preliminaries on Moebius maps and the circumcenter extension, and then in section 3 we prove the main theorem.

Preliminaries
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For details and proofs of the assertions made in this section we refer to [@biswas3], [@biswas5], [@biswas6], [@biswas7].

Moebius metrics and visual metrics
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Let $(Z, \rho_0)$ be a compact metric space of diameter one. For a metric $\rho$ on $Z$, the cross-ratio with respect to the metric $\rho$ is the function of quadruples of distinct points in $Z$ defined by $$[\xi, \xi', \eta, \eta']_{\rho} := \frac{\rho(\xi, \eta)\rho(\xi', \eta')}{\rho(\xi, \eta')\rho(\xi', \eta)}$$ A metric $\rho$ on $Z$ is said to be antipodal if it has diameter one and for any $\xi \in Z$ there exists $\eta \in Z$ such that $\rho(\xi, \eta) = 1$. We assume that the metric $\rho_0$ is antipodal. We say that two metrics $\rho_1, \rho_2$ on $Z$ are Moebius equivalent if for all quadruples of distinct points in $Z$, the cross-ratios with respect to the two metrics are equal. We let $\mathcal{M}(Z, \rho_0)$ denote the set of all antipodal metrics on $Z$ which are Moebius equivalent to $\rho_0$. For any $\rho_1, \rho_2 \in \mathcal{M}(Z, \rho_0)$, there exists a positive continuous function on $Z$ called the derivative of $\rho_2$ with respect to $\rho_1$, denoted by $\frac{d\rho_2}{d\rho_1}$, such that $$\rho_2(\xi, \eta)^2 = \frac{d\rho_2}{d\rho_1}(\xi)\frac{d\rho_2}{d\rho_1}(\eta) \rho_1(\xi,\eta)^2$$ for all $\xi, \eta \in Z$, and such that $$\frac{d\rho_2}{d\rho_1}(\xi) = \lim_{\eta \to \xi} \frac{\rho_2(\xi, \eta)}{\rho_1(\xi, \eta)}$$ for all non-isolated points $\xi$ of $Z$. Moreover, $$\left( \max_{\xi \in Z} \frac{d\rho_2}{d\rho_1}(\xi) \right) \cdot \left( \min_{\xi \in Z} \frac{d\rho_2}{d\rho_1}(\xi) \right) = 1$$ The set $\mathcal{M}(Z, \rho_0)$ admits a natural metric defined by $$d_{\mathcal{M}}(\rho_1, \rho_2) = \sup_{\xi \in Z} \log \frac{d\rho_2}{d\rho_1}(\xi)$$ The metric space $(\mathcal{M}(Z, \rho_0), d_{\mathcal{M}})$ is proper.

Let $X$ be a prope