# Construction of Regular Tetrahedron within Given Sphere/Lemma

## Lemma to Construction of Regular Tetrahedron within Given Sphere

In the words of Euclid:

It is to be proved that, as $AB$ is to $BC$, so is the square on $AD$ to the square on $DC$.

## 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:

Proposition $8$ of Book $\text{VI}$: Perpendicular in Right-Angled Triangle makes two Similar Triangles

and:

Proposition $4$ of Book $\text{VI}$: Equiangular Triangles are Similar

it follows that:

$BA : AD = DA : AC$
$BA \cdot AC = AD^2$
$AB : BC = EB : EF$

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.