carlosbr wrote: if $C_n$ does not belongs to $AB$, then $C_{n+1}$ is the circle of the triangle $\triangle{ABC_n}$. You probably mean "circumcentre"?

Consider $x_n=\frac{m\angle AC_nB}{\pi}$. Then the following reccurence holds: $x_{n+1}=2x_n$ if $x_n\le \frac 1 2$ and $x_{n+1}=1-x_n$ if $x_n\ge \frac 1 2$. Now the problem is a number theory one. The answer is $x_1\in \mathbb{Q}$ . Firstly it is obvious that if $2\mid q$ then, the denominator will not contain any $2$ after few steps when $x_n$ is multiplied by 2. Hence is suficient to examine the case when $(q,2)=1$. It is obvious that the denominator will be the same for all $x_i$. Denote by $p_n$ the denominator of $x_n$. If we prove that $p_n$ is eventually periodic, we are done, since the denominator is the same. But we can observe that at every step, $x_n\le 1$. Hence $p_n$ can take only finitely many values. Hence it is eventually periodic, since his reccurence depends only on the previous term. After some steps the angle $AC_nB$ can be same with $AC_{n+p}B$ but $C_n$ and $C_{n+p}$ can be simetric wrt $AB$. But is easy to see that, then $C_{n+2p}\equiv C_n$.

I forgot.. the case when $x_n\in \mathbb{R}\setminus\mathbb{Q}$. We can prove by induction that $x_{n+k}=a_kx_n+b_k$ with $a_k,b_k$ integers. Hence, in order to be periodic we must have, $x_n=x_{n+k}=a_kx_n+b_k$. Hence $x_n\in \mathbb{Q}$. Hence $x_1\in \mathbb{Q}$

is "segment bisector" mean perpendicular bisector?