Comparison of Sides of Five Platonic Figures/Lemma

Lemma to Comparison of Sides of Five Platonic Figures

In the words of Euclid:

But that the angle of the equilateral and equiangular pentagon is a right angle and a fifth we must prove thus.

(The Elements: Book $\text{XIII}$: Proposition $18$ : Lemma)

Proof

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.

$\blacksquare$

Historical Note

This proof is Proposition $18$ of Book $\text{XIII}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XIII}$. Propositions