For $m\ge 3$ and a triangulation $T$ of the $m$-gon there is a unique $k$ with $1<k<m$ such that $\{1,k\}$ and $\{k,m\}$ are both chords of $T$ or sides, and $T$ splits along $k$

Let $k$ be the least element of $\{2,\dots ,m-1\}$ such that $\{k,m\}$ is a diagonal of $T$ or the side $\{m-1,m\}$. This set is nonempty because $m-1$ belongs to it.

The chord $\{1,k\}$ is a side or lies in $T$. If it were a diagonal outside $T$, maximality would give a diagonal $\{p,q\}\in T$ crossing it, so $1<p<k<q$; if $q<m$ then $\{p,q\}$ would cross $\{k,m\}$, impossible, and if $q=m$ then $p<k$ would contradict the minimality of $k$.

Every diagonal of $T$ has both endpoints in $\{1,\dots ,k\}$ or both endpoints in $\{k,\dots ,m\}$. Indeed, if $\{p,q\}\in T$ had $p<k<q<m$, then it would cross $\{k,m\}$; and if $p=1<k<q<m$, then $q$ would satisfy the defining property of step 1.1 with $q<k$, again impossible.

The diagonals of $T$ with endpoints in $\{1,\dots ,k\}$ form a triangulation of the $k$-gon, and those with endpoints in $\{k,\dots ,m\}$ form a triangulation of the $(m-k+1)$-gon: they are pairwise non-crossing because they are diagonals of $T$, and they are maximal because any extra diagonal in one sub-polygon would also be a diagonal of the whole polygon and would not cross any member of $T$ by step 3.1. The index $k$ is unique, for if another index $\mathrm{\ell }$ with $1<\mathrm{\ell }<m$ had the same property, then the chords $\{1,k\}$ and $\{\mathrm{\ell },m\}$ would cross when $k<\mathrm{\ell }$, or the symmetric crossing would occur when $\mathrm{\ell }<k$.