Question

如图,△ABC中,∠B>∠C,AD⊥BC,AE平分∠BAC，求证∠DAE=1/2（∠B-∠C)

Answer 1

∵AE平分∠BAC，∴∠BAE=1/2∠BAC，
∵AD⊥BC，∴∠BAD=90°-∠B，
∴∠DAE=1/2∠BAC-(90°-∠B)
=1/2(180°-∠B-∠C)-90°+∠B
=90°-1/2∠B-1/2∠C-90°+∠B
=1/2∠B-1/2∠C
=1/2(∠B-∠C)。