G H ABCG. Neither can the base ABC be to the base DEF, as the pyramid A B C G to any solid greater than the pyramid DE FH. For, if it be possible, let this greater solid be Z. Because the base ABC is to the base D E F as the pyramid ABCG is to the solid Z. Therefore, by inversion, the base DEF is to the base A B C, as the solid Z is to the pyramid ABCG. But the solid Z is to the pyramid ABCG, as the pyramid DEFH to some solid (XII. Post. 2) less (V. 14) than the pyramid A B CG, because the solid Z is greater than the pyramid DEFH. Therefore, the base DEF is to the base ABC, as the pyramid B DEFH is to a solid less than the pyramid ABCG; the a M N T X CE V contrary to which has been proved. Therefore the base A B C is not to the base D E F, as the pyramid A B C G is to any solid greater than the pyramid DE FH. And it has been proved that neither is the base A B C to the base DE F, as the pyramid A B C G is to any solid less than the pyramid DEFH. Therefore, the base A B C is to the base D E F, as the pyramid ABCG is to the pyramid D E F H. Wherefore triangular pyramids, &c. Q. E. D. Polygonal pyramids of the same altitude, are to one another as their bases. Let the pyramids ABCDEM and F G H KLN be of the same altitude. The base ABCDE is to the base FG HK L, as the pyramid ABCDEM is to the pyramid F G H K L N. Divide the base A B C D E into the triangles A B C, A C D, and ADE, and the base F G H KL into the triangles FGH, FHK, and FKL; and let planes pass through ADM, ACM, FK N, and F HN. Therefore the pyramid ABCDEM contains as many triangular pyramids having the common vertex M, as the pyramid FGH KLN contains, having the common vertex N. A M N Because the triangle ABC is to the triangle FG H, as (XII. 5) the pyramid A B C M is to the pyramid FG HN; the triangle A CD is to the triangle F G H, as the pyramid A CDM is to the pyramid F G H N; and the triangle ADE is to the triangle FGH, as the pyramid ADEM is to the pyramid FG HN. Therefore the base A B C D E is to the base FG H, as the pyramid ABCDEM is to the pyramid F YG HN (V. 24, Cor. 2); and the base F G H KL is to the base FGH, as the pyramid FGHKLN is to the pyramid FGH N. But, by inversion, the base F G H is to the base F G H KL, as the pyramid FGHN is to the pyramid FG HKLN. Wherefore the base A BCDE is to the base FG H, the pyramid ABCDEM to the pyramid F G H N; and the base B C D F G H K FGH is to the base FGHK L, as the pyramid FG HN is to the pyramid FGHK LN. Therefore, ex æquali (V. 22), the base ABCDE is to the base FGH KL, as the pyramid ABCDEM is to the pyramid FGH KLN. Therefore polygonal pyramids, &c. Q. E. D. PROP. VII. THEOREM. Every triangular prism may be divided into three equal triangular pyramids. Let the prism ABCDEF be triangular. It may be divided into three equal triangular pyramids. F Join BD, E C, and CD. Because ABED is a parallelogram, and BD its diagonal, the triangle ABD is equal (I. 34) to the triangle EBD. Therefore the pyramid ABDC is equal (XII. 5) to the pyramid EBDC. Because FB is a parallelogram and CE its diagonal, the triangle ECF is equal (I. 34) to the triangle ECB. Therefore the pyramid ECBD is equal to the pyramid E CFD. But the pyramid ECBD has been proved equal to the pyramid ABDC. Therefore the prism ABCDEF is divided into three equal pyramids having triangular bases, viz. the pyramids ABDC, EBDC, and ECFD. Wherefore every A triangular prism, &c. Q. E. D. D E B COROLLARY 1.-From this it is manifest, that every pyramid is the third part of a prism having the same base and altitude. For if the base of the prism be any other figure than a triangle, it may be divided into prisms having triangular bases. COROLLARY 2.-Prisms of equal altitudes are to one another as their bases; because the pyramids upon the same bases, and of the same altitude, are (XII. 6) to one another as their bases, This proposition is the foundation of the rule for finding the solid content of a pyramid. PROP. VIII. THEOREM. Similar triangular pyramids, are to one another, in the triplicate ratio of their homologous sides. Let the pyramids ABCG and DEFH be similar and similarly situated. The pyramid ABCG has to the pyramid DEFH, the triplicate ratio of that which the side B C has to the homologous side EF. Complete the parallelograms ABCM, GBCN, and ABGK, and the parallelopiped B L. Complete also the parallelopiped EO. K L X O G H N R Because the pyramid ABCG is similar to the pyramid DEFH, the angle ABC is equal (XI. Def. 11) to the angle DEF, the angle GBC to the angle HEF, and the angle ABG to the angle DEH; and AB is (VI. Def. 1) to BC, as DE is to