A (Euclidean) polyhedral space is a metric space obtained by "gluing together" several (let's assume finitely many) Euclidean simplices (of varying dimensions) by identifying some faces via isometries. In other words, a polyhedral space is a simplicial complex where every simplex is equipped with a Euclidean metric and these Euclidean metrics agree on common faces. Such a space is naturally equipped with the quotient metric (i.e. the maximal metric that does not exceed the original Euclidean metric on every simplex); this makes it a geodesic metric space.
Including a simplex in a polyhedral space can make some distances smaller. For example, consider a circle made from two segments of different lengths: the distance between the endpoints of the longer one is smaller in the circle that in the original segment. Let me call a simplex in a polyhedral space convex if this does not happen to it (i.e. the distances induced from the polyhedral space on this simplex are the original Euclidean distances). It is obvious that simplices of the original triangulation can be subdivided into convex ones, isn't it? (For example, one can do sufficiently many barycentric subdivisions.)
Unfortunately, this obvious fact is not that easy to prove, and I was not able to locate any source where a proof can be found. (Although it is asserted without proof in several texts that I checked.) Well, I think I know a working plan of a proof, but it would be so long and full of nasty details... So here are my questions:
1) Does this fact (that every polyhedral space admits a subdivision into convex simplices) have a written proof somewhere (or maybe a slick proof that fits in MO)?
2) Is it true that this can be done by iterated barycentric subdivision?
ADDED. The existence of a desired subdivision follows from the fact that every polyhedral space admits a piecewise-linear path-isometric map to a Euclidean space. Unfortunately this fact does not seem to be written in citeable sources either. And it does not answer Q2 anyway.