The inner bisections of the $ \angle ABC $ and $ \angle ACB $ angles of the $ ABC $ triangle are at $ I $. The $ BI $ parallel line that runs through the point $ A $ finds the $ CI $ line at the point $ D $. The $ CI $ parallel line for $ A $ finds the $ BI $ line at the point $ E $. The lines $ BD $ and $ CE $ are at the point $ F $. Show that $ F, A $, and $ I $ are collinear if and only if $ AB = AC. $
1. **Given:**
   - The inner bisectors of $\angle ABC$ and $\angle ACB$ intersect at $I$.
   - The line through $A$ parallel to $BI$ intersects $CI$ at $D$.
   - The line through $A$ parallel to $CI$ intersects $BI$ at $E$.
   - The lines $BD$ and $CE$ intersect at $F$.

2. **To Show:**
   - $F, A$, and $I$ are collinear if and only if $AB = AC$.

3. **Using Desargue's Theorem:**
   - Consider triangles $ADE$ and $BIC$. By Desargue's Theorem, if $DE \parallel BC$, then the lines $AD$, $BE$, and $CI$ are concurrent.
   - Given that $AD \parallel BI$ and $AE \parallel CI$, we can infer that $DE \parallel BC$.

4. **Parallelogram Formation:**
   - Since $AD \parallel BI$ and $AE \parallel CI$, quadrilateral $ADIE$ is a parallelogram.
   - In a parallelogram, opposite sides are equal, so $AD = IE$ and $AE = ID$.

5. **Isosceles Trapezoids:**
   - Since $AD \parallel BI$ and $AE \parallel CI$, quadrilaterals $ADBE$ and $ADEC$ are isosceles trapezoids.
   - In isosceles trapezoids, the non-parallel sides are equal, so $AE = DB = DI$ and $AD = EC = EI$.

6. **Angle Congruence:**
   - Since $AD \parallel BI$ and $AE \parallel CI$, we have $\angle ADB = \angle DAE = \angle AEC$.
   - Therefore, triangles $ADB$ and $CEA$ are congruent by the Angle-Side-Angle (ASA) criterion.

7. **Equality of Sides:**
   - From the congruence of triangles $ADB$ and $CEA$, we get $AB = AC$.

8. **Conclusion:**
   - Since $F$ is the intersection of $BD$ and $CE$, and $A$ is the intersection of $AD$ and $AE$, and $I$ is the intersection of $BI$ and $CI$, $F, A$, and $I$ are collinear if and only if $AB = AC$.

\[
\boxed{AB = AC}
\]