Question

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 $\left(ABCD\right)$ $=$ ar $\left(PBQR\right)$.

Answer 1

Let us join $AC$ and $PQ$.
$\Delta ACQ$ and $\Delta AQP$ are on the same base $AQ$ and between the same parallels $AQ$ and $CP$.
$\therefore$ Area $\left(\Delta ACQ\right)$=Area $\left(\Delta APQ\right)$
$\Rightarrow$ Area $\left(\Delta ACQ\right)$-Area $\left(\Delta ABQ\right)$=Area $\left(\Delta APQ\right)$-Area $\left(\Delta ABQ\right)$
$\Rightarrow$ Area $\left(\Delta ABC\right)$=Area $\left(\Delta QBP\right)...\left(1\right)$
Since $AC$ and $PQ$ are diagonals of parallelograms $ABCD$ and $PBQR$ respectively,
$\therefore$ Area $\left(\Delta ABC\right)=\frac{1}{2}$ Area (||gm $ABCD$)...(2)
Area $\left(\Delta QBP\right)=\frac{1}{2}$ Area (||gm $PBQR$)...(3)
From equations(1),(2), and (3), We obtain
$\frac{1}{2}$ Area (||gm $ABCD$)=$\frac{1}{2}$ Area (||gm $PBQR$)
$\therefore$ Area (||gm $ABCD$)=Area (||gm $BPRQ$)