The points $M$ and $N$ are chosen on the angle bisector $AL$ of a triangle $ABC$ such that $\angle ABM=\angle ACN=23^{\circ}$. $X$ is a point inside the triangle such that $BX=CX$ and $\angle BXC=2\angle BML$. Find $\angle MXN$.
Source: Baltic Way 2010
Tags: geometry, circumcircle, angle bisector, geometry proposed
The points $M$ and $N$ are chosen on the angle bisector $AL$ of a triangle $ABC$ such that $\angle ABM=\angle ACN=23^{\circ}$. $X$ is a point inside the triangle such that $BX=CX$ and $\angle BXC=2\angle BML$. Find $\angle MXN$.