6. [9] Inside an isosceles triangle $A B C$, a point $K$ is marked such that $C K=A B=B C$ and $\angle K A C=30^{\circ}$. Find the angle $A K B$.

E. Bakayev
Answer: $150^{\circ}$. Solution 1. Construct an equilateral triangle $B C L$ (see the figure; points $A$ and $L$ are on the same side of the line $B C$). Points $A, C$, and $L$ lie on a circle with radius $B A$ and center at point $B$. Since $K$ lies inside triangle $A B C$, angle $A B C$ is greater than $60^{\circ}$, hence $L$ and $B$ lie on opposite sides of $A C$ and $L$ lies on the smaller arc $A C$. Therefore, the inscribed angle $C A L$ is half of the central angle $C B L$, which is $30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_92be13002631bb88c52bg-3.jpg?height=388&width=617&top_left_y=1645&top_left_x=1325)

Obviously, the point $K$ that satisfies the conditions of the problem is unique, so it coincides with the point symmetric to $L$ with respect to side $A C$. Thus, triangle $A K L$ is equilateral, and point $K$, like point $B$, lies on the perpendicular bisector of segment $A L$, from which $\angle A K B = \angle L K B = 180^{\circ} - 1 / 2 \angle A K L = 150^{\circ}$.

Solution 2. Let the height $B M$ of triangle $A B C$ intersect line $A K$ at point $O$ (see the figure). Then $\angle C O M = \angle A O M = 60^{\circ}$. Therefore, $\angle A O C = 120^{\circ}$ and $\angle C O B = 120^{\circ}$. Consequently, triangles $B O C$ and $K O C$ are congruent by two sides and the angle opposite the larger side (the so-called fourth criterion of triangle congruence). Therefore, $O B = O K$, meaning triangle $B O K$ is isosceles with an angle of $120^{\circ}$ at vertex $O$. Therefore, $\angle O K B = 30^{\circ}$, and $\angle A K B = 150^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_92be13002631bb88c52bg-3.jpg?height=351&width=557&top_left_y=2320&top_left_x=1366)

Solution 3. Construct an equilateral triangle $A C L$ on $A C$ such that points $L$ and $B$ lie on the same side of $A C$ (see the figure).

Draw the height $B M$ in triangle $A B C$, which is also the perpendicular bisector of side $A C$. Since $A L C$ is equilateral, point $L$ also lies on line $B M$. Additionally, draw the height $A N$ in triangle $A L C$. Since $A N$ is the angle bisector of $\angle L A C$, point $K$ lies on this line. Note also that $K$ lies on the same side of $B M$ as $A$, because due to $C K = C B$, it cannot lie inside triangle $B M C$; thus, $K$ lies on segment $A N$.

Notice that right triangles $B M C$ and $K N C$ are congruent by a leg and the hypotenuse (since $M C = A C / 2 = L C / 2 = N C$ and $B C = K C$). Therefore, first, $B M = K N$, and second, $B$ lies on segment $L M$ (since $B M = K N 2 \beta > \beta$, then $\angle A K C = 180^{\circ} - \beta$. Thus, $\angle A C K = \beta - 30^{\circ}$, from which $\angle K C B = \angle C - \angle A C K = (90^{\circ} - \beta) - (\beta - 30^{\circ}) = 120^{\circ} - 2 \beta$, $\angle B K C = 1 / 2 (180^{\circ} - \angle K C B) = 30^{\circ} + \beta$, $\angle A K B = 360^{\circ} - \angle B K C - \angle A K C = 360^{\circ} - (30^{\circ} + \beta) - (180^{\circ} - \beta) = 150^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_92be13002631bb88c52bg-4.jpg?height=496&width=500&top_left_y=89&top_left_x=1395)