Question

已知，如图CD是⊙O的切线，C是切点，直径AB的延长线与CD相交于D，连接OC、BC．                  

（1）写出三个不同类型的结论；                                                            

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

                                                                   

Answer 1

【考点】切线的性质．                                                                       

【专题】开放型．                                                                              

【分析】（1）CD是圆的切线可得出的有：OC⊥CD（切线的性质），CD²=DBDA（切线长定理），△BCD∽△CAD（弦切角定理），AB是圆的直角可得出的有∠ACB=90°（圆周角定理）等．只要正确的都可以；                

（2）由BD=OB可知，BC是直角三角形OCD底边上的中线，因此BC=OB=OD．因此三角形OBC就是个等边三角形，因此∠COB=60°，也就求出了∠D=30°，然后根据等边对等角，且外角为60°可在三角形OAC中求出∠A=30°，然后根据等角对等边即可得出CA=CD．                                                                           

【解答】（1）解：不同类型的结论有：                                                  

△BCD∽△CAD，                                                                              

OC⊥CD，                                                                                         

△ABC是直角三角形，                                                                            

OC²+CD²=OD²，                                                                                

CD²=DBDA，                                                                                     

∠ECD=∠OCA；                                                                               

                                                                                                          

（2）证明：∵CD是圆O的切线，                                                          

∴OC⊥CD，                                                                                      

∵OB=BD，                                                                                        

∴BC是直角三角形OCD斜边上的中线，                                                

∴BD=OB=BC=OC，                                                                           

∴△OBC是等边三角形，                                                                        

∴∠COB=60°，                                                                                 

∴∠D=90﹣60=30°；                                                                         

∵OA=OC，                                                                                       

∴∠A=∠OCA=30°，                                                                         

∴∠A=∠D，                                                                                     

即CA=AD．                                                                                       

【点评】本题主要考查了切线的性质，圆周角定理，等边三角形的性质等知识点的综合运用．