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.
Let AB=1, and center of nonagon O. By Law of Sines, AC of triangle ABC=1.87938... Circumradius of a nonagon is 1.4619... of the nonagon side. So, by Law of Sines, AE of triangle AOE=2.87938...
Therefore, this shows that AB+AC=AE since AB=1 plus AC=1.87938... , equals AE=2.87938...
|
|
Posted by CeeAnne
on 2004-11-28 18:21:21 |
Here, try this ...
