Problem 8. In a right triangle $ABC$ with a right angle at $C$, points $P$ and $Q$ are the midpoints of the angle bisectors drawn from vertices $A$ and $B$. The inscribed circle of the triangle touches the hypotenuse at point $H$. Find the angle $PHQ$.
Answer: $90^{\circ}$.

Solution. First, let's prove the following lemma.

Lemma. Let $ABC (\angle C=90^{\circ})$ be a right triangle, $I$ the intersection point of the angle bisectors $AM$ and $BK$, $S$ the midpoint of $KM$, and the inscribed circle of the triangle touches the hypotenuse at point $H$. Then the points $S, I$, and $H$ lie on the same line.

![](https://cdn.mathpix.com/cropped/2024_05_06_d5dd184c39392378f787g-3.jpg?height=318&width=604&top_left_y=1463&top_left_x=1323)

Proof. Let $D$ and $E$ be the projections of points $K$ and $M$ onto the hypotenuse $AB$ (see figure). Then $\triangle DKB = \triangle CKB$ by the hypotenuse and an acute angle, so $\angle DKB = \angle CKB$, which means $KB$ is the bisector of the external angle of $\triangle AKD$, and $AI$ is the bisector of its internal angle. Therefore, $I$ is the center of the excircle of $\triangle AKD$, so $DI$ is the bisector of angle $KDE$. Similarly, it can be shown that $EI$ is the bisector of angle $MDE$. Thus, $\triangle DIE$ is a right and isosceles triangle, meaning $IH$ is the perpendicular bisector of $DE$. The point $S$, being the midpoint of the lateral side of the right trapezoid $DKME$, also lies on the perpendicular bisector of $DE$. Therefore, the points $S, I$, and $H$ lie on the same line. The lemma is proved.

![](https://cdn.mathpix.com/cropped/2024_05_06_d5dd184c39392378f787g-3.jpg?height=317&width=640&top_left_y=2172&top_left_x=137)

Using the lemma, we will now prove that $\angle PHQ = 90^{\circ}$. Let $R$ be the midpoint of the hypotenuse $AB$. Then $PS$ and $RQ$ are the midlines of triangles $\triangle AKM$ and $\triangle AKB$, so $PS \parallel AK \parallel QR$ and $PS = \frac{1}{2} AK = QR$, meaning $PSQR$ is a parallelogram. Moreover, $\angle PSQ = 90^{\circ}$, as its sides are parallel to the sides of angle $\angle ACB = 90^{\circ}$. Therefore, $PSQR$ is a rectangle.

Let $\Omega$ be the circle circumscribed around the rectangle $PSQR$. From the lemma, it follows that $\angle SHR = 90^{\circ}$. Since $\angle SPR = \angle SHR$, point $H$ lies on the circle $\Omega$. Consequently, $\angle PHQ = \angle PRQ = 90^{\circ}$ (as inscribed angles subtending the diameter $PQ$).

Answer to variant 2: $90^{\circ}$.