Hi Edna.

If you draw a line between every two points on the polygon, you have drawn all the diagonals plus the sides.

Say you have an n-gon. Pick a corner to start at and draw lines to all the other corners. That's n-1 lines. Now pick the next point and draw a line to all the other corners (don't redraw the line to the first corner though - it is already drawn. How many is that? It is n-2.

So after two corners are done, we have n-1 + n-2 lines. And we keep going until the penultimate corner and we just draw 1 line to the last point. The last point already has lines to all other points, so it adds no lines.

What do we have now? n-1 + n-2 + n-3 + ... + 2 + 1. What does that add up to?

Let's add it to itself, then divide by two (because anything added to itself and then divided by two equals the original anyway, right?)

n-1 + n-2 + n-3 + ... + 2 + 1
= [ (n-1 + n-2 + n-3 + ... + 2 + 1) + (n-1 + n-2 + n-3 + ... + 2 + 1) ] / 2
= [ (n-1 + n-2 + n-3 + ... + 2 + 1) + (1 + 2 + 3 + ... + n-2 + n-1) ] / 2
= [ (n-1+1) + (n-2+2) + (n-3+3) + ... + (2+n-2) + (1+n-1) ] / 2
= [ n (n-1) ] / 2

It totals n(n-1)/2. To find that, I just moved the plus and minus terms around (this is allowed because of the commutative property of addition and subtraction).

But remember that this includes the sides, not just the diagonals. How many sides does an n-gon have? By definition, it has n sides. So we can subtract this from the total.

The number of diagonals is n(n-1)/2 - n.

Hope this helps,
Stephen La Rocque.