# $\mathrm{D}$ 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 $DE\|BC$.

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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.

Updated on 10-Oct-2022 13:41:59