Let's construct a triangle, given one of its angles $\alpha$, the angle bisector of this angle $f_{a}$, and the radius of the circumscribed circle $r$.

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Starting from the circumscribed circle, we construct the chord $BC$ (1st figure) corresponding to the given angle $\alpha$ at an arbitrary point $X$, which will be the side $a$ of the sought triangle.

![](https://cdn.mathpix.com/cropped/2024_05_02_cb9502f1d4d847bd86a8g-1.jpg?height=474&width=458&top_left_y=294&top_left_x=801)

1st figure

The angle bisector $f_{a}$ always bisects the arc $BC$ at point $F$, from which the side $BC$ is seen at an angle of $180^{\circ}-\alpha$. The distance $FA$ of the sought point intersects the side $BC$ at $D$. Although $DA = f_{a}$ cannot be used directly, the distance $FA$ can be constructed. Indeed,

$$
\triangle FDC \sim \triangle FCA
$$

because $\angle F$ is common, and $\angle DCF$ is equal to $\angle FAC$, as they are inscribed angles subtending equal arcs. Therefore,

$$
FD : FC = FC : FA
$$

or

$$
FC^2 = FA \cdot FD = FA \cdot (FA - f_{a})
$$

Based on this relationship, given $FC$ and $f_{a}$, $FA$ can be constructed as shown in the 2nd figure, using the theorem that the tangent segment $FC$ is the geometric mean between the two segments of the secant through $F$, $FA$ and $FA - f_{a}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_cb9502f1d4d847bd86a8g-1.jpg?height=325&width=360&top_left_y=1605&top_left_x=861)

2nd figure

Using the constructed ray $FA$, a circle centered at $F$ intersects the circumcircle of the triangle at point $A$. The second intersection point gives the mirror image.

Thus, there is always only one solution as long as $FA \leq 2r$, or (using the notation from the 1st figure)

$$
f_{a} \leq A'D' = A'B \cdot \cos \frac{\alpha}{2} = A'F \cos \frac{\alpha}{2} \cos \frac{\alpha}{2} = 2r \cos^2 \frac{\alpha}{2}
$$

In the case of equality, the triangle is isosceles ($AB = AC$), and if $f_{a} > 2r \cos^2 \frac{\alpha}{2}$, there is no solution.