Question

In the given figure, ar(DRC)=ar(DPC) and ar(BDP)=ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Answer 1

It is given that
$ar\left(\Delta DRC\right)=ar\left(\Delta DPC\right)$
As $\Delta DRC$ and $\Delta DPC$ lie on the same base DC and have equal areas, therefore, they must lie between the same parallel lines.
$\Rightarrow DC||RP$
Therefore, DCPR is a trapezium.

It is also given that
$ar\left(\Delta BDP\right)=ar\left(\Delta ARC\right)$
Then $ar\left(BDP\right)-ar\left(\Delta DPC\right)=ar\left(\Delta ARC\right)-ar\left(\Delta DRC\right)$
[since $ar\left(\Delta DRC\right)=ar\left(\Delta DPC\right)$]

$\Rightarrow ar\left(\Delta BDC\right)=ar\left(\Delta ADC\right)$

Since $\Delta BDC$ and $\Delta ADC$ are on the same base CD and have equal areas, they must lie between the same parallel lines.
$\Rightarrow AB||CD$
Therefore, ABCD is a trapezium.