The hypotenuse $AB$ of a right-angled triangle $ABC$ touches the corresponding excircle $\omega$ at point $T$. Point $S$ is symmetrical $T$ relative to the bisector of angle $C$, $CH$ is the height of the triangle. Prove that the circumcircle of triangle $CSH$ touches the circle $\omega$.
Let $G$ be the reflection of $H$ relative to the bisector of angle $C$ and let $M$ be the midpoint of $AB$. Let $a$, $b$, and $c$ be the respective side lengths of $ABC$.
It suffices to show that $(CGT)$ is tangent to $w$. Notice $G$ lies on $CM$. It is clear that $w$ is tangent to $BC$ at $T$ so it suffices to show that
$$MC\cdot MG=\left(\frac{c}{2}\right)\cdot \left(\frac{c}{2}-\frac{ab}{c}\right)=\frac{c^2}{4}-\frac{ab}{2}=\left(\frac{a-b}{2}\right)^2=MT^2$$
