Let $ABC$ be a non-degenerate triangle in the euclidean plane. Define a sequence $(C_n)_{n=0}^\infty$ of points as follows: $C_0:=C$, and $C_{n+1}$ is the incenter of the triangle $ABC_n$. Find $\lim_{n\to\infty}C_n$.
1. **Define the sequence and initial point:**
   Let $ABC$ be a non-degenerate triangle in the Euclidean plane. Define a sequence $(C_n)_{n=0}^\infty$ of points as follows: $C_0 := C$, and $C_{n+1}$ is the incenter of the triangle $ABC_n$.

2. **Show that the sequence is Cauchy:**
   To show that $(C_n)$ is a Cauchy sequence, we need to show that for every $\epsilon > 0$, there exists an $N$ such that for all $m, n > N$, $|C_m - C_n| < \epsilon$. This involves showing that the distance between consecutive points $C_n$ and $C_{n+1}$ gets arbitrarily small as $n$ increases. This step is omitted in the provided solution but is crucial for rigor.

3. **Angles of triangle $ABC_n$:**
   The angles of triangle $ABC_n$ are given by:
   \[
   \angle BAC_n = \frac{A}{2^n}, \quad \angle ABC_n = \frac{B}{2^n}, \quad \angle ACB_n = 180^\circ - \frac{A + B}{2^n}
   \]

4. **Law of sines in triangle $ABC_n$:**
   By the Law of Sines in triangle $ABC_n$, we have:
   \[
   \frac{BC_n}{\sin(\angle BAC_n)} = \frac{AC_n}{\sin(\angle ABC_n)}
   \]
   We are interested in the limit:
   \[
   \lim_{n \to \infty} \frac{\sin(A/2^n)}{\sin(B/2^n)}
   \]

5. **Simplify the limit using small angle approximation:**
   For small angles, $\sin(x) \approx x$. Therefore:
   \[
   \lim_{n \to \infty} \frac{\sin(A/2^n)}{\sin(B/2^n)} = \lim_{n \to \infty} \frac{A/2^n}{B/2^n} = \frac{A}{B}
   \]

6. **Determine the limit point:**
   The limit point $P$ on $BC$ satisfies $BP/PA = \angle B / \angle A$. This means that $P$ divides $BC$ in the ratio $A:B$.

The final answer is $\boxed{P}$, the point on $BC$ such that $BP/PA = A/B$.