Since $AMHN$ is a parallelogram, $AMNL$ is an isosceles trapezoid, so $AL \parallel MN$ Redefines $K$ as the second intersection of $AH$ and $(AMN),$ we will prove $LK$ passes through $O$ From $AL \parallel MN,$ combine with $AK \equiv AH$ is $A-$median of $AMN,$ $LK$ is $L-$ symmedian of $\triangle LMN$ We have: $\triangle BHN \sim \triangle CHM$ So $\frac{BN}{AM}=\frac{BN}{HN}=\frac{CM}{HM}=\frac{CM}{AN} \Rightarrow NA.NB=MA.MC$ So $M$ and $N$ have the same power wrt $(O),$ so $OM=ON$, combine with $AK, AO$ are isogonal wrt $\angle BAC$ $\Rightarrow O$ is the intersection of the tangent at $M, N$ of $(AMN),$ so $LK$ passes through $O$ $\blacksquare$

Here is my solution: Let us first prove a useful result: $A,B,C,L$ are concyclic (proof): Since $LN=AM=NH$, $LM = AN=MH$, one can show $\frac{MH}{MC}=\frac{NH}{NB}$ by showing that $\triangle NHB\sim\triangle MHC$ (by prove that $\angle NHB=\angle MHC$ and $\frac{NH}{HM}=\frac{HB}{HC}$, both are straight forward). Thus, $\frac{LM}{MC}=\frac{LN}{NB}$. One knows that $ALMN$ is an Isopope trapezoid, thus $\angle LNB=\angle LMC$. So $\triangle LMC \sim \triangle LNB$. Then we know $\angle BLC=\angle NLM$, since $\angle NLM=\angle BAC$, we know $\angle BLC=\angle BAC$. This leads to the conclusion. Now we prove $L,M,K,A$ are concyclic: starting from $O$, we draw perpendicular line to $BC$ intersecting $BC$ at $E$. Since $AH\parallel OE$, we know $\angle LKH=\angle LOE$. Let's compute $\angle LOE=\pi-\angle BOE-\angle BOL=\pi-\angle BAC-2\angle LCB$. On the other hand, let us compute $\angle LMC=\pi-\angle HMC-2\angle NML=\pi-\angle BAC-2\angle NML$. In first step, we already shown that $\triangle LMC\sim \triangle LNB$, this easily leads to $\triangle LNM\sim\triangle LBC$. Thus $\angle LCB = \angle LMN$. We have $\angle LMC = \angle LOE=\angle LKH$. Thus, $\angle LMA = \angle LKA$, which shows that $A,L,M,K$ are concyclic. Since $ALMN$ is isopope trapzoid, $A,L,M,N$ are concyclic. So $A,L,M,K,N$ are concyclic.

Attachments:

First of all note that $\triangle MHC \sim \triangle NHB$ $\implies \frac{MH}{MC} = \frac{NH}{NB}$ $\implies \frac{AN}{MC} =\frac{AN}{NB}$ $\implies AN.NB=AM.MC $ $\implies Pow(N,\odot (ABC))=Pow(M,\odot (ABC)) $ $\implies OM=ON$. Now note that $AH$ is median of $\triangle AMN \implies AO$ is symmedian of $\triangle AMN$(as $O$ & $H$ are isogonal conjugates.) But $OM=ON \implies $ $OM$ and $ON$ are tangents to $\odot(AMN)$. Also note that it follows easily from angle chasing that $ALMN$ is cyclic. Let $\odot(LMN)$ meet $AH$ in $K'$. All we have to prove is that $O,K',L$ are collinear. $HK'$ is median of $\triangle HMN$ and $\angle MK'N$ + $\angle MAN$ =180 $\implies K'$ is $H$-humpty point of $\triangle HMN \implies \frac{K'M}{K'N}=\frac{HM}{HN}=\frac{LM}{LN}$. So $LMK'N$ is harmonic quadrilateral $\implies$ tangents at $M,N$ to $\odot(LMN)$ and $LK'$ are concurrent. But $OM$ and $ON$ are tangents to $\odot(LMN)$. Thus $O,K',L$ are collinear as desired.

Sketch. Clearly, $AMHN$ is a parallelogram. From here, it easy to conclude that $ALMN$ is an isosceles trapezoid. Furthermore, since $\bigtriangleup MHC\sim \bigtriangleup NHB$ we conclude that $L$ is the center of spiral similarity carrying $\overline{BC}$ to $\overline{NM}$, thus $L$ lies on $(ABC)$, implying that $OM=ON$. Being $AH$ the $A$-median of $\bigtriangleup MAN$, we actually infer that $OM$ and $ON$ are tangent to $(ALMN)$. Because $$(P_{\infty},H;M,N)\overset{A}{=}-1$$we conclude that $OL$ and $AH$ meet each other at $(ALMN)$, as required. $\blacksquare$

Note that $\angle NAM = \angle MHN = \angle NLM$ and $LM = MH = AN$ so $ALMN$ is isosceles trapezoid. Note that $\angle LNA = \angle LMA$ and $\frac{LN}{NB} = \frac{HN}{NB} = \frac{HM}{MC} = \frac{LM}{MC}$ so $LNB$ and $LMC$ are similar so $\angle BLC = \angle NLM = \angle NAM = BAC$ so $L$ lies on $ABC$. Let $P,Q$ be midpoints of $AH$ and $BC$. Claim $: L,H,P$ are collinear. Proof $:$ Note that $O$ lies on perpendicular bisector of $AL$ so $OQ \perp MN \perp HL$ and Note that $AH = 2OP$ so $HP || OQ$ so $L,H,P$ are collinear. Note that $\angle LSH = \angle LOP = \angle 180 - \angle BOP - \angle LOB = \angle 180 - \angle A - 2\angle LCB = \angle 180 - \angle HMC - 2\angle LMN = \angle 180 - \angle HMC - \angle LMN = \angle LMC \implies LSA = \angle LMA \implies LMSA$ is cyclic so $LMSNA$ is cyclic.

Redefine $K$ as a second intersection of $(AMN)$ with $AH$. $LK$ is an $L$-symedian in $\triangle NLM$. Note that $\angle NHM+\angle BHC = \angle BAC + \angle BHC = \pi$, hence $H$ has an isogonal conjugate in $BNMC$, clearly it's $O$. It means that $\angle MNO = \angle BNH = \angle A = \angle CMH = \angle NMO = \angle NLM$, so $O$ is the intersection of tangents to $(NLM)$ in $N,M$, hence $O$ also lies on an $L$-symedian in $\triangle NLM$.