We will proceed by taking an inversion at $I$. Let $X^*$ be the image of $X$ under the inversion. Then $I$ is the orthocenter of $A^*B^*C^*$ and $IO^*$ is the line passing through $I$ and the circumcenter of $A^*B^*C^*$. Moreover, $IH^*$ intersects $IB^*C^*$ at $D^*$, and $M^*$ is the reflection of $I$ w.r.t. $B^*C^*$. Since $MD$ intersects $ABC$ at $E$, we have $IM^*D^*$ intersects $A^*B^*C^*$ at $E^*$, and so $E^*$ is the reflection of $D^*$ w.r.t. $B^*C^*$. These characterize all images but $H^*$, which we characterize below. Consider the reflection $H'$ of $H$ w.r.t. $BC$. Then $AH'$ is perpenticular to $BC$, and so $IA^*(H')^*$ is orthogonal to $IB^*C^*$. This shows that under the inversion of $IB^*C^*$, the circle $IA^*(H')^*$ maps back to itself. Therefore if $A'$ is the image of $A^*$ under the inversion of $IB^*C^*$, then $IA^*A'$ intersects $A^*B^*C^*$ again at $(H')^*$. Since $IHH'$ is orthogonal to $BC$, we know $H^*(H')^*$ passes throguh the circumcenter $O'$ of $IB^*C^*$. Therefore $H^*$ is the second intersection of $O'(H')^*$ and $IA^*A'$, showing that $H^*$ is the image of $(H')^*$ under the inversion of $IB^*C^*$. We have finally characterized $H^*$, and our goal is to prove that $IO^*$ is parallel to $H^*E^*$. To clear up the notations, we relabel $A^*,B^*,C^*,D^*,E^*$ as $A,B,C,D,E$; $I$ as $H$; $(H')^*$ as $X$; the circumcenter of $A^*B^*C^*$ as $O$; $M^*$ as $H'$; and $H^*$ as $P$. Let $P'$ be the reflection of $P$ with respect to $BC$. It suffices to show $DP'$ is parallel to $O'H'$. Let $XHO'$ intersects $ABC$ again at $Q$. Then $$\angle O'QX=\angle O'HX=\angle (O'H,AX)+\angle AXH.$$Since $AHA'X$ maps to itself under the inversion of $HBC$, we know that $O'H$ is tangent to $AHA'X$ at $H$, showing that $\angle AXH=\angle H'HO'$. Thus $$\angle O'QX=\angle (O'H,AX)+\angle H'HO'=\angle H'HX=\angle H'QX$$and so $O',H',Q$ are collinear. Since the image of $ABC$ under the inversion of $HBC$ is $BCP$ and the image of $XHO'$ under the inversion is $PH$, we know that $PH, O'H'$ and $BCP$ intersect at a point $R$. Note that $PHDR$ are now collinear, and the reflection of the line w.r.t. $BC$ is $P'H'E$. This shows that $PR,H'E$ and $BC$ are concurrent. Since $BCPR, BCH'E$ are concyclic, we know that $PRH'E$ are also concyclic. Lastly, since $PP'DE$ are concyclic, $$\angle PDP'=\angle PEP'=\angle PEH'=\angle PRH',$$showing that $DP'$ and $RH'$ are parallel to each other. We are thus done as $RH'O'$ are collinear.

Replace $H,O$ with any pair of isogonal conjugates $P,Q$. Apply inversion at $I$, and get a triangle $\triangle ABC$ with orthocenter $I$, a point $P$, the second intersection of $IP$ and $(IBC)$ is $D$, $M$ is the reflection of $I$ in $BC$, so $E$ is the second intersection of $(ABC)$ and $(IDM)$. We want to show $PE$ parallel to $IQ$, where $Q$ is a point which we'll characterize. First notice that $Q$ is concyclic with the $A, I$ and the reflection of $P$ in $AI$, so the midpoint $R$ of $PQ$ is on the Nine-point circle of $\triangle AIP$. Similarly $R$ is on the Nine-point circle of $\triangle BIP$ and $\triangle CIP$, so it is the Poncelet point of $ABCP$. This implies that if $X$ is the reflection of $I$ over $R$, $X \in (ABC)$ and $IPXQ$ is a parallelogram, so all we need to show is $X-P-E$. $X$ lies on the reflection of $(IAP)$ over $AP$, so $\angle (XP, XA)=\angle (AI, PI)$. By radical axes $ID, ME, BC$ concur, so $ME$ is the reflection of $IP$ over $BC$, which implies $\angle (AI,PI)=\angle (ME, MA) =\angle (XE, XA)$, and we are done. $\square$

Though the way I used was quite brutal and needed a lot of (?) properties, but it can solve the problem in 30 minutes. $\textbf{Definition}$:Let $\mathcal{C}$ be a circumconic, a point $X$ on it. For a point $P$, let the $\mathcal{C}$-cevian triangle of $P$ is $P_AP_BP_C$, then $P$, $P_AX\cap BC$, $P_BX\cap AC$, $P_CX\cap AB$ are collinear, and let it be $\textbf{Per}^\mathcal{C}_P(X)$. If $\mathcal{C}$ is circumcircle, we will omit it. Actually, we have $\textbf{Per}_I(E)=ID=IH$, so the Li4 point is H, and it implies that $\textbf{Per}_H(E)=HH=HX_{65}$, which means the steiner line of $E$ is $HX_{65}$. Hence, $E$ be $X_{108}$. There are many ways to finish the question, for instance consider the isoconjugate which makes $I$ maps to $H$, or prove $(IHE)$ passes through $F=(\infty_{IN})^*$ (By perfect hexagon, we have $\triangle FHI\sim\triangle FIO$), but here is my way to the destination. Since $\measuredangle(\textbf{Per}_I(E),BC))=\measuredangle AEI$, let $X$ be $X_{109}$. We have known that Steiner Line of $X_{109}$ is $IH$, so Li4 point is $I$, then $\textbf{Per}_I(X)=II=IO$. We have $\measuredangle(\textbf{Per}_I(X),BC))=\measuredangle AXI$. Now $(OI,IH)=\measuredangle(\textbf{Per}_I(X),BC))-\measuredangle(\textbf{Per}_I(E),BC))=\measuredangle AXI-\measuredangle AEI$. It's well known that $X_{108}, I, X_{102}$ and $X_{104}, I, X_{109}$ are collinear respectively. Hence $\measuredangle AXI-\measuredangle AEI=\dfrac{1}{2}(\widehat{AX_{102}}-\widehat{AX_{104}})=\dfrac{1}{2}\widehat{X_{102}X_{104}}=\measuredangle X_{102}EX_{104}=\measuredangle IEH$. Done.

