Question

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.

Answer 1

证明：（1）连结AD                             

            ∵∠DAC = ∠DEC  ∠EBC = ∠DEC

            ∴∠DAC = ∠EBC                 

            又∵AC是⊙O的直径 ∴∠ADC=90°   

              ∴∠DCA+∠DAC=90° ∴∠EBC+∠DCA = 90°

              ∴∠BGC=180°–（∠EBC+∠DCA) = 180°–90°=90°

              ∴AC⊥BH                        

          （2）∵∠BDA=180°–∠ADC = 90°  ∠ABC = 45°  ∴∠BAD = 45°

               ∴BD = AD

               ∵BD = 8  ∴AD =8              

               又∵∠ADC = 90°    AC =10

                 ∴由勾股定理 DC==  = 6

                 ∴BC=BD+DC=8+6=14             

                又∵∠BGC = ∠ADC = 90°    ∠BCG =∠ACD

                  ∴△BCG∽△ACD

                  ∴  =

                  ∴ =   ∴CG =            

                  连结AE   ∵AC是直径  ∴∠AEC=90°    又因 EG⊥AC

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

                  ∴CE = = 2                （10分）