In the following figure, AC // PS // QR and PQ // DB // SR.

Prove that :

Area of quadrilateral PQRS = 2 × Area of quad.ABCD.

From figure,

PQRS is a parallelogram.

Given,

AC // PS // QR and PQ // DB // SR.

∴ AQRC and APSC are also parallelograms.

We know that,

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

From figure,

△ ABC and || gm AQRC lie on same base AC and between same parallel lines AC and QR.

∴ Area of △ ABC = $\dfrac{1}{2} \times$ Area of || gm AQRC …………(1)

△ ADC and || gm ACSP lie on same base AC and between same parallel lines AC and PS.

∴ Area of △ ADC = $\dfrac{1}{2} \times$ Area of || gm ACSP …………(2)

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

⇒ Area of △ ABC + Area of △ ADC = $\dfrac{1}{2} \times$ Area of || gm AQRC + $\dfrac{1}{2} \times$ Area of || gm ACSP

⇒ Area of quadrilateral ABCD = $\dfrac{1}{2}$ (Area of || gm AQRC + Area of || gm ACSP)

⇒ Area of quadrilateral ABCD = $\dfrac{1}{2}$ Area of || PQRS

⇒ Area of || gm PQRS = 2 × Area of quadrilateral ABCD.

Hence, proved that area of quadrilateral PQRS = 2 × area of quad. ABCD.