如圖所示，點D、E分別是AB、AC的中點，點F、G分別為BD、CE的中點，若FG=6，則DE+BC=　　　　　...

問題詳情：

如圖所示，點D、E分別是AB、AC的中點，點F、G分別為BD、CE的中點，若FG=6，則DE+BC=　　　　　　，BC=　　　　　　．                                                                           

【回答】

12　， 　8　．                                                                                 

【考點】三角形中位線定理．                                                                  

【專題】計算題．                                                                              

【分析】根據中位線定理得：DE=BC，根據梯形中位線定理得FG=（DE+BC），由FG=6求得DE+BC的值即可．                                                                          

【解答】解：∵點F、G分別為BD、CE的中點，                                          

∴FG=（DE+BC），                                                                        

∵FG=6，                                                                                          

∴DE+BC=2FG=2×6=12；                                                                        

∵D、E分別是AB、AC的中點，                                                             

∴DE=BC，                                                                                     

∴DE+BC=BC+BC=BC=12，                                                                

∴BC=8．                                                                                           

故*為：12；8．                                                                            

【點評】本題考查了梯形的中位線與三角形的中位線的*質，是一道不錯的幾何綜合題．                 

知識點：平行四邊形

題型：填空題