Let $ABC$ be a triangle with incircle $(I)$, tangent to $BC$, $CA$, $AB$ at $D, E, F$ respectively. On the line $DF$, take points $M, P$ such that $CM \parallel AB$, $AP \parallel BC$. On the line $DE$, take points $N$, $Q$ such that $BN \parallel AC$, $AQ \parallel BC$. Denote $X$ as intersection of $PE$, $QF$ and $K$ as the midpoint of $BC$. Prove that if $AX = IK$ then $\angle BAC \le 60^o$.