Question

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

Answer 1

Given : $XY\parallel BC$ and $CF\parallel AB$

$\Rightarrow XF\parallel BC$ and $CF\parallel AB$

$\Rightarrow BCFX$ is a parallelogram.

$ar\left(△ACF\right)=\frac{1}{2}ar\left(BCFX\right)$ ... (1) ($△ACF$ and || gm $BCFX$ are on the same base $CF$ and between the same parallel $AB$ and $FC$)

$BE\parallel CY$ and $XY\parallel BC$ (given)

So, $BCYE$ is a parallelogram.

$ar\left(△ABE\right)=\frac{1}{2}ar\left(BCYE\right)$ ... (2) ($△ABE$ and ||gm $BCYE$ are on the same base $BE$ and between the same parallel lines $AC$ and $BE$)

$ar\left(BCFX\right)=ar\left(BCYE\right)$... (3) (Parallelogram $BCFX$ and $BCYE$ are on the same base $BC$ and between the same parallels $BC$ and $EF$.

From (1) , (2) and (3) ,

$ar\left(△ABE\right)=ar\left(△ACF\right)$