Let us join HF

In parallelogram ABCD,

AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel)

AB = CD (Opposite sides of a parallelogram are equal)

AD = BC

And AH || BF

AH = BF and AH || BF (H and F are the mid-points of AD and BC)

Therefore, ABFH is a parallelogram

Since,

ΔHEF and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF

Therefore,

Area of triangle HEF = * Area (ABFH) (i)

Similarly,

It can also be proved that

Area of triangle HGF = * Area (HDCF) (ii)

On adding (i) and (ii), we get

Area of triangle HEF + Area of triangle HGF = Area (ABFH) + Area (HDCF)

= [Area (ABFH) + Area (HDCF)]

Area (EFGH) = Area (ABCD)