Question

在三角形ABC中,AD是角平分线,M是BC延长线上一点,MN⊥AD于N,交直角AB于E，交AC于F。

求证：∠EMB+1/2（∠ACB-∠B）

Answer 1

∠EMB
= ∠AEM-∠B
= 90°-∠BAD-∠B
= 90°-(1/2)∠BAC-∠B
= 90°-(1/2)(180°-∠ACB-∠B)-∠B
= 90°-90°+(1/2)∠ACB+(1/2)∠B-∠B
= (1/2)(∠ACB-∠B)

Answer 2

∠EMB=1/2(∠ACB-∠B)