ments of the cylinder which are upon the segments of the circle cut off by AB, BC, CD, DA, are less than the parallelopipeds which contain them; therefore the prisms upon the triangles AEB, BFC, CGD, DHA, are greater than half of the segments of the cylinder in which they are; therefore, if each of the circumferences be divided into two equal parts, and straight lines be drawn from the points of division to the extremities of the circumferences, and upon the triangles thus made, prisms be erected of the same altitude with the cylinder, and so on, there must at length remain some segments *Lemma of the cylinder which together are less than the excess of the cylinder above the triple of the cone. Let them be those upon the segments of the circle AE, EB, BF, FC, CG, GD, DH, HA: therefore the rest of the cylinder, that is, the prism of which the base is the polygon AEBFCGDH, and of which the altitude is the same with that of the cylinder, is greater than the *1 Cor. triple of the cone: but this prism is triple* of the pyramid upon the same base, of which the vertex is the same with the vertex of the cone: therefore the pyramid upon the base AEBFCGDH, having the same vertex with the cone, is greater than the cone, of which the base is the circle ABCD: but this is impossible, because the pyramid is less than the cone, being contained within it; therefore the cylinder is not greater than the triple of the cone.

7.12.

Nor can the cylinder be less than the triple of the cone. Let it be less, if possible: therefore, inversely, the cone is greater than the third part of the cylinder. In the circle ABCD 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; and if, upon these squares there be erected parallelopipeds of the same altitude with the cone,

H

D

B

which are also prisms, the prism upon the square ABCD is the half of that which is upon the square described about the circle; for they are to one another as their bases; as are also the third parts of them: 32. 11. therefore the pyramid, the base of which is the square ABCD, is half of the pyramid upon the square described about the circle. But this last pyramid is greater than the cone which it contains; therefore the pyramid upon the square ABCD, having the same vertex with the cone, is greater than the half of the cone. Bisect the circumferences AB, BC, CD, DA in the points E, F, G, H, and join AE, EB, BF, FC, CG, GD, DH, HA: therefore each of the triangles AEB, BFC, CGD, DHA is greater than half of the segment of the circle in which it is. Upon each of these triangles erect pyramids having the same vertex with the cone: therefore each of these pyramids is greater than the half of the segment of the cone in which it is, as before was demonstrated of the prisms and segments of the cylinder; and thus dividing each of the circumferences into two equal parts, and joining the points of division and their extremities by straight lines, and upon the triangles erecting pyramids having their ver.tices the same with that of the cone, and so on, there must at length remain some segments of the cone, which together are less than the excess of the cone, *Lemma above the third part of the cylinder. Let these be the

segments upon AE, EB, BF, FC, CG, GD, DH, HA: therefore the rest of the cone, that is, the pyramid, of which the base is the polygon AEBFCGDH, and of which the vertex is the same with that of the cone, is greater than the third part of the cylinder: but this pyramid is the third part of the prism upon the same base AEBFCGDH, and of the same altitude with the cylinder: therefore this prism is greater than the cylinder of which the base is the circle ABCD: but this is impossible, because the prism is less than the cylinder, being contained within it: therefore the cylinder is not less than the triple of the cone. And it has been demonstrated that neither is it greater than the triple: therefore the cylinder is triple of the cone, or, the cone is the third part of the cylinder. Wherefore every cone, &c. Q. E. D.

PROP. XI. THEOR.

Cones and cylinders of the same altitude, are to one another as their bases.

Let the cones and cylinders, of which the bases are the circles ABCD, EFGH, and the diameters of their bases AC, EG, and KL, MN the axes of the cones or cylinders, be of the same altitude: as the circle ABCD to the circle EFGH, so shall the cone AL be to the cone EN.

If it be not so, the circle ABCD must be to the circle EFGH, as the cone AL to some solid either less than the cone EN, or greater than it. First, if it be possible, let it be to a solid less than EN, viz. to the. solid X; and let Z be the solid which is equal to the excess of the cone EN above the solid X; therefore the cone EN is equal to the solids X, Z together. In

the circle EFGH inscribe the square EFGH, therefore this square is greater than the half of the circle. Upon the square EFGH erect a pyramid of the same altitude with the cone; this pyramid shall be greater than half of the cone. For, if a square be described about the circle, and a pyramid be erected upon it, having the same vertex with the cone, the pyramid inscribed in the cone is half of the pyramid circumseribed about it, because they are to one another as their bases: * but * 6. 12. the cone is less than the circumscribed pyramid: therefore the pyramid of which the base is the square EFGH, and its vertex the same with that of the cone, is greater than half of the cone. Divide the circumferences EF, FG, GH, HE, each into two equal parts in the points O, P, R, S, and join EO, OF, fp, pg, GR, RH, HS, SE; therefore each of the triangles EOF, FPG, GRH, HSE is greater than half of the segment

of the circle in which it is. Upon each of these tri

⚫ Vertex is put in place of altitude which is in the Greek, because the pyramid, in what follows, is supposed to be circumscribed about the cone, and so must have the same vertex. And the same change is made in some places following.

angles erect a pyramid having the same vertex with the cone; each of these pyramids is greater than the half of the segment of the cone in which it is: and thus dividing each of these circumferences into two equal parts, and from the points of division drawing straight lines to the extremities of the circumferences, and upon each of the triangles thus made erecting pyramids having the same vertex with the cone, and so on, there must at length remain some segments of the cone which *Lemma are together less than the solid Z; let these be the segments upon EO, OF, FP, PG, GR, RH, HS, SE: therefore the remainder of the cone, viz. the pyramid of which the base is the polygon EOFPGRHS, and its vertex the same with that of the cone, is greater than the solid X; that is, the solid X is less than the pyramid. In the circle ABCD inscribe the polygon ATBYCVDQ similar to the polygon EOFPGRHS, and upon it erect a pyramid having the same vertex with the cone AL: and because, as the square of AC is 1. 12. to the square of EG, so is the polygon ATBYCVDQ to the polygon EOFPGRHS; and as the square of 2. 12. AC to the square of EG, so is the circle ABCD to 11. 5. the circle EFGH; therefore the circle ABCD is to

the circle EFGH, as the polygon ATBYCVDQ to the polygon EOFPGRHS. But as the circle ABCD to the circle EFGH, so is the cone AL to the solid X; and as the polygon ATBYCVDQ to the polygon * 6. 12. EOFPGRHS, so is the pyramid of which the base is the first of these polygons, and vertex L, to the pyramid of which the base is the other polygon, and its vertex N: therefore, as the cone AL to the solid X, so is the pyramid of which the base is the polygon ATBYCVDQ, and vertex L, to the pyramid the base of which is the polygon EOFPGRHS, and vertex N. But the cone AL is greater than the pyramid con14. 5. tained in it; therefore the solid X is greater than the