For example, let there be an arbitrary inscribed $\triangle A B C$ in circle $O$. Taking the midpoints of $\overparen{A B}$, $\overparen{B C}$, and $\overparen{C A}$ as $C_{1}$, $A_{1}$, and $B_{1}$ respectively, we obtain an inscribed $\triangle A_{1} B_{1} C_{1}$. Then, taking 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, we obtain another inscribed $\triangle A_{2} B_{2} C_{2}$. Thus, when $n$ approaches infinity, what is the trend of the three angles of $\triangle A_{n} B_{n} C_{n}$?
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, \[ A_{n}=\frac{1}{2}\left(\pi-A_{n-1}\right)=-\frac{1}{2}\left(A_{n-1}-\frac{\pi}{3}\right)+\frac{\pi}{3} \]
Iterating sequentially, we get \[ A_{n}=\left(-\frac{1}{2}\right)^{n}\left(A-\frac{\pi}{3}\right)+\frac{\pi}{3} \], so \[ \lim _{n \rightarrow \infty} A_{n}=\frac{\pi}{3} \]
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 close to an equilateral triangle.