Problem 1838

Let $T$ be the symmetry point of $A$ with regard to $BC$, $F$ be the projection of $A'$ onto $BC$, $M$ be the projection of $T$ onto $AC$.
Since $AC = CT$, we have $\angle TCM = 2\angle TAM$. Since
$$\angle TAM = \frac{\pi}{2} - \angle ACB = \angle OAB, \text{ we have}$$
$$\angle TCM = 2\angle OAB = \angle A'OB = \angle A'CF, \text{ and}$$
$$\angle TCH_A = \angle A'CF + \angle A'CT = \angle TCM + \angle A'CT = \angle A'CE.$$
Because $\angle CH_A T, \angle CMT, \angle CEA', \angle CFA'$ are right angles, therefore
$$\frac{CH_A}{CM} = \frac{CH_A}{CT} \cdot \frac{CT}{CM} = \frac{\cos \angle TCH_A}{\cos \angle TCM} = \frac{\cos \angle A'CE}{\cos \angle A'CF} = \frac{CE}{CA'} \cdot \frac{CA'}{CF} = \frac{CE}{CF},$$
i.e. $CH_A \cdot CF = CM \cdot CE$, so $H_A, F, M, E$ are on the same circle $\omega_1$.

Similarly, let $N$ be the projection of $T$ onto $AB$, then $H_A, F, N, D$ are on the same circle $\omega_2$. Since $A'FH_A T$ and $A'EMT$ are both right trapezoids, the perpendicular bisector of the segments $H_AF$ and $EM$ meet at the midpoint $K$ of the segment $A'T$, i.e. $K$ is the center of circle $\omega_1$, $KF$ is the radius of circle $\omega_1$. Similarly, $K$ and $KF$ are also the center and the radius of circle $\omega_2$, respectively. Thus, $\omega_1$ and $\omega_2$ are the same, $D, N, F, H_A, E, M$ are on the same circle. So $O_A$ is the midpoint $K$ of $A'T$, $O_AH_A \parallel AA'$.

Since $\angle H_CAO + \angle AH_CH_B = \frac{\pi}{2} - \angle ACB + \angle ACB = \frac{\pi}{2}$, we have $AA' \perp H_BH_C$, thus $O_AH_A \perp H_BH_C$, therefore $O_AH_A, O_BH_B, O_CH_C$ all pass through the orthocenter of $\triangle H_AH_BH_C$.