$g_{\alpha\beta}$, $g_{\beta\gamma}$, and $g_{\alpha\gamma}$ be the three main geodesics that make up the triangle (which connects $\alpha_0$, $\beta_0$, and $\gamma_0$). As before in Theorem \[hierarchy k-centered\], there are three cases:

1.  All three main geodesics have a curve in common.

2.  Any two of the main geodesics share a curve, but not the third.

3.  None of the main geodesics have common curves.

For the rest of the proof, note that if $v \in {\mathcal{C}}(S)$ is a non-domain separating curve, then $S \backslash v$ has one connected component with positive complexity, so by abuse of notation, we denote this component as $S \backslash v$. This means that every curve in ${\mathcal{C}}(S)$ not equal to $v$ intersects $S \backslash v$ so we can use the Bounded Geodesic Image Theorem on any geodesic that doesn’t contain $v$. Take two non-domain separating curve $v,w \in {\mathcal{C}}(S)$ such that $v$ and $w$ are disjoint. Then, because $\xi(S) = 3$, $S \backslash (v \cup w)$ has one connected component with positive complexity, and again we denote this component as $S \backslash (v \cup w)$. Furthermore, every curve in ${\mathcal{C}}(S)$ not equal to $v$ or $w$ intersects $S \backslash (v \cup w)$, so we may use 