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        1. (本題滿分9分)如圖所示,△ABC內(nèi)接于⊙O,AB是⊙O的直徑,點D在⊙O

          上,過點C的切線交AD的延長線于點E,且AE⊥CE,連接CD.

          (1)求證:DC=BC;  

          (2)若AB=5,AC=4,求tan∠DCE的值.

           

          (1)證明:連接OC················································································· 1分

          ∵OA=OC

          ∴∠OAC=∠OCA

          ∵CE是⊙O的切線

          ∴∠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的直徑  ∴∠ACB=90°

          ·························································· 5分

          ∵∠CAE=∠BAC  ∠AEC=∠ACB=90°

          ∴△ACE∽△ABC······················································································ 6分

              ∴   ······················································ 7分

           

          ∵DC=BC=3

          ····················································· 8分

           

          -----------9分         (其它解法參考得分)

           

          解析:略

           

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