题目

（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．  答案：（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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