The property I used is from permutation-chang.github.io and https://www.overleaf.com/read/gtthtdjsyxqd. However, there haven't been any English version yet, so the attachment is in Chinese.

The original idea of this problem comes form Here by me and Li4. So the following solution is based on the properties in the attachment. Consider an isoconjugate $\varphi: I\mapsto H$. Let $\mathcal{L}^{\varphi}$ be the image of the line at infinity under $\varphi$. Let $X$ be the the fourth intersection of the circumcircle and $\mathcal{L}^\varphi$. Then by $\textbf{Theorem 1.}$ take isogonal conjugate and $\varphi$. Let $P=I$ we have \[ \mathbf{Per}_{I}(X)=\mathbf{Per}^{\mathcal{L}^{\varphi}}_{H}(X)=IH \]However, by the assumption of the problem we have $\mathbf{Per}_I(E)=IH$. So we get $E=X$. On the other hand, consider isogonal conjugate and $\varphi$ as before. Now we take $P=H$ to get \[ \mathbf{Per}_{O}(E)=\mathbf{Per}^{\mathcal{L}^{\varphi}}_{I}(E)=OI \] Finally, use $\textbf{Proposition 9.}$ we get \[ \measuredangle (IH,IO)=\measuredangle(\mathbf{Per}_{I}(E),\mathbf{Per}_{O}(E))=\measuredangle HEI \]Then we're done.

We can replace $ O, H $ by any isogonal conjugate $ P, Q $ of $ \triangle ABC. $ From $ MI^2=MD \cdot ME $ we get $ \measuredangle MEI = \measuredangle (IQ,AI). $ i.e. the second intersection of $ EI $ with $ \Omega $ is the pole of the simson line of $ \triangle ABC $ with direction $ \perp IQ, $ so it lies on the circumconic of $ \triangle ABC $ passing through $ I, P $ and hence the Miquel point of $IPIQ $ lies on $ \odot (EIQ). $$\color{blue} ^{[1]}$ $ \qquad \blacksquare $ [1] Isogonal conjugate and circumconic (post #4)

Let $\triangle ABC$ be a triangle with incenter $I$ and circumcenter $O$. Let $R,A'$ be the reflection of $I,A$ across $BC,IO$ respectively, and $M,N$ be the minor and major midarc of $\overarc{BC}$. Lemma: $\triangle AIO \sim \triangle IRM$ and $R,M,A'$ are collinear.

Let $E'$ be the other intersection of $EH$ and $\Omega$. From the problem, since \begin{align*} \measuredangle OIH &= \measuredangle OIA + \measuredangle AIH \\&\overset{\text{Lemma}}{=} \measuredangle MRI + \measuredangle MID \\&= (\measuredangle A'MI + \measuredangle MIR) + \measuredangle IEM \\&= \measuredangle A'MA + \measuredangle AMN + (\measuredangle IEH + \measuredangle HEM )\\&= \measuredangle A'MN + \measuredangle IEH + \measuredangle E'EM. \end{align*}It suffices to show that $\measuredangle OIH = \measuredangle IEH$, or equivalently, $\measuredangle A'MN + \measuredangle E'EM = 0$ and $E'A' \parallel MN$. Let $M'$ be the reflection of $M$ across $BC$, and $S,X,F,W$ be $AM \cap BC$, $AH \cap \Omega$, $MD \cap AX$, and $M'H \cap AI$, respectively. Obviously, $M'H,XM,BC$ are concurrent at the point called $T$. By cross ratio-chasing, \begin{align*} (X,A';N,A)&=M(A,N;A',X) \overset{\text{reflection}}{=} M'(S,M;I,H) \\& = (S,M;I,W) \overset{H}{=} H(S,M;D,T) \\&= M(S,H;D,T) = (A,H;F,X) \\& \overset{E}{=} (A,E';M,X). \end{align*}Therefore, $(X,A_1;N,A) = (A,E';M,X)$. Since $XA \parallel NM$, hence, $E'A_1\parallel MN$ as desired.

Cute! By shooting lemma and angle chase we get $\measuredangle MEI = \measuredangle MIH$ so the dumpty point of $IHO$ lies on $(EHI)$. Now by Turkey TST 2023/5 we are done

Li4 wrote: Let $I$, $O$, $H$, and $\Omega$ be the incenter, circumcenter, orthocenter, and the circumcircle of the triangle $ABC$, respectively. Assume that line $AI$ intersects with $\Omega$ again at point $M\neq A$, line $IH$ and $BC$ meets at point $D$, and line $MD$ intersects with $\Omega$ again at point $E\neq M$. Prove that line $OI$ is tangent to the circumcircle of triangle $IHE$. Proposed by Li4 and Leo Chang. My generalization for this problem in this post.