证明：延长AB、CD交于点G
∵AD‖BC ∴GA/AB=GD/CD (1)
∵AE/EB=DF/FC ∴AE/EB+1=DF/FC+1 即 AB/EB=CD/FC ∴EB/AB=CF/CD
∴AB/CD=EB/CF (2)
∵AE/EB=DF/FC ∴EB/AE=FC/DF ∴EB/AE+1=FC/DF+1 即AB/AE=CD/DF
∴ AE/AB=DF/CD (3)
由（1）+（3）得 GE/AB=GF/CD
即 AB/CD=GE/GF (4)
由（2）和（4）得 EB/CF=GE/GF 即 GF/CF=GE/EB
∴EF‖AD‖BC