Two parallelograms are equal in area.

I tried this question by constructing a line $PD$ therefore forming two triangles $ADP$ and QDP but couldn't establish the congruency relation between the triangles. My approach was that if I have those triangles congruent then, $PQ = AD (CPCT)$ where $PQ$ and $AD$ is the individual side of parm. $ABCD$ and $PQRC$ respectively and we have $AD=BC$ & $PQ=RC$ (since, opposite side of parm. is equal) by joing $BD$ (forms diagnol) of parm. $ABCD$ and triangles $ABD$ and $BCD$ be equal in area similarly for parm. $PQRC$.

(may be my approach isn't on a right track or their are other ways to solve this particular question, please give it a try! )

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Referring to the following:-

Area of the red triangle = (1/2)[area of ABCD]

Area of the red triangle = (1/2)[area of PQRC]

Therefore, area of ABCD = area of PQRC

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A hint:

Extend the lines $A\vee B$ and $Q\vee R$ to the left util they intersect at $S$. Then bring the parallelogram $SDCP$ into the game.

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