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

Given:
Area (\(\triangle{DRC}\)) = Area (\(\triangle{DPC}\)).

As, \(\triangle{DRC}\) and \(\triangle{DPC}\) lie on the same base DC and have equal area, therefore, they must lie between the same parallel lines.

\(\because \) DC || RP
\(\therefore \) DCPR is a trapezium.

Also, Area (\(\triangle{BDP}\)) = Area (\(\triangle{ARC}\)).

\(\Rightarrow \) Area (\(\triangle{BDP}\)) - Area (\(\triangle{DPC}\)) = Area (\(\triangle{ARC}\)) - Area (\(\triangle{DRC}\))

\(\therefore \) Area (\(\triangle{BDC}\)) = Area(\(\triangle{ADC}\))

\(\because \) \(\triangle{BDC}\) and \(\triangle{ADC}\) lie on the same base CD and have equal areas, they must lie between the same parallel lines.

\(\therefore \) AB || CD
Hence, it is proved that ABCD is a trapezium.