PROPOSITION 12.
Similar cones and cylinders are to one another in the triplicate ratio of the diameters in their bases.
Let there be similar cones and cylinders, let the circles
ABCD,
EFGH be their bases,
BD,
FH the diameters of the bases, and
KL,
MN the axes of the cones and cylinders; I say that the cone of which the circle
ABCD is the base and the point
L the vertex has to the cone of which the circle
EFGH is the base and the point
N the vertex the ratio triplicate of that which
BD has to
FH.
For, if the cone
ABCDL has not to the cone
EFGHN the ratio triplicate of that which
BD has to
FH, the cone
ABCDL will have 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.
Let the square
EFGH be inscribed in the circle
EFGH; [
IV. 6] therefore the square
EFGH is greater than the half of the circle
EFGH.
Now let there be set up on the square
EFGH a pyramid having the same vertex with the cone; therefore the pyramid so set up is greater than the half part of the cone.
Let the circumferences
EF,
FG,
GH,
HE be bisected at the points
P,
Q,
R,
S, and let
EP,
PF,
FQ,
QG,
GR,
RH,
HS,
SE be joined.
Therefore each of the triangles
EPF,
FQG,
GRH,
HSE is also greater than the half part of that segment of the circle
EFGH which is about it.
Now on each of the triangles
EPF,
FQG,
GRH,
HSE let a pyramid be set up having the same vertex with the cone; therefore each of the pyramids so set up is also greater than the half part of that segment of the cone which is about it.
Thus, bisecting the circumferences so left, joining straight lines, setting up on each of the triangles pyramids having the same vertex with the cone, and doing this continually, we shall leave some segments of the cone which will be less than the excess by which the cone
EFGHN exceeds the solid
O. [
X. 1]
Let such be left, and let them be the segments on
EP,
PF,
FQ,
QG,
GR,
RH,
HS,
SE; therefore the remainder, the pyramid of which the polygon
EPFQGRHS is the base and the point
N the vertex, is greater than the solid
O.
Let there be also inscribed in the circle
ABCD the polygon
ATBUCVDW similar and similarly situated to the polygon
EPFQGRHS, and let there be set up on the polygon
ATBUCVDW a pyramid having the same vertex with the cone; of the triangles containing the pyramid of which the polygon
ATBUCVDW is the base and the point
L the vertex let
LBT be one, and of the triangles containing the pyramid of which the polygon
EPFQGRHS is the base and the point
N the vertex let
NFP be one; and let
KT,
MP be joined.
Now, since the cone
ABCDL is similar to the cone
EFGHN, therefore, as
BD is to
FH, so is the axis
KL to the axis
MN. [
XI. Def. 24]
But, as
BD is to
FH, so is
BK to
FM; therefore also, as
BK is to
FM, so is
KL to
MN.
And, alternately, as
BK is to
KL, so is
FM to
MN. [
V. 16]
And the sides are proportional about equal angles, namely the angles
BKL,
FMN; therefore the triangle
BKL is similar to the triangle
FMN. [
VI. 6]
Again, since, as
BK is to
KT, so is
FM to
MP, and they are about equal angles, namely the angles
BKT,
FMP, inasmuch as, whatever part the angle
BKT is of the four right angles at the centre
K, the same part also is the angle
FMP of the four right angles at the centre
M; since then the sides are proportional about equal angles, therefore the triangle
BKT is similar to the triangle
FMP. [
VI. 6]
Again, since it was proved that, as
BK is to
KL, so is
FM to
MN, while
BK is equal to
KT, and
FM to
PM, therefore, as
TK is to
KL, so is
PM to
MN; and the sides are proportional about equal angles, namely the angles
TKL,
PMN, for they are right; therefore the triangle
LKT is similar to the triangle
NMP. [
VI. 6]
And since, owing to the similarity of the triangles
LKB,
NMF, as
LB is to
BK, so is
NF to
FM, and, owing to the similarity of the triangles
BKT,
FMP, as
KB is to
BT, so is
MF to
FP, therefore,
ex aequali, as
LB is to
BT, so is
NF to
FP. [
V. 22]
Again since, owing to the similarity of the triangles
LTK,
NPM, as
LT is to
TK, so is
NP to
PM, and, owing to the similarity of the triangles
TKB,
PMF, as
KT is to
TB, so is
MP to
PF; therefore,
ex aequali, as
LT is to
TB, so is
NP to
PF. [
V. 22]
But it was also proved that, as
TB is to
BL, so is
PF to
FN.
Therefore,
ex aequali, as
TL is to
LB, so is
PN to
NF. [
V. 22]
Therefore in the triangles
LTB,
NPF the sides are proportional; therefore the triangles
LTB,
NPF are equiangular; [
VI. 5] hence they are also similar. [
VI. Def. I]
Therefore the pyramid of which the triangle
BKT is the base and the point
L the vertex is also similar to the pyramid of which the triangle
FMP is the base and the point
N the vertex, for they are contained by similar planes equal in multitude. [
XI. Def. 9]
But similar pyramids which have triangular bases are to one another in the triplicate ratio of their corresponding sides. [
XII. 8]
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,
U to
K, and from
E,
S,
H,
R,
G,
Q to
M, and setting up on each of the triangles pyramids which have the same vertex with the cones, we can prove that each of the similarly arranged pyramids will also have 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, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [
V. 12] therefore also, as the pyramid
BKTL is to the pyramid
FMPN, so is the whole pyramid of which the polygon
ATBUCVDW is the base and the point
L the vertex to the whole pyramid of which the polygon
EPFQGRHS is the base and the point
N the vertex; hence also the pyramid of which
ATBUCVDW is the base and the point
L the vertex has to the pyramid of which the polygon
EPFQGRHS is the base and the point
N the vertex the ratio triplicate of that which
BD has to
FH.
But, by hypothesis, the cone of which the circle
ABCD
is the base and the point
L the vertex has also to the solid
O the ratio triplicate of that which
BD has to
FH; therefore, as the cone of which the circle
ABCD is the base and the point
L the vertex is to the solid
O, so is the pyramid of which the polygon
ATBUCVDW is the base and
L the vertex to the pyramid of which the polygon
EPFQGRHS is the base and the point
N the vertex; therefore, alternately, as the cone of which the circle
ABCD is the base and
L the vertex is to the pyramid contained in it of which the polygon
ATBUCVDW is the base and
L the vertex, so is the solid
O to the pyramid of which the polygon
EPFQGRHS is the base and
N the vertex. [
V. 16]
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 of which the polygon
EPFQGRHS is the base and
N the vertex.
But it is also less: which is impossible.
Therefore the cone of which the circle
ABCD is the base and
L the vertex has not to any solid less than the cone of which the circle
EFGH is the base and the point
N the vertex 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 solid
O.
Therefore, inversely, the solid
O has to the cone
ABCDL the ratio triplicate of that which
FH has to
BD.
But, as the solid
O is to the cone
ABCDL, so is the cone
EFGHN 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 has not 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, as the cone is to the cone, so is the cylinder to the cylinder, for the cylinder which is on the same base as the cone and of equal height with it is triple of the cone; [
XII. 10] therefore the cylinder also has to the cylinder the ratio triplicate of that which
BD has to
FH.
Therefore etc. Q. E. D.