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

Theorem: Triangles on the same base and between same parallel lines are equal in areas. (1) Converse: Triangles with equal areas and on the same base lie between parallel lines. (2) Since, ar(DRC) = ar(DPC) and on the same base CD ⇒ DC is parallel to RP. [by (2)] So, DCPR is a trapezium. Given ar(BDP) = ar(ACR) Since, we know that ar(DPC) = ar(DRC), Therefore, on subtracting we get, ⇒ ar(BDP) - ar(DPC) = ar(ACR) - ar(DRC) ⇒ ar(DBC) = ar(ACD) and on the same base CD ⇒ AB is parallel to DC. [by (2)] So, DCBA is a trapezium.