连接fd、bf,fd与ab交与g,
af‖cd（bc),∠cde=∠eaf,∠ced=∠aef,ae=ed(e是ad的中点),△ced≌△aef,fe=ec,e是fc的中点.af=cd,
又af‖cd（bc),四边形afdc为平行四边形,df=ac,∠bdf=∠acb,又ab=ac
∠abc=∠acb,所以∠bdf=∠abc,
又ab=ac,df=ac（已证）,所以df=ab,bd为公共边,△abd≌△fbd,fb=ad
又af‖bc,四边形adbf为等腰梯形.