已知△ABC中,AB=AC直线DF交AB于点D,交BC于点E,交AC的延长线于点F,BD=CF,求证:DE=EF.
问题描述:
已知△ABC中,AB=AC直线DF交AB于点D,交BC于点E,交AC的延长线于点F,BD=CF,求证:DE=EF.
答
证明:作DG∥AC交BC于点G.∵DG∥AC,∴∠DGB=∠ACB,∵AB=AC,∴∠B=∠ACB,∴∠B=∠DGB=∠ACB,∴BD=DG,∠DGE=∠ECF,又∵BD=CF,∴DG=CF,在△DGE和△FCE中,∠DGE=∠ECF∠DEG=∠CEFDG=CF,∴△DGE≌△FCE(AAS...