lel to the base (30.), that is, to a pyramid BCDF, upon the same base BDF with the last, and having its vertex C in the same line CE, parallel to the base. Therefore the prism BEF is equal to the three pyramids ABCD, BCDE, and BCDF. But of these, the two last are each of them equal to the first A B C D, because they stand upon the same base B C D with it, and their vertices are in the same lines with it, viz., AE, FA, parallel to the base (30.). Therefore, the prism BEF is equal to three times the pyramid A B C D; and the pyramid ABCD is equal to a third part of the prism BEF. Therefore, &c. Cor. If a parallelopiped be described, having three of its edges the same with three conterminous edges of a triangular pyramid, the pyramid shall be equal to a sixth part of the parallelopiped; for the latter is divided by its diagonal plane into two equal prisms (22.), one of which stands upon the same base, and is of the same altitude with the pyramid, and is, therefore, by the proposition, equal to three times the pyramid. Scholium. This proposition might have been demonstrated somewhat more concisely. It will be found, however, that our demonstration applies equally to the following more general theorem :- If the upper part of a triangular prism (see the figure in the proposition) be cut away by a plane AEF, whether parallel to the base B CD, or otherwise, the remaining solid will be equal to the sum of three pyramids standing upon the same base BCD with the prism, and having for their vertices the upper extremities A,E,F of the diminished edges." planes VAC, VAD into triangular pyramids, having the common vertex V; and the prism, into as many triangular prisms, by the planes A Cca, ADda. And because the former have the same altitude, and stand upon the same bases with the latter respectively, they are equal to their third parts, each of each. Therefore, also, the sum of the former, that is, the whole pyramid V A B C DE, is equal to a third part of the sum of the latter, that is, to a third part of the whole prism Ad c. Therefore, &c. Cor. 1. The solid content of every pyramid is equal to one-third of the Cor. 1.). product of its base and altitude (29. Cor. 2. (Euc. xii. 5, 6.) Pyramids which have equal altitudes are to one another as their bases: and pyramids which have equal bases are to one another as their altitudes: also, any two pyramids are to one another in the bases and altitudes. (29. Cor. 2.). ratio compounded of the ratios of their Scholium. It was observed in the Scholium at i. 29. that the measurement of any plane rectilineal figure in general depends upon figure may be divided into triangles. In that of the triangle, because every such like manner, we may here remark, that the solid content of the pyramid leads to that of every other polyhedron; and that the latter may be found by adding together the contents of all the pyramids which constitute the polyhedron. These may be assumed with a common vertex, which may be either within the solid, in which case the several faces will be the bases of so many pyramids together making up the polyhedron, or at one adjoining faces will be divided into sides, of the angular points, in which case the and the others will be bases of the component pyramids. meet BC in E, and join D E. Then, because the triangles BCD, bcd are equiangular (12. and 15.), and therefore similar, BCD is to bed in the duplicate ratio (II. 42.) of BC to bc, or (I. 22.) BE, that is, in E the duplicate ratio of the same BCD to BED (II. 39.). Therefore the triangle BED is a mean proportional between BCD and bed (II. def. 11.) Now, the frustum is made up of three pyramids, namely, c B C D, Dbcd, and cb BD. And, of these the first has the same altitude with the frustum, and stands upon the base B C D; the second has likewise the same altitude with the frustum, and stands upon the base b cd; and the third, because c E is parallel (10.) to the plane B bd, is equal (30.) to the pyramid E6 BD, which has the same altitude, and stands upon the base BED, that is, upon a base which is a mean proportional between B C D and bcd. Next, let ABCDE, abcde be the two bases of the frustum of any pyramid, having the vertex V. B a E D Then, if the planes V A C, VAD be drawn cutting the bases in the straight lines AC, ac, AD, ad, the frustum in question will be divided into frustums of triangular pyra- A mids. And each of these, by what has been already