, 2023

Let the circumcircle of the triangle $O_1O_2O_3$ be $\Gamma$. Let the tangency point of $\omega_1$ and $\omega_2$ be $D$, $\omega_1$ and $\omega_3$ be $E$, $\omega_2$ and $\omega_3$ be $F$. Let the incenter of $O_1O_2O_3$ be $I$, then it's easy to see that $ID, IE, IF$ are perpendicular to the respective sides of $O_1O_2O_3$ since we have $|O_1D| = |O_1E|$, etc. The radical axis of circles: $\omega_1, \omega_2, \Gamma$ are concurrent, let's say at point $X$ where $Y, Z$ are defined similarly. Then $X$ lies on the perpendicular $IF$ since it is the radical axis of the tangent circles $\omega_1, \omega_2$. Let $O$ be the center of $\Gamma$ and $OO_1 \cap A_1B_1 = P$, then $OO_1 \perp A_1B_1$ and $X, D, P, O_1$ is cyclic. Therefore, $\angle DXO_1 = \angle DO_1O$ and similarly $\angle DXO_2 = \angle DO_2O$ and $XI$ is an angle bisector of the triangle $XYZ$, results for $Y$ and $Z$ follow similarly.