$H$ is the orthocenter. By simple angle chasing, we get that $\triangle AH_AB\sim\triangle AEA' \implies \triangle ABA' \sim \triangle AH_AE \implies$ $\angle AH_AE =\angle ABA' = \angle AHC$ by a simple angle chase. So $HC || H_AE$. Similarly, $HB || H_AD$. By another homothety, $\triangle ABC \sim \triangle ADE \implies AB || DE$. So triangles $BHC$ and $DH_AE$ are homothetic at point $A$. So we know all the angles of triangle $DH_AE$ now. The finish is now trivial. All we need to note is that $\angle O_AH_AH_B = A+(90-B)+C = 270-2B$. (Simple angle chase) Similarly, $\angle O_AH_AH_C = 270-2C$. Since now $\frac{\sin{O_AH_AH_B}}{\sin{O_AH_AH_C}} = \frac{\sin(270-2B)}{\sin(270-2C)}$ which is of course going to cyclically cancel out when we use Trig-Ceva.

we have that the triangles $ABC$ and $ADE$ are similar and coefficient of similarity is $ \frac{\cos A}{\sin B \sin C} $. Then $H_A$ be the orthocenter of the triangle $ADE$ and $ O_AH_A \perp H_BH_C $.

Let $X$; $Y$ be the intersections of $BC$ with $A'D$ and $A'E$. By cyclic quadrilateral $ADA'E$ we see $DE || BC$. Also we see $CAY$ must be isosceles since $\angle C'AE=90-C=\angle DEA'$ by simple angle chasing, and since $CE \perp A'Y$ then $E$ is the midpoint of $A'Y$. Similarly $D$ is midpoint of $A'X$ and so $A'EH_AD$ is a parallelogram. So $H_AE$ and $H_AD$ are heights of triangle $ADE$. Finally notice that $\angle (H_CH_B, BC)=B-C$ and $\angle(H_AO_A, BC) = C+90-B$ and so $H_AO_A \perp H_BH_C$. And so these lines concur because they are the heights of triangle $H_AH_BH_C$

easy for china tst $BFH_{B}\sim AA'B,$ $EF\left | \right |BC$ now easy by ceva in $H_{A}H_{B}H_{C}$

For the remainder of the proof, assume WLOG that $ O $ lies within $ \triangle{ABC} $ and that $ AB \ge AC. $ We proceed with complex numbers. WLOG assume that the circumcircle of $ \triangle{ABC} $ is the unit circle and let $ A, B, C, D, E, H_A, A' $ have complex coordinates $ a, b, c, d, e, h_a, x $ respectively. We easily find that the circumcenter of $ \triangle{BOC} $ has coordinates $ \frac{bc}{b + c} $ so we have that $ x $ satisfies $ \left(x - \frac{bc}{b + c}\right)\left(\overline{x} - \frac{1}{b + c}\right) = \frac{bc}{(b + c)^2}. $ Moreover because $ A' \in AO $ we have that $ x = a^2\overline{x}. $ Solving these two equations we easily find that $ x = \frac{a^2 + bc}{b + c}. $ Also, using the well-known formula for projections of points on to side lengths, we find that $ h_a = \frac{1}{2}\left(a + b + c - \frac{bc}{a}\right) $ and $ d = \frac{1}{2}\left(\frac{a^2 + bc}{b + c} + a + b - \frac{c(a^2 + bc)}{a(b + c)}\right) $ and $ e = \frac{1}{2}\left(\frac{a^2 + bc}{b + c} + a + c - \frac{b(a^2 + bc)}{a(b + c)}\right). $ Now it is trivial to verify that $ h_a + x = d + e $ so $ H_ADA'E $ is a parallelogram. This implies that $ \angle{H_ADE} = \angle{DEA'} = \angle{DAA'} = 90 - \angle{C} $ since $ ADA'E $ is clearly cyclic. Similarly, $ \angle{H_AED} = 90 - \angle{B} $ and therefore $ \angle{DH_AE} = 180 - \angle{A}. $ Now that we have all the angles in $ \triangle{DH_AE} $ an easy angle chase yields $ O_AH_A \parallel AO. $ But since $ O $ and $ H $ are isogonal conjuagates with respect to $ \triangle{ABC} $ and $ H_B, H_C $ are vertices of the pedal triangle of $ H $ with respect to $ \triangle{ABC}, $ it is well-known that $ OA \perp H_BH_C $ so we have that $ O_AH_A \perp H_BH_C. $ This implies that the three lines asked about by the problem concur at the orthocenter of $ \triangle{H_AH_BH_C} $ as desired.

