Problem 5. Inside the square $A B C D$, points $K$ and $M$ are marked (point $M$ is inside triangle $A B D$, point $K$ is inside $B M C$) such that triangles $B A M$ and $D K M$ are equal $(A M=K M, B M=M D, A B=K D)$. Find $\angle K C M$, if $\angle A M B=100^{\circ}$
Answer: $35^{\circ}$.

Solution. Note that triangles $A B M$ and $A M D$ are also equal by three sides. Thus, point $M$ lies on the diagonal $A C$ of the square, meaning $\angle M C D=\angle M A D=\angle M A B=45^{\circ}($ Fig. 1$)$.

In addition, from the equality of triangles $B A M$ and $D K M$, it follows that $\angle M K D=$ $\angle B A M=45^{\circ}$. Therefore, quadrilateral $M K C D$ is cyclic. Then $\angle K C M=\angle K D M=180^{\circ}-100^{\circ}-45^{\circ}=35^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5bfc2fb0dde28bf159fag-2.jpg?height=503&width=503&top_left_y=84&top_left_x=1938)

Fig. 1: to the solution of problem 5

## Criteria

4 p. A complete and justified solution is provided.

0 p. The problem is not solved or is solved incorrectly.

In other cases, sum the following criteria:

3 p. The cyclic nature of quadrilateral $M K C D$ is proven.

1 p. It is proven that point $M$ lies on diagonal $A C$.

0 p. The correct answer is provided.