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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 $LN = NA$.

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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.

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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.

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  • $\begingroup$ 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)? $\endgroup$ – user683941 Jun 23 at 9:40
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    $\begingroup$ @Adi03 If you consider the two equal angles on either side of the angular bisector, they should subtend equal chords/arcs. So, the point it meets the circumcircle should be the midpoint of that arc. If it's the midpoint of the arc, it must lie on the perpendicular bisector. That last step should be easy to prove. It's a pretty famous and useful property of angular bisectors. $\endgroup$ – Amit Rajaraman Jun 23 at 14:14

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