I think it's nice choice for juniors. Here is my solution. Firstly, we check the case if $AB=AK$. Then $AM$ coincides with perpendicular bisector of $BK$, so $BK\perp AM$. Moreover, because $BK$ is bisector and altitude in $\triangle ABL$, it's perpendicular bisector of $AL$. Thus, $NL=NA$. Now, $AB\not =AK$. Let circumcircle of $ABK$ meets $AM$ at $M'$. Then $KM'=M'B$, meaning that, if $M$ and $M'$ are distinct, $MM'$ is perpendicular bisector of $BK$ $\Longrightarrow AK=AB$, impossible. Thus, $M=M'$ and quadrilateral $BMKA$ is cyclic. Then \[ \angle KBA=\angle KMA=\angle NLA \] Quadrilateral $BLNA$ is also cyclic, and because $BN$ is bisector, we finally get $LN=NA$.

Because $BM=MK$ and $\angle BAM=\angle MAK$ ,after applying the Theorem of Sinuses in $\triangle ABM$ and in $\triangle MAK$ we see that $sin\angle ABM=sin\angle AKM$ so we conclude that either $\angle ABM=\angle AKM$ or $\angle ABM=180-\angle AKM$ which will mean that $BAKM$ is cyclic.Checking the first case we get $\frac{\angle B}{2}+\angle IBM=\frac{\angle B}{2}+\angle IBM+\angle C$ which means $\angle C=0$ so contradiction ,which confirms that $ABMK$ is cyclic.Moving further we get $\angle BAM=\angle BKM=\angle BNL$ so we conclude that $BANL$ is cyclic too.And the result comes by himself $AN=sin\frac{B}{2}\cdot 2R=AL$ where $R$ is the radius of the cercle passing through points $B,A,N$ and $L$. So we are done.

Ahiles wrote: Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ , where $L\in (BC)$ and $K\in (AC)$ . The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$. Remark. An easy and nice proposed problem ! Observe that always $AB>AK$ . Indeed, $m(\widehat{AKB})=\frac B2+C>\frac B2=$ $m(\widehat{ABK})$ $\implies$ $m(\widehat{AKB})>m(\widehat{ABK})\implies AB>AK$ . Otherwise, $AB>AK\Longleftrightarrow c>\frac {bc}{a+c}\Longleftrightarrow a+c>b$ what is truly always ! Quote: An easy extension. Let $AL$ be the angle bisector in the non-isosceles triangle $ABC$ , where $L\in (BC)$ and let $X\in (AL)$ be a point for which denote $K\in BX\cap AC$ . The perpendicular bisector of $BK$ intersects the line $AL$ at the point $M$. The point $N$ lies on the line $BK$ such that $LN\parallel MK$. Prove that $\frac {NA}{NL}=\frac {XA}{XL}\cdot\frac {IL}{IA}$ . Proof (similarly with frenchy's or Ahiles'). Is well-known that "in any nonisosceles triangle the intersection point between the bisector of an angle of triangle and the perpendicular bisector of the opposite side belongs to the circumcircle of the given triangle". Apply this property to the triangle $ABK$ and the vertex $A$ . Thus obtain that the quadrilateral $ABMK$ is cyclically $\implies$ $\widehat{BAM}\equiv\widehat {BKM}\equiv\widehat{BNL}\implies$ $\widehat{BAL}\equiv\widehat{BNL}\implies$ the quadrilateral $BANL$ is cyclically $\implies$ $\frac {NA}{NL}=\frac {\sin\widehat {NBA}}{\sin\widehat{NBL}}=\frac {\sin\widehat {XBA}}{\sin\widehat{XBL}}$ . Since $\frac {XA}{XL}=$ $\frac {BA}{BL}\cdot\frac {\sin \widehat{XBA}}{\sin\widehat{XBL}}=$ $\frac {IA}{IL}\cdot\frac {\sin \widehat{XBA}}{\sin\widehat{XBL}}$ obtain $\frac {\sin \widehat{XBA}}{\sin\widehat{XBL}}=\frac {XA}{XL}\cdot\frac {IL}{IA}$ . In conclusion $\frac {NA}{NL}=\frac {XA}{XL}\cdot\frac {IL}{IA}$ . Remark. In the particular case when $X:=I$ obtain the proposed problem. Another interesting particular case obtain such : denote $P\in AB\cap CI$ , $R\in AC\cap BI$ , $L\in BC\cap AI$ and $S\in PR\cap AI$ . Then for $X:=S$ obtain $NA=2\cdot NL$ . This results from the well-known relation $\frac {IA}{IL}=2\cdot\frac {SA}{SL}$ .

