Commentary on Proposition XI.28

Back to Proposition XI.28.

Simson properly observes that it ought to be proved that the diagonals of two opposite faces are in one plane, before we speak of drawing a plane through them. Clavius supplied the proof, which is of course simple enough.

Since EF, CD are both parallel to AG or BH, they are parallel to one another.

Consequently a plane can be drawn through CD, EF and the diagonals DE, CF are in that plane [XI. 7]. Moreover CD, EF are equal as well as parallel ; so that CF, DE are also equal and parallel.

Simson does not, however, seem to have noticed a more serious difficulty. The two prisms are shown by Euclid to be contained by equal faces—the faces are in fact equal and similar—and Euclid then infers at once that the prisms are equal. But they are not equal in the only sense in which we have, at present, a right to speak of solids being equal, namely in the sense that they can be applied, the one to the other. They cannot be so applied because the faces, though equal respectively, are not similarly arranged ; consequently the prisms are symmetrical, and it ought to be proved that they are, though not equal and similar, equal in content, or equivalent, as Legendre has it.

Legendre addressed himself to proving that the two prisms are equivalent, and his method has been adopted, though his name is not mentioned, by Schultze and Sevenoak and by Holgate. Certain preliminary propositions are necessary.

M E' D' E C' A' B' D A B C N

1. The sections of a prism made by parallel planes cutting all the lateral edges are equal polygons.

Suppose a prism MN cut by parallel planes which make sections ABCDE, A'B'C'D'E'.

Now AB, BC, CD, … are respectively parallel to A'B', B'C', C'D', … [XI. 16]

Therefore the angles ABC, BCD, … are equal to the angles A'B'C', B'C'D', … respectively. [XI. 10]

Also AB, BC, CD, … are respectively equal to A'B', B'C', C'D', … [I. 34]

Thus the polygons ABCDE, A'B'C'D'E' are equilateral and equiangular to one another.

2. Two prisms are equal when they have a solid angle in each contained by three faces equal each to each and similarly arranged.

Let the faces ABCDE, AG, AL be equal and similarly placed to the faces A'B'C'D'E', A'G', A'L'.

Since the three plane angles at A, A' are equal respectively and are similarly placed, the trihedral angle at A is equal to the trihedral angle at A'. [(3) in note to XI. 21] Place the trihedral angle at A on that at A'. Then the face ABCDE coincides with the face A'B'C'D'E', the face AG with the face A'G', and the face AL with the face A'L'. The point C falls on C' and D on D'.

K H L F G D C E A B K' H' L' F' G' D' C' E' A' B'

Since the lateral edges of a prism are parallel, CH will fall on C'H', and DK on D'K'.

And the points F, G, L coincide respectively with F', G', L', so that the planes GK, G'K' coincide.

Hence H, K coincide with H', K' respectively. Thus the prisms coincide throughout and are equal.

In the same way we can prove that two truncated prisms with three faces forming a solid angle related to one another as in the above proposition are identically equal.

In particular, Cor. Two right prisms having equal bases and equal heights are equal.

3. An oblique prism is equivalent to a right prism whose base is a right section of the oblique prism and whose height is equal to a lateral edge of the oblique prism.

Suppose GL to be a right section of the oblique prism AD', and let GL' be a right prism on GL as base and with height equal to a lateral edge of AD'.

L' K' H' M' G' D' C' E' A' B' L K H M G D C E A B

Now the lateral edges of GL' are equal to the lateral edges of AD'.

Therefore AG = A'G', BH = B'H', CK = C'K', etc.

Thus the faces AH, BK, CL are equal respectively to the faces A'H', B'K', C'L'.

Therefore [by the proposition above] (truncated prism AL) = (truncated prism A'L').

Subtracting each from the whole solid AL', we see that the prisms AD', GL' are equivalent.

Now suppose the parallelepiped of Euclid’s proposition to be cut by the plane through AG, DF.

Let KLMN be a right section of the parallelepiped cutting the edges AD, BC, GF, HE. Then KLMN is a parallelogram; and, if the diagonal KM be drawn,

△KLM = △MNK.

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

Now the prism of which the $\Delta\text{s}$ ABG, DCF are the bases is equal to the right prism on △KLM as base and of height AD.

Similarly the prism of which the $\Delta\text{s}$ AGH, DFE are the bases is equal to the right prism on △MNK as base and with height AD. [(3) above]

And the right prisms on $\Delta\text{s}$ KLM, MNK as bases and of equal height AD are equal. [(2), Cor. above]

Consequently the two prisms into which the parallelepiped is divided are equivalent.