证明：连接OC，BC，AC，∵DC切⊙O于C，∴OC⊥CD，∴∠D+∠COD=90°，∵AB为⊙O的直径，∴∠ACB=90°，∴∠CAB+∠B=90°，∵∠AEC=∠B，∴∠CAB+∠AEC=90°，∵ CE = BE ，∴∠COE=∠CAB，∴∠AEC+∠COE=90°，∴∠AEC=∠D．
