some $t\in (0,1)$, the point $q_{t}=(1-{t}) \cdot q_0 + {t} \cdot q_1$ is a point of the segment $\overline{u_{t}u'_{t}}$ and also a point of the segment $\overline{v_{t} v'_{t}}$. Since every node of $T$ only moves horizontally in $\langle \Gamma_0,\Gamma_1 \rangle$, we have that $q_0$ ($q_1$) lies in the strip delimited by the horizontal lines through $v_0$ and $v'_0$ (resp. through $v_1$ and $v'_1$). Since $\Gamma_0$ and $\Gamma_1$ are planar, $q_0$ and $q_1$ do not belong to the segments $\overline{v_0v'_0}$ and $\overline{v_1v'_1}$, respectively. It follows that $q_0$ is either to the left or to the right of the line $\ell_0^v$ that passes through $v_0$ and $v'_0$; analogously, $q_1$ is either to the left or to the right of the line $\ell_1^v$ that passes through $v_1$ and $v'_1$. Assume that $q_0$ is to the left of $\ell_0^v$; we claim that this implies that $q_1$ is to the left of $\ell_1^v$. The claim follows from the fact that, both in $\Gamma_0$ and in $\Gamma_1$, the two paths of $T$ from $u$ and $v$ to their lowest common ancestor $w$ are monotone in the $y$-direction (since $\Gamma_0$ and $\Gamma_1$ are strictly-upward), do not cross each other (since $\Gamma_0$ and $\Gamma_1$ are planar), and enter 