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 .