Geometry Problem about tangent lines Let S be the circumcenter of ABC. $A_0$ is the middle of arc BC not containing A, $C_0$ also the middle of arc AB without C. 
Let $S_1$ be a circle with center $A_0$, tangent to BC, $S_2$ with center $C_0$ and tangent to AB. Showing that the incenter I of ABC lies on the common extangent to $S_1$ and $S_2$.
I tried some angle chasing, considered common extangent of the circumcircle of ABC and both $S_1$ and $S_2$, tried Power of a Point theorem, polar bijection, but I don't know how to prove that the point I lies on the extangent. I'm really clueless how to proceed. I tried thise things bluntly but they seem to be irrelevant. How is it done?
 A: For now, we'll consider just the circle about $A_0$. 
Let $M$ be the midpoint of $\overline{BC}$ (and therefore also the foot of the perpendicular from $A_0$). Drop a perpendicular from $A_0$ to $\overline{AB}$ at $X$, and let $P$ be the point where this perpendicular crosses $\bigcirc A_0$. Finally, let $I$ be the incenter of $\triangle ABC$, so that $\overline{IA}$ and $\overline{IC}$ bisect $\angle A$ and $\angle C$.



*

*Since $\angle BAA_0$ and $\angle BCA_0$ subtend the same arc $\stackrel{\frown}{A_0B}$ of the circumcircle, they are congruent.

*By the Exterior Angle Theorem applied to $\angle I$ of $\triangle IAC$, we have 
$$\angle CIA_0 = \color{blue}{\angle IAC} + \color{red}{\angle ICA} = \color{blue}{\angle BCA_0} + \color{red}{\angle ICB} = \angle ICA_0$$ so that $\triangle A_0CI$ is isosceles, with $\color{violet}{\overline{A_0C}}\cong\color{violet}{\overline{A_0I}}$.

*As $\triangle AA_0X \sim \triangle CA_0 M$, we have
$$\frac{|\overline{A_0X}|}{|\overline{A_0A}|} = \frac{|\color{green}{\overline{A_0M}}|}{|\color{violet}{\overline{A_0C}}|}  \quad\to\quad \frac{|\overline{A_0X}|}{|\overline{A_0A}|} = \frac{|\color{green}{\overline{A_0P}}|}{|\color{violet}{\overline{A_0I}}|}$$
which shows that $\triangle A_0AX \sim \triangle A_0IP$, and therefore that $\overline{IP}\perp\overline{A_0P}$. 

*Therefore, the line through $I$ parallel to $\overline{AC}$ is tangent to $\bigcirc A_0$; likewise, this line is tangent to $\bigcirc C_0$ (not shown).
