证明：连结BE,CE
∵⊙O为△ABC的外接圆,
AE为⊙O的直径
∴∠ABE=90°=∠ACE
圆周角相等：∠ABC=∠AEC
∠AEB=∠ACB
AD⊥BC
∴Rt△ABD∽Rt△AEC
Rt△ABE∽Rt△ADC
∴AB/AE=AD/AC,AB/AD=AE/AC
∴AB·AC=AD·AE