8,9
[ Equilateral (regular) triangle ]

A line is drawn through vertex $C$ of the square $A B C D$, intersecting the diagonal $B D$ at point $K$, and the perpendicular bisector of side $A B$ at point $M$ ( $M$ is between $C$ and $K$ ). Find $\angle D C K$, if $\angle A K B=\angle A M B$.
Prove that triangle $A M B$ is equilateral.

## Solution

Since $\angle A K B=\angle A M B$, points $M, K, A$ and $B$ lie on the same circle. Therefore, $\angle M A B=\angle M K B=\angle A K B$ $=\angle A M B$, so $M B=A B$. Since $M B=M A$, triangle $A M B$ is equilateral. Thus, $\angle B K M=\angle A M B=60^{\circ}, \angle D C K=\angle B K M-$ $\angle K D C=15^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-02.jpg?height=640&width=541&top_left_y=1275&top_left_x=770)

## Answer

$15^{\circ}$.