．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E...

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                                

【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=，                                                                       

知识点：特殊的平行四边形

题型：选择题