Problem 7.2. (15 points) Given an isosceles triangle $A B C(A B=B C)$. On the sides $A B, B C, A C$ points $K, L, M$ are marked respectively such that $\angle A K M=90^{\circ}, \angle B L K=90^{\circ}$ and $K M=K L$. What is the measure of angle $C M L ?$
Answer: $90^{\circ}$.

Solution. From the condition, it follows that $\angle M K L + \angle L K B = 90^{\circ} = \angle L K B + \angle L B K$, from which $\angle M K L = \angle L B K$ (Fig. 1). Triangles $A B C$ and $M K L$ are isosceles with equal angles at the vertex, so the base angles are also equal: $\angle K L M = \angle M C L$. It remains to note that $90^{\circ} = \angle K L M + \angle C L M = \angle M C L + \angle C L M$, so $\angle C M L = 180^{\circ} - \angle M C L - \angle C L M = 90^{\circ}$.

Criteria

The largest suitable criterion is used:

15 p. Any complete solution to the problem.

5 p. The similarity (i.e., equality of corresponding angles) of triangles $A B C$ and $M K L$ is proven.

8 p. The problem is solved using the similarity of triangles $A B C$ and $M K L$, but their similarity is not proven.

0 p. Only the answer is provided.

![](https://cdn.mathpix.com/cropped/2024_05_06_58045c34aa4495fc9363g-3.jpg?height=502&width=483&top_left_y=92&top_left_x=485)

Fig. 1: to the solution of problem 7.2.