Prove, that every $(2n)$-gon have a diagonal which isn't parallel to any side of this polygon.
I was thinking about sth like: Let's suppose that ($2n-3$) diagonals from one point are parallel with ($2n-3$) sides of the polygonal. Then the task is to find $4$ other diagonals, which are not parallel with them (so we will have ($2n+1$) not parallel diagonals and only $2n$ sides, so one diagonal have to be not parallel to any side). But I don't know which one should I choose and if it`s a good idea at all.
If we assume the polygon to be convex, then it will be possible to do this with pigeon-hole, but I am not sure how (in my opinion probably by counting that there are more not parallel diagonals than sides, but there is the same problem - how to count them?).
Thanks for any help
I think I got it. There could me at most $(n-2)$ diagonals which are parallel to one side (because diagonals from vertex of this side could not be parallel to this side, so there is $(2n-2)$ vertexes left, but its $(2n)$-gon, so there must me one side from which there are no diagonals parallel). But this gives us at most $2n(n-2)=2n^2-4n$ diagonals which are parallel to any side of figure, and there is of course $n(2n-3)=2n^2-3n>2n^2-4n$ diagonals in $(2n)$-gon, so there must be diagonal which is not parallel to any side.