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

Answer 1

Given: A parallelogram $ABCD$ where $CP||AQ$ and $PBQR$ is a parallelogram.

To prove: ar$\left(ABCD\right)$=ar $\left(PBQR\right)$

Construction: Toin $AC$ and $PQ$

Proof:

For $\Delta ACQ$ and $\Delta AQP$

$\Delta ACQ$ and $\Delta AQP$ are on the same base $AQ$ and between the same parallels $AQ$ and $CP$
$\therefore Ar\left(ACQ\right)=Ar\left(APQ\right)$

[Triangles with same base and between the same base & between the parallels are equal in area.]

Subtracting $ar\left(ABQ\right)$ both sides we get:

$Ar\left(ACQ\right)-Ar\left(ABQ\right)=Ar\left(APQ\right)-Ar\left(ABQ\right)$\\ \Rightarrow Ar(ABC)=Ar(PBQ)\rightarrow (1)$

In parallelogram $ABCD,AC$ is diagonal

$\therefore \Delta ABC\cong \Delta ADC$[Diagonals of a parallelogram divides it into two congruent triangles]

$\Rightarrow Ar\left(ABC\right)=Ar\left(ADC\right)$[Area of congruent triangles is equal]

$\therefore Ar\left(ABC\right)=Ar\left(ADC\right)=\frac{1}{2}Ar\left(ABCD\right)\to \left(2\right)$

In parallelogram $PBQR,PQ$ is the diagonal

$\therefore \Delta PBQ\cong \Delta PRQ\Rightarrow ar\left(PBQ\right)=ar\left(PRQ\right)$

So, $ar(PBQ)=ar(PRQ)=\cfrac{1}{2}ar(PBQR)

\rightarrow (3)$

From $\left(1\right)$ we get,

$ar\left(ABC\right)=ar\left(PBQ\right)$

From $\left(2\right)$ and $\left(3\right)$ we get;

$ar\left(ABC\right)=\frac{1}{2}ar\left(ABCD\right)$

&

$ar\left(PBQ\right)=\frac{1}{2}ar\left(PBQR\right)\therefore \frac{1}{2}ar\left(ABCD\right)=\frac{1}{2}ar\left(PBQR\right)\Rightarrow ar\left(ABCD\right)=ar\left(PBQR\right)$

Hence proved.