Construction of Cube within Given Sphere

Proof

 * Euclid-XIII-15.png

Let $AB$ be the diameter of the given sphere.

Let $AB$ be cut at $C$ where $AC = 2 \cdot CB$.

Let $ADB$ be a semicircle on the diameter $AB$.

Let $CD$ be drawn from $C$ perpendicular to $AB$.

Let $DB$ be joined.

Let the square $EFGH$ be set out whose sides are all equal to $DB$.

From :
 * let $EK, FL, GM, HN$ be drawn from $E, F, G, H$ respectively perpendicular to the plane of the square $EFGH$.

Let $EK, FL, GM, HN$ be made equal to one of the straight lines $EF, FG, GH, HE$.

Let $KL, LM, MN, NK$ be joined.

The cube $FN$ can be seen to have been constructed which is contained by six squares.

Next it is to be demonstrated that the cube $FN$ can be inscribed in the given sphere.

Let $KG, EG$ be joined.

We have that $KE$ is perpendicular to the plane of the square $EFGH$.

So from :
 * $KE$ is perpendicular to the straight line $EG$.

Therefore $\angle KEG$ is a right angle.

Therefore also if $FK$ is joined, then $GF$ is perpendicular to $FK$.

For this reason, the semicircle described on $GK$ will also pass through $F$.

Similarly it will also pass through the remaining vertices of the cube.

Let $GK$ remain fixed, and let that semicircle be carried around and restored to the same position from which it began.

Thus the cube $FN$ has been inscribed in a sphere.

We have that:
 * $GF = FE$

and that $\angle F$ is a right angle.

Therefore from :
 * $EG^2 = 2 EF^2$

But:
 * $EF = EK$

Therefore:
 * $EG^2 = 2 \cdot EK^2$

Therefore from :
 * $GK^2 = EG^2 + EK^2 = 3 EK^2$

We have that:
 * $AB = 3 BC$

while:
 * $AB : BC = AB^2 : BD^2$

Therefore:
 * $AB^2 = 3 \cdot BD^2$

But it has been proved that:
 * $GK^2 = 3 EK^2$

while:
 * $KE = DB$

Therefore:
 * $KG = AB$

But $AB$ is the diameter of the given sphere.

Therefore:
 * the cube $FN$ can be inscribed in the given sphere

and:
 * the square on the diameter of the given sphere is three times the square on the side of the cube.