The observation we have just made proves, moreover, that no polyedron can have more than one polyedron symmetrical with it. For if upon a second base, a new polyedron were constructed symmetrical with the given one, the solid angles of this new polyedron would still be symmetrical with those of the given polyedron; hence they would be equal to those of the symmetrical polyedron constructed on the first base. Besides, the homologous faces would still be equal: hence those two symmetrical polyedrons constructed on the first and on the second base, would have their faces equal and their solid angles equal; hence they would coincide if applied to each other; hence they would form one and the same polyedron.

PROPOSITION III. THEOREM.

Two prisms are equal when a solid angle in each is contained by three planes which are similarly placed, and respectively equal in both.

Let the base ABCDE be equal to the base abcde, the parallelogram ABGF equal to the parallelogram abgf, and the parallelogram BCHG equal to bchg; then will the prism ABCI be equal to the prism abci.

For, lay the base ABCDE upon its equal abcde; these two bases will coincide. But the three plane angles, which form the solid angle B, are respectively equal to the three plane angles, which form the solid angle b, namely, ABC=abc, ABG abg, and GBC=gbc; they are also similarly situated: hence the solid angles B and b are equal, and therefore the side BG will fall on its equal bg. It is likewise evident, that by reason of the equal parallelograms ABGF, abgf, the side GF will fall on its equal gf, and in the same manner GH on gh; hence the upper base FGHIK will exactly coincide with its equal

fghik, and the two solids will be identical (1. VI.), since their vertices are the same.

Cor. Two right prisms, which have equal bases and equal altitudes, are equal. For, since the side AB is equal to ab, and the altitude BG to bg, the rectangle ABGF will be equal to abgf; so also will the rectangle BGHC be to bghc; and thus the three planes, which form the solid angle B, will be equal to the three which form the solid angle b. Hence the two prisms are equal.

PROPOSITION IV. THEOREM.

In every parallelepipedon the opposite planes are equal and parallel.

G

By the definition of this solid, the bases ABCD, EFGH are equal parallelograms, and their sides are parallel: it remains only to shew, that the same is true of any two opposite lateral faces, such as AEHD, BFGC. Now AD is equal and parallel to BC, because the figure ABCD is a parallelogram; for a like reason, AE is parallel to BF: hence (13. V.) the angle DAE is equal to the angle CBF, and the planes DAE, CBF are parallel; hence also the parallelogram DAEH is equal to the parallelogram CBFG. In the same way, it might be shewn that the opposite parallelograms ABFE, DCGH are equal and parallel.

Cor. Since the parallelepipedon is a solid bounded by six planes, whereof those lying opposite to each other are equal and parallel, it follows that any face and the one opposite to it may be assumed as the bases of the parallelepipedon.

Scholium. If three straight lines AB, AE, AD, passing through the same point A, and making given angles with each other, are known, a parallelepipedon may be formed on those lines. For this purpose, a plane must be extended through the extremity of each line, and parallel to the plane of the other two; that is, through the point B a plane parallel to DAE, through D a plane parallel to BAE, and through E a plane parallel to BAD. The mutual intersections of those planes will form the parallelepipedon required.

PROPOSITION V. THEOREM.

In every parallelepipedon, the opposite solid angles are symmetrical; and the diagonals drawn through the vertices of those angles bisect each other.

2

First. Compare the solid angle A (see preceding figure) with its opposite one G. The angle EAB, equal to EFB, is also equal to HGC; the angle DĂE DHE CGF; and the angle DAB=DCB=HGF; therefore the three plane angles, which form the solid angle A, are equal to the three which form the solid angle G, each to each. It is easy, moreover, to see that their arrangement in the one is different from their arrangement in the other: hence (23. V.) the two solid angles A and G are symmetrical.

Secondly. Imagine two diagonals EC, AG to be drawn both through opposite vertices: since AE is equal and parallel to CG, the figure AEGC is a parallelogram; hence the diagonals EC, AG will mutually bisect each other. In the same manner, we could shew that the diagonal EC and another DF bisect each other; hence the four diagonals will mutually bisect each other, in a point which may be regarded as the centre of the parallelepipedon.

PROPOSITION VI. THEOREM.

