∴∠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.