In bary we have $A_n=\frac{2}{3}C_{n-1}+\frac{1}{3}B_{n-1}$. Hence the centroids of all $A_nB_nC_n$ are equal (and equal to $(1/3,1/3,1/3)$) and rewriting as $A_n=\frac{1}{3}(A_{n-1}+B_{n-1}+C_{n-1})+\frac{1}{3}(C_{n-1}-A_{n-1})$ implies that, for each component separately, the largest absolute difference between a point's coordinate and $\frac{1}{3}$ decreases by a factor of at least $\frac{1}{3}$ when $n$ increases by $1$. Hence as $n \to \infty $, $A_n,B_n,C_n$ all go to the centroid of $A_0B_0C_0$, which is the centroid of all the $A_nB_nC_n$ as well and hence lies inside all of them. On the other hand, since $A_n,B_n,C_n$ all get arbitrarily close to the centroid, no other points can be contained in all $A_nB_nC_n$.

We have nested compact sets (closed triangles), so they have non empty intersection. We prove that they intersect in only one point. It's enough to prove that the three points $A_n, B_n, C_n$ converge to a common point $P$. Assume it's false. Then there exists some subsequences of $A_n,B_n,C_n$ (which for simplicity we denote again by $A_n, B_n, C_n$) that converge to points $A,B,C$ respectively and $\{A,B,C\}$ consists of at least 2 elements. Assume, first $A,B,C$ are distinct. Assume wlog that $\angle BAC\le 60^{\circ}$. Take $n$ large enough such that $A_n, B_n, C_n$ are close enough to $A,B,C$ respectively. Consider the next triangle $A_{n+1}B_{n+1}C_{n+1}$. Its side $B_{n+1}C_{n+1}$ is far enough from $A$ and so $A$ is outside $\triangle A_{n+1}B_{n+1}C_{n+1}$. But $A$ was a limit point of the sequence $A_n,n=1,2,\dots$, contradiction. Suppose now, $B=C\ne A$. Then $\angle B_nA_nC_n<60^{\circ}$ (it tends to 0 actually) and we apply the same argument. We proved that there is a unique point $P$ that's common for all the triangles. Now it remains to prove a small trifle - namely $P$ is in the interior of all the triangles. It was part of the Bulgarian text. Assume on the contrary $P$ is on some side, say $A_nB_n$, for some $n$. But it easily follows that $P$ is outside $\triangle A_{n+2}B_{n+2}C_{n+2}$ contradiction. That's it. Observe that the proof remains the same if we only require that $A_{n+1}, B_{n+1}, C_{n+1}$ are on the sides $B_nC_n, A_nC_n$ and $A_nB_n$ respectively, but not too close to the vertices $A_n,B_n,C_n$, for example the distance from $A_{n+1}$ to both points $B_n, C_n$ is greater than $\varepsilon\cdot |B_nC_n|$ for some fixed $\varepsilon>0$, and the same for $B_{n+1}$ and $C_{n+1}$ .