demonstrated, is equal to the sum of three pyramids, having the same altitude with the frustum, and for their bases respectively, the two bases of the frustum and a mean proportional between them. Therefore, their sum, that is, the frustum in question, equal to three pyramids having the same altitude, and for their bases the sum of the lower bases, the sum of the upper bases, and the sum of the means respectively. Now, the two former sums are the bases ABCDE, abcde. And the sum of the means is a mean between the bases ABCDE, abcde; for, the triangles ABC, ACD, &c. having to the triangles a bc, a cd, &c. the same ratio, each to each, viz. that of V A to v a2 (II. 42. and II. 32.) must have the same ratio also to their respective means (II. 27. Cor. 3. and is note*); wherefore the sum of the former, that is, the base ABCDE, has to the sum of the means the same ratio (II. 23. Cor. 1.), viz. the subduplicate ratio of ABC to abc, or of ABCDE to abcde; and accordingly (II. def. 11.) the sum of the means is a mean proportional between ABCDE and abcde. Therefore, in this case also, the frustum in question is equal to the sum of three pyramids, having the same altitude with it, and for their bases its two bases, and a mean proportional between them. Therefore, &c. have to the pyramid a bed the triplicate ratio of that which B C has to b c. Because the faces of the two pyramids are similar triangles, it is evident that *If a magnitude A be the first of three proportionals A, B, C, and if the same A be the first also of than b, C shall likewise be greater than c. three other proportionals A, b, c, then if B be greater For C is to c in the ratio which is compounded of the ratios of C to B, B to b, and b to c, that is, (because the first of these ratios is the same with that of B to A, and the last the same with that of A to b,) in the ratio which is compounded of the ratios of B to A, A to b, and B to b, or in the duplicate ratio of B to b. Wherefore because B is greater than b, C is likewise greater than c. Hence it follows, that if A, B, C be proportionals, and if A', B', C' be likewise proportionals, and if A' has a greater ratio to B' than A to B, A' shall also have a greater ratio to C' than A to C. For, if b be taken such that A' may have to b the same ratio as A to B, b will be greater than B' (II. 11.Cor. 3.); wherefore, if c be taken such that A', b, c may be proportionals, c will likewise be greater than C'; and therefore A' will have to C' a greater ratio than it has to c. But A' has to c the same ratio as A to C (II. 24.). Therefore A' has to C' a greater ratio than A to C. tudes A, B, C be proportionals, and likewise any other three, A', B', C', and if A is to C as A' to C', A is to B as A' to B'; for if A has not to B the same ratio as A' to B', then neither can A have to C the same ratio as A' to C'. This last is the theorem demanded in the And hence it is evident, that if any three magni text; and is evidently the converse of II. 27. Cor. 3., sufficiently apparent. from which alone, however, it may be considered as their corresponding edges are to one another in the same ratio, so that AB is to ab as B C to bc, and as BD to bd. And, because the plane and dihedral angles of the solid angle B are severally equal in order to the plane and dihedral angles of the solid angle b, these solid angles may be made to coincide. Therefore (27 Cor. 1.) the parallelopipeds, which have the edges BA, BC, BD, and ba, bc, bd, are to one another in the ratio which is compounded of the ratios of B A to ba, B C to bc, and BD to bd, that is, since these ratios are equal to one another, the triplicate ratio (II. 27. Cor. 2.) of that which BC has to.bc. But the pyramids ABCD, abcd are (31. Cor.) the sixth parts of these parallelopipeds, each of each, and therefore have to one another (II. 17.) the same ratio with the parallelopipeds. Therefore, the pyramid A B C D has to the pyramid abcd the triplicate ratio of that which B C has to bc. Otherwise: Take B E (II. 52.), a third proportional to BC, bc; and, again, BF a third proportional to bc, BE; so that BC may have to BF the triplicate ratio of that which it has to be (II. def. 11.): join D E, DF, AF; and, from A, a, let AP, ap be drawn perpendicular to the bases BCD, bed of the pyramids, each to each. Then, because the solid angles B, b may be made to coincide, and that in this case (5. Cor.) the triangles A B P,* abp will be