Wow. I feel so stupid. It took me so long to see the major idea. Solution Since $AL$ is the bisector of $\angle{KAB}$, the intersection of $AL$ with the circumcircle of $\triangle{AKB}$ is the midpoint of minor arc $KB$. This point also lies on the perpendicular bisector of $KB$ and therefore is $M$. Hence $AKMB$ is cyclic and $\angle{ABK}=\angle{AMK}=\angle{ALN}$. Therefore $ANLB$ is cyclic. Since $BK$ bisects $\angle{ABL}$, $N$ is the midpoint of minor arc $AL$ and therefore $LN=NA$.

See this: $ABMK$ cyclic, hence $\angle AMK=\angle ABK$, but $\angle ABK= \angle ABN$ and $\angle AMK= \angle ALN$, consequently $ABLN$ cyclic as well; with $BN$ angle bisector, done. Best regards, sunken rock

Dear Mathlinkers, for proving that ABLN is cyclic , we can use a converse of the Reim's theorem. Sincerely Jean-Louis

can somebody please post a figure?

A little late, but here's a diagram for anyone else wanting one. It includes the two important circumcircles. [asy][asy] unitsize(1.5cm); pair A, B, C, N, M, K, L; A=origin; B=(4.5,4); C=(7,0); L=extension(A, bisectorpoint(B,A,C), B, C); K=extension(B, bisectorpoint(A,B,C), A, C); M=extension(A, L, midpoint(B--K), bisectorpoint(B,K)); N=extension(B,K,L,L+M-K); draw(midpoint(B--K)--M); draw(A--B--C--cycle); draw(A--L); draw(L--N); draw(M--K); draw(B--N); draw(A--N); draw(circumcircle(A,B,K)); draw(circumcircle(A,B,N)); dot(M); label("A", A, SW); label("B", B, dir(60)); label("C", C, SE); label("L", L, NE); label("K", K, 2dir(310)); label("M", M, dir(320)); label("N", N, dir(300)); [/asy][/asy]

Dropping altitudes from $M$ gives congruent triangles that imply $(BLMC)$. $(BLMC)$ implies $(BNKC).$ Therefore, $\angle{NBK} = \angle{NKB}$

In triangle ABK, angle bisector of ∠A and perpendicular bisector of BK meet at M so ABKM is cyclic. ∠ABK = ∠KMA = ∠NLA ---> ABLN is cyclic. ∠LAN = ∠LBN = ∠NBA = ∠NLA ---> NL = NA.

