如图，E是矩形ABCD内的一个动点，连接EA、EB、EC、ED，得到△EAB、△EBC、△ECD、△EDA，设...

问题详情：

如图，E是矩形ABCD内的一个动点，连接EA、EB、EC、ED，得到△EAB、△EBC、△ECD、△EDA，设它们的面积分别是m、n、p、q，给出如下结论：                                                                 

①m+n=q+p；                                                                                     

②m+p=n+q；                                                                                     

③若m=n，则E点一定是AC与BD的交点；                                                

④若m=n，则E点一定在BD上．                                                               

其中正确结论的序号是（　　）                                                              

A．①③                       B．②④                       C．①②③                   D．②③④

【回答】

B【考点】矩形的*质．                                                                          

【分析】过E作MN⊥AB，交AB于M，CD于N，作GH⊥AD，交AD于G，BC于H，由矩形的*质容易*出①不正确，②正确；若m=n，则p=q，作AP⊥BE于P，作CQ⊥DE于Q，延长BE交CD于F，先*AP=CQ，再*△ABP≌△CFQ，得出AB=CF，F与D重合，得出③不正确，④正确，即可得出结论．                       

【解答】解：过E作MN⊥AB，交AB于M，CD于N，作GH⊥AD，交AD于G，BC于H，如图1所示：                  

则m=ABEM，n=BCEH，p=CDEN，q=ADEG，                                      

∵四边形ABCD是矩形，                                                                          

∴AB=CD=GH，BC=AD=MN，                                                                

∴m+p=ABMN=ABBC，n+q=（BCGH=BCAB，                                      

∴m+p=n+q；                                                                                     

∴①不正确，②正确；                                                                       

若m=n，则p=q，作AP⊥BE于P，作CQ⊥DE于Q，延长BE交CD于F，如图2所示：                

则∠APB=∠CQF=90°，                                                                           

∵m=BEAP，n=BECQ，                                                                      

∵m=n，                                                                                            

∴AP=CQ，                                                                                        

∵AB∥CD，                                                                                      

∴∠1=∠2，                                                                                      

在△ABP和△CFQ中，                                                                        

，                                                                         

∴△ABP≌△CFQ（AAS），                                                                   

∴AB=CF，                                                                                        

∴F与D重合，                                                                                    

∴E一定在BD上；                                                                              

∴③不正确，④正确．                                                                       

故选：B．                                                                                          

【点评】本题考查了矩形的*质、三角形面积的计算、全等三角形的判定与*质；熟练掌握矩形的*质，*三角形全等是解决问题的关键．                                                                          

知识点：特殊的平行四边形

题型：选择题