Problem

Source: 2019 Second Round - Poland

Tags: geometry, lengths, angles



Let $X$ be a point lying in the interior of the acute triangle $ABC$ such that \begin{align*} \sphericalangle BAX = 2\sphericalangle XBA \ \ \ \ \hbox{and} \ \ \ \ \sphericalangle XAC = 2\sphericalangle ACX. \end{align*}Denote by $M$ the midpoint of the arc $BC$ of the circumcircle $(ABC)$ containing $A$. Prove that $XM=XA$.