∴∠ADB =∞∠CBD (two straight lines are parallel and the internal dislocation angles are equal)

∫AE//CF

∴∠AED=∠CFB (two straight lines are parallel and the internal angles are equal)

∴△ADE∽△CBF

Once again: AD = BC

∴△ADE≌△CBF

∴BF=DE

∴BF-EF=DE-DF

Namely: BE=DF

& lt2> Proof: Let AD be handed over to BC at M point.

∴∠AMB=∠DMC (equal to vertex angle)

∫AB//CD

∴∠ ABC =∠ DCB, ∠bad =∠CDA (two straight lines are parallel with equal internal angles).

∴△ABM∽△DMC

AB = CD

∴△ABM≌△DMC

∴CM=BM

∵BE//CF can prove △ BEM △ CFM in the same way.

∴BE=FC

& lt3> Proof: Let AD be handed over to BC at M point.

∴∠AMC =∞∠BMD (equal to vertex angle)

∵∠ 1 =∠ 2 = 90.

∴△AMC∽△BMD

AC = BD

∴△AMC≌△BMD

∴CM=DM,AM=BM

∴CM+BM=DM+AM

Namely: AD=BC