Question

如圖，MN為⊙O的直徑，A、B是⊙O上的兩點，過A作AC⊥MN于點C，過B作BD⊥MN于點D，P為DC上的任意一點，若MN=20，AC=8，BD=6，則PA+PB的最小值是　　　　　　．                   

                                                                          

Answer 1

14　．                                                                                          

                                                                                                        

【考點】軸對稱-最短路線問題；勾股定理；垂徑定理．                                 

【專題】壓軸題；探究型．                                                                     

【分析】先由MN=20求出⊙O的半徑，再連接OA、OB，由勾股定理得出OD、OC的長，作點B關于MN的對稱點B′，連接AB′，則AB′即為PA+PB的最小值，B′D=BD=6，過點B′作AC的垂線，交AC的延長線于點E，在Rt△AB′E中利用勾股定理即可求出AB′的值．                                                                        

【解答】解：∵MN=20，                                                                        

∴⊙O的半徑=10，                                                                            

連接OA、OB，                                                                                  

在Rt△OBD中，OB=10，BD=6，                                                            

∴OD===8；                                                     

同理，在Rt△AOC中，OA=10，AC=8，                                                 

∴OC===6，                                                     

∴CD=8+6=14，                                                                                 

作點B關于MN的對稱點B′，連接AB′，則AB′即為PA+PB的最小值，B′D=BD=6，過點B′作AC的垂線，交AC的延長線于點E，                                                                                                

在Rt△AB′E中，                                                                                

∵AE=AC+CE=8+6=14，B′E=CD=14，                                                      

∴AB′===14．                                            

故答案為：14．                                                                            

                                                                          

【點評】本題考查的是軸對稱﹣最短路線問題、垂徑定理及勾股定理，根據題意作出輔助線，構造出直角三角形，利用勾股定理求解是解答此題的關鍵．