Here is a hybrid solution using projective, inversion and then ratio lemma to finish. Also, let me mention that the point $M$ in my solution is not the midpoint of $BC$, since it's useless anyway (so this a slight abuse of notation). Firstly let WLOG $AB>AC$ and define a few new points: let $Q$ be the A-HM point (the intersection point of $(BOC)$ and the symmedian), $AO \cap (ABC)=A', OH \cap (BOC)=H", AH" \cap (BOC)=M', OM' \cap BC=M, OQ \cap BC=Q'$ (and it is well-known that $AQ'$ is tangent to the circumcircle). The first part is to use projective geo: project $(AH, N\infty)=-1$ through $O$ onto the circle $(BOC)$ and then through $A$ onto $(BOC)$ to see the desired concurrency is equivalent to $(EQ, OM')=-1$. Now invert with respect to the circumcircle. Then $(BOC) \rightarrow BC, H \rightarrow H", Q \rightarrow Q', E \rightarrow E', M' (\in AH") \rightarrow M (\in (AOH)$, i.e. $\angle AOM=90)$, and we want to prove that $M$ is midpoint of $Q'E'$. We have that $OM \parallel AQ' \iff \frac {DO} {AO}=\frac{DM} {MQ'} $, so we want $\frac {AO} {DO}=\frac {ME'}{MD} \iff \frac {DA'} {DO}=\frac {DE'} {DM} \iff OM \parallel A'E'$ so we want $A'E'$ to be tangent to $(ABC)$. Now, we are left to do a bit more length bash. We define $f(Z)=\pm \frac{ZB} {ZC}$ and use the lemmas from mira74's ratio lemma handout (here: https://artofproblemsolving.com/community/c6h2357938p19166714) without proof. We have to prove that $f(E')=f(A')^2=(\frac {\cos \gamma} {\cos\beta})^2$. We have that $f(E')=f(E)f(O)=f(H')f(D)$ and it is easy to prove that the last one is $\frac{H'B}{H'C} \frac{DB}{DC} =\frac{HC}{HB} \frac{DB}{DC} =\frac{AC}{AB}\frac {\sin (90-\gamma)}{\sin (90-\beta)} \frac{AB}{AC}\frac{\sin (90-\gamma)}{\sin (90-\beta)}=(\frac{\cos \gamma}{\cos \beta}) ^2$ by using the classic version of ratio lemma and the equal segments $H'B=HC, H'C=HB$, so we are finally done.

Let $\Gamma = (ABC)$, $\omega = (BOC)$. Let $P = OD \cap \omega$, $K = OH' \cap \omega$, $F = OE \cap BC$, $Y = NO \cap BC$. Let $A' = AO \cap \Gamma$, and $X = NO \cap \omega$. It suffices to show $A, X, E$ collinear. Consider the inversion centre $O$ radius $OA$. $\Gamma$ is invariant, and $(BC, \omega)$ are swapped. Hence, we swap: \begin{align*} (A, A), (B, B), (C, C), (A', A'), (H', K), (D, P), (E, F), (X, Y). \end{align*}Furthermore, $((ODH'), KP)$ are swapped, and hence $E = \omega \cap (ODH') \Rightarrow F = BC \cap KP$. Now it suffices to prove $AOFY$ is cyclic. Claim 1: $AHA'F$ is cyclic. Proof: Note that $D$ lies on $BC$, the radical axis of $\omega$ and $\Gamma$. By inversion, $OP \cdot OD = OH' \cdot OK \Rightarrow DPKH'$ is cyclic by converse of Power of a Point (PoP). Thus, noting that $OM$ is the perpendicular bisector of $HH'$, $\measuredangle OHF = \measuredangle OHH' = \measuredangle HH'O = \measuredangle DH'O = \measuredangle OPK = \measuredangle OPF$. Hence, by converse of angles in a cyclic quad, $OHPF$ cyclic. $D = HF \cap OP \Rightarrow$ by PoP, $DH \cdot DF = DO \cdot DP = Pow_{\omega}(D) = Pow_{\Gamma}(D) = DA \cdot DA'$. Thus, $AHA'F$ cyclic by converse of PoP. $\square$ Claim 2: $AOFY$ cyclic. Proof: Take a homothety scale factor 2 from $A$. This maps $N \to H, O \to A'$. Thus, $NO \parallel HA'$. Now we have $\measuredangle FYO = \measuredangle HYN = \measuredangle YHA' = \measuredangle FHA' = \measuredangle FAA' = \measuredangle FAO$. Thus by converse of angles in a cyclic quad, $AOFY$ is cyclic. $\square$

