also a prism. Now we say that these two triangular prisms are equivalent to each other.

Indeed, if upon the edges GI, GE, GH, the parallelopiped GX be formed, the triangular prism EGI-CFH will be half of this parallelopiped (397); on the other hand, the prism AGH-FDE is also equal to half of the parallelopiped GX (406), since they have the same altitude, and the triangle AGH, the base of the prism, is half of the parallelogram GICH (168), the base of the parallelopiped. Therefore the two prisms E GI-CFH, AGH-FDE, are equivalent to each other.

These two prisms being taken from the pyramid S-ABC, there will remain only the two pyramids S-DEF, E-GBI; now we say that these two pyramids are equal to each other.

Indeed, since the following sides are equal, namely, BE = SE BG=AG = DE, EG = AD = SD, the triangle BEG is equal to the triangle ESD (43). For a similar reason, the triangle BEI is equal to the triangle ESF; moreover the mutual inclination of the two planes BEG, BEI, is the same as that of the two planes ESD, ESF, since BEG, ESD, are in the same plane, and BEI, ESF, are also in the same plane. If, then, in order to apply the one pyramid to the other, we place the triangle EBG upon its equal EDS, the plane BEI must fall upon the plane ESF; and, since the triangles are equal and similarly disposed, the point I will fall upon F, and the two pyramids will coincide throughout (384).

Therefore the entire pyramid S-ABC is composed of two triangular prisms AGF, GIF, equivalent to each other, and two equal pyramids S-DEF, E-GBI.

411. Corollary 1. From the vertex S let fall upon the plane ABC the perpendicular SO, and let P be the point, where this perpendicular meets the plane DEF, parallel to ABC; since SD=SA, we have SP=SO (408), and the triangle DEF= triangle ABC (218); consequently the solidity of the prism

AGH-FDE ABCX SO;
=

14

and the solidity of the two prisms AGH-FDE, EGI-CFH, taken together, is equal ¦ ABC × SO. These two prisms are less than the pyramid S-ABC, since they are contained in it; therefore the solidity of a triangular pyramid is greater than the fourth part of the product of its base by its altitude

412. Corollary 11. If we join DG, DH, we shall have a new pyramid D-AGH equal to the pyramid S-DEF; for the base DEF may be placed upon its equal AGH, and then, the angles SDE, SDF, being equal to the angles DAG, DAH, it is mani fest that DS will fall upon AD (364), and the vertex S upon the vertex D. Now the pyramid D-AGH is less than the prism AGH-FDE, since it is contained in it; therefore each of the pyramids S-DEF, E-GBI, is less than the prism AGH-FDE; therefore the pyramid S-ABC, which is composed of two pyramids and two prisms, is less than four of these same prisms. But the solidity of one of these prisms ABC × SO, and its quadruple ABC × SO; hence the solidity of any triangular pyramid is less than half of the product of its base by its altitude.

=

Fig. 215.

THEOREM.

413. The solidity of a triangular pyramid is equal to a third of the product of its base by its altitude.

Demonstration. Let S-ABC (fig. 215) be any triangular pyramid, ABC its base, SO its altitude; we say that the solidity of the pyramid S-ABC is equal to a third part of the product of the surface ABC by the altitude SO, so that

S-ABC ABC × SO, or = SOX

=

ABC.

If this proposition be denied, the solidity S-ABC must be equal to the product of SO by a surface either greater or less thanABC.

1. Let this quantity be greater, so that we shall have

S-ABC SOX (} ABC+M).

If we make the same construction as in the preceding proposition, the pyramid S-ABC will be divided into two equivalent prisms AGH-FDE, EGI-CFH, and two equal pyramids S-DEF, E-GBI. Now the solidity of the prism AGH-FDE is DEFX PO, consequently we shall have the solidity of the two prisms

AGH-FDE+EGI-CFH = DEF × 2PO, or = DEF × SO. The two prisms being taken from the entire pyramid, the remainder will be equal to double of the pyramid S-DEF, so that we shall have

2S-DEF = SO × (¦ ABC+M— DEF). But, because SA is double of SD, the surface ABC is quadruple of DEF (408), and thus

whence

ABC — DEF = ÷ DEF — DEF = } DEF ;

2S-DEF = SO × (} DEF + M),

or, by taking the half of each,

S-DEF SPX (} DEF+M).

It appears, then, that in order to obtain the solidity of the pyramid S-DEF, it is necessary to add to a third of the base the same surface M, which was added to a third of the base of the large pyramid, and to multiply the whole by the altitude SP of the small pyramid.

If SD be bisected at the point K, and if through this point a plane KLM be supposed to pass parallel to DEF, meeting the perpendicular SP in Q; according to what has just been demonstrated, S-KLM = SQ × (1⁄2 KLM + M).

