$\angle A B D=\angle C B E=\beta$

## Solution

Since

$$
\angle A B D=\angle A B C-\angle D B C, \angle C B E=\angle D B E-\angle D B C, \angle A B C=\angle D B E \text {, }
$$

then $\angle A B D=\angle C B E$. Points $A, D, C, E$ lie on a circle with center at point $B$ and radius equal to $A B$. Therefore, $\angle D A K=\angle C E K$.

Segment $B K$ is seen from points $A$ and $D$ at the same angle ( $\angle B A K=\angle B D K$ ). Therefore, points $A, D, K$ and $B$ lie on the same circle. Hence, $\angle D B K=\angle D A K$. Similarly, we can prove that $\angle C B K=\angle C E K$. Therefore,

$$
\angle A B K=\angle A B D+\angle D B K=\beta+\frac{\alpha-\beta}{2}=\frac{\alpha+\beta}{2}, \angle K B C=\frac{1}{2} \angle D B C=\frac{\alpha-\beta}{2}
$$

Thus,

$$
\frac{\angle A B K}{\angle K B C}=\frac{\alpha+\beta}{\alpha-\beta}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_c317f3f1f50d1d30304bg-32.jpg?height=425&width=694&top_left_y=1614&top_left_x=682)

Answer

$\frac{\alpha+\beta}{\alpha-\beta}$.
[ Segment visible from two points at the same angle ]

Two equal isosceles triangles $A B C$ and $D B E$ ( $A B=B C=D B=B E$ ) share a common vertex $B$ and lie in the same plane, with points $A$ and $C$ located on opposite sides of line $B D$, and segments $A C$ and $D E$ intersecting at point $K$. It is known that $\angle A B C=\angle D B E=\alpha<\frac{\pi}{2}, \angle A K D=\beta<\alpha$. In what ratio does the line $B K$ divide the angle $A B C$?