Let $AD$ be diameter in the circumcircle of $ABC$ and let $AO \cap BC=E, AO=R, \angle A=\alpha$. Let the tangents to the circumcircle of $ABC$ at points $B, C$ meet at $T$. Because $AO$ and $AH$, as well as $AM$ and $AN$ are isogonals we deduce that $\angle HAN= \angle MAO$. On the other hand $OM \cdot OT=OC^2=OA^2$ and also $KH \| AD$ thus $\angle OMA= \angle OAT= \angle HNA$. This means that the triangles $AOM$ and $AHM$ are similar so $NH/AH=MO/AO=\cos \alpha$ and $NH=AH \cos \alpha =2R \cos^2 \alpha$. Note that by symmetry $ON=OE=OD-DE=OD-HN=R-2R \cos^2 \alpha$. Now it remains to check that $OA^2+NH^2=ON^2+AH^2 \iff R^2+(2R \cos^2 \alpha)^2=(R-2R \cos^2 \alpha)^2+(2R \cos \alpha)^2$, which is equivalent to the perpendicularity of $AN$ and $OH$.

Here is a 100% synthetic solution. We start by proving a lemma. Lemma: The only points $K$ on line $AH$ with the property that $\angle{ABK}=\angle{ACK}$ are $K=A$ and $K=H.$ Proof: Let $B'$ be the symmetric of $B$ in line $AH.$ Then the angle condition translates as $K\in (AB'C).$ Hence $K$ takes at most two values. Since $A$ and $H$ clearly work, the lemma is proved. Note that $K$ is the symmetric of $A$ wrt $M$. Let $P=AO\cap BC$, $T$ the intersection of the tangent in $K$ at $(BKC)$ with line $BC$ and $H'=AP\cap KT.$ Since the reflection of $H$ in $M$ lies on $AO$, it's easy to see that $N$ and $P$ are symmetric wrt $M.$ $AN$ is the $A$-symmedian in $\triangle{ABC}$, so $KP$ is the $K$-symmedian in $\triangle{KBC}.$ Therefore $(H'P,H'T ; H'B,H'C)=-1.$ But $\angle{PH'T}=90^0$ (if $O'$ is the circumcenter of $\triangle{KBC}$, then $KT\perp KO'\parallel AO$), so $\angle{BH'A}=\angle{CH'A}.$ Consider the inversion in $A$ of radius $\sqrt{AB\cdot AC}$ composed with a reflection in the bisector of $\angle{BAC}.$ Denote by $H^*,K^*,M^*$ the images of $H',K,M$. $\angle{BH'A}=\angle{CH'A}$ gives $\angle{ABH^*}=\angle{ACH^*}$ and $H'\in AO$ gives $H^*\in AH.$ By the Lemma it follows that $H^*=H.$ Combining this with $\angle{AH'K}=90^0$, we obtain $HK^*\perp AK^*.$ (1) But $K^*$ is the midpoint of $\overline{AM^*}$ and $M^*$ is the second intersection of $AN$ with $(ABC)$, meaning that $OK^*\perp AK^*$. (2) Finally, from (1) amd (2) we obtain $AN\perp OH.$ This ends our proof.

I think it's much simpler . Let $X$ be on $BC$ such that $AX$ touches $(ABC)$. Clearly $OX \perp AN$ ($AN$ is the symmedian and so the polar of $X$). The condition tells us $HN \parallel AO$ and so $\angle AHN= \angle MOA = 180^0- \angle AXN $ and $AH \perp XN$, so $H$ is the orthocenter of the trinagle $\Delta AXN$. Now we have $XH \perp AN$, so $X,O,H$ are collinear and this line is perpendicular to $AN$.

Wow, @MilosMilicev is so good at geo! No wonder he got all the girls at IMO.

Can confirm, @MilosMilicev got me at the IMO.

