5
连结DF、OE,∵AD是直径,∴∠AFD=90°.又AB⊥BC,DC⊥BC,∴四边形BCDF是矩形.
∴BF=DC.由切割线定理得BE 2=BF·BA=1×4=4,BE=2.
∵OE⊥BC,DC⊥BC,AB⊥BC,∴CD∥OE∥AB.O为AD中点,
∴E为BC中点.∴BC=4.∴DF=4.在Rt△ADF中,AD=
=5.
5
连结DF、OE,∵AD是直径,∴∠AFD=90°.又AB⊥BC,DC⊥BC,∴四边形BCDF是矩形.
∴BF=DC.由切割线定理得BE 2=BF·BA=1×4=4,BE=2.
∵OE⊥BC,DC⊥BC,AB⊥BC,∴CD∥OE∥AB.O为AD中点,
∴E为BC中点.∴BC=4.∴DF=4.在Rt△ADF中,AD=
=5.