Wlog let $B > C$. Since $AH$ and $OD$ are parallel, and $AH/AI = 2R \cos A / (r/\sin(A/2)) = 2R \cos A / (R \cos A / \sin(A/2)) = 2 \sin(A/2) = BD/DO = ID/DO$, $AIH$ and $DIO$ are similar, so $\angle AHI = 90^\circ$. So we have $\angle HIO = 180^\circ - (B - C)/2$. Now consider the reflection of $O$ in $DE$, say $O'$. Then we have $\angle O'ED = \angle OED = 90^\circ - (B-C)/2 = 180 - \angle AED$, so $O'$ is on $AH$. So $DE$ is the perpendicular bisector of $OO'$, so the circumcenter of $OO'I$ is on $DE$. Now $\angle HO'O = \angle EO'D/2 = \angle EOD/2 = \angle EAD = (B-C)/2 = 180^\circ - \angle HIO$, so $HIOO'$ is cyclic, and thus the conclusion of the problem follows.

We use complex numbers with the assumption that $(O)$ is the unit circle. Let $A, B, C, D, I$ have complex coordinates $a^2, b^2, 1/b^2, -1, j$ respectively. Then $$e=\frac{-1}{a^2}\hspace{2 mm} \wedge \hspace{2 mm} j=\frac{-(ab^2+a+b)}{b}$$Let $R$ be the circumcenter of $\triangle OIH$. We have $$\overline{j}=-j\longleftrightarrow a^{2}b^{2}+a^2+b^2+2ab+1=0\hspace{2 mm} (\diamond)$$It's enough to show that $$\frac{\overline{r}+1}{r+1}=-a^2\longleftrightarrow j=\frac{-(a^2+1)(b^4+b^2+1)}{b^2(a^2-1)}$$Or $$a^{4}b^{2}+a^4+2a^{3}b+2a^{2}b^2+2a^2+2ab+b^2+1\stackrel{?}{=}0\hspace{2 mm} (\star)$$But $$(\star)=(a^2+1)\cdot (\diamond)=0.\hspace{3 mm} \blacksquare$$

parmenides51 wrote: Let $O, I, H$ be the circumcenter, the incenter, and the orthocenter of $\triangle ABC$. The lines $AI$ and $AH$ intersect the circumcircle of $\triangle ABC$ for the second time at $D$ and $E$, respectively. Prove that if $OI \parallel BC$, then the circumcenter of $\triangle OIH$ lies on $DE$. (Fedir Yudin) Cute! Let $L = AI \cap (ABC), K = LO \cap (ABC), S = KI \cap (ABC), G = \ell \cap BC$ (where $\ell$ is the perpendicular line from $I$ to $AI$). Then $AS$ is perpendicular to $BC$ since $\angle{IOL} = \angle{IOK} = 90$ implying that $ASLK$ is an isosceles trapezoid. Notice that $GI$ is tangent to $(BIC)$. If we let $S' = GL \cap (ABC)$, then $$GS'.GL = GB.GC = GI^2$$and $$\angle{LS'I} = \angle{LIG} = 90 = \angle{LS'K}$$thus $S' \equiv S$ and $G, S, L$ are collinear. Now observe that $$\angle{IGB} = \angle{IAH} = \angle{LKS} = \angle{SGB}$$hence $G,H,I$ are collinear. Let $T$ be the reflection of $I$ across $GL$. It suffices to show that $THIO$ is cyclic. To conclude, triangles $SIO$ and $SHI$ are similar via $\angle{SIH} = \angle{SKA} = \angle{SLI} = \angle{SOI}$ and $\angle{ISH} = \angle{OSI}$, thus $$\frac{OS}{ST} = \frac{OS}{SI} = \frac{OI}{IH}$$Combing the previous result with $\angle{OST} = \angle{OIH}$, we get that $\triangle OST, \triangle OIH$ are similar. Therefore $\angle{TOH} = \angle{SOI} = \angle{SIH} = \angle{TIH}$, as required. $\blacksquare$

I'll sketch a neat solution with Cartesian Coordinates: Let $O$ be the origin with positive $x$ axis in the direction of $OI$. Scale so the circumcircle is the unit circle and let $A=(a,\sqrt{1-a^2})$. Let $B=(b,-\sqrt{1-b^2})$ and $C=(-b,-\sqrt{1-b^2})$. WLOG $a>0$ (though we omit the case the square root is negative, that is obtuse triangles). We get $D=(0,-1)$ and $E=(a,-\sqrt{1-a^2})$. Reflect $E$ about $BC$ to get $$H=(a, \sqrt{1-a^2}-2\sqrt{1-b^2}).$$Now $I$ is the $x$ intercept of the line $AD$, that is $$I=\left(\frac{a}{1+\sqrt{1-a^2}},0\right).$$Using $DI=DB$ we get a relation between $a$ and $b$: $$(1+\sqrt{1-a^2})(1-\sqrt{1-b^2})=1.$$Now the perpendicular bisector of $OI$ has equation $x=I/2$ which intersects $DE$ at $F$ given by $$F=\left(\frac{a}{2(1+\sqrt{1-a^2})},\frac{-1-3\sqrt{1-a^2}}{2(1+\sqrt{1-a^2})}\right).$$Now it suffices to show that $OF^2=FH^2$, or for example (let $M$ be the midpoint of $IH$) the slopes $FM$ and $IH$ multiply to $-1$. Both approaches work out nicely using the relation between $a$ and $b$.