证明：∵AE切⊙O于点A，AB为⊙O的直径，∴∠BAE=90°，∠ACE=∠ACB=90°，∴∠ACE=∠BAE=90°． 又∵∠E=∠E，∴Rt△ECA∽Rt△EAB，∴ EC AE ＝ AE EB ，∴AE2=EB?EC．
