hence Also FHK : fhg :: FK × FH : ƒg × fh:: FH: fh (216). f FHG : FHK :: FG: FK or ƒ g. But the similar triangles FHG, fhg, give consequently FG:fg:: FH:ƒh; FHG: FHK:: FHK: fhg; 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, E-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 t 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 Fig. 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, GI, 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 evidently 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 the 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 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 III. 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 abcde 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 de, 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-a b c de, 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, without 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 abc de 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 mae, bae; in like manner, the pyramids N-ABC, n-a b c, being similar, the inclination of the planes NAC, BAC, is equal to that of the planes na c, bac; consequently, if we subtract the first inclinations respectively from the second, there will remain the inclination of the planes NAC, MAC, equal to that of the planes nac, mac. But, because the pyramids are similar, the triangle MAC is similar to mac, and the triangle NAC is similar to na c; therefore the triangular pyramids MNAC, mnac, have two faces similar each to cach, similarly placed, and equally inclined to each other; consequently the two pyramids are similar (425); and their homologous sides give the proportion Let P and p be two other homologous vertices of the same. polyedrons, and we have, in like manner, whence PN: pn:: AB : a b, PM: pm: AB: ab; MN: mn:: PN: pn:: PM: pm. Therefore the triangle PNM, formed by joining any three vertices of one polyedron, is similar to the triangle pnm, formed by joining the three homologous vertices of the other polyedron. Furthermore, let Q, q, be two homologous vertices, and the triangle PQN will be similar to pqn. We say also, that the inclination of the planes PQN, PMN, is equal to that of the planes p qn, pm n. For, if we join QM and q m, we shall always have the triangle QNM similar to q nm, and consequently the angle QNM equal to q nm. Suppose at N a solid angle formed by the three plane' angles QNM, QNP, PNM, and at n a solid angle formed by the plane angles qnm, qnp, pnm; since these plane angles are equal, each to each, it follows that the solid angles are equal. Whence the inclination of the two planes PNQ, PNM, is equal to that of the homologous planes p n q, p n m (359); therefore, if the two triangles PNQ, PNM, be in the same plane, in which case we should have the angle QNM = QNP + PNM, we should have, in like manner, the angle qnm=qnp+ pnm, and the two triangles qnp, pnm, would also be in the same plane. All that has now been demonstrated takes place, whatever be |