Question

如图，AD平分∠BAC点E.F分别在BD.AD边上，DE＝CD,EF＝AC,求证：EF∥AB

Answer 1

证明:延长FD到M,使握悔祥DM=DF;又DE=CD.则⊿CDM≌⊿EDF(SAS),∠EFD=∠CMD;CM=EF.
又EF=AC,则CM=AC,∠CAD=∠CMD.
又∠BAD=∠CAD,故∠BAD=∠CAD=∠前瞎CMD=∠EFD,得:EF/段搏/AB.

参考：http://zhidao.baidu.com/link?url=U1Y4UO2Fdm63eh2bVjFONKSqaw6Ub0FQR4j1WmduuEcjop4lDsBy7dWdNR51DnunsCm5qVVGLrZrU0m2ch-jUu7QpXAF8qww-1RkQO8iNXy

Answer 2

延长FD至G使DG=FD连接CG则⊿昌带CGD≌握唯⊿EDF∴∠CGD=∠EFD CG=EF ∵ EF=AC ∴AC=CG∴∠CAG=∠CGF∴∠CAG=∠EFD∵AD平分∠BAC∴段迅培∠BAD=∠CAD∴∠BAD=∠EFD∴EF∥AB

Answer 3

元芳