# Number of diagonals in a n-polygonal

I'm asked to find a proof that a polygonal with n edges has n(n-3)/2 diagonals

If you consider diagonals and edges all together, they represent the number of unordered couples of n points, that is C(n,2) = n*(n-1)/2.If you now subtract the number of edges (i.e. n ) you get that formula
Francesco G.
25 June 2014
Very clever indeed
ripemate
25 June 2014
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sharjil_lodhi
30 June 2014
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07:29 on 30/06/14
30 June 2014
Hi,This is Gem. My proof is as follows:Consider each node, there are n-3 diagonals (excluding adjacent nodes and itself, so n-3). There are n nodes in total. Hence n(n-3)But for each diagonal, we have counted twice the number (since each has 2 end nodes). So we divide above by 2.Hope it helped you.Gem
22:12 on 01/07/14
12 May 2020