Question

XY is parallel to the side BC of triangle ABC. If BE || AC and CF ||AB meet XY at E and F respectively, show that ar(ABE)=ar(ACF)

Answer 1

It is given that

$XY\parallel BC$ is perpendicular to $EY\parallel BC$

$BE\parallel AC$ is perpendicular to $\parallel CY$

$\therefore ,EBCY$ is a parallelogram.

It is given that

$XY\parallel BC$ is perpendicular to $XF\parallel BC$

$FC\parallel AB$ is perpendicular to $FC\parallel XB$

Therefore, $BCFX$ is a parallelogram.

Parallelograms $EBCY$ and $BCFX$ are on the same base $BC$ and between the same parallels $BC$ and $EF$.

Area$\left(EBCY\right)=$Area$\left(BCFX\right)$ ... $\left(1\right)$

Consider parallelogram $EBCY$ and $△AEB$

These lie on the same base $BE$ and are between the same parallels $BE$ and $AC$.

Area$\left(△ABE\right)=\frac{1}{2}$Area$\left(EBCY\right)$ ... $\left(2\right)$

Also, parallelogram $BCFX$ and $△ACF$ are on the same base $CF$ and between the same parallels $CF$ and $AB$.

Area$\left(△ACF\right)=\frac{1}{2}$Area$\left(BCFX\right)$ ... $\left(3\right)$

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