Question

三角形ABC中,AD平分角BAC,M是BC的中点,ME平行AD交AB于F,交CA的延长线于E,证明BF＝CE

Answer 1

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.