Construction of Regular Tetrahedron within Given Sphere/Lemma
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Lemma to Construction of Regular Tetrahedron within Given Sphere
In the words of Euclid:
(The Elements: Book $\text{XIII}$: Proposition $13$ : Lemma)
Proof
Let the figure of the semicircle be set out.
Let $DB$ be joined.
Let the square $EC$ be described on $AC$.
Let the parallelogram $FB$ be completed.
From:
and:
it follows that:
- $BA : AD = DA : AC$
Therefore from Proposition $17$ of Book $\text{VI} $: Rectangles Contained by Three Proportional Straight Lines:
- $BA \cdot AC = AD^2$
From Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base:
- $AB : BC = EB : EF$
But $EA = AC$ and so:
- $EB = BA \cdot AC$
and:
- $BF = AC \cdot CB$
Therefore:
- $AB : BC = BA \cdot AC : AC \cdot CB$
We also have:
- $BA \cdot AC = AD^2$
We have that $\angle ADB$ is a right angle.
- $DC$ is a mean proportional between $AC$ and $CB$
and so:
- $AC \cdot CB = DC^2$
Therefore:
- $AB : BC = AD^2 : DC^2$
$\blacksquare$
Historical Note
This proof is Proposition $13$ of Book $\text{XIII}$ of Euclid's The Elements.
It is suspected that this lemma is a later interpolation.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XIII}$. Propositions