设EF⊥AD于G∵AD平分∠BAC∴∠BAD＝∠CAD∵DE∥AC∴∠ADE＝∠CAD∴∠ADE＝∠BAD∴RtΔEDG≌RtΔEAG∴AG=GD∴RtΔFAG≌RtΔFDG∴∠FAG＝∠FDG又∵∠FDG＝∠B+∠BAD,∠FAG＝∠CAF+∠CAD∴∠B+∠BAD＝∠CAF+∠CAD (∵∠...