In $\square ABCD$,$A-Q-B$,$A-P-C$ and $A-R-D$. Given $\overline{PQ} \parallel \overline{BC}$ and $\overline{PR} \parallel \overline{DC}$. Prove that $\frac{AR}{AD} = \frac{AQ}{AB}$.

(N/A) In $\triangle ABC$,we are given that $\overline{PQ} \parallel \overline{BC}$.
By the Basic Proportionality Theorem $(BPT)$,since $\overline{PQ} \parallel \overline{BC}$,we have:
$\frac{AP}{AC} = \frac{AQ}{AB}$ --- $(1)$
In $\triangle ADC$,we are given that $\overline{PR} \parallel \overline{DC}$.
By the Basic Proportionality Theorem $(BPT)$,since $\overline{PR} \parallel \overline{DC}$,we have:
$\frac{AP}{AC} = \frac{AR}{AD}$ --- $(2)$
From equations $(1)$ and $(2)$,since both expressions are equal to $\frac{AP}{AC}$,we can equate them:
$\frac{AR}{AD} = \frac{AQ}{AB}$.
Hence,the result is proved.