Question

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

Answer 1

REF.IMAGE

$XY\parallel BC$

$\Rightarrow EN\parallel BC.$ & $BE\parallel CN$

In BCNE both pair of opposite are parallelogram

$\because$ BCNE is a parallelogram

parallelogram BCNE and $\Delta AEB$

area $\left(\Delta ABE\right)=\frac{1}{2}$ ar(BCNE)

BCFM is a parallelogram

$area\left(\Delta ACF\right)=\frac{1}{2}area\left(BCFM\right)$

parallelogram BCNE and BCFM are on the same base BC and between the same $\parallel$ BC and EF

area (BCNE) = area (BCFM)

$\Rightarrow$ area (ABE) = area (ACE)