（本题满分9分）如图所示，△ABC内接于⊙O，AB是⊙O的直径，点D在⊙O 上，...

（本题满分9分）如图所示，△ABC内接于⊙O，AB是⊙O的直径，点D在⊙O

上，过点C的切线交AD的延长线于点E，且AE⊥CE，连接CD．

（1）求证：DC=BC；

（2）若AB=5，AC=4，求tan∠DCE的值．

说明: 6ec8aac122bd4f6e

（1）证明：连接OC················································································· 1分 ∵OA＝OC ∴∠OAC＝∠OCA ∵CE是⊙O的切线 ∴∠OCE＝90° ·············································· 2分 ∵AE⊥CE ∴∠AEC＝∠OCE＝90° ∴OC∥AE ·················································· 3分 ∴∠OCA＝∠CAD ∴∠CAD＝∠BAC ∴ ∴DC＝BC ··························································································· 4分（2）∵AB是⊙O的直径 ∴∠ACB＝90° ∴·························································· 5分 ∵∠CAE＝∠BAC ∠AEC＝∠ACB＝90° ∴△ACE∽△ABC······················································································ 6分 ∴ ∴ ······················································ 7分 ∵DC＝BC＝3 ∴····················································· 8分 ∴-----------9分 (其它解法参考得分) 【解析】略

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