no 



M \THEMATICS. PLANE GEOMETRY. [BOOK VL PROP, xxvu. xxxin. 



cannot b) in a different straight lino from that of E G, 

 . . tht iKiralltlogramt art about the lame diagonal ; . ' . / 

 two rimilar parallelograms, etc. Q. E. D. 



rilOPOSITIOX XXVII. THBOEM. 

 Of all tht rectangle* contained by tht ngmentt of a giren 

 ArmgM line, the greatest it the rectangle of the two 

 tQVJil. parti of the line : that it, the square of half the 

 KML 



Thb to an obvlom corollary from Proportion V., Book II M already 



Mted it pa MS. The corre.|.ondln- Propoitinn of tuclid i. 



S!.^ complicated, and I'ropowuon. XXVlll., XXIX., .re .till 



M M. Plarfair. in bii edition of tht Klements, ha. therefore 



DUced these 'three Propositions by other, of a simpler character; 



3T exam pie U followed here; but the demonstration* of the 



Sit two pTopoaiOona will be found till Impler tban thoe of 



rtayfair. 



PROPOSITION XXVIII. PROBLEM. 



To divide a given straight line (AB), so that the rectangle 

 contained by itt segments may be equal to a given square 

 (C*), not greater than the square of half the line. 



.101 Bisect AB in D ;* then must AD be 

 either = C, or 70. If AD 

 = C, the tiling required is 

 done ; but if A D 7 C, draw 

 D F perpendicular to A B, t 

 * n I. and make D E = C, 

 as also D F = A D. With 

 centre D and radius D F de- 



A J) O. B scribe a circle ; the circum- 



ference must pass through the points A, B, because D A, 

 D F, DB are (by Const.) equal. Draw E P, parallel 

 to A B, meeting the circumference in P ; and draw P G 

 parallel to E D ; then G will be the point of division 

 required. 



. M i For (Const.) E G is a rectangle, .' . P G = E D ;* 

 but ED = C (Const) .'. PG=C: but AG-GB = 

 , 13 vi. P G,t .'. A G'G B = G, 2 .'. A B is divided in 

 G, so that A G'G B = C- ; which was to be done. 



PROPOSITION XXIX. PROBLEM. 



To prolong a given straight line (A B), so that the reetaru/Ie 

 contained by the segments between the extremities of the 

 given line, and the point to which it is prolonged, may 

 be equal to a given square (C 1 ). 



Bisect A B in D, draw B E perpendicular to it, and 

 make B E = C ; draw D E, and 

 with centre D and radius D E, 

 describe a circle meeting A B 

 prolonged in G, F ; then will 

 AG-GB = C 2 . Because D is 

 the centre of the circle F E G. 

 DF = DG ; and (Const.) 1) A 

 DB, .'. A F = B G. To each of these add A B, then 



w vi. FB=AG. ButFB-BG=BE 2 ,*.'. AG-GB 

 - BE = C ! , .'. A B is prolonged to G, so that AG'GB 

 = C* ; which was to be done. 



PROPOSITION XXX. PROBLEM. 

 To cut a given straight line (AB) in extreme and mean ratio. 



Divide A B in C, so that the rect- 

 angle AB'BC may be equal to A 1 



4 11 ii A C J ; t then because 

 AB'BC-AC 1 , .'. AB:AC::AC: 



17 vi. BC;*.'. AB is cut in extreme and mean 

 i Vet. vi. ratio ; t which was to be done. 



Nora. An example U here furnished of the way in which pain of 

 inritmm'niitrablf line* mar be found at pleasure. It waa proved 

 (Prop. XXIII., Book V.), that if two linei A B, A C, an (uch that 

 A Ii : AC :: AC: BC(= AB AC), then A B, A C, are incom- 

 mensurable. It would be ImpoMible, therefore, to eiprem the two 

 line* A B, A C. accurately by nunirrt ; and it U thus that a theory 

 of proportion, sufficiently comprehensive for the demandt of geo- 

 metry, eould never be rigorously established by aid of numbers 

 onlr : the baaei of the proposed triangles In Proportion I. of this 

 book, might, for aught we know to the contrary, be related as the 

 line* A B, A C above : and therefore could not be numerically 

 spewed. 



