressed by three letters, of which the letter that refers to the vertex is put lirtwijeii tho other two. Thus if the straight lines OA, OB, OC meet at tho point O, the angle contained liy tho straight lines OA, OB is named the angle AOB or BOA ; and the angle contained by OA, OC is named ih(! angle AOC or COA. Siniilavly the angle con- tained by OB, OC is referred to as the angle BOC or COB. But if there be only one angle at a i)oint, it may be expressed by a single letter, as the aiijle lit O. Of the two stiaight lines OB, OC shewn in tho adjoining figure, we recognize that OC is inure in- dined than OB to the straight line OA : this we express by saving that the angle AOC is greater than the angle AOB. Thus an angle must be regarded as having vuKjnilude. ^ . It should be observed that tho angle AOC is the sum of tho angles AOB and BOC ; and that AOB is the difference of tho angles AOC and BOC. The beginner is cautioned against supposing that the size of an angle is altered either by increasing or diminishing the length of its arms. [Auotlier view of an angle is recognized in nuiny branches of mathenuitics ; and though not emi)loyed by Euclid, it is here given because it furnishes more clearly than any other a conception of what is meant by the VKujnitude of an angle. Suppose that the straight line OP in the figure is capable of revolution about the point O, like the hand of a watch, but in the opposite direction ; and supjiose that in this way it has ])assed successively from the position OA to the positions occujiied by OB and OC. Such a line must have undergone more turninn in passing from OA to OC, than in passing from OA to OB; and consequently the angle AOC is said to be greater than the angle AOB.] O DKFINITIOXS. 3 ) straight I re not i.\ 10 vertex uf e angle. 7. When a straight lino standing on another straight line makes the adjacent angles equal to ono anothei-, each of the an- gles is called a right angle ; and the sti'aight line Aviiich stands on the other is called a perpendicular to it. N. All obtuse angle is an angle which is greater than one right angle, but less than two right anules. 0. An acute angle is an ;uig!o which is less than a rii^ht anule. \ urn of tho tbo uii''los size of au ngth of its lancbes of here given ion of what > OB; and ngleAOB.] O A [lu the adjoining figure the straight line OB may be supposed to have arrived at its present position, from tbo position occu- pied by OA, by revolution about tlie point O in eitlier of tire two directions indicated by tbo arrows : thus two straight lines drawii from a point may be considered as forniiiif,' tiro an-iles. (marked (i) and (ii) in tbo figure) of which the greater (ii) is said to be reHex. If the arms OA, OB are in the same ~ straight line, the angle I'ornnd by tbcni B on either side Ii cajled a straight angle.] 10. Any portion of a plane surfac(! bounded by one or more lines, straight or curved, is called a plane figure. The sura of the bounding lines is called the perimeter of the figure. Two figures arc said to be equal in area, when they enclose enuai portions of a plane surface. 11. A circle is a plane llguro contained by one line, which is called the circum- ference, and is such that all straight lines (h-awn from a certain point within the tigiirnto the circumference are equal to ono another : this point is called the centre of the circle. A radius of a circle is a straight line dravvn from tho centre to the circumference. 1—2 4 EUCLIU'S KI.EMENTH. 12. A diameter of .i cirelo is u strai^^ht line drawn throuf^li the centre, and terminated l)oth ways by the circumference. 13. A semicircle is the W'^uve bounded by a di;imeter of a circle and the part of the circumference cut ort' by the diametei'. [[. A segment of a circle is the iigure bounded by a strai-,'ht line and the part of the circumferenci! which it cuts ort". 1-j. Rectilineal figures are those which are bounded by straight lines. IG. A triangle is a plane figure bounded by thi'ee straight lines. Any one of the angular jioints of a triangle may be regarded as it3 vertex; and the opposite .side i.s then called the base. 17. A quadrilateral is a plane figure bounded by fvhioh has a riglit angle. Tlio sidn opposite to the ri^lit anK'lo in a right-angh il triaiiylo ia ca 'od the hypotenuse. 23. An obtuse-angled triangle is a triangle which has an obtuse an"lo. 21. An acute-angled triangle Is a triangle which lias three acute anirlcs. [It will bo Hoon liereafter (Book I. Proposition 17) that crmj tnmiiilc must /lan- at least two acute annles.] 25. Parallel straight lines are such as, being in the same plane, do not meet, iiowovor far thoy are i)roduct!d in either direction. 2G. A Parallelogram is a four-sided ligure which has its opposite sides pa- / rallel. /_ 27. A rectangle is a parallelogram which has one of its angles a right angle. 28. A square is a four-sided figure which has all its siJes equal and all its angles right angles. [It may easily be shewn that if a quadrilateral Jias all Its sides equal and one angle a right angle, then all its angles will be right angles.] 29. A rhombus is a four-sided figure which has all its sides equal, but its angles are not right angles. 30. A trapezium is a four-sided figure / which hixs two of its sides parallel, / 6 EUCLID'S KI.KMENTb. ft\ Tin: POSTULATES. Ill order to cffuttt tho constructions neceswary to tlio study of geomt'trv, it must be NUppDSfil tli.'it certuin iiiMcruuionts arn uvailable; but it has jilway.s bofu liekl that HUcli itistrmneiits ^*fi()ul(' 1)6 as fow in imnibt'i', and as simple in charneter as j.ossible. For th(» jnu'poses of tlu first Six r.ooks a Mrdi-iht rtihr and a paii "f coinpasMe!* are all tliat are needed; and in tlie i'ollow- ing Postulates, or requests, Kuelid demands tho iiso of sueh insti'Uiiients, and i»»suin(!s that they sufUeo, theoretieally as well fis praetieully, to eui ")' out the processes mentioned below. I'osTrLATES. Lot it be Lfi'anted, 1. That a straii^ht lino may bo drawn from any ono point to any otJier point. Wlien we draw a straiglit lino from the point A to tlie point B, we lire said to Join AB. 2. That ii^/iiiifi', that is to say, a, terminated straight line may be produced to any length in that straiglit line. 3. That a circle may be described from any centre, at any distance from that centre, that is, Avith a radius equal to any finite straight line drawn from the centre. It is important lo notice that tlie Postulates include no means of tlirert vn'(isii)r)iii'iit; lience the straight ruler is not supposed to bo (iradiuttcd ; and the compasses, in accordance with Euclid's use, are not to be employed lor triuisfn-ri)!'! distinices from oue part of a tigure to another. ox Tin: AXIOMS. The science of Oeouietrv is based upon certain simple state- ments, the truth of which is assumed at tho outset to be self- evident. These self-evident truths, called by Euclid Common Notiom, are now known as tite Axioms, Vff^i OKNPriAr, AXIOMS. > study of lonts arn itruiueiit.H meter as riihr and 1 10 I'lllloW- 3 of HUcll ly as well 1)\V. am' Olio oint B, wr> straight t line. centre, at ius equal o means of losed to be I's use, are of a lij'ure iplc state- :o be self- n Notions, The iicccssary charactoristics of an Axiom aro (i) 'J'liat it should hrt Holf-ei'tdvnt : that is, that its truth should 1)0 iuitiii'diately accoj)tod without ])ro.*' (ii) That it .slioui 1 ho ftnnhtmovtdl ; t'.i'it -s, that its trutli should not he dorivjihli' from aiiv othor l. iiii liion* siniplo than it.s(-H; (iii) That it shou,,i upply a hasis for tlm cstalilishmont of furtlicr truths. Thoso f;haractori:>iii'H may \\o. summod up in tlio following dofinition. Dki'IMTion, An Axiom i a Holf-evident truth, which neither rcquirns nor is capablo ol' pinnif, hut which serves as a f iunda- tion for future reasoning. Axioms aro of two kinds, general and f/cometncuL ( Joncral Axioms apply to nxttjiu'tudi'tt o/ alt kinth. ( ioometri- cal Axioms refer exchisively to (/eoiiii'trii'nl m'.cjnitiiil, h^ Mieli as have been already indicated in the definitions. CtEVKijal .Vxioms. ]. Things which ate ('(pial <<• iho same thing /ire equal (o om> another, L'. if ecjuals be added to equals, the wholes are equal. .'?. Jf e(]uals be taken from equals, the remainder.s are equal. 4. If equals be added to unequals, the \vhol< s aro un- e(iual, the greater sum being that whii-h includes the greater of the unequals. n. If (>(|uals be taken from unequals, the ren.-ainders are niuMpial, the greater remainder being that whiel' is left from the greater of the unequals. 0. Things whieh are doubh? of the same thing, or of equal things, are e(|ual to one another, 7. Things -svliich are halves of the same thing, )r of equal things, aro equal to one another. D.* The whole is greater than its part. * To proeorve the classification of ceneral and Keonietiical axi jms, we liavo placed Euclid's ninth axiom before the euihth. 8 Euclid's elements. (iiEO.METKICAL AxiOMS. 8. Magnitudes wl)ich can ho made to coincide with one another, are equal. Thif5 axiom afforda the ultimate test of the equality of two geome- trical magnitudes. It implies that any line, angle, or figure, may be supposed to be taken up from its position, and without change in size or form, laid down upon a second line, angle, or figure, for the purpose of comparison. This process is called superposition, and the first magnitude is said to be applied to tlie ntlicr. 10. Two straiglit linos cannot eiu;loso a space. 11. All right angles are equal. [The statement that all right angles are equal, admits of proof, and is therefore perhaps out of place as an Axiom.] 12. If a straight line meet two straight lines so as to make the interior angles on one side of it together less than two riglit angles, these straight lines will meet if con- tinually produced on the side on which are the angles whicli are together less than two right angles. That is to say, if the two straight lines AB and CD are met by the straight line EH at F and G, in such a way that the angles BFG, DGF are together less than two right angles, it is asserted that AB and CD will meet if continually pro- duced in the direction of B and D. [Axiom 12 lias been objected to on the double ground that it cannot be considered self-evident, and tliat its truth may be deduced from simpler principles. It is employed for the first time in the 2')th Pro position of Book I., where a short discussion of the difficulty will be The converse of this Axiom is proved in Book I. Proposition 17.] INTRODUCTORY. e with one two geome- ure, may be t change in jure, for the lagnitudc is ts of proof, es so as to etlier less eet if con- gles which at it cannot duced from ;e 2<)th Pro- iilty will be lition 17.] INTllODUCTORY. Plane Gcbnietry dccal.s with the properties of nil linos and figures that may bo drawn upon a plane .surfoce. iMiclid in his first Six Books confines himself to the properties ol straight hnes, rectilineal figures, and circles. The Definitions indicate the subject-matter of these books- t\w lostulates and Axio),)>i lay down the fundamental princii)les wliicli regulate all investigation and argument relatintr to this subject-matter. ° Euclid's method of exposition divides the sul)ject into a number of separate discussions, called propositions; each in-o- position, though in one sense compUte in itself, is derived from results previously obtained, and itself leads up to subsequent 2)ropositions. ^ Propositions are of two kinds, Problems and Theorems. A Problem proposes to effect some geometrical construction such as to draw some particular line, or to construct some re-' quired figure. A Theorem proposes to demonstrate some geometrical truth. A Proposition consists of the following parts : Tlio General Enunciation, the Particular Enunciation, tlie Construction, and the Demonstration or Proof (i) The General Enunciation is a preliminary statement, describing in general terms the purpose of the proposition. _ In a problem the Enunciation states the construction which it IS proposed to effect : it therefore names first the Data or things given, secondly the Quaesita, or things required. _ In a theorem the Enunciation states the proi)erty which it is proposed to demonstrate: it names first, tb^ Kv""f»i'=''''<' nr the conditions assumed; secondly, the Conclusion, or the asser- tion to bo proved. I 10 EUCLID'S ELEMENTS. (ii) The Particular Enunciation repeats in special terms the statement already made, and refers it to a diagram, which enables the reader to follow the reasoning more easily. (iii) The Construction then directs the drawing of snch .straight lines and circles as may bo required to eft'ect the purpose of a prol)lom, or to prove the truth of a theorem. (iv) Lastly, the Demonstration proves that tho object pro- posed in a pri)blem has been accomplished, or that tho property (Stated in a theorem is true. Euclid's reasoning is said to be Deductive, because by a con- nected chain of argument it deduces new truths from truths already proved or admitted. Tho initial letters q.E.f., placed at tho end of a problem, stand for Qnod erat Faciendum, vhirlt was to he done. The letters q.k.T). are appended to a theorem, and stand for Quod erat Demonstrandum, vbick ■n-as to he proved. A Corollary is a statement the truth of which follows readily from an established jiroposition ; it is therefore a[)pended to the proposition as an inference or deduction, which usually requires no further i)roof. The following symbols and abbreviations may be employed in writing out the propositions of Book L, though their use is not recommended to beginners. ••• for therefore. par' (or ii) for parallel, ,^ is, or are, equal to, par™ ?> parallelogram, ^ angle, sq. !) square, rt. L „ right angle. rectil. ^1 rectilineal. ''^ » triangle. st. line )) straight line, per p. „ perpendicular, pt. J> point ; and all obvious contractions of words, sucli as opp., adj., diag., &c., for opposite, adjacent, diagonal, &c. BOOK I. riiop. 1. 11 (cial terms 'am, which ig of such lio purpose object pro- 10 property 3 by a con- •oni truths I 2)ro1)lem, [ stand for )ws readily ded to the ly requires ' employed • use is not lie], lelograni, re, lineal, ght line, t; adj., diag., SECTION I. Pkopositiok 1. Pkoblkm. To describe an eqnilaterul iriam/Je on oint to draw a straiyht line equal to a fjiven sfrai(/ht line. Lot A be tlie given point, and BC tlio _i;iven straight lino. It is required to draw from the point A a straiglit line equal to BC. Constriiction. Join AB ; Post. 1. and on AB describe an etjuilateral triangle DAB. I. 1. From centre B, with radius BC, describe the circle CGH. Post. 3. Produce DB to meet the circle CGH at G. Post. 2. From centre D, Avith radius DG, describe the ciicle GKF. Produce DA to meet the circle GKF at F. Post. 2. Then AF shall be equal to BC. Proof. Because B is the centre of the circle CGH, therefore BC is equal to BG. J)ef. 11. And because D is the centre of the circle GKF, therefore DF is equal to DG ; Def. 11. and DA, DB, parts of them are equal ; Def. 19. therefore the remainder AF is equal to the remainder BG. Ax. 'X And it has been shewn that BC is equal to BG ; therefore AF and BC are each equal to BG. But things which are equal to the same tiling are equal to one another. Ax. 1. Therefore AF is equal to BC ; and it has been drawn from the given point A. q. e. f. [Thi3 Proposition is rendered necessary by the restriction, tacitly impo.sod by Euclid, that compasses shall not be used to transfer distances.l Book i. viiov. li. 13 equal to a ti[. therefore AC must fall along DF. And because AC is equal to DF, ^^!/J>- therefore tlie point C must coincide with the point F. Then B coinciding witlj E, and C with F, tlie base BC nmst coincide with the base EF; for if not, two straiglit lines would enclose a space ; which is impossible. j^^ 2q Thus the base BC coincides with the base EF, and is therefore equal to it. j ,. ,^ And tiie triangle ABC coincides witli tlie triangle DEF, and is therefore equal to it in area. ° Jx. S. And the remaining angles of the one coincide with the re- maining angles of tlie other, and are therefore equal to them, namely, the angle ABC to the angle DEF, and the angle ACB to the angle DFE. That is, the triangles are equal in all respects. (^.E.n. KoTK. It follow.s that two triauKleK wliich are equal in their several parts are equal also in omi ; but it sliould be observed that equality of area in two triangles does not necessarily iniijiy equalitv iu tlieir several parts : that is to say, triangles may' bo equal in lirci, without being of the same sJiapt'. Two triangles which are equal in all respects have i,h;iititif of form mid inafiintude, and are therefore said to be identically eaiial or congruent. j ^ < „ '^y<' /ojjo^ving application of I'roposition i anticipates the chief difficulty of Proposition 5. lu the equal sides AB, AC of an isosceles trian-le ABC, the ptunts X and Y are ta]:en, so that AX IS eciual to AY ; and BY and CX are joined. ►Shew that BY is equal to CX. In the two triangles XAC, YAB, XA is equal to YA, and AC is equal t.. AB ; H,/,,. that is, the two sides XA, AC are equal to the two sides YA, AB, each to each; and the_ angle at A, M-hich is contained by these sirles, la conunor! to both triangles : therefore the triangles are equal in all respects • BO that XC is equal to YB. 16 KUCLID'S ELEMENTS. Puoi'OsiTiox ;■). Theorem. 'The aiKjh's at the base of ati isosceles triangle are equal to one another: and if the equal sides be ■produced, the anyles on the other side of the base shall also be equal to one another. Let ABC l)u ;ui isoscelt'S triangle, liaviug tlic side AB ('(^ual to the side AC, and let the straight lines AB, AC be produced to D and E : then shall the angle ABC be equal to the angle ACB, and the angle CBD to the angle BCE. Construction. In BD take any point F; and from AE the greater cut off' AG equal to AF the less. i. 3. Join FC, GB. Then in the triangles FAC, GAB, FA is equal to GA, Constr. and AC is equal to AB, JJ^i/P- I also the contained angle at A is common to the two triangles ; therefore the triangle FAC is equal to the triangle GAB in all respects ; I. 4. that is, the base FC is equal to the base G B, and the angle ACF is equal to the angle ABG, also the angle AFC is equal to the angle AGB. Again, because the whole Af is equal to the whole AG, of which the parts AB, AC are ecjual, J^!/P' therefore the remainder BF is equal to the remainder CG. Proof Because BOOK I. I'UOr. 5. 17 'e are equal odiiced, the 'qual to one lie side AB ; AB, AC be angle ACB, Tlicn ill the two triangles BFC, CGB, BF is equal to CG, Proved. Ilcuuuse tlie line AD meets BC in E, nrovo that BE -EC. ' 3. ABCD is a rhombus and AC is joined: prove that the anglo DAB is e(iual to the angle DCB. 0. ABCD is a (piadrilatfriil having tlie opposite sides BC, AD ei|ual, and also the angle BCD e.iiial to tiie angbi ADC : iirove tliat BD is .'(inal to AC. ' 10. AB, AC are the eqiial sides of an isosceles triangle ; L, M, N an- the middle ](oints of AB, BC, and CA respcetivelv : prove tliut LM = MN. Prove also that the anglo ALM is eijual to the angle ANM. Definition-. Each ;. therefore the triangle DBC is equal in area to the trianf^le tJie part equal to the wliole ; wliich is absurd. Ax. 9. Therefore AB is not unequal to AC ; that is, AB is equal to AC. Q.E.D. Corollary. Hence if a triawjle is equiangular it is also equilateral. BOOK t. I'ltOI'. 7 t . 10 I till' equal Tkoi 'osniov 7. TiiKoKKji. On he mm. base, and '-. harin,, ,hW ddr. which a^t^^t /dnom those vheh are tenninatj at th. othn- .. /.v'v vqnnl to one another. J 3 0(|U)il to AB. I. 3. Constr. ;il to the //;yy>. B triangle 1.4. cl. Ax. 9. Q.E.D. dar it is side of t .n '''" '^"•", '""^ ^^' ••""' '"^ ^''- -^"'" scl«s AC, AD wind, are tenninat..(l at A, ,.,,ual to our S*;:; b"; "rr^ '''^'•'" f '"^ '^' ^^' ^^'-i- -- 1-- - natecl at B, ef|ual to one anotlior. /W/. 1. Construction. ,j,,i), cd. /Vofj/: Theu ill t}ie triangle ACD, Jjci-'ause AC is equal to ad' //,,,, tlun-efore the angle ACD is equal to the angl. ADC if,' ^'•Igl'e'eCD?^' '"''^' '''''' " ^''■"^'"" *'""^ ^^'^ I-'-t' ''-^' stiirf !nl' f r *'" '"':?^^ ^""^ '' ^"•^'^*^'- ^''-'"^ *'«« '-"'^'l^ BCD ; st.U^mme then ,s the angle BDC greater than tl.e angle Again, in the triangle BCD, because BC is equal to BD,' J/y„ therefore f IP an"lp pnr i- -.r^,, i - ii i -"!//'• l,nf u" -"^ '"^o'e BDC la equal to the angle BCD: i. 5 but It was shewn to be greater; which is impossible. 4..-' 20 Ki'if.in s kij:ment;'s. Casi-: 7r. AVIhh ou„ nf tlic vertices, as D, is within tlio other liianyli' aCB. A B Cousiruct'iuu. As before, join CD ; j>o^f ] •md produce AC, AD to E and F. Post "" 1 hen in the tria.i-le ACD, l.ecauso AC is erjual to AD, ll>m tlierefore the unoles ECD, FDC, on Wu^. other si(h; of tJm base, are equal to on(^ another. j -, But tlie an.^^le ECD is .-greater than its part, the aii^de BCD;" th(>.vfore the angh, FDC is also greater than tie -mvA^ still more then is the angle BDC greater than the anHe BCD. '^ Again, in the triangle BCD, heeause BC is eqn.-il to BD, //,,.; thereloi.ahe angle BDC is e.iual to tlu, angle BCD : lo but It has been shewn to be greater ; a\ hich^is impossible " 11..^ eas." n, wlueh the vertex of one triangle is on' a .sule ot tlie othei- iieeds mi demonstration • ■ ^^'i;;!;''^"''^^ ^^ cannot ],e equal to AD, and ro ,hhs of thn one runal. (,, luv ^idvsof ihevUwr, mch to each, and hacc likari.e their /,a.r, equal, then the awjle which is contained b,j the two ddcinf the one shxill be npad to the a>u,lc nddch is contained h„ the tico saUs ol the other. '^ BOOK I. PROP. 8. 21 D, is witliiii Po>st. 1. Post. 2. to AD, lliip. sidf! of the I. Ti. xni,'le BCD; the i\\VJ\^i 1 the Jiiude C5 3CD : I. 5. n})(jsyiblc. ;'lo is oil ;i JH ; .similarly ^•t» udi's of Ualned hij BA ACeuuSto tl.t "^"^ ^^^^^^^ '^^-^^ the two .ides BA to FO^ 1 A^ r '° '"'? ^°' ""^^ '■'^^■'' to each, numely BA to^ED, a.ul AC to DF, and also the base BC equal to the then shall tlie angle BAC he equal to the angle EDF. _ Proof, For if the triangle ABC he applied to tlu" nang e DEF, .so that the, point B n.ay be in E, and e straight line BC along EF ; fi t .,^'""' .''^^^'^"«^ BC is equal to EF, ![„., therefore the point C must coincide with tlie point F. ' ., p ,, Then, BC coinciding with EF it follows that BA and AC n,u.st coincide with ED and DF • fm- It not, they would have a different situation, as EG OF- tlH'H, on^the sanH3 base and on the san.e side of iTiw cont(:rini)ioi(s sides would be two triangles ha\ ing their e(|ual. „,, . , ^"-'t this is inii)os.sil)le. , - IW'I Tl " t^"' ^^' ^^ '"^'^^''^'^' ^^^th the sides ED, DF Hut Ls the angle BAC coincide.^ with the angle EDF, and is therefore equal to it. ° ' , -I.'". 6. oneumH^^T " ivnuv^l^^ may h^ made toa>iucid, ,nth Hence we are led t,. the iblluu iuy In.portanL Corollary. CoHoLLAifY. //• hi two tnanqh's the three ^hlrn o/' My line aro. ('nil,, J fn ihr fh , •; • ; , f ".5 OJ inf. tliut thr. tnan.jks are equal in all respects. 22 EUCLID'S ELEMENTS. u oi 1 101.. 7, AS Inch Irequently presents difficulty to a beginner. PnoposiTioN 8. Alternative Pkoof. then shall the angle BAG be equal to the an^le EDF For apply the tnangle ABC to the triangle DEF so tW R . iall on E, and BC alon" EF and so flmf ti, . , • . A . ^ "'^^ «ide of EF remote from D "'" ^'^'*'* ^ "'^-^ ^"^ «» the Let A'F? ho S '""'^ ^*'" •"" •"' ""^^^ BC ^« equal to EF Let A EF be the new position of the triangle ABC If neither DF, FA' nor DE, EA' are iS one straight line jom DA'. Case I. When DA' intersects EF. Then because ED is equal to EA', therefore the angle EDA' is equal to the angle EA'D Again because FD is equal to FA', therefore the angle F D A' is eq ual to the an-le FA D Hence the who e angle EDF is equil to the whoie a,^t EA'F • that IS, tlie angle EDF is equal to the angle BAG. ' ^^^^_^Two cases remain which may be dealt with in a similar n.anner: Cask II. When DA' meets EF produced, line^''' III. When one pair of sides, as DF, FA', are in one straight I. u. I. i>. 1 as it is indc- to a beifinner. ifles BA, AC equal to tlie :df. that B may •y be on the ; line, D. I. G. 0. I. o. EA'F: ir manner: ne straight BOOK r. PROP. 9. Proposition 9. Problem. 2.3 parts To bisect a ,lven angle, that is, to divide it into ttvo equal Let BAG Ije the given an,i,']e: it i.s required to bisect it. Construction. In AB take any point D; unci from AC cut offAE equal to AD. j ;j Join DE; }X' rf '• '''' ^^'^ '^'^"^ ''"^°*^ ^'■^'" A, describe an equi- Jateral trians to AB. Proof. For in the triangles DCF, ECF, f DC is equal to EC, ' Couxtr. iJccaiise -' '"^"^^ ^^ ^^ common to bfith ; |;ind the third side DF is equal to tlie third sid(! Therefore the angles DCF is equal to the aiigli^ ECF:' i. is. and these are adjacent angles. Put when a straight line, standing on another strai-dit iuK', makes the adjacent augies equal to one another, each <>t tlies(^ angh's IS called a rigjit angle; y>y- 7 therefore each of the angh^s DCF, ECF is a right angle." 'I licrcfore CF is at right angles to AB, and has been drawn from a point C in it. (^>.k.k. EXERCISE. lu the fjguro of iho above) proposition, slicxv that aioi pohit 1-0, or FC produccL], is ciuidistant IVom D and E. m 2G Euclid's elements. Pkoposition 12. Problem. nJn/rr ^VT/'^^'^^'^'' Verpendicular to a givm stmiaht hue oj unhmxted lenyth, from a given point vAont it. '^ Let AB be the oiven .st.-ai.rht line, which n.ay be pro- ted ,u ezther chrection, ,uk1 let C be the ,iven po'inr.^h. ^t^s;;:^:;^ :;f -^^- ^^' ^--^^ the drde pog, meeting AB fit F unci G. PoHt. 3. I. 10. Bisect FG at H ; Then shall the stmi^lit line CH be perpendicular to AB Join CF and CG. 1 roof. Then in the triangles FHC, GHC f FHisequaitoGH,' ' Constr. Because { '"^" "^ ^s common to both; and the third side CF is equal to the third side «. CG, beuig rada of the circle FDG : Def U therefore the ang e CHF is equal to the angl^ CHG -18 and these are adjacent angles. ' line nlkeHI.e "'f^\ ^^'\ ''""'"^^' ^^^ ^'^^'^'^^ «t^-ig''t iine makes the adjacent angles equal to one another each wh hTtwf ' ":r"!? '^ "«^''' •^">^'^' -^^ the straight Hne which stands on the other is called a perpendicular to it circle FDG. ^ mtersected in t\vo points by the BOOK I. Puoi>. 12. ivoh stmiyJtt out it. i;iy \m pro- point vvith- traiglit line oiu C take L-ircle FDG, Post. 3. I. 10. ir to AB. Coiistr. third side De/.U. 'HG; 1.8. V straiglit ther, eacli aiglit line r to it. the given Q. E. F. ted Ifngth, length in nts hy the KXEKCISKS ox PHOPOSITIONS 1 TO 12 to the base. ^ ^ °* ^"'^ ^^'^'^ i'^ pori)endicuIar basfW fnTLscStfln^^So t^;:^if ^°^\*^« -'-""*-« of tho are equal to one a^iothe"^ "^'^'^ 1'°'"^' °^ *he opposite sides. distl^tL'n^rexZSLVot-the'b/' '") ^^°T'^^ *"-«^« -« «1-- distant from the verlex! = '^'"'^ *^^* "'^^ **■« "^^^^ ^^^'^■ the angle AxV^t7uSoute\.n;^^^^^^^^ ''° ^"°"'^ ^^^^ -^ ^'-^ diagonaft^Lh ;^i*^^^^^^^^^^^^ '^"^"'^^'^ «^ '^ ^-^o-bus are bisected by the ^shevv that OA bleet: the ;n3: bZc':'"'''"' ^'"'-"^ "'"^^^ "^^'"^^ ^^ O = thesides^lln'^rlteli^';?;::,^^^^^ middle points of uiuudttial tiumglG is also equilateral. righVanler '^""' '^'' ^^''^^'""'^ °' '^^ ^•^^«™'^"-'' ^^^'^^ one another at poinSxtS^rartiki^irsouUt Ax"' "^ i-sceles tnangle ABC two are drawn inters^tintin'S': thew^^^ ^'^^^ '^ ^Y; and CX and BY (i) tlie triangle BOC is isosceles • •■■\ aS 'j'-'ii-'cts the vertical angle BAG • (1") AO'ifpioduced, bisects BC at right an-les In what case is this impossible ? I 28 kuclid's elements. Proposition- M]. Theorem. 1/ one straight line xduid iip,„i anothrir sfrairj/.f /inr^ fhrn thii adjiicnd aiKjIrs aJiall, be ,:i(Ii,'r /in, ri>f/U anohx, or totjrtlivr equal fo hro ri'jiif (hkjIp-^. n B Lf't the straight line AB stand upon tlio .straight line DC- tlion tho adjacent angles DBA, ABC shall he (M'tli.'r two i-i-dn, angles, or togetiier e(iual to two right angles. '^ C.\sE I. For if the angle DBA is equal to the an^jlo ABC, • aril of them is a right angle. "" i),.f -•' (Use it. Jhit if tlu' angh^ DBA is nr.t e(|nal 1o Ihc aiigh> ABC, from B draw BE at righl angl(>s to CD. i. ||, Proof, ^'ow Ihc aiigh' DBA is iiia(h' up of 1h.' two angles DBE, F_BA; to each of tlijse equals add the angh? ABC; then tlie two angles DBA, ABC are together equal to the three angles DBE, EBA, ABC. ' j,^., o. Again, the angle EBC is made up of the twoan Thv complement nf .„, :trntv. aiP'Iu is iu r/.^v-, / '^ nght anglo, ti.at is, the ...glo 1,- which Si I'sll^rf^'^^/:; ■ niLjle. Ill ht r^^^Zlr '"■""" '"■"' ^^"iPlementary, wh.n th.-ir suu. is a (ii) I'lif^supplement of ail aii<>-i(. Is ifs^A./^,./ /•. ^ »i, nta, ;, to th, sann' auH,/„s. «'itl,(',- ot tlip mfonor ()])iK,sitc .•u.-l.-s CBA, BAC? CovHtrnrfJon. P.is(.cfc AC at E • , ,<, .rom BE; ami j.nKlu.-o it to F, HK.Icin. EF ..,,,.1 j: Then ill tlic triiiiigh.s AEB, CEF, [ AEisequaltoCE,' ' ^<^,,,^^. rVonuse J, , ^lulEBtoEF; (y .; I also tho an-lo AEB is fqi,,-,! to <|,o ^prtif■,allv I opposite .'initio CEF; -1- tl..T.^>ro tl.. tmnslo AEB is oquul to tlu> triun^lo CEF in all respects: "^ so that the anole BAE is e,,nal to tho allele ECF '' '' I'-" tl.o ano^le ECD is .i,nvafer 1 hau its part, the an-h.' ECF- 1 .|-.vi....e he a,,ule ECD is ...eale,. than the auohTBAl ' tlKit ,s, the ;n.,le ACD is ^i^.eater than the ani,Wo BAC T.i a snn.lar way, if BC be bisected, and tho side AC produced to G, ,t n.ay be shewn that the an-de BCG k ATeatei- than the aiij/lo ABC. ^ But the an-le BCG is equal to the an-le ACD • ,15 therefore also the angle ACD is greater thaiTthe angle iBc." Q. K. D. a I i na ■ 0.1'trriiir intijlo, litii aiiijIi'.H. 1«' produced Jfl-c.-itcr t|i;iii I. 10. <'» BE. f. ;;. ('omlr. Cousfr. o vorticnllv I. i:l. isio CEF ill I. -1. ECF. •■mule ECF; Ul«' BAE; ;Ih BAC. !io side AC :!<' BCG is !D: 1.15. inr,de ABC, ....^ + Hii.-,ies BAC, ACB, as angles. " ' ^^^' '""^ ^"' "^^' J"«« ^»>«" two right q.E.u. KXtRCISES. of Miomir""" '"" '•'-"-"io" «" »» '" »hc.,v that it i. the come™ an.L »?Set":,:'4St.ir;h:;;;tv,'°iirp:,*^ «'"'"■ ioi^„.^Jr.:^^:;!„^-----rthe-£i:^^^^ 3—2 34 kuclid's elements. Proposition 18. Theorem. // one side of a triangle be r/reater than another, then the amjle opposite to the greater side shall he greater than the angle opposite to the less. '' Let ABC be a triangle, in which the side AC is greater triaii tne sule AB : then shall the angle ABC be greater than the an^rle acb p . Join BD. Iroof. Tlien in tlie triangle ABD, because AB is e(jual to AD, tlierefore the angle ABD is equal to the angle ADB. j. 5 But the exterior angle ADB of the triangle BDC is «ieaier than the interior opposite angle DCB, that is greater than the angle ACB. ' j jf Therrfore also the angle ABD is greater than the angle ACB- .still more then is tiie angle ABC greater tiian the ande ACB Q.E.D. 11 Euclid enunciated Proposition 18 iis follows: [For Exercise.s see page 38. J mother, then greater than 3 IS greater angle ACB. rt AD equal 1.3. ^DB. I. 5, ;le BDC is 1) tliat is, I. 16. ingleACB; the angle Q.E.D. ater angle rce ofdiffi- d in it and BOOK I. I'ROP, 19. 35 I'Roi'osiTiox Vd. Theorem. If one, angle of a triangle be greater than another, the the sue opposite to the greater angle shall be greater tha the side opposite to the less. A Let ABC be a triangle in which the angle ABC is -reater tlian the angle ACB : ^ i^ o'^-ittii then shall the side AC be greater than the side AB. Prooj. For if AC be not greater than AB It must be either equal to, or less than AB. ,. ^, ^, 15ut AC is not equal to AB, tor then the angle ABC would be equal to the angle ACB • i .1 but it is not. *llijp. f ^, , Neither is AC less tlwui AB • tor then the angle ABC would be less than the angle ACB ; 1. 1 S. but it is not : 'l/i/)) Therefore AC is neither equal to, nor less than AB. lliat IS, AC IS greater than AB. q.K.i). Note. The mode of demonstration used in this Pronosifinn ;. .r;" rc^tn'mutSf "^T""" '' '^ '^PP^-ble to carhf hie t ue and ? Pnn«r/ '^ exclusive suppositions must necessarily be ons in turn W/A '" '''''''"" *^1" ^^''^^^-^ "^ ^^^^' ^^ these supjos^^ Euclid enunciated Proposition 19 as follows: areJier .?/'"'"' T'^^! ""-^ ''''''^ ^''^'^"^^'^ '' subtended by the greater side, or, has the greater side apposite to if. [For Exercises see jmge 38,j 36 Euclid's elements. Proposition 20. Theorem. iki^lJr '''"' '^ ^^ "'^-'^Ole aretouetkeryreatcr tkan tk ., , „ r-<^fc ABC Lea tn'ande- ^Z S : "'^' ''-'' '' ''' '''^' ^^ ^^^*'-- ^-te. than the namely, ba, AC, shall bo ^n-eater than CB • AC, CB gieat(>i- than BA • .■ukI CB, BA greater than AC. CoHstvuction. Procluee BA to the T^.iJnf n , i • to AC. I^ '"'* '^' '"'iknig AD equal Join DC. J'rooj. Then in the tn'angl.. ADC, therefore tin; an.^le Arn i. I"'"/'* AC, 6'o;^s7r. Mnf +1, , '^'^'.^^^ ^'^ ^''li"il to theano],. ADC . "i J5Ut the angle BCD s <'reater fh-^,. fl T * ' therefoi-P ■.!.,. fi i ,>,'»aui than the angle ACD : Av <) iiitieioK also the angle BCD s «r,.eater tl...,.^l, . ,, '. , "^ Qicritt I tjian ij (J an-'le adp tl'at IS, than the angle BDC. ^ ' And in the triangle BCD because the angle BCD is -aviter th-.n fl therefore the^sicle BD i ?. e er a f ' "-f ' ^°''' ^^'•- i' JO tjic.icci than the side CB. i. 19 Tjut BA and AC are tog(-ther ei.ual to BD • therefore BA aiul an .... + i.i ^4"'^' culs of the side BC, let tlie two straight lines BD, CD be drawn to ;i point D within the triangle : then (i) BD and DC shall be together less than BA and AC • (11) the angle BDC shall be greater than the angle BAG. Construction. I'roduee BD to meet AC in E. J 'roof (i) In the triangh^ BAE, the two sides BA, AE are together greater than the third side BE : I. 20. to each of these add EC • then BA, AC are together greater than BE, EC. Ax. i. Again, in the triangle DEC, the two sides DE, EC are to- gether greater than DC : j oq to eaeh of these add BD ; then BE, EC are together greater than BD, DC. But it has been shewn that BA, AC are together greater than BE, EC : » » .still more then are BA, AC greater than BD, DC. (ri) Again, the exterior angle BDC of the trian-le DEC is greater than tlu; interior opposite angle DEC • - l(j Hiid the exterior angle DEC of the triangle BAE is -reater tlian the interior opjjosite angle BAE,^that is, than tlie a^iigle BAC ; 1 T stdl more tlu'ii t« Hi<' iM-rl" pnr . 38 Euclid's elements. KXERCISES ON TitorosiTioxs 18 and 19. 1. Tlic liypotem the greatest side of a right-angled triangle. 2. If two angles of a triangle are equal to one another, the sides also, which subtend the equal angles, are equal to one another. Tron 6 1 rove this indirectly by using the result of Prop. 18. i^n.-;?; n^^'/''''.^''?A°Jx''" isosceles triangle ABC, is produced to any point D ; shew that AD is greater than either of the equal sides. •1. If in a quadrilateral the greatest and least sides are opposite to one ano her, then each of the angles adjacent to the least side is greater than its opposite angle. nn/\ ^■\^ v'^^'f^ ^^^.' '^ ^^ '^ "°* erea^^v than AB, shew that hnL'pn ^ 1 ';,'^'''''To*''''"""^ **'^ ^'^^^^^ A and terminated by the base BO, is less than AB. "^ AOD ^^^J'^P' triangle, in which OB, OC bisect the angles ABC, ^Serthan OC^' ' '^ ^^ ^' ^'''''^''' ^^^"^ ^^' *''^" °^ '^ ON Pkoposition 20. the third side^'^^^'^"'''' "^ """^ *'™ '''^''' ''^ '' *"''"8le is less than nrP^;.Jln ^ ;i"^'^"'''^tf™l' if t^^'; opposite sides which are not parallel a e pioduced to meet one another; shew that the perimeter of the Se quaJrilatemr '""' '" ^°""'"^ '' ^''''''' '^"" '^'' P^"™^^^^' ^^ !>. The sum of the distances of any point from the three angular points ot a triangle is greater than half its perimeter. diagonal J^'' perimeter of a quadrilateral is greater than the sum of its 11 Obtain a proof of Proposition 20 by bisecting an angle by a straight hue which meets the opposite side. ^ ox Proposition 21. 12. In Proposition 21 shew that the angle BDC is greater than tlie angle BAC by joining AD, and producing it towards the base. 13. The sum of the distances of any point within a triangle from us angular points is less than the perimeter ot the triangle. BOOK I. I'uor. 22. 39 PjtoposiTioN 22. Pkohlem. ^ To describe a triamjh haviwj its sides equal to three is less than B Let A, B, C be the tliree given straiglit lines, of which any two are together greater than the third .h..llV' ''^'^"i?'^ ^"^ '^''''■^^' "" ^''"^^ «^ ^^1"-"1» the sides shall be equal to A, B, C. ^^Ta\Tnl- . ^^^-^ .^ ^'""'-^'^ ^^^^ ^^ tenninated at the point D, but unhmited towards E Make DF equal to A, FG equal to B, and G H equal to C. i ;5 In-oni centre F, Avith r.-,clins FD, describe the circle DLK ' J^iom centre G with radius GH, describe the circle MHK, cutting the former circle at K. Join FK, GK. Ihen shall the triangle KFG have its sides et.ual to the three straight lines A, B, C. ^ Proof. Because F is the centre of the cii-cle DLK, therefore FK is equal to FD : he/. 1 j but FD is equal to A; Covstr. therefore also FK is equal to A. Ax. 1. Again, Ijecause G is the centre of the circle MHk/ ' therefore GK is equal to GH : J)ef. Ij. but GH is equal to C ; Const r therefore also GK is equal to C. Ax 1 Tl.P..nf .1 /"f^f'G ^s equal to B. Comtr. li.e efore the triangle KFG has its sides KF, FG, GK equal respectively to the three given lines A, B, C q e ? 40 Euclid's elements. EXERCISE. On a ^'iveu base describe a triauglc, whoiio remaining sides sball be oqual to two Kivon straight lines. Point out bow tbe construction ails It any one ot the three given linos is greater than tbe sum of tne other two. PliOPOSITION- L'a. PuotiLEM. At a !/iveu point in a f/iven stravjU line, to make amjle equal to a (jicen anyle. an Lft AB be tl.e given straight line, uiul A the given i..)iut lu it ; iuid let DCE be the given angle. Tt is required to draw from A^'a straight line inakin.-- with AB an angle equal to tlie given angle DCE. * Construction. In CD, CE take any j^oints D and E; and join DE. From AB cut off AF equal to CD. i ?, On AF describe the triangle FAG, having the remaining sules AG, GF ('(jual respectively to CE, ED. i 22 Then .slw.U the angle FAG ^e equal to thJ angk; DCE.""" ^'>'('"J- h\ the triangles FAG, DCE, ( FA is equal to DC, Constr. J>ecause ^ and AG is equal to CE ; Constr. (and tlie base FG is equal to the base DE : Const,- theretore the angle FAG is equal to the angh; DCE. i .S* i hat IS, AG makes with AB, at the given point A, an angle equal to the given angle DCE. ' Q.E.f! BOOK I. I'Ror. 24. 41 sides shall be construction I the sum of Pkopositiox 24. ^ 7/ two triangles have two ,is, in wlildi tlie two sid ill" e«iu;i] to tlio two sid OS but tlie angle BAG greater tliaii tlie an-d' es ED, DF, each t o eacli. EDF tlieu sliall the base BC be greater than the b, Of tlie two side greater than DF. Else EF. s DE, DF, let DE be tliat wliich is not ConMructioH. At the point D, in the strai-lit 1 and on the same side of it (Mjual to the angle BAC. me ED, as DF, make the angle EDG I'roof Make DO and Tl e(iiial to DF Oi AC join EG, GF. 1. 23. I. 3. I )eLause ilso tl eont en in the triangles BAC, EDG, BA is equal to ED, and AC is equal to DG 7/ le contained aine Therefore tl e triangle BAC angle BAC h EDG : Constr. <'qual to the Cotistr. all res ipeets IS equal to the triangle EDG «o that the base BC is equal to the base EG. See note on the ucxt pa"c. Ill I. 4. 42 Euclid's ELiiMENis. Again, in tlie triangle FDG, because DG is equal to Df' tWforothe angle DFG is equal to the ;^^^^^^ but the angle DGF is greater than the aiUe EGF • therefore a so the angle DFG is greater than t fe an^le EGF • stall n,ore then is the angle EFG^greater than Z !^e Eg': And in the triangle EFG because the angle E^G is greater than the an-Ie EGF therefore the side EG is greater than the side EF : 'l9 I'ut EG was shewn to be equal to BC • ' ' * therefore BC is greater than EF. ' (^.k.u, * This condition was inserted by Simson to Pn«n,.» fi,n^ • xi 1 ins may be done as follows : Let EG DF, produced if necessary, intersect at K. Then, since DE is not greater than DF, a, n ''"*'' ^*^' s'"ce DE is not creater tlmn nr therefore the angle DGE is not great^^ tSe" a^n^fe DEG Lut the exterior angle DKG is greater than the ancle DEK • therefore the angle DKG is greater than ?l!e an^ie DG K Hence DG Is greater than DK. But DG isequal to DF • therefore DF is greater than DK. bo that tlic point F must fall below EG 1. 18, I. 1(). I. 19. 5. DGF, EGF; ngle EGF; ■ngle EGF. e EGF, :F; I. 19. hat, in the Without isps : for F uld require I condition Produced. G. K: 3K. 1. 18. 1. 1(>. I. 19. BOOK I. I-HOP. 24. Or the following method inuy be adopted. PitoposiTiox 2-1. [Alternativk Proof.] In the triangles ABC, DEF, let BA be equal to ED, and AC equal to DF, but let the angle BAC be greater than the angle EDF: then shall the base BC be greater than the >ase EF. . Fof apply the triangle DEF to tlie triangle ABC, so that D may fall on A, and DE along AB: then because DE is equal to AB therefore E must fall on B. ' And because the angle EDF is loss than the angle BAC therefore DF must fall bttw.en AB and AC i-.ct DF occupy the position AG. Cask I. IfG falls on BC: Then G must be between B and C- therefore BC is greater than BG. But BG is equal to EF : therefore BC is greater than EF P;o£f' ii"- ^{ ^ ^°^'* ""* fa" on BC. 43 which meet:: BC in K Join GK. Then in the triangles GAK, CAK, GA is equal to CA, and AK is common to both; land the angle GAK is equal angle CAK; therefore GK is equal to CK. But in the triangle BKG, th^'^^^TuT'''"^ '^'^T"' '^^" tl^« th"d side BG, I tlidt .s, BK, KC are together greater than BG ; therefore BC is greater than BG , or EF. q e d Because the two 1.9. Hyp. to the Conatr. I. 4. 20. 44 tUCLID's JatMEMS. ' '1 1 h I'KOI'USITION i>:.. TlIEuKKM. M.M / .. i... ./ ,/, ,„ ,. ,;,,^^ ,;^^ ^,,^,^,. ,^,;^,^ .^^^./ ^^^^^ . '/.. ..V /A«^ .r/.*cA /.,,,. ,h. ,,rmlr,' ba.,, shall bo. ,jrmterthan the awjle conknned by the corrc.pondiwj sides of the other. RA ;;!' ^^\ ^^V"' ^^^■^^•••'^''.^1<'« ^^J'ic-I. I.avo the two side. BA, AC (M,,iul t., 1l,(. two si BC would ]>e less than the base EF • 'i 04 liut it is not. ' j} ' Therefore the an,,le BAG is neither equal to, nor less than t he angle EDF ; that is, the angle BAG is greater than the angle EDF. q.k.d. KXERCISE. Hccoiding as AB is greater or less than AC. ' '}li,iunl sides a.v a./jnmu to <| nuai angles in tin; two trian-des. ^ ABC ACB e,,nal to t he < . o .•.ni^des DEF, DFE, ea, h to .al-h • and the snle BC e.jual to the side EF • then simll the triangle ABC l,e e^ual to the triungln DEF 111 all respects ; o '^■-r that is, AB shall 1..- e,|ual to DE, and AC to L^ and ti.e angle BAG shall l,e e,,ual to Mh- angle EDF I'or It A«3 be not equal to DE, one nuist he'-reater't h.o. the other. If possible, let AB he greater than DE Coyistrucflon. From BA eut oti" BG e.,nal to ED , " and joni GC. J'roo/: Then in the t\V(. triangles GBC, DEF, (GB is eijiial to DE, and BC to Ec also the contained angl- GBC is eq„al in'tiu cont/inieo angle DEF • // therefore the triangles are e.pml in all respects • ^4 so that the angle GCB is eqnal to the angle DFE " ' tlieietun- .lisu The angle GCB is equal to the angle ACB • Jr\ ' the part equal to the whole, which is inrpossible' Const)'. Hyp. 46 KUCLin'R KLEMKST8. Therefore AB is not unequal to DE, tluit is, AB is equul to DE. Hence in the triani,']es ABC, DEF, AB is e(iual to DE, Proved. also the contained anj,'!*' ABC is etjual to tlic contained angle DEF ; JI>IP- therefore i\\v triangles are ('(pial in all r(\spects : i. 4. so that the side AC is (Mjual t(» the side; DF ; and the angle BAC to the angle EDF. q e.d. Cask If. When the ecjual sides are opposite to eciual angles in the two triangles Let ABC, DEF be two triangles which have the angles ABC, ACB equal to the angles DEF, DFE, each to each, and the side AB ecjual to the side DE : then shall the triangles ABC, DEF be equal in all respects ; that is, BC shall be equal to EF, and AC to DF, and the angle BAC shall be equal to the angle EDF. « M BOOK I. I'ROI'. 26. 47 Provpd. lh,p. iml to tlio Ihjp. cts : I, 4. Q E.D. 'e to ('(jual the angles li to each, I respects ; ' DF, 3 EOF. ;. ... Proof. Conuli', Comtruction. Frou. BC .-ut ,AX BH <..,ual f. EF ••iikI join AH. TIhmi in the tiian,j,rj,..s ABH, DEF, fAB is ecjual to DE, , ■ 'uul BH to EF, ' tnuui . contained an •,'lo DEF • /, therefore the trian^Mc-s are e.jual in all respects ,4' soth theangh,AHBise,uaWothea.;;j^S^^^^^^ " '• Hut the angle DFE is equal to tho an-Ie APR ■ // therefore the ande AHB is ( (,u-il f, , . ' ^'^''^'' tl.ativtuexteric;^Hn^e:i^rt;:r^^ '-mter^ropp^it. angle ^whiehir^IU^^^ JheretoreBCisnot un,M,ual to EF that IS, BC is e(|ual to EF. H .nee a-, he triangles ABC, DEF, ( Ab ?s equal to DE, l»(!eause i . '"f'- '3< is equal to EF • i> , ulso tS-; .o.caimcl angle ABC is e.m.l to'th e I cont..uecI angl. def-^ '■'^ (^qual t<, the ^o that he sule AC is e.pml to the sid. DF and the angle BAC to the angle EDF. ' V.K.D. I. IC. Proved. n. K 48 KUCLIU S ELEilEXTb. OX THE IDENTICAL KQUALITV OF TUIAXGLES. At the closi; of the lir^t section of Book T., it is worth while to call special attention to those Propositions (viz. Props. 4, 8, 2(i) which deal with the identical equalitt/ of two triangles. The results of these Propositions may be suniniarized thus : Two triangles are equal to one another in all respects, when the following parts in each are equal, each to each. 1. Two sides, and the included angle. Prop. 4. ± The three sides. Proij. 8, Cur. .3. («) Two angles, and the adjacent side. \ (6) Two angles, and the side opposite one of - /Vo^;. 2(1. thoni. \ From this the beginner will periia|)s surnii.se that two tri- angles may be shewn to be vi{WK\ in all respects, when thcv have three parts equal, each to each ; but to this statement two obvious exceptions uuist be made. (i) AVhen iu two triangles the three anyles of one are e(pial to the thn-e aiKjI.'A of the other, each to each, it docs vot necessarily follow that the triangles arc equal in all respects. (ii) AVhen in two triangles two sides of the one are ecpial to two sides of the other, each to each, and one angle equal to one angle, these not being the angles included by the opial sides; the triangles are not necessarily equal in all res[)ects. In these cases a furtlier I'ondition nnist lie added to the hyixitliesis, before we can assert the identical equality of the two triangles. [See Theorems anil Exercises on iJook !., K\. l;5, Page \)-l.'] W'e observe that iu each of the three eases already proved of itleiitical eciuality in two triangles, namely in Proi)ositions 4, 8, 20, it is shewn that the triangles may be made to eoineide. ■iritk (me another: so that they are eipud in area, as iu all other resj)(!cts. Euclid however restricted himself to the use ef Proi). 4, when he required to deduce the equality in area of two triangles from the equality of certain of their parts. This restri(^tinn lia^ boo!! aha.ndoned in the present text-bo! si:, [See note to Pidp. ;M. | !l >l EXEKCISKS OX I'ltOl'S, 12 2fi 49 LES. worth while fops. 4, 8, 2(i) JS. rized thus : spects, wheu Pi'op. 4. I'rup. S, Cvi: ) hut two tri- '11 thcv liavii two oltvious 111(1 urc c(pi;il it (Iocs Ihit I'L'spijcts. le iire e(|ual gle equal to ciiuiil .si(l(,'s; .l(U'(l i(. tlic ility (if tlic Tago \)-i:\ 2iidy ])rovt;il o})ositi(jus 4, to eoincidi: a. as in all ;o tho iiso cf area of two it text-l>nok. EXERCISES o\ PRorosiTiox.s 12—20 i^osId.^ff H 1%'' A^r ^''" '^/r*'^'-' "^ ^^•^ ^""^le" '^t tlie base BC of an .in.As o^t .f;;ei;:.:s;;r ii ta BY iii'' f "^ ^^' ^"^- ^^-'«''* A and B: shew thit AX Jre./uil to BY '''"^''''^^ "i^'^" ^^ *"'■"'" Let AB be th(3 Kivon strait^'ht lino, ■iiiii p. t< tiu! givoji 2)oints. It is HMiuired to draw from P and Q (. a p,j,nt in AB, two stiai^lit line, tlmt shall l)c wiually incliiKj.r tu AB. Vomtniction. From P draw PH "■••• perpendicular to AB: produce PH to P, nudan. HP' cp.al to PH. Draw QP- nieetin,- AB in K. Then PK, QK «hall bo the rc.p.ued lu.e.s. ^Supply the pro(.f.J tw.^iv^inS;il-;---- •'. TlirouKli a r-iv.'ii i.oint dfiv 1 'f,--.; Ti,f r , •. oer- 4- -:i 60 EUCLID'S ELEMENTS. SECTION II. PARALLEL STKAIGIIT LINES AN1> PAKALLELOGUAMS. Uefinitiun. Parallel sti-aiifjit lines ai'o such as, beinGUAMS. ch US, being liey Jire pro- .hird «trai<'lit ctcridv angle, do of EF. j;lcs, ing iuigk's. Lsed upon his igiit lines so !in;e side of •aiglit lines, set on that iv less than > 'i and 6 are Lhat AB and liirka iipou it PUOPO.SITION- L'7. TilEOUKAl. CD ^^fV^'"' T"'^^'^' '■'"' ^^ ^'"<^ *''e tAvo stmi-d.t linos AB then shall AB and CD ),o parallel. ^»v;r./. Fo, if AB and CD be not parallel Jiut the ,mj,|,, AGH is e<|„,,l tn tlis nn-le GHK • ;/"'' .-nee t„e angle. AGH „„„ o'U a,,. ...tl^^tC;,,., 2^^': Ti.^ i? Winch IS nnijossihlc. ' ' '";t;i*C :' ■""^' '■" ''""'" "'"' •'"■.^ <—* -et ,..va,.ds tl.erc.fom tii,.y ,ue parallel. „, ,,, „_ 62 J.Utl.lD.S I.I.KMKMiS. Pjjopo.siTiox L'.s. Thkoukm. //' (> sfrahjlif Ilio\ /(ill'n,,/ /■ the liw; or If it ,ua.ke the inferior tnir/fes on the mme side tuyHther eipud lo two right ano/es, (hen the tico dravjht lines shtdl he jHiralle/. Let lliH strain-lit line EF cut tho two stcaii'lit lines AB CD ill G and H: niifl ' J'irsf, jot the oxterin)- aii,i(le EGB Ix' (>(|nal to t lie interior 0]>po,site anyle GHD: then shall AS and CD he parallel. Proo/. Becfiuse the an,i,d<^ EGB is equal to tlie an-le GHD- and because the .-inoh. EGB is .-ilso equal to th(> \(Mticallv op- posite angle AG H ; J_ j5 therefore the an.i^de AGH is equal to the ans^le GHd"; l)ut these are alternate aiiifles; tlierefore AB and CD are iiarallel. i. '21. q. K. ]). Secondlji, h't tlie two interior an<,dps BGH, GHD he to- ;,'ether equal to two ri^ht angles: then shall AB and CD Ijc parallel. Proof. JJecause tiie angles BGH, GHD are together equal to two right angles; " jj and because the adjacent anghvs BGH,.AGH a.-e also to<'eti.er equal to two light angles ; '"^j j -. tlierefore the angles BGH, AGH are together i^qual to' tlie two angles BGH, GHD. From these e(iuals take the coiiiinou aivle BGH- then the remainiiig angle AGH is equal to the lemainiiig aii an-le GHD; •'' .'1(1(1 to oaeJi the anu'lo BGH- tl'^Mi the angles AGH, BGH are tog.-ther greater th.-,n tho _ .-uiglos BGH, GHD. lint the adjacent anglos AGH, BGH are together e(,ual to two right angles; ' i 1 '] therefore the a'ngles BGH, GHD are together less than "two right anglos; therefore AB and CD moot towards B nnd D. Ax. 12. J5ut they never m.-.-t, .sinee they are narallel. Hyp. iiurcfore the angle AGH is not unequal to tlie angle GHD: that is, the alternate angles AGH, GHD are equal. {(her) 54 Euclid's elemknts. (ii) A'^nhu iH'CMiise tlio .-ui-lo AGH is ,.,{u...i f,, the v.vrtf- c.'illy opposiio ,'uij,'le EGB; i- .•ukI boeaiiso tho ;inj,de AGH is .mhiu] tw -1 (> na^ic GHD:' * thpn-fore the exterior angle £GB is equal i. the iiitenorm> l>osite angle G H D. ^ (iii) Lastly, the angle EGB is equal to the angle GHD- add to eacli the a?igle BG H ■ ilien the ,■ingl•^s EGB, BGH are together eqn,,,l to the an-Ies BGH, GHD. ^ ^ i' : Bhew that [I- 27.J 'ttiaitjht line 27 or 1. 28.J lines, and is the others. [I. 29.] t lines, each TzcpectlVSltf [I. 29."] It must 1)0 admitted that Euclid's twelfth Axiom i.s uu- satisfactory as the basis of a tlieo.y of parallel straight li, ei It cannot be regarded as either simple or self-evide.rt a t therefore falls short of the essential characteristics of an ax n noristhe difhcu ty entirely removed by considering it as a cr-" rollary to Proposition 17, of which it is the converse. I^lany .substitutes hav e been proposed ; but we need only notice here the system which has met with most general apiS .™d .::^r?unr:i:tai^:i^!'^^"'^ ^'^^^^^^^-■"' ^^^-^^ ^« i-^ to .fSstr^;;;^ ';:T'''"' ^'"'^'^ ''""' '''"'" '^ '"'' ^--^^"^ it iTt/;iWtll!'f.^"T" '"•' Payfair's Axiom; and thougii both s rniw ^^ "^ ?'"V objection, it is recom.uended as oti simper and more fundamental than that emnloved bv Luchd, and more readily admitted without proof ^ * ^ tlH. lirst part of Proi^sition 2!) is then given tluis. ^ PiioposiTiON 29. [Alterxativk Proof.J If a strmght line fall on two paralld straight lines, thenit shall make the alternate angles equal. Let the straight line EF meet the two parallel straight lines AB, CD at G and H : then shall the alternate angles AGH GHD be 0(]ual. ' For if the an-le AGH is not equal to tlie angle GHD: at G in the straight line HG make the angle HGP equal to the angle GHd' and alternate to it. i 93' Then PG and CD are parallel, i" 27 IkitAB and CD are parallel: il,m. therefore^the two intersecting straight'lines AG, PG are both parallel Tl,n, f .1 ^vhich is impossible. Tlayfaifs Axiom. that IS. the alternate angles AGH, GHD arc equal. Q.E.r. fttJ I.I.'l'LIDV, l.l.LMJiM. Proposition 30. Thkokkm. StniU/ht liii's wli'ii'h i; duo th '('• .saiiii' Hlrnlijlit I'tiiH cr H/ K / G/ •8 ■Q ill Lot tlio stmi^lit lilies AB, CD bo c.-ul stiviiiflit lino PQ : 1 l>.ii';iIlol to ilio thou shall AB an.] CD l)o parallol ivnv any straight lii .•111(1 PQ ill 1h(. points G, H, and K. lo EFoulfiii'' AB, CD. Pronj'. 'i'licii 1 moots thoiii. "vauso AB and PQ aro )>aralloI, and EF thei-ofore the an,^do AGK is o(jual to tlio allornato an,<,do GKQ. .Vnd bocanso CD and PQ aro pavallo], and EF moots tl thoroforo tlio oxtorior anglo GH opposite anglo HKQ. I. I'D. lem, D is equal to the interior I. 29 Thorofore the MngL3 AGH is oqu.al to the angl imd t Iioso are alternate augl o GHD es; therefore AB and CD aro jiarallol. ]. L'7, Q.i;.i). r...1^^w ^: 1 ^- f ''''^'''•'" ^^ '"^^^ <^°' t^>^ Proposition may I,o rslal hs led ,n a similar mamuT, thonKh in this case it scaredy neo.ls l.ioot; lor It IS uicouceivable that two straiglit lines, uhich 'do not ineetanint.nMeaiatestraiKlitliii... should nu^t or. Wotlu-r Pi„ (••'*? "^ this Pioj^ositioii may he readily deduced from riayfair's Axiom, of which it is tlie converse dnced Ihen there would be two intersecting straight lines loth parallel to a third strai-dit line: v.hich is impossible. Therefore AB and CD never ueet; that is, they are parallel. ' Hi mil fill I inn liOUK I. Iliol'. .31. ;,;' PnoposiTiox ."1. Pkodlk.m. Toilnnr a sfr„h,J,f H,,. th,-o,ujh a ,,;,;,, pnh.f p„rallrl tit (I. (jiri'u trodn(»' EA In F. Then sliall EF hx- parallel to BC. Proof Because the strai-ht line AD, nieetin-- the two stiaixl.t lines EF, BC, makes the .•dtein;.,te .•.ndes EAD ADC equal; ' /' , . (ousfr. theivfore EF is pai-.allel to BC; j. ^J iiiid it has been drawn through the given point A. ~ ' Q. K. F. KXKRCISES. 1. Any stmiKlit Ihio drawn parallel to Hip l.usn r.f nil isoscolos tnanjrle malvos equal an^'les with the sides. 