Let $AL$ and $BK$ be the angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects line $AL$ at $M$. The point $N$ is on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.
I am at a complete loss here, the hint says that I must first prove that $AKMB$ is a cyclic quadrilateral but I've no idea how to prove it or what to do once I have.
We know that if the angular bisector of $\angle P$ in $\triangle PQR$ intersects the circumcircle of $\triangle PQR$ at $S$, then $S$ is the midpoint of the arc $QR$ and lies on the perpendicular bisector of $QR$.
Here, as $AM$ is the angular bisector of $\angle A$ in $\triangle ABK$ and $M$ lies on the perpendicular bisector of $BK$, $M$ must be the midpoint of the arc $BK$ in the circumcircle of $\triangle ABK$. That is, $AKMB$ is cyclic.
$\angle ALN = \angle AMK = \angle ABK = \angle NBL$. Thus, $ABLN$ is also cyclic.
Now, the angular bisector of $\angle B$ intersects the circumcircle of $\triangle ABL$ at $N$, so $N$ must lie on the perpendicular bisector of $AL$ and thus, $LN=NA$.
This completes our proof.