如圖，E是矩形ABCD內的一個動點，連線EA、EB、EC、ED，得到△EAB、△EBC、△ECD、△EDA，設...

問題詳情：

如圖，E是矩形ABCD內的一個動點，連線EA、EB、EC、ED，得到△EAB、△EBC、△ECD、△EDA，設它們的面積分別是m、n、p、q，給出如下結論：                                                                 

①m+n=q+p；                                                                                     

②m+p=n+q；                                                                                     

③若m=n，則E點一定是AC與BD的交點；                                                

④若m=n，則E點一定在BD上．                                                               

其中正確結論的序號是（　　）                                                              

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

【回答】

B【考點】矩形的*質．                                                                          

【分析】過E作MN⊥AB，交AB於M，CD於N，作GH⊥AD，交AD於G，BC於H，由矩形的*質容易*出①不正確，②正確；若m=n，則p=q，作AP⊥BE於P，作CQ⊥DE於Q，延長BE交CD於F，先*AP=CQ，再*△ABP≌△CFQ，得出AB=CF，F與D重合，得出③不正確，④正確，即可得出結論．                       

【解答】解：過E作MN⊥AB，交AB於M，CD於N，作GH⊥AD，交AD於G，BC於H，如圖1所示：                  

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

∵四邊形ABCD是矩形，                                                                          

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

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

∴m+p=n+q；                                                                                     

∴①不正確，②正確；                                                                       

若m=n，則p=q，作AP⊥BE於P，作CQ⊥DE於Q，延長BE交CD於F，如圖2所示：                

則∠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與D重合，                                                                                    

∴E一定在BD上；                                                                              

∴③不正確，④正確．                                                                       

故選：B．                                                                                          

【點評】本題考查了矩形的*質、三角形面積的計算、全等三角形的判定與*質；熟練掌握矩形的*質，*三角形全等是解決問題的關鍵．                                                                          

知識點：特殊的平行四邊形

題型：選擇題