Question

（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

（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．

Answer 1

（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∽△AB C······················································································ 6分 ∴     ∴    ······················································ 7分 ∵DC＝BC＝3 ∴ ····················································· 8分 ∴ -----------9分          (其它解法参考得分)

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