Let $\vartriangle ABC$ be a triangle. Its incircle $\omega$ touches $ BC$, $CA$, $AB$ at points $A_0$, $B_0$, $C_0$ respectively. $AA_0$ meets $\omega$ at $A_1$. $A_2$ is the point of reflection of $A_1$ over $B_0C_0$. $B_1$, $B_2$, $C_1$, $C_2$ are dened similarly. Prove that the circumcentre of $\vartriangle A_2B_2C_2$ lies on the Euler line of $\vartriangle A_0B_0C_0$.
We show that $A_2$ lies on the circle with center $HG$. To do so, let $P$ be the midpoint of $BC$, and we claim that $PA_0$ is the reflection of $AP$ over $BC$. Indeed, note $(AA_0;BC)=-1$, so if $T=BB\cap CC$, we know that $(AA_0;T(AA_0\cap BC))=-1$, so thus we know that $BC$ is the angle bisector of $\angle APA_0$. In particular,
$$\measuredangle AA_0P=\measuredangle A_0AP+\measuredangle APA_0=2\measuredangle PAB-\measuredangle CAB+2\measuredangle APB=\measuredangle CBA-\measuredangle ACB$$Let $H’$ be the reflection of $H$ over $BC$. Then, $\measuredangle AA_0H’=\measuredangle ACH’=90^\circ+\measuredangle ACB-\measuredangle CBA$, so $\angle H’A_0M=90^\circ$. Reflecting over $BC$ and noting $BC$ is the angle of $\angle APA_0$, hence $PA_0$ maps to the $A$-median. Thus, $\angle HA_2G=90^\circ$, as desired.