Fig. 206.

sides is equal to the altitude of the prism. In every other case the prism is oblique, and the altitude is less than the side.

372. A prism is triangular, quadrangular, pentagonal, hexagonal, &c., according as the base is a triangle, a quadrilateral, a pentagon, a hexagon, &c.

373. A prism whose base is a parallelogram (fig. 206), has all its faces parallelograms, and is called a parallelopiped.

A parallelopiped is rectangular, when all its faces are rectangles.

374. Among rectangular parallelopipeds is distinguished the cube or regular hexaedron comprehended under six equal squares.

375. A pyramid is a solid formed by several triangular planes proceeding from the same point and terminating in the sides of a Fig. 196. polygon ABCDE (fig. 196).

Fig. 202.

The polygon ABCDE is called the base of the pyramid, the point S its vertex, and the triangles ASB, BSC, &c., compose the lateral or convex surface of the pyramid.

376. The altitude of a pyramid is the perpendicular let fall from the vertex upon the plane of the base, produced if necessary. 377. A pyramid is triangular, quadrangular, &c., according as the base is a triangle, a quadrilateral, &c.

378. A pyramid is regular, when the base is a regular polygon, and the perpendicular, let fall from the vertex to the plane of the base, passes through the centre of this base. This line is called the axis of the pyramid.

379. The diagonal of a polyedron is a straight line which joins the vertices of two solid angles not adjacent.

380. I shall call symmetrical polyedrons two polyedrons which, having a common base, are similarly constructed, the one above the plane of this base and other below it, with this condition, that the vertices of the homologous solid angles be situated at equal distances from the plane of the base, in the same straight line perpendicular to this plane.

If the straight line ST (fig. 202), for example, is perpendicular to the plane ABC, and is bisected at the point O, where it meets this plane, the two pyramids SABC, TABC, which have the common base ABC, are two symmetrical polyedrons.

381. Two triangular pyramids are similar when they have two faces similar, each to each, similarly placed, and equally inclined to each other.

Thus, if we suppose the angle ABC= DEF, BAC = EDF, ABS = DET, BAS = EDT, (fig. 203), if also the inclination of Fig. 203. the planes ABS, ABC, is equal to that of their homologous planes DTE, DEF, the pyramids SABC, TDEF, are similar.

382. Having formed a triangle with the vertices of three angles, taken in the same face or base of a polyedron, we can imagine the vertices of the different solid angles of the polyedrons, situated out of the plane of this base, to be the vertices of as many triangular pyramids, which have for their common base the above triangle; and these several pyramids will determine the positions of the several solid angles of the polyedron with respect to the base. This being supposed;

Two polyedrons are similar, when, the bases being similar, the vertices of the homologous solid angles are determined by triangular pyramids similar each to each.

383. I shall call vertices of a polyedron the points situated at the vertices of the different solid angles.

N. B. We shall consider only those polyedrons, which have saliant angles, or convex polyedrons. We thus denominate those, the surface of which cannot be met by a straight line in more than two points. In polyedrons of this description the plane of neither of the faces can, by being produced, cut the solid; it is impossible then, that the polyedron should be in part above the plane of one of the faces and in part below it; it is wholly on one side of this plane.

THEOREM.

384. Two polyedrons cannot have the same vertices, the number also being the same, without coinciding the one with the other.

Demonstration. Let us suppose one of the polyedrons already constructed, if we would construct another having the same vertices, the number also being the same, it is necessary that the planes of this last should not all pass through the same points as in the first; if they did, they would not differ the one from the other; but then it is evident that any new planes would cut the first polyedron; there would then be vertices above these planes and vertices below them, which does not consist with a convex polyedron; therefore, if two polyedrons have the same vertices, the number also being the same, they must necessarily coincide the one with the other.

Fig 204.

385. Scholium. The points A, B, C, K, &c. (fig. 204), being given in position to be used as the vertices of a polyedron, it is easy to describe the polyedron.

Take, in the first place, three neighbouring points D, E, H, such that the plane DEH shall pass, if there is occasion for it, through other points K, C, but leaving all the rest on the same side, all above the plane, or all below it; the plane DEH or DEHKC, thus determined, will be a face of the solid. Through one of the sides EH of this face, suppose a plane to pass, and to turn upon this line until it meets a new vertex F, or several at the same time F, I; we shall thus have a second face FEH or FEHI. Proceed in this manner, by making planes to pass through the sides of the faces, until the solid is terminated in all directions; this solid will be the polyedron required, for there are not two which can have the same vertices.

Fig. 205.

THEOREM.

386. In two symmetrical polyedrons the homologous faces are equal, each to each, and the inclination of two adjacent faces in one of the solids is equal to the inclination of the homologous faces in the

other.

