In an isosceles trapezoid $ABCD$, the base $AB$ is twice as large as the base $CD$. Point $M$ is the midpoint of $AB$. It is known that the center of the circle inscribed in the triangle $MCB$ lies on the circle circumscribed around the triangle $MDC$. Find the angle $\angle MBC$.
[img]https://cdn.artofproblemsolving.com/attachments/8/a/7af6a1d32c4e2affa49cb3eed9c10ba1e7ab71.png[/img]
1. **Identify the given information and set up the problem:**
   - $ABCD$ is an isosceles trapezoid with $AB = 2 \cdot CD$.
   - $M$ is the midpoint of $AB$.
   - The center of the circle inscribed in $\triangle MCB$ lies on the circle circumscribed around $\triangle MDC$.
   - We need to find the angle $\angle MBC$.

2. **Establish relationships and properties:**
   - Since $M$ is the midpoint of $AB$ and $AB = 2 \cdot CD$, we have $AM = MB = CD$.
   - In an isosceles trapezoid, the non-parallel sides are equal, so $AD = BC$.
   - Since $M$ is the midpoint of $AB$, $M$ is equidistant from $A$ and $B$.

3. **Use geometric properties and symmetry:**
   - Since $M$ is the midpoint of $AB$ and $AB = 2 \cdot CD$, $M$ lies on the perpendicular bisector of $CD$.
   - Therefore, $DM \parallel CB$ and $DM = CB$.

4. **Denote the angle $\angle MBC$ as $\alpha$:**
   - Since $DM \parallel CB$, $\angle MDC = \alpha$ by corresponding angles.

5. **Use the properties of the circles:**
   - The center of the circle inscribed in $\triangle MCB$ is the incenter, which lies on the angle bisector of $\angle MCB$.
   - The center of the circle circumscribed around $\triangle MDC$ is the circumcenter, which lies on the perpendicular bisector of $MD$ and $DC$.

6. **Relate the angles using the given circle properties:**
   - Since the incenter of $\triangle MCB$ lies on the circumcircle of $\triangle MDC$, the angle $\angle MOC$ (where $O$ is the incenter) must satisfy the relationship between the angles in the two triangles.
   - $\angle MOC = 180^\circ - \alpha$ (since it is the external angle of $\triangle MDC$).
   - The incenter angle $\angle MOC$ is also given by $90^\circ + \frac{\alpha}{2}$ (since it is the angle formed by the incenter in $\triangle MCB$).

7. **Set up the equation and solve for $\alpha$:**
   \[
   180^\circ - \alpha = 90^\circ + \frac{\alpha}{2}
   \]
   \[
   180^\circ - 90^\circ = \alpha + \frac{\alpha}{2}
   \]
   \[
   90^\circ = \frac{3\alpha}{2}
   \]
   \[
   90^\circ \cdot \frac{2}{3} = \alpha
   \]
   \[
   \alpha = 60^\circ
   \]

The final answer is $\boxed{60^\circ}$.