Question

如图,图1中三角形ABC是等边三角形,E为AC中点,F是线段BC延长线上一点,且CF=AE,连接BE,EF

Answer 1

1.因为△ABC为等边△,E为AC中点,因此∠EBC = 30° & AE = EC (等腰三角形顶角三线合一定律)因为AE = CF,AE = EC,因此EC = CF =>∠EFC = 30°△EBF中,因为∠EBC = ∠EFC = 30°,因此BE = EF

2.

在AC沿线上,取CG = AE,并连接FG因为∠2 = ∠1 = 60° (对顶角相等),CG = AE = CF,因此△CFG为等边△ => ∠CGF = 60° & FG = CG = CF因为 ∠A = ∠CGF = 60°,AE = FG,EG = EC+CG=EC+AE=AC=AB,因此△ABE全等于△GEF=> BE = EF