Question

设△ABC中∠ABC和∠ACB的角平分线分别交对边于点EF,若BE＝CF,求证AB＝AC

Answer 1

证法一 设AB≠AC,不妨设AB>AC,这样∠ACB>∠ABC,从而∠BCF=∠FCE=∠ACB/2>∠ABC/2=∠CBE=∠EBF.在△BCF和△CBE中,因为BC=BC,BE=CF,∠BCF>∠CBE.所以 BF>CE.(1) 作平行四边形BEGF,则∠EBF=∠FGC,EG=BF,FG=BE=CF,连CG,...