# 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