In $\Delta ABC$,$A-D-B$,$A-E-C$ and $BD = CE$. If $m\angle B = m\angle C$,prove that $\overline{DE} \parallel \overline{BC}$.

(N/A) Given: In $\Delta ABC$,$A-D-B$,$A-E-C$,$BD = CE$,and $m\angle B = m\angle C$.
Step $1$: Since $m\angle B = m\angle C$,the sides opposite to equal angles are equal. Therefore,$AB = AC$.
Step $2$: We can express the sides as $AB = AD + DB$ and $AC = AE + EC$ (since $A-D-B$ and $A-E-C$).
Step $3$: Substituting these into $AB = AC$,we get $AD + DB = AE + EC$.
Step $4$: Since $BD = CE$,we can substitute $CE$ with $BD$ in the equation: $AD + DB = AE + DB$.
Step $5$: Subtracting $DB$ from both sides,we get $AD = AE$.
Step $6$: Now,consider the ratios $\frac{AD}{DB}$ and $\frac{AE}{EC}$. Since $AD = AE$ and $DB = EC$,it follows that $\frac{AD}{DB} = \frac{AE}{EC}$.
Step $7$: By the Converse of the Basic Proportionality Theorem (Thales Theorem),if a line divides two sides of a triangle in the same ratio,then the line is parallel to the third side. Therefore,$\overline{DE} \parallel \overline{BC}$.