OMCC 2013 Problem 5

Let $ABC$ be an acute triangle and let $\Gamma$ be its circumcircle. The bisector of $\angle{A}$ intersects $BC$ at $D$ , $\Gamma$ at $K$ (different from $A$ ), and the line through $B$ tangent to $\Gamma$ at $X$ . Show that $K$ is the midpoint of $AX$ if and only if $\frac{AD}{DC}=\sqrt{2}$ .