Commentary on Proposition XI.37

Back to Proposition XI.37.

In this proposition it is assumed that, if two ratios be equal, the ratio triplicate of one is equal to the ratio triplicate of the other and, conversely, that, if ratios which are the triplicate of two other ratios are equal, those other ratios are themselves equal.

To avoid the necessity for these assumptions Simson adopts the alternative proof found in the MS. which Heiberg calls b, and also adopted by Clavius, who, however, gives Euclid’s proof as well, attributing it to Theon. The alternative proof proceeds after the manner of VI. 22, thus.

Make AB, CD, O, P continuous proportionals, and also EF, GH, Q, R.

K A B L C D M E F N G H Q R O P V S T

  1. Then, since

$$AB : CD = EF : GH,$$

we have, ex aequali,

$$AB : P = EF : R. \qquad \text{[V. 22]}$$

But

$$\text{(solid } AK\text{)} : \text{(solid } CL\text{)} = AB : P, \qquad \text{[XI. 33 and Por.]}$$

and

$$\text{(solid } EM\text{)} : \text{(solid } GN\text{)} = EF : R.$$

Therefore

$$\text{(solid } AK\text{)} : \text{(solid } CL\text{)} = \text{(solid } EM\text{)} : \text{(solid } GN\text{)}$$

  1. If the solids are proportional, take ST such that

$$AB : CD = EF : ST,$$

and on ST describe the parallelepiped SV similar and similarly situated to either of the parallelepipeds EM, GN.

Then, by the first part,

$$\text{(solid } AK\text{)} : \text{(solid } CL\text{)} = \text{(solid } EM\text{)} : \text{(solid } SV\text{)},$$

whence it follows that

$$\text{(solid } GN\text{)} = \text{(solid } SV\text{)}.$$

But these solids are similar and similarly situated; therefore their faces are similar and equal; [XI. Def. 10] therefore the corresponding sides GH, ST are equal.

[For this inference cf. note on VI. 22. The equality of GH, ST may readily be proved by application of the two parallelepipeds to one another, since, being similar, they are equiangular.]

Hence

$$AB : CD = EF : GH.$$

The text of the MSS. has here a proposition which is as badly placed as it is unnecessary. If a plane be at right angles to a plane, and from any one of the points in one of the planes a perpendicular be drawn to the other plane, the perpendicular so drawn will fall on the common section of the planes. It is of the nature of a lemma to XII. 17, where alone the fact is made use of. Heiberg observes that it is omitted in b and that the copyist of P knew other texts which did not contain it. From these facts it is fairly concluded that the proposition was interpolated. The truth of it is of course immediately obvious by reductio ad absurdum. Let the plane CAD be perpendicular to the plane AB, and let a perpendicular be drawn to the latter from any point E in the former.

C E D A G F B

If it does not fall on AD, the common section, let it meet the plane AB in F.

Draw FG in AB perpendicular to AD, and join EG.

Then FG is perpendicular to the plane CAD [XI. Def. 4], and therefore to GE [XI. Def. 3]. Therefore ∠EGF is right.

Also, since EF is perpendicular to AB, the angle EFG is right.

That is, the triangle EGF has two right angles: which is impossible.