mid VABC more than the sum of the prisms described about the pyramid ULMN exceeds the same pyramid VABC; and consequently the pyramid ULMN is greater than the sum of the prisms described about it, which is impossible (Prop. 65). The pyramids VABC, ULMN are, therefore, not unequal, that is they are equals to one another. Wherefore, pyramids that have, &c. which was to be proved.

PROP. LXVIII. THEOR.

Every prism having a triangular base may be divided into three pyramids that have triangular bases, and that are equal to one another.

Let ABCDEF (fig. 75) be a prism of which the base is the triangle ABC, and the triangle DEF is the side opposite to the base : the prism ABCDEF may be divided into three pyramids that have triangular bases, and that are equal to one another.

Join AE, EC, CD: the planes DEC, AEC divide the prism into three pyramids having triangular bases; and these three pyramids are equal to one another.

Because ABED is a parallelogram, of which AE is the diameter, the triangle ADE is equal to the triangle ABE (I. 34): therefore the pyramid of which the base is the triangle ADE, and vertex the point C is equal to the pyramid of which the base is the triangle ABE and vertex the point C (Prop. 67). But the pyramid of which the base is the triangle ABE, and vertex the point C, is the same solid as the pyramid of which the base is the triangle ABC, and vertex the point E; and this last pyramid is equal to pyramid of which the base is the triangle DEF and vertex the point C, these pyramids having equal bases ABC, DEF (Def. 15), and equal altitudes, the perpendiculars from E and C on the parallel planes ABC and DEF (Props. 28. 30). Therefore the three pyramids CEAD, EABC, CEDF are equal to one another. But the pyramids CEAD, EABC, CEDF make up the whole prism ABCDEF; therefore the prism ABCDEF is divided into three equal pyramids having triangular bases. Wherefore, every prism, &c.: which was to be proved.

Cor. 1. From this it is manifest that every pyramid is a third part of a prism which has the same base, and is of an equal altitude with it; for if the base of the prism be any other figure than a triangle, it may be divided into prisms having triangular bases.

Cor. 2. Pyramids having equal altitudes are to one another as their bases; because the prisms upon the same bases, and of the same altitude as the pyramids, are to one another as their bases (Prop. 62. Cor. 2).

Cor. 3. All pyramids are to one another in the compound ratio of their bases and altitudes; because the prisms having the same bases and having the same altitudes as the pyramids, and of which

these are the third parts, are to one another in the compound ratio of their bases and altitudes (Prop. 62. Cor. 3).

Cor. 4. Similar pyramids are to one another in the triplicate ratio, or as the cubes, of their homologous sides; because the similar prisms upon the same bases and having the same altitudes as the pyramids, and of which these are the third parts, are to one another in the triplicate ratio of the homologous sides of their bases (Prop. 62. Cor. 4).

PROP. LXIX. THEOR.

If from any point in the circumference of the base of a cylinder, a straight line be drawn perpendicular to the plane of the base, it will be wholly in the cylindric surface.

Let ABCD (fig. 76) be a cylinder, of which the base is the circle AEB, DFC the circle opposite to the base, and GH the axis; from E, any point in the circumference AEB, let EF be drawn perpendicular to the plane of the circle AEB; the straight line EF is in the superficies of the cylinder.

Let AGHD be the rectangle by the revolution of which the cylinder ABCD is described (Def. 23); then GH being at right angles to GA, the straight line which by its revolution describes the circle AEB, it is at right angles to all the straight lines in the plane of that circle which meet it in G, and it is therefore at right angles to the plane of the circle AEB; consequently AD, which is parallel to HG, is also at right angles to the plane of the circle AEB (Prop. 18) in every position of the rectangle AGHD in its revolution about HG. When, therefore, in the revolution of the rectangle AGHD, the straight line GA coincides with GE, AD being at right angles to the plane of the circle AEB will coincide with EF, which is at right angles to the same plane, because from the same point in a plane, there can be only one perpendicular to the plane upon the same side of it (Prop. 8). But AD is always in the superficies of the cylinder, for it describes that superficies; therefore EF is also in the superficies of the cylinder ABCD. Wherefore, if from any point, &c. which was to be proved.

