4. In a right triangle $ABC$, a height $BH$ is dropped onto the hypotenuse $AC$. Points $X$ and $Y$ are the centers of the circles inscribed in triangles $ABH$ and $CBH$, respectively. The line $XY$ intersects the legs $AB$ and $BC$ at points $P$ and $Q$. Find the area of triangle $BPQ$, given that $BH = h$.
Answer: $\frac{h^{2}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-13.jpg?height=477&width=891&top_left_y=869&top_left_x=662)

Solution 1. The lines $H X$ and $H Y$ are the angle bisectors of the right angles $A H B$ and $B H C$, hence

$$
\angle A H X=\angle B H X=\angle B H Y=\angle C H Y=45^{\circ} \text { and } \angle X H Y=90^{\circ} .
$$

The segments $H X$ and $H Y$ are in the same ratio as the radii of the inscribed circles of the similar triangles $A H B$ and $B H C$, so $\frac{H X}{H Y}=\frac{A B}{B C}$. Therefore, triangles $X H Y$ and $A B C$ are similar. Then

$$
\angle B A C=\angle H X Y=180^{\circ}-\angle P X H
$$

and quadrilateral $A H X P$ is cyclic. Therefore, $\angle Q P B=\angle A H X=45^{\circ}$, which means triangle $B P Q$ is isosceles. Note that triangles $P B X$ and $H B X$ are congruent. Indeed, side $B X$ is common, $\angle B P X=\angle B H X=45^{\circ}$, and angles $P B X$ and $H B X$ are equal since $B X$ is the angle bisector of $\angle P B H$. Then $B P=B H=h$ and $S_{B P Q}=\frac{B P^{2}}{2}=\frac{h^{2}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-13.jpg?height=482&width=894&top_left_y=1969&top_left_x=655)

Solution 2. Let $I$ be the incenter of triangle $A B C$. We will show that $I$ is the orthocenter of triangle $B X Y$. Let $B D$ be the angle bisector of triangle $C B H$, and $E$ be the intersection point of lines $A X$ and $B Y$ (see the figure). Then

$$
\angle D A E=\frac{1}{2} \angle B A C=\frac{1}{2}\left(90^{\circ}-\angle A B H\right)=\frac{1}{2} \angle C B H=\angle D B H
$$

Thus, in triangles $D A E$ and $D B H$, there are two pairs of equal angles. Therefore, their third angles are also equal, that is, $\angle A E D=\angle B H D=90^{\circ}$. Similarly, it can be verified that $Y I \perp B X$. Therefore, $B I \perp P Q$, so ray $B I$ is both the angle bisector and the altitude of triangle $B P Q$. Consequently, $\triangle B P Q$ is isosceles, from which $\angle B Q P=45^{\circ}$.

Furthermore, note that triangles $B H Y$ and $B Q Y$ are congruent, since they have a common side $B Y$ and two pairs of equal angles:

$$
\angle H B Y=\angle Q B Y \quad \text { and } \quad \angle B H Y=\frac{1}{2} \angle B H C=45^{\circ}=\angle B Q Y
$$

Therefore, $B P=B Q=B H=h$ and $S_{B P Q}=\frac{B Q^{2}}{2}=\frac{h^{2}}{2}$.