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
Very clever indeed
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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