X=reflection of O about BC. So X lies on BHC => 2A=BOC=BXC=180-A => A=60. AH=2RcosA=R=OX=O'X => K is midpoint of HX. since BHOC is cyclic, so as LKNM, midpoints of XB,XH,XO,XC.

Can anyone give us a good shape?

Similar to navi_09220114, but a bit more comprehensible ig. Let $\omega$ be the circle through points $B$, $C$ and $H$ and its centre is therefore $X$. Since $X$ lies on $(ABC)$ and $X$ lies on the perpendicular bisector of $BC$ by the radical axis of $(ABC)$ and $\omega$, we get that $X$ is the midpoint of arc $BC$ not containing $A$. Now, it is well-known that $XI=XB=XC=XH$, thus $\angle BHC=\angle BIC$, where $I$ is the incentre of $\triangle ABC$. Thus, we have $$180^\circ-\angle{BAC}=90^\circ+\frac{\angle BAC}{2}\implies \angle BAC=60^\circ,$$thus $\angle{BHC}=120^\circ=2\cdot \angle{BAC}=\angle{BOC}$, meaning $O$ also lies on $\omega$. Thus, $N$ is the midpoint of $OX$ by the radical axis again. Therefore, we have $$AH=2ON=OX=XO',$$where the first equality is well-known and since $AH\perp BC\perp OX\equiv O'X$, we have $AXO'H$ being a parallelogram, therefore $K$ is the midpoint of $XH$. Now do homothety centred at $X$ with factor $0.5$ and we obtain that $(BHOC)\rightarrow (LKNM)$. $\square$

Similar solution. $\angle BXH= 2\angle BCH=180-2B \implies 90-\angle C+\angle \frac{A}{2} =\angle XBH=\angle B$ So $\angle 90+\frac{A}{2}=180-\angle A \implies \angle A=60$ $\angle OXC=60, OXC$ is an equaliteral triangle. $\implies O \in (BHC)$ $AH \parallel OX$ and $AH=XO' \implies HK=KX$ Also $XN=\frac{XC}{2}$ because of $30-60-90$ triangle. $\implies XM=XL=\frac{XC}{2}=XN=XK$ $K,L,M,N$ are cyclic and $X$ is the circumcenter.