Demonstration. Let ABCDE (fig. 205) be the common base of the two polyedrons, M and N the vertices of any two solid angles of one of the polyedrons, M' and N' the homologous vertices of the other polyedron; the straight lines MM', NN', must be perpendicular to the plane ABC, and be bisected at the points m and n (380), where they meet this plane. This being supposed, we say that the distance MN is equal to M'N'.

For, if the trapezoid m M'N'n be made to revolve about mn, until its plane comes into the position of the plane m MN n, on account of the right angles at m and n, the side m M' will fall upon its equal m M, and n N' upon n N; therefore the two trapezoids will coincide, and we shall have MNM'N'.

Let P be a third vertex in the superior polyedron, and P' the homologous vertex in the other, we shall have, in like manner, MP= M'P', and NP = N'P'; consequently the triangle MNP, formed by joining any three vertices of the superior polyedron is equal to the triangle M'N'P', formed by joining the homologous vertices of the other polyedron.

If, among these triangles we consider only those which are formed at the surface of the polyedrons, we can conclude already that the surfaces of the two polyedrons are composed of the same number of triangles equal, each to each.

We say now that, if some of these triangles are in the same. plane upon one surface and form the same polygonal face, the homologous triangles will be in the same plane upon the other surface and will form an equal polygonal face.

Let MPN, NPQ, be two adjacent triangles, which we suppose in the same plane, and let M'P'N', N'P'Q', be the homologous triangles. We have the angle MNP = M'N'P', the angle PNQ = P'N'Q'; and, if we were to join MQ and M'Q', the triangle MNQ would be equal to M'N'Q', thus we should have the angle MNQ=M'N'Q'. But, since MPNQ is onc plane, we have the angle MNQ = MNP + PNQ; we have also

M'N'QM'N'P' + P'N'Q'.

Now, if the three planes M'N'P', P'N'Q', M'N'Q', are not confounded in one, they will form a solid angle, and we shall have the angle M'N'Q' < M'N'P' + P'N'Q' (356); therefore, as this condition does not exist, the two triangles M'N'P', P'N'Q', are in the same plane.

We hence infer that each face, whether triangular, or polygonal, in one polyedron, corresponds to an equal face in the other, and that thus the two polyedrons are comprehended under the same number of planes equal, each to each.

It remains to show that the inclination of any two adjacent faces in one of the polyedrons is equal to the inclination of the two homologous faces in the other.

Let MPN, NPQ, be two triangles formed upon the common edge NP in the planes of two adjacent faces; let M'P'N', N'P'Q', be the homologous triangles. We can conceive at No a solid angle formed by the three plane angles MNQ, MNP, PNQ, and at N' a solid angle formed by the three M'N'Q', M'N'P', P'N'Q. Now it has already been proved that these plane angles are equal, cach to each; consequently the inclination of the two planes MNP, PNQ, is equal to that of their homologous planes M'N'P', P'N'Q'′ (359).

Therefore in symmetrical polyedrons the faces are equal, each to each, and the planes of any two adjacent faces of one of the solids have the same inclination to each other as the planes of the two homologous faces of the other solid.

387. Scholium. It may be remarked that the solid angles of the one polyedron are symmetrical with the solid angles of the other; for, if the solid angle N is formed by the planes MNP, PNQ, QNR, &c., its homologous angle N' is formed by the planes M'N'P', P'N'Q', Q'N'R', &c. These last seem to be disposed in the same order as the others; but, as one of the solid angles is inverted with respect to the other, it follows that the actual disposition of the planes, which form the solid angle N' is the reverse of that which exists with respect to the homologous an gle N. Moreover the inclinations of the successive planes in the one are equal respectively to those in the other; therefore these solid angles are symmetrical with respect to each other. Sec

art. 360.

It will be perceived, from what has been said, that any polye dron whatever can have only one polyedron symmetrical with it. For, if there were constructed, upon another base, a new polyedron symmetrical with the given polyedron, the solid angles of this last would always be symmetrical with the angles of the given polyedron; consequently they would be cqual to those of the symmetrical polyedron constructed upon the first base. Moreover, the homologous faces would always be equal; whence these two symmetrical polyedrons, constructed upon the one base and upon the other, would have their faces equal and their solid angles equal; therefore they would coincide by superposition, and would make one and the same polyedron.

Fig. 200.

THEOREM.

388. Two prisms are equal, when three planes containing a solid angle of the one are equal to three planes containing a solid angle of the other, each to each, and are similarly placed.

Demonstration. Let the base ABCDE (fig. 200), be equal to the base abcde, the parallelogram ABGF equal to the patallelogram a bgf, and the parallelogram BCHG equal to the parallelogram bchg; we say that the prism ABCI will be equal to the prism abci.

For, let the base ABCDE be placed upon the base a b c d e, the two bases will coincide. But the three plane angles, which form the solid angle B, are equal to the three plane angles, which from the solid angle b, each to each, namely ABC=abc, AEG=abg,