We can show that $\angle A’C’B’=90^{\circ}$.

PS

let $CC'$ hits the circle $(ABC$) at $D$ ; $DAB$ is equilateral triangle then $\frac{B'C}{B'A} =\frac{BC}{BA} =\frac{\sin\angle BDC}{\sin 60^\circ} =\frac{CC'}{C'A} $ hence $C'B'$ is the bisector of $\angle CC'A$ More we know that the feet of the two in-bisectors of two angles concur with the ex-bsector of the third angle thus $C',A',B''$ are collinear and $(C,A;B',B'')=-1$ therefore $DC'$ is the ex-bisector of $ \angle CDA$ and $C'B'\perp C'A'$ RH HAS

Note that $BC$ is an external angle bisector in $\triangle ACC'$, so $A'$ is the center of the $A$-excircle in $ACC'$.Similarly, $B'$ is the center of the $B$-excircle in $\triangle BCC'$. Thus, $A'C'$ and $B'C'$ bisect $\angle BC'C$ and $\angle CC'A$, so $\angle A'C'B'=90^\circ$.