Here is my solution with complex numbers. It is easy to see that $A_bA_c\parallel EF$, where $D,E,F$ are foots. Let $M_a$ be the midpoint of $AH$. $O_a$ is on $AH$. Let $X$ be the intersection of $EF$ and $AH$. Use complex coordinates. Let $a,b,c$ be points $A,B,C$ on unit circle. We need to show that $\frac{O_a+O_b+O_c}{3}$ is of the form $k(a+b+c)$. By two sinus theorem in triangle $AHE$, $\frac{AX}{HX}=\tan B\tan C$. Since $M_a$ and $O_a$ are the circumcenters of similar triangles $AEF$ and $AA_bA_c$, $\frac{AM_a}{AO_a}=\frac{AX}{AH}$. From these equalities, we get $\frac{AO_a}{HO_a}=\frac{\tan B\tan C+1}{\tan B\tan C-1}$, so $$O_a=\frac{(\tan B\tan C+1)H+(\tan B\tan C-1)A}{2\tan B\tan C}=a+\frac{b}{2}+\frac{c}{2}+\frac{\tan A(b+c)}{2\tan A\tan B\tan C}$$Now we need to show that $\tan A(b+c)+\tan B(a+c)+\tan C(a+b)$, or equivalently $x=a\tan A+b\tan B+c\tan C$ is of the form $k(a+b+c)$. Borrowing a fact from barycentric coordinates, which is $H=(\tan A:\tan B:\tan C)$, we can say that $H=\frac{x}{\tan A+\tan B+\tan C}=a+b+c$, so $x=(a+b+c)(\tan A+\tan B+\tan C)$.

Here is a quick sketch of the synthetic solution (will add details later). First let $BA_c\cap CA_b= X$. Then show that $X$ lies on the radical axis of the circles $(O_b) $ and $(O_c) $ (just by angle chasing and writing powers). Then let $M_a$ be the midpoint of $O_bO_c$ and $A_bA_c\cap BC= A_0$. Show that $HM_a$ is perpendicular to the line $AA_0$ thus parallel to the line $O_aO$ (by taking harmonic ratios from $H$ and $A$ respectively). Then you can finally finish by looking at the triangle $O_aO_bO_c$.

another complex bash without all those annoying tangents. As usual, let $\triangle ABC$ be inscribed in the unit circle, so that $$|a|=|b|=|c|=1$$$$h=a+b+c$$$$g=\frac{a+b+c}3$$Now let line $A_bA_c$ intersect the unit circle at $X, Y$ respectively, so that $|x|=|y|=1$. Since $B, C, A_b, A_c$ are concyclic, we have $$\frac{(b-a_b)(c-a_c)}{(b-c)(a_b-a_c)} \in \mathbb{R} \implies \frac{(a-b)(a-c)}{(b-c)(x-y)} \in \mathbb{R} \implies a^2 = xy$$Since $A_b, A_c, H$ are collinear, we have $$h = x + y - xy\overline{h} \implies x + y = h + a^2\overline{h} = \frac{a^2b+a^2c+2abc+b^2c+bc^2}{bc}$$Then $$a_b = \frac{ab(x+y) - xy(a+b)}{ab-xy} = \frac{\frac{ab(a^2b+a^2c+2abc+b^2c+bc^2)}{bc}-a^2(a+b)}{ab-a^2} = \frac{b(a^2+ac+bc+c^2)}{c(b-a)}$$Similarly, $$a_c = \frac{c(a^2+ab+bc+b^2)}{b(c-a)}$$Now we find the coordinate of $O_a$ by shifting. Define the vectors $$a_b' = a_b-a = \frac{(b+c)(a^2+bc)}{c(b-a)}$$$$a_c' = a_c-a = \frac{(b+c)(a^2+bc)}{b(c-a)}$$We have $\overline{a_b'} = \frac{a_b'}{ab}$ and $\overline{a_c'} = \frac{a_c'}{ac}$. Then if $o_a' = o_a-a$, we have $$o_a' = \frac{a_b'a_c'(\overline{a_b'}-\overline{a_c'})}{\overline{a_b'}a_c'-a_b'\overline{a_c'}} = \frac{ba_c' - ca_b'}{b-c} = \frac{\frac{(b+c)(a^2+bc)}{c-a} - \frac{(b+c)(a^2+bc)}{b-a}}{b-c} = \frac{(b+c)(a^2+bc)}{(a-b)(a-c)}$$Then $$o_a = o_a' + a = \frac{a^3+abc+b^2c+bc^2}{(a-b)(a-c)}$$Similarly, $$o_b = \frac{b^3+abc+a^2c+ac^2}{(b-a)(b-c)}$$$$o_c = \frac{c^3+abc+a^2b+ab^2}{(c-a)(c-b)}$$Then the centroid of $\triangle O_aO_bO_c$ has coordinate \begin{align*} \frac{o_a+o_b+o_c}3 &= \frac{(b-c)(a^3+abc+b^2c+bc^2) + (c-a)(b^3+abc+a^2c+ac^2) + (a-b)(c^3+abc+a^2b+ab^2)}{3(a-b)(a-c)(b-c)} \\ &= \frac{2a^3b+2b^3c+2c^3a-2ab^3-2bc^3-2ca^3}{3(a-b)(a-c)(b-c)} \\ &= \frac{2(a-b)(a-c)(b-c)(a+b+c)}{3(a-b)(a-c)(b-c)} \\ &= \frac{2(a+b+c)}3 \end{align*}So in fact the centroid is the midpoint of segment $\overline{HG}$. $\blacksquare$

