^■B IMAGE EVALUAT90N TEST TARGET (MT-3) k A /, ^C> r^ ■^ *# I. S124 |M 1.1 « 12.0 lU SiUu L25 iU 12.2 1.6 P> /] >> <^1 >^ f — Ss.. '/ Hiotographic Sdences brporation G 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 m C\ \ signifie "A SUiVRE", le symbols "7 signifie "FIN". Lee cartee. planchee. tableaux, ate, peuvent Atre film^e * dee taux de reduction diffirents. Loraqua la document eet trop grand pour Hn reproduit m un soul cliche, il est film4 i partir da i'angle sup4rieur gauche, de gauche k droite. et de haut i* baa. on prenant le nombre d'Imegea n^caeaaira. Lea diagrammes suivants iHustrent la mithode. 1 2 3 4 6 \ \ ^ wm ■W p WWiW MMWiffi l\h I. <* uclid for '■ ^ iSino. dot; ^!^"^^ asurati< i^l Examples. *€ Algebra f amples. t JxC^ej to 1 Edition. ( i' igonome ,,:.rO' Examples. ir^'ey to r , r^ Crown 8 vo. B 3cKanics ^*^ amples. Th Ij^ Igebra fc . Witbnumei "I* ■ ^A^ey to tl and Schools A Treatise Edition. C 'I ^lane Trig With numei wmgm. s 3e EDUCATIONAL MATHEMATICAL WORKS BY > A I. TODHUNTEE, MX, P.ILS. ■ ■■ •* fl uclid for Colleges and Schools. New Edition fy},^ i8mo. cloth. 3«. 6!fHpi!P HfB •#• Key to the Plane Trigonometry for CoUegeeg and Schools. Crown 8vo. cloth, lot. 6d. A Treatise on Spherical Trigonometry for tlie U8e of Collegea and Schools. With numerous Examples. Third Edition. Crown 8vo. cloth. j{8. 6rf. A Treatise on Conic Sections. With numer- ous Examples. Fifth Edition. Crown 8vo. cloth, ft. 6d. A Treatise on the Differential Calculus. With numerous Examples. Sixth Edition. Crown Svo. cloth. ios,6 \ '.■■■ THE ELEMENTS OF EUCLID 1:..'^ V FOR THE USE OF SCHOOLS AND COLLEGES; COMPEISING THE PIEST SIX BOOKS AND P0BTI0N8 OF THE ELEVENTH AND TWELFTH BOOKS; WITH NOTES, AN APPENDIX, AND EXERCISES BY I. TODHUNTER M.A., F.R.S. NEW EDITION. SonOon : MACMILLAN AND CO. Toronto : COPP, CLARKE, AND CO. 1875 {A it Jiights reserved.] i mmmmtm^ i ^^''^mmm^mmm'mm^ 3B1 4 I Cambdlige: rRINTKD BY C. J. CLAY M. A AT THK UNIVKBSTTY FRKSS i I iJr / 4 PREFACE. I In offering to students and teachers a new editioc of the Elements of Euclid, it will bo proper to give some ac- count of the plan on which it has been arranged, and of the advantages which it hopes to present. Geometry may be considered to form the real founda- tion of mathematical instruction. It is true that some acquaintance with Arithmetic and Algebra usually precedes the study of Geometry; but in the former subjects a begin- ner spends much of his time in gaining a practical facility m the application of rules to examples, while in the latter subject he is wholly occupied in exercising his reasoning faculties. In England the text-book of Geometry consists of the Elements of Euclid ; for nearly every official programme of instruction or examination explicitly includes some portiou of this work. Numerous attempts have been made to find an appropriate substitute for the Elements of Euclid; but such attempts, fortunately, have hitherto been made in vain. The advantages attending a common standard of reference in such an important subject, can hardly be over- estimated; and it is extremely improbable, if Euclid were once abandoned, that any agreement would exist as to the author who should replace him. It cannot be denied that 1 wmmm-^-. VIU PREFACE. defecta and difficulties occur in tho Elements of Euclid, and that these become moro obvious as we examine the work more closely ; but probably during such examination the ^?onviction will grow deeper that those defects and diffi- culties are due in a great measure to the nature of the subject itself, and to the place which it occupies in a course of education; and it may be readily believed that an equally minute criticism of any other work on Geometry would reveal more and graver blemishes. Of all the editions of Euclid that of Robert Simson has been the most extensively used in England, ahd the pre- sent edition substantially reproduces Simson's; but his translation has been carefully compared with the original, and some alterations have been made, which it is hoped will be found to be improvements. These alterations, how- ever, are of no great importance ; most of them have been introduced with the view of rendering the language moro uniform, by constantly using the same words when the same meaning is to be expressed. As the Elements of Euclid are usually placed in tho hands of young students, it is important to exhibit the work in such a form as will assist them in overcoming the diffi- culties which they experience on their first introduction to processes of continuous argument. No method appears to be so useful as that of breaking up tho demonstrations into their constituent parts; this was strongly recommended by Professor Do Morgan moi^ than thirty years ago as a suitable exercise for students, and'the plan has been adopt- ed moro or less closely in some modern editions. An ex- cellent example of this method of exhibiting the Elements of Euclid will be found in an edition in quarto, published at the Hague, in the French language, in 1762. Two per- sons are named in the title-page as concerned in the work, PREFACE. Koenig and Blassiero. This edition bai served as the model for that which u now oflfered tx) the student; some slight modifications have necessarily been made, owing to the difference in the size of the pages. It will be perceived then, that in the present edition each distinct assertion in the argument begins a new linoj and at the ends of the lines are placed the necessary refer- ences to the preceding principles on which the assertions depend. Moreover, the longer propositions are distributed into subordinate parts, which are distinguished by breaks at the beginning of the lines. . This edition contains all the propositions which are usually read in the Univers-ties. ^fter the text wiU be found a selection of notes ; these are intended to indicate and explain the principal difficulties which have been noticed in the Elements of Euclid, and to supply the most important inferences which can bo drawn from the propo- sitions. The notes relate to Geometry exclusively; they do not introduce developments involving Arithmetic and Algebra, because these latter subjects are always studied in special works, and because Geometry alone presents suf- ficient matter to occupy the attention of early students. After some hesitation on the point, all remarks relating to Logic have also been excluded. Although the study of Logic appears to be reviving in this country, and may eventually obtain a more assured position than it now holds in a course of liberal education, yet at present few persons take up Logic before Geometry ; and it seems therefore premature to devote space to a subject which will be altogether unsuitable to the majority of those who use a work like the present. After the notes will be found an Appendix, consisting of propositions supplemental to those in the Elements of Jiluclidj it is hoped tbat a judicious choice has been made * PREFACE, from the abundant materials which exist for such an Ad- pendu. The propositions selected are worthy of notice on various grounds; some for their simplicity, some for their Tdne as geometrical facts, and some as being problems Which may naturally suggest themselves, brt of which the solutions are not very obvious. ITie work finishes with a collection of exercises. Geo- metrical deductions afford a most valuable discipline for a student of mathematics, especially in the earfepcriod o? Ills course; the numerous departments of analysis which subsequently demand his attention will leave him but little time then for pure Geometry. It seems however that the habits of mind which the study of pure Geometry tends to term, furnish an , ' antageous corrective for some of the evils resulting frc an exclusive devotion to Analysis, and it IS therefore desirable to engage the attention of beffin. ners with geometrical exercises. Many persons whose duties have rendered them familiar with the examination of lai-ge numbers of students in ^ementary mathematics have noticed with regret the frequ3nt failures in geometrical deductions. Several col- lections of exercises already exist, but the general com- plaint IS that they are too difficult. Those in the present volume may be divided into two parts; the first part con- tarns 440 exercises, which it is hoped will not be found beyond the power of early students ; the second pari; consists of the remainder, which may be reserved for practice at a later stage. Those exercises have.been principally selected from College and University examination papers, and have ^ been tested by long experience with pupils. It will be seen that they are distributed into sections according to the propositions in the Elements of Euclid on which they chieflv voi li. i~ "/ "^-' i^-"='*"*-»'""^j «*w ail aiigeu 111 order of difficulty, but It must sometimes happen, as is the case m I Ii an Ap^ notice on for their problems ^hich the 3s. Geo- ine for a )eriod of is which >ut little that the bends to 3 of the '^sis, and f begin- familiar cuts in ret the ral col- H corn- present rt con- found !onsists ;e at a elected i have ►eseen to the chiefly der of ) case s^i J*^j PHEFACK xi In the Elements of Enclid, that one example prepares the way for a 8et of others which are mnch easier Than Itself. It should be observed that the exercises relate to pnre Geometry; all examples which would find a more suitable place m works on Trigonometiy or Algebraical Ueometry have been carefully rejected. It only remains to advert to the mechanical execution of the volume, to which great attention has been devoted. Ihe figures will be found to be unusually large and dis- tinct, and they have been repeated when necessary, so that they always occur in immediate connexion with the corre- sponding text The type and paper have been chosen so as to render the volume as clear and attractive as possible The design of the editor and of the publishers has been to* produce a practically useful edition of the Elements of Mchd, at a moderate cost ; and they trust that the design has been fairly realised. Any suggestions or corrections relating to the work will be most thankfully received. I. TODHUNTER. St John's College, October i86a. i «u»j CONTENTS. PAGR Tntroductory Remarks ^y Book! I BookIL *"^"^^!^"!**!!!r!^!! sa Book III "" ' ,j Book IV ^.^!i"^"!'^!!"!""^!!"!!"!*!!!r*7' in BookV BookVI ''.ZZ'.''Z'. ,^3 Book xi ;;_;;;;■;;; ^;^ Book XII 244 Kotes on Euclid's Elements ^ -« * * 30 Appendix ^'^ * • 291 Exercises in Euclid ,40 i INTEODUCTORT EEMAEKS. The subject of Plane Geometry is here presented to the student arranged in six books, and each bo^k is subdivid^ into propositions. The propositions are of two kinds^^ W^ and .^.or^. In a problem something i, r^"^ to be^done, m a theorem some new principle is asserJed to A proposition consists of variova parta. We have fint the general enunciation of the problem or theorem; JZ example TojU^ribe an equilateral triangle on a ^iZ together le,. than two right angles. After the gene.^ enunciaUon follows the discussion of the proposition.^^ the enunciation IS repeated and applied to the particuT^; figure which IS to be considered ; as for example, Z^^^ be the git^en Hraight line: it U required So deocHbTan emlateral triangle on AB. 'rheZtructionS2Z Mows, which stetes the necessary stmight Imes and drdes wluch must be drawn m order to consUtute the ,ol^^ the problem, or to furnish assistance in the demonstZi^ of the theorem. Lastly, we have the demonstration TtsX «re*r2" ''' '-'''- '- "- '^^^' -^^i JeZtT i™l.l° r/'™^- is roauired; and binod ' --vi«vviVM iiiiu uumouairauon are com- xvi INTRODUCTORY REMARKS, Tho demonstration is a process of reasoning in which we draw inferences from results already obtained. These results consist partly of truths estiiblished in former propo- sitions, or admitted as obvious in commencing the subject, and partly of truths which follow from the construction that has been made, or which are given in the suppo^Uion of the proposition itself. The word hypotliem is used in the same sense as supposition. To assist the student in following the steps of tho reasoning, referetices are given to the results already ol> tamed which are required in the demonstration. Thus I. 5 mdicates that we appeal to the result established in the fifth proposition of the First Book; Constr. is sometimes used as an abbreviation of Construction, and Hyp, as an abbreviation of i^g/ioo^AmV. It is usual to place the letters q.b.p. at the end of the (Wscussion of a problem, and the letters q.e.d. at the end of the discussion of a theorem, q.e.p. is an abbreviation for quod erat faciendum, that is, which wa to he done; and Q.E.D. is an abbreviation for quod erat demonstrandum, that is, which was to be proved. in which L These er propo- I subject, itruction jpos^Ulon I used hi 3 of the eady o\y. rhus I. 5 d in the nietimes p, as an i of the e end of ition for ne; and andum, EUCLID'S ELEMENTS. BOOK I ;, DEFINITIONS. ^^ A msT U that which has no parts, or which has uo 2. A Une is length without breadth. 3. The extremities ofa line are points. breaV ''^"'^ " *^' "''''^ "^ '>'^r length and 6. The extremities ofa superficies are lines. bei4 tijSSe rtSl'ui: tf '" ""T^ »»y t'^o point, that BuperfidM. ^ "" between them lies whSUy in Mofwil'^LTXchm.iTli"*!"" «<:*'<>«»« to one same direction. "^^ together, but arc not in the 1 2 EUCLIUS ELEMENTS. 9. A plane rectilineal angle is the inclination of two straight lines to one another, which meet together^ but ar6\ not in the same straight line. Note, When several angles are at one point B, any one of them is expressed by three letters, of which the letter which is at the vertex of the angle, that is, at the point at which the straight lines that contain the angle meet one another, is put between the other two letters, and one of these two letters is somewhere on one of those straight lines, and the other letter on the other straight line. Thus, the angle which is contained by the straight hues ABy CB is named the angle ABC, or CBA ! the angle which is contained by the straight lines AB, DB 18 named the angle ABD, or DBA ; and the angle which IS contained by the straight lines DB, CB is named the an^le DBG, or CBD ; but if there be only one angle at a point, it may be expressed by a letter placed at that point; as the angle at ^. ^ 10. When a straight line standing on another straight line, makes the adja- cent angles equal to one another, each of the angles is called a right angle; and the straight line which stands on the - other is called a perpendicular to it. 11. An obtuse angle is that which is greater than a right angle. . ^ 12. An acute angle is that which 18 less than a right angle. • \ \ 1 :v n of two , but ar6\ ; B, any hich the IB, at the he angle > letters, I one of he other d by the >r CBA ! AB.DB ;le which kmed the Qgle at a at point; -A if ^SFINITIONS. ^ 13. A tenn or boundaiy is the extremity of anything, bouidarii^^ ^ *'^* ^^^^ ^ enclosed by one or morJ i^l^' A f'^'^^^e w a plane figure u'lu*'® drilateraj figures by four straight line.: foJLShK"^ ^^^ " Po'J«o»«, by mere thaa^ 24. Of three-sided figures, 1—3 ft III » EUCLms EISMMNTSL 25. An isosceles triangle \b that which ha« two sides equal : 26.^ A scalene trian,'le is that which has three unequal sides ; 27. A right^ngled triangle is that which has a right angle : [The side opposite to the right angle m a right-angled triangle is fre- quently called the hypotenuse.] XL 2®-..^'i obtuse-angled triangle is that which has an obtuse angle : 29. An acute-angled triangle is that which has three acute angles. Of four-sided figures, „ 30- .4 » vi «« iMeS^''p;:»<.«™ii'*T%°?/''"? «nd rhwiboid are not oftm it woV™Sv bo in™„-r»»"S'l.""***P^'«': »nd umversaur^pted] **''^*'««»t "^ "»8 regtrictioa wew POSTULATES.-+- let it be granted, point- to'Soa.eSJ:""*' "*' '^ ''""^ fro«» wj ona an A„3rL\*:S^^2*^„^«^sW ""e may be produced to •t 4 ^tf^^^StZi^'*"^'^'-"' -y «»"«^ 8 If EUCLWa BLEMENTS. AXIOMS.-^ to one a™ hfr7'^'^ ^"^ "^""^ ^ ^^"^ "^""^ "*^°« ^« ^V^ t 2. If equals be added to equals the wholes are equal, equS. ^^®^^" ^ ^^"^ ^^"» ®q»^« the remainders are unequal' ^"^^^^ ^ ""^^^ ^ unequals the wholes are ireLeJ^?"^" ^ ^^®" ^'^"* '^^^^^ *^® remaindeni onnfi *™"^ ^?i!^^ *^ ^°"^^® <>^ til© ««ne thlnff are equal to one another. * -^ 7. Thingd which are halves of the same thimr aro equal to one another. ^ ^"^ i. Mll^^tf^^^ '^^'''^^** ^^'^ ^"^ ^''^th^^ that 0. The whole is greater than its part 10. Two straight lines cannot enclose a space. 11. AU right angles are equal to one another. muirh\}{ * »t'?^8^^t JJne meet two straight lines, so as to make the two interior angles on the same side of if^V^ together less than two right angles, these straik. %^ being contmually produced, shall at length meSt f v ^^em which are the angles which are llss tha^two r^ht *' *^® distance AB, describe the wrcle BCD. [Postulau 3. 'f^ A^^J^^^^^ ^' ** *^® distance BA, describe tho circle ^G^. {Postulates. From the point Cy at which the circles cut one another, draw the straight Imes CA and Cj • to the points A and B, {Pott, 1. ^J?(7 shall be an equilateral triangle. • Because the point A is the centre of the circle BCD AC i^QCimX to AB, ; [DefinUicmU, And because the point B is the centre of the circle ACE, BC IS equal to BA. [Definition 15. But It has been shewn that CA is equal to AB\ therefore CA and CB are each of them equal to AB, But things which are' equal to tho same thing are equal to one another. ^ ^^^.^^ ^ - Therefore CA is equal to CB, Therefore CA, AB, BC are equal to one another. vvherefore the triangle ABC is equilateral, [Def. 24. and it is descnbed on the given straight line AB, q.b.f. 8 BUOLWS ELEMENTS. '^ PROPOSITION 2, PROBLBM. Oii^ZfgkrH^"' *" "'■<'«' " 'traigM linieguai to a line equal to M ^^ *^® Po«»* ^ a strai|ht From the point A to iSdraw the straight line ^^; ^p^J^Z and on it describe the eaui lateral triangle DAB [? i S'i ^^^^^'*^«*^>^* lines ^A J)B to E ^nd F, iPost.2, *rom the centre ^, at the dis- ^^i?; meeting />/^atGr.[i>o,,. 3 tance i>Gr describe the circle ^ArXme.t,tigi>^atZ.[Po.^3 -4// shall be equal to EC. SCU^l toX!"' ^ '» *"« «=«>tre of tho circle W^, And because the noint n i. fi.» ^ - lOejinition 15. ^X is equal to h^- " *^ <*»*'"« «f «»e cirole ©^z, ♦T" f ^' ^f ^"^ °f *"»«•» »«> equal ■ rS'f °" 'f Wore the reouunder AZ is 'e^ to thi^-^ det But it has been shewn that 5(7 is p„„,n.„ »„ U»*>» 8. one another. * ^ ^'^^ ^^^^ ^^^^g are equal to Therefore AZ is equal to BO. ^ ^^""^ ^' let ^5 and Obe the two given stmight lines, of which f BOOKL 8,4. 9 ^teSff ^J* j -g^l^ to out off ft^ ^B, fta the!SgM»ii ,^- ' [I. 2. a«d from the centre ^, at the astence AD, describe the Circle 2>^^ meeting ^^ At\ [Postulates. 4^ shall be equal to C. ^^^r?qud*to if' ^ " *« <*»«« of tho circte 2)^/; But C is equal to ^/> [D^nitim 15. line. ^7irrA& iltnT^zTc f^:t.tti!^. PROPOSITIOIf 4. THEOREM ^^s'XT^^^^ of th. one e,ml to tn. also have theXe^Z M^^ ^""-T ^^^^^*^^' ^^V '^l^ triangles shaUheeqZ andthlt ^"^^^ ^^^ '^' '^^ equaL each to ecu^hTameCLtZ f^^^^ Vt^^'' '^«^^ ^^ are opposite, ' ^«^^^y those to which the equal sides AB to DE, and i^ to ' ^^^^'^^^^^^^^'^an^ely, I>F, and the angle BAG a. / equal to the angle BDjr- the base i?(78hall be equal to the base E'i?; and the triangle ABC to the tri- angle J9^i^, and the other / angles shall be equal, each i w ^cii, tu vviiich the equal '^ ' "** *°S'® -4CS to the angle DF^ 10 EUCLIJys ELEMENTS, For if the triangle ABC be applied to the triangle DEF 80 that ihe point A may be on the point i>, and tho straight line AB on the straight line DE^ the point B will coincide with the point E^ because AB is equal to DK [Hyp. And, AB coinciding with />^,^(7willfalloni)^, because the angle BAG k equal to the angle EDF. [Hypothec, Therefore also the point (7 wiU coincide with the point F^ because ^ (7 is equal to DF Iff^othesisl mt the point 5 was shown to coincide with the point E therefore the base BC will coincide with the base EF; ' because, 5 coincidinff with E and C with F, if the base' BG does not coincide with the base EF, two straight lines wiB enclose a sp^ce ; which is impossible. [Axiom 10 Therefore the base BG coincides with the base EF and ia ^f^!^'*- [Axioms. I^^^Tn^l''^''}^' ^"^^^h ^.?^ ^^^^^id^ ^fcl^ the whole triangle 2)^i^, and is equal to it. [Axiom 8. ^n^l«^-!f ?t^®^f ^^®^ f *^® ^'^^ coincide with the other JnfM^^^ «^'^'*' ?^z.^^^.^^"^» ^ them, namely, the Sfle DFE ^""^ ' ^^ '^® ^ngleACB to the Wherefore, (/"^mjo triangles &c. q.e.d. PROPOSITION 5. THEOREM. TJie angles at the base of an isosceles triangle are eaijud LnZ «^^1^^^^; «^^./ the equal sides he produced the InotZr ""^ ^^ '^""^^ he equal to one Let ABC be an isosceles triangle, having the side A n equal to the sido AG. and let the S raight^lines AB AG ^e^a'^ri^*^^ Ti^^ *^^ '-^"^^^ ^^^«hall be equal to the angle AGB, and the angle CBD to the angle BCE, In BD tak^ pnv *v^in4^ rr atii from^^the greater cut off^Qf equal to^i^^the less, [1.3, t il ;■ and tho BOOK I. 6. 11 hypothesis, point -P, ypothesis, point jET. EF; base 5(7 ines will a?iom 10. ^, and is Axiom 8. le whole le other ely, the ^ to the *3 eqitaZ ced the ' to one deAB B,Aa qual to 7B. ss,ri.3. toid join Fa GB, Becauae^^is equal to^Qf, [Cmstr. and AB to AC, [Hypothms, the two sides ^^,^C'are equal to the two sides GA, AB, each to ewjh ; and they contain the angle FAG common to the two tnangles AFG, AGB; tiierefore the basei^Cis equal to the base GB md the triangle AFC to the triangle^ 6^5, and the remaining angles of the one to the remainini angles of the other, each to each, to which the equal sides are opposite, namely the angle ^Ci^ to the anffle A Br o«^ *i. ^^C'to the angle AGB. ^ ^' *"^ *^® a^ffJo • J. And because the whole AF is equal to the whole ^1^' of which the parts AB, AG are equa^ ^^^^^ole AG, the remainder BFis equal to the'remainder CGuZl; And FO was shewn to be equal to GB ; ^erefore the two sides -5J^, JPa are emial +A *!. * CG, GB, each to each ; ^^ *^ *^® *^« sid«8 andtheangb^j;(7was8hewntobeequalto 5a /"f f'"iS:'* 5^? ^^®° «^^^n that the who'lo an^lei 7?Ar IS equal to tl^ whole angle ACF ^ ^^ and^that the parts of these, the angles CBG, BCF are also therefore the remaininff anffle A nnia ^«,,^i a xi. 8ide of the base! ' ® the angles on the other Wherefore, ^A« an^/^, &c. q.b,^,, ^^^^ equESa?- ^^"^^ ^^^^y ^^^J-tHi^^Ie is also u EUCLIUS ELEMENTS. PROPOSITIONS THEOREM, U. 3. [Constructv^, nte to, the equal angles, shall he eoual to one another. ^^w* •««r®*^D^^ ^® * triangle, having the aide ^C7 shall be equal to the side AB nfl^^^^^^ "*** ^1"^^ *o ^A one of them must be greater than the other. ^J^^^S ^® *^® greater, and from it cutoff/)^ equal to ^C the less, and join 2>a ^en because in the triangles DEC, ACE I>B is equal to AC, * ' and EC is common to both, ^h to ^",^^' ^^ -« -l"*" to the two side, AC, CB, triangle />^C« e^A^h^e^S^^,*'-^- ^^. and tho the less to the greater; which is absnr^ , - .^ *' Therefon.^5 is not u^eqnrto ic- t^i ,v •» • ^^t"'- Wherefore, C/-.«.4S&o.-^p'*'^*"^'*""^'^'<'''^ equU^te^llS!^- ^''"* ^^'''^ equiangular triangle is also is equal to the angle ADO. fiT But the angle ACB is greater than the angle BCD [L 9 ^he^fore the angle ADC is also greater%han^Vl:gie much more then is the angle BDC greater than the angle Again, because BC is equal to BD, r^wx^- the angle BDCis equal to the angb BCD ^ '"^ iT But It h(^ been shewn to be greater; which is impossible.' But 'f one of the vertices as Dyhe withm the other triangle ACB, produce AC, AD to ^, >. . J^^^.^^ecause ^{7 is equal to ^-0,in the triangle ^GZ>, [ffyp, the angles BCD, FDC, on the other side of the base CD, are equal to one another. [r. 5. But the angle BCD is greater than the angle BCD, Wo™ the angle FDG U also greater than tSet^e mu^^ more then h the angle SDC greater than the angle Ag^,becaase5CU equal to 5i), trr,^,. . the aj,g,e ^z)(7 is eqJto the angle BCD. '^'^fiT The 1^" tT..*° "^ S^"'^'-' ^W"'" » impossible Of tte ne:Jrrdtir^^r *^-«"« ^ - » ^<»<^ Wherefore, OM ^^^ ^»i^ 5«^^ &c. q.e.d. PHOPOSITrnM- o mzTT^^T.*,.. -otnanglet hate two Hdet qfthe one eaual to ti^ tU otl^r, each to each, aSd have li^etMr Mtdet 14 EUCLWS ELEMENTS. hoiet equal, ih^ angle which u contained by the two HdeM ^t^'J^ 6. equa^^to the angle whichZ ZiZldb^ the two eidety equal to them, qfthe other, •*«*•««» oy A n^\n^^' f;^^u^® i*""^ triangles, having the two sides AB, ^C equal to the two s des DE Bl^t^sJh^^ l^T namely AJS to I>B^nd^AC ^o I^F^d^soZl hLTBG equal to^he base £F: the angle ^Jc7 shall be equd to ^e -n fw £J^® triangle ABChe applied to the triangle DEr 'f ^V:^^?,**^® straight line EF, the point C will also c^n cide with the point >, because BC is ^uaJ toEF rt^ liarAl"'.^''^^^''^'^ ^^' ^^ ^^ ^^ -^ cot ^^o?°» !f *K^^^ -^^ coincides with the base EF but tho 5?®* ^iVr^^ ?° pot coincide with the sides EIKFD^t have a different situation as EG, FG- then on thL «.?? base and on the same side of it «iere idU b? two S?anTe^ having their sides which are terminated at onreStI of the base equal to one another, and likewise theiS which are terminated at the other extremity. But this is impossible. rj l •art. e •>. l-fl«tom 8. Wherefore, ^^i^is equal to the angle EAF. [I. 8. h.. y?!^?®^'?^?//*^ ^^^/S r^c^eViw^ angle BAC u buected oy the straight line AF, q.e.p. PROPOSITION 10. PROBLEM, Jfo bisect a given Jinite straight line, that is to divide it into two equal parts. w*t^"«? »• Let ^5 be the given straight line : it is required to divide it into two equal parts. ' Describe on it an equilateral tnangle ABC, [i. 1. and bisect the angle AGB by the straight line CB, meeting AB at ^' [1. 9. AB shall be cut into two equal parts at the point B Because ^Cis equal to C7^, [Befin^tUmU. and 01) 18 common to the two triangles ACD, BCB ^Vr^h'; ^^' ^^ "^^ '^^ ^ *^" *^*^ «^^« ^^» ^A and tiie angle ACB is equal to the angle BCB ; [Cowft- [I. 4. therefore the baa** a tms. ^^i-.— 1 ■«■ i.- Wherefore ^A^ ^it?^ *^rat>A/ /m^ ^^ is divided *v>o eqtud parts at the point B. q.e.p. •«»''»«*« "•r^ " ww wy i y fwwiiiij uwy -i 16 EUCLIUS ELEMENTS. PROPOSITION 11 , PROBLEM. To draw a ttraigJU Urn at right angles to a giwn ttraight liney from a given point in the same. Let AE be the given straight line, and C the given Soint in it : it is required to raw from the point C a straight line at right anerles Take any point D in AC, and make CE equal to CD [J z OnDE describe the equilateral triangle DFE ' rr i and join CF. ' ■■ *• The straight line GF drawn from the given point CshaU be at right angles to the given straight line AB. Because DC is equal to CE, rrm,j,f^,^fi^ imd CF is common tS the two triangles DCF ^^^'T'"^' the two aides DC, CF are equal to the two sides EC CF each to each ; ' ^"^^ and the baae DF is equal to the base EF; [Dejinition 24, therefore the angle 2>C^is equal to the angle ECF- [1. 8. and they are acyacent angles. ' '. ^ «n"« ^atl?!,'*''^-^^* ?''®' Standing on another straight Ime. makes the adjacent angles equal to one another, each of the angles is caUed a nght angle ; [D^nitionlQ. therefore each of the angles DCF, ECF is a right angle. I' y}Tn'^{'''^ the given point C in thegiven straight hneAB, CFhas been drawn at right angles to AB. q.b,f. CoroUary By the help of this problem it may be shewn that two straight lines cannot «*«wu have a common segment. E If it be possible, let the' two straight fines ABC, ABD have the segment AB com- mon to both of them. From the point B draw BE at right angles to AB. Then, because ABC is a straight Ime, {Bypothem the angle CBE is efjual to the angle EBA. [D^ition la BOOK L 11, 12. 17 ll/i/potheais. Also, because ABD is a straight lino, the angle DBE is equal to the angle EBA. Therefore the angle DBS is equal to the angle CBE, [A^ 1 the less to the greater ; which is impossible uLJ^ I' PROPOSITION 12. PROBLEM, Take any point i) on the other side of ABy and from the centre C, at the distance CD, describe the circle EGF, meeting AB at F and G: [Postulate 3. Bisect FG at ^ [i. lo. and join C^. «i.oiTk® ^^^'^''^^t ^''ne CH drawn from the given point G shall be perpendicular to the given straight b^ne JJi Join CF, CG. Because i^^ is equal to JTWf/^ r/y , .. and //6 IS common to the two triangles FIIC GHC- t\t%tr;^^' ^^ ^^^ ^'l^ *« '^« twolSef f ^>^, and the ba^e GF is equal to the base GG ; IBejIniiim 15 therefore the angle CHFK, equal to the Lzl^CHG Xli and they are ac^acent angles. ' *■ ^LT^fhro^'-'*^'^^.^"''?^ '**"^""^ ^" ^'^other straight line makes the adjacent angles equal tf; one another each of tlm' &. Z^t^\t "?^^ ,?^^^' ^^^ *he straigSlhie 1 ch^ stands on the other is called a pernendjculai. to ii x^^^l Wherefore a perpendicular' CH has hem drmrl^Z^ 18 EUCLIUS ELEMENTS. \ if w PROPOSITION 18. THEOSEM, MfJ^hTj^^' ^^^""^ ^f *^/-ae>A/ line makes trith another itraight line on one side of it either are two rightanalZ or are together equal to ttoo right angles. ^ * Let the straight line AB make with the strakht h'ne CD on one side of it, the angles GBA, ABD • tS either are^two nght angles, or ail together eq^ to two'S? ^ B But if not. fro. the point E draw EE at r^^t^^ 'I toerofore the angles CEE, EBD are two right angleB-tw' NowtheMgle CBEh equal to the two ande. CBAARV. to each of tliese equaU add the angle^W- ^^'^>^^^' Agfa, the meU>l>BA is equal to the two angles^i; to each of these equals add the angle ABO- '^X'D^E,EStABG^' ^^' "^ ^^ to the three ^le^ret^iar^""^" "^^ shewn toie^Ji mTaBo' ""*'" ^^^' ^^^■'^*' ^"^ to the w?le» BntCBE, EBD are two right angles , ^^'^ '' therefore DBA, ABO are to. ■"*T BOOK /. 14, 15. 19 PROPOSITION 14. THEOREM. Jf, at a point in a straight line, two other ttraiaht h'^Am M^j/ri' two right angles, these two straiaht lines shall be in one and the same straight line. "^""'^^ '*"^ ntJXX&VD t^'^^^Sht line AB, let the two TisU MiS . ^ifZii £ •' d^^ together equal to two Fn?1f inf ^ '" theaamestraightlinewith CB. J ^« SdS^a'"'"" " *»^«-ki„g^^oqua. to^S Because AE]e e«"*»' ♦« *;*^ « •»•« • - each to each »d^_^^,i.i,arceq;:;,toth6twoti5i, CE,EF. M BOOKL 16,17. and the angle AEB is oqqal to the angle CEP hecauBothej are opposite ver- ' tical angles; £j j^ therefore the triangle AEB is equal to the triangle CEF ajd the remaining angles to' the remaining angles, each to each, to which the equal sides are opposite; fj, 4 therefore the angle BAE is equal to the angle ECF. But the angle ECD is greater than the angle ^C'JP.UW 9. © Therefore the angle ACD is greater than the anrfe BAE be J^dteTt^o^T^i^^^^^^ '^n^' an'd^^e'sj^^f ^ yfherefoTo,^ one side &o. <}.e.i,. ^-"^U-IC PROPOSITIOK 17. THEOREM Produce J56' to i>. >^ " tj^^^ Then because A CD is the exte- nor angle of the triangle ABC it 18 greater than the interior oppo- site angle ^^^ [I 10 To eachof these addtheangleJc/i Therefore the angles ACD ACB " are greater than the angles'^^a ACB. But «.e angles ACD, ACB ai-e together equal to two right tt^riXSes"^^^- ^^^'^^^ are together J'«,^' In the same mannAi* if »«««i.^ _i _^ V; -«(^A as alsa the an-les CAB, ABC, are togS MA less right angles. Wherefore, awy /too angles &c, Q.K.1>» 22 EUCLID'S ELEMENTS. PEOPOSmON 18, THEOREM. Because ^C is greater than Then,becanfk.^/)if is the ex- terior angle of the triangle BDC It IS greater than the interior oiv posite angle VOE. [i. jg But the ^hAVB h equal to the angle ABI>. n j because the side ^i) is equal to the side ^A fcL, Wore the angle AEI> is also greater than thi^^te Mu^ more then is the angle ABO greater ihan the angle Wherefore, within the trian.»:le : BDy DC shall be less than the other two sides BAj AC of the triangle, but shall contain an angle B£>0 greater than the mg\e BAG. Produce BD to meet ACatE. Because two sides of a triangle are greater than the third side, the two sides BA, AE of the triangle ABE are greater than the side -6.£?. [1.20; To each of these add J^C. * Therefore BA^ AC axe greater than BE, EC. Again ; the two sides CE, ED of the triangle GED are greater than the third side CD. [I. 20. To each of tlieso add DB. Therefore GE, EB are greater than CD, DB. But it has been shewn that BA. AC are greater than BE, EC; much more then are BA, ^C" greater than BD, DC. Again, because the exterior angle of any triangle is greater than the interior opposite angle, the exterior angle BDG of the triangle GDE is greater than the angle GED. , [I. 16. For the same reasdn, the exterior angle GEB of the tri- angle ABE is greater than the angle BAE. But it has been shewn that the togle BDG is greater than^ the angle (7^i?; much more then is the angle BDG greater than the angle BAG Wherefore, if from the ends &c. q.e.d. its B, C, I ^ :han the [I. 2O5 'ED are [I. 20. ter than angle is exterior le angle [I. 16. the tri- ter than^ le angle £001 I. 22. PROPOSITION 22. PROBLEM. t5 To make a triangle qf which ths sides shall he equal to three given straight lines, but any two whatever (/ thesfi must he greater than the third. Let AyB, C be the three given straight lines, of which any two whatever are greater than the third; namely, A and B greater than (7; ^ and C gieater than ^; and B and (/greater than A\ it is required to make a triangle of which the sides shall be equal to ^, ^, (7, each to each. Take a straight line DE terminated at the point 2), but unlimited towards E, and make DF equal to A, FG equal to B, and GH equal to C. [I. 3. Frum the centre F, at the distance FD^ describe the circle DKL. ' [Post. 3. Fi-om the centre (? at the distance GH, describe the circle JtiLKy cuttmg the former circle at JT. Join^^ JTG^. The triangle KFG shall have its sides equal to the three straight lines A, B, C. Because the point F is the centre of the circle DKL ^^isequaltoi^A [Definition 16. Therefore FK is equal to A. [^^^ j^ A ^in, because the point G is the centre of the circle HLK GH \% QcmaX to GK. rn.v? w Tr* T. X ^x,. , iLfefinttion 15. Therefore GK n equal to C. [^^ j, i^"i^Z^'A!% a*'°''" ^""'^ ^''' ^^' '^^ ="« ««"»' 26 BUCLms ELEMENTS. PROPOSITION 23. PROBLEM. pointLif ^d%ft tv^^^^^^^ ^ ^ ^-„ line AB, an angle eau^T fhl • ''" *^® ^^®° straight 7)0^. ' ^ ® ^"^ *^ *^^® g'ven rectilineal angle .In CD, CE take any points />, ig; and join DE. Make the triangle ^^G^ the sides of which shall be equal J?7^*^^S?^®® straight lines Ci>, i>iSr, ^(7;8othat^/' snail be equal U)CD,AG to CEy and ^G to DE. [i. 22. The angle /^(y shall be equal to the angle L>CE. an/frr ^^A^^ ^'^ ^^^^"^ *^ ^^' ^^> ^eh to each, therefore the angle EAG is equal to the angle DCE. [I. 8. lineABtZlfJf'^^J^^^ ^ *5 ^^^ ^^'^^'^ '^raight rectiflnJl"^^^^^^ ^^^' ^^ to th. giln PROPOSITION 24. THEOREM. 91068 oj the other, e(Kh to ewh,^hut the angle containedJ^i t' T/*f f ^-^ ^''^ "^ them greater thZth^Sec^ t^lfdjythetwo sides equal to them, of the 7C% Ce iild«^'^^i^%^£^>. *^? *ria,ngles, which have the tw6 each; ^^^^Ab^DeZ^A^ n^^^A^"^^ ^ ^AC^i than the lA^^: the^Se'l^V'sLITt \ BOOK /. 24, 26. 27 greater tlian the base EF, Of the two sides DE, DF, let DE be the side which is not greater than the other. At the point D in the straight line DE, make the angle EDG equal to the angle £AG, [1.23. aiidmake2)(yequalto-4{7or2>JF; [i 8 fxa^iom EG, GF, Because AB is equal to DE, [ffypothms. Bud AC to DG: rnJl* *- ^ » [Construction, the two sides BA, AC&re equal to the two sides ED, DG, each to each ; ' ' and the angle BA G is equal to the angle EDG ; [Cmstr, therefore the base BC is equal to the base EG, [I. 4. And because DG ia equal to DF, iConatructin. the angle DGF is equal to the angle DFG. [I. 5. But the angle DGFi» greater than the angle EGF, [Ax. 9. Therefore the angle DFG is greater than the angle EGF. Muc^ more then is the angle EFG greater than the angle ' lAxiom 9. And because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater angle is subtended by the greater side, [i 19, therefore the side ^G^ is greater than the side ^i^. But EG was shewn to be equal to BC; therefore BCk greater than EF y^herefore, if two triangles &c, Q.E.D. ^-.t> PPn'PncTnnTrkXT or BiE.^nffjy'yJ!^^ /i«»«^M^o sides of the one equal to tw^ 9%detqfthe other, each to each, 1m the b(mqf the one 28 EUCLIDS ELEMENTS. Bide^^^^^^'/f ^ w *Tu *^^^"^'?^ ^hich have the two sides AB, AC equal to the two sides DE, DF each to ^h, namely ^i?

^, and AC to DF huf the base BG greater than the base EF: the ang e BAG sS be greater thau the angle ^ ^^ For if not, the angle JiACiami be either equal to the angle EDF or less than the angle EDF. But the angle ^^(7 is not equal to the angle EDF, for then the base BG would be equal to the base ^i^- n a but it is not; ' rjj ;/* . * therefore the angle BAG'ib not equal to the angle TdfT' Neither is the angle BAG less than the angle EDF for then the base BG would be less than the base EF- FI 24 but it is not: rrr / . * therefore the angle BAGh not less than the angle EDF tt^^^^EDF"'"'' *^'' '''' ^""''^ ^^^ '' "^'y ^^'«i Therefore the angle BAG is greater than the angle EDF. , Wherefore, (/-^i^^o^riW/^/^&c. q.e.d. PROPOSITION 26. THEOREM. f^yl^softhe other, each to each, and one side equal to one side, namehj, either the sidet adjacent to the^otml ZtlCl T^tX^r^-r r^^^'^f '' ^"^ alius inlX otLr. ^'^^^^ ^^^^ one equal to the third angle of ths fmJtesAif^%rf^ ^'\ ^^ t"^"^Ie«> which have the angles ABC, BGA equal to the mglea^DEF, EFD, eacl^ L BOOK L 26. 29 to each, namely, ABC to DBF, and BCA to EFD ; and let tiiem haye also one side equal to one side ; and first let those sides be equal which are adjacent to the equal anirlea in the two triangles, namely, BG to EF: the other sides A^\ ^?. ^^"^ » ?'^^ ^ ®^^ namely, AB to DE, and AU to DF^ and the third angle BAC equal to the A third angle EDF, For if AB be not equal to DE, one of them must be greater than the other. Let AB be the greater, and make BG equal to DE, [I. 3. and join G^a Then because Gf^ is equal to 2)^, [Comtmctum^ f^d BO to EF; [Hypothem. the two sides GB, BCsltq equal to the two sides DE. EF each to each ; » * and the angle GBC is equal to the angle DEF ; [Bypothem. therefore the triangle GBC is equal to the triangle DBF, and the other angles to the other angles, each to each, to Which the equal sides are opposite ; [i. 4. therefore the angle GCB is equal to the angle BFE, But the angle BFE is equal to the angle ACB. [Hypothesia. Therefore the angle GCB is equal to the angle ACB, [Ax. 1. the less to the greater ; which is impossible. Therefore ^^ is not unequal to Z>^, that is, it is equal to it ; • and BC is equal to EF; [ffypothesis, mEF, tht tX^""^ ^"^ "' '^'^' '' *'^ '"^ ^''^* ~ -^o^- -i^"^ iJ3 wv|uu,i i/ii Luc* augte JJJiJ^'i \Hypoihm8» therefore the bas&^C is equal to the base DF. and tho third angle BAC to the third angle EDF. i;L4. 30 EUCLID'S elements: Next, let sides which are opposite to equal anrfes in each tnangie be e^ual to one Mother, namely. ZfiS I>E: hkewise lu this case the other sides shaU te^u^ ^h to each, namely, BG to EF mdACt^ n^^^TJ also the third angle I'ACe. Wherefore, if a straight line &c. q.e.d. PROPOSITION 28. THEOREM, If a straight line falling on two other straight lines, make the exterior angle equal to the interior and opposite angle on the same side of the line, or make the interior angles on the same side together eqtiol to two right angles^ the two straight lines shall be parcUlel to one another. 32 EUCLID'S ELEMENTS. Let the straight h'ne EF, which falls on the two straight mes AB, CD, make 'the exterior angle"j Mde, or make the mtenor angles on the same side BGtL Because the angle EGB is equal to the angle GHD, [Hyp, and the angle EGB ia also equal to the angle -4 Gf/?, [i. 15. therefore the angle AGH is equal to the angle GHD, [Ax.l. and they are alternate angles ; therefore AB is parallel to OD, mntl^'lL^^T.® *^? ^"^^^^ -^^^» ^^^ are together equal to two right angles, [ffypothesis. right aS'' ^^^ ''"'' ""^"^ *'^'*^'' '^^^^ *^ *^^ BG^Gni, ""^^'^ ^^^' ^^^ ^"^ ^^^^ *<^ *^« S. I^i^n^^^^® commonangle BGIT; therefore the remaining angle ^G^ZTisequal to the remaining angle GHD ; [Axiom 3. ajid they are alternate angles ; therefore ^^ ia parallel to C/>. p 27 [I. 27. Wherefore,^ a straight line &c. Q.E.D. h PROPOSITION 29. THEOREM, it^V^^fi^^M'"^^ '^^^^ ^'^'^^^ Pf^rcdlel straight lines. It makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite angle on the same side; and also the two interior angles on the same side together equal to two right angles. Let the straight line EF fall on the two parallel straight lines AB. CD • tho nif A,.nafo o«„i«„ .1 a « S JV^' ECB^lTf *"' T''^^^'^ irexreSorai^lS' M,GB, ihall be equal to the interior and opposite angle BOOK L 29. 83 on tijc same side, GIID, and the two interior angles on tlic same sidoi BQH, GHD, shall bo together equal to two right angles. ' For if the angle AGH bo not equal to the angle GHD, I one of them must be greater j than the other ; let the angle j AGH be the greater. Then the angle AGII ia greater than the angle GBD ; ; to each of them add the ando iBGH; ^ ^therefore the angles AGIIy BGII are greater than the \B,ug\Qa BGH, GHD. But the angles AGH, BGH are together equal to two nght angles; [1.13, therefore the angles BGH, GHD are together less than two right angles. But if a straight line meet two straight lines, so as to make the two interior angles on the same side of it, taken together, less than two right angles, these straight lines being continually produced, shall at length meet on that side on which are the angles which are less than two I right angles. [Axiom 12. \ Therefore the straight lines AB, CD, if continually pro- duced, will meet. I But they never meet, since they are parallel by hypothesis. Therefore the angle ^6?// is not unequal to tho angle GHD ; that is, it is equal to it. But the angle AGH is equal to the angle BGli. [I. 15. [Therefore the angle EGB is equal to the angle GHD. \Ax. 1. Add to each of these the angle BGH. Sj^S^Trrfc' ^^^^®^ ^^^> ^^^ ^^® ^^^ *o ^^^ anglca B{jrH, (xHD. UxiomX But the angles EGB, BGH are together equal to two nght angles. ^ rj jg iuuioiuru Liio angles jjixii, uHD are together equal to two nght angles. ^ [L/o»i 1. Wherefore, if a straight line &c. q.e.d. 34 EUCLID'S ELEMENTS, H PROPOSITION SO. THEOREM. Straight lines which are parallel to the tame ttraiaht line are parallel to each other-. Let AB, CD be each of them parallel to EF-, AB shall be parallel to CD. Let the straight Ime GHK mi AB, EF, CD. , Then, because GIIJC cuts the parallel straight lines AB^ EF, the angle AOIHb equal to the angle GHF. [I. 29. Again, because GK cuts the parallel straight lines EF, CD, the angle GHF is equal to the angle GKD. [I. 29* And it was shewn that the angle AGK\% equal to the atigle GHF. Therefore the angle A GK is equal to the angle GKD • N «. i and they are alternate angles \ ' ' therefore AB is parallel to CD. n 27 Wherefore, straight lines &c. q.k.d. PROPOSITION 31. PROBLEM. to l^tZ^r%t?Ur!r '''^"^' ^ ^^'^-^^-^ ^-«^^ Let ^ be the ^ven point, and BG the given straight line : it is required to draw a straight line th?oS tha point A parallel to the straighi line BC, '^^^^"S** ^^e In BC take any point ^ 2>, and join AD \ at the ^ - - A v point A in the straight line AD, make the angle DAE equal to the angle «— ^J>0\ ^ ^ [1.23. ^ and produce the straight line EA to F. i^JP shall be parallel to BC. BOOK I. 31, 32. 35 iVM ttraight ^ EF: AB -D [I. 27. nt parallel en straight irough the :h JT Because the straight line AD, which meets the two straight lines £C, EF, makes the alternate angles BAD ADC equal to one another, {C(yMtruction. EF is parallel to BC* [I 27. Wherefore the straight lineEAFit dratmi through the given point A, parallel to the given ttraight line BC. q.b.f. t PROPOSITION 32. THEOREM, If a side qf any triangle he produced, the exterior angle is equal to the two interior and opposite angles ; and the three interior angles qf every triangle are toge- ther equal to two right angles. Let ABC be a triangle, and let one of its sides BG be produced to D : the exterior angle ACD shall be equal to the two interior and opposite sugieB CAB, ABC - md *i^ .*^5f^ J.^\^^}P^ ^°^^®« ^^ *^e triangle, namely, ABC, BCAy CAB shall be equal to two right angles. Through the point Cdraw -^ C^ parallel to ^i?. [1. 3i. ^^\ JS Then, because -4 jB is par- allel to CE, and AC falls on them, the alternate angles BACy ACE are equal. [Again, because AB is parallel to CE, and BD falls on them, the exterior angle ECB is equal to the interior and opposite angle ABC. rj 29 I But the angle ACE was shewn to be equal to the angle j therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC. [Axiom 2. I To each of these equals add the angle ACB ; SlS« ^«*?® SJn^A^^' ^^^ ^^ ®^^ t^ t-e three angles CBA, BAC, ACB, ^ ^j^^^ 2. But the angles ACD, ACB are together equal to two right angles ; j-j jg therefore also the angles CBA, BAC, ACB are together equal to two right angles. [^^^ 1^ yfhQTQiovQ,ifasideqfan2ftria7igle&Q. q.e.d. 3^2 [I. 36 EVCLWS ELEMENTS. crquai to twice as many right angles with the angles at the SJnt ^°which 18 the common vertex ofthe triLigles" MthefisurelasX^ twice as many right angles . ;.^®^^"f every interior angle i^u' ^^*h»ts adjacent exterL angle ABD, is equal to two right angles j fj js il?Trf^T ""^ *^^ '"*«^^or angles of the figure, together with all Its exterior angles, are equal to n" STel^'reSMi'" ""^'^^ ^ the flffiirn t^„ ""° . i_?.? -"lua'. '9 all the interior anelfi. nf mi. ".. " ' ■'■■■a-- -^^'■i iTiui lour right andea m^fore aH the exterior angle's are'.i to W right BOOK L 33, 34. i- PROPOSITION 33. THEOREM. 37 The straight lines which jom the extremities of two equal and parallel straight lines towards the same paHs are also themselves equal and parallel, * Let ^jB and Cp be equal and parallel straight lines and let them be joined towards the same partfby the fnTpa^eT' ^^ "^^ ^^ '' ^^ ""^ ^^ «" be Iq^ Joined Then because AB ia par- allel to CD, [Hypothms, and BC meets them, the alternate angles ABC, BCD are equal. [f . 29. ' And because AB is equal to CD, [ffypotkesis. md BG 18 common to the two triangles ABC DCB - the two sides ^J5, BC are equal to the two sides DC CB each to each ; ' * ^ndjhe angle ABC was shewn t« be equal to the angle therefore the base ACis equal to the base BD and thA triangle ^5C7 to the trianglS BCD, andTe fther an^es are opptfte^^^^^ "''^ '^ '^'^^ '' ^^^^ *^^ equal sfdes therefore the angle ACB is equal to the angle CBD ImofAcT^^J^^ 't^^^K ^^"? ^^ "^^^^« *^^^ *^« straight imes^C, BD, and makes the alternate ana-lcs ACB CBD equal to one another, AC is parallel to BD. '[i 27 And it was shewn to be equal to it. Wherefore, (he straight lines &c. q.e.d. PROPOSITION 34. THEOREM „- ,.,,,. ,„„. ,_^ uiaaes ii inio two equal parts .S8 MUCLIiys ELEMENTS. :il sect it. ^"^^ ^^ *^^ diameter BG shall bi^ Because AB is parallel to CZ>, and BG meets them, the alternate angles ABG, BCD are equal to one an- ^^'^^^' [I. 29. And because^ (7 is parallel to ^i> and BG meets them, ine alternate angles AOTi mn « another. ^ -^^^^ ^"^^ ar« equal to one Therefore the two trianffles ATtr nnn i. . ^^' ^^• ^EG, EGA in the oneeaZifii::^ ^^l^ *^° ^n^^es the other, each to eacS aSn^side n^^'' ^^^' ^^^ ^^ two triangles, which is '-^l^Zt tot^^^^ ^ the ^^i^fZ^l^^T^ ert:S,and ^etAte^sfdSS^^^^^ angle (72)^. "'^^^^^ and the mgleBAG equal to the And because the angle ^j&tr is Pmioi+^i-v, , '•^'^^ and the angle CBI> to fhe angfe 7(7? *° *° ""^'^ ^^^' thewholeangIe^5i)iseq„altothewh;ieangle^Ci) Ur , An^the^^gle^^aha. been «hewn to ^^^^V^ ^Tqtl t*:„raSef " ""' ^°^'- °^ "^ parallelogram Also the diameter bisects the pamllelogram. tte twf h"'^.T1"' ^^' '^'^ ^Ceomaon, ScVr et" ^^' ^'^ '^<' ^'''^ 'o the two Jides 2>0, CB ZA^D^ ^""^ "^ "^ ^hefrn to be equal to the ttereforethetria„gle^5(7isequaltothetriangloi?<7i> ri . into two equaTpartl^ """'''''' ''''^ P^^aileiogram AGEB ^^^refore, the opposite stde^ &c. q.e.d. BOOK L 35. S9 PROPOSITION 35. THEOREM, Parallelograms on tTie same base, and between the same parallels, are eqttal to one another. Let the parallelograms ABCD, EBCF be on the same base BCy and between the sameparallels AF,BC: the paral- lelogram ABGD shall be equal to the parallelogram EBCF. If the sides AD, DF of the parallelograms ABCD, DBCF, opposite to the base BC, be teniiinated at the same point D, it is plain that each of the parallelograms is double of the triangle BDC', [I. 84. and they are therefore equal to one another. {Axiom 6. But if the sides AD, EF, opposite to the base BC of the parallelo- grams ABGD, EBCF be not terminated at the same point, then, because ABCD is a par- allelogram AD is equal to BC ; for the same reason ^i^is equal to BC', therefore AD is equal to EF-, [Aanom 1. therefore the whole, or the remainder, AE is equal to the whole, or the remainder, DF, [Axioma 2, 3. And AB is equal to DC ; [I. 34. therefore the two sides EA, AB are equal to the two sides FD, DC esuch to each ; and the exterior angle FDC is equal to the interior and opposite angle EAB ; [I, 29. therefore the triangle EAB is equal to the triangle ^^^' [I. 4. Take the triangle FDC from the trapezium ABCF, and from the same trapezium take the trians-le EAB. and the remainders are equal ; " [Axiom 3. that is, the parallelogram ABCD is equal to the parallelo- ^herefore, parallelograms on the same base &c. q.b.d. [I. 34. #"% 40 EUCLID'S ELEMENTS. V^^ifASf theS'i^Sram^ °" equal bases Join BE, Cir. . , Then, because i?(7 r ^M H isequaIto^(?,[//yp. ^O is equal to ^^; [Axiom 1. . _ and they are parallels, " "" ^ -d J^w,ed Wds the same parts by the st^Ss ?f .Ifflfcie^ot'r^^^^^^^^ -^^ -d selves equal and parallel '''e same parts are them- Therefore BE, Cffare hnfh «„ i , t^- 33. Therefore WcATpSeSm""^^'- . . And it is equal to ABCB h^t \i, [Definition. ba^^aau^etweell^f^a^rS^^^^^^^ to ttteS;r°" ""^ ^''-"«>°S-» m is eq^^ allo^^ri»f"^'<'^ ^^^^ is equal to the par- Wherefore, i>aro«(,j,ram&c. q.e.d. [^'«^'» 1. . PEOPOSmON37. TBEOREU M:X:,ZI'' ''"^'^'' '""' *^'-^- ^^ sa^ par. TiD^u*^® triangles .^J?(7, J^BG bo on the same base JiC, and between the same parallels ^Z),^a: the tri- *"^ir' "r*t'"^ oiiaii ue equal to the tnangle DBC, Produce ^2> both ways to the points .£;i^;'[Pc«^.l BOOK /. 37, 38. 41 ^een the same through B draw BE parallel to CL4, and through C draw CF parallel to BD, [I. 31, Then each of the figures EBGA, DBCF is a parallelo- &^^ ; {Definition, and EBCA is equal to DBCF, because they are on the same base BC, and between the same parallels BO, EF. [I. 35. And the triangle^i?C7ishalf of the parallelogram EBCA, because the diameter J ^ bisects the parallelogram; [I. 34. and the triangle DBC is half of the parallelogram DBCF, because the diameter DC bisects the parallelogram. [I. 34. But the halves of equal things are equal. [Axiom 7. Therefore the triangle ABC is equal to the triangle DBG. Wherefore, triangles &c. Q^d. PROPOSITION SsT THEOREM, Triangles on equal bases, and between the same par- allels, are equal to one another. Let the triangles ABC, DEF be on equal bases BC, EF, and between the same parallels 7?^, AD: the triangle ABC shall be equal to the triangle DEF, Produce^/) both ways to the points 9 h H ^Fl through B draw BG parallel to CA, and through F 6x2iVf FH paralleltoJ5^i>.[I.31. Then each of the figures GBCA, DEFH is a parallelogram. [Definition. And they are equal to one another because they are on equal bases BC, EF, and between the same parallels BF, GH, ^ Yi, 36. And the triangle ABC is half of the parallelogram GBCA, because the diameter ^^ bisects the parallelogram ; [I. 34. and the triangle DEF is half of the parallelogram DEFH^ because the diameter DF bisects the parallelogram. But the halves of equal things are equal. [Axicm. 7. Therefore the triangle ABC is equal to the triangle DEF, Wherefore, triangles &c. q.e.d. ■■f r 42 EUCLID'S ELEMENTS. tween the same parallels. ""^ '* ' *^^^ «^ail be b^ Join AD, ^2> shall be parallel to JBC. at H. ^'^^^^^ ^0 -^^^ meeting BD and join JE'a ^^' ^^' ^ ^^^-^^^^ ebI same parallels £C, AK ^^> ^^ between the But the triangle ABg\b eaual fn +!,« * • , ^^' ^7- the greater to the less • xchinh • • Uxiom 1. Therefore ^ T? i. T ' ^^ '^ impossible, t^rciore ^^ IS not parallel to BC. Btraig^JteXougHb^^^ > «^-"' t^^t no other' therefore AD is patlfel to ^a " "^''""'^ '^ ^^^ Wherefore, .^e,a/ ^^,-«^^^^^ ^^ q.e.d. PROPOSITION 40. r^^oi?^^. Let the equal triangles^^i^ S:f '""^^^^^^^^^^ ^^>m in the same straight lin^^rr^® "? ®^^*^ '^ases Bide of ,t : they shall be betweL^^^^^^^ t^' ^"^ ^? *^^ «^^« Join AD. "«iween the same parallels. ^i> shall be parallel to BF. ^^ ^ For ifit is not, through yf draw^t? parallel to "^i? meetme- I^n of ^ ,_ ^> 'J -ST-.- -^ — 1 • . r-r and jom G^^. Li. oi. BOOK L 40, 41. 43 Then the triangle ABC is equal to the triangle GEFy because they are on equal bases J3C, HF, and between the same parallels. [I. 33, But the triangle ABO is equal to the triangle DBF. [Hyp, Therefore also the triangle DBF is equal to the triangle ^^^> . [Axwrn-L the greater to the less; which is impossible. Therefore AGi% not parallel to BF, In the same manner it can be shewn that no other straight line through A but AD is parallel to BF ; therefore AD is parallel to BF, Wherefore, equal triangles &c. q.e.d. PEOPOSITION 41. THEOREM, If a parallelogram and a triangle he on the same hose and between the same parallels j the parallelogram shall be aotible of the triangle. Let the parallelogram ABCD and the triangle BBC he ^nJ'A^^^? ^^^® ,, F' ^^^ between the same parallels X . ' ^^^ : the parallelogram ABCD ghaU be double of tha tnangle BBC. Join AC, Then the triangle ABC is equal to the triangle BBC, because they are on the same base BC, and between the same parallels BC, AB. [I. 37. But the parallelogram ABCD is double of the triangle ABC, because the diameter ^C bisects the parallelogram. [J. 34. Therefore the parallelogram ABCD is also double of the tnangle BBC Wherefore, if a parallelogram &c. q.e.d. 44 EVCUD-S ELEMENTS. PEOPOSmON 42. PROBIEM. rectilineal angle, '' ^' angles equal to a given ?b.-Ul be equal to theVen trfe^ /%'?"''"«'«»«>•» that Its angles equal to Z>. "^""nsle ^i?c» and have one of Bi8ect5eat^:[I 10 ^'?n1if' ?''•^* "»« point f' J" JJ^e straight line £C, mkotheangleCfii^equa ' [I 23 through A draw ^V« parallel to^Candthrough O draw cr©. parallel to [I. 31 Therefore /-^Cr^yiaaWaUelogram. And, because 5^ is equal to ^C [Definition. iComtructim. they are-on eq^f basTil"!? ^ri''' ^^^. because parallels £(7, Jiff. ^^^ ■^^' -^^> and between the same Therefore the triano-le Ann.A^, - ^ ^- ^^• But tho pamllelogral FEGG f ''' triangle^^a AEG, because thl"f ?S S^"\''°"ble of the triangle the same parallels ^^^^ ^® ^° ^^' ""d between Wore the parallelog.. ^^^G is equal to the trL^le -<«H has one of its angles c;^^ equal to the givel:;,: Wherefore n >^^^^77 i ' Construction. "nicies CEEenualto thegtiinkTn '717 '^' "/ »'* \ BOOK I, 43, 44. 45 PEOPOSITION 43. TUEOREM, f The complements of the parallelograms which are about the diameter qf any parallelogram, are eqitai to one another. Let ABGDhQ a parallelo^am, of which the diameter is^C; md Eff GF parallelograms about AC, that 18, through which AC passes ; and BIT, KD the other paral- lelograms which make up the whole figure ABGD, and which are therefore called the complements: the comple- ment BK shall be equal to the complement KD, Because ABGD is a parallelogram, and AG its diameter, the triangle ABC is equal to the triangle ADC, [1. 34. Again, because AEKH is a parallelogram, and AK its diameter, the triangle AEK is equal to the triangle ^i- C [I. 34. For the same reason the triangle KGC is equal to the triangle KFC. Therefore, because the triangle AEK is equal to the tri- angle AHK, and the triangle KGC to the triangle KFC ; the triangle -4 ^iT together with the triangle KGC is equal to the triangle^ jfiT^ togetherwith the triangle KFC [Ax. 2. But the whole triangle ABC was shewn to be equal to the whole triangle ADC. Therefore the remainder, the complement BK is equal to the remainder, the complement KD, [Axiom 3. Wherefore, the complements &c. q.e.d. PROPOSITION 44. PROBLEM. To a given straight line to apply a parallelogram, fc/iic/i shall be equal to a given triangle, and ham one qf Its angles equal to a given rectilineal angle. 'Wm 46 mCLIlys ELEMENTS. tri«^Kd'ltg;:Sa'«-. «nd C the given to ap|,ly to the straight Tnl^'i'"**' »"S'r '' " reqSred the tr-angle C. and U^^g^,, ^'^^^ ««lW i^^?^':^:^A''2^rs%r'^''^ - *"« pa-ne,s two nght angles. * '"'^' "^-^-^ «•« together equal to two wS^"^,er^« ^^^' ^^ -e together le. V^" let them meet at K >f Produced , Therefore Z5i, equal to ^i.. i^"' "f ^ equal to the trianrfe O . f^- ^^ TWefor, Zir fa equal to the WaL n rc<««n..«». BOOK L 44, 46. 4r And because the angleGfJ9^isequal to theangley^^J/ [1 15 and Ukewise to the angle D ; [ConstL'tion. the angle ABM is equal to the angle D. [Axiom 1. Wherefore to the given straight line AB the paralldo^ gram LB is applied equal ta the triangle C, and having the angle ABM equal to the angle D, q.e.f. PEOPOSmON 45. PROBLEM. To describe a parallelogram equal to a given rectilineal Jigure, and having an angle equal to a given rectilineal angle* Let ABCD be the given rectilineal figure, and E the given rectilineal angle: it is required to describe a par- allelogram equal to ABCD^ and having an angle equal to E. L ♦k i^^ J^^,attd describe the parallelogram PIT equal to the triangle ADB, and having tlie angleW^ equa? to the and to the straight line GH apply the parallelogram GM eq^} t^ tL'e^aKr ^^^' ^^ ^^^^ *^« ^}ff The figure FKML^h^W be the parallelogram required.* ^^' ^^^ase the angle^is equal to each of the angles J'/r//, Add to each of these equals the angle KBG • therefore the ansrles PlTH. jrrtn ^^^ i\- ^, . , But^iT^^irG aretogetherequaltotworiglitangle8^29 thereforeJr^«,6^i,faretogethereq„altotworightangS yi** 48 EUCLID'S ELEMENTS, F G L ^ C IT "H M And because at the point H in the straight lino GH tim mnLli^'^^*^ • ^''^l' ^?' ^^' ^" «^« opposite sides of it make the adjacent angles together equal to two right angles KH IS m the same straight line with HM. ^ [in* Jr>l^"7?«^''"'u*^^f''''^^l'*""^^^ "^«e*« the parallels KM FG, the alternate angles MJUG, //^i^ are equal. [1.29 Add to each of these equals the angle UGL • hTJ^iigI '"^^'' ^^^^^' ^^^^' "'" "^^"^ '*" ^'^^ ^"^^^« ButMHGyHGLare together equal to two right angles- n29 therefore HGF, BGL are together equal toLo right LI!£. Therefore FG is in the same straight line with GL. [114 AndbecauseiT^is parallel to^6^,and^(3^toilfZ,[C7ona,v* KF IS parallel to ML ; FT qn and ^Jf, ^Z are parnllels ; [ConstLi^. therefore A'^is a parallelogram. [^,^,,,.^: iam ^1^"'" *^' ^^'^^^^ ^^^ ^« ^^^ to the parallelo- anA^\.r.4.*' 1 TN«^ [Construction, SfcSfeV^^- ^^^^ - equal to the Whole ■ori,^ - ^_ [Axiom 2. scnoea equal to the given rectilineal fwtirx A nrn ««^ halving the angle FKM equal to thX:7ntuE!^'^:l So Pm,al Jn .*P-P'y * Pai^ lelogram, which shall hafe an to a rfv^ r^wf"'®? rectilineal angle, and shall be equal !?..*. ^^®? rectilineal fignre ; namely, bv anolvm^ t„ ti,„ angle SSFL'H'LtiP*'*"*''"^''^"' «^"^ tS'ttelrst tri^ S soia ' ^^ *° ^'^ ^1"*' *" *e given angle ; BOOK I. 46. «9 r % PEOPOSITION 46. PROBLEM. To describe a square on a given straight line. Let AB be the given straight line : it is required to describe a square on AB. From the point A draw AG - at right angles to AB\ [I. ii. ^ and make AD equal to AB , [1. 3. through 2> draw DE parallel to d AB ; and through B draw BE parallel to AD. [i. 31. ADEB shall be a square. For ADEB is by construction a parallelogram ; therefore AB '\& equal to DE, and AD to BE. [i. 34. But AB is equal to AD. IConstructian. Therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral. Uxkw i. Jjikewise all its angles are right angles. For since the straight line AD meets the parallels AB DE the angles BAD, ADE are together equai to two right angles J ^ ^ '^^''^^ but ^^i) is a right angle ; IComtruction. therefore also ADE is a right angle. [Axiom 3. But the opposite angles of parallelograms are equal. [I. 34^ Therefore each of the opposite angles ABE, BED is a Therefore the figure ADEB is rectangular ; and it has been shewn to be equilateral Therefore it is a square. [Dejimtion 30, And it is described on the given straight line AB. q.b.f.* ^«^-^'^'''^^r^^' ^ *'"'""^ "*« ^emOMBtiauou ii 18 manifest that eyery parallelogram which has one right angle has all its angles right angles, . ' 50 EUCLID'S ELEMENTS. PEOPOSITION 47. THEOREM, -f In any right-angled triangle, the square which is de- scribed on the side subtending the right angle is eqiml to the squares described on the sides which contain the riaht angle. ^ V^^tA?,^^^ ^ right-angled triangle, having the right angle BAG : the square described on the side BG shall be equal to the squares described on the sides £A, AG. On BG describe the square BDEG^ and on BA, AG de- scribe the squares GB,HG', [1.46. through A draw AL parallel to BD or OE', [1.31. andjoin^^, i^a Then, because the angle i?^ (7 is aright angle, [Hypothesis. and that the angle BAG is also a right angle, [Definition 30. thetwo straight lines ^(7, AG, on the opposite sides of AB make with it at the point A the adjacent angles equal to two right angles ; - & m^^i therefore CA is in the same straight line with AG. [I 14 I^r the same reason, AB and AH^q in the same'straight Now the ajigle DBG is equal to the angle FBA. for each of them IS a right angle. . [iSoTlT Add to each the angle ABC. T Wore the whole angle DBA i» ^quaJ to the whole angle A J I. ., [Axiom 2. «^de?;r"o!tertot!fH^^' ^^ - ^'^^J^ ^? - - - " L-i^'i^'ifiidOn o\J, Md the . gle DBA is equal' to the angle FBC- «;jrefore the triangle ABD ia equ^ to the triangle [1.4. BOOK 1. 47, 48. 51 A ^^\ *^® pajftllelogi-am BL is double of the triangle ABD, because they are on the same baae BD, and between the same paraUels BD, AL, ' rj 41 And the square GB is double of the triangle FBa because ^^^.^-^B^r' "^ ^^' -^ »>^-- ^^^^^ But the doubles of equals are equal to one another. [Ax 6. Therefore the parallelogram BL is equal to the square GB «liAwn ?i!!f f?""® "'n''?®'*^ by joining AE, BK, it can be shewn, that the parallelogram GL is equal to the square CH. n::ferGB%T' '^'"' ^"^^^^ ^ ^^^^ *^^. *-^ ^^^ ^a olX f #^^ described on ^C, and the squares Therefore the square described on the side BC is equal to the squares described on the sides BAy AG. 4 ^ w Wherefore, in any right-angled triangle &c. q e d i PROPOSITION 48. THEOREM. n£fJht '^"^Z described on one of the sides of a tri- wl f ^ v^7/ ^^ *^f ^^^<^res descHbed on the other Two right tngle. contained lyy these two sides is a theYnWIe%'??hf '"V.^^i^^ ^^« ^^ *^« «i<^^« of thLthSmTjf 1/?'*''.^ ^"^ *^ ^^"^'•^^ described on angle ' ^ ^^^"^ ^^^^ '^''^^ ^« » ^'^S^^^i ^ From the point A xiraw ^Z> at nght angles io AG -, [i. n. and make AD equal to BA] [I. 3. and join i>C. _ Then because DA is equd* to BA, the square on DA is equal to the square on BA. To each of these add the square on AG. ^ I^^'^a'^ag'' ^'^'^'"®^ ^"^ ^^' "^^' ^'■^ ^^"^ **" *^® '^"'''''®« 4- .» 4U 52 EUCLID'S ELEMENTS, But because the angle DAG is a right angle, iCcyMtmcti^ the square on 2)Cis equal to the squares on DA, A G. [I. 47. And, byhypothesis,the square on -eCis equal to the squares on BA, AG. ^ Therefore thesquareonDCisequalto the square on BGU^X Therefore also the side DG is equal to the side BG. And because the side DA is equal to the side AB ; \Conztr, and the side AG \^ common to the two triangles DAG, BAG) the two sides DA, AG are equal to the two sides BA, AG, each to each ; and the base DG has been shewn to be equal to the base BG\ therefore the angle DAG is equal to the angle BAG. [I. 8. But DAG is a right angle ; [Construction, therefore also BAG is s, right angle. lAxiom 1. Wherefore, if the square &c. q.e.d. BOOK 11. definitions; ii«M^;n?I^^^f'^^^^?:''^^®'* parallelogram, or rectangle, is wud to be contained by any two of the straight lines which contain one of the rio-hfc ono-ips ^ wmcu .|>„!5V /"j^^^T Pa^'alloloffram, any of the parallelograms SlIedaGnoror *°-'"'"' ^'"^ *''^ *'"' complements, is BOOK 11. 1. 53 Thus the parallelogram HG^ together with the complements AF, FGj is the gnomon, which is more briefly expressed by the let- ters AGK, or EHG, which are at the opposite angles of the parallelo- grams which make the gnomon. PROPOSITIOlSr 1. THEOREM, ^ If there he two straight lines, one of which is divided into any number of parts, the rectangle contained hy the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of (he divided line. Let A and BG be two straight lines ; and let BG he divided into any number of parts at the points D, JE : the rectangle contained by the straight lines A, BG, shall be equal to the rectangle contained by A, BD, together with that contained by A, DE, and that contained by Ay EG, From the point B draw BF at right angles to BG ; [1. 11. and make BG equal to A ; [I. 3. through G draw GIT parallel to BG; and through I), E, G draw !>£:, EL, GB, parallel to BG. [I. 31. Then the rectangle BH is equal to the rectanarles BIT, DL, EH. ^n nr}LTn'^''^^ ^^/'Z^' ^^^ ^* ^» contained by rn ^n '^ f^'J^^^'}'^^ ^7 A ^A for it is contained by GB, BD, and GB is equal toi| ; ^ »^ Zb "f .^^"*^i»^ by A, m, because DK is equal to BG, which IS equal to A-, u 34 and in like manner EH is contained by A, EG. Therefore the rectangle contained by A, BG is eaual to the rectangles contained by ^,^Aandby^:2>!£^;Llby^^ y^^orefore,iftherebe two straight lines &c. q.e.d. M EUCLWS ELEMENTS, r B D i' B PROPOSITION 2. THEOREM, ^Jf ^ f^^^^Sfht line he divided into any two parts th^ ate together equal to the square on the whole line ' INoU. To avoid repeating the word m,ta,n,d too frequentfy, the^reotl^e contained by two straight Hnes AB, AC W|ometoie8 simply oaUed the rectangle And ^i^ is the rectangle contained hv BA AC fnr- if ,•- contained by DA, AC, of which DA is equal' to Si ' '* '' and CE is contained by AB, BC, for J5^ is equal to AB luSf^B *^^^^«ta«§^!e/^, AC, together with the rect- angle AB, BG, IS equal to the square on AB Wherefore, if a straight line &c. q.e.d. ' PROPOSITIONS. THEOREM. at tt'*St C^fhe'i^^tlf le't^t'-^^'r,?? *"" ^f^ BOOK II, 3, 4,. 55 On BC describe the square CDEB-^ produce ED to F, and through A a draw^ jPparallel to CD or BE. [1. 31. ~ Then the rectangle AEih equal to the rectangles AD, CE. But AE.is the rectangle contained by AB, BG, for it is contained hjAB, BE, of which BE ia equal to BC 'y and AD is contained by AC, CB, for CD is eaual to CB • and CE is the square on BC. " * Therefore the rectangle ^5, ^C^ is equal to the rectangle -40, 0/;, together with the square on BC. Wherefore, if a straight line &c. q.e.d. PROPOSITION 4. THEOREM. If a straight line he divided into any two parts, the square on the whole line is equal to the squares on th^ two parts, together with twice the rectangle contained hv the two parts. ^^ vff •i^o Let the straight line AB be divided into any two parts at the point C: the square on AB shall be equal tothn '^Il^jAc'^ci' *'^'*'" "''' *"^^^ the reXgle con! A tPJ?^ ^^ describe the square join BD; through C draw CGF parallel to AD or BE, and through G oxQ,vfHKi^2iYdMe\ioABoYDE. [1.31. Then, because CF is parallel to AD, and BD falls on them the exterior angle GGB is equal to the interior and opposite an- gle ADB ; v. 29. but the angle ADB is equal tc^e angle ABD, ri 5 because BA is equal to AD, being sides of a sqiaro ; therefore the angle CGB is equal to the nn^i e nna . r . . i Mid therefore the side CG is equal to the side CB. ' "n ' Q But CB is also equal to GK, and CG to BK - n \L therefore the figure WJT^ is equilateral * •aiE I'- A 56 IlUCLIiys ELEMENTS, i i [1. 34. and AxUm 1, A. ^ .B ther equal to two right Mirles ' ^^^ «« toge- Butirat7is a right angle " rr 7, <, • ^^ '*' Therefore G(75 is a right angle. ^ ' ^tl'^ '"" And therefore also the ang-Ieg GGir rirn ^ T" these are right angles ' ^^-^ opposite to For the same reason HF is also a square, and it is on the Ztui which IS equal to AG. [I 34 Wore^^^, cSTare the squares piemen^ "^T *' ''•"^''"'*"' ^'^ '« ^""J *<> *« com- And^^,.^;^irsSs*:srjr*-«'^^^'^^^ Therefore the four figures JW r^A^'^r. Spares on ^C7. ^^[Togf^f;^^^?;,^^-^-.-^;^ whicfir'thi^'q-^^i*?f J^l^« "P *•»« ^'«'J« figure ^2)^i?, Sc^no^lhShTwil^^^ oa Wherefore, C^a ,;ra«>A< ;;„, &, 7/^' ««• Corollary Frnm ^k«m . ' that parallellograms aboulllrdhS^^ '' ^^^^^^«t» hkewise squares. ' aumeter oi a square are ---x^oixiur^5. THEOREM. BOOK IL 6. 57 m: unequal parts, together with the square on the line between ttie points of section, is equal to the square on haif the line. mJ^/L*^J straight line AB be divided into two oqu.i S n ^vf ^'^^ ^? ^°^ '""^ *^« ^°^ ."al parts at the ^rn \ *^,V^*angle A£>,DB, together with the square on (7i>, shall be equal to the square on CB. On CB describe the ^ ^ ^ „ square CEFB\ [i. 46. 4^ C T) B ^^^ttJ^^' through i) draw I LI H />^(y parallel to C'i; or /?F- .K through ^draw ^ZJf paral- lel to 05 or i^jP; and through A draw ^^ parallel to CL jc g~T Then the complement Olf is equal to the complement to ?h\' who e i^^^^^^ ^^' *^^^^^^^^ *^« ^Me Ciif is equal But car is equal to ^X, [I 36 because ACU equal to CZ?. [flypoMm,.' Therefore also AL is equal to DF. [^^,<«» 1. to ApkSd C^^ ^^ ^^'' *''*'"«^''™ *« whole ^£is equal -R i. ATT' XI- * [Axiom 2. ^ualto i)i^^ rectangle contained by AD, DB, for 2>^ is and BF together with CZT is the gnomon CMG ; therefore thegnomonC^Gf isequal to the rectangle^D BB To^each of these add LG, which is equal to the square on rn, * - ^, til. 4, Corollary, and I. 34. Wherefore, if a straight line &c. q.e.d. thelnZ™ l^r^"''""" '\ " ""^^'f^^* *•>** *e difference of the squares on two unequal straight Unes AC, CD is eaual to the rectangle contained by.thir sum and differ^iw. 58 UCLIU8 ELEMENTS, PROPOSITION tf. THEOREM. ^ the straight UmwlZhU^^^ f^^^^ ''' part produced, "^ ^^ ^^^ A«(/'a«c? ^A^ togetW with the\qu^;^''*A>*^^^^ AD Db\ square on CD, 7 ^^^ ^® ®^^^^ ^ «io On CD describe the Bquaro (7^^/) j j-j ^g J^^";P^; ^iroiigh B draw ■^^G^ parallel to CE or i-> Ir *^^0S^I^ // draw iTXi^ parallel to ^2) or EF', and through ^ draw j^ v- .. ^^ parallel to CX or Z> J/; ^ * Then, because ^<7is equal to (7^ r^ '^V^* the rectangle AL is emial fr. +i.« x' , {.Hypothesis. but C^ is equal ioi^i;'' *° *'" '""^^'^ ^B; fi. 3s. therefore also AL is eqnal to /flf , , ^^- ^^■ J o each of these add Cilf- ' Umoml. therefore the whole ^i»f isennnl tr. n. But AM is the rSnT. .^T''°^^^•f^»••2• T^fore the rectangle ^i>, Z,^ ;, ,,^ ^'/'tL'SS T^eaeh of these add m, whicbis oq„al to the squalTo^i Therefore the rectangle An 7)7? li!' \^°'"<'"«7> a°d i. 3*. on GB, is eqnal to thi t„1„' f 5i,^*«*/r with the square But the gnomon CMG^nATr ? '^^ *^ ^^''^ ^fi"- ^^/•A ^hichTs «f fq^i^ein C^'^ "P *« '^'•"'^ fi^™ orjIl^tuJrotfXu^?^^^"^^'''- with t^ Wherefore, ir a # may be equal toC^; [pos^, 2. and I. 3. ^J^E'^? describe the square and construct two figures such as in the preceding propositions. Then, because CB is equal to BDy [Construction. and that CB is equal to 6?^, and BD to JTiV therefore GK is equal to ^iV. ' For the same reason PR is equal to RO. And becmise CB is equal to BD, and (?A^ to ^iV; the rect- omfSrelogram CO'.''"""' *^^"-'' '^« co„,p,e»e„te therefore also BN is equal to GR. j-^^.^ ^^ Therefore the four rectangles BN, CK, GR. RN&re eniml themCrr '' "^^ '' *^' ^'"^ ^^^ 4uaSuple of onHf [I. 34. [Axiom 1. Again, J)ecause CB is equal to ^i), and that HJJ is equal to BK, that is to CG, and that 05 is equal to GK, [ConstruciiorL [II. 4, Corollary, [I. 34. [1.34. BOOK IL 8. 61 parts, four ? and one of her part, it is made up ij two parts W, together [uax© on the B D K A [I. 34. [Axiom 1. ^, the rect- > rectangle [I. 36. nplements [I. 43. [Axiom 1. are equal of one of instruction. Corollary, [I. 34. [1.34. that IS to OP; [TL 4, Corollary, therefore CO is equal to OP, [Axiom 1. And because CO is equal to OP, and PR to /20, the rectangle AG'\% equal to the rectpngle MPy and the rect- angle PL to the rectangle /2i^. [I. 36. But i»fP is equal to PL, because they are the complements of the parallelogram ML ; [I. 43. therefore also ^G^ is equal to RF. [Axiom 1. Therefore the four rectangles AG, MP, PL, RF are equal to one another, and so the four are quadruple of one of them AG. And it was shewn that the four CK, BN, OR and RN are quadruple of CK-, therefore the eight rectangles which make up the gnomon AOH are quadruple of AJC. And because ^A' is the rectangle contained by AB, BO, for BK is equal to BO ; therefore four times the rectangle AB, BO is quadruple of^^. But the gnomon AOH was shewn to be quadruple of^A; Therefore four times the rectangle AB, BO is equal to the gnomon AOIL. [Axiom 1. To each of these add Xlf, which is equal to the square on AC. [II. 4, Corollary, and I. 34. Therefore four times the rectangle AB, BO, tof'ether with the square on -4(7, is equal to the gnomon AOH and the square XH. But the gnomon AOH and the square XH make up the figure AEFD, which is the square on AD. Therefore four times the rectangle AB, BO, together with the square on AC, is equal to the square on AD, that is to the square on the line made of AB and BC together. WhereforOj if a straight line &c. q.e.d. 62 EUCLWS ELEMENTS, I! H PROPOSITION 0. THEOREM. ^ \ ya straight line he divided into tiro equal and alto into two unequal parts, the squares on thi two uneauai parts are U>gether double of the square on fmlf Ih^uZ and of the squareon th^ li^te hetwL theZr^fttZ From the point C draw CE at right angles to ^ i?, [ r. 1 1 . and make it equal to AC or ^'^» [I. 3. and join EA, EB ; through D draw DF parallel to CE, and through E draw FG parallel *o ^-^ ; [I. 31. and join ^Z: Then, because AC is equal to CE, [Construction. the angle EACia equal to the angle AEC. [i 5 And because the angle ^C.^ is a right angle, [Construction. righ^^tr ''"^^'' ^^^ ^^^^'' *'^'"^'' "^"^ *^ r and they are equal to one another ; therefore each of them is half a right angle. f ?ighUngTe' ''^'''' '^'^ '^*^' ^^'^"^ ^^^' ^^^ i« l^alf Therefore the whole angle AFTB is a right angle. And because the angle G^^i^ is half a riffht an^lo an^ therefore the remaining angle EFG is half a right angle. Therefnrfi flio annrlA /nrrrrr i t J .1 - _ _ the side Ed is equal to the side GF. Vf^ Again, because the angle at B is half a right Migle, and the BOOK IL 9. 