已知,点A、F、E、C在同一条直线上,AF=CE,BE∥DF,BE=DF,求证:AB∥CD∵BE∥DF,∴∠AEB=∠CFD∵AF=CE,∴AF+FE=CE+FE,即AE=CF,在△CDF≌△ABE中,AE=CF,∠AEB=∠CFD,BE=DF.∴△CDF≌△ABE（SAS）,∴∠BAE=∠DCF∴AB∥CD...