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,
∴ Area (ΔHEF) = 1/2 Area (ABFH) ... (1)
Similarly, it can be proved that
Area (ΔHGF) = 1/2 Area (HDCF) ... (2)
On adding equations (1) and (2), we obtain
