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).]

Solution:

AC and PQ are joined.

Ar(△ACQ) = ar(△APQ) (On the same base AQ and between the same parallel lines AQ and CP)

⇒ ar(△ACQ)-ar(△ABQ) = ar(△APQ)-ar(△ABQ)

⇒ ar(△ABC) = ar(△QBP) — (i)

AC and QP are diagonals ABCD and PBQR.

∴,ar(ABC) = ½ ar(ABCD) — (ii)

ar(QBP) = ½ ar(PBQR) — (iii)

From (ii) and (ii),

½ ar(ABCD) = ½ ar(PBQR)

⇒ ar(ABCD) = ar(PBQR)