This question is edited following the comment of Joseph. He pointed out that the main object of the first version of this question is the cut locus.

Recall that the cut locus of a set $S$ in a geodesic space $X$ is the closure of the set of all points $p \in X$ that have two or more distinct shortest paths in $X$ from $S$ to $p$. http://en.wikipedia.org/wiki/Cut_locus

A simple lemma shows that, for a disk $D^2$ with a Riemannian metric, the cut locus of $D^2$ with respect to its boundary is a tree. A picture of such tree can be found on page 542, figure 17 of the article of Thurston "Shapes of polyhedra". The tree is white. http://arxiv.org/PS_cache/math/pdf/9801/9801088v2.pdf For an ellipse on the 2-plane, the tree is the segment that joins its focal points.

More generically for a Riemannian manifold $M^n$ with boundary, the cut locus of $\partial M$ should be a deformation retract of $M$. (I guess it is a $CW$ complex of dimension less than $n$.) To prove this lemma, notice that $M^n\setminus\operatorname{cut-locus}(\partial M^n)$ is canonically foliated by geodesic segments that join $X$ with $\partial M$.

I wonder if this lemma has a name or maybe is contained in some textbook on Riemannian geometry?

cut locusof the boundary of the disk? – Joseph O'Rourke Jul 14 '10 at 1:36medial axis. See the figure at the Wikipedia entry: en.wikipedia.org/wiki/Medial_axis . – Joseph O'Rourke Jul 14 '10 at 10:13