2. If IVon, any point in the bisector of an anfile a strai-'lit line is ••5. From a given jioint draw a stniiglit line that shall nia]c(> N\ith a jnvon straight line an angle equal to a given angle. 4. From X, a point in the base BC ol an isosceles trian<']e ABC a s^tnught hne IS drawn at right anghs to the base, cutting AB in Y and CA produced m 2 : shew tiie triangle AYZ is isoseeles. ' ; ■'*■ u\ tlie straiglit line which bisects an exterior angle of a triancrle IS parallel to the opposite si.le, shew that the triangle is i.sosce\ ^ 68 hUCLlU'ti ELEMENTS. Pkopositiov r>i>. Theorem. th.'H (i) tl,o exterior ungie ACD sl.ull he equal to th. sun, otthet' ''itmor opposite au-les CAB, ABC- (n) the three interior angles ABC. BCA, CAB .shall 1.0 tog(>ther e,,ual to two right angles. ComtrncHon. Through C (lr,uv CE p.-tralh^l to BA , .^ ''r^ti^hel''"^ '^^^'^"" °^ -'^^ ^^ -^ P-'^'H and AC tlK.etV,re the angle ACE is ecpntl to the alternate angle A^'ain, because BA and CE are parallel, and BD n.eets therf ' th.Te^ore the extcM.or angle ECD is i.^ual to he ,^^ "^^ opposite angle ABC. •"», jurdioi Therefore the whole exterior angle ACD is e„ual to" t"he sun, ot the two interior opposite angles CAB, ABC. (ii) Again, since the angle ACD is euual to the sun. of the angles CAB, ABC; , * to each of the.se equals add i „■ an-V BCA • ' '"''''''^* But the adjacent an,• , '• ^'A ^'<^'". ""si"^ as th. .is,,,-,, h"; lilies!' '""■" "" '"■■'":»' 'isi" ol' . i:. u. nuojv I. I'lun: :i2. CI s ftt tho hasfl Miicd by pi(.. I'llt HJlglfS of ^ilnrics (o rtrtUiueid igmv. iH it has f'l" ♦'(jii;il I- -iL'. er (.'(lutil les. tJif) iu- I- li-l.t \\ fur. tJi four y I'igiit i:. i). ^milulH''; 'V'^"'"' '"1"' ^'^ ''•" ^'-"••^' ^'- interior an^l.. and ^tlu, oxter,o.. an^l. arc together ,..,ual to two A^, 'I'l'i'i-oforu all till' ink-rior aii-hs u-itj, ..ii n /" !''' has sides. •' '" "'"^ '"^ ""' ''/-'"••" ''"li<-ivfon> all tho iiihM-ior an-h-s uiti, ..II H ""' T"'- " ° t^ K. 1>. kxi.;k(Msks on simsoVs cokoli.auiks, fA polygon is said to be regular wlini It l.-c ,.ii u • i f"i;,'lcs ("(lual.J °oUid,r \mi. n a u.i.s all itn sides niid all itn ■'.i^'/'^' 02 EUCLID'S ELEMENTS. Pkopositiox '.V,i. Theorem, The strai'iht lines ir/uch join the extre7nitu's of (wo eqanl and parallel stral/: Then because AB and CD are parallel, and BC meets them, therefore the alternate angles ABC, BCD are equal, i. l>D. Xow in the triangles ABC, DCB, {AB is ecjual to DC, T/^y,. Hnd BC is common to both; also the angle ABC is equal to the an-de theretore the triangles are equal in all respects; i. 4. so that the base AC is equal to the })ase DB,' and the angle ACB e(|ual to the angle DBC;' but these are alternate angles ; therefore AC and BD are i)ai'allel: i •); and it has been shi-wn that they are also equ;il. g. E. D. Dkimnitio.v. a Parallelogram is a four-.sid(!d ii-rure ^vll().se opposite sides are paiallel. " '■'>' of hvo equal KU'ts are thcin- itniiiflit linos; parts by tlio nilk-l. fall(;l, and BC i equal, i. 'I'd, B, iiup. to tlio angle Froi'ed. pects ; e DB, 1.4. DBC; I. -11 equal. . In a parallelogram the perpendiculars drawn from one pair of opposite angles to the diagonal which joins the other pair are equal. _ 14. If ABCD is a parallelogram, and X, Y respectively the middle points of the aides AD, BC ; siiew tiiat the hgure AYCX io a paraiieio- gram. (1(mI an applica- triiinjrlcs ABC, BC bisects the iitly established 'c, all its angles , the figure is a at, the figure is le angle a right cr. Hher, the figure l)isected by the 11 its angles are B diagonals are nt of a diagonal [josite sides, is of one equal to (/le of one equal \ all retijM'cts. des of one are mm one pair of lir are equal. rely the middle L i.s a paiaiieio- MISCELLANKOOT KXERCrSRS OX 8RCTI0XS I. AXD II. fig MLSCKLLANKOTTS 7:XERCISES OX SKCTIOXS r. AXD IT HtraighSe'""'"" ^' *^^" ''^'"'^^ «l^P-'^« -"«!- are in the same 3 In the fignre of Proposition U,. if AF is joined, show (i) that AFiseqimlto BC; .espeSL ''''"' '^^ '"'-^"^'^^^ ^^° '^-^ ^lual to the triangle CFA in all 4. ABC isa triancrlo rldit-an^lpd nt R nnri do • . ^ Bhow that the angle ACD is^ obtuse ' ^ '" P^'o^^^ced to D: 2 (/: 2f ^'''' '^'* ^" ^"^ '■"'^"'"^ P°^yS°" '^^ « «>^1<^« each angle contains — ~ right angles. <''. The angle contained by the bisectors of the ■uvlo^ nf ti,. i ^miulo ABC, s„ ,„;,( BP M dual to CQ, .!,„» llmt PQ i„ parallel to »n,l'l,,,M°,,';'DE°?c"" '"■"'>■'''"«''■■''. ,""* ft"' AB. BO arc oqual r)~2 SECTION III. TllK AREAS OK I'AHALLELOaKAM.S AND THIANGLKS. ITitherto when two ligures liavc been saiil to be equaL it has I'ci'ects^ '^ '"""^ ^■'-^'-'"^'■''■'^^''^ ^''l^'^l' t»^''^t i«> equal ill all In Section lir. of Eucli.r.s first Book, we have to consider the eqnahtv ni area of i>nnillelugnun,s and triangles which are not necessarily equal in all respects. [The ultimato test of equality, as wo liavo ah-eaily spen is afforded caiiy eanai cannot he n.^c:':^;.c;:::sriS::lsz^^ chanj^e ot torn.: hen.- the „,..thod of direct ..;>H7;o./^/SunsSl to the purpososot the present section. ' « is> uiisuueci We shall sec liowever from Euclid's proof of Pronocjltinn 'M fi,of two ti^ures which are not identically e,ual,n.y XSJ' 'be so Sdrt«l] '•'"' '^^'""' ^^^'^^ '' '' P«-^ble to Lfer the oqullity of Dkfinitioxs. _ 1. Th.> Altitude of a, pa,rallolo.T.-un with .-oforonco to a given Kido ...s baso, ,s tho porpondicnlar distance between tfic base and the opposite; side. 2. The Altitude of a triangle witli r(>fo.-onee to a given side as base, ,s the perpendicular distance of the opposite vertex troni the base. ^ ^ 67 lANCJLES. Ijc equal, it has is, equal in all ivo to consider gles which are Rocn, is afforded '"/// he made to h are not identi- indergoing f5ome fion is unsuited osition 3;', that ■ertheless be so the equality of •ffiM'ouco to a LUce between ce to a given the opposite BOOK I. I'UOP. 35. PimposiTiox 35. TiiiiOKK.M. vamll^^!7^Z • " '^" ''"""■ ^""^' "'"^ ^^''^^^^^" ^l^^ -"*'■ pai aueis, are equal ? ti area. A D F A D E F Let tlm parallelograms ABCD, EBCF be on tins ,s;uue Dase BC, and between the same parallels BO, AF • then shall the parallelogram ABCD be e.|ual" in area to the parallelogram EBCF. site^o'^tl ^' ^^ ^^'^ 'i'^''' ""^ ^''^' -^'"'^ purallelograms, oppo- poL D '''""'°" '' ^''' '■"■' ^«^'^'^"^^^ted at the sa!ne '"^"^0^'' '^ '^'' r--llelognuns is double of the therefore they are equal to one another. Ax. 0.' BC ^^3\ "^"^ 'I V'*' '''^'"' ^^' ^f"' "I'P"«^te to the base BC, aie not tenmnated at the same point: then because ABCD is a parallehjgram therefj^re AD is equal to the opposite side BC; i 3-1 and tor a snndar reason, EF is equal to BC ; tlierefore AD is equ;d to EF ' \ ,. \ Hence the whole, or remainder, EA is equal ta the Avi.ole' or remainder, FD. ' Then in the triangles FDC, EAB Bees ;ause FD is erpial to EA, ProvexL and DC IS e(jual to the opposite side AB, i 34 also the exterior angle FDC is eqmil to tlie interior I. opposite angle EAB. j oo therefore the triangle FDC is e,.,al 1o Ihe triai.gle EAB. Ti! l^roni the whole figure ABCF take the trian.-lo FDC • nnd from the same figure take the equal triangle EAB ; then the remainders are ^'ijual ■ j[x 3 'tr..?' ^^«^f '-^^"el^Sram ABCD is equal to the paralldo- V. K. D. W^^l SB EUCLID'a KLKMKNT.S. Pkopositiox 3G. Tjikouem. Pi frnlfefo(j ranis on equal basex, and betiveen the same farallelSf are equal in area. Ilyp. I. 34. Ax. 1. ]>t ABCD, EFGH bo paivilleloirraiiis on equal bases BC, FG, and Ix'tweeu the same parallels AH, BG : then shall tlie parallelograui ABCD be equal to tiie paral- leloifraiu EFGH. Construction. Join BE, CH. Proof. Then because BC is equal to FG ; and FG is equal to tlie opposite side EH ; tlierefore BC is etjual to £H: and they ai'e also parallel; -- ,ji ■ therefore BE and CH, which join them towards the same parts, are also (Mpial and parallel. i. 33. Tlierefore EBCH is a parallelo<,n-am. Def. 26. Now the parallelogram ABCD is equal to EBCH ; for they are on the same base BC, and between the same parallels BC, AH. I. 3.5. Also the parallelogiaiu EFGH is ecjual to EBCH; for they ar(> on the same base EH, and between the same parallels EH, BG. I. 35. Therefore the parallelogram ABCD is etjual to the paral- Ax. 1. Q. E. D. lelogram EFGH. From the last two Propositions we infer that : (i) ..1 jM rail el Of/ram is equal in area to a rectanyle of equal base and equal altitude. (ii) ParaUilvijrains on equal bases and of equal altitudes are equal in area. n the name n<»(>K I. j'Hoi'. ;J7. 69 (in) Of twopamlleloifra,n>t of equal altitadeH, that it the ifveater 'whirh h(fn the. greater hane. ; and of two paratlel\»ir(i,ii.'< on eipial buses, that is the greater' which has the arcatcr altmule. \\ buses BC, o tlie paral- Ilyp. H ; I. 34. Ax. 1. :ls ths same I. 33. Def. 26. CH; ;n the same I. 35. CH; ail tJie same I. 35. D the paral- x\.x. 1. Q. E. D. t: mjle of equal altitudes are PliOPOSITlON 37. TUKOKEM. Trianyleti on the same base, and between the same 'paral- lels, are equal in area. Let tlie triangles ABC, DBC be upon tlie aauiue base BC, and between the same parallels BC, AD. Then shall the triangle ABC be ecjual to the triangle DBC. Co II struct 1071. Through B draw BE i)aralkl to CA, to meet DA produced in E; I'yi through C draw CF parallel to BD, to meet AD produced in F.' Proof. Then, by construction, each of the ligures EBCA, DBCF is a parallelogram. j),,r ^)(j' And EBCA is equal to DBCF; for they arc. ..n the same base BC, and between the same parallels BC, L: . j •.;,_ Aiid the triangle ABC is half of the parallelogram EBCA, for the diagonal AB bisects it. i. ;;i. Also the triangle DBC is half of the parallelogram DBCF, for the diagoi al .■>C bisects it. i. 34. r.ut the halves of equ J hings are e{[ual ; J.r. 7. therefore the triangle ABC is "'^ual to the triangle DBC. ti.E.D. [For Exercises see page 73.] 70 KUCLID'S KI.KMENT8. PUOPO.SITIO.V lis. TjlKultKM. me piirallelsy lii Let the triangles ABC, DEF l,e on (.,,ual )m.ses BC EF and between tlie same pai-allels BF AD • ' ' then shall the triangle ABC be equal to the trian-^le DEF m.^'^7^., 1!T^' ' '-- ^° P-"^^ ^ CA, to through F draw FH parallel to ED, to meet AD produced in^H* defC!^ pZl:;;;iIg;.::r^^-"'^^^^"' '^-^ ^^ the^ures^cBCA, And GBCA is equal to DEFH • ^^' "^* tor they are on ecjual bases BC, EF and bpfw«on +i parallels BF, GH. ' '^^'^ween the same And the tz-iangle ABC is half of the parallelogran. GBCA,^' tor the diagonal AB bisects it. ^ :u Also the triangle DEF is half tlw:> i^n...,n i » , '^'"/•^ "•''Ji t"« parallelogram DEFH tor the diagonal DF Insects itl j 34 But the halves of equal things are equal • J r 7' therefore the triangle ABC is equal to the triangle DEF. From this Proposition we infer that Q.E.D. . ^ ji)^^ Tnan<;les on e. (i''e equal he greater lame base, r altitudp- .1'1U)P0.S1TI0N 3?'"' A^^'.PSC .vlm-I, stand on the san.o base BC u)hI on the same .side of it, bo o^ual in area : tlion shall they be between the same i.araHels • tliat IS, It AD be joined, AD shall be parallel to .BC. Comtruv.tUm. For if AD be not jiaralk-l to BC It possible, throu nil A draw AE parallel to BC i 3] nieetnig BD, or BD produced, in E. Join EC. rro^yf, KoNV tlie triangle ABC is equal to the triangle EBC r>ut the triangle ABC is e(,ual to the triangle DBC- //J>" therefore also the triangle DBC is equal to the triangb EBC ■ the whole equal to the part ; which is impossible. ' Therefore AE is not pai-allel to BC hnmlarly it can be shewn that no other st.'ai.d.t line through A, except AD, is parallel to BC. ^ Tlierefore AD is i)arallel to BC. V.K.n. From tliis Proposition it follows that : ^'gnal triangles on the mme base have equal altitudes. [For Exercises see page 73.] KUCl.li I.KMKWTtJ. ^. I'ltOPO.SlTION 10. TlIKC'llKM. Kiiiiid IriniHjIrtt, on etpiid hni^rn in ///,. ,sv,,/,^ Kfrnifffif, line ami OH the name side (/if, are between tlie same parallch. Let the ti'iangles ABC, DEF wiiioh stand on equal liases BC, EF, in the same straight line BF, and on the same side of it, be equal in area : then sliall they bo between the same parallels; that is, if AD be Joined, AD shall be parallel to BF. Consfrnrtion. For if AD be not parallel to BF, if possible, throiiuh A draw AG parallel to BF, i. 31. meeting ED, or ED produced, in G. Join GF. Proof . Now the triangle ABC is equal to the triangle GEF, for they are on equal bases BC, EF, and between the pame paralleln BF, AG. I. 38. But the triangle ABC is equal to the trio,ngle DEF: Ifi/p. therefore also the triangle DEF is equnl to the triangle GEF : the w!iOiM oqual to the part; wlueli is impossible. Thei-efore AG is not parallel to BF. Simil; through A, cxcpt AD, is j)arallel to BF Therefore AD is parallel to BF. Q.E.D. From this Proposition it follows that : (i) Equal triarujles oii equal bases hava equal altitudes, (ii) Equal triaiiijles of equal alUtudes have equal bases. avlv h can be shewn that no other straight lino K.VKKUtiKa ON lM(i>l*S). 37—40, 73 KM-.' ■('!,> !i; 'I'o.siTioNs 37 — 10. tiiiiht tine, ntllrh. \\v,i\ bases suiiie side Ms, :> BF. F, 1. 31. igleGEF, iU tho I. 38. IF: Ifijp. isrloGEF: ible. ight lino Q.E.D. titudes, I bases. DKKiMTioN |-:..i, I, ot' the tlirco 8ti-uii,'lit lines wliidi join the ui.-ul.u- puiuts (.f a tiiiiij,'!.! (,, the ji.iddl. points of the opposite skI. s call, d ix Median i-f the tniin-; IN Prop. 37. 1. 11', m I tiguiG ul' Prop. H7, AC and BD intersoci iu K, ,hew tJmt (_i) the triiiuyloH AKB, DKC ure C' ^ u.lm 'A. On the base of a givt-n triangk' construct a second irianulo ^lua in ar.'u to tho iiist, and having itb vortex in a given straight - given triangle 4. Describe an iscsceles trianglo equal in and standing on tlie aanie baae. ON Piiop. 38. 5. A triangle is divided by each of its w uns into two parts of equal area. ■' •' 0. A parallelogram is divided by its diuguiuilB into four trian-les 01 equal area. ° 7. Af ^ is a trianiive two sides of one respectively equal to two sides of the other, and the angles contained by those sides Lpidement- urij, the triangles are equal ni area. ON Prop. 39. 10. The straipht line which join^ the middle points of two sides of a tnunijle is parallel to tlie third .side. ^ j j AOC IS equal to the truu,^,t AD and CB are parallel. ON Puop. 40. MICROCOPY RESOLUTION TEST CHART (ANSI and ISO TEST CHART No. 2) 1.0 I.I 1.25 iti 150 UUl. Ill 2.8 II 3.2 m 114.0 1.4 II 2.5 2.2 2.0 1.8 1.6 A APPLIED IM/^GE Inc I65J East Main Street Rochester, New York 14609 USA (716) 482 - 0.W0 - Phone (716) 288 - 5989 - Fax 74 KUCLIi/S ELEMENTS. Pkoi'osition 41. Theorem. If a iKiralh'loijratn and a truiinjla be on the- natm hasp, and betwcjni tJie mine j/K«;-(t//f/.s', tlte jtand/elof/nmi shtdl he, double o/ the triautjk. L(!t tho parallologfiini ABCD, and the tnan;L,'le EBC be upon tlie .same base BC, and between the same parallels BC, AE: then shall the parallelogram ABCD be double of the triangle EBC. Construction. Join AC. Proof. Then the triangle ABC is ecpial to the triangle EBC,* for tliey ai'e on the same base BC, and between the same parallels BC, AE. i- 37. l>ut the parallelogram ABCD is double of the triangle ABC, for the diagonal AC l)ise(;ts the ])arallelogram. I..S4. Therefore the parallelogram ABCD is also double of the triangle EBC. Q.E.D. !« EXERCISF.S. 1. ABCD is a i)!ualleloKram, and X, Y are the middle points of tho sidos AD, BC; it' Z is any point in XY, or XY jtroiluued, shew tliat the triangle AZB is one quarter of the parallelo;,aani ABCD. 2. Describe a right-angled isosceles triangle equal to a given square. ?!. If ABCD is a parallelogram, an 1 XY any points in DC and AD respectively: .shew that the triangles AXB, BYC are equal in area. 4. ABCD is a parallelogram, and P is any point within it; shew that tlie Bum of the triangles PAB, PCD is equal to half the paral- lelogram. Ill I I BOOK I. I'KOl'. 42. 75 37 Proposition' 42. Problem. To describe ft iiaralldogram that ahaU he equal to a given triangle, and have one of its angles equal to a giveit, an,gle. 10. 23. 31. Lot ABC Ih! the mi\('n ti'i;iu;j;lo, unci D tlin ^dveii ;ni_!^le. It is required to descril)e a panillelogi'am equal to ABC, and liaving one of its angles e(i[ual to D. Construction. Bisect BC at E. At E in CE, make the angle CEF equal to D ; tlirougli A di-aw AFG pai-allel to EC ; and through C draw CG parallel to EF. Then FECG shall be the parallelogram required. Join AE. Proof. Now the triangles ABE, A EC are equal, for they are on equal bases BE, EC, and between the same parallels ; I. 38. therefore the triangle ABC is doubl(> of the triangle AEC. But FECG is a parallelogram by construction ; Def 26. and it is double of th< iangle AEC, for they are on the same base EC, uid between the same parallels EC and AG. i. 11. Therefore the parallelogram FECG is equal to the triangle ABC; and it has one of its angles CEF equal to the given angle D. Q. E. p. EXERCISES. 1. Describe a paralle'loffram equal to a piven square standing on the same base, and having an an^'le equal to half a right angle. 2. Describe a rhovubus equal to a given parallelogram and stand- ing on the same base. When does the construction fail? 76 EUCLID S KLEMEN-^S. Dkfinitiox. If in t]io rliacconal of a parallelogram any point is taken, and straight lines are drawn tlirough it parallel to the sides of the parallelo<,n-am ; then of the four parallelograms into which the wliol(> ligure is di\ided, tlie two tlii-ough which tlie diagonal passes are called Ppral- lelograms about tliat diagonal, and the other two, which with these make up the whole figure, are called the complements of the parallelograms about the diagonal. Thus? in tho figure given helow, AEKH. KGCF arc parallologrania about tiu! diagonal AC; and HKFD, EBGK are tlie complements of those parallelograms. Note. A parallelogram is often named by two letters only, these being placed at opposite angular points. Proposition 43. Theorem. The complements of the parallelograms about the diagonal of anp parallelogram, are equal to one another. Let ABCD he a parallelogram, and KD, KB the comple- ments of the parallelograms EH, GF about the diagonal AC: then shall the ' njilement BK be equal to the comple- ment KD. Proof. Because L^-" is a parallelogram, and AK its diagonal, tlieretore tlie tiiangu; AEK is etiual to the tria.igie AHK. i. 34. Tor a similar reason the triangle KGC is equal to the triangle KFC. Hence the triangles AEK, KGC are together equal to the triangles AHK, KFC. : ooK T. rnor. 44. •77 Djrani any irouij,h it ' tlie four iriod, tlie x1 Ppral- r'o, which ailed the onjil. andoprams jlements of But the whole triangle ABC is equal to the whole ti-iangle ADC, for AC bisects the parallelogram ABCD ; I. 'M. therefore tlie remainder, the complement BK, is equal to the remainder, the complement KD. Q.e.d, EXERCISES. In tbe figiu-e of Prop. 4B, i^rove that (i) The parallelogram ED is equal to the parallelogram BH. (ii) If KB, KD are joined, the triangle AKB is equal to the triangle AKD. only, these e diagonal he comple- gonal AC: lie comple- diagonal, AHK. I. 34. lal to the ual to the Proposition 44. Prorlem. To a fjivcn straight U)i(' to apply n parallelogram luJdch shall he equal to a given triangle, and have one of its angles equal to a given angle, F E Let AB bo the given straight lino, C the given triangle. and D til;' given angle. It is required to apply to the straight line AB a paral- lelogram equal to the triangle C, and having an angle equal to the angle D. Construction. On AB produced describe a parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D; i. 22 and I. 42*. through A draw AH parallel to BG or EF, to meet FG pro- duced in H. I. 31. Join HB. * This step of the construction is effected by first describing on AB produced a Lnaugle wlio^-ie sides are tubpecLively equal to fiiu.su of the triangle C (i. 22); and by then making a parallelogram equal to the triangle so drawn, and having an angle equal to D (i. 42). 78 euclid'h elements. Then bncuiso AH and EF aro parallel, and HF meets them, tluTefoi-(i the angles AHF, HFE an; together equal to two riiiht angles I. 29. lience thc^angles 6HF, HFE are together less th;in two riglit angles; therefore HB and FE ^vill mert ?£ produced towards B and E. ^■^''- 12. Produce them to n^f^et nt K. Through K draw KL parallel to EA or FH; i. 31. and produce HA, GB to meet KL in the points L and M. Then shall BL be the parallelogram required. Proof. Xow FHLK is a parallelogram, Constr. and LB, BF are the complements of the parallelograms about the diagonal HK: therefore LB is equal to BF. i. 43. But the triangle C is equal to BF; Constr. therefore LB is equal to the triangle C. And because the angle GBE is equal to the vertically oppo- site angle ABM, I- 1^- and is likewise equal to the angle D ; Constr. therefore the angle ABM is equal to the angle D. Therefore the parallelogram LB, which is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D. Q-E.F, HOOK I. l'ROI>. 45. 79 Piioposit;on 45. Pkoblkm. To describe a paralleloiirinti equal to a given, rectilineal figure, and having an angle equal to a given angle. K H M , Const)'. xralleloijrams Let ABCD 1)0 tlio given rectilineal i'.gui-e, and E the given angle. It is required to describe a panillelograni equal to ABCD, and having an angle equal to E. Suppose the given rectilineal tiguro to ho a quadrilateral. Construction. Join BD. Describe the pai'allelogi-ani FH (njual to tlio ti'iaiigle ABD, and having the angle FKH (squal to the angles E. I. 4l*. To GH apply the parallelogram GM, ecjual to the tiiangh^ DBG, and having the angle GHM equal to E. I. 44, Then shall FKML be the parallelogram required. l^roof. Because each of the angles.GHM, FKH is 0(jua] to E, therefore the angle FKH is ofpial to the angle GHM To each of these equals add the angle GHK then the angles FKH, GHK are together ec[ual to th(> angles GHM, GHK. But since FK, GH are paiallol, and KH meets them, therefore the angles FKH, GHK are toirether equal to two right an tries qui I. 29. therefore also the angles GHM, GHK ai-o together equal to two right angles : therefore KH, HM arc in the same straiuht Vnn\ i. 14, 11. K G 80 H M Again, hocuuso KM, FG Jiro jiai-allol, and HG nuM'ts tliom, ilmrcforo tlio altovii:>to an<,d(>s MHG, HGF aro o(iual : l. 21) to oach of tlicso equals adfl tlio an!j;l(! HGL; tlion tlio anglos MHG, HGL ato tog(>tli('i' oqual to the angles HGF, HGL. .Out l)ecause HM, GL are parallel, and HG meets them, then^foro the angles MHG, HGL ai'e together ecjual to two I'ight angles: I- 29. therefore also the angles HGF, HGL a)-e together equal to two I'ight angles : therefore FG, GL :u'(^ in the same straight line. I. 14. And hecause KF and ML are each parallel to HG, Constr. therefore KF is parallel to ML; I. 30. and KM, FL are parallel ; Com/r. therefor(^ FKML is a parallelogram. J)('/. 2G. And hecause the parallelogram FH is ecpial to the triangle ABD, Comlr. and the pa,r.allelogi'ani GM to the triangle DBG ; Consir. therefore the whole" parallelogram FKML is e(iual to tjie whole figure ABCD ; and it has the angle FKM ecjual to the angle E. By a series of similar steps, a parallelogram may lie eonstructed ecjual to a rectilineal figure of more than four sides. Q.E.F. BOOK I. PUOP. 4fi. 81 PllOPOHlTION 4<;. PUOIILKM. To (Irsrrihe a sqmtre on a given straiyht line. lU'ots tliom, |iial : T. 2!) OL; (» llio jiiiglps L'ets tlu'iit, er ('(|u;il to I. 29. her oquiil to lino. I. 14. HG, Const)-. I. 30. Consir. 11. ])ef. 2G. the tri.'iii<'lo CoitNfr. >BC ; ('onatr. Mjiial to the iiglo E. ;niin may 1)0 irc than four Q.E.F. I. 11. I. 3. I. 31. A B Lot AB be the given straight line : it is required to describe a square on AB. Const)'. From A draw AC at right angles to AB ; and ]u;ilqual to AB. Througli D (Inav DE i)ara]l('l to AB; and through B arnll('l()i,M'aiii BL is douMc of \\w triaiii^do ABD, for tlicy ''ii'o oil tin; saiiio baso BD, and Ix-twoon tho .saino jKirallcls BD, AL. I. 41. And tli(! K(|uar(! GB is douhlo of tho triaugh! FBC, for tlioy ar(^ on th(! sanio l)as(' FB, and botwooii tho same parallols FB, GC. I. 41. J Jut doulilf's of (>(jnals arc oqual • Ax. G. thcrofon; tho i)aralielo Iiypotonuso BC is ecjual to the sum of the sc^iiares dcsciibcd on tho two sides BA, AC. Q.K.n. NoTK. It is not npcesaary to the proof of this Proposition that tho three squiucs .should bo doscrihod crtenuil to the triangle ABC; mid Kinco cdcli Hciniire may bo drawn cither toirardu or mcu)/ from the triangle, it may be shewn that there are 2 :< 2 x 2, or eight, possible constructions. KXKKCISKS. lC, BAG IS a : line. i. 14. FBA, angle FBC. 1. In the liKuro of this Troposition, shew that (i) If BG, CH are joined, these straight lines arc parallel; (ii) Tlio jjoints F, A, K are in one straight lino; (iii) FC and AD are at right angles to one another; (iv) If GH. KE, FD are joined, the triangle GAH is equal tu the given triangle in all respects; and the triangles FBD, KCE are each equal in area to the triangle ABC. [See Ex. 9, p. 73.] 84 euclid'm KLKMKXTW. 2. Oil tilt! Huh'H AB, AC of (//(// triaii^'li' ABC, s(iiuu<'s ABFQ, ACKH lilt' (Icsciilii'd III til towiiid tin; tiiuiiL'l.', or liolh on llii) Kido I'Liiiutc! Iroiu it: sliiw lliut tlit.' stnu^.;hl liui's BH luul CG uro eiiiiiil. H, On Uio HJdi'H of liny triuii^'lc ABC, cMjuiliitcral triangk-H BCX, CAY, ABZ an; dtsi rilicd, nil "xtcriuiUy, or all towurdH the triaugle: hLow that AX, BY, CZ uio all iiiuul. 4. The xijuiiri- ilcstrilx'd on tin', dimjinutl of on DC is equal to the sum of the S([uares on CA, AD. ,.. And, by hypothesis, tl„> s.juare on BC is e-;iri'anged in reverse order, and the construction and proof presented in a synthetic form. 88 Euclid's elements. This unravelling of the conditions of a proposition in order to trace it back to some earlier i)rinciple on wliich it depends, is called geometrical analysis : it is tlie natural way of attack- ing most exurcises of a more difHcult type, and it is especially adapLetl to tlie soluti(jn o^ prdblenu-i. These directions arc so general that they caiuiot be said to amonnt to a vmtliud: all that can bo clauned for (jcomctrical Analysis is that it furnishes a mode of seurdiiixj for a swjijestion, and its success will necessarily depend on tlie skill and ingenuity with wliich it is employed : these may be expected to c(jmo witli experience, l»ut a thorough grasp of the chief Pro- positions of Euclid is essential to attamuig them. The practical application of tiiese hints is illustrated by the following examples. 1. Construct an hoacch't triinujit' havnid (I'lren the luiae, . ■y, I. 18. ry: ;ary. Q.E.P, xre not rrpial, f and unequal, 1. In a triangle ABC, if AU is not greater than AB, show that any straight line drawn through the viTtex A, and terminated by the liaso BC, is less than AB. 2. ABC '/,< n iriaiifilc, and tlir vertical anijle B^C in hi^ceted hy a stroiiiht line vhicli mt'et-t the base BC in X ; sln'ir that BA is ijreater tiian BX, and CA (jreater than CX. Jlenee obtain a proof of 1.20. 3. The iwrpendieular is the sltortis. straight line that can be drann from a ijicen point to a (jiren strainht line ; and of others, that which i.i nearer to the inrpendieuiur ix /<^s,s than the more remote ; and tiro, and only two equal straight lines co;, be drawn from .'e yiren point to the yiven straight line, one on each side of the perpendicular. 4. The sum of the distances of any point from the three angular points of a triangle is greater than half its perimeter. 5. The sum of the distances of any ])oint within a triancle from its angular poinis is less than the perimeter of the triangle. 91 KUCLIDH KLKMKNTK. i 0, The perimeter of u (luadrihituml i.s giLiiter than tli.i .sum of itrt diu^jonals. 7. The sum of the diagonals of a quadrilateral is loss usaa the sum of tin; four strai^'ht lines drawn from tlie au-'ular jiolnts to any given point. Prove this, and point out tlie exceptional ease. 8. Jii 5 lUi till! Hiiin of its .1 is lean auui the iliir jioint-s to any tuil ciise. <'/' "•.., ''"''' f"''' I^'"''''^'''^^ t" two other straight lines hiiewt 1. ih'l ; ''r ""-'"' ^""tained by eaeli pair are biseS bliew that the bisecting lines are i)iirallel. ^ •^^^'i-cteu, thpm i^f '""f \'' ^T '^''?" l^etween two parallels and terminated bv Problemb. point iifAR-"l^° ""'^ two given straight lines, and X is a given pSpL;Si^t d;S.mjn;Vf;;^?s'^' "^*^* '^ ^-^^ '^ ^'^-^^ ^ *^ E. so tha\ BD.ot EC ';;:;. I'^klTe^luar"^'^^" "^"^^ ^^^^'"-^ ^" ^ -'^ paraUel t?the'ba't"^Bc'';S ' 'T''''\ '' '}''''' ^ ^^'^'^''^^ ''»<^ ^E that DE ma; be';.;.S^ SsS^ o^feji:^';?^:^^'- "' ^ ^'"'^ "• '^ to tlie bn^^:3l:;::^;£' '--l"^-^ to draw a straight line parallel may be ecual to the i'ielSof Bold'cE." '^ "" ^' "" ^'^^ ^^ 7 I m KUCLIUS KLEMENT8. IV. OX rAlt.Vl.l,Kl,OGUAM.S. Si l'ni|i(isii ions ','/.', '.j\, .iiiil Uk; ilt'ductioiis I'roiu lUfst; l*r(>[)S. ,nvoii on p;ii^t! (j 1, iff 1. 'I'hi' slniiiilil liiii' ilniioi tliroimli fhf iiiidlh- point of a nidi' vf a triaittjie parallil to the Ixtnc, liisei'ts the remniniiiij aide. Jict ABC liu a A, and Z tlu! niiddlt! point of t!icsi,l('AB. Tlirouf'li Z, ZY isdniwn i-ar' tuBC ; (lioiisliiiU Y lie liic iniddlt'iiointof AC. Throuf;li Z draw ZX par' to AC. 1.8I. Tlicn in flu. a« AZY, ZBX, because ZY iind BC aiv ]).ir', .-. tlie z AZY- tlii,sectod at Y. 2. TIw sD-diiilit liiii' irJiicli jdiii.t the middle ^joints of two side.i of a ti'iaiiijle, is jxintllel to the tliird side. Let ABC be a A . imu Z, Y tlic middle points of llu! sidi's AB, AC: tlien slull ZYl.e par' to BC. l'i<)(bice ZY to V, making' YV t(iual to ZY. Join CV. Then in the A'' AYZ, CYV, ( AY^CY, Hijp. liccause andYZ^-^YV, Coimir. I and the / AYZ = tlie vert. opp. / CYV; .-. AZ^CV, and the / ZAY^the z VCY; hence CV is jiar' to AZ. But CV is equal lo AZ, that is, to BZ .'. CV is equal and i)ar' to BZ : ,•• ZV is e((na,l and par' to BC : that is, ZY is par- to BC. Hyp. I. 33. Q.E.n. I [A second ])roof of tin's proposition niav be derived from i. 38, 3t).] THKOUEJl.S AM) EXAMl'LEb UN HOOK I. 97 eso I'rops. (I nido of a I. 'JO, I. ;ii. y.K.i). sidi'ii of a I. '27. Hyp. T. B3. Q.E.I). 1 1.38,30.] f 3. The straitilit lint- which Joinn the middle points of two aides of a triaiKjle i.f cfunl to luilj the third side. •4. Shew that thr thiw slry the straiijiit line which Joins the middle points tf the Jther sides of the triut til' )Ui.'li C, the middle jioint of AB, a Hide of tho A ABK, CH liaH hvvn drawn paralii'l to tho side BK; .-. CH l)isocts AK: Ex. 1, p. WJ. that is, AHr HK; .-. XZ-ZY. Q.E.D. 11. 1/ three parnllcl stvaif/ht lines make equal intereepts on a fdurth strai'lht line trliieh vieets them, thei/ irill also nialce equal inter- cepts on anil other straiijht line lehieli meets them. 15. Kqual and parallel straight lines have equal projections on any otiter blraiijh' line. IC). AB is a {j;ivon strnij^'ht line hisoctod at 0; and AX, BY are lierpcndiculars drawn from A and B on any other straight line: shew tliat OX is equal to OY. 17. AB /.--• a iiiren straiijht line bisected at O : and AX, BY and OZ are j^erpendicuhtrs drawn to aiiij straight line PQ, which does not pass between A and B: shew that OZ /\- equal to half the sum of AX, BY, [OZ is said to be the Arithmetic Mean between AX and BY.] 18. AB is a given straiglit line bisected at O; and through A, B and O ])arallel st^raiirht lines are drawn to meet a given straigiit line PQ ill X. Y. Z : shew that OZ is o'lual t > half the sum, or half the differenre of AX and BY, according as A and B lie on the same side or ou opposite sides of PQ. r THKOKKMS AND KXAMI'LKH ON llooK I. 09 U the pro- B ^ Q K it line PQ: i;;CZ, BY I. ;u. =Q; I. -JR. I. 84. u A ABK, X. 1, !>. W'. Q.K.n. qiHtl iiiter- 'unn (in atiij kX, BY arp line: nhew JY «;«? OZ :',s' not. ])(l>iti AX, BY. BY.] rousli A, B rai^'ht line .)• half tlu! ^ 1'.). 7'« (//■/•/ equal to the an■ prorhired BCD is a teimsc of a '/potenuse. oiuliculars \v tliat the tlie feet of BC ; and ■qu'ctnu'hj : I' BAG. ratrn from 'ormetl arc '•rnrn from [lendlruhir of the two 'f 7. In a tri'iiinJi' if a jx'rprndicuJnr hi' drtiini from oiw I'.rlri'iin'lif of ihr hat^i' to tin' hiscctor of ilic rrrlinil tUKilf, (\) it will iiuihe icitli citJicrof tlie xidi'K nuitdiiiinn flic rcrticdl oii'ilc on oii'ilc cijiiol to luiJf tlic tiiiiii. of the aiii/ldf (It flic Inixe ; (ii) it vill mohc irith the ha.ie an miijlc equal tit half flic dijlcroice ■, that XL is half tlie diji'errnee (f the sides AB, AC. S. In ami triaufile the nnfile cnntaiiied hi/ the bisector of the rerticnl (imjlc and the pir/iendiciilar from the rertex to the base is equal to Iialf the diji'erenee of the ainjlcs at the base. [See Ex. H, p. o!).] 0. In a trianf;lc ABC the side AC is produc.d to D, and tbe anjj;les BAC, BCD are bisected by straigbt lines wliich meet at F; sliew that they contain an angle eipial to half the angle at B. 10. If in a right-angled triangle one of the acute angles is doulile of the other, shew that the hypotenuse is double of tiie minuter side. 1 1. If in a diagonal of a par.allelogram any two points equidistant from its extremities be joiiicd to the ojjposite angles, the iigure thus formed will be also a parallelogram. I 102 EUCLID'S KliElIKNTS. : ij , I'f ABC is a «iveii rquiljitcral triangle, and in the sides BC, CA AB the pouits X, Y, Z ai« taken respectively, so that BX, CY and AZ are all e(iual. AX, BY, CZ are now drawn, intersecting in P Q R- shew that the tiianKle PQR is eciuilateral. - > • 1^5. Tf in tlie RJdes AB, BC, CD, DA of a parallelogram ABCD tour points P, Q, R, 3 bo taken in order, one in each side,\so that AP BQ, CR, DS are all equal; shew that the tigure PQRS is a parallelo- gram. 14. In the figure of i. 1, if tJKs eirelcs intersect at F, and if CA and CB are produced to meet th(; circles in P and Q. respectively • shew that the pomts P, F, Q are in tlie same straight lino; and shew also that the triangle CPQ is equilateral. [Problems marked {*) admit of more than one solution.] 15. Through two given points draw two straight lines forming with a straight line given in position, an equilateral triangle. no. From a given point it is required to draw to two parallel straight lines two equal straight lines at right angles to one another. , ^^J- '^'I'l'ee given straight lines meet at a point; draw another straight line so that the two portions of it intercepted between the given lines may be eipial to one another. 18. From a given point draw three straight lines of given lengths so tha^ their extremities may be in the same straight lin.-. and inter- cept equal distances on that line. ' [See Fig. to i. 1(5.] 10. Use the properties of tlie equilateral triangle t(. i; isect a given finite straight line. "■b^vc.i 20. In a given triangle inscribe a rhombus, liaving one of its angles coincKhug with an angle of the triangle. VI. ox THK CONCLKUKN-Cl.; UF STHAUUIT LINKS I\ A TIUANGLK. Dkfixitioxs. (i) Tlireo or moro straight lines are said to l>e concurrent when they moot iu one point. (ii) Thre.! or more points ai-o said to l.o collinear when thov lie U[»()u one straight line. Wo here gi\o some propositions relating to the concurrence ot certain groups of straight lines drawn in a trian-le : the im- poi Uuicc ot these tlieoreuis will be more fully appreciated when the student is famdiar with Books in. and i\-. e« BC, CA, 2V find AZ 111 P, Q, R : ram A BCD 50 that AP, a parallelo- F, and if spectively ; line ; and on.] 3S forming e. .vo parallel ! another. L\v another itween the 3n ]en,£,'ths, and inter- ,'. to I. 10.] ect a given mo of its UIANCLE. ■e said to .'hen tbev acurrence : the iiu- ited when THKOKEMS AND KXAMl'LKtJ ON HOOK. 1. 103 4 1 1 I. TJie iwrpendiculam drawn to the aidci of a trian(jle-J'ro,ti their middle points are concurrent. Let ABC be a A, and X, Y, Z the middle poiiitH of its sides : then sliall the porp^ drawn to the sides from X, Y, Z be coneuneiit. i'voni Z and Y draw perp** to AB, AC; these perps, since they cannot be parallel, will meet at point O. Ax. 12. Join OX. Jt w required to prore tliat OX i^ pcrp. to BC Join OA, OB, OC. In the A^OYA, OYC, ( YA..YC, Because < and OY is common to both ; / also the / OYAr-^the z OYC, bcinj,' rt. l '. .-. OA- OC. Similarly, from the a" OZA, OZB, it may ho proved tliat 0A-=0E3. HenceOA, OB, OC are all e(i',ial. Again, in the a^ OXB, OXC ( BX ^CX, 1.4. n>jp. Decause 'and XO is common to both ; / alsoOB = OC: I'roced. '.-. the z OXB = the / OXC; i. 8. but tliene are adjacent / ** ; .-. they are rt. L«; /->''/• 7. that is, OX is perp. to BC. Hence the three perp'^ OX, OY, OZ meet in the point O. Q. K. ]). 2. Tlie bisectorii of the (Uiijleii (fn trianiile <.■;■.■ concurrent. Let ABC be a A . Bisect the / * ABC, BOA, by straij^ht lines which must meet at some point O. Ax. 12. Join AO. It /.v required to prore that AO bi-fcctn the I BAC. From O draw OP, OQ. OR perp. to the sides of tlu! a . Then in the a^OBP, OBR, ( the z OBP^the Z OBR, Because -' suid th.e z OPB=-the z ORB. bein^' rt. L », i and OB is common ; .-. OP^OR. I. 20 104 EUCLID'S ELEMKNTS. Similarly from tho A" OCP, OCQ, it may he shewn tliat OP=:OQ, .-. '^P, OQ, ORarcallciual.' Again in tlie a" ORA. OQA, jtlio ^"ORA, OQAun.Mf. l\ Bccanso'''"^'^ *'"' l'.vP"t'-'""w CA is J conmion, ^ aloORr^OQ; .1' roved. .-. the / RAO^thez QAO. p,^ jo^ ^^ ,,j That is, AO is tho hiscctor of tho / BAG. Plenee the bisectors of the three / « „,pet ^^ the point O. Q. K. I), |! w/.v( 10) 0/ tlu: tliinl ait/ilr ,7/-<' conrurn'iit. Let ABC he a a, of which tho sides AB, AC are ja'cxlucea to i'.ny points D and E. liisect the z ^ DBC, ECB by strai-lit lines which must meet at some point O. A.c. ]2. Join AO. It ?.s- rrqulrrd to proc; tlnit AO hiaccts the aunlc BAG. I'rom O draw OP, OQ, OR perp. to tho sides of tlie A . Then in the A« OBP, OBR, 'the I OBP^Uk! z OBR, Co„>>tr. Because-^ also the z OPB^tlie z ORB, bomt,' rt. L ", and OB is common ; •'• OP- OR. I. 20. Similarly in the a" OCP, OCQ, it may he shewn that OP:-:;OQ: .-. OP, OQ, OR are all equal. Af;ain in tlie a' ORA. OQA, r tile z« ORA, OQA are rt. L% ■ and tiie hypotenuse OA is common, ( als()OR = OQ; .-. the z RAO::=the z QAO. Because Proi'cd. Ex. 12, p. . Ey moans of this Corollary it may be shewn that in any triangle the shorter median bisects the greater side. [The point of intersection of the tlirec medians of a triangle is called the centroid. It is shown iu mechanics that a t'.ii;i triangular plate will balance in any position about this point centroid of a triangle is also its centrj of gravity.! therefore the lOR KUCLTDH ELKMKNTH. oppo. •). Tlir iicrpendicitlfirs drawn /mm the vertices of a triangle to the ositc sidfx are coHnirreiit. Ih'f. 20. I. 34. Let ABC I)y a a, aii, 1). 5i. l,p. lOH. C to thii Q.K.D. :ni.] from the s ortho- perpeii- THEOREM.S AND P:XAMPLKH ON BOOK I. 107 K VII. ON THE CONSTRUCTION OF TUIANGLES WITH GIVEN I'AUTS. No general rules can be laid down f(jr the solution _ of pro))lenis in this .section; but in a few typical cases we give constructions, wiiich the student will i'lnd little diiHculty in adapting to other (piestions of the same class. 1. CoHstrift a rliiht-nnijlcd tridinjlc, hdviiuj (liccn tlic lujpo'.ciiasc tuid the tiuiii of the reiiiauiiiin s/(/".s'. [It irt requiieil to construct a it. anj^ied A, having ilrt hypotenuse equal to the given straight line K, and the sum of its remaining sides equal to AB. From A draw AE making with BA an Z equal to half a rt. L . From centre B, with radius ecpud to K, de- sciibe a circle cutting AE in the points C, C. From C and C draw ])ndieular drawn In.m the vertex to tho base. [See Ex. 1, p. 8H.] I). Construct a trian;,'le, having given t!ie base, one of tho angles at the base, and the sum of tlie remaining side?. 10. Construct a triangle liaving given t!u< base, one of tlio angles at tlie base, and the dilterence of the remaining sides. 11. Cou.i)lied to rectilineal iiL,an'es Mill bo used as donotiug C(j[ualltij of area unless otherwiso statc'd. 1. HJiew that a pdrdllehxjrum is btsectrd hy airj siraiijlit line which i)as8es thromjli the middle j^oint of one of its diafjonals. [i. 2'.J, 2 O.J 2. Bisect a parallelogram by a straight lino drawn tluough a given point. "). Bisect a ]iarallelograni by a straight hue drawn perpendicular to one of its sides, X. Bisect a parallelogram by a straight line drawn parallel to a given straight line. 5. ABCD is a Irtipeziam in which the side AB is parallel to DC, ,S7/('((' titat its area is equal to tlie area of a piira'leloiirauL fornu'd laf dnncina throuah X, the middle puinc of BC, a straipht line i>araUel to AD. [I. 2«), 20,] 0. A trapezium is equal to a parallelogram whose base is half the sum of the parallel sides of the given ligure, and whose altitude is equal to the perpendicular distance between them. 7. ABCD is a trapezium in which the side AB is i)iirallclto DC; shew that it is doul)le of thr umlo formed by joining the extremities of AD to X, the middle poin.i of BC, 8. Rhew that a trapezium is bisected by the atniight line wliich joins the middle points of its parallel sides. [i- ii^-J )J 110 KICI.lhs KMOllCNT.S. on Til tho f..ll„win- K'l-oui) (.f KxorciNos tho in-nafs doucud chiefly l'r<)i.o,s.t.i<,us .37 una .Ts, ^,..1 tht- tw., c.mvor.so theoVem.s ^ iiLs AU I lul BD wliah j.mi tluur uxtiviiii(i,.,s an; namllcl shew that tho tnaiJKlu AXD is o^iiul to tlio tiiai.-lc BXC. 10. Jf two stiaiKlit lines AB. CD interspct i,t y ... ti ,t n 11. ABCD is u paialli-loKiaiM, and X any ),oint in the dia-'onal K^ li!t'oi!j " "'' ''■■"""''" ^BC, )iDC a,o .iual; iS AB^ir.^^^^ VyTn^''' '\'"^ ^' ® ^''" »"^Wlc." points of tho sicks AB, AC; show tliat it BQ and CR intds.ct in X tliotrianLdo BXC i <'.iuul to tho .inudiilatoml AQXR. [Seo Kx. n, p! mj ^ ^^^ '' in /,;';, /^;f'''^"''*l;^'V''''''^' "^^''" "''^'■^ '^f " qu'idrilateral be joined 11. Two trianf,'los of eciual area stand on tlio same base but on m he .same ha>.. ///(' triiDKili: OAC /n niiiiil tn tlir iliil'i'iriiir <;/ till! tridiiir OAB. Cask T. If O in without the l DAB and its opp. v. it. /L , tlien OA is willi- out tho par'" ABCD: thcrct'mo th(! pcrp. (hawn from C to OA is iijiial to the .suiii of the porp" drawn troiii B and D to OA. [Seo Ex. 'JO, p. W. 1 Now the A" OAC, OAD, OAB urn upon the sauic l)aM' OA; and tho altitude of tho a OAC witli rospcet to this hasc lias hwu sIkjwu to ht^ e(|ual to the sum of tho altitudes uf llip A'* OAD. OAB. Tlu loforo tho A OAC is (Miual to tho sum of the A" OAD, OAB. [See Ex. If,, p. 110. J <.'.i;.i>. i;). ABCD is a pnralloloRram, and throut,'h O. any point within it, straitiht hnos aro (hiuvii ])iuull( 1 to tho sidos of tlio parallolo;,'ranr, show that tho difloronoo of tlio piinillolo;;rann DO, BO is double of the triani^'le AOC. [See preceding theorem ^ii).] 20. The area of a ([uadrilateral is equal to the area of a triangle hnxlnl' two of iis sides oqual to tlu! diiigonals of the Kivon figure, and the includod angle equal to either of tho angles between tho dia- gonals. 21. ABC h (I triaiiiilc, tiiul D />• ,'Uj finuil u» an un^'lo (if liiu j,'ivtai triangli'. 'J.'{. Construct a trianylo eqUHl in area to a given tiiangli-, and having a given aititudu. '24. On a bast- of ^'ivc•n length construct a triangle equal to a given triangle, and having its vertex on a given straigiit line. 25, On a base of j^'iven length describe (i) an isosceles trianglu; (ii) a right-angled triangle, enual to a given triangle. '20, Construct a triangle iMiual to the stun or dilTerencu of two given triangles. [See Ex. Iti, p. lid. J 27. ABC is a ^'iven triangle, and X a idvi'n point: describe a triangle eimal to ABC, Imving its vertex at X, iuid its base in the .saniu straight line as BC. 2H. ABCD /s '/ ijiiitdriliiti'nil : mi tlir Ihim' AB coiistnict u tridiinlc niiKil in (nca to ABCD, and luiriti;! tlif aiKjIf |{l:\[s AND IN \MI'l,i;s oV HooK I. 11.T 11 M) a wivcn ;riiuif,'ldes, the angle contained by these : ides is acute; if greater, obtuse. 44 ABO i^ a triangle, right-angled at A; the si>les AB, AC are intersected bv a sliaight Hue PQ, and BQ. PC are jomed : shew that the sum of the squares on BQ, PC is equal to the sum of the squares en BC, PQ. 4.-. In a light-angled triangle four times the sum of the squares on the medians which bisect the sides containing the right angle is eijual to five times the square on the hypotenuse. 4('<. Describe a square whose area shall be three times that of a gveu si|uaie. 47. Divide a straight line into two parts such that tlie sum of their siiuares shall be equal to a given square. »i IX. ox LOCI. It is frequently required in tlic course of Tlane (ieonietiy to find the iiositioii of a point which Nitisfies given conditions. Now all ])robk'ins of tlii.s type hitherto considered have been found to 1)0 capable of deiinite dct(?rniinution, though some admit of more than one solution : tlus however will not be the case if (»ih/ one condition is given. For example, if we are asked to find n point which .shall be at a i^iven distance from ca given point, we obs(,'rve at once that the pro)>lem is imhienninate, that is, that it admits of an iiidetiiiite number of solutions; for the condition statnl is satisfied by any point on the cireumicrence of the circle described from the given point as centre, \yith _a radius equal to the given distance : moreover this condition is Hatisfted by no other pisint within, or without the c'uvh. i\gain, suppose that it is re.piircd to iiiid a point at a g distance from a given straight line. ivcu THKORKMS AXI) KX AM I'l.KS <)N I'.odK I. ii; to the sum B with four [Hill to five _^ Kcjiuiro on I'tS. the s(iiiares is iuaitti ; if AB, AC are : slu'W that the squarcrf the sqnavps right angle iiR'S that of tlie Hum of ieonietry to conditions. ha\o been .souio admit i the case if sked to tind fiveu point, :Ue, that is, lis ; for the rcumfcrenco itre, with a condition is le. t at a given Here, too, it is obvious that there are an iniiniie luuuber of such points, and that they lie on the two parallel straight lines which may be drawn on either side of the given stiaight line at the given distance iVom it : fnrtiici', no ])oint that is not on one or otiher of these parallels Nitisties the given londition. Hence we see that when one condition is assigned it is not sufficient to determine the position of a point absolutely, but it may have the ellect of restricting it to some definite line or lines, straight or curved. This leads us to the following definition. Dkfixition. The Locus of :i point satisfying an assigned condition consists of the line, lines, or part of a line, to which the point is thereby I'estr'cted ; provided that the condition is satisfied by every point on such line or lines, and by no other. A locus is sometimes defined as the path traced out by a point which moves in accordance with an assigned law. Thus the locus of a point, which is always at a given distance from a given point, is a circle of wliich the given point is t\w centi'e : and the locus of a point, which is always at a given distance from a given straight line, is a pair of i)ai'allel straight lines. We now^ see that in order to infer that a certain line, or system of lines, is the locus of a poiut, under a given coiulition, it is nec(;ssary to prove (i) that any point which fulfils the given condition is on tlie supposed locus ; (ii) that every point on the supposed locus satisfies the given condition. 1. Find the Uicn^ «/' */ point irliicli it< uliniti^ rijiildistiint fnnu two i/ii-en point.'i. Let A, B he the two given points. (h) lict P be any point eipiidistant fir.iu A and B, ko that AP = BP. lliscet AB at X, and join PX. Then in the a-^ AXP, BXP, ( AX = BX, Because and PX is common to both, ( alsoAP = BP, .-. tlio z PXA--:.the / PXB: and they arc adjacent / ■''; .-. PX is perp. to AB. .•. Any point which is equidistaMt from A and B is ou tlie straight line which l)isccts AB at ri<':ht angles. Con-'itr. Ihip. I. 8. no KUCLIDS KLKMKNTS. ij-i) Also overy point iu this line is equidistant from A and B. For let Q ho any point in this line. Join AQ, BQ. Thenin tliu a'' AXQ, BXQ, , AX:^BX, JJwause -) and XQ is common to both ; [also tlio / AXQ tho / BXQ, being rt. L *■ ; .•. AQ--BQ. That is, Q is tciuidirftant from A and B. 1.4. Hence we conclude that tlie locus of the point e(iuidistant from two given points A, B is the straight line whieli bisects AB at right angles. 2. To find Ihf hiciia of the middle jxdnt of a Ktrauiht line dimcn from a (jicen point to meet a (jiven straiglit line of unlimited lentjtli. B X -7- P^Qrfl Let A be the given iioint, and BC the given straiglit line of un- limited length. (a) Let AX be any !-traight line drawn tluxmgh A to meet BC, and let P be its iniddle iioint. Draw AF per]), to BC, and bisect AF at E. Join EP, and jjroduce it indetiiiitely. yince AFX is a A , and E, P the middle points of the two sides AF, AX, .•. EP is ])arallel to the remaining side FX. Ex, 2, p. \){\. .•• P is on the straight line which passes through the fi. red point E, and is parallel to BC. {(i) Again, every jwint in EP, or EP produced, iultils the required condition. For, in tliis straight line take any point Q. Join AQ, and jiroduce it to meet BC in Y. Then FAY is a A , and through E, the middle point of the side AF, EQ is drawn parallel to the side FY, . •. Q is the middle point of AY. Ex. 1, p. 06. Hence the required locus is the straight line drawn parallel to BC, and passing through E, the middle point of the perp. from A to BC. THEOREMS AND EXAMPLES ON BOOK I. ir ml B. 1.4. ;taiit from B iit ri^ht line (Irtnni lentjtli. 3. Find the locus of a jwint equidistant from two (ftven inter- secting straight lines. [See Ex. H, p. 4U.] 4. Fiud ihe locus of u point at a given radial distance from tlie circumference of a tj;ivi'U circle. 5. Find the locus of a point which moves so that the sum of its distances from two {^iveu intersecting straight lines of unlimited length is constant. 6. Find the locus of a point when the differences of its distances from two given intersecting straiglit lines of unlimited length is constant. 7. A straight rod of given length slides between two 8trai-;ht rulers placed at right angles to one another: tliiil the locus of its middle point. [See Ex. 2, p. 100.] 8. On a given base as hypotenuse right-angled triangles are described: lincl the locus of their vertices. ine of un- meet BC, esAF, AX, ;x, 2, p. ',)(). ed point E, 9. AB is a given straiglit line, and AX is the perpendicular drawn from A to any straight line passing through B: tind the locus of the middle point of AX. 10. Find the locus of the vertex of a triangle, when the base and area are given. 11. Find the locus of tlie intersection of the diagonals of a paral- lelogram, of which the base and area are given. 12. Find the locus of the intersection of the medians of a triangle described on a given base and of given area. he required ide AF, EQ :x. 1, p. 9(). illel to BC, A to BC. X. t)N THK INTERSECTION OF LOCI. It appears from various problems which have already been considered, that we iire often required to tind a point, the jHJsition of wliieli is suiyeet to two given conditions. The metliod of loci i.s very ti.seful in the «ohitiou of problems of this kind: for corresponding to each condition there will l)e a locus on which the recpiired point ntvist lie ; hence all points which are connnon to these two loci, tliat is, ah the ])oints of iutersevtion of the loci, will satisfy both the given conditions. 118 KUCI.lir.S i;i.K.\lK.\l;i. Example 1. To cuDKtrnct a triitnijle, haviiKj (jivdi tlw bane, the. altltudr, (iiid the leiujth of the inedinn which hi.' circle. Hence any points wliich are connnon to CD and tlie circle, satisfy both the given conditions: tliat is to say, if CD intersect the circle in E, F each of the points of intersection might be the vertex of the re(piired triangle. This supposes the length of the median Q to be greater than the altitude. Example '2. To jiiid a point e({uidiiit(int from tliree niren jJoints A, B, C, wliieli (ire not in tlie mine straigld line. (i) The locus of points equidistant from A and B is the straight lino PQ, which bisects AB at right angles, Ex. 1, p. 115, (ii) Similarly tho locus of points equidistant from B and C is the straight line RS which bisects BC at right angles. Hence tlie point connnon to PQ and RS must satisfy both con- ditions: tl'.at is to say, the ]>oint of intersection of PQ and RS will be equidistant from A, B, and C, These principles may also be used to ]iro\e the theorems relating to concurrency already j^'iven on page 103. Ex.\:mpm:. 'Jo prore thut the hiseclio:< (f the (iiinles if u iridiKjIe are concurrent. Let ABC bo a triangle. Bisect the / » ABC, BCA by straight lines BO, CO : these must meet at some point O. A.r. 1'2. Join OA. Then shall OA biseet tlie z BAC. Now BO is the loons of ])oiiits equi- distant from BC, BA; Kx. ii, p. -P.). .-. OP OR. Similarly CO is the locus of points equidistant from BC, CA. .-. OP-^OQ; hence OR OQ. .-. O is on the locus of points equidistant from AB and AC ; that is OA is the bisector of the l BAC. Hence the bisectors of the thrcn; z '^ meet at the point O. THKOUKMS AND KXAMl'I.KS oX I'.iinlC I. U!) (' bane, the hi' iilt.itiiilo ii ilisiaiico vith nuUus the circle, itersect the ! the vertex he median ifcii 2)oints he straight :. 1, p. 115. J and C is V liotli eon- ud RS will theoroms ' (I Iriduiilt; It nuiy Imppeii that the datii, nf the pn-liU'ni ure so reluteil to one anotiier that tlie resulting loei do not intersect: in this case the iirol)k'ni is ini[)ossilih'. For cxanii.le, if in Ex. 1, i-a-e IIS, the len-tli <|f the given meilian /,v lt's,i than tlie given altitude, tlie straiglit line CD will not he interseeted by the circle, and no triangle can fultil the conditions of the prol)lcni. If the length of the median Is rie coiintnict a tri(iii(jl,\ haviwi 'jivcn one auiiJc at the base and the length of the opposite side. Examine the relations irhieh must e.rist amonii tlie data in order that there man ''^' '""" *•'"'"' tions, one solution, or that the j'rahh'in maij he ii}q)ossible. 4. On the base of a given triangle construct a second triangle e(pml in area to the ihst, and having its vertex in a given straight line. 5. Construct an isosceles triangle oipud in area to a given triangle, and standing on the same base. C. Find a point which is at a given distance from a given point, and is equidistant from two given jiarallel straight lines. d AC: to. JiOOK Tl, fi! r i k\- Book 1 r. (IcjilsAvitli iIh- jii'c'i.s of i'('ctaiii;l('S .-iiid sijiiaros. J^KFIXITIOXS. 