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

Given:

\( \mathrm{XY} \) is a line parallel to side \( \mathrm{BC} \) of a triangle \( \mathrm{ABC} \).

\( \mathrm{BE} \| \mathrm{AC} \) and \( \mathrm{CF} \| \mathrm{AB} \) meet \( \mathrm{XY} \) at \( \mathrm{E} \) and \( F \) respectively.

To do:

We have to show that \( \operatorname{ar}(\mathrm{ABE})=\operatorname{ar}(\mathrm{ACF}) \).

Solution:

$BE \| AC$

This implies,

$BE \| CY$

$CF \| AB$

This implies,

$CF \| XB$

$XY \| BC$ and $CY \| BE$

Therefore,

$EYCB$ is a parallelogram.

$\triangle \mathrm{ABE}$ and parallelogram $EYCB$ lie on the same base $BE$ and between the parallels $B E$ and $A C$.

This implies,

$ar(\triangle \mathrm{ABE})=\frac{1}{2} ar(\mathrm{EYCB})$........(i)

$C F \| A B$ and $X F \| B C$

This implies,

$BCFX$ is a parallelogram.

$\triangle \mathrm{ACF}$ and parallelogram $BCFX$ lie on the same base $CF$ and between the parallels $A B$ and $CF$.

Therefore,

$ar(\triangle \mathrm{ACF})=\frac{1}{2}ar(\mathrm{BCFX})$..........(ii)

Parallelogram $BCFX$ and parallelogram $BCYE$ lie on the same base $BC$ and

between parallels $\mathrm{BC}$ and $\mathrm{EF}$.

Therefore,

$\operatorname{ar}(\mathrm{BCFX})=\operatorname{ar}(BCYE)$..........(iii)

From (i), (ii) and (iii), we get,

$\operatorname{ar}(\triangle \mathrm{ABE})=\operatorname{ar}(\triangle \mathrm{ACF})$

Hence proved.