PROP. LXX. THEOR.

A cylinder and a parallelepiped having equal bases and altitudes are equal to one another.

Let ABCD (fig. 77) be a cylinder, and EF a parallelepiped, having equal bases, viz. the circle AGB and the parallelogram EH, and having equal altitudes; the cylinder ABCD is equal to the parallelepiped EF.

If not, let them be unequal; and first, let the cylinder ABCD be

less than the parallelepiped EF; and let it be equal to the parallelepiped EQ, a part of EF cut off by the plane PQ parallel to NF. In the circle AGB inscribe the polygon AGKBLM that shall differ from the circle by a space less than the parallelogram PH (Supp. Prop. 8. Cor. 1), and cut off from the parallelogram EH, a part OR equal to the polygon AGKBLM: the point R will fall between P and N; because the parallelogram EH is equal to the circle AGB. On the polygon AGKBLM let a right prism AGBCD, of the same altitude as the cylinder, be constituted, by drawing from A, G, K,B,L,M, straight lines perpendicular to the plane AGB (Prop. 7), which prism will be less than the cylinder because it is within it (Prop. 69); and if through the point R a plane RS parallel to NF be made to pass, it will cut off the parallelepiped ES equal to the prism AGBC, because its base is equal to that of the prism and its altitude is the same (Prop. 62). But the prism AGBCD is less than the cylinder ABCD, and the cylinder ABCD is equal to the parallelepiped EQ, by hypothesis; therefore ES is less than EQ, and it is also greater, which is impossible. The cylinder ABCD, therefore, is not less than the parallelepiped EF. In a similar manner it may be shown that the cylinder is not greater than the parallelepiped EF. They therefore are equal. Wherefore, a cylinder and a parallelepiped, &c. which was to be proved.

Cor. 1. A cylinder and a prism having equal bases and altitudes are equal to one another; for each of them is equal to a parallelepiped of an equal base and altitude (Props. 62, 70).

Cor. 2. Cylinders having equal altitudes are to one another as their bases; for the parallelepipeds to which they are respectively equal are to one another as their bases (Prop. 57), that is as the bases of the cylinders. Therefore (Supp. Prop. 10) cylinders having equal altitudes are to one another as the squares of the diameters of their bases.

Cor. 3. For a similar reason (Prop. 58) cylinders are to one another in the ratio compounded of the ratios of their bases and altitudes. They are therefore to one another in the compound ratio of their altitudes and the squares of the diameters of their bases.

Cor. 4. Again, for a like reason (Prop. 59. Cor.), similar cylinders are to one another as the cubes of the diameters of their bases, or as the cubes of their altitudes.

If a cone be cut by a plane parallel to the base, the section will be a circle; and this section and the base will be to one another in the duplicate ratio of their distances from the vertex of the cone.

Let VAB (fig. 78) be any cone of which VC is the axis, and VD a perpendicular from V on the base AHB; and let the cone VAB

be cut by a plane EKF parallel to the base AHB: EKF is a circle, and EKF is to AHB as the square of VL to the square of VD.

Let VAB be a section of the cone passing through the axis VC and VD, the perpendicular from V on the base AHB, cutting the plane EKF in EGLF; and let VHC be a section of the cone by any other plane passing through VC, and cutting EKF in KG.

Then since EF and AB are parallel (Prop. 27) and likewise KG and HC, the triangles VAC and VEG, VHC and VKG, VCD and VGL, are similar.

Consequently

therefore, ex æquali

AC: CV:: EG: GV

CV CH: GV: GK
AC: CH EG: GK (V. 22);

but AC is equal to CH, therefore EG is equal to GK (V. A.).

