Question

已知：在△ABC中，以AC邊為直徑的⊙O交BC于點D，在劣弧上取一點E使∠EBC = ∠DEC，延長BE依次交AC于G，交⊙O于H.

(1)求證：AC⊥BH

(2)若∠ABC= 45°，⊙O的直徑等于10，BD =8，求CE的長.

 

 

 

 

 

 

 

 

 

 

 

Answer 1

證明：（1）連結AD                             (1分)

           ∵∠DAC = ∠DEC  ∠EBC = ∠DEC

           ∴∠DAC = ∠EBC                  (2分)

           又∵AC是⊙O的直徑 ∴∠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)

                 連結AE  ∵AC是直徑  ∴∠AEC=90°    又因 EG⊥AC

                 ∴ △CEG∽△CAE   ∴ =   ∴CE²=AC · CG = ´ 10 = 84

                 ∴CE = = 2               （10分）

 

解析:略