My solution: Since $ \angle DBA'=\angle BAA'+\angle AA'B=90^{\circ}-\angle ACB+90^{\circ}-\angle BAC=\angle CBA $ , so $ Rt \triangle BAH_A \sim Rt \triangle BA'D \Longrightarrow \triangle BAA' \sim \triangle BH_AD \Longrightarrow \angle H_ADA=90^{\circ}-\angle BAC \Longrightarrow DH_A \parallel A'E $ . Similarly we can prove $ EH_A \parallel A'D \Longrightarrow DH_AEA' $ is a parallelogram . Since $ A',H_A $ are symmetry WRT the midpoint of $ DE $ , so $ H_AO_A $ is parallel to $ A'O $ ( $ \because $ the circumcenter of $ \triangle A'DE $ is the midpoint of $ AA' $ ) $ \Longrightarrow H_AO_A \perp H_BH_C $ . Similarly $ H_BO_B \perp H_CH_A , H_CO_C \perp H_AH_B \Longrightarrow H_AO_A, H_BO_B, H_CO_C $ concur at the orthocenter of $ \triangle H_AH_BH_C $ . Q.E.D

It is easy to see homothety centred at $A$ sends $B \to D, C \to E, H \to H_A$ Thus $\angle O_AH_AH_B = B+(A+B-90)+A = 2A+2B-90$ Now by trig Ceva in $\triangle H_AH_BH_C$, we are done.

61plus wrote: 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 $BOC$ at $A'$. The projections of $A'$ onto $AB, AC$ are $D,E$, and $O_A$ is the circumcentre of triangle $DH_AE$. Define $H_B, O_B, H_C, O_C$ similarly. Prove: $H_AO_A, H_BO_B, H_CO_C$ are concurrent In fact 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 $\odot(BOC)$ and $\odot(BHC)$ swap under isogonal conjugation about $\triangle ABC$. Hence $\{A_0, A'\}$ is a pair of isogonal conjugates. Let $A_1, A_2$ be the projections of $A_0$ on lines $\overline{AB}, \overline{AC}$. Since isogonal conjugates share pedal circles, we see that $D,E,A_1,A_2, H_A$ 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, $\triangle HH_BH_C \mapsto \triangle A_0A_2A_1$ under homothety centered at $A$. Hence $\overline{H_BH_C} \parallel \overline{A_1A_2}$ and we're done. $\blacksquare$ Now it clear that $\overline{H_AO_A}, \overline{H_BO_B}, \overline{H_CO_C}$ concur at the orthocenter of $\triangle H_AH_BH_C$. $\blacksquare$

I have name orthocenter of $\Delta ABC$ as $K.$ Kindly adjust Claim: $BC\parallel DE$ Proof: Let $A''$ be the $A-$antipode WRT $(O)$. $\angle A''BA=\angle A''CA= 90^{\circ} \Leftrightarrow A''B\parallel A'D$ and $A''C\parallel A'C \Leftrightarrow \exists$ homothety sending $BC\leftrightarrow DE \Leftrightarrow BC\parallel DE \blacksquare$ Now it's also easy to see that $H_{A}\leftrightarrow H$ with the same homothety $\Leftrightarrow$ circumcenter of $(K_{A}DE)\leftrightarrow$ circumcenter of $(KBC).$ Now let circumcenter of $(HBC)$ be $O_{H}.$ $O_{H}H\parallel AO$ (Don't forget to see that $\square AOHO_{H}$ is a parallelogram!) $\Leftrightarrow H_{A}O_{A}\parallel AA' \Leftrightarrow AA'\perp H_{B}H_{C} \Leftrightarrow O_{A}H_{A}\perp H_{B}H_{C}$