Let $\Gamma = \odot(BOC)$; $P$ be antipode of $O$ in $\Gamma$; $U = AP \cap \Gamma \ne P$, $V = DU \cap \Gamma \ne U$; $X=AD \cap \Gamma \ne O$. $\boxed{\textbf{Part 1:} \text{ our problem is equivalent to points } O,H',V \text{ collinear}}$ Define $T = AE \cap \Gamma \ne E$. We want points $O,N,T$ to be collinear. Claim 1: $OH,ED$ concur at some point $S \in \Gamma$. Proof: Define $S = OH \cap \Gamma \ne O$. Note $$\angle OED = \angle OH'D = \angle OHD = \angle(OH,\text{tangent to } \Gamma \text{ at } O) =\angle OES $$It follows points $E,D,S$ are collinear, as desired. $\square$ [asy][asy] size(250); pair B=dir(-140),C=dir(-40),A=dir(115),O=(0,0),P=2*B*C/(B+C),H=foot(A,B,C),M=1/2*(B+C),N=1/2*(A+H),D=extension(A,O,B,C); path c=circumcircle(B,O,C); pair Hp=2*M-H,S=IP(H--10*H,c),E=IP(D--10*D-9*S,c),T=2*foot((O+P)/2,A,E)-E,U=2*foot((O+P)/2,A,P)-P,V=2*foot((O+P)/2,D,U)-U,X=foot(P,A,O); draw(A--B--C--A,purple); draw(H--A,orange); draw(S--O--P^^N--T^^A--X,brown); draw(A--E--P^^S--E--X,cyan); draw(A--P^^U--V,green); draw(O--V,cyan); draw(c,royalblue); draw(circumcircle(O,D,E),orange); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$O$",O,dir(80)); dot("$P$",P,dir(P)); dot("$H$",H,dir(-90)); dot("$M$",M,dir(-130)); dot("$N$",N,dir(120)); dot("$H'$",Hp,dir(80)); dot("$S$",S,dir(S)); dot("$E$",E,dir(50)); dot("$T$",T,dir(20)); dot("$U$",U,dir(-110)); dot("$V$",V,dir(V)); dot("$D$",D,dir(-110)); dot("$X$",X,dir(X)); [/asy][/asy] Now note that $$ (OA,OH ; ON,OM) = -1$$(just project onto $AH$ to see this). So $O,N,T$ collinear is equivalent to $$ -1 = (OA,OH ; OT,OM) $$But \begin{align*} (OA,OH ; OT,OM) &= (X,S ; T,P)_{\Gamma} \\ &\stackrel{E}{=} (X,D ; A , EP \cap AX) \\ &\stackrel{P}{=} (X, PD \cap \Gamma \ne P ; U,E)_{\Gamma} \\ &\stackrel{D}{=} (O,P ; V,S)_{\Gamma} \end{align*}So it suffices to show $$ (O,P ; V,S)_{\Gamma} = -1 $$This is equivalent to $$ OP \text{ bisects } \angle VOS $$which is further equivalent to $O,H',V$ collinear. $\boxed{\textbf{Part 2:} \text{ proving points } O,H',V \text{ are collinear}}$ Now we just focus on a very small part of the figure. Let $AP \cap BC = Y$. Since $$ (B,C ; D,H') \stackrel{V}{=} (B,C ; U, VH' \cap \Gamma \ne V)_{\Gamma} $$[asy][asy] size(200); pair B=dir(-140),C=dir(-40),A=dir(115),O=(0,0),P=2*B*C/(B+C),H=foot(A,B,C),M=1/2*(B+C),N=1/2*(A+H),D=extension(A,O,B,C); path c=circumcircle(B,O,C); pair Hp=2*M-H,S=IP(H--10*H,c),E=IP(D--10*D-9*S,c),T=2*foot((O+P)/2,A,E)-E,U=2*foot((O+P)/2,A,P)-P,V=2*foot((O+P)/2,D,U)-U,X=foot(P,A,O),Y=extension(A,P,B,C); draw(A--B--C--A,purple); draw(H--A--D,orange); draw(A--P^^U--V,green); draw(O--V,cyan); draw(c,royalblue); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$O$",O,dir(80)); dot("$P$",P,dir(P)); dot("$H$",H,dir(-90)); dot("$M$",M,dir(-90)); dot("$H'$",Hp,dir(70)); dot("$U$",U,dir(-110)); dot("$V$",V,dir(V)); dot("$D$",D,dir(-110)); dot("$Y$",Y,dir(-110)); [/asy][/asy] So $O,H',V$ collinear is equivalent to \begin{align*} (B,C ; D,H') = (B,C ; U,O) \\ \iff \frac{BD}{CD} \div \frac{BH'}{CH'} = \frac{BU}{CU} \div \frac{BO}{CO} \\ \iff \frac{BD}{CD} \cdot \frac{BH}{CH} = \frac{BU}{CU} \qquad \qquad (\star) \end{align*} Lemma 2: (Well Known) In a $\triangle ABC$, if points $W,Z$ lie on $B,C$ such that lines $AW,AZ$ are isogonal wrt $\angle BAC$, then $$ \frac{BW}{CW} \cdot \frac{BZ}{CZ} = \left( \frac{AB}{AC} \right)^2 $$ Proof: By Ratio Lemma, we have $$ \frac{BW}{CW} = \frac{AB}{AC} \cdot \frac{\sin \angle BAW}{\sin \angle CAW} \qquad , \qquad \frac{BZ}{CZ} = \frac{AB}{AC} \cdot \frac{\sin \angle BAZ}{\sin \angle CAZ} $$As $BW,CW$ are isogonal wrt $\angle BAC$, so $\angle BAW = \angle CAZ$ and $\angle CAW = \angle BAZ$. Now multiplying the two equations implies our Lemma. $\square$ Now we are ready to prove $(\star)$. Since $AH,AD$ are isogonal wrt $\angle BAC$, thus $$ \frac{BD}{CD} \cdot \frac{BH}{CH} = \left( \frac{AB}{AC} \right)^2 $$Now note that $P$ is the intersection of tangents at $B,C$ to $\odot(ABC)$. It follows $AP$ is the $A$-symmedian of $\triangle ABC$. Thus $AY,AM$ are isogonal wrt $\angle BAC$. Thus, $$ \frac{BY}{CY} = \left( \frac{AB}{AC} \right)^2 \div \frac{BM}{CM} = \left( \frac{AB}{AC} \right)^2 $$Now since $UP$ bisects $\angle BUC$, so it follows $$ \frac{BU}{CU} = \frac{BY}{CY} = \left( \frac{AB}{AC} \right)^2 $$So we have proven both sides $(\star)$ equal $\left( \frac{AB}{AC} \right)^2$. This completes the proof of the problem. $\blacksquare$

