Given a right triangle $ABC$ with $ \angle C=90^o$. On its hypotenuse $AB$ is arbitrary mark the point$ P$. The point $Q$ is symmetric to the point $P$ wrt $AC$. Let the lines $PQ$ and $BQ$ intersect $AC$ at points $O$ and $R$ respectively. Denote by $S$ the base of the perpendicular from the point $R$ on the line $AB$ ($S \ne P$), and let $T$ be the point of intersection of lines $OS$ and $BR$. Prove that $R$ is the center of the circle inscribed in the triangle $CST$. The point $Q$ is symmetric to the point $P$ wrt $AC$.

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