In $\triangle ABC$,$D$ is the mid-point of $AB$ and $P$ is any point on $BC$. If $CQ \parallel PD$ meets $AB$ in $Q$,then prove that $\operatorname{ar}(\triangle BPQ) = \frac{1}{2} \operatorname{ar}(\triangle ABC)$.

(A) is the mid-point of $AB$ and $P$ is any point on $BC$ of $\triangle ABC$. Given $CQ \parallel PD$ meets $AB$ in $Q$,we have to prove that $\operatorname{ar}(\triangle BPQ) = \frac{1}{2} \operatorname{ar}(\triangle ABC)$.
Join $CD$. Since the median of a triangle divides it into two triangles of equal area,we have:
$\operatorname{ar}(\triangle BCD) = \frac{1}{2} \operatorname{ar}(\triangle ABC) \quad \dots(1)$
Since triangles on the same base and between the same parallels are equal in area,we have:
$\operatorname{ar}(\triangle DPQ) = \operatorname{ar}(\triangle DPC) \quad \dots(2)$
[$\because$ Triangles $DPQ$ and $DPC$ are on the same base $DP$ and between the same parallels $DP$ and $CQ$]
Adding $\operatorname{ar}(\triangle DPB)$ to both sides of equation $(2)$:
$\operatorname{ar}(\triangle DPQ) + \operatorname{ar}(\triangle DPB) = \operatorname{ar}(\triangle DPC) + \operatorname{ar}(\triangle DPB)$
$\operatorname{ar}(\triangle BPQ) = \operatorname{ar}(\triangle BCD)$
From equation $(1)$,we know $\operatorname{ar}(\triangle BCD) = \frac{1}{2} \operatorname{ar}(\triangle ABC)$.
Therefore,$\operatorname{ar}(\triangle BPQ) = \frac{1}{2} \operatorname{ar}(\triangle ABC)$.