Fig. 218.

bases may be supposed to be situated in the same plane; and then the plane ab d produced will determine in the triangular pyramid a section f g h situated at the same altitude above the common plane of the bases; whence it follows that the section fgh is to the section abd as the base FGH is to the base ABD (408); and, since the bases are equivalent, the sections will be equivalent also. Consequently the pyramids S-a b c d e, T-f g h, are equivalent, since they have the same altitude and equivalent bases. The entire pyramids S-ABCDE, T-FGH, are equivalent, for the same reason; therefore the frustums ABD-d a b, FGH-h ƒ g, are equivalent; and consequently it will be sufficient to demonstrate the proposition enunciated, with reference merely to the case of the frustum of a triangular pyramid.

Let FGH-hfg (fig. 218), be the frustum of a triangular pyramid; through the points F, g, H, suppose a plane Fg H to pass, cutting off from the frustum the triangular pyramid g-FGH. This pyramid has for its base the inferior base FGH of the frustum; it has also for its altitude the altitude of the frustum, since the vertex g is in the plane of the superior base f g h.

This pyramid being cut off, there will remain the quadrangular pyramid gfh HF, the vertex of which is g, and the base fhHF. Through the three points f, g, H. suppose a plane fg H to pass, dividing the quadrangular pyramid into two triangular pyramids g-FƒH, gfh H. This last pyramid may be considered as having for its base the superior base gƒh of the frustum, and for its altitude the altitude of the frustum, since the vertex H is in the inferior base. Thus we have two of the three pyramids which compose the frustum.

It remains to consider the third pyramid g-Fƒ H. Now, if we draw g K parallel to ƒ F, and suppose a new pyramid K-F ƒ H, the vertex of which is K, and the base FƒH, these two pyramids will have the same base FfH; they will have also the same altitude, since the vertices g, K, are situated in a line g K parallel to Fƒ, and consequently parallel to the plane of the base; therefore these pyramids are equivalent. But the pyramid K-Fƒ H may be considered as having its vertex in ƒ, and thus it will have the same altitude as the frustum; as to its base FHK, we say that it is a mean proportional between the two bases FHG, fhg. Indeed the triangles FHK,ƒ h g, have the angle Ff, and the side FK=ƒg,

hence

Also

FHK : f h g :: FK × FH : ƒ g × ƒh :: FH:ƒh (216).

FHG: FHK:: FG: FK or ƒ g.

But the similar triangles FHG, fhg, give

consequently

FG:fg:: FH:ƒh;

FHG: FHK:: FHK : f h g ;

and thus the base FHK is a mean proportional between the two bases FHG, fhg; therefore the frustum of a triangular pyramid is equal to 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.

THEOREM.

423. If a triangular prism, whose base is ABC (fig. 216), be Fig. 216. cut by a plane DEF inclined to this base, the solid ABC-DEF, thus formed, will be equal to the sum of the three pyramids whose vertices are D, E, F, and the common base ABC.

Demonstration. Through the three points F, A, C, suppose a plane FAC to pass, cutting off from the truncated prism

ABC-DEF

the triangular pyramid F-ABC; this pyramid will have for its base ABC, and for its vertex the point F.

This pyramid being cut off, there will remain the quadrangular pyramid F-ACDE, of which F is the vertex, and ACDE the base. Through the points F, E, C, suppose a plane FEC to pass, dividing the quadrangular pyramid into two triangular pyramids F-AEC, F-CDE.

The pyramid F-AEC, which has for its base the triangle AEC, and for its vertex the point F, is equivalent to a pyramid B-AEC, which has for its base AEC, and for its vertex the point B. For these two pyramids have the same base; they have also the same altitude, since the line BF, being parallel to each of the lines AE, CD, is parallel to their plane AEC; therefore the pyramid F-AEC is equivalent to the pyramid B-AEC, which may be considered as having for its base ABC, and for its vertex the point E.

The third pyramid F-CDE, or E-FCD, may be changed in the first place into A-FCD; for the two pyramids have the same

base FCD; they have also the same altitude, since AE is parallel to the plane FCD; consequently the pyramid E-FCD is equivalent to A-FCD. Again, the pyramid A-FCD, or F-ACD, may be changed into B-ACD, for these two pyramids have the common base ACD; they have also the same altitude, since their vertices F and B are in a parallel to the plane of the base. Therefore the pyramid E-FCD, equivalent to A-FCD, is also equivalent to B-ACD. Now this last may be regarded as having for its base ABC, and for its vertex the point D.

We conclude, then, that the truncated prism ABC-DEF is equal to the sum of three pyramids which have for their common base ABC, and whose vertices are respectively the points D, E, F.

424. Corollary. If the edges are perpendicular to the plane of the base, they will be at the same time the altitudes of the three pyramids, which compose the truncated prism; so that the solidity of the truncated prism will be expressed by

or

ABC × AE+ ABC × BF+ABC × CD,

ABC × (AE+BF+ CD).

THEOREM.