63 angle FDB a right angle, for it is equal to the interior and opposite angle ECB ; [I- 29. therefore the remaining angle BFD is half a right angle. Tlierefore the angle at ./J is equal to tlie angle BFDy and the side DF is oquul v J ^ . • side DB. [I. 6. And because ^ C is r .. ' ^ OE^ [Cmstruction. the square on AG is equal to .lie square on CB ; therefore the squares on /v tB are double of the square on AG. 33 ut the square on ^^ is equal to the squares on AG, GBf because the angle AGE is a right angle ; [I. 47. therefore the square on ^jG' is double of the square on AG. Again, because EG is equal to GF, [Construction, the square on EG is equal to the square on GF; therefore the squares on EG, GF are double of the square on GF. But the square on ^i^is equal to the squares on EG, GF, because the angle EGF is a right angle ; [I. 47. therefore the squar c on EF is double of the square on GF. And GF is equal to CD ; [I. 34. therefore the square on ^jPis double of the square on GD. But it has been shewn that the square on AE is also double of the square on AG. Therefore the squares on AE, EF are double of the squares on AG, GD. But the square on ^jP is equal to the squares on AE, EF, because the angle AEF is a right angle. [I. 47. Therefore the square on AF is double of the squares on AG,GD. But the squares on AB, DF are equal to the square on AF, because the angle ADFi^ a right angle. [I. 47. Therefore the squares on AD, DF are double of the squares on -4(7, CD. And Di^is equal to DB ; therefore the squares on AD, DB are double of the squares on AG, GD. Whorofoii?, if a straight line &c. q.e.d. 4^ 64 EUCLID'S ELEMENTS, PROPOSITION 10. THEOREM, If a straight line he bisected, and produced to any pointy the square on the whole line thu^ produ>ced, and the square on the part of it produced, are together double of th square on half the line bisected and of the square on the line made up cfthe half and the part prodmed. Let the straight hne AB he bisected at (7, and pro- duced to £> : the squares on AD, DB shfll be together double of the squares on A (7, CD, From the point (7draw CE2X right angles to AB, [I. 11, and make it equal io AG or CB; [1.3. andjoin^^, EB ; through E draw EF parallel to ABy and through D draw DF psLrAlel to CE. [1.31. Then becaifse the straight line j&iPmeets the parallels EC, FD, the angles CEF, EFD are together equal to two right angles ; [j, 29. and therefore the angles BEF, EFD are together less than two right angles. Therefore the straight lines EB, FD will meet, if produced, [Axiom 12. IConsiruction. [1.5, [Construction. towards B, D, Let them meet at G, and join AG. Then because ACis ecjual to CE, the angle CEA is equal to the angle EAC; and the angle ACE is a right angle ; therefore each of the angles CEA, EAC is half a right angle. [1. 32. For the same reason each of the angles CEB, EBC ia half a right angle. a right angle. EBC is half a right angle, right angle, for they are verti- [L 15. but the angle BDG is a rigu;. angle, because it is equal to the alternate angle DCE ; [I. 29. therefore theremainingangl©i)aj5ishalf aright angle, [1.32, Therefore the angle AEB And because the an the angle DBQ is also ha callj opposite; d to any iced, and er doitbk he sqitare duced. and pro- together BOOK 11. 10. 65 lal to two [1. 29. ther less iroduced, Axiom 12. istruction. [1.5. istruction. ' a right [I. 32. is half it angle, re verti- [I. 15. equal to [I. 29. :le,[1.32. and is therefore equal to the angle DBG; therefore also the side BD is equal to the side DG, [i. e. Again, because the angle BGF is half a right' anele and the angle at F b. right angle, for it is equal to the opposite angle £CD ; ^j 3^ therefore theremaininganglei^^(5^ishalf aright angle, [1. 32.' and is therefore equal to the angle UGF; therefore also the side GF is equal to the side FB. '[1. 6. And because BO is equal to CA, the square on BC ia equal to the square on CA ; therefore the squares on BC, CA are double of the square on CA. ^ But the square on ^^is equal to the squares on BC, CA. [1. 47. Therefore the square on ^ JE^ is double of the square on AC, Again, because GF is equal to FB, the square on GF is equal to the square on FB ; therefore the squares on GFy FB are double of the square on FB. ^ But thesquare on EG is equal to the squares on GF, FB,[I.i7. Therefore the square on BG is double of the square on FB. And FB is equal to CD ; [I. 34] therefore the square on BG is double of the square on CD. But it has been shewn that the square on AB is double of the square on AC. Therefore the squares on AB, BG are double of the squares on ^(7, CD. But the square on AG' ia equal to the squares on AB, 5^- [I. 47. Therefore the square on AG is double of the squares on AO, OD. But the squares on AD, DG are equal to the square on Therefore the squares on AD, DG are double of the squares on ^(7, (72). And DG is equal to DB ; therefore the squares on AD, DB are double of the squares on A 0^ CD.. y^^horefore, if a straight line &c, q.b.d, ; 6 m- I 66 EUCLID'S ELEMENTS, PROPOSITION 11. PROBLEM, To divide a given straight line into two parts, so that the rectangle contained by the whole and one qf the parts may he equal to the square on the other part. Let ABhQ the given straight line : it is required to divide it into two parts, so that the rectangle contained by the whole and one of the parts may be equal to the square on the other part. On AB describe the square ABDC; [r. 46. bisect ^ (7 at ^; [I. lo. join BE ; produce CA to F, and make EF equal to EB ; [I. 3. and on ^^ describe the square AFGH. [1. 46. AB shall \q divided at H so that the rectangle AB, BH is equal to the square on AH, Produce GHio K. Then, because the straight line ^C is bisected at E^ and pro- duced to F, the rectangle GF, FA, together with the square on AEy is equal to the square on EF. [II. 6. But EF is equal to EB, ICotustructhn. Therefore the rectangle OF, FA, together with the square on AEj is equal to the square on EB. But the square on EB is equal to the squares on AE, AB, because the angle EAB is a right angle. [I. 47. Therefore the rectangle OF, FA, together with the square on AE, is equal to the squares, on AE, AB. Take away the square on AE, which is common to both • therefore the remainder, the rectangle C/;/!^, is equal to the square on AB, ' [Axiom 3. But the figure FE' is the rectangle contamed by CF, FA, for FG i» equal to FA; and AD is the square on AB ; therefore FK is equal to AD. Take away the common imrt A fC. and th— ?**>mgin??««» 1P}T ia equal to the remainder HD, " ' ""~ [AxUmi, ■\. BOOK IL 11,12. 67 f, »o that the parti quired to tained by he square with the [II. 6. nstruction. le square AE, AB, [I. 47. le square to both • } equal to [AxUm> 3. CFy FA, [Axiom 3. But HD is the rectangle contained by AB, BH, for AB is equal to BJ) ; and FH is the square on ^^H"; therefore the rectangle ^5, jff^is equal to the squareon AH. Wherefore the straight line AB is divided at H, so that the rectangle AB, BH is equal to the square on AH. q.b.f. PROPOSITION 12. THEOREM. In obtuse-angled triangles, if a perpendicular he drawn from either of the acute angles to the opposite side pro- duced, the square on the side subtending thedbtuse angle is greater than the squares on the sides containing thedbtuse angle, by twice the rectangle contained by the side on which, when produced, the perpendicular falls, and the straight line intercepted without the triangle, between the perpendicular and the obtuse angle. Let ABC be an obtuse-angled triangle, having the obtuse angle ACB, and from the point A let ^Dbe drawn perpendicular to BC produced : the square on AB shall be greater than tlie squares on AG, CB, by twice the rectangle BCy CD. Because the straight line BD is divided into two parts at the point C, the square on BD is equal to the squares on BC, CD, and twice the rectangle BC, CD. [IL 4. To each of these equals adt the square on DA. Therefore the squares c DD, DA are equal to the squares on BC, CD, DA, find twice xhe rectangle BC, CD. [Axiom 2. But the square or BA is equal to the squares on BD, DAi because the angle ": D is a right angle ; [i, 47. and the square or j .i is equal to the squares on CD, DA. \i. 47. Therefore the square jn BA is equal to the squares on BC, CA, and twice the rectangle BC, CD ; that is, the square on BA is greater than ti«e squares oa Wherefore, in obtuse-angled triangles &c. q.e,d. 6—2 68 EUCLID'S ELEMENTS, PROPOSITION 13. THEOREM, ^ In every triangle, the square on the aide subtending an acute angle, is less than the squares on the sides con- taining that angle, by ttcice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle, and the acute angle. Let ABC be any triangle, and the angle at B an acute angle; and on BC one of the sides containing it, let fall the perpendicular AD from the opposite angle : the square on AC, opposite to the angle B, shall be less than the squares on GB, BA, by twice the rectangle CB, BD. First, let AD faU within the triangle ABC, Then^ because the straight line CB IS divided into two parts at the point D, the squares on CB, BD are equal to twice the rectangle contained by CB, BD and the square on CD. [II. 7. To each of these equals add the square on i>-d[. Therefore the squares on CB, BD, DA are equal to twico the rectangle CB, BD and the squares on CD, DA. [Ax. 2. But the square on AB is equal to the squares on BD, DA, because the angle BDA is a right angle ; [I. 47* and the square on ^(7 is equal to the squares on CD,DA. [1.47. Therefore tie squares on CB, BA are equal to the square on AC and twice the rectangle CB, BD ; that is, the square on AC alone is less than the squares on CB^ BA by twice the rectangle CB^ BD, Secondly, let -42) fall without the triangle ABC* Then because the angle at D is a right angle, [Oomtructim, the angle ACB is greater than aright angle J " [I. 16. hiending idea con- lined by tercepted opposite an acute ), let fall le square than the to twice 4. [Ax. 2. 3D, DA, [I. 47. M.[1.47. e square BOOK II. 13,14. '%. 69 and therefore the square on AB is equal to the squares on AC J GBy and twice the rectangle BC, CD, [II. 12. To each of these equals add the square on BG. Therefore the squares on AB^ BG are equal to the square on AC, and twice the square on BCy and twice the rect- angle BG, CD. [Axiom 2. But because BD is divided into two parts at G, the rect- angle DB, BG is equal to the rectangle BG, CD and the square on -BC; [II. 3. and the doubles of these are equal, that is, twice the rectangle DB, BG is equal to twice the rectangle BG, CD and twice the square on BG. Therefore the squares on AB, BG are equal to the square on AG, and twice the rectangle DB, BG ; that is, the square on -4 (7 alone is less than the squares on AB, BG by twice the rectangle DB, BC Lastly, let the side AG he perpendicular to BG Then BG is the straight line between the perpendicular and the acute angle at B ; and it is manifest, that the squares on AB, BG are equal to the square on AG, and twice the square on BG. [I. 47 and Ax. 2. Wherefore, m every triangle &c. q.e.d. PROPOSITION 14. PROBLEM. To describe a square that shall be equal to a given recti- lineal Jigure. Let A be the given rectilineal figure : it is required to describe a square that shall be equal to A. Describe the rect- angular parallelog" *ii^ BvDBeqvLdil tothe re o tilineal figure A, [1. 4^. Then if the sides of it, BjE, BD are equal to one another, it is a square, and what was req"ired is now dona t 70 EUCLII/S ELEMENTS, it; But if they are not equal, produce one of them BE to r make EE equal -« w ^, to ED, [I. 3. And bisect BE at G J [I. 10. from the centre (r, at the distance GB, or GF, de- scribe the semi- circle ^^JP, and produce Djfi^ to ^. The square described on EH shall be equal to the given rectilineal figure A. ^ Join Gir. Then, because the straight line BF is divided into two equal parts at the point G, and into two unequal parts at the point Ey the rectangle BE, EF, together with the square on GE, is equal to the square on GF, [II. 5 But GF is c^qual to GH, Therefore the rectangle BE, EF, together with the square on GE, is equal to the square on GH. But the square on GH\% equal to the squares on GE, EH-JilAI, therefore the rectangle BE, EF, together with the square on GE, is equal to the squares on GE, EH, Take away the square on GE, which is common to both ; therefore the rectangle BE, EF is equal to the square on ^^' [Axkm 3. But the rectangle contained by BE, EF is the parallelo- gram BD, because EF is equal to ED. [Cir/»T^ . i* io «.« : J x_ a. 1 .i centre ^ — -.— ^. ^v xa aw-^uucu w> auu ids BOOK IIL 1. 73 Draw within it any straight line AB^ and bisect AB at Z) ; [1. 10. from the point D draw DG at right angles to AB ; [I. il. produce CD to meet the cir- cumference at Ey and bisect GE at F. [I. 10. The point F shall be the centre of the circle -4 J5a For if F be not the centre, if possible, let G be the centre ; and join GA, GD, GB, Then, because DA is equal to DB, [Construction. and DG is common to the two triangles ADG, BDG ; the two sides AD^ DG are equal to the two sides BD, DG, each to each ; and the base GA is equal to the base GB, because they are drawn from the centre G ; fi. Definition 15. therefore the angle ADG is equal to the angle BDG. [I. 8. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle ; [I. Definition 10. therefore the angle BDG is a right angle. But the angle BDF is also a right angle. [Construction. Therefore the angle BDGia equal to the angle BDFj [Ax. 11. the less to the greater ; which is impossible. Therefore G is not the centre of the circle ABO. In the same manner it may be shewn that no other point out of the line CB is the centre ; and since C^ is bisected at F, any other point in CB divides it into unequal parts, and cannot be the centre. Therefore no point but F is the centi e ; that is, F is the centre of the circle ABO: which was to be found. From yiis it is manifest, that if in a circle Corollary. a straight line bisect j^jfther at right angles, the centre of the circle is in the straight line which bisects the other. 74 EUGLIiyS ELEMENTS, PEOPOSITION 2. THEOREM. If any two points be taken in th^ circumference qf a circle^ the straight line which joins them shall fall within the circle. Let ABC be a circle, and A and B any two points in the circumference : the straight line drawn from ^ to ^ shall &11 within the circle. For if it do not, let it fall, if possible, without, as AEB. Find D the centre of the circle ABC ; [III. 1. and join DAy DB ; in the arc AB take any point F^ join DF^ and produce it to meet the straignt line AB at E. Then, becaiuse DA is equal to DBy [I. Definition 15. the angle DAB is equal to the angle DBA. [I. 6. And because AE^ a side of the triangle DAE^ is pro- duced to Bj the exterior angle DEB is greater than the interior opposite angle DAE. [i. 16. But the anglei>^^was shewn to beequal to the angle DBE; therefore the angle DEB is greater than the angle DBE. But the greater angle is subtended by the greater side ; [1. 19. therefore DB is greater than DE. But DB is equal to DF; [I. Definition 15. therefore DF is greater than DE, the less than the greater ; which is impossible. Therefore the straight line drawn from ^ to .5 does not fall without the circle. In the same manner it may be shewn that it does not fall on the circumference. Therefore it falls within the circle. "Wherefore, if any two points &c. q.e.d. PROPOSITION 3. THEOREM. If a straight line drawn throiMkdhe centre of a circle. &'9«M/* » ••# iiiyfii itiw •*» n Wfiic/h nwMi uoi poss i/trotu/h ih3 BOOK IIL 8. u ice qf a ' within oints in A U>B [1.6. is pro- ban the [I. 16. )DBE; BBE, 9; [1. 19. ition 15. greater; oes not oes not % circle^ uffh the centre, it ihall cut it at right angles; and if it eui it at right angles it thall bitect it. Let ABC be a circle ; and let CD, a straight line drawn through the centre, bisect any straight line^J5, which does not pass through the centre, at the point F: CD shall cut AB at right angles. Take E the centre of the circle ; andjoin^^, EB. [III.l. Then, because ^i^is equal to FB, [ffypothasv and FE is common to the two triangles AFE^ BFE ; the two sides AF, FE are equal to the two sides BF,^E, each to each ; and the base EA is equal to the base EB; [I. D. IMAGE EVALUATION TEST TARGET (MT-3) /. 1.0 I.I 11.25 m, lu no I: 9 2.2 £ LS i2.0 I 1.4 Hill 1.6 V] '/J >^ ^c^l ^^/ ^^^ ^%^^ M <$> ■^f/ Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. MS80 (716) 872-4503 •%^ #p; z ^ ^ o !^ 76 EUGLIUS ELEMENTS. [III. 1. PROPOSITION 4. TffEOnEM, If in a circle ttco straight lines cut one andtheVf ichich do not both past through the centre, they do not bisect one another. Let ABOD he a circle, and AC, BD two straight lines in it, which cut one another at the point ^, and do notl>oth pass through the centre : AC, BD shall not bisect ono another. If one of the straight lines pass through the centre it is piain that it cannot be bisected by the other which does not pass through the centre. But if neither of them pass through the centre, if possible, lot AE be e^ual to EG, and BE equal to ED. Take F the centre of the circle and join EF, Then, because FE, a straight line drawn through the centre, bisects another straight line A O which does not pass thiuugh the centre; [Hypothesis, FE cuts -4 C at right angles ; [III. 3. therefore the angle FEA is a right angle. Again, because the straight line FE bisects the straight line BD, which does not pass through the centre, \Hyp. FE cuts BD at right angles ; [IIL 3. therefore the angle FEB is a right angle. But the angle FEA was shewn to be a right angle ; therefore the angle FEA is equal to the angle FEB, lAx, 11. the less to the greater ; which is impossible. Therefore AG, BD do not bisect each other. Wherefore, (/* in a arcfo&c. q.e.d. PROPOSITIOi 5. THEOREM. If two circles cut one another, they shall not have the same centre. TiAr. ^.rtA f.nr/\ MiMnlAa «=s-wsp r-ii-»r VTTV «^i.VSwS /yT»rf -, •0^.0^ Mr ««*««( V22v S^^Xl.^^ j>X XL. BOOK IIL 6, 6. T7 pointa By C: they shall not haye the aime centre. For, if it lie possible, let E be their centre ; join EOy and draw any straight line EFO- meeting the circumferences at FmdG, Then, becaiise S is the cen- tre of the circle ABCy EC is equal to EF. [I. Bt^nitUm 15. Again, because E is the centre of the circle CDGy EC is eqml to EG, [I. D^nitwn 15, But EC was shewn to be equal to EF ; therefore EF is equal to EG, [Axiim 1. the less to the greater; which is impossibla Therefore E is not the centre of the circles ABC, CDG, Wherefore, ifttix) circles &c Q.E.D. PROPOSITION 6. THEOREM, If two circles touch one another internally^ they shall twt have the same centre. Let the two circles ABC, CDE touch one another inter- nally at the point C : they shall not have the same centre. For, if it be possible, let F be their centre ; join FC, and draw any straight line FEB, meeting the circum< ferences at E and B, Then, because F is the centre of the circle ABC, FC ia equal to FB. [I. Di^. 16. Again, because F is the centre of the circle CDE, FC is equal to FE, But -FC was shewn to be equal to FB ; therefore FE is equal to FB, the less to the greater ; which is impossible. Therefore F is not the centre of the circles ABC, CDE, [I. JD^nition IS, {Axiom h rj> «..^ ^.* j^^ M.^ ^ « ^ ; y >r«»KV v» '^•«ap %r. w If EUCLID'S ELEMENTS, PROPOSITION?. THEOREM. V .••^?^?^^*^*^ ^ ^^J^ *^ *^ diameter of a cirde which u not the centre, qf all the straight lines iM^Znhe drawn from this point to ths circuwf<^ZcT^ gZtJt ^tUtjn which the centre is, and tjL oth^ p^rtofZ n^ll ^^ ^f^^;. aM cf any others, tLwhihU nearer to thestraight line t-^hich posset through thecentre t^^ways greater than one mcriremote; 7ndfromm Ifr^JTr ^^^''f^}^ drawn to the drcumferlZ two •trawht lines, and only two, which are equal to one anl ther, one on each side qf the shortest lin^ Let ABCD be a circle and AD ita diameter in whiVii et any pint F be taken which is not the S ' ?et^ be the centre : of all the straight lines FB, FaFG &c tha? can be drawn from ^ to the drcumferencef Ef whkh S«^ *Jroiih E, shall be the neatest, and >i>, ffie oSer mt of the diameter AD, shall be the least • and of ^I others FB shall be greateV than FC, and FCi^FG Join BE, CE, GE, Then, because any two sides of a triangle are greater than the third side, p. 20. therefore BE, EF are greater thanJ5i^. But BE is equal to ^ JF j [l.lkf, 15. therefore AE, EF are greater than BF, that is, ^^ is greater than BF, Agdn, because BE is equal tb CE, {l, B^fUtian 15. and ^^18 common to the two triangles BEF, CEF- but the angle ^^^is greater than the angle CEF; therefore the base FB is greater than the base FC, [I. 24 tha?!^^'"® ""^^^ ** ""^y ^ '^^^ ^^** ^^ ^ greater Afirain. becaufM^ f^Jf wm i^M^^ i.\^^^ m^v •>* ^A -■ O* »»»*S«v. Therefore FA is the greatest^ and FD the least of all the g' raight lines from F to the circumference ; and FB is greater tl in jFC, and FC than jP(?. Also, there can be drawn two equal straight lines from the point F to the circumference, one on each side of the shortest line FD, For, at the point E, in the straight line EF, make the angle FEH equal to the angle FEG, [I. 23. and join ^iT. Then, because EG is equal to EH^ [I. D^nition 15. and EF is common to the two triangles GEF, HEF; the two sides EG, EF are equal to the two sides Eff, EF, each to each ; and the angle GEF is equal to the angle HEF ; ICmatr. therefore the base FG is equal to the base FH. [I. i. But, besides FH, no other straight line can be drawn from ^to the circumference, equal to FG. For, if it be possible, let FK be equal to FG. Then, because /X is equal to -F(t, ' Iffppotheni, and jPjfiTis also equal to FG, therefore Fff is equal to FJC ; [Axiom 1. that is, a line nearer to that which passes through the centre is equal to a line which is more remote ; which m impossible by what has been already shewn. Wherefore, if any point be taken &c. q.e.d. PROPOSITION 8. THEOREM, If any point he takevt, without a circle, and straight lines be drawn from it to the circumference, one qf which passes through the centre; of those which fall on the con- cave circumference, tJie greatest is that which parses through the centre, and of the rest, that which is nearer to the one passing through the centre is always greater man one mvre remote ; Out of those which fall on ih€ m EUCLXiyS ELEMENTS. eonvM circun^ference, the least it that between the point without the circle and the diameter; and c^ the rest, that which is nearer to the least is always less than one more remote; and from the same point there can he drawn to ths circumference two straight lines, and only two, which are equal to one another, one on ecmh side of the shortest line. Let ABC be a circle, and D any point without it, and from D let the straight lines DA, I>E, DF, DC be drawn to the circumference, of which D^ passes through the centre: of those which fall on the concave circumference AEFC, the greatest shall be DA which passes through the centre, and the nearer to it shall be greater than the more remote namely, 2>^ greater than DF, and DF greater than DG\ but of those which fall on the convex circumference GKLH the least shall be DG between the point D and the dia^ meter AG, and the nearer to it shall be less than the more remote, nan^ely, i>iriess than Z>Z, and DL less than DH, Take M, the centre of the circle ABC, [III. i. and join ME, MF, MC, MH, ^ ML,MK. Then, because any two sides of a triande are greater than the third side, [I. 20. therefore EM,MD bxq greater thsm ED. H. But^jlf is equal to AM-, \l.Def. 15. therefore AM, MD are greater than ED, that is, ^2> is greater than ED, Again, because EM is equal' to i^Jf, and MD'is common to the two triangles EMD, FMD ; the two sides EM,MD are equal to the two sides FM, MD, each to each ; but the angle EMD is greater than the angle FMD ; therefore the base ED is greater than the base FD. [1. 24. In the same manner it may be shewn that FD is ffreater than CD J BOOK IIL a SI Therefore DA is the greatest^ and DE gpreater than DF^ and DF greater than DC, Again, because MKy KD are greater than MD^ [I. 20. and MK is equal to MB^ [I. DeJinUi<^^ 16. the remainder KD is greater than the remainder GD, that is, GD is less than KD, And because MLD is a triangle, and from the points Jf, />y the extremities of its side MDy the straight lines MK, DK are drawn to the point K within the triangle, therefore MK, KD are less than MZ, LD ; [L 21. and MK\& equal to ML ; [I. Dejinitvm, 16. therefore the remainder KD is less than the remainder LD^ .In the same manner it may be shewn that LD is less than HD, Therefore DG is the least, and DiTless than DL, and DL less than DH, Also, there can be drawn two equal straight lines from the point D to the circumference, one on each side of the least line. For, at the point Jf, m the straight line MD^ make the angle DMB equal tc the angle DMK^ [I. 23, and join J9i?. Then, because MK is equal to MB, and MD is common to the two triangles KMD, BMD ; . the two sides JOf, MD are equal to the two sides BM^MD, each to each ; and the angle DMK ia equal to the angle DMB ; [Constr, therefore the base DK ia equal to the base DB, [I. 4. But, besides DB, no other straight line can be drawn from D to the circumference, equal to DK, For, if it be possible, let DN be equal to DK. Then, because DN is equal to DK, and DB is also equal to DK, therefore DB is equal to DN; [Axiom 1. that is, a line nearer to the least is equal to one which is more remote; which is impossible by what has been already shewn. WherfififtrA. »V/» t^fm i\ « Vk 6 mfu. 83 BUCLIiyS ELEMENTS. \i PBOPOSITION 9. THEOREM, If a point be taken within a circle^ from which then fall more than two equal straight lines to ttie circum- ference, that point is the centre of the circle. Let the point D be taken within the circle ABCy from which to the circumference there fall more than two equal stiuight lines, namely DA, DB, DC: the point D shaU be the centre of the circle. For, if not, let E be the centre ; join 2>^and produce it both ways to meet the circumference at F and & ; then FG is a diameter of the circle. Theuj because in FG, a diameter of the circle ABC^ the point D is taken, which is not the centre, DG is the greatest straight line from D to the circumference, and DG is greater than DB, and D3 greater than DA ; [IJI. 7, but they are likewise equal, by hypothesis j which is impossible. Therefore E is not the centre of the circle ABC. In the same manner it may be shewn that any other point than D is not the centre ; therefore D is the centre of the circle ABC. , y^YiQVQtoxQ^ if a point be taken &Q. q.e,d. PROPOSITION 10. THEOREM. One circumference of a circle cannot cut another at more than two points. If it be possible, let the circumference ABC cut the circumference DEF at more than two points, namely, at the points B, Gy F. Take K, the centre of the circle ABC, [HI. 1. and join KB, KG, KF. Then, because iT is the centre of the circle ABC^ ' BOOK IIL 10, U. u therefore KB, KG, KFmq all eqaal to each other. [I Dtf.iB fj^t wh?^? within the circle DEF, the point K is taken! from whic.^ to the circumference DBF f^more thwrt^ r^eWL^i)!^' ^^' ^^' ^^'^'^^ ^}^ But ^is also the centre of the circle ABG. [Cmstructin' Therefore the same point is the centre of two circles which cut one another; wrues which is impossible. pjjj ^ y^her-iore, one circun\ference &c, Q.a.D. PROPOSITION 11. THEOREM. nElT- ^t^^^^ ^^H^^ ^^ another internally, the straight ihrr^^ht^''''^ i^r ^''*^^^' ^''""ff prod Jed, shall fZ through the point qf contact. • "'^^ t^aiw oVJlf *'"''. ''i''^?* ^:?,^» ^-^^ *«"cJ> one another inter- . ^or, if not, let it pass otherwise, if possible, as FGDH, and join •a.F,AQ; Then, because AQ^ GF are greater than AF, [i. 20. and AF is equal to HF, [I. Bef. 15. therefore -4(y, G^i^, are greater than HF. Take away the common part GF; ^th^fore the remainder AG is greater than the remainder But -46? is equal to 2>(y. ft 7i^«-v tr Therefore the straight line which joins the Domts F a b^g^roduced, cannot pass otherwise th^ CglTthe that is, it must pass through A. , V iioQ mrdes «c. Q.K.D, ,f 6~? txri ,. #- 84 EUCLIUS ELEMENTS,- PROPOSITION 12. THEOREM, '^ Iftioo circlet touch one another externally^ the Hraight line which Joine their centres shall pan through the point qf contact. Let the two circles ABC, ADE touch one another ex- ternally at the point A ; and let F be the centre of the circle ABC, and G the centre of the circle ADE\ the straight line which joins the points F^ O, shall pass through the point A, For, if not, let it pass otherwise, if pos- sible, as FODO, and join FA, AG, Then, because i^ is the centre of the cir- cle ^^(7, /!ii is equal to FC; [LBef.lS, and because G is the centre of the circle ADE, GA is equal to GD ; therefore FA, AG are equal to FC, DG, lAxiom 2, . Therefore the whole FG is greater than FA, AG, But FG is also less than FA, AG; [I. 20. which is impossible. Therefore the straight line which joins the points F, G, cannot pass otherwise than through tht? point A, that is, it must pass through A. Wherefore, if two circles &c q.e.d. PROPOSITION 13* THEOREM. One circle cannot touch another at more points t/tan * wie, whether it touches it on the inside or outside, - For, if it be possible, let the circle EBF touch the circle ABC at more points than one ; and first on the inside, at the points B, D, Join BD, and draw GH\Aw^ Ing BD at rignt angles. [1. 10, 11. Then, because the two points ^, 2> are in the circum- ference of each of the circles, the straight line BD fidls within each of them ; Ifll. 2. BOOK III la a5 raigJU ' point ler ex- of the Hx the irough ctom 2. [I. 20. than' h the >n the bisect- 10, 11. ircum- rii. 2. ..J^ and thereforo the centre of each circle is m the straight hne OH which bisects BD at right angles ; [III. l, c^i, therefore G^iT passes through the point of contact [III. 11. But GH does not pass through the point of contact be- cause the points B, D are out of the line QH\ which is absurd. Therefore one circle cannot touch another on the inside at more points than one. Nor can one circle touch an- other on the outside at mo:e points than one. For, if it be possible, let the circle -4 C^ touch the circle ABG at the points -4, 6*. Join AC, Then, because the two points ^, (7 are in the circumference of the circle A GK, the straight line A G which joins them, falls within the circle AGK] [in. 2. but the circle AGK is without the circle ABC) [Hypothesta, therefore the straight line AGh without the circle ABG. But because the two points A, G are in the circumference of the chcle ABG, the straight line -4 C7 falls within the circle ABG \ [III. 2. which is absurd. Therefore one circle cannot touch another on the outsido at more points than one. And it has been shewn that one circle cannot touch another on the inside at more points than ono* "WheretoTo. one ctrcU i^-z o.b.i>* 8d EUCLID'S ELEMENTS. PROPOSITION 14. THEOREM, f ■ 1 f^-®" **'® centre of the and from E draw ^Z", ^-fi! per. pendiculars to ^^, 6'i»; fi,]2, and join ^^(jEO. Then, because the straieht Bne -BJf pajBing through the ^°*«'. «»ts the straight Itoe AB, whtch does not pass through the centre, at right angles, it aTso bisects it ; rrTT , StTtTv^" '^'^ *° ^^' ■«•» ^^ ^^ is equal to CG-^ ^ ^® *«"*'« »» ^». Srsq'S^r^T^l! «'1»'^« °« ^-S « eq^a to the re- ^d Wore the straight line EF is equal to th/s^ht But straight lines in a circle are said to be equaUy distant thi BOOK m. 14, IS. 87 Wore AB, CD are equally distant f Jtte ^^S" squares on EQ, ^^. ^^''^'^^'' ^^^ ^^» ^^ are equal to tho therefore the remaining rnu^e on WA K. ^ , T^f ***^' maining square on(?^f ^ ^'^ ^-^ ^* ®^^ *^ ^-^ re- and therefore the stnught Hue AF is equal to the^sShi to^5 was shewn to be double of ^4^, and CD double Therefore ^5 is equal to Ci). r, . ^ . Wherefore, .g^^a/,^rae>A^ //;,,, &e. q.e.d. "^ PROPOSITION 15. THEOREM ¥-^m ,rea^r[&i^^^ centre is 19 marer to the centre thanthllesT ' ^ ^''^''^^^ AD shall be^i^er thL ''*^*^^'"^^^^ isnot a diameter, and BC shall be greater than FQ, JpJ'^^J^^ ^®"*^® ^ draw »/?' J^ perpendiculars to 7.^^' [1.12. ^^^ioiiiEB,EG,FF, f^ j?S®"'l^2?^® ^^ « equal toJ9^,and^/)to^C;[I.i)e/.l5. 1 « 83 EUCLWS ELEMENT,^ [I. 20. -A^B but BE, ECuxe greater than BG; therefore also A J) is greater than BO, And, because BC is nearer to the centre than FG^ [Hypothesis, EH is less than EK, [III. Def, 5. Now it may be shewn, as in the preceding proposition, that BG 13 double of BIi , and FQ double of FK, and that the squares on EH, HB are equal to the squares ^^-^ are equal to two right angles; which is impossible. ^ [i 17 Therefore it must fall without the circle, as AE, .r^ot^^l ^,®*^T?r ^^ ^ivm^U line AE and the' circumfer- ence, io straight line can be drawn from the point A, which does not cut the circle. ^ •'^,*ruiuu For, if possible, let ^i^ be between them; and from the centre D draw DG perpendicular to AF\ [i. 12. let DG meet the circumference at H. Then, because the angle DGA is a right angle, [Comtmction. the angle DAG is less than a riffht angle; ^j «17 therefore DA is greater than i)^'. [I.19, But DA is equal to DH\ *ilS!f ^rif^.!?.?'«''t«'-««»' -DG", the less than the greater: 5^bSw^n"4i^n!f!?H ^^°-^ "^^^ ^" ^^^ fr^«> the point drclt circumference, so as not to cut the [I. D^nition 15. 9a EUCLID'S ELEMENTS. Wherefore, ths straight line &c. q.e.d. CoROLL^Y. From this it is manifest, that tae straight toe which IS drawn at right angles to the diameter of a circle from the extremity of it, touches the circle ; [III.iV.2. and that it touches the circle at one point only, because if it did meet the circle at two points it would fall witmn It. riTT 9 Also it is evident, that there can be but one straight line which touches the circle at the same point. PEOPOSITION 17. PROBLEM, "t To draw a straight line from a given point, either ^r^e ^^ *^ circumference, which shall touch a given nn^^\ M *^® ffi^en point A be without the given circle 5 iT* ^* w reqmred to draw from A a straight line, which snail touch the given circle. Take E, the centre of the circl©, [III. 1. and jom ^^ cutting the circum- ference of the given circle at D ; and from the centre E, at the distance EA, describe the circle AFG\ from the point 2> draw DFai right angles to jE^4 , [I. ii. and join ^/'cutting the circum- ference of the given circle at B ; jom AB, AB shall touch the circle BCD eqi^^'to^i^'"^ ^ '" *^^ ''^"*''^ ^^ *^^ ^^^^^^ ^^^» ^^ '« ,.,"''_., [I. Dtfinitwa 15. wtll to'^Z? -^ " *® "^"''^ «f *« arcle BOD, m is mu i? .1 ' . . , [^' L>efinit{(m 15. ^d tiie angle at JT is common to the two triangles AEB, ^f^^I^^u^ triangle ^JSr^ is equal to the triangle FED, whiwi,''*^®'' T^^^^ ^ *^® **<^«^ «^sies, each to each, to which the equal sides are opposite; .£1.4. BOOK IIL 17, la n \ therefore the angle ABE is equal to the angle FDE, But the angle FDE is a right angle ; ICimtLstum. therefore the angle ABE is a right angle. [AxUm 1. And EB is drawn from the centre; but the straight Ime drawn at nght angles to a diameter of a circle, from the extremity of it, touches the circle ; [in. le, CoroUary, therefore AB touches the circle. And AB is drawn from the given point A, q.b.p. But if the given point be in the circumference of the circle, as the pomt 2>, draw DE to the centre E, and DF at nght angles to2>^; then i)i^ touches the circle. [HI. 16, Cor. PROPOSITION 18. THEOREM. If a straight line touch a circle the straight line drawn from the centre to the point qf contact shall ^ perpen- dtculm' to the line toiiching the circle, ^ ^ ^ Let the straight line DE touch the circle ABC at the point Ci take I\ the centre of the circle ABC, and draw the straight line FG: FC shall be perpendicular to DE. For if not, let FG be drawn from the point F perpen- dicular to DEf meeting the cir- cumference at B, Then, because FGOb a right angle, [ffypothesis. FCG is an acute angle; [I. 17. and the greater angle of every triangle is subtended by the greater side ; [i. 19. therefore FCia greater than FG. But ^C is equal to i^5; rr 7)-^«v ^r therefore FB is greater than FG, the less than the greater • which IS impossible. »*" w*w greater, Therefore FG is not perpendicular to DE. .X J?t^x*^^_^?® ^^^^^ it maj be shewn that no other +k«tS'" "i;^?'"*" ^ ^f perpendicular to DE, but FC- therefore i^Cis perpendicular to 2>^. ' ^Wherefore, if s iiiraight line &e. q.e.d. 93 EUCLID'S ELEMENTS, PROPOSITION 19. THEOREM, If a ttraight line touch a circle, and from the point of contact a straight line he drawn at right angles to the touching line, th'. centre of the circle shall he in that line. Let the straight line DE touch the circle ABC at (7. and from C let CA be drawn at right angles to DE. the centre of the cu cle shall be in CA, For, if not, if possible, let F be the centre^ and join CF, Then, because DE touches the circle ABCf and FC is drawn from the centre to the point of contact, FC is perpendicular to DE; [III. is. thereforetheangle^C^isarightangle. But the angle AGE is also a right 5- a«gl©; »* [Constructim. therefore the angle FCE is equalto the angle ACE, [Ax. 11. the less to the gi-eater; which is impossible. Therefore i is not the centre of the circle ABC, }\ 5!?^ ^^^ manner it may be shewn that no other point out of tA IS the centre ; therefore the centre is in CA, Wherefore, if a straight line &c. q.e.d. PROPOSITION 20. THEOREM, "j" The angle at the centre of a circle is double of the angle at circun^ference on the same hase, that is, on the same arc. Let ABC be a circle, and BEC an angle at the centre, and BAC an angle at the cu-cumference, which have the same arc, BC, for their bas0; the angle BEC shall be double of the angle BAC, Join AE, and produce it to JP. - First let the centre of the circle be within the angle BAC, Then, because EA is equal to EB^ thj angle EAB is equal to the ttiifiio EBA I [I. 5, therefore the angles EAB^ EBA are double of the angle EAB, w. BOOK IJL 20, 21. 93 But the angle BEFia equal to the angles EAB^ EBA ; [1. 82. therefore the angle BEF is double of the angle EAB. For the same reason the angle FEC is double of the angle EAC. Therefore the whole angle BEG is double of the whole angle BAG, Next, let the centre of the circle be without the angle BAG. Then it may be shewn, as in the first case, that the angle FEG is double of the angle FAC^ and that the angle FEBy a pai t of the first, is double of the angle jP^jS, a part of the other; therefore the remaining angle BEG is double of the remaining angle BAG. Wherefore, the angle at the centre &c. q.b.d. PROPOSITION 21. THEOREM, -f The angles in the same segment of a circle are equal to one another. Let ABGD be a circle, and BAD, BED angles in the same segment BAED: the angles BAD, BED shall be equal to one another. Take F the centre of the circle ABGD. [III. 1. First let the segment BAED be greater than a semicircle. Join BFy DF. Then, because the Bugle BFD is at the centre, and the angle BAD is at the circumference, and that they have the same arc for their base, namely, BGD ; therefore the angle BFD is double of the angle BAD.II11.20, For the same reason the anirle BFD is double BED. )gl< Therefoire the angle BAD is equal to the angle BED, [Ak, T, ^ EUCLID 8 ELEMENTS. wmfcbe!''* ^"^ ^^^ritBAEDU not greater thJa Draw ^JF* to the centre, and pro- duce It to meet the circumference at C/,andjom CE, Thea the segment BAEC is greater than a semicircle, and there- fore the angles BAG, BEG in it, are equal, by the first casa. For the same reason, because the segment GAED is greater than a semicircle, the angbs GAD, GED are equal wglSiJ.^^ ""^""^^ ^«^^' ^^^ " ^^"«^ *<> *h« ^hole HTk * ' » [Axiom 2» wnerefore, , ^CX) are together equal to two right angles. ^heretorey t/ie opposite anfflea &c, qjld. PROPOSITION 23. TffBOJREM. On the same straight line, and on the same side of it there cannot be two similar segments of circles, not coinH- ctdmg with one another. If it be possible, on the same straight line AB, and on the same side of it, let there be two similar segments of circles ACB, ADB, not coinciding with one anottier. Then, because the circle A GB cuts the circle ADB at ttie two points A, By they cannot cut one another at any other point ; [III. 10. therefore one of the segments must fall within the other; let ACB fall within ADB-, draw the straight line BCD, and join AG, AD. Then, because ACB, ADB are, by hypothesis, similar segments of circles, and that similar segments of circles contam equal angles, [m. Definiticm 11. therefore the angle ACB is equal to the angle ADB ; that is, the exterior angle of the triangle ACD is equal to the mtenor tnd opposite angle ; which ia impossible. [116 '^^etQfoTQ, on the same straight line kc, q.e.d. 96 EUCLWS ELEMENTS. lii PROPOSITION 24, THEOREM. ^ Similar segmentt qf circies on equal straight linei ar€ equal to one another. Let AEB, CFD be similar segments of circles on the equal straight lines AB, CD ; the segment AEB shall be equal to the segment CFD. For if the segment r AEB be applied to the segment CFD, so that the t)oint A A may be on the point C;aiid the straight line ^jBon the straight line CD, the P^ ; [I. g. but DA Is equal to DC-. r^j^.^s^^t^rgfe'^*- therefore DB is equal to DC. ' [Axiom 1. BOOK III. 25. 97 [1.6. Therefore thethree straight lines r>A,DB,DCitaTb^%T*/^ \^*' «'«.<«f'an£ 18 common to the two triangles ADE CDE • ^hZt^,"^^' ^^"^ '^'^ **• the two sides CD.DE, tfem*U a";?Kj" "^ *<»"'<' ^ CWor e«M. of therefore the base J^^ is equal to the base EC. [i. 4, But EA was shewn to be equal to EB • therefore EB is equal to J^a ' [^^.^ j Therefore the three straight lines EA, EB.EGbxq all equal - and therefore ^ is the centre of the circle [m 9' fftri •' f ^' describe a circle ; this will pass through the dcScffi ' ''^''^^ ''^ ""^^^ ■^■^^^« ^ fieg^ent i^ *i. ^il^ ^* ^? evident, that if the angle ABD be irreater tZ^In^^^'^i^^S^'' centred fills wfthout tC^! T^^u^rFj 7»n\'^*''®'■®/*'^® ^®«» *^a° a semicircle ; but ^ feL^w^S^ ^ less than the anrfe BAD, the centre ^ fells within the segment ABO, whicb is therefore greater than a semicirola, ° ««.««» Wherefore, a segmen hat been described of Sic ^/^ circle being given, the circie ch it is a seament. qmv. ■MWMMMiMIMPM 98 EUCLWS ELEMENTS. I i li PROPOSITION 26. THEOREM, ^ In equal circlet, equal angles stand on equal aret, whether they be at the centres or circumferences. Let ABG. T>EF be equal circles ; and let BOC^ EHF be equal angleb in them at their centres, and BAG, EDF equal angles at their circumferences: the arc BKC shall be equal to the arc ELF, JobkBOyEF. Then, because the circles ABC, DEF are equal, {Hyp, the straight lines from their centres are equal ; [III. Dtf, 1. therefore the two sides BG^ GGare equal to i^e two sides Jg[^, ^/; each to each ; and the angle at G' is equal to the angle at H ; [Hypothesis, therefore the base BG is equal to the base EF. [I. 4. And because the angleat^ isequalto the angle sADy[Hyp, thesegment^^(7is similar to the segment EDF\ [IILD^/lll. and they are on equal straight lines BG, EF. But similar segments of circles on equal straight lines are equal to one another; ^ [IIL 24. therefore the segment BAG is equal to the segment EDF But the whole circle ABG is equal to the whole circle DEF; " [Hypothesis. therefore the remaining segment BICG is equal to the re- maining segment ELF ; [Axiom 3. therefore the arc BKG is equal to the arc ELF, BOOK HL 27. 89 I arctf \EHF \EDF CsbAU . 2)^. 1. ro sides [1.4. z>,[J5r«p; nes are [IIL 24. le circle tpothesis, the re- l-KOPOSITION 27. TBSORBM. BO^Eifi^^^ ^ f^"^ 1"'^ *°. BOOK III, 29, 30. 101 ^ PKOPOSmON 29. THEOREM, Tn equal circlet, equal arct are etibtended bv eaual straight Hnee. ^ ^ Let ABC, DEF be equal circles, and let BQC, EHF be equal arcs m them, and join BC, EFi the straight Uum BC ahall be equal to the straight line EF. Take K, Z, the centres of the circles, [ill. l. and join BK, KG, EL, LF. JJ«n» because the arc BOC is equal to the arc the angle BKC is equal to the angle ELF, [m. 27. And because the circles ABC, DEFfocQ equal, [Hypuhtm, the straight lines from their centres are equal; [III. D^. i. therefore the two sides BK, ^C7are equal to the two sides ^Z, ZjP, each to each ; , » and they contain equal angles ; therefore the base BC is equal to the base EF, [I. i. Wherefore, in equal circles &c q.e.d. PEOPOSmON 30. PROBLBm To bisect a given arc, that is, to divide it into two eaual parts, * If- ■ m 102 EUCLIL^S ELEMENTS. Let ADB be the given arc : it is required to Msect it Join AB ; bisect it at C ; [1. 10. from the point C draw CD at right angles to AB meeting the arc at D. [1. 11. The arc ADB shall be bisected at the point 2>. Jom AD, DB. Then, because AOis equal to CB, [Construction, and CD is common to the two triangles ACD, BCD; the two sides AC, CD are equal to the tT^o sides BC, CD, each to each t and the aiigle ACD is equal to the angle BCD, bjcause each of them is a right angle ; [Constructum. therefore the base ADi& equal to the base BD, [I. 4. But equal straight lines cut off equal arcs, the greater equal to the greater, and the less equcl to the less ; [III. 28. and each of the arcs AD, DB is less than a semi-circum- ference, because DC, if produced, is a diameter ; [III. 1. Cor. t)ierefore the arc AD is equal to the arc DB. : WheteioxQ the given arc U hiiected" at D. q.k.p. ' PROPOSITION 31. THEOREM. t. In a circle the angle in a semicircle is a right angle; hut the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle. Let ABCD be a circle, of which BC is a diameter and £i the centre ; and draw CA, dividing the circle into the segments ABC, ADC, and join BA, AD, DC: the angle m the semicircle BAC shall be a right aiigJo; but the angle in the segment ABC, which is greater than a ii BOOK HI, 31. 103 semicircle, shall be less than a right angle; and the angle in the segment ADC, which is less than a semicircle, shall be greater than a right angle. Join-4 J^,andproduce^-4 toi^. Then, because EA is equal to EB, [I, Definition 15. the angle EAB is equal to the angle EBA ; [I. 5. and, because EA is equal to EG, the angle EAC is equal to the angle EGA ; therefore the whole angle BAG is equal to the two anrfes, ABO, ACB. [AxiL 2. But FAG, the exterior angle of the triangle ABG, is equal to the two angles ABG, A GB ; [I. 82. therefore the angle jB^ (7 is equal to the angle FAG, [Ax. 1, and therefore each of them is a right angle. [T. D(f, 10. Therefore the angle in a semicircle BAGia a right angle. And because the two angles ABG^ BAG, of the triangle ABG, are together less than two right angles, [I. 17. and that BAG has been siiijwn to be a right angle, therefore the angle ABGi» less than a right angle. Therefore the angle in a segment ABG, greater than a semicircle, is less than a right angle. And because ABGD is a quadrilateral figure in a circle, any two of its opposite angles are together equal to two right angles; [111.22. thore/ore the angles ^J?(7, ADG are together equal to two right angles. But the angle ABG has been shewn to be less than a right angle; fliArAfnrA fliA anflrlA J T^d ia flfi*Aaf a«* fVion a «>in>lif onrvlA ■fv^^^e v«54¥Si r^ SJ O" Therefore the angle in a segment ADG, less than a semi- circle, is greater than a right angle. "Wherefore, the angle &a q.e.d. J It <1! 104 EUCLID'S ELEMENTS, I!' ^1 ' n ' \ i. I Corollary. From the demonstration it is manifest that if one angle of a triangle be equal to the other two, it is a right angle. For the angle adjacent to it is equal to the same two angles; [1.32. and when the adjacent angles are equal, they are right angles. [I. Definition 10. PEOPOSITION 32. THEOREM, If a straight line touch a circle, and from the point of contact a straight line he drawn cutting the circle, the angles which this line makes with the li?ie touching th& circle shall be equal to the angles which are in the alternate segments cfthe circle. Let the straight line EF touch the circle ABCD at the point B, and from the point B let the straight line BD be drawn,' cutting the circle : the angles wMch BD makes with the touching line EF, shall be equal to the angles in the alternate segments of the circle ; that is, the angle i>-ffi^ shall be equal to the angle in the segment BAD^ and the angle DBE shall be equal to the angle in the seg- ment -5C72). From the points draw BA at right angles to EF, [1. 11. and take anv point G in the arc BD. and jom AD, DC. CB, Then, because the straight line EF touches the circle ^^CZ> at the points?, [Hyp. and BA is drawn at rights angles to the touching line from the point of contact B, [Constructicm, therefore the centre of the circle fa in BA, [III. 19. Therefore the angle ADB^ being in a semicircle, is a right angle. [III. 31. Therefore the other two angles BAD^ ABD are equal to n right angle. ° ' ^ [I. z% But ^i?i^ is also a right angle. , [Consfructian, BOOK III. 32,33. 105 tnanifest r two, it mo two [1.32. po right lition 10« )oint of rclCf th& ing thJd Iternate 3CD at mQBl) > makes igles in angle BAD, the seg- Iruction, III. 19. a right III. 31. lal to 9 [1. 3a ruction^ Therefore the angle ABF is equal to the angles BAly ABD. ' From each of these equals take away the common ancle ABD; ** therefore the remaining angle 2>5i?' is equal to the remain- mg angle ^^i>, iAxiom3. which is m the alternate segment of the circle. And because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are together equal to two right angles. ^ Jjjj 22. But the angles DBF, DBE are together equal to two nght angles. |-j jg Therefwe the angles i>i?i?; DBE are together equal to the Ami the angle DBF has been shewn equal to the angle therefore the remaining angle DBE is equal to the re- maining angle BCD, £^^,.^ 3^ which is in the alternate segment of the circle. Wherefore, if a straight line &c. q.e.d. PROPOSITION 33. PROBLEM, ^ On a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle. Let AB be the given straight line, and G the given rectilineal angle: it is required to describe, on the given straight line AB a segment of a circle containing an anglo equal to the angle C. as First, let the angle C be a right angle. Bisect AB at F, fl. 10. and from the centre Fy at the distance jF!fi, describe the Bumicircle AH3, > Then the angle AHB in a semicircle is equal to. the right angle O, [III. 81, J 106 EUCLIiyS ELEMENTS, But if the angle G be not a right angle, at the point A, in the straight line ABy m"\Q the angle BAD equal to the angle C\ [1. 23. from the point A^ draw AE at right angles to ^Z>;[I.ll. bisect ^5 at -F; [1. 10. from the point F, draw FG at right angles to AB) [1. 11. and join GB, Then, because AF is equal to BF, [Const. and FG is common to the two triangles AFG, BFG\ the two sides ^F, FG are equal to the two sides BF^ FGi each to each ; and the angle ^4^6? is equal to the angle BFG ; [i. Definition 10. therefore the base ^6r is equal to the base BG ; [I. 4. and therefore the circle described from the centre G, at the distance GAy will pass through the point B, Let this circle be described; and let it be AHB. The segment AHB shall contain ah angle equal to the given rectilineal angle C. Because from the pomt A, the extremity of the diameter AE, AD is drawn at right angles to AE, [Construction. therefore AD touches the circle. [in. 16. Corollary. And ))ecause AB is drawn ft-om the point of contact A, the angle DAB is equal to the angle in the alternate segment AHB. . j-jll 32^ But the angle DAB is equal to the angle CI ICmstr. Therefore the angle in the segment AHB is equal to the angle C'. [Axiom 1. Wherefore, on the given straight line AB, the segment AHB qf a circle has been described, containing an angle equal to the given angle 0» q.e,p. BOOK III. 34,35. 107 PROPOSITION 84. PROBLEM. From a given circle to cut off a segment containing an angle equal to a given rectilineal angle. Let ABC be the given circle, and D the given recti- lineal angle : it is required to cut oflF from the circle ABG a segment containing an angle equal to the angle /). Draw the straight line JSF touching the circle ABC at the point 5; [III. 17. and at the point J5,in the straight line BF^ make the angle FBC equal to the angle D. [I. 23. The segment BA Cshall contain an angle equal to the angle £>, Bemuse the straight line FF touches the circle ABO, and BC is drawn from the point of contact B, [Constr. therefore the angle FBG is equal to the angle in the alternate segment BAG of the circle. [in. 32. But the angle FBG is equal to the angle D. [Comtmctiim. Therefore the angle in the segment BAG is equal to the ^«Sl«^- [Aadoml. ^ .^^^^^^^T^ from the given circle ABG, the segment BAL has hem cut off, containing an angle equal to the given angle D, q.e.p. ^ PROPOSITION 35. THEOREM, "f" If two straight lines cut one another ,-^.,,.^.„^ .^ vt.-s- ths rectangle contained by the segments of one oflhem shall be equal to the rectangle contained by the segments qr the other. 108 EUCLIU8 ELEMENTS. i Let the two strakht lines AC, BD cut one another at the point 4 within the circle ABCD : the rectangle con- tained by AE, EG shall be equal to 6 ^" « the rectangle contamed by BE, ED. If A Cmd BD bo*h pass through the centre, so that E is the centre, it is evident, since EA, EB, EC, ED are all equal, that the rect- angle AE, EG IS equal to the rect- angle BEy ED. n„f ?!?* Hi ^°^^ *^?^? ^^' P^«« *^^°"g^ t^e centre, and cut the other AG, which does not pass through the centre, at right angles, at the point E. Then, ifBD be bisected at F, F is the centre of the cu*cle ABGD- join AF. Then, J because the straight line BD which passes through the centre, cuts the straight line AG, which does not pass through the centre, at right angles at the Ppjpt E, [Bypothesis. AEiB equal to EG. [III. 3. And because the straight line BD is divided into two equal parts at the point F, and into two unequal parts at the pomt E, the rectangle BE, ED, together with the square on EF, is equal to the square on FB, [11. 6. that IS, to the square on AF. But the square on ^.Pis equal to the squares onAE,EF.[IA7. Therefore the rectangle BE, ED, together with the square on EF, is equal to the square^ on AE, EF. [Axiom 1. Take away the common square on EF; then the remaining rectangle BE, ED, is equal to the remaining square on AE, ' • that is, to the rectangle AE, EG. 4\.^^^1^ ^^!i^^% ^^^?^ P^®^®» through the centre, cut the other y4C/, which does not oass through the cGn*»^ at uie pomt ^, but not at right angles. " Then, if b3 be bisected at F, F is the centre of the circle ABGD: jom AF, and from iJ'draw FG perpendicular to AG, [1. 12. BOOK HI, 35. lOd )ther at Efle con- fcre, and centre, ttto two )arts at ith the [II. 6. F.[1.47. square awom 1. to the re, cut \iB& BCD', £1.12, Then AG\% equal to GC) [III. 3. therefore the rectangle AlE, EG, together with the square on EG, is equal to the square on AG, [II. 6. To each of these equals add the square on GF ; then the rectangle AE, EC^ to- gether with the squares on EG, GF, is equal to the squares on AG, GF. [Axiom 2. Biit the squares on EG, GF are equal to the square on EF; and the squares on AG, GF are equal to the square on ^^- [1. 47. Therefore the rectangle AE, EC, together with the square on EF, is equal to the square onAF, that is, to the square on FB. But the square on FB is equal to the rectangle BE, ED. together with the square on EF, [n. 6, Therefore the rectangle AE, EC, together with the square on EF, is equal to the rectangle BE, ED, together with the square on EF, Take away the common square on EF; then the remaining rectangle AE, EC is equal to the remaming rectangle BE, ED. [Axiom 3. Lastly, let neither of the straight lines AC, BD pass through the centre. Take the centre F, [III. 1. and through E, the intersection of the straight lines AC, BD, draw the diameter GEFH, Then, as has been shewn, the rectang^le GE, EHia equal to the rectangle AE, EC, and also to the rectangle BE, ED ; therefore the rectangle AE, EC is equal to the rectangle BE, ED. {Axiom 1. therefore, if two straight line« &c. q.b.d. 110 EUCLIiya ELEMENTS. ..f ^ PROPOSITION 86. THEOREM, ^ If from any point without a circle two straight line* be drawn, one qf which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part qfit without the circle, shall be eqtial to the square on the line which touches it. Let D be any point without the circle ABC, and let DCA, DB be two straight fines drawn from it, of which DCA cuts the circle and DB touches it: the rectangle AD, DO shall be equal to the square on DB, First, let DCA pass through the centre B, and join £B. Then BBD is a right angle. [III. 18. And because the straight line AC is bisected at B, and produced to D, the rectangle ^Z>,i>(7 together with the square on EC is eqiml to the square on BD, [II. 6. But BC is equal to BB ; therefore the rectangle AD, DC together with the 'square on JSB is equal to the square on BD. But the square on BD is equal to the squares^n BB, BD, because BBD is a right angle. [I. i7. Therefore the rectangle AD, DC, together with the square on BB is equal to the squares on BB, BD. Take away the common square on BB; then the remaining rectangle AD, DC is equal to the square on DB. ^ lAxiom 3. Next let DCA not pass through tue centre of the circle ABC; take the centre B; [HI. i. from B draw ^i^ perpendicular to AC; mdjomBB,BC,BD. fi,« ™?li^®???-^! *^® straight line ^^which losses through vsiS centre, cu*^ i,iie otraiuut line AC, which does not pass through the centre, at right angles, it also bisects it ; [III. 3. therefor© ^ F is equal to Fa [1. 12. bi BOOK ILL 86. xn And because the straight. line AC \& bisected at F, and produced to 2>, the rectangle AD^ DC, t(^ther With the square on FC, is equal to the square on F2), [II. 6. To each of these equals add the square on FE, Therefore the rectangle AD, DC together with the squares on CF, FB^ is equal to the squares on DF, FE. [Axum 2. But the squares on CF, FE are equal to the square on CE, be- cause CFE is a right angle ; [I. 47. and the squares on DF^ FE are equal to the square on DE. Therefore the rectangle AD, DC, together with the square on CE, is equal to the square on DE, 5\ But CE is equal to BE; therefore the rectangle AD, DC, together with the square on BE, is equal to the square on DM, But the square on DE is equal to the squares on DB, BE, because EBD is a right angle. [I. 47. Therefore the rectangle AD, DC, together with the square on BE, is equal to the squares on DB, BE. Take away the common square on BE ; then the remaining rectangle AD, DC is equal to the square on DB. Wherefore, iffrmi any point &c. Corollary. If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them without the circles are equal to one another; namely, the rectangle BA, AE is equal to the rectangle CA, AF; for each of them is ea.ual to the SQimre on the Straight Hne AD, which touches the circle. l^Axiom 8* Q.E.D. 112 EUCLID'S ELEMENTS. * PROPOSITION 87. THEOREM, \ If from any point without a circle there be drawn two straight lineSj one qf which cuts the circle^ and the other meets t/, and if the rectangle contained by the whole line which cuts the circle, and the part (fit without the circle, he equal to the square on the line which meets the circle, the line which meets the circle shall touch it. Let any point Z> be taken without the circle ABC^ and from it let two straight lines DCA, DB be drawn, of which DC A cuts the circle, and DB meets it; and let the rectangle AD, DC be equal to the square on DBi DB shall touch the circle. Draw the straight line DE, touching the circle -4-5C7; [III. 17. find FihQ centre, [III. l. and join FB, FD, FE. Then (the angle FED is a right angle. [iii. is. And because DE touches the circle ABC, and DCA cuts it the rectangle AD, DC is equ^ to the square on DE. [III. 36. But the rectangle AD, DC is equal to the square on DB. [Hyp. Therefore the square on 2>^is equal to the square on D^;[^«.l. therefore the straight line DE is equal to the straight line DB, And EFi% equal to BF-, [I. Definition 15. therefore the two sides DE, EF&re equal to the two sides DB, BF each to each ; and the base DFis common to^the two triangles DEF, DBF; therefore the angle DEF is equal to the angle DBF. [L 8. But DEF is a right angle ; - [Construction, therefore also DBF is a right angle. And BF, if produced, is a diameter; and the straight line which IS drawn at right angles to a diameter from the AY^^MItYklfv nf it- t-rxtt^l^g^r. tX^^ ^f_.1 rr-rr t y» >^, .-- II-.-, therefore 2>i5 touches the circle -45^ Wherefore, (//row a i?om<&c. q.e.d. BOOK IT. DEFINITIONS. 1* A BKOTiLiXEAL figure 18 said to be inscribed in another rectilineal figure, when all the angles of the in- scribed figure are on the sides of the figure in which it is inscribed, each on each. 2. In like manner, a figure is said to be described about another figure when all the sides of the circumscribedl figure pass through the angular points of the figure about which it & de- scribed, each through each. 3. A rectilineal figure is said to be inscribed in a circle, when all the angles of the inscribed figure are on the circumference of the circle. 4. A rectilineal figure is said to be descnbed about a circle, when each side of the circumscribed figure touches the circumference of tlie circle. . 6. In like manner, a circle is said to De inscribed in a rectilineal figure when the circumference of the circle touches each side of the figure. 8 114 EUCLmS ELEMENTS. 6. A circle is said to be described about a rectilineal figure, when the cir- cumference of the circle passes through all the angular points of the figure about which it is described. 7. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle. PEOPOSITION 1. PROBLEM. In a given circle, to place a straight line, equal to a . given straight line, which is not greater than the diameter qf the circle* ^ Let ABO be the given circle, and D the given straight line, not greater than the diameter of the circle: it is required to place in the circle ABC, a straight line equal to D. Draw BC, a diameter of the circle ABO. Then, if BO is e^ual to /), the thin^ required is done ; for in the circle ABC, a straight line is placed equal to L>. But, if it is not, BC is greater than D, [Hypothem. Make CE equal to D, [I. 3. and from the centre O, at |he distance CE, describe the circle AEF^ and join CA. Then, because is the centre of the circle AEP, CA is equal to CE-, " [I. Definition 15. but CE is equal to Z) ; [Construction, therefore CA is equal to D. [Axiom 1. Wherefore, in the circle ABO, a straight line CA is placed equal to the given straight line 2>, which is not greater than the diameter of the circle, q.e.f. BOOK IK a, 3. PROPOSITIGN 2. PROBLEM. ^ U5 F « £w^/i^ "^^^^ '^ tWcrjJd a triangle equiangular to a given trtangle* *-j ^®J ^^^. ^ *^« ?>von circle, and DBF the iriTif triangle: it is required to inscribe in the circle ABC a tnangle equiangular to the triangle D£F, Draw the straight line OAH touching the circle at the point -4 ; [III. 17. at the point A, in the straight line Aff^ make the BJngleHA Cequal to the angle 2>^^; [1. 23. and, at the point A^ in the straight line AG, make the angle GAB equal to the angle DFE ; md ioinBa ^iBCshall be the triangle required. Because GAH touches the circle ABC and Jr u drawn fn,m the point of contacts, ^^'''[Z.^,^ therefore the angle ffACit equal to the angle ABO in the alternate segment of the cirote. "* mi 82 But theangle SACa equal to the angle DEF. roLtr' Iherefore the angle ABC ia equal to the angle DSf. [Ax 1 angte°i)*l "*"'* '*^°'' *^ ^^^ ^^^ iB equal to tte 6 "6*»-«^^^. [I. 32, .lajMwiM 11 and 3. on^<^ zy-fi/>, aw<;? it ts inscribed m the circle ABC, q.e.f. / PROPOSITIONS. PROBLEM. 8—3 ^»wn triangif^ 116 EUCLID'S ELEMENTS. let ABC be the given circle, and DEF the given tri- angle : it is required to describe a triangle about the circle ABC, equiangular to the triangle DEF, Produce ^i^ both ways to the points ,x t^ G, Hi take JT the / \ 7s centre of the circle ABC) [HI. 1. from K draw any radius KB\ at the pomt K\ in the straight line KB. make the angle BKA equal to tne angle DEa, and the angle jB^C equal to the angle DFH-, [1. 23. ¥*4,*i*^??S'^.**^® J^^"^ -^» ^» ^> draw the straight lines LAM, MJ$N, NCL, touching the circle ABC [HI. 17. LMN shall be the triangle requir 3d. Because LM, MN, NL touch the circle ABC at the pomts A, B, Cy [Construction, to which from the centre are drr wn UTA, KB, KC, therefore theangle8atthepoints^,5,Carerightangles.[III.18. ^"i^ J>^a^se the four angles of the quadrilateral figure AMBKBxe together equal to four right angles, for it can be divided into two triangles, and that two of them KAM, KBM are right angles, tjierefore the other two AKB, AMB are together equal to two right angles. [Axhm 3. But the angles DEG, DEF are together equal to two right angles. ^ [I. 13. Therefore the angles AKB, AMB are equal to the angles DEGj DEF'f of which the angle AKB is equal to the angle DEG j [Ccmstr. therefore the remaining angle AMB is equal to the re- maining angle DEF, [Axiom 3. XI- _ iu i/iiO o&kUKj maiitier trie angle LTvM may be shewn to be equal to the angle DFE. Therefore the remaining angle MLN is equal to the remammg angle EDF. [i. 32, Axiom 11 and 3. BOOK IK 4. 117 i ^« J? n^^^® ^5^. ^r**«f ^^ ^^N is equiangular to ths trU angle DBF, and it %% described about the circle ABC, Q.B.F. PROPOSITION 4. PROBLEM. To inscribe a circle in a given triangle. Let ABC he the given triangle : it is required to inscribe a circle in the triangle ^^G ° ^ *"^ Bisect the angles ABC. ACB, by the straight lines BD, CD, meeting one another at the point and from D draw DE, DF, DG per- pendiculars to AB.BC, CA. [1. 12. Then, because the angle FBD is equal to the angle FBD, for the angle ABC is bisected by BBf [Construction. and that the right angle BED is a u c equal to the right angle BFJ); Uxiom Ih therefore the two triangles FBD, FBB have two angles of the one equal to two angles of the other, each to ewh; and the Bide BD, which is opposite to one of the equal angles m each, is common to both ; therefore their other sides are equal ; [i. 26 therefore BE is equal to BF. For the same reason Z)^ is equal to BF. Therefore BE is equal to BG. ' [Axiom 1. Therefore the three straight Imes BE, BF, BG are equal to one another, and the circle described from the centre D, at the distance of any one of them, will pass through the extremities cf the other two; "6"""" ?i?f o'* 7'" JT^ the straight lines AB, BC, CA, because ^frl^h^'lf/^ the Doints E, FG are right aiJgles/and the straight line which 18 drawn from the extremity of a dia- meter, at nght angles to it, touches the circle. [III. 16. Cor Therefore the straight lines AB, BC. CA do each of them ioucu ^"« errcie, and therefore the circle is inscribed in the triangle JLJUj. Wherefore a circle has been inscribed in the given frtangie, q.k.f. " • • ■ -V 118 EUCLID'S ELEMENTS, * PROPOSITION 5. PROBLEM, ^ To describe a circle about a given triangle. Let ABC be the given triangle ; it is required to do- scribe a circle about ABC ^^*«u w ao- Bisect AB, AC at the points /), E; [r. 10. from these points draw DF, EF, at right angles to ^^»^^; [I. 11. DF^ EFy produced, will meet one another; for if they do not meet they are parallel, therefore AB, AC, which are at right angles to them are parallel ; which is absurd : . let them meet at F, and join* FA ; also if the pomt F be not in BCy join BF, CF, Then, because AD is equal to BD, [Construction, and DF is common, and at right angles to AB, therefore the base FA is equal to the base FB, [I. 4. In the same manner it may be shewn that FC is equal to FA. Therefore FB is equal to FC; [Axiom 1. anji,^4, FB, FC are equal to one another. Therefore the circle described from the centre JP, at the distance of any one of them, will pass through the extre- mities of the other two, and will be described about the triangle ABC, Wherefore a circle has been described about the given triangle, q.e.p. Corollary. And it is manifest, that when the centre of the circle falls within the triangle, each of its angles is less than a right angle, each of them being in a segment greater than a semicircle ; and when the centre is m one of the sides or the triangle^ the anrfe opposite to this side, bemg m a semicircle, is a nght angle ; and when the centre :^!^^^- Ai.-al-ii*!iali BOOK IV. 5,6. 119 falls without the tnangle, the angle opposite to the side beyond which it la, being m a segment less than a semi- circle, IS greater than a right angle. [ill. 31. Therefore, conversely, if the given triangle be acute- angled, the centre of the cu-cle falls within it : if it be a nght-^gled tnangle, the centre is in the side opposite to the nght angle ; and if it be an obtuse-angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle. *^ PROPOSITION 6. PROBLEM. To inscribe a square in a given circle. Let ABGD be the given circle: it is reqmred to in- scnbe a square in ^^(72). Draw two diameters A (7, BB of the circle ABGD, at right an- gles to one another ; [III. 1, 1. 11. and join AB, BO, CD, DA. The figure ABGD shall be the square required. Because BE is equal to DE, for ^ is the centre; and that EA is common, and at right angles to -52); therefore the base BA is equal to the base DA [I 4 t^B a! or dT'' ''^^'' ^^' ^^ "^^ ^^'^ ""^ *^^^ ®^^ Therefore the quadrilateral figure ABGD is equilateral It IS also rectangular. ABGD %'*i^if oV'''^- ^^ ^^^°^ ^ ^^^^^ter of the circle ABGD, BAD is a semicircle; [C, and that -4(7 is equal to oacli of the two GH, FK, and that BD is equal to each of the two GF, HK, therefore GH, FK are each of tliem equal to GF, or UK; therefore the quadrilateral figure FGHK is equilateral. It is also rectangular. For since AEBG is a parallelogram, and AEB a right angle, therefore A GB is also a right angle. [i. 84. In ^e same manner it may be shewn that the angles at //, A, F are right angles ; I BOOK IV, a 121 therefore the quadrilateral figure FGHK\& rectangular. And it has been shewn to be equilateral ; therefore it is a square. Wherefore a square hm been described about the given circle, q.e.p. PROPOSITION 8. PROBLEM, To inscribe a circle in a given square. Let A BCD be the giyen square: it is required to in- scribe a circle in ABGD, Bisect each of the sides AB, AD at the points F^ E ; [I. lo. through E draw EH parallel to AB or DGy and through F draw FK parallel to AD or BC, [I. 81. Then each of the figures AK, KB, AH, HD, A G, GO, BG, GD is a right-angled parallelogram ; and their opposite sides are equal. [i. 34. And because AD is equal to AB, [I. Dejinition 30. and that AE is half of AD, and AF half of AB, [Constr. therefore AE is equal to AF, [Axiom 7, Therefore the sides opposite to these are equal, namely. FG equsl to GE, ^ ' [I. si' In the same manner it may be shewn that the straight hues GH, GKsre each of them equal to FG or GE, Therefore the four straight lines GE, GF, GH, GK are equal to one another, and the circle described from the centre G, at the distance of any one of them, will pass through the extremities of the other three ; and it will touch the straight lines AB, BC, CD, DA, because the angles at the points E, F, H, K are right angles, and the straight line which is drawn from the extremity of a diameter, at right angles to it, touches t*l®f ircl®- [III. 16. Corollary. Therefore the straight lines AB, BC, CD, DA do each of them touch the circle. Wherefore a circle has been inscribed in the given square. q.e.f. 122 EUCLIL^S ELEMENTS, PROPOSITION 9. PROBLEM. , To describe a circle about a given square. Let ABCD be the given square: it is required to desenbe a circle about ABCD. *^«4uu^ea lo Join AG, BD, cutting one an- other at E. Then, because AB is (i aal to ADy and AC is common to the two tri- angles BAC, L>AC; the two sides^^, A C are equal to the two sides 2)^, ^^each to each ; and the baae -SC7is equal to the base i>6'j therefore the angle BAG is equal to the angle DAC [I. 8. and the angle BAD is bisected by the straight line AC. A nn *5?^n "/?7!T''°®'' '* ""^^ ^® ^^^^ t^^at the angles im^BD AC. ^""^ «®^erally bisected by the straight Then, because the angle DAB is equal to the angle ABC and that the angle ^^5 is half the angle />^5, ' and the angle EBA is half the angle ABC, therefore the angle EAB is equal to the an^le EBA ; lAx 7 and therefore the side EA is equal to the side Eb! [I 6 liniVrTi?! J"^"T IV'^y ^® '^^^^ *^a<^ the 'straight imes -fiC, ^Z) are each of them equal to^^ or jE'^ Wherefore the four straight lines EA, EB, EC, ED are Z^« 1^''''? fi?^*5?^' ^'^^ *^« «^r^I® iescribed'from the centre A, at the distance of any one of them, will nass through the extremities of the other three, aiidwiK described about the square ABCD. ' ^qmr^^'^^qlr'! ''''''^' ^ ^^"^ described about the given To PROPOSITION 10. PROBLEM. -f describe an isosceles angles at the base double triangle, having each qf the (if the third angle. BOOK IV. 10. 123 Take any straight line AB^ and divide it at the point (7, so that the rectan- gle ABy BC may be equal to the square on A C; [II. 11. from the centre A, at the distance AB, describe the circle BDE, in which place the straight line BI> equal to AC, which is not greater than the diameter of the circle BDB; [iv. i. andjomD^. The triangle ABD shall be such as is re- guii-ed; that is, each of the angles ABD, ADB shaU be double of the third angle ^^i). . f^^^iiJP'j and about the triangle ACD describe the circle ACD, ^y 5 Then, because the rectangle AB, BC is equal to the square on AC, [Construction. and that Ada equal to BD, [ConstrucHon. therefore the rectangle AB, BC is equal to the square And, because from the point jB, without the circle ACD, two straight lines BCA, BD are drawn to the circumference, one of which cuts the circle, and the other meets it, and that the rectangle AB, BC, contained by the whole of the cutting line, and the part of it without the cVcle, is equal to the square on BD which meets it ; therefore thestraightline^2>touchesthecrrcle^C2>.[III.37. And, because BD touches the circle ACD, and DC is drawn from the point of contact D, therefore the angle BDCi» equal to the angle D AC in the alternate segment of the circle. [m. 32. To each of these add the angle CD A : ^}fJ^f^^J? }!!? ^^®^® *°^^® ^^-^ « equal to the two angles tDA, DAC f^aj^om 2. But the exterior angle BCD is equal to the angles CDA DAC, [I. 32. 