Let $ABC$ be a triangle with incircle $\omega$, incenter $I$ and circumcircle $\Gamma$. Let $D$ be the tangency point of $\omega$ with $BC$, let $M$ be the midpoint of $ID$, and let $A'$ be the diametral opposite of $A$ with respect to $\Gamma$. If we denote $X=A'M\cap \Gamma$, then prove that the circumcircle of triangle $AXD$ is tangent to $BC$.