Proving diagonals in convex polygons with pigeon hole principle

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allistair
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I got 6 problems that I needed to proove using the pigeon hole principle and I was able to solve 5 of them but this last one is giving me some problems.

In each convex polygon with 2*n vertices there is at least one diagonal that isn't parallel with either one of the sides of the polygon.

I would appreciate some help to point me in the right direction or maybe an example of a similar proof that uses the pigeon hole principle, thanks in advance
 
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i'm obligated to use the pigeon hole principle, i can't use anything else (i'm not sure what you mean by 'discrete math', or did you mean that i posted this in the wrong forum?)
 
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How many diagonals are there? How many can be parallel to a side?
 
I'm trying to find a function that gives the number of diagonals in funtion of the number of vertices but i don't see a connection both of them

i looked it up and apparently there is a formula for it, for a polygon with 2n vertices the number of diagonals is 2n*(2n-3)/2, i hope i'll be able to use this
 
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