4. Given an isosceles right triangle \(ABC\) with hypotenuse \(AB\). Point \(M\) is the midpoint of side \(BC\). On the smaller arc \(AC\) of the circumcircle of triangle \(ABC\), point \(K\) is chosen. Point \(H\) is the foot of the perpendicular dropped from \(K\) to line \(AB\). Find the angle \(\angle CAK\), given that \(KH = BM\) and lines \(MH\) and \(CK\) are parallel.
Answer: $22.5^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_05_06_2ae48c21b42f492bf703g-05.jpg?height=982&width=991&top_left_y=380&top_left_x=521)

First solution. Let $D$ be the intersection point of lines $A B$ and $C K$. Since quadrilateral $A B C K$ is cyclic, $\angle A K D = \angle A B C = 45^{\circ}$ and $\angle A K B = \angle A C B = 90^{\circ}$. Lines $H M$ and $C D$ are parallel, so segment $H M$ is the midline of triangle $C B D$ and, therefore, point $H$ is the midpoint of segment $B D$. Then, in triangle $K B D$, the median coincides with the altitude, making the triangle isosceles. Therefore,

$$
\angle K B D = \angle B D K = \frac{1}{2} \left(180^{\circ} - \angle B K D \right) = \frac{1}{2} \left(180^{\circ} - 90^{\circ} - 45^{\circ} \right) = 22.5^{\circ}
$$

Thus,

$$
\angle C A K = \angle C B K = 45^{\circ} - \angle K B D = 45^{\circ} - 22.5^{\circ} = 22.5^{\circ}.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_2ae48c21b42f492bf703g-05.jpg?height=717&width=759&top_left_y=1960&top_left_x=657)

Second solution. Let $O$ be the midpoint of the hypotenuse $A B$, then $O$ is the center of the circumcircle of triangle $A B C$ and, therefore, $A O = B O = K O$. Then $O M$ is the midline of triangle $A B C$ and, in particular, $O M$ is parallel to $A C$ and, thus, $\angle O M B = \angle A C B = 90^{\circ}$. Consequently, triangles $O M B$ and $O H K$ are congruent by two sides and the included angle $\left(B O = K O, \angle O M B = 90^{\circ} = \angle O H K \text{ and } B M = K H \right)$, and then $\angle K O A = \angle O B M = 45^{\circ}$. Therefore, triangle $A O K$ is isosceles with the vertex angle equal to $45^{\circ}$, so $\angle K A O = 90^{\circ} - \frac{45^{\circ}}{2} = 67.5^{\circ}$. It remains to note that

$$
\angle C A K = \angle K A O - \angle C A O = 67.5^{\circ} - 45^{\circ} = 22.5^{\circ}
$$