Question

The side AB of 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).

Answer 1

Let us join $AC$ and $PQ$.

$△ACQ$ and $△AQP$ are on the same base $AQ$ and between the same parallels $AQ$ and $CP$.

$\therefore$ Area $\left(△ACQ\right)=$ Area

$\left(△APQ\right)$

$\Rightarrow$ Area$\left(△ACQ\right)-$Area$\left(△ABQ\right)=$ Area$\left(△ACQ\right)-$Area$\left(△ABQ\right)$

$\Rightarrow$ Area $\left(△ABC\right)=$ Area $\left(△ABQ\right)$ ... $\left(1\right)$

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

$\therefore$ Area $\left(△ABC\right)=\frac{1}{2}$ Area of parallelogram $ABCD$ ... $\left(2\right)$

Area$\left(△QBP\right)=\frac{1}{2}$ Area of parallelogram $BPRQ$ ... $\left(3\right)$

From equations $\left(1\right),\left(2\right),$ and $\left(3\right)$, we obtain

$\frac{1}{2}$ Area of parallelogram $ABCD=\frac{1}{2}$ Area of parallelogram $BPRQ$

$\therefore$ Area of parallelogram $ABCD=$ Area of parallelogram $BPRQ$