In the given fig, $a r (D R C) = a r (D P C)$ and $a r (B D P) = a r (A R C)$ . Show that both the quadrilaterals $A B C D$ and $D C P R$ are trapeziums.

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Solution

$T h e o r e m :$ Triangles on the same base and between same parallel lines are equal in areas. .............(1)
$C o n v e r s e :$ Triangles with equal areas and on same base lies between parallel line ..............(2)
Since, $a r (D R C) = a r (D P C)$ and on same base $C D$
$\Rightarrow D C$ is parallel to $R P$ . [by (2)]
So, $D C P R$ is a trapezium.

Given $a r (B D P) = a r (A C R)$
Since, we know that $a r (D P C) = a r (D R C)$ ,
Therefore, on subtracting we get,
$\Rightarrow a r (B D P) - a r (D P C) = a r (A C R) - a r (D R C)$
$\Rightarrow a r (D B C) = a r (A C D)$ and on the same base $C D$
$\Rightarrow A B$ is parallel to $D C$ . [ by(2)]
So, $D C B A$ is trapezium.

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