Let $T$ be the intersection of the circumcircle of $BC_1B'$ with the line $B'C'$. From angle chasing, we get $\angle ABI=\angle C'AB$, $\angle BIC=120^{\circ}$, and $\angle AC'I=\angle AB'I=60^{\circ}$. Since $BB'TC_1$ is cyclic we have $\angle C'TC_1=\angle C_1BI=\angle C'AC_1$ therefore $ATC_1C'$ is cyclic which implies $$\angle ATB'=180^{\circ}-\angle C'TA=180^{\circ}- \angle C'CA=\angle AB_1B'$$, hence $AB'B_1T$ is cyclic which gives that $\angle B_1TB'= \angle B_1AB'=\angle B_1CC' \Rightarrow CB_1TC'$ is cyclic. $\blacksquare$