124 f EUCLWS ELEMENTS. Therefore the angle BDA Is equal to theangle^C72>. [Ax.l, But the angle BDA is equal to the angle DBA, [l. 5. because AD is equal to AB. Therefore each of the angles BDAy DBA, is equal to the angle BCD. [Axiom 6. And, because the angle DBO Is equal to the angle BCD, the side DB is equal to the side DC; [I. 6. but DB was made equal to CA ; ^ therefore CA is equaljto CD, [Axiom t, and therefore the angle CA D is equal to the angle CD A. [1. 5. 'Dierefbre the angles CAD, CD A are together double of the angle CAD. But the angle BCD is equal to the angles CAD, CD A. [1. 32. Therefore the angle BCD is double of the angle CAD, And the angle BCD has been shewn to be equal to each of the angles BDA, DBA ; therefore each of the angles BDA, DBA is double of the angle BAD, Wherefore an isosceles triangle has been described, having ea^h qf the angles at the base double of the third angle* q.e.p. I PROPOSITION 11. PROrLEM. To inscribe an equilateral and equiangular pentagon in a given circle. ^ ^ , Let ABCDE be the given circle: it is required to inscribe »» equilateral and equiangular pentagon in the -""V'^^'x^Jt '5H^\*^*^^ wia«gie» jbi:t±i, having each of the angles at C, H, double of the angle at F-, [IV. 10. !^,u \''''"*v ^^^^^. inscribe the triangle ACD, equian^ gular to the triangle FQH, so that the angle CAD may BOOK IV. 11. 125 bo equal to the an^le at Fy and each of the angles ACD, ADC equal to the angle at G or H; [IV. 2. and therefore each of the angles ACD, ADC is double of the angle CAD ; bisect the angles ACD, ADC by the straight \\m%CEjDB\ [1.9. and join AB, BC, AE, ED. ABCDE shall be the pentagon required. For because each of . the angles ACD, ADC is. double of the angle CAD, and that they are bisected by the straight lines CE, DB, therefore the five angles ADBy BDC, CAD, DCE, ECA are equal to one another. But equal angles stand on equal arcs ; [III. 26. therefore the five arcs AB, BC, CD, DE, EA are equal to one another. And equal arcs are subtended byequal straight lines ; [III.2ft therefore the five straight lines AB, BC, CD, DE, EA are equal to one another ; and therefore the pentagon ABCDE is equilateral It is also equiangular. For, the arc AB is equal to the arc DE ; to each of these add the arc BCD ; therefore the whole arc ABCD is equal to the whole fUTcBCDE. [Axicm2. And the angle AED stands on the arc ABCD, and the angle BAE on the arc BCDE, Therefore the angle AED is equal to the angle BAE. [IIL 27. For the same reason each of the angles ABC, BCD, CDE is equal to the angle AED or BAE ; ti ^_. XI- A A ■nryr\Tn •- • '-•ss And it has been shewn to be equilateral. Wherefore an equilateral and equiangular pentagon has been inscribed in the given circle. q.b.f. 126 EUCLID'S ELEMENTS, PROPOSITION 12. PROBLElff. J- a^'^gi:^ZT'"'"^'^ """^ **«.-«»l/«/«r pentagon Let ABODE be the given ctcle: it is rennired tn Let the angles of a pen- tagon, inscribed in the circle, by the last proposition, be at the points A, B, C, D, E so that the arcs AB, Bci CD DE, EA are equal M'*^- on ^^''"Jl'f ''^'^'''' ^"^ ^^' ^^ ^® ^"«^ *o th« square* of which the s<3(uare on /'Cis equal to the square on FB^ therefore the remaining snuarG on nir ia «. remaining square on jBJT, * " ^ ^erefore the straight line CK Ixum, line BIT. equal to the stndgh BOOK IV, 12. 127 And because FB is equal to FC, and FK'n common to the two triangles BFKj CFK\ the two sides BF^ FK are equal to the two sides. CF, FK^ each to each ; and the base BK was shewn equal to the base CK\ therefore the angle BFK'm equal to the angle CFK^ [I. 8. and the angle BKF to the angle CKF» [I. 4. Therefore the angle BFCia double of the angle CFK, and the angle BKC is double of the angle CKF, For the same reason the angle CFD is double of the angle CFL, and the angle OLD is double of the angle CLF. And because the arc BG is equal to the arc CJ>, the angle BFC is equal to the angle CFD ; [III. 27. and the angle BFC is double of the angle CFIT, and the angle CFD is double of the angle CFL ; therefore the angle CFK\a equal to the angle CFL, [Ax.*!, And the right angle FCKis equal to the right angle FCL, Therefore in the two triangles FCIT, FCL, there are two angles of the one equal to two angles of the other, each to each ; and the side FCy which is adjacent to the equal angles in each, is common to both ; therefore their other sides are equal, each to each, and the third angle of the one equal to the third angle of the other ; therefore the straight line CK is equal to the straight line CLy and the angle FKC to the angle FLC. [I. 26. And because CKis equal to CL, Xi^is double of OK, In the same manner it may be shewn that JSX" is double of BIC, And because ^JTis equal to CK, ai was shewn, and that HJSris double of BIT, and ZATdouble of CAT, therefore HK is equal to XJT. lAxLm 6, ^J^Jt®_^"*®, manner it may beshewn that GIL trJrij JriJLi mv eacii Oi them equal to UK or LJC; therefore the pentagon GHKLM\& equilateral ' It is also equiangular. f , 128 EUCLWS ELEMENTS. \ For, since the angle FKC is equal to the angle FLO, and that the ande HKL is double of the angle FkC^ and the angle KLM double of the angle FLG, as was shewn, therefore the angle HKL is equal to the angle KLm. [Axiom 6, In the same manner it may be shewn that each of the angles KHG, HQMy GML is equal to the angle HKL or KLM-, therefore the pentagon QHKLM is equiangular. *' And it has been shewn to be equilateral. Wherefore an equilateral and equiangular pentagon hat been described about the given circle, q.e.f. PBOPOSITION 13. PROBLEM, To inscribe a circle in a given equilateral and equi^ angular pentagon. Let ABODE be the given equilateral and equiangular pentagon: it is required to inscribe a circle in the pen- tagon ABODE, Bisect the angles BCD, CDE by the straight lines CF, DF] [I. 9. and from the point F, at which they meet, draw the straight lines FB, FA, FE. ^ Then, because BC is equal to DC, [Hypothesis, and CF is common to the two triangles ^6!F,i>Ci^; the two sides BC, CF are equal to the two uides DC^. OF. each to each ; and the angle BCF is equal to the angle DCF ; [Ctmstr. therefore the base BF is equal to the base DF, and th«i BOOK IV, \X 129 other angles to the other angles to which the equal sides are opposite; [Li. therefore the angle CBF is equal to the angle CDF, And because the angle CDE is double of the angle CDF, and that the angle CDE is equal to the angle CBA^ and the angle CDF is equal to the angle CBF, therefore the angle CBA is double of the angle CBF\ therefore the angle ABFis equal to the angle CBF; therefore the angle ABC is bisected by the straight line BF. In the same manner it may be shewn that the angles BAEj AED are bisected by the straight lines AF, EF, From the point F draw FG, FH, FK, FL, FM perpen- diculars to the straight lines AB, BC, CD, DE, EA. [1. 12. Then, because the angle FCB[ is equal to the angle Fcur, and the right angle FIIC equal to the right angle FKC\ therefore in the two triangles FHC, FKC, there are two angles of the one equal to two angles of the other, each to each; and the side FC, which is opi-osite to one of the equal angles in each, is common to both ; therefore their other sides are equal, each to each, and therefore the perpendicular FH is equal to the perpen- dicular FK. [I. 26. In the same manner it may be shevm that FL^ FM, FG are each of them equal to FH or FIT. Therefore the five straight lines FG, FH, FK, FL, FM are equal to one another, and the circle described from the centre F, at ^he distance of any one of them will pass through the extremities of the other four ; and it will touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right angles, [Construction. and the straight line drawn from the extremity of a dia- meter, at right angles to it, touches the cirdLe ; [III. 16. rrkA'..rv^^..£^ ««.^l^ ^Ci.1,^ -i. i~i.x i:^^. A n zj/nr ryrx T\ry jp j i>.UvXwAV£M wMwII Vi viiw wi>iUlgiiw iiliVH ^m-jU, jurXy, XJJL/^, JJ'jLt, jl^am. touches the circle. Wherefore a circle lias been inscribed in the given equilateral an d eqtiiangtUar pentagon. q.b.f. 130 EUCLID'S ELEMENTS, PROPOSITION 14. PROBLEM^ To describe a circle about a given equiiatercd and equi- angular pentagon. Let ABODE bo the given equilateral and equiangular pentagon : it is required to describe a circle about it. Bisect the angles BCD^ CDE by the straight lines OF, DF\ [1.9. and from the point F, at which they meet, draw the straight Unes FBy FA, FE. Then it may be shewn, as in the preceding proposition, that the angles CBA, BAE, AED are bisected by the straight lines BF, AF^EP. And, because the angle BCD is equal to the angle CDE, and that the angle FCD is half of the.angle BCD, and the angle FDC is half of the angle CDE, therefore the angle FCD is equal to the angle FDC; [Ax, 7. therefore the side FCia equal to the side FD. [i. 6. In the same manner it may be shewn that FB, FA, FE are each of thera equal to F(f or FD ; therefore the five straight lines FA, FB, FC, FD, FE are equal to one another, and the circle described from the centre F, at the distance of any one of them, will pass through the extremities of the other four, and will be de- scribed about the equilateral and equiangular pentagon ABCDE. Wherefore a circle has been described about the given equilateral and equiangular pentagon, q.e.p. PROPOSITION 16. PROBLEM, To inscribe an equilateral and equiangular hexagon Let ABCDEFhQ tht given cu-c :>: it is required to in- scribe an equilateral and equiangular hexagon in it. Find the centre G of the circle ABCDEF^ [III. i. BOOK IV. 15. 131 \Aviom, 1. and draw the diameter AGD • from the centre D, at the dis- tance DG^ describe the circle lomEG, (7Gf,and produce them to thepoints^, F\ and join^i?. BG, db, DE.EF, FA, The hexagon ABCDEF shall be equilateral and equiangular. For, because G i% the centre of the circle ABCDEF. GE is equal to GD; and because 2> is the centre of the circle EGCH^ DE is equal to DG; therefore GE is equal to DE, and the triangle EGD is equilateral ; therefore the three angles EGD, GDE, DEG are equal to one another. . [I. 5. clllJ^. But the three angles of a triangle are together equal to two right angles; ^ ^.32. m^^h^^ *^® a°S:i© ^^^ is the third part of two right n/?^ *^®u^xu? ?^^°®^ ^* ^^y ^® shewn, that the angle DGO IS the third part of two righi angles. «tri?M?rf«T» .^^^ J^"*^^^^* ^^"^ ^^ ^akes With the PmSf fn f^ ^-\}^^ adjacent angles J^(3^C; CG^i? together equal to two right angles, £1 13* ri'ght angles^ remaining angle CGB is the third part of two anot£^ *^® ^^^^^ ^^^' ^^^' ^^^ ^^^ ^^^^1 *^ o^« ^ ^^t'i^Vfl^r' '^"'^ '^' vertical ^opposite angles Therefore the six angles EGD, DGC, 0GB, BGA AGF FGE are equal to one another. ' 9—2 If I 132 EUCLID'S ELEMENTS, But equal angles stand on equal arcs ; [m. 26. therefore the six arcs AB, BG, CD, DE, ER FA are equal to one another. And equal arcs are subtended by equal straight lines ; [III.29. therefore the six straight lines are equal to one another, and the hexagon is equilateral. It is also equiangular. V For, the arc AF is equal to the arc ED j to each of these add the fktcABCD; therefore the whole arc FABCD is equal to the whole arc ABCDE; and thc^ angle FED stands on the arc FABCD, and the angle A FE stands on the arc ABCDE; therefore the angle FED is equal to the angle AFE. In the same manner it may be shewn that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED ; therefore the hexagon is equiangular. And it has been shewn to be equilateral ; and it is inscribed in the circle ABCDEF, Wherefore an equilateral and equiangular hexagon has teen inscribed in the given circle. q.b.f. Corollary. From this.it is manifest that the side of the hexagon is equal to the straight* line from the centre, that is, to the semidiameter of the circle. Also, if through the points "A, B, G, D, E, F, there be drawn straight lines touching the circle, an equilateral and equiangular hexagon will bo described about the circle, as may be shewn from what was said of the pentagon : and a circle may be inscribed in a given equilateral and equi- angular hexagon, and circumscribed about it, by a. method like that used for the pentagon. [III. 27. BOOK IV, 16. 133 PROPOSITION 16. PROBLEM. To inscribe an equilateral and equiangidarquindecagon in a given circle. Let ABCD be the given circle: it is required to in- scribe an equilateral and equiangular quindecagon in the circle ABCD. Let -4(7 be the side of an equilateral triangle inscribed in the circle ; [IV. 2. and let AB be the side of an equilateral and equiangular pentagon inscribed in the circle. [IV. H. Then, of such equal parts as the whole circumference ABGDF contains fifteen, the arc ABC, which is the third part of the whole, contains five, and the arc AB^ which is the fifth part of the whole, contains three ; therefore their difference, the arc J5(7, contains two of the same parts. Bisect the arc 5(7 at -^; [III. 30. therefore each of the arcs BE, EG is the fifteenth part of the whole circumference ABGDF. Therefore if the straight lines BE, EG be drawn, and straight lines equal to them be placed round in the whole circle, [IV. l. an equilateral and equiangular quindecagon will be in- scribed in it. Q.E.P. And, in the same manner as was done for the pentagon, if through the points of division made by inscnbing the quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon will be de- scribed about it ; and also, as for the pentagon, a circle may bo inaeribed in a sriyen p.niii1atf»rfl.! at^fl Annianfmlay quindecagon, and circunascribed^about it. BOOK V. DEFINITIONS. 2. A greater magnitude is said to be a mnUinU ^e fes3, when the greater is measured by the kss fci! samt JM^h'e 15 tW^tlr? hTi VVeT '^ when any equimultiples ;hatever of the fet and thlX'n' BOOK V, DEFINITIONS. 135 Wiien four magnitudes are proportionals it is usually expressed by saying, the first is to the second as the third is to the fourth. 7. When of the equimultiples of four magnitudes, taken as in the fifth definition, the multiple of the first is greater than the multiple of the second, out the multiple of the third is not greater than the multiple of the fourth, then the first is said to have to the second a greater ratio than the third has to the fourth ; and the third is said to have to the fourth a less ratio than the first has to the second. 8. Analogy, or proportion, is the similitude of ratios. 9. Proportion consists in three terms at least 10. "When three magnitudes are proportionals, the first Is said to have to the third the duplicate ratio of that which it has to the second. [The second magnitude is said to be a mean propov" tional between the first iisd the third.] 1 1 . When four magnitudes are continued proportionals, the first is said to have to the fourth, the triplicate ratio of that which^ it has to the second, and so on, quadruplicate, &c. increasing the denomination still by unity, in any num- ber of proportionals. Definition qf compound ratio. When there are any number of magnitudes of the same kind, the first is said to have to the last of them, the ratio which is compounded of the ratio which the first has to the second, and of the ratio which the second has to the third, and of the ratio which the third has to the fourth, and so on unto the last mag- nitude. For example, if A, B, C, D be four magnitudes of the same kind, the first A is said to have to the last i>, the ratio compounded of the ratio of A to B, and of the ratio of B to C, and of the ratio ofCtoD; or, the ratio of A to D is said to be compounded of the ratios of A to B,B to (7, and (7 to i>. And if^A^ has to B the same ratio that E has to Fj and B to C the sauie ratio that 6^ has to H ; and U to D the same ratio that JT has to L ; then, by this definition, A is said to have to D the ratio compounded of ratios which w-e the same with the ratios of E to F^G to H, and JT to X. / 136 EUCLIiyS ELEMENTS, I And the same thing is to be understood when it is more bneflj expressed by saying, A has to D the rati'o ^. poimded of the ratios o{Eil>F,GU, H, and 1" to Z. In Uke manner, the same things being supposed if if hM to iV_ the same ratio that^ Eas to i> ; & for i^ sake of shortness, M is said to have to N the ratio com! pomided of the ratios of ^ to ^, G to ^, and K^L. I.- 1^' f° Proportionals, the antecedent terms are said to S^e Shfr ""' '""*'"'= '" "^ *•"• oonsequeTtJ to Geomotera make use of the following technical words maSte*^'" ^^1? of "hanging eithef the ord^ or ufe pASnsji^ *~ '"^'' "^ *'"" "'oy <»n«n»e stiU to be altemata^''^w"f^' *"■ '^'r««»'^<'. ^7 permutation or aisemately; when there are four proportionals and it » S^'foX" v!*?6'"* " "^ "^^ ^'^'^ " theT^o^d'is to pro^rtio^ra i^iirftatt? j?:^ Tu^'z first as the fourth is to the third. V. K « «> the 15. Componendo, by composition : when there are fnn,. proportionals, and it is inferred, that the firrCtW Ju 5»e second, is to the second, as the third Zlf^Z with the fourth, is to the fourth. V. 18 ^S^ther 16. pividendo, by division ; when there are four wo iwrtionals, and it is inferred, that the excess of fhJ fi^«f above the second, is to the'second L the excLs of^^^^^^^ third above the fourth, is to the fourth, y.^^""^^ ^^ ^^^ 17. Convertendo, by conversion; when there arA fnnt. proportionals, and it is inferred, that the firs? is to iti 18. Bxmguali distantia, or ex wquo, from eanalitv nt distance; when there is any 'number of maSX Z^f uiiiu cvvo, aua to many others, such that thev are' nromr' tionals when taken two and two of each mnV t^^^?^'" inferred, that the first is to t^ last of the fi™t^nl '^ magnitudes, as the first is to the Ct of the othe^* BOOK V, DEFINITIONS, 137 Of this there are the two following kinds, which arise from the diflFerent order in which the magnitudes are taken, two and two. 19. Ex cBquaU. This teim is used simply by itself, when the first magnitude is to the second of the first rank, as the first is to the second of the other rank ; and the second is to the third of the first rank, as the second is to the third of the other ; and so on in order ; and the inference is that mentioned in the preceding definition. V. 22. 20. Ex (Bquali in proportioneperturhata seu inordinate, from equality in perturbate or disorderly proportion. This term is used when the first magnitude is to the second of the first rank, as the last but one is to the last of the second rank ; and the second is to the third of the first rank, as the last but two is to the last but one of the second rank ; and the third is to the fourth of the first rank, as the last but three is to the last but two of the second rank ; and so on m a cross order; and the inference is that mentioned in the eighteenth definition. V. 23. Afl TnnT*A AXIOMS. 1. Equimultiples of the same, or of equal magnitudes, are equal to one another. 2. Those magnitudes, of which the same or equal mag- nitudes are equimultiples, are equal to one another. 3. A multiple of a greater magnitude is greater than the same multiple of a less. 4. That magnitude, of which a multiple is greater than the same multiple of another, is greater than that other uiagmtude. 138 EUCLIiyS ELEMENTS, I fL II G B PKOPOSITION 1. THEOREM, If any number of magnitudes he equimultiples of as many, each of each; whatever multiple any one ojf them is of its part, the same multiple shall all the frst magni- tudes he of all the other. Let any number of magnitudes AB, CD be equimul- tiples of as rpany others E, F, each of each: whatever multiple AB is of E, the same multiple shall AB and CD together, be of ^ and i^ together. For, because AB\^ the same multiple of E, that CD is of JF*, as many magnitudes as there are in AB equal to E, so many are there in CD equal to F, Divide AB into the magnitudes AG, GB, each equal to E-, and CD into the magni- tudes CH, HD, each equal to F. Therefore the number of the magnitudes GH, HD, will be equal to the number of the magnitudes AG, GB. And, because AG i^ equal to E, and Cfl" equal to F, therefore AG and CH together are equal to E and F together ; and because GB is equal to E, and HD equal to F, therefore GB and HD together are equal to E and F together. iAxiom 2. Therefore as many magnitudes as there are in AB equal to E, so many are there in AB and CD together equal to E and F together. Therefore whatever multiple AB is of ^, the same multiple is AB and CD together, of ^ and F together. Wherefore, if any number of magnitudes &c. Q.E.D. PROPOSITION 2. THEOREM. Ai. V i\^ ^^^K ^^ i^^ ^^^^^ multiple of the second that the intra u oj t/16 fourth, and the fifth the same multiple of the second that the sixth is of the fourth; the first toge- ther with the fifth shall he the same multiple (tfthe second, JJiat the third together with the sixth U (if the fourth. H D F BOOK V. 2, 3. 139 pies of as of them is '8t magni' ) equimul- whatever ? and CD hat CD is F ? equal to lual to E ) multiple Q.E.D. B G £ Let AB the first be the same multiple of C the second, that DE the third is of F the fourth, and let BG the fifth be the same multiple of C the second, that EH the sixth is of i^ the fourth : AG^ the first together with the fifth, shall be the same multiple of G the second, that DH, the third together with the sixth, is of ^ the fourth. For, because AB is the same multiple of C that DE is of jP, as many magnitudes as there are in ^^ equal to (7, so _ many are there in DE equal to F. For the same reason, as many magnitudes as there are in BG equal to C, so many are there in EH equal to F. Therefore as many magnitudes as there are in the whole AG equal to (7, so many are there in the whole i)jy equal to F. Therefore ^^ is the same multi- ple of (7 that J9^ is of jP. Wherefore, if the first he the same multiple &c. q.e.d. Corollary. From this it is |>lain, that if any number of mag- nitudes AB, BG, GH be multi- ples of another G ; and as many DEj EK, KL bo the same mul- tiples of F^ each of each ; then the whole of the first, namely, AH, is the same multiple of C, that the whole of the last, namel v, Z>X,isofi^. I£ J 15 C K H C L F ' that the uiiple of Irst toge- ie second, irth. ~ •« PROPOSITION 3. THEOREM. If the first he the same multiple of tJie second that ths third is of the fourth, and if of the first and the third there he taken equimultiples, these shall be equimultiples, the one q/ the second, and the other of the fourth. 140 EUCLID'S ELEMENTS, £ A G C *u Vii:^ *]J®. ^I^^ ^? S*® «^™® multiple of B the second, that (7 the third ig of 2) the fourth ; Snd of A and C let the equimultiples EF and GH be taken: i&i^ shall be the same multiple of B that GH is of B. For, because ^i^ is the same multiple of A that GH is 01 O, [nypothests. as many ma^iitudes as ^ there are in jE'jP equal to ^, so many are there in G-S" equal to C. Divide EF into the magnitudes EK'y KF, ^ each equal to A\ and GH into the magni- tudes GL, LH, each equal to G. TherefoiPe the number of the magnitudes J5:^, ^i?; will be equal to the number of the magnitudes GL, LH. And because A is the same multiple of B that C is ^^^» [Hypothesis. and that E^is equal to A, and GL is equal to O; [Constr. therefore ^^is the same multiple of B that GL is of D. For the same reason XFis the same multiple of B that Therefore because EK the first is the same multiple of B the second, that GL the third is of Z) the fourth, and that ^i?^ the fifth is the same multiple of 5 the second, that LH the sixth is of D the fourth ; ^^ the first together with the fifth, is the same multiple of B the second that GH the third together with the sixth, is of £> the fourth. [y, 2. In the same manner, if there-be more parts in JET^'equal to A and m GH equal to (7, it may be shewn that EF is the same multiple of B that GH is of 2). [V. 2, (7or. Wherefore, if the first &c. q.e.d. PROPOSITION 4. THEOREM. .J.fJ^fi^^*^'^^ the same ratio to the second that the third has to the fourth, and if there be taken any equi- BOOK K 4. 141 i F C B G H multipUi whatever of the first and the thirds and dho any equimultiplet whatever of the second and the fourth^ then the multiple of the first shall have the same ratio to the multiple qf the second, that the multiple cf the third has to the multiple of the fourth. Let A the first have to B the second, the same ratio that C the third has to V the fourth ; and of A and C let there be taken any equimultiples whatever E and F, and of B and D any equimultiples whatever Okh^H: E shall have the same ratio to O tnat F has to H, Take of E and F any equi- multiples whatever K and Z, and of G and H any equimul- tiples whatever M and N, Then, because E is the same multiple) of A that JF'is of (7, and of E and F have been taken equimultiples K and L\ therefore K is the same mul- tiple of A that L is of C. [V. 3. For the same reason, M is the same multiple of -6 that N is of D. And because .4 is to J? as (7 is to 2), [Hypothesis. and of A and C have been taken certain equimultiples K and L, and of B and Z> have been taken certain equimultiples M and N; therefore if iT be greater than M, L is greater than N ; and if equal, equal ; and if less, less. But K and L are any equimultiples whatever of E and F, and iltf and iVare any equimultiples whatever of G and H\ therefore ^ is to G^ as i^ is to H. [V. D^nition 5, "Whereforey if the first SiC. q.e.p. Corollary. Also if the first have the same ratio to the second that the third has to the fourth, then any equi- multiples whatever of the first and third shall have the same ratio to the second and fourth; and the first and I P^. Definition 5. 142 EUCLID'S ELEMENTS. third Bhall have the same ratio to any equimultiDles what- ever of the second and fourth, "iuiupies wnat- .r,^^}V^ f n ^ -^ ""3: equimultiples whatever JT and L and of ^ and i> any equimultiples whatever G and ^" ' muSle if" tLt itrf-'b"' ''"'"■ '•"" ^ " *^« '«-« And because ^ is to 5 as C is to 2), lHypothe>U. and of ^ and C have been taken certain equimultioles IT miTlfi, ^"' ^ ^^^^ *^-^ *^- c^rt^e&f therefore if iT be greater than G, L is greater than H- and If equal,' equal ; and if less, less. ^ \yD^^t;^t H^!? n ^A k ^^® ^"y equimultiples whatever of JS: and F and Gf and ^are any equimultiples whatever of iand /^ therefore ^ is to ^ as ^is to 2). fV. i>^n^Y^ g.' In the same way the other case may be demonstrated. PROPOSITION 5. THEOREM. If one magnitude he the same multiple of another that a magnitude taken from the first is nfn J^J^l't^.T * i fif ^ xu 1 i ^^ *^® s^"^® multiple of the remainder Fn that the whole AB is of the whole CD, ^emainaer J^D, \\.2^ht^^i^^ T"® multiple of FD, that AE is of CF- tW^ore AE is the same multiple of CF that EG sk But ^^ is the same mulfcinlA nf nw +i,«+ >< » i^ ^i? ^« ' the^ore £isofi^;[F3(poiAe5w. B D E P therefore KH is the same multiple of F that GD is of jP ; therefore KH is equal to GD. [V. Axixm 1. From each of these take the common magnitude GH\ then the remainder C^is equal to the remainder HD. But (TiTis equal to F\ therefore HD is equal to F. C H \0(m9tvucti(m. r f f^ 144 euclid:s elements. G c H ]!^ext let GB be a multiple of E: HD shall b. [V. Axiom 1. From each of these take the common magnitude CHi then the remainder CKia equal to the remainder HD. And because CK is the same multiple of F that GB is ^^ -^» IComtruction, and that C7jr is equal to HD ; therefore HD is the same multiple of i^ that GB is of -K Wherefore, \ftwo magnitudes &a q.e.d. B D £ I r I PROPOSITION A. THEOREM. Jf the first of four magnitudes have the same ratio to tlie second that the third has to the fourth, thm, if the first he greater than the second, '^the third shall also he greater than the fourth, and if equal equal, and if less less. Tcke any equimultiples of each of them, as the doubles of each. Then if the double of the first be greater than the double of the second, the double of the ^hird is sreater tha.n tho double of the fourth. " [V. J)^nition 5, But if the first be greater than the second, the double of the first is greater than the double oi the second; BOOK V, A, B, 145 therefore the double of the third is greater than the double of the fourth^ and therefore the third is greater than the fourth. In the same manner, if the first be equal to the second, or less than it, the third may be shewn to be equal to tho fourth, or less than it. Wherefore, if the first &c. q.e.d. G H B D F PEOPOSITION B, THEOREM. If four magnitudes he proportionals, they shall also be proportionals when taken inversely. Let -4 be to J5 as C is to i> : then also, inversely, B shall be to ^ as /) is to 01 Take of B and D any equimul- tiples whatever E and F ; and of A and G any equimultiples whatever G and II. First, let B be greater than Gj then G is less than B. Then, because -4 is to J5 as <7 is to Z) ; [Hypothesis. and of A and G the first and third, G and // are equimultiples j and of B and D the second and fourth, B and F are equimultiples ; and that G is less than -£^ ; therefore // is less than F ; [V. Def. 5. that is, F is greater than H. Therefore, if B be greater than G, F is greater than H. In the same manner, if B be equal to G^ F may be shewn to be equal to H; and if less,- less. But B and F are any equimultiples whatever of B and D, and G and // are any equimultiples whatever of A a<" is to G. [V. Defmitim 5. Wherefore, if four magnitudes &c. q.e.d. 10 146 EUOLIjyS ELEMENTS. PROPOSITION C, THEOREM. \ BCD G F H If the fint he the same multiple of ths seeond, or the Mmepart of it, that the third is of the fourth, the first shall he to the second as the third is to the fourth. First, let A be the same multiple of B that C is of 2): A shall be to -S as C is to D. Take of A and O any equimultiples whatever E and F; and of B and D any equimultiples whatever G and H. Then, because A is the same midtiple of B that C is of .D ; [Hypothesis. and that E is the same multiple of ^ that ^is of C; [Constructian. therefore E is the same multiple of B that i^ is of Z); ' [V. 3. thatis,^andi^areequimultiplesofJ5and2). But'G^ and ff are equimultiples of B and D ;* [Construction, therefore if .& be a greater multiple of B than G is of ^, i^ is a greater multi- ple of Z) than ^ is of 2> ; that is, if E be greater than G, F is greater than H. In the same manner, if J? be equal to G, F may be shewn to be equal to H\ and if less, less. But E and F are any equimultiples whatever of A and G, and G and H are any equimultiples whatever of B and D ; [C^niti(m 5, Next, let A bo the same part of B that C is of 2): A shall be to jB as Cis to 3, For, since A is the same part of B that C is of i>, therefore B is the same multiple of A that J) is of C; therefore, by^he preceding case, B is to ^ a» 2> 1» to O' ; therefore, inversely, ^ is to J5 as C' is to D. Wherefore, if the first &c. q.b.d. D [y. B, BOOK V. A 7. 147 A B C D r PROPOSITION i). THEOREM, ^f the fir it he to the second as the third is to the fourth and Y the first be a multiple, or a part, of the second, the third shall be the same multiple, or the same part, of the fourth. ' *^ Let ^ be to j5 as C7is to 2>. And first, let ^ be a multiple oiB: C shall be the same multiple of 2>. Take B equal to A ; and what- ever multiple ^ or J? is of B, make F the same multiple of D, Then, because A is to ^ as C* W to D, [Hypothesis. and of B the second and D the fourth have been taken equimultiples ^ and -Fj [C(mitruction. therefore ^ is to ^ as C is to ^' [V. 4, Corollary. But A is equal to E ; [Construction. therefore Cis equal to i^. [V. A. And F is the same multiple of D that ^ is of -6 ; [Construction. therefore C is the sam3 multiple of D that A is of B. Next, let ^ be a part otB.C shall be the same part of i> For, because ^isto^asCistoD; [HypothesU. therefore, inversely, ^ is to -.4 a?:* Z> is to C [V B But ^ is a part of 5; ' [Hypotlesi^, that IS, 5 IS a multiple of A ; therefore, by the preceding case, D is the same multiple of Cj that is, C is the same part of 2> that A is of B, y^herefore, if the first &c. q.e.d. PROPOSITION 7- THEOREM. Equal magnitudes have the same ratio to the sams magnitude; and the same has the same ratio to equal *fi'(iffnziuaeSt 10—2 I 148 EUCLID'S ELEMENTS. D A G B Let A and B be equal magnitudes, and C any other magnitude: each of the magnitudes A and B shall have the same ratio to (7; and G shall have the same ratio to each of the magnitudes A and B. Take of A and B any equimultiples ■whatever D and E ; and of C any mul- tiple whatever F, Then, because D is the same mul- tiple of A that E is of 5, [Construction, and that -4 is equal to B ; [Hypothesis. therefore D is equal to E. \Y. Axiom 1. Therefore if D be greater than F, E is greater than ^; and if equal, equal; " " 6 F and if less, less. But D and E are any equimultiples whatever of A and B, and i^ is any multiple whatever of (7; [Construction. therefore A is to (7 as B is to (7. [V. D^. 5. Also C shall have the same ratio to A that it has to B. For the same construction being made, it may be shewn, as before, that D is equal to E. Therefore if F be greater than D, F is greater than E ; and if equal, equal ; and if less, less. But F is any multiple whatever of C, and D and E are any equimultiples whatever of A and B; [Construction. therefore Ciato A as C is to B. [V. Definitvm 5. Wherefore, eqtial magnitudes &c. q.e.d. PROPOSITION 8. THEOREM. Of uneqtud magnitudes^ the greater has a greater ratio to the same than the less has; and the same mag- nitude has a greater ratio to the less than it has to the greater. Let AB and BG be unequal magnitudes, of which AB is the greater; and let D be any other magnitude what- ever : AB shall have a greater ratio to D than BG has to D ; and D shall have a greater ratio to BG than it has to AB. BOOK r. 8. 149 f If the magnitude which is not the greater of the two AC. CB, be not less than D, take EF, FG the doubles oi AG, GB (Figure 1). But if that which is not the E Fig. r. greater of the two AG, GB, bo less than D (Figures 2 and 3), this magnitude can be multiplied, so as to become greater than i>, whether it be ^0 or GB, Let it be multiplied until it be- comes greater than i), and let the other be multiplied as often. L K ? Let EFhe the multiple thus taken o{AG, and FG the same multiple oiGB; therefore FF and FG are each of them greater than D. And in all the cases, take IJ the double of D, ^its triple, Fig. 2. Fig. 3. and so on, until the multiple ^ of i) taken is the first which ' '^ is greater than i^G'. Let X bo that multiple of D, namely, the first which is greater than FG ; and let K be tho multiple of Z> which is next x T less than L. Then, because L is the first multiple of Z) which is greater q b than FGj [Construction, L K H P Cl the next preceding multiple K is not greater than FG ; that is, FG is not less than K. And because FF is the same multiple of AC that FG is of CBf [Construction. therefore EG is the same multiple of AB that FG is of GB ; [V. 1. that is, EG and FG are equimultiples of AB and CB. E c- • B X K P 150 EUCLIUS ELEMENTS, lOomtruction, And it was shewn that EG is not less than iT ^ and EE is greater than D ; [C(mHmctvm. therefore the whole EG is greater than JT and i> together. But ^and D together are equal to Z; therefore EG is greater than L. But EG is not greater than L. And ^(y and FG were shewn to be equi- multiples of ^^ and BC j ^ and Z is a multiple of D. [Construction. Therefore AB has to Z> a greater ratio than BC has to JD. [v. Dejinitimi 7. Also, D shall have to J5C a greater ratio than it has to AB. i C For, the ,same construction being made. ^Ir^?^.^® s^®^'^' *hat L is greater than ^ ^ 6? but not greater than EG. I And Z is a multiple of Z>, [Construction. md EG and J^6^ were shewn to be equi- multiples of AB and CB. TWore i> has to BC a greater ratio than it has ^p., - ^ [V. I>^nition 7. Wherefore, qf unequal magnitudei &c. q.e.d. I PKOPOSITION 9. THEOREM. Magnitudes which have the same ratin m /x^ -«^ m^i^'nt*.^/^, ar. .^„«; And, because (7 is to .8 as (7 is to u4, [Hy^otheaU. and that F the multiple of the first is greater than E the multiple of the second, [Corw^rMcfion. therefore F the multiple of the third is grea4ier than 2> the multiple of the fourth. [V. Definition 5. But F is not greater than Z> ; IConstruction. [Constmctioru [V. Definition 6. [(7(WM are equimultiples of B and ^, and that E is less than D, therefore B is less than A, [y. iaj^wji 4. BOOK r. 11. 153 PROPOSITION 11. THEOREM, Ratios that are the same to the same ratio, are the same to one another. Let ^ be to i? as C is to 2), and let Cbe to 7> as ^ la to i^ : ^ shall be to J5 as j& is to F, H- A B- L- D- r- Take oi A^C, E any equimultiples whatever G,H,K\ and of By i>, F any. equimultiples whatever Z, J/, N. -* Then, because ^ is to B as (7is to J9, [Hypothesis. and that G and H are equimultiples of A and C, and L and M are equimultiples of B and D ; [Construction. therefore if G be greater than Z, J? is greater than M; and if equal, equal ; and if less, less. [V. Definition 6. Again, because (7 is to Z) as -^ is to F, [Hypothesis. and that H and K are equimultiples of C and Ey and Jf and N are equimultiples of Z> and F ; [Conserucfiow. therefore if If be greater than J/, if is greater than N; and if equal, equal ; and if less, less. [V. Definition 6. But it has been shewn that if G be greater than Z, H is greater than M ; and if equal, equal ; and if less, less. Therefore if G be greater than Z, K is greater than N; and if equal, equal ; and if less, less. And G and K are any equimultiples whatever of A and E, «,nd Z and N^ are anv eauimultipies whatever of B and ^. Therefore ^ is to 5 as i^ is to F. [V. iJtySniiton 5. Wherefore, ratios that are the safne &c. Q e,d. I 154 EUCLIUS ELEMENTS. PKOPOSITION 12. THEOREM, Tf anynurr^hor of moffnitudes be proportionals, as ons (If the ante . .. ?'^ ,, to its consequent^ so shall all the ante- cedents i c iu a!:, lis ccmsequents. Let any number of magnitudes A, ByaD, E Fh& proportionals ; namely as ^ is to i?, so let C be to i), and A- H- c- E- — M- F- ar.7^}%^l<^iP' ^ ^"7 equimultiples whatever 6?, ^, JT; and of B, D, F any equimultiples whatever Z, J»/, iV". ' ,.«7fh?f'/S®^^T ^ " *o.^ as C is to 2) and as JF is to /; and that (? //; ^ are equimultiples of A, a E, and Z, ifcT, iV* equimultiples of ^,i>,^; ' ' [C7on,.;u^o^ therefore if (? be greater than X, i7 is greater than M, fo°s ^ greater than iV^; and if equal, equal ; and if less' * [V. Definition 5. Therefore, if G^ be greater than Z, then G, ZT, JT together andlf'l^ss'"leM "" ^' '^ together; and if equal, equal; But G^, and G, H, K together, are any equimultiples whatever of^, and ^,C;Z? together; [v-l; and Z, and Z, ^, iV" together are any equimultiples what^ ever of 5, and 5, Z>, /'together. [y f Therefore as ^ is to J5, so are ^, C7, ZT together to Wherefore, i/*a«ynwm&^r&c. q.e.I). PROPOSITION 13. THEOREM. ,Jf}^fip^J^^^ the same ratio to the second which the third has to the fourth, hut the third to the fourth a greater BOOK r. 13. 155 ratio than the fifth to the sixths thejirst shall have to the second a greater ratio than the fifth has to the sixth. Let ^ the first have the same ratio to B the second that C the third has to D the fourth, but C the third a greater ratio to D the fourth than E the fifth to F the sixth: A the first shall have to B the second a greater ratio than E the fifth has to F the sixth. H- C D N- E For, because C has a greater ratio to D than E has to Fy there are some equimultiples of G and E^ and some equi- multiples of D and F, such that the multiple of C is greater than the multiple of />, but the multiple of E is not greater than the multiple of F. [V. Definition 7. Let such multiples be taken, and let G and H be the equi- multiples of V and E, and K and L the equimultiples of i>andi^; so that Q is greater than JT, but ^is not greater than L. And whatever multiple G is of (7, take M the same mul- tiple of A ; and whatever multiple K is of D, take N the same multiple of B. Then, because ^ is to -B as C is to D, [ffypothais. and M and G are equimultiples of A and C, and N and A" are equimultiples of B and 2> ; [Oomtructim. therefore if M be greater than N, G is greater than AT ; and if equal, equal ; and if less, less. [V. Definition 5. But G is greater than K; therefore M is greater than iV. But H is not greater than L ; [Construction, [Construction, J n^ J Tl- aau jlki uuu Jul iiiu uquimuibipies oi -d ana ^, ana iv ana ju are equimultiples of J5 and F\ [Constructum, therefore ^ has a greater ratio to B than ^ has to i^. Wherefore, if the first &c. q.e.d. 156 M i EUCLID'S ELEMENTS. Corollary. And if the first have a greater ratio to the second than the third has to the fourth, but tKird tj^T^ "1*^° to the fourth that th3 fifth hai to the sixth. It may be shewn, m the same manner, that the first has a greater ratio to the second than the fifth has to the sixth. PROPOSITION 14. THEOREM, *ir^ J^ ^rK!^l^ *^^ ^^^^ ^^^^^ io ihe second tJiat ths the third the second shall he greater than the four ^h- and \f equal, equal; and if' less, less. ' kx^.l'n:^ ft® S*?* ^^^® *^® sa^^e ratio to B the second Csih^^^''^ ^T *?u^ *^® ^^"^*h ^ if ^ be greaterXn leif f ^ ^^*®'' *^^° ^ ^ if ®^"^^' ^^"^1 i and if less" '^ 8 A B C D BCD A J3 C D li-or k' ' . . greater than G: ^shaU be greater tha« 2). For, because ^ is greater than a * \m,r.^ and 5 is any other magnitude; C^^Po^Aem. therefore ^ has to B a greater ratio than Chas to ^ [V 8 But ^ 18 to ^ as (7 is to D. ,„ \' ^' Thereibre Chas to 2>agreater ratio than C has tJ^^^^^^^ greatei ^tL^s Wsf "' *'^^ '' ^"^'^ *^^ -me has'the Therefore i> is less than ^; that is, B is greater than />!'' vJT^'T "^J^^ "^^^"^ *^ ^'- ^ «^" l>« equal to D. For, ^ IS to ^ as (7, that is A, is to i>. V.«o^Ae.M Therefore B is equal to D. i^VPoth^^^ BOOK r. 14, 13. 167 er ratio to It the third > the sixth, first has a the sixth. d that the 'ater than '*rth; and he second eater than nd if less, C D erthanZ). hypothesis, > B. [V. 8. hypothesis. B. [V. 13. le has the [V. 10. bhan D. il toZ). hypothesis. [V. 9. Thirdly, let A be less than C: B shall be loss than V. For, C is greater than A. And because C '^ to /) as ^ is to J5 ; [ffypothms. and C is greater \ i A; therefore, by c '. case, I> is greater than i? ; that is, J9 is 1 'ss t* -in 2>. Whei-efore, ir h jjirst &c. q.e.d. PROPOSITION 15. THEOREM, Magnitudes have the same ratio to one another that their equimultiples have. Let AB be the same multiple of C that DE is of F-. C shall be to F as AB is to DE, For, because AB is the same multiple of C that DE is of JP, [Hypothesis. therefore as many magnitudes as there are in AB equal to G, so d many are there in DE equal to F. Divide AB into the magnitudes ^(y, GH, HB, each equal to G\ and i>^ into the magnitudes DK, KL^ LE, each equal to F, Therefore the number of the mag- nitudes A G, GH, HB will be equal to the number of the magnitudes DKy KL, LE. And because AG^ GHy HB are all equal; [Consimction. and that DK, KL^ LE are also all equal ; therefore ^(^ is to DK as GH is to -£X, and as HB is to LE. [V. 7. But as one of the antecedents is to its consequent, so are all the antecedents to all the consequents. [V. 12. Therefore as u4(t is to DK so is AB to DE. But AG\^ equal to (7, and DK is equal to F, Therefore as (7 is to -F so is AB to DE, Wherefore, magnitudes &c. q.e.d. G H K '«L^. 158 EUCLID'S ELEMENTS. PEOPOSITION 16. THEOREM. fjV-^S^n ^'*^?»^«*^^' of the same kind he proportionals iMy shall also be proportionals when taken alternately. Let A, B, Cy p be four magnitudes of the same kind which are proportionals; namely, as ^ is to ^ so let be i«f 1 • ^l '] l^^*i ^® proportionals when taken alter- nately, that IS, A shall be to C as ^ is to D E- A— B- H- i! i I ! I T{?ke of ^ and B any equimultiples whatever E and F. and of C and 2) any equimultiples whatever G and H. Then, because E is the same multiple of ^ that F is of B, and that magnitudes have the same ratio to one another tnat their equimultiples have ; [v. 15. therefore -4 is to i? as ^ is to /: But ^ is to ^ as a is to i>. [HypotJiesk. Thereibre (7 is to 2) as JE7 is to F. [y. n. Again, because G and Zfare equimultiples of G and 2)! therefore C7 is to 2) as G^ is to iT. [v. 15. But it was shewn that (7 is to 2) as ^ is to i^. Therefore J^ is to Z' as (3^ is to ^. ' [y. n. But when four magnitudes are proportionals, if the first be greater than the third, the second is greater than the fourth ; and if equal, equal ; and if less, less. [V. 14. Therefore if E be greater than. G, F is greater than ZT* and if equal, equal ; and if less, less. But ^ and 2^ are any equimultiples whatever of A and An ^^® ^^ equimultiples whatever of G ^ ^' iComVmctym. Therefore ^ is to C as 5 is to 2). [V. Btfinitwn, 5. Wherefore, i/-/ottr w^w»7Mfl?^j? &a q,e.d= BOOK V. 17. 159 PROPOSITION 17. THEOREM. If magnitudes, taken jointly, he proportionaiSf they thall also be proportionals when taken separately; that isj if two magnitudes taken together have to one qf tJhem the same ratio which two others have to one of these, the remaining one of the first two shall have to the other the same ratio which the remaining one of the last ttvo has to the other qf these. ^ ^ Let JB, BE, CD, DF be the magnitudes which, taken jointly, are proportionals ; that is, let ^^ be to BE as CD is to DFi they shall also be prwortionals when taken separately; that is, AE shall be to EB as C7JFig to FD, Take of AE, EB, CF, FD any equimultiples whatever QH, HK^ X LM,MN', and, again, of EB, FD take any equimultiples whatever KX, NP. Then, because ^^is the same multiple of -4^ that HK\% of EB ; therefore GH is the same multiple of AE that GX is of AB. [V. 1. But GJI is the same multiple of AE that LM is of CF, [Conttr, therefore GlCis the same multiple of AB that LM is of CF. Again, because LMh the same multiple of CF that MN is of FD, therefore LM is the same multiple of CF that LN is of CD. j-y J SHS^^^ ^*! *®^ ^ ^® *^® same multiple of CF that CriL IS 01 AB. Therefore GIT is the same multiple of AB that ZiV is of VD ; tliat is, GITtLnd ZiV are eauimnUinlpa fif A Tt »»fi nT\ ^ "-•~*a vsj^,-s*vCT T»'i afA^fi^ «7£X\«& w'^t*^# H E- G A D F K H [Construction. 160 EUCLID'S ELEMENTS, II ; I . ! >I M Again, because UK is the same multiple ot EB that MN is of FDf and that KX is the same multiple of Eli that iVP is of FDj [Construction. therefore HX is the same multiple of EB that MP is of FI> ; [V. 2. ^ that is, IfX and 3fP are equimulti- ples of EB and FD. And because -4jB is to BE as CD is to DF, [Hypothesis. j; and that GX and ZiV are equimul- tiples ofAB and CZ>, and HX and 3zP are equimultiples of EB and H FD, thereforeiffi^iTbe greater than ^X, E ZiVis greater thanifcfP ; and if equal, equal ; and if less, less. [V. Bef. 5. But if txli be greater than XX, G A ^ «„ then, by adding the common mag- nitude HX to both, GX is greater than HX ; therefore also LN is greater than MP ; and, by taking away the common magnitude MN from both, LM is greater than NP. Thus if GHhe greater than XX, LM is greater than NP. In like manner it may be shewn that, if GH be equal to XX, LM is equal to NP ; and if less, less. But GH and LM are any equimultiples whatever of AE and CF, and XX and NP are any equimultiples whatever of EB and FD ; [Construction. therefore ^^ is to EB as CF is to PD, [V. Bejinition 6, Wherefore, if four magnitudes &c. q.e.d. ■^ II PROPOSITION 18. THEOREM. If magnitudes, taken separately, he proportionals, they shall also he proportionals when taken jointly; that is, if the first he to the second as the third to the fourth, the first and second together shall he to the second a* the third and fourth together to the fourth. BOOK r. 18. 161 EB that le of EB Mtruction. K M Let AE, EB, OF, FD be proportionals ; that ia, let ^t^ be to EB as CF is to FD: they shall also be propor^ tionals when taken jointly : that is, AB shall be to BE as (7i> is to i>i^. Take of AB^ BE, CD. DF any equimultiples whatever QH,HK,LM,MN\ and, again, of BE^ DF take any equimultiples whatever Then, because KO and NP are equimultiples of BE and DF, and that KH and iVJf are also equimultiples of BE and DF ; [Construction. therefore if ICO, the multiple of BE, be greater than ^jfiT, which is a multiple of the same BE, then iVP the multiple of DF is also greater than NM the multiple of the same DF] and if iTO be equal to Kff, NP is equal to NMi and if less, less. M'N from than NP. be equal latever of [multiples nstruction. definition 5. nals, they that is, {f mrth, the the third I B £ liff P First, let KO be not greater than KH\ therefore NP is not greater than NM, And because GH and HK are equimultiples oi AB and BE, [Construction. and that AB is greater than i?JB; therefore GH is greater than HK', [V. liciow ?. J but KO is n( *: greater than KH \ [ffynothesis. therefore {'lU w greater thau KO. In "'ke manner it may be 3h ^'n tbit LM is great' .^an NP. { Thus if KO be not greater than KII, fcheii G'^, the multiple of AB, m always greater than KO, uhe multiple of BE ; and Ukewisc LM,^lQ multiple of OD, is ^.reater hm NP, mv multiple of x^r ; n li -«r'y*^?:^?^:- rrr 162 EUCLID'S ELEMENTS. m V Next, let KO be greater than KH; therefore, as has been shewn, NP is greater than NM, And because the whole GH is the same multiple of the whole AB that HK is of BEj [Comtructwn. therefore the remainder GK is the same multiple of the remainder AE that GH is of AB ; [V. 5. which is the same that Z3f is of CD, [Construction, In like manner, because the whole LM is the same multiple of the whole CD that MN is of DF, [Construction. therefore the remainder LN is the same multiple of the remainder CF that LM is of CD. [V. 5. But it was shewn that LM is the same multiple of CD that GKiaofAE. Therefore GK is the same multiple of AE that LN is oiCF; that is, (r^and ZiV are equimultiples of ^i^ and CF. And because KO and NF are equimultiples of BE and DF'f [Construction. therefore, if from KO and NP there be taken Kff and NMf which are also equimultiples of BE and DF, [Constr. the remainders ffO and MP are either equal to BE and />^, or are equimultiples of them. Suppose that HO and MP are equal to BE and DF. Then, because ^-^ is to EB as CiP is to FDj [Hypothesis. and that GK and ZiV are equimultiples of ^i^ and CF; therefore GKia to EB as LN is to FD. [V. 4, (7or. But ^0 is equal to BE^ and JfP is equal to DF; [Hyp. therefore GKh to HO as LN io bU JXi. t. * [V. 6. 0| H G B P M Kl- 1> BOOK F. 18. 163 Oi H B £ 1 F A C N Tlierefore if GK be greater than HO^ LN is greater than MP \ and if equal, equal ; and if les«, less. [V. A . Again, suppose that HO and MP are equimultiples oi EB 21x16. FD, Then, because ^^ is to EB as Ci^ is to FD\ [Hypothesis. and that GK and ZiV are equimultiples of ^i^ and CF, and //O and MP are equi- multiples of EB and /7) ; therefore if GK be greater than HO^ LN is greater than MP J and if equal, equal ; and if less, less ; [V. Definition 5. which was likewise shewn on the preceding supposition. But if Gffhe greater than KOj then by taking the com- mon magnitude AjETfrom both, GK is greater than HO; therefore also ZiV is greater than MP; and, by adding the common magnitude NM to both, ZM is greater than NP, Thus if GH be greater than KO, LM is greater than NP. In like manner it may be shewn, that if GH be equal to JTO, LM is equal to NP ; and if less, less. And in the case in which KO is not greater than KH it has been shewn that GH is always greater than KO and also LM greater than NP, ' But GK And LM are any equimultiples whatever of AB and CD, an ^ KO and NP are any equimultiples whatever of BE and DF, iComtruction. therefore ^^ is to BE as CD is to DF. [V. Definition 5, y^h&cetorej if magnitudes &Q, q.e.d. 11—2 164 EUCLID'S ELEMENTS. . F PROPosmoisr 19. theorem. If a whole nmgnitude he to a whole as a magnitude taken from the first is to a magnitude taken from the other, the remainder shall he to the remainder as the whole w to the whole. Let the whole AB be to the whole CD as AE, a mag^ mtude taken from AB, is to CF, a magnitude taken from CX>: the remamder EB shall be to the remainder FD as the whole AB is to the whole CD. For, because AB is to CD as AE is to ^^i [Hypothesis, therefore, alternately, AB is to AE as CD IS to GF. [V. 16, And if magnitudes taken jointly be pro- portionals, they are also proportionals when taken separately; [v. 17. therefore EB is to AE as FD isto CF\ therefore, alternately, EB is to FD as AE is to CF. [V. 16. But AE is to CF as AB is to CZ) ; {hJ, i^GYQioYQEBhioFD 2isABhioCD. [V.11. Wherefore, if a whole &c. q.e.d. Corollary. If the whole be to the whole qa a mao^- mtude taJcen from the first is to a magnitude taken from the other, the remamder shall be to the remainder as the magnitude taken from the first is to the magnitude taken from the other. The demonstration is contained in the preceding, i. PROPOSITION E. . THEOREM. If four magnitudes he proportiondls, they shall also he proportionals hy conversion ; that is, the first shall he to ifs excess ahove the second as the third is to its excess above the fourth. D AT? Jlf ;^^.^? ^A^ ^ ^^ w to ^^'' ^B shall be to BOOK V, E, 20. 165 For, because AB is to BE as CD is to DF] ^ [Hypothesis, therefore, by diyision, AE is to EB as CF is to Fl>; [V. 17. and, by InTersion, EB is to AE as FD is to CF. [V. B. Therefore, by composition, AB is to AE as CD is to C/l [V. 18. Wherefore, if/our magnitudes &c. q.e.d. E B D- PROPOSITION 20. THEOREM, If there he three magnitudes, and other three, which have the same ratio, taken two and two, then, if thejirst be greater than the third, the fourth shall he greater than the sixth; and if eqwd, equal; and if less, less. Let A, B, C be three magnitudes, and D, E, F other three, which have the same ratio taken two and two ; that is, let u4 be to .B as Z) is to E, and let j5 be to C' as j& is to i^: it A be greater than C, D shall be greater than F; and if equal, equal ; and if less, less. First, let A be greater than C: D shall be greater than F. For, because A is greater than C, and B is any other magnitude, therefore A has to ^ a greater ratio than (7 has to J5. [V. 8. But ^ is to J5 as i) is to ^; [Hypothesis. therefore D has to J^ a greater ratio than C has to B. [V. 13. And because JB is to C as ^ is to F, [Hyp. therefore, by inversion, C is to B bb F i. to E. [V. B. And it was shewn that D has to ^ a greater ratio than C has to B ; therefore D has to ^ a greater ratio than i^hastojF; [V. 13, C^or. therefw^ D is greater than F, [V. 10. Abo 1 166 EUCLID'S ELEMENTS, 'A Secondly, let A be equal to C: D shall be equal to F. For, because A U equal to G, and B is any other magnitude, therefore -4 is to J5 as C is to 5. [V. 7. But ^ is to i? as 2) is to E^ [Hypothms, and C7 is to -5 as Fh to E, [Hyp. V. B, therefore D is to j& as -F is to J? ; [V. 11. and therefore D is equal to F, [V. 9. Lastlv, let A be less than C: D shall be less than F, For (7 is greater than A ; and, as was shewn in the first case, C7 is to jBasJf'isto^j ' and, in tlie same manner, ^ is to ^ as ^ is to 2); therefore, by the first case, F is greater than D ; that is, 2> is less than F, Wherefore, if there he three &c. q.e.i>. Ji 6 ABC I> E P M fS » PflOPOSITION 21. THEOREM. If there be three magnitudes, and other three, which have the mmf ratio, taken two and two, hut in a cross order th&mf the first he greater than the third, the -ClZi. '^^Jf ^T^'"* ^'^ ^^ ''^^^' ^^^ if ^ual, equal ; and if less, less. ^ «^»«*i Let A, B C be three magnitudes, and 2>, F, F other th^e, which have the same ratio, taken two W two. but n a cross order; that is, let ^ be to i? as ^ is to y and less ^ ' ^" ®*^"^^' ^"^""^^ ^ ^"^ '^ ^^' First, let A be creater than n- n oii, loan i^; ,11 k^ Ltf«».^A "w {^* w«««W« BOOK K 21. 167 For, because A is greater than (7, and B is any other magnitude, therefore A has to B & greater ratio than C has to B, [V. 8. But ^ is to 5 as ^ is to i^; [EypotJiesu. therefore E has to i^ a greater ratio than (7 has to ^. [V. 13. And because Z? is to C7 as Z> is to B, [Hypothesis, therefore, by inversion, (7 is to J5 as B is to D. [V. ^' And it was shewn that B has to F a greater ratio than C has to B; therefore J^ has to i^ a greater ratio than -S has to J? ; [V. 13, Cor, therefore F is less than D ; [V. 10. that is, JD is greater than F. Secondly, let A be equal ioC: J) shall be equal to F, For, because A is equal to (7, and B is any other magnitude, therefore ^ is to i? as G is to B. [V. 7. But ^ is to i? as J5^ is to i^ ; [Hyp, and C7is to 5 as jR? is to i) ; [ffyp.y- B, therefore ^ is to JF* as JS: is to i> ; [V. 11. and therefore D is equal to F. [V. 9. Lastly, let A be less than (7: D sliall be less than F. For C is greater than A ; and, as was shewn in the first case, C is to 5 as J^ is to 2) ; and, in the same mannerj i? is to ^ as Fisto^; therefore, by the first case, F'n greater than D ; that is, 2> is less than F. Wherefore, \f there he three &e. q.e.». A P B E C A B i D E F A B D E C F 168 ^K II 1 I'l EUVLWS ELEMENTS. PROPOSITION 22. THEOREM. If there he any number of magnitudes, and at manp others, which have the same ratio, taken two and two in order, the first shall have to the last of the first maa- mtudes the same ratio which the first of the others Aa« to the last. [This proposition is usually cited by the words ex a^aU.I First, let there be three magnitudes A, B, C, and other three JJ, E, F, which have the same ratio, taken two and two m order ; that is, let ^ be to 5 as i) is to E, and let B be to C/ as -E" is to Fi A shall be to Cas DiaioF. Take of A and D any equi- multiples whatever G and H; and of J^ and E any Equimul- tiples whatev er K and L ; and of G and F any equimul- A ]& d? tiples whatever M and N. Then, because ^ is to ^ as 2) ^ ^ "^ is to E', [ffypothesis. and that G, and IT are equi- multiples of ^ and D, and ^ and i equimultiples of BmidE; IComtructian. therefore 0^ is to iT as -S" is to For the same reason, -T is to Jf as Z is to N. And because there are three magnitudes G, K, M, and other three H, L, iV, which have the same ratio tken two anci two, therefore if G be greater than M, H is greater than iV; and if equal, equal ; and if less, less. [v. 20. ^^A ^ ^^A ?r ^^® ^"y equimultiples whatever of A and D, and ilf and i\r are any equimultiples whatever of Cand F. Therefore ^ is to C as i> is to F. [v. mnition 6. Next, let there be fou- magnitudes, A, B, C, />, and BOOK V. 22, 23. 169 A. B. 0. D. £. F. G. H. other four J?, F^ (?, -ff", which have the same ratio taken two and two in order ; nameiv, let 4 be to J5 as ^ is to F^ and i5 to C' as Z' is to Gf and C7 to D as Qi^io H.A shall be to Z> as JS; is to H. For, because A, B,G are three ma^itudes, and Ey Fy G other three, which have the same ratio, taken two and two in order, [ffypothesu. therefore, by the first case, -4istoCasJ^isto6r. But C is to 2> as G^ is to ^j [Hypothesis. therefore also, by the first case, .4 is to JO as -E^ is to H, And so on, whatever be the number of magnitudes. "Wherefore, y tJiere he any number &c. q.e.d. PROPOSITION 23. THEOREM. If there he any number of magnitudes, and as many others, which fiave the same ratio^ taken two and two in a cross order, the first shall have to the last of the first magnitudes the same ratio which the first of the others has to the last. First, let there be three magnitudes, A, B, G, and other three D, E, F, which have the same ratio, taken two and two in a- cross order; namely, let ^ be to -B as J^ is to -F, and B to (7 as Di^toE. A shall be to Cas i> is to F. Take of A, B, 2> any equimultiples whatever G, H, K', and of C, E, F any equimultiples whatever L, M,N. Then because G and JOT are equimultiples of A and B, and that magnitudes have the same ratio which their equimultiples have ; [V. 15. therefore ^ is to ^ as (r is to^. ABC a H L j> E r Aliu, lur bixo 5aiu6 ioEdVu, -S is to i^ as iHf is to iV. >.^. .^a.^ IMAGE EVALUATION TEST TARGET (MT-3) fe ^ a\ 1.0 I.I 1^128 |2^ US ^ ^" ja yi 12.2 U8 Pi a 14.0 1.8 11.25 ju 1.6 ^t— - 6" — ► ^ />lii!^ .%'"'■ /: ^ '/ r Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 ^ V iV \\ [V ^qU -^A^^ <^^<^^ ^J^ -C^' ^° J "^ m 170 BUCLliyS ELEMENTS i ABC ^5 1- K But ^ 18 to J9 as ^ is to ^» [ffypotkens. Therefore ^ is to ^ as 3/ 18 to iV: [V. 11. And because B isio C as J9 is to JS; [ffypothesia, and that IT and iT are equimultiples of B and D, and L and M are equimul- tiples of (7 and JS^; iComtr, therefore if is to Z as AT ia to Jf. [V. 4. And it has been shewn that (y is to ^ as Jf ia toiV. Theii since there are three magnitudes 6?, II, Z, and other three IT, M, N, which have the same ratio, taken two and two m a cross order; therefore if ^ be greater than X, JTis greater than JV; and If equal, equal ; and if less, less. [V. 21 . ^*i* ? ^"? ^*^® ^^^ equimultiples whatever of ^ and Z>, «ttd i and iy are any equimultiples whatever of G and jP; I [Y. Definition 5, A. B. C. D. E. F. G. H. therefore -4istoCasi>istojFf Next, let there be four magnitudes ^i' .4' ?' ^' *"^ ®^^®'" ^^^^^ ^» -^» ^» ^» which have the same ratio, taken two and two in a cross order ; namely, let ^ be to 5 as (? is to If, and 5 to C as i^is to 6?, and C' to X> as J^ is to jP; -4 shall be to 2> as J5: is to ^ xv^^5i^®^^?®.-^\^' Care three magnitudes, and i^, G,II other three, which have the samQ ratio, taken two and two m a cross order ; [ffypotkesis. therefore, by the first case, ^ is to C as i^ is to JK But C is to 2) as iE? is to i^; [ffypotheit^, therefore also, by the first case, :^ is to 2) as ^ is to ij. And so on, whatever be the number of magnitudes. ■•Tri I? uoi eioi o, V imre oe any number &c. Q.BLD. BOOK y. 24. 171 PROPOSITION 24. THEORSM, Tf the Jirtt have to the second the same ratio which the third has to the fourth^ and the fifth have to the second the same ratio which the sixth has to the fourth^ then the Jirst and fifth together shaU ^^ave to the second the same ratio which the third and sixth together have to the fourth. Let AB the first have to C the second the same ratio which DE the third has to jP the fourth ; and let BG the fifth haye to C the second the same ratio which EH the sixth has to J' the fourth : AG, the first and fifth together, shall have to C the second the same ratio which Dlf, the third and sixth together, has to J" the fourth. For, because -ffG^ is to C as 1?^ is to JF, [Hypothms. therefore, hy inversion^ C is to BG asFiBtoEff. [V. jB. And because AB is to C as DE is to-F, [Hypothms. B and C7is to J56? as -F is lo ^^; therefore, ex aequali, AB is to BG BsDEistoEH. [V. 22. And, because these magnitudes are proportionals, they are also proper- J^ tionals when taken jointly ; [V. 18. therefore ^G^ is to BG as DH is to EH. But BG u to C SiS EH is to JF; therefore, ex sequali, AG is to C&b DH is to i^." ' |. V. 22. V^here^QrOy if the first &c. q.b.d. CoBOLLARY 1. If the same hvpothesis be made as in the proposition, the excess of the first and fifth shall be to the second as the excess of the third and sixth is to the fourth. The demonstration of this is the same as that of the proposition, if division be used instead of composition. CoROLLART 2. The proposition holds true of two ranks of magnitudes, whatever be their number, of which each of the first rank has to the second magnitude the same ratio that the corresponding one of the second rank has to the fourth magnitude ; ns is manifest n £ I D [ffyp hesis. 172 EUCLIUS ELEMENTS, \ B £> H PEOPOSITION 25. THEOREM, Tffour magnitudes of the tame kind le proportionals the greatest and least qf them together shall he greater than the other two together. y^i^^er Let the four ma^itudes AB, CD, E, F be propor- tionals; namely, let AB be to CD asEktoF; ^d let AH bo the greatest of them, and consequently F the • [V. ^ V. 14 t^^th^^ ^ together shall be greater than CJ) and E Take AG equal to E, and Cff equal to F, Then, because AB is to CD a« ^ is to jP, iffypothesis. and that AG ia equal to E, and CH equal to F ; [Com^ruc^ion. therefore AB ia to <)!) sls AG is to (7£r. [V. 7, V. 11. And because the whole AB is to the whole CD as J Gf is to CH; the whole AB is to the whole CD. [y 19 But ^5 is greater than CD; Iffypothesi^. therefore BG is greater than Dff, j-y ^ And because ^ C is equal to E and Cff equal to i^; [Comtr to ethT/ ^^ ^^ ^ together are equal to Cff and i And if to the unequal maffnitudes BG, Dff, of which BG IS the greater, therele added e^ual maLE namely^(y and J- to BG, and C^ and^^ to D^en AB and ^together are greater than CD and E together. 'Wherefore, i/four magnitudes &c. q.e.d. I E BOOK VL DEFINITIONS. /I 1. Similar rectilineal figures are thQse which have their several angles equal, each to each, and the sides about the equal angles proportionals. 2. Reciprocal figures, namely, triangles and pajallelo- grams, are such as have their sides about two of their angles proportionals in such a manner, that a side of the first figure is to a side of the other, as the remammg side of this other is to the remaining side of the first. 3. A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less. 4. The altitude of any figure is i]xQ straight line drawn from its ver- tex perpendicular to the base. 174 EUCLIJyS ELEMENTS. \ H G B PROPOSITION 1. THEOREM. Triangles and parallelograms qf the same altitude are to one another as their bases. wr^nJ^ triangles ABV, ACD, and the parallelograms fS.nT.^rfK*^ ^?®/i***i;l®' °^™^^^' *^« perpendicular drawn from the pomt ^ to ^i): as the base Wis to the base CZ>, so shall the triangle ABC be to the triangle ACD. aikd the parallelogram EC to the parallelogram CF, Produce BD both ' take any number of straight lines BG, OH, each equal to BC^ and any number of straight lines DK, KL, each equal to' CZ); [1.3. andjoin^(?,^J?;^^, ALt Then, because CB, BG, GHsae all equal, [Construction, the triangles ABC, AGB, AUG are all equal. [I. 38. Therefore whatever multiple the base IIC is of the base BC, the same multiple is the triangle AHC of the tri- angle ABC For the same reason, whatever multiple ^he ^as© CL is of the base CD, the same multiple is the tiLdgle ACZ of the triangle A CD. And if the ba^e HC be equal to the baae CL, the irianrfe AHC 18 equal to the triangle ACL; and if the base HC be grater than the base CL) the triangle AHC is greater than the tnangle ACL ; and if less, less. [i. 33. Therefore, since there are four magnitudes, naiyisely. the two bases BC, CD, and the two triangles ABC, ACD ; and of the base BC, and the triangle ABC, the first and the third, any eouimultiples whatever have been taken nameW, the base HC and the triangle AHC; and of the base CD and the tnangle ACD, the second and the fouriJi. any eqmmultiDles whatever have been taken, namely, the base CL and the triande ACL • '* BOOK VI. 1,2. 175 itude are telo^ms endicular is to the X\qACD, strwtion, [I. 38. ^he base the tri- 7L is of ACL of triangle taaeffO greater [I. 88. ^ely, the ACD; irst and \ taken, I of the • fourth, ely, the and mnce it has been shewn that if the base M7be greater than the base CL, the triangle AHC is greater than the triangle ACL ; and if equal, equal ; and if less, less ; therefore as the base BC is to the base CD, so is the triangle ABC to the triangle ACD. [V. DefinUion 5. And, because the parallelogram CE is double of the triangle ABC, aud the panUlologram CF is double of the triangle ^Ci>; P- 41. and that magnitudes have the same ratio which their equi- multiples have ; [V* 15. therefore the parallelogram EC is to the parallelogram CF as the triangle ABC is to the triangle AVD. But it has been shewn that the triangle ABC is to the triangle ACD as the base BC is to the base CD ; therefore the parallelogram EC is to the parallelogram CF as the base BC is to the base CD. [V. 11. Wherefore, triangles &c. q.e.d. Corollary. From this it is plain that triangles and parallelograms which have equal altitudes, are to one an- other as their bases. For, let the figures be placed so as to have their bases in the same straight line, and to be on the same side of it; and having drawn perpendiculars from the vertices of the triangles to the bases, the straight line which joins the ver- tices is parallel to that in which their bases are ; [I. 33. because the perpendiculars are both equal and parallel to one another. [^* 28. Then, if the same construction be made as in the pro- position, the demonstration will be the same. PBOPOSITION 2. THEOREM. If a straight line he drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those sides produced, proportionally; and if the sides, or the sides produced, he cut proportionally, the straight hne which joins the points cf section, shall he parallel to the r*^ «viy««M«M/* mt/ia f\f the trianale. -JiV- w 176 EUCLIUS ELEMENTS, ; Lei DE be drawn parallel to BC^ one of the side* of the triangle ABO: BD shall be to i>^ aa CE is to EA, ■ Join BE, CD, Then the triangle BDE is equal to the triangle CDE, because they are on the same base £>E and between the same pat^lels DEj BC, [1. 37, And ADE is another triangle ; and equal magnitudes have the same ratio to the same magnitude ; ry^ 7^ therefore the triangle BDE is to the triangle ADE as the triangle C£>E is to the triangle ADE. But the triangle BDE is to the triangle ADE as BD is to DA ; because the triangles have the same altitude, namely, the perpendicular drawn from E to AB, and therefore they are to one another as their bases. [vi. 1. For the same reason the triangle CDE is to the triamrle ADE as CE is to EA. ^^ Therefore BD is to DA as QE is to EA. [V. 11. Next, lot BD be to i)^ as O^ is to EA, and ioin DE- 2>J: shall be parallel to j9a . , wiu jum x/^ . For, the same construction being made, because BD iaioDAG&CEia to EA, [ffypothesis, and as BD is to DA, so is the triangle BDE to the trj>angle ADE, ^yj j and as C7^ is to JB:^ 80 is the triangle ODE to the trianrie ADE: MM BOOK VL 2,8. 177 iideft of s, therefoi« the triangle BDE is to the triangle ADE as the triangle CDE is to the triangle ADE\ [V. 11. that is, the triangles BDE and CDE have the same ratio to the tiiangle ADE, Therefore the triangle BDE is equal to the triangle CDE, [V. 9. And these triangles are on tho same base DE and on the same side of it ; but equal triangles on the same base, and on the same side of it, are between the same parallels ; [I. 89. therefore DE is parallel to BC. Wherefore, \fa straight line £ic q.e.d. een the [1. 87. ie same [V. 7. 7 as the as BD ely, the khey are [VI. 1. triangle [V.ll. in DE: ^othesis, to the [VI. 1. triangle rXTT 1 PROPOSITION 3. THEOREM, -^ If the vertical angle of a triangle be bisected by a straight line which also cuts the bascy the segments of the base shall have the same ratio which the other sides of the triangle have to one another; and if the segments of the base have the same ratio which the other sides of the triangle have to one another f the straight line drawn from the vef'tex to the point of section shall bisect the vertical angle. Let ABC be a triangle, and let the angle BAG be bisected by the straight line AD^ which meets the base at D: BD shall be to DC as BA is to ^a Through C draw C!^ parallel to DA, [I. 31. and let BA produced meet CE at E. Then, because the straight line AC meets the parallels ADj EC, the angle A CE is equal to the alternate angle CAD; [1.29. but the angle CAD is, by hypothesis, equal to the anglo BAD; therefore the angle BAD is equal to the an^le ACE. [Ax, % 12 I7B EUCLID'S ELEMENTS. If A^Ain, because the straight line BAE meets the parallels ADy EC, the exterior angle BAD is equal to tiie interior and opposite angle AEC ; but the angle BAD has been shewn equal to the 9Xi%\QACE\ therefore the angle A CE is equal to the angle AEC\ [Axiom 1. and therefore AG is equal to ^^. [I. 6. And, because AD \a parallel to EG. [Constr. one of the sides of the triangle BCE^ therefore BD is to DC as BA is to AE ; but ^^ is equal to AG \ therefore BD is to DC ha .B^ is to AG, [I. 29. [VI. 2. [V. 7. Next, let BD be to 2)C as J5^ is to AG, and join AD: the angle j?^ (7 shall be bisected by the straight line AD, For, let the same construction be made. Then J52) is to 2X7 as 5-4 is to AC\ and BD is to /)(7as BA is to AE, because AlD is parallel to EC ; therefore BA is to -4(7 as BA is to ^i^ ; therefore ACSa equal to AE ; and therefore the angled JF(7 is equal to the angle -4(7^51 [L 6. But the angle -4^(7is equal to the exterior angle J5^i) ; [1. 29. andtlie angle JtCJ^isequalttfthealtemateangleC^Z); [1.29. thei-efore the angle BAD is equal to the angle CAD) [Ax. 1. that is, the angle BACk bisected by the straight line AD, "Wherefore, if the vertical angle &c. q.e.i). [Hypothem, [VI. 2. [Conatraction, [V. 11. [V. 9. BOOK VL A. 179 PROPOSITION A. THEOREM, Jf the exterior angle of a triangle, made hy producing one cf its aides, be bisected by a straight line which also cuts the base produced, the segments between the dividing straight line and the extremities of the base shall have t/ie same ratio which the other sides of the triangle Jiave to one another; and if the segtnents cf the base produced ham the same ratio which the other sides qf the triangle have to one anotlier, the straight line drawn fr&m the vertex to the point of section sIuUl bisect the exterior angle qfthe triangle. • Let ABC be a triangle, and let one of its sides BA be produced to E-, and let the exterior angle CAE he bisected by the straight line AD which meets the base produced at 2> : BD shall be to DC as BA is io AC, Through C draw CF parallel to AD, [I. 31. „ meeting AB at F, Then, because the straight line AC meets ^^ the parallels AD, FC, the angle A CF is equal to the alternate angle (Mi>;[1.29. but the angle CAD is, by hypothesis, equal to the angle DAE; * therefore the angle DAE is equal to the angle ACF. [Ax. 1. A^in, because the straight line FAE meets the parallels AD, PC, the exterior angle DAE is equal to the interior iind opposite angle AFC; , [I. 29. but the angle DAEhsLS been shewn equal to the Bugle ACF; therefore the angle ACF is equal to the angle AFC; [Ax. 1. and therefore AC ia equal to AF, And, because AD is paralled to FC^ one of the sides of the triangle BCF ; therefore BD is to DC as BA is to AF ; but ^i^is equal to AC; [1.6. IConstruction, [VI. 2. M..^.^^lf n T\ •_ i._ »^./n• SUviv4%/4v ju>Jy io kv X/v uS jlj^ 15 10 ^C/» [V r /. 12—2 I 180 EUCLW8 ELEMENTS. Next, lot BD be to DC as BA is to AC \ and join AD: the exterior an^le CAE shall be bisected oy the straight line ^2>. For, let the same con- struction be made. Then BD is to DC ta B A IB to AC 'f {Hypothesis. and BD is to 2>C as i?