Let $AK$ and $AT$ be the bisector and the median of an acute-angled triangle $ABC$ with $AC > AB$. The line $AT$ meets the circumcircle of $ABC$ at point $D$. Point $F$ is the reflection of $K$ about $T$. If the angles of $ABC$ are known, find the value of angle $FDA$.
1. **Reflecting Point \( D \) in \( T \):**
   - Let \( X \) be the reflection of \( D \) in \( T \). Since \( AT \) is the median, \( T \) is the midpoint of \( AD \). Therefore, reflecting \( D \) in \( T \) gives us point \( X \) such that \( T \) is the midpoint of \( DX \).

2. **Using the A-HM Point:**
   - The A-HM point is a special point in triangle geometry, often related to reflections and symmetries. Here, reflecting \( D \) in \( T \) to get \( X \) implies that \( X \) lies on the circumcircle of \( \triangle ABC \).

3. **Cyclic Quadrilateral \( ALKX \):**
   - Let \( L \) be the foot of the A-external angle bisector. By a well-known lemma in triangle geometry, the quadrilateral \( ALKX \) is cyclic. This means that the opposite angles of this quadrilateral sum to \( 180^\circ \).

4. **Angle Relationships:**
   - Since \( ALKX \) is cyclic, we have:
     \[
     \angle ALK + \angle AXK = 180^\circ
     \]
   - Also, since \( K \) is the reflection of \( F \) about \( T \), we have:
     \[
     \angle KXT = \angle FDA
     \]

5. **Calculating \( \angle FDA \):**
   - From the cyclic nature of \( ALKX \), we know:
     \[
     \angle KXT = \angle ALB
     \]
   - Since \( \angle ALB \) is related to the external angle bisector, we have:
     \[
     \angle ALB = \frac{B - C}{2}
     \]
   - Therefore:
     \[
     \angle FDA = \frac{B - C}{2}
     \]

The final answer is \( \boxed{ \frac{B - C}{2} } \)