过B作BG∥AC交EM的延长线于G.
∵BG∥AC,∠BGD＝∠CEM,∠GBM＝∠ECM,而AM＝CM,∴△BMG≌△CNE,
∴BG＝CE.
∵AD∥EM,∴∠BFM＝∠BAD,∠CEM＝∠CAD,而∠BAD＝∠CAD,∴∠BFM＝∠CEM.
由∠BGD＝∠CEM,∠BFM＝∠CEM,得：∠BGM＝∠BFM,∴BG＝BF.
由BG＝CE,BG＝BF,得：BF＝CE.