Molson Hart

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?

Sir, is there any chance I could get you to confirm whether or not my proof works? I'm not a very strong mathematician, but I spent quite a bit of time imagining hexagons in my head and I would feel great pride if I solved this problem. Thanks!

Very candidly, I simply do not understand your rounding argument. In your diagram, where do you get n-1? There are four seemingly parallel segments, two of which are sides (that's why I once mentioned a second side.) There are n-2 diagonals