Question

Question 4
In $\Delta$ ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (shown in figure), then prove that $ar\left(\Delta BPQ\right)=\frac{1}{2}ar\left(\Delta ABC\right)$.

Answer 1

Construction: Join PQ and CD.

Proof:
Since D is the mid-point of AB, CD is the median of$\Delta$ ABC.
We know that, a median of triangle divides it into two triangles of equal areas.

$\therefore \left(\Delta BCD\right)=\frac{1}{2}ar\left(ABC\right)$
$\Rightarrow ar\left(\Delta BPD\right)+\left(\Delta DPC\right)=\frac{1}{2}ar\left(ABC\right)$ ....(i)

Now, $\Delta$ DPQ and $\Delta$ DPC are on the same base DP and between the same parallel lines DP and CQ.

So, $ar\left(\Delta DPQ\right)=ar\left(\Delta DPC\right)$ ....(2)
On putting the value from equation (ii) in equation (i), we get;

$ar\left(\Delta BPD\right)+\left(\Delta DPQ\right)=\frac{1}{2}ar\left(\Delta ABC\right)$
$\Rightarrow ar\left(\Delta BPQ\right)=\frac{1}{2}ar\left(\Delta ABC\right)$