In $\triangle ABC,$ if $L$ and $M$ are the points on $AB$ and $AC,$ respectively such that $LM \parallel BC.$ Prove that $\operatorname{ar}(\triangle LOB) = \operatorname{ar}(\triangle MOC).$

(N/A) Given: $\triangle ABC$ with $LM \parallel BC,$ where $L$ lies on $AB$ and $M$ lies on $AC.$
Proof:
Since $\triangle LBM$ and $\triangle LCM$ are on the same base $LM$ and between the same parallels $LM$ and $BC,$ their areas are equal.
$\therefore \operatorname{ar}(\triangle LBM) = \operatorname{ar}(\triangle LCM)$
We can write these areas as the sum of two smaller triangles:
$\operatorname{ar}(\triangle LBM) = \operatorname{ar}(\triangle LOM) + \operatorname{ar}(\triangle LOB)$
$\operatorname{ar}(\triangle LCM) = \operatorname{ar}(\triangle LOM) + \operatorname{ar}(\triangle MOC)$
Substituting these into the equality:
$\operatorname{ar}(\triangle LOM) + \operatorname{ar}(\triangle LOB) = \operatorname{ar}(\triangle LOM) + \operatorname{ar}(\triangle MOC)$
Subtracting $\operatorname{ar}(\triangle LOM)$ from both sides,we get:
$\operatorname{ar}(\triangle LOB) = \operatorname{ar}(\triangle MOC)$
Hence proved.