In $\Delta ABC$,$\overline{AD}$ is a median. The bisectors of $\angle ADB$ and $\angle ADC$ intersect $\overline{AB}$ and $\overline{AC}$ at $E$ and $F$ respectively. Prove that $\overline{EF} \parallel \overline{BC}$.

(N/A) In $\Delta ABD$,$DE$ is the bisector of $\angle ADB$. By the Angle Bisector Theorem,$\frac{AE}{EB} = \frac{AD}{DB}$.
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $DB = DC$. Thus,$\frac{AE}{EB} = \frac{AD}{DC}$.
In $\Delta ADC$,$DF$ is the bisector of $\angle ADC$. By the Angle Bisector Theorem,$\frac{AF}{FC} = \frac{AD}{DC}$.
Comparing the two equations,we get $\frac{AE}{EB} = \frac{AF}{FC}$.
By the Converse of the Basic Proportionality Theorem (Thales Theorem) in $\Delta ABC$,since $\frac{AE}{EB} = \frac{AF}{FC}$,it follows that $\overline{EF} \parallel \overline{BC}$.