Really nice problem.....My solution with Complex Numbers Note that $P=AE \cap NO$, we need to show $BOPC$ is cyclic. Set $(ABC)$ as unit circle ,so that $|a|=|b|=|c|=|d|=1$. Also with suitable rotation $AO$ be real axis and $a=1$.Then $$m=\frac{b+c}{2} \wedge \bar{m}=\frac{b+c}{2bc}$$$$h=\frac{b+c+1-bc}{2} \wedge \bar{h}=\frac{bc+b+c-1}{2bc}$$$$2n=1+\frac{b+c+1-bc}{2} \iff n=\frac{b+3+c-bc}{4} \wedge \bar{n}=\frac{b+c+3bc-1}{4bc}$$$$h'=2m-h \iff h'=\frac{bc+b+c-1}{2} \wedge \bar{h'}=\frac{b+c+1-bc}{2bc}$$$$d= \frac{\bar{b}c-b\bar{c}}{(\bar{b}-\bar{c})-(b-c)} \iff d=\frac{b+c}{bc+1} \wedge \bar{d}=d$$ Now we are going to hardest part of problem which is computing $e$ $$BOEC \text{ is cyclic } \iff \frac{b(e-c)}{c(e-b)} \in \mathbb {R} \iff \frac{c(e-c)}{b(e-b)}=\frac{c(\bar{e}-\bar{c})}{b(\bar{e}-\bar{b})} \iff \bar{e}=\frac{e}{e(b+c)-bc} (\spadesuit)$$$$OEH'D \text{ is cyclic } \iff \frac{e(h'-d)}{h'(e-d)} \in \mathbb {R} \iff \frac{e(h'-d)}{h'(e-d)}=\frac{\bar{e}(\bar{h'}-\bar{d})}{\bar{h'}(\bar{e}-\bar{d})} \iff \bar{e}=\frac{e(bc-b-c-1)}{bc+b+c-1-2e(bc+1}(\clubsuit) $$ If we equalize $(\spadesuit)$ and $(\clubsuit)$ we get nice expression of $e$ $$e=\frac{bc-1}{b+c+2} \wedge \bar{e}=\frac{1-bc}{b+c+2bc}$$ Now we can compute $p$ $$p=\frac{n(e-\bar{e})}{n(1-\bar{e})-\bar{n}(1-e)} \text{ after simplifying } p=\frac{2bc(b+1)(c+1)}{(b+c)(b+c+3bc-1)}$$ It remais to show that $$BOPC \text{ is cyclic } \iff \frac{b(p-c)}{c(p-b)} \in \mathbb{R} \iff \frac{(1-c)(b^2+3b+3bc+c)}{(1-b)(c^2+3c+3bc+b)} \in \mathbb{R} \implies \text { equal to its conjugate }$$ So we are done!

