Biyankov Yu.A.

In an isosceles triangle \(ABC\), a point \(M\) is marked on the lateral side \(BC\) such that the segment \(MC\) is equal to the height of the triangle dropped to this side, and a point \(K\) is marked on the lateral side \(AB\) such that the angle \(KMC\) is a right angle. Find the angle \(ACK\).
Given that $CL = CM$ (see the figure). Therefore, right triangles $CLK$ and $CMK$ are congruent by hypotenuse and leg, and $CK$ is the angle bisector of $\angle LCM$. Let $\angle BAC = \angle BCA = \alpha$, then $\angle LCA = 90^\circ - \alpha$, and $\angle LCB = \alpha - (90^\circ - \alpha) = 2\alpha - 90^\circ$. Thus,

$\angle ACK = \angle ACL + \angle LCK = 90^\circ - \alpha + \frac{1}{2}(2\alpha - 90^\circ) = 45^\circ$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c271881b12104a3b803eg-10.jpg?height=463&width=463&top_left_y=1575&top_left_x=795)

## Answer

$45^\circ$.