解:(1)证明:连接OD,∵BC是⊙O的切线,∴OD⊥BC,∵AC⊥BC,∴OD∥AC,∴∠2=∠3;∵OA=OD,∴∠1=∠3,∴∠1=∠2,∴AD平分∠BAC;(2)解:∵BC与圆相切于点D.∴BD2=BE﹒BA,∵BE=2,BD=4,∴BA=8,∴AE=AB﹣BE=6,∴⊙O的半径为3.