Example 6 Suppose in circle $O$ there is an arbitrary inscribed $\triangle A B C$. Take the midpoints of $\overparen{A B}$, $\overparen{B C}$, and $\overparen{C A}$ as $C_{1}$, $A_{1}$, and $B_{1}$ respectively, to form an inscribed $\triangle A_{1} B_{1} C_{1}$. Then take the midpoints of $\overparen{A_{1} B_{1}}$, $\overparen{B_{1} C_{1}}$, and $\overparen{C_{1} A_{1}}$ as $C_{2}$, $A_{2}$, and $B_{2}$ respectively, to form another inscribed $\triangle A_{2} B_{2} C_{2}$. What is the trend of the three angles of $\triangle A_{n} B_{n} C_{n}$ as $n$ approaches infinity?
This is a ternary sequence problem. As shown in Figure 13-2, it is evident that
$$
A_{1}=\frac{1}{2}(B+C)=\frac{1}{2}(\pi-A)=-\frac{1}{2}\left(A-\frac{\pi}{3}\right)+\frac{\pi}{3}
$$

Similarly, $\quad A_{n}=\frac{1}{2}\left(\pi-A_{n-1}\right)=-\frac{1}{2}\left(A_{n-1}-\frac{\pi}{3}\right)+\frac{\pi}{3}$
Iterating sequentially, we get $\quad A_{n}=\left(-\frac{1}{2}\right)^{n}\left(A-\frac{\pi}{3}\right)+\frac{\pi}{3}$, so $\lim _{n \rightarrow} A_{n}=\frac{\pi}{3}$
Similarly, $\lim _{n \rightarrow \infty} B_{n}=\lim _{n \rightarrow \infty} C_{n}=\frac{\pi}{3}$
Therefore, when $n$ tends to infinity, the three interior angles of $\triangle A_{n} B_{n} C_{n}$ all tend to $\frac{\pi}{3}$. That is, $\triangle A_{n} B_{n} C_{n}$ becomes increasingly similar to an equilateral triangle.