Let $L$ be the midpoint of $A$ symmedian chord of $\triangle ABC$. It is sufficient to prove that $\angle ALH=90$.Let $AO \cap \odot(ABC)=R, AM \cap \odot(ABC)=Q, BB \cap CC=P$ and $AO \cap \odot(BOC)=G$ an $AL \cap \odot(ABC)=E$. Now do a $\sqrt{bc}$ inversion at $A$ followed by reflection along the $\angle A$ bisector. Let $D$ be the foot of the perpendicular from $A$ to $BC$. Note that $$L \mapsto K, H \mapsto G, D \mapsto R, N \mapsto Q$$. It suffice to prove that $\angle AGK=90$. We have given that $K,N,H$ collinear $\iff A,G,Q,L$ lie on a circle. Note that $AN.NE=BN.CN=HN.HK$$\implies A,H,E,K$ is conclyc. So $M=LG \cap AQ$. Hence $\angle AGL=\angle OGL=\angle OPL$ and $\angle LGQ=\angle LAR$ . Hence $\angle RGQ=\angle OPL+\angle QAL=\angle OMA=\angle RQA \implies R,G,K,Q$ lie on a circle. So $\angle AGK=\angle AQR=90$.

I guess this solution has not been posted yet... Reflect $N$ in $M$ to get point $P$.Observe that the reflection of $H$ in $M$ is $A$ antipode and reflection of $K$ in $M$ is $A$,so we have that $A$,$O$ and $P$ are collinear.Therefore $AD$ and $AP$ are isogonal in $\angle A$.Since $AP$ is parallel to $KH$,we have $\angle HAN$=$\angle HKA$.$(1)$ Now let $AN$ intersect $(ABC)$ in $Q$.$NQ\times NA=NB\times NC=NH\times NK$. So $HAKQ$ is cyclic. Therefore $\angle HKA=\angle HQA$ $(2)$. From $(1)$ and $(2)$,we conclude that $HA$=$HQ$ so $OH\perp AN$

Note that it's well known that $K$ is reflection of $A$ across $M$ so $ABKC$ is parallelogram. Note that $\angle CKH = \angle CBH = \angle CAH$ and $\angle CAN = \angle BAM = \angle CKM$ so $\angle HAN = \angle AKH$. Let $HM$ meet $ABC$ at $S$ Note that $HM = MS$ so $AHKS$ is parallelogram. Note that $\angle AOH = \angle AMB = \angle CMK$ and $\angle NAO = \angle NAM + \angle MAO = \angle NAM + \angle MKH = \angle NAM + \angle NAH = \angle MAH = \angle MKS$ and $KS || AH \perp BC$ so $\angle CMK + \angle MKS = \angle 90 \implies AN \perp OH$.

Consider $(ABC)$, $\Omega=(BHC)$, $\omega=(H,AH)$. Note that $N$ lies on radical axis of the first two circles. Note that $\Omega$ and $\omega$ intersect on the lines $AB, AC$ in the points $P, Q$ such that $PC=AC, QB=BA$. $\triangle ABC \equiv \triangle AQP$, so $AN$ is its $A$-median, and $HK$ is a perpendicular bisector of $QP$, so $N$ lies on the segment $PQ$, hence it's radical center of $(ABC),\Omega,\omega$, so $OH \perp AN$.

As said above it is well known that $K$ is reflection of $A$ across $M$ and $ABKC$ is a parallelogram. Realize that the intersection of $AM$ and $(BHC)$ is the $A-Humpty$ point and to prove the statement is enough to show that $AN$ $\cap$ $HO$ is the $A - Dumpty$ point. $\angle CKH = \angle CBH = \angle CAH$ and $\angle CAN = \angle BAM = \angle CKM$ so $\angle HAN = \angle AKH$. Let $AN$ intersect $(ABC)$ at $T$. Note that by PoP in $(BHC)$ and $(ABC)$ we get that : $AN.NT = HN.NK$ so $AHTK$ is cyclic so $\angle HAN= \angle HKT=\angle AKH$. so $HA=HT$.Its also well known that the line $OA-DUMPTY$ is the perpendicular bisector of $AT$. So our statement is proved.