PROPOSITION III. If the three faces of one trihedral angle be equal to the three faces of

another, each to each : then the dihedral angles of the former will be equal to those of the latter, each to each ; viz., those to which the equal faces are opposite. Case 1. Let the equal parts of the two figures lie in the same order

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in the two trihedral angles S. ABC, s. abc. Make SA, SB, SC, sa, sb, sc, all equal to one another; and draw the planes ABC, abc. Through S and s draw perpendiculars SP, sp to those planes ; and join AP, PB, ap, pb.

Then, since AS, SB, are equal to as, sb, (Constr.) and the angle ASB to asb (Hyp.), the base AB is equal to ab. In a similar manner BC is equal to bc, and CA to ca. Wherefore the two triangles ABC abc are in all respects equal.

Then (Prop. 11., Chap. iv.) P and p are the centres of the circles which circumscribe ABC, abc; and, therefore, their radii AP, PB, ap, pb are equal; and the triangle APB is in all respects equal to apb.

Let, now, the triangle ABC be applied to abc, so that AB, BC coincide with ab, bc: then the triangle ABC will coincide entirely with abc, and the point P with p (the bases AB, ab coinciding, and the sides being equal each to each). The perpendicular PS, therefore, will coincide with the perpendicular ps; and the vertex S will be in the line ps.

If it be denied that S coincides with s, let it be at some other point s' in the perpendicular ps; and join as', bs', cs'.

Then since the two sides cs', bs', are equal to CS, BS, and these again to cs, bs; the two triangles csb, cs'b have the two sides cs, bs, equal to the two cs', bs', and the base bc common. Whence the angle cs'b is equal to the angle csb.

But p being the centre of the circle about abc, the perpendicular from p falls between b and c; and hence (Prop. V., Chap. 111.) the angle cs'b is greater or less than csb, according as s' is in ps, or in its prolongation.

These two conclusions are contradictory; and hence the non-coincidence of S and s impossible. Whence S coincides with s, and the trihedral angles having their parts coincident, those parts are equal each to each ; viz., the dihedral angles opposite to the equal faces.

Case 2. Let the equal parts be ranged in reverse order in the two trihedral angles. Produce the three faces of S. ABC: their intersections SA', SB', SC' will be in continuous lines with SA, SB, SC; and

hence the angles A'SB', B'SC', C'SA' being opposite to ASB, BSC

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CSA, are respectively equal to them. Whence the faces of S. A'B'C' are also respectively equal to those of s. abc; and by the former case, the dihedral angles of S. A'B'C' are equal to those of s. abc, each to each, respectively.

But the dihedral angle (A'SB', B'SC') is opposite to (ASB, BSC), and therefore equal to it; and hence the dihedral angle (ASB, BSC) is equal to (asb, bsc). In the same manner the other dihedral angles are proved to be equal.



Tais Chapter is the foundation of the “ Mensuration of Solids," and has scarcely any other direct application in mathematics.

The general history of the method is this, as far as regards polyhedrons :

(1.) To investigate the conditions of equality of two parallelopipeds, or their proportionality to other data.

(2.) To extend the same to triangular prisms-sometimes even to include these amongst the former.

(3.) To divide every polyhedron into parallelopipeds or prisms (most commonly the latter), and then assign a rectangular parallelopiped equal to the sum of them all.

Of the curved solids, notice will be taken hereafter.

It will conduce to convenience of expression to make an unusual de. finitional distinction here, between a right parallelopiped and a rectangular one.

DEF. 1. A rectangular parallelopiped is one which has all its angles right angles and all its faces rectangles.

Der. 2. A right parallelopiped has for one pair of opposite faces any pair of equal parallelograms; but the other four faces rectangles.

This is in reality considering them as right prisms upon rectangular bases, and upon parallelogram bases respectively. The view is only changed from its somewhat simplifying the language of the exposition.


There is one important principle to be noticed here, in regard to the equality of solids bounded by plane surfaces. It came before us partially in the last Chapter, but it takes a prominent place in this.

It has been already shown that two trihedral angles may have all the component parts of the one equal to all the corresponding ones of the other, each to each: but that, except these parts succeed each other, in the same order of rotation, the two trihedral angles do not admit of supraposition. When they succeed in a reversed order of rotation, they have been called, as already explained, symmetrical trihedral angles. We can hardly deny the equality of two things, all the separate elements of which are the same, each in each ; merely because their order of succession is from right to left instead of from left to right. It was not, however, necessary to insist upon this in the discussion of solid angles, as the terms “ equal” and “ symmetrical” met the two cases, without any further inference being drawn. Here, however, where the volume of the space enclosed is concerned, the case is otherwise. We are compelled either to prove or assume that the volumes of space comprehended by the boundaries of two symmetrical parallelopipeds or triangular prisms are equal.