If the plane BDHF pass through two opposite and parallel edges BF, DH, it will divide the parallelepipedon AG into two triangular prisms ABDHEF, GHFBCD, which will be symmetrical with each other.

In the first place, those solids are evidently prisms; for the triangles ABD, EFH, having their sides equal and parallel, are equal; also the lateral faces ABFE, ADHE, BDHF are parallelograms; hence the solid ABDĤEF is a prism so likewise is GHFBCD. We are now to shew that those prisms are symmetrical.

On the base ABD, construct the prism ABDE/F/H' such that it be symmetrical with the prism ABDEFH. According to what has been already proved (2. VI.), the plane ABF/E/ must be equal to ABFE, and the plane ADH/E to

E

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ADHE: but comparing the prism GHFBCD with the prism

ABDH/E/F', we find the base GHF-ABD; the parallelogram GHDC, which is equal to ABFE, also equal to ABF/E/; and the parallelogram GFBC, which is equal to ADHE, also equal to ADH'E' therefore the three plane planes which form the solid angle G in the prism GHFBCD are equal to the three planes which form the solid angle A in the prism ABDH/E/F', each to each; and they are similarly arranged: hence those two prisms are equal, and might be made to coincide. But one of them ABDH/E/F/ is symmetrical with the prism ABHEF; hence the other GHFBCD is also symmetrical with that prism.

PROPOSITION VII. LEMMA.

In every prism ABCI, the sections NOPQR, STVXY, formed by parallel planes, are equal polygons.

:

F

H

Y

S

22

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For the sides ST, NO, are parallel, being the intersections of two parallel planes with a third plane ABGF; those same sides ST, NO, are included between the parallels NS, OT, which are sides of the prism hence NO is equal to ST. For like reasons, the sides OP, PQ, QR, &c. of the section NOPQR, are respectively equal to the sides TV, VX, XY, &c. of the section STVXY. And since the equal sides are at the same time parallel, it follows that the angles NOP, OPQ, &c. of the first section are respectively equal to the angles STV, TVX of the second. Hence the two sections NOPQR, STVXY are equal polygons.

A

N

E

B

Cor. Every section in a prism, if drawn parallel to the base, is also equal to that base.

PROPOSITION VIII. THEOREM.

The two symmetrical triangular prisms ABDHEF, BCDFGH, into which the parallelepipedon AG may be decomposed, are equivalent.

a

E

H

G

Through the vertices B and F, draw the planes Badc, Fehg at right angles to the side BF, and meeting AE, DH, CG the three other sides of the parallelepipedon, in the e points a, d, c towards one direction, in e, h, g towards the other: the sections Badc, Fehg will be equal parallelograms. They are equal (7. VI.), because they are formed by planes A perpendicular to the same straight line, and consequently parallel; they are parallelograms, because aB, dc, two opposite sides of the same section, are formed by the meeting of one plane with two parallel planes ABFE, DCGH. For a like reason, the figure BaeF is a parallelogram; so also are BFgc, cdhg, adhe, the other lateral faces of the solid BadcFehg; hence that solid is a prism (Def. 4. VI.); and that prism is right, because the side BF is perpendicular to its base.

This being proved, if the right prism Bh is divided, by the plane BFHD, into two right triangular prisms aBdeFh, BdcFhg; we are now to shew that the oblique triangular prism ABDEFH will be equivalent to the right triangular prism aBdeFh. And since those two prisms have a part ABDheF in common, it will only be requisite to prove that the remaining parts, namely, the solids BaADd, FeEHh are equivalent.

Now, by reason of the parallelograms ABFE, aBFe, the sides AE, ae, being equal to their parallel BF, are equal to each other; and taking away the common part Ae, there remains Aa Ee. In the same manner we could prove Dd-Hh. Next, to bring about the superposition of the two solids BaADd, FeEHh, let us place the base Feh on its equal Bad: the point e falling on a, and the point h on d, the sides eE, hH will fall on their equals aA, dD, because they are perpendicular to the same plane Bad. Hence the two solids in question will coincide exactly with each other; hence the oblique prism BADFEH is equivalent to the right one BadFeh

In the same manner might the oblique prism BDCFHG be proved equivalent to the right prism BdcFhg. But (3. VI. Cor.) the two right prisms BadFeh, BdcFhg are equal, since