In $\square ABCD$,$AB = AD$. The bisector of $\angle BAC$ intersects $\overline{BC}$ at $E$ and the bisector of $\angle DAC$ intersects $\overline{DC}$ at $F$. Prove that $\overline{EF} \parallel \overline{BD}$.

(N/A) In $\triangle ABC$,$AE$ is the angle bisector of $\angle BAC$. By the Angle Bisector Theorem,we have $\frac{BE}{EC} = \frac{AB}{AC}$.
In $\triangle ADC$,$AF$ is the angle bisector of $\angle DAC$. By the Angle Bisector Theorem,we have $\frac{DF}{FC} = \frac{AD}{AC}$.
Since it is given that $AB = AD$,we can substitute $AD$ with $AB$ in the second equation: $\frac{DF}{FC} = \frac{AB}{AC}$.
Comparing the two equations,we get $\frac{BE}{EC} = \frac{DF}{FC}$.
By the Converse of the Basic Proportionality Theorem (Thales's Theorem) in $\triangle BCD$,since $\frac{BE}{EC} = \frac{DF}{FC}$,it follows that $\overline{EF} \parallel \overline{BD}$.