425. Two similar triangular pyramids have their homologous faces similar, and their homologous solid angles equal.

Demonstration. The two triangular pyramids S-ABC, T-DEF g. 203. (fig. 203), are similar, if the two triangles SAB, ABC, are similar to the two TDE, DEF, and are similarly placed (381); that is, if the angle ABS = DET, BAS=EDT, ABC=DEF, BAC EDF, and if, furthermore, the inclination of the planes SAB, ABC, is equal to that of the planes TDE, DEF. This being supposed, we say that the pyramids have all their faces similar, each to each, and their homologous solid angles equal.

Take BG=ED, BH= EF, BI= ET, and join GH, G1, IH. The pyramid T-DEF is equal to the pyramid I-GBH; for the sides GB, BH, being equal, by construction, to the sides DE, EF, and the angle GBH being, by hypothesis, equal to the angle DEF, the triangle GBH is equal to DEF (36); therefore, in order to apply one of these pyramids to the other, we can evident.y place the base DEF upon its equal GBH; then, since the plane TDE has the same inclination to DEF as the plane SAB has to

ABC, it is manifest that the plane TDE will fall indefinitely upon the plane SAB. But, by hypothesis, the angle DET = GBI, consequently ET will fall upon its equal BI; and since the four points D, E, F, T, coincide with the four G, B, H, I, it follows that the pyramid T-DEF will coincide with the pyramid I-GBH (384).

Now, on account of the equal triangles DEF, GBH, the angle BGH=EDF = BAC; consequently GH is parallel to AC. For a similar reason, GI is parallel to AS; therefore the plane IGH is parallel to SAC (344). Whence it follows that the triangle IGH, or its equal TDF, is similar to SAC (347), and that the triangle IBH, or its equal TEF, is similar to SBC; therefore the two similar triangular pyramids S-ABC, T-DEF have their four faces similar, each to each. Moreover the homologous solid angles are equal.

For, we have already placed the solid angle E upon its homologous angle B, and the same may be done with respect to the two other homologous solid angles; but it will be readily perceived that two homologous solid angles are equal, for example the angles T and S, because they are formed by three plane angles which are equal, each to each, and similarly placed.

Therefore two similar triangular pyramids have their homologous faces similar, and their homologous solid angles equal.

426. Corollary 1. The similar triangles in the two pyramids furnish the proportions

AB: DE:: BC:EF::AC:DF::AS:DT:SB:TE::SC: TF; therefore in similar triangular pyramids the homologous sides are proportional.

427. Corollary 11. II. Since the homologous solid angles are equal, it follows that the inclination of any two faces of one pyramid is equal to the inclination of the two homologous faces of a similar pyramid (359).

428. Corollary ш11. If a triangular pyramid SABC be cut by a plane GIH parallel to one of the faces SAC, the partial pyramid IGBH will be similar to the entire pyramid SABC. For the triangles BGI, BGH, are similar to the triangles BAS, BAC, each to each, and similarly placed; also the inclination of the two planes is the same in each; therefore the two pyramids are similar.

429 Corollary IV. If any pyramid whatever SABCDE (fig. 214) Fig. 214.

be cut by a plane a b c d e parallel to the base, the partial pyramid S-a b c d e will be similar to the entire pyramid S-ABCDE. For the bases ABCDE, a b c d e, are similar, and AC, a c, being joined, it has just been proved that the triangular pyramid S-ABC is similar to the pyramid S-abc; therefore the point S is determined with respect to the base ABC, as the point S is determined with respect to the base a b c (382); therefore the two pyramids S-ABCDE, S-abcde, are similar.

430. Scholium. Instead of the five given things, required by the definition, in order that two triangular pyramids may be similar, we can substitute five others, according to different combinations; and there will result as many theorems, among which may be distinguished the following; two triangular pyramids are similar, when they have their homologous sides proportional.

For, if we have the proportions

AB: DE:: BC: EF:: AC: DF:: AS: DT:: SB: TE :: SC: TF Fig. 203. (fig. 203), which contain five conditions, the triangles ABS, ABC, will be similar to DET, DEF, and the disposition of the former will be similar to that of the latter. We have also the triangle SBC similar to TEF; therefore the three plane angles, which form the solid angle B, are equal to the three plane angles which form the solid angle E, each to each; whence it follows that the inclination of the planes SAB, ABC, is equal to that of the homologous planes TDE, DEF, and that thus the two pyramids are similar.

Fig. 219.

THEOREM.

431. Two similar polyedrons have their homologous faces similar, and their homologous solid angles equal.

Demonstration. Let ABCDE (fig. 219) be the base of one polyedron; let M, N, be the vertices of two solid angles, with out this base, determined by the triangular pyramids M-ABC, N-ABC, whose common base is ABC; let there be, in the other polyedron, the base abcde homologous or similar to ABCDE, m, n, the vertices homologous to M, N, determined by the pyramids m-a b c, n-a b c, similar to the pyramids M-ABC, N-ABC; we say, in the first place, that the distances MN, mn, are proportional to the homologous sides AB, ab.

Indeed, the pyramids M-ABC, m-a b c, being similar, the inclination of the planes MAC, BAC, is equal to that of the planes