Let's construct a triangle, given the angle $\alpha$, the sum of the two sides forming the angle, and the median extending from the vertex of angle $\alpha$.
Let $ABC$ be the sought triangle and $AD=k_{a}$ the given median. Extend the median $AD$ and measure the segment $DE=AD=k_{a}$ on it. On the extension of $EB$, measure the distance $BF=BE$. Since $ABEC$ is a parallelogram, we have

$$
\angle BAC = \alpha = \angle ABF
$$

and thus in the isosceles triangle $ABF$,

$$
\angle AFB = \angle FAB = 90^{\circ} - \frac{\alpha}{2}.
$$

Therefore, in triangle $AEF$,

$$
\angle AFE = 90^{\circ} - \frac{\alpha}{2}, \quad EF = b + c, \quad AE = 2k_{a}
$$

and this triangle can be constructed. Then, on $AF$, we draw the angle $\angle FAB = 90^{\circ} - \frac{\alpha}{2}$ at $A$, which intersects $EF$ at $B$. $BD$ is the line parallel to $EF$ drawn from $A$, intersecting the third vertex $C$ of the triangle.

Since the construction of triangle $AEF$ involves two sides and the angle opposite one of them, our problem has 2, 1, or 0 solutions.