如图所示，点D、E分别是AB、AC的中点，点F、G分别为BD、CE的中点，若FG=6，则DE+BC=　　　　　...

问题详情：

如图所示，点D、E分别是AB、AC的中点，点F、G分别为BD、CE的中点，若FG=6，则DE+BC=　　　　　　，BC=　　　　　　．                                                                           

【回答】

12　， 　8　．                                                                                 

【考点】三角形中位线定理．                                                                  

【专题】计算题．                                                                              

【分析】根据中位线定理得：DE=BC，根据梯形中位线定理得FG=（DE+BC），由FG=6求得DE+BC的值即可．                                                                          

【解答】解：∵点F、G分别为BD、CE的中点，                                          

∴FG=（DE+BC），                                                                        

∵FG=6，                                                                                          

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

∵D、E分别是AB、AC的中点，                                                             

∴DE=BC，                                                                                     

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

∴BC=8．                                                                                           

故*为：12；8．                                                                            

【点评】本题考查了梯形的中位线与三角形的中位线的*质，是一道不错的几何综合题．                 

知识点：平行四边形

题型：填空题