1. A Rectangle is ,i p.'ir.illclo^rj'iu avIucIi li;i,s one of its jiiiglcs ;l I'iylit angle. It should be rt'ineiiibcrcil that it' a ]iarallelogram has one rifjht aii,L;l(;, (/// its aiif,'les arc li.L^lit aiif^les. [l]x. 1, p. (jl.] L'. A i-pcf:ai),i,^l(i is said to l)o contained l)y any two of its sides \vliich form a ri';lit angles : for it is clear that Loth the form and magnitude of a r<'etangh' are fully determined Avhcn the lengths of two sueli sides are given. Thus llio ivi'taiifrlo ACDB is said ^ /() he contii'nii'il Ia- AB, AC: or Ijy CD, DB : aud it X and Y uio two st:',ii:lit lines equal rospoctively to AB and AC, then the rectan-,'le conlaiiu'd In- X and Y is oijnal to the rectangle contained hv AB, AC. [SeeEx. ]-2, p. Gl.] X After Pi'oposition M, we shall use the ahbreviatiou rei-f. AB, AC to denote the 'recta m/le eoyifa'med hij AB and AC. o. .Ill any parallelogram the ligiiro foi'med hy eitlier of the parallelogi-ams altout a diagonal together with the two complements is called ;!, gnomon. Thus the shaded ])orlioii of the annexed - fit,'ure, consisting of liie ])ari).lleiognini EH together witij the compleuienls AK, KC is G the fiiiDinon AHF. The other gnomon in tlie ilLrur(> is that whicli is made up ot' AK, GF and FH, uamely the gnomon AFH. INTHODUCTOHY. 121 I NTKODUCTOKY. .B B H A ..'.f.'itiifj Pure Geometry maki's no ust- of 'nvmhi'r to estiwato the magnitude of the lines, angles, and figures with Avhich it deals: heiiL'e it re(|uirL's no imitx (if 'inaijiiltudc such as the student is J'aniiliar with iu Arithnu'tie, For example, tliougli (ieometry is eoncerned with the relative lengths of straight lines, it does not seek to express those lengths in terms of i/ardx, feet, or imh<'i<: similarly it does not ask iunv many i>(/uaro i/,/ tiro Htral-jht (lina in (leonietry eon'e.si>und.s to Mfj jn'Diliict of tiro luiiiibci'.'^ in Arithinel ic or Algchra; ami that the ,yii(tre ((i-xciuhi'd on a alnii'flit Hki' eorrcsponds to the xjiiare of u niirnhcr. Aeconliiigly it will Ik- ibuud in tli(> course of Jiook I J. that several theorems relatim;' to the anus of rectan;,des and .sijuures are analogous to well-known algehraic al i'orniuhe.' Ill view of these priiiei|)les tlui roL'taiiyhi conlaiiieil Iiv two straight lines AB, BC is soiuetiines expressid in the foria of a jiroduct. as AB.BC, and the si|uure described on AB as AB-. This notation, together with the signs + and -, will he eniployed in the additional nialter ap[iended to this Ijook; hut it U'^not (uimittcd info Enflid'.-^ text hccause it is desirable in the first instance to emphasize the distinction lietweeii geonnitrieal ina up of tlio t\!. similarly tlio lig. DH is the rectangle contained by P, DB. .-. tlio rectangl(f contained by P, AB is ("(jnal to the sum of the rectangles contained by P, AC, by P, CD, and by P, DB. ^^'•-"• COKRKSPONDIXO A I.OKnUAICAL I'OIJMILA. In iiccordanco with tlio principles explained on pa^^e 12-2, the result of this proposition may be written tlius : P.AB = P.AC4-P.CD-i-P.DB. Now if the line P contains j* units of length, ami if AC, CD, DB contain «, b, c units resi>ectively, then AB=^'i + h + c, and we have j){a + h + c) ==pa +ph +pc. 121 KUCMD's I'.I.KMKNTS. I'KOJ f>sn'[()\ -J, in:oKK>f. //' n sh-auilif !!„,'. Is ul i„lo an>j two ports ihn nw,,-,'. on fin: >rhul.'. lino, is ejiuni to the sum of the rectannl.H C(>Hftn„>;l h,/ tli>- irhole lim ami cuch of the ports SI. Let tlic stiuiirlit lino AB 1)0 divided 2>iU'ts AC, CB : at C info (lie two tiicii shall ^. AF is tli(* roctani;-]o .'ontainod 1. for th(! % AF is contained hy AD, AC ; and"' A also the lii;. CE is tho rootaii'd Constr. )y AB, AC D=. AB , , I- contained by AB, BC tor the tig. CE is contained by BE, BC ; and BE.:.AB. tl 10 S(i. on AB the sum of the roots. AB, AC, and by AB, BC coutaiiuHl by Q.E.D. t'OUKi:sPOXI)I\(i Al,fii;i!l{AlCAL F OK.MULA. TIic result of tliis inoiiositioii niay Le wiitt AB-=:AB. AC i AB . BC. Lot AC contain , and wo have {a-{-hY-^{a + h)a-\-{a + b)h. MOOK II. I'lloP. .'J. 125 parts, thi', rectnrKjlfiH ilic two 11 of flio I. 4G. I. 31. Co)istr. , AC; AB ; , BC; -AB. lined ])y Q.K.L). PitOpOHITlON .".. TllKOKKM. //' (I tti'd'itiJtt Hill' iff (lividtnl into (till/ hm purfs, thfi. rPdaiujJi- eoiitdiiiril tnj tlo' vliole dud om' of tin' jiarfs is rqiuil ill (III' .si/iniir. on. flmt part fiiiji'flirr v'ilk tin' rii'tanyln coHfaiucd 1)1/ till' tiro jtarts. \ C B Let tli(^ str;iiglit line AB he divided at C into tlit! two p!irts AC, CB: then slinll tlie reet. contained by AB, AC Le e(|ual to Ww S(i. on AC toi,f('tlier with the reet. contained liy AC, CB. On AC de.sciil)e the scjuaie AFDC ; i. ICi. iind th)•on,^•h B diviw BE par' to AF, meeting,' FD produced in E. I. 31. Now the \\\f. AE is inadf3 up of the tigs. AD, CE ; and of tliese, the Jig. AE th<> icct. contained l)y AB, AC ; for AF AC ; and the tig. AD is the sq. on AC ; Constr. also tlio lig. CE is tlio reet. containcMl Ity AC, CB ; for CD AC. .-. tlie reet. contained liy AB, AC is e(|Ual to the sq. on AC together witli the reet. contained l)y AC, CB. Q.E.D. COllRESPONDrXG ALCJKBRAICAL FORMULA. This result may bo written AB . AC = AC'-'+ AC . CB. Let AC, CB ciaitaiii units of Icngtli respectively, tlieu f\B = a + h, !i,nil we luive [a + 1>) a — a- i nh, NoTK. It bboultl bo observed that Props. 2 aiul ;! are niwcial coxei of Prop. 1. '■ 120 kuci.id's ki.kmknts. l*H()l'OMlTH>N IIKOKKM. //' ii, t,tr'. u ilir'ulid into (n,;/ Urn yarlA, the snnare on thr' irhol,'. Jin,: w ,'qu„l In lite mm of the Hqaaren on thi tico ports to,jcf/o'r vUh t,nr,', tlo: r>'i'tav,jk contnimd hi) thf two ports. A C , B H / G K / Let tl.c stral.'^lit liiu' AB lu' dividt-d at C into tli<^ two p.irts AC. CB : tlifii shall tlH^ s(i. on AB t><' <'(Hial to lli<> sum ot the s(i ; .-. the an.ude ADB - tho nni,de ABD ; i. T). .-. tho iingle CGB - tho anglo CBG. .-. CB CG. I- 6. And tho opp. sides of tho par'" CK aro ccpial ; l. 34. .-. the fi.if. CK is etpiilateral ; and tho angle CBK is a riglit angle ; W- ~^' .-. CK is a square, and it is described on BC. i. 40, Cor. Similarly tho tig. HF is tho sq. on HG, that is, the sq. oil AC, , for HG the opp. side AC. i- o-i "<«>l< II. I'ltoj'. I, 127 purtti, the le HfjuartiH confoined into tlui nil of tlu! CB. I. 4G. 1 G. n. A^'/iin, tlm coinplcnciit AG (li.« cninplrnicnt GE. i. I.'l. But tim (iJ,^ AG tli(« r..,(. AC, CB ; for CG CB ••■ the rect. AC, CB. CoKOLLAMV. PamUrh,;/rams ah.<„f thn ilUuionaU of a ftrpinrf an: ffinniselvm sfruarfs. squari's. CK is tlie CG, AD, DB. 1. 29. I. 5. I. 6. T. 34. Def. 28. I. 4n, Cor. b is, the sq. I. 34 coRRKspoxniy j-oKnuAHAr. foumii.a. Tlio result of tliiH iniivortfint J'^'oposition ii.-iy l,o wiitton tli AB-'-.AC-' + CB-'-t-2AC . cB. 't AC = r/, and CB = /;; then AB = .'/ f /;, a lid wo have („ I- ijf ,-, „2 + ,;, J. 2„ h. US; ir. E. ii^p 128 KUCLTD'S ELKMKNT8. PUOPOSITIOX f). TlIEOUKM. //■ '- points of srction, are tnyether equal to the square on half the Hue. Let tho str;iio-li1 lino AB l»o divided oqually ut P, and unoqually at Q : tl.oi/tlio roct. AQ, QB and llic sc]. on PQ shall bo to- iLfotliei' oqual to the scj. oi\ PB. On PB dost'iil)o tlic squavo PCDB. Join BC. 46. 0( Through Q draw QE par' to BD, cutting BC in F. i. :U. Through F draw LFHG par' to AB. Througli A draw AG par' to BD. Now tlio oornplenionL PF - tho conq)lonicnt FD : i. l-"). to oai'h add tlio tig. QL; then the lig. PL the tig. QD. But Iho tig. PL -the iig. AH, for tliey ai'o par'"^ on id Itascs and hetweon the same pai''". .-. the tig. AH the tig. QD. To each add the lig. PF : 1|U 1. 'M\ thiMi the Hg. AF ^- the gnomon PLE. Xow the tig. AF tiie roct. AQ, QB, for QB QF ; •. tlie rei-t. AQ, QB tlui gnomon PLE. To eacli add tho s(i. on PQ, tiiat is, the lig. HE ; ll. 4. then the rect. AQ, QB with the sq. on PQ the gnomon PLE with the hg. HE the wliole tig. PD, which is the s(|. on PB. nnoic ir. i>nor. "i. 120 nequaUy, he, .square f equal to fit P, and all bp to- I. 46. F. 1. :U. D ; I. 1:5. > pfiv"" on 1. 'Mk Tliut is, the roct, AQ, QB and tlio .S(}. on PQ are to,f,'otlirr pqual to the S(|, on PB. q. i;.n. (.'ouOLLAltv. From lliis Proposition it follows that th,i diffi'i'cnct' of the squaviiti on tiro xlnibjlit /uirs is equal to th>' rectnmjle cojitalned by their nu.iu. and difference. P -H — Q B For let X and Y be the given ^ ,st. lines, of which X is the greater. Draw AP equal to X, and pro- Y (luce it to B, making PB equal to Y AP, that is to X. From PB cut off PQ equal to Y. Then AQ is equal to tlio sum of X and Y, and QB is equal to the dift'ercncc of X and Y. Now because AB is divided equally at P and unequally at Q, .-. the rect. AQ, QB with sq. on PQ-tbe sq. on PB; n. ;'). that is, the difference of the sqq. on PB, PQ = the rect. AQ, Qb', or, the difference of the sqq. on X and Y-the reet. contained by the sum and the difference of X and Y. COKilKSfONDLNG ALtiEBKAlCAL POR.MULA. This result may be written AQ.QB ! PQ2^PB-. Let AB = 2a ; and let PQ = h ; then AP and PB cach = ((. Also AQ ::;- a + b; and QB = « - /;. Hence we have {a + h){a-h) + lr = a\ or {a + h){a-h)==^a^-h'\ QF; HE; II. 4. the fiy. HE EXERCISE. In the (dnire fujnre aheic that AP is half tht> sum of AQ and QB ; and that PQ u half their ilifference. 0-! 130 Euclid's i:i,i;mi;nts. PUOPOSITION G. TlIKOREM. ]f a strahjht line is hUectM and produced to any point, flie redamjle c.o^dalncd b// the, irhnle line i/ins prodncod, ayid the part of it proilucrd, /(>;/r/Iirr ir!f/t the square on half the line bisected, is rqtud, to the square on the sfrai;/ht li'ne made vp of the hatfaud the part produced. Lot tlio stmi,i,'lit lino AB l.o l.is(>oto(l .it P, .'uul pvo- (liicod to Q : then tlio roct. AQ, QB aiui tlio r^q. on PB slifill be to- ^etlior equal to the sq. on PQ. On PQ describe the square pCDQ. t. 46. Join QC. Throui^di B draw BE par' to QD, meeting QC in F. I. 31. Througli F draw LFHG par' to AQ. Through A draw AG par' to QD. Now tlie conipleniont PF ^-= tlie comph'mont FD. i. 43. But tlie fig. PF = the tig. AH ; for they are par"'« on vqual bases and between the same par'^ !• 30. .-. the lig. AH the fig. FD. To each add the lig. PL; tlien tlie tig. AL ^ th.o gnomon PLE. Now the lig. AL^ the rect. AQ, QB, for QB-^QL; .•. the rect. AQ, QB r the gnomon PLE. To each add th,- scj. on PB, that is, the Jig. HE; then the root. AQ, QB with the sq. on PB : tht^ gnomon PLE with the fig. HE 1 the whole tig. PD, which is the squ.aro on PQ. That is, tlu^ rect. AQ, QB and the stj. on PB are together N equal to llie sq. on PQ. q.K.D. BOOK II. riior. 6. 131 coRUESi'ONi)i\(; ai-(;i:l!UAu;ai. formula. This result may Lc writtL-n AQ.QB + PB-^PQ-. Let AB-2a\ and let PQ-h- then AP uutl PB e;u:h = «. AlsoAQ-a + />; aiul QB-?>- «. lleuce we have {a + l))(h- a)-\'(i- = h-, or (h + (I ) {h - (i) — b- - a'-. 8 Definixion. It' a point X is taken in a .straight line AB, or in AB produced, the distances of the point of section from the ex- A ^ trcmities of AB are said to be the segments into which AB is divided at X. . In the former case AB is divided internally, in the latter case externally. B Thus in the annexed Ih^'u-es tlie segments into which AB is divided at X arc the Unes XA and XB. This definition enables us to include I'rops. 5 and (5 in a single Enunciation. //■ (t, t>tr(ii[iht Hue in hhected, and (i/so ilirldi'd {intenudlij or I'.v- tvrnnUij) into tico unequal aciinunitx, the rcctanijle contained hi/ the un- equal neiiments is equal to the dijlerenee of the squares un half the line, and on the line between tJie points of section. K.XKUCISi:. Bhew that the Enunciations of I'rops. 5 and may take the following form : Tlie reetan[ile contained hi/ two straiijld lines is etiiuil to the dij'er- cncc of the squares on half their sum and on half their diil'ercuce. [See Ex., p. I'i'J.] 132 KUCLID « KLKMKNTS. PllOl'OSITIOX 7. TllKOllKM. //■ (I titraiyhl line is Jicidtd into any l/co jmrts, tin' trnm (if the .squares on the ichole Hue and on, one of the parts is equal to tidco the reetainjle contained />// the whole and that jnirt, toijether with the square on the other ■part. A C B H / w / / / G r D 1 F c Let tlic straiylit line AB be divided at C into the two parts AC, CB : then sliidl tlie sum of the sijii. on AB, BC he eiiiud to twice the reet. AB, BC togc.'tlier witl) the scj. on AC. On AB descrilje tht^ scjuare ADEB. I. 16. oni BD. Througli C draw CF par^ to BE, meeting BD in G. i. 'i\. Through G draw HGK par' to AB. Now tlie eompkment AG -^ llie complement GE : i. 13. to each add the tig. CK: then the tig. AK the tig. CE. JJiit the lig. AK-^tlie. rect. AB, BC ; lor BK-BC. .-. tlie two tigs. AK, CE - twice the rect. AB, BC. Jiut the two tigs. AK, CE make up the gnomon AKF and the lig. CK : .-. tiie gnomon AKF with the tig. CK = twice the rect. AB, BC. To each udd the iig. H F," which is the sq. on AC : then the gnomon AKF with the tigs. CK, HF ^ twice the rect. AB, BC with the sq. on AC. Kow the S(iq. on AB, BC the tigs. AE, CK — the gnomon AKF with the tigs. CK. HF -twice the rect. AB, BC Avith the sq. on AC. HOOK 11. ruor. <. 133 or CUUKKSruNDINt! ALliKBHAlOAL FORMULA. The result of tliis propoKitiou may 1)<^ written AB' + BC--2AB. BC 1 AC-. Ul AB_((, ami BC-=/>; then AC =- <( - /a lleuce we have a- + h' - '^ob -!- (a - b)-, (a-hf^tr -'2(ib\b-. Thkuukm. M X G P K / Proposition t^. // a ^iMviht line he dlddrd into an// in:o purh, j'oin'^ times the rectangle contained by the ■whole line and one oj the parts, touether with the square on the other part, is cqaat to the square on the straight line which is nuide up oJ the whole and that part. [As this proposition is of little iuiportuuee we merely t;ive the ti{;mc, and the leading points in FAiclid's prooi.J °LetAB be divided utC. ^ C B D Produce AB to D, uuikiiig BD equal to BC. Ou AD desci'iljo the s(i[iuiie AEFD; :iud complete the construction us lu- dicuted in the ligure. Euclid tlien proves (i) that the rigs. CK, BN, GR, KO are all equal. (ii) that tiie rigs. AG, MP, PL, RF are all cqual^ Hence the eight riguies named aljo\e are four tunes tlie sum of the tigs. AG, CK ; tiiat is, four times the hg. AK ; that is, four times the rect. AB, BC. . , • But the whole rig. AF is made up of these eiglit figures, together with the rig. XH. Nvliich is the scp on AC: hence the scj. on AD four times tlie rect. AB, BC, together with the in{. on AC. Q.E.D. H L F B The accompanying tigure will .suggest a less cumbrous proof, which wo leave as an Exercise to the istudcnt. i I i i 11 i:U EUCMDS KliKMENTS. PhoPOSITION 'J. TllKOHK M. If a strah/Jd 11 m, is divided eqnalhj mid also tuiequaUy, the sum of the sqiuires o)i the tiro uivqnnl ixirts is twice the sum of the squares on, half the line and on the line between the 2)oints of section. / ^ ^^ r / / \ / /- \ / \ Q B Let the straight line AB Ik; divided ('(Hially at P, and unc([tially at Q : th(Mi sliall the siini of the sijiq. on AQ, QB bo twice the HUiii of the s([(|. oil AP, PQ. At P draw PC at rt. aii,t,de.s to AB ; and make PC e(ni;il lo AP oi- PB. Join AC, BC. Tliroui^di Q draw QD i)ar' to PC; and tlirotigh D draw DE })ar' to AB. Join AD, Then since PA = PC, .-. t lie angle PAC--= the angh; PCA. Antl since, in the triangle APC, the aiigh; APC is a rt. Ji'igltS dnistr. .-. the sum of thi; angles PAC, PCA is a rt. angle: i. ,'52. Iience each of tlie a.igles PAC, PCA is lialf a rt, angle. So ' o, each of thi; angles PBC, PCB is h.alf a rt. angle, .'. tlu; whole angle ACB is a rt. an e.vt. angle CED the int. opp. angle CPB, l. 21). .•. the ;uigh^ CED is a rt. angle : and tlie angle ECD is lialf a rt. angle. Pravrd. .•. .'ilso the angle EDC is li;ilf a. rt. anrh; and QB = u - h. 1 leuce we luive (,H.7,)2+(„ .h)■- = 2{,^'^-lr-). 13R KUCLID's Kl.fciMENTS. PUOI'O.SITIOX 10. TllKOUKM. //' It filrahjhl fine is hifU'cti'il and proiluwd to atiij point, till' si'iii, of ilia tn/iKtras on, the ii'lude liiir, IIiuh jn'odnccd, anil on, till', piirt prodnccd, is tioicn the siuu of tin: sijnnres on linlf the Hue bisected and on the line made Uj) of the half and the Ijart 2»'oduccd, c " ^ ^-~-~^- E D Let i\\v si. line AB he l)isf'ct»'cl at P, mikI jirodiucd to Q: then sliall the sum of the .scjij. oii AQ, QB be twice the suiu of the sqq. on AP, PQ. At P draw PC at riiirht anj^les to AB and make PC ecjual to PA or PB. Join AC, BC. I. 11. I. '6. Througli Q draw QD par' to PC, to meet CB produced in D I. 31. and through D draw DE par' to AB, to meet CP produced in E. Join AD. Then since PA -^ PC, Conslr. .'. the angle PAC - the angle PC A. 1.5. And since in the triangle APC, the angle APC is a rt. angle, .•. the sum of the angles PAC, PCA is a rt. angle. i. S'2. Hence each of the angles PAC, PCA is half a rt. angle. So also, each of the angles PBC, PCB is lialf a rt. angle. .". the whole angle ACB is a rt. angle. Again, the ext. angle CPB the int. opp. angle CED : I. 29, .'. the an!"le CED is a rt, anjile : and the angle ECD is lialf a rt. angle. .'. the ansxle EDC is half a rt. anijle. Proved. I. 32. the angle ECD the angle EDC ; EC - ED. I. 6. HUUK II. I'llOl'. 10. L train Diij point, laced, <(H(l cs oil luilf If and the ircd to Q : twice tilt! J. II. I. 0. l>rocluced I. 31. produced Const I'. 1. ^K I rt. angle, e. i.'aj. angle, t. angle. :d : I. 29. I'roi'ed. I. 32. I. G. the angle DQB \\w alt. angle CPB. .-. tin- ande DQB is a I't. angle 1. 2D. Also (he angl(! QBD tlie vert. (.pp. angUi CBP that is, the angle QBD is lialf a rt. angle. .-. the angl<^ QDB is half a it. angle : .•. tiie angle QBD thi^ angle QDB ; .-. QB^-QD. Now the Ml, on AP the stj. on PC ; for AP PC. lUit tlie s(|. on AC - the sum of tlu^ s(i(i. on AP, PC, for the aiig!e APC is a rt. angle. .', the s(i. on AC is twice the stj. on AP. 1. o. 1. 0. (>/^s■ t)\ 1. 1; CD is twice the s(i. on ED, tliat is. sq JSo also, the isq. on twice the sq. on the opp. side PQ. '• ^1- Now the S(iq. on AQ, QB the sqq. on AQ, QD - the sq. on AD, foi- AQD is a it. angle ; i . 1 < • the sum of the sqij. m\ AC, CD, for ACD is a rt. angle ; l. 17. :.-^ twice the S(|. on AP with twice the sq. on PQ. Prooed. 'J'hat is, the sum of the sijij. on AQ, QB is twice the sum of the sqci- on AP, PQ. *^-^'-^- CUUUKSPONDlNa ALGEBUAICAL I'UUMULA. The result of this propositiuii may be written AQ- + BQ- = 2(AP- + PQ-). LlL AB_2(/; and PQ = 6; then AP and PB cauh-*;. Also AQ:=^rt + ^; and BQ-:/y-'/. Hence we have {a^h)'--v{h~aY-^2{a--\-h-). EXERCISK. Shew that the enunciations of Props. <» and 10 may take the following lonn : The sum of the squares on two struhjht Jnw^ h equal to twice the sum of the squares on half their sum and on half their difference. 138 KUCIJD.S Kl.KMENTS. PllOPUSlTlON 11. PllOULKM. To (Ih'idti a fjiven hIi'u'kjIiI. line into two jmrls, hu t/iaf, fhe rcchitKjJe coifalned by (he irJiole and one part may be equal to the square on the other part. Let AB he tlii- given straiglit line. It is ro(iuii(.'(l to divide it into two parts, so that the leotanpjle contained l)y th(; wiiole and one part may bo e(i[ual to the square on the other part. On AB describe tlie square ACDB. i. K!. Bisect AC at E. I. 10. Join EB. Produci! CA to F, making EF e(jual to EB. I. 3. On AF describe the sijuare AFGH. I. Hk Then shall AB be divided at H, so that the rect. AB, BH is equal to the s([. on AH. Produce GH to meet CD in K. Then because CA is bisected at E, and produced to F, .-. the rect. CF, FA with the sq. on AE = the sq. on FE ii. G. = the sq. on EB. Constr. But the sq. on EB th(> sum of the .sqq. on AB, AE, for tlie anyle EAB is a rt. ant-le. i. 47. .-. the rect. CF, FA with the sq. on AE th(! sum of the S(jq. ni) AB, .AE. From these take the sq. on AE : then the rect. CF, FA = the sq. on AB. HOOK II. riioi'. 1 1, 139 l8, Hu thai ft may be ) th .'it the t may bo I. I. 46. 10. AB, I. ;5. I. 40. BH is (1 to F, E II. G. B. Cunstr. I. 47. i\n of tlio I'.wt the n\c\. CF, FA iUo lij,'. FK ; for FA FQ ; niul tho sq. on AB =-- tiu! ti<;. AD. ConHtr. .-. tho ti",'. FK: tlu) li-,'. AD. From those toko, th(! coiimion Hi,'. AK, thon tho i-oniiiiuing iig. FH (lio I'oiiiiiiiiiii;,' i\'ii th'' perp^'u- diniJitr } fit fall on it fi'tna th)', oppoalte unijl'\ and th> acute nni/lf. Let, ABC lie finy tri;m<,'lo luiviiij,' tlio anj^le .:t F-. nii acute /Li,;'.-^ ; Jiiul let AD \)o, tlm ]>( ip. dfawii from A to tlip opp. side. BC" hen sli.'i'i tlu' S(|. on AC be less than flic stiiii of tl sqq. oi AB, BO, l)y twioo tlio i'<'ct, CB, BD. 10 Now fall witliiii tli<> trianirlc ABC, ;is in F\>'. 1, or witliiHit it, as I't. Fi<'. 2. Rpcauso Un Fig. 1. BC is divicU'tl inio two jyaits at D, III lin F both BD is divided into two parts at C, (•(iHrs. the sum of the scjij. on CB, BD twice (lie rect. CB, BD with the .sq, on CD. II. 7 To each add the s(|. on DA. Then the sum of the sqcj. on CB, BD, DA --- twice the rect. CB, BD witli the sum of the sq debcribe a square that shall be cqical to a yiccio recti lineal Jtijnre. m nii^^'lit lino 311 may ha ppositc the 'I'X cniitdiii- !')• Of' tlld^ll' I'lite ntifih', 1/ tirii'c till' ction of till' by tlu lielp •diiiht line. n{^le a per- e rectangle Lse is e'lual gilt angles, sum of tliG jntained by 1(1, p. 101."] gbt angles, tiie squares ed by these 10, p. 101.] Let A l)e tlie given rectiliiiejil figure. 1 1 is I'etiuired to describe a square e(ju;il to A. J)escril)e tlie pur'" BCDE etjnal to the tig. A, and iiaving the angle CBE a I'iglit ang:e. J. 45. Tlien if BC = BE, the tig. BD is a squai'e ; and Avhat was recjuired is done. JUit if not, produce BE to F, making EF equal to ED; f. '». and ^ isect BF at G. i. 10. From centre G, with raduis GF, describe the semicircle BHF: produce DE to meet tlie semicircle at H. Then shall the s(i. on EH ])e etiual to the given fig. A. Join GH. Tlieu because BF is divided equally at G and unecjually iit E, .•. tlie rect. BE, EF with the S(j. on GE the sq. on GF il. 5. = the sq. on GH. But the sq. on GH = the sum of the sqq. on GE, EH ; for the angle HEG is a rt. angle. I. 47. .'. the rect. BE, EF with the sq. on GE = the sum of tlie S(j(j, on GE, EH. From tliese take the sq. on GE : then tlie rect. BE, EF = the sq. on HE. But the rect. BE, EF the fig. BD ; for EF ED; Cunstr. and the fiir. BD the OK II. 145 8. The difference of the squares ou two sides of a triauijle is ennal to twice the rectangle contained by the base and the intercept between the middle 2)oint of the base and' the foot of the perpendicular drawn from the vertical an(jle to the base. Let ABC he a triangle, and let P be the middle point of tho base BC: let AQ be drawn pcrp. to BC. Then shall AB-- AC- = 2BC . PQ. l-'irst, let AQ full within the trianf,Ie. Now AB-'= BQ- I QA-, i. .17. also AC-':-^QC-4 QA-, . • . AB2 - AC- - BQ- - QC- Ax. 3. = (BQ + QC) (BQ - QC) ii. .'5. ^BC.'iPQ Ex. 1, p. I'J'.). = 2BC.PQ. Q.K.D. The case in which AQ falls outside tlie triangle presents no difficulty. 5). The square on anij sliaiiiht line drawn from the vertex of an. isosceles trian EB in tlic direction remote from A; and on AF' describe tlie square AF'G'H' on the side remote from AB. Then AB will be divided exter- nally at H' as required ] ON II. ]'2 AND 13. 22. In a triangle ABC the angles at B and C are acute: if E and F are the feet of per]iendieulars drawn from the o])j)Osite angles to the sides AC, AB, shew that tlie sifuarc on BC is equal to the sum of the rectangles AB, BF and AC, CE. 2'\. ABC is a triangle right-angled at C, and DE is drawn from a point D in AC perpendicular to AB : shew that the rectangle AB, AE is equal to the rectangle AC, AD. tlie tlioorom c on half the the theorem ncMjiially, the to twice the ur times the / SCi'lioil, (Did N ,• t'licw that the i('(;tan{.;]e is e(iual to if X is the ian^'le whose t-aiigled. ial section at •ee times the TTIKOIIKMS AXn KXAMPT.KS O-V TOOK TT. 14; 24. In any trinnfjle the sum of the nqvnres on two sides is equal to twice the square on half the third 'aide tor/ether with twice the square on the median ivhich biscctx the third side. Let ABC be a triangle, and AP the median hisecting the side BC. Then shall AB- + AC' -^ 2 BP- ! 2 AP". Draw AQ perp. to BC. Consider the case in wliich AQ falls within Dip trian^^lo, but docs not coincide with AP. Then of the angles APB, APC, one miii-t ho obdiso, and the other acute: let APB be obtuse. Then in the a APB. AB-^BP- !- AP- }- 2 BP . PQ. ir. 12. Also in the a APC, AC- CP- i-AP- 2CP.PQ. 11. 1:^. ButCP-=BP, .'. CP-rr:BP-; and the rect. BP, PQ:.^the rcct. CP, PQ. Hence adding the above re^^ults AB-" + AC- = 2.BP- + 2.AP-. q.k.ii. Tlie student will have no difficulty in adapting this proof to the cases in which AQ falls without the ti'iangle, or coincides with AP. :lr «. ke EF' equal be the siiuare lividod exter- ate: if E and angles to the le sum of the I drawn from the rectangle 25. The Sinn of the squares on tlie sides of a ixirailclonram is equ'il to the sum of the squares on the diagonals. 20. In any (luadrilateral the squares on the diagonals are togo- thci- equal to twice the sum of the squares on the straiglit lines join- ing the middle points of opposite sides. [See Ex. '.I, p. 97.] 27. If from any point within a rectangle straight lines are drawn to the angular points, the sum of the squares on one pair of the lines drawn to opposite angles is equal to the sum of tlie squares on the other pair. 28. The sum of the squares on the sides of a quadrilateral is greater than the sura of the squares on its diagonals by four times the square on the straight line which joins the middle points of tln' diagonals. 29. O is the middle point of a given straiglit line AB. and from O as centre, any circle is described; if P be any point on its circum- ference, shew that the sum of the squares on AP, BP /,s- constant. WM 148 KHCLTn's KI,KMK\TS. 30. (iivcn tli(> liase of a tiiiUiKle, and tlic sum of tlio .sciuarcs on Ihc sides t'oiniiii^' tlio vcitical ani^lo; lind tho locus ol' tho vcitox. ;>1. ABC is ail isosceles trianj^lo in whicli AB and AC avo equal. AB ■■ produced beyond the base to D, so that BD is equal to AB. Sliew that tho s(iuiiio on CD is equal to tlie s(iuaie on AB together with twice the square on BC, 'iV2. In a right-an^'led triangle the sum of tho squares on the strniglit lines drawn from tho right angle to llie points of tri- seetion ot the h.yiiotenusc is equal to tlvi times the square on the line between the points of trisection. ;n. Three times the sum of iiio f ;r.aroH on the sides of a tri- angU is equal to four times tho sum ui' Jl'e squav.>3 on tli', medians. Hi. ABC is a triangle, and O *h' |)oiut of interjection of its medians : shew that AB- I BC- 1 CA---. ;} (O A- -I 0B-' -l OC-;. :5.j. ABCD is a iiuadrilatoial, and X tho middle point of the utraight iiio joinir,.': I'u' bisections of the diagonals ; with X as centre anil chek is dcscriUid. and P is any ]>oiiit upon tliis circle: shew that PA- -I- PB'-+ PC'- + PD"' is cor.^tiiHt, jeing equal to XA-^-rXB- + XC--f XD-'-i (XP-. 'M. The squares on the diagonals of a fcrapeziiim arc together equal to the sum of the squares on its two obliiiue sides, with twice the rectangle contained by its parallel sides. < PUOBLEMS. o7. Construct a rectangle equal to the difference of two squares. .38. Divide a given straight line into two parts so that the rect- angle contained by them nay be e(|ual to the s(juare described on a given straight line which is less than half the straight line to bo (livided. ;}'.). (liven a square and one ^ide of a rectangle which is equal to the square, lind the other side. 40. Produce a given straight line so that the rectangle contained by the whole lino thus in'uducisl and tlie part ]iio(liifi'd, may be equal to the S(|uare on another given line. 41. Produce a given straight line so that the rect;nig]e contained by the whole line thus produoid and the i(ivn line shall be equal to the square on the part inodiicuil. 4'J, l.»ivide a straight linr AB into two parts at C, such that tho rectangle contained by BC and anotlier line X v.<:\j be equal to tho square ou AC. •A-. 'J,- siiuarcs on ;ci'tox. ) uro equal, uiil to AB. ,B ti'Mther PAKT Tl. ires on tbe nts of tri- laro on the ?s of a tri- medians. jtion of its oint of the X as centre : shew tliat re together with twice vo squares. lat the rect- cvibcd on a t line to bo ich is equal e contained lay l)u equal :le contained be equal to nch that tlio eqiml to tho BOOK. iir. Vnmk 111. (Ic.'ils with the proportios of (Jirclos. 1>KFINITI0NS. 1. A circle is a })lann fiquro bounded by one line, wliii-li is called the circum- ference, and is such that all straiijlit HiK's drawn from a certain point ^vithin the fi. A diameter of a circle is a straiL;lit. line drawn thi()U which joins any two points on the <*ircumfoi'onc('. From these delinitions it may be seen that u chord of a circle, which docs not ]iass throngli the cpntre, divides the circiunl'dence into two nncqual arcs ; of those, the groatfr is called the major arc, and tlie loss the minor arc. Thus tlie major aic is f/ri'iitrr, and the minor arc less than the scniichcumtcrencc. The major and minor arcs, into wliich a cir- cumference is divided by a chord, are said to be conjugate to om^ another. S. Chords of a circlo are said to Ijc equidistant from the <;ontre, when the perppudieulars drawn to thoni fi'oin the centre ai'e equal : and one chord is said to l)e further from the centre than another, when tiie por- pondiculai- drawn to it fioni tiie centre is greater than the iK'rpendicular drawn to the otlier. 9. A secant of a circle is a straight line of indefinite length, whicii cuts the cii-c«nifei'ence in two points. \ / 1 0. A tangent to a circle is a straight line whicli meets the eireuniference, but being produced, does not cut it. Such a linc^ is said to touch the circle at a point; and the point is called tlie point of contact. id to 1)0 t'li joins DKFINITIONS. If a KL'cant, which cuts a circle at the points P and Q, j,'ra(luall,v changes its position in such a way that Prtiuains lixed, the point Q will ultimately uppinach the lixcd point P, until at lenj^'th tluso points may ho nuulo to coincide. Wlicn tlic! stiai^^lit line PQ rcaclios this limiting position, it hccomcs the t(tui 11. Circles are said to touch one another when they meet, Init do not cut one another. \ \ /( When each of the circles which meet is oulaide the other, they are said to touch one another externally, or to have external contact: when one of the circles is vithiii the otiser, they are said to touch one another internally, or to have Internal contact. 12. A segment of a circle is the fi<^ure bounded by a fhord and one of the two arcs into which the chord divides the circumference. / \ The chord of ;i segment is .sometimes called its base. ir)2 KUc'IJDti KI.KMKNTR. Ill 1:5. All angle in a segment is one foniu'd by two stnii,i,'lit lines drawn froin ;iny point in the arc of tin; segment to the extremities of its elioi'd. [It will be shewn in Proposition 21, that all anRlos in tlio pame Hcf^moiit of a circle arc equal.] 14 .\n angle at the circumference of ^ ( iii'lo i.i oiKf t'oiiiied hy stiai^ht lines drawn fi'im a point on the cii-cumference to tin' extremities of an arc: sut-h an an,<,de is said to Stand Upon tlie arc, which it subtends. 1 ."). Similar segments of circU's ai'c those which contain etjnal an^h-s. /-::i: in. A sector of a circle is a iigun* bounded by two radii and the arc inter- cepted betwe.'o them. Sv.MliOLS ' i» AimUKVIATlO.NS. / In addition to the syndxjls and abbreviations given on page 10, we ^ ill use Hu. fojlowin:,-. • /or circle, ''' /'or circumieri nee. ~7\N tlio Hame 'iveii on HOOK t'hoi'. 1. 153 PUOI'OSITION 1. PllOHLKM. Tofnd thu centre of a gli^en circle. I. 10. r. 11. I. 10. Lot ABC Ix' a xivcii cifclc: it is r(Miuiroil to liiid its ccutw. In tlio givou circle dniw .-iny choid AB, iiucl Insect AB at D. From D (h-aw DC at ri«,'ht angles to AB; ami i)ro(luco DC to u.cct the j"" at E aiul C. Bisect EC at F. 'I lien shall F be the centre ot the :•") ABC. First, the centre of the circle must he in EC : for 'f not, let tlu? centre be at a point G without EC. Join AG, DG, BG. Then ill the A^ GDA, GDB, / DA- DB, f.'fltixt)'. Because- find GD is common ; ( and GA -- GB, for by supposition ihey are nwlii; .•. the ^GDA the ^GDB; T. 8. .•. these an.£;les, being adjacent, are rt. angles. But the _ CDB is a rt. angle ; Constr. :. the _ GDB the z. CDB, A''- H. the part equal to the \vliole, wliirh is impu^sible. .•. G is not the centre. So it may be shewn that no point outside EC is the centre ; .•. the centres lies in EL, .-. F, the middle point of the dianut.r EC, must be the centre of the O ABC. Q-^-F- CoROLLAUY. T/ie strai'jla law xh'irh Lisccfs a chord of a rirch at r'n/ht an _ DBE; .". in the A DEB, the side DB, which is opposite the greater angle, is greater than DE Avhich is opposite the le.ss: i. |!», that is to say, DE is less than a radius of the circle ; .". E falls within the circle. >So also any other point between A and B may be shewn to fall within the ciicle. ". AB falls within the circle. Q. K. D. Pkfinition. a part of a cnrved line is said to be concavn to a point when, ()iitts, rotujh two cu pohitt;, (jiiits. n\\ v.Licli 10 circuni- hoids of u jf parallel light lines are equal, shew that ie on the the given HOOK III. I'HOP. 4. PrOPOSITIOX 4. TlIKOHKM. 157 If hi a circle two chords cut one anoflier, irhich. do not both jxiss throvtjh the centre, they cannot both be bisected at their j)oint of i) iter section. Let ABCD be a circle, and AC, BD two clioids Avliich intersect at E, but do not both pass tlirough the centre: then AC and BD shall not Ite both bisected at E. Case T. If one chord ))asses tlirougli the centre, it is a diameter, and the centre is its middle point; .•. it cannot bo bisected by tlie other choi'd, wliich by liypo- thesis does not pass through the centre. Case II. If n(!ither chord passes tlu'ougli tlie centre; then, if possible, let E be the middle point of both ; that is, let AE - EC; and BE = ED. Find F, the centr(! of tlu; circle: iii. 1. Join EF. Then, because FE, which passes through tlie centre, bisects the chord AC, Jh/I'- .'. tlu^ _ FEC is a rt. angk;. in. ."}. And because FE, which ])asses through the centre, bi- sects the el lord BD, Jlyp- .'. the _ FED is a i-t. aiigh;. .'. the 1. FEC =: tlie I. FED, tlic wdiole e. EXERCISES. ON riiorosrnoN' 1. 1. If a parallelogram can be inscribed in a circle, the jjoiut of intersection of its diagonals must be at the centre of the circle. 2. lleclangles are the only parallelograms that can be inscribed in u circle, ON Pkopobition 5. 3. Two circles, which intersect at one point, must also intersect at another. BOOK I If. I'KOl'. G. 159 e the same ,t C: u centre; F autl G. C, llup. iFC, JJyi). Ible. re. il. K. D. the point of circle. be inscribed ,1bo iutersect Phopositiox (5. Theorem. //■ two circles touch om anoth'r hdermdh), thnj cannot iiace the same centre. Lot the two 0^ ABC, DEC toucli at C; one another internally tiien they shall not have the s;nne centre. For, if jmssible, let the two circles have tl and let it \>v called F. Join FC; and from F draw any si. line; to meet tl le same centre : i(^ 0'''"at E and B. Then, because F is the centre of the ^ABC FB=-FC. J^i/J> \nd l)ecause F is the centre of the DEC, l/fj2> FE -. FC FB = FE: tlie whole equal to its part, whicl the two circles liave not the IS nnpossi sanje centre bit Q. E. JL». Note. From Propositions o aud G it is seen that circles, whose . ircumfer^nces liave any point in common, cannot be concentric, unless they comcide entirely. u^cumi,, poin'iTnToSon'' "^'"--^-f— « "^ -"--^ric circles can have no II. E. 11 i!| 160 EUCLIDS KLEMENTS. Proposition' Thkohe.m. If from anij point icltlnn, a circle trhich is not the centre, sfraujht lines are druivn to the circunifennce, the (/rented is that which passes throu(/h the centre ; and the least is that tuhich, xvJiOb produced hackmirds, passes throm/h the centre: and of all other such lines, that which is nearer to the f/reatest is nhcai/s (jreatcr than, one more remote: also two equal straight lines, and onlij tu:o, can, 0", draicn, from the yiraa 'poiut to the circii/nference, one on each side (f the diameter. A. Let ABCD l»o a c\w\e, within wliicli any point F is taken, which is not tlie centre: let FA, FB, FC, FG be drawn to the O'", of wliicli FA passes through E tlie centre, and FB is nearer than FC to FA, and FC nearer than FG : and let FD be the lino whi(;h, when produced backwards, passes through the centre: then of all these st. lines (i) FA shall be the greatest; (ii) FD shall be the least; (iii) FB shall be greater than FC, and FC greater than FG; (iv) also two. and oidy two, e(iual st. lines can be di'awn from F to the j^''. .loin EB, EC, EG. (i) Tlien inthe.'.FEB, the two sides FE, EB.ire together greater than tin; tliird side FB. liut EB EA, being radii of tlie circle; .'. FE, EA arti togetlier greater than FB; that is, FA is greater than FB. 1. -JO. BOOK Iir. I'ROP, UU ! t]t,e centre, 3 greatct't h "(ist is tJiat the centre ; arer to the n 0'! draii;u. )b each side ■j F is taken, 3 dl'UWU to 3, and FB is G : and let irds, passes FC jireater nes fan bo ire to^ctlu'r 1. 'JU. 3 . 8iniilarly FA may be shewn to be greater tlian any otliej- St. Inie drawn from F to tlie o''"; .*. FA is the greatest of all such lines, (ii) In the AEFG, the two sides EF, FG are together greater than EG ; and EG - ED, being radii of tln^ eirclc; • • ^f'/^ '^"'^ togetlier greater than ED. .'0. (iii) Becji.iiKn J Take away the common part EF; then FG is greater than FD. Similarly any other st. line drawji from F to the o™ may be shewn to be greater than FD. .'. FD is the least of all siuh lines. In the A^ BEF, CEF, ,. , , BE-CE, ni. D.'f. I. -because -j and EF is conmion ; (but the L BEF is greater than the l CEF; .'. FB is greater than FC. i. 21. Similarly it may be shewn that FC is greater than FG. (iv) At E in FE make the i. FEH equal to the _ FEG. I. ^r^. Join FH. Then in the A^ GEF, HEF, ( GE-HE, III. y>/: 1. Jjccause -l and EF is comiiion; (also the _ GEF -. the _ HEF; Conf other straight line can he di-awn troni F to the Q^*^ equal to FG. For, if possible, let FK - FG. Then, because FH - FG, Proved. .-. FK : FH, that is a line nearer (., FA, th.> gnvitest, is equal to a line which IS more remott^; wiiich is iuqmssible. Proved. .-.two, and only two, equal st. lines can bo drawn from ''^'^^"^O" g.E.D. 11-2 1^1 ^^1 ^m 162 kitcmd's ki,emknt.s. Proposition 8. Tiieokkm. If from unif point without a circle sfmif/ht lines are clrmni to the circumference, of thoxf vhich fall on the concave cir- cuviference, the yreatcst is that irhlch passes throu;/k the cenfrf : and more remote: lastlji, from the f/iren point there can he dnuot to the rircumference tivo, and only two, equal straight lines, 07ie on each side of the shortest Hue. Let BGD 1h! :i ciicle of which O is tlie centre; and let A he any point outsiile the cireh' : let ABD, AEH, AFG, be St. lines di-awn from A, of wliicli AD passes through C, tho cuiitr<', and AH is nearer than AG to AD : then of St. lines d)-a\vn from A to the concave C '"", (i) AD shall be the greatest, and (ii) AH greater tiian AG : and of st. lines drawn from A to the convex O*^'', (iii) AB shall l)e the least, and (iv) AE less than AF. (v) Also two, and only two, ecpial st. liiu.'s can hv diawii from A to the "'. Join CH, CG, CF, CE. (i) Then in the /:. ACH, the two sides AC, CH are together greater than AH: I- -O- Init CH CD, l)eiiig radii of the circle; .". AC, CD are together greater than AH: that is, AD is greater than AH. Similarly AD may be shewn to be greater than any other tit. line di-awn from A to the concave O*'*'; .". AD is the greatest of all such lines. are dratini ncave cir- row/h fill'' Hiiifereiice, troiKjh thf, he least is itVH to the les, one on \ ; and let H, AFG, b(! ugh C, tho eater tluiu li.-in AF. es can 1)0 C, CH are I. liO. any other (ii) Because J ( (1 iJooK HI. riior. 8. Til the A** HCA, GCA. HC GC, and CA is connnon: 163 iir. J)ef, 1. nit tlie _ HCA is greater than the _ GCA: AH is greater than AG. I. 24. (iii) In the A AEC, the two sides AE, EC are to^etlier gi'eater than AC : ''j ;)q l)utEC=:BC; III. 7>/:"l . . tlie remainder AE is greater tlian tlie remainder AB. Siniihirly any other st. line drawn from A to tlie convcv O'*' may be shewn to be greater than AB; .". AB is the least of all such lines. (iv) In the A AFC, because AE, EC are drawn from the extremities of the base to a point E within the triangle, .". AF, FC are together greater than AE, EC. * i! 21. ButFC = EC, ui.DpfA. .'. the remainder AF is great«-r than the remainder AE. (v) At C, in AC, make the _ ACM e(|ual to the .. ACE. Join AM. Then in the two / /^ ECA, MCA, ( EC-MC, j„. ])^f\ 1. r.ecause -j and CA is common ; ( also the z. ECA -.the ^ MCA ; Consfr. .•.AE=^AM; I. 4. and besides AM, no st. line can be drawn from A to the C'", equal to AE. For, if possible, let AK - AE : then because AM = AE, Proved. AM AK; that is, a 'inc nearer to the shortest line is equal to a hne which is mo. - remote : which is impossible. Proved. .'.two, and uriy two, equal st. lines can be drawn from A to the 0-. o,E.n Wliore aifi Ihe limits of that part of the circumference v.'hicli is concave to the point A? If)1 KUCLin's KI.KMKNTS. Obs. Of tlic followinR proposition Euclid pave two distinct proofs, tht' livHt of wliicli lias tlu; advantage of beiuK lUirrl. ■ l'l!OI'(»SITIOX ;t. TlMX)KKM. [FutST riiOOF.] //' from a p(>!id vnthin a clrch. more^ than two eqvnl Kti'diiiht I'nvx nni. hi' (/niwn lo the fiiriniifcrniri; thd point ■is till', ci'iilri' (>/' i/i'' circb\ Let ABC 1)0 Ji circlo, iuul D a point within it, from whicli inoiH tlian two v^\\Xi\\ st. lines aie drawn to the 0'^ namely DA, DB, DC: then D sli.iU be the centre of the circle. Join AB, BC : Mild bisect AB, BC at E and F respectively. i. 10. Join DE, DF, Then in the A*^ DEA, DEB, f EA - EB, Conxtr. liecause I and DE is connnon; I and DA DB; Ifup- :. the i. DEA- the _ DEB; 1. •"<. .-. these an<,des, being adjacent, ai'e li. angles. Hence ED, which bisects the <'hovd AB :it rt. angles, must pass th.rough the centre. m. 1. Cor. Similarly it may be shewn that FD passes through the centre. ,•. D, which is the only point common to ED and FD, must be the centre. (^.k.U. BOOK 111. rnoi'. !). 165 net proofs, irn eqval Jiitt point •cm whicli •^^ namely ]. 10. Constr. Iff/p. I. 8. les. ifijlos, must HI. I. Cor. irough the D and FD, Proposition 0. Tiieohkm. [Skcond Proof.] //' from a point tcithiit o. cirrln uion' thait, tiro etpud xlnriijlit Ihu's c/' tin: circle.. Lot ABC be a ciicle, and D a point witliin it, fi'oin wliicli more than two equal st. lines an; (lra,\vn to the ( /'', namely DA, DB, DC : then D shall be the centre of the circle. Foi', if not, suppose E to bo tho centre. Join DE, and produce it to meet the C/'" nt F, G. Then because D is a ])()int within the circle, not tho centre, and l)ecause DF passes through the centre E ; ,'. DA, which is nearer to DF, is greater thaii DB, whicli is more remote : Jii. 7. but this is impossible, since by hypothesis, DA, DB, aro equal. .*, E is not the centre of the circle. *.\.nd whei'o\(M' we .suppose the centre E to be, othei'- wise than at D, two at least of the st. lines DA, DB, DC may \ny shi'wn to be unequal, which is conti'ary to hypo- thesis. .•. D is the centre of the • ABC. tj.K.D. * NoTK. For example, if the centre E wore supposed to be within tlie an-() pointH. If possihlf, let DABC, EABC hv two circles, euttii;,Lf one iiiiothei- at more than two points, namely at A, B, C. Find ' .cntre of the O DABC, iii. I. II HA, HB, HC. 'J'ht'n since tlu' centre of tlic^ •> DABC, .■. H HB, HC are all etiual. iii. Jh-J'. 1. .Viid hec.-mse H is a point within the ro EABC, from which more than two e(iiial st. lines, namely HA, HB, HC arc drawn to tlte "', .'. H is th(^ centre of the ■ EABC : iir. 0. that is to say, the two circl(\s ha\e a conniK.'i centre H ; Imt this is impossible, since they cut one another, iii. ."». Therefore one circle cannot cut another in more than two points. Q.E.]). Note. This })roof is iinperfcct, because it as.suincs that the centre ')f the circle DABC nm.st fall itliin the circle EABC; wliorcas it may be conceived to fall either without the circle EABC, or on its circumference. Heiicu to make the proof complete, two ndditional cases are required- •I ♦ MICROCOPY RESOLUTION TEST CHART (ANSI and ISO TEST CHART No. 2) 1.0 I.I 45 la 2.8 1 3.2 m II 4.0 1.4 ||i£ 2.2 2.0 1.8 ^ APPLIED IfVHGE Inc '653 East Main Street Rochester, New York 14609 USA (716) 482 -0300 -Phone (716) 288 - 5989 -Fat ^-fc—a—reac:;-^. 168 euclid's elements. Proposition 11. Theorem. If two circles touch one another internaUn, the sfrai(/ht line \vhich joins their centref<, heAwj produced, shall pass throv'jh the 2)oinf <>/ contact. Let ABC and ADE l)e two circles wliicli touch ouo anotlier internally at A ; let F be the centre of the © ABC, and G the centre of the ADE: then .shall FG produced pass through A. If not, let it pass otherwise, as FGEH. Join FA, GA. Then in the z\ FGA, the two sides FG, GA are together greater than FA : i- 20. hut FA :^ FH, being radii of the O ABC : !/ijp. .'. FG, GA are together greater than FH. Take away the counnon part FG ; then GA is greater tlian GH. Hut GA- GE, behig radii of the ADE : 7f'i/p. .'. GE is greater than GH, the part gi-eater than the whole ; which is inipossil)le. .". FG, when produced, must pa.ss through A. Q.K.I). EXERCISES. 1. If tlie tllKtance between the ceiitroa of two circle.s is equal to the difference of their radii, then the ch'clcs must meet in one point, but in no other; that is, they must touch one anothtn-. 2. 7/" /((•'-' I'irrh'!^ wlione coitrcs are A and B touch one another internaliij, and a xtraiiilit line be drawn throuiih their point of contact, cuttiwj the circumferences at P andQi\ shew tliat the radii AP and BQ arc parallel. sfnrlf/Jd 'mil juiss inch ouo e ABC, ! together " I. 20. Hyp. Jfyp. Q.K.D. is e(iual to I ono point, )//(,' anuthi'r 5 of contact, AP and BQ ' . ]500K iiT. rnop. T2. Proposition- \2. Tiieork.m. 100 //' hco circles touch one another externallij, the stravjht cine tcliich joins their centres shall pass throuyh the 2^oint of contact. Let ABC iuul ADE be two circles wliicli toiicli ono jiuotlier exteniiilly at A; let F be the centre of the ('•) ABC, and G tlie centre of the ADE : then shall FG pass through A, If not, let FG pass otherwise, as FHKG. Join FA, GA. Then in tlie A FAG, the two sides FA, GA are together greater than FG : I. 20. but FA --r. FH, being radii of the © ABC ; Jlyp. and GA ^= GK, being radii of the ADE ; Jhjp. :. FH and GK are together gi-eater than FG; which is impossible. .'. FG nuist pass through A. Q F.D. KXKUrTSKS. 1. yihil the Jocuit of the c<'iitn'!< of all circles vliich touch 'i fiireii circle at a (jiven jtoint. 2. Find the hem of the centres of all circles of (jiven radim^, which touch a (jiven circle. 3. If the distance between the centres of two circles is equal to the sum of their radii, then the circles meet in one point, but in no other; that is, they touch one another. 4. If two circles whoae centrcti are A and B touch one another externalhj, and a xtrai'iiit line he drawn throunh their point of contact cutting the circumferences at P and Q.; shew that the radii AP and BGl lire parallel. 170 Euclid's elements. PhOI'OSITIOX L'k TlIEOKKM. Troo circles cannot touch oiic another at more than one 2Joinf, tvhether internally or externally. Fig. 2 D G If possible, let ABC, EDF l)e two circles which touch one another at more than one point, namely at B and D. Join BD; and draw GF, bisecting BD at rt. angles, i. 10, 11. Then, v/hether tlie circles touch one another internal Iv, as in Fig. 1, or externally as in Fig. 2, bec-ause B and D are on the O''"** of both circles, .*. BD is a chord of both circles : .'. the centres of both circles lie in GF, whicji bisects BD at it. angles. m. 1. Cor. Hence GF which joins the centres n-ast pass through a point of contact; m. H^ and 12. which is impossible, .since B and D are without GF. .". two circles cannot touch one another at more than one point. Q.E.D. Note. It must be observed that the proof ' - ^iven applies, by virtue of Propositions 11 and 12, to hoth the ■ • fif,'urfcs: we h.ve tlierefore omitted the separate discussion of Fig. .,, wliich finds a place in most editions based on Simson's text. BOOK III. PROP. 13. 171 ii one Q.E.D. EXERCISES ON PROPOSITIONS 1-13. 1. Describe a circle to pass throuj^h two given points and have its centre on a given straight line. When is this impossible ? 2. All circles which pass through a fixed point, and have their centres on a given straight line, pass also through a second fixed point. S. Describe a circle of given radius to touch a given circle at a given point. How many solutions will there l)c? When will there be only one solution? 4. From a given point as centre describe a circle to touch a given circln. How many solutions will there be? 5. Describe a circle to pass through a given point, and touch a given circle at a given point. [See Ex. 1, p. 109 and Ex. 5, p. 150.] When is this impossible? 0. Describe a circle of given radius to touch two given circles. [See Ex. 2, p. 169.] How many solutions will there be ? 7. Two i>arallel chords of a circle are six inches and eight inches in length respectively, and the perpendicular distance between them is one incli : find the radius. 8. If two circles touch one another externally, the straight lines, whicii join the extremities of parallel diutueters towards opposite parts, must pass through the point of oontac c. 9. Find the greatest and least straight lines which have one extremity on each of two given circles, which do not intersect. 10. In any segment of a circle, of all straight lines drawn at right angles to the chord and intercepted between the chord and the arc, the greatest is that which passes through the middle point of the chord ; and of others that which is nearer the greatest is greater than one more rem^ te. 11. If from any point on the circumference of a circle straight lines be drawn to the circumference, the greatest is that which passes through the cei»tre ; and of others, that which is nearer to the greatest is greater than (me more remote ; and from this point there can be drawn to the circumference two, and only two, eiiual straight lines. 172 1 I ,i i: KL'CLID.S KLKMKNTS. rilOPOSITIOX 14, TjIEORliJI. Equal chords in a circ/e on GC ; .". FA = GC; and doubles of these equals are ecuial ; that is, AB^ CD. (^.K.l), %/ V)f. I. -i7. EXEKCISES. 1. Find the locuit of the middle points of equal rZ/orJs of a circle. 2. If two chords of a cu'cle cut one another, and make equal anf,'lcs with the straight line which joins their point of intersection to the centre, they are equal. .'). If tico equal chords of a circle intersect, sJicn' that the segments of the one are equal respcctirehj to the seqinents of the other. i. In a given circle draw a chord which shall be ecjual to one given .straight line (not greater than the diameter) and parallel to another. 5. PQl is a fixed chord in a circle, and AB is any diameter : sIkjW that the sum or dilierenc(! of the perpeiuliculars let fall from A and B on PQ is constant, that is, the samoi for all positions of AB. 174 euci.id's elements. I'UOPOSITION IT). TlIKORKM. The (liameler Is the (jrealest rhord in a circJe ; awl of others, that which is nearer (u the centre is y renter than one more remote: conversely, the (jreater chord is nearer /<> thr cmtre t/um the less. Lot ABCD hi- .1 circle, of which AD is a diameter, and E the centre ; and let BC and FG he any two chords, whose l)t!rp. distances from the centre are EH and EK : then (i) AD sliall be ^n-eater than BC : (ii) if EH is less than EK, BC shall he greater than FG : (iii) if BC is greater than FG, EH shall be less than EK. (i) Join EB, EC. Th(!n in the .:. BEC, the two sides BE, EC are togethei iireater than BC : I. -20. III. JJef. 1. but BE AE, and EC - ED ; .•. AE and ED together are greater than BC ; that is, AD is greater than BC. Similarly AD may be shewn to be greater than any other chord, not a diameter. (ii) Let EH be less than EK; then BC shall be greater than FG. Join EF. Since EH, passing through the centre, is perp. to the chord BC, .*. EH bisects BC ; in. 3. HOOK iir. PROP. 15. 176 Is 'jrcfUer •iih'i: t/iitn that is, BC is double of HB. For a similar reason FG is doublo of KF. Now EB^EF, in. /)>'/. 1. .". tlio sq. on EB the sq. on EF. But the sq. on EB the scjq. on EH, HB; for the _ EHB is a rt. alible j j. 47. also the sq. on EF tiie sq(|. on EK, KF; for the L EKF is a rt. angle. .". the .se/!10. BOOK III. PROP. \C). 177 Bocond (by irst in to bo ur. I'nooF.] ficifiif'ter of circle: this point ;re, and AB be drawn , sliall cut gle ; 7////). 1. 17. 3; I. 19. nrele; y be shewn prodoced, in.i)t/10. (ii) Draw EH peq). t(» BF. i. 12. Then ill the *.EHB, because the _ E'!B is a rt. nM;^h\ .'. the _ EBH is less than a rt. antfle ; i. I 7. .'. EB is i,'ivat(M- than EH; i. 1*1. that is, EH is less than a radius of the ciirif: .'. H, a point in BF, is within the circle; .". BF must cut the circle. v. K. it. Propositiox l(i. TiiEOUKAr. [Eitlid's Proof.] The atmiyht line drawn at rvjht anglea to a diameter of (t circle at one of its extre/nities, in a tayuji'ut to the circle: and tio olhi'i' strai'jht line can be drawn throuyh this point so as not to cut the circle. Let ABC he a circle, of which D is tlu^ centre^ and AB a diameter; ](>t AE Ix^ dr.iwu at it. an<,'les to BA, at its extremity A: (i) then shall AE l)e a tangent to the circle. For, if not, let AE cut the circle at C. Join DC. Then in the A DAC, because DA ■--- DC, iir. /)>/'. 1, .'. the _ DAC the _ DCA. But th<} _ DAC i.^ a rt. angle; /f,/p, .". the _ DCA is a rt. angle; that is, two angles of the A DAC are together ecjual to two rt. angles; which is impossible. i, 17. Hence AE meets the circle at A, but being produced, does not cut it ; that is, A£ is a tangent to the circle, in. J)e/. 10. 123 178 Kt'CMp's KI,F;>fKNTS. (ii) vXIso tlirou^li A in> otiM'r stniiylit lims l»iit AE ran ho drawn so as not to cut the circl»». For, if possihle, let AF he anotlier st. lino drawn tl. ougli A so as not to cut the circle. From D draw DG porp. to AF; I. 12. and let DG meet the Q"" at H. TIh'H in tlio '.DAG, hocauso the _ DGA is a rf. an!.fl<', .'. Ilio _ DAG is loss tlian a rt. aii,i(lo; i. 17. .■. DA is «j;roattM- than DG. I. 11). But DA DH, III. !)>'/. 1. .". DH is <,'r('at(i' tlian DG, the part greater than tlie whole, which is impossible. .". no st, line can be drawn from the point A, so as not to cut the circle, except AE. Corolla RiKs. (i) A ttinycu.t ttmch's a circle at one point only. (ii) Thn'P. can he hut one tanr/ent to a circh>. at a (jii'en foint. I»ut AE t'Hii A II i\. OUgll I. 12. 