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:

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

(The Elements: Book $\text{XIII}$: Proposition $13$ : Lemma)


Proof

Euclid-XIII-13-Lemma.png

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:

$BA : AD = DA : AC$

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.

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

$DC$ 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 theorem is Proposition $13$ of Book $\text{XIII}$ of Euclid's The Elements.
It is suspected that this lemma is a later interpolation.


Sources