In $\square ABCD$,$P \in \overline{AB}$,$Q \in \overline{BC}$,$R \in \overline{CD}$,and $S \in \overline{AD}$ such that $\frac{AP}{AB} = \frac{CQ}{BC} = \frac{CR}{CD} = \frac{AS}{AD} = \frac{1}{3}$. Prove that $\square PQRS$ is a parallelogram.

(A) Given: $\frac{AP}{AB} = \frac{AS}{AD} = \frac{1}{3}$.
In $\triangle ABD$,by the Converse of Basic Proportionality Theorem $(BPT)$,since $\frac{AP}{AB} = \frac{AS}{AD} = \frac{1}{3}$,it implies $\frac{AP}{PB} = \frac{AS}{SD} = \frac{1}{2}$. Thus,$PS \parallel BD$ and $PS = \frac{1}{3} BD$.
Similarly,in $\triangle BCD$,since $\frac{CQ}{BC} = \frac{CR}{CD} = \frac{1}{3}$,we have $\frac{CQ}{QB} = \frac{CR}{RD} = \frac{1}{2}$. Thus,$QR \parallel BD$ and $QR = \frac{1}{3} BD$.
From these two results,$PS \parallel QR$ and $PS = QR$.
Since one pair of opposite sides of quadrilateral $PQRS$ is equal and parallel,$\square PQRS$ is a parallelogram.