Construction of Regular Tetrahedron within Given Sphere/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:

and:

it follows that:
 * $BA : AD = DA : AC$

Therefore from :
 * $BA \cdot AC = AD^2$

From :
 * $AB : BC = EB : EF$

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 :
 * $DC$ is a mean proportional between $AC$ and $CB$

and so:
 * $AC \cdot CB = DC^2$

Therefore:
 * $AB : BC = AD^2 : DC^2$