Answer :

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)

1/2 AD = 1/2 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 = 1/2 × Area (ABFH) ….. (i)

Similarly,

It can also be proved that

Area of triangle HGF = 1/2 × Area (HDCF) …… (ii)

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

Area of triangle HEF + Area of triangle HGF = 1/2 Area (ABFH) + 1/2 Area (HDCF)

Area (EFGH)= 1/2 [Area (ABFH) + Area (HDCF)]

Area (EFGH) = 1/2 Area (ABCD)

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