In parallelogram ABCD, P is a point on side AB and Q is a point on side BC. Prove that :

(i) △ CPD and △ AQD are equal in area.

(ii) Area (△ AQD) = Area (△ APD) + Area (△ CPB)

We know that,

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

(i) From figure,

△ CPD and || gm ABCD are on the same base CD and between the same parallel lines AB and CD.

∴ Area of triangle CPD = $\dfrac{1}{2}$ Area of parallelogram ABCD ……..(1)

△ AQD and || gm ABCD are on the same base AD and between the same parallel lines AD and BC.

∴ Area of triangle AQD = $\dfrac{1}{2}$ Area of parallelogram ABCD ……..(2)

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

Area of triangle CPD = Area of triangle AQD.

Hence, proved that area of triangle CPD = area of triangle AQD.

(ii) From part (i), we get :

⇒ Area of △ CPD = $\dfrac{1}{2}$ Area of parallelogram ABCD

∴ Area of || gm ABCD - Area of △ CPD = $\dfrac{1}{2}$ Area of parallelogram ABCD ……..(3)

From figure,

⇒ Area of || gm ABCD - Area of △ CPD = Area of △ APD + Area of △ CPB …………(4)

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

⇒ Area of △ APD + Area of △ CPB = $\dfrac{1}{2}$ Area of || gm ABCD ……..(5)

Since,

⇒ Area of △ ADQ = $\dfrac{1}{2}$ Area of || gm ABCD [From eqn. 2]

Substituting value of $\dfrac{1}{2}$ Area of || gm ABCD from above equation in equation (5), we get :

⇒ Area of △ APD + Area of △ CPB = Area of △ ADQ.

Hence, proved that area (△ AQD) = area (△ APD) + area (△ CPB).