Gauss's lemma

@needham2021visual page 274.

On a general surface, if we launch particles in all directions from a point $p$, travelling a distance $\sigma$ along geodesics, they will form a geodesic circle $K(\sigma )$ with radius $\sigma$ (this is a definition).

Lemma
The geodesic circle $K(\sigma )$ cuts its geodesic every geodesic radius at a right angle.
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Proof
Idea: suppose $p$ is placed along a geodesic ${\gamma }_1$ from $o$ at distance $R$. Consider also the point $q$ at a nearby geodesic ${\gamma }_2$ to the previous one and at the same distance $R$. If the angle were lesser than $\pi /2$ we could look for a point $r$ on ${\gamma }_1$ such that we have a geodesic triangle $rpq$ with the angle at $q$ being $\pi /2$. In this case

so it cannot be $d(o,q)=R$.
By a symmetry argument (exchanging the roles of $p$ and $q$) it can be shown that the angle cannot be greater than $\pi /2$.
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