C B 



20 VI. 



PROPOSITION XXXI. THEOREM. 

 In a nq!>t-angled triangle (ABC) the rectilineal figure 

 described upon the side (B C) opposite to the right angle, 

 it equal to the similar and similarly situated figure* 

 described upon the sides (A B, A C) containing the right 

 angle. 



For figure on AB : figure on AC : : AB S : AC ';* 

 therefore fig. on A B + fig. 

 on A C : fig. on A C : : A U' 

 + is v. +AC:AC.t 

 that is, fig. on AB + fi<_'. 

 on AC :tig. on AC: : 1!< 2 : 

 A C. But fig. on B C : 

 . fig. on A C : : B C 2 : A C 2 ; 

 ^ sv. consequently,* fig. 



on A B -f fig. on A C : fig. on A : : fig. on B C : tig. 



on A C, . '. the consequents being equal, the antecedents 



1 9V. Cor. are equal, t .'. fig. on A B -{-fig. on AC = 



fig. on B C ; . . in a right-angled triangle, <fec. Q. E. D. 



PROPOSITION XXXII. THEOREM. 



If tux> triangles (A B C, D C E) which have two sides 

 (B A, AC) of the one proportional to two sides (C D, 

 D E) of the other be joined at an angle, so as to have 

 their homologous sides (&.!&, DC, and A C, D E) parallel 

 to one another, the remaining sides (B C, C E) sliall be 

 in a straight line. 



Because A B is parallel to D C, and A C meets them, 

 the angle B A C = A CD; for a 

 similar reason C D E = A C D ; 

 .-. B A C = CDE. And be- 

 cause the triangles ABC, D C E 

 have an angle A = D, and the 

 sides about those angles propor- 

 tional, namely, B A : A C : : 

 .Hyp. CD :DE,* the triangle 

 + 6 VI ABC, DC E, are equiangular; f .'- anglo 

 A B C = D C E ; and it was proved that B A C = A C D, 



ACE=ABC + BAC. Add A C B to each, . . 

 ACE + ACB = ABC + BAC + ACB = two right 

 angles ; that is, at the point C, in A C, the two straight 

 lines CB, C E, on opposite sides of it, make the adjacent 

 angles A C E, A C B, together equal to two right angles ; 

 . . C B, C E are in a straight line ; .'. if two triangles, 

 &c. Q. E. D. 



PROPOSITION XXXIII. THEOREM. 



In equal circles (ABL, DEN), angles (BGC, EHF) 

 at the centres or (B A C, E D F) at the circumfei 

 are to one another as the arcs (B C, E F) upon which 

 they stand ; and so also are the sectors (BGC, EH F). 



NOT*. In the following demonstration, by "the angle B O L" must 

 be understood ' the sum of the angles B O C, C Q K, &c. ;" and by 

 " the angle E H N," " the earn of the angles E II K, F H M, &c." 



Take any number of arcs, C K, K L, each equal to 

 B C and any number F M, M N, each equal to E F ; 

 aud draw G K, G L ; H M, H N ; then the angles at G 



27 III are a 'l equal,*.', whatever multiple B L is 

 of B C, the same multiple is the angle B G L of B G C. 

 For a similar reason, whatever multiple E X is of E F, 

 the same multiple is the angle E H N of E H F ; also if 



K 



B L = E N, then BGL = EHN; ifBL7EN, then 

 BGL7BHN; an.l if H L z E N, tli-n i:<;L^ 

 i: II N. Consequently (Def. V., p. 573) B C : E F : : 

 BGC : E H F ; and since the angles A, D, are the 

 halves of BG C, EHF, each of each, . . Ii C : K F : : 

 A : D ; . . the anjles are as the arcs on which they stand. 



