I'm working through this paper and I'm hung up on Proposition $3.1$. To strip away the context of the problem and present it in another light: suppose there are two intersecting arcs $ab$ and $cd$ on the unit sphere of equal length $d$, then one of the arcs $ac$, $ad$, $bc$, $bd$ has length less than $d$. I'm not very keen on spherical geometry so this is a bit inaccessible to me.
I tried doing a (spherical) law of cosines argument but couldn't get anywhere. I had four equations but they didn't seem easy to manipulate. I also considered the following proposition (which I don't know to be true or not): suppose $d < ac$, then the angle opposite $ac$ must be greater than $\frac{\pi}{2}$. If this is the case, then the inner angles would all add up to something greater than $2\pi$ which would be a contradiction. This would say that one of them has to be $\frac{\pi}{2}$ or smaller, meaning that $ac \le d$ which (I assume) would nearly prove the result.
Thanks for any assistance!
With Will's help, I've put together a solution. It is not very brief but here goes. All arcs here are made by great circles.
By the spherical law of cosines, if $abc$ makes a triangle then
$$\begin{alignat}{3} & \cos(a) &&= \cos(b)\cos(c)+\sin(b)\sin(c)\cos(A)\\ \cos(b)\cos(c)-\sin(b)\sin(c)\le & &&\le \cos(b)\cos(c)+\sin(b)\sin(c) \\ \cos(b+c) = & &&= \cos(b-c)\end{alignat}$$
Hence $b-c\le a \le b+c$. Consider the diagram below.
Suppose arcs $\stackrel{\frown}{ab}$ and $\stackrel{\frown}{cd}$ have length $D$. Let the arc $\stackrel{\frown}{Od}$ have length $\eta$ and $\stackrel{\frown}{Ob}$ have length $\xi$. Then by the triangle inequality,
$$\begin{align}\lvert\stackrel{\frown}{bd}\rvert&\le \xi+\eta \\ \lvert\stackrel{\frown}{ad}\rvert&\le D-\xi+\eta \\ \lvert\stackrel{\frown}{ac}\rvert&\le D-\xi+D-\eta \\ \lvert\stackrel{\frown}{bc}\rvert&\le \xi+D-\eta\end{align}$$
Adding the first and third inequalities gives $\lvert\stackrel{\frown}{bd}\rvert+\lvert\stackrel{\frown}{ac}\rvert\le 2D$. So either one of $\lvert\stackrel{\frown}{bd}\rvert$ or $\lvert\stackrel{\frown}{ac}\rvert < D$ or $\lvert\stackrel{\frown}{bd}\rvert = D = \lvert\stackrel{\frown}{ac}\rvert$.
By symmetry, the same inequalities hold for the other two arcs. Suppose by contradiction that all of the outside arcs have length $D$, then by the triangle condition above we have that $\lvert\stackrel{\frown}{bc}\rvert-\lvert\stackrel{\frown}{Oc}\rvert\le \xi$ but since $\lvert\stackrel{\frown}{bc}\rvert = D$ and $\lvert\stackrel{\frown}{Oc}\rvert = D-\eta$, we see that $\eta\le\xi$. By symmetry, we also have $\xi\le\eta$ and so $\xi=\eta$.
By the first inequality above, $D-\eta\le\xi$. But since $\eta=\xi$, we see that
$$D-\xi\le\xi\le2D-\xi$$
and hence $\dfrac{D}{2}\le\xi$ and so by symmetry (with $D-\xi$), we see that $\xi=\dfrac{D}{2}$. Since all sides of the triangles in the drawing are equal, the angles $\angle dOa=\angle doB = \angle bOc = \angle cOa = \dfrac{\pi}{2}$ and so the triangles are right triangles. By the spherical Pythagorean theorem (which follows directly from the spherical law of cosines), we have that
$$\cos(\lvert\stackrel{\frown}{ad}\rvert) = \cos(\lvert\stackrel{\frown}{Od}\rvert)\cos(\lvert\stackrel{\frown}{Oa}\rvert).$$
However since $\lvert\stackrel{\frown}{ad}\rvert = D$ and $\lvert\stackrel{\frown}{Od}\rvert = \dfrac{D}{2} = \lvert\stackrel{\frown}{Oa}\rvert$, we have that
$$\cos\left(2\frac{D}{2}\right) = \cos^2\left(\frac{D}{2}\right).$$
Since $\cos(2x) = 2\cos^2x-1$, we see that $\cos^2\left(\dfrac{D}{2}\right) = 2\cos^2\left(\dfrac{D}{2}\right)-1$. This implies that $\frac{D}{2} = 0,2\pi$ but neither of these are admissible since we required that it be a triangle. Hence we conclude that at least one of the outside arcs has length less than $D$.