I use standard notations for triangles. Since $AL$ is the bisector of $\angle {BAC}$, we have $LM=LK$, so $KM$ is perpendicular to $AN$, that is to say, $[AKNM]=\frac 12\cdot KM\cdot AN$. In the cyclic quadrilater $AKLM$ we obtain $KM=AL\cdot \sin {\alpha}$. But $ABL$ and $ANC$ are similar, because $\angle{BAL}=\angle{NAC}=\frac{\alpha}2$ and $\angle{ABL}=\angle{ANC}=\beta$ and so $AL\cdot AN=b\cdot c$. Now substituting we have $[AKNM]=\frac 12\cdot KM\cdot AN=\frac 12\cdot AL\cdot \sin {\alpha}\cdot AN=\frac 12\cdot b\cdot c\cdot \sin {\alpha}=[ABC]$

$\angle BAN = \angle NAC = \angle NBC = \angle BCN$. Let the circumcircle of cyclic quadrilateral $AKLM$ meet $BC$ at $P$, again. $\angle KPL = \angle KAL = \angle CBN \Longrightarrow KP \parallel BN$ and $\angle LAM = \angle MPC = \angle PCN \Longrightarrow PM \parallel CN$. By area relation of parallel lines, we get $[BKP]=[KNP]$ and $[CMP]=[PMN]$. So, \[ \begin{array}{rcl}[ABC] &=& [AKPM] + [BKP] + [CMP] \\ &=& [AKPM] + [KNP] + [PMN] \\ &=& [AKNM].\end{array}\]

Let $P, Q$ be the projections of $N$ onto $AB, AC$ respectively. Since $PN\parallel KL, NQ\parallel ML$ we infer that $[KLN]=[KLP],\ [LMN]=[LMQ]$, hence $[AKNM]=[APL]+[AQL]$. With $[BLP]=[CLQ]$ ( from $\triangle BNP\cong\triangle CQN\implies BP=CQ$) we are done, because, if $AB>AC$, $[AKNM]= [APL]+[AQL] = [APL]+[BPL]+[AQL]-[CLQ]$ $=[ABL]+[ALC]=[ABC]$; if $AB<AC$ the proof is similar. Best regards, sunken rock

I don't think anyone has done it this way, so:

I have a question though, is working backwards like I did in my solution too "informal" for an olympiad proof?

Dear Mathlinkers, again the Reim's theorem avoid angles chasing... Sincerely Jean-Louis

since $AL $ is $A$-bisector then $ LK=LM\implies ALK \equiv ALM\implies AK=AM\implies KM \perp AN $ besides $KM=\sin A .AL$ so $2[AKNM]=AN.KM=AN.AL\sin A=AB.AC\sin A=[ABC]$

We use conventional notation, setting $BC=a, AC=b$, and $AB=c$. Also, let $\angle BAC=\theta$ and $AL=d$. Note trivially that $[ABC]=\frac{1}{2}bc\sin\theta$. We note that by the Angle Bisector Theorem, we have $BL=\frac{ab}{b+c}$ and $LC=\frac{ac}{b+c}$. Now by Power of a point, note that $LM=\frac{a^2bc}{d(b+c)^2}$. We now compute $[AKNM]=2[AKN]=2\left(\frac{1}{2}d^2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+\frac{a^2bc}{2(b+c)^2}\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right)=\frac{1}{2}d^2\sin\theta+\frac{a^2bc}{2(b+c)^2}\sin\theta$. It is well-known that $d^2=\frac{bc}{(b+c)^2}((b+c)^2-a^2)$ so we have that $[AKNM]=\frac{bc}{2(b+c)^2}((b+c)^2-a^2)\sin\theta+\frac{a^2bc}{2(b+c)^2}\sin\theta=\frac{bc}{2(b+c)^2}\sin\theta((b+c)^2-a^2+a^2)=\frac{1}{2}bc\sin\theta$ as desired.

WLOG, suppose $AB \le AC$, and let $P, Q$ be the projections of $N$ onto $AB, AC$ respectively. By Thales', we know $APNQ$ is cyclic, so the Spiral Center Lemma implies $N$ is the Miquel Point of $PQCB$, i.e. $NPB \overset{+}{\sim} NQC$. Now, since $N$ is the midpoint of arc $BC$, we have $NB = NC$, so the two aforementioned triangles are actually congruent. Since $AN$ is the perpendicular bisector of $KM$, we know $$\frac{[NLK] + [NLM]}{[AKLM]} = \frac{2 \cdot [NLK]}{2 \cdot [ALK]} = \frac{NL}{AL}.$$Now, observe that $$\frac{[BLK] + [CLM]}{[AKLM]} = \frac{\frac{BK}{AK} \cdot [ALK] + \frac{CM}{AM} \cdot [ALM]}{2 \cdot [ALK]}$$$$= \frac{BK + CM}{2 \cdot AK}.$$ Because $NPB \cong NQC$, we have $$AB + AC = (AP - BP) + (AQ + CQ) = (AP + AQ) + (CQ - BP) = 2 \cdot AP$$so $$\frac{BK + CM}{2 \cdot AK} = \frac{(AB - AK) + (AC - AM)}{2 \cdot AK} = \frac{AB + AC - 2 \cdot AK}{2 \cdot AK}$$$$= \frac{2 \cdot AP - 2 \cdot AK}{2 \cdot AK} = \frac{KP}{AK} = \frac{NL}{AL}.$$Now, it's easy to see this implies $$[NLK] + [NLM] = [BLK] + [CLM].$$Adding $[AKLM]$ to both sides finishes. $\blacksquare$ Remark: Embarrassingly, this problem took me over an hour because I tried to use Trig in my initial attempt at a length bash. Anyways, this length bash should mostly be natural, as working backwards from $$\frac{[BLK] + [CLM]}{[AKLM]} = \frac{BK + CM}{2 \cdot AK} \overset{?}{=} \frac{NL}{AL}$$is a fairly direct line of attack. The hardest step is substituting $$BK + CM = AB + AC - 2 \cdot AK$$but the rest of my solution follows smoothly afterwards.

Let $X,Y$ be the projections of $N$ onto $AB,AC$ respectively. From the fact that $L,N$ lie on the A-bisector we know that $KL=ML$ and $XN = YN$. Now \[[ABC] = [ACL] +[ABL] = \frac{(AB+AC)\cdot KL}{2},\]\[[AKNM]=AK \cdot XN,\]so it is sufficient to prove that $\frac{(AB+AC)\cdot KL}{2\cdot AK \cdot XN} = 1$. From Thales' theorem $\frac{KL}{XN} = \frac{AK}{AX}$, so \[\frac{(AB+AC)\cdot KL}{2\cdot AK \cdot XN} = \frac{(AB+AC)\cdot AK}{2\cdot AK \cdot AX} = \frac{(AB+AC)}{2 \cdot AX} = \frac{(AX \pm BX+AY \mp CY)}{2 \cdot AX} = 1\]as desired.

Consider projections $P,Q$ of $N$ onto $AB,AC$ respectively. Throughout the solution all areas are oriented. $$|NB|=|NC|,|NP|=|NQ|\implies NBP\cong NCQ$$$$|BP|=|CQ|,|LK|=|LM|\implies \text{area} (LBP)=\text{area} (LCQ).$$Finally we deduce $\text{area} (AKNM)-\text{area} (ABC)=\text{area} (LBP)+\text{area} (LQC)=0\text{ } \blacksquare$