Question

如圖，△ABC中，AB=AC，AD⊥BC，CE⊥AB，AE=CE．求證：                    

（1）△AEF≌△CEB；                                                                            

（2）AF=2CD．                                                                                 

                                                                        

Answer 1

【考點】全等三角形的判定與性質(zhì)；等腰三角形的性質(zhì)．                               

【專題】證明題．                                                                              

【分析】（1）由AD⊥BC，CE⊥AB，易得∠AFE=∠B，利用全等三角形的判定得△AEF≌△CEB；                

（2）由全等三角形的性質(zhì)得AF=BC，由等腰三角形的性質(zhì)“三線合一”得BC=2CD，等量代換得出結(jié)論．                

【解答】證明：（1）∵AD⊥BC，CE⊥AB，                                           

∴∠BCE+∠CFD=90°，∠BCE+∠B=90°，                                               

∴∠CFD=∠B，                                                                                 

∵∠CFD=∠AFE，                                                                             

∴∠AFE=∠B                                                                                     

在△AEF與△CEB中，                                                                       

，                                                                             

∴△AEF≌△CEB（AAS）；                                                                    

                                                                                                          

（2）∵AB=AC，AD⊥BC，                                                                     

∴BC=2CD，                                                                                      

∵△AEF≌△CEB，                                                                            

∴AF=BC，                                                                                        

∴AF=2CD．                                                                                      

【點評】本題主要考查了全等三角形性質(zhì)與判定，等腰三角形的性質(zhì)，運用等腰三角形的性質(zhì)是解答此題的關(guān)鍵．