三角形ABC,D点为BC的中点.E点为AC边上一点.连接AD,BE.相交于F点.且有AE=EF.求证：BF=AC.

延长AD到点G,使得：DG = DA .因为,DG = DA ,DB = DC ,所以,ABGC是平行四边形；可得：AC‖BG ,AC = BG .因为,AC‖BG ,所以,∠BGF = ∠EAF .因为,AE = EF ,所以,∠EAF = ∠EFA .因为,∠BGF = ∠EAF = ∠EFA = ∠BFG ,...