8.5. On the sides $B C$ and $C D$ of the square $A B C D$, points $M$ and $K$ are marked respectively such that $\angle B A M = \angle C K M = 30^{\circ}$. Find $\angle A K D$.
Answer: $75^{\circ}$.

Solution. From the condition of the problem, it follows that $\angle B M A=\angle C M K=60^{\circ}$, and then $\angle A M K=60^{\circ}$ (see Fig. 8.5a). Further reasoning can be done in different ways.

First method. Let $A H$ be the perpendicular 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 $A H=A B$. Using this equality, we get that the right triangles $A K H$ and $A K D$ are equal by hypotenuse and leg. Therefore, $\angle A K D=\frac{1}{2} \angle M K D=\frac{150^{\circ}}{2}=75^{\circ}$.

In the given reasoning, it is used that point $H$ lies on the segment $M K$, and not on its extension beyond point $K$. This is indeed the case, otherwise $A H$ would intersect side $C D$ at point $X$, but then $A H>A X>A D$, which contradicts the equality $A H=A D$. For the lack of this explanation, the student's grade should not be reduced.

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 $\angle A K D=\frac{1}{2} \angle M K D=75^{\circ}$.

Third method. Extend the segment $K M$ to intersect the line $A B$ at point $P$ (see Fig. 8.5b). 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), then $P B=A B$, 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 in a right triangle, $A M=2 B M$ 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 its angle at the base is $75^{\circ}$. Since $\angle M K D=150^{\circ}$ and $\angle M K A=75^{\circ}$, then $\angle A K D=75^{\circ}$.

Grading criteria:

+ The correct answer and a fully justified solution are provided

![](https://cdn.mathpix.com/cropped/2024_05_06_831a12603e07f8334797g-2.jpg?height=342&width=363&top_left_y=162&top_left_x=1672)

A

![](https://cdn.mathpix.com/cropped/2024_05_06_831a12603e07f8334797g-2.jpg?height=636&width=369&top_left_y=518&top_left_x=1666)

Fig. 8.5 b

- Only the answer is provided