Let $ABC$ be an acute angled triangle and $CD$ be the altitude through $C$. If $AB = 8$ and $CD = 6$, find the distance between the midpoints of $AD$ and $BC$.
1. Let \( M \) be the midpoint of \( AD \) and \( N \) be the midpoint of \( BC \).
2. Draw a line through \( M \) that is parallel to \( CD \) (which means perpendicular to \( AB \)) and let \( P \) be the intersection of the line and \( AC \).
3. Since \( MP \parallel CD \) and \( AM = MD \), we see that \( MP = \frac{1}{2}(CD) = 3 \) and \( AP = PC \).

   - Here, we use the property of midpoints and parallel lines in triangles. Since \( M \) is the midpoint of \( AD \) and \( MP \parallel CD \), \( MP \) is half the length of \( CD \).

4. Since \( AP = PC \) and \( BN = NC \), \( PN = \frac{1}{2}(AB) = 4 \).

   - This follows from the fact that \( N \) is the midpoint of \( BC \), so \( BN = NC \). Also, since \( P \) is the midpoint of \( AC \), \( AP = PC \).

5. It is easy to observe that \( \angle MPN = 90^\circ \) using \( AD \parallel PN \) and \( DC \parallel MP \).

   - Since \( MP \parallel CD \) and \( CD \) is perpendicular to \( AB \), \( MP \) is also perpendicular to \( AB \). Therefore, \( \angle MPN = 90^\circ \).

6. Finally, using the Pythagorean theorem in \( \triangle MPN \), we obtain:
   \[
   MN = \sqrt{MP^2 + PN^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
   \]

The final answer is \( \boxed{5} \).