We know that

Area of parallelogram PQRS = ½ (Area of parallelogram ABCD)

Construct diagonals AC, BD and SQ

From the figure we know that S and R are the midpoints of AD and CD

Consider △ ADC

Using the midpoint theorem

SR || AC

From the figure we know that P and Q are the midpoints of AB and BC

Consider △ ABC

Using the midpoint theorem

PQ || AC

It can be written as

PQ || AC || SR

So we get

PQ || SR

In the same way we get SP || RQ

Consider △ ABD

We know that O is the midpoint of AC and S is the midpoint of AD

Using the midpoint theorem

OS || AB

Consider △ ABC

Using the midpoint theorem

OQ || AB

It can be written as

SQ || AB

We know that ABQS is a parallelogram

We get

Area of △ SPQ = ½ (Area of parallelogram ABQS) ……. (1)

In the same way we get

Area of △ SRQ = ½ (Area of parallelogram SQCD) …….. (2)

By adding both the equations

Area of △ SPQ + Area of △ SRQ = ½ (Area of parallelogram ABQS + Area of parallelogram SQCD)

So we get

Area of parallelogram PQRS = ½ (Area of parallelogram ABCD)

Therefore, it is proved that ar (||gm PQRS) = ½ × ar (||gm ABCD).