... cos ∠B = q 10 2 CHAPTER 5. TRIANGLESSee also Problems 2 .18 , 2.26, 2.36, 2.44, 2.54, 4.46, 5.56, 7.45, 10 .3, 10 .77, 11 .3, 11 .5, 16 .7, 18 .9, 18 .12 , 18 .15 , 18 .17 -18 .20, 18 .22, 18 .38, 24 .1. §4. Triangles ... Since △AC 1 B2∼ △BC 1 A 1 and △AB 1 C2∼ △CB 1 A 1 , it follows that AB2·C 1 B =AC 1 · BA 1 and AC2· CB 1 = A 1 C · B 1 A. Hence,AB2AC2=AC 1 C 1 B·BA 1 A 1 C·CB 1 B 1 A= 1. 5.75. ... ACC 1 and BCC 1 we getAC 1 C 1 C=sin ∠ACC 1 sin ∠AandCC 1 C 1 B=sin ∠Bsin ∠C 1 CB,i.e.,AC 1 C 1 B=sin ∠ACC 1 sin ∠C 1 CB·sin ∠Bsin ∠A.Similarly,BA 1 A 1 C=sin ∠BAA 1 sin ∠A 1 AC·sin...