Commentary on Proposition XIII.13

Back to Proposition XIII.13.

LEMMA.

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

For let the figure of the semicircle be set out, let DB be joined, let the square EC be described on AC, and let the parallelogram FB be completed.

A B C D E F
Since then, because the triangle DAB is equiangular with the triangle DAC,
[VI. 8, VI. 4]as BA is to AD, so is DA to AC,
[VI. 17]therefore the rectangle BA, AC is equal to the square on AD.
And since, as AB is to BC, so is EB to BF, [VI. 1] and EB is the rectangle BA, AC, for EA is equal to AC, and BF is the rectangle AC, CB,
therefore, as AB is to BC, so is the rectangle BA, AC to the rectangle AC, CB.
And the rectangle BA, AC is equal to the square on AD, and the rectangle AC, CB to the square on DC, for the perpendicular DC is a mean proportional between the segments AC, CB of the base, because the angle ADB is right. [VI. 8, Por.]
Therefore, as AB is to BC, so is the square on AD to the square on DC. Q. E. D.

The Lemma is with reason suspected. Euclid commonly takes more difficult theorems for granted in the stereometrical Books. It is also clumsy in itself, while, from a gloss in the proposition rejected as an interpolation, it is clear that the interpolator of the gloss had not the Lemma. With the Lemma should disappear the words “as will be proved afterwards” (p. 469).

In the figure of the proposition, the semicircle really represents half of a section of the sphere through its centre and one edge of the inscribed tetrahedron (AD being the length of that edge).

The proof is in three parts, the object of which is to prove

  1. that KEFG is a tetrahedron with all its edges equal to AD,

  2. that it is inscribable in a sphere of diameter equal to AB,

  3. that AB $^2 = \frac{3}{2}$ AD $^2$.

To prove (1) we have to show

  1. that KE $= $ KF $= $ KG $= $ AD,

  2. that AD $= $ EF.

  1. Since HE $= $ HF $= $ HG $= $ CD, KH $= $ AC, and $\angle\text{s}$ ACD, KHE, KHF, KHG are right, $\Delta\text{s}$ ACD, KHE, KHF, KHG are equal in all respects; therefore KE $= $ KF $= $ KG $= $ AD.

  2. Since AB $= 3$ BC, and AB $:$ BC $= $ AB $\cdot$ AC $:$ AC $\cdot$ CB $= $ AD $^2 : $ CD $^2$, it follows that AD $^2 = 3$ CD $^2$. But [XIII. 12] EF $^2 = 3$ EH $^2$; and EH $= $ CD, by construction

Therefore AD $= $ EF. Thus EFGK is a regular tetrahedron.

  1. We now observe the usefulness of Euclid’s description of a sphere [in XI. Def. 14].

Producing KH ($= $ AC) to L so that HL $= $ CB, we have KL equal to AB; thus KL is a diameter of the sphere which should circumscribe our tetrahedron, and we have only to prove that E, F, G lie on semicircles described on KL as diameter.

E.g. for the point E, since AC $:$ CD $= $ CD $:$ CB, while AC $= $ KH, CD $= $ HE, CB $= $ HL, we have KH $:$ HE $= $ HE $:$ HL, or KH $\cdot$ HL $= $ HE $^2$, whence, the angles KHE, EHL being right, EKL is a triangle right-angled at E [cf. VI. 8].

Hence E lies on a semicircle on KL as diameter. Similarly for F, G. Thus a semicircle on KL as diameter revolving round KL passes successively through E, F, G.

  1. AB $= 3$ BC; therefore BA $= \frac{3}{2}$ AC. And BA $:$ AC $= $ BA $^2 : $ BA $\cdot$ AC $= $ BA $^2 : $ AD $^2$. Therefore BA $^2 = \frac{3}{2}$ AD $^2$.

If $r$ be the radius of the circumscribed sphere,

$$\text{(edge of tetrahedron)} = \frac{2\sqrt{2}}{\sqrt{3}}\cdot r = \frac{2}{3}\sqrt{6}\cdot r.$$

It will be observed that, although in these cases Euclid’s construction is equivalent to inscribing the particular regular solid in a given sphere, he does not actually construct the solid in the sphere but constructs a solid which a sphere equal to the given sphere will circumscribe. Pappus, on the other hand, in dealing with the same problems, actually constructs the respective solids in the given spheres. His method is to find circular sections in the given spheres containing a certain number of the angular points of the given solids. His solutions are interesting, although they require a knowledge of some properties of a sphere which are of course not found in the Elements but belonged to treatises such as the Sphaerica of Theodosius.

Pappus’ solution of the problem of Eucl. XIII. 13.

In order to inscribe a regular pyramid or tetrahedron in a given sphere, Pappus (III. pp. 142—144) finds two circular sections equal and parallel to one another, each of which contains one of two opposite edges as its diameter. In this and the other similar problems he proceeds in the orthodox manner by analysis and synthesis. The following is a reproduction of his solution of this case.

Analysis.

Suppose the problem solved, A, B, C, D being the angular points of the required pyramid.

Through A draw EF parallel to CD; this will make equal angles with AC, AD; and, since AB does so too, EF is perpendicular to AB [Pappus has a lemma for this, p. 140, 12—24], and is therefore a tangent to the sphere (for EF is parallel to CD, the base of the triangle ACD, and therefore touches the circle circumscribing it, while it also touches the circular section AB made by the plane passing through AB and EF perpendicular to it).

G D H L K M C E P A B F

Similarly GH drawn through D parallel to AB touches the sphere.

And the plane through GH, CD makes a circular section equal and parallel to AB.

Through the centre K of that circular section, and in the plane of the section, draw LM perpendicular to CD and therefore parallel to AB. Join BL, BM.

BM is then perpendicular to AB, LM, and LB is a diameter of the sphere. Join MC.

Then LM $^2 = 2$ MC $^2$, and BC $= $ AB $= $ LM, so that BC $^2 = 2$ MC $^2$.

And BM, being perpendicular to the plane of the circle LM, is perpendicular to CM, whence BC $^2 = $ BM $^2 + $ MC $^2$, so that BM $= $ MC. But BC $= $ LM; therefore LM $^2 = 2$ BM $^2$.

And, since the angle LMB is right, BL $^2 = $ LM $^2 + $ MB $^2 = \frac{3}{2}$ LM $^2$.

Synthesis.

Draw two parallel circular sections of the sphere with diameter $d'$, such that

$$d'^2 = \frac{2}{3}d^2,$$

where $d$ is the diameter of the sphere.

[This is easily done by dividing BL, any diameter of the sphere, at P, so that LP $= 2$ PB, and then drawing PM at right angles to LB meeting the great circle LMB of the sphere in M. Then LM $^2 : $ LB $^2 = $ LP $:$ LB $= 2 : 3$.]

Draw sections through M, B perpendicular to MB, and in these sections respectively draw the parallel diameters LM, AB.

Lastly, in the section LM draw CD through the centre K perpendicular to LM.

ABCD is then the required regular pyramid or tetrahedron.