Zaslavsky A.A.

Given a triangle $ABC$. On side $AB$ as the base, an isosceles triangle $ABC'$ is constructed externally with the angle at the vertex being $120^{\circ}$, and on side $AC$ an equilateral triangle $ACB'$ is constructed internally. Point $K$ is the midpoint of segment $BB'$. Find the angles of triangle $KCC'$.
Let's construct the parallelogram $B^{\prime} C^{\prime} B C^{\prime \prime}$. The composition of rotations around $C^{\prime}$ by $120^{\circ}$ and around $C$ by $60^{\circ}$ translates $B$ to $B^{\prime}$, and thus is a central symmetry with respect to $K$ (see the figure). Under this central symmetry, $C^{\prime}$ is translated to $C^{\prime \prime}$, which means that triangle $C C^{\prime} C^{\prime \prime}$ is equilateral. Since $K$ is the midpoint of $C^{\prime} C^{\prime \prime}$, $C K \perp C^{\prime} K$ and $\angle C^{\prime} C K=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-31.jpg?height=472&width=830&top_left_y=1699&top_left_x=617)

## Answer

$90^{\circ}, 30^{\circ}, 60^{\circ}$.