Question

如图，在△ABC中，AB=10，AC=8，BC=6，经过点C且与边AB相切的动圆与CB，CA分别相交于点E，F，则线段EF长度的最小值是　　　　　　．                                                                  

                                                                          

Answer 1

　4.8　．                                                                                           

                                                                                                        

【考点】切线的性质；垂线段最短；勾股定理的逆定理．                               

【分析】设EF的中点为P，⊙P与AB的切点为D，连接PD，连接CP，CD，则有PD⊥AB；由勾股定理的逆定理知，△ABC是直角三角形PC+PD=EF，由三角形的三边关系知，PC+PD＞CD；只有当点P在CD上时，PC+PD=EF有最小值为CD的长，即当点P在直角三角形ABC的斜边AB的高CD上时，EF=CD有最小值，由直角三角形的面积公式知，此时CD=BCAC÷AB，进而求出即可．                                                                                 

【解答】解：如图，设EF的中点为P，⊙P与AB的切点为D，连接PD，连接CP，CD，则有PD⊥AB；              

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

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

∴PC+PD＞CD，                                                                                

∵当点P在直角三角形ABC的斜边AB的高CD上时，EF=CD有最小值，                    

∴CD=BCAC÷AB=4.8．                                                                            

故答案为：4.8．                                                                                

                                                                          

【点评】此题主要考查了切线的性质，勾股定理的逆定理，三角形的三边关系，直角三角形的面积公式求解，得出CD=BCAC÷AB是解题关键．