Upon the base ABD make the prism ABD-E'F'H' symmetrical with the prism ABD-EFH. According to what has been demonstrated (386), the plane ABFE' is equal to ABFE, and the plane ADH'E' is equal to ADHE; but, if we compare the prism GHF-BCD with the prism ABD-HE'F', the base GHF is equal to ABD; the parallelogram GHDC, which is equal to ABFE, is also equal to ABFE', and the parallelogram GFBC, which is equal to ADHE, is also equal to ADH'E'; therefore the three planes, which form the solid angle G in the prism GHF-BCD, are equal to the three planes, which form the solid angle A in the prism ABD-HE'F', each to each; they are, moreover, similarly disposed in the two cases; therefore these two prisms are equal, and, being applied the one to the other, would coincide. But one of them is symmetrical with the prism ABD-HEF; therefore the other GHF-BCD is also symmetrical with ABD-HEF.

LEMMA.

395. In every prism ABCI the sections NOPQR, STVXY (fig. 201), made by parallel planes are equal polygons.

Demonstration. The sides NO, ST, are parallel, being intersections of two parallel planes by a third plane ABGF; these same sides NO, ST, are comprehended between the parallels NS, OT, which are sides of the prism; consequently NO is equal to ST. For a similar reason, the sides OP, PQ, QR, &c., of the section NOPQR are equal respectively to the sides TV, VX, XY, &c., of the section STVXY. Besides, the equal sides being also parallel, it follows that the angles NOP, OPQ, &c., of the first section are equal respectively to the angles STV, TVX, &c., of the second (344). Therefore the two sections NOPQR, STVXY, are equal polygons.

396. Corollary. Every section made in a prism parallel to its base is equal to this base.

THEOREM.

Fig. 201

397. The two symmetrical triangular prisms ABD-HEF, BCD-HFG (fig. 208), which compose the parallelopiped AG, are Fig. 208. equivalent.

Demonstration. Through the vertices B, F, perpendicular to the side BF, suppose the planes Bad c, Fehg to pass, meeting

the three other sides, AE, DH, CG, of the parallelopiped, the one in a, d, c, the other in e, h, g; the sections Bad c, Fehg, will be equal parallelograms. They are equal, because they are made by planes, which are perpendicular to the same straight line, and consequently parallel (397); they are parallelograms, because the two opposite sides of the same section a B, d c, are the intersections of two parallel planes ABFE, DCGH, by the same plane.

For a similar reason, the figure Ba e F is a parallelogram, as also the other lateral faces BF g c, cdhg, adhe, of the solid Badc-Fehg; therefore this solid is a prism (368); and this prism is a right prism, since the side BF is perpendicular to the plane of the base.

This being premised, if the right prism Bh be divided by the plane BFHD into two right triangular prisms a B d-h e F, Bdc-g Fh, we say that the oblique triangular prism ABD-HEF will be equivalent to the right triangular prism a Bd-he F.

Indeed, as the two prisms have the part ABD he F common, it is necessary only to prove that the two remaining parts, namely, the solids Ba AD d, Fe EH h, are equivalent to each other.

Now, on account of the parallelograms ABFE, a BF e, the sides AE, a e, being each equal to its parallel BF, are equal to each other; if, then, we take away the common part A e, we shall have A a = E e. It may be shown, in like manner, that D d = Hh.

Now, in order to apply the two solids BaADd, Fe EH h, one to the other, let the base Feh be placed upon its equal Bad; the point e falling upon a, and the point h upon d, the sides e E, h H, will fall upon a A, d D, each upon its equal, since they are perpendicular to the same plane Bad; consequently the two solids under consideration will coincide entirely, the one with the other; therefore the oblique prism BAD-HFE is equivalent to the right prism B a d-h Fe.

It may be demonstrated, in like manner, that the oblique prism BDC-GFH is equivalent to the right prism B d c-g Fh. But the two right prisms Bad-Fch, Bd c-g Fh, are equal to each other, since they have the same altitude BF, and their bases Bad, Bdc, are each half of the same parallelogram (389). Therefore the two triangular prisms BAD-HFE, BDC-GFH, equivalent to equal prisms, are equivalent to each other.

398. Corollary. Every triangular prism ABD-HFE is half of the parallelopiped AG, constructed upon the same solid angle A with the same edges AB, AD, AE.

THEOREM.

399. If two parallelopipeds AG, AL (fig. 209), have a common Fig. 209. base ABCD, and have also their superior bases comprehended in the same plane and between the same parallels EK, HL, these two parallelopipeds will be equivalent.

Demonstration. The proposition admits of three cases, according as EI is greater than EF, less, or equal to it; but the demonstration is the same for each; and, in the first place, we say that the triangular prism AEI-MDH is equal to the triangular prism BFK-LCG.

