In the given figure, XY || BC, BE || CA and FC || AB. Prove that : ar (ΔABE) = ar (ΔACF).

Parallelograms BCYE and BCFX stand on the same base BC and lie between the same parallels BC and XY.

ar (∥ gm BCYE) = ar (∥ gm BCFX) = x (let) …..(1)

Δ ABE shares the base BE with parallelogram BCYE and lies between the same parallels BE and AC.

ar (Δ ABE) = $\dfrac{1}{2}$ ar (∥ gm BCYE) = $\dfrac{1}{2}x$ …..(2)

Δ ACF shares the base CF with parallelogram BCFX and lies between the same parallels CF and AB.

ar (Δ ACF) = $\dfrac{1}{2}$ ar (∥ gm BCFX) = $\dfrac{1}{2}x$ …..(3)

From equation (2) and (3), we get :

ar (Δ ABE) = ar (Δ ACF)

Hence, proved that ar (Δ ABE) = ar (Δ ACF).