Shortest paths between points on the boundary of a  convex set Is this statement true, and, if so, does it have a name?

Given a bounded closed convex set, $C\subset \mathbb R^n$, let
  $C^{int}$ be the interior of $C$ and $C^{bd}=C-C^{int}$ be the
  boundary of $C$.  Given two points $x,y\in C^{bd}$, any "shortest"
  path between $x,y$ in $\mathbb R^n-C^{int}$ is entirely inside
  $C^{bd}$.

 A: I am not aware of any name for this property. However, it can be proved rather quickly. I won't deal with the existence of geodesic (some compactness argument may work, but it is not obvious to me which one), and only prove that any geodesic in $X := \mathbb{R}^n-C^{int}$ between points of $C^{bd}$ must lie in $C^{bd}$.
Let $x$, $y$ be in $C^{bd}$. Assume that there exists a geodesic $\gamma : [0,1] \to X$ between $x$ and $y$. If this geodesic does not lie in $C^{bd}$, then there must be some point $z = \gamma (t_0) \in \gamma ([0,1])$ which does not belong to $C^{bd}$.
By the Hahn-Banach theorem, I can find an hyperplane $H$ which strictly separates $C$ and $z$. By the intermediate value theorem, there must be some times $t_1$, $t_2$ with $0 < t_1 < t_0 < t_2 < 1$ and such that $\gamma (t_1)$ and $\gamma (t_2)$ both belong to $H$. The geodesic $\gamma$ restricted to $[t_1, t_2]$ must also be a geodesic.
The shortest path between two points in $H$ is a line. So, $z$ cannot belong to the geodesic between $\gamma (t_1)$ and $\gamma (t_2)$. We get a contradiction. Hence, the geodesic between $x$ and $y$ must lie in $C^{bd}$.
