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 equal parts.

Answer 1

$\because AC$ is a diagonal of the || gm $ABCD$
Refer image,

$\therefore ar\left(\Delta ACD\right)=\frac{1}{2}ar\left(gmABCD\right)$.......(1)
Now, in $\Delta AOP$ and $\Delta COQ$
$AO=CO$ [$\because$ Diagonals of || gm bisects each other]
$\angle AOP=\angle COQ$ [Vert, opp $\angle s$]
$\angle OAP=\angle OCQ$ [Alt. $\angle s;AB||CD$]
$\Delta AOP\cong$ $\Delta COQ$ [By ASA cong. Rule]
Hence, $ar\left(\Delta AOP\right)=$ $ar\left(\Delta COQ\right)$ [Cong. Area axiom]......(2)
Adding ar(quad.$AOQD)$ to both sides of (2), we get
$ar(quad.AOQD)+ar\left(\Delta AOP\right)=ar(quad.AOQD)+ar\left(\Delta COQ\right)$

$\Rightarrow ar(quad.APQD)=ar\left(\Delta ACD\right)$

But, $ar\left(\Delta ACD\right)=\frac{1}{2}ar||\left(gmABCD\right)$ [From (1)]

Hence, $ar(quad.APQD)=\frac{1}{2}ar\left(||gmABCD\right)$

So, $PQ$ divides the parallelogram into two parts of equal area.