If E, F, G and H are mid-points of the sides AB, BC, CD and DA, respectively of a parallelogram ABCD, prove that area of the quad. EFGH = $\dfrac{1}{2}$ area of || gm ABCD.

Parallelogram ABCD with E, F, G and H as mid-points of the sides AB, BC, CD and DA, respectively is shown below:

AH = $\dfrac{1}{2}$AD (As H is the mid-point of AD)

BF = $\dfrac{1}{2}$BC (As F is the mid-point of BC)

So, AH = BF and AH || BF (As AD || BC).

So, ABFH is a || gm.

Since, || gm ABFH and △EFH are on same base FH and between same parallel lines AB and FH so,

area of △EFH = $\dfrac{1}{2}$ area of || gm ABFH …….(i)

Similarly,

HD = $\dfrac{1}{2}$AD (As H is the mid-point of AD)

FC = $\dfrac{1}{2}$BC (As F is the mid-point of BC)

So, HD = FC and HD || FC (As AD || BC).

So, HFCD is a || gm.

Since, || gm HFCD and △HFG are on same base HF and between same parallel lines HF and DC so,

area of △HFG = $\dfrac{1}{2}$ area of || gm HFCD …….(ii)

Adding (i) and (ii) we get,

area of △EFH + area of △HFG = $\dfrac{1}{2}$ area of || gm ABFH + $\dfrac{1}{2}$ area of || gm HFCD

area of quad. EFGH = $\dfrac{1}{2}$ (area of || gm ABFH + area of || gm HFCD)

= $\dfrac{1}{2}$ (area of || gm ABCD).

Hence, proved that area of the quad. EFGH = $\dfrac{1}{2}$ area of || gm ABCD.