Question 223233
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Consider an *[tex \Large n]-gon.  You can't draw a diagonal from a vertex to itself, so that is one fewer than the number of vertices.  Then you can draw a line segment from a vertex to either of the adjacent vertices, but each of those would be coincident with a side and therefore not a diagonal.  That's two more fewer than the number of vertices.  All together, 3 fewer diagonals than the number of vertices.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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