, 2022

Take a point $X$ such that $\triangle XHI \sim \triangle XIO$. We first prove that $X$ lies on $\Omega$: take $Y, Z$ such that $\triangle XAY \sim \triangle XBZ \sim \triangle XHI$, then by spiral similarity, $\triangle IAH \sim \triangle OYI$. This tells us that $\angle OYI = \angle IAH = \angle OAI$, that is, $A, O, I, Y$ are concyclic. Similarly, $B, O, I, Z$ are concyclic. Let $P$ be the intersection of $AY$ and $BZ$; then again by spiral similarity, $\triangle AIB \sim \triangle YOZ$, so we obtain
$$\angle APB = \angle YAI + \angle AIB + \angle IBZ = \angle YOI + \angle AIB + \angle IOZ = 2 \cdot \angle AIB = \angle ACB,$$
that is, $P$ lies on $\Omega$. Then from $\angle AXB = \angle (AY, BZ) = \angle APB$ we get that $X$ also lies on $\Omega$.
Since $\angle XHI = \angle XIO$, $OI$ is tangent to $\odot(XHI)$, so we only need to prove that $E, X, H, I$ are concyclic. Note that $\angle DBM = \angle BEM$, so $MD \cdot ME = \overline\{MB\}^2 = \overline\{MI\}^2$, hence
$$\angle MEI = \angle DIM = \angle HIA = \angle IOY = \angle IAP = \angle MEP,$$
that is, $E, I, P$ are collinear. Therefore, from $\angle IEX = \angle YAX = \angle IHX$ we obtain that $E, X, H, I$ are concyclic.