How to show the free product of two hyperbolic groups is still a hyperbolic group? I saw from a paper which claimed that this is a easy consequence from the definitions, but I can't give a proof of it just by the definitions. So could you give me some ideas? Thanks!
 A: Hint 1: Let $S_1$ (resp. $S_2$) be a finite generating set of a hyperbolic group $G_1$ (resp. $G_2$). Then $S_1 \cup S_2$ generates $G_1 \ast G_2$. Moreover, given some $g \in G_1 \ast G_2$, every geodesic between $1$ and $g$ can be written as a sequence of elements
$$1, \ s_1^1, \ldots, \ s_1^1 \cdots s_{n_1}^1, \ldots, \ s_1^1 \cdots s_{n_1}^1 \cdots s_1^k \cdots s_{n_k}^k,$$
where, if we set $s^i = s_1^i \cdots s_{n_i}^i$, then $g=s^1 \cdots s^k$ is an alternating word in $G_1$ and $G_2$ and, for every $1 \leq i \leq k$, the path 
$$1, \ s^i_1,  \ldots, \ s^i_1 \cdots s^i_{n_i}$$
defines a geodesic in the Cayley graph of $G_1$ or $G_2$ with respect to the generating set $S_1$ or $S_2$ (depending on whether $s^i \in G_1$ or $s^i \in G_2$).
Hint 2: Let $\Delta$ be a geodesic triangle. Say that $1,g,h$ are the three corners of $\Delta$. Written $g$ and $h$ as alternating words, let $p$ denote their maximal common prefix. Then, using the previous description of the geodesics, notice that $p$ belongs to the three sides of $\Delta$.
