Comparison of Sides of Five Platonic Figures/Lemma

Proof

 * Euclid-XIII-18-Lemma.png

Let $ABCDE$ be a regular pentagon.

Let the circle $ABCDE$ be circumscribed around the pentagon $ABCDE$.

Let $F$ be the center of the circle $ABCDE$.

Let $FA, FB, FC, FD, FE$ be drawn.

The straight lines $FA, FB, FC, FD, FE$ bisect the vertices $A, B, C, D, E$ respectively.

The angles at $F$ form a total of $4$ right angles, and are equal.

Therefore one of them, for example $\angle AFB$ equals one right angle less a fifth.

Therefore $\angle FAB + \angle ABF$ consists of one right angle plus a fifth.

But:
 * $\angle FAB = \angle FBC$

and so $\angle ABC$ equals one right angle plus a fifth.