Points $C_1$ and $C_2$ lie on side $AB$ of triangle $ABC$, where the point $C_1$ belongs to the segment $AC_2$ and $\angle ACC_1= \angle BCC_2$. On segments $CC_1$ and $CC_2$ points $A'$ and $B'$ are taken such that $\angle CAA'= \angle CBB' = \angle C_1CC_2$. Prove that the center of the circle $(CA'B')$ lies on the perpendicular bisector of the segment $AB$.