In like manner it may be shown that any other straight line drawn from G to the circumference of the section EKF is equal to EG; therefore the section EKF is a circle, and G is its centre.

Since circles are to each other as the squares of their diameters (Supp. Prop. 10), and therefore as the squares of their radii (V. 15), the circle EKF is to the circle AHB as the square of EG to the square of AC; but by similar triangles

therefore, ex æquali therefore, alternando

EG: GV:: AC: CV

GV: VL:: CV: VD

EG VL: AC: VD (V. 22);
EG: AC: VL: VD (V. 16);

and therefore the square of EG is to the square of AC, as the square of VL to the square of VD (VI. 22). Consequently the circle EKF is to the circle AHB, as the square of VL to the square of VD. Wherefore if a cone be cut by a plane, &c. : which was to be proved.

PROP. LXXII. THEOR.

If a right cone and a cylinder have the same base and the same altitude, the cone is the third part of the cylinder.

Def. A right cone is a cone of which the axis is at right angles to the plane of the base, and which therefore may be described by the revolution of a right-angled triangle about one of its sides containing the right angle, which side remains fixed.

Let the cone ABCD (fig. 79) and the cylinder BFKG have the same base, the circle BCD, and the same altitude, the perpendicular from the point A upon the plane BCD, the cone ABCD is the third part of the cylinder BFKG.

For if not, let the cone ABCD be the third part of another cylinder LMNO, having its altitude equal to that of the cylinder BFKG, but its base LIM not equal to the base BCD; and first, let BCD be greater than LIM; therefore the cylinder LMNO less

than the cylinder BFKG; and consequently the cone ABCD less than the third part of the cylinder BFKG.

Then, because the circle BCD is greater than the circle LIM, a polygon may be inscribed in BCD, that shall differ from it less than LIM does (Supp. Prop. 8. Cor. 1), and which, therefore, will be greater than LIM. Let this be the polygon BECFD; and upon BECFD, let there be constituted the pyramid ABECFD, and the prism BCFKHG.

Because the polygon BECFD is greater than the circle LIM, the prism BCFKHG is greater than the cylinder LMNO, for their altitudes are equal and the prism has the greater base (Prop. 70. Cor. 1). But the pyramid ABECFD is the third part of the prism BCFKHG (Prop. 68. Cor. 1), and therefore it is greater than the third part of the cylinder LMNO. Now the cone ABECFD is, by hypothesis, the third part of the cylinder LMNO; therefore the pyramid ABECFD is greater than the cone ABCD; and it is also less, because it is inscribed in the cone; which is impossible. Therefore the cone ABCD is not less than the third part of the cylinder BFKG.

In like manner, by circumscribing a polygon about the circle BCD, it may be shown that the cone ABCD is not greater than the third part of the cylinder BFKG: therefore the cone ABCD is equal to the third part of the cylinder BFKG. Wherefore, if a cone and a cylinder, &c.: which was to be proved.

A series of cylinders all of the same altitude may be described about any cone, and another series of cylinders, all of the like altitude, may be inscribed in the cone, such that the sum of either series shall differ from the cone by a solid less than any given solid, however small.

Let VAB (fig. 80) be any cone; then a series of cylinders all of the same altitude may be circumscribed about VAB, and another series of cylinders all of the like altitude may be inscribed in VAB, such that the sum of either series shall differ from VAB by a solid less than any given solid S, however small.

Let VACB be a section of the cone by a plane passing through the axis VC perpendicular to the base. Divide the axis of the cone, VC into any number of equal parts in the points c1, C2, C3, C4; through these points draw straight lines parallel to AB the diameter of the base; and complete the series of circumscribed and inscribed rectangles in the triangle VAB, and through C, c1, c2, &c. draw Ce1, ice2, ice, &c. perpendicular to the bases of the circumscribed and inscribed rectangles: then if the circumscribed rectangles revolve about the sides Ce1, c12, c2o3, &c., and the inscribed rectangles revolve about the sides c1, c2i2, cig, &c., it is evident that they will