Triangulation of Lattice Polygons What's a neat proof that every lattice polygon can be split into elementary triangles?
By a lattice polygon I mean a polygon with vertices on some equidistant grid; by an elementary triangle I mean a lattice polygon on three vertices with no other boundary or interior points.
 A: We choose a graph $G$ with $v$ vertices and construct a $v$-vertex polygon either via triangulating it or erasing edges from a maximally planar graph of $n$-vertices. We then turn triangular regions into triangles through Fáry's theorem.
1) Let $G$ be the graph of a lattice polygon. The vertices $G$ form a closed path. Choose a face inside the closed path: if it is not triangular, then draw an edge between two unconnected vertices, if it is triangular, leave it alone. Repeat the process until all bounded regions of $G$ are triangular. We can't draw any more edges since a triangular region is formed by $K_{3}$.
A closed path bounding a nontriangular face never has edges between non-adjacent vertices, so we should always be able to draw one without making $G$ nonplanar.
2) We know that in a simple planar graph $e \leq 3v - 6$, so the maximum $e$ is $3v - 6$. This is possible only when $2e = 3f$ (this is the inequality from which the first one is derived, namely $3f \leq 2e$). If a face $f$ is triangular, then $3f$ is the same as counting every edge bounding it twice. So, to count every edge in the graph twice using faces, each must be part of a triangular region. 
We showed that a maximal planar graph with $v$ vertices can be triangulated with $3v - 6$ edges. Erase edges until only the desired polygon remains, which is composed of triangular regions.
The planar graphs constructed through (1) and (2) can be drawn with straight line segments by Fáry's theorem, turning the triangular regions into actual triangles.
