Angles+Distances

Point $D$ is on the $AC$ line over $A$ , such that $AD=AO$ . Note that $AO+AC=BC$ . From that $CD=BC$ . $O$ is the center of the circle, so $\angle OCD=\angle OCB$ . Since $CD=BC$ and $\angle OCD=\angle OCB$ , $\triangle CDO\sim\triangle COB$ . From that $\angle CDO=\angle CBO$ .

Since in a triangle the sum of any two angle is equal to the third angle's exterior angle and $\angle CAO=\beta =35°$ , $\angle ADO=\dfrac{\beta}{2}=\angle CDO$ . We know that $\angle CDO=\angle CBO$ , i.e. $\dfrac{\beta}{2}=\gamma$ . Therefore $\angle ABC=2\gamma =\beta =\boxed{35°}$