Let $M$ be an $n$-dimensional Riemannian manifold, $n \ge 3$.
Let $T$ be a geodesic triangle in $M$. Does there exist a (smooth) $2$-dimensional embedded submanifold $S \subset M$ such that $T \subseteq S$?
I am OK with $S$ being a manifold with boundary or corners.
I thought of taking $S$ to be something like the convex hull of $T$, but this concept is a bit tricky for general Riemannian manifolds...
Edit:
The local case where $T$ is contained in some convex neighborhood of a point $p$ is sufficiently interesting for my purposes.
However, even when $p$ is one of the vertices, the following naive attempt seems suspicious:
Let $\triangle pqr$, suppose $q=\exp_p(v), r=\exp_p(w)$.
Then the image of $(t,s) \mapsto \exp_p\big(t(sv+(1-s)w)\big)$, where $0 \le t,s \le 1$ is a surface. But it is not clear to me whether the edge $qr$ is contained in this image. (I think this is not always the case, e.g. when $M$ is a sphere. In particular $s \mapsto \exp_p(sv+(1-s)w)$ does not coincide with $qr$).