Let $ABC$ be a right triangle with right angle at $B$. Let $ACDE$ be a square drawn exterior to triangle $ABC$. If $M$ is the center of this square, find the measure of  $\angle MBC$.
1. **Identify the given information and draw the diagram:**
   - $ABC$ is a right triangle with $\angle ABC = 90^\circ$.
   - $ACDE$ is a square drawn exterior to triangle $ABC$.
   - $M$ is the center of the square $ACDE$.

2. **Determine the properties of the square $ACDE$:**
   - Since $ACDE$ is a square, all its sides are equal, and all its angles are $90^\circ$.
   - The center $M$ of the square $ACDE$ is equidistant from all its vertices.

3. **Analyze the position of $M$ relative to triangle $ABC$:**
   - Since $M$ is the center of the square, it lies on the perpendicular bisector of both $AC$ and $DE$.
   - The coordinates of $M$ can be found by averaging the coordinates of $A$ and $C$ (or $D$ and $E$).

4. **Use symmetry and geometric properties:**
   - Since $ACDE$ is a square, $M$ is also the midpoint of the diagonal $AE$.
   - The diagonal $AE$ of the square is at a $45^\circ$ angle to the sides of the square.

5. **Determine the angles involving $M$:**
   - Since $M$ is the center of the square, $\angle AMC = 45^\circ$ because $M$ lies on the diagonal $AE$.
   - $\angle MBC$ is the angle we need to find.

6. **Use the properties of the right triangle and the square:**
   - In right triangle $ABC$, $\angle BAC = 45^\circ$ and $\angle BCA = 45^\circ$ because it is an isosceles right triangle.
   - Since $M$ is the center of the square, $\angle AMC = 45^\circ$.

7. **Calculate $\angle MBC$:**
   - $\angle MBC$ can be found by considering the angles around point $B$.
   - Since $\angle ABC = 90^\circ$ and $\angle MBC$ is part of this right angle, we can use the fact that $\angle MBC$ is complementary to $\angle MCB$.

8. **Conclude the measure of $\angle MBC$:**
   - Since $\angle MCB = 45^\circ$ (as $M$ is the center of the square and $C$ is one of the vertices), $\angle MBC = 45^\circ$.

Therefore, the measure of $\angle MBC$ is $\boxed{45^\circ}$.