<4 is to AF; [VI. 2. therefore BA is to -4 C as BA is to AF; ^ [V. 11, therefore AC is equal to -4/*, [V. 9, and therefore the angle ACFi» equal to the angle AFC- [1.5. But the angle A FC is equal to the exterior angle DAE ; ll 29. and th^ angle A CFi& equal to the alternate angle CAD ; [1. 29. therefore the angle CAD is equa? to the angle DAE; [Ax, 1. that is, the angle CAE is bisected by the straight line AD, Wherefore, \ft?ie exterior angle &c. q.b.d. PEOPOSITION 4. fHEOREM, The sides about the equal angles qf triangles which are equiangular to one another are proportionals; and tJwse which are opposite to the equal angles are homologous sides, that is, are the antecedents or the consequents of the ratios. Let the triangle ^^Cbe equiangular to the tr^gle DCE, having the angle JBC equal to the angle DCE, and the angle ACB equal to the angle DEC, and consequently the angle 5-4(7equaltotheangle CD^-. the sides about the equal angles ofthe triangles Ji56', 2>(7^, shall be proportionals ; and those shall do the homolo- gous sides, which are oppo- site to the equal angles. Let the triangle DCE be placed so that its side CE may be contiguous to BC, and in the same straight 1? •a1_ .a Li. ii^. IJ BOOK ri. 4, 5. 181 Then tho angle BCA is equal to the angle CED; [ffyp, add to each the angle ABC\ therefore the two angles ABC, BCA are equal to the two angles ABC, CED ; [Axiom 2. but the angles ABC, BCA are togothor less than two right angles; [1.17. therefore the angles ABC, CED are together less than two right angles ; therefore BA and ED, if produced, will meet. [Axiom 12. ^et them be produced and meet at the point F, Then, because the angle ABC is equal to the angle DCE, [Hinpothesia, BF is parallel to CD ; P. iv, and because the angle ACB Is equal to the angle DEC, i^/yp. ^C7U parallel to i^j&. [1.28. Therefore FACD is a parallelogram; and therefore-4i^isequal toCi>,and^(7is equal to i^D. [1. 84. And, because AC iB parallel to FE, one of the sides of the triangle FBE, therefore BA is to AF taBCiaioCEi but A Fib equal to CD ; therefore i?^ is to CD as 5C is to CE; and, alternately, AB is to DC as DC is to CD. Ag^n, because CD is parallel to BFy * therefore BC is to CD as FD is to DD ; but FD is^equal to AC; therefore DC is to CD as ^C is to VE ; and, alternately, BC is to CA as CE is to ED, Then, because it has been shewn that AB is to BC as DC is to CE, and that DC is to CA as CD is to ED; therefore, ex sequali, BA is to ^C as CD is to DE. [V. 22. Wherefore, the tides &c. q.e.d. PROPOSITION 5. THEOREM, jy the sides of two triangles, about each of their angles, he proportionals, the triangles shall be equiangular to one another, andshall have thoseangles equalwhichareoppotite to the homologous sides. [VI, 2. [V. 7. [V. 16. [VI. 2. [V. 7. [V. 16. ^*t 183 EUCLIUS ELEMENTS. Let the trianglea ABG, DEF have their sides propor- tional 'so that AB is to BG as DE is to EF\ and BC to itly, iquali, BA CA ad ^i^ is to FD ; and, consequent to ^C as iE^/> is to i>i^: the trian le , iBG shall te equian- gular to the triangle DEF, and they shall have those angles equal which are opposite to the homologous sides, namely, the angle ABG equal to the angle DEF, and the angle EGA equal to tl angle EFD, and the angle BAG equal to the angle J5^2>i^.. -At the point E, in the straight line EF, make the angle FEG equal to the angle ABG\ and at the point F, in the stn. /^t line EF, make the angle / V equal to the angle EGA ;, [I. 23. therefore the remaining angle EOF is equal to the remain- ing angle BAG, Therefore the triangle ABG is equiangular to the triangle GEF) and therefore they have their sides opposite to the equal angles proportionals ; [VI. 4. therefore AB is to BG as GE is to EF, But AB is to 5(7 as DE is to EF: therefore DE is t6 EF as GE is to ^i^ ; therefore DE is equal to GE, For the same reason, DF is equal to GF. Then, because in the two triangles DEF, GEF, DE is equal to GE, and EF is common ; the tv/o sides i)JS; -^i^ are equal to the two sides GE, EF, each to eac!^ j and the base DF is equal to the base GF ; therefore the angle DEF is equal to the angle GEF., fl. 8, and the other angles to the other angles, each to each, to which the equal sides are opposite. [I. 4. therefore the angle DFE is equal to the angle GFE, and the angle EDF 1? equal to the angle EGF. [Hypothesis, [V. 11. [V. 9. BOOK ri 6,6. 183 And, because the angle VEF is equal to the angle OEF, and the angle GJSF is equal to the angle ABC, [Constt, therefore the angle ABC is equal to the angle DBF, [Ax. 1,^ For the same reason, the angle ACB is equal to the angle J>FBf and the angle at A is equal to the angle at 2>. Therefore the triangle ABC is equiangular to the triangle DBF, Wherefore, if the sid€s &Q, q.e.d. ♦ PSOPOSITIOK C. THEOREM, If two triangUs have one angle of the one equal to on§ angle qf the other, and the sides about the equal angles proportionals, the triangles shall he equiangular to one another, and shall have those angles equal which are op" posite to the homologous sides. Let the triangles ABC, DBF have the angle BAC in the one, equsd to the angle BDF in the other, and the sides s^bout those angles proportionals, namely, BA to AC as BD is to £>F: the triangle ABC shall be equiangular to the triangle DBF, and shall have the angle ABC equal to the angi . DBF, and the angle A CB equal to the angle BFB, At the point i>, in the straight line DF, make the angle FBG equal to either • of the angles BAC, jJiDF ; and at the point F, in the straight line DF, make the angle DFC equal to the angle ACB\ [I. 23. therefore the remaining angle at G is equal to the remam* ing angle at B, Therefore the triangle ABC is equiangular to the triangle DGF', < 1 _ i» trk M i ^ 1.^ A /^ r^ T\ i _ J. ^ T^ 7TI Ifi^teiuio JLijci. iS XiK* j£i.\J US \jfjLi io uU U£ t But BA is to AC2ABDS& to DF\ therefore J&2> is to BF as G^Z) is to DF\ therefore BB is equal to QD^ rrTT J [ffypothesiM, [V. 11. [V. ». 184 MUCLIUS ELEMENTS, And DF is common to the two triangles ETiF^ GDF\ therefore the two sides ED^ DF are equal to the two sides QDf J>Fy each to each ; and the angle EDF is equal to the angle QDF\ [Cmttr, therefore the basd EF is equal to the hase GFy and the triangle EDF to the trian^fle GDFf alid the re- mammg angles to the re- maining angles, each to each, to which the equal sides are opposite; p. 4. therefore the angle DFG is equal to the angle J)FE, and the angle at G is equal to the angle at E. Bat thiB angle DFG is equal to the angle ACB; [Comir. therefore the angle ACB is equal to the angie DFE, [Ax, 1. And the angle BAG is equal to the angle EDF; [ffypothestt. therefore the remaining angle at J? is equal to the remain- ing angle at ^. Therefore the triangle ABC is equiangular to the triangle DEF, Wherefore, if ttoo triangles &G. q.e.i>. PROPOSITION 7. THEOREM. ]f two triangles hate one angle q/" the one equal to one angle of the other, and the sides about two other angles proportionals; then, if each of the remaining angles he either lioS, or not less, than a right angle, or if one of them he a right angle, the triangles shall he equiangular to one another, and shall have those angles equal aihout which the sides are proportionals. Let the triangles ABC, DBF have one angle of the one equal to one angle of the other, namely, the angle other angles ABC, DEF, proportionals, so that AB is to BC as DE is to EF; and, first, let each of the remaining angles at C and F be less than a right angle : the triangle ^^G shall be equiangular to the truvngle DEF^ BXid sluill book: VI. 7. 185 hare the angle ASC equal to the angle DBF, and the , angle at C equal to the angle at F» For, if the angles ABC, DBF be not equal, one of them must be greater than the other. Let ABC be the greater, and at the point B, in the straight line AB, make the angle A3Q equal to the angle DBF, [1. 28. Then, because thd angle at A is equal to the angle at 2>, [Hyp, and the angle ABQ is equal to the angle DEF^ [Comtr, therefore the remaining angle AGB is equal to the re- maining angle DFE ; therefore the triangle ABG is equiangular to the triangle DBF, Therefore AB\&\ja BG9A DE is to EF. [VI. 4. But ABi&ioBCvA DE\& to EF) [ffypothuis, therefore ^ i5 is to 5C7 as ^^ is to -S^ ; [V. 11. therefore BC is equal to BG ; [V. 9. and therefore the mgle BCG is equal to the angle BGC. [1. 5. But the angle BCG is less than a right angle ; [Hyp, therv3fore the angle BGC is less than a right angle ; and therefore the adjacent angle AGB must be greater than a right angle. [!• 13. But the angle AGB was shewn to be equal to the angle fitF; therefore the angle at F is greater than a right angle. But the angle at JP' is less than a right angle ; [HypotheM, which is absurd. Therefore the angles ABC and DEF are not unequal ; that is, they are equal. And the angle at A is equal to the angle fc D; [ffypothesU. therefore the remaining angle at C is equal to the remain- ing angle at i^ ; therefore the tri ^gle ABC is equiangular to the triangle DEF. me EUCLIJyS ELEMENTS, [1.5. Next, let each of the angles at C and i^ be not left than a right angle : the triangle ABC shall be equiaiurular to the tri'ingle DEF, For, the same ccn- struction being made, it may be shewn in the same manner, that BO is equal to BG ; therefore the angle BCG is equal to the angle -56?C. But the angle BCQ is not less than a right angle ; \hZ therefore the angle BGC\% not less than a right angle; that is, two angles of the triangle BCQ are together' not Jess than two right angles ; which is impossible. [f. 17. Therefore the triangle ABC may be shewn to be equi- angular to the tnangle DBF, as in the first case. Lastly, let ^ne of the angles at (7 and i^ be a right Angle, namely, the angle at C\ the triangle ABC shall be equiangular to the triangle DBF. For, if the triangle ^^(7 be not equiangular to the triangle DBF, at the point B, in the straight line AB^ make the angle ABG equal to the ^-gle DBF, [I. 23. Then lu niiay be shewn, as in the first case, that BC is equal to BG ; therefore the angle BCG is equal to the angle jB6rC. [1. 5. * But the angle BCG is a right angle: [Hypothesis. B therefore the angle BGG is a right angle ; thatjs, two angles of the triangle BCG are to^rether equal i»o j**vo ngut angles; whicu is impossible. ^- [I. 17. Therefore the triangle ABC is equiangular to flie triangle DBF, therefore, if two trtangies&c. q.e.d. BOOK ri. 8. 187 PROPOSITION 8. THEOREM. "^ In a right^ngled triangle, if a perpendicular be drawn from the right angle to the base, ths triangles on each side of it are similar to the whole triangle, and to one another. Let ABC be a right-angled triangle, having the right angle BAG; and from the point A, let AD be drawn per- pendicular to the base BC: the triangles DBA, DAO shall be sunilar to the whole triangle A.BGt &i^d to one another. For, the angle BAG is equal to the angle BDA, each of them being a right angle, [Axiom 11. and the angle at B is common to the two triangles ABG, DBA ; therefore the remaining angle AGB is equal to the remaining angle DAB. Therefore the triangle ABG is equiangular to the triangle DBA, and the sides about their equad angles are propor- tionals; [VI. 4. therefore the triangles are similar. [VI. BefinitUm 1. In the same manner it may be shewn that the triangle DAG is similar to the triangle ABG. And the triangles DBA, DAG being both similar to the triangle ABG, are similar to each other. "Wherefore, in a right-angled triangle &c. q.e.d. Corollary. From this it is manifest, that the perpen- dicular drawn from the right angle of a right-angled triangle to the base, is a mean proportional between the segments of the base, and also that each of the sides is a mean proportional between the base and the segment of the base adjacent to that side. For, in the triangles DBA, DAG, BDhioDAfi&DAiiioDG) [VI. 4. J •_ At-. „1_. 4 -nn T\-n A BGia io BA &B BAia to BD; and in the triangles ABG, DAG, BCiaio OAa&GA is io CD. [VI. 4 [VI. 4. 188 EUCLIJyS ELEMENTS. PROPOSITION 9, PROBLEM. From a given straight line to cut off any part required. Let AB be the given straight line: it is required to cut off any part from it ^ From the point A draw a stn^ht line AGf making any angle with AJB\ in ^G^ take any point 2), and take A (7 the same mtdti^le of A 2>, that AB is of the part which is to be cut off from it ; join jBC. and draw DE parallel to it AE shall be the part required to be cut off. For, because ED is parallel to BC, one of the sides of the triangle ABC, therefore p2> is to 2>^ as J5^ is to J5^ ; and, by composition, CA is to AD as BA is to AE. [V. 18. But CA is a multiple of AD ; [Construction, therefore BA is the same multiple of AE ; [V. i>. that is, whatever part AD is of AC, AE is the same part of AB. Wherefore, /rom the given straight line AB, the part required has Ifeen cut off, q.e,p. [Construction, [VI. 2. PROPOSITION 10. PROBLEM. To divide a given straight line similarly to a given divided straight line, that is, into parts which shall have the same ratios to one another, that the parts of the given divided straight line have. Let AB be the straight line given to be divided, and ACihQ given divided straight line: it is required to divide AB similarly io AC. Let AChQ divided at the points D, E; and let AB, ^C be placed so as to contain any angle, and join BC ; through the point D, draw DF ^raliei to BC, and through the point E draw EG parallel to BC. [I. 31. AB shall be divided at the points F, a, similarly to AC, BObK VL 10, 11. 189 Through D draw DiTAT parallel to AB, [I. 81. Then each of the figures Fllf HB is a parallelogram ; therefore DHS& equal to FG, and HKis equal to GB, [1.34. Then, because HE is parallel to KG, IComtruetion, one of the sides of the triangle DKG^ therefore KH is to ITD as CE is to ED, [VI. 2. But KHiA equal to BG, and HD is equal to GF\ therefore BG is to (?i^ as C.S is to ED. [V. 7. Again, because FD is parallel to GE^ * [Con«en«? is to DA, Wherefore the given straight line AB is mvided simi* larly to the given divided airaigh^ line AC, Q.B.P. PROPOSITION 11. PROBLEM. To find a third proportional to two given straight lines. Let ABy AC be the two given straight lines: it is re- quired to find a third proportional to ABy AC Let ABy AC \m placed so as to contain any angle ; produce ABy ACy to the points 2>, E\ and make BD equal to AC\ [I. 3. join BCy and through D drav^j& parallel to BC. [I. 31. CE shall be a third proportional ioAByAC. For, because BC is parallel to DEy one of the sides of the triangle ADEy therefore AB is to BD as ^(7 is to (7^ ; but BD is equal to -4C7 ; therefore ^^ is to -4C as ^C7 is to Ci^. [V. 7, Wherefore to the two given straight lines AB, ACy a third proportional CE is/ound, q.b.f, [Construction, [VL2. [Construction. 190 EUCLIiys ELEMENTS. {CoMtruction, PROPOSITION 12. PROBLEM, lin^, ^"^ "^ '^''"'*'^* P'''>P^''^^<>^^ io three given Hraighi ■ Let ^, By C be the three given straiffht Kne«- a i» required to find a fourth proportional to^^^ c * _ Take two straight Unes, ' ' ■DE, DFy containing any an- gle JSTDJT; anTl in these make -gw equal to A, GE equal to ^, and jOiT equal to C; [1.3. join QH, and througa E draw ii'J'' parallel to GH, [j. 31. HF shall be a fourth propor- tional to ^, ^, (7. ^ For, Ifecause GHh parallel to ^i?; one of the sides of the triangle DEF, ' therefore DG h to GEbs DH is to HF, rvT 9 Stl^oV "^"^ *^ ^' ^^ ^^ ^^"^ '^ ^^ and S^t therefore :* is to i? as (7 is to ffl^. l<^<^ru.t^. Wherefore to the three given strainht li^^» a n n fourth proportional HF i^foZd t^l '' "^^ ^' ^' * PROPOSITION 13. PROBLEM. iSt.'^'''^'' '^"'^^''^PO^^^^^^eiween two given straight Place^5,J5Cin a straight line, and on ^(7 describe the semicircle ADC-, from tha point B draw J9i> at right MilifiOS to JiiL FT n iOgico tO ^1/'. [1. 11. BD shall be a mean propor- tional between AB and BG, I BOOK VL 13, 14 191 Join ^2>, i><7. Then, the angle ADC^ being in a semicircle, is a right angle; [III. 81. and because in the right-angled triangle ADC, DB [a dra^n from the right angle perpendicular to the base, therefore DB is a mean proportional between AB, BC, the segments of the base. [VI. 8, Corollary, Wherefore, between the two given straight lines ABg BC, a mean proportional DB is found, q.e.f. PBOPOSITION U. THEOREM. Equal parallelograms which have one angle qf the one equal to one angle of the other , have their sides about the equal angles reciprocally proportional; and parallelo- grams which have one angle of the one equal to one angle of the other, and their sides about tfte equal angles reci- procally proportional, are equal to one another. Let AB, BO be equal parallelograms, which have the angle FBD equal to the angle BBG : the sides of the parallelograms about the equal angles shall be reciprocally proportional, that is, DB shall be to BE9& GB is to BF, Let the parallelograms be j^aced, so that the sides DB, BE may be in the same straight line ; therefore also FB, BG are in one straight line. [1. 14. Complete the parallelogram FE. Then, because the parallelogram AB is equal to the parallelogram jB(7, lEypothem. and that FE is another parallelogram, therefore AB is to FE as BG is to FE. [V. 7. But AE is to FE as the base DB is to the base BE, [V 1. 1. and BG is to FE as the base GB is to the base BF; [VL 1. therefore DB is to BE 9^ is to BF, [V, 11. Idi EUCLII^S ELEMENTS, Kezt, let the angle FBD be eaual to the angle EBQ^ and let the sides about the equal angles be reciprocally Sroportional, namely, DB ^ BE ^ GB Sa \jQ BF: the parallelogram AB shall be equal to the parallelo- gram j9C7. For, let the same con- struction be made. Then, because DB is to BE as G^A is to BF, [ffypotkesis. and that DB is to BE as the parallelogram AB is to the parallelogram FE, [VI. l. and that GB is to BF as the parallelogram -BC7 is to the parallelogram^^; [VI. l. therefore the parallelogram AB is to the parallel(^ram FE as the ^rallelogram BC is to the parallelogram FE; [V. 11. therefore the parallelogram AB is equal to the parallelo- gram jBC [V. 9, Wherefore, equal parallelograms &c. q.e.d. II PROPOSITION 15. THEOJtEM. Equal triangles which have one angle qf the one equal to one angle qf the other, have their sides about the equal angles reciprocally proportional; and triangles which have one anqle of the one equal to one angle of the other, and their sides about the equal angles reciprocally pro^ portional, are equal to one another. . Let ABO, ADE be- equal triangles, which have the angle BAG equal to the angle DAl^x the sides of the triangles about the equal angles shall be reci- procally proportional ; that is, CA shall be to AD as EA is to AB. Let the triangles be placed so that the sides OA, AD may be in the same straight line, :ie EBG, ;iprocally BOOK VL 15, 16. 193 TypothestB, is to the [VI. 1. is to the [VI. 1. ?; [V.ll. )aranclO' [V.9, ne equal he equal » which he otheVf illy pr. Then, because the triangle ABC is equal to the trian- gle ADEt lffypoth€su» and that ABD is another triangle, therefore the triangle ABC is to the triangle ABD as the triangle ABE is to tho triangle ABD. [V. 7. But the triangle ABC is to the triangle ABD as the base CA is to the base AD, [VI. 1. and the triangle ADE is to the triangle ABD as the base EA is to the base AB; [VI. 1. therefore CA ia to AD &bEA is to AB. [V. 11. Next, let the angle BAC be equal to the angle DAS^ and let the sides about the equal angles be reciprocally, proportional, namely, CA to AD os EA is to ABi tlio triangle ABC shall be equal to the triangle ADE, For, let the same construction be made. Then, because CA is to AD as EA is to AB, [ffypcthesie, and that CA is to AD as the triangle ABC is to the triangle ABD, [VI. 1. and that EA is to AB as the triangle ADE is to the triangle ABD, [VI. 1. therefore the triangle ABC is to the triangle ABD as the triangle ADE is to the triangle ABD j . [V. 11. therefore the triangle ABCia equal to the triangle A DE. [V. P. Wherefore, equal triangles &c. QiE.d, PKOPOSITION 10. TBMOMM. If foar ttraight lines be^ proportionals, tlie rectangle contained by the extremes is equal to the redtangle con- tained by the means; and if the rectangle contained by the extremes be equal to the rectangle contained by thi mean^ ihe/our straight lines are proportionals. ;r 194 EUCLliyS ELEMENTS. i I F- Lot the four straight lines AB^ CD^ E, F, be propor- tionals, namely, let -4/? bo to <7Z> as -& is to JF": the rect- angle contained hy AB and F shall be equal to the rect- angle contained by pD and E, From the points Af C, draw A G, CH at right anglesto-4i?,Ci>;[I.ll. mekA A O equal to Fy and CiT equal to JS?; [1.3. and complete the paral- lelograms i5(7,i>ir. [1.31. Then, because ^^ is iaCDQ&Ei&ioFylHyp, and that E is equal to CH, and F is equal to nAG-f \C is to C, But B is equal to D ; [Construction, Therefore A iaio B bs B kto C [V. 7. Wherefore, if three straight lines &c. q.e.d. PROPOSITION 18. PROBLEM, ^ On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure^ Let ^i? be the given straight line, and CDEF the given rectilineal figure of four sides : it is required to de- scribe on the given straight line ^5, a rectilineal figure, similar and similarly situated to CDEF, Join DF\ at tho point A^ in the straight line A By make the angle BAG equal to the angle DCF\ and at the Doint B. in the Straight line A B, make the angle ABG equal toUi€angleCi>i^i[I,23» P4 BOOK VI. 18. 197 therefore the remaining angle AQB is equal to the reman- ing angle CFD^ and the triangle AGB is equiangular to the triangle CFD. Again, at the point B, in the straight line BG^ make the angle G^^^ equal to the angle FDE\ and at the point Q, in the straight line BG, make the angle BGH equal to the angle DFE ; [I. 23. therefore the remaining angle BHG is equal to the re- maining angle DEF, and the trian jle BHG is equiangular to the triangle DEF. Then, because the angle A GB is equal to the angle CFD^ and the angle BGH equal to the angle DFE) [Construction. therefore the whole angle AGH is equal to the whole angle GFE. [Axiom 2. For the same reason the angle ABH is equal to the angle CDE. And the angle BAG is equal to the a^igle DCF, and the angle BHG is equal to the angle DEF. Therefore the rectilineal figure ABHG is equiangular to the rectilineal figure CDEF, Also these figures have their sides about the equal angles proportionals. For, because the triangle BAG is equiangular to the triangle £>CF, therefore BA U to AG as DC is to OF. [VI. 4. And, for the same reason, ^6? is to GB as CF is to FD. and BG k to GH as DF is to FE ; therefore, ex sequali, ^6r is to GHuh CFk to FE. \Y. 22. In the same manner it may be she^vn that AB is to BH as CD is to DE. And GH is to HB as FE is to ED, [VI. 4. Therefore, the rectilineal figures ABHG and CDEF are equiangular to one another, and have their sides about the equal angles proportionals ; ciiUFcioio Xiiiiij ciic oiiiiiiiii tO O116 anOtiiei. l'^' i-, l/tjinition i. Next, let it be required to describe on the given straight line ABjSb rectilineal figure, similar, and similarly situated, to tlie rectilineal figure CDKEF of five sides. 198 EUCLID'S ELEMENTS. Join DEf and on the giren straight line AB describe, as in the former case, the rectilineal figure ABHGy similar, and similarly situated to the rectilineal figure CDEF of four sides. At the point B^ in the straight line BH^ make the angle HBL equal to the an- gle EDK\ and at the point H^ in the straight line BH^ make the an- gle BHL equal to the angle DEK\ [I. 23. therefore the remaining angle at L is equal to the remain- ing angle at K, Then, because the figures ABRG, CDEF are similar, the angle ABH is equal to the angle CDE\ [VI. Def. 1. and the angle HBL is equal to the angle EDK\ [Comtr, therefore the whole angle ABL is equal to the whole angle CDK^ ' [Axiom 2. For the same reason the whole angle GHL is equal to the whole angle FEK, Therefore the five-sided figures ABLHG and CDKEFqxq equiangular to one another. And, because the figures ABHG and CDEF are similar, therefore AB is to BH as CD is to DE\ [VI. Definition 1. but BH is to BL as DE is to DK; [VI. 4. therefore, ex sequali, AB is to BL aa CD is to DIT. [V. 22. For the same reason, GHia to HL as FE is to EK^, And BL is to LH as DK is to KE. [VI. 4. Therefore, the five-sided figures ABLHG and CDKEF are equiangular to one another, and have their sides about the equal angles proportionals ; therefore they are similar to one another. [VI. Definition 1. In the same manner a rectilineal figure of six sides •nnn \\A Af^anvf^ckA nn ft s^vsn Rtroifirht line^ similar and sinnlariy situated to a given rectilineal figure' of six sides; and so on. q.£.f. 3 describe, TGf Bimilar, CD£F of he remain- 3 similar, :VI. Def. 1. ; [Constr, the whole [Axiom 2. ^ual to the yKEFta^Q ire similar. Definition 1. [VI. 4. )K. [V. 22. [VI. 4. I CDKEF ides about !)e^i^ton 1, ' six sides imilar and six sides; BOOK VL 19. PROPOSITION 19, THEOREM. 199 Similar triangles are to one another in the duplicate ratio of their homologous sides. Let ABC and DEF be similar triangles, having the angle B equal to the angle E, and let ^5 be to BC as DE is to EF, so that the side BG is homolo- gous to the side EF: the triangle ABG shall be to the tri- angle DEF in the duplicate ratio of BG to EF, Take 56^ a third proportional to BG and J^^, so that BG may be to EF as ^i" is to BG ; [VL H. and join AG. Then, because AB is to 5C as DE is to .&i^, [ffypothesU. therefore, alternately, AB is to DE as J5C7 is to EF; [V. 16. but j5(7 is to EF as j^i^ is to BG ; [(7on8erMC««>n. therefore AB is to 2>iS; as EF is to J56? ; [V. 11. that is, the sides of the triangles ABG and DEF, about their equal angles, are reciprocally proportional ; but triangles which have their sides about two equal angles reciprocally proportional are equal to one another, [VI. 15. therefore the triangle ABG is equal to the triangle DEF, And, because BG is to EF as EF is to BGj therefore BG has to BG the duplicate ratio of that which BG has to EF. [V. Definition 10. But the triangle ABG is to the triangle ABG as BG is to BG ; [VI. 1. therefore the tnangle ABG has to the triangle ABG the duplicate ratio of that which BG has to EF, But the triangle -456r was shewn equal to the triangle DEF; therefore the triangle ABC has to the triangle DEF the duplicate ratio of that which BG has to EF, [V. 7. 'Whereforej similar triangles &G. q.b.d. Corollary. From this it is manifest, that if three 200 EUCLID'S ELEMENTS. straight lines be proportionals, as tlie first is to the tliird, so is any triangle described on the first to a similar and jsimilarly djscribed tiiangle on the second. ■it PROPOSITION 20. THEOREM. Similar polygons may he divided itito the same number of similar triangles^ having the satne ratio to one another tJmt the polygons have; and the polygons are to one anotJier m the duplicate ratio of their homologous sides. Let ABODE, FGHKL be similar polygons, and let AB be the side homologous to the side FG : the polygons ABCDE^ FGHKL may be divided into the same number of similar triangles, of which each shall have to each the same mtio which the polvgons have ; and the polygon ABCDE shall be to the polygon FGHKL in the duplicate ratio of ^^ to i'^^. Join BE, EG, GL, LH. Then, because the polygon ABCDE is similar to the poly- gon FGHKL, Ulypothem. the angle BAE is equal to the angle GFL, and BA is to AE as GF is to FL. [VI. Definition 1. And, because the triangles ABE and FGL have one angle of the one equal to one angle of the other, and the sides about these equal angles proportionals, therefore the triangle ABE is equiangular to the triangle J^<^L, [VI. 6. and therefore these triangles are similar ; [VI. 4. therefore the angle ABE is equal to the angle FGL, nooK VI. 20. 201 But, because the polygons are similar, [ffypothms, tlierefore the whole angle ABC is equal to the whole angle FGH; [VI. Definition 1. therefore the remaining angle EBGis equal to the remain- ing angle LGH. {Axiom 8. And, because the triangles ABE and FGL are similar, therefore EB is to BA as LG is to GF\ and also, because the polygons are similar, [Hypothesis » therefore AB is to BC as FG is to GH ; [VI. Definiti&n 1. therefore, ex sequali, EB is to BO as LG is to GH\ [V. 22. that is, the sides about the equal angles EBC and LGH are proportionals ; therefore the triangle EBC is equiangular to the triangle LGH] [VI. 6. and therefore these triangles are similar. ' 1TI« 4. For the same reason the triangle ECD is similar to the triangle LHK. Therefore the similar polygons ABODE, FGHKL may be divided into the same number of similar triangles. Also these triangles shall have, each to each, the same ratio which the polygons have, the antecedents h^m^ABE. EBO, EOD, and the consequents FGL, LGH, LHK-, and the polygon ABODE shall be to the polygon FGHKL in the duplicate ratio of ^i? to FG. For, because the triangle ABE is similar to the tri- angle FGL, therefore ABE is to FGL in the duplicate ratio of EB to LG. [VI. 19. For the same reason the triangle EBC is to the triangle LGH in the duplicate ratio of EB to LG. Therefore the triangle ABE m to the triangle FGL as the triangle EBCis to the triangle LGH. [v. 11. Again, because the triangle EBO is similar to the tri* angle LGH, therefore EBO is to LGH in the duplicate mtio of EO UiLH, j:VI. la, II 202 EUCLID'S ELEMENTS, For the same reason the triangle EGD is to the triangle LHK in the duplicate ratio of EG to LH. Therefore the triangle EBG is to the triangle LGH as the triangle EGD is to the dangle LHK, [V. 11. But it has been shewn that tlie triangle EBG is to the tri- angle LOH as the triangle ABE is to the triangle FGL. Therefore as the triangle ABE is to the triangle FGL, so is the triangle EBG to the triangle LGH, and the triangle EGD to the triangle LHK\ [V. 11. and therefore as one of the antecedents is to its consequent 80 are all the antecedents to all the consequents ; [V. 12. that is, as the triangle ABE is to the triangle FGL so is the polygon ABGDE to the polygon FGHKL. But the triangle ABE is to the triande FGL in the duplicate ratio of the side AB to the homologous side FG\ [VI. 19. therefore the polygon ABGDE is to the polygon FGHKL in the duplicate ratio of the side AB to tne homologous side i^6?. Wherefore, similar polygons &c. q.e.d. Corollary 1. In like manner it may be shewn that similar four-sided figures, or figures of any number of sides, are to one another in the duplicate ratio of their homo- logous sides ; and it has already been shewn for triangles ; therefore universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides. Corollary 2. liioAB and FG, two of the homologous sides, a third proportional M be taken, [YI. 11. M then AB has to Jlf the duplicate ratio of that which AB has to FG, [V. D^nition 10, BOOK VL 20, 21. 203 But any rectilineal figure described on -45 is to the similar and similarly descrioed rectilineal figure on FG- in tho duplicate ratio oiABio FGj [Corollary 1. Therefore as -45 is to ilSf, so is the figure on AB to tho figure on 2^6?; [V. 11. and this was shewn before for triangles. [VI. 19, Corollary, Wherefore, universally, if three straight lines be propor- tionals, as the first is to the third, so is any rectilineal figure described on the first to a similar and similarly described rectilineal figure on the second. PROPOSITION 21. THEOREM. Rectilineal figures which are similar to the same recti* lineal figure^ are also similar to each other. Let each of the rectilineal figures A and B be similar to the rectilineal figure C: the figure A shall be similar to the figure B. For, because -4 is similar to (7, [Byp. A is equiangular to (7, and A and C have their sides about the equal angles propor- tionals. [VI. Def. 1. Again, because B is similar to (7, [Hyp. B is equiangular to (7, and B and G have their sides about the equal angles proportionals. [VI. Definition 1. Therefore the figures A and B are each of them equian- gular to (7, and have the sides about the equal angles of each of them and of (7 proportionals. Ther^re A is equiangular to J?, [Axiom 1. nn/l A n-nA Z? VtnTTA 'l-VkAi«. oi^Ad nVk/^iif*. 4-na o/^llo1 OTtrviAa rw/V- tliiU. ^i UliVJ. -L» •_!£« T W VXiV4i Ci^JLVSS iX7JXr:W ViS-_- -w-V^^llsa VSXt^i.\."J jLTi \j— portionals; [V. 11, therefore the figure A is similar to the figure B. \Yl.Def. 1. Wherefore, rectilineal figures &c. q.e.d. 204 EUCLID'S ELEMENTS. PROPOSITION 22. THEOREM. If four straight lines he proportionals, the similar rec- tilineal figures similarly described on them shall also he proportionals; and if the similar rectilineal figures simi' larty described on four straight lines he proportionals, those straight lines shall he proportionals. Let the four straight lines AB, CD, EF, GH be pro- portionals, namely, AB to CD as EF is to GH-, and on AB, CD let the similar rectilineal figures KAB, LCD be simi- larly described ; and on EF, GU let the similar rectilineal figures MF, NH be similarly described: the figure KAB shall be to the figure LCD as the figure MF is to tiie ■figure NW, 8 P B ^0 AB and CD take a third proportional X, and to EF and GH a third proportional 0. [VI. 1 1 . Then, because ^^ is to CD as EF is t GH, [Hypothesis, and AB is to CD as CD is to X; [Constructim. and EF is to GH as GH is to 0-; [Construction. therefore CD h to X as GH is to 0. [V. 11. And AB is to CD as EF is to GH; therefore, ex sequali, ^J9 is to X as EFh to 0. [V. 22. But as AB is to X, so is the rectilineal figure KAB to the rectilineal figure LCD ; [VI. 20, Corollatnf 2. ^nd as EF is to O, so is the rectilineal figure MF to the rectilineal figure NH; [VI. 20, Corollary 2. BOOK VI. 22, 23. 205 therefore the figure KAB is to the figure LCD as the %are MF is to the figure NH. [V. 11. lar rec- also he Honals, be pro- on AB, )e siini- ;tilinea1 }KAB to tiie to EF "VI. 11. pothesis, Iruction. Wuciion. [V. 11. [V. 22. AB to ^llary 2. to the )llai^ 2. Next, let the figure KAB be to the similar figure LCD as the figure MF is to the similar figure NU', AB shall be to CD as EF is to GH, Make as ^5 is to CZ> so EF to P/J : [VI. 12. and on PB describe the rectilineal figure /ST?, similar and similarly situated to either of the figures MF, NH. [VI. 18. Then, because AB is to Ci) as EF is to Pi?, and that on AB^ CD are described the similar and simi- larly situated rectilineal figures KABy LCD, and on EF, PR the similar and similarly situated recti- lineal figures 3f/^, aS'/? ; therefore, by the former part of this proposition, KAB is to LCD as MF is to SB. But, by hypothesis, KAB is to LCD as 3fPis to NH; therefore MF is to SB as MF is to Nil; [V. 11. therefore SR is equal to NET. [V. 9. But the figures iSR and NH are similar and similarly situated, [Construetion, therefore PR is equal to GH, And because ^5 is to (7i> as EF is to PR, and that PR is equal to GH ; therefore ^^ is to CD as ^Pis to GH. ' [V. 7. Wherefore, if /our straight lines &c. q.e.d. PROPOSITION 23» TBMOREM. Paralielcgrame which are equiangular to one another have to one another the ratio which ie compounded qf the raiios ^ their sides. 206 EUCLIIfS ELEMENTS. u Lei the parallelogram ^C be equiangular to the paral- lelogram CF, having the apglo BCD equal to the angle ECQ\ the parallelogram AC shall have to the parallelo- gram CF the ratio which is compounded of the ratios of their sides. Let BC and CQ be placed in A straight line ; therefore DC and CE are also in a straight line; [i. 14. complete the parallelogram DG ; take any straight line K, and make -ST to Z as J5C is to (7Gf, and L to ilf as Z>(7is to CE\ [VI. 12. then the ratios of ^ to Z and of Z to Jf are the same with the ratios of the sides, namely, of BC to(7^aniiof2>CtoCiS^. But the ratio of .ST to iJf is that which is said to be com- pounded of the ratios of iT to Z and of Z to Jlf ; [V. Def. A, therefore ^ has to Jf the ratio which is compounded of the ratios of the sides. Now the parallelogram ^C is to the parallelogram CH M BC IS to CG; * ^I 1 but BC is to CG as ^ is to Z ; [Comtructlm] ttierefore the parallelogram AC is to the parallelogram CZTasjiristoZ. [V. 11. Agai^ the parallelogram CH is to the parallelogram CF asDCiaio CE; ' [VI. l. buti>C7isto(7ZrasZi8toiJf; ' [ConstructioT^. tiherefore the paraUelogrdm JOH is to the parallelogram CFasZisto J[f. [V. 11. Then, since it has been shewn that the parallelogram AC is to the parallelogram CH as K is to Z, and that the parallelogram CH is to the parallelogram CF as Z is to il!f, therefore, ex aeauali. the naraT1e^rtff«jm j/7 ?o +0. *i»^ 1 lelogram C7^as iTis toiJf. ""^ ■'"' "" ^= ^-^- Tf^Tl/ But ^ has to M the ratio which is compounded of the ratios of tlie sides; BOOK VI. 23, 24. 207 therefore also the parallelogram ^C has to the parallelo- gram CF the ratio which is compcunded of the ratios of the sides. \(here(oref parallelograms &c. <3.e.d. PROPOSITION 24. THEOREM, Parallelograms about the diameter of any parallelo- gram are similar to the whole parallelogram and to one another, . Let A BCD be a parallelogram, of which -4(7 is a diameter; and let EG and BIC be parallelograms about the diameter: the parallelograms EG and JuK shall be Bimilar both to the whole parallelogram and to one another. For, because DC and G^JP are parallels, A E B the angle ADC is equal to the angle A GF. [I. 29. q And because BCmd EF are parallels, the angle ABC is equal totheangle-4JS^i^.[I.29. And each of the angles BCD and EFG is equal to the opposite angle BAD^ [1. 34. and therefore they are equal to one another. Therefore the parallelograms ABCD and AEFG are equi- angular to one another. And because the angle ABC is equal to the angle AEF, and the angle BAC is common to the two triangles BACfm^^EAF, therefore these triangles are equiangular to one another; and therefore AB is to BC as AE is to EF. [VI. 4. And the opposite sides of parallelograms are equal to one another; [1. 34. therefore AB is to AD as AE is to .4^, Kad DC \%\xiCB2AGF\a to FE, and CD is to DA2i&FG\& to GA, [V.J, 208 EUCLIUS ELEMENT^'. Therefore the sides of the parallelograms ABCD and AEFG about their equal angles are proiwrtional, and the parallelograuis are therefore similar to one an- other. r,rr , . For the same reason the parallelogram ABCD is similar to the parallelogram FHGK, Therefore each of the pa- rallelofframs EG and HK is similar to BD ; therefore theparallelogram EQ is similar to the parallelogram HK. W herefore, parallelograms &c. q.e. d. [VI. D^nUion 1. A B ' PROPOSITION 1>5. PROBLEM. To describe a rectilineal figure which shall he similar to one given rectilineal Jigure and equal to another given rectilineal figure. Let ABC be the given rectilineal figure to which the figure to be described is to be similar, and D that to which It IS to bo equal: it is required to describe a rectilineal figure smiilar to ABC and equal to Z>, On the straight line i?6Mescribe the parallelogram BE equal to the figure ^5C. e ^xjr Oil the sti^aight line €E describe the parallelogram CM TX.}^%?' ^^ ^*^"*S *^« ^Si« ^"^^ eq»a» to the angle CBL CI. 45, CarolUwy, BOOK VI. 25, 26. 209 therefore BG and CF will be in one straight line, and LE and EM will be in one straight lino. Between 5(7 and C/'find a mean proportional OH, [VI. 13. and on (r^ describe the rectilineal figure KGHy similar and similarly situated to the rectilineal figure ABO, [VI. 18, KGH shall bo the rectilineal figure required. For, because BO is to GH as GH is to CF^ [Comfructum, and that if three straight lines be proportionals, as the first is to the third so is any figure on the first to a similar and similarly described figure on the second, [VI. 20, Cor. 2. therefore as BO is to CF so is the figure ABC to the figure KGH. But as BC is to CF so is the parallelogram BE to the parallelogram CM; [VI. 1. therefore the figure ABC is to the figure KGH as the pa- rallelogram BE is to the parallelogram CM. [V. 11. And the figure ABC ia equal to the parallelogram BE; therefore the rectilineal figure KGH is equal to the paral- lelogram CM. ' [V. 14. But the parallelogram CM is equal to the figure D ; [Constr, therefore the figure KGH is equal to the figure Z), [Axiom 1. and it is similar to the figure ABC. \ Construction. Wherefore the rectilineal figure KGH has been de- scribed similar to the figure ABC, and equal to D. q.e.p. PROPOSITION 26. THEOREM, If two similar parallelograms have a common angle^ and he similarly situated^ they are about the same diameter. Let the parallelograms ABCD, AEFG be similar and similarly si- tuated, and have the common angle BAD: ABCD and AEFG shall be about the same diameter. For, if not, let, if possible, the parallelogram BD have its diame- ter AHG in a different straight hne from AF, the diameter of the 14 210 EUCLID'S ELEMENTS. 4 parallelogram EQ\ let GF meet AHG at H^ and tlirough H draw -flX parallel io AD or BC. [1. 31. Then the parallelograms ABGD and AKIIQ are about the same diameter, and are therefore similar to one an- other; [VI. 24. therefore DA is to ^j5 as GA is to ^if. But because ABCD and AEFG are similar parallelo- grams, [ffypothesii, therefore DA is to AB as GA is to AE. [YI. Definition 1. Therefore 0^^ is to ^iT as GA is to ^^, [V. 11. that is, GA has the same ratio to each of the straight lines AJCahdAE, and therefore AKh equal io AE, [V. 9. the less to the greater ; which is impossible. Therefore the parallelograms ABCD and AEFG must have their diameters in the same , Lraight line, that is, they are about the same diameter. Whereforej if two similar parallelograms &c. q.b.d. PROPOSITION 30. PROBLEM. To cut a given straight line in extreme and mean ratio. Let AB be the ^ven straight line; it is required to cut it in extreme and mean ra^o. Divide AB ?A, the point C, so that the rectangle containefd by -j ^i?^jB<7 may be equal to the square A C B on AG. [11.11. Then, because the rectangle ABy BC is equal to the therefore AB is to ^(7 as ^ C is to CB, [VI. 17. Wherefore AB is cut in extreme and mean ratio at the point C. Q.e.p. \Y1. Definition Z. parallelo- Hypothtsii, Definition 1. [V. 11. aiglit lines [V. 9. 'FG must kat is, they C. Q.E.D. . an ratio, red to cut B lal to the ffistruciion, [VI. 17. ratio at efinition 3. BOOK VL 31. PEOPOSITION 31. THEOREM, 211 In any right-angled triangle, any rectilineal figure de- scribed on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle. Let ABC be a right-aftgled triangle, having th j right angle 5^(7: the rectilineal figure described on ^(7 shall be equal to the similar and similarly described figures on BA and CA. ^ Draw the perpendicular AD, :i. 12. Then, because in the right- angled triangle ABC, AD is drawn from the right angle at A, perpendicular to the base BCy the triangles ABDy CAD are similar to the whole triangle (7^-4, and to one another. [VI. 8. And because the triangle CBA is similar to the triangle ABD, therefore CB is to BA as BA is to BD, [VI. Befi 1. And when three straight lines are proportionals, as the first is to the third so is the figure described on the first to the similar and similarly descrU ed figure on the second; [VI. 20, Corollary 2. therefore as CB is to BD so is the figure described on CB to the similar and similarly described figure on BA ; and inversely, as BD is to BC so is the figure described on BA to that described on CB, [V. jB. In the same manner, as CD is to CB so is the figure described on CA to the similar figure described on CB, Therefore as BD and *GD together are to CB %o are the figures described on BA and CA together to the figure described on CB, [V. 24, But BD aiid CD together are equal to CB ; therefore the figure described on BC is equal to the similar and similarly described figures on BA and CA. [V. A, Wherefore, in any right-angled triangle &a q.e.d, . 14—2 212 EUCLID'S ELEMENTS. PEOPOSITION 32. THEOREM, If two triangles^ which have two sides of the one pro- portional to two sides of the other, he joined at one angle 90 as to have their homologous sides parallel to one anotlier. the remaining sides shall he in a straight line. Let ABC and DOE be two triangles, which have the two sides BA, AC proportional to the two sides CD, DE, namely, B A to AC as C£> is to DE; and let AB be parallel to DC and AC parallel to DE: BC and CE shall be in one straight line. For, because AB is parallel to DCf [Hypothesis, and ^<7 meets them, the alternate angles BAC, ACD are equal; [I. 29. for the same reason the tangles ACDyCDEwcoeqml; therefore the angle ^^C7is equal to the angle CDE. [Ax. 1. And because the triangles ABC, DCE have the angle at A equal to the angle at D, and the sides about these angles proportionals, namely, BAio AC as CD is to DE, [ffyp, therefore the triangle ABC is equiangular to the triangle DCE] |-yj Q^ therefore the angle ABC is equal to the angle DCE. And the angle BAC was shewn equal to the angle ACD ; therefore the whole angle ACE is equal to the two ano-les ABC and BAC ^ [Axiom 2. Add the angle ACB to each of these equals ; then the angles ACE and ACB are together equal to the Biiglea ABC, BAC, ACB. But the angles ABC, BAC, ACB are together equal to two right angles ; [I. 32. fciieixiorc tiiG angles ACE and ACB are together equal to two nght anglcH. ' And since at the point C, in the straight line AC the two straight Imes BO^ CE which are on the opposite sides e one pro- ' one angle le anoUier^ \ have the I CD, DE, et AB be I CE shall 9E. [Ax. 1. 9 angle at ese angles 5^, [Hyp, e triangle [VI. 6. :je. eACD; .wo angles [^ariom 2. lal to the ' equal to [I. 32. ' equal to B AC, the >site sides BOOK VI, 32, 33. 213 of it, make the adjacent angles ACE^ ACB together equal to two right angles, therefore BC and CE are in one straight line. [1. 14. "Wherefore, if two triangle$ &c. q.e.d. t PROPOSITION 33. THEOREM, In equal circles, angles, whether at the centres or at the circumferences, have the same ratio which the arcs on which they stand have to one another; so also have the sectors. Let ABC2iXidL DEF be equal circles, and let BOC and EHF be angles at their centres, and BAC and EDF angles at their circumferences : as the arc BC is to the arc EF so shall the angle BGC be to the angle EHF, and the angle BAC to the angle EDF', and so a£o shall the sector BGChe to the sector EHF, Take any number of arcs CE". KL, each equal to BC, and also any number of arcs FM, MN each equal to EF) and join GK, GL, HM, HN, Then, because the arcs BC, CK, KL, are all equal, [Cmstr* the angles BGC, CGK, KGL are also all equal ; [III. 27. and therefore whatever multiple the arc BL is of the arc BC^ the same multiple is the angle BGL of the angle BGC. ~ ' For the same reason, whatever niultiple the arc EN is of e same multiple is the angle EHN of the the arc EF^ angle EHF. EUCLID'S ELEMENTS. And if the arc BL be equal to the arc EN^ the angle BGL ig equal to the angle EUN \ [in. 27. and if the arc BL be greater than the arc EN^ the angle BQL is greater than the angle EHN\ and if less, less. Therefore since there are four magnitudes, the two arcs J5Q EF^ and tne two angles BQC^ EHF\ ■ and that of the arc BG and of the angle BGG have been taken any equimultiples whatever, namely, the arc BL and the angle BGL ; and of the arc EF and of the angle EHF have been taken any equimultiples whatever, namely, the arc EN and the angle EUN; and since it has been shewn thTit if the arc BL be greater than the arc EN, the angle BGL is greater than the angle EHN\ and if equal, equal ; and if less, less ; therefore as the arc BG is to the bvo EF, so is the angle BGG to the angle EHF. [V. Definition 5, But as the angle BGG is to the angle EHF, so is the angle BAG to the angle EDF, [V. 15. for each is double of each ; [III. 20. therefore, as the arc BG is to the arc EF so is the angle BGG to the angle EHF, and the angle BAG to the angle E£>F, ^ Alan aa 4-ha oK/t R/Y ia 4-/\ 4^1^/% nt»n "PT^ r,n. c1^»11 BGG be to the sector EHF. Join BG, GK, and in the arcs BG, GK take any points X, 0, and jom BX, XG, GO, OK, BOOK VI. S3. 215 [III. 27. the angle 3, less. , the two Then, because in the triangles BGC^ CGK^ the two sides BG^ GC are equal to the two sides CG^ GKj each to each ; and that they contain equal angles ; [III. 27. therefore the base BC is equal to the base CJT, and the triangle BGCia equal to the triangle CGIl, £1. 4. [lavo been J BL and een taken ^ and the e greater the angle the angle ejinition 5, so is the [V. 15. [III. 20. :he angle ^he angle ly points And because the arc BC is equal to the arc CJCy [Cmstr. the remaining part when BG is taken from the circun^- ference is equal to the remaining part when CK is taken from the circumference ; therefore the angle BXC ia equal to the angle OOK". [III. 27. Therefore the segment BXC is similar to the segment COX; [III. Definitum 11. and they are on equal straight lines BCy CX, But similar segments of circles on equal straight lines are equal to one another; [III. 24. therefore the segment BXC is equal to the segment COX. And the triangle BGC was shewn to be equal to the triangle CGX; therefore the whole, the sector BGC, is equal to the whole, the sector CGX, lAxiom 2. For the same reason the sector XGL is equal to each of the sectors BGC, CGX. In the same manner the sectors EHF, FHM, MHN may Therefore whatever multiple tlio arc BL is of the arc BC, the same multiple is the sector BGL of the sector BGC\ 216 EUCLID'S ELEMENTS. and for the aame reason whatever multiple the arc EN is of the arc ^i^, the same multiple is the sector EHN of the sector £dliF, And if the arc BL be equal to the arc EN the sector BGL is equal to the sector EHN\ Mid if the arc ^Z be greater than the arc EN, the sector BGL IS greater than the sector EHN-, and if less, less. ^InC^Epifr.y.f'r ' ' ^^^r magnitudes, the two arcs 2/0, hJ^^ and the two sec .rj BGC^ EIIFi and that of the arc BG and of the sector BGG have been taken any equimultiples whatever, namely, the arc BL and the sector ^(5=Z ; and of the arc EF and of the sector EHF have been taken any equimultiples whatever, namely, the arc EN and the sector Z!^iV; and since it has been shewn that if the arc BL be greater than the arc Z^iV; the sector BGL is greater than the sector EHN\ and if equal, equal ; and if less, less ; therefore as the arc ^^ is to the arc EF, so is the sector BGGio the sector EHF . [v. D^mtim 5. Wherefore, in equal circles &c. q.e.d. * BOOK VI. B, PROPOSITION B. THEOREM. 217 Jfthe vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle con-- talned by the segments of the base, together with the square on the straight line which bisects the angle. Let ABC be a triangle, and let the angle BAG he bisected by the straight line AD: the rectangle BA.AG shall be equal to the rectangle BD^ DC, together with tiio square on AD. Describe the circle ACB about the triangle, [IV. 5. and produce ^Z> to meet the circumference at E^ and join EG. Then, because the angle BAD is equal to the angle EAGf [Hypothesis, and the angle ABD is equal to the angle AEG, for they are in the same segment of the circle, [m. 21. therefore the triangle BAD is equiangular to the triangle EAG. Therefore BA is to ^i> as EA is to AG; [VI. 4. therefore the rectangle BA, AG is equal to the rectangle EA,ADj [VI 16. that is, to the rectangle ED, DA, together with the square oh AD. [IL3, But the rectangle ED, DA is equal to the rectansrle BD,DC; [III. 35. therefore the rectangle BA, AG is equal to the rectangle BD, DG, together with the square on AD. ^^hGreiovejif the vertical angle &o, q.e.d. 218 EUCLWS ELEMENTS, \ ', i PROPOSITION C. THEOREM. 4- If from the vertical angle of a triangle a straight line he drawn perpendicidar to the base, the rectangle contained hythe Bides of the triangle is equal to the rectangle con- iained by the perpendicular and the diameter qfthe circle described about the triangle, * Let ABO be a triangle, and let AD he the perpen- dicular from the angle A to the base BC: the rectangle BAf AC shall be equal to the rectangle contained hj AD and the diameter of the circle described about the triangle. Describe the circle ^(7^ about the triangle ; [IV. 5. draw the diameter AE, and join E€. Then, because the right angle BDA is equal to tho angle ECA in a semi- circle ; [III. 31. and the angle ABD is equal to the angle A EC, for they are in the same segment of the circle ; [IH. 21. therefore the triangle ABD is equiangular to the triangle AEC. Therefore BA is to AD as EA is to AC; [VL 4. therefore the rectangle BA, AC is equal to the rectangle EAy AD. Tvi. iQ^ Wherefore, if from the vertical angle &c. q.b.d. ^th PROPOSITION D.. THEOREM. The rectangle contained by the diagoiials of a qmdri. lateral figure inscribed in a circle is equal to both the rectangles contained by its opposite sides* LOOK VL D. 219 night line contained %ngle con- 'the circle e perpen- rectaugle jd by AD i triangle. [III. 21. ) triangle [VI; 4. rectangle •VI. 16. qvadri' both the Let ABGD be any quadrilateral figure inscribed in a circle, and join AG,BD'. the rectangle contained by AC, BD shall be equal to the two rectangles contained by AB.GDm^hyAD.BG, M Make the angle ABE equal to the angle DBG; [I. 23, Ad to each of these J^uals the angle EBD, ^en the angle ABD is equal to the angle EBG [Axiom 2. And the angle BDA is equal to the smgleBGE, for they are in the same seg- ment of the circle ; [111.21. therefore the triangle A BD is equiangular to the triangle EBG. Therefore ^Z> is to DB aaEGisioGB) therefore the rectangle ADj GB is equal to the rectangle DB.EG. [VI. 16. Again, because the angle ABE is equal to the angle ])J3Q [Consttntction, and the angle BAE is equal to the angle BDG, for they are in the same segment of the circle ; [HI. 21. therefore the triangle ABE is equiangular to the triangle DBG. Therefore BA is to AE as BD is to DG; [VI. 4. therefore the rectangle BA, DG is equal to the rectangle AE.BD. [VI. 16. But the rectangle ADy GB haa been shewn equal to the rectangle DB, EG; therefore the rectangles AD, GB and BA, DG are together equal to the rectangles BD, EG and BD, AE ; that is, to the rectangle 5Z>, ^O: ^^ [II. 1. Wherefore, the rectangle corUained &c. q.£.d. [VI. 4. BOOK XI. DEFINITIONS. 1. A SOLID is that which has length, breadth, and thickness. ^ 2. That which bounds a solid is a superficies. 3. , A straight line is perpendicular, or at right angles, to a plane, when it makes right angles with every straight line meeting it in that plane. 4. A plane is pei-pendicular to a plane, when the straight lines drawn in one of the planes perpendicular to the common section of the two planes, are perpendicular to the other plane. 6. The inclination of a straight line to a plane is the acute angle contained by that straight line, and another drawn from the point at which the first line meets the plane to the point at which a perpendicular to the plane drawn from any point of the first line above the plane, meets the same plane. 6. The inclination of a plane to a plane is the acute* angle contained by two stl-aight lines drawn from any the same point of their common section at right angles to it, one in one plane, and the other in the other plane. 7. Two planes are said to have the same or a like inclination to one another, which two other planes have, when the said angles of inclination are equal to one 8. Parallel planes are such as do not meet one another though produced. BOOK XL DEFINITIONS. 221 dth, and it angles, ' straight ^hen the icular to mdicular le is the another eets the lie piano le plane^ le acute any the les to it, •r a like es have, to one another 9. A solid angle is that which is made by more than two plane angles, which are not in the same plane, meeting at one point. 10. Eoiial and similar solid figures are such as are contained by similar planes equal in number and magni- tude. [See the Notes^ 11. Similar solid figures are such as have all their solid angles equal, each to each, and are contained by the same number of similar planes. 12. A pyramid is a solid figure contained by planes which are constructed between one plane and one point above it at which they meet 13. A prism is a solid figure contained by plane figures, of which two that are opposite are equal, similar, and par- allel to one another ; and the others are parallelograms. 14. A sphere is a solid figure described by the revolu- tion of a semicircle about its diameter, which remains fixed. 15. The axis of a sphere is the fixed straight line about which the semicircle revolves. 16. Tlie centre of a sphere is the same with that of the semicircloo 17. The diameter of a sphere is any st -aight line which passes through the centre, and is terminated both ways by the superficies of the sphere. 18. A cone is a solid figure described by the revolution of a right-angled triangle about one of the sides containing the right angle, which side remains fixed. If the fixed side be equal to the other side containing the right angle, the cone is called a right-angled cone ; if it be less than the other side, an obtuse-aneled cone : and if greater, an acute-angled cone. 19. The axis of a cone is the fixed straight line about which the triangle revolves.. 222 EUCLID'S ELEMENTS. I 20. The base of a cone !s the circle described by tliat tide containing the right angle which revolves. 21. A cylinder is a solid figure described by the revo- hition of a right-angled parallelogram about one of its sidca which remains fixed. 22. The axis of a cylinder is the fixed straight lino about which the parallelogram revolves. 2.3. The bases of a cylinder are the circles descnbed by the two revolving opposite sides of the parallelogram. 24. Similar cones and cylinders are those which have their axes and the diameters of their bases proportionals. 25. A cube is a solid figure contained by six equal squares. 26. A tetrahedron is a solid figure contained by four equal and equilateral triangles. 27. An octahedron is a solid figure contained by eight equal and equilateral triangles. 28. A dodecahedron is a solid figure contained by twelve equal pentagons which are equilatend and equi- angular. 29. An icosahedron is a solid figure contained by twenty equal and equilateral triangles. A. A parallelepiped is a solid figure contained by six quadrilateral figures, of which every opposite two are parallel. PROPOSITION 1. THEOREM, One part of a straight Itm cannot be in a plane, and GgiOifiCf- part iCiihoUt it. If it be possible, let AB, part of the straight line ABCf be in a plane, and the part EC without iU BOOK XL 1,2. 223 Then slnco tho straight line ABh in the plane, it can be produced in that plane; let it be produced to D ; and let any plane pass through the straight line^ 2>,and be turned about until it pass through Uie point C. Then, because the points B and C are in this plane, the straight line BO is in it. [i. Definitim 7. Therefore there are two straight lines ABCy ABD in the same plane, that have a common segment AB\ but this is impossible. [I. ll, Corollary* Wherefore, one part of a straight line &c. q.e.d. PROPOSITION 2. THEOREM, TiDO straight lines which cut one another are in one plane; and three straight lines which meet one another are in one plane. Let the two straight lines AB, CD cut one another at E: AB and CD shall be in one plane; and the three straight lines EC, CB, BE which meet one another, shall be in one plane. Let any plane pass through the straight line EBy and let the plane be turned about EB, produced if necessaiy, until it pass through the iwint C Then, because the points E and C are in this plane, the straight line EC is in it ; [I. Definition 7, far the same reason, the straight line BC is in the same plane ; and, by hypothesis, EB is in it. Therefore the three straight lines EC, CB^ BE are in one plane. m.-,*- A 73 JJUiV .£S.JJ are; \y'J^ CwX V AAA vl«W therefore AB and CD are in one plane. Wherefore, two straight lines &c. q.e.p. £XI. 1, 224 EUCLID'S ELEMENTS, PEQPOSITION 3. THEOREM, If two planes cut one another their common section w a straight line. Let two planes A By BG cut one another, and let BD be their common section : BD shall be a straight Ime. If it be not, from B to D, draw in the plane AB the straight line BED, and in the plane BG the straight line BFD. [Postulate 1. Then the two straight lines BEDy BFD have the same extremities, and therefore Include a space be- tweei^ them ; but this is impossible. [Axiom 10. Therefore BD, the common section of the planes AB and BG cannot but be a straight lino. "WheretoYOj if two planes &c. q.eld. \ B c E[ r j>\ A i I PROPOSITION 4. THEOREM, If a straight line stand at right angles to each of two straight lines at the point of their intersectiony it shall also he at right angles to the plane ichich parses through them, that is, to the plan^ in which they are. Let the straight line EF stand at right angles to each of the straight lines AB, CD, at E, the X)oint of their interscc'tion : EF shall also be at right angles to the plane passing through AB, GD. Take the straight lines AEy EBy CEy ED, all equal to one another ; join AD, GB; through E draw in the Diane in which are AB. GD any straight line cutting AD at (9, and GB at H-, and from any point F in EF draw FA, FG, FD, FG, FHyFB. on section id let BD , line. 3 AB and ch of two •%, it shall ?5 through !es to each BOOK XL 4. 225 Then, because the two sides AE, ED are equal to the two sides BEy EC, each to each, [Construction, and that they contain equal angles AED, B^^C; [I. 16. therefore the base AD is equal to the base BC, and the angle DAE is equal to the angle EEC. [I. 4. And the angle AEG is equal to the angle BEIT; [1. 15. therefore the triangles AEG, BEH have two angles of the one equal to two angles of the other, each to each ; and the sides EA, EB adjacent to the equal angles are equal to one another; \Comirvi:tion, therefore EG is equal to EH, and AG is equal to ^^' [I. 26. And because EA is equal to EB, [Construction, and EF is common and at right angles to them, [Hy^othem, therefore the base ^i^ is equal to the base BF, [I. 4. For the same reason CF is equal to DF. And since it has been shewn that the two sides DA. AF^VQ equal to the two sides CB, BF, each to each, and that the base DF is equal to the base CF; therefore the angle Z^^i' is equal to the angle CBF. [T. 8. Again, since it has been shewn that the two sides FA, AG are equal to the two sides FB, BH, each to each, and that the angle FAG is equal to the angle FBH\ therefore the base FG is equal to the base FH, [I. 4. Lastly, since it has been shewn that GE is equal to HE, and EF is common to the two triangles FEG, FEH ; and the base FG has been shewn equal to the base FH\ therefore the angle FEG is equal to the angle FEH [I. 8. Therefore each of these angles is a right angle. [I. Btjln. 10. In like manner it may be shewn tliat EF makes right angles with every straight line which meets it in the plan© passing through AB, CD. Therefore EF is at right angles to the plane in which are AB,CD. [XI. Dejinition 3. therefore, if a straight line &c. q.e.d. 15 ! * 228 EUCLID'S ELEMENTS. PROPOSITION 6. THEOREM, If three straight lines meet all at one point, and a ^trainM line stand at right angles to each qf them at that %ini^heZTs^^^^^ lines shall be in one and the same plane, i Let the straight line AB stand at «g\* a«g^®« *^ f'^ of thrstraight lines BG. BD, BE, at B the IK)mt wh^e they meet xBC, BD, BE shall be m one and the same plane. For, if not. let, if possible, BJ) and BE be in one plane, and 5(7 without it ; letaplane pass through AB and BG\ the common section of this plane with the plane in which are BD and BE is a straight line ; \^l., 3. let this straight line be BF, Then the three straight lines ^ . , AB, BG, BF are all in one plane, namely, the plane wmcn passes through ^5 and ^C. , "*xt,^ And because ^|BtandB at right angles to eg of^^e straight lmesi?/>,-S A , K u therefore it is at right angles to the plane passmg through them * therefore it makes right angles with eveiT straight Jline meeting it in that plane. [XL Definitvm d. But BF meets it, and is in that plane ; therefore the angle ABF is a right angle. But the angle ABC is, by hypothesis, a right angle ; therefore the angle ABC is equal to the angle ABF ; [Ax. 11. and they are in one plane ; which is impossible. lAxwm 9. Therefore the straight line BG is not without the plane m therefore the three straight lines BG, BD, BE are in and the same plane. NV herefore, if three straight lines &c. Q.E.D. one nnty and a hem at that nd the same gles to each point where id the same plane which each of the [Hypothesis. sing through [XI. 4. straight line [. Definition 3. angle ; 3. [Axiom 9. t the plane in BOOK XL 6. PEOPOSITION C. THEOREM. 227 If two straight lines he at right angles to the same plane, they shall be parallel to one another* Let the straight lines ABj CD be at right angles to the same plane : AB shall be parallel to CD, Let them meet the plane at the ^^ points Bf i5 ; join BD ; and in the plane draw DJE at right angles to BD ; [1. 11. make DB equal to AB ; [L 3. andjom^j&,-4^,^i>. Then, because AB is perpendicular to the plane, [Hypothesis. it makes right angles with every straight line meeting it in that plane. [XI. Def. 3. But BD and BE meet AB, and are in that plane, therefore each of the angles ABDj ABE is a right angle. For the same reason each of the angles CDB, CDE is a right angle. And because AB is equal to ED, [Construction, and BD is common to the two triangles ABD, EDB, the two sides AB, BD are equal to the two sides ED, DB, each to each ; and the angle ABD is equal to the angle EDB^ each of them being a right angle ; [Axiom 11. therefore the base AD is equal to the base EB. [I. 4. Again, because AB is equal to ED, [Construction, and it has been shewn that BE is equal to DA ; therefore the two sides AB, BE are equal to the two sides EDy DAf each to each ; and the base AE is common to the two triangles ABE, EDA; thersforfl thfi aticlft A HJH ia eoual to the anfflc EDA. FT. 8^ E are m one ■ ^^^ ^^^ ^^^^^ ^^^ jg ^ ^^^ii angle. E.D, therefore the angle EDA is a right angle, that is, ED is at ridit angrlcs to AD^ l5-2 22S EUCLIUS ELEMENTS. But ED is also at right angles to each of the two BD^ CD; therefore ED is at right angles to each of the three straiffht lines BD AD, CD, at the point at whicli they meet ; therefore these three straight lines are all in the same plane. [XI. 5. But AB is in the plane in which are BD, DA ; [xi. 2. therefore AB, BD, CD are in one plane. And each of the angles ABD, CDB is a right angle ; therefore AB is parallel to CD. [X. 28 y^herefore, if twb straight lines &c^ q.e.d. I PROPOSITION 7. THEOREM. ir two straight lines he parallel, the straight line drawn from any point m one to any point in the other, is in the same plane with the parallels. ' Let ^5, CD be parallel straight lines, and take any pomt E m one and any point F in the other: the straight line which joms E and F shall be in the same plane with the parallels. ^ For, if not, let it be, if pos- sible, without the plane, as EGF; andinthe planed ^Ci>, in which the parallels are, draw the straight line EMF from E to F. Then, since ^^(yi?' is also a straight line, [Hypothesis. "' the two straight lines EGF, EHF include a space between them ; which is impossible. |;^^,v,m 10. Therefore the straight line ioininn" fVsp tv^infa t? ^.r-. j i? \^ not without the plane in which the parallels AB, CD are; therefore it is in that plane. Wherefore, if two straight lines &c q.b.d* [I 28. Hne drawn r, is in the I take any le straight plane with L e between [Axiom 10. ' ] XT •- CD are; BOOK XL 8. PROPOSITION 8. THEOREM. 229 //"/tro straight lines he paraUel, and one of them he at right angles to a plane, the other also shall he at right angles to the same plane, ^f k^^ ^?h^^ ^^ *^? parallel straight lines; and let one I nt"^ . . X®. ^^ r^^*^* a"^^®s to a plane : the other CD shall be at right angles to the same plane. Let AB, CD meet the plane at the points Bj D ; join BD ; therefore AB, CD, BD are in one plane. [xi. 7. In the plane to which AB is at right angles, draw DE at right angles to ^Z); [i. n. make DE equal to ABi [l. 3, and join BE, AE, AD. Then, because AB is at right angles to the plane, [Hypothesis. it makes right angles with every straight line meeting it m that plane ; [XI. Definitwn 3. therefore each of the angles ABD, ABE is a right angle. And because the straight line BD meets the parallel straight lines AB, CD, the angles ABD, CDB are together equal to two right angles. ■" [1,2,^. But the angle ABD is a right angle, [H^/^otliesis] therefore the angle CDB is a right angle ; that is, CD is at right angles to BD. And because AB is equal to ED, [Construction. and BD is common to the two triangles ABD, EDB; the two sides AB, BD are equal to the two sides Ed\dB each to each ; * and the angle ^^i) is equal to the angle EDB, each of wiuiii uoiug it ligiit aiigie ; [AxUm 11. therefore the base AD is equal to the base EB. [i. 4. Again, because AB is equal to ED, [Cmstruction. and BE has been shewn equal to DA, 230 EUCLID'S ELEMENTS. tbo two sides AB, BE are equal to the two sides ED, DA, each to each ; and the base ^^ is common to the two triangles ABE^ EDA ; therefore the angle ABE is equal to the angle ADE. [I. 8. But the angle ABE is a right angle ; therefore the angle ADE is a right angle ; that is, ED is at right angles to ^2>. But ED is at right angles to BDj [ConM, therefore ED is at right angles to the plane which passes through BD^ DA, [XI. 4. and therefore makes right angles with every straight line meeting it in that plane. [XI. Definition 3. But OD is in the plane pa,ssing through BD, DA, because all three are in the plane in which are the parallels AB, CD; therefore ED is at right angles to CD, and therefore CD is at right angles to ED. But CD was shewn to be at right angles to BD ; therefore CD is at right angles to the two straight lines BD, ED, at the point of their intersection D, and is therefore at right angles to the plane passing through BD, ED, [XI. 4. that is, to the plane to which AB is at right angles. "Whereioro, if two straight lines &c. q.e.d. PROPOSITION 9. THEOREM. Two straight lines which are each of them parallel to the same straight line, and not in the same plane with it, are parallel to one another. Let AB and CD bo each. of them parallel to EF, and not in the same plane with it : AB shall be parallel to CD. In EF take any point G ; in the plane passing through EF and A B. draw from G t he straight , angles to EF; and in the plane passing through EF and CD, draw from G the line GH at right BOOK XL 9,10. 231 straight line GK at right angles to EF. [t. 1 1. Then, because EF is at right angles to GH and ""^^ {Construction, EFia at right angles to the plane -ff(?jr passing through them. jXI 4. And EF is parallel to AB ; [Hypothesis, therefore AB is at right angles to the plane HGK. [XI. 8. For the same reason CD is at right angles to the piano ncref ore AB and CD are both at right angles to the plane HGK, ° Therefore AB is parallel to CD. [xi. 6. therefore, if two straight lines &c. Q.E.D. PROPOSITION 10." THEOREM. If two straight lines meeting one another be parallel to two ot/iersthat meet one another, and are not in the same plane tcith the frst two, the first two and the otJier two shatl contain equal angles. Let the two straight lines AB, BC; which meet one an- other, be parallel to the two straight lines DE, EF, which meet one another, and are not in the same plane with AB, BC: the angle ABC shall be equal to the angle DEF. Take BA, BC, ED, EFsdl equal to one another, and join AD, BE, CF, AG,DF. > y^^» Then, because AB is equal and parallel to DE, therefore AD is equal and parallel to ^^' [1. 33. For the same reason, CF is equal auu parallel to HJH. Therefore are each of them equal and parallel to BE. Therefore AD is parallel to CF, [XL Ok 23^ EUCLID'S ELEMENTS. and AD is also equal to CF% [Axiom 1. Therefore -4(7 is equal and parallel to DF. [I. 33. And because AB^ BG are equal to DE, EFj each to each, and the base ACh equal to the base DF, therefore the angle ABC is equal to the angle DEF. [I. 8. Wherefore, if two straight lines &c. ♦ PROPOSITION 11. PROBLEM. To draw a straight line perpendicular to a given plane, from a given point without it. Let A be the given point without the plane BH: it is required to draw from the point A a straight line perpen- dicular to the plane BH, Draw any straight line BC in the plane BH, and froni the point A draw AD perpendicular to i5(7. [1.12. Then if AD be also perpen- dicular to the plane BH, the thing required is done. But, if not, from the poipt D draw, in the plane BH, the straight line DE at right angles to BG, [1. 11. - and from the point A draw -4i^ perpendicular to DE, [1. 12. AF shall be perpendicular to the plane BH, Through F draw 6?jSr parallel to BG. [I. 31. Then, because BG is at right angles to ED and DA, [(7omTr. BG is at right angles to the plane passing through ED and DA. [Xi. 4. And GH is parallel to BGi [Construction, yivenplane, I BH: it is line perpen- BOOK XL 11, 12. 233 > DE, [1. 12. [I. SI. 'yA, [Comtr. irough ED [XI. 4. Construction, therefore GIT is at right angles to the plane passing through ED and DA ; [XI. 8. therefore GH makes right angles with erery straight line meeting it in that plane. [XI. Definitim 8. But ^iP meets it, and is in the plane passing through ED and DA ; therefore Gffis at right angles to ^i^, and therefore AF is at right angles to GIT, But AF is also at right angles to DE; [Constructwn, therefore AF is at right angles to each of the straight lines Gil and DE at the point of their intersection ; therefore ^i^^is perpendicular to the plane passing through GH and DEy that is, to the plane BH, [XI. 4. Wherefore, /row the given point Ay without the plane BHy the straight line AF has been drawn perpendicular to tJie plane, q.e.p. PROPOSITION 12. PROBLEM, To erect a straight line at right angles to a given plane, from a given point in the plane. Let A be the given point in the ^ven plane : it is re- quired to erect a straight line from the point A, at rigb^j angles to the plane. From any point B without the plane, draw BC perpendicular to the plane ; [XI. 11. and from the point A draw AD parallel to BG, [I. 31. AD shall be the straight line re- quired. / — For, because ^7>and BC ore / two parallel straight lines, [Comir. ^ — and that one of them BG is at right angles to the given plane, o [Construction, the other -42> is also at right angles to the given plane. [XI. 8, Wherefore a straight line has been erected at right a?i». gles to a given plane, from a given point in it, q.e.p. 231 I EUCLWS ELEMENTS, PROPOSITION 13. THEOREM, From the fame point %n a given plane, there cannot U two straight lines at right angles to the plane, on the same side of It; and there can he hut me perpendicular to a plane from a point mthout the plane. For, if it be possible, let the two straight lines AB AG be at riffht angles to a given plane, from the same point A m the plane, and on the same side of it. Let a plane pass through BA, AC; the coinmon section of this with the given plane is a straight line; [XI. 3. let this straight line be DAE, Then the three straight lines AB, AC, DAE are all m one plane. And because CA is at right angles to the plane, [ffypothes^is. it makes right angles with every straight line meeting it in the plane. fXI. Bejitdtion 3. But DAE meets CA, and is in that plane ; therefore the angle C7^J^ is a right angle. For the same reason the angle BAE is a right angle. Therefore the angle CAEh equal to the angle J5^it:; [Ax.!!, and they are in one plane ; which is impossible. f^^.,^ g^ Also, from a point without the plane, there can be but one perpendicular to the plahe. another^^''^ ^""^^ ^® '^^' *^®^ ^^^^ ^® P^^"®^ ^ ^"® which is absurd. ^herefore^from the same point &c, q.e.d. BOOK XL 14,16. 