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. Show that
ar(ABCD) = ar(PBQR).
[Hint : Join AC and PQ. Now compare ar(ACQ) and ar(APQ).]

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Solution

AC and PQ are joined.
$A r (△ A C Q) = a r (△ A P Q) (On the same base AQ and between the same parallel lines AQ and CP)$
$\Rightarrow a r (△ A C Q) - a r (△ A B Q) = a r (△ A P Q) - a r (△ A B Q)$
$\Rightarrow a r (△ A B C) = a r (△ Q B P) — (i)$
AC and QP are diagonals ABCD and PBQR.
$∴ a r (A B C) = \frac{1}{2} a r (A B C D) — (i i)$
$a r (Q B P) = \frac{1}{2} a r (P B Q R) — (i i i)$
From (ii) and (ii),
$\frac{1}{2} a r (A B C D) = \frac{1}{2} a r (P B Q R)$
$\Rightarrow a r (A B C D) = a r (P B Q R)$

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