Question

求解答八年级数学几何题已知：如图,AD//BC,E是线段CD的中点,AE平分角BAD,求证：BE平分角ABC

Answer 1

延长AE、BC,相交于点F.

已知,AD‖BC,∠DAE = ∠BAE ,DE = EC ,

可得：∠BFA = ∠DAE = ∠BAE ,AE = EF ,

所以,BA = BF ,BE是等腰△BAF底边上的中线,

可得：BE平分等腰△BAF的顶角∠ABF,

即有：BE平分∠ABC .