Indeed, since AE is parallel to BF, and HE to GF, the angle AEI= BFK, HEI= GFK, HEA GFB. Of these six angles the three first form the solid angle E, and the three last the solid angle F; consequently, since these plane angles are equal, each to each, and similarly disposed, it follows that the solid angles E, F, are equal. Now, if the prism AEM be applied to the prism BFL, the base AEI being placed upon the base BFK, these two bases, being equal, will coincide; and, since the solid angle E is equal to the solid angle F, the side EH will fall upon its equal FG. Nothing further is necessary in order to show that the two prisms will coincide throughout; for the base AEI and its edge EH determine the prism AEM, as the base BFK and its edge FG determine the prism BFL (388); therefore these prisms are equal.

But, if from the solid AL we take the prism AEM, there will remain the parallelopiped AIL; and, if from the same solid AL we take the prism BFL, there will remain the parallelopiped AEG; therefore the two parallelopipeds AIL, AEG, are equivalent.

THEOREM.

400. Two parallelopipeds, which have the same base and the same altitude, are equivalent.

Demonstration. Let ABCD (fig. 210) be the common base of Fig. 210. two parallelopipeds AG, AL; since they have the same altitude,

Fig. 210.

Fig 211.

their superior bases EFGH, IKLM, will be in the same plane Moreover, the sides EF, AB are equal and parallel, as also IK, AB; consequently EF is equal and parallel to IK; for a similar reason, GF is equal and parallel to LK. Produce the sides EF, HG, also LK, MI, till they shall, by their intersections, form the parallelogram NOPQ; it is evident that this parallelogram will be equal to each of the bases EFGH, IKLM. Now, if a third parallelopiped be supposed, which, with the same inferior base ABCD, has for its superior base NOPQ, this third parallelopiped will be equivalent to the parallelopiped AG (399); since, the inferior base being the same, the superior bases are comprehended in the same plane and between the same parallels GQ, FN. For the same reason, this third parallelopiped will be equivalent to the parallelopiped AL, therefore the two parallelopipeds AG, AL, which have the same base and the same altitude, are equivalent.

THEOREM.

401. Every parallelopiped may be changed into an equivalent rectangular parallelopiped having the same altitude and an equivalent base.

Demonstration. Let AG (fig. 210) be the proposed parallelopiped; from the points A, B, C, D, draw AI, BK, CL, DM, perpendicular to the plane of the base, and we shall thus have the parallelopiped AL equivalent to the parallelopiped AG, and of which the lateral faces AK, BL, &c., will be rectangles. If, then, the base ABCD is a rectangle, AL will be the rectangular parallelopiped equivalent to the proposed parallelopiped AG.

But, if ABCD (fig. 211) is not a rectangle, draw AO, BN, each perpendicular to CD, also OQ, NP, each perpendicular to the base, and we shall have the solid ABNO-IKPQ, which will be a rectangular parallelopiped. Indeed, the base ABNO and the opposite base IKPQ are, by construction, rectangles; the lateral faces are also rectangles, since the edges AI, OQ, &c., are each perpendicular to the plane of the base; therefore the solid AP is a rectangular parallelopiped. But the two parallelopipeds AP, AL, may be considered as having the same base ABKI, and the same altitude AO; consequently they are equivalent; thereFig. 210, fore the parallelopiped AG (fig. 210, 211), which was first changed into an equivalent parallelopiped AL, is now changed

211.

into an equivalent rectangular parallelopiped AP, which has the same altitude AI, and of which the base ABNO is equivalent to the base ABCD.

THEOREM.

402. Two rectangular parallelopipeds AG, AL (fig. 212), Fig. 212. which have the same base ABCD, are to each other as their altitudes AE, AI.

Demonstration. Let us suppose, in the first place, that the altitudes AE, AI, are to each other as two entire numbers, as 15 to 8, for example, AE may be divided into 15 equal parts, of which AI will contain 8, and through the points of division x, y, z, &c., planes may be drawn parallel to the base. These planes will divide the solid AG into 15 partial parallelopipeds, which will be equal to each other, having equal bases and equal altitudes; we say equal bases, because every section of a prism MIKL, parallel to the base, is equal to this base (395), and equal altitudes, because the altitudes are the divisions themselves A x, xy, yz, &c. Now, of these 15 equal parallelopipeds 8 are contained in AL; therefore the solid AG is to the solid AL as 15 is to 8, or in general as the altitude AE is to the altitude AI.

Again, if the ratio of AE to AI cannot be expressed in numbers, we say still, that the proportion is not the less true, namely, solid AG solid AL:: AE : AI.

:

For, if this proportion does not hold, let us suppose that

solid AG solid AL:: AE: AO.

Divide AE into equal parts, each of which shall be less than IO; there will be at least one point of division m between I and O. Let P be the parallelopiped which has for its base ABCD and for its altitude Am; since the altitudes AE, A m, are to each other as two entire numbers, we shall have

But AO is greater than Am; it is necessary, then, in order tha. this proportion may take place, that the solid AL should be greater than P; on the contrary it is less; consequently it is impossible that the fourth term of the proportion