∴∠GBC= 1/2 ∠ABC，∠GCB= 1/2 ∠BCD

In parallelogram ABCD, AB is parallel to CD.

∴ ∠ABC+∠BCD= 180

∴∠gbc+∠gcb= 1/2(∠abc+∠bcd)=90

△GBC，∠ g = 180-(∠ GBC+∠ GCB) = 90。

Similarly, it can be proved that the other three internal angles of the quadrilateral EFGH are all 90.

So the quadrilateral EFGH is a rectangle.