Point $O$ lies in the interior of $\Delta PQR$. Given $O-A-P$,$O-B-Q$,$O-C-R$,$\overline{AB} \parallel \overline{PQ}$ and $\overline{BC} \parallel \overline{QR}$. Prove that $\overline{AC} \parallel \overline{PR}$.

(N/A) Given: Point $O$ lies in the interior of $\Delta PQR$. $O-A-P$,$O-B-Q$,$O-C-R$,$\overline{AB} \parallel \overline{PQ}$ and $\overline{BC} \parallel \overline{QR}$.
To prove: $\overline{AC} \parallel \overline{PR}$.
Proof:
In $\Delta OPQ$,since $\overline{AB} \parallel \overline{PQ}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{OA}{AP} = \frac{OB}{BQ} \quad \dots (1)$
In $\Delta OQR$,since $\overline{BC} \parallel \overline{QR}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{OB}{BQ} = \frac{OC}{CR} \quad \dots (2)$
From equations $(1)$ and $(2)$,we get:
$\frac{OA}{AP} = \frac{OC}{CR}$
In $\Delta OPR$,since $\frac{OA}{AP} = \frac{OC}{CR}$,by the converse of the Basic Proportionality Theorem $(BPT)$,we have:
$\overline{AC} \parallel \overline{PR}$.