Let $ABCD$ be a rectangle of sides $AB = 4$ and $BC = 3$. The perpendicular on the diagonal $BD$ drawn from  $A$ cuts $BD$ at point $H$. We call $M$ the midpoint of $BH$ and $N$ the midpoint of $CD$. Calculate the measure of the  segment $MN$.
1. **Identify Coordinates and Diagonal:**
   - Let the coordinates of the rectangle be \( A(0, 0) \), \( B(4, 0) \), \( C(4, 3) \), and \( D(0, 3) \).
   - The diagonal \( BD \) can be represented by the line segment from \( B(4, 0) \) to \( D(0, 3) \).

2. **Equation of Diagonal \( BD \):**
   - The slope of \( BD \) is \( \frac{3 - 0}{0 - 4} = -\frac{3}{4} \).
   - The equation of the line \( BD \) is \( y = -\frac{3}{4}x + 3 \).

3. **Perpendicular from \( A \) to \( BD \):**
   - The slope of the perpendicular line from \( A \) to \( BD \) is the negative reciprocal of \(-\frac{3}{4}\), which is \(\frac{4}{3}\).
   - The equation of the perpendicular line passing through \( A(0, 0) \) is \( y = \frac{4}{3}x \).

4. **Intersection Point \( H \):**
   - To find \( H \), solve the system of equations:
     \[
     \begin{cases}
     y = -\frac{3}{4}x + 3 \\
     y = \frac{4}{3}x
     \end{cases}
     \]
   - Substitute \( y = \frac{4}{3}x \) into \( y = -\frac{3}{4}x + 3 \):
     \[
     \frac{4}{3}x = -\frac{3}{4}x + 3
     \]
     \[
     \frac{16}{12}x + \frac{9}{12}x = 3
     \]
     \[
     \frac{25}{12}x = 3
     \]
     \[
     x = \frac{36}{25} = 1.44
     \]
     \[
     y = \frac{4}{3} \times 1.44 = 1.92
     \]
   - Thus, \( H \left( \frac{36}{25}, \frac{48}{25} \right) \).

5. **Midpoint \( M \) of \( BH \):**
   - Coordinates of \( B \) are \( (4, 0) \).
   - Coordinates of \( H \) are \( \left( \frac{36}{25}, \frac{48}{25} \right) \).
   - Midpoint \( M \) is:
     \[
     M = \left( \frac{4 + \frac{36}{25}}{2}, \frac{0 + \frac{48}{25}}{2} \right)
     \]
     \[
     M = \left( \frac{\frac{100}{25} + \frac{36}{25}}{2}, \frac{48}{50} \right)
     \]
     \[
     M = \left( \frac{136}{50}, \frac{24}{25} \right)
     \]
     \[
     M = \left( \frac{68}{25}, \frac{24}{25} \right)
     \]

6. **Midpoint \( N \) of \( CD \):**
   - Coordinates of \( C \) are \( (4, 3) \).
   - Coordinates of \( D \) are \( (0, 3) \).
   - Midpoint \( N \) is:
     \[
     N = \left( \frac{4 + 0}{2}, \frac{3 + 3}{2} \right)
     \]
     \[
     N = (2, 3)
     \]

7. **Distance \( MN \):**
   - Use the distance formula:
     \[
     MN = \sqrt{ \left( 2 - \frac{68}{25} \right)^2 + \left( 3 - \frac{24}{25} \right)^2 }
     \]
     \[
     MN = \sqrt{ \left( \frac{50}{25} - \frac{68}{25} \right)^2 + \left( \frac{75}{25} - \frac{24}{25} \right)^2 }
     \]
     \[
     MN = \sqrt{ \left( \frac{-18}{25} \right)^2 + \left( \frac{51}{25} \right)^2 }
     \]
     \[
     MN = \sqrt{ \frac{324}{625} + \frac{2601}{625} }
     \]
     \[
     MN = \sqrt{ \frac{2925}{625} }
     \]
     \[
     MN = \sqrt{4.68}
     \]
     \[
     MN = 2.16
     \]

The final answer is \( \boxed{2.16} \).