EF; that is, the sides about the equal angles are proportionals. Therefore the parallelogram B M is similar to the parallelogram EP: for the same reason, the parallelograms BN and ER are similar to the parallelograms B K and EX. Therefore the three parallelo- A grams BM, BN, and B K are similar to the three EP, ER, and EX. But the three B M D E F BM, BN, and BK, are equal and similar (XI. 24) to the three which are opposite to them; and the three EP, ER, and EX are equal and similar to the three opposite to them. Therefore the solids BL and EO are contained by the same number of similar planes; and their solid angles are (XI. B) equal. Therefore the solid BL is similar (XI. Def. 11) to the solid E O. But similar parallelopipeds have the triplicate (XI. 33) ratio of that which their homologous sides or edges have. Therefore the solid BL has to the solid EO the triplicate ratio of that which the side BC has to the homologous side EF. But the solid BL is to the solid EO as (V. 15), the pyramid ABCG is to the pyramid DEFH; for the pyramids are each the sixth part of the solids, since the prisms which are halves (XI. 28) of the parallelopipeds are triple (XII. 7) of the pyramids. Therefore, likewise the pyramid ABCG has to the pyramid DEFH, the triplicate ratio of that which B C has to the homologous side of EF. Wherefore similar triangular pyramids, &c. Q. E. Ď. COROLLARY.-From this it is evident, that similar polygonal pyramids are to one another in the triplicate ratio of their homologous sides. For they may be divided into similar pyramids having triangular bases; because the similar polygons, which are their bases, may be divided into the same number of similar triangles homologous to the whole polygons. Therefore, as one of the triangular pyramids in the first polygonal pyramid is to one of the triangular pyramids in the other (V. 12), so are all the triangular pyramids in the one to all the triangular pyramids in the other; that is, as the one polygonal pyramid is to the other. But one triangular pyramid is to its similar triangular pyramid, in the triplicate ratio of their homologous sides. Therefore, the one polygonal pyramid has to the other, the triplicate ratio of that which one of the sides of the one has to the homologous side of the other. This proposition is the foundation of the rule for the mensuration of similar solids and all regular polyhedrons. The bases and altitudes of equal triangular pyramids are reciprocally proportional: and triangular pyramids, of which the bases and altitudes are reciprocally proportional, are equal to one another. First, let the pyramids ABCG and DEFH, be equal to one another. The base ABC is to the base DEF, as the altitude of the pyramid DEFH is to the altitude of the pyramid A B CG. Complete the parallelograms AC, AG, GC, DF, DH, and HF; and the parallelopipeds BL and EO. Because the pyramid ABCG is equal to the pyramid DE FH, and the solid BL is sextuple (XI. 28 and XII. 7) of the pyramid ABCG and the solid EO sextuple of the pyramid DEFH. Therefore the solid BL is equal (V. Ax. 1) to the solid E O. But the bases and altitudes of equal parallelopipeds are reciprocally proportional (XI. 34). Therefore as the base B M to the base EP, so is the altitude of the solid EO to the altitude of the solid B L. But the base B M is to the base E P, as (V. 15) the triangle A B C is to the triangle DEF. Therefore the triangle ABC is to the triangle DEF, as the altitude of L N X the solid EO is to the altitude of the solid B L. K B F D E Next, let the bases and altitudes of the pyramids A B C G and DEFH be reciprocally proportional. The pyramid ABCG is equal to the pyramid DEFH. The same construction being made, because the base ABC is to the base D E F, as the altitude of the pyramid D E F H is to the altitude of the pyramid ABCG; and the base A B C is to the base DE F, as the parallelogram B M is to the parallelogram E P. Therefore the parallelogram B M is to the parallelogram EP, as the altitude of the pyramid DEFH is to the altitude of the pyramid A B C G. But the altitude of the pyramid DEFH is the same with the altitude of the parallelopiped EO; and the altitude of the pyramid ABCG is the same with the altitude of the parallelopiped BL. Therefore the base B M is to the base EP, as the altitude of the parallelopiped E O is to the altitude of the parallelopiped B L. Therefore the parallelopiped BL is equal to the parallelopiped EO (XI. 34). But the pyramid ABCG is the sixth part of the solid BL, and the pyramid D'EFH is the sixth part of the solid EO. Therefore the pyramid ABCG is equal (V. Ax. 2) to the pyramid DEF H. Therefore the bases, &c. Q. E. D. A cone is the third part of a cylinder, having the same base and altitude. Let a cone and a cylinder have the same base, viz., the circle ABCD, and the same altitude. The cone is the third part of the cylinder, that is, the cylinder is triple of the cone. If the cylinder be not triple of the cone, it must either be greater than the triple, or less than the triple. First, let it be greater than the triple. Inscribe the square A B CD in the circle: this square is greater (XII. 2) than the half of the circle A B CD. Upon the square A B C D` erect a prism of the same altitude with the cylinder; this prism is greater than half of the cylinder. For, let a square be described about the circle, and let a prism be erected upon the square, of the same altitude as the cylinder. Because the inscribed square is half of that circumscribed; and upon these square bases are erected parallelopipeds, viz., the prisms of the same altitude. Therefore the prism upon the inscribed square ABCD is the half of the prism upon the circumscribed square, for they are to one another (XI. 32) as their bases. But the cylinder is less than the prism upon the circumscribed square. Therefore the prism upon the inscribed square ABCD is greater than half of the cylinder. Bisect the arcs AB, BC, CD, and DA in the points E, F, G, and H; join AE, EB, BF, FC, CG, B H E F D GD, DH, and HA: and each of the triangles AEB, BFC, CGD, and DHA is greater than the half of the segment of the circle which contains it (XII. 2). Erect prisms upon each of these triangles, of the same altitude with the cylinder; each of these prisms is greater than half of the cylindric segment which contains it. For, if through the points E, F, G, and H, parallels be drawn to A B, BC, CD, and D A, and parallelograms be completed upon these straight lines, and parallelopipeds be erected upon the parallelograms; the prisms upon the triangles A E B, BFC, CGD, and D H A, are the halves of the parallelopipeds (XII. 7, Cor. 2); and the cylindric segments, which are upon the circular segments, cut off by A B, BC, CĎ, and DA, are less than the parallelopipeds which contain them. Therefore the prisms upon the triangles AEB, BF C, CGD, and DHA, are greater than half of the cylindric segments which contain them. Therefore, if each of the arcs be bisected, and straight lines be drawn from the points of bisection to the extremities of the arcs, and upon the triangles thus made, prisms be erected of the same altitude with the cylinder, and so on; there will at length remain some cylindric segments which together are less (XII. Lem. I) than the excess of the cylinder above the triple of the cone. Let these cylindric segments be those upon the circular segments AE, EB, BF, FC, CG, GD, DH, and H A. Therefore the rest of the cylinder, that is, the prism upon the polygon AEBFCGDH, is greater than the triple of the cone. But this prism is triple (XII. 7, Cor. 1) of the pyramid, having the same base and vertex with the cone. Therefore the pyramid upon the base AEBFCGD H is greater than the cone upon the circle ABCD; but it is also less, for the pyramid is contained within the cone, which is impossible. Therefore the cylinder is not greater than the triple of the cone. Let it be H Nor can the cylinder be less than the triple of the cone. less, if possible; therefore the cone is greater than the third part of the cylinder. In the circle A B C D inscribe a square; this square is greater than the half of the circle. Upon the square ABCD erect a pyramid, having the same vertex with the cone; this pyramid is greater than the half of the cone; because, as was before demonstrated, if a square be described about the circle, the square ABCD is the half of it. If upon these squares there be erected parallelopipeds or prisms of the same altitude with the cone, the prism upon the inscribed square A B C D is the half of that which is upon the circumscribed square; for they are to one another as their bases (XI. 32) ; and so are also the third parts of these solids. Therefore the pyramid upon the inscribed square ABCD is half of the pyramid upon the circumscribed square. But the latter pyramid is greater than the cone which it contains. Therefore the pyramid upon the inscribed square ABCD is greater than the half of the cone. Bisect the arcs AB, BC, CD, and DA, in the points F, F, G, and H, and join AE, E B, BF, FC, CG, GD, DH, and HA. Each of the triangles AEB, BF C, C G D, and DHA is greater than half of the segment of the circle which contains it. Upon E G F |