, 2024

Let us define $B_a$, $B_c$, $C_a$, $C_b$ similarly. Let the circle passing through $B$, $C$, $A_b$, $A_c$ be $\omega_a$ and let us define $\omega_b$, $\omega_c$ analogously. We will start with a lemma that we will use repeatedly.

Lemma: Let $ABC$ be a triangle, $O$ be its circumcenter and $AO \cap BC = D$. Let $E$, $F$ be points on the lines $AB$, $AC$ respectively such that $DB = DE$ and $DC = DF$. Then we have $BC \parallel EF$.

*Proof:* Let the midpoints of the segments $BE$ and $CF$ be $K$ and $L$, respectively. Since $\angle AKL = \angle ADL = \angle B$.

Let $BA_c \cap CA_b = X$.

*Claim 1:* $X$ lies on the radical axis of the circles $\omega_b$ and $\omega_c$.

*Proof:* Let $\omega_c \cap CA_b = C_1$ and $\omega_b \cap BA_c = B_1$. We will prove that $B$, $C$, $B_1$, $C_1$ are concyclic which implies the claim. Since $\angle C_a C_1 C = \angle C_a C_b C = \angle A$ we have that $A$, $B$, $C_a$, $C_1$ are concyclic. Using the lemma, we have $A_b C_b \parallel B_a B_c$ hence $A_b$, $C_b$, $A$, $C$ are concyclic and $B$ lies on the radical axis of the circles $\omega_c$ and $(AA_b C_1 C_a)$ therefore $B$, $C_1$, $C_a$ are collinear. Hence we have $\angle BC_1 C = 180^\circ - \angle A$ and $\angle BB_1 C = 180^\circ - \angle A$ can be obtained similarly. $\square$

Let $M_a$ be the midpoint of the segment $O_b O_c$ and let $O$ be the circumcenter of the triangle $ABC$.

*Claim 2:* $HM_a \parallel OO_a$.

*Proof:* $A$ lies on the radical axis of the circles $\omega_b$, $\omega_c$ and from Claim 1 we have $AX \perp O_b O_c$. Let $BE$, $CF$ be the altitudes of the triangle and $A_b A_c \cap BC = T$. Let the second intersection of the line $HM_a$ with the circle $(AFHE)$ be $R$ and the foot of the perpendicular from $H$ to the line $AX$ be $S$. Since the points $O_a$, $O_b$, $O_c$ lie on the altitudes of the triangle $ABC$, we have $(O_b, O_c; M_a, \infty) = -1$ and projecting this to the circle $(AFHE)$ from the point $H$ we obtain $(E, F; R, S) = -1$. Then projecting this from $A$ to the line $BC$ we have $(C, B; AR \cap BC, AX \cap BC) = -1$ hence we must have $AR \cap BC = T$. Since $R$ lies on the circle with diameter $AH$ we obtain $HM_a \perp AT$. Finally, since the line $AT$ is the radical axis of the circles $\omega_a$ and $(ABC)$ we have $OO_a \perp AT$ which implies

$$HM_a \parallel OO_a.$$ $\Box$

Now, let $G_1$ be the centroid of the triangle $O_aO_bO_c$, we will prove that $G_1$ is the midpoint of $HG$. Triangles $HM_aM_c$ and $OO_aO_c$ have sides that are parallel to each other, hence they are similar and the similarity ratio is equal to $\frac{O_bO_c}{M_bM_c} = 2$. Therefore the intersection of the lines $O_aM_a$ and $HG$ divides both segments with the ratio $2 : 1$ and this point is in fact $G_1$ since $O_aM_a$ is a median, and it is also the midpoint of $HG$ hence we are done.