In $\Delta ABC,$ the bisectors of $\angle B$ and $\angle C$ intersect $\overline{AC}$ and $\overline{AB}$ at $D$ and $E$ respectively. If $\overline{DE} \parallel \overline{BC},$ prove that $\Delta ABC$ is an isosceles triangle.

(N/A) Given: $\overline{BD}$ is the bisector of $\angle B$,so $\angle EBD = \angle DBC = \frac{1}{2} \angle B$.
$\overline{CE}$ is the bisector of $\angle C$,so $\angle DCE = \angle ECB = \frac{1}{2} \angle C$.
Since $\overline{DE} \parallel \overline{BC}$,by alternate interior angles,$\angle EDB = \angle DBC$.
Thus,$\angle EDB = \angle EBD$,which implies $\Delta EBD$ is isosceles with $EB = ED$.
Similarly,$\angle DEC = \angle ECB$,so $\angle DEC = \angle DCE$,which implies $\Delta DCE$ is isosceles with $DC = ED$.
Therefore,$EB = DC$.
In $\Delta ABC$,since $\overline{DE} \parallel \overline{BC}$,by the Basic Proportionality Theorem (Thales Theorem),$\frac{AE}{EB} = \frac{AD}{DC}$.
Since $EB = DC$,we have $AE = AD$.
Adding $EB$ to both sides: $AE + EB = AD + DC$,which gives $AB = AC$.
Since two sides are equal,$\Delta ABC$ is an isosceles triangle.