If E, F, G, and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar(EFGH) = 1/2 ar(ABCD)

Given: E, F, G, and H are respectively the mid-points of the sides of a parallelogram ABCD.
To Prove: ar(EFGH) = 1/2 ar(ABCD)
Construction: H and F are joined.
Proof: AD || BC and AD = BC (Opposite sides of a parallelogram)
=> 1/2 AD = 1/2 BC
Also, AH || BF and DH || CF => AH = BF and DH = CF
H and F are mid points
Thus, ABFH and HFCD are parallelograms.
Now, △EFH and ||gm ABFH lie on the same base FH and between the same parallel lines AB and HF.
∴ Area of △EFH = 1/2 ar(ABFH) — (i)
Also, Area of △GHF = 1/2 ar(HFCD) — (ii)
Adding (i) and (ii), Area of △EFH + area of △GHF = 1/2 ar(ABFH) + 1/2 ar(HFCD)
=> Area of EFGH = Area of ABFH
=> ar(EFGH) = 1/2 ar(ABCD)