8,9  [ relationships of linear elements of similar triangles

The height of a right triangle, dropped to the hypotenuse, divides this triangle into two triangles. The distance between the centers of the inscribed circles of these triangles is 1. Find the radius of the inscribed circle of the original triangle.

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Let $ABC$ be the original triangle, $CD$ its height dropped to the hypotenuse, $O_{1}$ and $O_{2}$ the centers of the inscribed circles of triangles $ADC$ and $CDB$ respectively, $r_{1}$ and $r_{2}$ their radii, and $r$ the radius of the inscribed circle of triangle $ABC$.

Triangles $ACD$ and $CBD$ are similar to triangle $ABC$ with coefficients $AC / AB$ and $BC / AB$ respectively, so from the equality $AC^{2} + BC^{2} = AB^{2}$ it follows that $r^{2} = r_{1}^{2} + r_{2}^{2}$.

In the right triangle $O_{1}DO_{2}$, the legs are equal to $r_{1} \sqrt{2}$ and $r_{2} \sqrt{2}$, so

$1 = \left(r_{1} \sqrt{2}\right)^{2} + \left(r_{2} \sqrt{2}\right)^{2} = 2 r_{1}^{2} + 2 r_{2}^{2} = 2 r^{2}$

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-19.jpg?height=684&width=435&top_left_y=1&top_left_x=820)

## Answer

$\frac{1}{\sqrt{2}} \cdot$