Therefore, EF= 12AC, FG= 12BD, GH= 12AC, HE= 12BD,

In trapezoidal ABCD, AB=DC,

So AC=BD,

∴EF=FG=GH=HE，

The quadrilateral EFGH is a diamond.

Let AC and EH intersect at point m,

In △ABD, e and h are the midpoint of AB and AD, respectively.

EH∨BD，

Similarly, GH∑AC,

∵AC⊥BD，

∴∠BOC=90，

∴∠EHG=∠EFG=90，

∴ quadrilateral EFGH is a square.

② connect eg.

In the trapezoidal ABCD,

E and g are the midpoint of AB and DC respectively,

∴EG= 12(AD+BC)=3.

At Rt△EHG,

∫EH2+GH2 = EG2，EH=GH，

∴EH2=92, that is, the area of the quadrilateral EFGH is 92.