连接AC∵AD是直径∴∠ACD=∠ACE=90°∵CB=CE，那么∠E=∠CBE∠CBE=∠D∴∠E=∠D∵∠E=∠D，∠ACD=∠ACE=90°AC=AC∴△ACE≌△ACD(AAS)∴∠CAE=∠CAD即∠CAB=∠CAD那么BC=CD∴弧BC=弧CD即C是弧BD中点
