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 $\text{ar}(ABCD) = \text{ar}(PBQR)$.
[Hint: Join $AC$ and $PQ$. Now compare $\text{ar}(ACQ)$ and $\text{ar}(APQ)$.]

(A) We have a parallelogram $ABCD$. $AB$ is produced to $P$.
$CB$ is produced to $Q$ and parallelogram $PBQR$ is completed.
Let us join $AC$ and $PQ$.
Since $ABCD$ is a parallelogram and $AC$ is its diagonal,
$\text{ar}(\Delta ABC) = \frac{1}{2} \text{ar}(ABCD) \quad \dots(1)$
Since $PBQR$ is a parallelogram and $PQ$ is its diagonal,
$\text{ar}(\Delta PBQ) = \frac{1}{2} \text{ar}(PBQR) \quad \dots(2)$
Since $\Delta ACQ$ and $\Delta APQ$ are on the same base $AQ$ and between the same parallels $AQ$ and $CP$,
$\text{ar}(\Delta ACQ) = \text{ar}(\Delta APQ)$
Subtracting $\text{ar}(\Delta ABQ)$ from both sides:
$\text{ar}(\Delta ACQ) - \text{ar}(\Delta ABQ) = \text{ar}(\Delta APQ) - \text{ar}(\Delta ABQ)$
$\text{ar}(\Delta ABC) = \text{ar}(\Delta PBQ) \quad \dots(3)$
From $(1), (2),$ and $(3)$,we get:
$\frac{1}{2} \text{ar}(ABCD) = \frac{1}{2} \text{ar}(PBQR)$
Therefore,$\text{ar}(ABCD) = \text{ar}(PBQR)$.