Problem
Find the diameter of a circle inscribed in three sides of a right triangle $ABC$ ($\angle B$ is a right angle).
$$ $$
$$ $$
$\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$
$$ $$
$$ $$
$$ $$
$$ $$
$$ $$
$$ $$
$$ $$
$$ $$
$$ $$
$$ $$
Solution
Let the lengths of the three sides be $AB=c$, $BC=a$, and $CA=b$, respectively. From the figure, if the area of a right triangle is $S$, then
$$S=\frac{ac}{2} — [1],$$
If the diameter of the inscribed circle is $x$, then from Heron’s formula,
$$S=\frac{1}{2}×\frac{x}{2}×(a+b+c) — [2],$$
From [1] and [2],
$$\frac{ac}{2}=\frac{1}{2}×\frac{x(a+b+c)}{2},$$
$$x=\frac{2ac}{a+b+c}.$$
(Answer) $x=\frac{2ac}{a+b+c}.$
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.50; p.311.