3. The three points $A, B, C$ form a triangle. $A B=4, B C=5 A C=6$. Let the angle bisector of $\angle A$ intersect side $B C$ at $D$. Let the foot of the perpendicular from $B$ to the angle bisector of $\angle A$ be $E$. Let the line through $E$ parallel to $A C$ meet $B C$ at $F$. Compute $D F$.
Answer:
$1 / 2$

Since $A D$ bisects $\angle A$, by the angle bisector theorem $\frac{A B}{B D}=\frac{A C}{C D}$, so $B D=2$ and $C D=3$. Extend $B E$ to hit $A C$ at $X$. Since $A E$ is the perpendicular bisector of $B X, A X=4$. Since $B, E, X$ are collinear, applying Menelaus' Theorem to the triangle $A D C$, we have
$$
\frac{A E}{E D} \cdot \frac{D B}{B C} \cdot \frac{C X}{X A}=1
$$

This implies that $\frac{A E}{E D}=5$, and since $E F \| A C, \frac{D F}{D C}=\frac{D E}{D A}$, so $D F=\frac{D C}{6}=\frac{1}{2}$.