Question

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

                                                                          

Answer 1

14　．                                                                                          

                                                                                                        

【考點(diǎn)】軸對(duì)稱(chēng)-最短路線問(wèn)題；勾股定理；垂徑定理．                                 

【專(zhuān)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(zhǎng)，作點(diǎn)B關(guān)于MN的對(duì)稱(chēng)點(diǎn)B′，連接AB′，則AB′即為PA+PB的最小值，B′D=BD=6，過(guò)點(diǎn)B′作AC的垂線，交AC的延長(zhǎng)線于點(diǎn)E，在Rt△AB′E中利用勾股定理即可求出AB′的值．                                                                        

【解答】解：∵M(jìn)N=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(diǎn)B關(guān)于MN的對(duì)稱(chēng)點(diǎn)B′，連接AB′，則AB′即為PA+PB的最小值，B′D=BD=6，過(guò)點(diǎn)B′作AC的垂線，交AC的延長(zhǎng)線于點(diǎn)E，                                                                                                

在Rt△AB′E中，                                                                                

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

∴AB′===14．                                            

故答案為：14．                                                                            

                                                                          

【點(diǎn)評(píng)】本題考查的是軸對(duì)稱(chēng)﹣?zhàn)疃搪肪�問(wèn)題、垂徑定理及勾股定理，根據(jù)題意作出輔助線，構(gòu)造出直角三角形，利用勾股定理求解是解答此題的關(guān)鍵．