in the same plane, and on account of the parallelism of AP, ap (5.) will be equiangular (I. 15.), A P is to ap (II. 31.) as A B to ab, or as B C to bc. Again, because, in the triangles DB C, dbc, DB is to db as BC to be; that is, as bc to BE; the triangles DBE, dbc are equal to one another (II. 41.) But DBE is to DBF (II. 39.) as BE to BF; that is, as BC to bc. Therefore, dbc is to DBF as BC to bc. But B C is to bc as AP to a p; therefore, dbc is to DBF as AP to ap, and consequently (II. 11. Cor. 2.) the ratio, which is compounded of the ratios of dbc to DB F, and a p to AP is a ratio of equality. But the pyramid abcd is to the pyramid ABFD in a ratio which is compounded of the ratios of dbc to DBF, and ap to AP (32. Cor. 2.); therefore, the pyramid ABFD is equal to the pyramid a bcd. Now, the pyramid A B CD is to the pyramid ABFD, as the *The lines BP, bp are omitted in the figure, any two similar pyramids, and let B C,bc be homologous edges: the pyramid V ABCDE shall be to the pyramid vabcde in the triplicate ratio of that which B C has to bc. of the solid angle V are severally equal Because the plane and dihedral angles to the plane and dihedral angles of the solid angle v, these solid angles may be the bases ABCDE, abcde will be made to coincide; and, if this be done, parallel (15.), because the faces V AB and vab, VBC and vbc, &c. being similar, their sides AB and ab, BC and bc, &c. will be parallel (I. 15.). Hence it appears that the planes V AC angles with the adjoining faces of and vac, VAD and vad make equal the pyramids; and that the triangles VAC and vac, as also VA D and vad, that the triangles A B C, ACD, ADE are similar. And it is easy to show are respectively similar to the triangles abc, acd, ade (II. 32.). Therefore, the pyramids in question are divided by those planes into similar triangular pyramids VABC and vabc, VACD and vacd, VADE and vade (def. 10.). And, since the homologous edges of cd, DE and de, are to one another in these pyramids, BC and bc, CD and the same ratio; and, by what has been *These demonstrations are respectively analogons to those given of 11. 42. in page 66. The second, it is obvious, admits of considerable abridgment, by reference to 32. Cor. 2.; for, since the pyramids are to one another in the ratio which is compounded of the ratios of their bases and altitudes, and that the bases are to one another in the duplicate ratio of two homologous edges (II. 42.), and the altitudes have to one another the same ratio with those edges, it follows that the pyramids are to one another in the triplicate ratio of their homologous edges. And in this form the demonstration applies equally to the general case in which the bases are similar polygons. homologous edges: the polyhedron ACG shall be to the polyhedron a cg in the triplicate ratio of that which BC has to bc. In the first place, it is evident that the corresponding edges of the polyhedrons are to one another each to each in the same ratio, because they are homologous sides of the similar faces of the polyhedrons. Let the polyhedrons be so placed as to have two corresponding edges AB, ab parallel to one another, and two corresponding faces ABCD, abcd adjoining to those edges likewise parallel. Then, because the angles ABC, abc are equal to one another, BC is likewise parallel to be (15.); and for the like reason CD, DA are parallel to cd, da respectively. Again, since AB is parallel to ab, a plane may be made to pass through AB parallel to the plane abhg (15.), and that plane is no other than AB HG, because the dihedral angles at A B and a b are equal of one another (17. Scholium, 3. and 11.Cor.1.); and for the like reason the other similar faces of the two polyhedrons lie in parallel planes. Therefore, also, any two corresponding edges being the common sections of such planes, are parallel (12. Cor.). Join AC and ac, AH and ah, AK and a k, AL and al; that is, join the points A, a, with all the angular points of their respective figures, to which they are not already joined by the edges terminating in A, a. Then, because the triangles 'A B C, abc, are similar (II. 32.) the angles BAC, bac are equal to one another; but the planes BA C, bac are parallel, and A B is parallel to ab; therefore, also (15.), AC is parallel to ac: and, for the like reason, AH is parallel to a h. Again, because A F and FK are parallel respectively, and in the same ratio, to af and fk, the