For a javascript rendered 3D model of the path of configurations reported by Ogievetsky and Shlosman in arXiv:1805.09833 see this page on my website.
The configurations in this path are composed of two "three-pole teepees" pointed in antipodal directions. This is also true of the Figure 3 configuration in the question. But a crucial difference is that in Ogievetsky and Shlosman's configurations, the two teepees are both right-handed. In Figure 3, the two teepees have different handedness.
You can set up the calculation of OS in the case of opposite-handed teepees. Just switch $\delta$ to $-\delta$ for $D$, $E$, and $F$ in their Equation (5). Now solving for $d_{AB}=d_{AD}=d_{BD}$ is even easier. From $d_{AD}=d_{BD}$ I get $\kappa=0$, and from $d_{AB}=d_{AD}$, I get $$s+\frac{3(s-1)}{t+4-3s}+\frac{s-s^2}{s+t}=0\text,$$ where $s=(\sin\phi)^2$ and $t=(\tan\delta)^2$.
There are two real solutions for $t$ when $0\le s\le \tfrac13$, and then the solutions become complex for $\tfrac13<s\le1$. All the real solutions give $d_{AB}=d_{AD}=d_{BD}\le1$, with equality only for the double solution $s=\tfrac13$, $t=1$, which is Figure 3.