△ABC内接于⊙O,AD⊥BC,D为垂足E是弧BC的中点,求证：∠EAO=∠EAD

Question

△ABC内接于⊙O,AD⊥BC,D为垂足E是弧BC的中点,求证：∠EAO=∠EAD

Answer 1

证明：
连接OB,则
∠AOB=2*∠ACB,∠OAB=∠OBA
∵AD⊥BC
∴∠OAB
=(1/2)(180°-∠AOB)
=90°-(1/2)∠AOB
=90°-∠ACB
=∠DAC
∵E是弧BC的中点
∴∠EAB=∠EAC
∴∠EAO
=∠EAB-∠OAB
=∠EAC-∠DAC
=∠EAD
得证