If E, F G and H are respectively the midpoints of the sides AB, BC, CD and AD of a parallelogram ABCD, show that ar(EFGH) ar(ABCD).

Construct line HF as shown and construct perpendiculars EJ and GK on HF as shown

The line HF divides the parallelogram ABCD into two parallelograms ABFH and parallelogram HFCD

Consider parallelogram ABFH

EJ is the perpendicular distance between AB and HF therefore EJ is the height of parallelogram ABFH and also EJ is height of ΔEFH

Area of ΔEFH = × HF × EJ

But area of parallelogram ABFH = HF × EJ

Therefore, area of ΔEFH = × area of parallelogram ABFH …(i)

Consider parallelogram HFCD

GK is the perpendicular distance between DC and HF therefore GK is the height of parallelogram HFCD and also GK is height of ΔGFH

Area of ΔGFH = × HF × GK

But area of parallelogram HFCD = HF × GK

Therefore, area of ΔGFH = × area of parallelogram HFCD …(ii)

Add equation (i) and (ii)

⇒ area of ΔEFH + area of ΔGFH = × area of parallelogram ABFH + × area of parallelogram HFCD

⇒ area of ΔEFH + area of ΔGFH = × (area of parallelogram ABFH + area of parallelogram HFCD) …(iii)

From figure area of ΔEFH + area of ΔGFH = area of parallelogram EFGH and

area of parallelogram ABFH + area of parallelogram HFCD = area of parallelogram ABCD

therefore equation (iii) becomes

area of parallelogram EFGH = × area of parallelogram ABCD

hence proved