(Dragomir Grozev) 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. $\square$

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. $\square$