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

(i) area of ∆CPD = area of ∆AQD

(ii) area of ∆ADQ = area of ∆APD + area of ∆CPB.

Parallelogram ABCD with point P on the side AB and Q on the side BC is shown below:

∆CPD and || gm ABCD are on the same base CD and between the same parallels lines AB and CD.

∴ Area of ∆CPD = $\dfrac{1}{2}$ Area of || gm ABCD ……(1)

∆AQD and || gm ABCD are on the same base AD and between the same parallel lines AD and BC,

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

From (1) and (2),

Area of ∆CPD = Area of ∆AQD.

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

(ii) From part (i) we get,

Area of ∆CPD = $\dfrac{1}{2}$ Area of || gm ABCD

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

From figure,

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

From (3) and (4) we get,

⇒ Area of ∆APD + Area of ∆CPB = $\dfrac{1}{2}$ Area of || gm ABCD

Since,

Area of ∆ADQ = $\dfrac{1}{2}$ Area of || gm ABCD (From eq 2)

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

Hence, proved that area of ∆ADQ = area of ∆APD + area of ∆CPB.