已知△ABC内接于圆O，AO是半径，AD⊥BC于点D，求证：∠BAD=∠OAC

垂径定理啊
提示 思路：∠BAD=∠OAC → △ABE≌△ACE → 作BC中点E → AD⊥BC

证明：延长AO交圆O于E,连接CE
∵AD⊥BC
∴∠ABC ∠BAD＝90
∵∠AEC、∠ABC所对应圆弧都为弧AC
∴∠AEC＝∠ABC
∵直径AE
∴∠OAC ∠AEC＝90
∴∠OAC ∠ABC＝90
∴∠OAC＝∠BAD