Question 145
Let $ABC$ be a triangle with incircle $\omega$, incentre $I$ 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 $BC$.