For now let's consider the points $A_0$, $A_1$ and $A_2$. Since $\triangle ABC$ is an acute triangle, then the point $O$ lies inside $\triangle ABC$, hence the point $A_2$ lies on the line $AC$ (Fig. 40). Let's prove that the quadrilateral $OA_0CA_2$ is inscribed. Indeed, since $A_2O \parallel BC$ we have
$$\angle OA_2A = \angle BCA = 90^\circ - \angle A_0AC = 90^\circ - \frac{1}{2} \angle A_0OC = \angle OA_0C.$$

Now let $X$ be the second intersection point of circumscribed circle of quadrilateral $OA_0CA_2$ with the line $CH$. Let's prove that the points $A_0$, $B$, $A_1$ and $X$ lie on the same circle. Indeed,
$$\angle A_1BH = 90^\circ - \angle A = \angle C_0CA = \angle AA_0C_0,$$
that is quadrilateral $A_0BA_1H$ is inscribed. Also $\angle A_1BH = 90^\circ - \angle A = \angle C_0CA = \angle AA_0C_0$

$$\angle HXA_0 = 180^\circ - \angle A_0XC = \angle A_0OC = 2\angle A_0AC = 2\angle A_0BC = \angle A_0BH,$$
since because of the fact that the points $H$ and $A_0$ are symmetric with respect to the line $BC$.

Now let's prove that the line $A_1A_2$ through a point $X$. This follows from that fact that
$$\angle A_1XA_0 = 180^\circ - \angle A_1BA_0 = \angle A_0CA_2 = 180^\circ - \angle A_2XA_0.$$
Now we have
$$\angle OXH = 180^\circ - \angle OXC = \angle OA_2A = \angle C, \text{ and also } \angle OXT = \angle OCA = 90^\circ - \angle B.$$
So we proved that if $X$ is an intersection point of $CC_0$ and $A_1A_2$, then $\angle OXT = 90^\circ - \angle B$, $\angle OXH = \angle C$.

Fig. 40

Now let $Y$ be the intersection point of $AA_0$ and $C_1C_2$. Analogously one can prove that $\angle OYT = 90^\circ - \angle B$, and $\angle OYH = \angle A$. Consider the points $X, H, Y, O$. Clearly $\angle YHX = \angle A$. Then, since $\angle OXH + \angle XHY + \angle HYO = 180^\circ$, the points $X, Y, O$ lie on one line. Hence, we have $\triangle XHY$ with the point $O$ on the side $XY$. This triangle is similar to $\triangle ABC$, moreover since $\angle TXY = \angle TYX = 90^\circ - \angle B$, then in this similarity the ray $XT$ corresponds to the ray $CO$, and the ray $YT$ to $AO$. Hence, the point $T$ corresponds to the point $O$, so $T$ is the center of circumscribed circle $\triangle YHX$. Thus we have $\angle THX = 90^\circ - \angle XYH = 90^\circ - \angle A = \angle HCA$, that is $HT \parallel AC$.