E, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.

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

AH || BF and and DH || CF

⇒ AH = BF and DH = CF (H and F are mid points)

∴, ABFH and HFCD are parallelograms.

We know that, ΔEFH and parallelogram ABFH, both lie on the same FH the common base and in-between the same parallel lines AB and HF.

∴ area of EFH = 1/2 area of ABFH — (i)

And, area of GHF = 1/2 area of HFCD — (ii)

Adding (i) and (ii),

area of ΔEFH + area of ΔGHF = 1/2 area of ABFH + 1/2 area of HFCD

⇒ area of EFGH = area of ABFH

∴ ar (EFGH) = 1/2 ar(ABCD)