Geodesic quadrangles in CAT($0$) spaces

Question

I am trying to show that any geodesic quadrangle $Q$ in any CAT($0$) space $X$ has a comparison quadrangle in $\mathbb{R}^2$ (same definition as for triangles). One can split $Q$ in two triangles $T_1$ and $T_2$ and consider the quadrangle $\overline{Q} \subseteq \mathbb{R}^2$ obtained gluing the comparison triangles $\overline{T_1}$ and $\overline{T_2}$ along the "diagonal", which we call $\overline{d}$ (note that, in general, we would not obtain the same quadrangle $\overline{Q}$ if we had chosen the other "diagonal" in $Q$). Pick $p,q \in Q$, wlog one in $T_1$ and one in $T_2$. If the straight line segment between $\overline{p}$ and $\overline{q}$ crosses $\overline{d}$, then we call the intersection point $\overline{g}$, its "preimage" $g$, and we obtain the inequality $$d_X(p,q) \le d_X(p,g) + d_X(g,q) \le d_{\mathbb{R}^2}(\overline{p},\overline{g}) + d_{\mathbb{R}^2}(\overline{g},\overline{q}) = d_{\mathbb{R}^2}(\overline{p},\overline{q}),$$ and everything is fine. If, however, $\overline{Q}$ is not so nice and we do not get the intersection point $\overline{g}$, then the above reasoning does not work. I would appreciate any hint on how to adapt the proof for the general case.