Proof: By Euclid
(related to Proposition: Prop. 13.15: Construction of Cube within Given Sphere)
- Let the diameter $AB$ of the given sphere be laid out, and let it have been cut at $C$ such that $AC$ is double $CB$.
- And let the semicircle $ADB$ have been drawn on $AB$.
- And let $CD$ have been drawn from $C$ at right angles to $AB$.
- And let $DB$ have been joined.
- And let the square $EFGH$, having (its) side equal to $DB$, be laid out.
- And let $EK$, $FL$, $GM$, and $HN$ have been drawn from (points) $E$, $F$, $G$, and $H$, (respectively), at right angles to the plane of square $EFGH$.
- And let $EK$, $FL$, $GM$, and $HN$, equal to one of $EF$, $FG$, $GH$, and $HE$, have been cut off from $EK$, $FL$, $GM$, and $HN$, respectively.
- And let $KL$, $LM$, $MN$, and $NK$ have been joined.
- Thus, a cube contained by six equal squares has been constructed.
- So, it is also necessary to enclose it by the given sphere, and to show that the square on the diameter of the sphere is three times the (square) on the side of the cube.
- For let $KG$ and $EG$ have been joined.
- And since angle $KEG$ is a right angle - on account of $KE$ also being at right angles to the plane $EG$, and manifestly also to the straight line $EG$ [Def. 11.3] - the semicircle drawn on $KG$ will thus also pass through point $E$.
- Again, since $GF$ is at right angles to each of $FL$ and $FE$, $GF$ is thus also at right angles to the plane $FK$ [Prop. 11.4].
- Hence, if we also join $FK$ then $GF$ will also be at right angles to $FK$.
- And, again, on account of this, the semicircle drawn on $GK$ will also pass through point $F$.
- Similarly, it will also pass through the remaining (angular) points of the cube.
- So, if $KG$ remains (fixed), and the semicircle is carried around, and again established at the same (position) from which it began to be moved, then the cube will have been enclosed by a sphere.
- So, I say that (it is) also (enclosed) by the given (sphere).
- For since $GF$ is equal to $FE$, and the angle at $F$ is a right angle, the (square) on $EG$ is thus double the (square) on $EF$ [Prop. 1.47].
- And $EF$ (is) equal to $EK$.
- Thus, the (square) on $EG$ is double the (square) on $EK$.
- Hence, the (sum of the squares) on $GE$ and $EK$ - that is to say, the (square) on $GK$ [Prop. 1.47] - is three times the (square) on $EK$.
- And since $AB$ is three times $BC$, and as $AB$ (is) to $BC$, so the (square) on $AB$ (is) to the (square) on $BD$ [Prop. 6.8], [Def. 5.9] , the (square) on $AB$ (is) thus three times the (square) on $BD$.
- And the (square) on $GK$ was also shown (to be) three times the (square) on $KE$.
- And $KE$ was made equal to $DB$.
- Thus, $KG$ (is) also equal to $AB$.
- And $AB$ is the radius of the given sphere.
- Thus, $KG$ is also equal to the diameter of the given sphere.
- Thus, the cube has been enclosed by the given sphere.
- And it has simultaneously been shown that the square on the diameter of the sphere is three times the (square) on the side of the cube.
- (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
