6. In triangle $A B C$, angle $A$ is equal to $50^{\circ}, B H$ is the altitude. Point $M$ on $B C$ is such that $B M=B H$. The perpendicular bisector of segment $M C$ intersects $A C$ at point $K$. It turns out that $A C=2 \cdot H K$. Find the angles of triangle $A B C$.
Solution. Draw a perpendicular to $BC$ from point $M$. Let $K^{\prime}$ be the point of its intersection with $AC$. Then the segment $MK$ is the midline in $\triangle K^{\prime}MC$, that is, $K^{\prime}K = KC$. From the fact that $HK = KC + AH$, we get $AH = HK^{\prime}$. Right triangles $BHA$ and $BHK^{\prime}$ are equal by two legs, $BHK^{\prime}$ and $BMK^{\prime}$ are equal by a leg and hypotenuse. Then

$$
\angle B = 3 \angle AHB = 3 \cdot (90^{\circ} - 50^{\circ}) = 120^{\circ} \text{.}
$$

The remaining angle $C$ is

$$
180^{\circ} - 50^{\circ} - 120^{\circ} = 10^{\circ}
$$