of $A$ until an opposite vertex is reached) and let $d_n$ be the vertex of $A_n$ (for $n \in \operatorname{\mathbf{N}}$) such that $d_n \to d$. Denote also by $\tilde{d}$ the limit of $(\varphi_n d_n)$. By Proposition \[proposition:opposite\], $c_n$ is opposite $d_n$ for sufficiently large $n$. We can also assume by passing to a subsequence that $(\varphi_n c_n)$ converges to some $\tilde{c}$. If $\tilde{c}$ is opposite $\tilde{d}$, then we can immediately complete the proof using Lemma \[lemma:opposite\].

![Illustration of Proposition \[proposition:centerok\].[]{data-label="picture:centerok"}](figure3.eps)

If, on the contrary, $\tilde{c}$ is not opposite $\tilde{d}$, then we cannot proceed in this way. In this case, pick a sequence $e_n \to e$ with $e_n$ (resp. $e$) adjacent to $d_n$ (resp. $d$) but not contained in $A_n$ (resp. $A$). It is easy to show that such a sequence exists, using for example Proposition \[proposition:projections\]. Since $e$ is almost opposite $c$, $e_n$ is almost opposite $c_n$ for sufficiently large $n$ (Proposition \[proposition:almost-opposite\]). For these $n$, there is a gallery from $d_n$ to $c_n$ of length $m = \operatorname{\mathrm{diam}}\Delta$ passing through $e_n$. Let $d_n = e_0^{(n)}, e_1^{(n)} = e_n, \ldots, e_{m-1}^{(n)}, e_m^{(n)} = c_n$ be the vertices of this gallery, as illustrated in Figure \[picture:centerok\]. After passing to subsequences, we 