235 PROPOSITION 14. THEOREM, Planes to which the same straight line is perpendicular are parallel to one another. Let the straight line ABhQ perpendicular to each of the planes CD and EF\ those planes shall be parallel to one another. For, if not, they will meet one toother when produced ; let them meet, then their com- mon section will be a straight line; let GH be this straight line ; in it take any pomt K, and join AK.BK. Then , because AB\% perpen- dicular to the plane EF, VEyp, it is perpendicular to the straight line BK which is in that plane ; [XI. Definition 3. therefore the angle ABKi^ a right ang'e. For the same reason the angle BAK is a right angle. Therefore the two angles ABK, BAK of the triangle ^^^are equal to two right angles; which is impossible. [I. 17. Therefore the planes CD and EF, though produced, do not meet one another ; that is, they are parallel. [XI. Deration 8. Wherefore, planes &c. q.b.d. PROPOSITION 15. THEOREM, T P tni./\ at^nitnin'kt InnnAa nnh'i/o'h nvtaaf nvta nvtrifhov TiA inn.irnJ.J.ol Ji.J fK/V cvt ixt-yitrv z-Zffzra ct.TZ-v---r^ rrv.~.".- --r '- ; j <, • to two other straight lines which meet one am)ther, hut are not in the sam>e plane with the first two, the plane pass- ing through these is parallel to the plane passing through the others. li^ 236 EUCLID'S ELEMENTS. Lot AB, BGy two straight linos which moot one another, be parailol to two other straight lines DE, EF. which meet one another, but are not in the same plane with AB, BC- the piano passing through AB, BC, shaU be parallel to the plane passing through DEy EF, From the point B draw BG perpendicular to the plane pass- mg through DE, EF, [XI. 11. and let it meet that plane at G ; through G draw GH parallel to ED, and 6?^parallel to EF. [1.31. Then, because BG is per- pendicular to the plane passing through DEy EF, [Comtructi(m. it ma^es right angles with eyery straight line meeting it m that plane; \X.l. DejinUi^i Z, but the straight lines GHmdi GlCmeei it, and are in that plane ; therefore each of the angles BGH and BGK is a right angle. ^ Now because BA is parallel to ED, {Hypothem. and GHh piu-allel to ED, [Comtructum, therefore BA is parallel to GH ; [xi. 9. therefore the angles ABG and BGH are together equal to two right angles. ° |-j^29. And the angle BGHY^tLS been shewn to be a right angle; therefore the angle ABG ih a right angle. For the same reason the angle ^56? is a right angle. Then, because the straight line GB stands at ri-ht Section i?" ^ ^"'^ ^'"'' ^^' ^^' ^* ^^'^ ^^^^^^ therefore GB is nAmenrlinnin*. +« +i»^ ^i ^ . ., BA BC * ""* "-•-"«* -v wiw piiiiiu j^assing liirougii DE EF '^ ^^ P^'l^^^^c^ar to the plane passing through ' * [Cimstruction, ie another, ^hich meet I AB, BC: allel to the meeting it 'Jejinitwn 3. ire in that is a right HypothesU. onstructifOn, [XI. 9. her equal [I. 29. t angle; gle. at right ' point of g tiirough [XI. 4. J through nstruction. BOOK XL 16,16. 237 But planes to which the same straight line is perpendicular are parallel to one another; [XI. 14. therefore the plane passing through AB, BC is parallel to the plane passing through DEy EF, Wherefore, ^two straight lines &c. Q.E.ii PROPOSITION 16. THEOREM. Jf two parallel planes be cut by another plane^ their cotnmon sections with it are parallel. Let the parallel planes ABj CD be cut by the plane EFHGj and let their common sections with it be £E. GH: EF shall be parallel to Gil. For if not, EF and GH^ being produced, will meet cither towards F^ H, or towards E^ G. Let them be pro- duced and meet towards F, H at the point K. Then, since EFK is in the plane AB, every point in EFK IS in that plane ; [XI. 1. therefore K is in the plane AB. For the same reason K is in the plane CD. Therefore the planes AB, CD, being produced, meet one an- other. But they do not meet, since they are parallel by hypothesis. Therefore EF and GHy being produced, do not meet to- wards F, H. In the same manner it may be shewn that they do not meet towards E, G. But straight lines which are in the same plane, and which being produced ever so far both ways do not meet are therefore EF is parallel to GH. "^hexeioTQf if two parallel planes ^ Q.E.D, G^ 238 EUCLWS ELEMENTS. PKOPOSmON 17. THEOREM, If two itraiglU lines he cut hy parallel planes^ the(f thMl he cut in the same ratio. Let the straight linos AB and CD be cut by the pa- rallel planes 6?^, KL^ MN, at the points A, E, JB, and CyF.D: AE shall be to EB as C'i^is to FD. Join AC,BD,AD) let ' AD meet the plane KL at the point X; and join EXy XF» Tlien, because the two parallel planes KL, MN are cut by the plane EBDX, the common sections EX, BD are parallel ; [XI. 16. and because the two pa- rallel planes GH, KL are cut by the plane AXFG^ the common sections AG, XF are parallel. [XT. 16. And, because EX is parallel to BD, a side of the triangle ABD, therefore AE is to EB as AX is to XD. [yi. 2. Again, because Xi^is parallel to ^(7, a side of the triangle ADC, therefore AX is to XD as CF is to FD. [VI. 2. And it was shewn that AX is to XD as AE is to EB ; therefore AE is to EB as CF is to FD. [V. 11. Wherefore, if two straight lines &c. q.e.d. PROPOSITION 18. THEOREM. If a straight line he at right angles to a plane, every plune which passes through it shall he at right angles to that plane. Let the straight line AB be at right angles to the plane CK\ every plane which passes through AB shall be ftt right angles to tiie plane CK. BOOK XL 18, la 239 ane9^ they )y the pa- B^ By and ie of the CVI. 2. 10 triangle [VI. 2. )JSB; [V. 11. me, every angles to 68 to the ? shall be 0- Let any plane DE pass through AB^ and let (7j& be the common section of the planes DEy CK\ [XI. 3. take any point F in OEy from which draw FQy in the plane DE, at right angles to CE. [1. 11. Then, because AB is at right angles to the plane ^■'^> [Hypothesis. therefore it makes right angles with every straight line meeting it in tliat plane ; [XI. Befinitwn 3. but CB meets it, and is in that plr e ; therefore the angle ABF is a right angle. But the angle GFB is also a right angle j ICcyMtmctim, therefore FG is parallel to AB, [i. 28. And AB is at right angles to the plane CK] [HypoihesU. therefore FG is also at right angles to the same plane. [XI. 8. But one plane is at right angles to another plane, when the straight lines dra\\Ti in one of the planes at right angles to their common section, are also at right angles to the other plane ; [XI. Definition 4. and it has been shewn that any straight line FG drawn in tbe plane DE, at right angles to CE, the common section of the planes, is at right angles to the other plane CK\ therefore the plane DE is at right angles to the plane CK, In the same manner it may be shewn that any other plane which passes through AB h Q,i right angles to the plane CK. Wherefore, if a straight line &c. q.e.d. PROPOSITION 19. THEOREM, If two planes which cut one another he each of them perpendicular to a third plane, their common section shall be perpendicular to the same 2:)lane. 240 EUCLID'S ELEMENTS. Lfet the two planes BA, BC be each of them peippen- dicular to a third plane, and let BD be the common section of the planes BA^ BG\ BD shall be perpendicular to the third plane. For, if not, from the point />, draw in the plane BA^ the straight line DE at right angles to ADy the common section of the plane BA with the third plane ; [1. 11. and from the point 2), draw in the p>lane BC, the straight line DF^X right angles to CD, the common section of the plane BC with the third plane. • [1. 11. Then, because the plane BA is perpendicular to the third plane, IHypofhesk. and DE is drawn in the plane BA at right angles to ^i> tlieir common section ; [Comtructwn, therefore DE ia perpendicular to the third plane. [XI. D^. 4. In the same manner it may be shewn that DF is per- pendicular to the third plane. Therefore from the point D two straight lines are at right angles to the third plane, on the same side of it; which is impossible. [XI. 13. Therefore from the point D, there cannot be any straight line at right angles to the third plane, except BD the com- mon section of the planes BA, BC\ therefore BD is perpendicular to the third plane. Wherefore, if two planes &c. q.e.d. PROPOSITION 20. THEOREM, If a solid angle he contained hy three plane angles, any two qfthem are together greater than the third. Let the solid^ angle^ at_^ be contained by the three plane angles BAO^ CAJJ, uAii: any two of them shall be together greater than the third. If the angles BAC, CAD, DAB be all equal, it ia evident that any two of them are greater than the third. BOOK XL 20. 241 1 peippen- >n section ar to the hypothesis. )8 to AZ> nstruction, F is per- )a are at ie of it; [XI. 13. f straight the com- ghsj any he three shall be iial, it 18 third. [1.3. If they are not all equal, let BAG be that angle which is not less than either of the other two, and is greater than one of them, BAD. At the point -4, in the straight line BA^ make, in the plane which passes through BA, ACy the angle BAE equal to the angle BAB ; [1. 23. make AE equal to ^i> ; through E draw BEG cutting AB, AG Sit the points B, G: andjoin i>5,Z>C: Then, because -4i> is equal to AE, [Comtruction. and ABh common to the two triangles BAD, BAEf the two sides BAj AD are equal to the two sides BAy AE, each to each ; and the angle BAD is equal to the angle BAE; [Constr, therefore the base BD is equal to the base BE. [I. 4. And because BD, DG are together greater than BG, [1. 20. and one of them BD has been shewn equal to BE a part of BG, therefore the other DG is greater than the remaining part EG. And because ^2> is equal to AE, [Comtruction. and AG is common to the two triangles DAG, EAG, but the base DG is greater than the base EG; therefore the angle DAG is greater than the angle EAG. [1. 25. And, by construction, the angle BAD is equal to the angle BAE; therefore the angles BAD, DAG are together greater than the angles BAE, EAG, that is, than the angle BAG. But the angle BAGi& not less than either of the angles BAD. DAG I therefore the angle BAG together with either of the other angles is greater than the third. Wherefore, if a solid angle &c. q.£.]>. 16 f'' (I i ii II ^42 EUCLIUS ELEMENTS. PROPOSITION 21. THEOREM. Every solid angle is contained by plane angles^ which are together less than four right angles. First let the solid angle at A be contained bv three plane angles BACy CAB, DAB: these three shall be together less than four right angles. In the straight lines ^5,^(7, ^Z> v take any points B, OJ Z>, and join BGf CDy DB. Then, because the solid angle at B is contained by the three plane angles CBA, ABD, DBG, any two of them are together greater than the third, {XI. 20. therefore the angles CBA, ABD are together greater than the angle DBG. For the same reason, tho angles BCA^ AGD are together greater than the angle DCB, and the angles CDA^ ADB are together greater than the angle BDG. . Therefore the six angles CBA, ABD, BGA, AGD, GDA, ADB are together greater than the three angles DBG, DGB,BDG; but the three angles DBG, DGB, BDG are together equal to two right angles. [I. 32. Therefore the six angles GBA, ABD, BGA, AGD, GDA, ADB are together greater than two right angles. And, because the three angles of each of the triangles ABC, ACDy ADB are together equal to two right angles, * [1. 32. therefore the nine angles of these triangles, namely, the angles CBA, BAG, ACB, AGD, GDA, CAD, ADB, DBA, DAB are equal to six right angles ; and of these, the six angles GBA, ACB, AGD, GDA, A T\T> T\T> A ..-^^ ,-.^««J--^_ Al X _;™l,i. ^m»~lr»„ therefore the remaining three angles BAG, GAD, DAB^ which contain the solid angle at A^ are together less than four right anglen. is, les, which bv three shall be BOOK XI. 21. 543 5 together r than the 72), CDA, les D£C, bher equal [I. 32. 7i>, CDA, ) triangles two right [I. 32. imely, the D, ADB, 7), CDA, D, DABi less than Next, let the solid angle at A be contained by any number of plane angles BAG, CAD, DAE, EAF, FAB: these shall be together less than four right angles. Let the planes in which the an- gles are, be cut by a plane, and let the common sections of it with those planes be BC, CD, DE, EF, FB, Then, because the solid angle at B is contained by the three plane angles CBA, ABF, FBC, any two of them are together greater than the third, [XI. 20. therefore the angles CBA, ABF are together greater than the angle FBC For the same reason, at each of the points C, D, E, F, the two plane angles which are at the bases of the triangles having the common vertex A, are together greater than the third angle at the same point, which is one of the angles of the polygon BCDEF, Therefore all the angles at the bases of the triangles at^ together greater than all the angles of the polygon. Now all the angles of the triangles are together equal to twice as many right angles as there are triangles, that is, as there are sides in the polygon BCDEF\ [I. 32. and all the angles of the polygon, together with four right angles, are also equal to twice as many right angles as there are sides in the polygon ; [I. 32, Corollary 1. therefore all the angles of the triangles are equal to all the angles of the polygoti, together with four right angles. [Ax. 1. But it has been shewn that all the angles at the bases of the triangles are together greater than all the angles of the polygon ; therefore the remaining angles of the tHangles, nameiv, those at the vertex, which contain the solid angle at A, uie together less than four right angles. 16-3 li^^ H BOOK XII. LEMMA. If , from the greater of ttDd uneqtml magnitudes there he taken more than its Ivalfy and from the remainder more than its half and so on, there shall at length re- main a magnitude less than the smaller qf the proposed magnitudes. Let AB and C be two unequal magnitudes, of which AB\% the greater: if from AB there be taken more than its half, and from the remainder more than its half, and so on, there shall at length remain a magnitude less than (7. For G may be multiplied so as at length to become greater than AB, A Let it be so multiplied, and let DE its multiple be greater than AB^ and lot DE be divided into DFy FG^ GE, each equal to C From AB take Bfff greater than its half, and from the remainder AH take BJC Sweater than its half, and so on, until there ) as many divisions in ^5 as in BE ; and let the divisions in AB be AK, KH, HB, and the divisions in BE be BF, FG, GE, K H G B c is Then, because DE is greater than AB; and that EG token from BE is not greater than its half: but BH taken from AB is greater than its half ; therefore the remainder BG is greater than the remainder AH* BOOK XIL 1. 245 ides there emainder length re- proposed , of which nore than ilf, and so than G, G h i its half: remainder Again, because DO is greater than AH\ and that OF is not greater than the half of DO, but HIC is greater than the half oi AH', therefore tlie remainder DF ia grr \ter than the remainder AK, But DjPIs equal to C; therefore (7 is greater than AK; that is, AK is less than C. q.e.d. And if only the halves bo taken away, the same thing may in the same way Ij den mstrated. PROPOSITION 1, THEOREM, Similar polygor t inscribed in circles are to one another as the. squares on uieir diameters. Let ABODE, FGHKL be two circles, and in them the similar polygons ABODE, FGHKL-, and let BM, GNhQ the diameters of the circles: the polygon ABODE shall be to the polygon FGHKL as the square on BM is to the square on GM. 3om AM, BE, FN, GL. Then, because the polygons are similar, , therefore the angle BAE is equal to the angle GFL, and BA is to AE as OF is to FL. [VI. D^nitim 1. Therefore the triangle BAE is equiangular to the triangle GFL; [VI. 6. therefore the angle AEB is equal to the angle FLO But the angle AEB is oqual to the angle AMB, ai angle FLO is equal to the angle FNG ; [I therefore the angle AMB is equal to the angle FNG the 246 EUCLID'S ELEMENTS. And the angle BAM is equal to the angle GFN tor each of them is a nght angle, [uj. ^^^ Therefore the remaining angles in the triangles AMB, I'JSU are equal, and the triangles are equiangular to one another ; ^ ® therefore BA is to BM as GF is to GN, [VI. 4. and, alternately, BA is to GF as BM is to GN\ [Y. 16. therefore the duplicate ratio of BA to GF is the same as ,the duphcate ratio of BM to GN. [V. Bejinitim 10, V. 22. . ^Jut the polygon ABODE is to the polygon FGHKL in the duphcate ratio of BA to (9i^ ; [vi. 20. and the square on BM is to the square on GN in the du- plicate ratio of BM to GN ; [vi. 20. therefore the polygon ABODE is to the polygon FGHKL as the square on BMis to the square on GN [V. 11. Wherefore, similar polygons &c. q.e.d. PROPOSITTOF 2. THEOREM, Circles are to one an<^ther as the squares on their diameters. Let ABOD, FFGH be twp circles, and BD, i^// their diameters: the circle ABOD shall be to the circle EFGH as the square on BD is to the square on FH, fi."" 5" ^"^^ C"x;ic ^iBCD io to some space either less than the circle EFGH, or greater than it. First, if possible, let it be as the Mrcle ABOD is to a space S less than the circle EFGH, BOOK XIL 2. 247 In the circle EFGH inscribe the square EFOH. [IV. 6. This square shall be greater than half of the circle EFGff, For the square EFGH is half of the square which can be formed by drawing straight lines to touch the circle at the points E, Fj G, H -, and the square thus formed is greater than the circle ; therefore the square EFGH is greater than half of the circle. Bisect the arcs EF^ FG, GH, HE at the points K^ and join EK, KF, FL, LG, GM, MH, HN, NE, Then each of the triangles EKF, FLG, GMH, HNE shall be greater than half of the segment of tlie circle in which it stands. For the triangle EKF is half of the parallelogram which can be formed by drawing a straight line to touch the circle at K, and parallel straight lines through E and jP, and the parallelogram thus formed is greater than the segment FEK-, therefore the triangle EKF is greater than half of the segment. > And similarly for the other triangles. Therefore the sum of all these triangles is together greater than half of the sum of the segments of the circle in which ii i.__j Again, bisect EK, KF,&c. and form triangles as before ; then the sum of these triangles is greater than half of the sum of the segments of the circle in which they stand. 248 EUCLID'S ELEMENTS. If this process be continued, and the triangles b^ sup- posed to be taken away, there will at length remahi seg- inents of circles wliich are together less than the excess of the circle EFG7T (•K.ve the space S, by the preceding rri^^t^f^ ^}l^ segments EK, KF, FL, LG, GM, MH, It IS, IV£, be those which remain, and which are together less than the excess of the circle above S; i *j!SJSf°''® ***® ^®s* of ^^^ '--*"«^<' namely the polygon EKFLGMHN, is greater than the space^^. ^™ In the cu-cle u45C7i) describe the polygon ^X5(?(7Pi)i2 similar to the polygon EKFLGMHN, Fi^prr^!rJ^^^^^^ ^^^OCPDR is to the polygon ^KFLQMHN as the square on BD is to the square on *^' [XII. 1. that IS,, as the circle ABGD is to the space S. [Hyp., V. 11. 5 n^n®- ^^T^T ^i^^^^^fj^? i» less than thJ circle ABGD m which it is inscribed, , therefore the polygon EKFLGMHN is less than the space S\ j-y j^ but it is also greater, as has been shewn ; which is impossible. Therefore the square on BD is not to the square on Jf'/Z^a w Til ^^^ '* may be shewn that the square on Z'// IS not to the square on BD as the circle EFGII is to any space less than the circle ABGD. ^ ^« « w Nor is the square on BD to the square on >i7 as ^%r^^i^^ ^-5(7/> IS to any space greater than the circle For, if possible, let it be as the circle ABGD is to a space T greater than the circle EFGff. ^^ Then inVAlSJolw +l»/» ar»iin«A «». TCTT S- J._ xl_ _ - _ -^ _> - "" '"^'j ","•'' ~-i-=^-- y" -i^^-f i3 %,%} wie square Oil Jji/ as the space T is to the circle ABGD, ?.^/?^rr*^® *P^^ ^ ^^ *^ *'*® ^^^^^ ^SOD so is the circle Jiif UM to some space, which must be less than the circle BOOK xn. 2. 249 laiii seg- jxcess of receding together polygon CPDR polygon uare on [XII. 1. ., V. 11. ) circle tan the [V. 14. lare on an the lare on ?is to FH as > circle I space >ii BD ABGDf because, by hypothesis, the space T is greater than the circle J^/'G^if. [V.14. Therefore the square on FH is to the square on BD as the circle EFGH is to some space less tluui the circle ABCD', which has been shewn to be impossible. Therefore the square on BD is not to the square on FH ^ the circle ABCD is to any space greater tnan the circle EFGH, And it has been shewn that the square on BD is not to the square on FH as the circle ABCD is to any space less than the circle EFGH, Therefore the square on BD is to the square on FHfUi the circle ABCD is to the circle EFGH Wherefore, circles &c. q.b.d. circle I circle NOTES ON EUCLID'S ELEMENTS. Thb article Eucleidea in Dr Smith's Dictionary of Greek and Roman Biography wm written by Professor De Morgan ; it contains an account of the works of Euclid, and of the various editions of them which have been published. To that article we ^fer the student who desires full information on these subjects. Perhaps the only work of importance relating to Euclid which has been published since the date of that article is a work on the Porisms of Euclid by Chasles ; Paris, i860. Ehclid appears to have lived in the time of the first Ptolemy, B.C. 3^3—^83, and to have been the founder of the Alexandrian mathematical school. The work on Geometry known as The Element! of Euclid consists of thirteen books; two other books have sometimes been added, of wliioh it is supposed that Hypsicles was the author. Besides the Elements, Euclid was the author of other works, some of which have been preserved and some lost. We will now mention the three editions which are the most valuabb for those who wish to read the Elements of EucHdin the original Greek. 0) The Oxford editioii in folio, published in 1703 by David Gregory, under the title Ed^XdSou tA trutdfieva. " As an edition of the whole of Euclid's works, this stands alone, there being no other in Greek." De Morgan. (1) Euclidis Elementorum Lihri aex priores...edidit Joannes Gulwlmua Canierer. This eflition was published at Berlin in two volumes octavo, the first volume in 1824 and the second in 1825 It contains the first six books. oUhe j^/ewcn<« in Greek with a Latm Translation, and very good notes which form a mathema- ticaJ commentary on the subject. (3) Midis Elementa ex optimis libris in umm tironum Graxe edita ah Brntsio Ferdlnando August This edition was published at Beriin in two volumes octavo, the first volume in 1826 and the second in 1829. It contains the thh-teen books of the UemenU in Greek, with a coUection of various readings. NTS. Oreek and Morgan ; it the various t article we )e subjects, aclid which rork on the it Ptolemy, .lexandrian ffn as The tber books k Hypsicles 3 author of ome lost, e the most Liclid in the by David an edition e beine: no It Joannes •lin in two d in 1825. jek with a mathema- 1. tlronum lition was volume in 1 books of readings. NO TES ON EUCLID'S ELEMENTS. 261 A third volume, which was to have contained the remaining works of Euclid, never appeared. *' To the scholar who wants one edition of the El<»ments we should decidedly recommend this, AS bringing togett. • . ' that has been done for the text of Euclid's greateat worV. 7)e Morgan. An edition of t \^ of Euclid's works in the original hu long been promisee' byT'uDner the well-known Gernan publrsher, as one of hie serie>>' > mpact editions of Greek and Latin authors ; but we behove there is no hope of its early appearance. Robert Simson's edition of the ElemenU of Euclid^ which we have in substance adopted in the present work, differs con- siderably from the original. The English reader may ascertain the contents of the original by consulting the work entitled The Elements of Euclid with dissertations .. .by Jams Williamson. This work consists of two volumes qusxto ; the first volume was published at Oxford in 1781, and the second at London in 1788. Williamson gives a close translation of the thirteen books of the Elements into English, and he indicates by the use of Italics the words which are not in the original but which are required by our language. Ainong the numerous works which contain notes on the Elements of Euclid we will mention four by which we have been aided in drawing up the selection given in this volume. An Examination of the first six Books of Euclid's Elements by William Austin... Oxford, 1781. Euclid's Elements of Plane OeomMry with copious nofe«...by John Walker. London, 1827. The first six hooks of the Elements of Euclid with a Common' tary... hy Dionysius Lardner, fourth edition. London, 1834. Short supplementary remarks on the first six Books of Euclid^i Elements, by Professor De Morgan, in the Companion to the Almanac for 1849. We may also notice the following works: Geometry, Plane, Solid, and Spherical,... London 1830; this forms part of the Library of Useful Knowledge. TMor^mes et ProhUmes de G4om4trie Elem4ntaire par Eugine Catalan. . . Troisiime edition. Paris, 1 85 8. For the History of Geometry the student i» i-efefred to Montucla's Histoire des MatMmatiques, and to Chasles's Aper^u historique sur Vorigine et le dev4loppement des M^thodes en Gio- m4trie... I 252 NOTES ON THE FIRST BOOTC ^ Definiiions. Tlie first seven definitions have given rise to con- siderable discussion, on which however we do not propose to enter. Such a discussion would consist mainly of two subjects, both of which are unsuitable to an elementary work, namely, an exami> nation of the origin and nature of some of our elementaiy ideas, and a comparison of the original text of Euclid with the substitu- tions for it proposed by Simson and other editors. For the former subject the student may hereafter consult Whewell's History of Scientific Idem and Mill's Logic, and for the latter the notes in Camerer's edition of the Elements of Euclid. Wo will only observe that the ideas which correspond to the words pointy line, and surface, do not admit of such definitions as will really supply the ideas to a person who is destitute of them. The sp-called definitions may be regarded as cautions or restric- tions. Thus a jpoint is not to be supposed to have any size, but only position; a line is not to be supposed to have &ny breadth or thickness, but only length ; a surface is not to be supposed to have any thickness, but only length and breadth. The eighth definition seems intended to include the cases in which an angle is formed by the meeting of two cuj'ved lines, or of a straight line and a curved line ; this definition however is of no importance, as the only angles ever considered are such as are formed by straight lines. The definition of a plane rectilineal angle is important ; the beginner must carefully observe that no change is made in an angle by prolonging the lines which form it, away from the angular point. Some writers object to such definitions as those of an equi- lateral triangle, or of a square, in which the existence of the object defined, is assumed when it ought to be demonstrated. They would present them in such a form as the following: if there be a triangle having three equal sides, let it be called an equilateral triangle. Moreover, some of the definitions are introduced prematurely. Thus, for example, take the definitions of a right-angled triangle and an obtuse-angled triangle ; it is not shewn until I. 17, that a triangle cannot ha;t> both a right angle and an obtuse angle, and so cannot be at the same time right-angled and obtuse- angled. And before Axiom 1 1 has been giveiK it is concAivaUd BUCLIUS ELEMENTS. ^53 rtn<'>Aiva! that the same angle may be greater than one right angle, and ■loss than another right angle, that is, obtuse and acute at the same time. The definition of a square assumes more than is necessary. For if a four-sided figure have all its sides equal and one angle a right angle, it may be shewn that all its angles are right angles ; or if a four-sided figure have all its angles equal, it may be shewn that they are all right angles. PostulaieQ. The postulates state what processes we assume that we can effect, namely, that we can draw a straight line between two given points, that we can produce a straight line to any length, and that we can describe a circle from a given centre with a given distance as radius. It is sometimes stated that the postulates amount to requiring the use of a ruler and compasses. li must however be obsei'ved that the ruler is not supposed to be a graduated ruler, so that we cannot use it to measure off assigned lengths. And we do not require the compasses for any other process than describing a circle from a given point with a given distance as radius ; in other words, the compasses may be supposed to close of tiiemselves, as soon as one of theb points ia removed from the paper. Axioma, The axioms are called in the original Cowrrum Notions. It is supposed by some writers that Buclid intended his postulates to include all demands which are peculiarly geo- metrical, and his common notions to include only such notions as are applicable to all kinds of magnitude as well as to space magnitudes. Accordingly, these writers remove the last three axioms from their place and put th^m among the rvQstulates ; and this transposition is supported by some manuscripts and some versions of the Elements. The fourth axiom is sometimes referred to in editions of Euclid when in reality more is required than this axiom ex- presses. Euclid says, that if A and B be unequal, and C and D equal, the sum of A and C is unequal to the sum of B and D, What Euclid often requires is something more, namely, that if A be greater than B, and C and D be equal, the sum of A and C is greater than the sum of B and D. Such an axiom as this is requu-ed, for example, in I. 17. A similar remark applies to the fifth axiom. In the eighth axiom the words "that is, which exactly fill 254 NOTES ON II i l.it « the original Greek. They are objectionable, because lines and angles are magnitudes to which the axiom may be applied, but they cannot be said io fill space. On the method of superposition we may refer to papers by Professor Kelland in the Transactions of the Boyal Society of Edinburgh, Vols. xxi. and xxiil. The eleventh axiom is not required before T. 14, and the twelfth axiom is not required before I. -29 ; we shall not consider these axioms until we arrive at the propositions in which they are respectively required for the first time. The first book is chiefly devoted to the properties of triangles and parallelograms. We may observe that Euclid himself does not distinguish between problems and theorems except by using at the end of the investigation phrases which correspond to Q.E.P. and q.b.d. respectively. I. 2. This problem admits of eight cases in its figure. For it will be found that the given point may be joined with either end of the given straight line, then the equilateral triangle may be described on either side of the straight line which is drawn, and the sides of the eqdlaterai triangle which aw produced may be produced through eUher extremity. These various cases may be left for th« exercise of the student, as they present no difficulty. Ther^^will not however always be eight different straight lines obtained which solve the problem. For example, if the point A falls on 5(7 produced, some of the solutions obtained coincide; this depends on the fact which follows from I. 32, that the angles of all equilateral triangles are equal. I. 5' "Join FC' Custom seems to allow this singular ex- pression as an abbreviation for "draw the straight line i^(7," or for ''join i? to C by the straight line i?(7." In comparing the triangles BFC, CGB, the words "and the base BO is common to the two triangles BFC, COB'' are usually inserted, with the authority of the original. As however these words are of no use, and tend to perplex a beginner> we have followed the example of some editors and omitted them. A corollary to a proposition is ah inference which may b© deduced immediately from that proposition. Many of the corol- lariea in the £lementt Mra not in the original text^ but mtro- duced hv the editors, EUCLID'S ELEMENTS. 255 It ha"? been suggested to demonstrate I. 5 by mpei'poiitwn* Conceive the isosceles triangle A BO to be taken up, and then re- placed so that AB falls on the old position of AC, and ^(7 falls on the old position of AB. Thus, in the manner of I. 41 we can shew that the angle A BC is equal to the angle A CB. I. 6 is the converse of part of I. 5. One proposition is said to be the converse of another when the conclusion of each is the hypothesis of the other. Thus in I. 5 the hypothesis is the equality of the sides, and one conclusion is the equality of tha Angles; in I. 6 the hypothesis is the equality of the angles and the conclusion is the equality of the sides. When there is more than one hypothesis or more than one conclusion to a pro- position, we can form more than one converse proposition. For example, as another converse of I. 5 we have the following: if the angles formed by the base of a triangle and the sides pro- duced be equal, the sides of the triangle are equal; this pro- position is true and will serve as an exercise for the student. The converse of a true proposition is not necessarily true ; the student however will see, as he proceeds, that Euclid shews that the converses of many geometrical propositions are true. * I. 6 is an example of the indirect mode of demonstration^ in which a result is established by shewing that some absurdity follows from supposing the required result to be untrue. Hince this mode of demonstration is called the reductio ad oMurdur^. Indirect demonstrations are often less esteemed than direct de- monstrations ; they are said to shew that a theorem is true rathet than to shew why it is true. Euclid uses the reductio ad absur- dum chiefly when he is demonstrating the converse of som# fomler theorem ; see T, 14) i9j a^j 40. Sottie remarks on indirect demonstration by Profctvsor Syl- vester, Professor .s )e Morgan, and Dr Adamson will be found in the volumes of tb< f ^rloaophical Magazine for 1852 and it* 5 3. I. 6 is no', required by Euclid before he reaches II. 4 ; so that I. 6 might '.41 ' *novtd from its present place and demonstrated hereafter i tinner .^ ays if we please. For example, I. 5 might be placed after 1. 18 md demonstrated thus. Let the angle ABO be equal to the angle AOB', then the side AB shall be equal to the side A 0. For if not, one of them must b'^ Treater than the oth ; suppose AB greater than A 0. Then the a,ngle J OB h p-eater ^hi^. There are two other cases ; for the straight line A G may pass through B or G, or it may fall outside BO; these cases may be treated in the same manner as that which we have considered. I. 8. It may be observed that the two triangles in I. 8 are equal in all respects; Euclid however does not assert more than the equality of the angles opposite to the bases, end when he requires more than this result he obtains it by using I. 4. I. 9. Here the equilateral triangle DEF is to be described on the side remote from A, Ijecause if it were described on the taTM side, its vertex, jP, might coincide with -4, and then the e0nairuv;tluu would fitil* EUCLIU8 ELEMENTS. 257 tothbsis. Or 3. Bisect the . Then the f I. 0,6. le two might lich has been ides i 5, AC ;he base BQ L to the angle iangle ABC, onterminous, (. Let GBO • corresponds il to BO, the In the same [. Therefore 5(7(7, that is, A G may pass ;ase3 may be cjonsidered. es in I. 8 are )rt more than •^id when he 1.4. be described jribed on the md then the I. II. The corollary was added by Simson. It is liable to serious objection. For we do not know how the perpendicular BE is to be drawn. If we are to use I. 1 1 we must produce AB^ and then we must assume that there is only one way of pro« ducing ABj ic9 otherwise we shall not know that there is only one perpendicular; and thus we assume what we have to demonstrate. Simson's corollary might come after I. 13 and be demon- strated thus. If possible let the two straight lines ABC, ABD have the segment AB common to both. From the point B draw any straight line BE. Then the angles ABE and EBC are equal to two right angles, by I. 13, and the angles ABE imdi EBD are also equal to two right angles, by I. 13. Therefore the an- gles ABE and EBC are equal to the angles ABE axA EBD. Therefore the angle EBC is equal to the angle EBD] which is absurd. But if the question whether two straight lines can have a com. /non segment is to be considered at all in the Elements, it might occur at an earlier place than Simson has assigned to it. For example, in the jfigure to I. 5, if two straight lines could have a common segment AB, and then sep: ate at B, we should obtain two different angles formed on the other side of BC hy these produced paiiis, and each of them would be equal to the angle' BCG, The opinion has been maintained that even in I. i, it is tacitly assumed that the straight lines A C and BC cannot have a common segment at C where they meet ; see Camerer's Euclid, pages 30 and 36. Simson never formally refers to his corollary until XI. i. The corollary should be omitted, and the tenth axiom should be extended so as to amount to the following ; if two straigLi lines coincide in two points they must coincide both beyond and between those points. I. 12. Hero the straight line is said to be of unlimited length, in order thai wo in&y ensure that it shall meet the circle. Euclid distii^guishes between the terms at right angles and perpendicular. He uses the term at right angles when the straight line is drawn from a point in another, as in I. 11 ; and he uses the term perpendicular when the straight Une is drawn from a point without another, as in I. n. This distinction howevei? is often disregarded by modern writers. I. 14. Here Euclid first requires his eleventh axiom. F<* 17 i5S NOTES ON In the flemon&i-a^ion we have the angles A BO and J^^'eqiaal to two right angles, and also the angles ABG and ABD equal to two right angles ; and then the former two right angles are equal to the latter two right angles by the aid of the eleventh axiom. Many modern editions of Euclid howayer refer only to the first axiom, as if that alone were sufficient; a similar remark applies to the demonstrations of I. 15, and I. 24. In these cases we have omitted the reference purposely, in order to avoid per- plexing a beginner; but when his attention is thus drawn to the circumstance he will see that the first and eleventh axioms are both used. We may observe that errors, in the references witli respect to the eleventh axiom, occur in other places in many modem edi- tions of Euclid. Thus for example in III. i, at the step "there- fore the angle FBB is equal to the angle ODB^' a reference is given to the first axiom instead of to the eleventh. Tjiere seems no objection on Euclid's principles to the fol- lowing demonstration of his .eleve":th axiom. Let ^^ be at right a'hgles to DAC at the point A, and EF at right angles to JIEG at the point E: then shall the anglea BAG ftnd FEG be equal* B D K H E G Take any length A 0, and make ^i), ER, EG all equal to A C. Now apply HEG to DA C7, so that H may be on D, and HG on J)Cy and B and F on the same^ide oiJ)Q\ then G will coincide with (7, and E with A, Also EF shall coincide with AB ; for if not, suppose, if possible, that it takes a difierent position as AK. Then the angle BAK is equal to the angle HEF^ and the angle CAK to the angle GEF) but the angles EEF and GEF are equal, by hypothesis; therefore the anj^les DAK and CAK are equal But the angles DAB and CAB are also equal, by hypothesis; is |^»«a%er vuan uuc aagla iJAJL \ uuati* 4^1^^ >, .1^ /Kin ?^'eq]aal to D equal to !S are equal e eleventh jfer only to ilar remark these cases avoid per- awn to the axioms are I respect to lodem edi- ep "there- eference is to the fol- I, and EP 2gles£AC Q lal to A C. id JfG on 11 coincide IB; for if )n as AK. the angle are equal, ire equal ^pothesis; BUCLWS ELEMENTS. 259 fore the angle DAB is greater than the angle CAK. Much more then is the angle DAK greater than the angle CAK, But the angle DAK was shewn to be equal to the angle CAK] which is absurd. Therefore EF mus^ coincide with AB\ and therefore the angle FEO coincides with the angle BAC^ and ia equal to it. I. 1 8, I. 19. In order to assist the student in remembering which of these two propositions is demonstrated directly and which indirectly, it may be observed that the order is similar to that in I. 5 and 1. 6. ' . I. 20. ** Proclus, in his commentary, relates, that the Epi- cureans derided Prop. 20, as being manifest even to asses, and needing no demonstration ; and his answer is, that though the truth of it be manifest to our senses, yet it is science which must give the reason why two sides of a triangle are greater than the third: but the right answer to this objection against this and the 2rst, and some other plain propositions, is, that the number of axioms ought not to be increased without neces- sity, as it must be if these propositions be not demonstrated.'* Simson. I. -21. Here it must be carefully observed that the two straight lines are to be drawn from the ends of the side of the triangle. If this condition be omitted the two straight lines will not necessarily be less than two sides of the triangle. I. 32. "Some authors blame Euclid because he does not demonstrate that the two circles made use of in the construction of this problem must out one another: but this is very plain from the determination he has given, namely, that any two of the straight lines DFj FG, GHy must be greater than the third. For who is so dull, though only beginning to learn the Elements, as not to perceive that the circle described from the centre F, at the distance FD, must meet FU betwixt P and H, because FD is less than FH; and that for the like r«won, the circle de- scribed from the centre (7, at the distance 6'^...must n^et DG betwixt D and G', and that these circles must meet one another, because FD and GH are together greater than FG T* Simson. The condition that B and C are greater than A, ensures that the circle described from the centre G shall not fall entirdy within the circle described from the centre F; the condition that A 4Uid a are greater than C> ensures thai< iue ulrule utwcriuea 17—2 260 NOTES ON I from the centre F shall not fall entirely within the cirde de- sorlbed from the centre 0\ the condition that A and C are greater than jB, ensures that one of these circles shall not fall entirely without the other. Hence the circles must meet. It is easy to see this as Simson says, but there is somethirg arbi- trary in Euclid^a selection of what is to be demonstrated and what is to be seen, and Simson*s language suggests that he was really conscious of this. > I. 24. In the construction, the condition that DE is to be the side which is not greater than the other, was added by Simson ; unless this condition be added there will be three cases to consider, for F may fall on EO, or alove EQ, or helow EG. It may be objected that even if Simson's condition be added, it ought to be shewn that F will fall Idow EO, Simson accordingly says "...it is very easy to perceive, that DQ being equal to DP, the poiht is in the circumference of a circle described from the centre D at the distance DF, and must be in that part of it which is above the straight line EF, because DG falls above DF, the angle EDO being greater than the angle EDF.'*'* Or we may shew it in the following manner. Let M denote the point of intersection of DF and EG. Then, the angle DHG is greater than the angle DEG, by I. 16; the angle DEG is not less than the angle DGE, by I. 19; therefore the angle DHG is greater than the angle DGH. Therefore DH is leas than DG, by I. 20. Therefore DH is less than DF. If Simson's condition be omitted, we shall have two other cases to consider besides that in Euclid. If F falls on EG, it is obvious that EP is less than EG. If F falls dbovc EG, the sum of DF and EP is less than the sum of DG and EG, by I. ai ; and therefore EP is less than EG, I. 26. It will appear after I. 32 that two triangles which have two angles of the one equaLto two angles of the other, each to each, have also their third angles equal. Hence we are able to include the two cases of I. 26 in one enunciation thus ; if two triangles have all the angles of the one respectively equal to all the angles of the other, each to each, and have also a side of the one, opposite to any angle, equal to the side opposite to the equal angle in the other, the triangles shall he equal in all respects. The first twenty-six propositions constitute a distinct section EUCLWS £Ll^M£JjN2)b\ 261 cirde de- ind C are M not fall eet. It is hirg arbi- l and what ■was really £^ is to be added by three cases mEG. It I added, it ccordingly ual to BPf d from the part of it above DF, )r we may i point of is greater t less than is greater , by I. 20. two other £0, it is }, the sum I. 21 ; and ^les which (ther, each e are able us ; if two to all the f the one, inal angle let section of the first Book of the Elements. The principal results are those contained in Propositions 4, 8, and 26 ; in each of these Propositions it is shewn that two ii'iangles which agree in three respects agree entirely. There are two other cases which will naturally occur to a student to consider besides those in Euclid ; namely, (i) when two triangles have the three angles of the one respectively equal to the three angles of the other, (2) when two triangles have two sides of the one equal to two sides of the other, each to each, and an angle opposite to one side of one triangle equal to the angle opposite to the equal side of the other triangle. In the first of theso two cases the student will easily see, after reading I. 29, that the two mangles are not necessarily equal. In the second case also the triangles are not necessarily equal, as may be shewn by an example; in the figure of I. 11, suppose the straight line FB drawn; then in the two triangles FBE, FED, the side FB and the angle FBC are common, and the side FE is equal to the side FDy but the triangles are not equal in all respects. In certain cases, however, the triangles will be equal in all respects, as will be seen from a proposition which we shall now demonstrate. If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles opposite to a pair of equal sides equal; then if the angles opposite to the other pair of equal sides he both acute, or both oUvm, or if one pf them he a right angle, the two triangles are equal in all respects. Jjei ABC md DEF he two triangles; let AB he equal to DE, and BC equal to EFf and the angle A equal to the angle D. First, suppose the angles C and F acute angles. If the angle B be equal to the angle E, the triangles A BO, DEF are equal in all respects, by I, 4. If the angle B be not* equal to the angle E, one of them must be greater than the other; suppose the angle B greater than the angle E, and make the angle ABG equal to the angle E Then the triangles ABG, DEF are equal in all respects, by I. 26; therefore BG is equal to EF, and the angle EGA is equal to the angle EFD. But the angle EFD is acute, by hypothesis ; therefore the angle BGA is (icute. Therefore the angle BGC is obtuHo, by I. 13. But it nas 2G2 NOTES ON I 11 been shewn that BG is equal to EP\ and EP is equal to BC^ by hypothesis ; therefore BQ is equal to BO. Therefore the an- gle BGG is equal to the angle BCG, by I. 5 ; and the angle BOG is acute, by hypothesis; therefore the angle ^G'C is acute. But BGC was shewn to be ob- tuse j which is absurd. Therefore the angles ABC DBF are Tnc'^^nTv' *^** '1'.**"',^ *'' '*1"*^- '^^^^^^^^ t^« triangle. ABiJ, DBF are equal in all respects, by I. 4. Next, suppose the angles at C and F obtuse angles. The demonstration is similar to the above. angle C7, If the angle B be not equal to the angle E, make the ' lit nil *ngle ABG equal to the angle E. Then it may be shewn, as before, that BG is equal to BC, and therefore the angle BGC is equal to the angle BCG, that is, equal to a right angle. There- fore two angles of the triangle BGC are equal to two right angles ; which is impossible, by I. 17. Therefore the angles ABO and DEF are not unequal; that is, they are equal. Therefore the triangles ABC, DEF are equal in all respects, by I. 4. If the angles A and D ar^ both right angles, or both obtuse, the angles and F must be both acute, by I. 17. If AB is less than BO, and DE less than EF, the angles at and F must be both acute, by I. 18 and I, 17. The propositions from I. 27 to I. 34 inclusive may be said to constitute the second section of the first Book of the Elemmf^, They relate to the theory of parallel straight lines. In I. 29 EucUd nees for the first time his twelfth axiom. The theory of parallel straight lines has always been considered the great difficulty o* Siciuoatary georuetry, and many attempts have been made EUCLID'S ELEMENTS. 263 in overcome this difficulty in a better way than Euclid has done. We Bhall not give an account of these attempts. The student who wishes to examine them may consult Camerer's Euclid^ Grer- gonne's Annaks de MatMniatiqrieB, Volumes XV and xvi, the work by Colonel Perronet Thompson entitled Oeometry without Axioms, the article Parallels in tho English Cydopadia, a me«- moir by Professor Baden Powell in the second volume of the Memoirs of the Ashmolean Society, an article by M. Bouniakofaky in the Bulletin de VAcadSmie Imp^riale, Volume v, 8t P^ters- bourg, 1863, articles in the volumes of the Philosophiccd Magazine for 1856 and 1857, and a diasertation entitled Sur un point de Vhistoire de la GSoirUtric chez les Grecs par 4. /. -fir. Vincent. Paris, 1857. Speaking generally it may be said that the methods which differ substantially from Euclid's involve, iu the first place an axiom as difficult as his, and then an intricate series of proposi- tions ; while in Euclid's method after the axiom is once admitted the remaining process is simple and clear. One modification of Euclid's axiom has been proposed, which appears to diminish the difficulty of the subject. This consists in assuming instead of Euclid's axiom the following ; two inter- secting straight lines cannot be both ■parallel to a third straight line. The propositions in the Elements are then demonstrated as in Euclid up to I. 28, inclusive. Then, in I. 29, we proceed with Euclid up to the words, ''therefore the angles BGH, GHD are kss than two right angles." We then infer that BGH and GHD must meet: because if a straight line be drawn through (7 so as to make the interior angles together equal to two right angles this straight line will be parallel to CD, by I. 28 ; and, by our. axiom, there cannot be two parallels to CD, both passing through G. This form of making the necessary assumption has been recommended by various eminent mathematicians, among whom may be mentioned Play fair and De Morgan. By postponing the consideration of the axiom until it is wanted, that is, until after I. 28, and then presenting it in the form here given, the theory of parallel straight lines appears to be treated in the easiest manner that has hitherto been proposed. I. 30. Here we may in the same way shew that \t AB and EF are each of them parallel to CD, they are parallel to each other. It has been said that the case considered in the text is so obvious as to need no demonstration ; for n AB and CD cau IMAGE EVALUATrON TEST TARGET (MT-3) 4r A r/. 1.0 I.I 11.25 U HI £ U£ 12.0 b I. 12.2 1.8 14 11116 V] W ^. '/ Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, NY. 14580 (716) 873-4503 •NT d^ o h ^";^^ v\ J ^ o ■X !V» mp 264: NOTES ON n«ver meet EF, which Kea between them, they cannot ineet one another. I. 34. The coroUarieg to I. 34 were added by Simson. In the second oorollaiy it ought to be stated what is meant by an exterior angle of a rectiUneal figure. At each angular point let ime of the sides meeting at that point be produced; then the exterior angle at that point is the angle contained between this produced part and the side which is not produced. Mhtr of tne sides may be produced, for the two angles which can thus be obtained are equa!, by I. 15. The rectilineal figures to which Eu- clid confines himself are those in which the angles all face inwards; we may here however notice another class of figures. In the accompanying diagram the angle AFC f aces outwards, and it is an angle less than two right angles ; this angle however is not one of the interior angles of the figure AEDCF, We may consider the corre- eponding interior angle to be the excess of four right angles above the angle AFC; such an angle, greater than two right angles, is called a re-entrctnt angle. Tho Jirst of the corollaries to I. 32 is true for a figure which has a re-entrant angle or re-entrant angles; but the aecond is not. I. 31. If two triangles have two angles of the one equal to two angles of the other each to each they shall also have their third angles equal. This is a very important result, which is often required in the Elements. The student should notice how this result is established on Euclid's principles. By Axioms 11 and 2 one pair of right angles is-equal to any other pair of right angles. Th'jn, by I. 32, the three angles of one triangle are together equal to the three angles of any other triangle. Then, by Axiom 2, the sum of the two angles of one triangle is equal to the sum of the two equal angles of the other • and then, by Axiom 3, the third angles are equal. After I. 32 ^e can draw a straight line at right angles to ft given straight line from its extremity, without producing th« given straight line. Let is be the given straight line. It in required to draw from A a straight line at right angles to A B.. ABO. EUCLIirS ELEMENTS. 265 OvlAB describe the equilateral triangle ABO. Produce BO to i>, so that OD may be equal to {7J3. 3 om AD. Then ^i> shall be at right angles to AB. For, the angle CADia equal to the angle CDA^ and the angle CaB is wqual to the angle OBA^ by I. 5. There- fore the angle BAD is equal to the two anj^les il J9i>, BDAy by Axiom a. Therefore the angle BAD is a right angle, by L 32. The propositions from I. 35 to I. 48 inclusive may be said to constitute the third section of the first Book of the Elemmtt, They relatd to equality of area in figures which are not neces- sarily identical in form. I. 35. Here Simson has altered the demonstration given by Euclid, because, as he says, there would be three cases to con- sider in following Euclid's method. Simson however uses the third Axiom in a peculiar manner, when he first takes a triangle from a trapezium, and then another triangle from the same trapezium, and infers that the remainders are equal. If the demonstration is to be conducted strictly after Euclid's manner, three cases must be made, by dividing the latter part of the demonstration into two. In the left-hand figure we may suppose the point of intersection of BE and DC to be denoted by O. Then, the triangle A. BE is equal to the triangle DCF\ take away the triangle DOE from each; then the figure ABGD is equal to the figure EOCF ; add the triangle QBO to each ; then the parallelogram A BCD is equal to the f rallelogram EBOF, In the right-hand figure we have the trian^^e AEB equal to the triangle DPO; add the figure BE DC to each; then the parallel- ogram A BCD is equal to the parallelogram EBOF, The equality of the parallelograms in I. 35 is an equality of area, and not an identity of figure. Legendre proposed to use the word equivalent to express the equality of area, and to restrict the word eqtml to the case in which magnitudes admit of super- position and coincidence. This distinction, however, has not been generally adopted, probably because there are few cases in which any ambiguity can arise ; in such cases we may say es* pecially, equal in area, to prevent misconception. Cresswell, in his Treatise of Geometry , has given a demon- ii4-»«4-: ^e T ^ at ««■> u:»i. .v.^^- 4.1.. i. i.u.. 11-1 .«.,. V,, o,i — *— ^Sd JM««« 4Mr4*WA^^^4ft »»>*.i»«i# 266 NOTES ON divided tnto pain of pieces admitting of luperpoMtioii ahd coin* cidence ; see also his Preface, page x. I. 38. An important case of I. 38 is that in which the tri- angles are on equal bases and have a common vertex. I. 40. "We may demonstrate I. 40 without adopting the in- direct method. Join BD, CD. The triangles DBC and JDEF are equal, by I. 38; the triangles il^Cand DEF are er al, by hypothecs; therefore the triangles DBC and ABC ate equsAf by the first Axiom. Therefore AD ia parallel to BC, by I. 39. Philosophical Mdgazine, October 1850. I. 44. In I. 44, Euclid does not shew that AH and FQ will meet. "I cannot help being of opinion that the construe- lion would have been more in Euclid's manner if he had made GH equal to BA and then joining HA had proved that HA was parallel to OB by the thirty-third proposition." WiUianuon, I. 47. Tradition ascribed the discovery of I. 47 to Fytha- goraA Many demonstrations have been given of this cele- brated proposition ; the following is » Fi/thafforische Lehnatz. , .,ZweyU. .,A usgabe, Mainz. 1 82 1. BUCLtXfS JELEMENTS. 267 i^d coin* i the tri- ig the in- ad DBF r al, by jqual, by y I- 39. and FO ionstruc- %d made JLl was Pytha- lis oele- resting. i-angled iof the [MMntion res may a third are col* \edJ)er THE SECOND BOOK. The second book is devoted to the investigation of relatiolit between the rectangles contained by straight lines divided mto 83gments in various ways. When a straight line is divided into two parts, each part is calted a segment by Euclid. It is found convenient to extend the meaning of the word segment, and to lay down the following defi- nition. When a point is taken in a straight line, or in the f straight line produced, the distances of the point from the ends of the straight line are called segments of the straight line. When it is necessary to distinguish them, such segments are called in- ternal or aOemaly according as the point is in the straight ^'ne. or in the straight line produced. The student cannot fail to notice that there is an analogy between the first ten propositions of this book and some element- ary facts in Arithmet: and Algebra. Let A BCD represent a rectangle which is 4 inches long and 3 inches broad. Then, by draw- ing straight lines parallel to the sides, the figure may be divided into 12 squares, each square being described on a side which represents an inch in length. A squal*e described on a side measuring an inch is called, for shortness, a square inch. Thus if a rectangle is 4 inches long and 3 inches broad it maybe divided into 12 square inches; this is expressed by saying, that its area is equal to 12 square inches, or, more briefly, that it oontains n square inches. And a similar result is easily seen to hold m all sunilar cases. Suppose, for example, that a rectangle is 11 feet long and 7 feet broad; then its area is equal to 12 times 7 square feet, that is to 84 square feet; this may be expressed briefly in common language thus; if a rectangle measures la feet by 7 it contains 84 square feet. It must be carefully observed that the sides of the rectangle are supposed to be measured by the same unit of length. Thus if a rectangle is a yard in length, and a foot and a half in breadth, we D c A B MIP 268 NOTES ON must express each of these dimensions in terms of the siime unit; we may say that the rectangle measures 36 inches by 18 inches, and contains 36 times 18 square inches, that is, 648 square inches. Thus universally, if one side of a rectangle contain a unit of length an exact number of times, and if an adjacent side of the rectangle also contain the same unit of length an exact number of times, the product of these numbers will be the number of square units contained in the rectangle. ^ Next suppose we have a sqtiare, and let its side be 5 inches in length. Then, by our rule, the area of the square is 5 times 5 square inches, that is 25 square inches. Kow the number 25 is called in Arithmetic the square of the number 5. And universally, if a straight line contain a unit cf length an exact number of times, the area of the square described on the straight line is denoted by the square of the number which denotes the lengUi of the straight line. Thus we see that there is in gene al a connexion betweeti the product of two numbers and the rectangle contained by two straight lines, and in particular a connexion betv/een the square of a number and the square on a straight line ; and in consequence of this connexion the first ten propositions in Euclid's Second Book correspond to propositions in Arithmetic and Algebra. The student will perceive that we speak of the square de* scribed on a straight line, when we refer to the geometrical figure, and of the square of a number when we refer to Arithmetic. The editors of Euclid generally use the words "square described upon** in I. 47 and I. 48, and 'terwards speak of the square of a straight line. Euclid himself retains throughout the same form of expression, and we have imitated him. Some editors of Euclid have added Arithmetical or Alge* braical demonstrations of the propositions in the second book, founded on the connexion we have explained. We have thought it unnecessary to do this, because the student who is acquainted with the elements of Arithmetic and Algebra will find no diffi- culty in supplying such demonstrations himself, so far as they are usually given. We say so far as they are usually given, because these demonstrations usually imply that the sides of rectangles can always be expressed eaxictly in terms of some unit of length; whereas the student will find hereafter that this is not the case, owing to the existence of what are technically called Lncam/manturahle macmitudecL. We dn not enher on this subiectt EUCLID'S ELEMENTS. 269 ame unit; 1 8 inches, ire inches, a unit of ide of the aumber of of square y ; inches in JB 5 times e number ' 5. And an exact le straight motes the tweeti the I by two ! square of Qsequence 's Second :ebra. quare de* cal figure, letic. The described square of lame form or Alge* md book, 6 thought cquainted I no diffi- r as they \lly given, I sides of some unit this is not klly called s subiect. as it would lead us too far from Euclid's Elements of Qeometry^ with which we are here occupied. The first ten propositions in the second book ci Euclid may be arranged and enunciated in various ways ; we will briefly indicate this, but we do not consider it of any importance to dis- tract the attention of a beginner with these diversities. II. 1 and II. 3 are particular cases of II. i. II. 4 is very important; the following particular case of it should be noticed ; the square described on a straight line made up of two equal straight lines is equal to four times the square described on one of the two equal straight lines, II. 5 and II. 6 may be included in one enunciation thus ; the rectangle under the sum and difference of two straight lines is equal to the difference of the squares described on those straight lines; or thus, the rectangle contained by two straight lines together with the square described on half their difference^ is equal to the square described on half their sum. II. 7 may be enunciated thus; the square described on a sira^ht line which is the difference of two othtr straight lines is less than the sum of the squares described on those straight lines by twice the rectangle contained by those straight lines. Then from this and II. 4, and the second Axiom, we infer that the square described on the sum of two straight lines, and the square described their difference, are together double of the sum of the squares described on the straight lines; and this enunciation includes both II. 9 and n, 10, so that the demonstrations given of these pro- positions by Euclid might be superseded. II. 8 coincides with the second form of enunciation which we have given to II. 5 and II. 6, bearing in mind the particular case of II. 4 which we have noticed, II. 1 1. When the student is acquainted with the elements of Algebra he should notice that II. 11 gives a geometrical con- struction for the solution of a particular quadratic equation. II. n, II. 13. These are interesting in connexion with I. 47; and, as the student may see hereafter, they are of great import" ance in Trigonometry ; they are however not required in any of the parts of Euclid's Elements which are Usually read. The converse of I. 47 is proved in I. 4S; and we can easily shew that converses of II. 12 and II. 13 are true. Take the following, which is the converse of II. 11; if the i^Uari deieribtd 9» one side of a triangle he greater than the mni 270 NOTES ON of the tquarti described on the other two ndes^ the angU opposite to the first side 'Is obtuse. For the angle cannot be a right angle, since the square de* scribed on the first side would then be equal to the sum of the squares described on the other two sides, by I. 47 ; and the angle cannot be acute, since the square described on the first side would then be le^s than the sum of the squares described on the other two sides, by II. 13; therefore the angle must be obtuse. Similarly we may demonstrate the following, which s the con- verse of II, 13; if the square described on one side of a tricmgle be less than the sum of the squares described on the othe. two sides, the angle opposite to the first side is acute. II. 13. Euclid enunciates II. 13 thus; in actUe-angled tri- angles, ^, \ and he gives only the first case in the demonstration. But, -as Simson observes, the proposition holds for any triangle ; and accordingly Simson supplies the second and third cases. It has, ^however, been often noticed that the same demonstration is applicable to the first and second oases ; and it would be a great improvement as to brevity and clearness to take these two cases together. Then the whole demonstration will be as follows. Let ABC be any triangle, and the angle at B one of its acute angles ; and, if ilC7 be not perpendicular to BC, let fall on BOy produced if necessary, the perpendicular AD from the opposite angle: the square ou AC oppiosite to the angle B, shall be less than the squares on CB, BA, by twice the rectangle HB, BD, - B C First, suppose AC not perpendicular to BC. The squares on CB, BD are equal to twice the rectangle CB, BD, together with the square on CD. [II. 7. To each of these equals add the square on DA. Therefore the squares on CB, BD, DA are equal to twice the rectangle CB, BD, together with the squares on CD, DA, But the square cin ^^ is eau&I to the sauaree on BD. DA. mUCLWS ELEMENTS. 271 U oppotiit iquare de- im of the 1 the angle first side >ed on the ) obtuse. 8 the con- « tricmgle two tides f \ngled tri' »nstratioD. r triangle ; cases. It stration is be a great two cases Hows, me of its ]et fall on from the e £, shall rectangle A f rectangle twice the and the square on J C is equal to the squares on CD, DA, because the angle BDA is a right angle. [1-47. Therefore the squares on CB, BA are equal to the square on AC, together with twice the rectangle CB, BD \ that is, the square on AC alone is less than the squares on GBy BAf by twice the rectangle CB, BD. Next^ suppose A C perpendicular to BC. Then BC is the straight line intercepted be- tween the perpendicular and the acute angle at B. And the square on ^1 fi is equal to the squares on AC, CB. [I. 47. Therefore the square on AC ia less than the squares on AB, BC, by twice the square on BC. II. 14. This is not required in any of the parts of £uclid't Elementi which are usually read; it is included in VI. aa. THE THIRD BOOK. The third book of the Elements is devoted to properties of circles. Diflferent opinions have been held as to what is, or should be, mcluded in the third definition of the third book. One opinion is that the definition only means that the circles do not cut in the neighbourhood of the point of contact, and that it must be shewn that they do not cut elsewhere. Another opinion is that the definition means that the circles do not cut at all; and this seems the correct opinion. The definition may therefore be pre- sented more distinctly thus. Two circles are said to touch inter- nally when their circumferences have one or more common points, and when every point in one circle is within the other circle, except the common point or points. Two circles are said to touch externally when their circmuferences have one or more common points, and when every point in each circle is without •the other circle, except the common point or points. It is then shewn in the third Book that the circumferences of two circles which touch can have only one common point. A straight line which touches a circle is often called a tan- gent to the circle, or briefly, a tangent. It is veiy convenient to have a word to denote a portion of 272 NOTES ON the boundary of a circle, and accordingly we use the word o*ic. Euclid himself uses circumference both for the 'whole boundary and for a portion of it. III. I. In the construction, DO is said to be produced !e jB; this assumes that D is within the circle, which Euclid demon* Btrates in III. 7, III. 3. This consists of two parts, each of wHch is the con- Terse of the other; and the whole proposition is tiie conyerse of the corollary in III. i. III. 5 and III. 6 should have been taken together. They amount to this, if the circumferences of tmo circles meet at a point ikey cannot have the sam£ centre, so that circles which have the same centre and one point in their circumferences common, must coincide altogether. It would seem as if Euciid had made three cases, one in which the circles cut, one in which they touch internally, and one in which they touch externally, and had then omitted the last case as evident. III. 7, III. 8. It is observed by Professor De Morgan that in III. 7 it is assumed that the angle FEB is greater than the angle FBOf the hypothesis being only that the angle DFB is greater than the angle DFC; and that in III. 8 it is assumed that JT falls within the triangle DLM, and E without the triangle J)MF. He intimates that these assumptions may be established by means of the following two propositions which may be given in order after I. 21. The perpendicular is the shortest straight line which can he dravm from a given point to a given straight line / and of others that which is nearer 'to the perpendicular is less than the mme remote, and the converse; and not more than two equal straight lines can he drawn from the given point to the given straight line, cm on each side of the perpendicular. Every straight line drawn from the vertex of a triangle to the hose is less than the greater of the two sides, or than either of them ^ they he equal. The following proposition is analogous to III. 7 and III. 8. If any point he taken on the circumference of a circle, of all the straight lines which can he drawn from it to the circumference, the greatest is that in which the centre is j and of any others, that which is nearer to the straight line which passes through the centre is always greater than one more remMe; and from the same peint there can he dravm to the circumference two straiaht Urns, and EUCLID'S ELEMENTS. 27a word o*ic. boundary odueed U id demon* B the con- »nvexlM of er. They at a point have the Qon, must lade three ley touch had then trgan that than the J DPB ia i OMUIMd 6 triangle stablished r be given ch can he \ of others the mort il straight light line, rigle to the er of them 1 III. 8. 