triangles AFK, afk are similar (15. and II. 32.), and their planes parallel; wherefore, as before, AK is parallel to a k; and for the like reason the triangles AD L, adl are similar and in parallel planes, and A L is parallel to al. And because the triangles ACL, acl have the three sides of the one parallel to the three sides of the other, each to each, they are equiangular (15.), and lie in parallel planes: the same also may be said of the triangles AK L, a kl, and A H K, a hk. Therefore, the pyramids into which the solid A C G is divided by the planes ACL, ADL,AFK, AHK, AKL are similar to the pyramids into which the solid. acg is divided by the planes acl, adl, afk, ahk, akl, each to each: for it has been shown that the faces in these planes are similar, each to each, and their other faces are similar, because they are similar faces, or similar parts of similar faces, of the polyhedrons; also the dihedral angles made by any two similar and adjoining faces are equal (17. Scholium, 3.), because their planes are parallel. And the former pyramids are to the latter (34.) in the triplicate ratios of their homologous edges, each to each; that is, in the triplicate ratio of B C to bc, because their edges have to one another the common ratio of BC to bc. Therefore, also, the sum of the former pyramids, that is, the polyhedron ACG, is (II. 23. Cor. 1.) to the sum of the latter, that is, to the polyhedron a cg, in the same ratio, viz., the triplicate ratio of B C to b c. Therefore, &c. Cor. Similar polyhedrons are to one another as the cubes of their homologous edges (27. Cor. 2.). SECTION 4.-Problems. In the following problems it is assumed that a sphere may be described from any centre, and at any distance rom that centre: also, a plane is consi with the radius CD, describe a sphere DEF. If this sphere touch the plane in D, C D is the perpendicular required (8. Cor.); if not, take K, the centre of the circle D G H in which it cuts the plane, and join CK; CK will be the perpendicular required (8. Cor.). Therefore, &c. Another method has been seen in prop. 7. Cor. It appears from the demonstration, that, if a point be equally distant from the three angles of a triangle, it must lie in a perpendicular to the plane of the triangle, which passes through the centre of the circumscribing circle. Scholium. Had the intersection of the spherical surfaces, in this problem, been considered at a greater length, the analogy of the construction to that employed in the Another method has been seen in first method of Book i. prop. 44, might prop. 7. PROP. 37. Prob. 2. To draw a straight line perpendicular to a given plane A. B. from a given point C in the same. G From the centre C, with any radius CD, describe, in the plane AB, the circle DE F. In the circumference of this circle, take D B Now, let CG be the perpendicular required. From the centre D, with any radius greater than D C, describe a sphere, and let its surface cut the perpendicular CG in G; and join GD, GE, G F. Then, because GE and GF are each of them equal to GD (8.), if spheres be described from the points E and F, each with a radius equal to DG, their surfaces will like. wise each of them pass through the point G. And G is the only point on that side of the plane A B, which is found at the same time in the surface of each of the three spheres. For, every point, as G, which is at equal distances from the three points D, E, F, must lie in the perpendicular which passes through the centre of the circle DEF, because (I. 7.) the angles G C D, GCE, GCF are equal to one another, and have been lost sight of in the detail. The subject is, however, of sufficient interest to merit the attention of the student; and, as it has not hitherto found a place in this treatise, the present is, perhaps, the most proper for its consideration. 1. Two spheres will, 1°, cut one another, or 20, touch one another, or 3°, one of them fall wholly without the other, according as the distance between their centres is, 1°, less than the sum, and greater than the difference of their radii, or, 2°, equal to the sum, or to the difference of their radii, or, 3°, greater than the sum, or less than the difference of their radii. This is easily inferred from what has been already demonstrated with regard to the intersection of circles; for it is evident that the sections of two spheres, made by a plane passing through their centre and through any |