Let the $B$ and $C$ tangents intersect at point $L$, which lies on $(BOC)$ such that $OL$ is a diameter. Let $A'$ be the antipode of $A$ on $(ABC)$ and suppose $A'A' \cap BC = T$. Redefine $E$ as $OT \cap (BOC)$, I'll show that it lies on $(OH'D)$. Note that $A'H'$ is an altitude in the right triangle $DA'T$ Note that by power of point from $T$, we have that $$TD \cdot TH' = TA'^2 = TB \cdot TC = TO \cdot TE$$so $E$ indeed lies on $(OH'D)$. Further, since $TO \cdot TE = TA'^2$ in right triangle $OA'T$, we have that $A'E$ must be an altitude, and consequently, $L,A'$ and $E$ must be collinear. Now suppose $NO$ meets $BC$ at $Z$. Invert about the circumcircle to get that the problem is now equivalent to having points $A,O,Z,T$ concyclic. Since $N$ is the midpoint of $AH$ and $O$ is the midpoint of $AA'$, we have that $ZO$ is parallel to $HA'$ and so $\frac{DO}{DZ} = \frac{DA'}{DH}$. Also, $AHA'T$ is cyclic due to the right angles, giving that $DH \cdot DT = DA \cdot DA'$. We are now ready to finish the problem. Observe that $$DA \cdot DO = DA \cdot \frac{DZ \cdot DA'}{DH} = DZ \cdot \frac{DA \cdot DA'}{DH} = DZ \cdot \frac{DH \cdot DT}{DH} = DZ \cdot DT$$implying that $AOZT$ is indeed cyclic, as desired. $\blacksquare$

Attachments:

Let $AO$ meets the circumcircle of $\triangle ABC$ the second time at $A_1$ (as a anti-diametral point of $A$), and $OH$ meets the circumcircle of $\triangle BOC$ the second time at $T$. It suffices to show that $\angle TON = \angle TEA$. We have that $TO$ is the angle-bisector of $\angle BTC$ and $OB^2=OH\cdot OT$, i.e. $OH\cdot OT = R^2$, where $R$ is the radius of the circumcircle of $\triangle ABC$. This gives us that $OA_1^2=R^2=OH\cdot OT$ and $OA_1$ should be a tangent to the circumcircle of $\triangle THA_1$, therefore, \[ \angle TA_1O=\angle OHA_1=\angle HON. \]Note that the last equality holds by $ON\| A_1H$. Thus, $\angle TON = \angle HON = \angle OA_1T=\angle TA_1A$. On the other hand, we have \[ \angle OED=\angle OH'D=\angle OHD = \angle BOH+\angle OBH = \angle BOT + \angle OCB = \angle OCT = \angle OED, \]that is $E$, $D$, $T$ are collinear. Now \[ ED\cdot DT = CD\cdot DB = A_1D\cdot DA \]and we can see that $AEA_1T$ is a cyclic quadrilateral, which implies $\angle TEA=\angle TA_1A$. Hence $\angle TON = \angle TA_1A=\angle TEA$, and we are done!

