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

解析