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.

1) All nine interior angles are 140 degrees.
2) Triangle ABC is isosceles, so angle BAC=BCA=20 degrees. Same reasoning for congruent triangle BCD.
3) BD=AC and is parallel to AE, from steps 1 and 2.
4) Angle BAE= BAC+CAD+DAE; =20+20+20=60 degrees.
5) Angle DEA=60 degrees also, by similar reasoning.
6) Draw line from B parallel to DE, intersecting AE at a point P.
7) BDEP is a parallogram with angles of 120 and 60.
8) EP=BD=AC (See step 2), and AP=BP=AB (triangle APB is equilateral because all its angles=60 degs.)
9) AE=EP+AP. Substituting from step 8, AE=AC+AB.

Posted by sizz on 2004-11-27 12:14:21