Let $\mathbb H^2$ be the hyperbolic space and consider a hyperbolic triangle $(v_1,v_2,v_3)$ and a point $v \in \mathbb H^2$ as showed in the figure : 
Let $\mu_i$ is the hyperbolic distance from $v$ to $v_i$ and $\varepsilon_i$ is the hyperbolic distance from $q_i$ to $v$.
Proposition : Let $v$ within $T$. For each $i=1,2,3$, let $G_{i}$ be the geodesic passing through the points $v$ and $v_{i}$, and let $q_{i}$ denote the intersection of $G_{i}$ with the edge of $T$ opposite to $v_{i}$ . Then the hyperbolic barycentric coordinates of $v$ are: $$ \phi_{i}=\frac{\sinh \varepsilon_{i}}{\sinh \left(\varepsilon_{i}+\mu_{i}\right)}. $$
In the proof, the author cited that it suffices to consider the case $i=1 $ : $$v=\frac{\sinh \varepsilon_{1}}{\sinh \left(\varepsilon_{1}+\mu_{1}\right)} v_{1}+\frac{\sinh \mu_{1}}{\sinh \left(\varepsilon_{1}+\mu_{1}\right)} q_{1}$$.
Edit : I think the author used the uniqueness of barycentric coordinates and this remark : On the geodesic boundary $\left(v_{1}, v_{2}\right)$, for example, we have, for $v \in\left(v_{1}, v_{2}\right)$, $$ v=\frac{\sinh \left(c-d_{\mathbb{H}^{2}}\left(v_{2}, v\right)\right)}{\sinh (c)} v_{1}+\frac{\sinh \left(d_{\mathbb{H}^{2}}\left(v_{1}, v\right)\right)}{\sinh (c)} v_{2}, $$ where $c$ is the geodesic edge lengths between $v_{1}$ and $v_{2}$. The problem remains now to show that : $$\sum_{i=1}^3 \frac{\sinh \varepsilon_{i}}{\sinh \left(\varepsilon_{i}+\mu_{i}\right)}=1.$$