1.17. I. 11). m. />/: 1. possible, so as not to rcle at (me at a (jivpn HOOK 111. I'uor. 17. ritOl'OSITluN 17. ruuni.K.M. 7o t/nnr a tiiinjvid to a cii'clts from a yiwn. jminl t'ltluf uH, or irifhout f/ie circiiui/ernru'f. Fig. 1 Fig. 2 III. 1. Lnt BCD Ui the «,'iveii circle, aiul A tln^ '/wen point: it is reciuired to draw from A a tangent to tlie (• CDB. Ca.sk J. If the j,dveu point A i.s on the /''. Find E, tl»e centre of the circle. Join EA. At A draw AK at rt. anj^des to EA. i. H. Then AK Ijeing perp. to a diameter at one of its extrmnitiesj is a tangent to the circle. m i(;] Casp: it. If tlie given point A is without the o"'. Find E, the centre of the circle; m. 1. and join AE, cutting the BCD at D. Fi'om centre E, with radius EA, d(!scribe the (• AFG. At D, draw GDF at rt. angles to EA, cutting the OAFG at F and G. j I j Join EF, EG, cutting the ©BCD at B .and C. Join AB, AC. Then both AB and AC shall be tangents to the ©CDB. For in the A» AtB, FED, { AE=-: FE, being radii of the ©GAF; Because /and EB - ED, being radii o' tlie © BDC ; (and the included angle AEF is common'; .*. the .L ABE = the ^ FDE, j 4 11 I 180 EUCLID'S ELEMENTS. .. :t But the L. FDE is a rt. .iiitflc, Conslr. .'. tho L ABE is ;i rt. angle ; hence AB, being drawn at rt. angles to a diameter at one of its extremities, is a t.uigent to tlie 0BCD. in. IG. Similarly it may be shewn that AC is a tangent, q. e.f. CoROLLAiiY. If two tangents are draicn to a circle fro'ni an external 'point, then (i) they are equal; (ii) they snhtoid f'qual angles at the centre : (iii) they make equal angles tvith the straight line which joins the given point to the centre. For, in the above figure, Since ED is perp. to FG, a chord of the ,:; FAG, .•.DF = DG. 111. o. Then in the A« DEF, DEG, DE is connnon to lioth, Because - and EF= EG ; m. JJef. 1. and DF : DG ; Proved. .-. the L DEF ^- the _ DEG. i. t^. Again in the A' AEB, AEC, j AE is connnon to both, Because - and EB = EC, [and the _ AEB :^ the _ AEC: Proved. :. AB - AC : I. 4. and the ^ EAB^ the _ EAC. q.e.d. KoTE. If the given point A is within the circle, no solution isi possible. Hence wc see that this problem admits of ticti solutions, one solu- tion, or no solution, according as the given point A is without, on, or tinthin the cirnnmferencc of a circle. For a simpler method of drawing a tangent to a circle from a given point, see page 202. Conslr. eter at one III. IG. :ent. Q. E. F. ', circle from 'hey subtend anyles tc'dh 3 centre. FAG, HI. 6. III. Def. 1. Proved. I. ^. Proved. I. 4. Q.E.D. no solution is ions, one solu- vithout, oil, or e from a given BOOK IIT. PROP. 18. PkOPGSITIOX 18. TlIEOUEM, 181 The slraiyht line drawn from the centre of a circle to the point of contact of a tanyent is perpendicular to the tangent. G E Let ABC 1)0 Ji ciirlo, of wliich F i.s the centre; and let tlie st. line DE touch tlie circh; at C : then shall FC he perp. to DE. For, if not, suppose FG to he perp. to DE, r. \'l. and let it meet the 0™;it B. Then in the A FCG, l)ecause the L FGC is a rt. angle, Jfi/j). .'. the L. FCG is less than a rt. angle : I. 17. .'. the L. FGC is greater than tlie l. FCG ; .". FC is greater than FG : i. lU. hut FC -^ FB ; .'. FB greater than FG, the part greater than tlie whole, which is inipossihle. .'. FC cannot be otherwise than per}), to DE : that is, FC is perp. to DE. t^K.i). EXERCISES. 1. Draw a tan^'cnt to a circle (i) parallel to, (ii) at right aiiglus to a given straight line. 2. TdiiiiL'iits draicn to a circle from tlie c.rtrcinitics of n. diaincdn' are jmrallcl. 8. Circles which touch one another intcrnaUij or e.iii'riiaUij hare a common tangent at their point of contact. 4. In two concentric circla ant/ chord of the outer circle irJiidi toucbex the inner, i.t hixcrfed at the -point of contact. r». In tico concentric circles, alt chords of the (niter circle which toucli the inner, are equal. 182 Euclid's elements. Proposition' 19. Theorem. y/ic St might line drawn perpendicular to a tangent to a circle from the point of contact passes through the centre. D C E Let ABC be n circle, and DE a tangent to it at tiie j)oint C ; and let CA be drawn perp. to DE : then shall CA pass thrcugh the centre. For if not, suppose tlie centre to be outside CA, as at F. Join CF. Then because DE is a tangent to the circh', and FC is tlrawn from the centre F to the point of contact, .'. the _ FCE is a rt. angle. iii. 18. But the ;_ ACE is a rt. angle ; ^fgi'- .'. the _ FCE the _ ACE ; the part equal to the whole, which is impossible. .*. the centre cannot be otherwise than in CA; that is, CA passes through the centre. Q.E.D. EXERCISES OX THE TANGENT. Propositions 10, 17, l.S, 19. 1. The centre of any circle which touches two intersectiuff straiifht lines muat lie on the bisector of the an{/le hettveen them. 2. AB and AC are two tan-rentH to a circle wliose ceutre is O; shew that AG bisects the chord of contact BC at right angles. BOOK III. I'Ror. 19. 183 anyeat to a centre. le point C ; , as at F. ', and FC HI. 18. Hup. sible. DA; Q.E.D. iwj straight !eutre is O; ties. 3. If two circles are concntric all tangents drawn from points on the circumference of the outer to the inner circle are tMjual. 4. The diameter of a circle bisects all chords which are parallel to the tangent at cither extremity. 5. Find the locim of the centres of all circles which touch a tjlvcn straight line at a given point. 6. Find *he locus of the centres of all circles which touch each of two w "r straight lines. 7. Fuh rhe lociiK of the centres of all circles which touch each of two interstctuuj strnifiht lines of unlimited leni/th. 8. Describe a circle of given radius to touch two given straight lines. y. Through a given point, within or without a circle, draw a chord e(iual to a given straiglit line. In order that the problem may be possible, between what limits must the given line lie, wheu the given point is (i) without the circle, (ii) within it? 10. Two parallel tangents to a circle intercept on any third tan- gent a segment which subtends a right angle at the centre. 11. //( ani/ quadrilateral circumscrilx-d about a circle, the sum of one pair of opposite sides is equal to the sum of the other pair. 12. Any parallelogram which can be circumscribed about a circle, must be equilateral. 13. If a quadrilateral be described about a circle, the angles sub- tended at the centre by any two opposite sides are together equal to two right angles. 14. AB is any chord of a circle, AC the diameter through A, and AD the perpendicular on the tangent at B: shew that AB bisects the angle DAC. 15. Find the locus of the extremities of tangents of fixed length drawn to a given circle. 16. In the diameter of a circle produced, determine a point such that the tangent drawn from it shall be of given length. 17. In the diameter of a circle produced, determine a point such that the two tangents dj-awn from it may contain a given angle. 18. Describe a circle that shall pass through a given point, and touch a given straight line at a given point. [See page 183. Ex. 5.] 19. Describe a circle of given radius, having its centre on a given straight line, and touching another given straight line. 20. Describe a circle that shall have a given radius, and touch a given circle and a given straight line. How many such circles can be drawn ? 184 Euclid's elkmknts. PilOPOSITIOX 20. TlIKOKEM. The angle at the centre of a circle is double of an. ayiyle at the circunfereace, standiwj on tin', saim arc. Fig. 2 I. n. Let ABC l»e a circle, of wliicli E is tlie centre; and let BEC be an angle at tlie centre, and BAG an angle at the O^'", standing on the same arc BC : then shall the _ BEC be double of the L BAC. Join AE, and produce it to F. Cask I, When tlie centre E is within the angle BAC. Then in the A EAB, because EA := EB, .'. the _ EAB the _ EBA ; .'. the sum of the _ ^ EAB, EBA twice the i EAB. But the ext. _ BEF:-:the sum of the l. " EAB, EBA- i. o'l. .'. the _ BEF-^ twice the l EAB. Similarly the _ FEC- twice the _ EAC. .". the sum of the ^^ BEF, FEC^twice the sum of the ^3 EAB, EAC; that is, the :_ BEC =- twice the l BAC. Case II. When the centre E is without the _ BAC. As before, it may be shewn that the _ FEB = twice the l FAB • also the s. FEC ^^ twice the i. FAC; .-. the difference of the _ « FEC, FEB -:= twice the difference of the L « FAC, FAB : that is, the l BEC - tv.ice the _ BAC. y.E.D. n)i anyle BOOK III. rnop. 21. Note. If the arc BFC, on which the angles stand, is greater than a semi-circumference, it is clear that the angle BEC at the centre will be reflex: but it may still be shewn as, in Case I., that the reflex t BEC is double of the L BAC at the G'®, standing on the same arc BFC. 185 ; ;iud let t the O '", le BAC. I. T). . EAB. BA; I. o-l. BAC. le^ FAB; le^ FAC; lifi'creiico Vi.E.D. Proposition 21. Theorem. Amjles in the saine segment of a circle are equal. A L(;t ABCD bo ;i circle, and let BAD, BED be angles in tlie same segment BAED: then sliall the _ BAD - tlie l. BED. Find F, the centre of the circle. in. 1. Case 1. When the segment BAED i.s gi'eater than a semicircle. Join BF, DF. Then the L. BFD at the centre ~ twice the i. BAD at \\w O"*-", standing on the same arc BD: ill. 20. and similarly the l BFD ^- twice the L. BED. iii. 20. .*. the L BAD ^- the L BED. Case IT. When the si^gment BAED is jiot gie.iter th.an a semicircle. 186 EUCLID'S KLEMENTa. Join AF, juxd produce it to meet the C" at C. Join EC. Then since AEDC is a seniicirclo; .•. the se,i,niieut BAEC is i,'r('ater than a semicircle: .'. the Z BAG = tlie ._ BEC, in this so<,Mnent, Cnstlier = two rt. angles. ' j. ;}•_>. .". the _^ ADC, ABC together r- two I't. angles. Similarly it may he shewn that th(^ _ >^ BAD, BCD together =.- two rt. angles. Q. K. D. EXERCISES. 1. If a circle can be described about a i)aralIelograra, tlie parallelogram must be rectangnlar. 2. ABC is an isosceles triangle, and XY is drawn parallel to llio base BC: shew that the four points B, C, X, Y lie on a circle. 3. If one side of a quadrilateral iii-^crihed in a circle js produced, the exterior (uujh in. enual to the opposite inltrior unale of the quadri- lateral, K 4 BOOK III. PROP. 22. 189 Proposition 22. [Alternativo Proof.] liOt ABCD bo a quadrilatoral inscribod in tho <^ ABC : then shall tho / "ADC, ABC tu^;cth(n-^. two rt. anj^'lcs. Join FA, FC. Then tho L AFC at tho centre ^ twiec tlie L ADC at tho C'', standing on tlio name aro ABC. m. -JO. Also tlie reflex anf^lc AFC at die centre = twice the z ABC at the C", standing' on the name arc ADC. iii. '20. Hence the; /« ADC, ABC are together halt' the sum of the z AFC and tlie reflex an^le AFC ; but these make up four rt. angles: 1. 1."). Cor. 2. .•. the L ** ADC, ABC together = two rt. angles. g.K.n. Dkfimtiox. Four or more points througli which which HOOK III. I'ltdi". 24. Puoi'osrnox IM. Thkokkm. li>l Simi/ar s,';,ni^„fH of n,Wf'.^ on eqttnl ,'/innl.^ ,n'n ,'nHtd to OHO. o.nothi'i: Let AEB find CFD be .sin.ilur s(.^rn,„ut.s on c.u.il chords AB, CD: ^ ^ then shall the segment ABE - the segment CDF. For if the segment ABE l.e .•ippli.-d to the segment CDF so that A tails on C, and AB falls along CD; then since AB CD, .". B must coincide with D. .-. the segment AEB must coincide with the segment CFD • tor it not, on the same chord and on the same side of it tliere would be two simihir segments of circles, not co- mculuig with one another: which is impossible. m 2\\ the segment AEB -^ the segment CFD. Q. E. D. KXERCISKS. 1. Of two segments standing on tlie same clionl, the greater segment contains the smaller angle. b'emei 2 A segment of a circle stands on a chord AB, and P is anv point on the same side of AB as the segment: shew that the angle APBs greater or less than the angle in the segment, according as P is within or without the segment. ^ „ f'v ^\!^' -^ ""'f /^"' *"''''"'' P""''^'' "f f''e sides of a Uiunnle, nvd X V.S- the Joot o; the perpendicular let fall from one vertex on the oppimte sale :>du'w that the four poiuts P. Q, R, X are concuclic. [See page 9b, Ex. 2: also page 100, Ex. 2.] 4. Use the preceding exercise to shew that the middle points of the . /J irliick tin; [/inn tire in a pitrt. L-'t ABC lit' Jill Jirc of fi ciiilc: it, is rtM|uir('(l in '('SiTilic the wiiolc " nf wliidi tin- afc ABC is !i }»uft. Ill tlui >/i\('\\ ;irs, .". tlic c'('iitr(f of tli(^ ciivU' lies in DF. ill. !. ('"r Again, lu't-ausf EF bisects tlu; chord BC at it. an,i,'lt'K, .". tlie centre of tiie circle lies in EF. ill. 1. Cor. .-. the centra' of the ciivh' is F, the only point coiiiiiioii to DF, EF. l{enc(i the ( )"' of a circle descrihed from centre F, with radius FA, is that of which the gWcn arc is a part, v- 1'. F- * NoTK. Euclid gave this proposition a somewhat differoiit form, as follows: A scyment of a circle bcliKj (!ntr(> F, with )art. ^^ i;. F. different form, chrli' (if which 1 HOOK llf. VHov, 2«. VJ.i I'llofosiTlON ♦Jd. 'J'j,K(,itKM. i-ot ABC, DEFl.r,..,ua)ciivI,.samll..t^'' - ;in(I GG - HF, (•'11(1 (he _ BGC th<> _ EHF- • ". BC^EF. A-aiii, hivause tlH' _ BAC the _ EDF //.> . the si^.uH'ut BAG is similar to the se-ii,eu;, EDF , 1 ,- IIL O' f' ].') •111(1 t.icy ai'e ou o(iual clumU BC, EF; J/i/p. I. I. 7 ••• the .segment BAG : : tiie se^^u.cut EDF ,j, 04 5iit the whole OABC .. the whole r., def- ' l;l IllllHr u/^,,. .,,.,. i. 1-,,,^ ,1 . . > •• 1 he I'diiaiiiiu"- so'fiMf'iif- Ricr^ +l.„ • • ' r. ••.^. llf lit BKC the l■elllalmn,^'sem, lent "aiv BKC tlK'aiv ELF. ELF. I [I'ov an Altornativo Proof and E.\ •xcrciscs see ij]). 197, !!)«.] 13-2 194 EUCLID'S ELEMENTS. Proposition 27. Theorem. In equal circles the angles, whether at th^^f^l'J '^'' cWcui,^fereiices, ivhich stand on equal arcs, shall he equal. Let ABC, DEF be equal circles, and let the arc BC = tlie arc EF : then shall the _ BGC - the _ EHF, at the centres; and also the _ BAG = the _ EDF, at the 0"^ If the _« BGG, EHF are not equal, one must he the greater. If possible, let the _ BGC be the greater. At G, in BG, make the ^ BGK equal to the _ EHF. i. -J. Tiien because in the equal ©' ABC, DEF, the ' BGK = the _ EHF, at the centres; Constr. :. the arc BK-the arc EF. ni- -o. But the arc BC - the arc EF, J^I/P- • the arc BK:=the arc BC, a part equal to the whole, which is impossible. .•. the _ BGC is not unequal to the _ EHF ; that is, the _ BGC - the _ EHF. A,ul since the _ BAG at the O^ is half the _ BGC at the 11 1« ^ \J» centre, and likewise the _ EDF is half the _ EHF, . tbo _ BAC - the ^ EDF. Q. E. D. [For Exercises see pp. 197, 108.] .1 I3UOK: III. i-Hop. 28. 195 centres or the U be equal. Pkoi'u«itio.v 28. Tiieouk.m. Ta equal circle, the arc, which are cut ol}' b., ,nacd '^<^^;/' ill, J ""-» e centres ; eO'''^ e must be the eater. 3 _EHF. I. 23. DEF, tres ; Constr. III. 26. lll/p. iipossible. B _EHF; le _ BGC at the III. 20. ! _EHF, F. Q.E.D. Let ABC, DEF be two equal circles nuc let the chord BC = the chord EF- ti-Hsh;d tlH.nwMorarcBAC.theiuajorarc'EDF- .tud the nuuor a,,, bgc = the minor arc EHF. Find K a.ul L tlu; centres of the 3^ ABC, DEF • ,„ and join BK, KC, EL, LF. Then because the ^ ABC, DEF are equal, . . their radii are equal. Hence in the A" BKC, ELF, ,. ( BK = EL, l>e('ause -j KC = LF, (and BC = Ef'; •'• the _ BKC = the _ ELF : • '. tile arc BGC ^ the are EHF ; and these are the minor ai-cs.' Jh,t the whole O- ABGC = tiie whole Q- DEHF- . . the remaining arc BAC = the remaining arc EDF and these are the niujor aixs. q.^.j,. I. N. III. 2G. I [For Exercises see p^j. l'J7, l'J8. j 196 Euclid's elements. Thopositiox "29. Theokkm. hi, rqnnl clrch's lite c/ionh; vhich cut <>/ eqind HOPosiTioNs LH), 27. 1. //■ liro chiinls of a circli' arc jxinilli'l, tln'ij Intercept eqitdl (ires. 2. Tlio straight linos, whicli join tlie (■xtrcniitics of two VAinal arcs of !i circle fowards the same parts, arc parallol. ;>. Ill II rirele, or in eijudi rirele at the eentrcK are cuiial. !tijr>i are equal if tlieir (//(///<'.< (il (I I'cs, sh((Il EF. s. F, 111. J III. '2t (■(|u;ii circles; •ejisou, J'ruvcd. I. 4. m, E. D. 'ccpt cqiiitl iircK. OS of two o(iiial K.xtHci.sK.s ox raoi's. 28 2!). 197 i>. If tiro chonh iutrmpct irifJiin n n;,'„i > 4i ''••'«n'a;:,:,:::;::;!cJ:';T;,^^^^^ ;•'«•'•■. ;'"■-/ «.™. „„ „„„„ llinj cut „,r. IlKcamiJcrnce Inj III,. ,l/jl,.,:„a of Ihe „„.» ■•»-«»,.»« »,. ., 4,^,;i;;;:,:t;;;r ^";,:/;--;;; ^;^ «- m,. '•ne ..f tlie circles: shew that v ion ^ l \, "^ t'Hcnniforcncc „f .1.0 tn„.,.,„ XYZ i„ ,on„, „t tlK^o^^f .t Iri^.a^^S" ""«'" "'■ OX I'Kdl'OSITION.s •>S -TO c(iual. ^ '^*'^^' (") to\\ards opi^osite parts, arc st..i"ht '^^^xk^'^^'7''^'':^ of two c,„al circles two to the chord QY. "" '''''''' "'''^'^ t^'^ chord PX is equal the .traiKlit linos which join^^^,,^.J u irtr'^f^^^l'^ '^'''' '^'^'' are equal. •* »^xtunuties towards the same i)arts "ir^M m™.^Q i^S^rionStwI. ■'?;' °^ '"■";""'"«" '^ °"^ tlml BP-BQ. iinimuM Ijy tlu- circumferences: sliew i" .:;>;;;\i?;;-= b;;;%';;;:;;^- ;;;s-;?='-".* ^bc, / /'■ //;(■//• (iii'jli'i 198 KUCLUJ'S KLEMENTS. Note. Wu luive given Euclid's cleinonstrations of Propositions ?.(j, 27, 28, 29 ;_ but it sliould bo noticoil that all these propoaitious also admit of direct proof by the method of ttiq)er2)onition. To illustrate this method wo will ai)i)ly it to Tropositiou 2(i. Pkoposition L'»). [Alternative Proof.] In ciiual circlcx, tin' arcs wliick suhtend eqiidl ditijlcs, tclicthcr at till' coitrcs or circumferences, shall be equal. Let ABC, DEF bo equal circles, and let the / « BGC. EHF at the cent/es be equal, and consequently the z » BAG, EDF at the o''"-'" e-nient ABC is !,n'<'ater, and the segment is ADC less than a seniicifcle: then (i) tlu> any-le in tin' semicircle BAC shall he a rt. aiigh- (ii) the angle in the segment ABC shall be less than a I't. angl(; ; (iii) the angle in \\\v segm A ABC, becauso tlin two _ ^ ABC BAC -i.c togetlun- less than two rt. an-Ics; ' j j!;. I'rnrv,/. e'liK'nt ABC, is .'ind of these, the i BAC is a rt. an. KXKHCISES. J. AvircJe described on the Jnjpotrints,- of a rl<'l,t;unilrd triaini/,- "■^ duunctrr, parses thnmih the opposite .n,„ul,,r poh'i. ' '' ^tvithMinT^.'.'"'/'^ i-IKht-an^'lod triangles is describcl upon a given points ly-potonuse: Imd the locu. of the oppo.iite angului 3. A straiglit ml of given Icngtii slide. l)etween f,vo ^tvih'hi nSiL'its'. "' "«'" ""«'"» "' "- »■"">-- «- t,;c',;,:„f;,"t' a''»m.«;.;'''iS'i;s/;;/igf-'- *««■ "« "- .>.™.t« p, b. q . r,. A circle is described on ou(> of the e.Ruil sides of an isoscele- inn^ icfl,?^^ f 'If ''^''r ' ^'''':' ',"^"''""' '■"'^^'»^^' t'^" ^^ianiet..- of the imiei IS e(nial to the radius „f the outer. Sliew that -inv ,.hr>v,1 ,,f the out..r cucle drawn f^om the point of coAS, Hseci^^ y ] ' circumierence of the inner circle. i^'teinci o} tiic . 7. Circles described on any two sides of a trian-1,. as di'uueters nitersect on the third side, or the third side produced ^''""^^"^ or wiii^ur;;:^ drcS.:!^r" "'^" "^^ ^^^^'^ ^'-^^^ ^^ -*'-'> ••»' !) Describe a s-iuare c-iual to the difference of two given squares. 10. Through one of the points of intersectirni of two circles draw a chord of one circle which .sJkiU be bisected by the other. 11 On a given straight line as base a system of Ofiuilateral four dSS:" " '""""'= ''"' "" ^"^"« ^^^ "- inte;sisi;n'^ S; 202 Euclid's ki.kmknts. NoTK 1. The extension of Proposition 'JO to straiijht unci vejlex iln^'l(.'S fhinislu.'s a simple iilternative proof of the tirst theorem contained in Proposition lil, viz. The (iiKjlc ill (I scinicirch' /s (f ri'ilit s; i.' ~U','. .'. the _^ DBE, DBF t(),!4etli(T Uw. _« BCD, BAD: and of tlicsc Oti. II ijli'i'H stniKiht Hill' to (fi'HCt'ihe a aetjui' itt n <' /'(•/'' ic/i'k'/i, .s/iiilf I'onJaln _ C. i. '2:». I'roiii A draw AE at rt. angles to AD. I. 11. Bisect AB at F; I. 10, jind fi-oni F draw FG at rt. angles to AB, cutting.!; AE at G. Join GB. Then in tlu^ .^ AFG, BFG. ( AF BF, CiHislr. r.ecause -: and FG is coninioii, (and tli(^ _ AFG^=tlio _ BFG, lieing rt. angles; .•. GA GB : T. I. .'. the circle desci'ilied from ceiiti'(> G, with I'jidius GA, will pass through B. Describe this circ](!, and call it ABH: then the segment AHB shall contain an angle equal to 0. Because AD is drawn at rt. angles to the radius GA fioni its extremity A, .". AD is a tangent to the circle; iir. 10. and from A, its point of contact, a choid AB is drawn; .". t!ie :_ BAD::=tlie angle in the alt. segment AHB. ill. '.VI. lUit the .1 BAD =- the 1 C : Comtr. .'. tlie .angle in the .segment .AHB — the C. .'. AHB is tlie segment required. <,>. k.f. 20(; la'Cl.ins l.I.K.MKNTS. No IK. Til the |)iirticiiliir casi' wlicn tlie ({ivon unglo C U a rt. an' tlie s(■^,'lll(■llt ic(|iiiiv(l will lie tlio H<'iiii<'ircl(! (It'scril)t'(l on tlio jjivcii Ht. Hiiu AB; lot' tho. uuahi in u Hciniciielo in a it. aii^l"'. iii. :n. -^s B KXKHCISKS. [Tho following exorcises ilopond on tho corollary to rroitosition "21 ^,'ivt•^ on pa^'o iHj, nuint-ly Till' liiciiHi)/ the vi'rticfs of iriituiih'^ which Ktuiiil on the sdiiic hasf mid hurt' a (fivi'ii rcrtiml iniiili', m thr (ire of the xeijmcnt xtditdiiiif on this hase, miil rnntdiiiiiKj an umjie cqiiiil to iltc ijivcn (itiiile. Kx<'rcises 1 and 2 afford ^ood illustrations of the nictliod of find- ing required points hy the Intersect inn of Loci. See page 117.] . 1. Dencril/c n ti-itimjle on a tiireii huac, h.arinij u ^ivn vertical iiiiiile, II 11(1 liiiriiiii its vertex mi ii ijiren straiijiit line. 'J. Ciiiistriict 11 triiniiiic, huviinj ijicca the base, the vertical amjle (iiiil (i) one otiier side. (ii) the altitude. (iii) the leii'itli of the inedinn which hisect/i the hiise. (iv) the point at which the jwrpendicular from the vertex meets the base. ii. Constrnct a triangle having ijiven the base, the vertical anole, and the point at which the base is cat bij the bisector of the vertical amjle. [Let AB be the base, X tlie f,'iven point in it, and K the given angle. On AB dcseribe a segment of a circle containing an angle eipial to K; coin])lote the c™ by drawing the arc APB. Bisect the arc APB at P: join PX. and i)rodiiec it to meet the C'-^ at C. Then ABC shall be the required triangle.] 4. Construct a t' 'i/le hariufj ijivcn the base, the vertical angle, and the sum of the rei,. ninii sides. [Let AB be the given base, K the given angle, and H the given line equal to th(! sum of the sides. On AB describe a segment containing an angle equal to K, also another segment containing an angle cciual to half the z K. From centre A, with radius H, descirihe a circle cutting the last drawn segment at X and Y. .Join AX (or AY) cuttin<' the first segment at C. Then ABC shall be the required triangle.] /'•, Construct a triangle liaving given the base, the vertical angle, and the dilTereu!.-e of the remaining sides. I Proposition 'J I ))i (III- utmic ham' cut xtdiKlittji un Hill'. fui'thoil of Hiid- iij,'e 117.] . ijir'-u viTtic'il 1 I'ertical autjle xixe, ram the vertex vertiraf unfile, of the rertii'dl 1(1 K the f,'iven ninj^ an iin^le Bisect the arc C. Tlicn ABC vertical angle, \ the given line loiit containin-^ an angle ec|Uiil !S(!ribe a circle 'or AY) cutting Bd triangle] vertical angle, HOOK iir. I'ltoi'. 34. VUDPO.SITION .'51. Tu<»HliK.M. 20" From, a t/irr)!, circ/n tu nit (>[)' n, niyiwii/ ir/iic/t shall coiihVH' (in aufff'' rfputl, to <« yiven oikjIc. Let ABC lie llu! ifivcii cii-rln, jiiul D tlio •;iv«'ii aii.i,'Ii': it is r«'(iiiiii'(l to cut dll" from the • ABC u .sc^niciit Mhitli sliiill coiitiiiii an uiighi «M[Uiil to D. 'Yiikv any jxiiiit B on {\n\ "', and at B di-aw the tangent EBF. jii. 17. At B, in FB, iMak(i the _ FBC ((jual to tlic „ D. I. '-'.'». i'liL'ii the .segment BAC .shall contain an anghj ecjual to D. r.ccau.se EF is a tangent to tlie circle, and fj-oni B, its 1)oint of contact, a chord BC is drawn, .". the _ FBC the angh; in tlie alternate .segment BAC. III. '.VI. lint the _ FBC = the ^ D; Constr. .'. th(! angle in the segnuuit BAC ~ th(^ _ D. Jlence fi'om the given (:, ABC a segment BAC has been ■cut ott', containing an angle equal to D. q.v.. v. EXEijcisns. 1 . The chord of a given .segment of a circle is produced to a llxci' lioint: on this straight line .so produced draw a segment of a circle similar to the given segment. 2. _ Through a given point without a circle draw a straight line that will cut off a segment capable of containing an angle equal to a .givcu angle. .- t iJ U, F U 208 kuclid's klemexts. Pkopositiox .•}."). Theorem. IJ f):-(> eliards of a circle cut one nnother, the recianyJe cnntidui-(l In/ i},,' sajnteids of one shall he equal to ihe recl- anijle contained by ihe sefjments of the other. ACBD, cut (iiic jinothur Let AB, CD, two chords of tli(! at E: then sIkiU the rect. AE, EB the rect. CE, ED. Fiiul F tlio centre of the GACB: iii. 1. From F draw FG, FH pcn-p. ivspecti\-ely to AB, CD. i. vi. Join FA, FE, FD. 'Hicii Itec'uise FG is drawn from tlie centre F perp. to AB, .". AB is hisected at G. m. 3. For a similar reason CD is bisected at H. Again, because AB is divided equally .-it G, and unecpiully ;it E, .'. the rect. AE, EB with the .S(i. on EG ^ the S([. on AG. ll. :>. To each of these eipuds add the sq. on GF; then the rect. AE, EB with tlu; S(i(|. on EG, GF tli(! sum of tiie .sq(i. on AG, GF. But the sqcp on EG, GF the s(}. on FE; j. 47. and the sq([. on AG, GF the scj. on AF; for the angles at G are rt. angles. .'. the lect. AE, EB with the sq. on FE th<^ .si]. on AF. Similarly it may be shewn that the rect. CE, ED with the s([. on FE .the S(i. on FD. But the sq. on AF = the K(|. on FD; for AF^^^ FD. ,'. the rect. AE, EB with the s(}. on FE - the rect. CE, ED with the sq. on FE. From these equals take the sq. on FE: then the I'ect. AE, EB^the rect. CE, ED. q. E. D. 1300K HI. I'liur. 35. •2m CoKOLl.AKY. //' tliroiKjIt (I. jlxnd point %oillu)r a ci/r/r aitij niunhir vj clxmh are dnivm, tka rcctjdiyli's cutUained Ixj tJieii' sc(jiaiuds arc idi i-qtud. NoTK. The folluwini,' special cases uf tliis jjiopositiDn deservo notice. ji) when tlio given chords botli pass tlirough tlic centre: (ii; wlien one chord passes through the centre, and cuts thu otlier at right angles : (iii) wlien one chord passes through the centre, and cuts the otlicr ohH(iuely. In i^ach of these cases the genoial proof iei|uires some inoditlca- tion, which may be left as an exercise to the student. ]:xi;i!(:i.si;s. I. Two str(ii(/ht linen AB, CD iiilerscrl at E, sn Unit llo' nrtdiiijli' AE, EB /s fiiudl, t(i tilt' rcvtainile CE, ED: ulicic Hint the four points A, B, C, D are conci/clic. '2. The rectangle contained by the segments of any diord drawn tlirough a given point witliia a circle is eijual to the square on half the shortest chord which nuiy be drawn through that point. .'5. ABC is a triangle right-angled at C; and from C a jjcrpcn- dicuiar CD is diawn to the hy))otenuse : shew tluit the s(iuare on CD is e(iual to the rectangle AD, DB. ■1. ABC is a triangle; and AP, BQ the i)erpendiculars droppi-d from A and B on the opposite sides, intersect at O: shew that' tiie rectangle AG, OP is equal to the rectangle BO, OQ. 5. Two circles intersect at A and B. and tlu'ough any point in AB their common chord two chords are drawn, one in each circle; shew that their four extremities are concyclic. (\. A and B are two points within a circle such that the rectangle contained by the segments of any chord drawn through A is equal to tin; rectiiugle ctmtained by tlie segments of any chord through B: shew that A and B arc equidistant from tlie centre. 7. //■ tlironiih E, a iwint without a circle, two ficcmita EAB, ECD or,' (Innrit: shew that the rectawile EA, EB is ((iiuil to the reetaioile EC, ED. [Proceed as in III. fi;";, using II. (>.] 8. Through A, a point of intersection of two circles, two straight lines CAE. DAF are drawn, each passing through a centre and teiini- nated l)y tlie circumferences: shew that the rentangbj CA. AE is ecuul to the rectangle DA, AF. ^ I 14-2 210 Krci.lDS KI.KMKNTS. PHOPOSITION .")(;. THKOKKM. If fmia u)iy point ivif/ionf a circh; a tamjeHf a, id n swaut be draioi, (Jieii the v'danfjle contahwd h;/ f.Jw. ichoJe. .secant and the part of it withuat the circle shall be equ(d to the square on the tanyent. Let ABC ))e a ciivlc: and fmm D a ixtiiit Avitliout it, let tIi(M(^ 1)0 drawn the secant DCA, and the tangent DB. then the lect. DA, DC shall be equal to the scj. eijuals tak(^ tli(> s([. un EB: then the rect. DA, DC ~ th(^ sq. on DB. t^K.j). NoTi:. This proof may eiisily 1m? adapted to the case when tlio secant passes through tho ccutio of the circle. BUOIC III. TUOl'. '.W. 211 rn(/enf tnid a hi/ till', tell oh; II he equal to •itliout it, let it DB: iCJ. (Jll DB. Ill . 1. I. 12. lire, is perp. IH. n. ?d to D J on FD. 11. G. EF: ^ tll(3 SflfJ. on s a rt. angle; s a rt. angle; )!■ EBD i: a ill. 18. . on EB, BD. aso when tho CoROLLAK^'. //' from a given point ttithouf ct. DA, DC the S(|. on DB: tlieu siiall DB he a tangent to the circle. From D draw DE to touch the oABC: iii.lT. let E he the point of contai-t. Find the centre F, and join FB, FD, FE. iji. 1. Then since DCA is a secant, and DE a tan,<,'ont to th(> cin-h', .'. the rect. DA, DC tiie S(|. on DE, iii. WC). r.ut, by hypothesis, the nnt. DA. DC the S(|. on DB; .". tlu^ S([. on DE the sc]. on DB, ." DE = DB. Hence in the .*." DBF, DEF. [ DB = DE, l'.ecaus(> -' and BF - EF; [ and DF is conunon; .'. the _ DBF =^-- the _ DEF. lUit DEF is ;i rt. angle ; .'. DBF is also a rt. angle; and since BF is a radius, .*. DB touches the ©ABC at the point B J'rnri'd. III. D'f I. T. S. III. IS. Q. K. D. NOTE OX TllK MKTIIOI) OK LIMITS AS AI'l'LIKU TO TANGKNCY. Euclid (Icdnrs ii tani^'cnt to a circln ns n .-ilrdijilit line which virctx the c ire II )ii/i' re )!(■<• , hut iiciuij produce d, docs not cut it: and from tliis ddinitiou he deduces tlie fundanientul theoioiu tliat a tanj^'ciit is per- pendicular to the radius drawn to the point of contact. Prop. 1(>. But this result may also be established by the Method of Limits, which rej^ards the tan<,'ent as tiic ultiiuiitc 2>o>iitiou of a xecant when H)i tiro ^)o/»^s• of interKcetiou irith tlw. cireinufereiiee nre hroutilit into roin- eiiienee [See Note on page I.jIJ: and it may be shewn that every tlieorcm relatinj,' to the tanj,'eut may be derived from some nioie t,'i:neral i)roposition relatinj,' to the seeant, by considering the ultimate case when the two points of intersection coincide. 1. To 2»'ore hij the Metliod of Limitii that a taixjoit to n circle is at riiiltt (ini/len to the radius drawn to tlie point cf coiitnct. Let ABD be a circle, whose centn; i-i C; and PABQ a secant cutting tiit; ■ "• in A and B ; and let P'AQ' be the limiting position of PQ when the jioiut B is i)r()uglit into coincidence with A: then shall CA be perp. to P'Q'. Bisect AB at E and join CE: then CE is perp. to PQ. in. .'5. Now let the secant PABQ change its position in such a way liiat while the ])oint A remains llxcd, the point B eou- tiiiually approaches A, and ultimately coincides witli it ; tlu'U. howcrcr iieiir B approaches to A, the st. line CE is always ])erp. to PQ, since it joins the centre to the middle point of tlu; chord AB. But in the limiting position, when B coincides with A, and the secant PQ becomes the tangent P'Q', it is clear that the point E will also coincide with A; and the ijerpemlicular CE becomes the radius CA. liencG CA is perp. to the tangent P'Q' at its point of contact A. y, K. 1). NoTi:. It follows from Proposition 2 that a straif/ht Hue cannot cut tlie circumference of a circle at more than two points. Now when the two i^oints in which a secant cuts a circle move towards coinci- dence, the secant ultimately becomes a tangent to the circle: we infer therefore that a tangent cannot meet a circle other\vi.-.e than at its point of contact. Thus lAiclid's definition of a tangent may bo deduced from that given by the Method of Limits. i-SI 214 kuclid's KLKMKNTS. 2. liij thix Mi'thod ]>n)j)i)sitioii ;}2 win/ h,- drrireil as a .yin-inl cafe from Proposition 21. For l<'t A and B be two points on the C" of the ABC; and let BCA, BPA be any two an^^'lfs in the segment BCPA : then the / BPA .the z BCA. m. -j]. Produce PA to Q. Now let the point P continually approaeli the tixed point A, and ultimately coincide with it; then, lioircvcr lu-ar P iikdi approaeli to A thez BPQ::::thez BCA. III. 21. But in the limiting position wluii P coincides with A, and the secant PAQ hecoiuos the tangent AQ' It is clear that BP will coincide wiUi BA ' and the /. BPQ ])econics the /. BAQ' ' Hence the / BAQ'_- tlie i BCA, in the alternate segment. o. ,:. ,.. Q' Q The contact of circles may be treated in a .similar manner by adoptmg the following definition. ^ Definition If one or other of two intersecting circles alters its position in such a way that the two points of intersection continually approach one another, and ultimately coincide ; in the limitin-^ posi- tion they are said to touch one another, and the point in which the two points of intersection ultimately coincide is called the point of KXA.VPLKS ON LI.MIT.S. 1. Deduce Proposition l;i from the Corollary of Proposition 1 ant: Propo.sition 8, '■ ' 2. Deduce Propositions 11 and 12 from Ex. 1, page l;"i6. .'5. Deduce Proposition (5 from Proposition '>. 4. Deduce Proposition i;j from Proposition 10. r>. Shew that a straight line cuts a circle in two different points T^vo coincident points, or not at all, according as its distance from flie centre is less than, e(pial to, or greater than a radius. niitv of the diameter? " " •' 9. From Ex. 1, ])age 100, deduce the corresponding property of a triangle inscribed in a circle. bi i^ .ju.a TUKOUEM.S AND i;:L.UirLL.S ON BuuR 111. 215 t xitccuil Cdie THE(JREMS AND EXA^SIPLKS OX liOOK III. I. ox TIIK CKNTRK AND CHORDS OK A OIHOLK. Sec rr()])ositi(>ns 1, 3, 14, 1">, 2"). 1. .1// circles which piiax tliroiHili a jlrt'd point, (iml liar,- their (•('litres on (I (jiren xtnii(iht line, pans ulxo tiinnnih a aeeoiid fixed point. Let AB be the given st. lino, and P the given point. From P draw PR perp. to AB ; and produce PR to P', making RP' e(inal to PR. Tlion all circles which pass through P, aiul liave llieir centres on AB, si'.all pass also through P'. For let C be the centre of uni/ one of these circles. Join CP, CP'. Theninthe A'CRP, CRP' j' CR is common, I'.ecause 3 andRP-RP', Constr. I and the Z CRP-the / CRP', being xL angles; .-. CP-CP'; I- >• .-. the circle whose centre is C, and whicli ]iasses throngli P, nuist pass also through P'. But C is the centre of (un/ circle of the system ; .-. all circles, which pass through P, and have their centres in AB, pass also through P'. ^- '•■ ^'• 2. Describe a circle that shall pass through three (jiven points not in the same straight line. 2 If; la'ci.iit's i:t.K.Mi:\'is. H. Di'scril,.; ii ciivlo tliat slmll imss tluDUKii two -iv.'ii points ami iiavi) Its centre in a Kiven stiui^'lit Ynw. When is tlii.s iuipu.sHiblo? i. .l)escnl)(; a circle of f^iv.-ii radius to pass tlirougli two L'iveii ]ioiiits. When IS this iiiipossiWo? p. ABC is an isosceles trian^'le; anase, or the base proauced, at X and Y.' t»Ii(_'\\ 11 lilt D A C Y. <; If two circU.s which intersect aic cut h.y a strai^-ht line paiallel totlieconnnon diord, shew that, the parts of it iuferc.i.te.l l)etwoen the circumferences arc iMjual. 7 U two ..irclos cut one anotlier. any iw.j sliai-lu lines drawn ihrouKh a point ot section inakin- equal an-lcs with the common .•honl_, and terminated by the circumferences, are c.iual. WW V> |i. l-)('». I 1 • L -'• • »-» ♦ l„ ''^,; 1^^" *''•' .'■"■!.'^''' .'■''' ""*; ■'"^'f'"'^'- "'■ '--'ll •-fnii-ht lines drawn through a point ot section and terminated by the circumferences, tho greatest IS that wlucli is parallel to the line joining tho centres. ',). Two circles, Avhose centres are C and D, intersect at A, B- and throu^di A a stia.^dit hne PAQ is drawn terminated by the cucum erences: >t PC QD intersect at X, shew tliat the angle PXQ IS equal to the angle CAD. " ^ 10. Through a point of section of two circles which cut one bPect'el f7 '' "^T"^' ^T ^^'"^''^"^^^-^ I'y "'- circumferences and bisected at tlie point ot section. 11- AB is a fixed diameter of a circle, whose centre is C; and rem P, any point on th.circunderence, PQ is drawn j.erpcndicular to AB; sh.'w that the bisector of t'le angle CPQ always intersects the circle in one or other of two fixed points. 12 Circles are described on the sides of a (luadrilateral as diameters: shew that the common chord of any two consecutive circles IS parallel to the common chcnd of the other two FFx '( p. 1)7.] ■ ■■ ' i;3. Two e(iual circles touch one another externallv, and tliroueh the point of contact two chords are drawn, one in clich circle at right angles to each other: shew that the straight Ihie ioinin" their otiier extremities is equal to the diameter of either circle. 14. Straight lines arc drawn from a given external point to the cireumference ot a circle : find the locus of their middle points. [IjX. ii, I). \U.j 'I'HKi>i!i:ms a.-d kn.\.mi'i.i:s on iiouk m. '2V '11 puints and upossiblo? b'h two {,'ivcii c A, as Cf litre, I, at X and Y. Rtmi;;lit lino t iitfcrccptcd linos ilniwii tllu troiiimou !il. [Kx. I'i, lines drawn fc'itincos, tho oiitroH. ect at A, B; iitod by tin; ! angle 'PXQ licli cut one it'renecs and •e is C; and urpendicular iitorsects tlio Irilatoral as consocutivu .vo. [\']x. ;», uid through :h circle, at oining tlieir loint to the Idle iJoints. )posite sides uf AB, any segments lit II. ON THK 'I'AN(;i;NT AM) TIIIC CONIACI" t)l' »II!CM;.S. !^t!0 i'r()!)o^iti(.iis II, I-', I's 17, 1^, l'.». 1. All e(iual chords placed in a given eirclii touch a llxed coticen- ti ie ciii'lc. 2. If I'roiu an external jioint two tangents are drawn to a circle, the angle contained by them is double the angle contained by the chord ol' contact and the diameter drawn through one of the points of contact. ;{. Two circles touch one another externally, and through the )-)int of contact a straight line is drawn lerndiiated by the clrciini- ierence.^: shew that the tangents at its extrenuties are i>aiallel. ■1. Two circles intersect, and through (Uie jioint of section any straight line is drawn terminated by the ciicumferenccs: shew that tlui angle between the tangents at its extremities is equal to the angle between the tangents at the point of section. .'). Show that two parallel tangents to a circle intercept on any third tangrnt a segment which subtends a right angle at the centre. 0. Two tangents are drawn to a given circle fiom a fixed ext(.'rnal point A, and any third tangent cuts them jjroduced at P and Q: shew that Pd subtends a constant angle at the centre of the circle. 7. In aiiij ijudih-llateral cirntimcrihed about a circle, tlie .sum of one imir of o/'ponite sides is equal to the sum of the other imir. H. //■ ///(,' i<\nn of one puir of opposite sides of a quadrilateral is equal to' the sum of the other pair, shew that a circle maij be in^criheil in tliejiiiure. [Bisect two adjacent angles of the llgure, and so describe a circle to toucli three of its sides. Then prove indirecdly by nieanu of the last exercise that this circle nuist also touch the fourth side.] '.). Two circles touch one another internally: fhew that of all chords of the outer circle which touch the inner, the greatest is that which is perpendicular to the straight line joining the centres. 10. Ju any triangle, if a circle is described from the middle point of one side as" centre and with a radius equal to half the sum of the other two sides, it will touch the circles desciibed on these sides as diameters. 11. Through a given point, draw a straight line to cut a circle, so that the part intercepted by the circuiuference may hi; equal to a given straight line. Jn order that the problem may be jjossihle, between what limits must the given line lie, when the yiveu point is (i) without the circle, (ii) within it '{ I 218 KUCLin's KLKMKXTS. .sl.ovv that tl>.. tan«(.nts to tlu-.n ut tho points wli.r. tliry cut a « ■ u j.|i.ull..] stra>,],t lin. all toud. a fixod circlo, whose contri isthe '^^WeW „,^J^ V *''•', "i"-''*^" ,*""*'^* "•"" 'i'"'*-'"''- intoinally, an.l any third crclul.e,l..sonh,.,lt.,nH,inKlH,th; tlwn thn sn.n of tho distancoR o Ji'cortant? "^ "'■*■'" '""" ^'"^ ^"^"^'"^ '"■ ^'^^' ^^^" ^'iven circle; ^,.„^"^^•^''''"}/''^^'"'"^"'' ''"'"^^ ^•"•'' ^'"i<^ tlio pairs ..f tan<'onts drawn in.m tlion. to a -ivcn cin-le contain a constant anj^dc. .rlvo,?;;..''?"'^ ''' '7'"^ "";'' ^''''*' ^''" tHiiKcnts drawn from it to two fmpossible-'' '""•' '"'" " ^''" ""■''" "^'"'"''* ^''"-'- ^^''»''" ^« ^'"'^ K). If three cirtd.'s touch one another two nnti two: prove that Tin: Common Twokxts to Two C nicM:s. 17. 'J'o (Irair a n iiiiniiin tuihinit to tiro rirrli First, if tlio given circles are external t intersect >) one anotlier. or if tl lev Let A he tho c«;.itre of tl ^'router circle, and B tl of the less. 10 le centre 1- roin A, with radius e(iual to tlie dill™ of the radii of tl «iven circles, descrihe a circle: and from B draw BC to touch the last drawn circle. Join AC, and produce it to meet the j^'ieater of the f,'iven circles at D. TlirouKh B draw the radius BE direction. par' to AD. and in the sanio then DE shall i Jo in I' ir since AC M' a coiinii.in tangent to the t he ( lift'" hetwcen AD and BE. wo n'ivcn ciieles. and CD DE CD BE: is jiar' to BE; But since BC is a t is equal and imr' to CB. angent to the circle at C. hence eacli of tl the I ACB is a rt. anL'l le aiigl(>s at D es at U ana t is a rt. angL. •■• DE is a tangent to both circles.. Count r. Coiistr. I. m. III. 18. I. 2!). ' cut a f,'iv<'ii 1 in the piven 111 any third distances of ^'iv(•^ circles >f tanj^cntM )iii it to two Vlieji is this ; prove tiiat ict an; Con- or if they 1 tlie sanio cs. < 'oiistr. ('(iiistr. I. 38. iir. IS. I. 2!>. THKOIIKMS AM) KXAMl'I.KS o\ llooK III. Ill) J I follows from liyjiothesis that the jjoiut B is outside tiie circlo UHcd in the construetion : .-. two tanf,'ents siurh as BC may always he drawn to it from B ; lien. a to it from B. Hence tico more connnon tar ,'ents may he drawn to the j,'iven circles : these will he found to pass l)etwcen the j^iven circles, and art: called the transverse common tangents. Tims, in general, four common tangents may he drawn to two given circles. The student should investigate for himself the mnnherof conuuon tangents %vhich may la; drawn in the following special cases, noting in each case where the general construction fails, or is modified : — (i) When the given circles intersect : (ii) When the given circles have external contact : (iii) When the given circles have internal contact: (iv) When one of the given circles is wholly within the other. 18. Drair tlic diirct connnon fmiijciits to tiro equal circles. lit. If the two direct, or the two transverse, common tangents aro drawn to two cirwles, the parts of the tangents intercepted be- tween the points of contact are equal. 20. If four crramon tangents are drawn to two circles external to one another; shew that the two direct, and also the two transverse, tangents intersect on the straight line which joins the centres of the circles. 21. Two given circles have external contact at A, and a direct common tangent is drawn to touch them at P and Q: shew that PQ subtends a riglit angle at the point A. 22. Two circles have external contact at A, and a direct conimou tangent is drawn to touch them at P and Q : shew that a circle described on PQ as diameter is touched at A l)y the straight line which joins the centres of the circles». 'itl 220 i:rcr,it>s i;M;Mi;\rs. M _.{. 1 wo ciiclfs wliDsc ei'iitu's iiii! C und C liavo external conttict III A. an< a .liicct coinni(.ii tan>,'.'iit is drawn to tom-h tlicm at P and Q: sliow that the bisectors of tlio aii^l-s PCA, QCA mr.-t at ri^'ht aiiKlrs in PQ. And if R {a tli.,. poiut of inUMsectio,. of tho buseators, kIiow thut RA is also a cotimion lan-ent to the ciielo-;. 24. Two circles liave external contact at A, and n direct common tun^^.nt IS drawn to t- u,'!, tiuMu at P and Q: shew that tho 8(iuar« on PQ IS ciual to tlie rcctau-;lt; contained by the diameters of tho circles. . '-'.->. Draw a tan-ent to a -iven circle, so that tho part of it nterceptcd h,v anotiier «iven drclc may l>c equal to h ^dven strai-ht hno. When IK this imi)ossibk>.' " 'JC. Draw a secant to two f,Mven circk's. so tliat th.> parts of it intercepted by tiie circnniferencos may be equal to two j^iven straight J'nom.Kiih (IN Tamikxcv. The lollowinj,' nxeivises an- M.lved l,v thr Methwd of Jnter- seetK!!! oi l.oci, exiilaiiKMl on pjim; J |7. The student should l.eyiu hy making hinLself iUmiliar with till) loHowing loci. ^^^^0) T/w locu.^ of thr rn,tn:< of ,,■/>..•/.'.• which pax, throu,,h lira ^jircn (ii) The /,Hw/.s' o/- the cclrcs of circles which touch a <,ivcn Htmi,,ht line ut a tiireit point. ' (ill) The hens of the rcntrcx of circles which touch student sh.niM investigate the iini and relations among tlie data, in order tluit the irrobleni mav jKiSSlhlo. ^ its be 27. Describe a circle to touch three given straight lines. L.iv.':;^sf JSJ' r' "^ ';''"'''•*" ^"''- ^'"'""^'^ ^ ^'^^^" l'"i"t and touch a given stiaight line at a given point. 29. Describe a circle to puss through a .dv<-n pniut, and to-ch i given circle at a given point. ^ ' '' B'^WBtor. j; -m. £. Et_ ■m.imi.Ms AM) i;\A.\iiM,i;s u.\ ikmik iir. 221 riial contact tlicm lit P yA uiiM't at tioii of tlio cilL'loS. cct conunon ; tho 8i|uan) OtflH of tllij luiit of it III .stiui^,'lit piuts of it oil HtiiUKht I mI' Intcr- iiiliar with 7/ tiri) ijimi U'ti ntniijiht 'en circle nt '•cii Hlrniijltt ivi'u circle, ticu (fircit th(! liiuitH la Uiay be s. 1(1 touch !l id tortch il 80. Dcseriljo ft circU' of yiven riuUin to pu s lhi'oii;,'li n kIvi'IJ lioiiit, aiul touch u giveu HtmiKlit Hint. ;;!. describe ii ch'clo of given uuVma to touch two given ciivlts. .'<2. DiMcrilii' a circle of -jivcn nulius to tcjucli two given htniight lines, .'i;{. l)t'S(iil)c !v ciiclc of given nulius to Inuch u given circle juul ii given straight line. :M. Describe two circles of i^iven radii to to'ich i.ne unotlier and u given Ktrtiight line, on the same side of it. ;$.'». If a circle touches ii given circle and a given straight line, -^hcw that tile points of contact and an extremity of the diameter of tiie given circle at jight angles to the given line are' coUinear. H(). To tlcxvriln' a circle to touch a ;iii>'ii circle, ,iii(l nl.-n li> hmch a llircn -itroifilit line at a i/iveit jxiint. Let DEB bo the given circle, PQ the given st, line, and A the given point in it : it is rtMjuired to describe a circle to touch the • DEB, and also to toucii PQ at A. At A draw AF iierj). to PQ : i. 1 ! . then the centre of the required circle must lie in AF. m. I'.i. Find C, tl'.o centre of the DEB, III. 1. .lid draw a diameter BD perp. to PQ: P join A to one extremity D, cutting the o™ at E. .Toin CE. and jiroduci^ it to cut AF at F. Then F is the centre, and FA the radius oC he reipiired ciicie. . [^"I'Ply t^ie proof: and shew thai a secoi.' lution is obtained bv joining AB, and producing it to meet tlio c'"' ■ also distinguish between the j,;itiire of the contact of the circles, when PQ cuts, touches, or is without tho given circle.] .37. Describe i ele to touch a given straight line, and to toucli a given circle at a ^iven point. 38. Describe n circle to touch a given circle, ha\e ii eeutie in a given straight line, and pass through a given point in that .straight line. [I'lir ofchor pi'ublom.s of tho .sauie elas.s .see page 23r>.] Q 9n(> EUCLID'S p:LEMKNTa 'K ?' Uktjiocional Ciuclks. l)i;KiNiTin\. Circles which iiiter.soct at a- point, so that the two taii<,'cuts at that point are at riglit anj^duH to one anotlier, are .-said to be orthogonal, or to cut one another ortho- gonally. 89. In two interHectiug ch-oles the angle between the taogents at one point of intersection is equal to the angle between tho tangents at the other. 40. If two circh's cut one another ortliononaUi/, the taufjent to each circle at a imiiit of intcracetion icill pasn tliroiKjh the centre of the other circle. 41. If two circles cut one another orthotjonalhj, the .square on the (lixtance between their centres is equal to the sum of the squares on (heir radii. 42. Find the locus of the centres of all circles which cut a given circle orthogonally at a given point. 43. Describe a circle to pass through a given point and cut a given circle orthogonally at a given point. III. ON ancjlj:s i\ secmknts, and angles at the CENTRES AND CIKCUMKERENCES OF CIRCLES. See Propositions 20, -21, 22 - 2(5, '27, 28, 2t); 31, 32, 33, 34.. 1. If two cJiords intersect u-itliiii a circle, theij form an anijle equal' to that at the centre, subteuded by lialf tlie sum of (he ares they cut (|//'.. Let AB and CD be two chords, intersecting at E within the given i- ADBC: then shall the / AEC be equal to the angle at the centre, subtended by half the sum of the arcs AC, BD. Join AD. Then the ext. z AEC = the sum of the int. opp. Z'EDA, EAD; that is. the sum of the z 'CDA, BAD. l)Ut the /"CDA, BAD are the angh's m the G'« subtended by the aios AC, BD ; .-. their sum = half the sum of the angles at tiie centre subtended by the same arcs; or, the z A EC = the angle at the centre subtended by half the sum of the arcs AC, BD. o. K, i>. it, so tliat tho one iuiotiiei", jther ortho- 1 the taugentu }ii tlio tangents the idiifjcnt to h the centre of , THEORKMS AND EXAMPLES ON BOOK HI. 23 2. If two chords when inoduced intersect outside a circle, they form lut angle equal to that at the centre subtended by half the difference at the arcs they cut off. 3. The sum of the arcs cut off by two chords of a ch-cle at riglit angles to one another is equal to the semi-circumferonce. 4. AB, AC are any two chords of a circle ; and P, Q are the middle points of the minor arcs cut off by them: if PQ is joined, cutting AB and AC at X, Y, shew that AX AY. ij. If one side of a quadrilateral inscribed in a circle is produced, the exterior angle is equal to the opposite interior angle. 6. If two circles intersect, and any straight lines are drawn, one through each point of section, terminated by the circumferences; shew that the chords which join their extremities towards the same parts are parallel. 7. ABCD is a quadrilateral inscribed in a circle ; and the opposite sides AB, DC are produced to meet at P, and CB, DA to meet at Q: if the circles circumscribed about the triangles PBC, QAB intersect at R, shew that the points P, R, Q are coUinear. 8. If a circle is described on one of the sides of a right-angled triangle, then tlie tangent drawn to it at the point whore it cuts the liypotenuse bisects the other side. 9. Given three points not in the same straight line : shew how to find any number of points on the circle which passes through them, without finding the centre. 10. Through any one of throe given points not in the same straight lino, draw a tangent to tlie circle which passes through them, without finding the centre. 11. Of two circles which intersect at A and B, the circumference of on(! passes througli the centre of the other : from A any straiglit liiuf is drawn to cut the first at C, the second at D ; shew that'CB - CD. 12. Two tangents AP, AQ are drawn to a circle, and B is the middle ])oint of the are PQ, convex to A. Sliew tliat PB bisects tlie angle APQ. 13. Two circles intersect at A and B ; and at A tangents are drawn, one to each circle, to meet the circumferences at C and D : il CB, BD are joined, shew that the triangles ABC, DBA are equiangular to one anotlu'r. 14. Two segments of circles are uescribed on the same chord and on the same ide of it ; tlie extremities of the common chord are joined to any jjoiiit on the are of the exterior segnirnt ; .slicw tliut the ;U(; intercepted on the interior segment is i-onstaiit. I H. E. 15 224 Euclid's klements. 15. If a series of triangles are drawn standing on a fixed base, and having a given vertical angle, shew that thi; bisectors of the verti- cal angles all pass through a fixed point. 1(1. ABC is a triangle inscribed in a circle, and E the middle ])oint of the arc subtended by BC on tlie side remote from A: if through E a diameter ED is drawn, shew that the angle DEA is half the dilferen('(> of the angles at B and C. [See Ex. 7, p. 101.] 17. If two circles touch each other internally at a point A, any chord of the exterior circle which toucb;s the inteiior is divided at its point of contact into segments which subtend equal angles at A. is. If two circh.'s touch one another internally, and a straight line is drawn to cut them, the segments of it intercepted between the circumferences subtend equal angles at the point of contact. ! ThK Oni'HOCENTnE OF A TllIAXfil.K. 19. The licrpendiculars draicn from the lyrtirpn of a trhiiKjh' to tlip npjwsite sidm are concurrent. In the A ABC, let AD. BE be the perp'' drawn from A and B to the oppo- site .sides; and let them intersect at O. Join CO; and produce it to meet AB at F. It /s rt'qxirrd to shcir that CF h prrp. to AB. Join DE. Then, because the z '^ OEC. GDC are rt. angles, JI'.IP- :. the points O, E, C, D are concyclic : .-. the Z DEC = the z DOC, in the same segment; = the vert. opji. / FOA, Again, because the Z " AEB, ADB are rt. angles, Hyp. .: the points A, E, D. B are concyclic : .-. the z DEB-the z DAB, in the same segment. .-. the sum of the Z " FOA, FAOr^^the sum of the z • DEC, DEB -a rt. angle: l^llp- :. the remainin" z AFO = art. angle: i. H*i. that is, C.-' is perp. to AB. Hence the three perp^ AD, BE, CF meet at the point O. Q. k. p. (For an Alternative Proof see page 10(1. J THKORKMS AND EXAMPLKS UN BOOK It I. 22;') in a lixed base, ovs of the verti- [ E tho mlddlfi lote IVoin A: it' -lo DEA i^ half 1. 101.) t a point A, any is di villi id at its ngles at A. , and a straight )ted between tho )ntact. of a triiUHih' to D C ^gmcnt; Hyp. gment. Z" DEC, DEB HlljK e : I. 3y. 5int O. Q. E. P. C.J Definitions. (i) Tho iiiti-rsec;tiou of tho porpendicnlirs diMwii iVoiu tlu' vortiofts of :i, tnnu^do to tlui opposite sides is ciUcd its ctho- centre. (ii) 'I'ho triiuiglo ftn-nu'd l.y joiiiin;^' tlio find, of th.- ppri.cndi- • nlars is (inllcd ihr. pedal or orthocentric triangle. 20. In (III (initc-diKjlfd trioiiiilc tlic pirju'iuliculdrii ilniwn from till' vi'rtici'A to the opposite )!ides bi.'^cct tli<; oii'ilc^ of tlic pcilal tri'aiuilc l/iroiiiili irliicli tliey pti>is. In the acute-angled a ABC, let AD, BE, CF be the porp" drawn from tho vertices to tlie opposite sides, meeting' at the orthoeentre O; and let DEF be tile pedal triangle : then shall AD, BE, CF bisect respect- ively the z » FDE, DEF, EFD. For, as in the last theorem, it may be shewn that the points O, D, C, E are con cyclic ; .-. the z ODE = the i OCE, in the same segment. Similarly the points O, D, B, F are concyclic ; .