Question

The diagonals of a parallelogram $ABCD$ intersect at a point $O$. Through $O$, a line is drawn to intersect $AD$ at $P$ and $BC$ at $Q$. Show that $PQ$ divides the parallelogram into two parts of equal area.

Answer 1

Step 1 . Explaining the given terms

According to the given details

The diagonals of a parallelogram $ABCD$ intersect at a point $O$.

Through $O$, a line is drawn to intersect $AD$ at $P$ and $BC$ at $Q$ and we have to prove

$ar$ (parallelogram $PDCQ$) $=$ $ar$ (parallelogram $PQBA$).

Step 2. Proof

Since $AC$ is a diagonal of parallelogram $ABCD$ and diagonal of parallelogram bisect it in two triangle of equal area.

$\therefore ar\left(\Delta ABC\right)=ar\left(\Delta ACD\right)=\frac{1}{2}ar\left(\right||gmABCD)\dots \left(1\right)$

From $\Delta AOP$ and $\Delta COQ$

$AO=CO$ [Since, diagonals of a parallelogram bisect each other,]

$\angle AOP=\angle COQ$ [Vertically opposite angles]

$\angle OAP=\angle OCQ$ [Alternate interior angles]

$\Rightarrow \Delta AOP\cong \Delta COQ$ [By $ASA$ Congruence Rule]

$\therefore ar\left(\Delta AOP\right)=ar(\Delta COQ$ [ As Congruent figures have equal areas]

So,

$ar\left(\Delta AOP\right)+ar\left(\right||gmOPDC)=ar\left(\Delta COQ\right)+ar\left(\right||gmOPDC)$

$\Rightarrow ar\left(\Delta ACD\right)=ar\left(\right||gmPDCQ)$

$\Rightarrow \frac{1}{2}ar\left(\right||gmABCD)=ar\left(\right||gmPDCQ)$ [From eq. $\left(i\right)$]

Now we have ,

$ar\left(\right||gmPQBA)=ar\left(\right||gmPDCQ)$

$\Rightarrow ar\left(\right||gmPDCQ)=ar\left(\right||gmPQBA).$

Hence, Proved $ar$ (parallelogram $PDCQ$) $=$ $ar$ (parallelogram $PQBA$).