• 32. 11.

+ Lemma

cone.

having the same vertex with the cone; this pyramid is
greater than the half of the cone; because, as was before de-
monstrated, if a square be described about the circle, the
square ABCD is the half of it: and if upon these squares
there be erected solid parallelopipeds of the same altitude
with the cone, which are also prisms, the prism upon the
square ABCD is the half of that which is upon the square de-
scribed about the circle; for they are to one another as their
bases *; as are also the third parts of them: 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
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 those pyramids is
greater than the half of the segment of the
cone in which it is, as was before demon-
strated 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 ex-
tremities by straight lines, and upon the triangles erecting
pyramids having their vertices 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
above the third part of the cylinder: let these be the seg-
ments 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 it is also
less, for it is contained within the cylinder; which is impos-
sible: therefore the cylinder is not less than the triple of the
cone. And it has been demonstrated, that neither is it greater

:

E

B

F

BOOK XII. PROP. XI.

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.

PROPOSITION XI.

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

Let the cones and cylinders, of which the bases are the circles ABCD, EFGH, and he axes KL, MN, and AC, EG the diameters of their bases, 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, 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 circumscribed about it, because they are to one another as their bases: but the cone

T

C F

H

R

骨

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.

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.

6. 12.

Lemma.

* 1. 12.

a 2. 12.

* 11. 5.

* 6 12.

⚫ 14.5.

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 triangles, 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 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. 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 to the square of EG, so is the polygon ATBYCVDQ to the polygon EOFPGRHS; and as the square of AC to the square of EG, so is the circle ABCD to 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 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 pyra

A

N

H

V

C EK

R
G

T

mid 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 contained in it; therefore the solid X is greater than the pyramid in the cone EN: but it is less, as was shewn; which

* 14.5.

is absurd: therefore the circle ABCD is not to the circle EFGH, as the cone AL to any solid which is less than the cone EN. In the same manner it may be demonstrated, that the circle EFGH is not to the circle ABCD, as the cone EN to any solid less than the cone AL. Nor can the circle ABCD be to the circle EFGH, as the cone AL, to any solid greater than the cone EN. For if it be possible, let it be so to the solid I, which is greater than the cone EN: therefore, by inversion, as the circle EFGH to the circle ABCD, so is the solid I to the cone AL: but as the solid I to the cone AL, so is the cone EN to some solid, which must be less than the cone AL, because the solid I is greater than the cone EN; therefore, as the circle EFGH is to the circle ABCD, so is the cone EN to a solid less than the cone AL, which was shewn to be impossible; therefore the circle ABCD is not to the circle EFGH, as the cone AL is to any solid greater than the cone EN. And it has been demonstrated, that neither is the circle ABCD to the circle EFGH, as the cone AL to any solid less than the cone EN; therefore the circle ABCD is to the circle EFGH, as the cone AL to the cone EN: but as the cone is to the cone, so is the cylinder to the cylinder, because the cylinders are triple of the cones, each of each: therefore as the 10. 12. circle ABCD to the circle EFGH, so are the cylinders upon them of the same altitude. Wherefore cones and cylinders, &c. Q. E. D.

15.5.

PROPOSITION XII.

THEOR.-Similar cones and cylinders have to one another, the See N. triplicate ratio of that which the diameters of their bases have.

Let the cones and cylinders of which the bases are the circles ABCD, EFGH, and the diameters of the bases AC, EG, and KL, MN the axes of the cones or cylinders, be similar: the cone of which the base is the circle ABCD and vertex the point L, shall have to the cone of which the base is the circle EFGH and vertex N, the triplicate ratio of that which AC has to EG.

For if the cone ABCDL has not to the cone EFGHN, the triplicate ratio of that which AC has to EG, the cone ABCDL must have the triplicate of that ratio to some solid which is

less or greater than the cone EFGHN. First, let it have it to a less, viz. to the solid X. Make the same construction as in the preceding proposition, and it may be demonstrated the very same way as in that proposition, that the pyramid of which the base is the polygon EOFPGRHS, and vertex N, is greater than the solid X. Inscribe also in the circle ABCD, the polygon ATBYCVDQ similar to the polygon EOFPGRAS, upon which, erect a pyramid having the same vertex with the cone and let LAQ be one of the triangles containing the pyramid upon the polygon ATBYCVDQ, the vertex of which is L; and let NES be one of the triangles containing the pyramid upon the polygon EOFPGRHS of which the vertex is N;

15.5.

• 6.6.

and join KQ, MS. Then, because the cone ABCDL is similar * 24 Def.11. to the cone EFGHN, AC is * to EG as the axis KL to the axis MN; and as AC to EG, so is AK to EM; therefore as AK to EM, so is KL to MN; and alternately, AK to KL, as EM to MN: and the right angles AKL, EMN are equal: therefore, the sides about these equal angles being proportionals, the triangle AKL is similar to the triangle EMN. Again, because AK is to KQ, as EM to MS, and that these sides are about equal angles AKQ, EMS, because these angles are, each of them, the same part of four right angles at the centres K, M, therefore the triangle AKQ is similar to the triangle. EMS. And because it has been shewn, that as AK to KL, so is EM to MN, and that AK is equal to KQ, and EM to MS, therefore as QK to KL, so is SM to MN; and therefore, the sides about the right angles QKL, SMN, being proportionals, the triangle LKQ is similar to the triangle NMS. And because of the similarity of the triangles AKL, EMN, as LA is to AK, so is NE to EM; and by the similarity of the triangles AKQ, EMS, as KA to AQ, so ME to ES: therefore, ex æquali*, LA is to AQ, as NE to ES. Again, because of the similarity of the triangles LQK, NSM, as LQ to QK, so NS to SM; and

6. 6.

• 22.5.