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        1. 已知,如圖CD是⊙O的切線,C是切點,直徑AB的延長線與CD相交于D,連接OC、BC.                  

          (1)寫出三個不同類型的結(jié)論;                                                            

          (2)若BD=OB,求證:CA=CD.                                                          

                                                                             


          【考點】切線的性質(zhì).                                                                       

          【專題】開放型.                                                                              

          【分析】(1)CD是圓的切線可得出的有:OC⊥CD(切線的性質(zhì)),CD2=DBDA(切線長定理),△BCD∽△CAD(弦切角定理),AB是圓的直角可得出的有∠ACB=90°(圓周角定理)等.只要正確的都可以;                

          (2)由BD=OB可知,BC是直角三角形OCD底邊上的中線,因此BC=OB=OD.因此三角形OBC就是個等邊三角形,因此∠COB=60°,也就求出了∠D=30°,然后根據(jù)等邊對等角,且外角為60°可在三角形OAC中求出∠A=30°,然后根據(jù)等角對等邊即可得出CA=CD.                                                                           

          【解答】(1)解:不同類型的結(jié)論有:                                                  

          △BCD∽△CAD,                                                                              

          OC⊥CD,                                                                                         

          △ABC是直角三角形,                                                                            

          OC2+CD2=OD2,                                                                                

          CD2=DBDA,                                                                                     

          ∠ECD=∠OCA;                                                                               

                                                                                                                    

          (2)證明:∵CD是圓O的切線,                                                          

          ∴OC⊥CD,                                                                                      

          ∵OB=BD,                                                                                        

          ∴BC是直角三角形OCD斜邊上的中線,                                                

          ∴BD=OB=BC=OC,                                                                           

          ∴△OBC是等邊三角形,                                                                        

          ∴∠COB=60°,                                                                                 

          ∴∠D=90﹣60=30°;                                                                         

          ∵OA=OC,                                                                                       

          ∴∠A=∠OCA=30°,                                                                         

          ∴∠A=∠D,                                                                                     

          即CA=AD.                                                                                       

          【點評】本題主要考查了切線的性質(zhì),圓周角定理,等邊三角形的性質(zhì)等知識點的綜合運用.

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