$D$ and $E$ are points on sides $AB$ and $AC$ respectively of $\Delta ABC$ such that $\operatorname{ar}(DBC) = \operatorname{ar}(EBC)$. Prove that $DE \parallel BC$.

(N/A) Given: In $\Delta ABC$,$D$ and $E$ are points on $AB$ and $AC$ such that $\operatorname{ar}(DBC) = \operatorname{ar}(EBC)$.
To prove: $DE \parallel BC$.
Proof:
We are given that $\operatorname{ar}(DBC) = \operatorname{ar}(EBC)$.
Both triangles $\Delta DBC$ and $\Delta EBC$ lie on the same base $BC$.
We know that if two triangles have the same base and equal areas,they lie between the same parallels.
Therefore,the line segment $DE$ must be parallel to the base $BC$.
Hence,$DE \parallel BC$.