[asy][asy] import olympiad; unitsize(50); pair A, B, C, Q, P, R, S, M, O; // Coordinates A=(0,0); B=(3,0); C=(3.4,3.5); Q = extension(A, bisectorpoint(C,A,B), B, C); P = extension(B, bisectorpoint(C,B,A), A,C); R = extension(midpoint(B--P), bisectorpoint(B,P), A, Q); S = extension(B, P, Q, Q+R-P); M = foot(R,P,B); O = circumcenter(A,P,B); // Drawing draw(A--B); draw(B--Q); draw(Q--C); draw(A--P); draw(P--C); draw(A--Q); draw(B--S); draw(midpoint(B--P)--R); draw(R--P); draw(Q--S); draw(circumcircle(S,Q,B)); draw(circumcircle(A,P,R)); draw(B--R); draw(A--S); dot(O); draw(O--P, dashed); draw(O--B, dashed); draw(M--O); // Labels label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, N); label("$Q$", Q, NE); label("$P$", P, N); label("$R$", R, N); label("$S$", S, NW); label("$M$", M, N*dir(30)); label("$O$", O, NE); [/asy][/asy] Claim: Quadrilaterals $APRB$ and $ASQB$ are cyclic. Proof: Let $R'$ be the point which bisects minor arc $BP.$ We will show $R'=R.$ We start with the angle bisector of $\angle BAC.$ Note that equal angles subtend equal arcs, so the angle bisector of $\angle BAC$ obviously intersects point $R'.$ Now, let's consider the perpendicular bisector of $BP.$ Let $M$ be the midpoint of $BP$ and $O$ be the circumcenter of $\triangle APB.$ Notice $\angle PMO = \angle BMO = 90^\circ$ by the perpendicular bisector condition and $OM=OM,$ so by RHS Congruence, $\triangle PMO \cong \triangle BMO.$ Thus, we know $\angle POR = \angle BOR,$ which implies arcs $PR$ and $BR$ are equal, as equal central angles are subtended by equal arcs. Therefore, we have shown $OR$ bisects minor arc $BP,$ (so $R$ is the midpoint of minor arc $BP$) implying the perpendicular bisector of $BP$ lies on $R=R',$ so $APRB$ is cyclic. Now, with this information, let's take a look at quadrilateral $ASQB.$ Recall that $QS \parallel RP.$ So, $\angle BAR = \angle BPR = \angle BSQ.$ Thus, $ASQB$ is cyclic, and our claim is proven. Claim: Segments $QS$ and $SA$ are equal. Proof: Recall that $BP$ is the angle bisector of $\angle ABC,$ so $\angle ABS = \angle SBQ$ (Note $S$ is simply an extension of $BP$). Furthermore, if two central angles in a circle are equal, the corresponding chords are also equal in length. Thus, we have $QS=SA,$ as desired.

@above, you need to redefine $Q\mapsto L, P\mapsto K, M\mapsto R, N\mapsto S$. Let $AL$ intersect the circumcircle of $\triangle ABK$ at $M'$. Since $\angle BAK$ is inscribed, angle bisector $AL$ meets the circumcircle of $\triangle ABK$ at the midpoint of minor arc $BK$. Since $M'$ is the midpoint of minor arc $BK$, it must also be on the perpendicular bisector of $BK$. But since the perpendicular bisector of $BK$ intersects $AL$ at $M$, we must have $M'=M$. Therefore, quadrilateral $BAKM$ is cyclic. Hence $\angle ABK = \angle AMK$. But since $KM \parallel LN$, we must also have $\angle ABK = \angle AMK=\angle ALN$. So quadrilateral $LBAN$ must also be cyclic. Since $BN$ is a perpendicular bisector and $\angle ABL$ is inscribed, it follows that $N$ is the midpoint of $\overarc {AL}$. Thus, we have $LN=NA$ as desired.

Since $M$ lies on the angle bisector of $\angle ABK$ and on the perpendicular bisector of $BK$, we obtain that $ABMK$ is cyclic (indeed $AB \neq AK$, else $\angle AKB = \angle ABK = \angle CBK$, which is impossible since $\angle AKB$ is an external angle in triangle $CBK$). Hence $\angle ABN = \angle ABK = \angle AMK = \angle ALN$ (the latter from $MK \parallel LN$), so $ABLN$ is cyclic. Since $BN$ bisects $\angle ABL$, we obtain $AN = NL$, as desired.

Back to the problem. The diagram is given below. Since $DM$ is the perpendicular bisector of $BK$, $BM=MK$. So, by our Lemma, $ABMK$ is cyclic as $\angle BAM=\angle MAK$. Notice that, $BC \nparallel AC$ which implies $\angle AKB\ne \angle KBC=\angle ABK\implies AB\ne AK$. Since $MK\parallel NL$, $\angle BNL=\angle BKM=\angle BAM=\angle BAL\implies ABLN\text{ is cyclic}$. Therefore, \[\angle LAN=\angle LBN=\angle ABN=\angle ALN\implies AN=LN\]and we are done.

Redefine $N$ as intersection of perpendicular bisector of $AL$ with $BK$, then clearly from the definitions and angle bisector of midpoints of arcs (alternatively just ghost point) we get that $ANLB, AKMB$ are cyclic and now Reim's just finishes.