The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Figure).
Show that ar (ABCD) = ar (PBQR).
[Hint : Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]

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.