∴AE∥CF,AB=CD，

E is the midpoint of AB, f is the midpoint of CD,

∴AE=CF，

∴ Quadrilateral AECF is a parallelogram,

∴AF∥CE.

De sigma BF can be obtained in the same way,

∴ quadrilateral FGEH is a parallelogram;

(2) When the parallelogram ABCD is rectangular and AB=2AD, the parallelogram EHFG is rectangular.

∫E and f are the midpoint of AB and CD respectively, AB=CD,

∴AE=DF and AE∑df,

∴ Quadrilateral AEFD is a parallelogram,

∴AD=EF，

If ∵AB=2AD and E is the midpoint of AB, then AB=2AE,

So AE=AD= 12AB,

At this time, EF=AE=AD=DF= 12AB, ∠ EAD = ∠ FDA = 90,

∴ Quadrilateral ADFE is a square,

∴eg=fg= 12af,af⊥de,∠egf=90，

At this point, the parallelogram EHFG is a rectangle.