Problem

Source:

Tags: geometry, incenter, circumcircle, geometry proposed



On the circumcircle of the acute triangle $ABC$, $D$ is the midpoint of $ \stackrel{\frown}{BC}$. Let $X$ be a point on $ \stackrel{\frown}{BD}$, $E$ the midpoint of $ \stackrel{\frown}{AX}$, and let $S$ lie on $ \stackrel{\frown}{AC}$. The lines $SD$ and $BC$ have intersection $R$, and the lines $SE$ and $AX$ have intersection $T$. If $RT \parallel DE$, prove that the incenter of the triangle $ABC$ is on the line $RT.$