AB=CD,BE=DF,AF=CE求证BE‖DF

如图,AB=CD,BE=DF,AF=CE.说明(1)BE=DF(2)AB//CD,BE//DF.   已知：如图,BE⊥AC,DF⊥AC,垂足分别是点E、F.AF=CE,BE=DF.求证：AB=CD 在四边形ABCD中,AB‖CD,AB⊥CD,∠AEB=∠CED.F为BC的中点,求证AF=DF=0.5（BE+CE) 如图AB=CD AF平行DF CF平行BE 求证AF=DE 已知:点A、F、E、C、在同一条线上,AF=CE,BF平行DF,BE=DF.求证:AB平行CD 已知四边形ABCD,延长AB到E使BE=DF,连接CE、AF.求证:AF=CE 如图所示,已知CE垂直AB于E,DF垂直AB于F,AF=BE,AC=BD.求证：CE＝DF 已知,如图,AB//CD,AD//BC,点E,F分别在AD,BC上,且BE=DF.求证AF=CE 如图,已知AB＝CD,DF＝BE,AF＝CE,求证:∠CDF. 已知：如图,AB＝CD,AE＝CF,BE＝DF.求证：AF∥CE 如图 已知AB=CD,AE=DF,CE=BF.求证：AF=DE 如图,ab=cd,ae=df,ce=fb,求证af=de