Do general Riemannian manifolds satisfy the SAS (side-angle-side) postulate?

Question

By triangle I have in mind something where all sides must be length minimizing geodesics, i.e. geodesic segments. In particular the distances between their endpoints will be the length of the geodesic. Cf. this question.

Question: Assume $M$ is a connected and geodesically complete smooth Riemannian manifold. Then given points $A$, $B$, $C$, $A'$, $B'$, $C'$, and $d(A,B) = d(A',B')$, $d(B,C)=d(B',C')$, and $\angle (A,B,C) = \angle(A',B',C')$, can we conclude that $d(A,C)=d(A',C')$, $\angle(B,C,A)=\angle(B',C',A')$, and $\angle(C,A,B)=\angle(C',A',B')$?

Question 2: Same as the above but assume $M$ is additionally simply connected. (E.g. no tori.)

Because this is probably a basic and well-known result, a pointer to a reference would suffice.

Background/motivation/what I've considered so far: The SAS postulate is true in absolute geometry (i.e. Euclidean or hyperbolic), cf. this question. And apparently it is also true in spherical geometry using the correct notion of "triangle" or "side", i.e. length-minimizing geodesics, cf. this question.

So that covers all of the ($2$-dimensional, but the proofs probably don't depend on the dimension) Riemannian manifolds of constant curvature. But those are a very special class, so it's not clear to me that it would generalize.

There would seem to potentially be topological obstructions in general, i.e. due to the Riemannian manifold being mildly "pathological", hence my restriction to connected and geodesically complete (= complete metric space by Hopf-Rinow theorem) smooth Riemannian manifolds. (To be honest those are the only Riemannian manifolds I am interested in anyway.)

But even for "non-pathological" Riemannian manifolds, it still seems like the most possible obstruction to generalizing a result that holds for those with constant curvature is topological. E.g. with a torus, if the hypothetical missing thirds segment "passes through the donut hole" for $A$ and $C$, but not for $A'$ and $C'$, then I could imagine this failing.

But would that actually be a valid counterexample? And if so, is lack of simple connectedness the only possible obstruction?

A "global" topological obstruction (such as lack of simple connectedness) being the only possible obstruction seems conceivable to the extent that Riemannian manifolds are "locally" or "infinitesimally" Euclidean. But of course all of this is mostly hand-waving and speculation on my part.

Related questions:
Riemannian Triangles Third Side
Bounding angles in Riemannian triangles with bounded sides

Answer 1

This will pretty much never be true outside of a constant curvature situation. For example, you could take $\mathbb{R}^2$ with a metric that is flat on some ball $U$ and has constant curvature $-1$ on some other ball $V$. If you then take $A,B,C\in U$ and $A',B',C'\in V$ which form the same angle and lengths at $B$ and $B'$ (with the entire triangles contained in $U$ and $V$), the other side and angles of the triangles will not be the same (since one is a Euclidean triangle and the other is a hyperbolic triangle, so for instance the angles cannot be the same since the angles of the Euclidean triangle add up to $\pi$ and the angles of the hyperbolic triangle add up to less than $\pi$).

In general, the SAS postulate can be thought of as a sort of homogeneity statement: the reason you would expect it to be true is that there is an isometry taking $B$ to $B'$ and the sides of angle $ABC$ to the sides of angle $A'B'C'$. So it is saying in some sense that your space looks the same at every point and from every direction. When you have some random Riemannian metric that is different everywhere, there's no reason to expect something like this to be true.