Question

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．                                                          

                                                                   

Answer 1

【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．