In Fig.9.29, `a r\ (B D P)\ =\ a r\ (A R C)`and `a r\ (B D P)\ =\ a r\ (A R C)`. Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Explanation

We have to prove quadrilaterals `ABCD` and `DCPR` are trapeziums.

It is given that `Area (triangleDRC) = Area (triangleDPC)`

As `triangleDRC` and `triangleDPC` are lying on the same base `DC` having equal areas, therefore, they must lie between the same parallel lines.
According to Theorem 9.3:
Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.
`DC || RP`
Therefore, `DCPR` is a trapezium.
It is also given that `Area (triangleBDP) = Area (triangleARC)`

Now, subtract `ar (triangleDPC)` fromm `ar (triangleBDP)` and `ar (triangleDRC)` from `ar (triangleARC)`
`ar (triangleBDP) - ar (triangleDPC) = ar (triangleARC) - ar (triangleDRC)` [Since, `ar (triangleDPC) = ar (triangleDRC`)]`ar (triangleBDC) = ar (triangleADC)`

Since `triangleBDC` and `triangleADC` are on the same base `CD` having equal areas, they must lie between the same parallel lines. (According to Theorem 9.3)
`AB || CD`

Therefore, `ABCD` is a trapezium.