\leq 2M$.

Consider the geodesic triangle $v_{\alpha\beta}^{+1}v_{\alpha\gamma}^{-1}v_{\beta\gamma}^{+1}$ in ${\mathcal{C}}(S \backslash v)$. We now have the picture in ${\mathcal{C}}(S \backslash v)$ as in Figure \[2 links are thin\]. By Theorem \[curve hyp\], the inner triangle is $17$ centered, call this center $z$. Combining Theorem \[curve hyp\] and Lemma \[centered to thin\], the outer three triangles are $17*4$-thin. Therefore $z$ is at most $17*5 + 2M = 285$ away from each of the geodesics in the hierarchy triangle $\alpha\beta\gamma$ whose domain is ${\mathcal{C}}(S \backslash v)$.

This all implies that $\alpha\beta\gamma$ is 285-centered at $\{v, z\}$.

**Case 2**: Assume that at least two main geodesics share a common curve, but there is no point that all three main geodesics share the same curve. First assume there is only one such shared curve. Without loss of generality assume that $g_{\alpha\beta}$ and $g_{\alpha \gamma}$ share the curve $v$. Then we can consider a new triangle with the main geodesics forming the triangle $v\beta_1\gamma_1$, see Figure \[Case 2\]. This new triangle has no shared curves so is covered by Case 3.

Now assume there is more than one shared curve between the main geodesics. By definition of a geodesic, for any two main geodesics that share 