$XY$ 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 $\operatorname{ar}(ABE) = \operatorname{ar}(ACF)$.

(N/A) We have a $\Delta ABC$ such that $XY || BC$,$BE || AC$,and $CF || AB$.
Since $XY || BC$ and $BE || AC$ (given $BE || AC$ and $E$ lies on $XY$),we have $BCYE$ as a parallelogram.
Now,the parallelogram $BCYE$ and $\Delta ABE$ are on the same base $BE$ and between the same parallels $BE$ and $AC$.
Therefore,$\operatorname{ar}(\Delta ABE) = \frac{1}{2} \operatorname{ar}(BCYE) \quad -(1)$
Again,since $CF || AB$ and $XY || BC$,we have $BCFX$ as a parallelogram.
Now,$\Delta ACF$ and parallelogram $BCFX$ are on the same base $CF$ and between the same parallels $AB$ and $CF$.
Therefore,$\operatorname{ar}(\Delta ACF) = \frac{1}{2} \operatorname{ar}(BCFX) \quad -(2)$
Also,parallelogram $BCFX$ and parallelogram $BCYE$ are on the same base $BC$ and between the same parallels $BC$ and $EF$.
Therefore,$\operatorname{ar}(BCFX) = \operatorname{ar}(BCYE) \quad -(3)$
From equations $(1)$,$(2)$,and $(3)$,we get:
$\operatorname{ar}(\Delta ABE) = \operatorname{ar}(\Delta ACF)$.