如图，在矩形ABCD中，∠ADC的平分线交BC于点E、交AB的延长线于点F，G是EF的中点，连接AG、CG． ...

问题详情：

如图，在矩形ABCD中，∠ADC的平分线交BC于点E、交AB的延长线于点F，G是EF的中点，连接AG、CG．                                            

（1）求*：BE=BF；                                                                         

（2）请判断△AGC的形状，并说明理由．                                              

【回答】

【解答】（1）*：∵四边形ABCD为矩形，                                              

∴AB∥CD，AD∥BC，                                                                            

∴∠F=∠CDF，∠ADF=∠BEF，                                                             

∵DF平分∠ADC，                                                                            

∴∠CDF=∠ADF，                                                                            

∴∠F=∠BEF，                                                                                  

∴BE=BF；                                                                                         

（2）解：△AGC为等腰直角三角形，理由如下：                                          

如图，连接BG，                                                                               

由（1）可知BE=BF，且∠FBE=90°，                                                      

∴∠F=45°，                                                                                      

∴AF=AD=BC，                                                                                 

∵G为EF中点，                                                                               

∴BG=FG，∠EBG=45°，                                                                         

在△AGF和△CGB中，                                                                           

，                                                                                 

∴△AGF≌△CGB（SAS），                                                                   

∴AG=CG，∠AGF=∠BGC，                                                                   

∴∠BGF+∠AGB=∠AGB+∠AGC，                                                         

∴∠AGC=∠BGF=90°，                                                                           

∴△AGC为等腰直角三角形．                                                                 

知识点：特殊的平行四边形

题型：解答题