Triangulation of a simple polygon (elementary proof?)

Let $C$ be a simple (Jordan) polygon in the Euclidean plane. I would like to prove the existence of a triangulation of $C$.

This seems possible if we assume the Jordan curve theorem. Can we prove it directly without using the theorem? Even better, is there an elementary proof?

Here's an elementary proof, but it will simultaneously prove the triangulation theorem and the Jordan curve theorem for simple (non-self-intersecting) polygons by concurrent induction. $\def\nn{\mathbb{N}}$ $\def\rr{\mathbb{R}}$ $\def\tri{\triangle}$ $\def\seg#1{\,\overline{\!#1\!}\,}$ $\def\less{\smallsetminus}$

Notation

Let "$int(\seg{AB})$" denote the interior of the segment $\seg{AB}$, that is without its endpoints.

Let "$int(C)$" to denote the interior of the simple closed curve $C$ when it exists.

Let "$ext(C)$" to denote the exterior of the simple closed curve $C$ when it exists.

Proof

For each simple polygon $C$ let $P(C)$ be the assertion that all the following hold:

1. $C$ partitions the plane into two path-connected regions called its interior and exterior.

2. The interior and exterior of $C$ are not connected by a path not intersecting $C$.

3. Every point on $C$ is on the boundary of both the interior and exterior of $C$.

4. The exterior of $C$ is connected to $\infty$ by a path not intersecting $C$.

5. $C$ with its interior can be partitioned into triangles.

Clearly $P(C)$ is true for any triangle $C$.

Take any polygon $C$ with more than $3$ sides.

By induction we can assume that $P(C')$ for any polygon $C'$ with fewer sides than $C$.

Let $Y$ be the bottommost vertex of $C$ and $X,Z$ be its left and right adjacent vertices in $C$.

If $C$ does not pass through $int(\tri XYZ) \cup int(\seg{XZ})$:

  Let $C'$ be the polygon derived from $C$ by removing the vertex $Y$.

  Then $C'$ has an interior $I$ and exterior $E$ and $\seg{XZ}$ is on the boundary satisfying $P(C')$.

  Also $Y$ and hence $int(\tri XYZ)$ is connected to $\infty$ by a path avoiding $C'$.

  Thus $int(\tri XYZ)$ is in the exterior of $C'$.

  Let $I' = I \cup int(\seg{XZ}) \cup int(\tri XYZ)$, and let $E' = E \cap ext(\tri XYZ)$.

  Then $I',E',C$ are disjoint and $I' \cup E' \cup C = \rr^2$ and $I',E'$ are each path-connected.

  Also $I'$ is not connected to $E'$ by a path avoiding $C$ because otherwise:

    Let $F$ be such a path connecting $I'$ to $E'$ avoiding $C$.

    If $F$ ever passes through $int(\tri XYZ)$:

      $F$ must eventually exit $int(\tri XYZ)$ because $F$ ends in $E'$.

      Let $M$ be the last point where $F$ exits $int(\tri XYZ)$ [which exists by continuity of $F$].

      Then $M$ is in $int(\seg{XZ})$.

      Let $F'$ be the path derived from $F$ by removing the section from its start point to $M$.

      Then $F'$ connects $I$ to $E$ avoiding $C'$ because $F'$ starts across $\seg{XZ}$ from $int(\tri XYZ)$.

    Similarly if $F$ ever passes through $int(\seg{XZ})$.

    Therefore $I$ and $E$ are connected by a path avoiding $C'$, which gives a contradiction.

  And every point on $C$ is in $( C' \less \seg{XZ} ) \cup \seg{XY} \cup \seg{YZ}$, and hence on the boundary of $I'$ and $E'$.

  And $E'$ is trivially connected to $\infty$ by a path avoiding $C$.

  And $C \cup I' = ( C' \cup I ) \cup \tri XYZ$ can clearly be partitioned into triangles.

If $C$ passes through $int(\tri XYZ) \cup int(\seg{XZ})$:

  Let $V$ be the point on the curve $C$ that is in $int(\tri XYZ) \cup int(\seg{XZ})$ and is furthest from $\seg{XZ}$.

