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

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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.

Updated on 10-Oct-2022 13:42:00