China Team Selection Test

Let $T$ be the symmetry point of $A$ with regard to $BC$, $F$ be the projection of $A^{\prime }$ onto $BC$, $M$ be the projection of $T$ onto $AC$. Since $AC=CT$, we have $\angle TCM=2\angle TAM$. Since $$\angle TAM=\frac{\pi }{2}-\angle ACB=\angle OAB,\text{ we have}$$ $$\angle TCM=2\angle OAB=\angle A^{\prime }OB=\angle A^{\prime }CF,\text{ and}$$ $$\angle TC{H}_A=\angle A^{\prime }CF+\angle A^{\prime }CT=\angle TCM+\angle A^{\prime }CT=\angle A^{\prime }CE.$$ Because $\angle C{H}_{A}T,\angle CMT,\angle CE{A}^{\prime },\angle CF{A}^{\prime }$ are right angles, therefore $$\frac{C{H}_A}{CM}=\frac{C{H}_A}{CT}\cdot \frac{CT}{CM}=\frac{\cos \angle TC{H}_A}{\cos \angle TCM}=\frac{\cos \angle A^{\prime }CE}{\cos \angle A^{\prime }CF}=\frac{CE}{C{A}^{\prime }}\cdot \frac{C{A}^{\prime }}{CF}=\frac{CE}{CF},$$ i.e. $C{H}_A\cdot CF=CM\cdot CE$, so $H_A,F,M,E$ are on the same circle ${\omega }_1$.



Similarly, let $N$ be the projection of $T$ onto $AB$, then $H_A,F,N,D$ are on the same circle ${\omega }_2$. Since $A^{\prime }F{H}_{A}T$ and $A^{\prime }EMT$ are both right trapezoids, the perpendicular bisector of the segments $H_{A}F$ and $EM$ meet at the midpoint $K$ of the segment $A^{\prime }T$, i.e. $K$ is the center of circle ${\omega }_1$, $KF$ is the radius of circle ${\omega }_1$. Similarly, $K$ and $KF$ are also the center and the radius of circle ${\omega }_2$, respectively. Thus, ${\omega }_1$ and ${\omega }_2$ are the same, $D,N,F,H_A,E,M$ are on the same circle. So $O_A$ is the midpoint $K$ of $A^{\prime }T$, $O_A{H}_A\parallel A{A}^{\prime }$.

Since $\angle H_{C}AO+\angle A{H}_C{H}_B=\frac{\pi }{2}-\angle ACB+\angle ACB=\frac{\pi }{2}$, we have $A{A}^{\prime }\perp H_B{H}_C$, thus $O_A{H}_A\perp H_B{H}_C$, therefore $O_A{H}_A,O_B{H}_B,O_C{H}_C$ all pass through the orthocenter of $△H_A{H}_B{H}_C$.