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

【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