59th Ukrainian National Mathematical Olympiad

For now let's consider the points $A_0$, $A_1$ and $A_2$. Since $△ABC$ is an acute triangle, then the point $O$ lies inside $△ABC$, hence the point $A_2$ lies on the line $AC$ (Fig. 40). Let's prove that the quadrilateral $O{A}_{0}C{A}_2$ is inscribed. Indeed, since $A_{2}O\parallel BC$ we have $$\angle O{A}_{2}A=\angle BCA=90^{\circ }-\angle A_{0}AC=90^{\circ }-\frac{1}{2}\angle A_{0}OC=\angle O{A}_{0}C.$$

Now let $X$ be the second intersection point of circumscribed circle of quadrilateral $O{A}_{0}C{A}_2$ with the line $CH$. Let's prove that the points $A_0$, $B$, $A_1$ and $X$ lie on the same circle. Indeed, $$\angle A_{1}BH=90^{\circ }-\angle A=\angle C_{0}CA=\angle A{A}_0{C}_0,$$ that is quadrilateral $A_{0}B{A}_{1}H$ is inscribed. Also $\angle A_{1}BH=90^{\circ }-\angle A=\angle C_{0}CA=\angle A{A}_0{C}_0$

$$\angle HX{A}_0=180^{\circ }-\angle A_{0}XC=\angle A_{0}OC=2\angle A_{0}AC=2\angle A_{0}BC=\angle A_{0}BH,$$ 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 $A_1{A}_2$ through a point $X$. This follows from that fact that $$\angle A_{1}X{A}_0=180^{\circ }-\angle A_{1}B{A}_0=\angle A_{0}C{A}_2=180^{\circ }-\angle A_{2}X{A}_0.$$ Now we have $$\angle OXH=180^{\circ }-\angle OXC=\angle O{A}_{2}A=\angle C,\text{ and also }\angle OXT=\angle OCA=90^{\circ }-\angle B.$$ So we proved that if $X$ is an intersection point of $C{C}_0$ and $A_1{A}_2$, then $\angle OXT=90^{\circ }-\angle B$, $\angle OXH=\angle C$.

Fig. 40

Now let $Y$ be the intersection point of $A{A}_0$ and $C_1{C}_2$. 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 $△XHY$ with the point $O$ on the side $XY$. This triangle is similar to $△ABC$, 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 $△YHX$. Thus we have $\angle THX=90^{\circ }-\angle XYH=90^{\circ }-\angle A=\angle HCA$, that is $HT\parallel AC$.