Question

如圖，已知△PDC是⊙O的內(nèi)接三角形，CP=CD，若將△PCD繞點P順時針旋轉，當點C剛落在⊙O上的A處時，停止旋轉，此時點D落在點B處．                                                                     

（1）求證：PB與⊙O相切；                                                                  

（2）當PD=2，∠DPC=30°時，求⊙O的半徑長．                                    

                                                            

Answer 1

【考點】切線的判定與性質；全等三角形的判定與性質；旋轉的性質．                 

【專題】探究型．                                                                              

【分析】（1）連接OA、OP，由旋轉可得：△PAB≌△PCD，再由全等三角形的性質可知AP=PC=DC，再根據(jù)∠BPA=∠DPC=∠D可得出∠BPO=90°，進而可知PB與⊙O相切；                                               

（2）過點A作AE⊥PB，垂足為E，根據(jù)∠BPA=30°，PB=2，△PAB是等腰三角形，可得出BE=EP=，PA=2，PB與⊙O相切于點P可知∠APO=60°，故可知PA=2．                                           

【解答】（1）證明：連接OA、OP，OC，由旋轉可得：△PAB≌△PCD，                   

∴PA=PC=DC，                                                                                 

∴AP=PC=DC，∠AOP=∠POC=2∠D，∠APO=∠OAP=，                   

又∵∠BPA=∠DPC=∠D，                                                                       

∴∠BPO=∠BPA+=90°                                                      

∴PB與⊙O相切；                                                                             

                                                                                                          

（2）解：過點A作AE⊥PB，垂足為E，                                                

∵∠BPA=30°，PB=2，△PAB是等腰三角形；                                           

∴BE=EP=，（6分）                                                                          

PA===2                                                                          

又∵PB與⊙O相切于點P，                                                                     

∴∠APO=60°，                                                                                 

∴OP=PA=2．                                                                                    

【點評】本題考查的是切線的判定與性質、全等三角形的判定與性質及圖形旋轉的性質，能根據(jù)題意作出輔助線是解答此題的關鍵．