如图，在正方形ABCD中，AD=5，点E、F是正方形ABCD内的两点，且AE=FC=3，BE=DF=4，则EF...

问题详情：

如图，在正方形ABCD中，AD=5，点E、F是正方形ABCD内的两点，且AE=FC=3，BE=DF=4，则EF的长为（　　）                                                                             

A．                           B．                       C．                           D．

【回答】

D【考点】正方形的*质；全等三角形的判定与*质；勾股定理；等腰直角三角形．                  

【分析】延长AE交DF于G，再根据全等三角形的判定得出△AGD与△ABE全等，得出AG=BE=4，由AE=3，得出EG=1，同理得出GF=1，再根据勾股定理得出EF的长．                                           

【解答】解：延长AE交DF于G，如图：                                                   

∵AB=5，AE=3，BE=4，                                                                         

∴△ABE是直角三角形，                                                                         

∴同理可得△DFC是直角三角形，                                                           

可得△AGD是直角三角形，                                                                     

∴∠ABE+∠BAE=∠DAE+∠BAE，                                                           

∴∠GAD=∠EBA，                                                                            

同理可得：∠ADG=∠BAE，                                                                    

在△AGD和△BAE中，                                                                       

，                                                                             

∴△AGD≌△BAE（ASA），                                                                   

∴AG=BE=4，DG=AE=3，                                                                       

∴EG=4﹣3=1，                                                                                 

同理可得：GF=1，                                                                            

∴EF=，                                                                       

故选D．                                                                                             

【点评】此题考查正方形的*质，关键是根据全等三角形的判定和*质得出EG=FG=1，再利用勾股定理计算．                                                

知识点：特殊的平行四边形

题型：选择题