Problem

Source: iran tst 2014 third exam

Tags: geometry, circumcircle, trigonometry, ratio, symmetry, geometry proposed



The incircle of a non-isosceles triangle $ABC$ with the center $I$ touches the sides $BC$ at $D$. let $X$ is a point on arc $BC$ from circumcircle of triangle $ABC$ such that if $E,F$ are feet of perpendicular from $X$ on $BI,CI$ and $M$ is midpoint of $EF$ we have $MB=MC$. prove that $\widehat{BAD}=\widehat{CAX}$