BOOK n. PROP, vi. x.] MATHEMATICS. PLANE GEOMETRY. 



561 



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t 4 IT. Cor. to each of these add L G or C l) 2 ,t .' . C M G 

 + LG= AD-DB + CD 2 : butCMG + LG = CF = 

 C B 2 , .-. A D-D B + C D 2 = C B 2 ; . . if a, straight line, 

 <tc. Q.E. D. 



COR. From this it is manifest, that the difference of 

 the squares of two unequal lines, A C, C D, is equal to 

 the rectangle contained by their sum and difference ; that 

 is, A C 1 C D 2 = A D DB, where AD = AC + CD, 

 andDB = CB CD = AC CD. 



PROPOSITION VI. THEOREM. 



If a straight line (AB) be bisected (in C), and prolonged 

 to any point (D), the rectangle contained by the whole 

 line (AD) thus prolonged, and the prolonged part (BD), 

 together with the square of (C B) half the line bisected, 

 is equal to the square of (C D), the line made up of the 

 hlf and the prolonged part : that is, A D "D B + C B 3 

 = CD J . 



46 I. Upon C D describe the square OF:* draw 

 jv c n n DE, and through B draw 



B H G parallel to C E or 

 D F, and through H 

 draw K L M parallel to 

 + 311. AD, orEF;t 

 also through A draw A K 

 parallel to CL or D M. 

 Then because A C=CB, 

 .381. AL = CH; 

 butCH = HF,t.'. AL 

 + 43 I. = H F : to each of these add CM, . . AM 

 gnomon C M G ; but A M is the rectangle A D 'D B, 

 4ii.cor. for DM = DB, .'. CMG = ADDB: 

 + 4 n. cor. add LG or CB't to each of these, . . C M G 

 + L G = A D D B + C B 1 ; but CMG+LG = CF = 

 C D, 1 . . A D D B + C B 2 - C D 2 ; . . if a straight line, 

 <tc. Q. E. D. 



PROPOSITION VII. THEOREM. 



If a straight line (A B) be divided into any two parts (in C), 

 the squares of the whole line, and of one of the parts, 

 are equal to twice the rectangle contained by the whole 

 and that part, together with the sruare of the other part : 

 that ia, A B J + BC 2 = twice A B'BC + A C 1 . [See 

 Diagram, Prop. IV.] 



48 I. Upon A B describe the square AE,* and con- 

 struct the figure as in the preceding propositions. Then 

 t 48 1. because AG = GE,t if C K be added to 

 each, AK = CE,.-.AK + CE= twice A K. But 

 A K + C E = gnomon AKF + CK, /. AKF + CK 

 = twice A K ; but twice A K = twice A B'B C, for 



4 ii. Cor. B K = BC,' /. A K F + C K = twice AB 

 BC. To each of these add H F or AC 2 , .-. AKF + 

 C K + H F = twice AB'BC+AC 2 : but AKF + CK 

 + H F = A B 1 + B C 2 , . . A B 2 + B C 2 = twice AB B C 

 + A C ; .-.if a straight line, <tc. Q. E. D. 



PROPOSITION VIII. THEOREM. 



If a straight line (A B) be divided into any two parts 

 (in C), four times the rectangle contained by the whole and 

 one of the parti (B C), together with the square of the other 

 part (A C), t* equal to the square of the. line (AD = A.B 

 + B C), made up of the whole and that part : that is, 

 four times A B B C + A C 2 = A D 2 . 



I- Prolong A B to D, so that B D = C B ;* 

 + 48 I. and upon A D describe the square A F ;t and 



construct two figures such as in the preceding. Then 

 because C B=BD, and CB = G K, and BD = KN,* 

 841. .-.GK = KN. For a like reason, PR=RO; 

 and because CB=BD, and GK = KN, .-.CK = BN, 

 t SB i. and G R = R N.f 

 But C K = R N, Deing com- 

 431. piemen ts,* .-. BN 

 = G R, . . the four rectangles, 

 B N, C K, G R, R N, are 

 equal, and are therefore qua- 

 druple of C K. Again, be- 

 cause CB = B D, and B D = 

 t 4 11. cor. B Kt = C G, 

 Mi- and C B =-- G K* = 



+ 4 II. Cor. (!P ( +. .-. CG = 

 VOL. I. 



G P, and because C G=G P, and P R = RO, .' . A G= 



.361. MP,*andPL=RF. But M P = P L, being 



+ 431. complements, f .'. AG=RF.'.the four, AG, 



MP, PL, RF, are equal, and are together quadruple AG. 



