# Task 9.4

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 the angle $A K D$.

## Number of points 7

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# Answer:

$75^{\circ}$

## Solution

## First Solution

If we find angle $A K D$, it is only present in triangle $A K D$, and we do not know another angle in this triangle.

All the information in the problem is above line $A K$. Therefore, we need to move upwards.

And we need to find angle $A K M$. Triangle $A K M$ is also not a good one; it is unlikely to expect properties of equilateral or even isosceles triangles from it. Therefore, we extend angle $A K M$ to another triangle: we extend $K M$.

It will intersect $A B$ at some point $P$. In this triangle, we immediately note that angle $P$ is $30^{\circ}$. Therefore, triangle $A P M$ is isosceles. Let's try to express the lengths of the sides of triangle $A K P$. For this, we denote the side of the square as $a$. Then triangles $P B M$ and $A B M$ are equal, so $A P = 2a$. Next, $P M = A M = 2 B M$ (the leg opposite the $30^{\circ}$ angle is half the hypotenuse) and $M K = 2 M C$.

Then $P K = 2(B M + M C) = 2a$. As expected, triangle $A P K$ is isosceles, angle $K$ in it is $75^{\circ}$, and the required angle is $75^{\circ}$.

## Second Solution

$K A$ turned out to be the bisector of angle $M K D$. $M A$ is the bisector of angle $B M K$. These are the external angles of triangle $M K C$. It remains to see that $A C$ is the bisector of the internal angle $C$ of this triangle. Therefore, point $A$ is the excenter of triangle $M C K$. We draw the diagonal $A C$ in the square. Then point $A$ is the intersection of the bisectors of the internal angle $C$ of triangle $M C K$ and the external angle $M$ of the same triangle. That is, point $A$ is the excenter of this triangle. Consequently, the line passing through the third vertex of this triangle and point $A$ will also be the bisector of the external angle $K$ of this triangle. Again, we get the same $75^{\circ}$.

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