-. the z. ODFnrtlio z OBF, in the samo segment. Rut the I OCE rr the z OBF, each being the comp' of the / BAG .■. the z ODErT:the z ODF. Similarly it may be shown that tlie z ^ DEF, EFD are bisected bv '. .njOLLARY. (i) Every two sidefi of the peihil triaiiqJe are equallii :''h-d to that side of the original trianyle in irhirli they meet. i'orthe z EDC = thecomp'of the z ODE -the comp' of the z OCE = the / BAC. Similarly it may be shewn that the z FDB-tlie z BAC .-. the z EDO =. the z FDB:-the z A. In like manner it may be proved that the z bEC=the z FEA==the z B, and the z DFB = the z EFA--.the z C, M ,^°'*°^':t''''- ^,"^ P'' ^'''"'^^''' °^^' A^l"' DBF are equiangular to one another and to the triangle ABC. Note. If the angle BAC is obtuse, then the perpendiculars BE. CF t'lsect externally the corresponding angles of the pedal triangle. 15-2 i iC 22G kucud's elemknts. \'- erticca circle, irhicli 21 In any triangle, if the perpendiculars drau-n from the r 0)1 'he opposite sidc^ are produced to meet the circumscribed then eacli xide bisect.-> that portion of the line perpendicular to it lies hetu-een the orthocentre and the circumference. Let ABC 1k' a triangle in which tlio pei'ijon- (liouhirs AD, BE are drawn, intcrsoctin?^ at O tlu' (irthocentro; and let AD he produced to meet the ' .'■'■ of the circunisoriljiiifj; circle at G : then shall DO- DG. Join BG. Then in the two a" OEA, ODB, Ok; / OEA-tho z ODB, beinp^ rt. anodes; imd the Z EOA -^ the vert. opp. / DOB; .-. the remaining Z EAO = tho renuuning 15ut the z CAGr^the z CBG, in the same segment; .-. the Z DBO=:the z DBG. Tlien in the a'DBO, DBG, (the z DB0=-the z DBG, r.eeause 'the z BDO-^the z BDG, / and BD is common; .-. DO---DG. Vrot'cd. I. 20. Q. K. n. «>•> In an acute-anahd triannlc the three sidex are the external his^cTor^of thean!]lc>i of the pedal triani of a triangle (UHliin orthocentre are each equal to the circle circumscribed about the triangle. 2(5 D E are taken on the circumference of a Kemicircle described on a 'given straight Hue AB • the chords AD BE and AE, BD intersect (produced if necessary) at F and G: sh,>w that FG is per- liendicular to AB. .>r, /^B'^D 1'= a pavall'^opraui: AE and CE are drawn at right angleato A B, and CB respectively: shew that ED, if produced, will be ncrpendiculav to AC, om the rerticea }scribed circle, 'hir to it which I'rnreiJ. I. 20. Q. K. 1>. ire Die external n ohttise-nnpled iiternal hii^ector^ ), t<]ieir that the then muj one of " triiniffle vJioxe cea of a triangle cus of X, and shew liuit it passes through B. [Ex. 'J, p. 21(3.] 48. Two circles intersect at A and B, ancV through P, any point on the circumference of one of them, two straight lines PA, PB are drawn, and produced if necessary, to cut tlie other circle at X and Y: find the locus of the intersection of AY and BX. 4l). Two circles intersect at A and B ; HAK is a fixed straight line drawn tb ough A and terminated by the circumferences, and PAQ is any other straight line similarly drawn; find the locus of the intersection of HP and QK. 1.1 230 KUt'LIU's KLKMENTS. .jU. I wo segments of ciicles ftie on Uio same cliord AB uiul on the same side of it; and P and Q are any points one on eacli air • tind the locus of the intersection of the bisectors of the angles PAQ^ PBQ. " rA Two circles intersect at A and B ; and through A any stral'dit line PAQ IS drawn terminated by tlie circumferences: lind tlie locus of the middle point of PQ. Ui\ -AIlSCKI-LANKOL-S ExAMI'LKS 0\ ANCil.KS IX A ClliCI.K. r,2 ABC is'ii triangle, and circles are drawn through B, C, cutting the sides m P Q P , Q', . . , : .shew that PQ, P'Q' . . . ure parallel to one anotlier and to the tangent drawn at A to the circle circumscribed about the triangle. 53. Two circles intersect at B and C, und from any ])oint A, on the circumlerciice ot one of them, AB. AC are drawn, and j)roduced if necessary, to meet the other at D and E: shew that DE is parallel to the tangent at A. '■ 54 A secant PAB and a tangent PT are drawn to a circle from an external ponit P; and the bisector of the angle ATB meets AB at C : shew that PC is ecpial to PT. ^Jin -^''T *" ^'''"'*/^ ?'- ^I^'^ circumfereiu^e of a circle two chords AB, AC are drawn, and also the diameter AF: if AB, AC are produced to meet the tangent at F m D and E, shew that the triangles ABC, ALD are equiangular to one another. r>(;. O is any point within a triangle ABC, and CD, OE, OF are drawn perpendicular to BC, CA, AB resi.ectivelv : shew that the angle BOC is equal to the sum of the angles BAC, EDF. ',7. If two tangents are drawn to a circle frt)m an external iioint shew that they contain an angle cpial to the ditierence of the an.'le- in the segments cut off by the chord of contact. _ 08. Two circles intersect, und through a point of section a straight line IS drawn bisecting the angle between the diameters through that point : shew that this straight line cuts off similar segments from the two circles. ^ r,„iiit,, ami fmu >i Jur, panif in. the vommim chord produml t,n,;i,'nU arc dnncii U,'nll the (•//r/t'.v; Hhcw thut the poiiiLs of contact lie on a circle which cuts all the (fiven circles orthnionallij. 7. All circlet, which pa^.^ throniht line oj uideiuute leniith cnllin;, AB or AB produced at' ri,,ht anoles • <>'.>>l/'''}^ CD aji.,-ed sh uiiht line of indefniite leu.th; AP ,., a„ii 8t,-aifiht line drawn throu lu A to meet CD at P- and in J\P a pond Q is taken such that the rectaiojle AP, AQ />■ constant: find the locus of Q. 1-J. Two circles intersect orthogonally, and tangents are drawn irom any point on the circumference of one to touch the other : prove that tlie first circle ])asses through the middle point of the chord of contact ot the tangents. [Ex. 1, p. i;:j.'5.] I ^'?" A^ «t'""'''i-cle is des.;ribed on AB as diunietei, and any two cliords AC, BD are drawn intersecting at P : shew that AB- = AC . AP + BD . BP. Iti. 'I'wo circles_intersect at B and C, and the two direct oimnuii lungents AE and DF are drawn : if the eonmion chonl is produced to meet the tangents at G and H, shew that GH- = AE-'-|- 3C-. 17. If from a point P, without a circle, PM is drawn perpendicular to a diameter AB, and also u secant PCD, shew that PM- = PC . PD + AM . MB. THKORKMH AND KXAMPI.KH OS HOOK III. SB& 18. Tliii!i> ciicli's iiitiMNeet at D, uiul tlioir otlit r points of inter- section aro A. B, C ; AD cuts tlio ciiclo BDC at E, and EB, EC cut the riicU's ADB, ADC rt'sjtoctivcly at F una Q : shew tluil the | ints F, A, Q lire <•(.,. .iit'rvr, iiml f, B, C, Q (•((iicvi-lic. 19. A semicircle is dPscriheU i>n ii ^^iv.'u diameter BC, ami from B and C any two clioid^t BE, CF mo drawn intors».'eting within tho Hoiiiieiiclt' at O; BF and CE :ui- jnodiicrd to meet at A: s)u>\v that the Hiiia of tlH^Hqiiaivs on AB, AC in ('i|iml fo hviec tlio Hquare on the tall^^ont from A toncthcr witJi tho Hfjuarc on BC. 20. X and Y aro two hxod poiiitH in the dian.eter of a circlo oqnidiHtant fioni tlio centre C : tlnon^di X any chord PXQ is drawn, and itH oxtivmities are joined to Y ; sliew tiiiit tl\() Kiini of l\n\ Hiinares on tho sides of tiie triangle PYQ is constant. jHee i). MY. Kx. 21,1 I'llOIlT.KMS 0\ TaNOKNTY. 21. To (h'xcr'the " chdi' to panx thvoinih tint ijir n jutiiits diid tn touch a ijivfu Htrn^i'lit Jiiu . / Let A and B tc the glv r* points, and CD tlie pive x Hi , line: H it is nHpiired to u.scribe ;■ circlo to pasa ♦^^hrouRh A im-, '.> nnd to fonch CD. Join BA, and prodnee i* t'> nioi'! '' CD at P. I)escriI)o n s(|naro oipial to tlie C P Q root. PA, PB; ii. U. and from PD (or PC) cnt off PQ ofpial to a side of this sqnare. Through A, B and Q descriljo a circle. Kx. 1, ]>. I.'C. Then since the rect. PA, PB^the sq. on PQ, .-. the ABQ touches CD at Q. m. :>7. V. K. V. Note, (i) Since PQ may he taken on cither side of P, it is clear that there are in f^eneral two solutions of the problem. (ii) Wlien AB is parallel to the f,'iven lino CD, the above method is not appiicabie. In tiiis case a simple coiistrnction foiiows from in. 1, Cor. and jii. 10- and it will he found that only one solution existfi m 236 KUCMD S KLKMKNTS. ■ . Is' {■ t ! i i •11 i'j ii 'ii 11 22. To describe a rircle to p,i^s through two qiven points and lo touch a gicen circle. Lot A and B be tlie givpn points, and CRP tlie given circle : it is required to describe a circle to pass througli A and B, and to touch the >•. CRP. . , Throu;,'h A and B de- scribe aiuj circle to cut the _ given circle at P and Q. "' ^: — r":' 'B Join AB, PQ, and pro- '~n duce them to meet at D. ^ From D draw DC to touch the -iven circle, and let C be the point of contact. ' Then the circle described through A, B, C will touch tlie "iven cncle. n ' •" For, from the ©ABQP. the rect. DA, DB = the rect DP DQ- and from the ■ PQC, the rect. DP, DQ-the sq. on DC; ' ui .30 .-. the rect. DA, DB = th.' sq. on DC : .-.DC touches the ABC at C. m .37 But DC touches the © PQC at C ; Co'mtr .-.the ©ABC touches the given circle, and it passes throu'di tlie given pouits A and B. '"' n t. t./ ■ ^^"^^-^ (i) ^i»^e V'o tangents may be drawn from D to the given cn-cle, it follows that there will be two solutions of the problem. • ^aI '^^^? general construction fails when the straight line bisect- ing AB at right angles passes through the centre of the given circle- the problem then becomes symnietrical, and the solution is obvious ' 2:',. 7'.) tlcscrihe a circle to p„^x throwih a niveu iwiiit touch two (liven straight lines. Let P be the given point, and AB, AC the given straight lines: it is required to describe a circle lo pass through P and to touch AB, AC. Now the centre of every circle ^yhich touches AB and AC must lie on the bisector of the z BAC. Ex. 7, p. is;;. Hence draw AE bisecting the I BAC. From P draw PK perp. to AE, and produce it to P'. making KP' equal to PK. ' lUUl to THEOKEMS AND EXAMPLES 0\ BOOK III. 237 pohitu (Hid ./A the point tlie {?iven DQ: III. .'ifi. III. 37. Conntr. ongh tlio Q.E.F. D to tho problem. lie bisect- en circle : r>l)vioiis. it (Uid to 6 Then every circle which has its centre in AE, and passes through P, must also pass throuf^h P'. Ex. 1, p. 2ir>, Hence the prolileni is now reduced to drawing' a circle through P and P' to touch fither AC or AB. Ex. 21, p. 2'^'>. Produce P'P to meet AC at S. Describe a square equal to the reet. SP, SP'; n, 14. and cut off SR equal to a side of the Hiiuare. Describe a circle thioush the points P', P, R; then since the rect. SP, SP'=-the sq. on SR. Constr. .*. the circle touches AC at R ; lu. H7. and since its centre is in AE, the bisector of the / BAC, it may be shewn also to touch AB, g. k, f. Note, (i) Since SR may be taken on either side of S, it follows that there will be two solutions of the problem. (ii) If the Riven straight lines are parallel, the centre lies on the parallel strai^dit line mid-way between them, and the construction proceeds as before. 24. To describe a rircle to touch tiro f]iren straight linen and a ijiven circle. Let AB, AC bo the two given st. lines, and D the centre of the given circle ; it is required to describe a circle to touch AB, AC and the circle G whose centre is D. Draw EF, GH par' to AB and AC respectively, on the sides remote from D, and at distances from them equal to the radius of the given circle. Describe the 0MND to touch EF and GH at M and N, and to pass through D. p.^^^ 23 p •J3(; Let O be the centre of this circle. • • - > 1 • - •Tom CM, ON, CD meeting AB, AC and the given circle at P Q an(i R. ' Ao'^l'^" ^^"'f'*^ described from centre O with radius OP will toucli AB, AC and the given circle. For since O is the centre of the M N D, .-. OM = ON ^OD. ButPM = QN = RD; Comtr, .-. OP = OQ-OR. ■■• ■'' ;;"'cle described from centre O, with radius OP, will pass through Q and R. re And since the z ' at M and N are rt. angles, m 1h .-. the / » at P and Q are rt. angles" ' / 29 ,-. the V) PQR touches AB and AC. i! •■ 238 kuclid's klkmknts. AjuI siuee R, the point iu which the oirolos meet, is nn tlie line of centres OD, . '. the PQR touches tlie given circle. q. k. r. Note, There will be two solutions of this problem, since two circles may he drawn to touch EF, GH and to pass throuph D. ittitj i 3 25. To ilt'xcrihi' a circh' tn ])as>< throiiali a rjivini paint ntul toiirJi / DHE is a rt. angle, being in a semicircle. in. 31. also the angle at F is a rt. angle; Const r. .-. the points E, F, G, H are concvclic : .-. the rect, DE, DF-the rect. DH, DG : m. .'{li. but the rect. DE, DF==the rect. DK, DP : i'ouatr. ,-. the rect. DH, DG = the rect. DK, DP : .•. the point H is on the PKG. Let O be the centre of the r^ PHG. Join OG, OH, OH. Then OG and DF are par', since they are both perp. to AB ; and DG meets them. .-. thezOGD = thezGDC. t. 2'.), But since OG = OH, and CD-CH, •. thezOGHn^thezOHG ; and the z CDH -.tiie Z CHD : • t>ia .' OMn ti-.o.-r.Mn. .-. OH and CH are in one st. line, the PHG touches the given "j DH! (), K. F, THEOREMS AND EXAMPLES ON BOOK HI. 2:30 tlip liiif> of Q. K. F. iiuce two 1 D. NoxK. (i) Since two circles may be drawn to pass through P, K and to toucli AB, it follows that there will be two solutions ui the present problem. (ii) Two more solutions may be obtained by joining PE, and )irocfeding as before. TJie student should examine the nature of the contact between the circles in each ease. :.2I.p. 2:ir.. 2(5. Describe a circle to pass througli a given point, to touch a given straight line, and to have its centre on another given straight line. 27. Describe a circle to pass through a given point, to touch a given circle, and to have its centre on a given straight Inic. 28. Describe a circle to pass through tw > given points, and to intercept an arc of given length on a given circie. 2'.). Describe a circle to touch a given circle and a given slraiglit line at a given point. ;{0. Describe a circle t(j toucli two given circles and a given straight line. ON MAXIJ[A AM) MINIMA. 111.31. Const r. III. .■{(». I 'oust r. AB; T. 211. ;hd '?. K. r. We gatlier fi'oiu the Theory of Loci tliat the position of an angle, line or figure is capable under suitable eoutUtions of 'f^radual change ; and it is usually found that change of position involves a < orrespunding and gradual change of JixigaitiuP. I'nder these circumstances wo may be recpiired to note if any situations exist at which the magnitude in (piestion, aftei' increasing, begins to doci'ease; or after decreasing, to increase;: in such situations the ^[agnitu(lo is s/iid to liave reached a Maximum or a Minimum value; for in the former case it is greater, and in the latter case less than in adjacent situations on either side. In the geometry of the circle and straight line we only meet with such cases of continuous change as admit of ii>ie transition from an increasing to a decreasing state - or vice versa— so that in all the problems with which we ha\e to ileal (where a single circle is involved) there can be only one Maximum ;nid on(> MininiuTn the Maximum being the CTcatc^^t, and the -Minimum being the least value that the variable magnitude i^s capable of taking. II. E. 16 240 Euclid's klements. h! 1 llms <>f ia any particular i)roblem; but by means of It a situation is suggested, which on further examination may bu shewn to give the maximum or minimum value sought for. For example, suppose it is required I.O determine (he greatest stnwjht lu>c thai, may he dmira nerpen- dioidar to the chord of a srnnujnt of a circle and intercepted between the chord and the arc: we immediately anticipate that the greatest perpendicular is that which occupies a si/mmetrical ])osition in the figure, namelv tne peri)cndiciuar \\hidi passes through the middle point of the chord; and on lurther examination this may be i)roved to be the case by means of i. 19, and i. .34. Again we ai'c able to lind at what point a geometrical magni- tude, varying under certain conditions, assumes its Maximum or Minimum value, if Ave can discover a construction for drawing the magnitude so that it may have an assigned value: for we may then examine between what limits the assigned value must le in order that the construction mav be possil)le; and the higher (.r lower limit will give the Maximum or .Minimum souglit lor. It was pointed out in the cliai)ter ..ii the intersection t>f i.oci, [see page 119] that if under certain conditions existing anion"- the data, two solutions of a problem are ])o.ssible,and under otluT conditions, no solution exists, there will always be some inter- mediate condition under which one and oah one distinct solution IS possible. Under these circumstances this single or limiting soluti(Mi will always be found to correspond to the maximum or minimum Aalue of the magnitude to be constructed. 1. For example, suppose it is re(iuircd to diviilc a ijivcn straipht line .so thot the nrtaiif/h' coHtaim'd hi the two ncf) mentis mail be a muximiim. We may lirst attempt to divide th- Hiv..ii strai-ht linn so that the recuuiHle coiitaintHl by its segments may have a niven area— that is be equal to the square on a given straight line. ' THEOREMS AND EXAMrLKb ON UOOK 111. 241 1 maximuui the magni- tural thero- ir when, in u/mmetrical case. fill or posi- euoiigli to y iiieuns of ion may be t for. iiitercepted ulicular is re, namely )iiit of the .1 to be the cal uiagui- iximuiu or )r drawiiiLT 10 ; for we ■ahio must ; and the .Miiiimmu DU of iioci, iig among iider f)ther niie inter- 3t sohitioii 5 sohition minimum ni'd hy the so tliat the % — that is, Let AB bo the given straight line, and K the bide of the given s{|uare : K it is required to divide thti st. hne AB at a jwint M, so that the reet. AM, MB may be eciual to the mi un K. Adopting a construction su<-'gosted l^y ii. 1-1, describe a semicircle on AB; and at any i)oint X in AB, or AB produced, draw XY perp. to AB, and equal to K. Through Y draw YZ par' to AB, to meet tlie arc of tlie senheirele at P. Then if the peip. PM is drawn to AB, it nuiy be siiewn after the manner of ii. 14, or hy iii. ;^5 that the rect. AM, MB-the sq. on PM. — the sq. on K. So that the rectangle AM, MB increases as K increases. Now if K is less than the radius CD, then YZ will meet tlie arc of the semicircle in two ])oints P, P'; and it follows that AB may he divided at two points, so that the rectangle contained hy its segments may be equal to the s(iuarp on K. If K increases, the st. line YZ will recede from AB, a.iu the points of intersection P, P' will con- tinually approach one another; until, when K is eoual to the radius CD, the St. line YZ (now in the position Y'Z') will nu'ct the are in tiro coincident jtoiiita, that is, will touch the semicircle at D; and there will be only one solution of the problem. If _K is greater than CD, the straight line YZ will nut meet the semicircle, and the problem is impossible. Hence the greatest h-ngth that K may have, in order that the con- struction may be possible, is the radius CD. .•. the rect. AM, MB is a maxinmm, when it is equal to the S(iuare on CD; that is, when PM coincides with DC, and consequently when M is the middle point of AB. (\b». The special feature to be not'.'cd in this problem is that the maximum is found at the transitional point between (wo solutions and uo solution : that is, When the two Boiutioua coincide and become identical. 16-2 242 kucmd's ki,kmknts 1 1 ^2^^^^^ » ^B^^B^'^ i ■ ili] i^ n 'I'lio fulluwing extiini»lo illu.stmte.s the .same point, I ?i J^^J'."^^ ^^ whatiMint in a given straight line the unnle mbtrndid (>!/ the Iinejotnui;/ two given points, whieh are on the san'ie xide of the given atratgiit line, is . I A and B, ides of AB: THKORKMH Wr KXAMPT-FS OX BOOK Tlf. hence two polntfi in CD may be found Buch tliat tlio anj^'le sul'tended by AB at each of thoni is greater tlian the an^'le subtended at any other point in CD on tho sumi' xidr oj' AB. Wo add t,\vo moro examples of eonsi(lf>ra})le iniportance. :{. In a straifiht line of indejinite lenr/th find a point such that the Sinn, of its distances front two tiiven jtoints, on tlir stiim' sidi' of tlw iiirrn linr, shall he a niinintiini. Let CD be the given st. Hue of indefinite h^nj^'th, and A. B tlie ^iveii ^B ])oints on tlie same side of CD : it is required to find a point P in CD such that the sum of AP, PB is A a minimum. f '~v^ Draw AF porp. to CD; I and produce AF to E, makiiij^' FE ' t(]ual to AF. C F , P Q o •Toin EB, cuttiuK CD at P. j /' - ' •loin AP, PB. 1 ,'''- ' Tlien of ail lines drawn from A ^ an. I B to a jxiint in CD, the sum of AP, PB shall be tlie least. For, let Q be any other i)oint in CD, ./ / Join AQ, BQ, EQ. Now in the a' AFP, EFP, ( AF-EF, J^eeanse [and FP is common; Constr. /and the z AFP-the i EFP, bcinf' rt. auo shewn that AQ ' EQ. Now in the A EQB, the two sides EQ, QB are to.rether m-ealer ihan EB; hence, AQ, QB are together greater than EB, that is, greater than AP, PB. Similarly the sum of the st. lines drawn from A and B to anvotb(>r lioint in CD may be shewn to be greater than AP. PB. .-. the sum of AP, PB is a mininnun. 0. K. n. NoTK. It follows from the above proof that tlu' I APF-rthe z EPF , 4 the /. BPD. i/jr," Thus the sum of AP, PB is a minimum, when these lines arc I'lfialli/ inclined to CD, i1 (I u KHmn's KLKMKNTS. !i «■ 4. is i>;-i feet distant from the bridge. 10. From a given point P without a circle whose centre is C, draw a strai{,'ht line to cut the circumference at A and B, so that the triangle ACB may be of maximum area. 17. Shew that the greatest rectangle which can bo inscribed in a circle is a square. 18. A and B are two fixed points without a circle : l"ind a point P on the circumference such that tlie sum of the squares on AP, PB may he a minimum. [See p. 147, Ex. 24.] 2in KroiJD a KLKMENTS. r.t. A segmcul i»f 11 circlf is described on tlu- diurd AB : find a iM.iiit C oil itH aro ho that tlie sum of AC, BC may lie a maxiiiium. 20. Of all tiiaiiffh'x Hint cnn bp hini-rihnt in u ri,vh that which h(iK the t/n-ntt'st pcvimetvv ix ('(jitilntfml. yi. Of all triau!ilen that can he irnvrilit;! in n ,,ii;u circlf that vhich has the ffrcntcst area is equilateral. 22. Of all triaiifile.t that ctni be iim-ribetJ in a uiven trianqle that irhirh htiH the least perimeter is the triaiuile faniied'bif joininq'the feet I'J the perpeiKliculars draini from the vertices' oti opposite side's. 2:5. Of all rectangles of given area, the square lias the least neri- meter. '■ 2f- ^' • '• liianglft of maximum area, havluK its angles •'ciiial ,0 those 'A a yxven triangle, and its sides passing thiuiigh (hie,, given points. ° Vr. HAHDKIl MISf'KLL.WKOIs KXAMPLFS. 1. AB is a diaineter , a g.v ,u circle : and AC, BD, two chords on the same side ot AB, intersect at E: shew fliat the cireh- which passes through D, E, C cuts the given circle orthrigonally. 2. Two circles whose centres are C and D intersect at A and B and a straight line PAQ is drawn through A and terminated hv the circuinlerences: prove, that (i) the angle PBQ~the angle CAD (ii) the angle BPC = the angle BQD. ^. Two chords AB. CD of a circle whose centre is O intersect at right angles at P : shew that (i) PA- 4 PB- + PC- + PD-'=: 4 (radius)-, (ii) AB- + CD' + 40P- 8 (radiusj-. 1. Two i)arallel uingents to a circle interc i,l on any third tangent ,i ])ortu)n which is so divided at its point of contact that the rectangle contained hy its two parts is e(pial to the s.iuare on the ra(hiis. 5. Two equal circles move between two straight lines iilaced at light angles, so tiiat each straight lii- is touched by one circle and the two circles touch one another . 1 the locus of the point ot contact. ' G. AB is a given diaui ter of a rifcle, and CD is any parallel chord: it any jx.iiit X in ,B is joined to the e.;..remities of CD •iUGw thai XC- + XD-=XA2+XB-. i! AB: tiuil a xiinuiii. that which chrlf that •/(/H///f thai ilifl thrfcft U'K, loaHt pori- its aufjtles Lwo nhonln rcle \vliic)i I A and B, ted by the iteraoct at finy third 5t that thi! re on the los ])l;iced >iie eirclp, the point y parallel \ IIOOK III. 247 I 7. PQ is a fixed elioid in a riivli>, ui, 1 PX. QY anv two parallel fhorda throu-'li P uiid Q: shew tliiit XY IuiicIu'm u (i\.d eoncentric circle. H, Two ( inul circles intersect at A and B; ami fium C anv point on the cireumt('reno(( of < of flicni a i»erp(nidiculur is drawn'to AB, meeting the otiior circle u > and O' : shew that eitht;r O or O' is fli.. nrtlinceiitrn (»f the triangi >BC. DistinKuish hetweeu the two oases. 9. Three eipial cii les jmss throuf^'h tlic "^amo i)o;nt A, and their other points of intersection an; B. C. D : shew tliut of tiie four lioinfs A, B, C, D, each is the orthoceutre of tlie triangle formed hy joiiiinj,' tJM) other three. 10. From a fjiven point without a (urcle draw a straJKlit line to the concave ciieunit'erence ho as to be bisected by tlie convex circumference. When is this problem impossible? 11. Draw a 8traif,'ht line cutting,' two concentric circles n th-r I lie chord intercepted by tlie circumference of the greater circle nia\ be double of the chord intercepted by tht less. 12. ABC is a trianf?le inscril)ed in a circle, and A', B', C are the middle ])oints of the ares subtended by the sides (remot(s from the opposite vcitiees): find the relation between the angles of the two triauKles ABC, A'B'C ; and prove that the pedal triau;,'].- of A'B'C is equiangular to the triangle ABC. 13. The oj)posite sides of a quadrilateral inscribed in a circle are produced to meet: hew that the bisectors of the two angles s(» formed are perpend, Jar to one another. 14. If a quadrilateral can have one circle inscribed in it, and another circumscribed about it; shew that the straight lines joining the opposite points of contact of the inscribed circle are perpenrlieular lo one another. !.■>. (liven the base of a tiiangle and the sum of the remaining sides; find the locus of the foot of the perpendicular from imo extremity of the base on the bisector of the exterior vertical angle. 10. Two circles touch each other at C, and btraight lines are drawn through C at right angles to one another, meeting tin; circles at P, P' and Q, Q' respectively: if the straight line which joins the centres is terminated by the "ircumferenccs at A and A', shew that P'P= hQ'Q- A'A-. 1.. Two circles out one another orthogonally nt A and B; P is any point on the arc of one eireie inteicepledby ilie other, and PA, PB are produced to meet the circumference of tin econd circle ut C and D : shew that CD is a diameter. ■ 248 RtlCUD'S KLEMKNTS. Ml MM FT! If! i '!3! IS. ABC is a tiiant,'Ie, iiiid from niiy j)uiiit P pcrpc'iidiculara PD, PE, PF art' dniwn to tin- sides: if S,, S.„ S., nn- tin- i-t'iifrrs of tlio circlos (;irc'iiiiiscrilM(| aliuiit tlii> triiiii;,'rfs ERF, FPD, DPE, hIk'w tlmt tlu) triiiiifjli! SiSjS,, i.s t'i|iiiaii;;iilar to tin triiiiij,d«' ABC,' mill that tho Hiiie.s of tlic oin' arc respuctivcdy half of llic sides of thu • ithor, ]i>. Two taiif,'(>nts PA, PB aro dniwii froiri an <'xt.Tnal point P to a <,'iv('ii cirtdc, and C is tlii' middle jiuint of lln' chord of contact AB: if XY is any clKird thron^di P, slm- that AB bisects tho angkt XCY, 20. Given the sum of two straifjht lines and tho rcctan;,dt) con- tidncd 1>y thoni (etjnal to a given S(innre) : find the lines, 21. (iivon the sum uf the .sf|uares nn two stiaii^ht lines and the rcclangle contained hy tlieni : lind the lines. 22. (Jiv 1 the snm of two straight lines and the sum of tho R(|liare8 on tl»em : lind the linos. 2:5. (liven tlic diircreuci- lirtween two straight lines, and the root- angle contained liy thein : find the lines, 24, Given the stun or dill'crence of two straight lines and tho difforenco of their .squares : lind the lines. 25. ABC is a triangle, and the internal and external hiscctora of the angle A meet BC, and BC i)rodiict.'d, at P and P': if is the mi.ldlo point of PP', .shew that OA is ii tangent to tliu circle circinu. Hcribed about tho triangle ABC. 2(5. ABC is a trian^de, and fioni P, any point on tho cinum- fercnco of tlic circle circuniscril)cd alnmt it, perpendiculars aro drawn to the hides BC, CA, AB meeting the circle again in A'. B' C- prove that ' ' (i) the triangle A'B'C is ideniicrJly equal to the triangle ABC, (ii) A A', BB', CC arc parallel. 27. Two equal circles intersect at lixed points A and B, and from any point in AB a peipendiciJar is drawn to meet tlie circumferences on tho same side of AB at P und Q: shew that PQ is of constant length. 28. Tho straight lines which join the vertices of a triano'le to the centre of its circumscribed circle, are perpendicular respectively to the .sides of the jjedal triangle. 29. P is any point on the circumference of n circle circumscribed about a triangle ABC ; and peri)endiculars PD. PE nie an^Ic Imtwci'ii AP and the dinmotcr of the cii-funisi iil)t'd circle throii^di A. 31. Shew that the circles circumscrihed about the four trian;,'le.s formed liy two pairs of intersectinf,' straij^ht lines meet in a point. 32. Shew lliat the orthorentres of the four triangles formed by two jiairs of intersecting straight lines are colliiicar. angle con- 's ami the im of th(! I the reot- s and the isoctorfl of O is till! (( circuui- cucum- aro drawn '. B', C; ,'lo ABC. and from nifcrcnces ' constant igle to the •ely to the imseribed :n from P le circunx- Os TIIR CoNHTnUCTION OP TllIANni,KH. ^:\. Given the vertical nnulo, one of tho sides ccmtainlng it, an.l the lengtli of tho prrppiidiciilar from tho vi-rtcx on the bas.;: construct t\w triangle. 34. (riven the feet of the i)(>rpondiculars drawn from the vortiees on the opposite sides : construct the triangle. 3,>. (liven the base, the altitude, and the radius of the circum- scrilK'd circle: construct the triangle. 30. Given the base, the vertical angle, and the sum of the squares on the sides containing the vertical angle ; construct the triangle. 37. (riven the base, the altitude and the snin of the squares on the sides containing the vertical luv^Ur. constiuct tin; triangle. .3S. Given the base, the vertical angle, and the difference of the squares on the sides containing the vertical auKle: construct the tri- jin<,'le. 3'.». Given the vertical angle, and the lenj,'ths of the two m(;dians drawn from the extremities of the base: construct tlie triangle. 40. Given the base, the vertical angle, and the dilTerenco of the angles at the base: construct the triangle. 41. Given tha base, and the position of the bisector of the vertical angle : construct the triangle. 42. Given the base, the vertical a-^gle, and tlie length of the bisector of the vertical angle : construct tne triangle. 43. (riven the perpendicular from the vertex on the base, the bisector of the vertical angle, and the median which bisects the base : construct the triangle. 44. Given tlie hispctov of the vertical angle, the median bisect- nig the base, and the dift'ereuce of the angles at the base : construct the triangle. li Mir ROOK (V. Book IV. consists onlin'ly of pichl-iiis, doaliiio- svitli vfinou.s ivctilmoal ti;;aire,s in ivlution to the (•irclps'~'wliicli iwiss tliroiioli thcif ■•ni,o;ular points, or mo toiiolied bv tlicir sules. •' nr,FINITIO\S. 1. A Polygon is a iPctilinoal fi-un' hounded l,v n.oro tii.'in tour sides. A Pnly.«>T)U of _/7/v sides is e;illed M NW" sides „ seven sides M ''/17//7 sides V ff^ii sides ., iiveJve sides )> fifteen sides a Pentagon, Hexagon, Heptagon, Octagon, Decagon, Dodecagon, Quindecagon. 2. A Polygon is Regular Avlien all its sides ai-e eriual, and all its angles ai-e (Mjual. o • ). A rectilineal figure is said to he inscribed in a circle, wiicn ail its angular l)oiuts are on the circunifej-ence of the cu'cle: Jiiid a, circle is saifl to he circumscribed about a i-ectilineal figure, mIicu the circuni ference of the circle i)asses thi-ougli all the Jiiigular points of the figure. ^. A rectilineal figure is said to he circumscribed about a circle, when each side of the figure is a tangent to the cii'cle : .iiul a circle is said to be inscribed in a recti- lineal figure, when the ci/cumference of the circle is touched hy each side of the figure. A straight line is said to l)e placed in a i). ir<.'l< ?%■•- ■■•" ■.-> .^ciMi iw m- jt/j.ai;cu ill a <'ll'<'|(i Us extremities are on the circumference of the circle,' ^\ heu BOOK IV. PRor. ]. ■les wJiifli ! by their )>v iiioro >n, n, an, ?on, :agon. 251 PUOI'OSITIOX Pkoblem. cu Jit a !/io'>t rlrde to jilace a chord eqmd to a yii,^.^ slraighl lute, which is not greater than the diameter of the circle. b-t ABC l)(j t\w. given circle, ;iud D the gi\en .stniiuht line not greater than tiie diameter of the circle : It is re(iuired to place in the ©ABC a chord ecjual to D. Draw CB, a diameter of the © ABC. Then if CB -^ D, the thing required is done. J^ut if not, CB must lie greater than D. From CB cut cti" CE ('(fual to D : and from centre C, with radius CE, describe the cutting the given circle at A. Join CA. Then CA shall l»e the chord I'tMjuired. For CA -: CE, l)eing radii of the AEF : and CE - D : .'. CA - D. J[yp, I. ;j. AEF, Contilr, . Problem. _ hi a yiv,n circle to inscribe a Iriawjle eqaianun/nr lo a given Inauyle. ^ ^ ! ! H *ii;! :M Ia'I ABC be the givou uirdc, and DEF the given triau-lr- It IS reqiunul to inscrihc in tlie ^j ABC a tnan-h' cuiiaiKmlar to tlie A DEF. n I o At any point A, on the ^^ of the ^ABC, draw the tangent GAH. ,,, t- ° III. 1 i. At A make tJK! _ GAB e(iual to tlie _ DFE ; i. l';{. and make the _ HAC (Miiial to the _ DEF. i. 20. -loin BC. Then ABC shall be the triangle re(|uired. lieeause GH is a tangent to the -ABC, and from A its point ot eontact the ehord AB is drawn, .•. the _GAB ihv. „ ACB in tlie alt. segment: in. ;}•' .-. the _ACB- the .DFE. (Jonstr. .Similarly the _ HAC the _ ABC, in the alt. segment- .-. the _ ABC -the _ DEF. Cou,,lr. Hence the third _ BAC the third _ EDF for the three anghvs in eaeh triangle are together enual to two rt. angles. i •'•' .•.the.;, ABC is e.juiangular to the .:. PEF, and it'ls msenbcd us ihc -sABC; V- K. F. BOOK IV. PUOi'. 3. 253 i;i\v till! III. 17. I. li.S. I. 2:3. Proposition ',i. Puoblem. AhoHi a fjiven circle to circumscribe a triauak egui- anynlar to a given trianyle. G E H Mb JU. 1. I. L';;. H' K. F. Let ABC bo the given circle, uml DEF iUr. -iveii ^ It IS re(,inred to circumscribe ubout the 0ABC a triangh^ i-quuingular to tlie .:. DEF. ^ • Produce EF both ways to G and H. Pind K tlie centre of the • ABC, and draw any radius KB. At K make the _ BKA equal to the _ DEG ; ^^ and make the _ BKC e(|ual to tiie _ DFH ' riirou^h A, B, C draw LM, MN, NL perp. to KA, KB, KC riien LMN shall be tlie triangle i-equired. Because LM, MN, NL are drawn pei-p. to radii at theii extnnnities, .•. LM, MN, NL ai-e tangents to the ciivle. ju. K). And because the four angles of tlu; (,uadrilateral AKBM together four rt. angles; j. 3,,. ^. ^,_ and of these, the _ -' KAM, KBM, are rt. ang' 's; Cundr . . the ^« AKB, AMB, together- two rt. angles. J5ut the _« DEG, DEF together -two rt. angles; i. !;{ •". the _ « AKB, AMB - tlii! _ « DEG, DEF ; and of these, the _ AKB - the _ DEG ; ' Conf,(r .". the I. AMB -the _ DEF. Simihiriy it may !)(> shewn that the _ LNM ^ the _ DFE .'. the third _ MLN - tlm third „ EDF. i. 3l\ .'. the /I LMN is e.piiangular to the ^ DEF and it is circumscribed about tJie (.:;ABC. ,. ,. ,.' 254 euclid'« klements. Mi l> I' H 1*K01'0S1TI()N I. PkOHLEM. To iuiscribe a circle in a (/icen IriaH'jfc Ifill^ Let ABC be the ifiveii ti-iii]i<'le : It IS re(|uirecl to inscribe a circle in tlio zlABC. Diset-t tlie _ '^ ABC, ACB by the st. lines Bl, CI, which intersect at I. ,^ «j_ From I draw IE, IF, IG perp. to AB, BC, CA. i. I'J. Then in the /.» EIB, FIB, ( the _ EBI : the L FBI ; Conslr. I5ecause -jand the _ BEI .the l BFI, being rt. angles; ( and Bl is connnon ; ■■• IE IF. I. 20. Similarly it may be shewn that IF ■ IG. .*. IE, IF, IG are all ecjual. Ki-om centre I, with radius IE, describe a circle: this cii'cle must pass through the points E, F, G ; and it will be inscribed in the A ABC. Koi- since IE, IF, IG are radii of the ;•; EFG ; and since the l. "" at E, F, G are rt. angles ; .". the C!.'EFG is touched at these points by AB, BC, CA : Jll. 10. .'. the .V EFG is inscribed in th(! ..ABC. Q. K. V. NoiK. i'lxtm page lo;} it U accu tlmt il' Al be jolucd, Uiou AI biswtsj the angle BAC. BOOK IV. PUOl'. 4. 255 Hence it foUowH that the bisectors of the angles of a tiiomjle are concurrent, the iMint of intersection hvimj t)ic centre of the invcrihetl circle. The centre of the circle inscribed in a triangle is sometimes called its In-centre. DllKIN'lTION. , CI, whicli r. S). 1. IL'. Constr. I. L'tJ. irc'lo : F, G; BC, CA ; m. IG. <^>. K. 1'. cd, tiien A I A fiivlt' wiiii'li Toiiciics <»ii(^ .sitl<' of a triaiiijh' and tlu' other two sidi's jnodiiccil is said to be an escribed circle of tlie triiinglc. To draw an escribed circle of a i/icen trianijlc. Let ABC bu the \i,\\vn triangle, of which the two sides AB, AC are produced to £ and F: it is reciuired to describe a circle touching BC, and AB, AC produced. JJisect the z » CBE, BCF by the st. lines BIj, Cli, which intersect at I,. i. 'J. From Ij draw IjC, IjH, IjK perp. to AE, BC, AF. I. 12. Then in the A" I^BG, I,BH, I'x^cause - the z liBG = the z IjBH. Constr. and the z l,GB = lhe'z l,HB, being rt. angles; also l,B is common ; l,G l,H. Similarly it may be shewn that IjH = l,K; .". I,G, l,H, IjK are all equal. From centre I, with radius l,G, describe a circle: this circle must pass through the points G, H, K : and it will be an escribed circle of the a ABC. For since IjH. I,G, \^K are radii of the (•; HGK, and since the angles at H, G, K are rt. angles, . the GHK is touched at these points by BC, and by produced : .'. the G GHK is an escribed circle of the a ABC. It is clear that every triangle has three escribed circles. AB, AC Note. From page 104 ii, is seen that if Al, be joined, then Alj bisects the angle BAC: hence it follows that The bisectors of iico e.vterior angles of a triangle and tiie bisector of the third angle are concurrent, the point of intersection being the centre of an escribed circle. 2f>(! Euclid's ktjcmrnts. F'ropositton "». Prohlrm. To i'ircu7nnrrihp, a flrcle. about a glceii triangle. 'lb Let ABC be the jajiven triangle : it is recjuii-ed to circumscribe a circle about the A. ABC. Draw DS bisecting AB at rt. angles ; i. 11. and draw ES bisecting AC at rt. angles; llitMi since AB, AC are neither par', nor in the same st. lino, .". DS and ES must meet at some point S. Join SA ; ,ind if S 1)e not in BC, join SB, SC. Then in the A" ADS, BDS. j AD BD I '.realise -/and DS is connnon to lioth ; (and tiie l. ADS = the _ BDS, l)eing rt. /mgles ; .•. SA -- SB. Siiuilaj'ly it may l)e shewn tliat SC - SA. .". SA, SB, SC are all equal. From centre S, with ladius SA, describe a circle: this circle must pass through the points A, B, C, and is Ihei'efoi'o circumscribed about the A ABC. 9. R. P. It follows that (i) when the centre of the circumscribed circle falls ir'tthiu tlie triangle, each of its angles must be acute, for (vich angle is then in a segment greater than a semicircle : (ii) when the centre falls on one of the, mhs of the triangle, the angle opposite to this side must be a right angle, for it is the angle in a scnnicircle: agle.. HOOK TV. PROP. 5. e A ABC. I. 11. niP St. lino, S. 267 (iii) when the centre falls vrithont the triangle, the angle opposite to the side beyond which the centre falls, must be obtu.se, for it i.s the angle in a segment less than a semicircle. Therefore, conversely, if the given triangle he acute-angled, the centre of the circnnmirihed circle falh uithin it: if it be a right-angled triangle, the centre falh on the hypotenuse ; if it be an obtuse-angled triangle, the centre falls without the triangle. ■^rfnn\T Pt^^.l?^ it is seen that if S be joined to the middle point ot BC, then the jommg hne is perpendicular to BC. H<'rtcf the perpemUciilars drawn to the sidenof a triangle from their middle pomtH are concurrent, the jmint of intersection hehiq' the centre of the circle circumscribed about the triannle. The centre of the circle circumscribed about a triauRle ia some- times called its circum-centre. E.XRRCLSRS. t. antfle.s : circle : , C, and is i}. R. P. circle falKs acute, for emicircle : ides of the he a right On the Inscribep, CiRcuMacRiBEP. and EscRiBKn CrRcnRs or a Trianolk. 1. An equilateral triangle is inscribed in a circle, and tangents are drawn at its vertices, prove that (i) the resulting figure is an equilateral triangle: (ii) its area is four times that of the given triangle. fi,,v^i ?^'?"?f. '^ '''\''}^, *° *°"<''^ t^^o parallel straight lines and a third straight line winch meots them. Shew that two such circles can be drawn, and that they are equal. 3. TiHir.yes which have equal bases and equal vertical analeg hare equal ctr-vnMribed circles. •' 4. I is the centre of the circle inscribed in the triangle ABC and I, IS the centre of the drcle which t-uches BC and AB, AC produced: shew that A, I, Ij are ccllinear. o. If the imcribed and , rcK^scribei' circles of a triangle are con- centric, shew that the triangle i. a^Uateral; and that the diameter of the circumscnhed circle is dmibk • r -if of the inscribed circle. 6. ABC is a triangle; and I. S are the centres of the inscribed and circurascnbed circles; if A, I, S are collinear, shew that AB- AC. 17-2 258 KUCLID'h KTiKMKNTS. 7, The 8um of the diameters of the inscrihed and circnmacribod circles of a right-anj^'lod triangle is equal to the sum uf the sides containing the right angle. 8, If the circle inscribed in a triangle ABC touches the sides at D, E, F, show that the triangle DEF is acute-angled; and express its angles in terms of the angles at A, B, C. n. If I is the centre of the circle inscribed in the triangle ABC, and I, the centre of the escribed circle which touches BC; shew that I, B, 1^, C are concyclic. 10. In any triangle the difference of two sides is equal to the dif- ference of the segments into which the third side is divided at the ])()int of contact of the inscribed circle. 11. In the triangle ABC the bisector of the angle BAC meets the base at D,and from I tlio centre of tlio inscribed circle a perijcndicular IE is drawn to BC : shew that the angle BID is equal to the angle CIE. 12. In the triangle ABC, I and S nro the centres of the inscribed and eircumseiil)od circles: show tliat IS subtends at A an angle equal to JiiiU" the difference of the angles at the base of the triangle. la. In a triangle ABC, I and S are the centres of the inscribed and circumsciibed circles, and AD is drawn perpendicular to BC: shew that Al is the bisector of the angle DAS. 14. Shew that the area of a triangle is equal to the rectangle contained by its semi -perimeter and the radius of the inscribed circle. ir.. The diagonals of a quadrilateral ABCD intersect at O: shew that the centres of tlie circles circumscribed about the four triangles AOB, BOC, COD, DOA are at the angular points of a parallelogram. Ifi. In any triangle ABC, if I is the centre of the inscribed circle, and if Al is produce I to meet the circumscribed circle at O ; shew that O is the centre of the circle circumscribed about the triangle BIC, 17. Given the base, altitude, and the radius of the circumscribed circle; construct the triangle. 18. Describe a circle to intercept equal chords of given length on three given straight lines. r.). In an equilateral triangle the radii of the circumscribed and escribed circles are respectively double and treble of the radius of the inscribed circle. 20. Three circles whose centres are A, B, C touch one another externally two by two at D, E, F: shew that the inscribed circle of the triangle ABC is the circumscribed circle of the triangle DEF, 1 ircnmscribod (.)!' tlic sides the Bides at i express its L-ianRle ABC, 3 ; shew tliat lal to the dif- ividod at the \C meets the )cri)endiculiir le angle CIE. tlie inscribed 1 angle equal ngle. the inscribed nilar to BC: the rectangle cribed circle. t at O : shew 'our triangles arallelograni. icribed circle, O ; shew that nglc BIC, :ircuniscribed reu length on mscribed and radius of the I one another ibed circle of ;le DER nooK IV. vnov. H. 259 Puoi'usiTiuN 0. Phoblkm. To inscribe a square in a f/ircn circ/e. Let A BCD bo the given circle : i< is ivcjuircd to inscribe a sfiuare in the ;•, ABCD. Find E the centre of the circle : m. | find diuw (wo diameters AC, BD pei'p. to one another, i. 11. Join AB, BC, CD, DA, Then the lig. ABCD shall Iks tlici S(|uare required. For in tho A^ BE A, DEA, f BE - DE, Because - and EA is conunon ; and tlie _ BEA = tin* _ DEA, ]iein<' rt. angles ; .". BA- DA. J, J f:>iniihir]y it may be shewn that CD ^ DA, and that BC _ CD. .". the iig. ABCD is equilateral. And since BD is a diameter of the ABCD, .". BAD is a semicircle; .". the _ BAD is a rt. ani^le. m, 31. ►Similarly the other an^^les of the fig. ABCD are rt. angles. .". the iig. ABCD is a square, and it is inscribed in the given circle. y. K. !■. [For Exercises see page 203. J 260 KUCLID'H KLEMKNT8. 'Ill' ;i Proposition 7. Problem. To circumscribe a square about a r/ivoi circle. Q A F B E J H K Let ABCD be the <^iven circle : it is required to circumscribe a square about it. Find E the centre of the ABCD : ill. 1. and draw two diameters AC, BD perp. to one auoth«M\ i. 1 1. Through A, B, C, D draw FG, GH, HK, KF perj). to EA, EB, EC, ED. Then tlie fig. GK shall be the square requircil. Because FG, GH, HK, KF are drawn perp. to radii at their extremities, .'. FG, GH, HK, KF are tangents to the circle, ill. IG. And because the .:. ** AEB, EBG are both rt. angles, Constr. :. GH is par' to AC. 1. 28. Similarly FK is par' to AC : and in like manner GF, BD, HK are par'. Hence the ligs. GK, GC, AK, GD, BK, GE are par'^ .•. GF and HK each ~ BD ; also GH and FK each AC : but AC - BD ; .". GF, FK, KH, HG are all equal : that is, the tig. GK is equilateral. And since the iig. GE is a par'", .•. the L. BGA - the l BEA ; i. .M. but the A BEA is a rt. angle ; Coimtr. .'. the .1 at G is a rt. angle. Himilarly the l " at F, K, H are rt. angles. .". the Iig. GK is a square, and it has been circumscribed about the ©ABCD. i^.E.F. le. t it. 111. 1. ii-. 1.11. to EA, EB, -ed. i at their . III. IG. s, Co7ist): 1. 28. ar"'\ 1. HI. Coiiatr, uiusuribod . E.F. BOOK IV. I'HOP. S. S61 PlioPOSITlON ?<. t'jluULEM. To inncriOe a circle in, a yicYii i^i/ntn A E D B H K Lt;t A BCD be the given square : it is required to inscribe a «nrcl<' in the s»i. ABCD. Bisect tlu! sides AB, F and E. i. lO. Tiu-ough E draw EH lui B or DC : i. 31. and thiuugh F draw FK i)ar' to At BC, meeting EH at G. Jsow AB -= AD, being the .sides of a square ; and their halves are ecjual ; Counlr. ■'■ AF AE. Jj: 7. Hut the lig. AG is a par'"; Count t: .'. AF~GE, and AE - GF; .•. GE OF. Siinihirly it may be shewn that GE GK, and GK GH : .'. GF, GE, GK, GH are all equal. From centre G, with radius GE, describe a circle; this circle must pass through the? points F, E, K, H : and it will be touched l)y BA, AD, DC, CB; iii. 16. for GF, GE, GK, GH are i Hi ; and the angles at F, E, K, H are rt. angles. i. 29. Hence the 0FEKH is inscribed in the sq. ABCD. y. K. F. [For ExercisL's see p. 20;}.] i'i ''•^^^^^Hl f ^W^^^^^^H ;;h^^^| .^I^^^H 'TW MICROCOPY RESOLUTION TEST CHART (ANSI and ISO TEST CHART No. 2) 1,0 1 i\M 2.8 2.5 lliM s;2 1^ mil 9.9 1 I.I 1 2.0 1.8 1.25 1.4 1.6 ^ APPLIED IfVHGE In. ^r"_ 'f"^ Eost Main Street ~.JS Rochester, New York 14609 USA '-= (716) 482 - OJOO - Phone =i (716) 2H8 - 5989 - Fax 262 EUCLID'S ELEMENTS. Propositiox 9. Problem. 7'o circumscribe a circle about a given square. i I n\t I. ])ef. 28. I. D<'f. L'8. I. 8. Let A BCD be the given square : it is required to circumscribe a circle about the sq. ABCD Join AC, BD. intersecting at E. Then in tlie A ^ BAG, DAC, ( BA = DA, Because -< and AC is connnou ; I and BC - DC ; .'. tlie L BAC the i. DAC : that is, the diagonal AC bisects the l. BAD. Similarly the remaining angles of the square are bisected by the diagonals AC or BD. Hence each of the l ^ EAD, EDA is half a rt. angle ; .'. the z. EAD = the /.EDA: .•. EA -- ED. I. G. Similarly it may be shewn that ED EC, and EC = EB. .'. EA, EB, EC, ED are all equal. From centre E, with radius EA, describe a circle : this circle must pass through the points A, B, C, D, and is therefore circumscribed about the sq. ABCD. Q. e.p. BOOK IV. PROP. 9. 263 IP Definition. A rectilineal figure about which a circle may be described is said to be Cyclic. EXERCISES ON PROPOSITIONS G — 9. 1. Jf a circlf can he inscribed in a quadviJateral, shew that the sum of one pair of oppoxite sides is equal to the sum of the other pair. 2. If the snm of one pair of opposite sides of a quadrilateral is equal to the sum of the other pair, shew that a circle may be inscribed in the figure. [Bisect two adjacent angles of the figure, and so describe a circle to touch three of its sides. Then prove indirectly by means of the last exercise that this circle must also touch the fourth side.] 3. Prove that a rhombus ami a square arc the onlij parallehirirams in ichich a circle can be inscribed. 4. All cyclic parallelograms are rectangular. 5. The greatest rectangle irhich con be inscribed in a given circle is a square. G. Circumscribe a rhombus about a given circle. 7. All squares circumscribed about a given circle are equal. 8. The area of a square circumscribed about a circle is double of the area of the inscribed square. 0. ABCD is a square inscribed in a circle, and P is any point on the arc AD : Shew that the side AD subtends at P an angle three tiv as great as that subtended at P by any one of the other sides. 10. Inscribe a square in a given square ABCD so that one of iu angular points should be at a given point X in AB. 11. In a given square inscribe the square of minimum area. 12. Describe (i) a circle, (ii) a square about a given rectangle. 13. Inscribe (i) a circle, (ii) a square in a given quadrant. 14. ABCD is a square inscribed in a circle, and P is any ]inii)t on the circumference ; shew that tbo sum of tlie squares on PA, PB, PC, PD is double the square on the diameter. [See Ex. 24, p. 147.] I 264 EUCLID'S ELEMENTS. PiioPosiTioN 10. Problem. To describe att isosceles trianyle havlny each of the awjles (It the base douhle of the third angle. Ti^kc any straight line AB. Divide AB at C, so that the rect. BA, BC -tlie sq. on AC. II. 11 From centre A, witJi radius AB. describe the ©BDE ; 1/. i. IV. 5. Constr. Constr. III. 37. and in it place the chord BD equal to AC. Join DA. Tlien ABD shall be the triangle required. Join CD ; and about the AACD cir-cumscribe a circle. Then the rect. BA, BC ^ the t (j. on AC - the sq. on BD. Hence BD is a tangen^ to tlie 0ACD : and from the point of contact D a chord DC is drawn ; .•. the _ BDC ^. the _ CAD in the alt. segment, in. 32. To each of these equals add the ;_ CDA : then the whole _ BDA ~i\\e sum of t' ^ CAD, CDA. But the ext. _ BCD - the sum of the _ ^AD, CDA; i. 32. .'. the _ BCD ^-the _. BDA. And since AB - AD, being iv.Jii of the BDE, .'. the L DBA ^the _ BDA : i. o. .". the L. DBC - the _ DCB; BOOK IV. PROP. 10. 265 I!' .'. DC^DB; 1. 6. that is, DC = CA : Constr. .'. the _ CAD - the l. CDA ; i. 5. .'. tlie sum of tlie _ " CAD, CDA - twice the angle at A. But the _ ADB - tlie sum of tlie _ ** CAD, CDA ; Proved. each of the _ ** ABD, ADB - twice the aiijjrle at A. (i. E. F. EXERCISES ON PROPOSITION 10. 1. 0. 1. In au isosceles triangle in which each of the angles at the base is double of the vertical angle, shew that the vertical angle is one-fifth of two right angles. 2. Divide a ri>jht angle into jive equal parts. 3. Describe au isosceles triangle whose vertical angle shall be three times either anj/le at the base. Point out a triangle of this kind in the figure of Proiwsitiou 10. 4. In theji;iure of Proposition 10, if the two circlea intersect aC F, slmo that BD'= DF. 5. 1)1 the Jujure of Proposition 10, sliew tliat tJie circle ACD /s etiual to the circle circumscribed about the triangle ABD. 6. In the figure of Proposition 10, if the two circles intersect at F, shew that (i) BD, DF are sides of a regular decagon inscribed in the circle EBD. (ii) AC, CD, DF are sides of a regular pentagon inscribed in the circle .aCD. 7. In the figure of Proposition Id, shew that the centre of the circle circumscribed about the triangle DBC is the middle point of the arc CD. 8. In the figure of Proposition 10, if i is the centre of the circle inscribed in the triangle ABD, and I', S' the centres of the inscribed and circumscribed circles of the triangle DBC, shew that S'i — S'i'. 266 EUCLID'S ELEiMENTS. \ 1 it 1( i' Proposition 11. PRonLEM. To inscribe a regular pentagon in a given circle. A Let ABC bo ii miveu circle : it is required to inscribe a regular pentagon in the G) ABC. Describe an isosceles AFGH, having each of the angles at G and H double of the angle at F. iv.*10. In the 0ABC inscribe the AACD e(|niangular to the AFGH, jy 2. so that each of the _ ^ ACD, ADC is dou])](i of the l CAD. Bisect the l"" ACD, ADC bv CE and DB, which meet the O'^at E and B. ' I 9^ Join AB, BC, AE, ED. Then ABCDE shall l)e tlie JV(jiiired regular pentagon. Because each of the _ ** ACD, ADC twice the l CAD ; and because the l " ACD, ADC are bisected by CE, DB,' ■ the tive ^ ^ ADB, BDC, CAD, DCE, ECA are all equal. .". the five arcs AB, BC, CD, DE, EA ai-e all equal. . the five chords AB, BC, CD, DE, EA are all equal .". the pentagon ABCDE is e(|uilateral. Again the arc AB ^the ai'c DE ; to each of these eciuals add the arc BCD : the whole arc ABCD - the whole arc BCDE : O III. 20. III. 2l». Proved. hence the angles at the 0'=*' which stand upon these equal arcs are equal; ju 07 that is, the l AED ^ the _ BAE. In like manner the remaining angles of the pentagon may be shewn to l)e equal ; '^ .". the pentagon is equiangular. Hence the pentagon, being botli equilateral and equi- angular, IS regular; and it is inscribed in the ©ABC. q.e.p. nooK IV. PROV. 12. 267 Proposition 12. TuomjiM. To circumscribe a reyulnr 2>6>itayon about a yiveu clrch. liocause Let ABCD he the given circu; ; it is i-equired to circumscribe a regular pentagon about it. Inscribe a regular pentagon in the 0ABCD, iv. 11. and let A, B, C, D, E be its angular points. At the points A, B, C, D, E draw GH, HK, KL, LM, MG, tangents to the circle. in. 17. Then .ihall GHKLM be the required regular pentagon. Find F the centre of the ©ABCD ; ni. 1. and join FB, FK, FC, FL, FD. Then in the two A« BFK, CFK, BF = CF, being radii of the circle, and FK is common : and KB KC, being tangents to the circle from the same point K. iii. 17. Cor. :. the L BFK ^ the L CFK, I. 8. also the l. BKF = the u CKF. i. 8. Cor. Hence the l. BFC ^ twice the l. CFK, and the L BKC ~ twice the i. CKF. Himilai-ly it may be shewn that the l. CFD = twice the l CFL, and that the z. CLD = twice the L CLF. But since the arc BC - tl:e arc CD, iv. .