In the given figure, ^@D^@ is the midpoint of side ^@AB^@ of ^@\ | Math Question and Answer

Answer:

Step by Step Explanation:

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 $Q$ .
Let us join $CD$ and $PQ$ .
We know that a median of a triangle divides it into two triangles of equal area.

In $\Delta ABC$ , $CD$ is a median. $\begin{aligned} \therefore \space & ar(\Delta BCD) = \dfrac { 1 } { 2 } ar(\Delta ABC) \\ \implies& ar(\Delta BPD)+ar(\Delta DPC) = \dfrac { 1 } { 2 } ar(\Delta ABC) &&\ldots \text{(i)} \end{aligned}$
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: $\begin{aligned} &ar(\Delta BPD) + ar(\Delta DPQ) = \dfrac { 1 } { 2 } ar(\Delta ABC) \\ \therefore \space & ar(\Delta BPQ) = \dfrac { 1 } { 2 } ar(\Delta ABC) \end{aligned}$

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