ows from Proposition \[qisom\] that $F : X \to Y$ is an isometry. $\diamond$

Proof of main theorem
=====================

Let $X, Y$ be complete, simply connected Riemannian surfaces of pinched negative curvature $-b^2 \leq K \leq -1$, and let $f : \partial X \to \partial Y$ be a Moebius homeomorphism with circumcenter extension $F : X \to Y$. All the tools are now in hand for the proof of the main theorem:

[**Proof of Theorem \[mainthm\]:**]{} As mentioned in the previous section, for any $x \in X$ there exists a probability measure $\mu_x$ on $\partial X$ with support contained in $K_x$ such that $\mu_x$ is balanced at $x \in X$, and $f_* \mu_x$ is balanced at $F(x) \in Y$. As shown in [@biswas6], this is equivalent to the fact that the convex hull in $T_x X$ of the compact $\{ \overrightarrow{x\xi} : \xi \in K_x \}$ contains the origin of $T_x X$ and the convex hull in $T_{F(x)} Y$ of the compact $\{ \overrightarrow{F(x)f(\xi)} : \xi \in K_x \}$ contains the origin of $T_{F(x)} Y$. By the classical Caratheodory theorem on convex hulls, since $X$ is of dimension two this implies that there exists $1 \leq k \leq 3$ and distinct points $\xi_1, \dots, \xi_k \in K_x$ and $\alpha_1, \dots, \alpha_k > 0$ (all depending on $x$) such that $$\alpha_1 \overrightarrow{x\xi_1} + \dots + \alpha_k \overrightarrow{x\xi_k} = 0$$ and $\alpha_1 + \dots + \alpha_k = 1$. Since the vectors $\overrightarrow{x\xi_i}$ are non-zero we must have $2 \leq k \leq 3$. Now if any two of the vectors $\overrightarrow{x\xi_i}, \overrightarrow{x\xi_j}$ for some $i \neq j$ are linearly dependent, then since they are distinct unit norm vectors we must have $\overrightarrow{x\xi_i} = -\overrightarrow{x\xi_j}$, hence $\xi_j = i_x(\xi_i)$. Thus $\xi_i, i_x(\xi_i) \in K_x$, and it follows from Lemma \[antisom\] that $F$ is an isometry and we are done. In particular if $k = 2$ then we are done. Thus we may as well assume that for any $x \in X$, there exist distinct points $\xi_1, \xi_2, \xi_3 \in K_x$ and $\alpha_1, \alpha_2, \alpha_3 > 0$ (all depen