Question

XY is a line parallel to side BC of a triangle ABC. If BE||AC and CF||AB meet XY at E and F respectively,
show that $ar\left(\Delta ABE\right)=ar\left(\Delta ACF\right)$

Answer 1

It is given that
$XY||BC\Rightarrow EY||BC$
$BE||AC\Rightarrow BE||CY$
Therefore, EBCY is a parallelogram.

Also,
$XY||BC\Rightarrow XF||BC$
$FC||AB\Rightarrow FC||XB$
Therefore, BCFX is a parallelogram.

Parallelograms EBCY and BCFX are on the same base BC and between the same parallels BC and EF.
$\Rightarrow$ ar (EBCY) = ar (BCFX) ...(1)

Consider parallelogram EBCY and $\Delta AEB$
These lie on the same base BE and are between the same parallels BE and AC.
$\Rightarrow ar\left(ABE\right)=\frac{1}{2}ar\left(EBCY\right)..\left(2\right)$

Also, parallelogram BCFX and $\Delta ACF$ are on the same base CF and between the same
Parallels CF and AB.
$\Rightarrow ar\left(\Delta ACF\right)=\frac{1}{2}ar\left(BCFX\right)..\left(3\right)$

From equations (1), (2), and (3), we obtain ar $\left(\Delta ABE\right)=ar\left(\Delta ACF\right)$