Let $ABC$ be a triangle, with $A'$, $B'$, and $C'$ the points of tangency of the incircle with $BC$, $CA$, and $AB$ respectively. Let $X$ be the intersection of the excircle with respect to $A$ with $AB$, and $M$ the midpoint of $BC$. Let $D$ be the intersection of $XM$ with $B'C'$. Show that $\angle C'A'D' = 90^o$.