I was reading the paper "Planar separators" by Alon, Seymour and Thomas (available on the first author's webpage). They consider a planar triangulation, that is, a maximally planar graph $G$ drawn in the plane so that the boundary of every face is a triangle. They construct a cycle $C$ on vertices $v_0, \ldots , v_{2k-1}$ such that when $D$ denotes the subgraph of $G$ induced by $C$ and the vertices on the inside of $G$ (in the fixed drawing), $C$ is *geodesic* in $D$: For any two vertices of $C$, their distance in $D$ is the same as in $C$. Then they say

There are $k+1$ vertex-disjoint paths of $D$ between $\{ v_0, \ldots, v_k \}$ and $\{ v_k, \ldots, v_{2k-1}, v_0 \}$. For otherwise, by a well-known form of Menger's Theorem for planar triangulations, there is a path of $D$ between $v_0$ and $v_k$ with $\leq k$ vertices.

What's more, they take these paths $P_0, \ldots, P_{k}$ so that $P_i$ joins $v_i$ to $v_{2k-i}$.

Unfortunately, I do not know the mentioned "form of Menger's Theorem for planar triangulations" and do not see how to derive it, given that it appears to include the notion of distances and provides a linkage from each $v_i$ to $v_{2k-i}$. A search on google has also not been successful.

Does anybody know the statement they refer to? Has it appeared somewhere explicitly? If not, can someone indicate how to prove it?