The height dropped from the vertex of the right angle to the hypotenuse divides the triangle into two triangles, each of which has an inscribed circle. Find the angles and the area of the triangle formed by the legs of the original triangle and the line passing through the centers of these circles, if the height of the original triangle is $h$.

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Consider the rectangle formed by the intersection of the lines containing the legs of the original triangle and the radii of the circles drawn to the points of tangency with these legs.

## Solution

Let $A B C$ be the given triangle, $\angle C=90^{\circ}, C D$ be its altitude, $O_{1}$ and $O_{2}$ be the centers of the circles inscribed in triangles $A C D$ and $B C D, r_{1}$ and $r_{2}$ be their radii, $P$ and $Q$ be the points of tangency with sides $A C$ and $B C, L$ and $K$ be the points of tangency with side $D C, M$ and $N$ be the points of intersection of the line $O_{1} O_{2}$ with sides $A C$ and $B C, F$ be the point of intersection of the lines $\mathrm{PO}_{1}$ and $Q O_{2}$.

Then $P C Q F$ is a rectangle, $P F=C Q, Q F=C P$. Therefore,

$$
\begin{aligned}
& F O_{1}=P F-r_{1}=C Q-r_{1}=C K-r_{1}, \\
& F O_{2}=F Q-r_{2}=C P-r_{2}=C L-r_{2} .
\end{aligned}
$$

Since $C K+r_{2}=C L+r_{1}=C D$, then $C K-r_{1}=C L-r_{2}$. Therefore, $\angle F O_{1} O_{2}=45^{\circ}$. Consequently,

$$
\angle C N M=45^{\circ}, C N=C Q+r_{2}=C K+r_{2}=C D=h .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f3f7c9453eb6f3f37ea7g-36.jpg?height=335&width=760&top_left_y=1885&top_left_x=655)

## Answer

$45^{\circ} ; 45^{\circ} ; 90^{\circ} ; \frac{h^{2}}{2}$.