Let the circumcenter of triangle $ABC$ be $O$. $H_A$ is the projection of $A$ onto $BC$. The extension of $AO$ intersects the circumcircle of $\triangle BOC$ at $A'$. The projections of $A'$ onto $AB$ and $AC$ are $D$ and $E$, respectively. $O_A$ is the circumcenter of triangle $DH_AE$. Define $H_B$, $O_B$, $H_C$, and $O_C$ similarly. We aim to prove that $H_AO_A$, $H_BO_B$, and $H_CO_C$ are concurrent.

To prove this, we first establish that the concurrency point is the orthocenter of $\triangle H_AH_BH_C$. Let $A_0$ be the reflection of $A$ in $\overline{BC}$.

Claim: $\overline{H_AO_A} \perp \overline{H_BH_C}$.

Proof: Note that the circumcircles $\odot(BOC)$ and $\odot(BHC)$ are isogonal conjugates with respect to $\triangle ABC$. Hence, $\{A_0, A'\}$ is a pair of isogonal conjugates. Let $A_1$ and $A_2$ be the projections of $A_0$ onto lines $\overline{AB}$ and $\overline{AC}$, respectively. Since isogonal conjugates share pedal circles, the points $D$, $E$, $A_1$, $A_2$, and $H_A$ are concyclic. Note that $H_A$ is also the circumcenter of $\triangle AA_1A_2$, hence $\overline{H_AO_A} \perp \overline{A_1A_2}$. Clearly, $\triangle H_AH_BH_C$ maps to $\triangle A_0A_2A_1$ under a homothety centered at $A$. Therefore, $\overline{H_BH_C} \parallel \overline{A_1A_2}$, and we have $\overline{H_AO_A} \perp \overline{H_BH_C}$.

Thus, it is clear that $\overline{H_AO_A}$, $\overline{H_BO_B}$, and $\overline{H_CO_C}$ concur at the orthocenter of $\triangle H_AH_BH_C$.

The answer is: \boxed{\text{The lines } H_AO_A, H_BO_B, \text{ and } H_CO_C \text{ are concurrent at the orthocenter of } \triangle H_AH_BH_C.}