证明:
∵AD⊥BC
∴∠CAD=90°-1/2∠C
∵∠A=180°-∠B-∠C,AE平分∠BAC
∴∠CAE=90°-1/2∠B-1/2∠C
∴∠DAE=∠CAE-∠CAD=(90°-1/2∠B-1/2∠C)-(90°-1/2∠C)=1/2(∠C-∠B)
证明:
∵AD⊥BC
∴∠CAD=90°-1/2∠C
∵∠A=180°-∠B-∠C,AE平分∠BAC
∴∠CAE=90°-1/2∠B-1/2∠C
∴∠DAE=∠CAE-∠CAD=(90°-1/2∠B-1/2∠C)-(90°-1/2∠C)=1/2(∠C-∠B)