# Convex Hull = Boundary + Segments (follow up)

My question is related to this question, but even less general. Suppose $$B \subseteq \partial A$$ is a subset of the boundary of a compact convex set $$A \subseteq \mathbb{R}^n$$ such that the property mentioned in the question holds for the convex hull of $$B$$, $$conv B$$. i.e., suppose the boundary of $$convB$$ consists only of $$B$$ and segments that join two points from $$B$$. Can we, in that case, say that all points in $$convB$$ are convex combination of at most two points from $$B$$?

We can even assume $$B$$ is a connected subset of $$\partial A$$ if that makes this property hold.

For context, these are all ways I'm exploring to solve my original question here.

As always, I'm most open to counterexamples which show this is false.

Let $$A$$ be a tetrahedron (in $$\Bbb R^3$$) and $$B$$ its 1-skeleton, i.e., the "wireframe" made of $$A$$'s edges and vertices. This males $$\operatorname{conv}(B)=A$$ and as every point on a triangular face of $$\partial A$$ is on a line segment between two points on its boundary, $$B$$ has the required property (and is conneted). However, every line segment between points in $$B$$ lies in such a triangular face, i.e., interior points of $$A$$ are not convex combinations of only two points of $$B$$.