证明:连接BE,
∵AE是直径,∴∠ABE=90°,
∵AD⊥BC,∴∠ADC=90°=∠ABE,
∵∠E=∠C,
∴ΔABE∽ΔADC,
∴AB/AD=AE/AC,
∴AB*AC=AD*AE.
(注:结论不是AB*BC=AD*AE)