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Answer :
Construction: Let us join AC and PQ.
\(\triangle{ACQ}\) and \(\triangle{AQP}\) are the same base AQ and between the same parallels AQ and CP.
\(\therefore \) Area \(\triangle{ACQ}\) = Area \(\triangle{AQP}\)
\(\Rightarrow Area (\triangle{ACQ} ) - Area (\triangle{ABQ} ) = Area (\triangle{APQ} ) - Area (\triangle{ABQ})\)
\(\Rightarrow \) Area \((\triangle{ABC}\)) = Area (\(\triangle{QBP}\)) ...(i)
\(\because \) AC and PQ are diagonals of parallelograms ABCD and PBQR, respectively
\(\therefore \) Area (\(\triangle{ABC}\)) = (\(\frac{1}{2} \) ) Area (ABCD) ...(ii)
Also, Area (\(\triangle{QBP}\)) = (\(\frac{1}{2} \) ) Area (PBQR) ...(iii)
From equations (i),(ii) and (iii), we get,
(\(\frac{1}{2} \) ) Area (ABCD) = (\(\frac{1}{2} \) ) Area(PBQR)
\(\therefore \) Area (ABCD) = Area(PBQR)
Hence, proved.