E, F, G and H are the mid-points of the sides of a parallelogram ABCD. Show that area of quadrilateral EFGH is half of the area of parallelogram ABCD.

Since, H and F are mid-points of AD and BC respectively.

∴ AH = $\dfrac{1}{2}AD$ and BF = $\dfrac{1}{2}BC$

Since, ABCD is a parallelogram.

∴ AD = BC and AD || BC (Opposite sides of parallelogram are equal)

⇒ $\dfrac{AD}{2} = \dfrac{BC}{2}$ and AD || BC

⇒ AH = BF and AH || BF.

Since, one pair of opposite sides are equal and parallel.

∴ ABFH is a || gm.

We know that,

The area of a triangle is half that of a parallelogram on the same base and between the same parallels.

Since, || gm ABFH and triangle HEF are on the same base FH and between the same parallel lines HF and AB.

∴ Area of △ HEF = $\dfrac{1}{2}$ Area of || gm ABFH ……….(1)

Since, || gm HFCD and triangle HGF are on the same base FH and between the same parallel lines HF and DC.

∴ Area of △ HGF = $\dfrac{1}{2}$ Area of || gm HFCD ……….(2)

Adding equations (1) and (2), we get :

⇒ Area of △ HEF + Area of △ FGH = $\dfrac{1}{2}$ Area of || gm ABFH + $\dfrac{1}{2}$ Area of || gm HFCD

⇒ Area of quadrilateral EFGH = $\dfrac{1}{2}$ (Area of || gm ABFH + Area of || gm HFCD)

⇒ Area of quadrilateral EFGH = $\dfrac{1}{2}$ Area of || gm ABCD.

Hence, proved that area of quadrilateral EFGH is half of the area of parallelogram ABCD.