Let two mutually disjoint geodesics $L_1$ and $L_2$ in $\Bbb H$ be given, where $\Bbb H$ is an upper half plane in $\Bbb C$ with hyperbolic metric $\rho$. Then $\rho(z,w)$ is strictly convex on $\{(z,w)\in\Bbb C^2\mid z\in L_1,w\in L_2\}$. Equivalently, for every $z,z'\in L_1$ and every $w,w'\in L_2$, the inequality $$\rho(z_0,w_0)\leq{\rho(z',w')+\rho(z,w)\over 2}$$ holds, where $z_0$ and $\omega_0$ are the middle-points of $z$ and $z'$ and of $w$ and $w'$ (w.r.t the hyperbolic distance), respectively.
Proof. It suffices to show the inequality when $z\neq z'$. Let $L$ be the geodesic on $\Bbb H$ passing through $z_0$ and $w_0$. Let $\hat{z}',\hat{z}$ and $\hat{z}_0$ be points symmetric to $z,z'$ and $z_0$ respectively, with respect to $w_0$ (in the sense of the hyperbolic geometry). (The following figure replaces $\Bbb H$ by the unit disk and assume $w_0 =0$).
By the definitions, $\rho(z,w) = \rho(\hat{z}',w')$, and it is clear that
$$\rho(z',\hat{z}')\leq\rho(z',w')+\rho(w',\hat{z}').$$
On the other hand, let $\gamma_1$ and $\gamma_2$ be the elliptic elements of $\mathrm{Aut}(\Bbb H)$ of order two which have $z_0$ and $w_0$, respectively, as fixed points. $\color{red}{\text{Then the composition $\gamma_2\circ\gamma_1$ is a hyperbolic element with the axis $L$, and sends $z_0$ and $z'$ to $\hat{z}_0$ to $\hat{z}'$, respectively}}$. $\color{red}{\text{Hence we have $2\rho(z_0,w_0) = \rho(z_0,\hat{z}_0)<\rho(z',\hat{z}')$.}}$
The above red statements don't make sense to me. Especially, in the first statement, it says the composition of elliptic elements is hyperbolic. If $\gamma_1,\gamma_2$ are hyperbolic elements, we can conjugate in $\mathrm{Aut}(\hat{\Bbb C})$ so that $\hat{\gamma}_1:z\mapsto e^{i\theta}z$ and $\hat{\gamma}_2:z\mapsto e^{i\eta}z$ for some $\theta,\eta\in\Bbb R$. The square of the trace of $\gamma_2\circ\gamma_1$ can't exceed $4$ so can't be hyperbolic element. Even if the first red statement is true, I can't understand why it implies the second statement.