5. The circle inscribed in the right triangle ABC touches the legs CA and CB at points P and Q, respectively. The line PQ intersects the line passing through the center of the inscribed circle and parallel to the hypotenuse at point N. M is the midpoint of the hypotenuse. Find the measure of angle MCN.
Answer: $90^{\circ}$.

Instructions. Let the center of the inscribed circle be denoted by I. The points of intersection of the line,

![](https://cdn.mathpix.com/cropped/2024_05_06_cd3075ae5b9d36827637g-1.jpg?height=257&width=578&top_left_y=1647&top_left_x=2143)
parallel to the hypotenuse and passing through I, with the legs AC and BC, are denoted by K and T, respectively. We will assume that point N lies on the extension of segment KT beyond point K (see figure). It is easy to see that IPCQ is a square. PQ is the axis of symmetry of the square, and point N lies on it, which means that triangles NPI and NPC are symmetric and, therefore, equal. We have $\angle NCP = \angle NIP$. Angles NIP and ABC are equal as acute angles with respectively parallel sides. Angles MBC and MCB are equal, as angles at the base of an isosceles triangle (property of the median drawn to the hypotenuse). Combining the obtained angle equalities, we have $\angle MCB = \angle NCA$. Adding angle ACM to each of them, on one side we get a right angle, and on the other - $\angle NCM$. Hence the answer.

Criteria. Only answer: 0 points.

Noted that IPCQ is a square: 1 point.

Proved the equality of triangles NPI and NPC: 2 points.