Let $H$ be the orthocenter of triangle $ABC$ and we know that $\overline{AO}$ and $\overline{AH}$ are isogonal. Notice that $$\angle AED=\angle AA'D=90-\angle BOA=\tfrac{1}{2}\angle AOB=\angle ACB$$so $\overline{BC}\parallel\overline{DE}.$ Also, $\triangle ABH_A\sim\triangle AA'E$ by AA so $\triangle ABA'\sim\triangle AH_AE$ by SAS. Thus, $\overline{HC}\parallel\overline{H_AE}$ as $$\angle EH_AA=\angle A'BA=180-\angle BAO-\angle OA'B=\angle AHC.$$Hence, $\triangle HBC\sim\triangle H_ADE$ and \begin{align*}\angle O_AH_AH_B&=\angle DH_AC-\angle DH_AO_A+\angle EH_AC+\angle CH_AH_B\\&=\beta+\gamma-\beta+(90-\beta)+\alpha\\&=90+\alpha+\gamma-\beta\\&=270-2\beta\end{align*}and Trig Ceva on triangle $H_AH_BH_C$ finishes. $\square$

Notice $\triangle ABH_A \sim \triangle AA'E$, so by spiral similarity and angle chasing we can conclude $\overline{HC} \parallel \overline{H_AE}$. This is enough to imply that $H_A$ is the orthocenter of triangle $ADE$, so $$\angle O_AH_AH_B = \angle A + (90^\circ - \angle B) + \angle C = 270^\circ - \angle B.$$The result follows by multiplying cyclically and trig Ceva.

Note that, $$\begin{aligned} \angle ABH_A & = 180^{\circ} - (\angle BAC + \angle BCA)\\ & = \angle OCB +\angle HAC\\ & = \angle AA'B +\angle BAA'\\ & = \angle DBA' \end{aligned}$$Hence, $\triangle BAH_A \sim \triangle BA'D$ $$\frac{BA}{BA'} = \frac{BH_A}{BD}\rightarrow \frac{BA}{BH_A} = \frac{BA'}{BD}$$Thus, $ \triangle BDH_A \sim \triangle BA'A \implies \angle BAA'= BDH_A$ $$\begin{aligned} \angle AA'E + BA'D &= 180^{\circ} - (\angle A'AE + \angle DBA')\\ &= 180^{\circ} -(\angle BAH_A + \angle ABH_A )\\ &= 90^{\circ} \end{aligned}$$$$\begin{aligned} \angle H_ADA' + BA'A &= 90^{\circ} - \angle BDH_A + \angle BDH_A\\ &= 90^{\circ} \end{aligned}$$$$ \therefore \angle H_ADA' + DA'E = 180^{\circ} \implies H_AD // EA'$$Similarly, we can say that $H_AE//DA'$ $\therefore H_A,D,A',E$ is a parallelogram We know that $A'$ is the reflection of $H_A$ across the midpoint of $DE$ (i.e the intersection point of $H_AA'$ and $DE$) . Let $O_{A_1}$ be the circumcenter of $\triangle DEA'$. Then $O_{A_1}$ is the midpoint of $AA'$ and $O_AO_{A_1}$ passes through the midpoint of $DE$. Hence, $H_AO_A // AA'$ $AA'\perp H_BH_C \implies O_AH_A \perp H_BH_C$ We know that an orthocenter exists for a triangle. Consider $\triangle H_AH_BH_C$. $O_AH_A \perp H_BH_C$. Similarly, $ O_BH_B \perp H_CH_A, O_CH_C \perp H_AH_B$. Hence all of them intersect at one point, namely the orthocenter of the orthic triangle.