## Task A-1.2.

Given is a right triangle $A B C$ with a right angle at vertex $C$. Let $N$ be the foot of the altitude from vertex $C$, $M$ the midpoint of the hypotenuse, and $L$ the intersection of the angle bisector of the right angle with the hypotenuse. If the measure of angle $\varangle L C N$ is $15^{\circ}$, determine the measure of angle $\varangle M C L$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_f93036d069fffd92d333g-02.jpg?height=745&width=1239&top_left_y=998&top_left_x=337)

Let $\alpha$ be the measure of angle $\varangle B A C$.

From the right triangle $A B C$, we have that $\varangle C B A=90^{\circ}-\alpha$.

Since point $M$ is the midpoint of the hypotenuse $\overline{A B}$ of the right triangle $A B C$, it follows that $M$ is also the center of the circumscribed circle of triangle $A B C$.

Therefore, triangle $M C A$ is isosceles, so $\varangle A C M=\varangle M A C=\alpha$.

From the right triangle $B C N$, we have that $\varangle N C B=\alpha$.

We conclude that $\varangle N C B=\varangle A C M$.

Since $L$ lies on the angle bisector, we have $\varangle A C L=\varangle L C B$. 1 point

From the last two equalities, we conclude that $\varangle M C L=\varangle L C N=15^{\circ}$.

Note: The last step in the solution can be replaced by two smaller steps: calculating the angle $\alpha$ (considering that $\varangle A C L=\varangle L C B=45^{\circ}$), and finally calculating the unknown angle $\varangle M C L$ from $\alpha$. There are two possibilities for the measure of this angle: in the case where $|A C|>|B C|$, the points $A, M, L, N$ and $B$ are located on the hypotenuse in that order, and $\alpha$ is $30^{\circ}$; in the case where $|A C|<|B C|$, the points $B, M, L, N$ and $A$ are located on the hypotenuse in that order, and $\alpha$ is $60^{\circ}$. For a full score, the solution does not need to comment on both cases (only one case is sufficient), nor does it need to prove the order of points $M$, $L$, and $N$ on the hypotenuse.