Proposition 12

Let there be similar cones and cylinders, let the circles ABCD and EFGH be their bases, BD and FH the diameters of the bases, and KL and MN the axes of the cones and cylinders.

I say that the cone with circular base ABCD and vertex L has to the cone with circular base EFGH and vertex N the ratio triplicate of that which BD has to FH.

For, if the cone ABCDL does not have to the cone EFGHN the ratio triplicate of that which BD has to FH, then the cone ABCDL has that triplicate ratio either to some solid less than the cone EFGHN or to a greater.

First, let it have that triplicate ratio to a less solid O. Inscribe the square EFGH in the circle EFGH. Therefore the square EFGH is greater than the half of the circle EFGH.

Now set up on the square EFGH a pyramid with the same vertex as the cone. Therefore the pyramid so set up is greater than the half part of the cone. Bisect the circumferences EF, FG, GH, and HE at the points P, Q, R, and S, and join EP, PF, FQ, QG, GR, RH, HS, and SE.

Therefore each of the triangles EPF, FQG, GRH, and HSE is also greater than the half part of that segment of the circle EFGH about it.

Now set up on each of the triangles EPF, FQG, GRH, and HSE a pyramid with the same vertex as the cone.

Therefore each of the pyramids so set up is also greater than the half part of that segment of the cone about it.

Thus, bisecting the circumferences so left, joining straight lines, setting up on each of the triangles pyramids with the same vertex as the cone, and doing this repeatedly, we shall leave some segments of the cone which are less than the excess by which the cone EFGHN exceeds the solid O.

Let such be left, and let them be the segments on EP, PF, FQ, QG, GR, RH, HS, and SE. Therefore the remainder, the pyramid with the polygonal base EPFQGRHS and vertex N, is greater than the solid O.

Now inscribe in the circle ABCD the polygon ATBUCVDW similar and similarly situated to the polygon EPFQGRHS, and set up on the polygon ATBUCVDW a pyramid with the same vertex as the cone.

Let LBT be one off the triangles containing the pyramid with polygonal base ATBUCVDW and vertex L, and let NFP be one of the triangles containing the pyramid with polygonal base EPFQGRHS and vertex N. Join KT and MP.

Now, since the cone ABCDL is similar to the cone EFGHN, therefore BD is to FH as the axis KL is to the axis MN.

But BD is to FH as BK is to FM, therefore BK is to FM as KL to MN. And, alternately BK is to KL as FM is to MN.

And the sides are proportional about equal angles, namely the angles BKL and FMN, therefore the triangle BKL is similar to the triangle FMN.

Again, since BK is to KT as FM is to MP, and they are about equal angles, namely the angles BKT and FMP, for whatever part the angle BKT is of the four right angles at the center K, it is the same part as the angle FMP of the four right angles at the center M. Then, since the sides are proportional about equal angles, therefore the triangle BKT is similar to the triangle FMP.

Again, since it was proved that BK is to KL as FM is to MN, while BK equals KT, and FM equals PM, therefore TK is to KL as PM is to MN. And the sides are proportional about equal angles, namely the angles TKL and PMN, for they are right, therefore the triangle LKT is similar to the triangle NMP.

And since the triangles LKB and NMF are similar, therefore LB is to BK as NF is to FM. And since the triangles BKT and FMP are similar, therefore KB is to BT as MF is to FP. Therefore, ex aequali, LB is to BT as NF is to FP.

Again, since the triangles LTK and NPM are similar, therefore LT is to TK as NP is to PM, and since the triangles TKB and PMF are similar, therefore KT is to TB as MP is to PF. Therefore, ex aequali, LT is to TB as NP is to PF.

But it was also proved that TB is to BL as PF is to FN. Therefore, ex aequali, TL is to LB as PN is to NF.

Therefore in the triangles LTB and NPF the sides are proportional. Therefore the triangles LTB and NPF are equiangular, hence they are also similar.

Therefore the pyramid with triangular base BKT and vertex L is similar to the pyramid with triangular base FMP and vertex N, for they are contained by similar planes equal in multitude.

But similar pyramids with triangular bases are to one another in the triplicate ratio of their corresponding sides.

Therefore the pyramid BKTL has to the pyramid FMPN the ratio triplicate of that which BK has to FM.

Similarly, by joining straight lines from A, W, D, V, C, and U to K, and from E, S, H, R, G, and Q to M, and setting up on each of the triangles pyramids with the same vertex as the cones, we can prove that each of the similarly arranged pyramids also has to each similarly arranged pyramid the ratio triplicate of that which the corresponding side BK has to the corresponding side FM, that is, which BD has to FH.

And one of the antecedents is to one of the consequents as all the antecedents are to all the consequents, therefore the pyramid BKTL is to the pyramid FMPN as the whole pyramid with polygonal base ATBUCVDW and vertex L is to the whole pyramid with polygonal base EPFQGRHS and vertex N, hence the pyramid with base ATBUCVDW and vertex L has to the pyramid with polygonal base EPFQGRHS and vertex N the ratio triplicate of that which BD has to FH.

But, by hypothesis, the cone with circular base ABCD and vertex L also has to the solid O the ratio triplicate of that which BD has to FH, therefore the cone with circular base ABCD and vertex L is to the solid 0 as the pyramid with polygonal base ATBUCVDW and vertex L is to the pyramid with polygonal base EPFQGRHS and vertex N. Therefore, alternately the cone with circular base ABCD and vertex L is to the pyramid contained in it with polygonal base ATBUCVDW and vertex L as the solid O is to the pyramid with the polygonal base EPFQGRHS and vertex N.

But the said cone is greater than the pyramid in it, for it encloses it. Therefore the solid O is also greater than the pyramid with polygonal base EPFQGRHS and vertex N. But it is also less, which is impossible.

Therefore the cone with circular base ABCD and vertex L does not have to any solid less than the cone of with circular base EFGH and vertex N the ratio triplicate of that which BD has to FH.

Similarly we can prove that neither has the cone EFGHN to any solid less than the cone ABCDL the ratio triplicate of that which FH has to BD.

I say next that neither has the cone ABCDL to any solid greater than the cone EFGHN the ratio triplicate of that which BD has to FH.

For, if possible, let it have that ratio to a greater solidO. Therefore, inversely, the solid O has to the cone ABCDL the ratio triplicate of that which FH has to BD. But the solid O is to the cone ABCDL as the cone EFGHN is to some solid less than the cone ABCDL.

Therefore the cone EFGHN also has to some solid less than the cone ABCDL the ratio triplicate of that which FH has to BD, which was proved impossible.

Therefore the cone ABCDL does not have to any solid greater than the cone EFGHN the ratio triplicate of that which BD has to FH.

But it was proved that neither has it this ratio to a less solid than the cone EFGHN. Therefore the cone ABCDL has to the cone EFGHN the ratio triplicate of that which BD has to FH.

But the cone is to the cone as the cylinder is to the cylinder, for the cylinder with the same base as the cone and of equal height with it is triple the cone. Therefore the cylinder also has to the cylinder the ratio triplicate of that which BD has to FH.

Therefore, similar cones and cylinders are to one another in triplicate ratio of the diameters of their bases.