(related to Proposition: Prop. 13.13: Construction of Regular Tetrahedron within Given Sphere)

- Let the diameter $AB$ of the given sphere be laid out, and let it have been cut at point $C$ such that $AC$ is double $CB$ [Prop. 6.10].
- And let the semicircle $ADB$ have been drawn on $AB$.
- And let $CD$ have been drawn from point $C$ at right angles to $AB$.
- And let $DA$ have been joined.
- And let the circle $EFG$ be laid down having a radius equal to $DC$, and let the equilateral triangle $EFG$ have been inscribed in circle $EFG$ [Prop. 4.2].
- And let the center of the circle, point $H$, have been found [Prop. 3.1].
- And let $EH$, $HF$, and $HG$ have been joined.
- And let $HK$ have been set up, at point $H$, at right angles to the plane of circle $EFG$ [Prop. 11.12].
- And let $HK$, equal to the straight line $AC$, have been cut off from $HK$.
- And let $KE$, $KF$, and $KG$ have been joined.
- And since $KH$ is at right angles to the plane of circle $EFG$, it will thus also make right angles with all of the straight lines joining it (which are) also in the plane of circle $EFG$ [Def. 11.3] .
- And $HE$, $HF$, and $HG$ each join it.
- Thus, $HK$ is at right angles to each of $HE$, $HF$, and $HG$.
- And since $AC$ is equal to $HK$, and $CD$ to $HE$, and they contain right angles, the base $DA$ is thus equal to the base $KE$ [Prop. 1.4].
- So, for the same (reasons), $KF$ and $KG$ is each equal to $DA$.
- Thus, the three (straight lines) $KE$, $KF$, and $KG$ are equal to one another.
- And since $AC$ is double $CB$, $AB$ (is) thus triple $BC$.
- And as $AB$ (is) to $BC$, so the (square) on $AD$ (is) to the (square) on $DC$, as will be shown later [see lemma].
- Thus, the (square) on $AD$ (is) three times the (square) on $DC$.
- And the (square) on $FE$ is also three times the (square) on $EH$ [Prop. 13.12], and $DC$ is equal to $EH$.
- Thus, $DA$ (is) also equal to $EF$.
- But, $DA$ was shown (to be) equal to each of $KE$, $KF$, and $KG$.
- Thus, $EF$, $FG$, and $GE$ are equal to $KE$, $KF$, and $KG$, respectively.
- Thus, the four triangles $EFG$, $KEF$, $KFG$, and $KEG$ are equilateral.
- Thus, a pyramid, whose base is triangle $EFG$, and apex the point $K$, has been constructed from four equilateral triangles.
- So, it is also necessary to enclose it in the given sphere, and to show that the square on the diameter of the sphere is one and a half times the (square) on the side of the pyramid.
- For let the straight line $HL$ have been produced in a straight line with $KH$, and let $HL$ be made equal to $CB$.
- And since as $AC$ (is) to $CD$, so $CD$ (is) to $CB$ [Prop. 6.8 corr.] 3, and $AC$ (is) equal to $KH$, and $CD$ to $HE$, and $CB$ to $HL$, thus as $KH$ is to $HE$, so $EH$ (is) to $HL$.
- Thus, the (rectangle contained) by $KH$ and $HL$ is equal to the (square) on $EH$ [Prop. 6.17].
- And each of the angles $KHE$ and $EHL$ is a right angle.
- Thus, the semicircle drawn on $KL$ will also pass through $E$ [inasmuch as if we join $EL$ then the angle $LEK$ becomes a right angle, on account of triangle $ELK$ becoming equiangular to each of the triangles $ELH$ and $EHK$ [Prop. 6.8], [Prop. 3.31] ].
- So, if $KL$ remains (fixed), and the semicircle is carried around, and again established at the same (position) from which it began to be moved, it will also pass through points $F$ and $G$, (because) if $FL$ and $LG$ are joined, the angles at $F$ and $G$ will similarly become right angles.
- And the pyramid will have been enclosed by the given sphere.
- For the diameter, $KL$, of the sphere is equal to the diameter, $AB$, of the given sphere - inasmuch as $KH$ was made equal to $AC$, and $HL$ to $CB$.
- So, I say that the square on the diameter of the sphere is one and a half times the (square) on the side of the pyramid.
- For since $AC$ is double $CB$, $AB$ is thus triple $BC$.
- Thus, via convertion, $BA$ is one and a half times $AC$.
- And as $BA$ (is) to $AC$, so the (square) on $BA$ (is) to the (square) on $AD$ [inasmuch as if $DB$ is joined then as $BA$ is to $AD$, so $DA$ (is) to $AC$, on account of the similarity of triangles $DAB$ and $DAC$.
- And as the first is to the third (of four proportional magnitudes), so the (square) on the first (is) to the (square) on the second.] Thus, the (square) on $BA$ (is) also one and a half times the (square) on $AD$.
- And $BA$ is the diameter of the given sphere, and $AD$ (is) equal to the side of the pyramid.
- Thus, the square on the diameter of the sphere is one and a half times the (square) on the side of the pyramid.
- (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"