Question

△ABCD中,AB＞AC,∠AEF=AFE,延长EF于BC的延长线交于G,求证∠G=1/2（∠ACB-∠B） 提示∠AE

Answer 1

已知,∠AEF = ∠AFE ,

而且,∠AEF+∠AFE+∠A = 180° ,

即有：2∠AEF+∠A = 180° ,

可得：∠AEF = (1/2)(180°-∠A) = (1/2)(∠ACB+∠B) ；

所以,∠G = ∠AEF-∠B = (1/2)(∠ACB+∠B)-∠B = (1/2)(∠ACB-∠B) .