Comparison of Sides of Five Platonic Figures/Lemma
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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