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.
(In reply to
Trying to put a picture into words... by Alec)
Alec's proof is imaginative in its construction, but to be strict, it is not really a proof. When he says:
>A mildly confusing geometric demonstration…
>Make 4 identical nonagons, and Label the vertices in each from A to I. I have the horizontal bases all as GH in my orientation. Line the nonagons up so that G in the 2nd one is coincident with C in the 1st, H in the 3rd is coincident with F in the 2nd, and B in the 4th is coincident with G in the 3rd. This will make I in the 4th coincident with E in the 1st one.
The last statement, while it is true, does not follow formally (it is not proved by the previously statements).
re: Trying to put a picture into words...
