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: (XII. 6.)
but 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 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; (Lemma.)

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; (x11. 1.) and as the square of AC to the square of EG, so is the circle ABCD to the circle EFGH; (XII. 2.)

therefore the circle ABCD is to the circle EFGH, as the polygon ATBYCVDQ to the polygon EOFPGRHS: (v. 11.)

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: (XII. 6.)

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 contained in it; therefore the solid X is greater than the pyramid in the cone EN: (v.14.) but it is less, as was shewn; which 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; (v. 14.)

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, (v. 15.) because the cylinders are triple of the cones, each of each, (XII. 10.) therefore as the circle ABCD to the circle EFGH, so are the cylinders upon them of the same altitude.

Wherefore, cones and cylinders of the same altitude are to one another as their bases. Q.E.D.

PROPOSITION XII. THEOREM.

Similar cones and cylinders have to one another the 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, if possible, 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 EOFPGRHS, 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, and join KQ,

MS.

Then, because the cone ABCDL is similar to the cone EFGHN, AC is to EG as the axis KL to the axis MN; (x1. def. 24.) and as AC to EG, so is AK to EM; (v. 15.) 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:

(vi. 6.)

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. (vI. 6.) 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. (v. 22.)
Again, because of the similarity of the triangles LQK, NSM,
as LQ to QK, so NS to SM;

and from the similarity of the triangles KAQ, MES,

as KQ to QA, so MS to SE:

therefore, ex æquali, LQ is to QA, as NS to SE: (v. 22.) and it was proved that QA is to AL, as SE to EN: therefore, again, ex æquali, as QL to LA, so is SN to NE: wherefore the triangles LQA, NSE, having the sides about all their angles proportionals, are equiangular and similar to one another: (VI. 5.)

and therefore the pyramid of which the base is the triangle AKQ, and vertex L, is similar to the pyramid the base of which is the triangle EMS, and vertex N,

because their solid angles are equal to one another, and they are contained by the same number of similar planes: (XI. B.)

but similar pyramids which have triangular bases have to one another the triplicate ratio of that which their homologous sides have; (XII. 8.) therefore the pyramid AKQL has to the pyramid EMSN the triplicate ratio of that which AK has to EM.

In the same manner, if straight lines be drawn from the points D, V, C, Y, B, T to K,

and from the points H, R, G, P, F, O to M,

and pyramids be erected upon the triangles having the same vertices with the cones,

it may be demonstrated that each pyramid in the first cone has to each in the other, taking them in the same order, the triplicate ratio of that which the side AK has to the side EN;

that is, which AC has to EG:

but as one antecedent to its consequent, so are all the antecedents to all the consequents; (v. 12.)

therefore as the pyramid AKQL to the pyramid EMSN, so is the whole pyramid the base of which is the polygon DQATBYCV, and vertex L, to the whole pyramid of which the base is the polygon HSEOFPGR, and vertex N:

wherefore also the first of these two last-named pyramids has to the other the triplicate ratio of that which AC has to EG:

but, by the hypothesis, the cone of which the base is the circle ABCD, and vertex L, has to the solid X, the triplicate ratio of that which AC has to EG;

therefore, as the cone of which the base is the circle ABCD, and vertex L, is to the solid X, so is the pyramid the base of which is the polygon DQATBYCV, and vertex L, to the pyramid the base of which is the polygon HSEOFPGR, and vertex N:

but the said cone is greater than the pyramid contained in it; therefore the solid X is greater than the pyramid, the base of which is the polygon HSEOFPGR, and vertex N: (v. 14.)

but it is also less; which is impossible:

therefore the cone, of which the base is the circle ABCD and vertex L, has not to any solid which is less than the cone of which the base is the circle EFGÍ, and vertex N, the triplicate ratio of that which AC has to EG.

In the same manner it may be demonstrated, that neither has the cone EFGHN to any solid which is less than the cone ABCDL, the triplicate ratio of that which EG has to AC.

Nor can the cone ABCDL have to any solid which is greater than the cone EFGHN, the triplicate ratio of that which AC has to EG.

For, if it be possible, let it have it to a greater, viz. to the solid Z: therefore, inversely, the solid Z has to the cone ABCDL, the triplicate ratio of that which EG has to AC:

but as the solid Z is to the cone ABCDL, so is the cone EFGHN to some solid, which must be less than the cone ABCDL, (v.14.) because the solid Z is greater than the cone EFGHN; therefore the cone EFGHN has to a solid which is less than the cone ABCDL, the triplicate ratio of that which EG has to AC, which was demonstrated to be impossible:

therefore the cone ABCDL has not to any solid greater than the cone EFGHN, the triplicate ratio of that which AC has to EG: and it was demonstrated, that it could not have that ratio to any solid less than the cone EFGHN;

therefore the cone ABCDL has to the cone EFGHN, the triplicate ratio of that which AC has to EG:

but as the cone is to the cone, so the cylinder to the cylinder; (v. 15.) for every cone is the third part of the cylinder upon the same base, and of the same altitude: (XII. 10.)

therefore also the cylinder has to the cylinder, the triplicate ratio of that which AC has to EG.

If a cylinder be cut by a plane parallel to its opposite planes, or bases; it divides the cylinder into two cylinders, one of which is to the other as the axis of the first to the axis of the other.

Let the cylinder AD be cut by the plane GH parallel to the opposite planes AB, CD, meeting the axis EF in the point K,

and let the line GH be the common section of the plane GH and the surface of the cylinder AD.

Let AEFC be the parallelogram in any position of it, by the revolution of which about the straight line EF the cylinder AD is described; and let GK be the common section of the plane GH, and the plane AEFC.

And because the parallel planes AB, GH are cut by the plane AEKG,

their common sections AE, KG, with it, are parallel: (x1.16.)

wherefore AK is a parallelogram,

and GK equal to EA the straight line from the centre of the circle AB: for the same reason, each of the straight lines drawn from the point K to the line GH may be proved to be equal to those which are drawn from the centre of the circle AB to its circumference, and are therefore all equal to one another;

therefore the line GH is the circumference of a circle of which the centre is the point K: (1. def. 15.)

therefore the plane GH divides the cylinder AD into the cylinders AH, GD;

for they are the same which would be described by the revolution of the parallelograms AK, GF about the straight lines EK, KF: and it is to be shewn, that the cylinder AH is to the cylinder HC, as the axis EK to the axis KF.