解: ∵AD ⊥BC
∴∠BDA=90 °
∴∠BAD=90 °-∠B
又∵AE 平分∠BAC
∴∠BAE=∠BAC
=(180°-∠B-∠C)
∴∠EAD=∠BAD-∠BAE
=90°-∠B-(180°-∠B-∠C)
=90°-∠B-90°+∠B+∠C
=∠C-∠B
=(∠C-∠B)
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