In $\square ABCD$,$P \in \overline{AB}$,$Q \in \overline{BC}$,$R \in \overline{CD}$,and $S \in \overline{DA}$ such that $\frac{AP}{PB} = \frac{AS}{SD}$ and $\frac{CB}{QB} = \frac{CR}{RD}$. Prove that $\overline{PS} \parallel \overline{QR}$.

(N/A) $1$. In $\triangle ABD$,we are given $\frac{AP}{PB} = \frac{AS}{SD}$. By the Converse of the Basic Proportionality Theorem (Thales Theorem),$PS \parallel BD$.
$2$. In $\triangle BCD$,we are given $\frac{CB}{QB} = \frac{CR}{RD}$. This can be rewritten as $\frac{BQ}{QC} = \frac{DR}{RC}$. By the Converse of the Basic Proportionality Theorem,$QR \parallel BD$.
$3$. Since $PS \parallel BD$ and $QR \parallel BD$,by the property that lines parallel to the same line are parallel to each other,we conclude that $PS \parallel QR$.