连接OC，AC∵C是AD弧的中点∴∠ABC=∠CBE∵AB是直径，CE⊥BD∴∠ACB=∠CEB=90°∴△ABC∽△CBE∴∠OAC(∠BAC)=∠BCE∵OA=OC∴∠OAC=∠OCA=∠BCE∵∠OCA+∠BCO=90°∴∠BCE+∠BCO=90°即∠OCE=90°∴OC⊥CE∴CE是⊙O的切线