'He, of all vrnference, )hers, that the centre ame point lines, and only two, which are equal to one another, one on ^tch tide f^ tk4 ffreatest line. The first two parts of this proposition are contained In III. 15 ; ail three parts might be demonstrated in the manner of III. 7, and they should be demonstrated, for the third part ii really required, as Mre shall see in the note on III. 10. III. 9. The point £ might be supposed to fall within the angle A DC. It cannot then be shewn that DC is greater than J)B, and BB greater than DA, but only that either DO or DA is less than DB ; this however is sufficient for establish* ing the proposition. Euclid has given two demonstrations of III. 9, of which Slmson has chosen the second. Euclid's other demonstration ii as follows. Join D with the middle point of the straight lint AB; then it may be shewn that this straight line is at right angles to AB ; and therefore the centre of the circle must lie in this straight line, by III. i, Corollary. In the same manner it may be shewn that the centre of the circle must lie in the straight line which joins D with the middle point of the straight line BC. The centre of the circle must therefore be at D, because two straight lines cannot Lave more than one commmi point. III. 10. Euclid has ^ven two demonstrations of III. 10, of which Simson has chosen the second. Euclid's first demonstra> tion resembles his first demonstration of III. 9. He shews that the centre of each circle is on the straight line which joins K with the middle point of the straight line BG, and also on the straight line which joins K with the middle point of the straight line BH ; therefore K must be the centre of each circle. , The demonskation which Simson has chosen requires some additions to make it complete. For tlie point K might be sup- posed to fall without the circle DEF, or on its circumference, or within li'f and of these three suppositions Euclid only considers the last. If the point K be supposed to fall without the circle^ DEF we obtain a contradiction of III. 8 ; which is absurd. If the point K be supposed to fall on the circumference of the circle DEF we obtain a contradiction of the proposition which wo have enunciated at the end of the note on III. 7 and III. 8; .which is absurd. What is demonstrated in III. 10 is that the circumferences of two circles cannot have more than two common |>oints; there it 10 ■■ 274 NOTES ON nothing in th« demonstrfttion which aaaumei that the droiei cut one another, but the enunciation refers to tliis case only because it is Rhewn in III. 13 that if two circles touch one another, their circumferences cannot have more than one common point. III. II, III. 1 3. The enunciations as fpven by Simson and others speaJc of the point of contact ; it is however not shewn nntU III. 13 that there is only one point of contact. It should be observed that the demonstration in III. 1 1 will hold even if D and M be supposed to coincide, and that the demonstration in IIL n will hold even if C and D be supposed to coincide. We mny combine III. il and III. la in one enunciation thus. If two circlet touch one another their circumferencee cannot have a common point out of the direction of the straight line which joins the centres, III. II may be deduced from III. 7. For Off is the least line that can be drawn from to the circumference of the circle whose centre is /', by III. 7. Therefore OH is less than (?i, that is, less than OD; which is absurd. Simikrly III. la may be deduced from III. 8. III. 13. Simson observes, ''As it is much easier to imagine that two circles may touch one another within in more points than one, upon the same side, than upon opposite sides, the figure of that case ought not to have been omitted ,* but the construction in the Greek text would not have suited with this figure so well, because the centres of the circles must have been placed near to the circumferences; on which account another construction and demonstration is given, which is the same with the second part of that which Gampanus has translated from the Aralic. where, without any reason, the demonstration is divided into two parts." It would not be obviorfs from this note which figure Simson himself supplied, because it is uncertain what he means by the "same side" and "opposite sides." It is the left-hand figure In the first part of the demonstration. Euclid, however, seems to be quite correct in omitting this figure, because he has shewn in III. 1 1 that if two circles touch internally there cannot be a point of contact out of the direction of the straight line which joins the centres. Thus, in order to shew that there is only one point of contact, it is sufficient to put the second supposed point of contact on the direction of the straight line which joins the EUVmy^S MLEMEN'm 21b cei...j». Accordingly in hia own dcmonatration Euclid ^^on- iinus himself to the right-hand figure ; and he ihc vi that this case cannot exist, because the straight line BD would be a diameter of both circles, and would theref(re be bisected at two different points ; which is absurd. Euclid might have used a s.milar method for the second pari of the proposition ; for as there cannot be a point of contact out of the straight line joining the centres, it U obvimuly impoaibU that there can be a second point of contact when the circles touch extemallj. It is easy to tee this ; but Euclid preferred a method in which thero is more formal jasoning. We may observe that Euclid's mode of dealing with the contact of circles has often been censured by commentators, but apparently not always with good reason. For ( •'^mple. Walker gives another demonstration of III. 13; and •'ays that Euclid's is worth nothing, and that Simson fails ; for it is not proved that two circles which touch cannot have any arc common to both circumferences. But it is shewn in III. 10 that this is impos- sible; Walker appears to have supposed that fll. 10 is limited to the case of circles which cut. See the note on III. 10. III. 17. It is obvious from the construction in III. 17 that two straight lines can be drawn from a given external point to touch a given circle ; and these two straight lines are equal in length and equally inclined to the straight line which joins the given external point with the centre of the ^^iven circle. After reading III. 31 the student will see that the problem in III. 17 may be solved in another way, as follows: describe a circle on ^ j^ as diameter; then the points of intersection of this circle with the given circle will be the points of contact of the two stndght lines which can be drawn from A to touch the given circle. III. 18. It does not appear that III. 18 adds anythmg- to what we have already obtained in III. 16. For in III. 16 it is shewn, that there is only one straight lino which touches a given circle at a given point, and that the angle between this straight line and the radius drawn to the point of contact is a right angle. ' III. 20. There are two assumptions in the demonstration of in. 20. Suppose that A is double of B and C double of D; then in the first part it is assumed that the sum of A and (7 is double ot the sum of B &nd J>, and in the second part it is as- 18—2 276 NOTES ON Rumed that the difference of A and C is double of the difl^rencd of B and D, The former assumption is a particular case of V. i| And the latter is a particular case of Y. 5^ An important eztersion may be given to III. ^o by intro- ducing angles greater l.ian cwo right angles. For, in the first figure, suppose we draw the straight lines BP and CF^ Then, the angle BE A is double of the angle BFA, and the angle CEA ig double of the angle CFA ; therefore the sum of the angles BE A and CEA is double of the angle BFC. The sum of the angles BEA and CEA is greater than two right angles ; we will call the sum, the re-entrant angle BEC, Thus the re-entrant angle BEC is double of the angle BFC. (See note on I. 32). If this extension be used some of the demonstrations in the third book may be abbreviatod. Thus III. a i may be demonstiratcd without making two cases ; III. 11 will follow immediately from the fact that the sum of the angles at the centre is equal to four right Angles; and III. 31 will follow immediately from III. ao. III. a I. In III. 11 Euclid himself has given only the first rase; the second case has been added by Simson and others. In either of the figures of III. 2 1 if a point be taken on the same side of BD as A , the angle contained by the straight lines which join this point to the extremities of BD is greater or less than the angle BAD, according as the point is within or without the angle BAD; this follows from I. i. We shall have occasion to refer to IV. 5 in some of the remaining notes to the third Book ; and the student is accord- ingly recommended to read that propodition at the present stage. The following proposition is very important. 1/ any numler of triangles he constructed on the same base and on the same side of it, with equal vertical angles, the vertices will all lie on the cir- cumference of a segment of tt circle. For take any one of these triangles, and describe a circle round it, by IV. 5 ; then the vertex of any other of the triangles must be on the circumference of the segment containing the assumed vertex, since, by the former part of this note, the vertex cannot be without the circle or within the circle. Ill- 11^ The converse of III* 11 is true Jind very ini" portant; namely, if two opposite angles of a quadrilateral U together equal to two right angles, a circle may he circumscrthed apptU the guadrilaterah for, let A BCD denote the quadrila* EUCLWS ELEMENTS. 277 of V. I, jy intro- the first Then, gle CEA e angles a of the we will i-entrant n I. 32). Lhe third •nstratcd ely from 1 to four [II. 20. the first L others, bhe same es which than the he angle ) of the I ancord- present ' numbtf \ame side I the civ' a circle triangles ling the le vertex rery im* ateral hi mscrihed ^uadrila* teral. Describe a circle round the triangle ABC, by IV. 5. Take any point E, on the circumference of the segment cut oflf by AC, and on the same side of ^C as i) is. Then, the angles at B and E are together equal to two right angles, by III. ^^ ; and the angles at B and J) are. together equal to two right angles, by hypothesis. Therefore the angle at E is equal to the angle at D, Therefore, by the preceding note D is on the cir- cumference of the same segment as E. III. 32, Tha converse of III. 32 is true and important; namely, if a straight line meet a circle, and from the point qf meeting a straight line he dravm cutting the,cirde, and the ofiigle between the two straight lines be equal to the angle in th£ alternate segment of the circle, the straight line tohiah meets the circle shaU touch the circle. .-.it This may be demonstrated indirectly. For, if possible, sup- pose that the straight Une which meets the circle does not touch it Draw through the pomt of meeting a straight line to touch the circle. Then, by III. 3« ^^^ ^^^ hypothesis, it will follow that two different straight lines pass through the same point, ^ make the same angle, on the same side, with a third straight line which also passes through that pouit; but this is unpoa- wbiO' ... ..i. L III. 35, III. 36. The following proposition constitutes a large part of the demonstrations of III. 35 and III. 36. // any point be taken in the base, or the base produced, of an xsoscelet tnangle, the rectangle contained by the segments of the base u equal to the difference of the square on the straight line joining this poi-'. to the vertex and the square on the side of the triangle. This proposition is in fact demonstrated by Euclid, without using any property of the circle ; if it were enunciated and de- monstrated before III. 35 and III. 36 the demonstrations of these two propositions might be shortened and simplified. The following converse of III. 35 and the Corollary of III. 30 may he noticed. // two straight lines AB, CD intersect atO, and the rectangle AO, OB be equal to the rectangle CO, OD. the circum- ference of a circle mil pass through the four points A, B, C, D. For a circle may be described round the triangle ABC, by iv, 5; ana men iu luay wc sn^-ti «!-.--.^ j-, -„- — III. 35 or the Corollary of III. 3(5 that the circumference of tha circle will also pass through D. 278 NOTES ON THE FOURTH BOOK. 'to fourth Book of the Elementa consists entirely of problems Ihe first hve propositions relate to triangles of any kind • the wmainmg propositions relate to polygons which have aU 'their sides aqua^ and all thet angles equal. A polygon which has all its sides equal and aU its angles equal is called a regular polygon. I Y. 4. By a process similar to that in IV. 4 we can describe a circle which shall touch one side of a triangle and the other two sides produced. Suppose, for example, that we wish to d^cnbe a circle which shaU touch the side BC, and the sides A ^^ ^^ Produced: bisect the angle between AB produced and BO, and bisect the angle between -4 C7 produced m^ BC- then the point at which the bisecting straight lines meet will be th J centee of ths required circle. The demonstration wiU be similar to thali m IV. 4. A circle which touches one side of a triangle and the other w ^^ Pro^^^ced, is called an escribed circle of the triangle. We can also describe a triangle equiangular to a given tri- angle, and such that one of its sides and the other two sides produced shaU touch a given circle. For, in the figure of IV. 3 suppose iliT produced to meet the circle again j and at the point of mtersection draw a straight line touching the circle; this straight hne with parts of iNT^ and NO, wiU form a triangle, which will be equiangular to the triangle MLN, and therefore equiangular to the triangle EDF-, and one of the sides of this triangle, and the other two sides produced, will touch the given circle. IV. 5. Simson introduced into the demonstration of IV < the part which shews that 2)7^ and EF ynW meet. It has also been proposed to shew this in the following way: join BE- then the angles EDF and DEF i^^ together less than the angles ABF and AEFy that is, they are together less than two right angles; and therefore DF and EF will meet, by Axiom 12. This assumes that ADE and AED are acute angles ; it may how. ever be easily shewn that DE is parallel to BC, so that the taiangle ABE is equiangular to the triangle ABO: and we must A OB may be acute angles. IV. 10. The vertical angle of the triangle fn IV. 10 is easily seen to be the fifth part of two right angles ; and as it EUCLW8 ELEMENTS. 27d may b« biieoied, we can thus divide a right angle geometrically into five equal parts. ^ It follows from what Is given in the fourth Book of the Elements that the circumference of a circle can be divided into 3, 6, n, 14, ... . equal parts ; and also into 4, 8, 16, 32, .... equal parts ; and also into 5, 10, ao, 40, ... . equal parts; and also into 15, 30, 60, no, equal parts. Hence also r^ular polygons having as many sides as any of these numbers may be inscribed in a circle, or described about a circle. Thi« however does not enable us to describe a regular polygon of any assigned number of sides ; for example, we do not know how to describe geometrically a regular polygon of 7 sides. It was first demonstrated by Gauss in 180 1, in his Dtsqwr- sitimes ArithmeticcBf that it is possible to describe geometrically a regular polygon of a" + 1 sides, provided a* + 1 be a prime num- ber; the demonstration is not of an elementary character. As an example, it follows that a regular polygon of 17 sides can be described geometrically ; this example is discussed in Catalan's ThSorimes et ProhUmes de Qiomitrie EUmentaire. For an approximate construction of a regular heptagon ieo the Philoiopkical Magazine for February and for AprU, 1864. THE FIFTH BOOK. The fifth Book of the Elements is on Proportion, Much has been written respecting Euclid's treatment of this subject; besides the Commentaries on the Elements to which we have already referred, the student may consult the articles Jlaiio and Proportion in the English Cyclopoedia, and the tract on the Connexion of Number and Magnitude by Professor De Morgan. The fifth Book relates not merely to length and space, but to any kind of magnitude of which we can form multiples. V. Def. I. The word part is used in two senses in Geometry. Sometimes the word denotes any magnitude which is less than another of the same kind, as in the axiom, the whole w greater than its part. In this sense the w )rd has been used up to tha presetit poiiiii, out lu viie uitn jdook. xjucuu, uoiiuiics vuw ..vi«. --«• a more restricted sense. This restricted sense agrees with that which i-^ dven in Arithmetic and Algebra to the term aliquoi partf or W the term submultiple. ■1 ^m NOTES ON V. Ikf. 3. SimsQu considers that the definitions 3 and 8 are "not Euclid's, but added by some unskilful editor." Other com- mentators also hare rejected these definitions as useless. The last ii»ord of the third definition should be quanluplieity, not qwrniiy; so that the definition indicates that ratio refers to the nwmher of times which, one magnitude contains another. See De Morgan's Differential and Jntegml CalculuSy page 18. V. D(f. 4. This definition amounts to saying that the quan- tities must be of the same kind. V. J)ef, 5. The fifth definition is the foundation of Euclid's doctrine of proportion. The student will find in works on Alge- bra a comparison of Euclid's definition of proportion with the Simpler definitions which are employed in Arithmetic and Algebra. Euclid's definition is applicable to incommensurable quantities, aa well as to commensurable quantities. We should recommend the student to read the first propo- eitiod of the sixth Book immediately after the fifth definition of the fifth Book ; he will there see how Euclid applies his defi- iii^^^ion, and will thus obtain a better notion of its meaning and im- jportance. Comjpound Ratio. The definition of compound ratio was supplied by Simson. The Greek text does not give any defini- tion of compound ratio here, but gives one as the fifth definition of the sixth Book, which Simson rejects as absurd and useless. V. Defs, 18, 19, 20. The definitions 18, 19, 20 are not pre- sented by Simson precisely as they stand in the original. The last sentence in definition 18 was supplied by Simson. Euclid does not connect definitions 19 and 20 with definition 18. In 19 he defines ordinate proportion, and in 20 he defines pertiirbate proportion. Nothing would be lost if EucUd's definition 18 were entirely omitted, and the term ex cequali never employed. Euclid «mploys such a term in the enunciations of V. 20, 21, 22, 23; but it seems quite useless, and is accordingly neglected by Simson and others in their translations.- The axioms given after the definitions of the fifth Book are not in Euclid; they were supplied by Simson. ... -.„.„^....„„_ „j s„_ „.j^j. ^ypg. mjgni; OQ aiviaed into four sections. Propositions i to 6 relate to the properties of equi- multiples. Propositions 7 to 10 and 13 and 14 connect the aoUon of tiie ratio of magnitudes with the ordinary notions of EUCLW8 ELEMENTS. 281 knd 8 Are iiher com- 8S. The ictty, not >r8 to the SeeDe he qaan* Eudid'a on Alge* with the Algebra, itities, a» * propo- nition of his defi- fandim- Eitlo was y defini- efinition seless. not pre- a. The Euclid 18. In ir^rbate 18 were Euclid 12, 23; ^Simson 00k are uto four Df equi- lect the tions of grealfTt equal, end 2«m.. Pjoppflitions 11, 11, 15 and 16 may bo considered as introduced to shew that, if four quantttUa of Me $ame kind h^ pr,<^rt*onal$ they toill alio he proportionals when taken alternately. The remaining, propositions shew that mag- nitudes are proportional by composition, by division, and ex asquo. In this division of the fifth Book propositions 13 and 14 are supposed to be placed immediately after proposition 10; and they might be taken in this order without any change in Euclid's demonstrations. The propositions headed Aj B, C, D, E were supplied by Simson. V' i» ^> 3» 5> ^« These are simple propositions of Arithmetic, though they are here expressed in terms which make them ap- pear less familiar than they really are. For example, Y. i *' states no more than that ten acres and ten roods make ten times as much as one acre and one rood.*' De Morgan, In V. 5 Simf>on has substituted another construction for that given by Euclid, because Euclid's construction assumes that we can divide a given straight line into any assigned number ci equal parts, and this problem is not solved until YI. 9. Y. 18. This demonstration is Simson's. We will give here Euclid's demonstration. Let il^ be to jFJ5 as Ci'' is to FD: AB shall be to BE as CD iaUi DP, For, if not, AB will be to BE as CD is to some magnitude less than DFy or greater than DP. First, suppose that ^.iS is to BE as CD is to DQ, which is less than DP, Then, because ^J5 is to 5 J? as (7Z) is to DQ, therefore il J^ is to EB as CQ is to GD. [ Y. 1 7. But AE is to EB as CP is to PD, [Hypothesis. therefore CG is to GD as CP is to PD, [Y. 1 1. But CG is greater than CP; therefore GD is greater than PD. But GD is less than PD ; which is impossible. In the same manner it may be shewn that AB is not to BB as CD is to a magnitude greater than DP, Therefore AB\fi%oBEM CD is to DP. The objection urged by Simson against Euclid's demonstra- tion is that ''it depends upon this hypothesis, that to any three magnitudes, two of which, at least, are of the same kind, there F G B D [Hypothesis. [Y14. 282 NOTES ON F p may be a fburUi proportional : . . Euclid doei not demon- strate li nor does he shew how to find the fourth proportional before the lath Proposition of the 6th Book. . . . ." The following demonstration is given by A^stii in his Excml nation of the first svx hoohs of Euclid's ElmenU. Let AE be to EB asCF iatoFB: AB shall be to ^i? as CD is to DP, For, because ^^ is to ^^ as CF is to FB therefore, alternately, AE is to CF as EB is to FD, ry^ jg And as one of the antecedents is to its con- sequent so is the sum of the antecedents to the sum of the consequents ; lY ii therefore as ^^ ia to i^T) so are AE md EB together to CF and FJ) together, that ^s, ^5 is to CD as EB is to PD. Therefore, alternately, AB ia to EB as CD is to FB. [V i6 • Z' i""^* A J^^ ^?* "^^P ''' *^® demonstration of this proposition IS "take AO equal to E and Cff equal to F^>; and here a refer- ence is sometimes ^iven to I. 3. But the magnitudes in the proposition are not necessarily straight Unes, so that this refer- ence to I. 3 should not be given; it must however be assumed that we can perform on the magnitudes considered, an operation similar to that which is performed on straight Unes in 1. 3. Since tiie fifth Book of the Elements treats of magnitudes generaUy, and not merely of lengths, areas, and angles, there is no reference made m it to any proposition of the first four Books. wl,?r^*'^-!?'^' 'T f*^P°«i*i«°« '•^lati^ff to compound ratio, which he distinguishes by the letters F, G, H, JT; it seems how! ever unnecessary to reproduce them as they are now rarely read and never required. , '' B D THE SIXTH BOOK. The sixth Book of the Elements consists of the appUcationof the ^eory of proportion to establish properties of geometrical A 7^J ^•^* '; ^^^ *^ important remark bearing on the first definition, see the note on VI. 5. VI. Lef a. The second definition is useless, for Euclid makes no mention of reciprocal figures. EUCLIJD^S ELEMENTS, 283 VI. Def. 4. The fourth definition is strictly only applicable to a triangle, because no other figure has a point which can be exclusively called its vertex. The altitude of a parallelogram is the perpendicular drawn to the base from any point in the op- posite sidel VI. 1, The enunciation of this important proposition is open to objection, for the manner in which the sides may be cut is not sufficiently limited. Suppose, for example, that AD is double of DB, and CE double of EA ; the sides are then cut proportionally, for each side is divided into two parts, one of which is double of the other ; but DE is not parallel to BG. It should therefore be stated in the enunciation that the segments terminated at the vertex of the triangle are to he homologout term^ in the ratios, that is, are to he the antecedents or the consequents of the ratios. It will be observed that there are three figures corresponding to three cases which may exist ; for the straight line drawn pa* rallel to one side may cut the other sides, or may cut the other sides when they are produced through the extremities of the base, or may cut the other sides when they are produced through the vertex. In all these cases the triangles which are shewn to be equal have their vertices at the extremities of the base of the given triangle, and have for their common base the straight line which is, either by hypothesis or by demonstration, parallel to the base of the triangle. The triangle with which these two triangles are compared has the same base as they have, and has its vertex coinciding with the vertex of the given triangle, VI. A, This proposition was supplied by Simson, VI. 4, We have preferred to adopt the term "triangles which are equiangular to one another," instead of ** equiangular triangles," when the words are used in the sense they bear in this proposition. Euclid himself does not use the term egydan- gular triangle in the sense in which the modem editors use it in the Corollary to I. 5, so that he is not prevented from using the term in the sense it bears in the enunciation of VI. 4 and else- where ; but modem editors, having already employed the term in one sense ought to keep to that sense. In the demonstrations, where Euclid uses such language as "the triangle ABC is equi- angular to the triangle BEP^'* the modem editors sometimes adopt it, and sometimes change it to "the triangles ABC and J>EF are equiangular." . In VI. 4 the manner in which the two triangles are to be I 284 NOTES ON ;tly > be described ; their bases an same straight line and contiguous, their vertices are to be on the same side of the base, and each of the two angles which have a common vertex is to be equal to the remote angle of the other triangle. By superposition we might deduce YI. 4 immediately from VI. 2. VI. 5. The hypothesis in VI. 5 involves more than is di- rectly asserted; the enunciation should be, '^if the sides of two triangles, taken in ordeTj about each of their angles .....;" that is, some restriction equivalent to the words taken in wder should be introduced. It is quite possible that there should be two triangles ABCy DEF, such that AB \% io EC as DE is to EP, and BG to CA as DF is to ED, and therefore, by V. 11, AB to AC OB DF is to EF; in this case the sides of the triangles about each of their angles are proportionals, but not in the same order, and the triangles are not necessarily equiangular to one another. For a numerical illustration we may suppose the sides of one triangle to be 3, 4 and 5 feet respectively, and those of another to be 12, 15 and 20 feet respectively. Walker, Each of the two propositions VI. 4 and VI. 5 is the converse of the other. They shew that if two triangles have either of the two properties involved in the definition of similar figures they will have the other also. This is a special property of triangles. In other figures either of the properties may exist alone. Eor example, any rectangle and a square have their angles equal, but not their sides proportional; while a square and any rhombus have their sides proportional, but not their angles equal. VI. 7. In "VI. 7 the enunciation is imperfect ; it should be, " if two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals, so thai the sides subtending the equal angles a/re homologous; then if each ....." The imperfection is of the same nature as that which is pointed out in the note on VI, 5. Walker, The proposition might be conveniently broken up and the essential part of it presented thus : if two triangles have two sides of the one proportional to two sides of the other, and the angles opposite to one pair of homoiogov,s sides equal, the angles which are opposite to the other pair of homologous sides shaU either be eqtuiU, or be together equal to two right angles, For, the angles included by the proportional sides must be /. EUCLWS ELEMENTS. 2^5 I bd in the be on the ch have % the other tely from ban is di* ies of two • •••.• > }, in (yrder should be DE is to )y V. 23, I triangles the same Etr to one the sides I those of I converse ler of the ;ures they triangles. >ne. For )qual, but rhombus hould be, 3 angle of ionals, so \; then if e as that and the two sides the angles which are be equal, must be A either equal or imequal. If they are equal, then since the tri- angles have two angles of the one equal to two angles of the other, each to each, they are equiangular to one another. We have therefore only to consider the case in which the angles in- eluded by the proportional sides are unequal. Let the triangles ABO, I>EF have the angle iXA equal to the angle at D, ^and ABXo £C as DE is to EF, but the angle ABC not equal to the angle DEF: the angles AC£ and DFE shall be together equal to two right angles. Tor, one of the angles A BC, DEF must be greater than - - the other ; suppose A BC the greater ; and make the angle A BO equal to the angle DEF. Then it may be shewn, as in VI. 7, that BG is equal to BC, and the angle BGA equal to the angle EFD. Therefore the angles ACB and DFE are together equal to the angles BGC and A OB, that is, to two right angles. Then the results enunciated in VI. 7 will readily follow. For if the angles A CB and DFE are both greater than a right angle, or both less than a right angle, or if one of them be a right angle, they must be equal. VI. 8, In the demonstration of VI. 8, as given by Simson, it is inferred that two triangles which are similar to a third triangle are similar to each other; this is a particular case of VI. 21, which the student should consult, in order to see the validity of the inference. VI. 9. The word part is here used in the restricted sense of the first definition of the fifth Book. VI. 9 is a particular case of VI. 10. VI. 10. The most important case of this proposition is that in which a straight line is to be divided either internally or ex- ternally into two parts which shall be in a given ratio. The case in which the straight line is to be divided iniemalhf is ^ven in the text ; suppose, for example, that the given ratio is that of AE to. EC', then .4^ is divided at in the given ratio. Suppose, however, that AB is to be divided exlemoUly hi & ^ven ratio; that is, supTK>se that AB is to be produced so that the whole str' * ht line made up oi AB and the part produced may be io the p^rt produced in a.given ratio. Let the given ratio / 286 NOTES ON ym' b« that oiACUi CE. Join EB ; through C draw a straight liii« parallel io EB; then this straight line will meet ^J9, produced through Bf at the required point. VI. II. This \a a particular case of VI. 12. ^^I. 14. The following is a full exhibition of the iteps which lead to the result that FB and BG are in one straight line. The angle DBF is equal to the angle QBE I [Hypothe^k, add to each the angle FBE; therefore the angles DBF, FBE are together equal to the angles QBE, FBE. lAxiom 2. But the angles DBF, FBE are together equal to two right angles; [I. 13. therefore the angles OBE, FBE are together equal to two right angles; {Axiom 1. therefore FB and BG are in one straight line. [I. 14. VI. 15. This may be inferred from VI. 14, since a triangle is half of a parallelogram with the same base and altitude. It is not difl&cult to establish a third proposition conversely connected with the two involved in VI. 14, and a third propo- sition similarly conversely connected with the two involved in VI, 15. These propositions are the following. Equal parallelograms which have their aides reciprocally prO" portional, have their angles equal, each to each. Equal triangles which have the sides alout a pair of angles reciprocally proportional, have those angles equal or together equal to two right angles. We will take the latter proposition. Let ABC, ADE be equal triangles ; and let CA be to AD M AE ia to AB: either the angle BAG shall be equal to the angle DAE, or the angles BAG and DAE shall be together equal to two right angles. [The student can constrtict the figure for himself.] Place the triangles so that GA and AD may be in one straight line; then if EA and ^-B are in. one straight line the angle BAG is equal to the angle DAE, [I, j^. If EA and -4jB are not in one straight line, produce BA through A to F, so that AF may be equal to AE; join DF and EF» Then becauae GA ia to AD ?!B AE is ta A B r.'7'wn.'}/,.'^-»s's and -42? is equal to AE^ [Gonstruction* therefore GA is to AD as AF iaioAB, [V. 9, V. 11. Therefore the triangle DAF is equal to the triangle BAG. £VI. i*. EUCLID'S ELEMENTS. 287 produced )p8 which ae. ypothetis, !ie angles Axiom 2. wo right [I. 13. wo right Axiom I. [I. 14. triangle inversely i propo- olved in »% pi'O' f angles ler eqiial ) to AD 1 to the er equal straight :le BAP [I. 15. through Iruction, V. II. TI. 16, But the triangle DAE is equal to the triangle BA 0. [ffypotheiU. Therefore the triangle Dil-ff is equal to the triangle Dili^. [Ax. i. Therefore EF is parallel to AD. [I. 39" Suppose now that the angle DAE la greater than the angle DAF. Then the angle CAE is equal to the angle AEF, [I. 29. and therefore the angle CAE is equal to the angle AFE, [I. 5. and therefore the angle CAEia equal to the angle BAC. [I. ag. Therefore the angles BAC and DAE are together equal to two right angles. Similarly the proposition may be demonstrated if the angle DAE is less than the angle DAF. VI. 16. This is a particular case of VI. 14. VI. 17. This is a particular case of VI. 16. VI. 22. There is a step in the second part of VI. 22 which requires examination. After it has been shewn that the figure SR is equal to the similar and similarly situated figure NH, it is added "therefore PB is equal to OH.'' In the Greek text reference is here made to a lemma which follows the proposition. The word lemma is occasionally used in mathematics to denote an auxiliary proposition. From the unusual circumstance of a reference to something following, Simson probably concluded that the lemma could not be Euclid's, and accordingly he takes no notice of it. The following is the substance of the lemma. If PR be not equal to GH, one of them must be greater than the other; suppose PR greater than GIT. Then, because SR and NH are similar ^res, PR is to PS as Gffia to GN. [VI. Definition i. But PR is greater than Gff, [Hypothesis, therefore PS is greater than GN. [V. 14. Therefore the triangle RPS is greater than the triangle HGN. [I- 4» ^a^o»* 9- But, because SR and NH are similar figures, the triangle RPS is equal to the triangle lIGN; [VI. 20. which is impossible. Therefore PR is equal to GH. ... -.-.■. r. r »/ . ^__ riT\ _„,! /3Z7^««' VI. 23. m uuo ngure 01 v x. z^ suppuisu jjs^- »«« «« -~S5 Sinzi.4 Then the triangle BCD is to the triangle GCE as the parallelo- gram -4(7 is to the parallelogram CF. Hence the result may be extended to triangles, and we have the following theorem, 288 NOTES ON trumglet which have one angle of the om equal to cni an^U of the other, have to one anoUier the ratio which is compounded of the ratio* of their sides. Then YI. 19 is an immediate consequence of this theorem. For let ABC and DEF be similar triangles, so that AB UioBC as i)^ is to EF] and therefore, alternately, .45 is to DJ? as BC is to EF, Then, by the theorem, the triangle ABC has to the triangle DEF the ratio which is compounded of the ratios o( AB to BE and of BG to EF, that is, the ratio which is compounded of the ratios of BC to EF and of BC to EF. AuJ, Irum the definitions of duplicate ratio and of compound ratio, it Tollows that the ratio compounded of the ratios of BC tr ■i'f and of BC to EF is the duplicate ratio of BC to EF. VI. 25. It will be easy for the student to exhibit in detail the process of shewing that BC and CF are in one straight line, and also LE and EM ; the process is exactly the same as that in I* iit ^y whieh it is shewn that KU and MM are in one straight line, and also FO and OL. It seems that VI. 2$ is out of place, since it separates pro- positions so closely connected as VI. 24 and VI. 26. We may enunciate VI. 25 in familiar language thus: to make a figure which shall have the form of one figure and the size of another, VI. 76, This proposition is the converse of VI. 24 ; it might be extended to the case of two similar and similarly situated parallelograms which have a pair of angles vertically opposite. We have omitted in the sixth Book Propositions 27, 28, 29, and the first solution which Euclid gives of Proposition 30, as they appear now to be never required, and have be6n condemned as useless by various modern co- .nei.wvt/^ra; see Austin, Walker, and Lardner. Some idea of tbo nr/'r^f )f these f repositions may be obtained from the foll6wi^»^ ..wateiiient of the problem pro- posed by Euclid in VI. ag. AB is a, given straight line; it lias to be produced through 5 ta a point 0, and a parallelogram described on ^ subject to the following conditions ; the paral- lelogram is to be equal to a given rectilineal figure, and the parallelogram on the baae BO which can be cut off by a ^traight line through £ is to be similar to a given paralielo* gram. VI. 31. This proposition seems of no use. Moreover the enunciation is imperfelem pro- e; it lias llelogram the paral- and the [)ff by a parallelo* (over the produced through D to a point P^ such that DP is equal to DB j and join CP, Then the triangle CDP will satisfy aU the conditions in Euclid's enunciation, as well as the triangle CDE\ but CP and CB are not in one straight line. It should be stated that the bases must lie on corresponding sides of both the parallels; the bases CP and BO do not lie on corresponding sides of the parallels AB and DC, and so the triangle CDF wild not fulfil all the conditions, and would therefore be excluded. • VI. 33. In VI. 33 Euclid implicitly gives up the restriction, which he seems to have adopted hitherto, that no angle is to h% considered greater than two right angles. For in the demon- stration the angle BQL may be any multiple whatever of the angle BQC, and so may be greater than any number of zighi angles. VI. By Cf D. These propositions were introduced by Simson. The important proposition VI. D occurs in the MryciXif "Zivraiii of Ptolemy. THE ELEVENTH BOOK. In addition to the first six Books of the Elements it is usual to read part of the eleventh Book. Eor an account of the contents of the other Books of the Elements the student is referred U the article Bucleides in Dr Smith's Dictionary of Greek and Roman Biography , and to the article Irrational QtMm- tUies in the English Cyclopaedia. We taay state briefly that Books VII, VIII, IX treat on Arithmetic, Book X on Irra- tional Quantities, and Books XI, XII on Solid Geometry. XI. Def. 10. This definition is omitted by Simson, and justly, because, as he shews, it is not truo that solid figures contained by the same number of similar and equal plane figures are equal to one another. For, conceive two pyramids, which have their bases similar and equal, but have different altitudes. Suppose one of these bases applied exactly on the other; then if the vertices be put on opposite sides of the base a certain solid is formed, and if the vertices be put on the same side of the base another solid is formed. The two solids thus formed are con* but they are not equal. It will be observed that in this example one of the solids hai A re-entrant solid angles see page 954. It is however true thM 19 290 NOTES ON two ewwix Bolid figures are equal if tbey are contained by equal plane figures similarly airanged; see Catalan's Thiorimes et Prchlhnet de G4om4trie EUmentaire. This result was first demon- strated by Cauohy, who turned bis attention to the point at^the request of Legendre and Malus; see the Journal de VEcole Polytechniqiie, Cahier i6. XI, Def. 26. The word tetrahedron is now often uf jd to denote a solid bounded by any fuur triangular faces, that is, a pyramid on a triangular base ; and when the tetrahedron is to be such as Euclid defines, it is called a regular tetrahedron. Two other definitions may conveniently be added. A straight line is said to be parallel to a plane when they do not meet if produced. The angle made by two straight lines which do not meet is the aingle contained by two straignt lines parallel to them, drawn through any point. ^I, 21, In XI. 21 the first case only is given in the ori« ginal. In the second case a certain condition must be intro- duced, or the proposition will not be true; the polygon BCDEF must have no re-entrant angle. See note on I. 32. The propositions in Euclid on Solid Greometry which are now not read, contain some very important results respecting the volumes of solids. We will state these results, as they are ofben of use; the demonstrations of them are now usualiy given as examples of the Integral Calculus. We have already explained in the notes to the second Book how the area of a figure is measiured by the number of square inches or square feet which it contains. In a similar marner the volume of a solid is measured by the number of cubic inches or cubic feet which it contains ; a cubic inch is a cube in which each of the faces is a square inch, and a cuUc foot is similarly defined. *■ The volume of a prism is found by multiplying the number of square inches in its base by the number of inches in its altitude ; the volame is thus expressed in cubic inches. Or we may multiply the number of square feet in the base by the number of feet in the altitude ; the volume is thus expressed in cubic feet. By the base of a prism is meant either of the two equal, timilar, and parallel figures of XI. Definition 13; and the altitude of the prism is the perpendicular distance between these two planes, n prispi volum A of a ] base X Yi stude] on ^i T^ are ve tio'iis. and is thetiiv Li EUCLIUS ELEMENTS, 291 I by equal ■st demon- tint at the de I'Ecole n uf jd to that is, a iron is to Iron. n they do >t meet is HLf drawn n the ori« be intro- 1 BCBEF The rule for the volume of a prism involves the fact that prisjM qn equal loHt and between the same paralleli we emtal in volume. A parallelepiped is a particular case of a prism. The volume of a pyramid is one third of the volume of a prism on the same base and having the same altitude. For an account of what are called the Jive regular solidt the student is referred to the chapter on Polyhedrons in the Treatise on Spherical Trigonometry, THE TWELFTH BOOK. Two propositions are given from the twelfth Boole, as they are very important, and are required in the University Examina- tioas. The Lenmia is the first proposition of the tenth Book and is required in the demonstration of the second propoiritioii of the twelfth Book. ^ ^ nrhich are ecting the they are w usualiy ond Book of square larner the inches or Lich each similarly e number hes in its I. Or we se by the Dressed in )f the two ; and the 'een these 19-2 APPENDIX. of i] This Appendix consists of a collection of important pro- positions which will be found useful, both as aflfording geometrical exercises, and as exhibiting results which are often required in mathematical mvestigations. The student will have no diflBculty in drawing for himself the requisite figures in the cases where they are not given. \ ! f 1. The sum of the squares on the. sides of a triangk is equal to twice the square on half the ba>e^ together with twice the square on the straight line which joi?i8 the vertex to the middle point qfthe base. Let ABC be a triangle ; and let D be the middle point of the base AB» Draw CE perpendicular to the base B E meeting it at JE; then JS may be either in AB or ia AB produced. First, let E coincide with D; then the proposition follows immediately from I. 47. Next, let E not coincide with D; then of the two angles ADC and BDC^ one must be obtuse and one acute. Suppose the angle ADC obtuse. Then, by II. 12, the square on AC m equal to the squares on AD, DC, toge- ther with twice the rectangle AD, DE) and, by II. 13, the square on BC together with twice the rectangle BD, DE is equal to the squares on BD, DC. Therefore, by Axiom 2, the squares on AC BC-, toerether with twice the rectansle BD, DE are equal to the^squares on AD„ DB, and twice the square on DC, together with twice' the rectangle AD,DE. But -4Z) is equal to Z>5. Therefore the squares on AC, BCBTe equal to twice the squares on AD, DC, 294 APPENDIX. ^hth K '*^^ f^^^^- ^"^^^^^^ 'Within a circle, the angle tchichthey include w measured by haHfthe mm of the in. terceptea arcs, v •« •'•- ioinAD"^ ^^^^^ ^^ ^^ ^^ ^^ * ^"^^® intersect at JET; The angle J[^(7 is eanal to the angles ADE, and DAE, by I. 32; that is, to the angles standing on the arcs AG and BD. Thus the angle AEG is equal to an angle at the cir- cumference of the circle stand- ing on the sum of the arcs AG and BD\ and is therefore equal to an angle at the centre of the circle standing on half the sum of these arcs of i:'!:^'GB'^^^j^.^^ '' "^'^^'^ ^^ ^^ «^e sum *h. ^' V *'^?- ^t%^ produced intersect without a circle the angle which th^ include is measured by Mf the difference of the intercepted arcs, ^ ^ Let the chords AB and CD of a circle Y»rodiinAi1 I'n tersect at E ; join AD ^-^ "* » circle, produced, m- difference Of AG and BD-, and is therefore equal to an angle at the centre of the circle standing on half the ^er" ence of these arcs. ^ ^ oiner- APPENDIX, 295 , 4. To draw a straight line which thali touch two given circles. Let A be the centre of the greater circle, and B the f^'iu ^U^^ ^®^ ^^^^^^' ^^^^ centre ^y and radius equal to the difference of the radii of the given circles, desCTibe a circle; from B draw a straight line touchmg the circle 80 described at G, Join AC and produce it to meet the circumference at 2>. Draw the radius BE parallel to AD, and on the same side of ^^; and join DE. Then DE shall touch both circles. See I. 33, 1. 29, and III. 16 Corollary. Since two straight lines can be drawn from B to touch the described circle, two solutions can be obtained ; and the two straight lines which are thus drawn to touch the two given circles can be shewn to meet AB, produced through B, at the same point. The construction is applicable when each of the given circles is without the other, and also when they intersect. When each of the given circles is without the other we can obtain two other solutions. For, describe a circle with A t\r% j» ^^_i. J __ Ji 1 X_ J-1- ^ i,^ -i» J ■» _ J.. A '«. OS M> \;wiivri; uiiu. ^OiUiiis \;i|Uti>i &u wiio ouui OI wio lauli 01 the given circles ; and continue as before, except that BE and -42) will now be on opposite sidea of AB. The two straight lines which are thus drawn to touch the two given ciix;les can be shewn to intersect AB at the same point 296 APPENDIX. 5. To describe a circle which shall pass throttgh three given points not in the same straight line* This ia solved in BucUd IV. 6. 6. To describe a circle which shall pass thrmigh two given points on the same side of a gieen straight line, and touch that straight line. Let A and B be the given points ; join AB and pro* duce It to meet the given straight line at C, Make a square equal to the rectangle CA, GB (II. 14), and on the ffiven straight line take CE equal to a side of this square. Describe a circle through A, B, E (6); this will be the circle required (III. 37). Since E can be taken on either side of (7, there are two solutions. The construction fails if AB is parallel to the given Straight |ine. In this case bisect AB at D, and draw DG at nght angles to AB, meeting the given straight line at C, Then descnbe a circle through A, B, (7. ^ 7. To describe a circle which shall pass through a given point and touch two given straight lines. Let A be the given point; produce the given straight hues to meet at B, and join AB. Through B draw a straight line, bisecting that angle included by the given Straight lines within which A lies; and in this bisecting on one of the given straight lines, meeting it at />; with centre a and radius CD, describe a circle, meeting AB, produced if necessary, at E. Join CE; and through A draw a straight hne parallel to CE, meeting BC, produced if APPENDIX. 297 necessary, at F, The circle described from the centre Fj with radius FA, will touch the given straight lines. For, draw a perpendicular from F on the straight line BD, meeting it at G. Then CE is to FA as BC is to BF. and CD is to jP(5^ as BG is to BF (VI. 4, V. 16). There- fore GE is to i^^ as GD is to FG (V. 11). Therefore CE is to CD as /!4 is to ^G^ (V. 16). But CE is equal to CD', therefore FA is equal to FG (V. ^). If A is on the straight line BC we determine E as before ; then join ED^ and draw a straight line through A parallel to ED meeting BD produced if necessary at G ; from G draw a straight line at right angles to BG, and the point of intersection of this straight line with BC^ produced if necessary, is the required centre. As the circle described from the centre (7, with the radius (7Z>, will meet AB at two points, there are two solutions, If A is on one of the given straight lines, draw from A a straight line at right angles to this ffiven straight lino; the point of intersection of this straight line with either of the two straight lines which bisect the angles made by the given straight lines may be taken for the centre of the required circle. If the two given straight lines are parallel, instead of drawing a straight line BG to bisect the angle between them, we must draw it parallel to them, and equidistant from them. 298 APPENDIX. 8. To describe a circle which shall touch three given straiglu lines, not more than two qf which are parallel. Proceed as in Euclid IV. 4. If the given straight Imes form a triangle, four circles can be described, namely, one as in Euclid, and three others each touching one side of the triangle and the other two sides produced. If two of the given straight lines are parallel, two circles can be described, namely, one on each side of the third given straight lino. 9. To describe a circle which shall touch a given circle^ and touch a given straight line at a given point. Let A be the given jpoint in the given straight line and G be the centre of the given circle. Through C draw a Btipight Ime perpendicular to the given straight line and meeting the circumference of the circle at B and Z>, of which D is the more remote from the given straight line. Join AD^ meeting the circumference of the circle at K From A draw a straight line at right angles to the given straight line, meeting CjE: produced at F. Then /'shall be the centre of the required circle, and FA its radius. For the angle AEF is equal to the angle CED (1. 15); and the angle EAF is equal to the angle CDE (I. 29); therefore the angle AEF is equal to the angle EAF\ therefore -4 F is equal to EF (I. 6). APPENDIX, 299 ch three hich are ight lines mel^, one e side of If two B8 can be rd given a given n point. ght line, C draw iH line, In a similar manner another solution may be obtained by joining AB. If the given straight line falls without the given circle, the circle obtained by the first solution touches the given circle externally, and the circle obtained by the second solution touches the given circle internally. If the given straight line cuts the given circle, both the circles obtained touch the given circle externally. 10. To describe a circle which sJicUl pass through two given points and touch a given circle. Let A and B be the given points. Take any point C on the circumference of the given circle, and describe a circle through A, B, G. If this described circle touches the given circle, it is the required circle. But if not, let D and 2), straight circle at ) to the iJPshall ma (1.15); (I. 29); EAF; ^5 and*^n ?^'"* °/ interaection of the two circles. Let Wl^srifl A^' ^ ^*" ^ *"« reS^VSl! Kn P®'"® ^^^ *^^ solutions, because two straight lines can be drawn from E to touch the given circle. *.oor "x? ^^^76*V' ""^ which bisects AB at right ansfles passes through the centre of the given ch-cle, the con- f1 *1SV'^''/^'' ^^. ""^^ ^^ ^'•^ P^^"e'- I« this case ^ must be determined by drawing a straight line paraUel to AB so as to touch the given circle. p*nui«* 300 APPENDIX. I 11. To describe a circle which $haU touch two given straight lines and a given circle. Braw two straight lines parallel to the given straight lines, at a distance from them equal to the radius of the given circle, and on the sides of them remote from the centre of the given circle. Describe a circle touching the straight lines thus drawn, and passing through the centre of the given circle (7). A circle having the same centre as the circle thus described, and a radius equal to the excess of its radius over that of the given circle, will be the re- quired circle. Two solutions will be obtained, because there are two solutions of the problem in 7 ; the circles tiius obtained toujch the given circle externally. "We may obtain two circles which touch the given circle internally^ by drawing the straight lines parallel to the given straight Imes on the sides of them adjacent to the centre of the given circle. 12. To describe a circle which shall pass through a given point and touch a given straight line and a given circle. We will suppose the given point and the given straight line without the circle ; other cases of the problem may be treated in a similar manner. Let A be the giveft point, and B the centre of the given circle. From B draw a perpendicular to the given straight line, meeting it at (7, and meeting the circum- ference of the given circle at D and JS, so that D is be- tween B and C. Join JSA and determine a point Fin UA, produced if necessary, such that the rectangle EA^ EF may be equal to the rectangle EC, ED ; this can be done hj describing a circle through A\ (7, D, which will meet EA at the required point (111. 36, Corollary). Describe a circle to pass through A and F and touch the given straight line (6) ; this shall be the required circle. APPENDIX. 301 two given For, let the circle thus described touch the giren straight line at G ; join EO meeting the given circle at H, ni straight ius of the from the ching the he centre centre as ;he excess >e the re- 3 are two obtained ven circle the given he centre hrough a I a given i straight n maybe e of the ;he given D circum- D is be- F in EA, EA, EF be done nil meet escribe a 1 straight and join DH. Then the triangles EHD and ECG are similar: and therefore the recUngle JB'C, ED is equal to the rectangle EGy EH (III. 31, VI. 4, VI. 16). Thus the rectangle EA^ EF is equal to the rectangle EH^ EG ; and therefore ^ is on the circumference of the described cu-cle (III. 36, Corollary). Take K the centre of the described circle; join KG.KH, and BH. Then it may be shewn that the angles KHG and EHB are equal (I. 29, I. 6). Therefore KHB is a straight line; and therefore the described circle touches the given circle. Two solutions will be obtained, because there are two solutions ol the i)roblem in 6 ; the circles thus described touch the given circle externally. By joining DA instead of EA we can obtain two solu- tions in whiwi the circles described touch the given circle iateFually. B02 APPENDIX. 13. To describe a circle which shall totich a given straight line and two given circles. Let A be the centre of the larger circle and B the centre of the smaller circle. Draw a straight line parallel to the given straight lino, at a distance from it equal to the radius of the smaller circle, and on the side of it remote from A. Describe a circle with A as centre, and radius equal to the difference of the radii of the given circles. Describe a circle which shall pass through B, touch exter- nally the circle just described, and also touch the straight line which has been drawn parallel to the given straight Ime (12). Then a circle having the same centre as the second described circle, and a radius equal to the excess of its radius over the radius of the smaller given circle, will be the required circle. Two solutions will be obtained, because there are two solutions of the poblem in 12 ; the circles thus described touch the given circles externally. We may obtam in a similar manner circles which touch the given circles internally, and also circles which touch one of the given circles internally and the other exter- nally. 14. Let A he the centre of a circle, and B the centre of a larger circle; let a straight line be drawn tottching the former circle at C ^nd the latter circle at D, and meeting AB produced through A at T. From T draw any straight line meeting the smaller circle at K and L aTid the larger circle at M and N ; so that the five letters T, K, L, M, N are in this order. Then the straight lines AK, KC, CL, LA shall be respectively parallel to the straight lines BM, MD, DN, NB; and the rectangle TK, TN shall be equal to the rectanale TL. TM. and equal to the rectangle TO, TD, " ' " Join ACi BD, Then the triangles 7!4Cand TBD are APPENDIX, 303 a given efmlangular ; and therefore TA is to 7!5 a» -4C7 is to BD (VI. 4, V. 16), that is, as -4 JS: is to BM. d B the narallel al to the k remote d radius 1 circles, h exter- straight straight I as the e excess Q circle, are two ascribed ;h touch h touch r exter* J centre mucking D, and r draw and L, J letters ht lines to the ctangle 72>are Therefore the triangles TAIC and TBM are similar (VI. 7); therefore the angle TAK is equal to the angle TBM; and therefore ^A is parallel to B3f. Similarly AL is parallel to BN. And because AIC is parallel to BM and ^C parallel to BDy the angle CAK is equal to the angle DBM\ and therefore the angle CLK is equal to the angle DNM (III. 20) ; and therefore CL is parallel to i>iV. Similarly 6^ is parallel to />ili/. Now TM is to TD as TD is to TN (III. 37, VI. 16); and TM\B to TD as TK is to TC (VI. 4); therefore TK is to TC Q& TD is to 7W; and therefore the rectangle TK, TN ia equal to the rectangle TC, TD, Similarly the rectangle TL, TM is equal to the rectangle TC, 2D, If each of the given circles is without the other we may suppose the straight line which touches both circles to meet AB at T between A and B, and the above results will all hold, provided we interchange the letters ^ and Z ; so that the five letters are now to be in the following order, Z, K, T, My N The point ris caiied a centre qfsim^Uwk of tlio twa circles, 304 APPENDIX. 16. To describe a circle which shall pass through a given point and touch two given circles. Let A be th^ centre of the smaller circle and B the centre of the larger circle ; and let E be the given point. Draw a straight line touching the former circle at C and the latter at i), and meeting the straight Ime AB pro- duced through A, at T. Join TE and divide it at'/ so *^^* S;?.'*®m?"^^®. 2!£r i!F may be equal to the rectangle and touch either of the given circles (10); this shall be the required circle. APPENDIX, 305 throttgh a and B the iven point. at C and ABr pro- t at Jrso rectangle »-» 3 -rt . JU OMXX £' all be the For suppose that the circle is described so as to touch the smaller given circle ; let ^ be the point of contact; we have then to shew that the described circle will also touch the larger given circle. Join TG, and produce it to meet the larger given circle at H. Then the rectangle TG, TH\% equal to the rectangle TC, TD (14); therefore the rectangle TG^ TH ia equal to the rectangle TE, TF\ and therefore the described circle passes through H, Let be the centre of this circle, so that OGA is a straight line ; we have to shew that OHB is a straight line. Let TG intersect the smaller circle again at K\ then AK is parallel to BH{14) ; therefore the angle AKT is equal to the angle BUG ; and the angle A KG is equal to the angle AGK, which is equal to the angle OGH^ which is equal to the angle OHG. Therefore the angles BHG and OHG together are equal to AKT and AKG together ; that is, to two right angles. Therefore OHB is a straight line. Two solutions will be obtained, because there are two solutions of the problem in 10. Also, if each of the given circles is without the other, two other solutions can be obtained by taking for T the point between A and B where a straight line touching the two given circles meets AB. The various solutions correspond to the circiun- stance that the contact of drcles may be external or internal. 16. To describe a circle which shall tovjch three given circles. Let A be the centre of that circle which is not greater than either of the other circles; let B and O be the centres of the other circles. With centre -6, and radius equal to the excess pf the radius of the circle with centre B over the radius of the circle with centre Ay describe a circle. Also with centre C, and radius equal to the excess of the raniiia of the circlfi with opntTfi fJ ovftr t.hf* i^diiis fsf fiia circle with centre A, describe a circle. Describe a circle to touch externallv these two described circles and to pass through ^ (15). fhen a circle having the same centre as tlie last described circle, and having a radius equal to 20 306 APPENDIX, the excess of its radius over the radius of the circle with centre A, will touch externally the three given circles. In a similar way we may describe a circle touching internally the three given circles, or touching one of them externally and the two others internally, or touching one of them internally and the two others externally. 17. In a given indefinite straight line it is required to find a point sttch that the sum of its distances from two given points on the same side of the straight line shall he the least possible. Let A and B be the two given points. From A draw a perpendicular to the given straight line meeting it at C\ and produce AG to D so that CD may be equal to AG. Join DB meeting the given straight line at E, Then E shall be the required j^oint. For, let J^be any other point in the given straight line. Then, because AG ^r equal to DC, and EG is common to the two triangles ACE, DGE\ and that the right angle AGE is equal to the right angle DCE\ therefore AE is equal to DE. Similarly, ^i^ is equal to DF, And the ...... ^C T\T7t s~J T^Ti i_ A XI Tjr^ /T /-.yvX _ At- __._^_^- SUa*a vi Jt^S: Mfilva. JL' JiJ Ao^rUtiwvI' viiciii XJ JLi (i, Zv /" I ilioi OlOl the sum of AF and FB is greater than BD\ that is. the sum of ^i^ and FB is greater than the sum of DE and EB\ therefore the sum of AF and FB is greater than the sum of ^^ and EB, circle with drcles. 3 touching le of them ling one of s required mces from *aight line m A draw ig it at C\ lal U> AG. , Theni? APPENDIX. 307 18. The perimeter qf an isosceles triangle is less than that qf any other triangle qf equal area standing on the same base. Let ABC be an isosceles triangle; AQG any other tri- jj q angle equal in area and stand- " ing on the same base .^(Z Join BQ ; then BQ is paral- lel to -4(7(1. 39). And it will follow from 17 that the sum oi AQ and QG is greater than the sum of AB and BC. 19. If a polygon he not equilateral a polygon may he found of the same number qfsides^ and equal in area, hut havitig a less perimeter. *aight line. jommon to ight angle )re AU is And the j-i- -^_^_^- liat is. the r DE and JHter than For, let CJp, DE be two a<\jacent unequal sides of the polygon. Jom CE. Through D draw a straight hne parallel to CE, Bisect CE at Z ; from L draw a straight lSf„^^J^^J^* ?°&^®® J^ ?? meeting the straight line drawn turoaftii u at K, luen uy removing from the given poly* gon the triangle CDE and applying the triangle GKE, we obtam a polygon having the same number of sides a« the given polygon, and equal to it in area, but havmg a less perimeter (18). * £0—2 308 APPENDIX. 20. A arid B are ttco given points on the fame tide qf a given straight line^ and AB prodttced meets the given straight line at Q\ of all points in the given straight line on each side of C, it is required to determine thai at which AB subtends the greatest angle. Describe a circle to i)ass through A and 5, and to touch the given straight line on that side of C which is to be considered (6). Let D be the point of contact: JD shall bi the required point. For, take any other point E in the given straight line, on the same side of (7 as i) is ; draw EAj EB ; then one at least of these straight Imes will cut the circumference ADB. Suppose that BE^cvAr the circumference at F] join AF. Then the angle AFB is equal to the angle ADB (III. 21); and the angle AFB is greater than the ande AEB (1. 16); therefore the angle ADB is greater than the angle AEB. Q1 A yf M// K fiiv^A 4-/»m/\ rtim^n^ i^xviT/M / a aj^i* / It >l/y% n and AB i* drawn and produced both ways so as to divide the whole circumference into two arcs; it is required to determine the point in each qf these arcs at which AB subtends the greatest angle^ APPENDIX. 809 ante tide qf 9 the given ■raight line ine that at Describe a circle to pass through A and B and to touch tlie circumference considered (10) : the point of contact will be the required point. The demonstration is similar to that in the preceding proposition. B, and to which is to x>ntact: D ;raight line, *; then one cumference ^;iom AF. 3 (III. 21); EB (1. 16); gle AEB. ti. it, ^•i'fVT/i? . aw to divide required to which AB 22. A and B are two given points withoiU a given circle ; it is required to determine the points en the cir- cumference qf the given circle at which AB subtends the greatest and least angles. Suppose that neither AB nor AB produced cuts the given cwcle. Describe two circles to pass through A and B, and to touch the given ch-cle (10) : the point of contact of the circle which touches the given circle externally will be the point where the angle is greatest, and the point of contact of the circle which touches the given circle internally will be the point where the angle is least. The demonstration is similar to that in 20. If AB cuts the given circle, both the circles obtained by 10 touch the given circle internally ; in this case the angle subtended by ^^ at a point of contact is less than the angle subtended at any other point of the circumference of the given circle which is on the same side of AB. Hero the angle is greatest at the points where AB cuts the circle, and is there equal to two right angles. If AB produced cuts the given circle, both the circles obtained by 10 touch the given circle externally ; in this case the angle subtended by ^5 at a point of contact is greater than the angle subtended at any other point of the circumference of the given circle which is on the same side of AB. Here the angle is least at the points where AB produced cuts the circle, and is there zero. 310 APPENDIX, 2S. Jy there he four magnitudes such that tfie first U to the second as the third is to the fourth; then shall the first together with the second be to the excess of the first above the second as the third together with the fourth is to the excess oftJie third ab&ce th^ fourth. For, the first together with the second is to the second as the third tf^^other with the fourth is to the fourth (V. 18), Therefore^ aii : rj itely, the first together with the second is to the third together with the fourth as the second is to the fourth (Y. 16). Similarly, by V. 17 and V. 16, the excess of the first above the second is to the excess of the third above l^e fourth as the second is to the fourth. Therefore, by V. 11, the first together with the second is to the excess of the first above the second as the third together with the fourth is io the excess of the third above the fourth. i f |j j 1 i '1 t j ! 1 i 1 ' 24. The straight lines drawn at right angles to the sides of a triangle from the points qf bisection qfthe sidet meet at the sanrie point. Let ABChQ a triangle; bisect J5(7at 2), and bisect GA at E ; from D draw a straight line at right angles to BG^ and from E draw a straight line at rigat angles to CA\ let these straight lines meet at G^ : we have then to shew that the straight line which bisects AB Q.i right angles ^so passes through G, From the triangles BDG and APPENDIX. 311 CDG we can shew that BG is equal to CG ; and from the triangles GEG and AEG wfe can shew that CG is equal to AG ; therefore BG is equal io AG. Then if we draw a straight line from G to the middle point of AB we can shew that this straight line is at right angles to ^^: that is the line which bisects AB at right angles passes through G» 25. The straight lines drawn frmn the angles of a triangle to the points of bisection qf the opposite sides meet at the same point. Let ABO be a triangle ; bisect BG at D, bisect CA at ^; and bisect AB at F-j join BE and CF meeting at G; Join ^6J^ and GD: then AG md GD shall lie in a straight line. The triangle BEA is equal to the triangle BEG, and the triangle GEA is equal to the triangle GEG (I. 38); therefore, by the third Axiom, the triangle BGA is equal to the triangle BGO, Similarly, the triangle CGA is equal to the triangle 0GB, Therefore the triangle BGA is equal to the triangle CGA. And the triangle BGB is equal to the triangle OGD (1. 38) ; therefore the triangles BGA and BGD together are equal to the triangles VGA and CGU together, merefore the triangles BGA and BGD together are equal to half the triangle ABC. Ilierefore G must fall on the straight line AD ; that is, AG and GD lie in a straight line; 312 APPENDIX. 26. The straight lines which bisect the angles qf a triangle meet at the same point. Let' ABC be a triangle; bisect the angles at J? and G by straight lines meetmg at G ; join AG ; then AG shall bisect the angle at ^. ,, From G draw GD perpendicular to BCy GE perpen- dicular to GA, and GF perpendicular to AB, From the triangles BGF and BGD we can shew that GF is equal to GD ; and from the triangles CGE and CGD we can shew that GE is equal to GD ; therefore GF is equal to GE, Then from the triangles AFG and AEG we can shew that the angle FAG is equal to the angle EAG, The theorem may also be demonstrated thus. Produce AG to meet BG at H. Then AB is to BH as AG is to GH, and AG is to Cff as ^G^ is to GH (VI. 3); there- fore AB is to BH as ^(7 is to CHCV. 11) ; therefore AB is to ^C as BH is to*(7^ (V. 16); therefore the straight line AH bisects the angle at A (VI. 3), 27. Let two sides of a triangle he produced through the base; then the straight lines which bisect tJie two exterior angles thus formed, and the straight line which bisects the vertical angle of the triangle, meet at the same This may be shewn like 26: if we adopt the second method we shall have to use VI. A, APPENDIX. 313 glea qf a ; J? andC 28. The perpendiculars drawn from the angles of a triangle on the opposite sides meet at the same point. Let ABC be a triangle; and first suppose that it is not obtuse angled. From B draw BE perpendicular to CA \ AG shall ? perpen- thew that 'JGE md efore GF md AEG bhe angle Produce AG is to i); there- 3fore AB i straight through t/ie two fie which the same 9 second from (7 draw OF perpendicular to AB; let these perpen- diculars meet at G; join AG, and produce it to meet BG at D : then AD shall be perpendicular to BC, For a circle will go round AEGF {Note on III. 22) ; there- fore the angle FAG is equal to the angle FEG (III. 21). And a circle will go round BCEF (III. 31, Note on III. 21) ; therefore the angle FEB is equal to the angle FOB. Therefore the angle BAD is equal to the angle BGF. And the angle at B is common to the two triangles BAD and BGF. Therefore the third angle BDA is equal to the third angle BFC (Note on I. 32). But the angle BFG is a right angle, by construction ; therefore the angle BDA is a right angle. In the same way the theorem may be demonstrated when the triangle is obtuse angled. Or this case may be deduced from what has been already shewn. For suppose the angle at A obtuse, and let the peipendicular from B on the opposite side meet that side produced at E, and let the perpendicular from G on the opposite side meet that side produced at Fi and let BE and GF be produced to meet at G. Then in the triangle BGG the perpendiculars BF and GE meet at A ; therefore by the former case the straight line GA produced will be perpendicular to BG* 3U APPENDIX. i n !i 29. 1/ from any point in the circumference qf the circle described round a triangle perpendiculars he drawn to the sides of the triangle^ the three points qf intersection are in the same straight line. ■ Let ABC be a triangle, P any point on the circum- ference of the circumscribing circle; from P draw PD, P£, PF perpendiculars to the sides BCy CA, AB respec- tively \ DyEjF shall be in the same straight line. [We will suppose that P is on the arc cut off by AB, on the opposite side from (7, and that E is on CA produced through A ; the demonstration will only have to be slightly modified for any other figure.] A circle will go round PEAF {Note on III. 22) ; there- fore the angle PFE is equal to the angle PAE (III. 21) But the angles PAE and PAC are together equal to two right angles (I. 13); and the angles PAG and PBC are together equal to. two right angles (III. 22). Therefore the angle PAE is equal to the angle PBC\ therefore the angle PFE is equal to the aiigle PBG. Again, a circle will go round PFDB {Note on III. 21) ; therefore the angles PFD and PBD are togetliier equal to two right angles (III. 22). But the angle PBD has been shewn equal to the angle PFE. TTierefore the angles PFD m^ PFE are together equal to two right andes, Therefore EFim^ FD are in the same straight line. APPENDIX, 315 CB qf the he drawn tersection 3 circum- raw PD, 80. ABC 18 a triangle^ and O it the point qf inter- section of the perpendiculars from A, B, on the opposite sides of the triangle: the circle which passes through the middle points of A, OB, 00 will pass through the feet of the perpendiculars and through the middle points qf the sides qfthe triangle. Let Z), E, F be the middle points of OA, OB, 00 respectively ; let G be the foot of the perpendicular from A on BC, and H the middle point of BC, Jrespec- j ABf on )roduced ) slightly ); there- in. 21). 1 to two '^(7 are herefore )fore the [II. 21) ; er equal BJD has le angles ) angles, llien OBG is a right-angled triangle and E is the middle pomt of the hypotem\se OB ; therefore EG is equal to EO; therefore the angis EGO is equal to the angle EOG. Similarly, the m^lQ EGO is equal to the angle FOG. Therefore the angle FGE is equal to the angle FOE. But the angles FOE and BAG are together equal to two right angles; therefore the angles FGE and BAO are together equal to two right angles. And the angle BAG is equal to the angle EDF, because ED, DF are parallel to B A, AC (VI. 2). Therefore the angles FGE and EDF are together equal to two right angles. Hence G is on the circumference of the circle which passes through D, E, F /TIT- J- TTT «r^S \U.\iJi& Oi6 iii. ZZy. Again, FH is parallel to OB, and EH parallel to 00; therefore the angle EHF is equal to the angle EGF> Therefore ^is also on the circumference of the circle. 