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$ \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.