In the given figure, if PQRS is a parallelogram

Given PQRS is a parallelogram, so PQ || SR and PS || QR. Also, AB || PS.

To prove $O C \| S R$

Proof in $\triangle O P S$ and $\triangle O A B$, $P S \| A B$

$\angle P O S=\angle A O B$ [common angle]

$\angle O S P=\angle O B A$ [corresponding angles]

$\therefore \quad \Delta O P S \sim \Delta O A B \quad$ [by AAA similarity criterion]

Then, $\frac{P S}{\cap AR}=\frac{O S}{\cap QR}$ $\ldots$ (i)

In $\triangle C Q R$ and $\triangle C A B$, $Q R\|P S\| A B$

$\angle Q C R=\angle A C B$ [common angle]

$\angle C R Q=\angle C B A$ [corresponding angles]

$\therefore$  $\triangle C Q R \sim \Delta C A B$ 

Then, $\frac{Q R}{A B}=\frac{C R}{C B}$

$\Rightarrow$ $\frac{P S}{A B}=\frac{C R}{C B}$ .....(ii)

[since, $P Q R S$ is a parallelogram, so $P S \equiv Q R$ ]

From Eqs. (i) and (ii), 

$\frac{O S}{O B}=\frac{C R}{C B}$ or $\frac{O B}{O S}=\frac{C B}{C R}$

On subtracting from both sides, we get

$\frac{O B}{O S}-1=\frac{C B}{C R}-1$

$\Rightarrow$ $\frac{O B-O S}{O S}=\frac{C B-C R}{C R}$

$\Rightarrow$ $\frac{B S}{O S}=\frac{B R}{C R}$

By converse of basic proportionality theorem,

$S R \| O C$

Hence proved.