Question

如图：在△ABC中,∠A=60°,∠B,∠C的平分线BE,CF相交于点O.求证：OE=OF,BF+CE=BC

Answer 1

楼主的图坑人啊,E、F标反了

在BC上截取BG＝BF,连结OG

∵BF＝BG,∠OBF＝∠OBG,OB＝OB

∴△OBF≌△OBG

∴∠BOF＝∠BOG,OF＝OG

∵∠BOC＝180°－∠OBC－∠OCB＝180°－1/2(∠ABC＋∠ACB)＝180°－1/2(180°－∠A)＝120°

∴∠BOG＝∠BOF＝180°－∠BOC＝60°

∴∠COG＝60°＝∠BOF＝∠COE

∵∠OCG＝∠OCE,OC＝OC

∴△COG≌△COE

∴CG＝CE,OG＝OE

∴BC＝BG＋CG＝BF＋CE,OE＝OF