$ \mathrm{D} $ and $ \mathrm{E} $ are points on sides $ \mathrm{AB} $ and $ \mathrm{AC} $ respectively of $ \triangle \mathrm{ABC} $ such that ar $ (\mathrm{DBC})=\operatorname{ar}(\mathrm{EBC}) $. Prove that $DE\|BC$.
Given:
\( \mathrm{D} \) and \( \mathrm{E} \) are points on sides \( \mathrm{AB} \) and \( \mathrm{AC} \) respectively of \( \triangle \mathrm{ABC} \) such that ar \( (\mathrm{DBC})=\operatorname{ar}(\mathrm{EBC}) \).
To do:
We have to prove that $DE\|BC$.
Solution:
ar \( (\mathrm{DBC})=\operatorname{ar}(\mathrm{EBC}) \)
This implies,
$\triangle DBC$ and $\triangle EBC$ are equal in area and have the same base $BC$.
Therefore,
$\triangle DBC$ and $\triangle EBC$ lie between the same parallel lines.
This implies,
$DE \| BC$
Hence proved.
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