证明:(1)∵AB=AC
∴△ABC是等腰三角形,∠ABC=∠ACB
∵点A、B、C、D在⊙O上
∴∠ACB与∠ADB是园周角且同弧AB
∴∠ACB=∠ADB,即∠ABC=∠ADB
∵在△ABE和△ADB中,∠ABC=∠ADB,∠BAD=∠DAB
∴△ABE∽△ADB
(2)连接OA
∵点A、B、C、D在⊙O上,AB=AC
∴OA垂直平分BC
∵AE=1/2ED,FB=1/2BD
∴AD=3/2ED,DF=3/2BD
即AD/ED=DF/BD=3/2
∵在△DAF和△DEB中,
∠ADF=∠EDB,AD/ED=DF/BD
∴△DAF∽△DEB
∴∠F=∠EBD
∴BE∥FA
∴OA⊥AF
∵OA是⊙O的半径
∴AF是⊙O的切线