I found a very interesting solution using mixtilinear circles

Lemma 1

Proof

Lemma 2

Proof

In this problem, let $AO$ meets $(BOC)$ at $X$ then we can notice that $E$ is the tangential point of $X-mixtilinear\,\,incircle$ and $(BOC)$ and $A$ being the $X-$excenter Therefore by two lemmas, the problem ends

I have a easier prove: $AO$ meet $(ABC)$ at $A'$,$AE,AD$ meet $(BOC)$ at $F,J$ As $OM//AH,\frac{AO}{OA'}=\frac{HM}{MH'}=1$ $A'H'$ is perpendicular to $BC$ $OE$ meet $BC$ at $K$ As $\angle OKD=180^o-\angle EOJ-\angle ODK=\angle ODC-\angle EOJ=\angle EJD$ So we have $KEDJ$ is concyclic As $ON//A'H$ We only need that $OF//A'H$ As $OE·OK=OD·OP=OB^2=OA^2$ We have $\triangle OAE\sim\triangle OKA$ So $\angle AKH=\angle AKO+\angle DKE=\angle DAE+\angle EJD=\angle JEF=\angle JOF$ So $OF//A'H\Leftrightarrow\angle HA'O=\angle JOF\Leftrightarrow\angle HA'O=\angle AKH\Leftrightarrow AHA'K is concyclic$ As $DA·DA'=DB·DH=DO·DJ$ So $\frac{DA'}{DJ}=\frac{DO}{DA}=\frac{DM}{DH}$ So $A'M//JH$ As $\angle KJO=\angle OED=\angle OH'D=\angle OHD$ We get $KJHO$ is concyclic So $KOMA$ is concyclic As $\angle KA'A=\angle KMO=90^o=\angle KHA$ So $KA'HA$ is concyclic We get what we desired.

VicKmath7 wrote: Here is a hybrid solution using projective, inversion and then ratio lemma to finish. Firstly let WLOG $AB>AC$ and define a few new points: let $Q$ be the A-HM point (the intersection point of $(BOC)$ and the symmedian), $AO \cap (ABC)=A', OH \cap (BOC)=H", AH" \cap (BOC)=M', OM' \cap BC=M, OQ \cap BC=Q'$ (and it is well-known that $AQ'$ is tangent to the circumcircle). Why OM' intersects BC at M?

Let $A'$ be the antipode of $A$. $OE\cap BC=G,OD\cap (ABC)=K,OF\cap BC=L$ $\textbf{Claim:} \ AHA'G$ is cyclic. \[\angle OKG=\angle DEO=\angle DH'O=\angle OHD=\angle OHG\]$\textbf{Claim:} \ GA'$ is tangent to $(ABC)$. \[DA.DA'=DB.DC=DO.DK=DG.DH\]Thus $\angle AA'G=\angle AHG=90$ $\textbf{Claim:} \ A,O,L,G$ are cyclic. By inversion centered at $O$ with radius $OB$, $\overline{AEF}\leftrightarrow (AOLG)$. $\textbf{Claim:} \ HA'\parallel OF$ \[DG.DH=DA.DA', \ DO.DA=DL.DG\implies \frac{DH}{DL}=\frac{DA'}{DO}\]$\textbf{Claim:} \ O,F,N$ are collinear. $ON\parallel HA'\parallel OF$ thus $N,O,F$ are collinear as desired.$\blacksquare$