If we proceed thus to form a series of pyramids, the sides of which decrease in the ratio of 2 to 1, and the bases in the ratio 4 to 1, we shall soon arrive at a pyramid S-a b c, the base of which a b c shall be less than 6M. Let So be the altitude of

this last pyramid; and its solidity, deduced from that of the preceding pyramids, will be

Sox ( a b c + M).

It

But M > a b c, and consequently a b c + M > } abc. would follow, then, that the solidity of the pyramid S-a b c is greater than Sox abc; which is absurd, since it was proved in the preceding proposition, corollary 11, that the solidity of a triangular pyramid is always less than half of the product of its base by its altitude; therefore it is impossible that the solidity of the pyramid S-ABC should be greater than SO × ABC.

2. Let S-ABC be equal to SO × (ABC-M); it may be shown, as in the first case, that the solidity of the pyramid S-DEF, the dimensions of which are less by one half, is equal to SPX (DEF—M);

and, by continuing the series of pyramids, the sides of which decrease in the ratio of 2 to 1, until we arrive at a term S-a bc, we shall, in like manner, have the solidity of this last equal to

But, as the bases ABC, DEF, KLM..... ab c, form a decreasing series, each term of which is a fourth of the preceding, we shall soon arrive at a term a b c equal to 12M, or which shall be comprehended between 12M and 3M; then, M being either

Fig. 214.

equal to or greater than a b c, the quantity abc-M will either be equal to or less than abc; so that we shall have the solidity of the pyramid S-a b c either equal to or less than

Sox abc;

which is absurd, since, according to corollary 1 of the preceding proposition, the solidity of a triangular pyramid is always greater than the fourth of the product of its base by its altitude; therefore the solidity of the pyramid S-ABC cannot be less than SO × ABC.

We conclude, then, according to the enunciation of the theorem, that the solidity of the pyramid SABC= SO × ¦ ABC, or = ABC × SO.

414. Corollary 1. Every triangular pyramid is a third of a triangular prism of the same base and same altitude; for ABC × SO is the solidity of the prism of which ABC is the base and SO the altitude.

415. Corollary II. Two triangular pyramids of the same altitude are to each other as their bases, and two triangular pyramids of the same base are to each other as their altitudes.

THEOREM.

416. Every pyramid S-ABCDE (fig. 214) has for its measure a third of the product of its base by its altitude.

Demonstration. If the planes SEB, SEC, be made to pass through the diagonals EB, EC, the polygonal pyramid S-ABCDE will be divided into several triangular pyramids, which have all the same altitude SO. But, by the preceding theorem, these are measured by multiplying their bases ABE, BCE, CDE, each by a third of its altitude SO; consequently the sum of the triangular pyramids, or the polygonal pyramid S-ABCDE will have for its measure the sum of the triangles ABE, BCE, CDE, or the polygon ABCDE, multiplied by SO; therefore every pyramid has for its measure a third of the product of its base by its altitude.

417. Corollary 1. Every pyramid is a third of a prism of the same base and same altitude.

418. Corollary 11. Two pyramids of the same altitude are to each other as their bases, and two pyramids of the same base are to each other as their altitudes.

419. Scholium. The solidity of any polyedron may be estimated by decomposing it into pyramids, and this decomposition may be effected in several ways; one of the most simple is by means of planes of division passing through the vertex of the same solid angle; then we shall have as many partial pyramids as there are faces in the polyedron excepting those which contain the solid angle from which the planes of division proceed.

THEOREM.

420. Two symmetrical polyedrons are equivalent to each other, or equal in solidity.

Demonstration. 1. Two symmetrical triangular pyramids, as S-ABC, T-ABC (fig. 202), have each for its measure the prod- Fig. 202. uct of the base ABC by a third of its altitude SO or TO; therefore these pyramids are equivalent.

2. If we divide, in any manner, one of the symmetrical polyedrons in question into triangular pyramids, we can divide the other polyedron, in the same manner, into triangular pyramids symmetrical with the former (382); but the triangular pyramids in the one case and the other being symmetrical, are equivalent, each to each; therefore the entire polyedrons are equivalent to each other, or equal in solidity.

421. Scholium. This proposition seems to result immedi ately from a former (386), in which it was shown that, with respect to two symmetrical polyedrons, all the constituent parts of the one are equal respectively to those of the other; still it was necessary to demonstrate it in a rigorous manner.

THEOREM.

422. If a pyramid is cut by a plane parallel to its base, the frustum which remains, after taking away the smaller pyramid, is equal to the sum of three pyramids, which have for their common altitude the altitude of the frustum, and whose bases are the inferior base of the frustum, its superior base, and a mean proportional between these bases.

Demonstration. Let S-ABCDE (fig. 217) be a pyramid cut Fig. 217. by the plane a b d parallel to the base; let T-FGH be a triangular pyramid, whose base and altitude are equal or equivalent to the base and altitude of the pyramid S-ABCDE. The two