In the given figure,if $PQRS$ is a parallelogram and $AB \parallel PS$,then prove that $OC \parallel SR$.

(N/A) Given: $PQRS$ is a parallelogram,so $PQ \parallel SR$ and $PS \parallel QR$. Also,$AB \parallel PS$.
To prove: $OC \parallel SR$.
Proof: In $\triangle OPS$ and $\triangle OAB$,since $PS \parallel AB$:
$\angle POS = \angle AOB$ (Common angle)
$\angle OSP = \angle OBA$ (Corresponding angles)
Therefore,$\triangle OPS \sim \triangle OAB$ (by $AAA$ similarity criterion).
Thus,$\frac{PS}{AB} = \frac{OS}{OB}$ ..........$(i)$
In $\triangle CQR$ and $\triangle CAB$,since $QR \parallel PS \parallel AB$:
$\angle QCR = \angle ACB$ (Common angle)
$\angle CQR = \angle CAB$ (Corresponding angles)
Therefore,$\triangle CQR \sim \triangle CAB$ (by $AA$ similarity criterion).
Thus,$\frac{QR}{AB} = \frac{CR}{CB}$.
Since $PQRS$ is a parallelogram,$PS = QR$. Substituting this,we get:
$\frac{PS}{AB} = \frac{CR}{CB}$ ..........$(ii)$
From equations $(i)$ and $(ii)$:
$\frac{OS}{OB} = \frac{CR}{CB} \Rightarrow \frac{OB}{OS} = \frac{CB}{CR}$.
Subtracting $1$ from both sides:
$\frac{OB}{OS} - 1 = \frac{CB}{CR} - 1$
$\frac{OB - OS}{OS} = \frac{CB - CR}{CR}$
$\frac{BS}{OS} = \frac{BR}{CR}$.
By the converse of the Basic Proportionality Theorem $(BPT)$,$SR \parallel OC$ (or $OC \parallel SR$). Hence proved.