如图,AD平分∠BAC点E.F分别在BD.AD边上,DE=CD,EF=AC,求证:EF∥AB_百度...答:证明:延长FD到M,使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=∠CMD=∠EFD,得:EF//AB.参考:http://zhidao.baidu.com/link?url=U1Y4UO2Fdm63eh2bVjFONKSqaw6Ub0FQR4j1WmduuEcjop4lDsBy7dWdNR51DnunsCm5...