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 Fig. 9.26). Show that ar (ABCD) = ar (PBQR).

[Hint: Join AC and PQ. Now compare ar (ACQ) and ar (APQ)]

Let us join AC and PQ

ΔACQ and ΔAQP are on the same base AQ and between the same parallels AQ and

CP.

Area (ΔACQ) = Area (ΔAPQ)

Area (ΔACQ) − Area (ΔABQ) = Area (ΔAPQ) − Area (ΔABQ)

Area (ΔABC) = Area (ΔQBP) (i)

Since AC and PQ are diagonals of parallelograms ABCD and PBQR respectively,

Area (ΔABC) = Area (ABCD) (ii)

Area (ΔQBP) = Area (PBQR) (iii)

From equations (i), (ii), and (iii), we obtain

Area (ABCD) = Area (PBQR)

Area (ABCD) = Area (PBQR)