Solved with Om245. Very good problem. We might have slightly overcooked this but a solution is a solution. So, we define the points $A'$ - the $A-$antipodal point , $K$ the orthocenter and $H_A$ the reflection of the orthocenter across side $BC$. Further, let $O'$ be the second intersection of the $A-$diameter with $(BOC)$ and $Y$ be the second intersection of line $\overline{H'E}$ with $(BOC)$. Our first few claims will deal with locating $H'$ and $E$. We start off by noting that, \[\measuredangle EYO' = \measuredangle EOD = \measuredangle EH'C\]so it follows that $\overline{YO'} \parallel \overline{BC}$. A direct consequence of this observation is that $HH'O'Y$ and $H_AA'O'Y$ are both isosceles trapezoids. Now with these observations in hand, we can show our first key claim. Claim : Lines $\overline{H_AM}$ and $\overline{H'E}$ are parallel. Proof : We first show that $\overline{HO'}$ and $\overline{KA'}$ are parallel. To see why, let $L$ be the midpoint of $AK$. Then, it is clear that $LO \parallel KA'$. Now, we consider a $\frac{\sqrt{AB \cdot AC}}{2}$ inversion centered at $A$. This will then clearly map $H$ to $O$ and vice versa. Thus, $(BOC)$ maps to the nine-point circle of $\triangle ABC$. Thus, since $L$ lies on this nine-point circle and the $A-$altitude it follows that $L$ maps to $O'$ and vice versa. This means, $\overline{LO} \parallel \overline{HO'}$ from which it indeed follows that $\overline{HO'}$ and $\overline{KA'}$ are parallel as desired. Considering a reflection across the perpendicular bisector of $BC$ proves the claim. Now, let $E'$ be the reflection of $E$ across the perpendicular bisector of $BC$. By symmetry it then follows that $E'$ lies on $(BOC)$ and $\overline{HO'}$. We can then show the following result. Claim : Points $A$ , $E'$ , $H_A$ and $O'$ are concyclic. Proof : For this, simply note that due to the reflections, \[HE' \cdot HO' = H'E \cdot H'Y = H'B \cdot H'C = HB \cdot HC = HA\cdot HH_A\]which clearly implies the desired claim. Now we are almost done. Let $S = \overline{AH} \cap \overline{H'E}$. Then, we can make the following final piece of observation. Claim : Quadrilateral $ASA'E$ is cyclic while quadrilateral $SH_AH'A'$ is a parallelogram. Proof : The first part is easy since, \[\measuredangle A'AS = \measuredangle O'AH_A = \measuredangle O'E'H_A = \measuredangle A'EY = \measuredangle A'ES \]which implies that $ASA'E$ is indeed cyclic as claimed. For the next part, note that since we previously showed $H_AM \parallel H'E$, and $M$ is the midpoint of $HH'$, it follows that $H_A$ is the midpoint of $HS$. But then, \[SH_A=H_AH = A'H'\]and since $\overline{SH_A} \parallel \overline{A'H'}$ it follows that $SH_AH'A'$ is indeed a parallelogram. Now, with all these results in hand, we are left with a direct angle chase. Let $R = \overline{AE} \cap \overline{NO}$. Note that, \[\measuredangle ORA = \measuredangle NOA + \measuredangle OAE = \measuredangle HA'A + \measuredangle A'AE = \measuredangle HA'A + \measuredangle A'SH' =\measuredangle HA'A + \measuredangle H_AH'Y = \measuredangle HA'A + \measuredangle O'HA' = \measuredangle HO'O = \measuredangle EYO\]from which it follows that $R$ lies on $(BOC)$ as we indeed set out to show.

A very nice and elegant problem, probably one of my favorite ones so far! My solution is identical to mathuz's above, but I've spotted some beautiful properties in the problem, so let me name them: 1) $AO \cap (BOC) = L, BC \cap OE = J, OH' \cap (BOC) = K \Rightarrow J,K,L$ are collinear. 2) $OH \cap (BOC) = T, AO \cap (ABC) = P \Rightarrow T,P,J$ are collinear 3) $OM \cap (BOC) = G \Rightarrow G,P,E$ are collinear 4) $H'EJK$ is cyclic, and $P$ is it's tangent