Let $M$ be the intersection point of the medians $AD$ and $BE$ of a right triangle $ABC$ ($\angle C=90^\circ$). It is known that the circumcircles of triangles $AEM$ and $CDM$ are tangent. Find the angle $\angle BMC.$

1. **Identify the given elements and properties:**
   - \( \triangle ABC \) is a right triangle with \( \angle C = 90^\circ \).
   - \( AD \) and \( BE \) are medians intersecting at point \( M \).
   - The circumcircles of \( \triangle AEM \) and \( \triangle CDM \) are tangent.

2. **Use properties of medians and centroid:**
   - The centroid \( M \) of a triangle divides each median into a ratio of \( 2:1 \).
   - Since \( \triangle ABC \) is a right triangle, the centroid \( M \) lies at \( \left( \frac{a}{3}, \frac{b}{3} \right) \) where \( a \) and \( b \) are the lengths of the legs \( AC \) and \( BC \) respectively.

3. **Circumcircles tangency condition:**
   - The tangency of the circumcircles of \( \triangle AEM \) and \( \triangle CDM \) implies that the angle between the tangents at \( M \) to these circles is \( 90^\circ \).

4. **Calculate the angles:**
   - Let \( O' \) and \( O'' \) be the circumcenters of \( \triangle AEM \) and \( \triangle CDM \) respectively.
   - The tangency condition implies \( \angle EMO' + \angle CMO'' = 90^\circ \).

5. **Use the properties of the right triangle and medians:**
   - In \( \triangle ABC \), the medians \( AD \) and \( BE \) intersect at \( M \), the centroid.
   - Since \( \angle C = 90^\circ \), the medians \( AD \) and \( BE \) are perpendicular to each other at \( M \).

6. **Determine \( \angle BMC \):**
   - Since \( M \) is the centroid and the medians are perpendicular, \( \angle BMC \) is formed by the intersection of the medians.
   - Therefore, \( \angle BMC = 90^\circ \).

The final answer is \( \boxed{90^\circ} \).