Do Equidistant points form a Geodesic in the Torus? Consider the Torus $T \subseteq \Bbb R ^3$ seen as a revolution surface, around the $y$ axis, of a circle in the $xy$ plane given by the equation $(x-R)^2+y^2=r^2$, with $R>r>0$. We name $d_l$ the intrinsic metric of the torus, a.k.a. the geodesic distance. Given two points $p,q \in T$, I want to know if the set of equidistant points
$$\{v \in T : d_l(v,p)=d_l(v,q) \}$$
contains the image of a geodesic curve. I tried to make some calculations, but things got messy pretty fast. Acording to this, let $F:[0,2\pi]\times[0,2\pi] \rightarrow T$ be the map
$$F(\phi,\theta)=(\ (R+r \cdot \cos\theta)\cdot \cos\phi\ , \ (R+r \cdot \cos \theta)\cdot \sin\phi \ , \ r \cdot \sin\theta\ )$$
If $h$ is a real number with $|h|\leq R+r$ and $\sigma(t)=F(\mu(t),t)$ is a curve where
$$\mu '(t)= \frac{r\cdot h}{(R+r\cdot \cos t)\cdot \sqrt{(R+r\cdot \cos t)^2-h^2}}$$
Then the reparametrization of $\sigma$ by arc leangth is a geodesic. Moreover, every geodesic can be obtained this way (at least locally). With this at hand, I tried to calculate the distance between two points in the torus, say $F(0,0)$ and $F(\phi_0, \theta_0)$. If $\sigma$ is as above, then
$$\Vert \sigma '(t)\Vert = \frac{r\cdot (R+r\cdot\cos t)}{\sqrt{(R+r \cdot \cos t)^2-h^2}}$$
Where $h$ is...some smart number that depends on $\phi_0,\theta_0$ (and I don't know how to calculate). Anyway, I coludn't get much further. This expression probably can't be integrated and I don't know how to go on.
 A: Here is a class of counter-examples:
Suppose that $M$ is a (not necessarily complete) connected Riemannian surface such that any two points of the universal cover of $M$ can be connected by a unique geodesic. Then the following are equivalent:
(a) the bisector of every geodesic segment in $M$ contains a (nondegenerate) geodesic subsegment.
(b) $M$ has constant curvature.
For instance, this applies if $M$ is a complete Riemannian manifold of curvature $\le 0$.
This can be derived from Theorems 46.1 and 47.4 in
Busemann, Herbert, The geometry of geodesics, Mineola, NY: Dover Publications (ISBN 0-486-44237-3). x, 422 p. (2005). ZBL1141.53001.
In particular:
Suppose that $M$ is an arbitrary Riemannian surface and $m\in M$ any  point. Let $U$ be a convex neighborhood of $m$ in $M$ (it always exists). Then $U$ has constant curvature if and only if for every two points $p, q\in U$ the bisector of $pq$ contains a nondegenerate subsegment contained in $U$.
In other words, the statement you are looking for always fails locally except for metrics of constant curvature.
Actually, Busemann works in a much greater generality of G-spaces (which are metric spaces satisfying certain conditions).

The following is a reasonable guess in view of Busemann's results (it's probably known to be true):
Suppose that $M$ is a connected Riemannian surface. Then the following are equivalent:
(a) the bisector of every geodesic segment in $M$ contains a (nondegenerate) geodesic subsegment.
(b) $M$ has constant curvature.
The direction (b)$\Rightarrow$(a) is well-known and easy to prove. Even more: Under the constant curvature assumption, the bisector is a geodesic graph in $M$.
