证明:(1)连接OD.
∵AD∥OC,
∴∠BOC=∠OAD,∠COD=∠ODA,
= ;
(2)由(1)∠BOC=∠OAD,∠OAD=∠ODA.
∴∠BOC=∠ODA.
∵∠BOC+∠ADF=90°.
∴∠ODA+∠ADF=90°,
即∠ODF=90°.
∵OD是⊙O的半径,
∴CD是⊙O的切线.
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