KöMaL 2023 B5305

Let $A_1$ and $A_2$, respectively, denote the points lying closer to $B$ and closer to $C$ that divide side $BC$ of a triangle $ABC$ in a 1:2 ratio. Define points $B_1$ and $B_2$ on side $CA$, and points $C_1$ and $C_2$ on side $AB$ in an analogous way. Prove that the centroid of triangle $ABC$ lies on the line joining the common points of the circumscribed circles of triangles $A_1B_1C_1$ and $A_2B_2C_2$.