Any solution?

By definition, $A_{n+1}$ lies on the internal angle bisector of $\angle B_nA_nC_n$ and $B_{n+1}$, $C_{n+1}$ on the external angle bisector. Thus $A_{n+1}A_n \perp B_{n+1} C_{n+1}$ for each $n$, i.e. $A_n$ is the foot of the perpendicular from $A_{n+1}$ to $B_{n+1}C_{n+1}$. Hence if all coordinates of $A_{n+1}$, $B_{n+1}$, $C_{n+1}$ are rational for some $n$, the the coordinates of $A_n$ are rational as well, since it is a rational function of these (see for example here). Analogously, $B_n$ and $C_n$ have rational coordinates. By backward induction, we see that $A_0,B_0,C_0$ must have rational coordinates. But then the area of $\triangle A_0B_0C_0$ would be rational as well, which is not true. Hence the answer is $\boxed{\textnormal{no}}$.

Thx $ $