Molson Hart

Assume every diagonal is parallel to a side. Number of unique diagonals for a 2n gon is (2n-3)*n. Sides will have an average of (2n-3)*n/(2n) parallel to it. You can simplify and then round this up to say that some side has n-1 diagonals parallel to it. Including the side, ...

...we have n-1+1 parallel line segments, each with a pair of unique vertices (2n total) that are shared with the 2n gon. Let's now attempt to draw the shape, starting with the parallel side, connecting it to the diagonals. 1 side, connect to the first diagonal (+2 sides), second

Just for once, trying to read across several tweets, "...we have n-1+1 parallel line segments, each with a pair of unique vertices (2n total)" is suspect: you try to include the second side which you know nothing about

I was on phone my phone. I had no choice at the time of writing. If it helps I'm happy make an image and tweet it. I don't understand what you're saying. Let n = 3, there's an average of 9/6 = 1.5 diagonals parallel to each side. Round it up to 2, for one side.

Just read what you wrote (the sentence I quoted). As I see it: 2n vertices is all you have; n parallel segments joining 2n vertices necessarily include 2 sides. Just try to consolidate your tweets when you have time

I am not sure if you can have n-1 diagonals parallel to a side

I can explain that with greater clarity. (2n-3)/2 is the average of unique diagonals parallel to each side. It is not a whole number. There must be some side that has more than (2n-3)/2 unique diagonals parallel to it for the average to be lower. Round (2n-3)/2 to n-1. Clear now?