•. the L BFC the ^ CFD : ill. and the halves of these angles are equal, that is, the ^ CFK - the z_ CFL. 11. 27. i " 268 KUOLID'S ELRMRNTS. Q Then in the A** CFK, CFL, the L CFK = the L CFL, Proved. BeoauRO \ and tlie l FCK tlie l FCL,bfnn,i;rt..iiij4le,s, ui. 18. and FC is comnion ; .'. CK : CL, I. 'JO. and the l FKC =- the l FLC. Hence KL is double of KC ; similai-ly HK is double of KB. And since KC ~ KB, ill. 17. Cor. :. KL=:HK. In the same way it may be shewn that every two con- secutive sides are equal ; .'. the pentao;on GHKLM is equilateral. Again, it has been proved tliat the l FKC = the :_ FLC, and that the l ^ HKL, KLM are respectively double of these angles : .•. th(^ _ HKL = the _ KLM. In the same way it may be shewn that every two con- secutive angles of the figure are equal ; .'. the pentagon GHKLM is equiangular. .'. the pentagon is i-egular, and it is circumscrilied about the 0ABCD. Q.E. P. CoROLLARV. Similarly it viay he proved thai if tangents are drawn at the vertices of any regular polygon inscribed in a circle, they rc'dl form another regidar polygon of the same species circumscribed about the circle. [For Exercises see p. 270.J 1 nooK TV. PROP. 13. 209 1 PuoposiTiox 1.'). Phohlkm, 1 7V) hisi-rihe a clri'ff in, o, (/Ircn rcgiiltir pfi)itaqnn. I A I G/<^^-^^^S^ 1 I Proved. hV^ \"""~"/- 1 s,ui.l8. V / \f ■ I. 26. C K D Let ABODE be the given regulju* pentagon : Ml eof KB. it is icfiuired to inscribe a circle within it. II 17. Cor. 1 Mispct two consecutive l^ BCD, CDE by CF and DF ? which intei'sect at F. T 1) ■ M' '^1 wo con- Join FB ; 11'' 1 and draw FH, FK perp. to BC, CD. T. 12, if 1 Then in tlie A' BCF, DCF, ' ' ^1 L FLC, of these BC : DC, Hyp. JVcause - and CF is common to both ; ^^^1 and the l BCF == the l DCF ; f'onnfr. .". the Z.CBF -:the .lCOF. i. 4. i 1 wo con- But the _ CDF is half an angle of tl.e regulai" pentagon : .', also the ^ CBF is half an angle of the regular pentagon : that is, FB bisects the l ABC. 1 rl about So it may lie shewn that if FA, FE were joined, these '1 Q. E. F. lines would bisect the i_ ** at A and E. 1 tmigenffi Again, in the A*" FCH, FCK, :| ribed in ( the L FCH = the z. FCK, Consfr. . MM he same Because < and the l FHC -the i. FKC being it. nngles ; 1 ( also FC is common ; ' /H • •. FH = FK. I. 20, ■ tH Similarly if FG, FM, FL be drawn perp. tn BA. AE. ED, H it may be shewn that the tive perpendiculars diawn from F - ' H to the sides of the pentagon are all equal. 1 u 270 kuclid'h ki.kmionts. A From centre F, witli radius FH, lU'serilx! a circle; lliis circle* must pass tlirougii the points H, K, L, M, G ; and it will 1)0 touclied at those points by the sides of tiie pentagon, for the :_ ** at H, K, L, M, G ar(! rt. l '. Constr. .', the 0HKLMG is inscribed in the given pentagon. (^. e. i'. (JoKOiiLAKV. 7'Af bisectors of the anyles of a reynlar pcutayou meet eiidiculavs therefore a idiculars as BOOK IV. NOTE ON REGULAR POLYGONS. 275 2. If a polygon inscribed in a circle in equilateral, it is also equiangular. Let AB, BC, CD be consecutive sides of an equilateral polygon inscribed in the © ADK; then shall this polygon be equiangular. Because the chord AB = the chord DC, Hyp. .*. the minor arc AB = the minor arc DC. in. 28. To each of these equals add the arc AKD : then the arc BAKD=rthe arc AKDC; .•. the angles at the C^o, which stand on these equal arcs, are equal ; that is, the z BCD = the z ABC. iii. 27. Similarly the remaining angles of the polygon may be shewn to be equal: .•. the polygon is equiangular. q.e.d. 3. If a polygon inscribed in a circle is equiangular, it is also equilateral, proinded that the number of its sides is odd. [Observe that Theorems 2 and 3 are only true of polygons inscribed in a circl'^. The accompanying figures are sufficient to shew that otherwise a polygon may be equilateral without being equiangular, Fig. 1; or equiangular without being equilateral. Fig. 2.] Fig. I Fig. 2 r^^.y for we are enabled to divide the Note. The following extensions of Euclid's constructions Eegular Polygons should be noticed. By continual bisection of arcs, circumference of a circle, by means of Proposition G, into 4, by means of Proposition ir>, into 3, by means of Proposition 11, into 5, 10, 20,. by means of Proposition 16, into 15, 30, GO, 8, IG,..., G, 12,..., 2 . 2", . . . equal parts ; 3 . 2",... equal parts; 5 . 2",... equal parts; 15 . 2", . , . equal parts. Hence we can inscribe in a circle a regular polygon the number of whose sides is included in any one of the formula) 2 . 2", 3 . 2", 5 . 2", 15 . 2", 11 being any positive integer. In addition to these, it has been shewn that a regular polygon of 2" + l sides, provided 2" + l is a prime number, may be inscribed in a circle. 276 Euclid's elrments. I t EXERCISES ON PROPOSITIONS 11 — IG. 1. Express in terms of a right angle the magnitude of an angle of the following regular polygons : (i) a pentagon, (ii) a hexagon, (iii) an octagon, (iv) a decagon, (v) a quindccagon. 2. The angle of a regular pentagon is trisected by the straight lines which join it to the opposite vertices. 3. In a polj-gon of n sides the straight lines which join any angular point to the vertices not adjacent to it, divide the angle into a - 2 equal parts. 4. Shew how to construct on a given straight line (i) a regular pentagon, (ii) a regular hexagon, (iii) a regular octagon, 5. An equilateral triangle and a regular hexagon are inscribed in a given circle ; shew that (i) the area of the triangle is half that of the hexagon ; (ii) the square on the side of the triangle is three times the square on the side of the hexagon. 6. ABODE is a regular pentagon, and AC, BE intersect at H : show that (i) AB = CH = EH. (ii) AB is a tangent to the circle circumscribed about the triangle BHC. (iii) AC and BE cut one another in medial section. 7. The straight lines which join alternate vertices of a regular pentagon intersect so as to form another regular pentagon. 8. The straight lines which join alternate vertices of a regular polygon of 71 sides, intersect so as to form another regular polygon of n sides. If 9! = G, shew that the area of the resulting hexagon is one-third of the given hexagon. 9. By means of iv. IG, inscribe in a circle a triangle whose angles are as the numbers 2, 5, 8. 10. Shew that the area of a regular hexagon inscribed in a circle is three-fourths of that of the corresponding circumscribed hexagon. 1 i;: H la u'i i,n angle of le straight 1 join any angle into u* octagon, iscribed in THEOREAtS AND EXAMPLES ON BOOK IV. ;on; times the sect at H : about the f a regular f a regular polygon of Dne-third of ngle whose i in a circle I hexagon. I. ON THE TRIANGLK AM) ITS CHICLES. 1 D E F arc the points of contact of the inscribed circle of the triangle ABC. and D„ E^, F, the points of ^»f "''^ «//^%Xo'f« circle, which touches BC and the other sides produced: a, b, c denoe the lengths of the sides BC, CA, AB; s the semi-penmeter of the triangle, and r, ri the radii of the inscribed and escribed circles. Prove the folloicing equalities:— (i) (ii) (iii) AE = AF==s-n, BD= Bf=s-b, CD = CEt:^s-r. AE, = AFi=^s CDi = CEi-^ BDi=BFi ,s- - h, S -C. (iv) CD = BDi and BD^CDj. (V) EEi^FFj — a. (vi) The area of the A ABC =r»=rj(s-«). iii 278 EUCLID'S ELEMENTS. 2. In the friaurjie ABC, I u the centre of the inscribed circle, and I,, lo, I., the centres of the escribed circles touchimj respectively the sides BC, CA, AB and the other sides jivoduccd. Prove the folloiving properties : — (i) The points A, I, I, are collinear; so are B, I, I.,; and C, I, I,,. (ii) The points L, A, I., are collinear; so are I,, B, I,; and anothei joi (iii) The triannlrs BI,C, CI,A, Al.jB are equiangidar to one _ _ _ (iv) The triangle \^\.,\.^h equiannular to tlie triamile formed bu joining the points of contact of the inscribed circle. " ' (v) Of the four points I, I,, l„, I,, each is the orthocentre of tlie triangle whose vertices are the other three. (vi) The four circles, each of which passes throunh three of the points \, \^, L, I.,, are all equal. ' THEOREMS AND EXAMPLES OX BOOK IV. 279 3. With the notation of page 277, shew that in a triangle ABC. If the angle ut C is a right angle, r = s - c ; )\ — s-h; r» ~s - a ; 7:^ = s. 4. With the figure given on page 27*'3, shew that if the circles whose centres are I, Ij, T,, I;, touch BC at D, Dj, D^, D.j, then (i) DD.= D,D., = i. (ii) DD.,= DiD.-c. (iii) D.^D.y-^b + c. (iv) DD^^b~c. 5. Shew that the orthocoitre and verJce.t of a trhnujle are the centres of the inscribed and escribed circles of the jicdtil tria)itilc. [See Ex. 20, p. 225.] 6. Given the base and vertical angle of a trian;ile,find the locus of the centre of tJie inscribed circle. [See Ex. 30, p. 228.] 7. Given the base and vertical angle of a triangle, find the locus of the centre of the escribed circle which touches the base, 8. Given the base and vertical angle of a triangle, shew that the centre of the circumscribed circle is fixed. 9. Given the base BC, and the vertical angle A of a triangle, fintl the locus of the centre of the escribed circle which touches AC. 10. Given the base, the vertical angle, and the radius of the inscribed circle ; construct the triangle. 11. Given the base, the vertical angle, and the radius of the escribed circle, (i) which touches the base, (ii) which touches one of the sides containing the given angle ; construct the triangle. 12. Given the base, the vertical angle, and the point of contact with the base of the inscribed circle; construct the triangle. 13. Given the base, the vertical angle, and the point of contact with the base, or base produced, of an escribed circle ; construct the triangle. 1-1. From an external point A two tangents AB, AC are drawn to a given circle ; and the an;^'le BAC is bisected by a straight line which meets the circumference in I and I,: shew that I is the centre of the circle inscribed in the triangle ABC, and Ij the centre of one of the escribed circles. 15. I is the centre of the circle inscribed in a triangle, and Ij, lo, I- the centres of the escribed circles ; shew that llj, IL, II3 arc bisected by the circumference of the circumscribed circle. 16. ABC is a triangle, and l^, ly the centres of the escribed circles which touch AC, and AB respectively: shew that the points B, C, \„, ij lie upon a circle whose centre is on the circumterence of the circle circumscribed about ABC. m 280 Euclid's elements. PI- !|H 17. With three given points as centres describe three circles touching one another two by two. How many sohitions will there be? 18. Two tanf,'cnts AB, AC are drawn to a given circle from an external jioint A; and in AB, AC two points D and E are taken Ro that DE is eciual to the sum of DB and EC: shew that DE touches the circle. 19. Given the perimeter of a triangle, and one angle in magnitude and position : shew that the opposite side always touches a fixed circle. 20. Given the centres of the three escribed circles ; construct the triangle. 21. Given the centre of the inscribed circle, and the centres of two escribed circles ; construct the triangle, 22. Given the vertical angle, perimeter, and the length of the bisect jr of the vertical angle; construct the triangle. 23. Given the vertical angle, perimeter, and altitude ; construct the triangle. 24. Given the vertical angle, perimeter, and radius of the in- scribed circle ; construct the triangle. 25. Given the vertical angle, the radius of the inscribed circ'e, and the length of the perpendicular fron; the vertex to the base ; construct the triangle 26. Given the base, the difference of the sides containing the vertical angle, and the radius of the inscribed circle ; construct the triangle. [See Ex. 10, p. 258.] 27. Given a vertex, the centre of the circumscribed circle, and the centre of the inscribed circle, construct the triangle. 28. Ill a triangle ABC, I is the centre of the inscribed circle ; shew lliat thn centres of the circles circumscribed about the triangles BIC, CIA, AIB lie on the circumference of the circle circumscribed about tli(i given triangle. 29. In a triangle ABC, the inscribed circle touches the base BC at D ; and r, r^ are the radii of the inscribed circle and of the escribed circle which touches BC : shew that BD . DC. 30. ABC is a triangle, D, E, F the points of contact of its inscribed circle ; and D'E'F' is the pedal triangle of the triangle DEF : shew that the sides of the triangle D'E'F' are parallel to those of ABC. 81. In a triangle ABC the inscribed circle touches BC at D. Shew that the circles inscribed in the triangles ABD, ACD touch one anothei'. luee circles iUtherebe? ile from an [ are taken DE touches 1 magnitude fixed circle. )nstruct the 3 centres of Qgth of the ; construct I of the in- ribed circ'e, ) the base ; taining tlie instruct the 10, p. 258.] cle, and the nrcle ; shew angles BIC, ;ribed about base BC at ;he escribed its inscribed DEF : shew ABC. BC at D. > touch one THEOREMS AND EXAMPLES ON BOOK IV. 281 O.N THE NiNK-roiNTS ClUCLE. 32. In any trimifili' the vihJdle polnt.'^ of the s/Jcs, the feet of the perpendiculars drawn from the vertices to the opposite sides, and the middle points of the 'lines joining the orthocentre to the vertices are concijcUc. In the A ABC, let X, Y, Z be the middle points of the sides BC, CA, AB ; let D, E, F be the feet of the ))orp'* drawn to these sides from A, B, C ; let O be the orthocentre, and a, /3, 7 the middle points of OA, OB, OC: then shall the nine points X, Y, Z, D, E, F, a, /3, 7 be concyclic. Join XY, XZ, Xa, Ya, Za. Nowfrom the a ABO, since AZ = Z B, andAa^aO, /////). .-. Za is par to BO. Ex. 2, p. •)•;. And from the A ABC, since BZ = Z A, andBX = XC, Hup- :. ZX is par' to AC. But BO makes a rt. angle with AC ; .-. the Z XZa is a rt. angle. Similarly, the z XYa is a rt. angle. .-. the points X, Z, a, Y are concyclic : that is, a lies on the C'"'" of the circle, which passes through X, Y, Z ; and Xa is a diameter of this circle. Similarb" it may be shewn that (i and 7 lie on the C" of the circle which passes through X, Y, Z. Again, since aDX is a rt. angle. Hyp. .'. the circle on Xa as diameter passes through D. Similarly it may be shewn that E and F lie on the circumference of the same circle. .-. the points X, Y, Z, D, E, F, a, |3, 7 are concyclic. q.e.d. From this property the circle which passes through the middle points of the sides of a triangle is called the Kine-Foints Circle ; many of its properties may be derived from the fact of its being the circle circumscribed about the pedal triangle. Hyp. 1.29. I 282 kuclid's klkments. 3ii. To prove that (i) the cottn' of the nine-points circle is the middle point of the straight line ivhich joins the orthocentre to the circumscribed centre : (ii) the riidins of the nine-points circle is half the radius of the circumscribed circle : (in) the centroid is coUinear with the circumscribed centre, the nine-jjoints centre, and the orthocentre. In the A ABC, let X, Y, Z be the niidcUo points of the sides; D, E, F tiie foot of tlic porp"; O the ortho- ccntro; S and N tlie centres of the circumscribed and nine-points circles respectively. (i) To prove that N is the middle point of SO. It may bo shewn that the perp. to XD from its middle point bisects SO; Ex. 14, p. 98. Similarly the perp. to EY at its middle point bisects SO : that is, these perp" intersect at the middle point of SO: And since XD and EY are chords of the nine-points circle, .-. the intersection of the lines which bisect XD and EY at rt. angles is its centre: ni. 1. .•. the centre N is the middle point of SO. (ii) To prove that the radius of the nine-points circle is half the radius of the circumscribed circle. By the last Proposition, Xa is a diameter of the nine-points circle. .-. the middle point of Xa is its centre: but the middle point of SO is also the centre of the nine-points circle. {Proved.) Hence Xa and SO bisect one another at N. Then from the a" SNX, ONa ( SN = ON, Because] andNX^Na, (andthe zSNX = the zONa; .-. SX = Oa = Aa. And SX is also par' to Aa, .-. SA = Xa. But S A is a radius of the circumscribed circle ; and Xo i-z a diavictcr of the nine-points circle ; .*. the radius of the nine-points circle is half the radius of the circum scribed circle. I. 15. 1.4, I. 33. y^ ille potnt of ibed centre: adius of tite ! centre, the A yfj \ ^ 1 I f 1 7 "- V /... " \ /jN^ o\e D C SO: ,'cle, it rt. angles in. 1. rcle is half Jints circle. oints circle. {Proved.) 1.15. I. 4. I. 33. he circum- THEOREM8 AND KXAMl'LI'-S ON HOOK IV. 283 (iii) To prove that the centroid is collinear with points S, N, O. Join AX and draw a,'/ par' to SO. Lot AX lUL'ct SO at G. Then from the a AGO, since Aa = aO and ay is par' to OG. .-. fii'j-^-uG. Ex. i:i, p. 98. And from the A Xaq, since aN - NX, and NG is par' to a;/, .-. i;CT0UY. TIio first four books of Euclid deal with tlio .absolute equality or inequality of (Geometrical magnitudes. In tho Fifth Book magnitudes are compared by considering their ratio, or relative greatness. The meaning of the words ratio and proportion in their simplest aritlimetical sense, as contained in tho following defini- tions, is probably familiar to the student : The ratio of one number to another is the multiple, part, or parts that the first number is of the stjcond; and it mai/ therefore be measured bi/ the fraction of v:hich the first number ts tho numerator and the second the denominator. Four numbers are in proportion when the ratio of the first to the second is equal to that of the third to the fourth. But it will be seen that these definitions are inapplicable to Geometrical magnitudes for the following reasons ; (1) Pure Geometry deals only with concrete magnitudes, re- presented by diagrams, but not referred to any common unit in terms of which they are measured: in other words, it makes no use of number for the purpose of comparison between difterent magnitudes. (2) It commonly happens that Geometrical magnitudes of the same kind are incommensurable, that is, they are such that it is impossible to express them e.vacthj in terms of some common unit. For example, we can make comparison between the side and diagonal of a square, and we may form an idea of thcu' relative great- ness, but it can be shewn that it is impossible to divide either of thein into equal parts of which the other contains an exact nunther.^ And as the magnitudes we meet with in Geometry are more often incom- mensurable than not, it is clear that it would not always be possible to exactly represent such magnitudes by numbers, even if reference to a common unit were not foreign to the principles of Euclid. Tt IP. therefore necessary to establish the Geometrical Theory of Proportion on a basis quite independent of Arithmetical principles. This is the aim of Euclid's Fifth Book. i I 2Hfi KUCr,ID's KLKMKNTS. 111!' Wo Khali nnploy tlu! following notation. Capital letters, A, B, C, ... will be used to dcuoto tho maRnitudcB tlienisclvcs, not diiij .uiiiin'rinil or (itficbntintl iniutninrit of tlunii, and small lettciH, in, ii, p,... will lie usid to dcnott; whole niiiulxTH. AIho '* .ill ho nsHiuueil that niiiUi)ilifati((ii, in tiic Kenst; of repeated uiilion, cat) he ajipiiod to any nia^jnitiido, so that j;t . A or «(A will luiioto tiio ni**gniMide A taken in times. The symbol > "ill ho n^^ed t' = or < B ; and conversely. (2) HiA -I- «iB + ... = ;« (A + B +.. .). (a) If A - B, then 7hA - iuB-= m (A - B). (I) wA + »A i-...-(/«|-n + ...) A. (5) If in --^ It, then ?// A - itA ~- (in - n) A. (()) VI . /(A = inn . A - nm . A -- ii . ?// A. ',]. The Ratio of one magnitude to anotiior of ti»e sanu) kind is the relation wliidi the llrst bears to the second in respect of qiiantnpUritij. The ratio of A to B is denoted tliu.';, A : B; and A is called the antecedent, B the consequent of the latio. The term quaiitiipliciti/ denotes the capacity of the fir.st magnitude to contain the second with or without renuiinder. If the magnitudes are commensurable, their quantuplicity may be expressed nnincvicallii by observing what multiples of the two magnitudes are equal to one another. Thus if A — ma, and B — mi, it follows that nA niB. In this case in A = — B, and the quantuplicity of A with respect to B is the arith- metical fraction — . \ *..: DEFINITION,-^. 287 ma(^nitudos /■ thi'iii, and iImts. AIho of rrpf'jitcd \ or iiiA will , 1111(1 • : fi or Itiplo of t\ imiubur of But if till! iimniiitiulcs luo iiicoiiiuiciitjiirublo, no nuiltiplo of tlu; fust cull In i'(|iml to any multiple ttf tlio hccoiuI, and tluioforo tho i|uantiiplicitv of onn with iiHpcct to tlio other eannot exuetly ho cxpreHsed niiiiui ically : in tliis case it in deteiiniiied l)y examining' how tho niul|i|ilif< of oiio niaj^'nitudo uiu distiihuted among the multiplcH of the otiier. Thus, let all llie niiiltipi of A lif I'diiikmI, th ^m' extending,' («/ illji III turn; also let all Ihi' iiii , tuples of B he foiineu ;,;;il placed in tlieir proper oi Icr of iiia^^'iiiliide aMioii!,' llir multiples of A. 'riiis foriim thu relative s. ile of the two nui; iiiludes, and the (piantiiplieity of A with re8i)ect to B is cHtimated UyexuminiiiK how tlie multipleH of A are distributed anions tliose of B in their n lativo scale. In other words, the ratio of A to B is known, if for all integral valueH of in wo know the multiples iiB und (n + i) Q lutwcen whicli mA licH. Itiple of u iuiiiV)er of Hclf-ovldent. < B ; anil In tho caso of two given maL'nitudos A and B, tiic relative scale of multiples is definite, and is ditTiicnt fn.ni that of A to C, if C ditYers fmm B by any ma^'iiitude however small. lor let D he tlio dilTerenco between B and C ; tiien liowevei Hinall D maybe, it will lie possible to Und a number in such tliut ;//D -A. In tiiis ease, iiiB und mC wmild differ by u mu^,'nitude greater than A, und therefore cioiild not lie iietweeii Hk; .same two niiilliples of A ; ho that after a certain point tho relative Hcale of A and B would dilTor from that of A and C. f the same second in ; and A is itio. ;t magnitude magnitudes iniiitcricalhf Miual to one In this case is the arith- [It is worthy of notice that we can always estimate, the ari hmctical ratio of two iucommeusurublc magnitudes ivitltin any rcqiui d degree of accuracy. For suppose that A und B are incommensurable; divide B into iii eijuul jiiirts eueli equal to /•<, so tliut B _: /H^i, where m is an iih'gcr. Also suppose (i is contained in A moro than n times and les. than (« + l) tunes; then A B ^und<(^:^i)^. /"/i 111^ ' tiiiit is, lies between - and D VI III so that differs from - by a (luantity less than - O III, • "^ III And since ■\e can choose ji (our unit of measurement) us small as we please, iii ca t be made us great us we please. Hence - can bo made as small as w III please, and two integers n and m can be found whose ratio will exproK.- that of a and b to any rciiuirod degree of accuracy.] U. L. 19 288 EDCLIUS ELEMENTS. It 1. The ratio of one luagnitude to another is equal to that of a third magnitude to a fourth, when if any equi- multiples whatever of the antecedents of the ratios art; taken, and also any equimultiples whatever of the con- sequents, the multiple of one antecedent is greater than, equal to, or less than that of its consequent, according as the nuiltiple of the otlier antecedent is greater than, ecjual to, or less than that of its consecjuent. Thus the ratio A to B is equal to that of C to D when mC > - or < nD according as viA > - or < uB, whatever whole numbers '/a and ii may be. Again, let m be any whole number whatever, and u another whole luimber determined in kucIi a way tliat either iii/K is eciual to iiB, or 7H,A lies between «B and (« + l) B; then the definition asserts that the ratio of A to B is equal to that of C to D if mCuD when ;)(A=^hB; or if mC lies between uD and (n + l) D when m/K lies between uB and (!M-1)B. In other words, the ratio of A to B is equal to that of C to D when the multiples of A are distributed among those of B in the same manner as the multiples of C are distributed among those of D. 5. When the ratio of A to B is ec^ual to that of C to D the four magnitudes are called proportionals. This is ex- pressed by saying " A is to B as C is to D", and the proportion is written A : B : : C : D, or A : B : D. A and D are called the extremes, B and C the means; also D is said to be a fourth proportional to A, B, and C. Two terms in a proportion are said to be homologous when they are both antecedents, or both consequents of the ratios. [It will be useful here to compare the algebraical and geometrical definitions of proportion, and to shew that each may be deduced from the other. According to the geometrical definition A, B, C, D aie in propor- tion, when mC> = ofD according as ;hA> = ' "'I'l^iplving both sides by '" , ^^c obtain viA _ )/(C iiB " nD '" lience from tlie nature of fiactions, mC:- ^ ^-iiD according as ?;tA> :;= <»B, which is tiie geometrical test of proportion. Given that «tC> :_ D ; tlien it will be possible to find some fraction -- vvhicli lies between them, n and ,ii being positive integers. A „ B'^in (1); C n b"m {•■^h l''ioi>i (1), vhA>«B; and these contradict the hypothesis. Therefore ^ and ^ are not unequal ; that is, -^ = g ; which provea the proposition.] G. Tlie ratio of one nia-iiitude to anotiun- is greater tiian that of a third magnitude to a fourth, M-hen it is liossible to find equimultiples of the antecedents and equi- niultiples of the consequents such tliat while the multiple ot the antecedent of tlie lirst ratio is greatcu- than, or equal to, that of Its consequent, the multiple of the anteced«.w of the second is not greater, or is less, than that of its consequent. 19-2 llenee uud •3;j 290 KUCLIDS ELEMKNTS. This dctiiiitiun iisserls tluit il whole numbers m aucl n can be foiina Buch that wliile viA is (greater than uB, iiiC is not pi'cater than nD, or while )«A--»B, mC is loss than iiD, then the ratio of A to B is greater than that of C to D. 7. It" A is VA\mi\ to B, ilw. r.itio of A to D is called a ratio of equality. it' A is gi'eatei' tii.iii B, tlic; ratio of A to B is (.•ailed a ratio of greater inequality. if A is less than B, the ratio of A to B is called a ratio of less inequality. 8. Two ratios arc saitl to be reciprocal when the ante- cedent and consequent of one are the conse(iuent and ante- cedent of the other respectively; thus B : A is the reciprocal of A : B. I). Three niagnitud(>s of tlu; same kind ai'c; said to be pro])ortionals, Avlien the ratio of tlu; first to the second is equal to that of the second to the third. Thus A, B, C ;u'e proportionals if A : B :: B : C. B is called a mean proportional to A and C, and C is called a third proportional to A and B. 10. Three or more magnitudes are said to be in con- tinued proportion when the ratio of the lirst to the second is equal to that of the second to the third, and the ratio of the second to the third is equal to that of the third to the fourth, and so on. 11. When there are any nuadxu" of magnitudes of the same kind, the hrst is said to have tbvaiciil treatment of ratio and proportion when applied to geometrical magnitudes cannot be consideied exact, it will periiaps be useful here to summarise in algebraical form the principal theorems of proportion contained in Book V, The student will then perceive that its leading propositions do not introduce new ideas, but merely supply rigorous ))roofs, based on the geometrical definition of proportion, of results already familiar in the study of Algebra. We shall only here give those propositions which are afterwards referred to in Book VI. It will be seen that in their algebraical form many of them are so simple that they hardly require proof. Summary op Prin'cipal Theorems of Book V. Phoposition 1. Jiatios which are equal to the same ratio are equal to one another. That is, If A : B-X : Y and C : D-rX : Y; then A:B-C:D. Paoi'osrnoN ;i. If four vuuinituilr.^ arc proportioiialt, tlit'ij arc aho proportionals ivlieu taken iurcry-elij. That is, if A : B:^C : D, then B:A=:D:C. This inference is referred to as invertendo or Inversely. PuorosiTioN 4. (i) Equal waiinitUiJen hare tlic name ratio to the i: .'//.-' y.aiiie ratio tn equal iiiapnitudes. For if A-^B, then C:A^C:B. ortion when pxact, it will ;he principal nt will then w ideas, hut definition of ehra. e afterwards braical form )f. K V. le another. proportionah y. mannitudi'. inptiifudcs. SUMMARY OF PRINCIPAL THEOREMS OF BOOK V. Proposition C. 'J93 (i) Miiiinitudeii ichich have thf same ratio to thf same maqnitude are equal to one another. That is. if A : C---B : C, tlien A-B. (ii) Those iiuiiiuiludi-s to irliieh tlir savie mafinitiide has the .lame ratio are equal to one aitotlwr. That is. if C : A C : B. then A B. Vhopostiion H. Maiinitudes have the same ratio to otic another which their equi- multiples have. That is, A : B. wA : mB. where vi is any whole nnmbei-. I'liOPOSTTTOX 11. If four maiinitudes of the same kind are i)rnportinnaU, thei/ are aho proportionals when talen alternateli/. If then shall For sijice u- 1 • 1 B , A B C B .-. mnltiplyuiK by ^ , we liav.' g • ^ " ^ " C ' A : B C D, A : C B D. A C B " D' that is, A _ B c~b' or A : C = B : D. This inference is referred to as alternando or alternately. r jo if 294 i:urr,nvs klkmknts. I'lJOPOHITION 12. If (oiji iinnihfr of HKiiinititdcK of the Hniiir hind aif proportionaU, then tin oiip itf thi' (iiiti'c('(}r)itK /.v to its coiisciincut, xo ix thr xiini of thi- (iiifeccdeiits to thr smii of the coiinciiui'iits. Let A : B -C : D--E : F-=...; then shall A : B - A + C j- E i ... : B-f D + F + .... ACE For put oach of tlio oiiual vntios , _ , ^ ,. . . oqnal to /; ; B D F thfn A=--Bk, C D/,, E-F/,-. :/• ACE A + C + E-!-... _B/,'-l-D/, + F/,- ^ " B-!-D i F-i ... ' B + D + F . "~B D F .-. A : B = A + C + E i... : B + D i F i .... This inference is sometimes referred to as addendo. Proposition liJ. (i) If four nuifinitudoa are proportioiudx, the sum of the fi rut and seroiid. in to the second as the sum of the tliird and fourth is to the fourth. Tiet A : B = C : D, then shall A+B : B = C + D : D. For since A C B D' A , C , .•.B + 1 = D + 1 = tliat is, A+B C+D B ~ D ' or A + B:B = C + D:D. This inference is referred to as componendo. (ii) f four magnitudes^ are proportionals, the difference of the first and seco)id is to the second as the difference of the third and fourth n to the fnirtJi. That is, if A : B = C : D, then A~B:B-C~D:D. The ]n'oot' is similar to that of the former case. This inference is referred to as dividendo. SUMMARY OP rnTXOTT'AL TIIKOUKMS OF nooK V, onr. 95 ■oportioniil>i, ' tilllll of tlic hr fimt ami > tite fiuirlli. of the first fourth if. to Puoi'OfilTION 14. If there are two sets of wafpiitiules, such that the first !s to the second oj thejirst set as the first to the second of the other .set, and the serond to the third of the frst set as the second 'to the third of the other and so on to the last nuiimitiide : then the frst is to the last' of the I'rit set as thejirst to the last of the other. First lot thoro ho. tliroo maRnitudos. A, B, C, of ono set, and tluw, H. Q, R, ot anotlioi' set, and let and then shall For since A : Br-P: Q, B :C-Q : R; A :C-P : R. A P , B Q B^Q-^'>'^C^R = A B that is, or B C A : C P Q Q • R' P r' P : R. Similarly if A B B-P:Q, C=.Q: R, L: M = Y : Z; it can hfi proved that A : M - P : 2 . This inference is referred to as ex sequali. Conni.LARV. If and then shall For since or A : B-P : Q, B : C-R : P; A : C-R : Q. A P , B R B"^Q'""^^C-^P A B P R ■ B • C " Q • P • A_ R ■ C Q' A : C-R : Q. 296 buclid'8 elements. Hhoposition 15. Kf A : B = C : D, and E : B-F : D; thfn nhall A i-E : B-.C 1 F : D. For Hinco AC , E F B~D'""'* B D = . A i E 1 F B " ' D ' tliat is, A 1 E : B..C IF : D. t)iat Ib, Tropobitiox 1(5. I-'et A : B=:C : D; Ihen slmll tho duplicate ratio of A : B he Iqmd to (,]„■ du,,li„at. ratio Let X be a third proportional to A, B; so that A : B-B : X; B_A ■ X b' B A_A A X • B B ' b • A _ A2 X B~-' But A : X is the duplicato ratio of A : B ; .-. the duplicate ratio of A : B - A- : B-'. But since A : B = C ; D; AC ■ B D ' A'-' _ C-' ■ B-' D^' Of A- : B2_^C-^ : D- ; that is, the duplicate ratio of A : B - tho duplicate ratio of : D. Converxehi, let the duplicate ratio of A : B be eaual to fbp f\»r<\\ cate ratio of C : D ; " • d ■»« equal lo the rtupli- then shall A : B = C • D for since a-j . b*=:C- • D- ••• A:B-C:D.' PROOFS OP THE PROPOSITIONa OF BOOK V, 297 •qua I; and ioate ratio 3 : D. the dnpli- PrOOFS op TIIK PitOPOSITIOVS OP r.OOK V. nKRtVFn FROM TIIK (iKOMKTIJK^AL OKFINITIOX oF PliOPORTION. Ohs. Tlio Projiositions of Book V. are all theorpiiis. Proposition 1. Ratios whh'h arc equal to the. same ratio are equal to one. another. Let A : B ;: P : Q, aiul also C : D :: P : Q; tliou sliall A : B :: C : D. For it. is evident that two scalers or arrangements of multiples which agree in every respect with a third scale, will agree with one another. Proposition 2, //■ tii-o ratios are equal, the antecedent of the second is (/reater than, equal to, or less than its consequent according as the. antecedent of the first is (/reater than, equal to, or less than its coiisequert!. Let. then accord in u- A : B ;: C : D. C> == or < D. as A > or < B. This follows at once from Def. 4, by taking 711 and n each equal to unity. 29H KIiri.in'H KI.KMFNTS. I'KOI'OSITION .".. //• iH^o rs of A aro distributed .■iinon.,' thoso of B in the same manner as the multiples of C a)'e anionic thos(f of D; therofom also, the multij)l(-s of B are distributed amonc those ot A m the same manner as the nudtinles of D are among those of C. That is, B : A :: D : C. NoTK. Tliis proposition ia Rometimps cnunriated thus and the inference is referred to as Invertendo or Inversely. ■ Proposition 4. Equal magnitudes have the same ratio to the same rnaq- tvUnde; and the same magnitude has the same ratio to eaual 7)iagnitHd('S. ' Let A, B, C be three magnitudes of t*.r^ same kind, and let A be equal to B : then shall A : C : : B : C »^Tid C : A : : C : B. Since A: B, their nndtiples are identical and therefore are (hstn])uted m the same way among the multiples of C. ■. A : C :: B : C, /W 4. .•. also, invertendo, C : A :: C : 3. y 3 I'ROOFB t)F THK I'Uol'OHITlUNS Of HuoK \ 299 I'UUI'OSITION r». 0/ Itro luuqiuil motjuitudes, (/,>' f/nttt-ir /m.^ - /rmfer ratio to a third mnynilade than tho. h;,sN htus ; nth. same magnitiule has a ,/reater ratio to the Om of two maanitiidea than %t has to the greater. ^''<''*6A k't A I)(! > B; then shall a : C Ik; ;- B : C. Since A B, it will lu> possil)lo to iincl m such that mA exceeds j/iB hy a magnitude greater than C ; lieuce if mk lies between uC and (u i 1)C, iiiB ^ /iC. and if m^ - nO, then wtB < nC\ .*. A : C > B : C. Sbcondli/y let B be < A ; then shall C : B Ije > C : A, For taking //* and u as before, ViC > 7«B, while nC is not > inIK ; Uej: G. .". C : B-C : A. I^K/'- <^. V, 3. PUOI'OSITION (i. Mayiiitudes which hacc the stone ratio to the mine inag- uitnde are equal to one another; and (hose to which the mine magnitude has the same ratio are equal to one another. First, let A : C :: B : C; then shall A- B. For if A > B, then A : C :- B ; C, and if B>A, then B : C ^ A : C, v. 5. which contradict the hyjjothesis; .■. A- B. I 300 KUcLID's Kf.KMKNTH. iSe.cowilii, let C : A :: C : B; ( Im'Ii slijill f^ _ B_ neauist! C : A :: C ; B, .'., inn- I'll' utfi), A : C : : B : C A B, liy till) Hrst ji.iit ot' tin; in-vo(. V. '6. ■ jiii 8| I'lWPOHlTlON 7. y/tff^ naii/nUnUn wliirh lias a /iii/inie is the less of the two. ^''irsf, let A : C ))(-. B : C; then shall a he > B. For it" A B, then A : C :: B : C, M Jiich is coiitrnry to the; liypothcsis. Ami if A <- B, thru A : C -- B : C; whitli is contrary to the Jiyjiothesis; .'. A> B. Secoudlij, l(!t C A be>C : B; tlion shall ., ),(. < b. For if A^ B, thfii C : A : : C : B, which is contrary to the hyi)othesis. And if A,. B, then C : A -. C : B; whidi is contrary to the hypothesis; .•. A < B. V. J. V. 5. V. 4. V. y. 1^^ V. ;{. ivo; a)i(l i (ltd It it V. i. V. n. V. 4. V. 5. 1 KOOKS UK TIIK I'UUl'OSilTlONti OF HOOK V. I'UUI'OHITION ^. 301 Mivjniftulis haci' (hi miw: ratio ht ow; auollitt' w/iirh their tqniianlfiiih-i hare. hot A, B 1m' two iiui,i,'iiitu(l<'s; tl't'll «IihII a . ^ ;: ln^ : inB. Jf y>, q lie any two wliol(^ iiuiiilx-rs, tlu'ii in , ;>A :• or ■ //* . yB accordiiii,' !is y;A:- or -78. lint m . ]>A J) . niA, tuul j)i . rjQ ij.iiiB: .'. i> . mA -.- or A > oi' -yB; .'. A : B :: niA : iiiB. L(!t A : B :: C : D. Tlicn since A : B :: ///A : mB, Dej\ 1. Coil. and C : D : : uO '. nD .'. ill A : mB :: nC : uD. Phoposition !). V. 1, // two ratioH an', equal, and any eqaimultipUs 0/ tlm nutecedents and also of' the consfqnruts are taken, the. mnltiple 0/ the first anffcnlfitt has to that of its consfqiuud the, saiiui ratio as the laalliple of the other antecedent has to that of its consequent. Let A : B :: C : D; then shall mA -. uB :; mC : wD. Let p, q ])v any two whole numbers, then because A : B : : C : D, ptn . C > = or <7/t. D iiccording as y>m . A > ~ or < 17/t . B, /^il. that is, J) . iiiC > - or -- q , nD, according as p . niA ~^ ~ or C D; .•. A:- C. iSiuiilarly it juay \n) slicwii tliat if B < D, th(>u A < C, and if B D, then A - C. V. o. V. /, Pkopositiox i J . // four niatjitiludes of the same kind are proportionals, theij are also proportionals when taken alternately. Let A, B, C, D be four uiagnitudcs of the same kind sucli that A : B :: C : D; then shall A : C : : B : D. Lucause A : B :: nilK : /mB, v. 8. and C : D :: wiC : ?iD; .'. 9//A : /mB :: nC : ?iD. v. L .". niA > - or -, nC according as »tB >- or - or <>/B, so is ?;/C> or < uD, and 7nE> oi- < ?^f, .'. so IS nl^ + viC + 7/iE + . . . > or < mB + uD + nF + . . . or m (A + C + E +. . .) > ^ or < w(B -f D + F + . . .) i and m and ti are any whole numbers; .". A : B : : A + C + E + . . . : B + D i F + . . . . Def. 4. NoTK. This inference is usually referred to as addendo. Proposition' l.",. If four mnynitudes are proportiona/s, the sum or difer- ence oj the first and second is to the second as the sum or difference of the third and fourth is to the fourth. theix shall Let A : B : : C : D : A + B : B :: C f D : D, aiul A ~ B : B : : C '- D • D. Tf m be any whole nund)er, it is possible to find another nundHM- n such that mA ,iB, or lies between uB and (n -I- 1 ) B, .". //iA i niB mB -r ,iB, or lies between niB + 7iB and mB + {n + 1) B. 20 U. K. 304 kuclid'8 elements. l)ut '))}^ ! '>mB ^ wi(A + B), and inQ + nE -^ {m + ?i) B ; .'. 7>a(A + B) : {ni + ii) B, or lies betwoiMi {m + vO B and {ill + 7i + 1 ) B. Also because A : B : : C : D, .'. mC~vD, or lies between uD and (?M- 1)D; i)^/! 4. .". m{C + D) : {ni + ?t) D or lies l)etween {in + n) D and (7/^ 4- 7?, + 1 ) D ; that, is, the multiples of C i- D are distributed anioiif^ tliose of D in the same way as the multiples of A + B amon» those of B; .•. A4- B ; B :: C I D : D. 1 n the same way it may be pi'oved that A - B : B : : C - D : D, or B - A : B : : D - C : D, according as A is > ov < B. NoTF. Tlicso inferences are referred to as componeudo .and dlvi- deudo respectively. Propositiox 11. Tf there are ttoo sets of magnitud''s, such (hat the first is to the second of the first set as the first to the second of the other set, and the second to the third of thf first srt as the second to the third of the other, and so on to the last nnu/ni- tiufe: then the first is to the laM of the first set as the first to the last of the other. First, let thei'e be three ma,ia(/ui- le first to f one set M^ow, if ;//A ?> nC, I lion ink : 7hB - ?iC : mB; and .", ///P> //R, Similarly it may he sliewu that 711P or < ;/R, according as mA or < ?yC, .". A : C :: P : R. V. 7. Dr/: 4. SecoiuUi/, let there he any numhei' of magnitudes, A, B, C, ... L, M, of one set, and the same numlx-r P, Q, R, ...Y, Z, of anothei' set, such that A : B : : P : Q, B : C :: Q : R, L : M : : Y : Z ; then shall A : M :: P : Z. For A : C :: P : R, and C : D : : R : S ; .". hy the Ih'st cas(^ A : D ; : P : S. and so on, until iinally A : M - P : Z. Note. Tliis infoi'oncc^ is I'ofc^rrcd to jis ex jequall. C/OROLLARV. If A : B : : P : Q, and B : C : : R : P ; then A : C :: R : Q. Proved. Ifyp. ;. y, Cor. V. 0. V. 3. Propo.sitio:: 1 ."i. If A B : : C : D, anf! E B ; : F : D; then shall A f E B : : C f F : D For since E B : : F : D, . , invertcudo. B E : : D : F. ex Also A : B :: C : D, f^'iwili, A : E ;: C : F, Ifljp. V. 3. 20-2 30G KUCLTDS KLEMENTS. .". , romponendo, A + E ; E Ai,'ain, E ; B .". , p:c a'qnali, A • E : B C + F : F. F : D, C + F : D. V. 13. Hyp. v. 14. Proposition 1(1. ]f two ration are equal, their dujiHcatp. ration are equal ; and converselj, if the dup/irate ratios of two ratios are equal, the ratios themselves are equal. Let A : B : : C : D ; tlien sliall the duplicate ratio of A to B be equal to that of C to D. Let X be a thii'd proportional to A and B, fuid Y a third propoi'tional to C and D, so that A : B :: B : X, and C : D :: D : Y; then because A : B :: C : D, .'. B : X :: D : Y; .*. , ex requali, A : X :: C : Y. But A : X and C : Y are respectively the duplicate ratios of A : B ond C : D, Ihf ]:]. .'. the duplicate ratio of A : B - that of C : D. Conversely, let the duplicate ratio of A : B - that of C : D; then shall A : B :: C : D. Let P be such that A : B : : C : P, .". , invertendo, B : A :: P : C. Also, by hypothesis, A : X :: C : Y, .■. , ex cvquali, B : X :: P : Y; l»ut A : B :: B : X, .-. A : B :: P : Y; v. 1. .". C : P :: P : Y; v. 1. that is, P is the mean proportional between C and Y. .•. P--D, .'. A : B :: C : D. 1^ V. 13. Hyp. V. 14. BOOK VI. Definitions. 1. Two rectiliiuMl Hi^furcs arc said to !>»' equiangular when the angles of the lirst, taken in order, .'iiv, equal respectively to tliose of tin; .second, taken in order. Each angle of the first figure is said to correspond to the angle to which it is equal in the second tigure, and sides opposite to corresponding angles are called corresponding sides. 'J. Uectilineal figures are said to be similar when tht-y are e(i[uiangular and have the sides about tlu; e(}ual angles proportionals, the corresponding sides being homologous. [.See Def. 5, page '2^X.\ Thus tlic two quadi'ilateral.s ABCD, EFGH arc similar if thu angles at A, B, C, D are respec- « lively equal to these at E, F, G, H, ^^ and if the following proportions ^^-''"'^ \ hold ^-^^ \ F AB : BC :: EF : FG, A\ \ ^-'-^\ BC : CD:: FG : GH, \ \ ^\ \ CD: DA :: GH : HE. ^- ^ DA : AB :: HE : EF. H 3. Two figures are said to have their sides about two of their angles reciprocally proportional when a side of the first is to a side of the second as the remaining side of the second is to the remaining side of the first. \. A straight line is said to be dixided in extreme and mean ratio when the whole is to the greater .seK()I'. I, .?()n ciud tli(! l);ist! HC ; jukI of tlii^ coiiscc^ueuts, any (njui- iiiultiplex li;i\<5 l)Oeii taken, iiaiiit'ly the A ACM and the base CM ; and since it has been shewn that the A AHC is greater than, equal to, or less than the A ACM, according as HC is greater than, eciual to, or less than CM; .'. the four original magnitudes are proportionals, \, l)ij'. I. that is, the A ABC : tlu^ A ACD :; the hase BC : the base CD. <;. i;.l). CouoLLARV. The areas of paraUcloyvitinn of the, saian altitude are to urn; anot/fr as their bases. Let EC, CF b<^ par"" of the; same altitude; then shall the par'" EC : the par'" CF :: BC : CD. Join BA, AD. TJien th(^ A ABC : the i\ACD :. BC : CD; Proved. but the par'" EC is double of the A ABC, and the par"' CF is double of the A ACD; .'. the par'" EC : the par'" CF :: BC : CD. v. 8. NoTK. Two stiaiglit lines are cut proportionally wlien the KCg- luciits of one line aie in the sauie ratio an the corresponding sogmentti of the other. [See definition, page 13].] Ffg.i A X B A Fig. 2 B Y D D Thus AB and CD are cut proportionally at X and Y, It' AX : XB :: CY : YD. And the same definition applies etpially whether X and Y divide AB, CD internally ay in Fig. 1 or externally as in Fig. 2. 310 KUCLID'S loLKMENTS Pkoposition l*. Tjieokkm. _ ff a stmufhl line he drauu,. pa,;, lid in our sole of a tnau.jlr, H shall ml the ofh.r sales, or those sides produced, jirojiorlionalli/: Converselii }/ the sides or the sides prodiwed he cat pro. porhonalhj, I hr straight line whirh joins the points o/' section, shall he pa rail rl to the rcniainin.j side of the trianalk I.et XY 1.C (Iniwn pa,' (.. BC, one of tlu, sidrs oF Ihu ABC I then .shall BX : XA :: CY : YA. 'Unn BY, CX. The., (Ih- BXY^the A CXY, Wma un (In- saine l.ase XY and hetwceu the same pavaliels XY, BC; , w; and AXY i.« another ti'ian,<,Hc; •■• Hie BXY : the ,\AXY ;: the .CXy": the , AXY v 1 i.uttho A BXY : tlie l.AXY :• E/ : XA, " v/ l' and the A CXY : the , ,AXY : : CY : YA. • '. BX : XA :: CY : YA. ^ | Converselii, h't BX : XA :: CY : YA, a.id k-L XY he joined- then .shall XY he par' to BC. As het'ore, join BY, CX. \\y hyi)othesis BX : XA ::'cY : YA ; but BX : XA :: the , \ BXY : the AXY vi J and CY : YA :: the I.CXY : tlie .^AXY; .'. the BXY : the .a AXY :: the ._ CXY : tlie / AXy' v 1 • •. the .A BXY -the .:.CXY; v"(i' and they are triangles on the .same ha.se and on tlie same f^ide or it, .'. XY ifi par to BC. i ^tj sii/e of (I. rut, jH'ti >f S'Xfloil., 1^ s of llu. l);i.sc XY I. ;57. :y. v. I. VI. I. N. 1. ' joiiiL'd; r', VI. J. / Y. V. J. y. (i. It' same 1. au. S;.E.D. BOOK A'l. I'KOP. 2. 311 KXKKCISKS. 1. Shew that t\ery 'l" ^'^'"V "" ''*'"'*' ^ '" ""' ''""'I'K'" I'li^t^ uf two triangles ACB ^DB. straight lines are drawn parallel to AC, AD, meeting BC BD at F, «j : shew that FG is parallel to CD. Do'^'^i" a triangle ABC the straight line DEF meets the sides gC, CA. AB at the i).)ints D, E, F respectivelj*. and it makes equal angles with AB and AC: prove that BD : CD :: BF : CE. 5. Jf th(! hisector of the angle B of a triangle ABC meds AD at riglit angles, shew that a line through D parallel to BC will biseet (). From B and C, the extremities of the base of a triangle ABC, lines BE, CF are drawn to the opposite sides so as to intersect oil the median from A: shew that EF is parallel to BC. 7. From P, a given point in the side AB of a trian-de ABC draw a straight line to AC produced, so that it will be bisected by BC. H. Find a point within a triangle such that, if straight lines be drawn from it to the three angular points, the triangle will be divided into three equal triangles. 312 KircMDS KliK.MKN'I'S. I'UOi'OSlTlOX .). 'rilKOUKM. /////'• rtrtlrcd atKjl'- of a h'tdnyle he binected by astmiyht linn ivhkli cuts the basi; the ftuf/iiicnfs of the base shall ha ce lo one another the same ratio as the reinain'uir/ sides of the triaiu/lc: Converseli/, if the base be dirided so that its sef/nieuls have to one another the same ratio as the remaining sides of the triangle hare, the straight line d rami from the vertex lo the point of sretion shall bis>rt the rvtieal angle. Ill llu! _ ABC let tlic _ BAG lie bist'ctud \>y AX, wliicli laeots tho lja.se at X ; then shall BX : XC :: BA : AC. Through C Jiuw CE par' to XA, to meet BA produeed Then hecause XA and CE are [jar', .'. the L BAX '--^ the int. opji, _ AEC, i. L"J. .uid the _ XAC = tho alt. _ ACE. i, 29. But the _ BAX ---the ^XAC; J/i/p. .". the _AEC- the i.ACE; AC AE. I.e. Again, because XA is })ai-' to CE, .-i side of the A BCE, .". BX : XC :: BA : AE; vi. 2. that is, BX : XC :: BA : AC. ' a xti'niyht sliuU havi; idfs of the. ( sefftneulti i(j sides of i vertex lu HOOK VI. 1>H01'. :\. Convcraelu, U-l BX : XC :: BA : AC; iiml Irt AX he j then .sliiill the _ BAX _ l XAC. For, with the same construction as l)efoic', because XA is pai" to CE, /i side of tlie l\ BCE, .'. BX : XC :: BA : AE. I Jut l)y liypothesis BX ; XC :: BA : AC; .". BA : AE :: BA : AC; .•. AE:-AC; ". the :. ACE- the _ AEC. Hut because XA is par' to CE, .'. the ^ XAC the alt. _ ACE. and tht! ext. _ BAX tii" int. opp. _ AEC; .'. tht; _ BAX -the _ XAC. 313 oiiied: VI «( \ 1 1 r» 1. L"J 1. L"J l,>. K. I). AX, wliich produced r. 31. EC, 1. I'D. I. 29. Hyp. I. G. A BCE, A'l. L'. KXKKCLSKS. A rxi' A^i^'J^ '^^'^^ ,^^ °* ^ triungle ABC is bisected at D, luul the anclfs ADB ADC JUL" busected by tlie Btrui«ht lines DE, DF, meeting AB, AC at E, F respectively: bliew that EF is parallel to BC. -• Apply Proposition 3 to trisect a given linito straight line. ;i. If the line bisecting the vertical angle of a triangle be divided uito parts which are to one another as tlie base to tlie sum of the sides, the point of division is the centre of the inscribed circle. •1. A- BCD is a (luadriiateral: slunv that if the bisectors of the angles A and C meet in the diagonal BD, the bisectors of the angles B and D will meet on AC. 5. Cumtruct a truni/*'■ s'kIi' (if II tridii'ili' f>'' proiliii'iul, iinc(('(l by AX wliieli meets the hii«c profhieed iit X : then sliall BX : XC :: BA : AC. Thruujfh C chaw CE par' to XA, and let CE meet BA at E. 'L'lieii because AX and CE are par', .'. the ext. _ FAX - tlio int. opp. _ AEC, and the _ XAC -^ the alt. _ ACE. JUit the _ FAX the _ XAC ; .-. the _ AEC- the _ ACE: .-. AC- AE. Agaui, because XA is par' to CE, a side of the . .-. BX : XC :: BA : AE; that ife. BX : XC : : BA : AC. I. 31. I. 2\}. Hyp. I. G. BCE, ( 'onstr. VI. 2. U'uiK \ r, I'lUM'. A. Ml: //n: exf trior ii'li ruts till'. or (Did till' iced ho re fo s of thti iri- •i:rfi\i: fo ihc Convurxdy, lot BX : XC ;: BA : AC. mikI let AX \»> joiii.-d then slmll tli.> FAX the ,. XAC. For, with the siDiio constriU'tion Jis Ix-fote, lu'cuuse AX is jwir' to CE, ;i side of tlic / BCE, ". BX : XC :: BA : AE. lint hy hypothi'sis BX : XC :: PA : AC; .•. BA : AE :: BA : AC; .-. AE AC, . . the _ ACE the _ AEC. I5ut Ijeeuuse AX is par' to CE, .'. the j_ XAC = the alt. _ ACE, and the ext. _ FAX tlie int. opji. „ AEC ; t. 29. .". the _ FAX the :. XAC. (/. IM). N r. '1. v. !. I. 'i. and let tin; jts the base I. :u. Propositions ;J luul A imiy l)fi botli included in one einniciution as follow.s : // till' iiiti'i-ior or c.vterlor rriticiil (utf it. meeting AB in C and D • shew that '\C : CB :: AD : DB. 2. From a point A straight linos are drawn making the angles BAC, CAD, DAE, each equal to half a right angle, and tin n are cut by a straight liise BODE, which makes BAE an isosceles triangle: .shew that BC m DE is a mean proportional between BE and CD. 3. By nu uns of Propositions S and A, inove that the straight lines bisecting one angle of a triangle internallv, and the other t^Nvo e-Uernally, are concurrent. 31G Euclid's elements. Proposition I. Theohkm. If two triangles he equicmgiihir to one another, the sides about the equal angles shall he proportionals, those sides v?hi('h are opposite to equal angles being homologous. B C E Let the A ABC be equiangular to the ADCE, having the L ABC equal to the l DCE, tlie l BCA equal to the l CED, and consequently the l CAB equal to the l EDC: i. 32.' then shall the sides about these equal angles be propor- tionals, namely AB BC and AB BC CA AC DC CE DC ED, DE. the Let the ADCE l)e placed so that its side. CE may l)o (contiguous to BC, and in the same straight line with it. Then because the _ « ABC, ACB are together less than two rt. ancfles, "^ . i - 1 , I. i ( . and the ^ACBr-^the _DEC; Hyp, -'^ ABC, DEC are together less than two rt. angles; .•. BA and ED will meet if produced. Ax. 12.' I^et them })e produced and meet at F. Then because the l ABC ^ the l DCE, .*. BF is par' to CD; and l)ecause the l ACB :^ tlu^ l. DEC, .'. AC is par* to FE, .'. FACD is a par'"; .*. AF=CD, and AC FD, r. iU Jrgp. I. 2S. Jfyp. I. '2K BOOK VI. VROP. 4. 317 er, the sides those sidt'tt Jiaving tho the Z.CED, !: I. 32. bo propor- DE may l)o vvitli it, f less chiin I. 17. Hyp. rt. angles ; .i.r. 12. Hyp. I. 2S. I. 28. I. U. Again, l)ecause CD is par' to BF, a .side of the A EBF, ■. BC : CE :: FD : DE; vi.' 2. but FD - AC; .". BC : CE :: AC : DE; and, (dternatehj, BC : CA :: CE : ED. v. 1 1 Again, because AC is par' to FE, a side of the A FBE, .'. BA : AF :: BC : CE; vj* o. but AF :==CD; .". BA : CD :: BC : CE; and, cdternately, AB : BC :: DC : CE. Also BC : CA :: CE : ED; ."., ex a'.qnali, AB : AC :: DC : DE. V. 11. ProveAl. V. 14. Q. K. D. [For Alternative Proof see Page 320.] EXKRCISRS. 1. If one of the parallel sides of a trapezium is double the other, siiew tliat the diagonals intersect one another at a point of trisection. ' .1 ?•■*. ^^r^^'^rfi''^ -^P "^ ^ triangle ABC any point D is taken : shew .at If AD, DC AB BC are bisected in E, F, G, H respectively. then EG is equal to HF. i . . 3. AB and CD are two parallel straight lines; E is the middle pomt of CD ; AC and BE meet at F, and AE and BD meet a G .shew that FG is parallel to AB. 4. ABCDE is a regular pentagon, and AD and BE intersect in F • shew that AF : AE :: AE : AD. <-'■••-. .),„?"r J"" *^t,^«"^'^„°^ ^- 1^ «'^«^^ that EH and GF are parallel, and that FH and GE will meet on CA produced. 0. Cliords AB and CD of a circle are produced towards B and D respectively to meet in the point E, and through E, the line EF is drawn parallel to AD to meet CB produced in F. Prove that EF is a mean proportional between FB and FC. *->-.. u I 318 KUCTilBH KLKMKNTS. Proposition Theorem. //■ t)ip. tilths of tiro trianylex, taken in order about each of their ((n(/lfis, be pi'oportionals; the triawih's shall be equi- angular to one another^ havinf/ those angles equal which are opposite to tJie homologous sides. so tlwit L(^t the A*" ABC, DEF l»ave their sides proportionals DE EF EF, FD, AB : BC BC : CA :ni(1 foiisequently, ex (tquali, AB : CA :: DE : FD. Tlien sliall tlie triangles be equiangular. At E in FE make the i. FEG equal to the l ABC; and at F in EF make the _ EFG equal to the _ BCA; T. 2:'>. tlien the remaining _ EG F - the remaining i. BAC. I. .'^2. .•. the A GEF is equiangular to the A ABC; .-. GE : EF :: AB : BC. But AB : BC :: DE .-. GE : EF :: DE .-. GE-DE. Similarly GF DF. Then in the triangles GEF, DEF ( GE-DE, Because <^ GF = DF, (and EF is common; EF: EF; VI. 4. irijp. V. 1. .-. the and the and the _ EGF lUit the .-. the Similarly, the .1 GEF the LDEF, _GFE the _ DFE, the _ EDF. _GEF :tl!(^ _ABC; _ DEF = the _ ABC. i. EFD ^- the u BCA, I. «. Const r thoiil each of all be equi- il m/iich are BOOK VI. TROP. (i. foportionals, ir. .ABC: BCA; T. 23. _ BAG. I. Wl. i^BC; vr. 4. Hyp. V. 1. I. s. Cons f I 310 .'