∴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.