题目

已知：在△ABC中，以AC边为直径的⊙O交BC于点D，在劣弧上取一点E使∠EBC = ∠DEC，延长BE依次交AC于G，交⊙O于H.(1)求证：AC⊥BH(2)若∠ABC= 45°，⊙O的直径等于10，BD =8，求CE的长.          答案：证明：（1）连结AD                             (1分)           ∵∠DAC = ∠DEC  ∠EBC = ∠DEC           ∴∠DAC = ∠EBC                  (2分)           又∵AC是⊙O的直径 ∴∠ADC=90°    (3分)             ∴∠DCA+∠DAC=90° ∴∠EBC+∠DCA = 90°             ∴∠BGC=180°–（∠EBC+∠DCA) = 180°–90°=90°             ∴AC⊥BH                        （5分）          （2）∵∠BDA=180°–∠ADC = 90°  ∠ABC = 45°  ∴∠BAD = 45°              ∴BD = AD              ∵BD =8  ∴AD =8                 （6分）              又∵∠ADC =90°    AC =10                 ∴由勾股定理 DC== = 6                ∴BC=BD+DC=8+6=14              (7分)               又∵∠BGC = ∠ADC =90°    ∠BCG =∠ACD                 ∴△BCG∽△ACD                 ∴ =                  ∴ =   ∴CG =             (8分)                 连结AE  ∵AC是直径  ∴∠AEC=90°    又因 EG⊥AC                 ∴ △CEG∽△CAE   ∴ =   ∴CE2=AC · CG = ´ 10 = 84                 ∴CE = = 2               （10分） 解析:略 

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