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.

I haven't looked at any of the other comments so forgive me if I repeat someone.
Each internal angle of a regular nonagon is 140 degrees given by the formula: (180)(n-2) where n is the number of sides.
Next Draw triangle ABC (A, B, and C are the points of the nonagon.) I assumed each side was 1 unit. With the law of cosines:
AC^2=1^2+1^2+(2)(1)(1)(cos140)
AC=sqrt[2-2cos140]
AC=1.879385
AB (assumed to be 1) + AC=2.879385

Now for AE.
First I figured the length of a circumradius by drawing a triangle with its vertices at the center, A, and B. The angle between the 2 circumradii is 40 (360/9). Assuming again that AB=1, we have an isosceles triangle. Bisect the center angle and AB with an altitude and do some trig to find that the circumradius=(1/2*sin20). Now do a similar procedure to find AE. Draw 2 circumradii (one to A, one to E) the angle between them is 160 (360*4/9). Do the same thing with this isosceles triangle as the last one and find AE to be (sin(80)/sin(20)) or 2.879385 (the same as AB + AC)

Posted by Sam on 2004-12-12 21:07:20