Draw a regular nonagon. If we label the vertices of the polygon sequentially, starting from A, we can draw three line segments: AB, AC, and AE.

Show that AB + AC = AE.

By drawing radii from the centre (say O) to A and B the distance AB can be calculated by creating two right angle triangles by bisecting angle AOB to the centre of AB.

AB = 2.r.sin (AOB/2)

AC = 2.r.sin (AOC/2)

AE = 2.r.sin (AOE/2)

It is easy to show that AOB = 40 degrees (360 degrees divided by the 9 internal angles), AOC = 80 degrees and AOE = 160 degrees.

So if sin20 + sin40 = sin80 (which it does) then AB + AC = AE.

This might be a bit backward, possibly inside-out.  Please forgive me if it doesn't qualify as mathematical proof.

Posted by Fletch on 2004-11-25 17:30:12