4. Point $M$ is the midpoint of the leg $A B$ of an isosceles right triangle $A B C$ with a right angle $\angle B$. The bisector of angle $\angle A$ intersects the circumcircle of the triangle at point $L$. Point $H$ is the foot of the perpendicular dropped from $L$ to the line $A C$. Find the angle $\angle A M H$.
Answer: $112.5^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_05_06_2ae48c21b42f492bf703g-02.jpg?height=999&width=996&top_left_y=346&top_left_x=513)

First solution. Let $D$ be the intersection point of lines $A C$ and $B L$. Since quadrilateral $A B L C$ is cyclic, $\angle A L B = \angle A C B = 45^{\circ}$ and $\angle A L C = \angle A B C = 90^{\circ}$. Additionally, $\angle D A L = \frac{1}{2} \angle B A C = 22.5^{\circ}$. Then,

$$
\angle A D L = \angle A L B - \angle D A L = 45^{\circ} - 22.5^{\circ} = 22.5^{\circ}
$$

Therefore, triangle $A L D$ is isosceles and its altitude $L H$ is also a median. Thus, $A H = H D$ and segment $H M$ is the midline of triangle $A B D$, in particular, lines $H M$ and $B D$ are parallel, and hence, $\angle A M H = \angle A B L$.

Since $A L$ is the angle bisector of $\angle B A C$ and quadrilateral $A B L C$ is cyclic, $\angle C B L = \angle C A L = 22.5^{\circ}$. Therefore,

$$
\angle A M H = \angle A B L = \angle A B C + \angle C B L = 90^{\circ} + 22.5^{\circ} = 112.5^{\circ}.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_2ae48c21b42f492bf703g-02.jpg?height=734&width=763&top_left_y=2032&top_left_x=652)

Second solution. Let $O$ be the midpoint of the hypotenuse $A C$, then it is the center of the circumcircle of triangle $A B C$ and, therefore, $C O = L O$. Consequently, triangle $A O L$ is isosceles and $\angle O L A = \angle O A L = \frac{1}{2} \angle B A C = 22.5^{\circ}$. Angle $\angle L O H$ is an exterior angle for triangle $A O L$, so $\angle L O H = \angle O L A + \angle O A L = 22.5^{\circ} + 22.5^{\circ} = 45^{\circ}$. Thus, triangle $L H O$ is isosceles and right-angled. Segment $M O$ is the midline of triangle $A B C$ and, therefore, it is parallel to side $B C$. Hence, triangle $A M O$ is isosceles and right-angled. Consequently, triangles $A M O$ and $L H O$ are congruent by two angles and a side $(A O = L O)$, and then $M O = O H$ and triangle $M O H$ is isosceles with an exterior angle of $45^{\circ}$. Therefore, the angles at its base are each $22.5^{\circ}$. It remains to note that $\angle A M H = \angle A M O + \angle O M H = 90^{\circ} + 22.5^{\circ} = 112.5^{\circ}$.