Let $D, E, F$ be the feet and let $AH\cap (ABC)=K_A$. We have $\angle HK_AB=\angle ACB=\angle AA_bH$ so $(A_bBK_AH)$ is cyclic. Let $T_A$ be the reflection of $H$ over $O_a$. Now, $2|AO_a|=\dfrac{|AA_b|}{\sin{AA_cA_b}}=\dfrac{|AH||AK_a|}{|AB|\sin{ABD}}=\dfrac{|AH||AK_a|}{|AD|}$. Then, $|HT_A|=2|AH|-2|AO_a|=|AH|\left(\dfrac{2|AD|-|AK_a|}{|AD|}\right)=\dfrac{|AH|^2}{|AD|}$. Then, $|AT_A|=|AH|-|HT_A|=|AH|\left(\dfrac{|AD|-|AH|}{|AD|}\right)=\dfrac{|AH||HD|}{|AD|}$. Hence, $|AT_A||AD|=|AH||HD|$. Similarly, $|BT_B||BE|=|BH||HE|$ and $|AH||HD|=|BH||HE|$ so $$\dfrac{|AT_A|}{|BT_B|}=\dfrac{|BE|}{|AD|}=\dfrac{|AB|\sin{BAC}}{|AB|\sin{ABC}}=\dfrac{|BC|}{|AC|}$$ Then, $\dfrac{|AT_A|}{|BC|}=\dfrac{|BT_B|}{|AC|}$. Similarly, the ratio $\dfrac{|CT_C|}{|AB|}$ is equal to these. Now, we can finish the proof by the following lemma. Lemma: Given a triangle $ABC$ with orthocenter $H$. Given points $X, Y, Z$ on $AH, BH, CH$ such that $$\dfrac{|AX|}{|BC|}=\dfrac{|BY|}{|CA|}=\dfrac{|CZ|}{|AB|}$$Then, the centroids of $\triangle ABC$ and $\triangle XYZ$ coincide.

Hence, the centroid of $\triangle T_AT_BT_C$ is $G$. Then, the centroid of $\triangle O_aO_bO_c$ is the midpoint of $[HG]$.

hmm this solution seems new

Interesting. Let $A'$, $B'$, and $C'$ be the reflections of $A$ over $O_A$, and similar which lie on $(AA_BA_C)$ and similar circles. We show the centroid of $A'B'C'$ is $H$, which finishes by vectors, as \[\frac{A'+B'+C'}{3}=\frac{2(O_A+O_B+O_C)-A-B-C}{3}\]so \[\frac{O_A+O_B+O_C}{3}=\frac{A'+B'+C'+A+B+C}{6}=\frac{3H+3G}{6}=\frac{H+G}{2}\] First, $A'A_B \perp AB$ and $A'A_C \perp AC$, with other similar perpendicularities holding. Thus $CC_B \parallel AA'\parallel B'B_C$. So it suffices to show that $AA'$ or the $A$-altitude bisects $B_CC_B$ which is immediate from $\angle HB_CC_B=\angle A =\angle HC_BB_C$.

Cute!

Interesting Problem Let $AD$, $BE$, and $CF$ be altitudes. Let $H_a$ be the antipode of $A$ in $(AA_bA_c)$ and define $H_b$ and $H_c$ similarly. As $BCEF$ is cyclic we must have $EF\parallel A_cA_b$ and as $H$ is the antipode of $A$ in $(AEF)$ we must have that $AEHF$ is homothetic to $AA_cH_aA_b$. The key claim in the problem is that $H$ is the centroid of $H_aH_bH_c$. It is sufficient to show that $HH_a$ bisects $H_bH_c$. As $H_bB_c\perp BC$ and $H_cC_b\perp BC$ it is sufficient to show that $HH_a$ bisects $B_cC_b$. We are done though as $\angle HB_cC_b=\angle BAC=\angle HC_bB_c$. Finally $$\frac{O_a+O_b+O_c}{3}=\frac{A+H_a+B+H_b+C+H_c}{6}=\frac{2}{3}(A+B+C)$$