316 APPENDIX. Similarly, the two points in each of the other dides of the trianjBfle ABC may be shewn to be on the circum- ference of the circle. The circle which is thus shewn to pass through these nine points may bo called the Nine points circle: it has some curious proi>erties, of which we will now give two. The radius of the Nine points circle is half <\f the radius of the circle described round the original triangle. For the triangle DBF has its sides respectively halves of the sides of the triangle ABG^ so that the triangles are similar. Hence the radius of the circle describecT round DEF is half of the radius of the circle described round ABC. ^ If ^ he the centre of the circle described round the triangle ABC, the centre of the Nine points circle ii the middle point cf SO. Jt)r HS is at right angles to BG, and therefore parallel to GO. Hence the straight line which bisects HG at right angles must bisect SO. And H and G are on the circum- ference of the Nine points circle, so that the straight line which bisects HG at right angles must pass through the centre of the Nine points circle. Similarly, from the other sides of the triangle ABC two other straight lines can be obtained, which pass through the centre of the Nine points circle and also bisect SO. Hence the centre of the Nine points circle must coincide with the middle point of SO. We may state that the Nine points circle of any triangle touches the inscribed circle and the escribed circles of Uie triangle : a demonstration of this theorem will be found in the Plane Trigonometry, Chapter xxiv. For the history of the theorem see the Nouvelks Annales de Mathema- ttques for 1863, page 562.*. 31. If two straight lines bisecting two angles of a tri- angle and terminated at the opposite sides be equal t/ie oiseeted angles shall be equal. Let ABC be a triangle; let the straight line BD bisect tHe angle at B, and be terminated at the side ^C; and let the straight line CE biaent tho murle a+. C oti/1 k^ +..^_ mmated at the side AB ; and let the Straight line ^ibe equal to the straight Ime CE\ then the angle at B shall bo equal to the angle at C, APPENDIX. 317 sides of circum- g^h these 3: it has two. f of the riangle, ly halves igles are d round d round und the 'le u the parallel at right circum- ght line ugh the le other can be e points le Nino SO. triangle of the ) found history ithema' fa tri" laly tJie ) bisect C; and BDhe hall be For, let BD and CE meet at ; then if the angle OBC be not equal to the angle OCB^ one of them must bo greater than the other ; let the angle OBC be the greater. Then, because GB and BJD are equal to BC and CJE, each to each; but the angle CBD is greater than the angle BCE; therefore CD is greater than BE (I. 24). . On the other side of the base BC make the triangle BCF equal to the triangle CBE, so that BF may be equal to CE, and CF equal to BE (I. 22) ; and join DF. Then because BF is equal to BD^ the angle BFD is equal to the angle BDF, And the angle OCD is, by hy- pothesis, less than the angle OBE ; and the angle COD is equal to the angle BOE; therefore the angle ODC is greater than the angle OEB (I. 32), and therefore th© angle ODC is greater than the angle BFC. Hence, by taking away the equal angles BDF and BFDy the angle FDC is greater than the angle DFC; and therefore OF is greater than CD (I. 19) ; therefore BE is greater than CD. But it was shown that CD is greater than BE; which is absurd. ThAypfnre tb**- asi^les OBC and OCB are not uneQuaL. kud therefore tl that is, they are equal equal to the angle ACB. angle [For the history of this theorem see Lady'i and Gen- tleman* 9 Diary for 1859, page 88.] I I i; t 318 APPENDIX. 32 If a quadrilateral figure does not admit ofhavina fJZfj.'fr^''^ roMwc? e^, the sum of the rectangles col iztin'eViy%rz%:^^^ " ^^^^^^^ ^^^^ ^^ ^-^-^^^ o^n^f i/l^^^ ^® .* ,, shall 3 J)BC, hen the (VI. 4); fore the I to the DBA. B A3D CJS &s DA is ith the E, BD I to the id J?C. (I. 20); «*A/tf._ ?. 33. jfy* ^A^ rectangle contained by the diagonals qf a quadrilateral be equal to the mm of the rectangles con- tained by the opposite sides, a circle can be described round the quadrilateral. This is the converse of VI. i>; it can be demonstrated indirectly with the aid of 32. 34. It is required to find a point in a given straight line, such thai the rectangle contained by its distances from two given points in the straight line may be equal to ths rectangle contained by its distances from two other given points in the straight line. Let A, B, Of D be four given points in the same straight line: it is required to find a point in the straight line, such that the rectangle contained by its distances from A and B may be equal to the rectangle contamed by its distances from G and D, On AD describe any triangle AED ; and on CB de- scribe a similar triangle CFB, so that CF is parallel to AE, and BFio DE\ Join EF, and let it meet the given For, CE is to OA as OF is to 00 (VI. 4) ; therefore OE is to OF as OA is to 00 (V. 16). Similarly OE is to Oi?^ as OD is to OB. Therefore OA is to OC as Oi> is to OB (V. 11). Therefore the rectangle OA, OB is equal to the rectangle 00, OD. 320 APPENDIX. The figure will vary slightly according to the situation 9f the four given points, but corresponding to an assigned situation there will be only one point such as is reqmred. For suppose there could be such a point P, besides the point which is determined by the construction given above ; and that the points are in the order A, G, 2), B^ O. P. Join PE, and let it meet CF, produced at G ; join BG. Then the rectangle PA, PB is, by hypothesis, equal to the rectangle PC, PD ; and therefore PA is to PC as PD is to PB. But PA is to PC as PE is to PG (VI. 2) ; there- fore PD is to P^ as PE is to P^ (V. 1 1) ; therefore BG is parallel to DE. But, by the construction, BF is parallel to ED\ there- fore BG and BF are themselves parallel (I. 30) ; which is absurd. Therefore P is not such a point as is required. « ON GEOMETRICAL ANALYSIS. 35. The substantives ancdysis and synthesis, and the corresponding adjectives analytical and syntJietical, are of frequent occurrence in mathematics. In general analysis means decomposition, or the separating a whole into its parts, and synthesis means composition, or making a whole out of its parts. In Geometry however these words are used in a more special sense. In synthesis we begin with results already established, and end with some new result; thus, by the aid of theorems already demonstrated, and problems already solved, we demonstrate some new theo- rena, or solve some new problem. In analysis we begin witn assuming the truth of some theorem or the solution of some problem, and we deduce from the assumption con- sequences which we can compare with results already esta- blished, and thus test the validity of our assumption. 36. The propositions in Euclid's Elements are all ex- hibited synthetically ; the student is only employed m ex- amining the soimdness of the reasoning by which each successive addition is made to the collection of geometrical truths already obtained ; and there is no hint given as to the mfl.nner in which ^h^^ -nrv^rknaifir^na xxTt^wo. <^ma^»in1K* Aia covered. Some of the constructions and demonstrations appear rather artificial, and we are thus naturally mduced to enquire whether any -rules can be discovered by which we may be guided easily and naturally to the investigation , of new propositions. 3 situation 1 assi^ed \ required, esides the tion given ; join BG. [ual to the asP2)is 2) ; there- refore BQ D\ there- ; which is luired. I. > and the 'ol, are of 1 analysis Q into its g a whole vords are egin with }w result; Eited, and lew theo- we begin olution of ition con- )ady esta- 3n. •e all ex- ed in ex- ich each ^metrical '^en as to APPENDIX. 321 v4X£3** istrations induced by which stigatiou 37. Geometrical analysis has sometimes been described in language which might lead to the expectation that directions could be given which would^ enable a student to proceed to the demonstration of any proposed theorem, or the solution of any proposed problem, with confidence of success ; but no such directions can be given. Wo will state the exact extent of these directions. Suppose that a new theorem is proposed for investigation, or a new problem for trial Assume the truth of the theorem or the solution of the problem, and deduce consequences from this assumption combined with results which have been already established. If a consequence can be deduced which contradicts some result already established, this amounts to a demonstration that our assumption is inad- missible ; that is, the theorem is not true, or the problem cannot be solved. If a consequence can be deduced which coincides with some result already established, we cannot say that the assumption is inadmissible ; and it may happen that by starting from the consequence which we deduced, and retracing our steps, we can succeed in giving a syn- thetical demonstration of the theorem, or solution of the problem. These directions however are very vague, be cause no certain rule can be prescribed by which we are to combine pur assumption with results already established ; and moreover no test exists by which we can ascertain whether a valid consequence which we have drawn from an assumption will enable us to establish the assumption itself. That a proposition may be false and yet furnish consequences which are true, can be seen from a simple example. Suppose a theorem were proposed for investi- gation in the following words ; one angle of a triangle is to another as the side opposite to the first angle is to the side opposite to the other. If this be assumed to be true we can immediately deduce Euclid's result in I. 19 ; but from Euclid's result in I. 19 we cannot retrace our steps and establish the proposed theorem, and in fact the proposed theorem is false. Thus the only definite statement in the directions can be deduced from an assumed proposition which con- tradicts a result already established, that assumed propo^ sition must be false, 21 322 APPENDIX. 38. We may mention, in particular, that a consequence would contradict results already established, if we could shew that it would lead to the solution of a problem already given up as impossible. There are three famous problems which are now admitted to be beyond the power of Geometry; namely, to find a straight line equal in length to the circumference of a given circle, to trisect any given angle, and to find two mean proportionals between two given straight lines. The grounds on which the geometrical solution of these problems is admitted to be impossible cannot be explained without a knowledge of the higher parts of mathematics ; the student of the Elements may however be content with the fact that innumerable attempts have been made to obtain solutions, and that these attempts have been made in vain. The first of these problems is usually referred to as tlie Quadrature of the Circle. For the history of it the student should consult the article in the English Cyclo- pcBdia under that head, and also a series of papers in the Athenceum for 1863 and subsequent years, entitled a Bydget of Paradoxes y by Professor De Morgan. For approximate solutions of the problem we may refer to Davies's edition of Hutton's Course of Mathe- matics, Vol. I. page 400, the Lady's and Gentleman^s Diary for 1866, page 86, and the Philosophical Magazine for April, 1862. The third of the three problems is often referred to as the Duplication of the Cube. See the note on VI. 13 in Lardner's Euclid^ and a dissertation by C. H. Biering en- titled Historia Problematis Culn Duplicandi..JE.9imi\dd, 1844. We will now give some examples of Geometrical cuia- lysis. 39. From two givers points it is required to draw to the same point in a given straight line, two straight lines equally inclined to the given straight line. Let A and B be the given points, and CD the given straight line. Suppose AE and EB to be the two straight lines equally inclined to CD. Draw BF perpendicular to CD. and produce AE and BF to meet at Q. Then the an APPENDIX. 323 msequence • we could a problem *ee famous the power u in length ; any given itween two feometrioal impossible the higher ments may e attempts (e attempts >rred to as y of it the lish Cyclo- pers in the entitled a m we may of Maths- rentleman^s Magazine Drred to as n VI. 13 in Biering en- ^..Haunise, 3trical ana* to draw to aight linei > the given aight lines lar to CD, i the angle ^£u j^^^ ^ *?? ^^'® ^^^^ ^y hypothesis ; and the angle A£a is equal to the angle Dm (L15). Hence the triangles BEF and GUF are equal in aU respects (I. 26); therefore FG is equal to FB. ^ ^ ' * r..!^!^ ^T^^ ^^^^l ^^^ ^® ^^y synthetically solve the Ho r;n?hT/.-^p"T"^^'^r ^ ^^^ ^^ produce In^ jA -11 w^^x^SV^® equal to FB-, then join AG, and ^ ^ wiU intersect CD at the required i)oint. ...rfihnT^h ^''^'^ "^ gif^en straight line into two parts Tli* ' Merence of the squares on the parts may he equal to a given square. Let^^ be the given straight hne, and suppose C the required point. ^ K. G 5 Then the difference of the KSl ""^'/^^ ^^ ??. ^« ^ ^^ ^^^^1 *o a given square. But the difference of the squares on AG md BC hlqvli iLtlr^'i^^^^ .^''''^'''^^^y *^^^*^ «^^ and difference; Wherefore this rectangle must be equal to the given square ?rir^ ^^' *^? ^^"^^^"^ synthetical solutSn oHs fhlAiff * ^^^^a^gle equal to the given square (1. 46); then of tht^!;r^^^ ^^ ^^ ^^ ^" b^ Val to the ride of the i^ctangle adjacent to AB, and is therefore known. ^ kno^"^ of -^^and CB is known. Thus AC and CB «:,i* il^^^l^. that the given square must not exceed the -iuarw vu ^x/, m order that the problem may be possible7 of thffwn^? ^"^^ positions of (7, if it is not specified which othlr^ w ^^'"^"ts AC and CB is to be grater than th© other; but only one position, if it is specified. ! 21—2 m APPENDIX. In like manner we may solve the problem, to produce a given straight line so that the square on the whole straight line made up of the given straight line and the part produced, may exceed the sqwwe on the part pro- ducediyy a given square, which is not less than the square on the given straight line. Tlie two problems may be combined in one enunciation thus, to divide a given straight line internally or exter- nally so that the difference of the squares on the segments may he equal to a given square, ^ 41. To find a point in the circumference of a given segment of a circle, so that the straight lines which join the point to the extremities of the straight line on which the segment stands may be together equal to a given straight line. Let ACBhe the circumference of the given segment, and suppose G the required point, so that the sum of AC and CB is equal to a given straight Ime. , ^ ^^ Produce ^(7 to i> so that CD may be equal to CB; and join DB, '' Then AD is equal to the given straight line. And the ande ACB is equal to -the sum of the angles CDBmd CBD (I. 32), that is, to twice the angle CDB (1. 5). There- fore the angle ADB is half of the angle in the given seg- ment Hence we have the following synthetical solution. Describe on AB a segment of a circle contammg^an angle equal to half the angle in the given segment, vv liu A as centre, and a radius equal to the given straight line. de8cri()e a circle. Join A with a point of intersection of this cffcle and the segment which has been described; tins APPENDIX. 325 ) produce he whole ? and the part pro- he square lunciation or exter- i segments if a given 7hich join on which ) a given Q segment, ;um of AG lal to CB) I. And the CDB and 5). There- f'ven seg- solution. \g an angle With A So raight line, 5rsection of 3ribed;this joining straight line will cut the circamference of the given segment at a point which solves the problem. The given straight line must exceed AB and it must not exceed a certain straight line which we will now deter- mine. Suppose the circumference of the given segment bisected at E : join AE, and produce it to meet the cir- cumference of the described segment at F, Then AE is equal to EB (III. 28), and EB is equal to EF for the same reason that CB is equal to CD. Thus EA, EBy EF are all equal ; and therefore E is the centre of the circle of which ADB is a segment ^III. 9). Hence AF is the longest straight line which can be drawn from A to the cir- cumference of the described segment; so that the given straight line must not exceed twice AE» 42. To describe an isosceles triangle having each of the angles at the hose double of the third angle. This problem is solved in IV. 10 ; we may suppose the solution to have been discovered by such an analysis as the following. Suppose the triangle ABD such a triangle as is required, so that each of the angles at B and D is double of the angle at A* Bisect the angle at D by the straight line DC. Then the angle ADC is equal to the angle at A ; therefore CA is equal to CD. The angle CBD is equal to the angle ADB, by hypothesis ; the angle CDB is equal to the angle at A ; therefore the third angle BCD is equal to the third angle ABD (I. 32). Therefore BD is equal to CD (I. 6) ; and therefore BD is equal to ^(7. Since the angle BDC is equal to the angle at A^ the straight line BD will touch at D the circle described round the triangle ACD {Note ow III. 32). Therefore the rectangle AB, BC is equal to the square on BD (III. 36). Therefore the rectangle AB^ BG is equal to the square ou^C/. Therefore AB is divided at C in the manner required in II. 11. , Hence the synthetical solution of the problem is evident. 326 APPENDIX, 43. To inscribe a square in a given triangle, Lei ABC he tho given triangle, and suppose DMfG the required square. Draw AH perpen- dicular to BC, and ^JT parallel to 5C; and let J5^ produc- ed meet AK at K. Then BG is to GF qaBA isto^JT, and BG is to GDs^aBA isto-4Jy(VI. 4). But GFiB equal to GD, by hypothesis. Therefore J5^ is to ^iT as BA is to Aff (V. 7, V. 11). Therefore AH is equal to AK (V. 7). Hence we have the following synthetical solution. Draw Alt parallel to BCj and equal to AH; and join Bl^, Then BIC meets ^(7 at one of the comers of the required square, and the solution can be completed. 44.^ Through a given point between two given straight lines, it is required to draw a straight line, sue that the rectangle contained by the parts between the given point and the given straight lines may be equal to a given r xtangle. Let P be the given point, and AB and AC the given straight lines ; suppose MPN the required straight line, so that the rectangle MP, PN is equal to a given rectangle. rroduce AP to Q, so that the rectangle AP, P§ may be equal to the given rect- angle. Then the rectangle JfP, PN is equal to ttie rectangle AP, PQ. Therefore a circle will go round AMQN {Note on III. 35). Therefore the angle PNQ is equal to the angle PAM (III. 21). Hence we have the following syntheticalsolution. Pro- duce AP to ^, so that the rectangle -4i^, jTC^ may be equal to the given rectangle; describe on PQ a segment of a circle containing an angle equal to the angle PAM) join P with a point of intersection of this circle and AC; the straight line thus drawn solves the problem. APPENDIX. 327 ^ 46. In a given circle it is required to inscribe a tri- angle so that two sides map pass through two given points, and the third side be parallel to a given straight line, C ; D /(VI.4). r, V. 11). )n. Draw r. Then d square, I straight that the ooint and xtangle. B ^o round PNQk on, Pro- ] may be segment le PAM; smdAC', Let A and B be the given points, and CD the given straight hne. Suppose PMN to be the required triangle inscribed in the given circle. Draw iV;^ parallel to ^J5; join EM, and produce it if necessary to meet AB at J?*. If the point F were known the problem might be con- sidered solved. For ENM is a known angle, and therefore the chord EM is known in magnitude. And then, since F is a known point, and EM is a known magnitude^ the posi- tion of M becomes known. We have then only to shew how F is to be determined. The angle MEN is equal to the angle MFA (I. 29). The angle MEN is equal to the angle MPNJlll. 21). Hence MAF and BAP are similar triangles ( Vl. 4). Therefore "n/r A i^ x^ A r:i ^^ T> A i^ 4-^ A T> rnu^^^^^^^ 4-Uo *uxA4-n«.^lA ■^'^^JL AtJ V\> ^M.J,,' C«k» J-r^JL AM W ^JLJi. • ^AA^A^A^^A^^ %t»*\J MA, AP is e^ual to the rectangle BA, AF (VI. 16). But since ^ is a given point the rectangle MA, AP is known j and AB is known; thus AF is determined. 3^8 APPENDIX. 46 In a gioen circle it is required to inscribe a tri^ angle so that the sides may pass through three given points. Let Ay B, Che the three given points. Suppose PMN to be the required triangle inscribed in the given circle. — Draw NE parallel to ABj and determine the pomt i^ as in the preceding problem. We shall then have to de- scribe in the given circle a triangle EMN so that two of its sides may pass through given points, -F and ^, andtho third side be parallel to a given straight line AB, ini3 can be done by the preceding problem. This example and the preceding are taken from the work of Catalan already cited. The present problem is sometimes called Castillon's and sometimes Cramers; the history of the general researches to which it has given rise will be found in a series of papers in the Mathematunan, Vol. III. by the late T. S. Davies. r\Tcr T r\nr 47. A loms consists of all the points which satisfy cer- tain conditions and of those points alone. Thus, for exam- ple, the locus of the pomts which are at a given distance APPENDIX. 329 he a tri' 'ee given %qPMN circle. e point F lvo to de- lat two of V, and the iB, This from the >roblem is Yier's; the given rise dmaticiafii satisfy cer- for exam- (n distance from a given point is the surface of the sphere described from the given point as centre, with the given distance as radius; for all the points on this surface, and no other points, are at the given distance from the given point If we restrict ourselves to all the points in a fixed plane which are at a given distance from a given point, the focus is the circumference of the circle described from the given point as centre, with the given distance as radius. In future wo shall restrict ourselves to loci which are situated in a fixed plane, and which are properly called plane loci. Several of the propositions in Euclid furnish good exam- pies of loci. Thus the locus of the vertices of all triangles which are on the same base and on the same side of it, and which have the same area, is a straight line parallel to the base ; this is shewn in I. 37 and I. 39. Again, the locus of the vertices of all triangles which are on the same base and on thd same side of it, and which have the same vertical angle, is a segment of a circle de- scribed on the base ; for it is shewn in III. 21, that all the points thus determined satisfy the assigned conditions, and it is easily shewn that no other points do. "We will now give some examples. In each example we ought to shew not only that all the points which we indi- cate as the locus do fulfil the assigned conditions, but that no other points do. This second part however we leave to the student in all the examples except the last two; in these, which are more difficult, we have given the complete investigation. 48. Required the locus of 'points which are equidis- tant from two given points. Let A and B be the two given points; join AB-, and draw a straight line through the middle point of AB at right angles to -4-5 ; then it may be easily shewn that this straight line is the required locus. 49. Required the locus of the vertices of all triangles on a given lose AB, such that the square on the side ter- minated at A may exceed ilie square on t/ie side iermi- nated at B, by a given sqtmre. Suppose C to denote a point on the required locus ; from C draw a perpendicular on the given base, meeting it, pro- mi 390 APPENDIX, dnced if necessary, at D, Then the square on AC \& equal to the squares on AD and CD, and the square on BC is equal to the squares on BD and CD (I. 47) ; therefore the square on AC exceeds the square on BC by as much as the square on AD exceeds the square on BD. Hence i> is a fixed point either in AB or in AB produced through B, (40). And tne required locus is the straight line drawn thi;ough D, at right angles to AB, 60. Required the locm qf a point such that the straight lines draton from it to touch two given circles may he equal. Let A be the centre of the greater circle, B the centre of a smaller circle ; and let P denote any pomt on the re- quired locus. Since the straight lines drawn from P to touch the given circles are equal, the squares on these str£light lines are equal. But the squares on PA and PB exceed these equal squares by the squares on the radii of the respective circles. Hence the square on PA exceeds the square on PB, by a known square, namely a square equal to the excess of the squard on the radius of the circle of which A ie the centre over the square on the radius of the circle of which B is the centre. Hence^ the required locus is a certain straight line which is at right angles to AB (49). This straight line is called the radical axis of the two circles. If the given circles intersect, it follows from III. 36, that the straight line which is the locus coincides with the produced parts of the common chord of the two circles. 61. Required the locus qf the middle points qf alt the chords of a circle which pass through a fixed point. Let A be the centre of the given circle; B the fixed APPENDIX. 331 8 equal BC'm ore the i as the ) Dis a i?.(40). hi^ougn traight may he ) centre the re- n P to Q these indP^ radii of exceeds square le circle adius of equired ogles to the two III. 36, vrith the cles. f qf all mint. ho fixed point; let any chord of the circle be drawn so that, pro- duced if necessary, it may pass through B. Let P be the middle point of this chord, so that P is a point on the re- quired locus. The straight line AP is at right angles to the chord of which P is the middle pomt (III. 3) ; therefore P is on the circumference of a circle of which AB is a diameter. Hence if .S be within the given circle the locus is the dr- cumfcrence of the circle described on AB as diameter ; if B be without the given circle the locus is that part of the circumference of the circle described on AB as diameter, which is within the given circle. 62. is a Jixed point from which any straight line is drawn meeting a Jixed straight line at P ; in OP a point Q is taken such that OQ is to OP in a Jixed ratio: determine the locus of(^. We shall show that the locus of Q is a strdght line. For draw a perpendicular from O on the fixed straight line, meeting it at C7 ; in OC take a point Z> such that OD is to OG in the fixed ratio ; draw from any straight line OP meeting the fixed straight line at P, and in OP take a point Q such that OQ is to OP in the fixed ratio; jom B QD. The triangles ODQ and OOP are similar (VI. 6); therefore the angle ODQ is equal to the angle OOP, and is therefore a right angle. Hence Q lies in the straight line drawn through D at right angles to OD, MP 332 APPENDIX. 63. U a fixed point from which any straight line is drawn meeting the circumference of a fixed circle at P ; in OP a point Q is taken such that OQ w ifo OP *w a fixed ratio: determine the locals of Q,, We shall shew that the locus is the circumference of a circle. For let C be the centre of the fixed circle; in OC take a point D such that OD is to OG in the fixed ratio, and draw any radius CP of the fixed circle ; di-aw DQ .parallel to CP meeting OP, produced if necessary, at Q, Then the triangles OOP and ODQ are similar (VI. 4), and therefore OQ is to OP as OD is to 00, that is, in the fixed ratio. Therefore Q is a point on the locus. And DQ is to CP in the fixed ratio, so that DQ is of constant length. Hence the locus is the circumference of a circle of which D is the centre. 64. There are four given points A, B, C, D in a straight line; required the locus qf a paint at which AB and CD subtend equal angles. Find a point in the straight line, such that the rect- angle OA, OD may be equal to the rectangle OB^ OC (34), and take OK such that the square on OK may be equal to either of these rectangles (II. 14) : the circumference of the circle described from as centre, with radius OK^ shall be the required locus, [We will take the case in which the points are in the followmg order, O, A, B, C, D.] For let P be any point on the circumference of this APPENDIX. 333 raight line ircle at P ; in a fixed irence of a ^ in OC take ratio, and ^Q .parallel Then the I therefore fixed ratio. ) is to CP bh. Hence ih i> is the ), D in a \chich AB t the rect- % 00 (34), )e equal to jnce of the Ky shall be are in the ice of this circle. Describe a circle round PADy and also a circle round PBO\ then OP touches each of these circles (III. 37) ; therefore the angle OP A is equal to the angle PDA, and the angle OPB is equal to the angle PGB (III. 32). But the angle OPB is equal to the angles OP A and APB together, and the angle PGB is equal to the angles CPD and PDA together (I. 32). Therefore the angles OP A and APB together are equal to the angles CPD and PDA together; and the angle OP A has been shewn equal to the angle PDA ; therefore the angle APB is equal to the angle CPD, "We have thus shewn that any point on the circumference of the circle satisfies the assigned conditions; we shall now shew that any point which satisfies the assigned conditions is on the circumference of the circle. For take any point Q, which satisfies the required con- ditions. Descnbe a circle round QAD, and also a circle round QBG. These circles will touch the same straight line at Q\ for the angles AQB and GQD are e^ual, and the converse of III. 32 is true. Let this straight line which touches both circles at Q be drawn ; and let it meet the straight line containing the four given points at B. Then the rectanrie EA^, BD is equal to the rectangle RB, EC; for each is'equal to the square on BQ (III. 36). Therefore B must coincide with (34) ; and therefore BQ must be equal to OK. Thus Q must be on the circumference of tlie circle of which is the centre, and OK the radius. .*' I 334 APPENDIX, 56. Required the locus of the vertices of all the tri- angles ABU 7JDhich stand on a given hose AB, and have the side AC to the side BC in a constant ratio. If the sides AG and BG are to be equal, the locus is the straight line which bisects AB at right angles. We will suppose that the ratio is greater than a ratio of equal- ity; so that ^(7 is to be the greater side. Divide AB at D so that AD is to DB in the given ratio (VI. 10) ; and produce AB to E, so that AE is to EB in the given ratio. Let P be any point in the required locus ; join PD and PE. Then PD bisects the angle APB, and PE bisects the angle between BP and AP produced. Therefore the angle DPE is a right angle. Tlierefore P is on the ch-cumference of a circle described on DE as dia- meter. We have thus shewn that any point which satisfies the assigned conditions is on the circumference of the circle described on DE as diameter ; we shall now shew that any point on the circumference of this circle satisfies the as- signed conditions. Let Q be any point on the cu*cumference of this circle, QA shall be to QB in iihe assigned ratio. For, take O the centre of the circle ; and join QO. Then, by construction, AE is to EB as AD is to DB, and therefore, alternately AE\% to AD as EB is to U±i\ therefore the sura of AE and AD is to their difference as the sum of EB and DB is to their difference (23) ; that is, twice ^O is to twice DO as twice DO is to twice BO ; therefore ^0 is to />0 as DO is APPENDIX. 335 f the tri- ind have I locus is S to B0\ that is, AO is to OQ as QO is to 05. Therefore the triangles AOQ and QOB are similar triangles (VI. 6); and therefore ^^ is to Qi? as QO is to J50. This shews that the ratio of -4 Q to BQ is constant; we have still to shew that this ratio is the same as the assigned ratio. We have already shewn that ^O is to DO as DO is to BO] therefore, the difference oi AO and DO is to DO as the difference of DO and BO is to BO (V. 17); that is, AD is to DO as BD is to BO ; therefore ^i) is to BD as i)0 is to B0\ that is, AD is to 2>5 as QO is to jBO. This shews that the ratio of QO to BO is the same as the assigned ratio. fles. We of equal- Lven ratio bo EB in •ed locus ; [PB, and produced, jfore P is E as dia- bisfies the the circle r that any IS the as- his circle, ike Othe istruction, ternatelyj^ ra of AE nd DB is ice DO as > as DO is ON MODERN GEOMETRY. 66. We have hitherto restricted ourselvea'tO Euclid's ' Elements, and propositions which can be demonstrated by strict adherence to Euclid's methods. In modem times various other methods have been introduced, and have led to numerous and important results. These methods may be called semi-geometrical, as they are not confined within the limits of the ancient pure geometry; in fact the power of the modem methods is obtained chiefly by combining arithmetic and algebra with geometry. The student who desires to cultivate this part of mathematics may consult Townsend's Chapters on the Modern Geo- metry of the Pointy Line, and Circle. Wo will give as specimens some important theorems, taken from what is called the theory of transversals. Any line, straight or curved, .which cuts a system of other lines is called a transversal ; in the examples which we shall give, the lines will be straight lines, and the sys- tem will consist of three straight lines forming a triangle. We will give a brief enunciation of the theorem which we are about to prove, for the sake of assisting the memory in retaining the result ; but the enunciation will not be fully comprehended until the demonstration is completed.; mua m 336 APPENDIX. 57 If a straight line cut the sides, or the sides pro- duced, of a triangle, the product qf three segments %n order is equal to the product of the other three segments. Let ABC he & triangle, and let a straight line be drawn cutting the side BG at D, the side CA at ^, and the side AB produced through B at F, Then BD and DC are called segments of the side BC, and CE and EA are called ^gm^ntLi the side CA, and also AF and FB are called segments of the side AB. Through A draw a straight line parallel to BC, meeting , DF produced at H. Then the triangles CED and EAH are equiangiilar to one another; therefore AH la to CD as AE is to EC (VI. 4). Therefore the rectangle AH, EC is equal to the rectangle CD, AE {yi. 16). . Again, the triangles FA H and i^-52>are equiangular to one ai^other ; therefore AHisU>BD&s FA is to FB (VI. 4). Therefore the rectangle AH, FB is equal to the rectangle BD, FA (VI. 16). Now suppose the straight lines represented by numbers in the manner explained in the notes to the second Book of the Elements. We have then two results which we can ex- press arithmetically: namely, the product AH. EC is equiU to the product CD.AE; and the product AH.FBiseqmi to the product BD.FA. Therefore, by the principles of arithmetic, the product AHEC. BD.FA is equal to the ^^rodnct AH. FB.OJJ.AJij^ and therefore, by the "principles of arithmetic, the proaucw BD . CE. AFis equal to the product DC.EA. FB. This is the result intended by the enunciation dven above. Each product is made by three segments, one fi-orn APPENDIX, 337 ide9 pro- nents in be drawn . the side DC are are called are called \ meeting liar to one 1 (VI. 4). rectangle angular to W (VI. 4). ) rectangle y numbers id Book of we can ex- ?(7 is equal ''B is equal e product 3.CD.AE, iie prOuUcu tion given 3, one from every side of the triangle : and the two segments which ter- minated at any angular point of the triangle are ne ^er in Uie same product. Thus if we begin one i)roduct with the seg- ment BDf the other segment of the side J9(7, namely Du, occurs in the other product ; then the segment CE occurs in the £rst product, so that the two segments CD and CE, which ternmiafe at C, do not occur in the same product ; and so on« The student should for exercise draw another figure for the case in which the transversal meets all the sides prodticedj and obtain the same result. 68. Conversely, it may be shewn by an indirect proof that if the product BD.CE. AF be equal to the product DC.EA.FBj the three points />, E^ ^ lie iu the same straight line. . 59. J/ three straight lines be drawn through the angular points of a triangle to the opposite sides, and meet at the same point , the product of three segments in order is equal to the product of the other three segments. Let ABC be a triangle. From the angular points to the opposite sides let the straight lines AOD, BuE, GOF be (h-awn, which meet at the point O: the product ^i^.52>.(7^shall be equal to the product FB.DC.EA, For the triangle ABD is cut by the transversal FOG^ and therefore by the theorem in 67 the following products are equal, AF, BC, DO^ and FB, GD, OA. Again, the triangle ACD is cut by the transversal EOBj and therefore by the theorem in 67 the following products are equal, AO,DB. CE and OD. BG.EA. 22 338 APPENDIX. Therefore, by the principles oiT arithmetic, the following products are equal, AF. BC , DO . AO . DB . CE and FB.CD,OA.OD,BG. EA. Therefore the following products are equal, AF.BD.CE and. FB.DC.EA, We have supposed the point O to be within the tri^jgle ; iii O be without the triangle two of the points i>, E, F will fall on the sides produced. 60. Conversely, it may be shewn by an indirect proof that if the product AF. BD . CE be equal to the product FB.DC.EA, the three straight Unes AD, BE, CF meGi at the same point. ^ 61. We may remarlj that in geometrical problems the following terms sometimes occur, used in the same sense as in arithmetic ; namely arithmetical progression, geometri- cal progression, and harmonical progression. A proposi- tion respecting. harmonical progression, which deserves notice, will now be given. 62. Let ABC he a triangle; let the angle A he hisected hy a straight line which meets BC at D, and let the ex- terior angle at A he 'bisected hg a straight line whichmeets BC, produced through C,atE: then BD, BC, BE shall be in harmonical progression. following CE and following APPENDIX. 339 For BD is to DC as BA is to ^C (VI. 3) ; and BE is to EG &B B A is to AC (VI, A). Therefore BD is to DO as BE is to EC{V.l 1) Therefore BD is to 5J^ as DC is to £'(7(V. 16). ThuR of the three straight lines BD,BC, BEy the first is .o tun third as the excess of the second over the first is to the excess of the third over the second. Therefore BD, BC, hE are in harmonical progression. This result is sometimes expressed by saying that BE is divided harmonically at D and C, .EA. I triangle; , Ey F will rect proof le product , CjPmeet ►blems the le sense as geometri- L proposi^ L deseryes he "bisected let the ex- hich meets ; BE shall I: 22—^ EXERCISES IN EUCLID. I. 1 to 15. describe an isosceles each of the sides equal to a given On a given straight line tri- straight angle having ^""^2 In the figure of I. 2 if the diameter of the smaller circle is the radius of the larger, shew where the given ^fnt ^nd the vertex of the constructed tnangle will be situated. ^.^^^^ ^^^^ ^^^^^ at right Ml- gles! any ^nt irelther of them is equidistant from the '"'Ttl^e'^^^'^ABC and ACB '^ the base of^ isosceles triangle be bisected by the straight Imes BD, CD. shew that DBG will be an isosceles tnaggle. 'tBAGS^ a triangle h^.ving the angle B double of the angle A, If BD bisects tho angle B and meets AC at D, '""Z 'tX. t^fofVf -if FC and BG meet at H ''^iXZl^o'lriffFC and BG meet at H, shew that AH bisects the angle BAG. 8. The sides .4 J5, ^i>ofa quadrilateral ^5 CDm^ eaual and the diagonal ^C^^ bisects the angle BAD: shew ffthe sides CT IndGD are equal, and that the diagond >1C bisects the angle ^(7/>. , q ^6rj9 ADB are two triangles on the same side of ^i^^'such that AG is equal to ^A a^\ -^^ J^ X .^ ^a, Mid ^4/> and 5(7 intersect at 0: shew that the in- a*^nria /fi /I « is isosceles. " "lO iJhe opposite angles of a rhombus are equal. 11. A di£«on 1 of a rhombus bisects each of the angles tlu-ough which it passes. EXERCISES IN EUCLID. 341 ogceles tri- en straight bhe smaller I the given igle will be it right an- t from the base of an , lines JBI>f e. ouble of the tAAC&iD, meet at H meet at H, ABCD are BAD\ shew the diagonal same side of » is equal to that the tri- 12. If two isosceles triangles are on the same base the straight line joining their vertices, or that straight line produced, will bisect the base at right angles. 13. Find a point in a given straight line such that its distances from two given points may be equal. 14. Through two given points on opposite sides of a given straight line draw two straight lines which shall meet in that given straight line, and include an angle bisected by that given straight line. 15. A given angle BAC is bisected ; if GA is produced to G and the angle BAQ bisected, the two bisecting lines are at right angles. 16. If four straight lines meet at a point so that the opposite angles are equal, these straight lines are two and two in the same straight line. I. 16 to 26. 17. ABC is a triangle and the angle A is bisected by a straight line which meets BC at D ; shew that BA is greater than BD, and CA greater than CD. 18. In the figure of I. 17 shew that ABC and ACB are together less than two right angles, by joining A to any point in BC. 19. ABCD is a quadrilateral of which AD is the longest side and BC the shortest ; shew that the angle ABC\% greater than the angle ADC^ and the angle BCD greater wian the angle BAD, 20. If a straight line be drawn through A one of the angular points of a square, cutting one of the opposite sides, and meeting the other produced at F, shew that AF is greater than the diagonal of the square. 21. The perpendicular is the shortest straight lino that can be dravm from a given point to a given straight line; and of others, that which is nearer to the perpen- dicular is less than the more remote; and two, and only two, equal straight lines can be drawn from the given point to the given straight line, one on each side of the perpen- dicular. 22. The sum of the distances of any point from the three angles of a triangle is greater than half the sum of the sides of the triangle. 342 EXERCISES IN EUCLID. 23. The four sides of any quadrilateral are together arf>-,i^T than the two diagonals together. tM The two sides of a triangle are together greater thaa i dee the straight line drawn from the vertex to the middle point of the base. 26. If one angle of a triangle is equal to the sum of the other two, the triangle can be divided into two isosceles ■ triangles. , 4 26. li the angle C of a triangle is equal to the sum of the angles A and B, the side ^-5 is equal to twice the straight line joining (7 to the middle point of ^i?. f 27. Construct a triangle, having given the base, one of the angles at ttie base, and the sum of the sides. * 28. The perpendiculars let fall on two sides of a tri- angle from any point in the straight line bisecting the angle between them are equal to each other. ^ 29. In a given straight line find a i)oint such that the ^ perpendiculars drawn from it to two given straight lines fihall be equal. ..,,,. , - 30. Through a given point draw a straight Ime such f that the perpendiculars on it from two given points may be on opposite sides of it and equal to each other. 31. A straight line bisects the angle A of a triangle ABC; from B a perpendicular is drawn to this bisecting straight line, meeting it at />, and BD is produced to meet AC or AG produced at E : shew that BD is equal to DE. 32. ABj AG are any two straight lines meeting at A: through any point P draw a straight line meeting themat E and F, such that AE may be equal toAF, 33. Two right-angled triangles have their hypotenuses equal,' and a side of one equal to a side of the other : shew that they are equal i|i all respects. r. 27 to 31. 34. Any straight line parallel to the base of an iso- sceles triangle makes equal angles with the sides. 35. If two straight lines A and B are respectively parallel to two others G and />, shew that the incUnatiou oi A to Bis equal to that of C to i>. 36. A straight line is drawn terminated by two parallel Btraight lines; through its middle point any Lcraight line u JSXERGISES IN EUCLID, 343 together 5r greater 3X to the le sum of ) iifosceles i the sum twice the • ise, one of 3 of a tri- the angle h that the light lines line such its may be a triangle \ bisecting )d to meet al to DE. tiug at A : ;i\iQm.2XE fpotenuses ,her: show drawn and terminated by the parallel straight lines. Show that the second straight line is bisected at the middle point' of the first. 37. If through any point equidistant from two parallel straight lines, two straiglit lines be drawn cutting the pa- rallel straight lines, they will intercept equal portions of these parallel straight lines. 38. If the straight line bisecting the exterior angle of a triangle be parallel to the base^ shew that the triangle is isosceles. 39. Find a point B in a given straight line CD, such that if AB be drawn to B from a given point A, the angle ABC will be equal to a given angle. 40. If a straight line be drawn bisecting one of the angles of a triangle to meet the opposite side, the straiglit fi lines drawn from the point of section parallel to the other sides, and terminated by these sides, will be equal. 41. The side BC of a triangle ABC is produced to a point i>; the angle ACB is bisected by the straight lino CE which meets AB sit E. A straight line is drawn through E parallel to BC^ meeting AG at F^ and tho straight line bisecting the exterior angle ACD at Q. Shew that EF'\% equal to FG. 42. AB is the hypotenuse of a right-angled triangle /^ ABG\ find a point Z> m AB such that DB may be equal/ " to the perpendicular from D ovlAC. 43. ABC is an isosceles triangle : find points D^ Ein it the equal sides ABy AC such that BB^ BE, EC may air ' be equal. 44. A straight line drawn at right angles to BC the base of an isosceles triangle ABC cuts the side AB si D and GA produced at E-. shew that AED is an isosceles triangle. of an iso- espectiveljr clinatiou of ;wo parallel ightlineiii I. 32. ^ 45. From the extremities of the base of an isosceles triangle straight lines are drawn perpendicular to the sides ; shew that the angles made by them with the base are each equal to half the vertical angle. 46. On the sides of any triangle ABC equilateral tri- angles BCD, GAE, ABFaxe described, all external: shew that tiie straight lines AD, BE, CF are all equal. 344 FXERCrSES IN EUCLID. 47. What is the magnitude of an angle of a regufar octagon? 48. Through two given points draw two straight lines forming with a straight line given in position an equilateral triangle. 49. If the straight lines bisecting the angles at the base of an isosceles triangle be produced to meet, they will contain an angle equal to an exterior angle of the triangle. 60. A is the vertex of an isosceles triangle ABC, and BA is produced to Z>, so that AD h equal to BA ; and DC\s drawn : shew that BCD is a right angle. 61. ABC is a triangle, and the exterior angles at B and C are bisected by the straight lines BD, CD respec- tively, meeting at D : shew that the angle BDC together with half the angle BAC make up a right angle. , 62. Shew that any angle of a triangle is obtuse, right, or acute, according as it is greater ihan, equal to, or less than the other two angles of the triangle taken together. 53. Construct an isosceles triangle having the vertical angle four times each of the andes at the base. 64. In the triangle ABC the side BC is bisected at E and AB at G ; AE is produced to ^ so that EF is equal to AE. and CG is produced to -fiT so that GK is equal to CG : shew that FB and HB are in one straight line. 66. Construct an isosceles triangle which shall have one-third of each angle at the base equal to half the vertical angle. 56. ABj AC are two straight lines given in position: it is required to find in them two points P and Q, such that, PQ being joined, AP and PQ may together be equal to a given stt^ght line, and may contain an angle equal to a given angle. 67. Straight lines are drawn through ihe extremities of the base of an isosceles triangle, making angles with it on the side remote from the vertex, each equal to one-third of one of the equal angles of the triangle and meeting the sides produced: shew that three of the triangles thus formed are isosceles. * 68= AEB^ GED are two straisrht linfts intersectins' at E ; straight lines AC^ DB are drawn forming two triangles AGE, BED; the angles ACE, DBE are bisected by the straight lines CF, BF, meeting at F. Shew that the angle CFB is equal to half the gum of the angles EAC, EDB. i roguFar s^ht lines [uilateral )S at the they will iriangle. IBC, and 3A'f and les at B > respeo- together se, right, 0, or less 'ether. ) vertical jtud at .E ' is equal equal to ne. lall liare e vertical position: I Qy such be equal I equal to jmities of rith it on 6-third of eting the rles thus triangles ;d by the the angle EDB. EXERCISES IN EUCLID. 845 59. The straight line joining the middle point of the hypotenuse of a right-angled triangle to the right angle is equa' l-^ '^alf the hypotenuses 60. ' om the angle ^ of a triangle ABC a perpen- dic" ' .» ^awn to the opposite side, meeting it, produced if nocoas, '', at Z>; from the angle B a perpenclicular is dra^/n *■ pie opposite side, meeting it, produced if neces- sary, nt £1 : shew that the straight lines which join JD and E to the middle point oi AB are equal. 61. From the angles at the base of a triangle perpen- diculars are drawn to the opposite sides, produced if neces- sary : shew that tho straight line joining the points of inter- section will be bisected by a perpendicular drawn to it from the middle point of tho base. 62. In the figure of 1. 1, if C and H be tho points of intersection of the circles, and AB be produced to meet one of the circles at K^ shew that CHK is an equilateral triangle. 63. The straight lines bisecting the angles at the base of an isosceles triangle meet the sides at I) and E: shew that DE is parallel to the base. 64. AB^ -4 (7 are two given straight lines, and P is a given point in the former: it is required to draw through P a straight line to meet AC 9i Q^ so that the angle APQ may be three times the angle AQP. 65. Construct a right-angled triangle, having given the hypotenuse and the sum of the sides. 66. Construct a right-angled triangle, having given the hypotenuse and the difference of the sides. 67. Construct a right-angled triangle, having given the hypotenuse and the perpendicular from the right angle on it. 68. Construct a right-angled triangle, having given the perimeter and an angle. 69. Trisect a right angle. 70. Trisect a given finite straight line. 71. From a given point it is required to draw to two parallel straight lines, two equal straight lines at right angles to^each other. 72. Describe a triangle of given perimeter, having it? angles equal to those of a given triangle. 346 EXERCISES IN EUCLID, I. 33, 34. 73. If a quadrilateral have two of its opposite sides parallel, and the two others equal but not parallel, any two of its opposite angles are together equal to two right angles. . . 74. If a straight line which joins the extremities of two equal straight lines, not parallel, make the angles on the same side of it equal to each other, the straight line which joins the other extremities will be parallel to the first. 75. No two straight lines drawn from the extremities of the base of a triangle to the opposite sides can possibly bisect each other. 76. If the opposite sides of a quadrilateral are equal it is a parallelogram. , ., , i 77. If the opposite angles of a quadrilateral are equal it ife a parallelogram. 78. The diagonals of a parallelogram bisect each other. 79. If tha diagonals of a quadrilateral bisect each other it is a parallelogram. , ". 80. If the straight line joining two opposite angles of a parallelogram bisect the angles the four sides of the pa- rallelogram are equal. 81. Draw a straight line through a given point such that the part of it intercepted between two given parallel straight lines may bo of given length. 82. Straight lines bisecting two adjacent angles of a parallelogram intersect at right angles. 83. Straight lines bisecting two opposite angles of a parallelogram are either parallel or coincident. ' 84. If the diagonals of a parallelogram are equal all its angles are equal. , ^^ - „ 85. Find a point shch that the perpendiculars let fall from it on two given straight lines shall be respectively equal to two given straight lines. How many such points are there ? 86. It is required to draw a straight lino which shall be equal to one straight line and parallel to another, and be terminated by two given straight lines. 87. On the sides AB, BC, and CD of a parallelogram ABCD three equilateral triangles are described, that on jtiU fc<)war(l» toe same parts as ipnu puniiiuii/giiiiii, auu, iiiiu^u on ABf CD towards the opposite parts: shew that the EXERCISES IN EUCLID, 347 te sides any iwo JO right IS of two I on the le which rst. remities possibly equal it re equal ;h other, ch other mgles of the pa- int such parallel ;les of a jles of a lal all its \ let fall Dectively h points ich shall p, and be lelogram that on that the distances of the vertices of the triangles onAB, CD from that on BC are respectively equal to the two diagonals of the parallelogram. 88. If the angle between two adjacent sides of a paral- lelogram be increased, while their lengths do not alter, the diagonal through their point of intersection will diminish. 89. Ay Bf C are three points in a straight line, such that AB is equal to BC: shew that the sum of the perpen- diculars from A and C on any straight line which ddes not pass between A and C is double the perpendicular from B on the same straight line. 90. If straight lines be drawn from the angles of any parallelogram perpendicular to any straight line which is jutside the parallelogram, the sum of those from one pair of opposite angles is equal to the sum of those, from the other pair of opposite angles. 91. If a six-sided plane rectilineal figure have its op- posite sides equal and parallel, the three straight lines jom- mg the opposite angles will meet at a point. 92. ABj AC are two given straight lines; through a given point E between them it is required to draw a straight line GEH such that the intercepted portion GH shall be bisected at the point E. . „ , 93. Inscribe a rhombus within a given parallelogram, so that one of the angular points of the rhombus may be at a given point in a side of the parallelogram. 94. ABCD is a parallelogram, and E^ F, the middle points of AD and BC respectively ; shew that BE and DF will trisect the diagonal AC I. 35 to 45. 95. ABCD is a quadrilateral having BC parallel to AD ; shew that its area is the same as that of the parallelo- gram which can be formed by drawing through the middle point of DC a straight line parallel to AB, 96 ABCD is a quadrilateral having BC parallel to AD, E is the middle point of DC; shew that the triangle AEB is half the quadrilateral. ^ , , x. 97. Shew that any straight line passing through the «^:^;iia »/^{*l4■ rxf fVio rHoTne+Ar of a narallelocram and termi- nated by two opposite sides, bisects the parallelogram. 348 EXERCISES IN EUCLID. ii 98. Bisect a* parallelogram by a straight line drawn through a given point within it. / 99. Construct a rhombus equal to a given parallelo- 100. If two triangles have two sides of the one equal to two sides of the other, each to each, and the sum of the two angles contained by these sides equal to two right an- gles, the triangles are equal in area. 101 A straight line is drawn bisecting a parallelogram ABGD and meeting AD at E and BG at Fi shew that the triangles EBF and CED are equal. 102. Shew that the four triangles into which a paral- lelogram is divided by its diagonals are equal in area. 103 Two straight lines AB and CD intersect at E, and the triangle AEC is equal to the triangle BED: shew tha^i^ds parallel to ^i>. . . o • 104 ABCD is a parallelogram; from any point F in the diagonal BD the straight lines PA, PC are drawn. Shew that the triangles PAB and PCB are equal. 105 If a triangle is described having two of its sides equal to the diagonals of any quadrilateral, and the in- cluded angle equal to either of the angles between these diagonals, then the area of the triangle is equal to the area of the quadrilateral. - ^ ^f 106. The straight line which joins the middle points ot two sides of any triangle is parallel to the base. 107 Straight lines joining the middle points of ad- jacent sides of a quadrilateral form a parallelogram. 108 D Eare the middle points of the sides AB, AC of a triangle, and CD, BE intersect at -f: shew that the triangle BFC is equal to the quadrilateral ^/>^^. 109. The straight 4ine which bisects two sides of any triangle is half the base. 110 In the base AC ot ta tnangle take any point D ; bisect AD, DC, AB, BC at the points^, F, G,H respec- tively : shew that EG is equal and parallel to FH. 111. Given the middle points of the sides of a tnangle, construct the triangle. i. . . i 112 If the mi(? die points of any two sides of a tnangle be ioined, the triangle so cut off is one quarter of the whole. bisected at the points E, F-, a perpendicular is drawn from A to the opposite side, meeting it at D, Shew that the EXERCISES IN EUCLID, 349 drawn rallelo- 5 equal of the ^ht an- logram w that , paral- i. b at E, ': shew t P in drawn. :s sides the in- Q these he area oints of of ad- iB.AG hat the of any [)int D\ respec- Angle, triangle e whole. BG are Tin from jhat the angle FDE is equal to the angle JBAC Shew also that AFDE is half the triangle ABC 114. Two triangles of equal area stand on the same base and on opposite sides: shew that the straight line joining their yertices is bisected by the base or the base produced. 115. Three parallelograms which are equal in all re- spects are placed with their equal bases in the same straight line and contiguous ; the extremities of the base of the first are joined with the extremities of the side opposite to the l^se of the third, towards the same parts : shew that the portion of the new parallelogram cut off by the second is one half the area of any one of them. 116. ABGD is a parallelogram; from D draw any straight line DFG meeting BC at J" and AB produced at G\ draw AF and CG: shew that the triangles ABF, CFG are equal. . 117. ABCis a given triangle: construct a triangle^ of equal area, having for its base a given straight line AD, coinciding in position with AB. 118. ABC is a given triangle: construct a triangle of equal area, having its vertex at a given point in BC aud its base in the same straight line as AB. 119. ABGD is a given quadrilateral: construct ano- ther quadrilateral of equal area having AB for one side, and for another a straight line drawn through a given point in CD parallel to AB» 120. ABGD is a quadrilateral: construct a triangle whose base shail be in the same straight line asABj vertex at a £ ven point P in CD, and area equal to thn* of the j^ iven quadrilateral. 12? ABC is a given triangle: construct a t langle of eq>U£ area, having its base in the sam» strain ine as^-S, and its vertex in a given straight line pavali^ji to AB. 122. Bisect a given triangle by a straight line drawn through a given point in a side. 123. Bisect a given quacirilateral by a, straight line c>awn through a given angular point. 124. If through the point withm a parallelogram ABGD two straight -J les are drawn puii^ilel to the sides, .uid the Darallelograms OB aivX OD are equal, the point O la 111 wliO iiitil^OuiM. Ji.\J» m 350 EXERCISES IN EUCLID, I I. 46 to 48. 125 On the sides AG, EC of a tmngle^5(7, squares AGDE, BCFH axe described: shew that the straight '"^^e'^^Thf fq^rron^^^^ side subtending an acute an- gle of 'a trian^e is less than the squares on the sides rontaininsr the acute angle. ,. , , ^ containing ^^ ^ ^^^^^ ^.^^ subtending an obtuse an- gle of a triangle is greater than the squares on the sides containing the obtuse angle. +„-o„crl« bn leas 128 If the square on one side of a triangle bo less than the squares on the other two sides, the angle contained by Siese sides is an acute angle; if greater, an obtuse ^"^129 A straight line is drawn parallel to the hypotemn^e of a right-angled triangle, and each of the acute ajgles is joined with the points%here this straight line mters^ts the sides respectively opposite to them: shew that the squares on tfie joining straight lines are together equal to the square on the hypotenuse and the square on the straight line drawn parallel to it. j, n n t% *\.^ «« 130. If any point P be joined to A, B, C, D, the an» gular points of a rectangle, the squares on PA and PO are together equal to the squares on P^ ana ±^JJ. 131. In a right-angled triangle if the square on one of the sides containing the right angle be three times the square on the other; and from the right angle two straight lines be drawn, one to bisect the opposite side, and the other perpendicular to that side, these straight lines divide the right angle into three equal parts. , 132. If ABC be a triangle whose angle A is a right angle, and BE, GF be drawn bisecting the opposite sides respectively, shew that four times the sum of the squares on BE BXid GF is equal to five times the square on BG. 133. On the hypotenuse BG, and the sides C7/1, AB of a right-angled triangle ^56^, squares BDEG, AK ana AG\re described: shew that the squares on i>g^and J^Jf arc togeiJier equal to five times the square on i^O. /■' EXERCISES IN EUCLID, 351 squares straight ate an- te sides iuse an- le sides bo less »ntained obtuse )otein!iie ingles is itersects ;hat the equal to straight . the an- IPC are m one of imes the straight and the es divide 3 a right nte sides 5 squares nBC. A, AB of AF, and and EF > II. 1 to 11. V 134. A straight line is divided into two parts ; shew ^that if twice the rectangle of the parts is equal to the sum of the squares described on the parts, the straight line is ,^isected. . 135. Divide a given straight line into two parts such that the rectangle contained by them shall be the greatest possible. // 136. Construct a rectangle equal to the difference of two given squares. 137. Divide a given straight line into two parts such that the sum of the squares on the two parts may be the least possible. 138. Shew that the square on the sum of two straight lines together with the square on their difference is double the squares on the two straight lines. 139. Divide a given straight lino into two parts such that the sum of their squares shall be equal to a given square. 140. Divide a given straight line into tvfo parts such that the square on one of them may be double the square on the other. 141. In the figure of II. 11 if Cffhe produced to meet i?i^ at L, shew that CL is at right angles to BF. 142. In the figure of II. 11 if i?^ and CH meet at O, shew that AO is at right angles to CH. 143. Shew that in a straight line divided as in II. 11 the rectangle contained by the sum and difference of the parts is equal to the rectangle contained by the parts. II. 12 to 14. 144. The square on the base of an isosceles triangle is e^cial to twice the rectangle contained by either side and by the straight line intercepted between the perpendicular let fall on it from the opposite angle and the extremity of the base. 145. In any triangle the sum of the squares on the sides is equal to twice the square on half the base together ...:i.l. X :__ J.1 XT _i.»~J^UX 1^>^A A-mTknrn -PK/^rM fK/k vertex to the middle point of the base. 3^2 EXERCISES IN EUCLID. 1 ifi ABC is a triangle having the sides^^and AC f if ip is produced beyond the base to Dm that TA""- ' iLf/to/« shew that the square on CD is equal '^'^ f 4^' The sum of the squares on the sides of a paral- lelogSm I eqSd to the^ sum of the squares on the ^Tr^' The ba.e of a triangle is given and is bisected by the ctntre of a given fcle : if the vert^^^^^^^ of the circumference, shew that the sum ot tue squares uu the two sides of the triangle is invariable. ^.^^Q^als geiSof the dkmeters of a Parallelogram as a^n^e P"'"l52"'tn'^T"he'diameter of a circle take two points C and^^Slfyistant from^^^ cen^re^d rom any jKnnt ^^:JZ irS^'arXS Sal to the squares °" fas '"toic the br«e of a triangle to D such that IqS^rron'^Xof thS'gle by the rectangle coutW by the «e«n«"te °f fte b^. ^^^^^ ^„ ^^0 hypotenuse Br of k rtehSd triangle ^SC: shew that the squares o^DA ILHo are together equal to the squares on EA and AB. . . . ,_ •., —i,:^!, /^ ;a o *«iif. angle. 156. ABU is a triangie f J^'T'''/^ SerDenddculaf to and DE is drawn from a pomt D m AO perpenaicmar EXERCISES IN EUCLID. 353 md AG so that IS equal square a, paral- on the ected by ny pomt uares on Liagonals 8 on the sides, of inter- t centre, ^ht lines four an- teral arc s by four e middle \ points C any point that the 8 squares such that al to the AD will from the } than the contained lypotenuse lie squares res on EA !'«"ht Q-Tigle. Lidiculaf to AB: shew that the rectangle ABj AE is equal to the rectangle AG^ AD, 157. If a straight line be drawn through one of the angles of an enuilateral triangle to meet the opposite side produced, so that the rectangle contained by the whole straiffht line thus produced and the part of it produced is equal to the square on the side of the triangle, shew that the square on the straight line so drawn will oe double the square on a side of the triangle. 158. In a triangle whoso vertical angle is a right ande a straight line is drawn from the vertex perpendicular to the base : shew that the square on this perpendicular is equal to the rectangle contained by the segments of the base. 159. In a triangle whose vertical angle is a right angle a straight line is drawn from the vertex perpendicular to the base : shew that the square on either of the sides adja^ cent to the right angle is equal to the rectangle contained by the base and the segment of it adjacent to that side. 160. In a triangle ABC the angles B and G are acute : if E and F be the points where perpendiculars from the opposite angles meet the sides AG^ AB, shew that the square on BG is equal to the rectangle AB, BF, together with the rectangle AG, CE. 161. Divide a ^iven straight line into two parts so that the rectangle contamed by them may be equal to the square described on a given straight line which is less than hedf the straight line to be divided. III. 1 to 15. 162. Describe a circle with a given centre cutting a given circle at the extremities of a diameter. 163. Shew that the straight lines drawn at right anglcii to the sides of a quadrilateral inscribed in a Circle from their middle points intersect at a fixed point. 164. If two circles cut each (ither, any two parallel straight lines drawn through the points of section to cut the circles are equal. 165. Two circles whose centres are A and B intcisect *^^,; through G two chords DGE and FGG are drawn cMurtlly inclined to AB and terminated by the circles; shew that DE and FG are equal. . 354 EXERCISES IN EUCLID. I 166. Through either of the points of intersection of two given circles draw the greatest possible straight line" tei-minated both ways by the two circumferences. 167. If from any point in the diameter of a circle straight lines are drawn to the extremities of a parallel chord, the squares on these straight lines are together equal to the squares on the segments into which the diameter is divided. 168. A and B are two fixed points without a circle PQR ; it is required to find a point P in the circumfer- ence, so that the sum of the squares described on AP and BP may be the least possible. 169. If in any two given circles -which touch one an- other, there be drawn two parallel diameters, an extremity of each diameter, and the point of contact, shall lie in the same straight line. , 170. A circle is described on the radius of another circle as diameter, and two chords of the larger circle are drawn, one thi-ough the centre of the less at right angles to the common diameter, and the other at right angles to the first through the point where it cuts the less circle. Shew that these two chords have the segments of the one equal to the segments of the other, each to each. 171. Through a given point within a circle araw the shortest chord. 172. O is the centre of a circle, P is any point in its circumference, PN a perpendicular on a fixed diameter: shew that the straight line which bisects the angle OPN always passes through one or the other of two fixed points. 173. Three circles touch one another externally at the points A^ B^ C\ from A^ the straight lines AB, AC are produced to cut the circle BC at D and B: shew that BE is a diameter of BG, and is parallel to the straight line joining the centres of the other circles. 174. Circles are described on the sides of a quadri- lateral as diameters : stew that the common chord of any adjacent two is parallel to the common chord of the other two. 175. Describe a circle which shall touch a given circle, have its centre in a given straight line, and pass through a given point in the given straight line. srsection of traight lino' s. of a circle * a parallel ^ether equal diameter is )ut a circle I circumfer- on AP and ich one an- I extremity II lie in the of another 5r circle are bt angles to iglesto the rcle. Shew 3 one equal e araw the )oint in its . diameter: mgle OPN xed points. Qally at the By AG are >w that DE traight line f a quadri- lord of any f the other ^ven circle, s through a EXERCISES IN EUCLID, III. 16 to 19. ' 353 t.J^i' ^^^ *x** *Y° tangents can be drawn to a circle length *''''" ™' P"'"** "^ **' *''«y "^o «f «q"»l line\'Jw''chTS'clrcr ^''" "'"^^^ "»« " "''^S''^ BtraStlinoXtJXtvt'cir^^lo'' ^"" *"^«" «'«' ^ a ^ 3/f^^%&nt°Vrat^rrt:Vet= ference shall be of given length. circum- 1 ^?^' J^T^^ circles have the same centre- show ^ht^k oil cho^ of the outer circle which toucHhe toer drcle a^ *i ^^^'.L .^^^^^^Ji a given point draw a straight line so thif the part mtercepted by the circumference of a ^vLdrl diameter""^ ^ ^"'^ '*'"^^^* "'^^ "^* greatir tha^ the 182. Two tangents are drawn to a circle at the oddo- site extremities of a diameter, and cut off from a ffi tangent a portion ^^: if C' be the centre of the cMe shew that ACBib^ ri^ht angle. 183. Describe a circle that shall have a given radius and touch a given circle and a given straight line. ^vi V • i?i r® ^^ ai;^^" *^ *^^^^ a given circle and a g^ven straight line. Shew that the points of contact are a ways m the same straight line with a fixed point in the circumference of the g:iven circle. circl^^* "^^^^ * straight line to touch each of two given pie p'^y^^^^\1 ^^^%.tZrt\^, flia/?^: ^/^^ a straight line cutting two given circles so Si feng^ths '''^^''^^^^^ ^^*^^" *he circles shall havQ 188. A quadrilateral is described so that its sides aUfaci^rexSaAJS^"^"'" ^*° "*> ^«-"''«<* *J3-2 356 EXERCISES IN EUCLID, iqo ABD, ACE aro two straight lines toucliinff a circle at B and V, and if DE be joined DE'i^ equal to liD ftud GE together: shew that DE touches the circle. 191 If a quadrilateral be dsscribod about a circle the ancrles 'subtended at the centre of the circle by any two opposite sides of the figure are ti^gcther equal to two ng t ang es^^ ^^„ ^^ ^ ^.^^^^ ^^ ^ .^^^^ ^ ^^^^^ ^^ ^^^j^ ^^^ when produced are cut by a straight line wh?ch touches U^^^ circle: shew that the tangents drawn from the points of section are parallel to each other. ^,Vni^a. 193 A straight line is drawn touching two circles: show that the chords aro parallel which join the points of conLt and the points where the straight line through the centres meets the circumferences. 194 If two circles can be described so that each touches the other and three of the sides of a quadrilateral figure then the difference between the sums of the opposite sides is doublo the common tangent drawn across the quad- " ^195 ' AB is the diameter and G the centre of a sem^ circle • shew that O the centre of any circle inscribed in the semicircle is equidistant from C and from the tangent to the semicircle parallel io AB. . • vi. i;„^„ 196 If from any point without a circle straight lines be drawn touching it, the angle contained by the tangents is double the angle contained by the straight line joining i\iQ points of contact and the diameter drawn through one ^^ V97?* A quadrilateral is bounded by the diameter of a circle, the tangents at its extremities, and a third tangent: shew that its f rea is equal to half that of the rectangle con- tained by the diametefr and the side opposite to it. 198. If a quadrilateral, having two of its sides parallel, be described about a circle, a straight line dmwn through the centre of the circle, parallel to either of the two paral- lel sides, and terminated by the other two sides- shall be equal to a fourth part of the perimeter of the figure. 199 A series of circles touch a fixed straight hne at a fixed point: shew that the tangents at the points where they cut a parallel fixed straight line all touch a fixed circle. onn Of all gtrnight lines which can be draw- from two given 'points to meet in the convex circmnfei ace ot a EXERCISES IN EUCLID. 357 idling a \ to BD >• ircle the any two , to two ch other chcs the )omts of circles : points of )ugh the lat each Irilateral opposite he quad- f a semi- 3ribed in ) tangent ight lines tangents le joining ough one eter of a tangent: Lngle con- 8 parallel, I through two paral- (. shall be ire. ^ht line at ttts where ited circle. from two lice of a given circle, the sum of the two is least which make equal angles with the tangent at the point of concourse. 201. C is the centre of a given circle, CA a radius B a point on a radius at right angles to CA ; join AB and produce It to meet the circle again at /), and let the tan- ffent at D meet CB produced at E: shew that BDE is an isosceles triangle. 202. Let the diameter BA of a circle be produced to * * so *«at -^P equals the radius ; through A draw the tangent ^£rZ>, and from P draw PEC touching the circle at 6 and meeting the former tangent at E\ join BC and produce it to meet AED at D ; then wiU the triangle IJIlC be equilateral. ° III. 20 to 22. 203. Two tangents AB, AC are drawn to a circle- ^ IS any point on the circumference outside of the triangle ABt .-shew that the sum of the angles ABD and ACD 204. P, Q are any points in the circumference^ of two segments described on the same straight line AB, and on the same side of it ; the angles PAQ, PBQ are bisected by the straight lines AR, BR meetmg at R : shew that the angle ARB is constant. ^ r?^^'j S^^ segments of a circle are on the same base AB, and P is any point in the circumference of one of the segments; the straight lines APD, BPC are drawn meet- ing the circumference of the other segment at Z> and C- ^(7 and BD are drawn intersecting at Q. Shew that the angle AQB is constant. 206. APB is a fixed chord passing through P a point of intersection of two circles AQP, PBR; and QPR is any other chord of the circles passing through P: shew that AQ and RB when produced meet at a constant angle. j?^I^ ^05 is a triangle; Cand D are points in BO and AO respectively, such that the angle ODC is equal to the angle DBA : shew that a circle may be described avuiiu. itii\j qU.ciU.riiab6i ai xt-UKJJJt IMAGE EVALUATION TEST TARGET (MT-3) V- ^ // // «5^:#.^^ <^^.«!>. 1$ t/. .

