In figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that, Area of ∆ABP + area of ∆DPC = Area of ∆APD.

We know that,

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

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

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

From figure,

Area of ||gm ABCD = Area of ∆APD + Area of ∆ABP + Area of ∆DPC

Dividing the above equation by 2 we get,

$\dfrac{\text{Area of ||gm ABCD}}{2} = \dfrac{\text{Area of ∆APD}}{2} + \dfrac{\text{Area of ∆ABP}}{2} + \dfrac{\text{Area of ∆DPC}}{2}$

From (i),

$\text{Area of ∆APD} = \dfrac{\text{Area of ∆APD}}{2} + \dfrac{\text{Area of ∆ABP}}{2} + \dfrac{\text{Area of ∆DPC}}{2} \\[1em] \text{Area of ∆APD} - \dfrac{\text{Area of ∆APD}}{2} = \dfrac{\text{Area of ∆ABP}}{2} + \dfrac{\text{Area of ∆DPC}}{2} \\[1em] \dfrac{\text{Area of ∆APD}}{2} = \dfrac{\text{Area of ∆ABP}}{2} + \dfrac{\text{Area of ∆DPC}}{2} \\[1em] \text{Area of ∆APD} = \text{Area of ∆ABP} + \text{Area of ∆DPC}$

Hence, proved that Area of ∆APD = Area of ∆ABP + Area of ∆DPC.