Let the vertices of n-gon have numbers (from 1 to n). Consider the first vertex. We can draw (n-3) diagonals from it: to the vertices with numbers 3, 4, …, (n-1). From the second vertex we can draw (n-3) diagonals too: to the vertices 4, 5, …, n. From the third vertex we can draw (n-4) diagonals: to the vertices 5, …, n. From the fourth vertex we can draw (n-5) diagonals: to the vertices 6, …, n. According to the same logic: From the (n-2)th vertex we can draw 1 diagonal too: to the vertex n. From the (n-1)th and n-th vertices we can’t draw any diagonal. So the total number of diagonals is: (n-3) + (n-3) + (n-4) + (n-5) + … + 2 + 1 = (n-3) + (n-2)(n-3)/2 = n(n-3)/2
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