如图,在△ABC中,AB=7,AC=11,点E是BC的中点,AD是∠BAC的平分线,EF‖AD...答:解:延长FE到G,使EG=EF,连接BG.∵EG=EF;EB=EC;∠BEG=∠CEF.∴⊿BEG≌⊿CEF(SAS),BG=CF;∠G=∠CFE,得BG∥AC.∵EF∥AD;AD平分∠BAC.∴∠BHA=∠G=∠CFE=∠CAD=∠BAD,得BH=BA=7;且四边形AFGH为平行四边形,HG=AF.则:AC=AF+CF=AF+BG=AF+(BH+HG)=AF+(7+HG)=7+2AF.即11=7...