J 358 EXERCISES IN EUCLID, ". 208. ABCD is a quadrilateral inscribed in a ciitile, and the sides ABj CD when produc3d meet at : shew that the triangles AOC, BOD are equiangular. 209. Shew that no parallelogram except a rectangle can be inscribed in a circle. , 210, A triangle is inscribed in a circle: shew that the sum of the angles in the three segments exterior to the triangle is equal to four right angles. 211. A quadrilateral is inscribed in a circle: shew that the sum of the angles in the four segments of the circle exterior to the quadrilateral is equal to six right angles. 212. Divide a circle into two parts so that the angle contained in one segment shall be equal to twice the angle contained in the other. 213. Divide a circle into two parts so that the angle contained in one segment shall be equal to five times the aiigle contained in the other. 214. If the an^le contained by any side of a quadri- lateral and the adjacent side prodiiced, be equal to the opposite angle of the quadrilateral, shew that any ijide of the quadrilateral will subtend equal angles at the opposite angles of the quadrilateral. 216. If any two consecutive sides of a hexagon inscribed in a circle bo respectively parallel to their opposite sides, the remaining sides are parallel to each other. 216. A, Bj C, D are four points taken in order on the cu^umferenco of a circle ; the straight Imes AB, CD pro- duced intersect at P, and AD. BC at Q : shew that the straight lines which respectively bisect the angles APC, AQC are perpendicular to each other. 217. If a quadrilateral be inscribed in a circle, and a straight line bo drawn making equal angles with one pair of opposite sides, it will jnake equal an^es with the ofiier pair. 218. 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 arc perpendicular to each otker. TIL 23 to 30. 219. The straight lines joining the extremities of the chords of two equal arcs of a circle, towards the same partS; are parallel to each other. EXERGlSEiS IN EUCLID. 359 bhat the to the e angle aes tho quadri- to the dide of pposite Scribed 3 sides, on the Z> pro- lat the APG, and a le pair Bother bed in at the of the of the 9 partS; 220. The straight lines in a circle which loin the a* treaties of two parallel chords are equal to edSrother. ^iU AB 18 a common chord of two cirHp« . flii.»iT».k rt two'?tL-ItSi'^^°*/-!? *«r«»'fe"nce of a circle 223. The straight lines bisecttoir any anirle of a n„oJ«- lateral inscribed in a circle and the opK exterior'SSri^" meet in the circumference of tho circfe «^«e"0"" Mgle, ^ • ^?*' ll-® ■? * diameter of a circle, and D is a ffiven point on the circumference, such that the arc^l^iifC than half the art: DA : dr^,. a chord AF^one ride^ tri^'^; i^o'stiiSt ifnet'lV'di^^'rrtoPtM* opposite sides at P anH o jIt • ^ f to meet the &e%t™^htlifej^il|^m Z^ir^tV '^^ length in all triangles oa tie s^t baS^ll« In'tf ^^ vertical aiwles eqiml to O. ' ^^ "*™'S one'X'STS::^^.? narssf rT'^'' terminated by the circles the sUfSfrifJ^-'^®.'*'^?''' ext^ti^Jth «ie oUi^? A ^^^Z i^^l^d" angk!io^i'e^fit^raLgM"^:nd'r"^''= *<' jl tlf-indf fiV s^aifhT linfs-te tt'eftly fe to ^5; any circle passing tlirough Ami ^ m^te tte» st^ught lines at Z and Jf. Shew that if AB bTt^twZ AL and AM the sum of AZ and ^ilf is constent. ,575^5 Ks.^1"""^^ '"'^^^"•^ diCnr?j?kd^ii 229. AOB and (702) are diameters of a cir^A nf ,n'«k* ^g^s.to each other; E is a point i^the arfic^^^^^ EFa IS a chord meeting CODsit F, and drp/ ' ' - 360 EXERCISES IN EUCLID. diroction that EF is equal to the radius. Shew that the arc BQ is equal to three times the arc ^^. 230. The straight lines which bisect the yertical angle? of all triangles on the same base and on the same side of it» and having equal yertical angles, all intersect at the same point 231. If two circles touch each other internally, any chord of the greater circle which touches the less sha'l be divided at the pomt of its contact into segments whicu subtend eq^ angles at the point of contact of the two circles. ii III. 31. 232. Right-angled triangles are described on the same hypt>tenuse: shew that the angular points opposite the hypotenuse all lie on a circle described on the hypotenase as diameter. 233. The circles described on the equal sides of an isosceles triangle as diameters, will intersect at the middle point of the base. . . 234. The greatest rectangle which can be mscnbed in a circle is a square. 236. The hypotenuse -4^ of a right-angled triangle ABC is bisected at D, and EDF is drawn at right angles to ABf and £>E and DF are cut oflf each equal to I) J ; CE and CF are joined: shew that the last two straight lines will bisect the angle C and its supplement respec- tively. 236. On the side AB of any triangle ABC as diameter a circle is described; EF is a diameter parallel to BC: shew that the straight lines. EB and FB bisect the interior and exterior angles at B. 237. If AD, CE be drawn perpendicular to the sides BC, AB of a triangle ABC, and DE be jomed, shew that the angles ADE and ACE are equal to each other. 238. If two circles ABC, ABD intersect at A and B, and AC, AD be two diameters, shew that the straight line CD will pass through B. 239. If be the centre of a circle and OA a radius and a circle be described on OA as diameter, the drcum- EXERCISES IN EUCLID. 361 jthat iho iX anglef ) side of \ at the lly, any $88 shaU bs whicii the two he same site the )otena8e )9 of an ) middle ribed' in triangle \i anglos to-D^; straight respec- liameter to BG: interior he sides lew that i and B, straight a radius drcum- [fS!.? ^^'^^ "^ll "^ ^*^c* any chord drawn through It from A to meet the exterior circle a rivtn TnirR'i^S ?w"?^ *?^^^^^. ^^^^'» «*™''g^* line at cifen S^i ?K^ *^^ the tangents drawn to it from two guen pomts m the straight Ime may be parallel. fA.^^\ . ®f?"^® a *^"*^1® ^i*Ii a given radius touching a given straight line, such that the tangents dr^wn to it iiom Uo given pomts in the straight lin? may ^^raUel. t^o^LJ^ ^^ *^® ^^^^ at the base of any triangle J^rpendiculars are drawn to the opposite sides, produced Jh^TentTe^te '^ ' Perpendicular cSwn to it from An ♦ti\- ^^i^ ^ diameter of a circle; B and C'are points on the circumference on the same side of AD i 2, pSn- itw?/rJS ^ ^" ^^ P^d^eed tough Cr meks TaTi?: shew that the square on AD is greater thai^ the sum of the squares on AB, BC, CD, by twfce the rectangle Sc, CK ^« ?i?^* .-^^1? *^® dir- eter of a semicircle, P is a point am%m'^1^''''T^ ^^ ^« perpendicular to ^^V on ^ p' »p *^ diameters two semicircles are described, and ^SopJ\fu *'^^^® l^**®*" circumferences at ^, B: shew that Q/2 w:U be a common tangent to them. 245. ^^, ^ C are two straight lines, B and are sriven points m the same; BD is dmwn pe ^dicular toTa ^d 2)^ perpendicular to AB; in Uke manner GFis drawi' Kl to^'^a ^^' "^^ ^^ *^ ^^- Shew that ^(^^ 246. Two circles intersect at the points A and B from which are drawn chords to a point Cil one of the drcui^ ferences, and these chords, pWuced if necessary, ciT the other circumference at D and U: shew that thi'straight jI^^v.'"^^^ ^^ "^?* angles that diameter of the circle ABC which passes through C. ^^J^^^' ^l » right angles io AB, 250. Two equal circles touch one another externally, and through the point of contact chords are drawn, one to each circle, at right angles to each other: shew that the straight line joining the other extremities of these chords is equal ana parallel to the straight line joining the centres of the circles. 261. A circle is described on the shorter diagonal of a rhombus as a diameter, and cuts the sides ; and the points of intersection are joined crosswise with the extremities of that diagonal: shew that the parallelogram thus formed is a rhombus with angles equal to those of the first. 252. If two chords of a circle meet at a right angle withih or without a circle, the squares on their segments are together equal to the squares on the aiameter. III. 32 to 34. 253. B is a point in the circumference of a circle, whose centre is G\ PA, a tangent at any point P, meets CB produced at A, and PD is drawn perpendicular to CB : shew that the straight line PB bisects the angle APD. 254. If two circles touch eauh ( lier, any straight line drawn through the point of contact will cut off similar seg- ments. 265. ABiA any chord, and ^2> is a tangent to a circle at A, DPQ is any straight line parallel to AB, meeting the circumference at P and Q. Shew that the triangle PAD is equiangular to the triangle QAB. 266. Two circles ABDHj ABG, intei-sect each other at the points A^ B ; from B a straight line BD is drawn in the one to touch the other ; and from A any chord what- ever is drawn cutting the circles at O^ &nd. II: shew that BG is parallel to Dff, 267. Two circles intersect at -4 and B. At A the tangents ACj AD are drawn to each circle and terminated JEXEROISES m EUCLIB. 363 if the tersect AB, Bmally, rn, one w that f these ing the tal of a points lities of formed t angle gments , whose Bts GB bo CB\ ?D. :ht line lar seg- a circle aeeting triangle h other rawnin i what- 3w that A the ninated bv the circumference of the other If Cli nn v^ u* j shew that AB or AB DroduPftT 4f L^f ' ^^J^ Joined, angle CBD produced, if necessary, bisects the a chord .nA J^^'^^y^PT* ^" *^® Circumference of a circle . 260. AS is any chord of a circle, P any noint nn rta cjrenmference of the circle ; PiJf is a WpeSffir on i J and IS produced to meet the circle at Q iWdATiU^ pen,™,d,cular to the tangent at P: she; ttat the tri^^le WAM la equiangular to the triangle PAQ ^nangw righTUl^ertotSrXr^ p'isT^i^(?„t''«' ^^ "* ^hewlh^l ll^ta^e^^it ^r ""** '' ^' ' '^^'•^^ verlS anS'tnd'the dS '"'^ ^^°° *« •-«' *« ver^^4 aS^t"(p:Sft » ft^ti^-^t from the vertex to the middle point of the base. , 265. Having given the base and the vertical ande of a S^'. ^ ^"^''^^^ ^^ *^^ gi-eatest whfn it U .. ?^^*- ^''T ^ ^^!®'^. ?^"^* ^ without a circle whose centre is draw a straiglit line cutting the circle at the Sf ^' "^ ^^""^ *^^ *'^* ^0(7 may be the gr^t^? 267. Two straight lines containing a constant ande always pass through two fixed pointe, their position beL pthermse unrestricted: shew that the straight line bisect fixed ointe ^^^^ ^^^^^ through .one or other of two 268. Given one angle of a triangle, the side opposite mmm mm 864 EXERCISES IN EUCLID, ii, and the »um of the other two sidea, coustmct the triangle. III. 35 to 37. 269 If two circles cut one another, the tangents drawn to the' two circles from any pomt in the common chord produced are equal. , j. ^ r, 270, Two circles intersect at .4 and 5 : shew that AB produced bisects their common tangent. 271 If AD, CE are drawn perpendicular to the sides BC AB of a triangle ABC, shew that the rectanglo con- tained by BC and BD is equa) to the rectangle confcamed by-S^and-B^. 272 If through any point in the common chord of two circles which intersect one another, there be drawn any two other chords, one in each circle, their four exi^remities shaU all lie in the circumference of a circle. ^73 From a given point as. centre describe a circle cutting a given straight line in two points, so that ^he rect. SSe conteined by their distances from a fixed pomt m the stniight line may be equal to a given square. 274 Two circles ABCD, EBCF, having the common tamrents AE and DF, cut one another at 5 and C7 and Sechord BC is produced to cut the tangents at G and /T: shew that the square on OH exceeds the square on AE or DF by the square on BC 275 A series of circles intersect each other, and are such that the tangents to them from a fixed pomt are S"hew that the straight lines oining the two pomts S^i^tersection of each pair will pass through this point. - 276. A BCis a right-angled triangle ; from any point D in the hypotenuse BC a sti^ight line is d'J^ J* "g^* anries to i^ meeting CA at E and BA produced at F: RhfwtC the square on DE is equal to tlie diflference of hrr^tngles Sd, DCf^ AE ^^^ ;ries'«^^/>T^^^ on DF is equal to the sum of the rectangles BD, DO and 977 It is required to find a point in the gtraight line wWoh touches a Srcle at the end of a 8>™a.d'^«^';> «»«^ ««Lt when a straight Une is drawn from this point to. the other exteemity ol the diameter, the rectangfe contamed EXERCISES IN EUCLID. 365 met the bs drawn }ii chord that AB ^he sides iglo con- on^^aihed d of two any two ties shall a circle the rect- nt in the common d C, and } and H: >n AE or , and are )oint are FO points point. iny point 1 at right ed at F'. erence of tie square , 2)(7and aight line Bter, such nt to the contained hj the part of it without the circle and the part within the circle may be equal to a given square not greater than tbuit on the diameter. IV. 1 to 4. 278. In IV. 3 shew that the straight lines drawn through A and B to touch the circle will meet. 279. In IV. 4 shew that the straight lines which bisect the angles B and C will meet. 280. In IV. 4 shew that the straight line DA will bisect the angle at A, > 281. If Sie circle inscribed in a triangle ABC touch the sides AB, AG fii the points D, E, and a straight line be drawn from A to the centre of the circle meeting the circumference at Gy shew that the point G is the centre of the circle inscribed in the triangle ADE. 282. Shew that the straight lines joining the centres of the circles touching one side of a triangle and the others produced, pass through the angular points of the triangle. 283. A circle touches the side JBG of a triangle ABO and the other two sides produced : shew that the distance between the points of contact of the side BG with this circle and with the inscribed circle, is equal to the differ- ence between the sides AB and AG 284. A circle is inscribed in a triangle ABC, and a triangle is cut off at each angle by a tangent to the circle. Shew that the sides of tbo three triangles so cut off are together equal to the sides of ABC, 285. D is the centre of the circle inscribed in a tri- angle BAG, and AD is produced to meet the straight line drawn through B at right angles to BD at O : shew that O is the centre of the circle wm^ touches lie side BG and the sides AB, ^(7 produced. 286. Three circles are described, each of which touches one «de of a triangle ABC, and the other two sides pro- duced. If 2) be the point of contact of the side BG, E that of AG, and F that of AB, shew that AE is equal to i?2>. BF to GE, ^di CD io AF, 287. Describe a circle which shall touch a given circle and two given straight lines which themselves touch the ;^ven circle. ■Mi •^■filpWWfTppp ^ 366 EXERCISES IN EUCLID. 288. If the three points be joined in which the drclo isiflcribed in a triangle meei/S the aides, shew that the r«N stilting triangle is acute angled. 289. Two opposite sides of a quadrilateral are toge- fiher equal to the other two, and each of the angles is less than two right angles. Bhew that a circle can be inscribed in the qua£ilateral. 290. Two circles HPL, KPM\ that touch each other externally, have the common tangents HK^ LM\ IlL and KM beinff jomed, shew that a cu'cle may be inscribed in the quadmatend HKML. 291. Straight lines are drawia from the angles of a triangle to the centres of the opposite escribed circles: shew that these straight lines intersect at the centre, of the inscribed circle. 292. Two sides of a triangle whose perimeter is con- stant are ffiven in ptosition: shew that the third side always touches a certain circle. • 293. Gi^en the base, the vertical angle, and the radius of the inscribed circle of a triangle, construct it IV. 6 to 9. '?34. In IV. 6 shew that the perpendicular from F on BG will bisect BG. ' - 295. If DE be drawn parallel to the base BC of a triangle ABCy shew that the circles described about the triangles ABC and ADE have a common tangent. 296. If the inscribed and circumscribed circles of a triangle be concentric, shew that the triangle must be equilateral. 297. Shew that if the straight line joining the centres of the inscribed and circumscribed circles of a triangle i^asses through one of^its angular points, the triangle is isosceles. 298. The common chord of two circles is produced to any point P ; PA touches one of the circles at -4, PBG is any chord (rf the other. Shew that the circle which passes through A^ By and C touches the circle to which PA is a tangent. 299. A quadrilateral ABCD is inscribed in a circle, and AD, BG are produced to meet at E: shew that the circle described about the triangle ECD will have the tangent at ^ parallel to ^i?. ipwfy ;he drclo b the rtii' \Te toge- es is less inscribed ch other i/Zand »ibed in les of a circles : re. of the r is con- lird side 16 radius )m F on BC of a bout the iles of a must be Q centres triangle iangle is duced to PBOis ch passes hPA is a circle, that the have the qa* EXERCIJSES IN EUCLID. mi «».^^S ^®*^^® * ^"J^'e which rf.aU touch a given straight line, and pass through two given points. "w^'gui ^ J^L- ^°*^"> \ <2r«^» wWch shall pass through two S Kength '^ ^''"^ * ^^''" "^'^'^*^*' linel cho^! ; 302. Describe a circle which shall have its centre in a given straight hue and cut off from two given st^gh? Imes chords of equal given length. otnugni 303. Two triangles have equal bases and equal v<,rti<^l angles: shew that the radius of the circumsSg cS of one tnangle is equal to that of the other. ^ 304. Describe a circle which shall pass throuirh two given points, so that the tangent drawn toit from fnothw given point may be of a givenlength. anoiner trianrf; ABO^.f^jTr' "^ i*** ^•"''^ inscribed in tho tnangle AliC, and ^O be produced to meet the circnm scnbed crcle at r, shew that rs, FO, and 5?C^3i 309. The opposite sides of a quadrilatorol inscribed in a cmsle are produced to meet at > and ft ^dSn? the tnang^so formed without the quadrilateral dS i^ :^'tt:^ut[ tf i;^^hfatgt -«'^' ^^^^^^ A\?' ^^^^^}^ * semicircle, ^5 beinir the diametfir and the two chords AD, BG intersect at I- shewThtt ff righTanSef '"'^^ ''"^^ ^""^^^ ^^" cut thrfon^^'at m 'Jmm 368 EXERCISES IN EUCLID, ^ -812. Tlie diagonals of a given qnadrilateral ^BCD intersect at 0:Xw that the centres of the «rde. de- scribed about the triangles OAB, OBG, OCD, OVA, wiU lie in the angular points of a parallelogram. 313. A circle is described round the triangle ABC\ the tangent at C meets AB produced at i> ; the circle whose centre is D and radius DC cuts AB vX E, shew that -&C bisects the angle ^CA * 314. AB, AC are two straight lines given in position; i?C is a stmight line of given length ; i>, E are the midd e «Ainfa nf AB AG- DF, EF&re drawn at right angles to 7^Acisvei^yely^ 'shew that ^^ wUl be constant for all positions oi BC. 316 A circle is described about an isosceles triangle ABC in which AB i. equal to AC', from - .a stm^K*;* Mne Is drawn meeting the base at i> and the circle at ^: shew that the circle which passes through J5, Z>, and E, touches AB, 316. ^C is a chord of a given circle ; B and -D are two ri^en points in the chord, Soth within or ^oth ^^^„«* ihe cfJcle: If a circle be descri^bed to Pass throug^^^ ^d J) and touch the given circle, shew that AB and CD «ubtend'equal angles at the point of contact 317. A and B are two points wi;thin a circle : find*^ point P in the circumference such that if PAH, ^^^ Srawn meeting the circle at H and JT, the chord HK shaU be the greatest possible. 318. The centre of a given circle is equidistant ftom two ffiven itraight lines : describe another circle which shall tZKese two straight lines and shal cut off from the given circle a segment containing an angle equal to a given angle. 319 is the centre -of the circle circumscribing a triangi; ABC', A E, F the feet of ^^^J P^^PfJ^^^^^/'S^ A, B, C oh the opposite sides : shew that OA, OB, OO are respectively perpendicular to EF, FD, DE. 320 If from any point in the circumference of a given circle straight lines be drawn to the four angular pmnte of an inscriW square, the sum of the sqmires on the four straight Unes is double the square on the diameter. *ippllli Hi EXERCISES IN EUCLID. 369 ABCD ABC', e circle ': shew tosition ; I middle ngles to bant for triangle straight ie at E'. and Ef d 2) are without h B and and CD find the P^iTbe STiTBhaU ant from [lich shall from the a given bribing a lars from ?, OC are )f a given iar points 1 the four r. de JriL'^boV'.'' cWe':^"'^'* '^"^^^ . «m«re can b, 322. Describe a circle about a given rectangle. 323, If tengents be dra^vn through the etlremities of IV. 10. 324. Shew thai: the aiirie ACD in the figure of I V. 10 18 equal to three times the angle at t.^ vertex of the triangle. ■ . 325. Shew that in the figure of IV. 10 there are two mangles which possesr the required property : si n that mere is also an isosceles triangle whose equa» angles are each one third pari; of the third angle. K «» are 326. Shew that the ba«e of the triangle in IV. 10 is ?S -I "5®x.**^ 5 ^^^^^ pentagon inscribed in the smaller circle of the figure. 327. On a given straight Ime as base describe an isoa-^ celes triangle having the third angle treble of each of the angles at the base. 328. In the figure of IV. 10 suppose the two circles to cut again at E: then DE is equal to DC, 329. If A be the vertex and BD the base of the con- structed triangle in IV. 10, D being one of the two point* of mtereev^tion of the two circles employed in the construc- tion, and -fe. the other, and AE be drawn meeting BD pro- duced at G, shew that GAB is another isosceles triamrle of the same kind. ® 330. In the figure of IV. 10 if the two equal chords of the smaller circle be produced to cut the larger, and these points of section be joined, another triangle will be formed having the property required by the proposition. 331. In the fi^re of IV. 10 suppose the two cm;les \a cut again at E\ join AE, CE, and produce AE, BD to meet at G : then CDGE is a parallelogram. 332. Shew that the smaller of the two circles employed in the figure of IV. 10 is equal to the circle described round the required triangle. 370 BXEBOISES IN EUCLID. ' 333 In the figure of IV.'IO if AF be the diameter of the^Lle? circlefSrF is equal ^ a radius of the circle which circumscribes the tnangle BbJJ. . IV. 11 to 16. ^34 The straight Uues which connect the angular winte of a regular pentagon which are not adjacent mter- Ktt «*' a^lar^poi^ of another regular pentagoiu v\'^ ABODE is a regular pentagon ; join Al> ana BE^^^ lit BE meet Aclt F; Sew that ^(7 is equal to t!ie sum of ^^ and -5/^. ... -36 Shew that each of the triangles made by jommg ,e extremities of adjoining sides of a regular Pentagon {s lew than a third and greater than a fourth of the whole area of the pentagon. 337. Shew how to derive a re^ar hexagon from an equUateral triangle inscribed in a circle, and from the con- ttZt^n shew that the side of the hexagon equals the w^S of the circle, and that the hexagon is double of the triangle. . , , i 338. In a given circle ms-ribe a tnangle whose angles are as the numbers 2, 8, 8. -, irr nn 339 If ABGDEF is a regular hexagon, and AOy BJJ, CE DF EA, FB be joined, another hexagon is formed whose area is one third of that of the former. 340. Any equilateral figure which is inscribed in a <4rcle is also equiangular. VI. 1,2. M Shew that one of the triangles in the figure of ^V lo'is a luean proportional between the other two. * 342. Throy,^ D, any ^mt in ^^^^^l^J^-^ ^AfA^FiB ame^ proportional between tiie tri- angles FBD, EDC. EXERCISES IN EUCLID. 371 343* Peipendiculars are drawn from any Doint within Mnilt SL*^™*.'?"'^? » J™"8rle """h that If straight Mde^rSi^Sl'? t° tt" ^"^ "^"S"'""- points the fai- ^S^tS^*"" .'?t2.t'""ee equal trianglk. iK A ^^■*' ^-^"®' straight lines are drawE mrallfil t/» ^A i^cD^^ ^^' ^^ ** ^' ^ ■■ *«^^^t ^ t'ti^ ii„J*^" f™™ "'y PO'^'t in the base of a triansle straight ^orof'^rSif^'f ^ *? *^« "'«'«« = *ew tha? theX* BC^t/fn.^ n ^^}%' ?^I '^^*' •>»« parallel to fae S i?- .W tw"**! ^p?t £ ; join S^ and^ meet- t^le^'ii^ "*"«'* -^^^ " ^"^ to the Jir^:.^^?S i' ii triMigle; any straight line paraUel to irTp ifjii/i ?? mee^ ^^if iJfl *^' if ^^be produced it wiU bisect £G ♦^ « t xu *"" *"*«* of a quadrilateral figure be oai^fil mt^ the other sides, or those sides pr^uced, proportiW a iriv«n rwlf o'f **J?^fK'i' '" "paired to draw from lin^ to i^^/i'^ the side ^5, or ^i? produced, a straight hne to AC, or reproduced, so that it may be bisected by VI. 3, A. Hgi^'^gii^toMTK^sri^f^.v/ri,^ 24—2 372 EXERCISES IN EUCLID. are drawn and produced to cut the circle at F atid G\ shew that the quadrilateral CFDG las any two of its adjacent sides in the same ratio as the remaming two. 354. Apply VI. 3 to solve the problem of the trisec- don of a fimte straight line. 355. In the circumference of the circle of which AB'ii a diameter, take any point P; and draw P(7, PD on opposite sides of AP^ and equally inclined to it, meeting AB at Cand D : shew that ACh to BCab AD is to BD, 366. AB is a straight line, and D is any point in it : determine a point P in -4 J9 produced such that PA is to PB as DA is to DB. 357. From the same point A straight lines are drawn making the angles BAC, CAD, DAE each equal to half a right angle, and they are cut by a straight line BGDE, which makes BAE an isosceles triangle : shew that BCov DEAb a mean proportional between BE and CD. 358. The angle ^ of a triangle ABC is bisected by AD which cuts the base at i>, ana is the middle point of BC: shew that OD bears the same ratio to OB that the difference of the sides bears to their sum. 359. AD and AE bisect the interior and extenor angles at ^ of a triangle ABC, and meet the base at i> and J?; and is the middle point of BC: shew that OB is a mean proportional between OD and OE. 360. Three points D, E^ F in the sides of a triangle ABCheing joined form a second triangle, such that any two sides make equal angles with the side of the former at which they meet : shew that ADj BE^ CF are at right angles to BCy CAy AB respectively. •VI. 4 to 6. 361. If two triangles be on equal bases and between the same parallels, any straight line parallel to their bases will cut off equal areas from the two triangles. 362. AB and CD are two parallel straight lines ; E is the middle point of CD ; AC and^j^ meet at P, and AE and BD meet at O- ; shew tnat i-xx is paraii^i vo ^msm 363. A,B,C are three fixed points in a straight line ; any straight lino is drawn through G ; shew that the per- pendiculars on it from A and -S are in a constant ratio. EXERCISES IN EUCLID, Silf3 o* ? ul J^*^® perpendiculars from two fixed points on a straight line passing between them be in a given ratio, tho straight line must pass through a third fixed point. 365. Find a straight line such that the perpendiculara e^h o«?S *^''^® ^'^^'^ ^''^''^^ ^^^^^ ^e in a Wen ratio to fi.«f ^*1: ^^i^^^l^ .? ?i^®" P^i"^* ^^^^ a straight lino, so that the parts of it Intercepted between that point ind perpendiculars drawn to the straight line from two other given points may have a given ratio. 367. A tangent to a circle at the point A intersects tMnlP'^fwi,*^"^"?*' ^* ^' ^' *^« A of contaTof which i^th the circle are i>, E respectively; and BE CD minf i «f^ ^""li ?. ^""^ ^^^^ P^i"*« ' -^^ «^d ^^ are fixed Sf^M 1"^^^.* t^' '.^^y «*^^^* J^e is drawn from P to meet ^^at M, and a straight line is drawn from Q ^mI^oJ^^ ""''^^^^ ^?^^^'' «l^ew that the S of Hr5i ^uJ^^^}^h ^°^ *^enee shew that the straight Ime through Jf and i\r passes through a fixed point. ^ dfe9. Shew that the diagonals of a quadrilateral, two of whose sides are parallel and one of them double of the other, cut one another at a pomt of trisection, ^•^?^^V ^-^u^^^ ^fiP *^^ P^^"*^ ^^ *^e circumference of a circle of which C is the centre ; draw tangents at ^ and 5 meeting at ^ ; and from A draw AN perpendicular to CE : shew that BTis ioBCBsBNisto NA 371. In the sides AB, AC o{ a triangle ABC are nl?"«*/7^ P'^'^'H ^> ,^' ^"^""^ *^at ^i) is equal to CE: jp ^S^^.® produced to meet at F: shew that AB is to ^c7as EF IS to 2)^. 372 If through the vertex and the extremities of the base of a triangle two circles be described intersecting each other m the base or base produced, their diameters are proportional to the sides of the triangle. 373. Find a point the perpendiculars from which on the sides of a given tnangle shall ue in a given ratio. *w^^"^* ^^"^A^'A^^ *^^ adjacent sides of a rectangle, two similar triangles are constructed, and perpendicular^ are drawr to AB AC from the angles which they subtend, intersectu.ar at the point P. If AB, AC be homologoiS 874 EXERCISES IN EUCLID. sides, shew that P is in aU cases in one of the dlagonalb of ^''a^^fTn'the fi^re of 1.43 shew that if ^G^ and J»fl be produced they wUl meet on ^(7 produced. V ^P5 W (7Q2> are parallel straight hnev^^^ JP is to PjB as DQ is to QC: shew that tho 8tr^«*»t Uii^ PQ. ^a BD, pn)duced if necessary^ will meet atj^ point : shew also that the straight lines PQ, AD, BO, pro- duced if necessary, will meet at a point. ^-^„«ed 377. ACB is a triangle, and the side -4(7 P P«>d^ced to ^80 that CD is equal to AC, and ^^ is JO^ned^/ «J^ straight line drawn parallel to ^B cyi\^i^^fi^^^AC,CB, and W the pointe of section stra ght ines be dra^ mraM to /)5, shew that these straight Imes will meet AB at points equidistant from its extremities. 378. If a circle be described touching externally two giv6n circles, the straight line passing through the pomts of contact will intersect the straight line passmg through the cc tres of the given circles at a fixed pomt. 379. D is the middle point of the «ide ^C? of a tn- angle ABC, and P is any point m AD ; through P the stilight lin^s BPE, CPF are drawn meeting the other sides at E, F: shew that J^Pis parallel ioBC 380 AB is the diameter of a circle, E the middle point of thf radius OB ; on AE, EB^ diameters c^rdes S^ described; PQL is a common tangent fleeting ^ circles at P and Q, and AB produced at L : shew that i?X is equal to the radius of the smaller circle. „„ ^81 ABCDE is a regular pentagon, and AV, Bt^ inters^t iS : sifew that a^^side of the pentagon is a mean proportional b^^^^^^^^^ ^ p,i,^ in a?trai4t line par^lel to AB ; ^^-^.g^^ ^^^i^^ B and P2> and QC meet at /9j shew that Ri:^ is parallel 383. ^ and 5 are two given points ;^C and BD are perpendicular to a given straight line ({^'x^^^^^^ Intersect at E, and J^Pis perpendicular to CD . shew that >i w and «P mftke eaual angles with ijV. ^ „^^ "~ ^ft4 From the angular points of a paraiieiograui^-i^t/^ pe^ndicu'S.^':^^ d^wn o; the f iagona^^^ E,F,G,H respectively : shew that hFQK is a paraueio- gram similar to ABCD% EXERCISES IN EUCLID. 875 IFH \y and raight b at a 7, pro- duced if any o,cA drawn [ meet lly two points iirough f a tri- P the 3 other middle circles ing the VN that 2), BE a mean 3 points meet at parallel 5i>are and 5(7 lew that them at )arallelo« 385. If at a given pomt two circles intereect, and their centres Ue on two fixed straight lines which piSs through i?J^lrS?*? '^®'' that whatever be the magnitude of X tSS ?.i5t7 ''•T^T" **H?^?*^ ^" always meet in one <5 two fixed straight Imes which pass through the given point^ VI. 7 to 18. o JS^' *^^.*r?v ^''*^l®i? *°^^^ ®^^ ®*^er, and also touch a given straight Ime, the part of the straight line between the points of contact is a mean proportional between the diameters of the circles. ♦k ??Z' ?^^!.^® h W^^ ^^^ ^^ ^ circle into two parts. So •ven rati? ^ ^^ ^^^^ ^^ ^ ^^ ^**^' ^ * 388. In a given triangle draw a straight Ime parallel to one of the sides, so that it may be a mean proportional between the segments of the base. f fv vu« 389. ABC is a triangle, and a perpendicular is dra^ from^ to the opposite side, meeting ii at D between B ^Sl ^ •j^??iJ'i°** ^^^-^ ^® ^ "^«»» proportional between BD and CD the angle BAOh a right angle. 390. ABC is a triangle, and a perpendicular is drawn D /^^'^J^"® opposite side, meeting it at D between ^ S;"'^ x' -^^j?®? *^a* i^ -^^ is a mean proportional between BD and BG, the angle ^^C is a right angle. 'L ^l' . ^ ^^ **^® c®^*^® ^^ * ci^'C^e, and ^ any point within It • GA IS produced, through ^ to a point B such that the radiusis a mean prooortional between GA and CB\ shew i???/ , ^J^Jt^ P^*°* ^^ **^® cu-cumference, the angles UFA and GBP are equal. 392. . is a fixed point in a given straight line OA, and a circle of given radius moves so as always to be touched by OA ; a tangent OP is drawn from to the circle, and m OP produced PQ is taken a third propor- tional to OP and the radius: shew that as the circle moves along OA, the point Q will move in a straight /??A rJ^^ ^''^^ parallel straight lines touch a circle, and SPT 13 another tangent cutting the two former tan- gents at S and Z, and meeting the circle at P \ shew 11 376 EXERCISES IN EUCLID. that the rectangle SP, PT is constant for all positions ^ 394. Find a point in a side of a triangle, from which two straight Imes drawn, one to the opposite angle, and the o?ier ^raUel to the ba^e, shall cut oflf towards the vertex and towards the base, equal triangles. 396. ACBisc^ triangle having a right angle at G ; from A a straight line is drawn at right angles to AB, cutting 5(7 proceed at E; from B a straight line is ^awn at right%les to AB, iutting AG produced at^: shew that the triangle EC^ is equal to the tnangie ACB. . _ „ . ^ 396. The straight line bisecting the angle ABC of the triangle ABC meets the straight Imes drawn througi M^parallel to BC and AB respectively, at E and F: shew that the triangles CBE, ABF are oqud. 397. Shew that the diagonals of any quadrilateral fiffire inscribed in a circle divide the quadrilateral mto four triangles which are similar two and two ; and deduce the theorem of IIT 35. ^ 398. AB, CD are any two chords of a curcle passmg tiirough a point 0; JEJ^is^any chord paraM to 0^ ; jom CE, DF meeting AB at the points G and ^, and DE,Ci meeting AB at the pomts AT and L : shew that the rect- angle OG, OH is equal to the rectangle OK, OL. 399. ABCD is a quadrilateral in a circle ; the straigljt lines CE, DiS^which bfsect the angles ^CB,M>B cniBD and ^C at-J^and G respectively : shew that EF is to EG as ED is to EC. 400 From an angle of a triangle two straight hues are drawn 'one to any point in the side opposite to the an^le, and the other to the circumference of the circumscnbmg circle, so as to cut from it a segment containmg an angle equal to the angle contained by the first drawn line and the side which it meets: shew that the rectangle con- tamed by the sides of the triangle is equal to the rectangle contained by the straight lines thus drawn. 401 The vertical angle C of a triangle is bisected by a gfraigbt line which meets the base at i>, and is produced to a point E, such that the rectangle contained by ox." bmi CE 18 equal to the rectangle contained by /6 and OJi: shew that if the base and vertical angle be given, the posh tion of .^ is invariable. EXERCISES IN EUCLID. 377 , which indthe vertex 7; from cutting awn at sW that [BC of ;hrough and?: ^lateral ral into deduce passing B\ join J)E, CF he rect- jf^^' ^'A^^^^ '5 inscribed in a right-angled trianrfo wnuse AB of the triangle: shew that the area of S^ square is equal to the rectanrio AD, BE ^ 403. ABCD is a parallelogram: from B a sfrftiViif hne IS drawn cutting the diagoK^ at ^, Ihl s'dTS^ rec^rie^^'i;^"- ^^ ??^.r^ ^' ^' «hew ttt fhe rectangle £Fy FGia equal to the square on BF. iurJl^' }' * ^}^^S^^ line drawn from the vertex of an isoscel^ triangle to the base, be produced to meet th^ circumference of a circle describel about the tnWle there, tangle contmncd by the whole line loproB ^d the part of it between the vertex and the Ce "hiUI trian^^:! *^' '^^"'^ "" '^^'' '^ *^^ ^^^ sides of the i^Ju^' '^Y^ ^}^isht lines are drawn from a point A to touch a circle of which the centre is JS; the poiSts oAn tact are joined by a straight Ime which cuts iS at #• Td stra^hf^ ^T*^^ ^.''^''^^ '^ described: shew t^t the m«pf U nlT ^'^''" r*^^"^"^^ ^ ^ *^«c^ this circr will meet it on the circumference of the given circle. Ill straight cut BD 8 to ^6? lines are le an^le, iscribmg an angle line and igle con- rectangle 3ted by a produced ■^ ■»">■ _ ' and CB: the post' VI. 19 to D. nf^tl^L f"" i^^sceles triangle is described having each ande«Tlr "t *^\t>as«. ^^uble of the third angTef if ?ho ^tej^t^^^^^^^^^^^ ^n the 5?oS 407. Any regular polygon inscribed in a circle is « mean proportional between the inscribed and circuSil^d regular polygons of half the number of sides ''''^^""^"^^ arc pT;allel *^^ ^^"'^ ""^ ^^' ^^ '^'^ *^"* ^^ ^^^ ^^ ..J^?: ,.^^^<*e a triangle into two eoual mrt^i hv « -uaiguu ime at ript angles to one of the sides." " ' "' ' the dLiite r^f ir^flf *"r^^^' t""^ *° ^«^ ^«^*Jier in are simUa? "" ""^ *^^'^ ^'^^^ «^^^ t^a* t^^ triangles 378 EXERCISES IN EUCLID. 411. Through a giyen point draw a chord in a given circle so that it shall be divided at the point in a given ratio. 412. From a point without a circle draw a straight line cuttmg the circle, so that the two segments shall be equal to each other. 413. In the figure of II. 11 shew that four other straight lines, besides the given straight line are divided in the required manner. 414. Construct a triangle, having given the base, the vertical angle, and the rectangle contained by the sides. 415. A circle is described round an equilateral triangle, and from any point in the circumference straight lines are drawn to the an^Iar points of the triangle: shew that one of these straight lines is equal to the other two together. , 416. From the extremities B, C of the base of an isosceles triangle ABC, straight lines are drawn at right angles to AB^ AG respectively, and intersecting at D: shew that the rectangle BCy AD is double of the rectangle AB, DB, 417.. ABC is an isosceles triangle, the side AB bemg equal to -46' ; .F is the middle point oiBG-, on any straight line through A perpendiculars FG and CE are drawn : shew that the rectangle AC, J^F is equal to the sum of the rectangles FC, EG and FAy FG. thesti angle] that ai angle. , 422 withou find wl 423 at ape distanc perpen the cen by the ing thrt 424. straight straight 425. two plai i>draw F: shei isperpe 426. and to a line join to the fc XI. 1 to 12. 418. Shew that equal straight lines drawn from a given point to a given plane are equally inclined to the plane. 419. If two straight lines in one plane be equally in- clined to another plane, they will be equally inclined to the common section of these planes. 420. From a point A a perpendicular is drawn to a plane meeting it at B ; from B a perpendicular is drawn on a straight line in the plane meeting it at (7: shew that ACia perpendicular to the straight line in the plane. 40 1 ji li/y is ft triansfle t the i>©n>endicula!^ from A and B on the opposite sides meet at D ; through D a straight line is drawn perptendicular to the plane of the triangle, and E is any point in this straight line : shew that 427. vertices . AB.AG GED : s] angles at to be on "428. another 1 tended b; not in thl angles su tenor an§ 429. EXERCISES IN EUCLID. 379 I ^¥J\LL v^ a. geometrical construction for drawing a 18 perpendicular to AB f^^i^ucou u necessary. XI. 13 to 21. vertl^ /^i !^*5l ta TS« T "^ *"»??««»!<»«, whoso tofe on oW^sUe s"§es of 5!?' '^«'««' *"??<»*"« ^ <««» -8^ 429. From the exwemiUes of the two paraUel straight im 880 EXERCISES IN EUCLID. lines AB^ CD parallel straight lines Aa, Bb, Ce^y Dd aro drawn meeting a plane at a, bjCjd: shew that ^jB is to CD as ab to cd. 430. Shew that the perpendicular drawn from the ver- tex of a regular tetrahedron on the opposite face is three times that drawn from its own foot on any of the other faces. 431. A triangular pyramid stands on an equilateral base and the angles at the vertex are right angles ; shew that the sum of the perpendiculars on the faces from any point of the base is constant. 432. Three straight lines not in the same ^lane inter- sect at a point, and through their point of intersection another straight line is drawn within the solid angle formed by them: shew that the angles which this straight lino makes with the first three are together less than the sum, but greater than half the sum, of the angles which the first thf^e make with each other. 433. Three straight lines which do not all lie in one plane, are cut in the same ratio by three planes, two of which are parallel: shew that the third will be parallel to the other two, if its intersections with the three straight lines are not all in the same straight line. 434. Draw two parallel planes, one through one straight line, and the other through another straight hue which does not meet the former. 435. If two planes which are not parallel be cut by two parallel planes, the lines of section of the first two by the &st two will contain equal angles. 436. From a point A in one of two planes are drawn AB at right angles to the first plane, and ^(7 perpendicular to the second plane, and meeting the second plane at B, C: shew that BG is perpendicular to the line of intersection of the two planes. . t. n i 437. Polygons formed by cuttmg a pnsm by parallel planes are eqiml. " 438. Polygons formed by cutting a pyramid by parallel planes are similar. 439. The straight Ime PBbp cuts two parallel planes at B, 6, and the points P, p are equidistant from the planes ; PAa^pcUdxa other straight lines drawn from Ir'yp to cut the planes : shew that the triangles ABC, abc are equal. 440. Perpendiculars AE, BF are drawn to a plane from tTi A perp with tlK 441. shew thai the sides 442. radii AP, circumfer to^AS: 443. ; sum of th extremitie gram. 444. ] gonal the i 445. 1 both of its 446. I angle be pi downward^ be obtainec j447. A igiveu tttraii ■given straig ■"'^Pand i Dd aro 4.B is to I tho ver- is thrco tier faces. unilateral BS ; shew from any ne inter- lersection ie formed bight lino the sum, L the first ie in one 3 of which el to the ght lines 3 straight hich does ut by two ro by the re drawn ►endicular >at J5,(7: section of y parallel )y parallel lei planes le planes ; ' , p to Cut ) equal. a plane EXERCISES m EUCLID. 381 from two points .4. B q.Kntr'^ ,'f . « i . - wit^ti;: given piano U peSc^uil."!^!'' «^i>«^n I. 1 to 48. she^thaUife';,S."of p7'^« t^^^K^*^ "= the sides of the triangk "^^ » 'ess fljan the sum of ^n^the^sL^„TS::r^.tee^%5'--X^^ ^i. both^^its^^a^Sri^ffiy:^^^'' » bisected by ""'• l^'l^Z^ih 'Jl*^^,^-^ "Mes of the tri- assummg any proposition beyond ^-447. ^18 a given point, and » i» |6;.eastraJgntHne: it ii requi^edto c stra5?ht Bne, a straight ' o-i 7 wis ^»Vi£iJ» ili (% to the A J- and P^ may be equal to a given length, 'hat the sum 382 EXERCISES IN EUCLID: I " 448. Shew that by saperpo«ti«m the fiwt case of I. 2G may be immediately demonstrated, and also tu© wcond case with the aid of 1. 16. . . , C r *v 449 A straight line is drawn termmated by one of the sides of an isosceles triangle, and by the other side pro- duced, and bisected bv the base : show that the straight lines Wins intercepted between the vertex of the isosceles triangle and this straight line, ai-e together equal to the two eqaal sides of the triangle. . ^ ,^ ... , -^^ . 450 Through the middle pomt M of the base BC of a triangle a straight line DME is drawn, so as to c x off equal parts from the sides AB, AC, produced if necepsary • • shew that BD is equal Ui GE. , ..v 451. Of all parallelograms which can be formed with diameters bf given lengths the rhombus is the greatest 452 Shew from I. 18 and I. 32 that if the hypote- nuse SCqS 9. right-angled triangle ABC be bisected at D, thk ADy BD, CD are all equal. 453. If two equal straight lines intersect each other any where at right angles, the quadrilateral formed by joining their extremities is equal to half the square on either straight line. 464. Inscribe a parallelogram in a given triangle, in such a manner that its diagonals shall intersect at a given point within the triangle. 455. Construct a triangle of given area, and having two of its sides of given lengths. 466. Construct a triangle, having given the base, the difference of the sides, a^d the difference of the angles at the base. . . , , ,. .x • 457. AB, AC are two giv^^ fVrai^ht lines: it is re- quired to find iu AB a point P, m..n t at if PO be drawn perpendicular to AC, the sum tf AP ai-d AQ may be equal to a given straight line. . , - xv 468. The distance of the vertex of a tnangle from the bisection of its base, is equal to, greater than, or less than half of the base, according as the vertical angle is a nght, aa acute, or an obtuse angle. 469. If in the side§ of a giver square, at equal distances ^^-_. J.I.. f.^.~ «.M>»ii1n«. ■r\i^infJa fmit* nfliAr r^Ain^A be taken. ^_^„ w , contained by the straight lines i^di joinlhem,' shall ^so be a square. one on each side, the figure 461. CA\ tJ BCtki, 462. triangle 463. isoscelet ther wit 464. be bisec duced t distance its distal 465. from on< given po; 466. and CB CFBG I cent 8id( whose a4 nals of tl lina 467. angles; i shew thai with the ABCD, 468. . m&ABG passing th if necessa centre F quadrilate i^ MIO BUU 469. ^ ^EXERCISES IN EUCLID. 383 460. On ft glyen straight line as base, congtnict a tri- angle, haring g,vr,n the dilerence of the 8 dT!md a Soit through which one of the sides h to pass. ^ •Y^^i ^-^^ ^^ * triangle in which BA is irrnafAr ^han ther with it a single isosceles triangle. ^ K« k-^^ X ¥^**1? ®^ ^^^ ®9^^ sides of an isosceles trianirl« 466. Determme the locus of a point whoM Hiafan/w* S?Si W """"^ " ^'""'^ '"* Ccrtrnttth^? uTirl^^ii t ^P'^«^\^^ ^B is bisected at Qand on ^^ ??i?»^ " diagonals any two parallelograms 'J^(7^ ^d cenf ridrarr^^^'.^'^it^^^^^^ who^ a^a^ wK^l "^- ^'^^x ^? ^'^^^ ^^ ^^ completed, and also that ris^f fc litS.^'' ""%?? ^^ ^^ : Bhew th^t thTdi^S' nate of these latter paraUelograms are in the same str^ght o««.ffo^* ^'^^^ ^* ? rectangle of which ^, C are ODDosito angles; i^ is any point in BC and F is any^nt ^^^ shew that twice the area of the ivimttl^ApTifJti: ' wi|i^,^he rectangle BE^ D^t nZ'Vif'^'^l and^^Vl^^;i^^-?^'*i;^''^*^^"^^^« o^ the same base I'b^T'^i ^7^J<>^^«i;^ thronghTand b^^^^ ^^^^HhF^^f '^ necessary: shew that the 4iwania,ierai^/&/>i^has the sum of two of its s^^'^o ^—^^ i^ Mio sum of the other two. " • '"'" --i^-s 469. Two straight lines AB, AC are given in positicm: 384 EXERCISES IN EUCLID, \\\i:' m. ¥• it is required to find in AB a point P, such that a perpen- dicular being drawn from it to AC, the straight line AP may exceed this perpendicular by a p'-oposed length. 470. Shew that the opposite sides of any equiangular hexagon are parallel,andthatanytwosideswhichareadiacent are together equal to the two to which they are parallel. 471 From D and E, the comers of the square 3DEG described on the hypotenuse BGoi a fig^it-angled triangle ABC, perpendiculars DM, EN are let fall on AC, AB respectiyely: shew that AM is equal to AB, and AJS equal to AC, 472. AB and AC are two given straight lines, Mid P is a given point : it is required to draw through P a straight line which shall form with AB 2^di AC the least possiole triangle. 473. ABC is a triangle in which C is a right an^le : she^ how to draw a straight Ime parallel to a given straight line, so as to be terminated by CA and CB, and bisected by^J5. 474. ABC is an isosceles triangle having the angle at B four times either of the other angles; AB is produced to i> so that BD is equal to twice AB, and CD is joined : shew that the triangles ACD and ABC are equiangular to one another. 476. Through a point K within a parallelogram ABCD straight lines are drawn parallel to the sides: shew that the difference of the parallelograms of which KA and AC are diagonals is equal to twice the triangle BKD, 476. Construct a right-angled triangle, having given one side and the diiference between the other side and the hypotenuse. 477. The straight 'lines AD, BE bisectuig the sides BC, AC of a triangle mtersect at Q: shew that AQ is double of GD. 478. BAC'i^ a right-angled triangle ; one straight line is drawn bisecting the right angle A, and another bisecting the base BC at right angles ; these straight lines mtersect equal to DA, ' 479. OtlAG the diagonal of a square ABCD, a rhom- bus AEFC is described of the same area as the square. and that 4 ang!< on 1 verte whicl 4f allele 48 squar 48 AEi that t the qi CFE. 48^ angle ; meets line w! shew 1 ABDl by a sti 486. BC, G sfcraighj is equal 487. AG, Bi Z; ZO mdB£ and JiTI ADEB of the p{ 488. shew ths through point. 4P.Q the same tercept e to their I perpen- meAP [angular idjacent rallel. BDEG triangle 1(7, AB nd AN les, and gh P a he least b an^le: straight bisected angle at ►reduced i joined : igular to iABCD bew that mdKC ag given and the the sides b AG is light line bisecting mtersect if D^ is , a rhom- j square, EXEHCISES m EUCLID. 085 vertex at »: shw that the I^r^f^f.^^T'^''' **» which bisects the a.ng\e SAG ''^^"*^° »*'^«lit line sq«^- ^"""^^ " ^l"^--* of given magnitude in a given ^ that the triangle £Fg' hhiMthf^ wters^ at /•; shew the qnadrilate^ il'J'^is eanlun "%*^'* f 4<^' '"'^ «»* Gf J or ^DjP ^-^^-^ " equal to either of the triangles Mglet the fn^gle A fa Sted'T^ "'f 'i''S>« O a right mwtsWat^ and «.« in^u^c^' ^. *''^''* «»« which line which metoVat^f^l tud^A' ^ f*'^*''* ^hew^^hat the triangle AOru't^^^^^^l^ by»sLiSMnll'to\r'^?ts*'il5fe""'^' Ve divided 486 3 7?r^n jr^Tfr, ^ Which will coincide. 50; C>,1fd b^tw^f^the™^!^''"^^?"^ ?° ^i-^ >«»«« f» fe£? "- &rt' t-lh^lfai 1^ and JT^ at ^ and »?;1S^?- , *° ^^' *°d meeting r& of the paralleIo^aSi™'cg- " ^"^"^ '° "^« »•«» ^l^nghtheintersoc^tionofthrdrgoCffbita^^^^^^^ I the same parall^^llfewXt^rSs oftL''^. •'!'"«.«» Wr ^ '«"^''' "f -y "'-filial ttete •s 386 EXERCISES IN EUCLID. I hi n 490 In a right-angled triangle, right-angledit J, if the ridi iob^ double of the side AB, tie angle B w more than double of the angle C. . ,. j 491, Trisect a parallelogram by straight lines drawn ftrougii one of its angular points. 4<»2 AHK is an equilateral triangle t^^CX* is a ,hombis.f S^to of which IS equal to a side of the triMigle InrttTsid ric and C/)^ which pass through H and ^re^Sctivelyf shew that the angle A of the rhombus is ten-ninths of a right angle. ^ 493. Trisect a given triangle by straight hues drawn from a riven point in one of its sides. 494 In the figure of I. 85 if two diagonals be. drawn tofte iwo pJr^Ue^grams respective y, «»«JXZlb^ tremitv of the base, and the intersection of the diagonals De jS^rith the intersection of the ^des or sito produ«.d^ in the figure, shew that the joinmg straight Ime will bisect Ute base. II. ItoU. 4<».5 Produce one side of a given triangle so that the wctoile Sned by this sidl and the produced part !^^ equTto the difference of the squares on the other *''°496 **"Produce a given straight line so that the sum of the w^reron the given straight line and on the part wodS^iay te equal to twice the rectangle contained by ?he wMe XightVe thus produced and the part pro- *"*^7 Produce a given straight lino so that the sum of the squarson the W "traight line and on the whole straiglt Une thus pi'oduced may be eqvml to t^«« *^ rectangle contained by the whole straight line thus pro- duced and the part produced. ,~„t„„-viB . 498. Produce a given straight line so that the jectang'^ contetaed by the whole straiglit line thus produced and the . ^rt S^u^ may be equal to the square on the given straiglit line. . ,. „ _ ..i.j x^. ^i« sii/>^ 499. Describe an isoscmusi ootuac-angivu "^7*0^- --^-^ that the square on the largest side may be equal to three times the square on either of the equal sides. 600. Find the obtuse angle of a triangle when the EXERCISm IN EUCLID. p»7 the rectangle of the sS ""^ "^** contaimn| it, by whe?^he^rrt,^o"|S sX^o^jf S"'-" »«^«- equal to a given quantity *^ °' *^® rectan^e is whenLte:* ^f';*;^''^^^^^^^ %8jve„ sqnan, w equal to a given qimntUy "' ' '"^^ "^ ""« rectangle givenV^ris'SXhlhaifoftre.^^ '°''-"''«J •» » , 604. Bivideagiv^stmi^hJi? H^'P"*'l"a>-e. the squares on th? who We ^Z "'**' '''° P'*'^ "«> that "nay be together douWe of the ' „are on tC *lf ^^ P"'^ 605, Two rectaneles ha™ f?..?! ° *"® "t^er part, meters: shew that thiv are I1?»T^ ^l^ *""* «9'»' Peri- the sum ofi>.4 andV^Wu^I^ ♦!* P^'"* »"* that jP^ : shew that the locus of ^ c^Lw^lS !r.»^ ^^ ""d lines through the cent^A «f Ti, , "• the two straisrht sides. ^ "'® "'"^'^^ Of the rectangle paraUel to^fa III. a to 37. ^5i>isequaltotheaSffi"''''' *'" *'"'" *''^ ^"^'« diculars from fh^ anglefl^gl:' o".".^ -?^' ^"^ *''« ^T^"- middle point of the tSrd 8M« thfP?.!'! f,!'^^'' and A" the ^i^^are each equal to 1 ' '" "■*' *« ''"S'^s -f-^^. t»„"*t /'•" " * diameter of a circle • ^f o«j ^ n two chords meeting the tan<,Ant „f d ^^ ^M ^-O are spectivelv • sIipu, S«; ti, ^ , at ^ at ^r and if re- 4al. ^- '"*'' ^^"^ the angles i^Gi: and rj)§ are 25-2 388 J^XEBCISES IN EUCLID. 613. Shew that the four straight lines bisectmg the angles of any quadrilateral form a quadrilateral which can be inscribed in a circle. x • i 614. Find the shortest distance between two circles which do not meet. . a ^ •* • 615. Two circles cut one another at a point A : it is required to draw through A a straight line so that the extreme length of it intercepted by the two circles may be equal to that of a given straight line. 616 If a polygon of an even number of sides be in- scribed in a circle, the sum of the alternate angles together with two right angles is equal to as many right angles as the figure has sides. 517 Draw from a given point in the circumfet^nce of a circle, a chord which shall be bisected by its point of mter- section with a given chord of the circle. ' 518 When an equilateral polygon is described about a circle it must necessarily be equiangular if the number of sides be odd, but not otherwise. 619 AB is the diameter of a circle whose centre is G, and DCE is a sector having the arc DE constant ; join AEy BD intersecting at P; shew that the angle APB is 620 If any number of triangles on the same base BC, and on the same side of it have their vertic^ angl^ equal and perpendiculars, intersecting at D, b© diuwn from B and h orthe opposite sides, find the locus of 2> ; and shew that all the straight lines which bisect the angle BDt pass through the same point. x, . i 621. Let O and (7 be any fixed points on the cu-cum- ference of a circle, and OA any chord; then if AVhe loined and produced to B, so that OB is equal to OA, the locus of 5 is an equal circle. , t, r. r „ 622. From any point P in the diagonal BD ol a parallelogram ABCD, straight lines PE PF, PG, PH are drawn perpendicular to the sides ABj i/0, Cx/, jja . shew that ^P is parallel to GH. t. ^ ^i- „ ^x^Ack 523 Through any fixed point of a chord of a circle oztf. J^|"^^fa", _^ „u^^ +u«* +!.« affoifyVif. lines from other ciiof as ui e ai av. u ; «"- •• vis^- vi-v^=^--. ~<^~'', , .» the middle point of the first chord to the middle pom the others will meet them all at the same angle. P24. ABC is ft straight line, divided at any poi of EXERCISES IN EUCLID, 389 f^X^^L^^k^^^ ^^ ^^^ ^^« «'°»«^r segments of cimes, having the common chord BD- CD anH An^^L and V»b;? oi^'- ' -?-^> ^^are joined: shew that i^i^ ^?^' A » ^r«» /'^ght line a part o^|v|" Cfth fl.- 5-" t-^ stja^ht me and two circles are given- find drawn to the circles are of equal length. ""Agents 62& In a circle two chords of ffivcn lenirth j.™ ,l«.w» so as not to intersect, and one of thesis Sin posfc the opposite extremities of the chords nrni^-^i?? ' straight lines intersecting within the drde- shew^Tf f^^ locus of the point of intersectiL ^U be a porUon^f the uin^r/afa-uio^ts^^ :^t^a :s r«Id !?i \ externally and internally r^sSeTy a? fA^F ^' *' '"='" ^-^^ ■« double 7thl 530. C is the centre of a circle, and CP is a nemen ^cnlaronaehord^P^: shew that tho sum of (^^Jd fi?i ^^^^ ''•'en C-P is equal to ^P. ** 631. ^5, ^(7 CZ> are tnree adjacent «!d^_>«»f ® the perpendicular from the vertex, and the difiference of the b^e angles. ^^^ ^^ ^ ^^^ ^^ same side of it are described two se^ents of circl^; P is any point in the circumference of one of the seg- ments, and the straight line BP cuts the circumference of the other segment at Q : shew that the angle P-4Q is equal to the angle between the tangents at A, 535 AKL is a fixed straight line cutting a given circle at ^ and X; APQ, ABS are two otlier ptrait^,^* lines making equal angles with ^JTZ, and cutting the cTrcle at P,Q and B, S: shew that whatever be the posi- ZTofAPQfindAM the straight line Jommff the mid- die points of PQ and BS always remams parallel to itself. . 536 If about a quadrilateral another quadrilateral can be described such tlat every two of its adjacent sides are equaUv inclined to that side of the former quadrilateral whiKets them both, then a chrcle may be described About the former quadrilateral. 537. Two circles touch one another internally at the Doint ^ : it is required to draw from A a straight hue S that the part of it between the circles may be equal tr^given stmight line, which is not greater than the difference between the diameters of the circles. 538 ABCD is a parallelogram ; AE is at right angles to aI and a^ is at right angles to CB : shew that ED, if woduced will cut AG oi right angles, produced, wmc ^^ ^^ °^ j^t ^f \t"^«i^^,r'^r A\o^Uv is let fall on the opposite side : shew that the rect- SeTcLt^^^^^^^ segments into which each perpen- SarTdiVided bV the point of intersection of the ttree are equal to each oth^^^^^ ^^ ^^^ ^^^ ^^ ^ ^^^ bisected bv two straight lines on which perpendiciJars are dmwn from the vertex: shew that the straight line which SSses through the feet of these perpendicular will be Urallel to the base and wiUbisocUlie^s^^^^^^^ _^^^, ^^ ^ 541. m a given circio iiisciii/*/ » ixy-^vaAif^zv v-i--^- -^ ^Ti"K=^^-angled triangle ABG perpendicular, ad; SElro let fall Sn BC, CA respectively; crdes EXERCISES IN EUCLID. sai described on AC, BG qa diameters meet BE. AD rean«fl. tiTely at F^Q and H, K: shew that F, cfffKUe^S^ circumference of a circle. > -* «« "u wio 643. Two diameters in a circle are at right anirles- from their extremities four parallel straight linS m« e^r ' St^"" ^''^^ '"''''*® *^® circumferfnce ^to f^ f.r.A^h7P^' "^ ^"^^^^^^^ Ppint of a semicircular arc^^A ^iifS^'' T "^^W ?^*^*"'^ ***^ diameter at D, and ti£ rilateii AEBC ''^'^"^ '''' ^^ ^ ^^"^^ *^^ ^"^^' of whth >! « ^^VTn '"''^''^V * ParaUelogram is described ?L Sif £.w^ ^£are adjacent sides: find the locus of the middle points of the diagonals of the parallelogram. o\.^^^:.f d.^ " * ^^^. ^^^^ ^^ * ^'^^^^ ^C' is a moveable S Sw^oTV^'^^^' a parallelogram is described of which AB and AG are atflacent sides: determine the greatest possible length of the diagonal drawn through A. y II R^^? ®^^^^ *^^^^^®3 '^^ Placed at such a distance apart that the tangent drawn to either of them from the centre of the other is equal to a diameter, shew that they will have a common tangent equal to the radius. 648. Find a point in a given circle from which if two tongents be dmwn to an equal circle, given in position, the chord joinmg the pomts of contact is equal to the chord tJh. ^Pi cmile formed by joininer the points of inter- secfaon of the two tangents produced; and determine the hmit to the possibility of the problem. ^ , ^^ *"» ni. ^i^'/y"^^ ^^ * diameter of a circle, and AF is any chord; G is ^ny point in AB, and through G a straight line is drawn at nght angles to AB, mieting AF, pro- duced if necessary at G, anc' meeting the circumference at ^'a "5^ *l\at the rectangle FA, AG, and the rectangle JJA, At, and the square on AD are all equal 660. Construct a triangle, having given the base, the veri;ical angle, and the length of the straight line drawn from the vertex to the base bisecting the vertical angle. «P ^r>^ given Circle: find a pomt P such that if AP, ^Z*, OP meet the circumference at D, E, F resnectivelv the arcs DE, EFm^y be equal to given ar<^ ^^^P^^*^®*^' 892 EXERCISES IN EUCLID. 652 Find the point in the circumference o^ a given circle, the sum of whose distances from two given straight lines at right angles to each other, which do not cut the circle, is the greatest or least possible. 553. On the sides of a triangle segments of a circle are described internally, each containing an angle equal to the excess of two right angles above the opposite angle of tiie triangle : shew that the radii of the circles are equal, that the , J7, i?* respectively: shew that J)E and DF are equaUy inclined to -3[i>. 555. The points of contfUJt of the inscribed arcle of a triangle are joined; and from tlie angular points of the triangle so formed perpendiculars are drawn to the opposite sides : shew that the triangle of which the feet of these perpendiculars are the angular points has its sides parallel to the sides of the original triangle. 666. Construct a triangle having given an angle and the radii of the inscribed and circumscribed circles. 657 Trian/les are constructed on the same base with equal vertical angles ; shew hat the locus of the centres of the escribed circles, each of which touches one of ttie sides externally and the other side and base produced, is an arc of a circle, the centre of which is on the circumference of the circle circumscribing the triangles. 558. From the angular points A, B, (7 of a triangle perpendiculars are drawn on the opposite sides, and ter- circumscribmg circle : if L be the point of intersection ot the perpendiculars, shew that LD, LE, LF are bisected by the sides of the triangle. EXERCISES IN EUCLID, d§3 5fi9. ABCDE is a re^lar pentagon ; join A C and BD intersecting at : shew that AO isTqual to DO Mid that the re^ctangle ^C CO is equal to the Squar? on^a ^* Ua ^ ^' ^ ^*'?'^^* ^"^® ^^ *^^ Siven length mores so that Its ends are always on two fixed straight lines (7P. CQ • straight lines from P and ^ at right angles to CP and CO respectively intersect at R ; jperpendiculars from P and S Wi nf^ ^/respectively intersect at S: shew that the aU7 ^^^ ^ ^^® ^^'^^^^ ^^^°ff *heir common centre 1.^^?^* I^^g^\*-a^?led triangles are described on the same hy^tenuse : shew that the locus of the centres of the in- scribed circles is a quarter of the circumference of a circle o: which the common hypotenuse is a chord 662. On a given straight line AB any triangle ACB is described; the sides AC, BC are bise^ed afd stra^ht lines drawn at nght angles to them through the points of bisection to intersect at a point D; find the locus 5f 2) ♦!»« o«^] *"A*T* ^ *"«?gh having given its base, one of the angles at the base, and the distance between the centre of the mscnbed circle and the centre of the circle touchm^ the base and the sides produced. ij«/^t* ^?«<^"^® ^ ^^cle which shall touch a given straight line at a given point, and bisect the circunSerence of a given circle. ^* ^^^' A^^^^^l ^ ^}^^^^ ^^^^^ »^a" pass through a given point and bisect the circumferences of two given circles 566. Within a given circle inscribe three equal circles touching one another and the given circle. 667. If the radius of a circle be cut as m II ll the greater segment will be the side of a regular decagon m- scnbed in the circle. 668. If the radius of a ch-cle be cut as in II n the square on its greater segment, together with the squari on the radius^ is equal to the square on the side of a reeular pentagon inscribed in the circle. ^^ 1- ^P^-xv.^t°°^ the vertex of a triangle draw a straight line to the base so that the square on the straight line mav bejequal to the rectangle contained by the segments of the 670. Four straight lines are drawn in a plane forming four trmngles; shew that the circumscribing circles of tnese triangles all pass through a common pomt. .Jl.___. 394 EXERCISES IN EUCLID. 671 The perpendiculars from the anglea A aud B of a triangle on the opposite sides meet at D j the circles de- icribid round J/Wand 2)^(7 cut ^.J5 or AB produced at the points Em&Fi shew that AE is equal to BF, 572 The four circles each of which passes through the centres of three of the four circles touching the sides of a triangle are equal to one another. , 673. Four circles are described so that each may touch internally three of the sides of a quadrilateral : shew that a circle may be described so as to pass through the centres of the four circles. 574 A circle is described round the triangle -45(7, and from any point P of its circumference perDcndiculars w^ drawn to BG, CA, AB, which meet the circle agam at Te, F: shew that the triangles ^J^^-and^^^ are equal iniafi respects, and that the straight lines AB, BE, CF are parallel. 675 With any point in the circumference of a given circle as centre, describe another circle, cutting the former at AaadB; from B draw in the described circle a chord BD equal to its radius, and join AD, cutting the given circle at Q : shew that QDia equal to the radius of the given circle. 676. A point is taken without a square, such that sti-aight lines being drawn to the angular points of the square, the angle contained by the two extreme straight lines is divided into three equal parts by the other two straight lines : shew that the locus of the pomt is the cir- cumference of the circle circumscribing the square. 577 Circles are inscribed in the two triangles formed bv drawing a perpendicular from an angle of a triangle on the opposfte side ; and analogous circles are described m relation to the two other like perpendiculars: shew that the sum of the diameters of the six circles together with the sum of the sides of the original triangle is equal to twice the sum of these perpendiculars. 578 Three concentric circles are drawn in the same Diane: draw a suai^uu imic, oucii vii-- ----- -z. ^.r^ Tv" "j Ween the inner and outer circumference maybe bisected at one of the points at which the straight Imo meets the middle circumference. EXERCISES m Eirczw. 896 [OormAllf.. a diameter, and P any point in the circum- ference of a circle; AP and BP are joined and produced rf necessary; from any point Cin AB t, straigfit line is * ^ ^i nght angles to AB meeting AP at 3 and ^P ^JtT? *^^,*"? circumference of the circle at F; shew that CZ> IS a third proportional to CIS and CF. 580. Ay B, C^are three points in a straight line, and D fu^Tu f ^^'''^ -i^ ^?^ ^^ subtend equal angles ; shew that the locus of D is the circumference of a circle. 681. If a straight line be drawn from one corner of a square cutting oflF on^fourth from the diagonal it will cut off one-third from a side. Also if straight lines be drawn similarly from the other comers so as to form a square, this square will be two-fifths of the original square 682. The sides AB.AG of a given triangle ABC are produced to any points 2>, JS, so that DJE is parallel to BO. 1 he straight line i>^ is divided at Fso that 2)Pis to FF Bs BDisto CF: shew that the locus of Pia a straight 683. A, B, G are three points in ordc^r in a straight line : find a point P in the straight lino so that PB may be a mean projportional between PA and PC, ' 684. Ay B are two fixed points on the circumference of a given circle, and P is a moveable point on the circum- ference; on PB is taken a point D such that PD is to 4 x^ \ constant ratio, and on PA is taken a point F such that PF is to PB in the same ratio : shew that DE always touches a fixed circle. 685. ABC is an isosceles triangle, the angle at A being four times either of the others : shew that if BC be bisected at D and F, the triangle ADF is equilateral. 586. Perpendiculars are let fall from two opposite ang:Ies of a rectangle on a diagonal: shew that they will **™,*?® .of the fomeJ Zts the common chord 5(7 at £, and the <='~">«^ ference of tho other circle at : shew that the angles EPO and DPO are equal for all positions of F. 691 ABC ABF are triangles on the same baae m the ratio of two to one -, ^J^ and ^f produced meet the "iito at i) and E : in FB a part FG is cut off equal to ^A.an« !fiG is Wsfctcd Tt : shew that BO i^U>^E»sDF^ to ^^m. A is the centre of a circle, ""d """^^^ 5'^'=! passes through A and cute the ^°P^%^^f^tnAtnmD adhord of the latter circle meeting BC at E, ana trom iv J^dZmDF^DG tangents to the former circle: shew n9^'\''!iricr tr'fides^of a triangle, are taken ^A,E-A% AC. re produced to F,f^^^J%^f^ w nnnol to ^D and CO equal toAE; BG, tJ' s'^, J™"?';' mS at If: shew that Sie triangle FHG is equal to the *™Str Ky^triig?ftlc if BD be^i^ken equal to one^otrth of To, and^C'^ one-fourth of AG, the stm^ht Une drawn from'C thi-ough the «tersect»n of 5^ and AD will divide AB into two parts, which are m the ratio oi "^M^.^lnv rectaineal figure is inscribed in a dj^le: shew that by'bisectmg the arcs fd draj"n« tengents^ the points of bisection parallel to the sideg of the reou EXERCISES IN EUCLID. S»7 — — • IK ^P^'^'T^can fo»-»'^a rectilineal figure circumfcribinir the circle equiangular to the fortner. "^nomg ^Jfl' i^i"* •* T*° proportional between two similar right-angled tnangles wfach have one of the sides contain- ing the riffht angle common. 597. In tho sides AC, EC of a triangle ABC points n and ^ are taken, such that CD and C'^ are respectivelV tho third parts of AG and BC-, BD and^yXdmwn interBecting at 0: shew that EG and DG^^Zl^^X the fourth parts of ^^ and BD, fvvwvwj A«/.h^n;i,.?^^ P T ^^?";S*^*^ of fc'^o circles which touch each other externally at C7; a chord AD of the former chord BF of the latter, when produced, touches the former at G?: shew that the rectangle contained by-4i> and BF IS four times that contained by DE and FQ, ^Jr^' I^^ T^l?^ intersect at A, and ^^0 is drawn meeting them at i? and (7; with ^, (7 as centres arTZ ?^^f *7'^- ?r''^*^^ ,®^^^ ?^ ^^^^^ intersects one of the fornier at right angles : shew that tliese circles and the circle whose diameter is BC meet at a point. HiviH«« A^^R^^.^ a regular hexagon I'shew that BF divides AD m the ratio of one to three. 601. ABC, DEF are triangles, ha^^'no- f,T»A ««^ie j -rtnnl areas of the tnaiigles are as ^0 to i);S?. „„ ^''^•i. l^-*^' -^^® *¥ ?"">*» at which the inscribed and Zw twifTi? ^'"'^'' r "^"^^ ^<^ "f » triangle ifl^, tZ ^H^^t}^ .produced to cut the escriled circle again at !>, then IfP is a diameter. o„/n" • '''il* "P^l® ^ ?/ * *™"e'e ^-SC-is a right angle, n^ n'^ *''^ ^°''* "*^.^'''; PeT>«ndieular from A on 1(7 .Ss^aXTa^feSf. "" ^^' ^^'= ^''^ ^''^^ «>« ^604. If from the point of bisection of any given arc of a circle two straight lines be drawn, cutting tIeThord of ?onl'.n^^i *^' ^i^^r Terence, the four poin^ts of Ltersec. tion shall also lie m the circumference of a circle in«r>«w ^'-'^ f''^^^^ ""^ a triangle ABC is touched by the W+wT^^ ^^ ^\ ^^^ ^^ *^^ ^««"l>ed circle at ^: f h!li^^* *i^® i*®;?*^^^'' contained by the radii is equal to the rectangle AD, DB and to the rectangle AE EB 898 EXERCISES IN EUCLID. 606. Shew that the locus of the pi^f «_^^i»„fj Btr^ght lines parallel to the base of a tnaBgU and termi- nftted bv its siaes is a straight lino. , v • 607 A p^lelogra ^ is^nscribed in a tri^gle, haying one side on^base of tho triangle, aijd the a(^acont sid^^ wrallel to a fixed direction : shew that the locus of the Kectkn of the diagonals of the para Uelogram is a straieht line bisecting the baiie of the tnangie. ^08 On a given straight line ^J5 aa hypotenuse a rishTanffled triangle is described ; and from -4 and B rtSi^Ks a^ dmwn to bisect the opposite sides ; shew that the locus of their intersection is a circle. 609 From a given point outside two g ven circlos whiXdoL'^meStfdiwa^str^ght line b-^ that^^^^^^^^^^^ tions of it intercepted by each circle shall be respectively ^Tf In a S'tri^gle inscribe a rhombus which shall have one of its angular points coincident with a pomt in the base, and a side on that base. y^ ^l r*. 611. ABC is a triangle having a right ai-|le at C7; ABDE is the square d. >cribed on the hypotenuse ; ±, (^, Ji arf Kin?s o? intersection of the diagonals of the squares on the hypotenuse and sides : shew that the angles DCE, G^FjEf are together equal to a right angle. MISCELLANEOUS. 612. is a fixed point from which any straight Une is drawiT meeting a fileS straight line at P; in OP a point Q I taken sulh that the rectangle OP, OQ is constant; shew that the locus of a is the circumference of a circle. 613 is a fixed point on tho circumference of a circle, from wHch any straiglit line is drawn meeting the circum- fe?^ce at P; in OP a point Q is taken such that the rS^ToP.OQ is constant : shew that the locus of Q la * ^^ei'f^ ThTopposite sides of a quadrilateral inscribed in _ JlT^^v.^^ I/f.A^.ni>d mfiftt at P and 0: shew that the squarTonWis eciual to the sum of tte squares OJJ m tangents from P and Q to the circle. EXERCISES IN EUCLID. 399 equal in arearh"ai?tte:^»t4ffi'''''^ at 1 and^/^' l^"® *''° *?1«?»t» *» a c^e, touching it equal to the tangeVt drawn from ^to the drSe ^ "^ the stra.-ght line jdlg The ST Jo bteX'^' ?^*? farther "^ '"^ -=-'- /J^P, i^ra*?e1,1^'t^l1 thr^f tfe'tnVX'ptro? fee^stf It^-JhS So^a-.^''^'' '"^^^ °- o^«>- WuVUeTi- drawn ■thrnri,*n'''^* ''"^t^^ a* ^ and 5, and CBD is through 5*'?'l£flT?'^'='!la>\t<' ^^.*» ""eet the circles! intSn7o^f • ^^^ ,'"■? '* ^'^"" bisecting either the nl r« ° 'f *"■ *"S^'^ between ^Cand AD, and mee^ 623. Divide a triangle by two straight lines into three 400 EXERCISES IN EUCLID. Y«tTte which, Tvhen properly arranged, shall form a paralle- ^^^rB^'t;:^t^tl^'^Tp U any point: fhe Wang^^pl^ls equal to thesum of thetnangles P^5 -^ef5^%o^cX^r -H^ih:^ ana a^traig^^^line ^n/^nVia drawn which meets one circle at ^ and JJ, ABijDI^ ^*. S 'j iJ o«/i thfiir common chord at G\ tewthtUlJe^rr^^'XtV^^^^^^^^ recWe 5C, C/) is to the rectangle AC, CE. THE END. SiSSSSSiTHSSSSTSVrWTiXiriS. u«v«.«T r.«» paralle- V point: ifference he angle md that BsPAB ight line and D, d at C: ^ as the BUIITT X'RSHil.