. tlie reimiiiiiiii,^ _ FDE the remainini,' _CAB; i. ;)2. that is, the ADEF is equiangulai- to the a ABC. <^> K.J), Phoi'ositiox 0. TiiKoin:.M. 7/ two trimu/les hare one avyh', of the one eKil (<, one amjle of the other, and the sides nbo2U the equal anyles pro- portionals, the triauyles shall be similar. I. 2."). I. 32. VI, i. Tu the .1^ ABC, DEF let tlie _ BAC -= the _ EDF, ami let BA : AC :: ED : DF. Then shall the A'' ABC, DEF be shiiilai'. At D ill FD make the _ FDG equal to one of the _ *• EDF, BAC at F in DF make the _ DFG equal to the _ ACB; .". the reuiaininif _ FGD the remaining _ ABC. Then the A ABC is e((uiangular to tiie \ DGF .'. BA : AC :: GD : DF. But BA : AC :: ED : DF; .". GD : DF :: ED : DF, .•. GD-ED. TJieu in the A« GDF, EDF, { GD:^ED, -, and DF is connnon ; [and the _GDF=-the _EDF, .•. the A^ GDF, EDF are equal in all respects, i. 4 so that the A EDF is «'quiangular to the A GDF; but the A GDF is eiiuiangular to the ABAC; Constr .'. the A EDF is ecpiiangular to the ABAC; their sides about the equal angles are proportionals, vr. 4 Beeause Constr. aco ctuuuu uie equal angles are propori that is, the A^ ABC, DEF are simihi (^ i;. I). n. K 21 320 kucmd's ki.k.mknts. ^G^E 1, From Definition 2 it, is soon thfit tiro conditions arn necessary for similarity of rectilineal IJ^'ures, namely (1) the ligures must be e(|Uian^,'ular, and (2) tL(! sides about the equal angles must be proportionals. In the case of trimKjIi'x we learn from l'ro])s. 4 and Tj that each of these conditions follows from the otlier : '' - ■ ' ever is not necessarily the case with rectilineal ligures of " three sides. this how- more than m JN()TK 2. We have given Euclid's demonstrations of I'rojiositions 1, o, (■) ; but these propositions also admit of easy proof by the method 01 Kuperposition, As an illustratiun, we will apply this method to I'ropositiou -1. Pkoposition 1. [Altj;i!nati\k Pjjoof.) ]f two t rid unit's be etiuUnnjiiltir to one miothcr, the. tiidcs ohout the equal auiiJes .shall l>c propnrtiunah, thoae side.s which are vppoMtc to equal au(jh'fi beiini humoluffous. B Tict tlu; A ABC 1)0 equiangular to the a DEF, having the / ABC equal to the I DLF, the / BCA equal to the / EFD, and conse- quently the I CAB equal to the / FDE: 1.32. then shall the sides about these e(iual angles be proijortionals.' '"" Apply the A ABC to the a DEF, so that B falls on E and BA along ED: then BC will fall along EF, since the / ABC = the z DEF. Let G and H be the points in ED and EF, on which A and Join GH. Then because the z EG H^ the / EDF, .-. GH is ])ari to DF: .-. DG : GE:: FH : HE; .-. , coniponnulo, DE:GE::FE:HE, :.,alterHatclii, DE : FE : GE : HE, that is, DE : EF :: AB : Bc'. Similarly by ai)plying the a ABC to the a DEF, so that the point C may tall on F, it may be ))roved that Et :Fi--;;BC;CA. .-., ex a'quali, DE : DF :: AB : AC, ii. 1;. i». lll/p. C fall. JI>ip. 1:5. 11. somlitions nrn (1) tlio liguros 111 aiigltjs niust from Pro))H. 4 licr : this liow- of more than if ri'opositioii:-; by the luctliocl position -1. ^\1 s/V/l'.-i- (iJiout titc. ire vppotiitc to If,' tho z ABC D, and consc- I. ^2. lortionals. on E unci BA DEF. lli/p. A iuul C fall. Hyp. V. 1:5. V. 11. ihat the point y. K. D. 1500K VI. IM{')1'. 7, PllOPOSITIOX 7. TllKORK.M. 321 // Uco tni'. and the ,:1'^ are equiangular and therefore sii;ii!ar. vi. \\ JJut if the _ BAG is not equal to the _ EDF, one of them must be the greater. Let the ^ EDF be greater than the _ BAC. At D in ED make the _ EDF' e(iual to the _ BAC. i. •';; Then the A« BAC, EDF' ai'o e(]uiangular, Conslr, .". BA : AC :: ED : DF'; but BA : AC :: ED : DF; .*. ED : DF :: ED : DF', .". DF=DF', .'. the .^DFF'--:the i. DF'F. But the i, ^ DF'F, DF'E are supplementary, .'. the .:_ " DFF', DF'E arc supplementary: that is, the _ '^ DFE, ACB are supplementaiy. g. K. 1). 21-2 VI. I. Jfyp. V. 1. I. 13. 822 kuclid's elements. COKOLLAHIES TO PltOPOSITlOX 7. Three cases of tbis tlicorem deserve special attention. It has been proved that if the angles ACB, DFE are not supplc- viciitiiri/, tliey are anud : and we know that of angles which are supplementary and unequal, one must be acute and the other obtuse. Hence, in addition to the hypothesis of this theortni, (i) If the anyiles ACB, DFE, opposite to the two homologous sides AB, DE are both acute, both obtuse, or if one of them is a riglit angle, it follows that these angles are etjual ; and therefore the triangles are similar. (ii) If the two given angles are right angles or obtuse angles, it follows that the angles ACB, DFE must be both acute, and therefore equal, by (i): so that the triangles are similar, (iii) If in each triangle the side opposite the given angle is not less than the other given side; that is, if AC and DF are not less than AB and DE respectively, then the angles ACB, DFE cannot be greater than the angles ABC, DEF, respectively; therefore the angles ACB, DFE, are both acute; hence, as above, they are ecjual ; and the triangles ABC, DEF similar. BOOK vr. PROP. 7. 323 K n. are not su2)plc- yand unequal, »o homologous e, or if one of obtuse anf,'lf' s, be both acute, n angle is not iC and DF are n lan the angles th acute; KXKKflSKS. ON Propositions 1 to 7. 1. Shew that the diagonals of a trapezium cut one another in the same ratio. 2. If three straight lines drawn from a point cut two parallel straight lines in A, B, C and P, Q, R respectively, prove that AB : BC :: PQ : QR. .'!. From a point O, a tangent OP is drawn to a given circle, and OQR is drawn cutting it in Q and R ; shew that OQ : OP :: OP : OR. 4. If two triaunlrn are on equal buses and hetween the same parallels, anij struhjht line parallel to their bases will cut off equal areas from the two triangles. '). If two straiijht lines PQ, XY intersect in a point O, so that PO : OX :: YO : OQl, prove that P, X, Q, Y are concyclic. 0. On the same base and on the same side of it two equal triangles ACB, ADB are described; AC and BD intersect in O, and through O lines paraliel to DA and OB are drawn meeting the base in E and F. Shew that AE = BF, 7. BD, CD are perijendieular to the sides AB, AC of a triangle ABC, and CE is drawn perpendicular to AD, meeting AB in E • shew that the triangles ABC, ACE are similar. 8. AC and BD are drawn perpendicular to a given straight line CD from two given points A and B ; AD and BC intersect in E and EF IS perpendicular to CD : shew that AF and BF make equal angles with CD. i o 9. ABCD is a parallelogram; P and Q. are points in a strai"ht hne parallel to AB ; PA and QB meet at R, and PD and QC meet"at S : shew that RS is i)arallel to AD, 10. In the sides A B, AC of a triangle ABC two points D, E are taken such that BD is equal to CE ; if DE, BC produced meet at F, shew that AB : AC :: EF : DF. _ 11. Find a point the perpendiculars from ivhich on the sides of a given triangle shall be n a given ratio. rX'CMU'.S KLEMKNTS. in: i'Koi'osirroN s. TiiKoitivM. fii. (I, rn/Jif-(in(ih;l IrhiiKilt' if a pn'ppndirn/ar he draion hyiii the rii/ht (Hi;//e to the ht/poteniDie, the Iriaur/les on each aid'- vf It urn isi/ni/irr to the whole trinnijle am} t<> one another. B DC Let ABC l»c a 1 riuii,!;lt' ri.i,dit.-iui'j;lc(l nt A, and let, AD 1)0 ])('!■]). to BC: liicii shall ili(« / -^ DBA, DAC ^>o. similar to i]\o \ABC and to Olio auotlicr. h\ the A" DBA, ABC, tlio L BDA r-. tlio L. BAG, boin;,' rt. angles, and tho _ ABC i.s common to both; .". the remaining ^ BAD = t]ie remaining i. BCA, i, 32. tliat i.s, the A'* DBA. ABC a-e equiangular: .'. they ai'e similar. vi. 4. In the ,sani(i Avay it may be proved that tiie A^ DAC, ABC are similar. Heni-c^ the A'^ DBA, DAC, lieing e(iuiangular to the same A ABC, are ecjuiangular to one another; .'. they are similar. vi. 4. Q.E.D. Coitoi.LAr.v. .l](>cause the A« BDA, ADC are similar, .•. BD : DA :: DA : DC; .•iiid because the A^ CBA, ABD ai'c similar, .'. CB : BA :: BA : BD; niid because the A" BCA, ACD are similar, .". BC : CA :: CA : CD. EXERCISES. 1. Prove that the liypotenuse is to one side as tlio second side i.s to tlie i)erpeiuhenlai'. 2. .S7/^'(f tluit the nnliux of a circle iii a menu prnpnrfinii'iJ between the sefiwriits of ani/ taiuicut beticcea itn imiut of contact and a pair oi'iniralh'l taiigents. BOOK VI. I'lior. 9. 325 r bo t/i'dii'ti les on, eorh nic another. let, AD 1)(! BCA, I. 32. ir; VI. 4. B A** DAC, o t]io same VI. 4. Q. K. D. similar, ?cond side i.s niitij hetwoeu and a jxiir "DKFixrnox. A loss inat(iiituclo is .said to bn a sub- multiple of a .<;roat(M', wlion tlic less is coiitainod an i\i'(tcf. miiiiIxT of times in tlio jri'eiiter, [Hook V. Dcf. L',] Pijoi'osiTiox ii. Pi!or.ij:.M. Fi'n)ii (I, fi'tron Hti'di'jht lino to rvt off dii;/ rrtpfired snb- 'iii'tdtiple. A F B Let AB 1)(* tlio 'Avow straight line. Tt is i('(|ini-e(l to cut (iH'a certain submultiplo from AB. I'lom A draw a straiqlit line AG of indefinite lengtji making any angle with AB. In AG take ;uiy j)oint D; and, by cutting off successive })arts each etjual to AD, make AE to contain AD as many times MS AB contains the i-eijuired subnudtii)le. •loin EB. Tlirougli D draw DF piii'' to EB, meeting AB in F. Then shaM AF be tiie retjuired submultiple. Because DF is par' to EB, a side of tlu' *. AEB, .-. BF : FA :: ED : DA; vi. 2. ."., roinponendo, BA : AF :: EA ; AD. v. 1'). But AE contains AD the required nund)er of times; Conxtr. .'. AB coiitains AF the re(|uii'ed nund)er of times; tliat ir-, AF is the retjuired submultiple. g. K. F. 1. 9 KXERCISES. Divide a straiglit line into five equal parts. Give a geometrical construction for cutting ufT L\vo-s(;voutlis of a given struit^ht line. 820 kuclid'h klemknth. m PliOl'OSITIOV 10. PitOIII.KM. y'o (/irh/r a .- a i)ar"'; .". DH FG, and HK GB. i. .Vl. Xow sinc(i HE is par' to KC, a side of tli(^ .'. DKC, .'. KH : HD :: CE : ED. vi. 2. Jiut KH BG, and HD GF; .". BG : GF :: CE : ED. y. 1. Again, becau.so FD is par' to GE, a, side of the A AGE, .". GF : FA :: ED : DA, yj. 2. and it lias been shewn tliat BG : GF :: CE : ED, ."., cv O'qnnJi, BG : FA :: CE : DA: \. \A. .". AB is divided similarly to AC. V-K. P. \ i,- 1 . i f , EXERCISE. Divide a straifiht line intfriuilly ami extn-nalhj in a qiven ratio. Is IS aliva lis possible ? tliis always possible It (/ivii/iting DK in F. i. .".1. Then shall HF be a thiid pi-oiK)rtional to A and B. Because GH is par' to EF, a side of tlu^ ADEF; .'. DG : GE :: DH : HF. But DG A; and GE, DH each B; i'omtr .-. A : B :: B : HF; that is, HF is a third proportional to A .uid B. (•i. r;. F K.XKIICISKS. 1. AB Ih a (lianniter of a circle, and through A any straight line is drawn to cut the circumference in C and tlie tangent at B in D : shew that AC is a third proportional to AD and AB. 2. ABC is an isosceles triangle having eaih of the anglfs at the base double of the vertical angle BAC ; the bistctor of the angle BCA meets AB at D. Shew that AB, BC, BD are three proportionals. 3. Two circles intersect at A and B ; and at A tangents are drawn, one to each circle, to meet the circumferences at C and D : shew that if GB, BD are joined, BD is a tliird proportional to CB, BA. 328 KUCLID'h KLKIIKNTH. l*IIOl'{>.SITIO\ iL'. PllOHLKM. Tii ft ml a Jon rlh iri'opoHtOixil In f/n-''' ;/!f/i s/rdh//// luias. ABC 6 ~ G r~x Lft A, B, C )«( (lie tlircf <,'i\rii sti-.ULflit, lines, it is i-oquircd to iiiul ;i fourth ])roi)()rtinii;il to A, B, C. Take two st^li,^'llt lines DL, DK cont.iiniu.ir iiiiy .uiii»ortional to A, B, C. TJocauRo GH is par' to EF, a side of the '> DEF; .". DG : GE :: DH : HF. vi. 2, But DG A, GE B, and DH O: Const r. .■. A : B :: : HF; tiiat is, HF is a fourtii i)ropoi-tiu)ial to A, B, C. f}.V..V. KXKfJC'ISKS. Annn^*" [™"\ P;. "'".'" ?'■ ^''*^ an-ular points of a parallelo-rnin A13CD. a HtrniKht \nw is drawn nuctiii;,' AB at E and CB at F ■ shew that CF IS a ionrtli ])r()iK)rtionaI to EA, AD, ami AB. 2. In a trianffle ABC the bisector of llic vertical auKl.. BAG mef^ts tlie base at D and the circumference of the circumscribed circle at E: shew that BA, AD, EA. AC are four proportionals. a. Fi-om a point P tangents PQ, PR are drawn to a circle wh().<.e centre is C, and QT is drawn pcrpendieulnr to RC iirodu! !'d • uhew that QT is a lourtJi proportional to PR, RC, and RT. Ijlll IllHiS, cs. A, B, ^. ) B; I. :;. 1. :n, B, c. »EF; vr. 2. ( 'oust I'. q. ]■:. I'. i'allolo;:min at F ; shew mir\a BAC ribcd circle ircio wliose IJooK VI. I'Unr. 13. PitOl'OStTION 1;;. PlIoiU.KM, n2a lilies Tojh/ti It iitran jirnpordoihtf hefwrt')! 'kd tjin u slrniglit Let AB, BC he the two given s(rai;:lit lines. 7t is re<|nir'e(l to lind ;i uHviii proportiouMl between tlieni. Place AB, BC in a straig'lit line, and on AC (lescriKr the seinicirt-lc ADC. From B draw BD at rt. angles to AC. r. 1 I. TIkmi shall BD he a mean proportional hetwem AB and BC. Join AD, DC. Now the ;_ ADC heitv, n ' semieirele is a rt. an<,de; in. ?A. and l)e(Miise in the ri^ht-.ui ;led /.ADC, DB is drawn from the rt. angle ju'rp. o ' ito In ^ -itenusc' .'. the A.jr:: DBC are similar ; A'l. S. .•. -J . ljD ;■ BD : BC; that is, BD is a, inc'ii; pi'oportional between AB and BC. KXKliCIHKS. 1. If from Olio mi'ilo A of a paiallolo>,'ram a .'itrai<,'lit lino 1)o drawn cutting,' the clia^fnal in E and tlic sides in P, Q. show that AE is a moan ]iroportional botweon PE and EQ. 2. A, B, C are three points in order in a straight line: find a poi)it P in the strui^'ht lino so tliat PB may be a mean proportional bc'twoen PA and PC. 3. Tlio diamotcr AB of a semieirele is divided at any point C, and CD is drawn at ri^'lit uncles to AB nicctin^' the circinnl'eronce iu D ; DO is drawn to liie centre, and CE is pi riKuidicnlar to OD : shew that DE iy a third proportional to AG and DC. f 330 EUCLID'S ELEMENTS. ■1. AC is the diameter of a semicircle on which a point B is taken so that BC is ciiual to the radius : shew that AB is a mean propor- tional between BC and the sum of BC, CA. ;">. A is any point in a semicircle on BC as diameter; from D any point in BC a p(ir])endicular is drawn meeting AB, AC, and tlie cir- cumference in E, G, F respectively; shew that DG is a third propor- tional to DE and DF. 0, Two circles touch externally, and a common tangent touches them at A and B : prove that AB is a mean proportional between the '■' " "" "' "■ ' [See Ex. 21, p. 210.] diameters of the circles. 7. If a straight line be divided in two given points, determine a third point such that its distances from tlie extremities may be ]iroportional to its distances from the given points. 8. AB is a straight line divided at C and D so that AB, AC, AD are in continued proportion; from A a line AE is drawn in any direc- tion and equal to AC ; shew that BC and CD subtend equal angles at E. 9. In a given triangle draw a straight line parallel to one of the sides, so that it may be a mean proportional between the segments of the base. 10. On the radius OA of a quadrant CAB, a semicircle ODA is described, and at A a tangent AE is drawn ; from O any line ODFE is drawn meeting the circumferences in D and F and tlie tangent in E : if DG is drawn perpendicular to OA, shew that OE, OF, CD, and OG are in continued proportion. 11. From any point A, in the circumference of the circle ABE, as centre, and with any ra. :s-^ r 332 kuclid's elements. Pj:ui'osrnox 15. Tnj:uuK.M. Triii'ii'jlas which are equal in area, and ichich have one (171.1/ fe of the one equal to one aiaj/e of the other, have their tiiil';s about the equal angles reriproralt ;/ proportional: Co'Hverselij, trlan;/les ivliic.h have one amjle of the o)ie equal to one amjle of the other, and the sides about thet^c amjles reciprocal t ij ]>roj)orl(oniii about, thcMC I. (I li;i\c the e ; ingles be line st. line; I. li. ), n>/p, lADD: VI. J. V. 1. EAD, ^ABD, A'l. 1. lABD, lAU^: V. I. (^. K. 1). nooK VI. fuoi'. 15. 333 KXKUCLSJCS. ON PllOrcSITIONS 11 AND 1.". 1. Pdnillclofiniiiix irliicli an' enua} in arm aud triilch h,ne their sides rcciiirucaUij propiirtiomil, have their luujles respect ivelij equal. 2. Triaitiiles irhich are e(]iial in area, and irhieh have the ■iides ahinit a pair of angles reciprocaUij proportional, hare those aiu/les equal or supplemcntarij, ;i. AC, BD are the diaKoiials of a tmpeziuiii wOiicli intersect in O ; It the side AB is parallel to CD, use I'rop. 15 to prove that the triangle AOD is eciual to the triangle BOC. 4. From the extremities A, B of the hypotenuse of a ri<'ht- angled triangle ABC lines AE, BD are drawn perpenlieular to AB. and nie.'ting BC and AC pruihiced in E and D respectively : employ Prop. 15 to shew that the triangles ABC, ECD are eciual in area. 5. On AB, AC, two sides of any triangle, squares are described externally to the triangle. Jf the squares are ABDE, ACFG shew that the triangles DAG, FAE are equal in area. (1. ABCD is a parallelogram; from A and C any two parallel straight lines are drawn meeting DC and AB in E and F resi)ectively EG, which 's ])arallel to the diagonal AC, meets AD in G : shew Ih'at the triangles DAF, GAB are equal in area. 7. Describe an isosceles triangle equal in area to a given triangle and having its vert.Jal angle equal to one of the angles of the given triangle. 8. Prove tliat the eiiuilateral triangle described on the hypotenuse of a right-angled triangle is eO.Srno\ is. J'lHUU.KM. On (t ijivca HtraiyJit line to (h'scrlhc (i rcct'dineal fyrire shnilar and siinilarl/j situiued to a (/icen rectUuieal Jhjure. Let AB be tlie ^aveii .st. line, and CDEF tlie given rootil. figure: lirst siii)pos(! CDEF to be a (luadrilateral. It is ifHiuired to «le.sciil)e on the st. line AB, a rectil. tigure similar and similarly situated to CDEF. Join DF. At A in BA make the _ BAG ecjual to the _ DCF, I. 23. and at B in AB make the ^ ABG e(]ual to tlie ^ CDF; .". the remaining ^ AG B = the remaining i. CFD; 1.32. and the AAGB is equiangular to the ACFD. Again at B in GB make the _ GBH equal to the :. FDE, and at G in BG make the _ BGH eijual to the i_ DFE; i. 23. .". the remaining _ BHG the remaining l DEF ; i. 32. and tiie .\ BHG is ('(juiangular to the A DEF. Tlien sliall ABHG be the retiuired figure. (i) To j)r()ve that tlie ([uadriJaterals are equiangular. Because the _AGB -tlK; _ CFD, and the _ BGH tli(^ _ DFE; Conslr. :. the whole _AGH. the whole _ CFE. Ax. 2. Similarly tlie _ ABH the _ CDE ; and tlie ;ingles at A and H are respectively e(|ual to the angles at C and E ; ' ' Conxt,-. .'. the iig. ABHG is ('(luiangular to tlie fig. CDEF. 338 i# Euclid's elemknts. (ii) To pi-ovH that tlio riundi-ilatprals liavo the si < ii-h thitl llw iiohiiinus liar, : ,nnl Ho- pnlnijoiin itn-. to oun aiuttlin- ht llii; diipl'iralv rii(n, ,,/ ll,eir Imiuitloijovs s'uir ll'S. Li!; abode, FGHKL 1)(' siiiiil.ir iioIv-mus, ;,u.1 let AB bo tho sU]() Jjoinoloyous to FG ; thou (i) the poly^irons in;.y he divided info the; suiiu^ mii.iber of simil/ir t> [ ii^u_. (ii) these trifui,<,de.s sliall have each (,, ,.,irl, tlie same ratio Hiat the pi'lyifous h.-ive- (iii) the poly-ou ABODE "^sliall Ik- fot'he lH)lygou FGHKL in the rlui)He;ite ratio of AB to FG. .ioiii EB, EO, LG, LH. (i) Then because the |)olvo]y-ou FGHKL, y/_^^^ .'. tlie _ EAB til.- _, LFG, and EA : AB :: LF : FG ; \ i. Di-f. 2. .•. the A EAB \< similar (o tlie .".LFG ; vi. 0. .'. the _ ABE the _ FGL. liut, because the poly.ijjoiis aiv simihir, Jfi/p. .'. the _ABC the _ FGH, vi. nf/ J. .•. the remainiii;^ _ EBC the j'eiiiainin-- _ LGH. And Ijecause the .'.^ ABE, FGL aiv similat-, Prnvrd. .'. EB : BA :: LG : C.r- and because the polygons are simihir, 7/yn .•. AB : BO . FG GH; vi. J)e/ 2, .'.. p.e i»juali, EB : 1,0 :: LG • GH, v. 14 that ih, llie sides abou ihe eqii, i ^ -^ ebO, LGH are proportiouais ; .•. the A EBO is similar to Hie _LGH. vi. G. •"W* ck Ui c'lch tlitit iin: tinittliir ill. and let AB Ix; sauu; miiiiber ai'Ii the sniiio )Iygoii FGHKL similar to tlio \ I. ih'f. '1. "G ; VI. G. VI. Ih'f. -2. _ LGH. iiilaf', Privi'd. VI. 7>y: 2. V. 14. LGH arc H. VI. G. IJOOK VI. I'ltdl'. 20. \) . Tn tlio saiiH! way it Jiiay hv pioxcd tliat tho .". ECD is similiir to tlu? ALHK. .'. tlni poly.L^'oiis I, ivt; bcon dividid into tlic! saiii" iiiiiiilici' ot" siiiiilai- triangles. (ii; A;i,'aiu, because the . ABE is .similar to tlic . FGL, .". the . ABE is to tlitj \FGL in die duplicate ratio of EB : LG; vi. I'J. ■ and, in Iik»' manner, ^ the A.EBC is to the . ,LGH in the dui.licafe ratio of EB to LG; .-.tho A ABE : the .6, PGL :: the .lEBC : (he ..LGH. v. 1. Tn like maimer it can be shewn tliit tho AEBC : th(^ .\LGH :: the A EDO : tho /.LKH, .•. tho A ABE : tho A FGL :: tlie AEBC : tho ..LGH :; th(! EDO : tho .'.LKH. I hit when any number of ratios are e»jual, as each ante- cedent is to its consoj)orfti»tti's, as thejirsf IS to the third, so is any rectUiiieal Ji(/vn' iliscribed on tJn' jlrx 'o a similar and similarly describe' rectilinad frpa-'i Vi, "s Kr.KMKNTH. ' ProPOSITIOV 21. TllKOHKM. l:ym,n',dju,>,res vhirl, ar.: shnUar to U,.- ,,unn rrclL Limaljujnr,; are nfso HUuUar to carh ot/in: L(,t ,.Hc-h of tlu. rvdilineul tij^ures A nnd B be similar to C- tlien sluill A l)e siinihir to B. For l)e<-au.se A is .similar to c, //„yy .-. A i.s e,iu.aujr„l,i,. to C, and the .side.s about ti.eir equal ' alleles arc proi.oi-tionals ^j 7W '^ A. K. D. the aniun recti- .siinilfu- to C; II UP. t tlieir e(jual VI. J)e/. 2. ' , . ^^m t their e(jual VI. Dfj: -2. > C, and ha\e I to the cor- t tlu.'ir equiil V. 1. HOOK VI. I'KOP. 22. ThOI-OHITIO.,' 2L'. J'ilKOKKM. 343 V y'^«'' utrnit/'it /iiirn be proporfioufff 'rtirniinf, IhjvrtH bi' nunifdi'/// dfsnihrd on lite, /ir'ni Hitt/ Hfcuiul, ami (t/si) (I jut'ir on. the third the, siitti/dr nntt d/nllttr/f/ dincribrd flifum on. the second, titt (t, rccfitinofd Jii/iirn on. the third is to thn fiimi/nr and siniilnr/i/ descrifml Jitjnre on the/ourth, the/our utrai'jlit tin, s sh
ut AB : CD :: .'. CD : X :: .'., ex (f.quali, AB : X :: But AB : X :: the tig. KAB and EF : O :: the tig. MF : Constr. GH 0. EF : GH ; Hyp. GH 0, V. 1. EF : 0. V. 14. : the fig. LCD, VI. 20, Cor. the fiir. NH; the fig. KAB : the fig. LCD : : the fig. MF : the fig. NH. 1. 344 KUC'LlD'rS iaK.MKNT^S. M A - G H C R tlie %. MF : the fig. NH; GH. Conversely, let the iig. KAB : the iig. LCD then sludl AB : CD :: EF To AB, CD, .'ukI EF tako a fourth ])ropni-tioTial PR- vi 10 and on PR descriho tlu^ li.i,.. SR Mniilar and siniilarly situatcnj to cither of tliehgs. MF, NH. yj j,.. Then because AB: CD ::EF: PR, Constr. . ., by the toniHM- part of the proposition, the tig. KAB : the %. LCD :: the ti<^ MF Jut the fig. KAB : tlie tig. LCD :: the tig. MF th(' fig. SR :: th(^ tig. MF the hiT. SR ^ th(! li<.-. NH the fig. MF tlio fig. SR. t]io fig. NH. J/y/>. tlie fig. NH, V. 1. And since the figs. SR and NH aTo similar and similar situated, GH*. ^»'ow AB .". AB PR CD CD EF EF PR; GH. Constr. . E. D. Jujiurs ore equal thnr homolooom sides are equal. The i.-oof is easy and may be left as an exercise for the student )|.:finitjox. A\ hen there are any nuinher of magnitudes ot the same kind, th(> first is said to have to the' last the ratio compounded of the ratios of the first t(, the second of I.e second to the third, and so on up to the ratio of ihe last Ijut one to tiie l;ist niay-nitude [B look V. Def. 1l>.] R the fig. NH; l1 PR: VI. 12. lariy .situatod VI. 1^. Coiuitr. ig. SR. ig. NH. 7/v//;. ig. NH, V. I. ncl .similiii'lv Consir. ii. E. D. ihirb/ Hitmited Tlio i)roof is magnitudes the last t\w K' scfoiid, of ratio of tlu; V. Dcf. 12.] ha BOOK vr. I'Rop. 23, 34; Proposition '1\\. Theokkm. ParnlMofjrams which tire ^'(pUaatjular to one auolhe/r have to our. another the ratio which is compoimded of the nilios oj their sides. Let the par'" AC Ix^ e; then DC and CE are also in a st. line. i. 14.' Complete the par'" DG. Take any st. line K, and to BC, CG, and K Hud a fourth proportional L- VF l-> and to DC, CE, and L take a fourth proportional M ; then BC : CG :: K : L, and DC : CE :: L : M. iJut K : M is the ratio compounded of the ratios K : L and L : M, v. z;,-/." ii'. tiiat IS, K : M is the ratio compounded of the ratios BC : CG and DC : CE. Now th(* par"' AC : th(^ par'" CH :: BC : CG vi. 1. : : K : L, Coustr. and tlie par'" CH : (he par"' CF :: DC : CE vi. 1. ,. , :: L : M, Constr. ..,rx (cquah, the par'" AC : the par'" CF :: K : M. v. II. ]5ut K : M is the ratio compounded of the ratios of the sides' .-. the par'" AC has to the par'" CF the ratio compounded' of tlie ratios of the sides. q.e.u. KXKKCI8E. 'f!;-^ areas of Lvvo triang!;.. or parallelogmius arc to one anoiiier iiUitudt's '^^'"l'""'"^'-''^ "* t'»' »'it>o« «f tboli- bases and of their t iii 34(5 tCCLIDS KLKMENTS. PnoposiTioN 24. Thkorem. P(ir((Udo(jratns about a / aiuj parallelograia ore nimlhtr to the ivhole jxiral/e/o^/ram and to one anothrr. J. 29. I. 29. Let ABCD be u par'" of wliicli AC is a diagonal; and let EG, HK be par"" about AC: then sliall the pai"" EG, HK be similar to the par'" ABCD, and to one another. For, because DC is par' to GF, .". the _ ADC -the _ AGF; and because EC is par' to EF, .•. the £.ABC = the ^AEF; iind each of the j. " BCD, EFG is equal to the opp. _ BAD, .•. the /. BCD the _ EFG ; [i. ;U. .*. the pai'"' ABCD is equianj^nilar to the pai-'" AEFG. .Vgain in the A'' BAC, EAF, because the _ ABC = the _ AEF, 1.29. and the _ BAC is connnon; .'. A*" BAC, EAF are equiangular to one anothei-; J. 32. .•. AB : BC :: AE : EF. vi. 4. But BC = AD, and EF .^ AG ; i. 34. .•. AB : AD :: AE : AG; and DC : CB :: GF : FE, and CD : DA :: FG : GA, .•. the sides of th(^ par'"' ABCD, AEFG almut their equal angles are proportional; .'. the par'" ABCD is similar to th(^ par'" AEFG. vi. Dcf. 2. Tn the same way it may be proved that the par'" ABCD is similar to the ])ai-'" FHCK, .'. each of the par"'- EG, HK is similar to the whole i)ar'" : .". the par'" EG i.=5 aimilar to the par"' HK. vi. 2i. Q. E. D. BOOK VI. PROP. 25. 347 y i>aralleh)fjraiti one anutJwr. (liugonal ; the ]»;ii-"' ABCD, J. L'9. 1. L'D. he opp. _ BAD, [i. ;u. par'" AEFG. -, 1. 2i am about the mine diayonul. A G D Let tho par"" ABCD, AEFG !).> .siinibir aiul similarly situatofl, and Jiave the coiuukiu aiii^le BAD: then shall these par'"' be about the same dia,. .■. GA : AK :: GA : AE; .'. AK AE, which is impossil)]e; .'. AC must pass through F; that i.s. the pai"'^ BD, EG are about the sann; diagonal. Q.K. I). HOOK vr. I'Rop. 30. 34J) (}h> BKFixrnox. A .st,-aio-],t ]ino is .said to l.o divi.lrd in extreme and mean ratio, wlu-n the whole is to the J.-Jute^ segment us the greater .segiueut is to the less. '' [Book VI. Def. 4.] PuoposiTiov ;10. Problem. ro ./tV/./. n, gh-.n xirahjht line In. extreme and mean ratio. cut FG, or FG C 3 , _ I^f't AB l)e the given St. line- It IS required to divide it iu extreu.e and mean ratio. tl!e't Tag. "" '" '^"' ''" '""'' ""'' ^^ "'■'^^' ^'^^ "i"'^^ *'^ Then ])ecause the rect. AB, BC the s(]. on AC, .-. AB : AC ;: AC : BC. ' yr. j; V. E. F. EXERCISES. ^\. a S^I'^i/S^FXr^^lS^ir^Sl!;:-' *? ;::-; divided iu extreme and mean i.ill^;:^rf-^iAr£^^^^^^^ ! 350 Euclid's klemen'ts. Proposition- .'^1. Tiikoukm. In a 7'l. but BC = the sum of BD, DC; .'. the tig. P^the sum of the tigs. R and Q. Q. I . r. NoTK. This proposition is a Reneralization of the 47tli Prop, of Book I. It will be a useful exercisi; for the stnilent to deduce the general theorenx from the particular case with tlie aid of prop. 20, Cor. 2. mre described similar and iny the ru/ht ich BC is the irly dt'sci'ibod tigs. Qaiid R. VI. 8. R,vi P. 20, Cor. V. 2. . P; R, Q tig. P; V. IT). lid Q. Q. I . P. le 47tli Prop, of t to deduce tlm .id of J'rop. 20, KXKRCrSES OX PROl', .31, EXERCISES. .351 r;„l.; „n '\"«Jt'^"-'>«'Vnangle if a perpendicular he drawn from tho Kl. a. gle to the opposite side, the sefrn.ents of tlie hypotenuse are in the duplicate ratio of the sides containing the right angle. t. /f ^^' "I Proposition .31, the figure on the hypotenuse is eniml o the given triangle, the figures on the other two sides are eacli em a to one of the parts into ^vhi(■h the triangle is divided hv the per e dicular from tlie right angle to the hypotenuse. " ' r . \- v^v^ '"•'^- ^\ '"'•' '»^'*^''^n« «f the triangle ABC which meet in u . It XY be joined, compare the areas of the triangles AGB, XGY. 4. Shew that simUar trlaiujU: an- to one another i„ the duplicate ratio qt (i) enrrespoialnai medians, (n) the radii of their ini-rihed eireles, (m) the radii of their cireuniserihed eireles. ' , ■''^' ^ ^IF^J""- ^^"^ Vedal trmni;]e of the triangle ABC; i.rove that the tmngle ABC is to the triangle DBF in the duplicate ratio of AB to oU. Hence shew that the fig. AFDC : the a BFD :: AD- : BD-. tl.nf R J^nn^"'' ?.^ of a triangle ABC is produced to a point D such that BD : DC in tlie duplicate ratio of BA : AC. Shew that AD is a mean proportional between BD and DC. 7. Bisect a triangle by a line drawn parallel to one of its sides. 8. Shew how to draw a line parallel to the base of a triangle so as to form with tlie other two sides produced a triangle double of the given triangle. '.». If through any point within a triangle lines be drawn from the angles to cut the opposite sides, the segments of anv one side will iiaye to each other the ratio compounded of the ratios of the segments ot the other sides. 10. Draw a f a.dght line parallel to the base of an isosceles tri- angle so as to cut c'^ -.x 'liangle which has to the whole triangle the latio of the base to a t'dj. 11. Through a given point, between two straight lines containing a given angle, draw a line which -.hall cut oG a triangle equal to a given rectilineal ligure. Obs. The .32nd Proposition a ^i .en by Euclid is de- lective, and as it is never applied, v. e have omitted it. II. E. 23 352 KCCLIDS KI.KMKNTS. Pkopositiox :\'.\. Thkokkm. Ill eqiKil circfcs, (iiiyles, wlntlier nt (In; coitn's or the, cii'- cHiiij't'nnccn, hace tlie .same ratio as the arcs on uhich they stand: so also have the sectors. m 1 Lot ABC and DEF 1»«^ ('(lual ciiclcs, and let BGC, EHF be angles at tlic ccntirs, and BAG and EDF angles at tlio Q*'''*; then shall (i) the _ BGG : th(i _ EHF :: the are BC : the arc EF, (ii) the _ BAG : the _ EDF :: tlu^ aie BG : the arc EF, (iii) tiie sector BGG : the seetoi- EHF :: the arc BG : the arc EF. Along the ()"' of the (mABG take? any number of ai'cs CK, KL each eciual to BG; and along the 6'*' of the 0DEF take any nunibei' of aics FM, MN, NR eacli e(|ual to EF. .loin GK, GL, HM, HN, HR. (i) Tlien the _^ BGG, GGK, KGL are all equal, for they stand on tlu^ eijual arcs BG, GK, KL: ill. 27. .*, tlie L. BGL is the same nmltiplc of tlie _ BGG that the arc BL is of the arc BG. Similai-ly the _ EHR is the same multiple of the ^ EHF that the arc ER is of the arc EF. And if the arc BL ^^ tlu! arc ER, the _ BGL the _EHR; and if the arc BL is great been tiiken, nfunely the _ BGL :ind the fire BL; find of the coiiKC(|nent.s finy eiiuinuiltiplcs lifive been taken, namely the _ EHR find th(^ fire ER : find it has been proved thfit the _ BGL is giefiter than, e([UfiI to, or less tlifin (he _ EHR fieri )r(liii,t( jisBL is greater tlmn, ('(jUfil to, or less tlmn ER; .". the four nuignitudes are proportiontils ; v. ])(-/, 4 that is, the _ BGC : the _ EHF :: the arc BC : the arc EF. (ii) And since the _ fiGC - twice the _ BAC, in. 20. and the _ EHF twice the _ EDf'; .-. (h.) _ BAC : the ^ EDF : : the arc BC : the arc EF. v. 8. (iii) Join BC, CK; fuul in th(^ arcs BC, CK t;ike any points X, O. Join BX, XC, CO, OK. Then in the A* BGC, CGK, [ BG. CG, iJecfiuse - GC-GK, [and the l BGC r:.the i. CGK; .•. BC CK; and the A BGC the A CGK. And becfiuse the fiic BC the fire CK, .'. the remaining arc BAC th(i remaining fire CAK. .'. the ^BXC^the ^COK; iir. 27. .•. tli(^ segment BXC is simihir to the segment COK; in. Dp/. find they stfind on equfil chords BC, CK; .•. the segment BXC-. the segment COK. iii. 24. And the A BGC - the A CGK; .". tlie sector BGC -- the sector CGK- 23-2 III. 27. T. 4. Constr. 354 EUC[ in'-^ KI.EMKXTS. («iii(! Snnilfirly it nuiy be shewn that the sectors BGC, CQK, KGL fire .-ill equal; niul likewise the sectors EHF, FHM, MHN, NHR are all equal. .". the sector BGL is the same multiple of the sector BGC that th(^ arc BL is of the arc BC ; and the sector EHR is the same multiple of the sector EHF tliat the arc ER is of the arc EF: And if the ai-c BL ^^ the arc ER, the sector BGL -- the sector EHR: Proced. and if the arc BL is greater than the arc ER, the sector BGL is greater than the sector EHR: and if the arc BL is less than the arc ER, the sector BGL is less than the sector EHR. Now since there are four magnitudes, namely, the sec- tors BGC, EHF and the arcs BC, EF; and of the antecedents any equimultiples have been taken, namely the sector BGL and the arc BL; and of the consequents any equimultiples have been taken, namely the sector EHR and the arc ER : and it has-' been shewn that the sector BGL is greater than, equal ir\ o? less than the sectoi- EHR according as the arc BL L; ;-r.'\ater than, ecjual to, or less than the arc ER; .". iK>r four magnitudes are proportionals; v. Bef. 4. that is, the sector BGC ; the sector EHF :: the arc BC : the arc EF. Q. K. D. ^ii i if HOOK VI. PROP. H. 355 "D lioposniox 15. Thp:o"km. , BGC, CGK, re all equal, sector BGC sector EHF R: J'rored. i ER, EHR: :r, HR. ely, the sec- antecedents sector BGL :juimultiples le arc ER : reater than, ling as the e arc ER; ; y.De/.i. the arc EF. q. K. D. // f/n: ve.rtk-'\seril«' a circ](! al)out tJM' dC, iv. 5. and produce AD to nifct the ' in E. .loin EC. Then in tlie A" BAD, EAC, because the _ BAD - the _ EAC, Jl,n, .■•nd th. ABD = th( _ AEC in the same segment; in. 'i>l.* . . the ^ BAD is ('(juiangular to the A EAC. i ;Jl'' • ". BA : AD :: EA : AC; ai 4 .•. the rect. BA, AC - the rect. EA, AD, vi. lo! - the rect. ED, DA, with the sij. on AD. Jiut the rect. ED, DA = the rect. BD, DC: in 35 .-. tlie rect. BA, AC - the rect. BD, DC, with tl • s,i. on' AD.' (^ K. \). KXKKCISK. ,..1, "u*)!^ vertical anf,'le BAC be externally Lisccted by a stiaiKht line Ar'fl ff"' ^v^T '" °' «^^ew ^''".t liit^ >t.cia„frle contained by BA, MICROCOPY RESOLUTION TEST CHART (ANSI and ISO TEST CHART No. 2) 1.0 I.I 1.25 "- IIIIIM Hill t 9 !r iiiiM I:: m L4 II 2.5 2.2 2.0 1.8 1.6 ^' x^PPLIED IIVMGE Inc 165 J East Main Street Rochester. New York 14609 (716) 482 - 0300 - Phone (716) 288 - 5989 - Fax USA 356 EUCLID'h ELKMEX'x.i. Pkopositiox C. Theohkm. If from the vprtical aixjle (fa tnawjh a strahjltt line he Jrawn perpevdlcular to the Ihiup, the rectnm/Ie contained by the sides of the triangle shall be equal to the rectan(/le con- tained bij the 'perpendicular and the diameter of the circle described about the trian. slra'ujlit line he, le co)itainp.d hy i rectangle copi- er of the circle le porp. from A root, contained nscribed about EC. IV. ;>. semicircle ACE, <,'uient; III. 21. EAC; 1.32. VI. 4. AD. VI. IG. (i.K.I). BOOK Vr. PROP. B. rROrOSITIOX J). TlIKOIlKM. 357 2'hi', rectan.yla coutttlaed Ity tlie dhujonah of a (juddri- lateral inscribed in a circle is equal to the sum of the two rectanyles contained by its opposite sides. Let ABCD be a (juadrilateral inscribed in a circle, and let AC, BD be it.s diagonals: i\nm tlio rect. AC, BD sliall be equal to tlie sum of tlio rect- angles AB, CD and BC, AD. Make the _ DAE ecpial to tlie _ BAC; 1.23. to each add the _ EAC, then the _ DAC the _ BAE. Then in the .::\« EAB, DAC, the _ EAB -= the _ DAC, and ihe _ ABE the _ ACD in the same segment; iir. 21. .'. the triangles ai-e e()uiaiigular to one another ; i. 32. .•. AB : BE :: AC : CD; .■. the rect. AB, CD - the rect, AC, EB. Again in the A"* DAE, CAB, tiie /. DAE the _ CAB, Co?istr. and the „ ADE-the :. ACB, in the same segment, III. 21. .". the triangles are equiangular to one another ; i. 32. .•. AD : DE : : AC : CB ; vi. 4. .'. the rect. BC, AD ^ the rect. AC, DE. vi. 10. But the rect. AB, CD - the rect. AC, EB. Proved. .'. the sum of the rects. BC, AD and AB, CD ^: the sum of the rects. AC, DE and AC, EB ; that is, the sum of the rects. BC, AD and AB, CD = the rect. AC, BD. u. 1. Q.E.L. VI. 4. VI IG. I: In^ 368 EucLin'.s kli:mi:nt.s. added by Robert Himson. Prop. D is iisually known as I'tolemy's theorem, iuul it is the pur- iicuiar case ot the lollowing more general theorem : The rectauole contained hy the dimjouaU of a quadrilateral is irs, than the .van oj the rectaunh; amtained hi its opposite xidrs, vuless a circle can be cnruiuscribed about the quadrilateral, in which case it i,s equal to that .lani. KXKHCISKS. 1. ABC is an isosceles triangle, and on the base, or base pro- auced any pouit X is taken : shew that the circumscribed circles of the triangles ABX, ACX are equal. ARP ?"'"'^\*^^ extremities B, C of the base of an Isosceles triangle ABC, straight lines are drawn perpendicular to AB, AC respectivety tbei"ctardeAB'''DB' '"^''''^ *^'''* *^^« ^'''^'=ta'^«l«^ BC, AD is double of • i'!' ^^1 *'^^,,^^i'^«"»al8 of a quadrilateral inscribed in a circle rae at ■ig It angles the sum ot the rectangles of the opposite ddes is double tHe area ot the figure. Rn"*i:- ^^^aV"" "; 'l^'^'-V'il'i^^i'^l inscribed in a circle, and the diagonal Tn^^le'DcfcB ' ' '"'*''"^'^' ^^' ^^ '' ^"^"'^^ *" ^^'''^ J If the vertex A of a triangle ABC be joined to any point in t e base, it will divide the triangle into two triangles such that their cucumscribed circles have radii in the ratio of AB to AC. (5. Construct a triangle, having given the base- the vertical angle, and tlie rectangle contained by the sides. ^ l,oI\i '^r?^ *"^"\^'^'^%"f equal area are inscribed in the same circle : ^ c\v that the rectangle contained bv any two sides of the one is to he rectangle con ained by any two sides of the other as the base of the second is to the base of the first. ,„>;.?; i ^ fi'''^'' •''' '^^'f ^^'^^^ ^■o""^! an e.iuilateral triangle, and from any point in the circumference straight lines are drawn to the angular !». ABCD is a quadrilateral inscribed in the antrle ABC : if the points A and C are fixed .e, and BD bisects ,>(• fl,-> ,.;,. ,1 1 D • "~ '-".""" V '-"^ ^ ""^ "Acu ui; the circumference of the ciicle and B is variable m position, shew th.t the sum of AB and BC has a constant ratio to BD. ill Euclid, but wove n, aud it is the par- IHadrilaterul in ic.ss ixite tii(h's, uitli'xs (I in which case it. v',s base, or base pro- mscribcd circles of n Isosceles triangle B; AC respectively, ', AD is double of d in a circle r.re at site sides is double J, and the diagonal 3 equal to the rect- d to any point in les such that iheir to AO. the vertical angle, 11 the saine circle : •i of the one is to ler as the base of igle, and from any I'll to the angular it lines is equal to e, and BD bisects the circumference t the sum of AB THEOREMS AND EXA:\rPLKS (^X r.OOK VT. I. ON HARMONIC SI'XrriON. 1. 7'(» divide, a (jivrn ntniifiht liiu' iittcrnaUij and I'.vtcnidUij xo that its seijmentu may be in a fiivcn ratio. H- K ^^--6/' L M A b Q M Let AB bo the given st. line, and L. M two otlicr st. lines wliicli deterniine the given ratio: it is nujuir. '} to divide AB internally ant? externally in the ratio L : M. Through A and B draw any two par' st. lines AH, BK. From AH cut oiY Aa equal to L, !ind from BK cut off Bb and B// each equal to M, Bh' being tal«.'n in the ,^ainc direction as .s«i"t to a harmonic laiigo js said to be a harmonic pencil. transvemh'''"^'^ ^""^ '^''''''" *"" '"' "" ''•^'^'''" "^ ^"''' ^'^ ^''^'^^^ ^^ 4. A Hvsteni of four .straight lines, no three of which are eoncurrent, is c;Uled a complete quadrilateral. i hesc straight hues will intersect two anhannunic rmiyc, one o^, each sfraUi Urn (/(L- 7>o/,,/ tun em. also yq, Bb, Qp wtU be eoncicrrenl. ,,J) ,F''' ^/^f''«i 5 to prove that in « comvlefc oundrihlrml hi nm:h tne three dmyonuls are drawn, the straight' line joinim, any nair '!/ opposac rerl.rs is cut harvmucally by the 'other twodiagZals! ^ i t cullod a range. one puir aro liar- r, it i.s .said to bo a point is called a ex ol' tlu; j)eiu;il, 't to a hariaonic f lines is called a I'ec of wliich arc 'WO in .six points, •oe strai'dit lines '.'/ of (I /Kiniiiiilic '«>/• Coitrcrselif '•ct interiialhj and ■ mnrfcN, one on the strnlfiht linrs- S ; then will Qq ^, P, B, Q and ' line {the points , Qq vill be con- (iiiii.ilTihjlcral in. ioinintj any 2>air diagonals. THKORKMS AXn EXAMPLKS 0\ HOOK VI. 363 II. 0\ TKNTUKS OF SIMILARITY AND SIMIUTUDK. 1. If an;/ tn-o unequal simitar fi(iure.-i are plaeed no that their homohufoHH sides are parallel, the lines joining eorresponiting points in the tn-o figures meet in a point, whose distanees from any tno corre- sponding j)nints are in the ratio of any pair of homologtms sides. Let ABCD, A'B'C'D' be two Hiniilar figures, and let them be placed so that their hoinolo;,'oiis sides are parallel; namely, AB, BC, CD, DA parallel to A'B', B'C, CD', D'A' respectively: then shall AA', BB', CC, DD' meet in a point, whose distances from any two corresponding points shall be in the ratio of any pair of homologous sides. Let AA' meet BB', produced if necessary, in S. Then because AB is par' to A'B'; Hyp. .•. the A^ SAB, SA'B' are equiangidar; .-. SA : SA'--AB : A'B'; vx. 4. .-. AA' divides BB', externally or internally, in the ratio of AB to A'B'. Similarly it may be shewn that CC divides BB' in the ratio of BC to B'C. But since the figures are similar, BC : B'C'=AB : A'B'; .-. AA' and CC divide BB' in the same ratio; that is, A A', BB', CC meet in the same point S. In like manner it may be proved that DD' meets CC in the point S. .-. AA', BB', CC, DD' are concurrent, and each of these lines is divided at S in the ratio of a pair of homologous sides of the two figures. Q. E. D. CoR. If any line is drawn through S meeting any pair of homolo- gous sides in K and K', the ratio SK : SK' is constajit, and equal to the ratio of any pair of homologous sides. Note. It will be seen that the lines joining corresponding p.ints are divided externally or internally at S according as the con-esp ond- in" sides are drawn, in the saine or in opposite rlireetirms, Tn eit)ier case the point of concurrence S is calied a centre of similarity of the two figures. I 1 i uwm I '* 364 i: I re r, id's i:r,K.\n:NTs. 'itfl 1 1 ■ ! Mr ! 1 -. A cmiiion hninnif STT' to two cireleA whm cn,lrc.< ,nr C C drawn m.'cUm, th.'sr two rirrh, in p. Q, „„,; p'' q' y',,i,,,,, ' ;■"■ the r,ulii CP, CQ .h.ll /. /v.^.v/Zn'/v innnmtoc'p'c'i' Al.. Cor. 1. It lias heou proved that SC : SC = CP : CP'- thus the external common tangents to the two circles meet at a noint S ^^JIch divides the line of centres eternally in the ratio of the Tui Similarly it inay be shewn that the transverse common tangents mtL of thc'Si. ' '^"" '^'' ^'''' °^' '''''''' "^*"'"^lly i" t''^^ Cor, 2. CC'is divided harmonically at S and S'. Dkfinition The points S and S' which divide* externally and .ntcvnaily he line of centres of two circles la tlic ratio of Jheir rad are called the external and internal centres of similituae respectively! e cciitren arc C, C, in/ stniinht line in '', Q', rrxiu'ctirrh/, dlil to C'P', C'Q'. eh he cqittll to t!h! I'Q'. it iinglo, III. IH. lar; VI. 7. ill' to T'P', T'Q' r; m. 37. VI. IG. ST' y. K. ]i. is meet at a point ratio of tlie radii. common tangents internally in tlie le externally anil itio of their radii tude respectively. THKOllKMS AND KX.VMl'LES OX HOOK VI. KXAMl'l.r.S. 305 1. Inscribu :•. nipiaro in a j,'iven triaiiiJtU). 2. In a given triangle iiiseril)e a triangle similar iind similarly situated to a givt.'n triangle. ;{. InHcribc a miuaio in a given sector of circle, so that two iiiigular points slial'. he on Hki uro of the Hector and the other two (111 tlie hounding radii. •t. /// ///(• I't'imr I'll pnift' 278, //' Dl iiic'ls the iiiscrilx'd circle in X, kIicw Unit 'a, X. D| oiiit to tlio II pitiiit. i.s<-!ill(' polar : also LM is the st intersect the full without it: tho tJiut^'L'ut at al Property of 'J. 1/ A (iiiil P iirr liny tiro pnhilx, and if tin- jmliir <>>' ^ iri'h rrxitert in nnij riirlf iinni^i'ii tliroiKjk P, tln'ii tlw pular of P inuxt juihh llirouijh A. lii't BC be Mie jwlar of tlie jioirit A witli iTspcct *-.) a circle whose ctntie is O, and l(!t BC pass thiouj'h P: tlioii shall tlu! i)oliir of P pass tliroiij,'h A. Join OP; and from A draw AQ pevp. to OP. We shall shew that AQ is the polar of P. Now since BC is tlie polar of A, .•. the /. ABP is a rt. aiifjle; ])>•/. 2, paKo 'JCtO. Hiid the /: AQP is a rt. anK'le: Cnitxtr. .". tho four points A, B, P. Qareconcyclic; .-. OQ.OP-OA.OB lii. ;i(i. — (radius)-, for CB is the polar of A: .-. P and Q arc inverse points with respect to the given cii And since AQ is i)erp. to OP, .'. AQ is the ))olar of P. That is, the polar of P parses Ihrouyh A. A similar proof applies to the case when the given juti without the circle, and the polar BC cuts it. cle. y. j;. 1'. nt A is ii. To itrovc that the locus of tJie intersection of tatijieiitK dm (I cirrie (it the e.rtreiiiities of nil chorda which piiKx tliroiKjlt ,. ••<•-' that IS, HK passes tlirough A. .-. P IS the i)oiut of intersection of tangents drawn at the ex- tremities ot a chord passing througli A. of A From I. and 11. we conclude that the required locus is the polar JsoTE If A IS wuhuiit the circle, the theorem demonstrated in lart I. ot the above proof still holds good ; but the converse theorem lJ^nV ^^""*l,V>"^ ■!?'• "" ^'""^^' "^ ^^- 1''"^' it" A i« without the ncle the polar BC will mtersc^ct it; and no point on that part of he polar which is withm the circle can be the point of intersection of tangents W- We now see that (i) The Polar, >f an external pninf with irxpcct to a circle is the chord <>J contact oj tamjoits draini I'roiii it. (ii) The Polar of ,ni internal point h the locus of the intersections 01 (an<,ents draicn at the extremities of all chords which pass through tharn,ii f'"^ ^'"'"''"J" l'"i"^ on the circuiuforencu /.• the tamjent at jents from P, Ex. I. 11. 23B. Cuimtr. yclic. ingles. III. 22. L'unstr. fA; Co list r. Siitisfips the given oiiit on it. Diiiw )icl of oontiicl HK rough A; for P \a Ex. 2, p. :i(J7. ilnuvu at the ux- ociis is tlic i)oI;ir ilonionstratcd in converse theorem ■ A is without tile on that part of of intersection of to a circle is tlic ( tin: intcr.si'cfi<)ii.'< 'dell 2>(i>i''i throiujii is tlic tamjciit at THi:oKEM8 AM) EXAMl'LKa ON BOOK VI, m,) H p aI K Q 3 AHB. Ex. 1, \). 'CM. The following theorem is known as the Harmonic I'ropcrty of Pole and I'olar. '1. Anij straiiiht line th-nwn throuiih a point is cut lianiumicallij 1/1/ the point, its j)()l(ir, and the riiruiujerence of the circle, Let AHB be a circle, P the given ])oint and HK its polar; let Pinjb be any straight line drawn through P meeting the polar at q and the o™ of the circle at a and 1/ : then shall P, ((, 7, 1> be a harmonic range. In the case here et)nsiclered, P is an external point. Join P to the centre O, and let PO cut the o™ at A and B : let the ])ohir of P cut the (/" at H and K, and PO at Q. Then PH is a tangent to tlie .-. From tho similar triangles OPH, HPQ, OP : PH-.PH : PQ, .. PQ. PO=PH- ^ Pa . Pb. :. the points O, Q, ((, 6 are concyclic: .-. the ^({QA^the z ahO =:tlie /. 0((/^ t. .1. — the z OQ.I), in the same segment. And since QH is '"n'p. to AB, .-. the Zr/QH= /6QH. .". Q.q and QP are the internal and exiernal bisectors of the Z aQ.1): .'. P, a, (/, /> is a harmonic range. Ex. 1, p. ;-J()0. Tlie student should investigate for himself the case when P is an internal point. Converseli/, it inai/ he shewn that if throutih a fixed point P nvy secant is drawn ciittiiifi the eirciniiference of a (jiveii circle at a and b, and if q is the hannonic conjit()ate at P with, res2>ect to a, b; then the locus of q is the polar of P with respect to the u circle tliii.t e.iii'h side is t le [lobir of tho o[»l)ositc vertex is «aid to lie self-conjugate with rcs[»cct to the circle. 24- .J - ^ - -~- - -- -J ■,s 870 KL'CI.IDS KI.K.MKNTS. !. i-..\a.\ii>ij:s o\ |'(»f,h and i'olak, 1. Til,- atniijilit line, wltich Joins (iii;i two points is tin' jiohir with rrsprrt to a (jicen circlr of the point of in'tcrscction of their polors. 2. The point of intersection of an;/ two strdiifht lines is the pole of the Ktraii/ht line which joins tlieir poles. ;•{. Find the locus of the poles of „ll straiijht lines which pass tlirouijli (I iiiren point. ■1. Find the locus of the jxiles, with respect to a P with respect to the circle. !). (iiren a circle, and a fixed point O on its circumference ■ P is nny point on the circle: find the locus of the point inverse to P with respect to any circle whose centre is O. _ 10. Given two points A and B, and a circle whose centre is O snewthat. the rcctaniile contained hi/ OA and the perpendicular from B on the polar of A is equal to the rectaniilc contained hij OB and the perpendicular from A on the polar of B. 11. Four points A. B, C, D are taken in order on the circumference oj a circle: DA, CB intersect at P, AC, BD at Q and BA. CD /// R- snew that the triangle PQR is self.conju.iatc with respect to the circle.' 12. dive a liuf'ui' construction for lindinq the polar of a qiren point with respect to a ijiren circle. Hence find a linear cdnstrnction jor draw mil a taniient to a circle from an external point. 13. //■ a triaiiiile is self-conjuI('A1, AXIS. 1. To lind the locits (f points from irliich the ttiinii'iits dniirn tii lirn (jiren circles are eiiiial. l'i«. * l-'i". 2 Let A and B bo the centres of the fj;ivon circles, wlioso radii are n and b; and let P bo any point such that the tan}.'ent PQ drawn to the circle (A) is equal to the tan. (B), (C). A) and (B); J [if p. '^) and (C), Tfup. ,nd (C), ronf^'h O. hat O is witli lU tiio tliree circles lit these common ly by III. 35. iical axes of three Let A and B be the centres of tlie given circles: it is required to draw their radical axis. If the given circles intersect, then the st. line drawn tlu'ongh their jioints of intei'section will be the radical axis. [Ex. 1, Cor. ]>. :i7'2.\ But if the given circles do not intersect, describe Jiiiy circle so as lo cut them in E, F and G, H ; Join EF and HG, and ])roduce them to meet in P, Join AB; and from P draw PS porp. to AB. Then PS shall be the radical axis of the o» (A), (B), Definition. If cadi pair of circles in a given sy.steiu have the same riulical axis, the circles are said to Ijo co-axal. i;xa:mpi,ks. 1. ,S7;('/(' tlint till' radlt'iil o.ris of tiro rirrlr.^ hi.'^rrt^ tiiii/ one of their eoiinuoii taiKjeiits. 2. If taiiiieuta are draini to tiro circles from any point on their radical a.vifi ; xheir titat a circle deacrihcd n-itli this point an centre and (1111/ one of the tangoita as radius, cuts Indh the (jiren. circles ortho- ijonallij. ;}. O is the radical centre of tliree circles, and from O a tangent OT is draicn to any one of tlieni: slieir that a circle ichose centre is O and radius OT cuts all the given circles orthogonally. 4. If three circles touch one another, taken tico and tn-o, shew tliat their common tangents at the points of contact are concurrent, . f 374 KlXl.ins K I, KM KM'; '■>. //' circJcH (lie (h'xrrihrd on tlir t/ii-i (liaiiu'ter, their radical ceiitrr /,v the orfli 'I'e sides of (I tridini/e tu C. All rircl 'es irhii-h ocriitrr of the frianpl. pai^x thromih a fixed point and rat cirele orthnfionnllii, pass thriwuh a second' li. red pa I point. a III re, I 7. Find the locus of the centres of all circles which pass throiujh . . . ,, is tnhen on the radical axis: shew /L/V, J T '/''•''•'/ ^''''''^ point as centre n-ith radius e.,u' l e ta,n^'£^^^ ""' [These fixed points are called the Limiting Points of the si,sten,.] po£; iJiHiTi::; t--^ i;;t Vz ::;:'z r^-"" '"■ form a harmonic range. ^ '' ""^ '""^ "■' ''''""•''•''• ,w;/;„vXrr„,r;^:r;,;-;;;!';; :,;:i:t" """" "•" "■■■ -»" "' 1 ;; term, S-,,„„e cl , L' ^ it "2i the XXl '"""f '""".v "' to an„ther will be ,len„ted ,, I fr,")^ °ff "^^j ''r 1 "'?'"'"*■ tl,e n„„.e,.at„,- unci the ..econdthe^le^mii^ /''"'•■'' "'° '""' '» i I lie triaiiplf. lit (lint cut (I i/iiro It. hick 2>tiss tlir' of (,)!,• /.,( f.fif ^0 .supi)ose that nunieric-illy in 2 such .segment lich the tir«L j.s THEOHE.MS AM) KXAMTIKS ON liOOK VI. M7.") V. OX TRAXSVF.USALS. l)i;i'iXTTlo\. A sti'aighi line (h-awn to cut. a ui\cii svsTcni ot lliii's is calleil a transversal. 1. If three concurrent straii/lit liiies are drawn from the iiniiiilar points of a triangle to meet the opposite sides, tlien the product of three alternate segments taken in order is enual to tlie product of the other tliree ncjiments. Let AD, BE, CF Le drawn from tho vertices of the a ABC t > intersect at O, and cut the opposite sides at D, E. F: then shall BD . CE . AF:= DC . EA . FB. By similar triangles it may be shewn that BD : DCr=the alt. of aAOB : the alt. of a AOC; BD_ aAOB '■ DC ~ A AOC "' similarlv. and CE_ aBOC EA a BOA AF_ aCOA FB " A COB ' I\rnltiplying tliese ratios, we have BD CE AF DC ■ EA ■ FB or, BD .CE. AF -DC. EA 1; FB. Q. 1'.. 1>. The converse of tliis theorem, which may he proved indirectly, i.-s very important : it may he enunciated thus : If three straight lines arawn from the vertices of a trianple cut the opposite sides so that the product of three alternate scfimeut's taken in order is equal to the product of tlie other three, tiien the tliree straight lines are concurrent. That is, if BD . CE . AF= DC . EA . FB, then AD, BE, CF arc concurrent. 1,1 HI Hill M I 376 KUCMd'.S Kl.K.MKNTS. i V Hi i [, ^^^^H' '' ^ ' ' i P r>in„l to thr product of th. otiur tinrr .,o>>,nitt "'" '" CA^ AB^omIT.'' ^-'T^^"' T''^ y^ ''' ^'■^"^vorsal moot thr ,.;,]... BC V.M, Mt,, oi tlicHc sides produced. ,at, D. E F- thon sliiill BD.CE.AF DC. EA FB ^I'lMAV AH par' to BC. incrtin- tl.o tmnsvcr.sul at H.' 'I'lien from tho siiuihir a' DBF. HAF, BDHA FB AK • DCE. HAE, CE EA • DC HA '■ BD CE EA FB • DC AF' BD.CE.AF DC.EA .>B BD.CE.AF^DC.EA.FB. «lnc(!d, as in Fi^'. 2. ^ '8- i > oi all tbreo .sides i^ro- The converse of this Theorem may be proved indirectly iiiid from tlip similar a' •• , l)y miiltiplioatioii, lli.'it is, or, 1, thi' xiih-A produced, ^^•< ftil,-cit ill ortlcr /s b\ U'Ct tllC Si(l(>,q BC. FB. H. MISOKLLANKOUS EXAMl'I.KS ()\ HOOK VI. DKKIN'ITIONS. .T (i) I two ti'iuiigk's iiro sm-li that tlii'on struiglit lines joiniiii,' corresjiiiuJing vertices ure eoncniTeiit, tliey are M.iid to he co^ polar. (ii) If two triiuigh'.s are siicli that the pohits of intersection of rorresponding .sides aiv collinear, ihey iire said to he co-axial, 'I'ifKllliKMK ro hi: I'KnVKli liy 'ri!A\SV|-.l!SAr,S. 1. 'J'lw stniiiiht liiicx which Jain I lie rcrticru nf n trininilc to tlic jutiiitu of contact of the iiixcriticil circle (or aiiii of the three cscrihed I Ircles) are concurrent. 2. The middle points of the diniionah oj' a complete iinadrilaleral lire col I incur. ii. Co-polar triani/les are also co-a.vial : and coiircrseh/ co-n.rinl Irian (/lex are also co-polar. 