### In the given figure, $D$ is the midpoint of side $AB$ of $\Delta ABC$ and $P$ is any point on $BC$. If $CQ||PD$ meets $AB$ in $Q$, prove that $ar(\Delta BPQ)$ is equal to $\dfrac { 1 } { 2 } ar(\Delta ABC)$. A C B Q D P

1. We are given that $D$ is the midpoint of side $AB$ of $\Delta ABC$ and $P$ is any point on $BC$.
Also, $CQ||PD$ meets $AB$ in $Q$.
2. Let us join $CD$ and $PQ$.
In $\Delta ABC$, $CD$ is a median. \begin{aligned} \therefore \space & ar(\Delta BCD) = \dfrac { 1 } { 2 } ar(\Delta ABC) \\ \implies& ar(\Delta BPD)+ar(\Delta DPC) = \dfrac { 1 } { 2 } ar(\Delta ABC) &&\ldots \text{(i)} \end{aligned}
4. But, $\Delta DPC$ and $\Delta DPQ$ being on the same base $DP$ and between the same parallels $DP$ and $CQ$, we have: \begin{aligned} ar(\Delta DPC) = ar(\Delta DPQ) &&\ldots \text{(ii)} \end{aligned} Using $\text{(i)}$ and $\text{(ii)}$, we get: \begin{aligned} &ar(\Delta BPD) + ar(\Delta DPQ) = \dfrac { 1 } { 2 } ar(\Delta ABC) \\ \therefore \space & ar(\Delta BPQ) = \dfrac { 1 } { 2 } ar(\Delta ABC) \end{aligned}