In $\Delta ABC$,points $D$ and $E$ lie on $\overline{AB}$ such that $A-D-E-B$ and $AD = EB$. $P \in \overline{AC}$ and $Q \in \overline{BC}$ such that $\overline{DP} \parallel \overline{BC}$ and $\overline{EQ} \parallel \overline{AC}$. Prove that $\overline{PQ} \parallel \overline{AB}$.

(N/A) In $\Delta ABC$,points $D$ and $E$ lie on $\overline{AB}$ such that $A-D-E-B$ and $AD = EB$.
Since $A-D-E-B$,we have $AE = AD + DE$ and $BD = DE + EB$.
Given $AD = EB$,we have $AE = AD + DE$ and $BD = DE + AD$,so $AE = BD$ ... $(1)$.
In $\Delta ABC$,since $\overline{DP} \parallel \overline{BC}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{AD}{DB} = \frac{AP}{PC}$ ... $(2)$.
In $\Delta ABC$,since $\overline{EQ} \parallel \overline{AC}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{BE}{EA} = \frac{BQ}{QC}$.
Since $AE = BD$ and $AD = EB$,this becomes $\frac{AD}{BD} = \frac{BQ}{QC}$ ... $(3)$.
From $(2)$ and $(3)$,we get:
$\frac{AP}{PC} = \frac{BQ}{QC}$.
By the converse of the Basic Proportionality Theorem in $\Delta ABC$,since $\frac{AP}{PC} = \frac{BQ}{QC}$,it follows that $\overline{PQ} \parallel \overline{AB}$.