Let's construct a triangle, given one of its angles $\alpha$, the length of the median $s_{a}$ starting from the vertex of this angle, and the angle $\varepsilon$ formed by the medians $s_{a}$ and $s_{b}$.
Consider the $\Delta ABC$ constructed (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_ca417d9860f8905cf9b9g-1.jpg?height=510&width=658&top_left_y=225&top_left_x=728)

Let the reflection of the vertex $A$ over the midpoint $A_{1}$ of the side $BC$ be $A'$. In the resulting parallelogram $AB A' C$, the diagonal $AA' = 2AA_{1} = 2s_{a}$ is seen from vertex $B$ at an angle of $180^{\circ} - \alpha$.

The construction process: We start with the segment $AA' = 2s_{a}$, and mark the point $S$ on it such that $AS = \frac{1}{3} AA'$. We construct the arc of a circle over $AA'$ corresponding to the viewing angle $180^{\circ} - \alpha$. The ray through $S$ that forms an angle $\varepsilon$ with $AA'$ intersects the viewing arc at point $B$. The construction of point $C$ is trivial ($A_{1}C = BA_{1}$). Since $\varepsilon$ can be measured in two ways, we generally get two solutions: $\Delta ABC$ and $\Delta AB^{*}C^{*}$. The two solutions coincide if $\varepsilon = 90^{\circ}$.