hypothesis equal to the given solid Z; therefore, the excess of the circumscribed, above the inscribed prisms, is less than the given solid Z. But the excess of the circumscribed prisms above the inscribed, is greater than their excess above the pyramid ABCD, because ABCD is greater than the sum of the inscribed prisms. Much more, therefore, is the excess of the circumscribed prisms above the pyramid, less than the solid Z. A series of prisms of the same altitude has therefore been circumscribed about the pyramid ABCD exceeding it by a solid less than the given solid Z.

PROPOSITION XVI. THEOREM.

Pyramids that have equal bases and altitudes are equal to one another.

Let ABCD, EFGH, be two pyramids that have equal bases BCD, FGH, and also equal altitudes, namely, the perpendiculars drawn from the vertices A and E upon the planes BCD, FGH. The pyramid ABCD is equal to the pyramid EFGH.

If they are not equal, let the pyramid EFGH exceed the pyramid ABCD by the solid Z. Then, a series of prisms of the same altitude may be described about the pyramid ABCD that shall ex

A

divided, namely, HT, TU, UV, VE, and through the points T, U, and V, let the sections TZW, UZX, V@Y, be made parallel to the base FGH. The section NQL is equal to the section WZT (II. 14); as also ORI to XEU, and PSM to YoV; and therefore, also, the prisms that stand upon the equal sections are equal (II. 10, Cor. 1), that is, the prism which stands on the base BCD, and which is between

the planes BCD and NQL, is equal to the prism which stands on the base FGH, and which is between the planes FGH and WZT; and so of the rest, because they have the same altitude; wherefore, the sum of all the prisms described about the pyramid ABCD is equal to the sum of all those described about the pyramid EFGH. But the excess of the prisms described about the pyramid ABCD above the pyramid ABCD, is less than Z; and therefore the excess of the prisms described about the pyramid EFGH above the pyramid ABCD, is also less than Z. But the excess of the pyramid EFGH above the pyramid ABCD, is equal to Z, by hypothesis; therefore, the pyramid EFGH exceeds the pyramid ABCD, more than the prisms described about EFGH exceed the same pyramid ABCD. The pyramid FFGH is therefore greater than the sum of the prisms described about it, which is impossible. The pyramids ABCD, EFGH, therefore, are not unequal, that is, they are equal to one another.

PROPOSITION XVII. THEOREM.

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

Let there be a prism of which the base is the triangle ABC, and let DEF be the triangle opposite to the base. The prism ABCDEF may be divided into three equal pyramids having triangular bases.

F

E

Join AE, EC, CD; and because ABED is a parallelogram, of which AE is the diameter, the triangle ADE is equal (Pl. Ge. I. 34) to the triangle ABE; therefore the pyramid of D which the base is the triangle ADE, and vertex the point C, is equal (II. 16) to the pyramid, of which the base is the triangle ABE, and vertex the point C. But the pyramid of which the base is the triangle ABE, and vertex the point C, that is, the pyramid ABCE is equal to the pyramid A

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DEFC, for they have equal bases, namely, the triangles ABC, DFE, and the same altitude, namely, the altitude of

the prism ABCDEF. Therefore the three pyramids ADEC, ABEC, DFEC, are equal to one another. But the pyramids ADEC, ABEC, DFEC, make up the whole prism ABCDEF; therefore the prism ABCDEF is divided into three equal pyramids.

COR. 1. From this it is manifest that every pyramid is the third part of a prism which has the same base, and the same 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 of equal altitudes are to one another as their bases; because the prisms upon the same bases, and of the same altitude, are (II. 10, Cor. 1) to one another as their bases.

PROPOSITION XVIII. THEOREM.

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 superficies.

Let ABCD 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 F be the point in which EF meets the plane DFC opposite to the base; join EG and FH; and let AGHD be the rectangle (II. Def. 13), by the revolution of which the cylinder ABCD is de- D scribed.

Now, because GH is 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 A circle AEB. But EF is at right angles to

с

B

the same plane; therefore EF and GH are parallel (I. 6), and in the same plane. And since the plane through GH and EF cuts the parallel planes AEB, DEC, in the straight lines EG and FH, EG is parallel to FH (I. 14). The

figure EGHF is therefore a parallelogram, and it has the angle EGH a right angle; therefore it is a rectangle, and is equal to the rectangle AH, because EG is equal to AG. Therefore, when in the revolution of the rectangle AH, the straight line AG coincides with EG, the two rectangles AH and EH will coincide, and the straight line AD will coincide with the straight line EF. 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.

PROPOSITION XIX. THEOREM.

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

Let ABCD be a cylinder, and EF a parallelopiped having equal bases, namely, the circle AGB and the parallelogram EH, and having also equal altitudes; the cylinder ABCD is equal to the parallelopiped EF.

D

QSF

If not, let them be unequal; and first, let the cylinder be less than the parallelopiped EF; and from the parallelopiped EF let there be cut off a part EQ by a plane PQ parallel to NF, equal to the cylinder ABCD. In the circle AGB inscribe the polygon AGKBLM that shall differ from the circle by a space less than the parallelogram PH, and cut off from the parallelo

G

M

B

K

E

PRN

H

gram EH, a part OR equal to the polygon AGKBLM. The point R will fall between P and N. On the polygon AGKBLM let an upright prism AGBCD be constituted of the same altitude with the cylinder, which will therefore be less than the cylinder, because it is within it; and if through the point R a plane RS parallel to NF be made to pass, it will cut off the parallelopiped ES equal to the prism AGBC, because its base is equal to that of the prism, and its altitude is the same. But the prism AGBC is less than the cylinder ABCD, and the cylinder ABCD is equal to the parallelopiped 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 parallelopiped EF; and in the same manner it may be shown not to be greater than EF.

PROPOSITION XX. THEOREM.

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

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

If not, let the cone ABCD be the third part of another cylinder LMNO, having the same altitude with the cylinder BFKG, but let the bases BCD and LIM be unequal; and first, let BCD be greater than LIM.

G

A

N

M

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 (Pl. Ge. Quad. 6), and which, therefore, will be greater than LIM. Let this be the polygon BECFD; and upon BECFD let there be stituted the pyramid ABECFD, and the prism BCFKHG.

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con

Because the polygon BECFD is greater than the circle LIM, the prism BCFKHG is greater than the cylinder LMNO, for they have the same altitude, but the prism has the greater base. But the pyramid ABECFD is the third part of the prism (II. 17) BCFKHG; 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. And