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, $$LN=NA$$.
• How exactly do we know that $S$ lies on the perpendicular bisector of $QR$? Is that a theorem or something? Can I get a proof of it (because I'll have to mention one)? – user683941 Jun 23 at 9:40