Question

如圖，在△ABC中，AB=10，AC=8，BC=6，經(jīng)過點C且與邊AB相切的動圓與CB，CA分別相交于點E，F(xiàn)，則線段EF長度的最小值是　　　　　　．                                                                  

                                                                          

Answer 1

　4.8　．                                                                                           

                                                                                                        

【考點】切線的性質(zhì)；垂線段最短；勾股定理的逆定理．                               

【分析】設(shè)EF的中點為P，⊙P與AB的切點為D，連接PD，連接CP，CD，則有PD⊥AB；由勾股定理的逆定理知，△ABC是直角三角形PC+PD=EF，由三角形的三邊關(guān)系知，PC+PD＞CD；只有當(dāng)點P在CD上時，PC+PD=EF有最小值為CD的長，即當(dāng)點P在直角三角形ABC的斜邊AB的高CD上時，EF=CD有最小值，由直角三角形的面積公式知，此時CD=BCAC÷AB，進(jìn)而求出即可．                                                                                 

【解答】解：如圖，設(shè)EF的中點為P，⊙P與AB的切點為D，連接PD，連接CP，CD，則有PD⊥AB；              

∵AB=10，AC=8，BC=6，                                                                       

∴∠ACB=90°，PC+PD=EF，                                                                   

∴PC+PD＞CD，                                                                                

∵當(dāng)點P在直角三角形ABC的斜邊AB的高CD上時，EF=CD有最小值，                    

∴CD=BCAC÷AB=4.8．                                                                            

故答案為：4.8．                                                                                

                                                                          

【點評】此題主要考查了切線的性質(zhì)，勾股定理的逆定理，三角形的三邊關(guān)系，直角三角形的面積公式求解，得出CD=BCAC÷AB是解題關(guān)鍵．