Question

如图，已知△PDC是⊙O的内接三角形，CP=CD，若将△PCD绕点P顺时针旋转，当点C刚落在⊙O上的A处时，停止旋转，此时点D落在点B处．                                                                     

（1）求证：PB与⊙O相切；                                                                  

（2）当PD=2，∠DPC=30°时，求⊙O的半径长．                                    

                                                            

Answer 1

【考点】切线的判定与性质；全等三角形的判定与性质；旋转的性质．                 

【专题】探究型．                                                                              

【分析】（1）连接OA、OP，由旋转可得：△PAB≌△PCD，再由全等三角形的性质可知AP=PC=DC，再根据∠BPA=∠DPC=∠D可得出∠BPO=90°，进而可知PB与⊙O相切；                                               

（2）过点A作AE⊥PB，垂足为E，根据∠BPA=30°，PB=2，△PAB是等腰三角形，可得出BE=EP=，PA=2，PB与⊙O相切于点P可知∠APO=60°，故可知PA=2．                                           

【解答】（1）证明：连接OA、OP，OC，由旋转可得：△PAB≌△PCD，                   

∴PA=PC=DC，                                                                                 

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

又∵∠BPA=∠DPC=∠D，                                                                       

∴∠BPO=∠BPA+=90°                                                      

∴PB与⊙O相切；                                                                             

                                                                                                          

（2）解：过点A作AE⊥PB，垂足为E，                                                

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

∴BE=EP=，（6分）                                                                          

PA===2                                                                          

又∵PB与⊙O相切于点P，                                                                     

∴∠APO=60°，                                                                                 

∴OP=PA=2．                                                                                    

【点评】本题考查的是切线的判定与性质、全等三角形的判定与性质及图形旋转的性质，能根据题意作出辅助线是解答此题的关键．