连接AD、AC、BC,⊙A中,△ACD是等腰三角形,∠D=∠ACF；在⊙O中考查△ABC和△ACF:：∠B=∠D=∠ACF,∠BAC=∠CAF是两三角形公用角,所以△ABC∽△ACF,AB/AC=AC/AF,即AC²=AF·AB,把半径AC换成AE就是AE²=AF·AB.