Let $ABC$ be a triangle with $A',B',C'$ are midpoints of $BC,CA,AB$ respectively. The circle $(\omega_A)$ of center $A$ has a big enough radius cuts $B'C'$ at $X_1,X_2$. Define circles $(\omega_B), (\omega_C)$ with $Y_1, Y_2,Z_1,Z_2$ similarly. Suppose that these circles have the same radius, prove that $X_1,X_2, Y_1, Y_2,Z_1,Z_2$ are concyclic.