Construct a triangle given an angle, the side opposite the angle and the median to the other side (researching the number of solutions is not required).
1. **Given Data and Definitions:**
   - Let \( AB \) be the given side.
   - Let \( \alpha \) be the given angle opposite to side \( AB \).
   - Let \( m_c \) be the given median to the other side.
   - Let \( M \) be the midpoint of \( AB \).

2. **Constructing Triangle \( ABC \):**
   - Construct triangle \( ABC \) such that \( \angle CAB = 90^\circ \) and \( \angle CBA = 90^\circ - \alpha \).
   - This construction ensures that \( \angle ACB = \alpha \).

3. **Constructing the Circumcircle \( \omega_1 \):**
   - Construct the circumcircle \( \omega_1 \) of triangle \( ABC \). The circumcircle passes through points \( A \), \( B \), and \( C \).

4. **Constructing Circle \( \omega_2 \):**
   - Construct circle \( \omega_2 \) with center \( M \) (the midpoint of \( AB \)) and radius \( m_c \) (the given median).

5. **Finding the Intersection Point \( D \):**
   - Let \( D \) be the point of intersection of circles \( \omega_1 \) and \( \omega_2 \).

6. **Constructing Triangle \( ABD \):**
   - The triangle \( ABD \) is the required triangle. Here, \( AB \) is the given side, \( \angle ADB = \alpha \) is the given angle, and the median from \( D \) to \( AB \) is \( m_c \).

To verify the construction:
- The point \( D \) lies on the circumcircle \( \omega_1 \), ensuring that \( \angle ADB = \alpha \).
- The point \( D \) also lies on the circle \( \omega_2 \), ensuring that the distance from \( D \) to \( M \) (the midpoint of \( AB \)) is \( m_c \), which is the given median.

Thus, the triangle \( ABD \) satisfies all the given conditions.

The final answer is the triangle \( \boxed{ ABD } \).