Let $M$ be the midpoint of $AB$. We will show that both $MX$ and $MN$ are parallel to $BC$.

To show that $MX$ is parallel to $BC$, we note that $BMX$ is an isosceles triangle with apex $M$. After all, $MX$ and $MB$ are the radius of the circle. It follows that $\angle MXB = \angle MBX$. Since also $\angle MBX = \angle XBC$, we have that $\angle MXB$ and $\angle XBC$ are alternate interior angles, and so $MX$ and $BC$ are parallel.

To show that $MN$ is parallel to $BC$, we note that triangle $ABC$ and $AMN$ are similar, since $\frac{|AN|}{|AC|} = \frac{1}{2} = \frac{|AM|}{|AB|}$ and $\angle BAC = \angle MAN$. It follows that $\angle AMN = \angle ABC$ and because these are corresponding angles $MN$ and $BC$ are parallel.

It follows that $MX$ and $MN$ are in fact the same line, and that line is also the line $XN$. Hence, $XN$ is parallel to $BC$.