[ Properties of bisectors, concurrency $]$

On the sides $B C$ and $C D$ of the square $A B C D$, points $M$ and $K$ are marked such that $\angle B A M=\angle C K M=30^{\circ}$. Find $\angle A K D$.

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From the condition, it follows that $\angle B M A=\angle C M K=60^{\circ}$, and then $\angle A M K=60^{\circ}$ (see the left figure). We can reason in different ways.

First method. Let $A H$ be the perpendicular dropped from vertex $A$ to $M K$. Then the right triangles $A M B$ and $A M H$ are equal by hypotenuse and acute angle, from which $A H=A B$. If point $H$ lies on the segment $M K$, then the right triangles $A K H$ and $A K D$ are equal by hypotenuse and leg. Therefore, $\angle A K D = 1 / 2 \angle M K D = 75^{\circ}$.

If $H$ lies on the extension of segment $M K$ beyond point $K$, then segment $A H$ intersects side $C D$ at point $X$. But then $A H > A X > A D$, which contradicts the equality $A H = A D$.
![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-06.jpg?height=990&width=1262&top_left_y=848&top_left_x=412)

Second method. The diagonal $C A$ of the square is the bisector of the internal angle of triangle $C M K$, and the ray $M A$ is the bisector of its external angle, so vertex $A$ is the center of the excircle of this triangle. Therefore, $K A$ is also the bisector of the external angle of triangle $C M K$, so $\angle A K D = 1 / 2 \angle M K D = 75^{\circ}$.

Third method. Extend segment $K M$ to intersect line $A B$ at point $P$ (see the right figure). Then $\angle P M B = \angle C M K = \angle A M B$. Therefore, the right triangles $P M B$ and $A M B$ are equal (by leg and acute angle), that is, $A P = 2 a$, where $a$ is the side of the given square, and $P M = A M$.

By the property of the leg opposite the $30^{\circ}$ angle, $A M = 2 B M$ and $M K = 2 M C$. Therefore, $P K = P M + M K = 2(B M + M C) = 2 B C = 2 a$.

Thus, triangle $A P K$ is isosceles with an angle of $30^{\circ}$ at vertex $P$, so the angle $M K A$ at its base is $75^{\circ}$, and

$\angle A K D = 180^{\circ} - 30^{\circ} - 75^{\circ}$.

## Answer

$75^{\circ}$.

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