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?