Let $ABC$ be an acute-angled triangle with $AB =4$ and $CD$ be the altitude through $C$ with $CD = 3$. Find the distance between the midpoints of $AD$ and $BC$
1. **Identify the given information and define the points:**
   - Triangle \(ABC\) is acute-angled.
   - \(AB = 4\).
   - \(CD\) is the altitude through \(C\) with \(CD = 3\).
   - Let \(M\) be the midpoint of \(AD\).
   - Let \(N\) be the midpoint of \(BC\).

2. **Extend \(BA\) to a point \(K\) such that \(AK = BD\):**
   - Since \(AB = 4\), and \(AK = BD\), we have \(BD = 4 - AD\).
   - Let \(AD = x\), then \(BD = 4 - x\).
   - Therefore, \(AK = 4 - x\).

3. **Calculate the length \(DK\):**
   - Since \(D\) is the foot of the altitude from \(C\) to \(AB\), \(DK = AD + AK = x + (4 - x) = 4\).

4. **Calculate the length \(KC\):**
   - Since \(C\) is on the altitude \(CD\), and \(CD = 3\), we use the Pythagorean theorem in \(\triangle DKC\):
     \[
     KC = \sqrt{DK^2 + CD^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5.
     \]

5. **Determine the midpoints \(M\) and \(N\):**
   - \(M\) is the midpoint of \(AD\), so \(M\) divides \(AD\) into two equal parts.
   - \(N\) is the midpoint of \(BC\), so \(N\) divides \(BC\) into two equal parts.

6. **Use the homothety argument:**
   - The problem has a homothety with a scale factor of \(\frac{1}{2}\) with the problem in the link.
   - The distance between the midpoints \(M\) and \(N\) is half the distance between \(A\) and \(K\).

7. **Calculate the distance between \(M\) and \(N\):**
   - Since \(M\) is the midpoint of \(AD\) and \(N\) is the midpoint of \(BC\), the distance between \(M\) and \(N\) is half the distance between \(A\) and \(K\).
   - The distance between \(A\) and \(K\) is \(5\), so the distance between \(M\) and \(N\) is:
     \[
     MN = \frac{5}{2} = 2.5.
     \]

The final answer is \(\boxed{2.5}\).