In the adjoining figure, ABCD is a parallelogram. Any line through A cuts DC at a point P and BC produced at Q. Prove that : ar (ΔBPC) = ar (ΔDPQ).

We know that,

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

Since, triangle APB and parallelogram ABCD are on the same base AB and between the same parallels AB and DC.

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

Since, triangle ADQ and parallelogram ABCD are on the same base AD and between the same parallels AD and BQ.

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

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

⇒ Area of △ APB + Area of △ ADQ = $\dfrac{1}{2}$ Area of ||gm ABCD + $\dfrac{1}{2}$ Area of ||gm ABCD

⇒ Area of △ APB + Area of △ ADQ = Area of ||gm ABCD …..(3)

From figure,

⇒ Area of △ APB + Area of △ ADQ = Area of quadrilateral ADQB - Area of △ BPQ ……(4)

From equation (3) and (4), we get :

⇒ Area of quadrilateral ADQB - Area of △ BPQ = Area of ||gm ABCD

⇒ Area of quadrilateral ADQB - Area of △ BPQ = Area of quadrilateral ADQB - Area of △ DCQ

⇒ Area of △ BPQ = Area of △ DCQ

⇒ Area of △ BPQ - Area of △ PCQ = Area of △ DCQ - Area of △ PCQ

⇒ Area of △ BCP = Area of △ DPQ.

Hence, proved that area of △ BCP = area of △ DPQ.