  Then $V$ is clearly a vertex of $C$, because it cannot be an interior point of an edge of $C$.

  And $\seg{YV}$ does not intersect $C$, because such an intersection would be further from $\seg{XZ}$ than $V$.

  Let $U,W$ be the left and right adjacent vertices of $V$ with respect to $\vec{YV}$.

  Let $C_1,C_2$ be the polygons derived from $C$ by splitting at $\seg{YV}$ with $C_1$ passing through $X$.

  Then $C_1$ has an interior $I_1$ and an exterior $E_1$ satisfying $P(C_1)$.

  And $C_2$ has an interior $I_2$ and an exterior $E_2$ satisfying $P(C_2)$.

  Then the path from $Y$ to $\infty$ avoiding $C_1 \less {Y}$ can be modified to connect $int(\seg{YZ})$ to $\infty$.

  Thus $int(\seg{YZ})$ is in the exterior of $C_1$.

  Also $int(\seg{UV})$ is connected to $Y$ by a path sufficiently close to $\seg{YV}$ on the left of $\vec{XYVW}$.

  And the left side of $\vec{XYV}$ near $Y$ is in the interior of $C_1$ because the right side is connected to $\infty$.

  Thus $int(\seg{UV})$ is in the interior of $C_1$.

  If $C_1$ passes through $W$:

    $C_2 \less \seg{YV}$ is a path avoiding $C_1$, and hence $int(\seg{UV})$ is in the exterior of $C_1$.

    Thus the interior and exterior of $C_1$ are not disjoint, which gives a contradiction.

  Therefore $C_1$ passes through $U$.

  Let $I' = I_1 \cup I_2 \cup int(\seg{YV})$, and let $E' = E_1 \cap E_2$.

  Then $I',E',C$ are disjoint and $I' \cup E' \cup C = \rr^2$ and $I',E'$ are each path-connected.

  Also $I',E'$ are not path-connected because otherwise [similar to the above argument]:

    Let $F$ be such a path connecting $I'$ to $E'$, which must not intersect $C$.

    If $F$ passes through $int(\seg{YV})$:

      Let $M$ be the last point of $F$ that is in $\seg{YV}$ [which exists by continuity of $F$].

      Then $M$ is in $int(\seg{YV})$.

      Let $F'$ be the path derived from $F$ by removing the section from its start point to $M$.

      Then $F'$ connects $I_1 \cup I_2$ to $E_1 \cap E_2$ not intersecting $C_1 \cup C_2$.

    Therefore some path avoiding $C_1$ connects $I_1$ to $E_1$ or $I_2$ to $E_2$, which gives a contradiction.

  And every point on $C$ is in $C_1 \cup C_2 \less int(\seg{YV})$, and hence is on the boundary of $I'$ and $E'$.

  And $E'$ is trivially connected to $\infty$ by a path avoiding $C$.

  And $C \cup I' = ( C_1 \cup I_1 ) \cup ( C_1 \cup I_2 )$ can clearly be partitioned into triangles.

Notes

There are small details that need to be filled in, but they are all easy local arguments. For example in all the places where we need to construct some path that is "near" enough to something, it can be done rigorously by first obtaining a $d > 0$ such that every vertex in $C$ is at least $d$ away from any edge that is not incident to it.

Also, the above proof can be shortened a bit by factoring, but I'm leaving it exactly as I obtained it.

I Found this .....

Here http://www.cs.uu.nl/docs/vakken/ga/slides3.pdf The work seems to be authored by prof. dr. M.J. van Kreveld and dr. W.G. van Toll at Universiteit Utrecht.

It's not hard to add the details

1. If there is a boundary point inside UVW (or on UW) then there nust be a vertex inside UVW (or on UW). Otherwise the side that contains the point would extend to intersect UV or VW or U or W any of which contradicts the polygon being simple.
2. Having chosen the vertex furthest from UW (any of them if there is more than one equidistant) construct a line through it parallel to UW intersecting UV at X and VW at Y. If there were a boundary point in VXY then by the same argument as (1) there would need to be a vertex. This would contradict the chosen vertex being furthest from UW. Since the inside of VXY is therefore clear of boundary points then a line from V to the chosen vertex does not cross the boundary.