解答：证明：连接AC，∵EA切⊙O于A，∴∠EAB=∠ACB．∵ AB = AD ，∴∠ACD=∠ACB，AB=AD．于是∠EAB=∠ACD．又四边形ABCD内接于⊙O，∴∠ABE=∠D．∴△ABE∽△CDA．于是 AB CD = BE DA ，即AB?DA=BE?CD．∴AB2=BE?CD．
