Given a circle, construct a triangle inside it, where one angle is given and the median from another angle's vertex is also given.
Let the given circle be $k$, its center $O$, and its radius $r$. Let the triangle $ABC$ be one that meets the given conditions, where the angle at $A$ is equal to the given angle $\alpha$, and the segment connecting the midpoint $B_1$ of side $AC$ to $B$ is equal to the given length $s_b$ of the median.

![](https://cdn.mathpix.com/cropped/2024_05_02_8e8f5ef7ce49d1f6809fg-1.jpg?height=578&width=528&top_left_y=318&top_left_x=793)

According to the relationship between inscribed and central angles, $\angle BOC = 2\alpha$. Therefore, by measuring $\alpha$ on both sides of an arbitrary radius $OF$ at $O$, the points where these new rays intersect $k$ are $B$ and $C$. Then, $A$ can only be on the arc $i$ of $BC$ that does not contain $F$, and $B_1$, which is the image of $A$ scaled down by a factor of 2 from the center $C$, can only be on the image of $i$ scaled down by the same factor, $i_1$ (which is, in other words, the arc of the Thales circle above the diameter $OC$ on the side of the line $BC$ that contains $A$, since the scaling places the center of $i_1$ at the midpoint of $CO$, and $C$ is also a point on $i_1$).

On the other hand, $B_1$ will also lie on the circle $k_b$ centered at $B$ with radius $s_b$. Therefore, the position of $B_1$ is determined by the intersection of $i_1$ and $k_b$; finally, $A$ is determined by the intersection of the line $CB_1$ with $k$.

The correctness of the construction is evident; the number of solutions is 2, 1, or 0, depending on the number of common points between $i_1$ and $k_b$.