XY is a line parallel to side BC of a ∆ ABC. If BE || AC and CF || AB meet XY at E and F, respectively. Show that

ar (∆ ABE) = ar (∆ ACF).

Given XY is a line parallel to BC of a ∆ ABC.

If BE || AC and CF || AB meet XY at E and F, respectively

To prove ar(∆ABE) = ar (∆ACF)

Proof Since, a∆ABC in which XY || BC,

also BE || AC, i.e. BE|| CY and CF || AB, i.e. CF || XB

Now, since EY || BC and CY || BE

So, EYCB is a parallelogram.

Since, ∆ABE and parallelogram EYCB lie on the same base BE and between the same parallel lines BE and CA.

∴ ar (∆ABE) = 1/2 ar (EYCB) …(i)

Again, CF||BX and XF||BC

∴ BCFX is a parallelogram.

Since, ∆ACF and parallelogram BCFX lie on the same base CF and between the same parallel lines BX and FC.

∴ar (∆ACF) = 1/2 ar (BCFX) … (ii)

Now, parallelogram BCFX and parallelogram EYCB lie on the same base BC and between the same parallel lines BC and EF.

∴ ar (BCFX) = ar (EYCB) …(iii)

From Eqs. (i), (ii) and (iii), we get

ar (∆ACF) = ar(∆ABE)