题目

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． 答案：D【考点】正方形的性质；全等三角形的判定与性质；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】此题考查正方形的性质，关键是根据全等三角形的判定和性质得出EG=FG=1，再利用勾股定理计算．                                                

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