I have never seen a satisfactory proof of this, without the adoption of considerations extraneous to those formally or virtually admitted by Euclid. Neither, though I have long thought about it, have I been more successful than my predecessors.

We have, then, no alternative but to assume as an axiom :

“ If the same triangular prism be equal to one triangular prism and symmetrical to another, these latter prisms contain equal volumes of space.”

The case of the parallelopiped can be proved from this, or rather it is a mere corollary. Indeed we are entitled to lay down as a working form of this axiom that:

Symmetrical triangular prisms and symmetrical parallelopipeds contain equal volumes of space, each to each.

It is deemed proper to draw the student's especial attention to this, rather than by passing it over, to leave him under the belief that his demonstration by supraposition of the equality of symmetrical figures was more rigidly complete than it actually was.

It may be remarked that the reason of this difficulty (that suprapotion should be universally applicable in the case of plane rectilineal figures, and not in solids bounded by planes) is thus explained. The uniformity of the plane renders a rectilineal figure capable of being turned over, so as to enable the figures traced on it to be applied one to the other in the same order though originally given in a reversed one. All the parts of both figures being in the coincident planes, there is no part of the one which may not coincide with a correspondent part of the other. The same, indeed, may be done with any face of the one of two symmetrical prisms and the corresponding face of the other; but this gives the prisms themselves on opposite sides of that common plane, and consequently not in a state of coincidence or supraposition.

Further, if we conceive the figures compared to be successively fitted in a hollow or matrix, it might simplify the conception of this solid supraposition or substitution.

PROPOSITION I. If two triangular prisms have one solid angle of the one equal or symmetrical to one solid angle of the other, and the three contiguous edges equal, they will have equal volumes.*

(1.) Let the triangular prisms ABCDEF and A'B'C'D'E'F' have the three plane angles at A equal to the three plane angles at A', and in the same order; and let the three edges at A be equal to the three edges at A': then these prisms will contain equal volumes.

For if A' be placed on A, A'B' on AB and the plane A'B'C' on ABC, the side A'C' will coincide with AC and the point C' with the point C, since the angle B'A'C' is equal to BAC, and the side A'C' to the side AC.

Also because the trihedral angle at A' is equal to that at A, the line A'D' will coincide with AD; and because A'D' is equal to AD the point D' will coincide with D.

Again, since D'A' and A'C' coincide with DA and AC, and ADEB, A'D'E'B' are equiangular parallelograms, the lines D'E', E'B' will coincide with DE, EB, and the point E' with the point E.

In the same manner the faces F'D'A'C' and FDAC will coincide.

Whence also the face F'D'E' will coincide with FDE, and F'C'B'E' with FCBE.

The boundaries of the two prisms therefore coinciding, they “ fill: the same space," and are therefore equal.

(2.) Let the two prisms be symmetrical : then by the axiom, they are of equal volume. We give the figure for illustration.

* Whatever three conditions are given for defining the equality of the solid angles, the equality of those angles justifies our assumption of any other three that may be more convenient for immediate use. We shall avail ourselves of this fact to avoid circumlocution in the details of our demonstratious..

In these the angle B'A'C' is turned from the side A'B' in a reversed order to that in which BAC is turned from AB. If the face A'D'E'B' were made to coincide with ADEB, the prisms would lie on opposite sides of the plane of the paper (a plane common to both prisms), and coincidence would not take place. The line A'D' would not generally coincide with AD; indeed only when the planes D'A'B', DAŽ were perpendicular to the planes B'A'C", BAC respectively, or (which implies the same thing) when the angles DAB, DAC, D'A'B', D'A'C' were all equal.

If the base A'B'C' be reversed 80 as to apply altogether to ABC, the prisms fall on opposite sides of that base; and hence equality by coincidence cannot be inferred. The axiom alone can justify the inference of equality of volumes.

Or again, if the planes of the bases be made continuous, symmetry of the figures results, but no coincidence.

The same conclusions follow in whatever way we dispose of the Ek figures for the purpose of comparison.

PROPOSITION II. Two parallelopipeds which have one trihedral angle of the one equal or symmetrical to that of another angle in the other; and the edges uniting in those angles equal, each to each corresponding one, will be of equal volume.

The proof is so similar to that in the preceding proposition, that the student ought to draw it up for himself.

PROPOSITION III. A plane drawn through two opposite edges of a parallelopiped divides its volume into two equal parts.

For it is easily to be shown (by the student) that the parallelopiped is thereby divided into two symmetrical triangular pyramids; and hence the inference by means of the axiom.

PROPOSITION IV. Any plane drawn through the centre of a parallelopiped divides it into two parts which are of equal volume.

This is a slight extension of the preceding one, and it is left for the student to prove that the parts are symmetrical, and hence of equal volume.

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