4. The six centres of .similitude of three circles lie three hij three on four straiijht lines. I MISCELLAXEOUH EXAMPLES OX BOOK VF. Q. E. I). nitlior meet two 1 tliree .sides pro- ii'ectly : Ic and the third t the product oi' e product of the ig to the concur- ■y tao niethod of ortant theorems 1. Through D. any point in the base of a trian;^!.' ABC. straight lines DE, DF are drawn parallel to the sides AB, AC, and meeting; the sides at E, F: sluw that the triangle AEF is a mean ]iroportional between the triangles FBD, EDC. 2. If two triangles have one angle of the one equal to one .THgle of the other, and a second angle of tlie one supplementary to a second angle of the otlier, then the sides about the third angles are proportional. H. AE bisects the vertical angle of the triangle ABC and meets the base in E; show that if circles are described about the triangles ABE, ACE, the diameters of these circles are to each other in 'the same ratio as the segments of the base. 4. Through a fixed point O dr.avf a :,traight line so that the parts intercepted between O and the perpendiculars drawn to the straight line from two other lixed points may have a given ratio, KLCi.lDs i;i,i:.\ii:\is. •). Tlif aii','1.. A of II (liaiiL'lt. ABC is l.isoft.wl l... ao BC „, D. HM.l AX is tin. .n.-lian ll..t , . BC s u th^, vn V""' rl.« sa.no ratu, to XB as tl... ditTcenc. of tl?. sid.is has^.I'lheif s^ n'r ;;;i;;.^M.ointof bc. t^.. ob i.';; li;:' ^.i^.iirsji.^ ^^li Q •* \ i s 1 HiH fixeil points; AB fiii,l CD ai.. lix,,! „..pi1],.1 any st.a..i,t lin. is dnuv,. f.ou. P to .ne't A^'h^^m! 7. P aud H. C ,s til.. ini(l.l],. point of an a.r of a ci.clo who^e chord is ■ny point m tl.o vouiiv^ate an: : shew that AD ) DB AB: D IS DC : : AB .- AC. 1>. Ill the trian-Ie ABC the sid.; AC is douhl,. „f Rr rf r^n contained by the sc'nients of fl,.. l,vv.lv; '^ ^^'^ rectangle «u,„ Of the ..^cansi^c;;;-,,:!, t:;.i.i!'C™?»:f Ih'rx"' '° *"- 12. D is a point in the side AC of the trianrrip ARr „v, i ir • Doint in AB If Rn nc a- ■ -i ^ i-'/^ •'''anf,'ie ABC and E is a 13. If the perpendiculars from two fixed poiuta on n cf.. • ^ , dm,mHor,bea „bo„t ABD. AOD are as AD to Ca'" "'" "'"'"^ -Uj MISl'KM.ANKCMS liXAMIM.KS oN ItiioK \ I. ;;7i) il l)V AD mooting ilifw tlijit XD liiis 1ms to tlieir siiiii. trianjilp iiiteinally w tlmt if O is tlK' oiml hctwi'on OD III!' fixed punillt'l to iiioct AB at M. M uicctiiij,' CD at . and tlit'iice shew »«h a fixed point. whose chord is of BC. If CD, n(!t'tin^' AB in D \CD, ABC, CDE 1(1. AB, CD ar« two diameters of tiie ,Iiew lluit DC is a mean Itroportional between DE and DF. ir -in it ill 380 i:i'ii.ii>s i;i,|.:mi.;nt; 11: m 2(5. If tlirouKh tl)f ini p.,i„t of tlin 1 lini>I)t' (liiiwii iiittTscct iiK <»ii.> nidi, (,f tlm triiiii^'lr. tli.. oti ms(( of II trliiiiKli' uin HM.l th,. linn <|i(i\vii immllcl to llio I)iim; 1" HT plfldlK ci'vifl. (1 1 uiiinoiiiciilly, If iKiiii tiitlii'i 1 loiu til.) voitox, it will I ^■tiiiK tilt Mic.lmn from tl )uw aiiKl.' of H tWiiiiKli! a lii„. |„. ,1 niwu JlHl drawn 'ii(--!i!i\ uuaJlel P^Mho.base Irom tl.o vnlfx. it will !.,. diNidod ;... tl,o .nlr,„„I ,„„1 .xtcruu, i,c»lm-» of ll,o Isli ACB AB'^Ar.^ !« 'I «!;•■" point nuts.d,. th.. un-U,. tnnu..d hvtw<.Kiv..n lines tho problem admit otT " ^^"'' """'^ ''^"'"^'^"'^ ^^^^'^ the A SCT : tlu) a ACB :: the a ACB .- tlu; a CMP. ;;."). ABC is a triaii<,di,' hiN(:ril..,-d iu a circi.. An af i- ilisidod ' an iiii.^'lt' uiid its 1 B, nrid C is an flifw tliiit CP. CQ ACB. ('. driiw a Htriii},'Iil A imuiikl to iIk; I'.v two ^ivfu lines P sni'h tiiiit (ho ■ AC luay Juive a '. draw a stnu^'ht at point and tlio y .solutiuMH does unibur of circles y point of this 'ii! other circles ; ! intersections of le shall bo equal. Juehing it at P; »ve that Ih' jxiint (if till- ribcd ; PQL i.s a iid AB ]iiuduuLd li r circle. «7. The vertical aiijfU; C of a triangle in bisecttd hv a straiud aiKHt line Winch incetH tli»' Imse at D, and is pioducid to a jioiiit E, hiicI. that the reetan^-ie contiii 1 l),v CD and CE is eiinul to the rectaiiKh' contained hy AC and CB: hIk-w that if the haHe and ve-tical anclo he given, the position of E is invai able. 38. ABC is an iHOHcelew tiiiui-i.' having tlie bai-i' aiiK'les at B ■■•■d C each d(.i hie of the vertical angle: if RE and CD bisect tlie l»aso angles and lueet tlio opposite hWps in E and D. shew tliat DE divides tlio triangle into tigni to BC. who 'SO ralio is e.iiini lo that of AB :«;». If AB. tln! diameter of a sei nicircle. l:e bisected in C and on AC and CB cintles be dcscriijcd, and in tin space between the tf cncmnferenceH a circle be inscribed, shew that its diameter will to that of the etiual circles in the ratio of two to tl irei- nee, •M. O is the centre of a circle inscribed in a ipiadiilateru! ABCD. a lini EOF is drawn an. I making ecpial angles with AD and BC. ami lii.M'tit.v rhein_ in E and F respectively: shew that the triangles AEO, BQF are similar, and that AE : ED -CF : FB. 41. Froin the last exercise deduce the following; I'l circle of a triangle ABC touches AB it. F; XOY is drawn tl ic inseribeil centre making e(inal angles wilh AB and AC, and meeting tl n'oui'ti ll !U lb ill X and Y resi)ectively : shew that BX : XF . AY :"yC 12. In.scribe a square in a given .semicircle. •13. Inscribe a square in a given Hegnient of a circle. 11. Describe an eiiuilateral triangle e(iual to a given isoscelen triangle. -i.j. Describe a square having given the dlfferenco between a diagonal and a side. •It). Given the vertical angle, the ratio of the sides containing it, ;t;id the diameter of the circumscribing circle, construct the triangle. ' ■17. Given the vertical angle, the line bisecting the base, and the angle the bisector makes with the base, construct the triangle. IH. In a given circle inscribe a triangle so that two sides mav pass through two given points and the third side be parallel to a given straight line. 49, In a given circle inscribe a triangle so that the sides may pass through three given points. '"'■ f 3H2 KLc ),I1)".S KI.KMKNTS. • \ 'w. : M '"■'' {""»■ P^""t« in a Htmight line, and O is sucii R point ,n It tluit tlif u-ctauf,' e OA, OY is equal t.. the reotan^-le OB, OX: It a circle be described with centre O and radius e.iual to a n.eani)ro,.ortional between OA and OY, sheNv that at every point on this circle AB and XY will subtend equal an<,'les. r.l p is a h^xed point and OP is any line drawn to meet a fixed straij,dit hne in P; it on OP a point Q is taken so that OQ to OP is a constant ratio, liiid the locus of Q. o2 O is a fixed point, and OP is any line drawn to meet the rOQ r'op *" ^'^^■^^'"■f ^' '" P= ^t' on OP a point Q is taken so tluit OQ to OP is a constant ratio, find the locus of Q. /i:! If from a f,'iven point two straight lines are drawn includin-' a given angle, and having a fixed ratio, find the locus of the extremity uUme ot them when the extremity of the other lies on a fixed straight (bp'lte J,^i' '^«*^-^'f t 1'"^ PAB, two points A and B are marked and the fine PAB is made to revolve round the fixed extremity P C is -i fixed point hi the j.lanc in whidi PAB revolves; prove tliat if CA i Trl'"' •^,*;","''^ ''"'^ *'"-' parallelogram CADB be completed, the locus ot D will be a circle. ^ _o5. Find the locus of a point whose distances from two fixed points are in a given ratio. of] Find the locus of a point from which two given circles sub- tend the same angle. r,7. Find the locus of a point such tli.it its distances from two intersecting straight lines are in a given ratio. r.„lS;i n'^" *^;.«^,"-'"i-^ on page 'AU, sliew that QT, PT' meet on the ladical axis ot the two circles. r)9. ABC is any trian.trle, and on its sides erniilatoral triangles are described externally : if X, Y, Z ...e the centres of their inscr bed circles, shew that the triangle XYZ is e.iuilaten.i. '^uiueu t!0. If S, I are the centres, and R, r the radii of the circumscribed and inscribed circles of a Inangl,., and if N is the centre of its nine- points circle, prove tliat (i) 6l-=:R--2Rr, (ii) NI = .\R-r. Establish corresiiondinHproiurties for the escribed circles anrM.once piove that (li(- nuie-pomts euvle touehes the inscribed and escribed cir(«les of a triangle. "Sk ; line, and O is such I) tlie rcctaujjile OB, I milius equal to a lat at every point on iwu to meet a fixed ) that OQ to OP ia SOLID (;K()iMi:TIiY drawn to meet the loiat Q is taken so of Q. me drawn iucludin;^' cus of the extremity ;s on a fixed strai''ht 1 B are marked and :tremity P. C is a ; prove that if CA be comi)leted, the ces from two fixed ) given circles sub- distances from two , PT' incut on the atoral triangles are i>i' their inscribed f tlie circumscribed centre of its nine- circlo.s, and lience •ibcd and escribed KL^CLll). ]}(){)Iv XI. J)ki-initioxs. From the Deiinitions of IJook I. it v.ill Ik, .vnieinlHnrd tiiat (i) A line is tliut Avliich lia.s lenr/th, without l.readth 01* tiiicknoss. (ii) A surface is that whieli has feu!/(h and hrcaJf,) without tinckiu'ss. ' To these definitions we have now to add : _ (iii) SpaC3 is that winch has le>i, brearffL and thickness. Thus a line is said to bo of one dimension ; a surface is said to be of two dimensions • and space is said to be of three dimensions. The Propositions of Euclid's Eleventh IJook here oi\,.„ establish the first principles of the geometry of spare ov solid geometT!,. They deal with the properties of strai'-dit hues which are not all in the same plane, the relations which straight lines bear to planes which do not contain those lines, and the relations which two or more planes bear to one another. Unless the contrary is stated thr> strai.i^dit lines are supposed to be of indetiriite len-th, and the planes of infinite extent. Solid geometry then proceeds to discuss the pi-onertie,s of solid figures, of surfaces which are not planes, and of lines wluch can not be drawn on a plane surface I:, a II. E. 25 384 EUCLin'S ELEMENTS, Lines and Planes. J. A str;ii, i v« 2j-2 m! ( 1 ;t3 3H« I'.ITMD's KI.KMKX'l'S. fi. One plane is perpendicuUir to another plane when r///// straiglit Ime drawn in „„(. of the planes perpcntlicukr to til ... plane . . i perpen to tlie eoninion section is also perpendicular to the other r.l.n, r n ,• =^^^J^''''!'f /'R."^'^^ t],G i,]ane EB is perpendicular to the piano CD, t any stnxight hno PQ, drawn in the plane EB at right planrCD. "'™'"°" """*'''" ^^' '' ''■'" ^* "S''* ^"8^^^ *° *^^ 7. The inclination of a plane to a plane is the acute an,£(le contained by two straight lines drawn from any point in tlie common section at right angles to it, one in one plane and one in the other. Thus in the adjoininp; figure, the strai;-dit lino AB is the com- mon H"ction of tlie two inter- sectin;,' planes BC, AD; and from Q, (I, I If point in AB, two straight lines QP, QR arc drawn perpendicular to AB, one in each plane: then the inclination of the two planes is measured by the acute angle PQR. Note This dellnition assumes that the angle PQR is of constant magnitude whatever point Q is taken in AB : the truth of id^Jcl assumption is proved in Proposition 10. dlh^ltlSgle^""™''^ ^'^ "'' intersection of two planes is called a It may be lu-oved that two planes when the dihedral angle formed bv t perpendicular to 1 is a right angle. ■;;< DKFINITIONS. 38i lotlier plniie when Ties perpondiculrtr cul.ir to the other peipendicular to the he plane EB at right right angles to the plane is the acute 'n from any point to it, one in one PQR is of constant the truth of which planes is called a 8 Parallel planes are suc-li .-is do not meet wh.'n pro- duced. ' 9. A straigl)t line is parallel to a plane if it dno^ „„t meet tlie plane wlien produced. 10. The angle between two straio-ht liiu^s which do not nieet is the angle contained hy two infer^^rtino strai-dit lines respectively parallel to th«> iw., non-intersectin<.- Jin's Thus if AB and CD are two straight lines which do not meet, and «ih, be are two intersecting lines parallel respectively to AB and CD ; then the angle between AB and CD is measured by the angle abc. 11. A solid angle is that which is made by three or more plane angles which lia\e a common a ertex, but are not in the same plane. A solid angle made by Hirer, plane angles is said to be trihedral ; if made by more than three, it is said to be polyhedral. A solid angle is sometimes called r. comer. 12. A solid figure is any i)ortion of space boujided by one or more surfaces, plane or curved. These surfaces are called the faces of the solid, and llie inter- sections of adjacent faces are called edges. 7 .(nil mlar to one another t angle. 7 I 388 kucltd'h klements. \ ; i POLYHEDHA. i:?. A polyhedron is ;i :;o]i(l fimiro Itouiulcd l)v pIjuio iaecs. 0/«. ^ A ]>lnPo roctilinoal y tliree pairs of paral](>l jil.-nie faces. Fi?. 2. IIIh A parallelepiped may be rectangular as in lig. 1, or oblique as Jn tig. 2. PEPINITIONS. 389 )unfl('fl l)y piano liavc thrre Hides; 'Iron must at loast iiust at least have . by plane facos, lilar and equal ices are para 11 p- ism. A prism is adjacent parallel- rwise the prism is "0 })oundofl l)y 1, or obiitpie as Jn IG. A pyramid is a solid li<,'ure bounded by plane faces, of wliich one is a poly<,'on, and the rest are triangles having as bases the sides of the polygon, and as a oom- nioii vertex- some point not in the yilnne of the polygon. The polygon is called the base of the pyramid. A pyramid having for its base a rt'qnJar polygon is said to bo right when the vertex lies in the straight line diawn perpendicular to the base from its central point (the centre of its inscribed or cir- cumscribed circle). 17. A tetrahedron is a pyramid on a triangular base : it is thus con- tained hy/onr triangular faces. 18. Polyhedra are classiiied according to the iiuiaber of their y*ac(?6' ,• thus a hexahedron has six faces; an octahedron has right faces ; a dodecahedron has tivelrc faces. 19. Similar polyhedra are such as have all their solid angles e(jual, each to each, and are bounded by the same number of similar faces. 20. A Polyhedron is regular when its faces are similar and equal regular polygons. ' 1 . .•590 KLTI.IDs KI,J..\IK\TS. 21. It Will 1... proved (hvo. jKige 4l>0) th.-it IIhmv caii only \n'jn-t' rc^ailar polyliPflru. 'I'licy arc (Icliiird jis follows. (i) A regular tetrahedron is u solid fii^urp bounded by four pluiK* faces, Avliicli are e(iual and ecjui- lateral trlanijles. (ii) A cube is a .solid ii,i?iire bounded l)y sU plaiie faces, which Hi'e equal i!(juare6. (lii) A regular octahedron is a solid ii'^uvo. bounded by eij/it plane faces, wiiich tire equal and equilateral trianyles. (i\ ) A regular dodecahedron is a solid iitfure bounded by twelve plane faces, Avhich are equarand regular ]>enta(jons. HKKlNtTroN.s 391 that ihcic can (v) A regular icosahedron is a solid liguro bounded ))y tuwufi/ j)lane fac«!.s, Avinch aii; ('(luai and equ ilateral trianyks. Solids ok ]Ievolution. 22. A sphere is a solid HjLfure described by th(> revo- lution of a semicircle about its diameter, "svliich remains iixed. The axis of the sphere ia the fixed straight line about whicli the Bemicircle revolves. The centre of the sphere is the same as tlie centre of tlie semi- circle. A diameter of a sphere is any strai^lit lino whicli jjasses throu"h the centre, and is terminated bolli wa\s hv tlie surface of the sphere. 23. A right cylinder is a solid figure described by the levolution of a rectangle about one of its sides which re' ains fixed. The axis of the cylinder is the fixed straight line about which the rectangle revolves. The bases, or ends of the cylinder are the circular faces described by the two revolving opposite si les of the rectau'Ie. 392 KUCLID'H ELEMENTH. 24. A right cone is .-i solid liuuiv ch'.scribed by the rpvolution of a nj,'lit aii. Note. This proposition scarcely needs proof, for the truth of it J^JdTplanr' ""n-diately from the definiLns' of a BtraS^L line It r.hould be observed that tlie method of proof used in thi>^ nnd the next proposition rests upon the following Horn. annQ f!"""^ f ""'{'«''^^ ^^^''«* t»rn8 ahout a ,fixed straight line as an axis, it can be made to pass through any point in space. \ % I P :i\u Krci, Ill's I I.KMKNI's. Puoi'osiTiox 2. Tiii.;.)ui.:m. Au,/•. 25' M. mother (ire in ohh ch each pair intpr^ Hook xr. I'Hoi'. ;}. 305 ii/raii//U linr. Pkoi'ohitiov ;{. Tmkorem. "i; aiH>ihrr ih'u' romiiioit itecfiou in a ^rsf'ct at E; : AB and CD at B le plane. \B; until it passes (lis piano, J^'f. T) and xi. 1. >lane. the plane wliich I)lane. q. e. d. can he made io ^iiiea. and a given point Ax. p, 3D3. ight lines; xi, 2. "-•'''; XI. 2. i>ip«. I. Ih-f. 25. >?^^ fx ; \ ""^ e) '■F ; Aj ^ ^^ Let the two planes XA, CY cut one another, and l(«t BD be then' con rnon section : then shall BD he a st. lino. For if not, from B to D in the plane XA draw the st. line BED j and in the plane CY draw the st. line BFD. Then the st. lines BED, BFD have the same extremities; • ". they inclu(l(> a spaci;; liut this is impossible. ••• the common section BD cannot be otherwise than a st. line. Q. E. \h Or, more briefly thu.s — l^et the planes XA, CY out one another, and let B and D be two points in their common section. Then because B and D are two points in the planer XA . . the St. line joinin^r b, D lies in that plane, i. Def 5 And because B and D are two points in the plane Cy' . . tlie St. line joinin.,^ B, D lies in that plane. Hence the st. line BD lies in both planes, Ti • ^^ ^^ therefore their common section. linj''' ''''"""''" '**'^'*^"" oi" <^lie two planes is a strahjU Q.E.D. i| 396 Euclid's elkmextk. Pkopohitiov 1. TiiEOHKM. [Alternative Proof.] // a tilmv/ht Uiui is perpettdicular to each of two straiyht Hues at their 2>oiiU of iutersectioti, it shall also be perpen- dicular to the plane in which thej lie. L<^t tlici stniiglit ]in(^ AD ])(^ perp. to cacli of tlie st. lines AB, AC at A their point of intersection: then shall AD be perp. to the plane in which AB and AC lie. Produce DA to F, niakiniLj AF equal to DA. Diaw any st. lino BC in the plane of AB, AC, to cut AB, AC at B and C; and in the same plane draw through A any st. line AE to cut BC at E. It is re([uirod to prove that AD is porp. to AE. Join DB, DE, DC; and FB, FE, FC. Then in the .\'* BAD, BAF, because DA ::^:: FA, Conatr. and the connuon side AB is perp. to DA, FA : .-. BD:.-BF. I. 4. SiniiL'irly CD= CF. Now if the ABFC be turned about its base BC until the vertex F comes into the plane of the A BDC, then F will coincide with D, since the conterminous sides of the triangles are equal, i. 7. .'. EF will coincide with ED, that is, EF = ED. BOOK XL PROP. 4. 397 I. 8. Hence in tlic, a '' DAE, FAE, since DA, AE, ED FA, AE, EF respectively, .". the I DAE -= the L. FAE. That is, DA is perp. to AE. Similarly it may be shewn that DA is perp, to every st. line v/hich meets it in tlie pbine of AB, AC ; .". DA is perp. to this plane. q.e.d. fl r St. line AE to cut Propositiox 4. Theouem. [Euclid's Proof.] If a stra'ujht hue in perpe)idicnlar to ench of iwo straight lines at their point of intersection, it shall also he perpen- dicular to the j^lune in lohich they lie. Let the st. line EF be perp. to each of the .st. lines AB, DC at E their point of intersection : then shall EF be also perp. to the plane XY, in which AB and DC lie. Make EA, EC, EG, ED all equal, and join AD, BC. Through E in the plane XY draw any st. line cutting AD and BC in G and H. Take any pt. F in EF, and join FA, FG, FD, FB, FH, FC. Then in the a*" AED, BEC, because A E, ED =- BE, EC respectively, Constr. and the i. AED =-- the £ BEC ; i. 1.5. .'. AD =:^ BC, and the _ DAE = the _ CBE. i. 4. 398 KUCLID's ELIiMKNTS. till In the ,\ -^ AEG, BEH, b('Oiius(( tlio _ GAE --■ the _ HBE, Jintl tlie _ AEG ^^ the l. BEH, and EA = EB; .". EG ■--- EH, and AG = BH. Again in the a » FEA, FEB, because EA = EB, and the common side FE is perp. to EA, EB; .-. FA. FB. Similarly FC= FD. xVgain in the A« DAF, CBF, because DA, AF, FD -^ CB, BF, FC, respectively, .•. tlie :_ DAF ^ the i. CBF. And in the a'' FAG, FBH, because FA, AG, = FB, BH, respectively, and the ^ FAG =- the ^ FBH, .'. FG = FH. I'roved. I. 15. Constr. r. 26. Jl;ip. I. 4. I. S. Proved. I. 4. Lastly in the a « FEG, FEH, because FE, EG, GF = FE, EH, HF, respectively, .". the 1. FEG the :lFEH; that is, FE is perp. to GH. r. 8. Similarly it may be shewn that FE is perp. to every St. line which meets it in the plane XY, .'. FE is perp. to this plane. xi. Def. Q.E.D. KuoK XI. vnov. f). 3!)y , Proved. I. 4. perp. to evciy PUOPOSITIOV 5. THE0U1:.M. // a strahjlit line is prrpendindar to cark of thrca con- current stvaUjJd lines at their point of intersection, these three straight lines shall he in one plane. Lot the straight lino AB be perpendicaiLii- to e;ich of the .straight lines BC, BD, BE, at B il section ; u'lr j)oint of inter- then shall BC, BD, BE be m one ane let XY be the plane whicli passes tlu'oui^h BE BD • xi. 2 and, if possible, su])pose tliat BC is not in t!iis phme. Let AF be the plane which passes through AB, BC • ion section of the two planes XY, AF be tl and let the comn St. line CF. 10 XI. 3. Tl len since AB is perp. to BE and BD, and since BF is in tl .•. AB 10 same plane as BE, BD, IS also perp, to BF. But AB is perp. to BC; XI. 4. , ^ -j;^. - . __ , J!yp. the ^« ABF, ABC, Avhich are in the same plane, are both rt. angles ; which is impossible. .'. BC is not outside the plane of BD, BE : that is, BC, BD. BE ar<» \\\ ?'.?'.*> ')]j me C^.K.f). I H. E. 26 400 EUCLID'S ELEMENTS. PkOPOSITIOX G. TlIEOUEM. Iftroo sfirriglit lines are perpendicular to the sarna 2)lane, tluuj sJiall hi: /Ktrallcl to oiio. avothor. jirf Lut the st. Hues AB, CD be perp. tc tlie plane XY tlien sliall AB and CD be par'.* Let AB and CD meet the plane XY at B and D. J on I BD and in tho. plane XY draw DE perp. to BD, mak equal to AB. m'' DE .1 oni BE, AE, AD. Tl len sinre AB is perp. to tlie plane XY Hyp. AB is also perp. to BD and BE, which meet it in that ])lane that is, the ^ '^ ABD, AGE ai-e rt. ancfk XI. Drf. 1. Similarly the l. " CDB, CDE are rt. anspectively, Comtr. and the _ ABD ^ the ^ EDB, being rt. angles ; .•. AD = EB. I. 4. Again in the a " ABE, EDA, because AB, BE ED, DA, respectively, and AE is conauon ; .•. tie _ABE=-the ^ EDA. i. 8. * NoTi:. In order to shew that AB and CD are parallel, it is Jiocessary to i)rovc that (i) tliey are in tiie same plane, (ii) the anglea ABD. CDB, are supplementary. BOOK XI. PROP. 7. 401 to the same ^j/«we, But the .L. ABE is a rt. angle ; Proved. .'. the z_ EDA is a rt. angle. But tli<^ _ EDB is a i-t. angle by construction, and the _ EDC is a rt. angle, since CD is perp. to the plane XY. ,f^^^ Hence ED is perp. to the three lines DA, DB, and DC; .'. DA, DB, DC are in one plane. xi. 5. But AB is in the plane which contains DA, DB; xi. 2. ". AB, BD, DC are in one plane. And each of tlie _ -< ABD, CDB is a rt. angle; U>/p. .'. AB and CD aie par'. i. 28. t^.E.D. ProPOSITIOX 7. TlIEOllEM. _ _ If two strahjht lines are iiaraJhl, the ,strai',s KLKMKNTH. Proposition .s. Thkoukm. ^ If (wo Htraiyht lines are parallel, and if one of then zs rerpendirnhr to a plane, then the other shall aha he per ■penilicnlnr to ilir same 'plane. m Let A B, CD be two par' st. lines, of whidi AB is pern to the plane XY: ^ ^' tiien CD shall also be perp. to the same plane. Let AB and CD meet the plane XY at the points B, D. Join BD ; and 'a the plane XY draw DE i)erp. to BD, makin<' DE equal to AB. ^ Join BE, AE, AD. Then because AB is perp. to the plane XY, II mu :. AB IS also perp. to BD and BE, wliich meet it in that l'^''''''' , . Xl.IJrfl. that IS, the /. « ABD, ABE are rt. angles. Now in the A"* ABD, EDB, because AB, BD - ED, DB, respectively, Constr. and the /_ ABD ;::^ the l EDB, being rt. angles; .'. AD-EB. ° ' I, 4. Again in the A^ ABE, EDA, because AB, BE=rED, DA respectively, and AE is common ; .'. the L ABE - the l. EDA. i. 8. BOOK Xr. I'1501'. 8. uy.] Hut tlin z. ABE is a rt. angle ; /'roved .'. tlio £. EDA is a rt. aii,oflo : that is, ED is porp. to DA. But ED is also perp. to DB : Comtr. .'. ED is porp. to tli(^ piano containiii'^r db, DA. xr 4 And DC is in this plane ; for 1)oth DB and DA are in the plane of the par'^ AB, CD. xr. 7 XI. /)>•/: 1. .'. ED is al.so porp. to DC ; that i.s, the ^ CDE is a rt. ando. Airain since AB and CD arc par', and since tlu? l ABD is a rt. angle .". the 1.CDB is also a rt. ;uiglf\ .'. CD is porp. Ijoth to DB and^DE; .•• CD IS also perp. to the plane XY, which contains DB, DE. , Q. i;. r>. I. 29. which AB is perp. KXKUCI.SKS. , making DE equal f,.n,h„^'^^f Porpenaicuki- h tho loMst straight h'nG that can be drawn trom ail external point to a plane. 2. Equal straight lines drawn from an external point to a plane aiv equally niclined to tho perpendicular drawn from that point to the plane. i ^^iul uu 3. Shew that two pbservations with a spirit-level arc sufficient to flWu-rnV-r'' 1^J=^"« i« l^"7^o»tal: and prove that for this purpose Itie two positions of the level must not bo parallel. 5 Shew how to determine in a Riven strai^dit line tho i.r.int %hid. IB equidistant iro.n two fixed points. When is this im- possible r miln./.l/'' ^^?'\f^ -'"^ '\ ^'^?"'^ ^^ "" I^'^"'^' ■'^'^"^ ^^i^it auy plane passing through the given straight line will have with the given plane a common section which is parallel to the given strair-ht line 7 404 EUCLID'S ELEMENTS. I'hOI'OSITIOX 0. TlIEOllEM. Tim straight Ihicn lohich are 2^firalld to a third straight line are 2^araUel to one another. Let the st. linos AB, CD l)e eacli jjar' to the st. line PQ : then .shall AB be par' to CD. I. If A B, CD and PQ are in one jjlane, the propo.sition has already been proved. j 3q IT. But if AB, CD anri PQ arc not in one plane, in PQ take any point G ; and from G, in the plane of the pai" AB, PQ, dr.uv GH perp. toPQ; j^ also from G, in the plane of the par'' CD, PQ, draw GK perp. to PQ. ill Then because PQ i.s perp. to GH and GK, Constr. .-. PQ is perp. to the plane HGK, which contains them. XI. 4. But AB is par' to PQ ; //y.;. ^ .". AB is also perp. to the i:»lane HGK. xi. 8. Similarly, CD is perp. to the plane HGK. Hence AB and CD, being perp. to the same plane, are par' t^ one another. ^^ g Q.E.D. ItOOlv XI. I'UOI*. 10. 405 M. 'o a third straight Q e St. lino PQ e propositioia has I. 30. ! plane, ^B, PQ, draw GH I. 11. ' CD, PQ, draw I. 11. cl GK, Constr. contains them. XI. 4. Hyp. n HGK. XI. 8. le HGK. le plane, are par' Xi. 6. Q.E.D. PROPOSITIOX 10. TlIKOREM. If licn intersr.dinr/ fttrnuiht lines avp rmpecfivelj/ 2Hira1le.l to two ilh. r infersrctiiu/ f .J.). I. XI. 9. I. 33. 1. 8. Q.E.D. 4()(i I (TCMD's i:ri;Mi;NTs. if| II ]'|{OPOSITIO\ I I. PltOMLKM. Tn ,/r,n,^ a s/mH,hf, line pn'pnidlmlar to a uivca nfa jTom a (jivnn point (nitsi,/>; it. ne r Let A 1)(' (li(« ,oi\(Mi point ou(si(| BC ' i l"") TlHMi if AD is also porp. to the piano XY, what 'was i-cfiuiri'd IS dono, lint if not, from D drav; DE in tlio piano XY perp. *° ^"' T. 11. and from A draw AF jh rp. to DE. i. 12. Then AF shall bo jK-.-p. to the plane; XY. Throu,i,di F draw FH par' to BC. i .31 Now because CD is porp. . DA and DE, Con'sir. .". CD IS perp. to the plane containini;- DA, DE. xi. 4. And HF is par' to CD ; .•. HF is also perp. to the piano containing DA, DE. xi. 8. And sinoo FA meets HF in this plane, .'. the L HFA is a rt. ani^do ; xi. Drf. 1. tliat is, AF is perp. to FH. And AF is also pei-p. to DE ; Co7,Mr. . . AF IS perp. to tlio plane containing FH, DE ; tiiaf; is, AF is perp. to the piano XY. q.e.f. ili'tiin a sfi'(ii(//it line perpeniflrii/tir f'> a. (flri'n pfmn'. froiii. it fflvoi point ill, fhn plnnr. B Lot A bo tlio given |»f»iiit in tlio piano XY. It is required to draw fi'oiii A a st. lino poi-p. to the piano XY. From any point B outsido tho piano XY draw BC perp. to tho plane. xr. 1 1. Then if BC passes througli A, what was required i.s dono. Ijut if not, from A draw AD par' to BC. i. 31. Then AD shall bo tho perpendicular re(juired. For since BC is perp. to tho piano XY, Constr, and since AD is ])ar' to BC, Constr. :. AD is also pcM-f). to the jdauo XY. XI. S. ti.K.F. Kxi:itcisi:s. 1. Equal straif^'ht linoH drawn to inrct a i)Ianc from a jroint without it are equally inclined to the plain'. 2. Find tho locus of tho foot of tho iicrpondicular drawn from a given point upon any plant! which 2>assus through a given straight line. ?}. From a given point A a perpendicular AF is drawn to a plane XY; and from F, FD is drawn perpendicular to BC, any line in that plane: shew that AD is also perpendicular to BC. I If 408 euclid'h klkmknth. PUOPOHITIOV 1.^. TheOKKM. Only nvP prrpendlrnlar ran be drawn fn n oivm vfans from a ytven point either in th- plane or outsid. iL ' Case I. Let the givon point A be in tJ.e given i>l„.e XY • X;Vl.o1^! ' '' '^" ^^^P^- "'' ^" '- ^'^- '■- A lo let \t.sJ^Z ot^tT "'"''' '"^^^ ^^ '-^'"^ AC; and and XY. ' '■"'"""•'' "■"^'"^^ '^^ ^''^^ Pl''^"es DF Then the st. lines AB, AC, AE are in one plane • BA .^"f '^^"•'^"^''^BA is porp. to the plaiie XY, Ifyn . . BA IS Hlso porp. to AE, which n^oets it in this plane f that IS, the :_ BAE is a rt. a.igk, . • fl,. s ^/^''''^^''^''y, the L. CAE is a rt. angle. . , Which is impossible. Cask IT Let the given point A l,e outside the plane XY Ihen two perp^ cannot be drawn from A to tlfe lan7- for If there could be two, they wou .d be par' xV 6 which is absurd. ' q.^.j,' HOOK xr. riior. 14. 409 r> n given plane idti it. veil plane XY; iwn fioiu A to and AC ; and the planes DF A I. 3, 3 plane. XY, Ifi/p. this pl.ine ; XI. Def. 1. mo, are equal to the plane e plane XY. the j)lane ; par. XT. 6. Q.E.D. Proposition- 11. Thkorf.m. Pfnnen to lohlfh the. same straight lino, ia perpendicular are. jyarnlld to on/: another. Let the st. line AB bo porp. to each of the pianos CD, EF : then shall those pianos bo par'. For if not, they will meet when produced. " possible, let the two planes meet, and let Hui st. line uH be their common section. xi. 3. In GH take any point K ; and join AK, BK. Tlu^n because AB is perp. to the piano EF, .'. AB is also perp. to BK, which meets it i' !iis plane; XI. 7>y: 1. that is, the l AEK is a rt. angle. Similarly, the BAK is a rt. an«i;lo. .'. in the A KAB, tlx' - BK, BAK are together equal to two i*t. angles ; which is impossible. I. 17. .". the planes CD, EF, though prodnced, do not nii'ot : that is, they are par'. q.e.d. i! ■> 410 kix'lid'.s i:j,kmI':\ts. Proposition 15. Tiikoiiem. //• two inf.'rf^rcHnci strahihf, lines are parallel respectively to two other xnfrr.jtlu:, straight lines vMch are not in the same plane uutk them, then the plane containing the Jn.t patr shall he parallel to the plane contavrdng the second jJiXbl % -\ ^■t lino/Dr''Fp'''''V''f' ^"^ ^''' '•^?P'^^«vely par' to the AB, BC : ' '""^ ''""^ "' *''" ^'""^^^ P^-'^"« '^« tlu'u slKill tlio piano containing AB, BC l)o par' to the piano contannng DE, EF. ' co tne From B draw BG porp. to tho piano of DE, EF • xr 1 1 ••iiid lot it meet that plane at G.' Through G draw GH, GK par' respectively to DE, EF. i 31 _ Thou because BG is perp. to the plane of DE, EF . BGjs also perp. to GH and GK, which n.eet it in that 4\ ' 4. • IP, XI- J)ef. 1. tliat IS, each of tho z. « bGH, BGK is a rt. angle. Now l)ecauso BA is pat'' to ED, and because GH is also par' to ED, .". BA is par' to GH. And since the ^ BGH is a it. angle ; .'. the L ABG is a rt. angTe. ' Snnilarly the c CBG is a it. an^le Hyp, Constr. XI. 9. Proved. I. 29. HOOK XI. I'Uor. If). 411 tl re3j)ecttvely are not in mtaining tho, n(j the second Then since BG is perp. to each of the st. lines BA, BC, .". BG is perp. to the i)laue containing tlieni. xi. 4. But BG is also perp. to the plane of ED, EF : Constr. that is, BG is pcu'p. to tlu; two j)]anes AC, DF : .'. these planes are par'. xi. 1-4. Q.K.D. Proposition 1G. Tiikoukm. If two parallel planes are cut hi/ a third plane their common sections with it shall he ^>ar«7/e^. K p.ir' to tlie 10 plane as par' to the F; xr. II. , EF. I. 31, E, EF, it it in that xr. Be/. 1. xngle. Hyp. Constr. XI. 9. Proved. I. 29. N / / / 'F c H N \ D b N G B Let the par' planes AB, CD be cut l)y the plane EFHG, and let the st. lines EF, GH Le their conuuon sections with it : then shall EF, GH be par'. For if not, EF and GH will meet if produced. If possible, let them meet at K. Then since the whole st. line EFK is in the plane AB, xi. 1. and K is a point in that line, .'. the point K is in the plane AB. Similarly the point K is in the plane CD. Hence the planes AB, CD when produced meet at K ; whicn is impossible, since they are par'. Hyp. .'. the st. lines EF and GH do not meet; and they are in the same plane EFHG ; .". they are par'. i. Def. 25. i > i if 412 EUCLID'a ELEMENTS, Proposition 17. TiiEojiEM. Straight lines which are cut h>j parallel planes are cut projwrtionally. ru ^u, tV^: 1^""' ^^' ^^ ^^ '^^ ^'y ^^'^ *^^^e« P^-^r' planes GH, KL, MN ;i;, the points A, E, B, and C, F, D : then shall AE : EB :: CF : Fd! Join AC, BD, AD ; and let AD meet the plane KL at the point X : join EX, XF. Then because the two par' planes KL, MN are cut by the plane ABD, ■^ :. the common sections EX, BD are par'. xi. 16. and because the two par" planes GH, KL are cut by the phine DAC, '' .. the common sections XF, AC are par'. xi. IG. And because EX is par' to BD, a side of the A ABD, .*. AE : EB :: AX : XD. 'vi. Again because XF is par' to AC, a side of the A DAC .•. AX : XD :: CF : FD. Hence AE : EB :: CF : FD. y. 1 Q.E.D. Definition. One plane is perpendicular to another plane, wheii any straight line drawn in one of the planes perpendicular to their common section is also perpendicular to the other plane. [Book xi. JJef. 6.] VI. 2. VI. 2. ? 2Jluues are cut ree par' planes D : )iiit X : N are cut by u-'. XI. 16. •e cut by the ir'. XI. IG. ! AABD, VI. 2. he ADAC, VI. 2. V. 1. Q.E.D. ■ to anotlier of the planes perpendicular k XI. JJej: G.] BOOK XI. PROP. 18. PuoposiTioN 1(S. Tjikorem. 413 If a sfrair/ht line is jicrptnuCicHhir to a jdnw, then every plane v)hi<'h passes throvijh the straiyld line is also 2>^'>'P^^*'' dicular to the (jioai, plane. Let the st. line AB be perp. to the plane XY ; and let DE be any plane passing through AB : then shall the plane DE be perp, to the plane XY. Let the st. line CE be the common section of the planes XY, DE. XI. 3. From F, amj point in CE, draw FG in the plane DE perp. to CE. i. 11. Then because AB is perp. to the plane XY, Uyp- .'. AB is also perp. to CE, which meets it in that plane, XI. D>f. 1. that is, the <:. ABF is a rt. angle. But the z. GFB is also a rt. angle ; Constr. .". GF is par' to AB. I. 28. And AB is perp. to th(! plane; XY, ^^!/J'- .'. GF is also perp. to the plane XY. xi. 8. Hence it has been shewn that any st. line GF drawn in the plane DE perp. to the common section CE is also perp. to the plane XY. .". the plane DE is perp. to the plane XY. xi. Uef. 6. Q.E.D. EXKIICISE. Sliew that two planes are perpendicular to una another when the dihedral angle formed by them ix a riijht awjle. 414 KUCIJDS KI.K.MKNTS. Proposition li). Tiikouicm. _ // two intersecting pianos are mch prrpendlcnlar to a tlnrdpl'ine, their common section shall also he perpendicular to that plane. Lot eacli of the planes AB, BC be perp. to tlie plane ADC, and let BD be their common section : then sliall BD be perp. to the plane ADC. For if not, from D draw in the plane AB the st. line DE perp. to AD, tlie common section of the planes ADB, ADC : and from D draw in the plane BC the st. line DF perp' to DC, the common section of the planes BDC, ADC. Then because the plane BA is perp. to the jilane ADC, and DE is drawn in the plane BA peip. to AD the comnfon section of these planes, Constr :. DE is perp. to the plane ADC. xi. Def. 6. Similarly DF is perp. to the plane ADC. .-. from the point D two st. lines are drawn perp. to the plane ADC ; which is impossible. xi 13 Hence DB cannot be otherwise than perp. to tlie plane ADc! (J. IC.l), tl. 'rpendictdar to a he 2^eri)endicxilar •p. to the plane 3 ADC. 3 the st, line DE les ADB, ADC : I. 11. it. line DF perp. >C, ADC. 3 plane ADC, Hyp. AD the common Co7istr. DC. XI. Def. 6. ADC. vii perp. to the XI. 13. )thej)lane ADC. Q.i:.]). BOOK xr. inioi'. 20. Proposition 20. Theorem. 415 Of the three plane tiagUs lohiclt form a tvlhednd awjle, any two are together greater than the iJiird. Let the trihedral angle at A be formed by the tliree plane L * BAD, DAC, BAC : then shall any two of them, such as the l ' BAD, DAC, be together greater than the third, the L BAC. Case I. If the l BAC is less thai\, or equal to, either of the ^^ BAD, DAC; it is evident that the L^ BAD, DAC are together greater than the _ BAC. Case II. But if the l. BAC is greater than either of the ^« BAD, DAC; then at the point A in the plane BAC make the L BAE equal to the /. BAD ; and cut off AE equal to AD. Through E, and in the plane BAC, draw the st. line BEC cutting AB, AC at B and C : join DB, DC. Then in the a « BAD, BAE, since BA, AD = BA, AE, respectively, Consir. and the L BAD = the L BAE ; Constr. .-. BD = BE. I. 4. Again in the A BDC, since BD, DC are together greater than BC, n. E. and BD = BE, .'. DC is greater than EC, I. 20. Proved' 27 416 Euclid's km^mknts. And in the a *^ DAC, EAC, Jjecause DA, AC = EA, AC respectively, but DC is greater than EC ; .". the i. DAC is greater than the ^ EAC. But the ^ BAD = the z. BAE ; . . the two i. « BAD, DAC are together greater than'the BAG, Q.E.D, Constr. Proved. I. 25. Constr. Pkoposition 21. Theouem. livery {convex) ,oUd anyh is formed b>/ phme amil which are (oyrthrr /rss thauj\mv rhjU anyles. ~y BSC^ csn^ n^'c!''^ """''' ''^ ^ ^"^ ^°"'''''^ ^'y ^^'"^ I'^'''"^ z. « ASB, then shall tlie sum of these plane angles be less than four rt. angles. % .ely, Constr. ', Proved. - EAC. I. 25. Constr. ater than tho Q.E.D. // 2)lane anyles piano z.»ASB, ! less than four iJooK xr. PROP. 21. 417 For let a i)liUH; XY intorsect all the arms of tho plane angles on the same side of tlu; vertex at tho points A, B, C, D, E: and let AB, BC, CD, DE, EA be tho common sections of the plane XY with tho planes of the several angles. Within the polygon ABCDE tak(! any point O ; and join O to each of the vertices of tho polygon. Then since the /_ ^ SAE, SAB, EAB form the trihedral angle A, .'. the L. ^ SAE, SAB are together greater than the _ EAB ; XI. 20. that is, the z. "* SAE, SAB are together greater than the l. " OAE, OAB. Similarly, the L. * SBA, SBC are together greater than tlie _ " OBA, OBC: and so on, for each of the angular points of the polygon. Thus by addition, tlie sum of the base angles of the triangles whose vertices are at S, is greater than tho sum of tho base angles of the triangles whose vertices are at O. But these two systems of triangles are ecjual in numb(>r ; .'.the sum of all tho angles of tho one system is equal to the sum of all the angles of tho other. It follows that tho sum of the vertical angles at S is less than the sum of the vertical angles at O. But the sum of the angles at Q is four rt. angles; .'.the sum of the angle's at S is less than four rt. an_, 'Ior. Q.K.D. }\ote. Tliis proposition was not f^ivon in this form by Euclid, who established its t)'uth only ni the case of trilicilral ansb'^- '-I'be above domoiiFtration, however, ap])lies to all eases in which the polygon ABCDE is convex, but it must be observed that vrithout thisi condition the proposition is not necessarily true. A solid angle is convex when it lies entirely on one side of each of the infinite planes which p ss through its plane angles. If this is the case, the polygon ABCDE will have no rc-eiitnint angle. And it i.i clear that it would not be possiliic to apply xi. 20 to a vertex at which a re-entrant angle existed. 27-2 1 1' 41H KUri.ID's KI.KMKNTS. KxKKcisKs OX Book XI. 1. Equal 8triiii,'ht lines drawn to u plane from a point without it liavn cciiial projections on tliat plane. 2. 1 1' S is the centre of the circle circumscrihed about tlic triangle ABC, and if SP is drawn pcrjK iidicnlar to the plane of the triangle, hhew l!;at any point in SP is equidistant from the vertices of the tiian;,le. ;>. Find the locus cf p.)int,s in space equidistant from three Rivon ])(iints. 4. From FiXauij)!!! 2 deduce a i)ractical method of drawing n ])erpeudicular from a given i)oint to a ])l.ane, having given ruler, compasses, i;nd a straight rod longer than the required perpen- dicular. 5. Give a geometrical construction for drawing a straight line equally inclined to tluee slraigiit lines which meet in a point, but are not in the same plane. (J. In a (jauche quadrilateral (that is, a quadrilateral whose sides tue not in the H;ime jilane) if the mi'Mle points of adjacent sides are joined, the ligure tiius formed is a parallelogram. 7. AB ;ind AC are two straight lines intersecting at right angles, and from B a perjx'ndicular BD is drawn to thei)lane in which they arc : shew that AD is perpimdicular to AC. 8. If tvs-o intersecting planes are cut by two parallel planes, the linos of soiiti.ni of the first pair with each of the second pair contain equal angles. 9. If a straight line is parallel to a plane, shew that any plane passing tin ou'-'li tiic given straight line will intersect the given i)lano m a line of section which is parallel to the given line. 10. Two intersecting planes pass one through each of two parallel straight lines ; shew that the common section of the planes is parallel to the given lines. 11. If a straight line is parallel to each of two intersecting planes, it is also parallel to the common section of the planes. 12. Through a given point in space draw a straight line to inter- sect each of two i;iven straight Hues which are not in the same plane. ].•!. If AB, BC, CD are straight lines not all in one plane, shew that a plane which j)as:-es tluough the middle point of each one of them is parallel both to AC and BD. 11. From a given point A a perpendicular AB is drawn to a plane XY; and a second perpendicular AE is drawn to a straight line CD in the i)lane XY : shew that EB is perpendicular to CD. oiii a point without j(l about tlic triangle me of tho triangle, tho vertices of tlie nt from three {»ivon ithod of drawing a Having given ruler, e retjuired pcrpen- .ing a straight line ; iu a point, but are ■ilatcral wliose sides f adjacent Hides are ;ing at right angles, lane in which they parallel planes, tho second pair contain ihew that any jjlane jct the given piano ine. 3ugli each of two ection of the planes intersecting planes, mes. ;raight line to inter- e not in the same in one plane, shew ; of each one of them AB is drawn to a rawn to a straight uliciilar to CD. KXKIICIHKS (IN' lUtoK XI. -Ill) 1."), From a i)oint A two pcrpendii-ulius AP, AQ iin> ilnuvn (hk; to each of two intersi'cting pianos : shew that tin; ooimukih sc-i-tiou of these planes is perpendicular to tho plane of AP, AQ. 10. From A, a i)oint in one of two given intersecting iilanoa, AP is drawn perpendicular to tho first plane, and AQ pi rixnoicular to the second : if these perpendiculars meet tlie secoii I i)lane at P and Q, shew that PQ is perpendicular to the common section of tho two planes. 17. A, B, C, D arc four points not in one plane, shew that tho four angles of the gauche (luadrilateral A BCD are together less than four right angles. 18. OA, OB, DC ire three straight lines drawn from a given point O not in the same ])lane, and OX is another straight lino within the solid angle formed by OA, OB, OC : shew that (i) the sum of the angles AOX, BOX, COX is greater than half the sum of tho angles AOB, BOC, COA. (ii) the sum of the angles AOX, COX is lesri than tho sum of the angles AOB, COB. (iii) the sum of the angles AOX, BOX, COX is less than the sum of the angles AOB, BOC, COA. 10. OA, OB, 00 arc three strai^'lit lines formin;,' a ;-oli 1 angle at O, and OX bisects the plane angle AOB; show that tho unglu XOC is less than half the sum of the angles AOC, BOC. 20. If a point be equidistant from the angles of a right-!in;j;Ied triangle and not in the plane of the triangle, tlic line joining it with tho middle point of the hypotenuse is perpendicular to tho plane of tho triangle. 21. The angle which a straight line makes with its projection on a plane is less than that wliich it makes with any other straight lino which meets it in that plane. 22. Find a point in a given piano sueh that the sum of its distances from two given points (not in the plane but on tho same bide of it) may be a minimum. 23. If two straight lines in one plane arc ciiually inclined to another plane, they will be eipially inclined to tho eoinin;)n section of these planes. 24. PA, PB, PC are three concurrent straight lines each of which is at right angles to the other two: PX, PY, PZ are perpen- diculars drawn from P to BC, CA, AB respectively. Shew that XYZ is the pedal triangle of the triangle ABC. 25. PA, PB, PC are three concurrent straight lines each of which is at right angles to the other two, and from P a perpendicular PO i.i drawn to the piano of ABC : shew that O is tho orthocentre of the triangle ABC. 420 kuclid'h klkmknth. tmin rHIOOIlEMS AND KXAMPLi'LS ON liUOK XL. Orfixitions. (i) Linoa wliicli nrn drawn on a piano, or throuRli which a piano may be uiadcj to pasn, am said to Ijo co-planar. (ii) The projection of a line on a piano is tlio locus of the feet of perpendiculars drawn from all points iu the giveu lino to the plane. TiiEORKM 1. Tlie jmyectioii of a strauiht line on a plane h itself a stnmjht line. •' Lot AB 1)0 tliof^ivcn st. lino, and XY the given piano. From P, any point in AB, draw Pp porp. to tho piano XY • It is ro.pnrod to ^-hew that tlie locus of p is a st. line. From A and 8 draw IKa, Bh porp. to the plane XY Now since Art, P^), B/> are all porp. to the plane XY, .•. they are par'. And since these par'» all intersect AB, •'• tiioy are co-plauar. the point p IS in the common section of the planes A/;, X Y • that is, p is in the st, line ah. ' ' ' xr. (5. XI. 7. But j() is any point in the projection of AB .-. the projection of A B is the st. line ah. ' Q.B.O. ' HOOK XI. tlirnuRli wliich a r. Iio locus of tlio feot given lino to the I a 2)Ian<; h itself a en piano. 10 plane XY a St. line. ane XY lano XY, xr. 0. XI. 7. men Ah, XY; AB, «'>. Q.B.D. THEOHKMH AND EXAMPLKH ON nOOK XT. 421 Thkokkm 2. Draw / '« p(ira!lel,'pipM are concurrnit and fucdf ^^^'^' '•' " P"'"""' ^^ '''^'"'^ ^^^'°' CD'A'B' are opposite I'or fiuioo the planes DA', AD' nre par', and tliH plane DB nie.ts tlicin, .-. the common sections AB and DC aropar'. Similarly AD and BC are luir'. .-. thf! fi;;. A BCD is a pai'» andAB = DC; alsoAD=BC Def. XI. 10. X. 84. Similarly each of the facos of tlie paiP'^'' i^ a nar'" ■ so that thoudKes AB, CD', B'A', DC aro equal and par'- Then in the opp. faces ABCD. C'D'A'B', we hav(> AB = CD' and BC :-- D'A' ■ P,-nv^ i and since AB, BC a... respectively par' to CD' D'A' A the z ABC = tliti z'C'D'A'- ' ' v, io .-. the par- ABCD .: the par- C'D'A'B' identically. P. 64, Ex 11 bisect^oile alc>t£'""^'^ ^^'' ""'' ^^'' °°' «^"^" ^« concurrent and Join AC and A'C. iMiimdcil hy tliito trf parallt'lograiiiH, III. //•(.' i't)ii(:uirrnt and 'A'B' are opposite he opposite faces >•'. Def. par' XX. 10. 1.34. par'" ; 1 and par' : kewise AC, BD', Proved. XI. 10. • r. 64, Ex. 11. 3 concurrent and 'I'lIKOKKMS AMI KXAMI'I.KS n\ IIOOK XI. IL'.*? Then sincii AC is equal and j)Rr' to A'C, .-. tlie t\iJ, ^CA'C isii]mi'"; .•. its tliai?ouaU MA', CC hi i tt tiii" atiotlu'r That is, AA' passt-s through O, tha niitMle point ( CC . Similarly if BC and B'C wore joined, the fl^;. BCB'C would bo a par'" ; .•. the diagonals BB', CC hiweet one another. That is, BB' also passis lllI()U^'h O the middle point of CC. Hiniilarly it may be shewn that DD' passes through, and in bisected at,'0. ^^ "• Thkohkm 4. '/'/(c xtftiijilit //;/<'.•.• irliirli join the vrrtir.i'H of' a dtm- hedroii to (ha ct'nlioUh of tlu' opponitt: J'ucis an' concuiifut. I.ot A BCD 1)0 a tetrahedron, and let it, . .7,., 1^3, y^ he the centroids of the faces opposite respt>(tivt iy to A, B, C, D. Then shall A//, , B./.,, Cry.,, D;u be concurrent. Take X the middle point of the edge CD ; then (I. and '/., niu?^t lie; respectively in BX and AX, ' " so that BX ::^ :5 . Xr/, , 1'. lOr,, Ex. 4. and AX-:5 . X//./. .-. //,f/.j is pat' to AB. And A//,, B//., must intersect one another, since they are both in the plane of the'A AXB : let them intersect at the point G. Then by similar A«,, AG : G//. = AB : n^ti, = AX : X7.. -. 3 : 1.' .-. B/7., cuts A//, at a point G wliosc distance from f/i~\ . A.7,. Similarly it may be shewn that Ci/., and Dg^ cut Af/, at tiie sarao point; .•. these lines are concurrent. g.K.n. 424 kuclid's elements. li i*o 7'"'^°"^^^ '"'• . (') V, SAB are similar .-. ab : AB = S6 : SB = bc : BC, for the a-^ Qbc, BCo are similar. Or, ah : bez=/\B : BC. In like manner, be : ('(/-BC : CD. And so on. .-the n<^H abed. ABCD are equiangular, and liavc their sides about the equal angles proportional. .•. they are similar, (ii) From S draw S.rX perp. to the planes abed, ACCD and nieetmg them at .r and X. Then shall fig. abed -. fig. ABCD = S.c- : SX^. Join a.r, AX. Then it is clear that tho a'' Sax, SAX are similar. And the tig. abed : fig. ABCD = *76- : AB- vi OQ --aS'' : AS-', = S.c- ; SX-, Q.E.D. drawn parallel to cate ratio of their ion formed by a r. ir'. hi to DA. 3, BC, XI. 10. are equiil to the ir, BCc arc similar. lavc tlieir sides ihcd, ADCD and SX2. imilar. ^'- VI. 20. >-, ^". Q.K.D. DEFINITIOX. I^EFINITIOX. 425 A polyhedron is regular when its facen are similar and equal regular polygons. Theorem 6. There cannot he more than five re v_v n-D and F = /j. So that E-V=:F-'2, or, E-|-2 = F + V. This is known as Eider's Theorem. • >; Miscellanp:ous Examples ox fSoLiD Geometky. 1. The projections of parallel straight lines on any plane arc parallel. Ar^'"]-, ^' "'' '"^'"^ ^'' ^^''^ *^'° projections of two parallel straight lines AB, CD on any jilane, shew that AB : CD~ah : cd. 3. Draw two parallel pianos one through each of two straight hnes which do not intersect and are not parallel. 4. If two straight lines do not intersect and arc not parallel on what planes will their projections bo parallel? 5. Find tlie locus of the middle point of a straight line of constnnt hngth whose extremities lie one on eiichof two nnn-interscctiiig straight hues, having directions at right angle,^ to one another, ° t o/ edges, and V of ;ogotlici' the facc^ ibi.T of edKcs, and tion, anil that thu vertices us cdyes, ;s and oiw edge in Pecond face wo 1 tliu number of vertices and livo re on adding the [gcs by one iiioro laced in liosition, 's nor vertices ; tEOMETRY. n any plane arc Uel straight lines I of two straight 3 not i^arallcl, ou t line of constant jrsccliug straiglit MISCKLLANKOUK KXAMI'LKS ON SOMP GEOMKTHY. 429 6. Thiee points A, B, C are taken one on each of the conter- minous edges of a cube: prove tliat the angles of the triangle ABC are all a" te. 7. If a parallelepiped is cut by a i)lane which intersects two pairs of opposite faces, the common sections form a parallelogram. 8. The square on the diagonal of a rectangular parallelepiped is etjuiil to the sum of the sipiares on the three edges conterniinoua witli thu diagonal. 9. The square on the diagonal of a cube is three times the square on one of its edges, 10. The sum of the squares on the four diagonals of a parallele- piped is equal to the sum of the squares on the twelve edges. 11. If a perpendicular is drawn from a vertex of a regular tetra- hedron on its triangular base, slicw that the foot of the perpendicular will div'de each median of the base in the ratio 2 : 1. 12. Prove that the perpendicular from the vertex of a regular tetraliedron upon the oi)2)osite face i;; three times that drop2)ed from its foot upon any of the other faces. 1^. If A P is the perpendicular drawn from the vertex of a regular tetrahedi'on upon the opposite face, slicw that i3AP- = 2(/-, where a is the length of an edge of the tetrahedron. 11. The straight lines whicli join the middle points of opposite edges of a tetrahedron are concuiTont. 15. If p tetrahedron is cut by any plane parallel to two opposite edges, the section will be a parallelogram. IG. Prove that the shortest distance between two ojiposite edges of a regular tetrahedron is one half of tlic diagonal of the square on an edge, 17. In a tetrahedron if two pairs of oi)posite edges are at right angles, then the third pair will also be at rii,'ht angles. 18. In a tetrahedron whose o[)p()site edges are at right angles in pairs, the four i^erpendiculars drawn from the vertices to tlie opj)OKito faces and the three shortest distances between opposite edges are concurrent. 19. In a tetrahedron whose opposite edges are at right angles, the sum of the squares o.i each pair of opposite edges ^s the same. 20. The sum of the squares on the edges of any tetrahedron is four times the stmi of the S(iuares on the straight lines which join the middle points of opposite edges. 430 Krci.lKS KI.I'.MKNTS. 21. ^u liny tetrahedron the plane which bisectn a dihedral an^le divideH ;'ie opposite ed(,'e into Hcf^nicnts which aro proportional to the ureas of the faces meeting; at that edge. 22. If !,lie auf^les at one vertex of a tetrahedron are all right angles, and tlie opjiosite I'ace is eciuilateral, shew tliat tlie sum of the pcrnerdieulars dii)]i])ed from any point in this face upon the other thre? faces is constant. L'lJ. Shew that the polygons formed hy cutting a prism by i)arallcl planes are ecpial. 24. Three straight lines in space OA, OB, OC, nte )untually it right angles, and their lengths are n, h, c : e.xpress 'he tui-fi of tha triangle ABC in its simjdest form. 2.5. Find the diagonal of a regular octahedron in u ims u^ one of its edges. 2(5. Shew how to cut a cube by a plane so that the lines of section may form a regular liexagor. 27. Shew that every section of sphere by a plane is a ciicle. 2H. Find in terms of the ]( ;igtli of au tJge the radius of a R)>here inscribed in a regular tetrahedron. 2!* ''Uid the locus of points in a given plane at which .a straight line of >l\eii ]■:;'.. gth and p'rsition subtends a right angle. 30. A l!.':evi j r.int O is joined to any point P in a given piano which (loef. 1 o' ." imain O; on OP a point Q is IrAen such that the rectangle Of-, OQ is constant: shew that Q lies on a li.\cd sphere. ctH !i (lihedml nn^le ! proportional to the L'clrou are all right that tl\(; sum of tho i'ace upon tliu other ; a prism by parallel OC, :i(H )iuitually 'it. ross I he iw/a of tha )n in uniis oi one i>f it the lines of section ilane hi a circle. he radit.H of a Hjiliere le at which a btraight t anule. t P in a given piano la.\on snch that the on a fixed sphere. f