In $\triangle ABC$, the angles are $\alpha, \beta, \gamma$. Let $K$ be the center of the circumcircle of the triangle. What angle does the line $CK$ form with the side $AB$?
Assuming that $\alpha>\beta$ does not harm the generality. Let the line $C K$ intersect the side $A B$ at $D$, the circumcircle at $E$ for the second time, and denote the sought angle by $\delta$.

Considering the position of point $D$, generally three cases must be distinguished:

(1) $\alpha<90^{\circ}$, but $\alpha-\beta<90^{\circ}$ and $\alpha-\beta>90^{\circ}: D$ is on the extension of segment $A B$ beyond $A$ (2nd figure - Vö. $d$ ) with the boundary). Regarding case (1), the angle $\delta$ is an acute angle (since $\alpha>\beta$) and is the angle at $D$ in $\triangle A D C$ (1st figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_892c4cb6a99aa45d648fg-1.jpg?height=626&width=590&top_left_y=454&top_left_x=754)

In the right triangle $C A E$,

$$
\angle A E C = \beta,
$$

as the inscribed angles subtending the common arc $\widehat{A C}$, and thus as the complementary angle

$$
\angle A C E \equiv \angle A C D = 90^{\circ} - \beta
$$

Therefore, from $\triangle A D C$,

$$
\begin{gathered}
\delta = 180^{\circ} - \alpha - (90^{\circ} - \beta) = 90^{\circ} - \alpha + \beta = \\
= 90^{\circ} - (\alpha - \beta)
\end{gathered}
$$

We arrive at the same result in case (2) as well, by the same reasoning.

![](https://cdn.mathpix.com/cropped/2024_05_02_892c4cb6a99aa45d648fg-1.jpg?height=537&width=678&top_left_y=1580&top_left_x=710)

In case (3) - when $\alpha - \beta > 90^{\circ} - \mathrm{in} \triangle A D C$ (2nd figure)

$$
\begin{aligned}
\delta = & 180^{\circ} - (180^{\circ} - \alpha) - [180^{\circ} - (90^{\circ} - \beta)] = \\
& = \alpha - 180^{\circ} + 90^{\circ} - \beta = (\alpha - \beta) - 90^{\circ} .
\end{aligned}
$$

Therefore, we can state generally that when $\alpha > \beta$,

$$
\delta = \left|90^{\circ} - (\alpha - \beta)\right|
$$

Boundary cases:
a) $\alpha = \beta, \delta = 90^{\circ}$ ( $C K$ is the axis of symmetry of the isosceles triangle);
b) $\gamma = 90^{\circ}$ (i.e., $A B$ is a diameter) in which case $\alpha = 90^{\circ} - \beta$, and thus $\delta = 2 \beta$;
c) $\alpha = 90^{\circ}, \delta = \beta (C K \equiv C B, D \equiv B)$;
d) $\alpha - \beta = 90^{\circ}, \delta = 0 (C K \| A B)$.

Remark: The solvers - almost without exception - dealt exclusively with case (1).