This is a followup to my last question - Can we prove that simple polygons can always be split in half (vertex-wise) by diagonals?

Is there a constant natural number K for which the following is true:

For any simple polygon with more than 3 vertices there always exists a diagonal which:

- is inside the polygon
- doesn't intersect with any edges
- splits the polygon in two polygons in such a way that the difference between their vertex counts is smaller than K