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.

(N/A) Given: $A$ parallelogram $ABCD$ where diagonals $AC$ and $BD$ intersect at $O$. $A$ line $PQ$ passes through $O$ such that $P$ lies on $AD$ and $Q$ lies on $BC$.
To prove: $\operatorname{ar}(\quad. APQB) = \operatorname{ar}(\quad. PQCD) = \frac{1}{2} \operatorname{ar}(\|gm ABCD)$.
Proof:
In $\triangle AOP$ and $\triangle COQ$:
$1$. $AO = CO$ (Diagonals of a parallelogram bisect each other).
$2$. $\angle AOP = \angle COQ$ (Vertically opposite angles).
$3$. $\angle OAP = \angle OCQ$ (Alternate interior angles,as $AD \parallel BC$).
Therefore,$\triangle AOP \cong \triangle COQ$ by the $ASA$ congruence rule.
This implies $\operatorname{ar}(\triangle AOP) = \operatorname{ar}(\triangle COQ)$ (Congruent areas axiom).
Now,add $\operatorname{ar}(\quad. OPCD)$ to both sides:
$\operatorname{ar}(\triangle AOP) + \operatorname{ar}(\quad. OPCD) = \operatorname{ar}(\triangle COQ) + \operatorname{ar}(\quad. OPCD)$
$\operatorname{ar}(\quad. APCD) = \operatorname{ar}(\triangle OPD + \triangle OPCD) = \operatorname{ar}(\triangle OCD + \triangle OPCD) = \operatorname{ar}(\triangle OCD + \triangle COQ) = \operatorname{ar}(\triangle OCD + \triangle AOP) = \operatorname{ar}(\quad. APQD)$.
Since $AC$ is a diagonal,$\operatorname{ar}(\triangle ADC) = \frac{1}{2} \operatorname{ar}(\|gm ABCD)$.
By adding $\operatorname{ar}(\triangle AOP)$ to $\operatorname{ar}(\triangle ADC)$ and subtracting $\operatorname{ar}(\triangle COQ)$,we find that $\operatorname{ar}(\quad. APQD) = \frac{1}{2} \operatorname{ar}(\|gm ABCD)$.
Similarly,the other part $\quad. PQCB$ also has an area equal to $\frac{1}{2} \operatorname{ar}(\|gm ABCD)$.