The three points $A, B, C$ form a triangle. $AB=4, BC=5, AC=6$. Let the angle bisector of $\angle A$ intersect side $BC$ 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 $AC$ meet $BC$ at $F$. Compute $DF$.
1. **Identify the given elements and their relationships:**
   - \( A, B, C \) form a triangle with sides \( AB = 4 \), \( BC = 5 \), and \( AC = 6 \).
   - The angle bisector of \( \angle A \) intersects \( BC \) at \( D \).
   - The foot of the perpendicular from \( B \) to the angle bisector of \( \angle A \) is \( E \).
   - The line through \( E \) parallel to \( AC \) meets \( BC \) at \( F \).

2. **Use the Angle Bisector Theorem to find \( BD \) and \( DC \):**
   The Angle Bisector Theorem states that the angle bisector of an angle in a triangle divides the opposite side into segments proportional to the adjacent sides. Therefore,
   \[
   \frac{BD}{DC} = \frac{AB}{AC} = \frac{4}{6} = \frac{2}{3}
   \]
   Let \( BD = 2x \) and \( DC = 3x \). Since \( BD + DC = BC \),
   \[
   2x + 3x = 5 \implies 5x = 5 \implies x = 1
   \]
   Thus, \( BD = 2 \) and \( DC = 3 \).

3. **Determine the coordinates of \( D \):**
   Since \( D \) divides \( BC \) in the ratio \( 2:3 \), and \( B = (0,0) \) and \( C = (5,0) \),
   \[
   D = \left( \frac{2 \cdot 5 + 3 \cdot 0}{2+3}, \frac{2 \cdot 0 + 3 \cdot 0}{2+3} \right) = \left( \frac{10}{5}, 0 \right) = (2,0)
   \]

4. **Find the coordinates of \( E \):**
   Since \( E \) is the foot of the perpendicular from \( B \) to the angle bisector \( AD \), and \( AD \) is the angle bisector, \( E \) lies on \( AD \). Given that \( AD \) is a line, we need to find the equation of \( AD \). However, for simplicity, we can use the fact that \( E \) is the midpoint of \( BG \) where \( G \) is the reflection of \( B \) across \( AD \).

5. **Determine the coordinates of \( F \):**
   Since \( E \) is the midpoint of \( BG \) and \( F \) is on \( BC \) such that \( EF \parallel AC \), we can use the similarity of triangles. Given that \( \triangle BEF \sim \triangle BGC \),
   \[
   \frac{BF}{BC} = \frac{BE}{BG} = \frac{1}{2}
   \]
   Therefore, \( BF = \frac{1}{2} \times 5 = 2.5 \).

6. **Calculate \( DF \):**
   Since \( D \) is at \( (2,0) \) and \( F \) is at \( (2.5,0) \),
   \[
   DF = |2.5 - 2| = 0.5
   \]

The final answer is \( \boxed{0.5} \)