Let $ABC$ be a triangle with $AB \neq AC$. The incircle of $ABC$ touches the sides $BC$, $CA$, $AB$ at $X$, $Y$, $Z$ respectively. The line through $Z$ and $Y$ intersects $BC$ extended in $X^\prime$. The lines through $B$ that are parallel to $AX$ and $AC$ intersect $AX^\prime$ in $K$ and $L$ respectively. Prove that $AK = KL$.