In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$,$M \in \overline{AD}$ and $N \in \overline{BC}$. If $\overline{MN} \parallel \overline{AB}$,prove that $\frac{DM}{MA} = \frac{CN}{NB}$.

(N/A) $1$. Given: In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$. $M$ is a point on $\overline{AD}$ and $N$ is a point on $\overline{BC}$ such that $\overline{MN} \parallel \overline{AB}$.
$2$. Since $\overline{AB} \parallel \overline{CD}$ and $\overline{MN} \parallel \overline{AB}$,it follows that $\overline{MN} \parallel \overline{CD}$.
$3$. Draw diagonal $\overline{AC}$ intersecting $\overline{MN}$ at point $P$.
$4$. In $\triangle ADC$,since $\overline{MP} \parallel \overline{DC}$,by the Basic Proportionality Theorem (Thales Theorem),we have $\frac{DM}{MA} = \frac{CP}{PA}$.
$5$. In $\triangle ABC$,since $\overline{PN} \parallel \overline{AB}$,by the Basic Proportionality Theorem,we have $\frac{CP}{PA} = \frac{CN}{NB}$.
$6$. From the two equations,since both are equal to $\frac{CP}{PA}$,we conclude that $\frac{DM}{MA} = \frac{CN}{NB}$. Hence proved.