如图,△ABC中,AD平分∠BAC,BE⊥AC于点E,交AD于点F.试说明∠AFE=二分之一(∠ABC+∠C).
问题描述:
如图,△ABC中,AD平分∠BAC,BE⊥AC于点E,交AD于点F.试说明∠AFE=二分之一(∠ABC+∠C).
答
∠AFE=∠BFD
∠BFD=∠ABF+∠BAD
1/2(∠ABC+∠C)=1/2(180-∠A)=90-1/2∠A=∠ABF+∠BAD
所以∠AFE=1/2(∠ABC+∠C)