And it was proved that the four, C K, B N, G R, R N, are 



quadruple C K, . ' . the eight rectangles, which compose 



the gnomon A O H, are quadruple of A K ; and because 



AK = AB-BC, for BK = BC, .-. four times A BB C 



= four times AK : but it was proved that A O H = four 



times AK, .-. fourtimes ABB C = AO H. Toeachof 



4 it. Cor. these add XH or AC-',* .'. four times 



AB-BC+AC 2 = AOH + XH; but AOH + XH 



-AD 1 , .-. four times AB BC + AC 1 = AD J ; .-. if 



a straight line, <fec. Q. E. D. 



NOTE. For another and shorter demonstration of toil proposition, 

 >ee the remarks at the end of Book II. 



PROPOSITION IX. THEOREM. 



If a straight line (A B) be divided into two equal, and also 

 into two unequal parts (in C and D), the squires of the 

 two unequal parti are together double of the square of 

 half the line (A C), and of the square of the line (G D), 

 between the points of section : that is, A D 1 + D B s = 

 twice A C* + twice C D 2 . 



11 1. From C draw C E at right angles to A B ;* 

 + si. and make it = AC or CB.t Draw E A, EB; 



and through D draw D F parallel to C E, and through F 



31 1. draw F G parallel to B A :* draw also A F. 

 Then because A C = C E, the angle EAC = AEC;t 



+ S I. and because A C E is a right angle, the two 

 others, AEG, E A C, must together make one right 



32 1. angle ;* and as they are equal, each must be 

 half a right angle. For a like reason, each of the angles, 

 C E B, E B C, is half a right angle, . . A E B is a right 

 angle. And because G E F is half a right angle, and 



E G F a right angle, for it is = 

 + 291. ECB, t .'. the re- 

 maining angle E F G is half a 

 right angle, . . G E F = E F G, 

 .61. and .-. EG=GF.* 

 Again, because B is half a right 

 angle, and F D B a right angle, 

 + 291. for it is = ECB, t 

 .* . the remaining angle, B F D, is half a right angle, . . 



61 B = BFD, and .-. DF = DB. And because 

 AC = CE, .-. AC 2 =CE 2 , .'. A C" + C E-' = twice 

 AC 2 ; but AE = AC 2 +CE 2 , because A C E is a right 



+ 47 i angle,t ' A E 2 = twice A C ! . Again, because 

 EG = GF, ..EG J = GF 2 , .-. E G 2 + G F 2 = twice 



471 OF 8 : but EF J = EG 2 + GF 2 ,* .-. E F* = 

 + S4I twice GF 2 ; and G F = CD,t .-. E F 2 =twice 



CD' :but AE 2 = twice AC 1 , .-. AE 2 + EF 2 = twice 



.47 1. AC s +twice CD 2 : but A F 2 = A E 2 + EF 2 ,* 



because A E F is a right anle, . . A F 2 = twice A C 2 + 



twice CD 2 : but A ]) 2 + D F' 2 = A F 2 , because A D F is 



+ 47 1. a right angle, t . . A D 2 + I > F 2 = twice A C 2 



+ twice CD 2 ; and D F = DB, .'. A I) 2 + DB 2 = u>ice 



A C 2 + twice C D : ; . . if a straight line, &c. Q. E. D. 



PROPOSITION X. THEOREM. 



If a straight line (A B) be bisected (in C), and prolonged 

 to any point (D), the square of the whole line thus pro- 

 longed, and the square of the part of it prolonged (D B), 

 are together double the square of (A C) half the line 

 bisected, and of the square of (C D) the line made up of 

 the half and the prolonged part : that is, A D' + D B' 

 = twice A C 2 + twice C D 2 . 



u i. From C draw C E at right angles to A B,* 

 + 31. and make it = A C or C B ;t and draw A E, 



E B. Through E draw E F parallel to A b, and through 



si I. D draw D F parallel to CE.* Then because 

 E C, F D are parallels, the angles CEF+EFD= two 



+ 29 1. right angles,t . ' . B E F + E F D are Uss than 

 two right angles, . . E B, F D, if prolonged, will meet 



AX. 12. towards B, D.* Let them meet in G ; and 

 draw A G. Then because A C = C E, the angle C E A 



5 I. =E A C,t and A C E is a right angle, .'. CE A, 



32 1. E A C are each half a right angle.* For a like 



4 o 



D 



