Let $\triangle ABC$ be a triangle satisfying $AB<AC$. Let $D$ be a point on the segment $AC$ such that $AB=AD$. Let then $X$ be a point on the segment $BC$ satisfying $BD^2=BX\cdot BC$. Let the circumcircles of the triangles $\triangle XDC$ and $\triangle ABC$ intersect at $M \neq C$. Prove that the line $MD$ goes through the midpoint of the arc $\widehat{BAC}$ of the circumcircle of $\triangle ABC$.