PROPOSITION 19.
Between two similar solid numbers there fall two mean proportional numbers; and the solid number has to the similar solid number the ratio triplicate of that which the corresponding side has to the corresponding side.
Let
A,
B be two similar solid numbers, and let
C,
D,
E be the sides of
A, and
F,
G,
H of
B.
Now, since similar solid numbers are those which have their sides proportional, [
VII. Def. 21] therefore, as
C is to
D, so is
F to
G,
and, as D is to E, so is G to H.
I say that between
A,
B there fall two mean proportional numbers, and
A has to
B the ratio triplicate of that which
C has to
F,
D to
G, and also
E to
H.
For let
C by multiplying
D make
K, and let
F by multiplying
G make
L.
Now, since
C,
D are in the same ratio with
F,
G, and
K is the product of
C,
D, and
L the product of
F,
G,
K,
L are similar plane numbers; [
VII. Def. 21] therefore between
K,
L there is one mean proportional number. [
VIII. 18]
Let it be
M
Therefore
M is the product of
D,
F, as was proved in the theorem preceding this. [
VIII. 18]
Now, since
D by multiplying
C has made
K, and by multiplying
F has made
M, therefore, as
C is to
F, so is
K to
M. [
VII. 17]
But, as
K is to
M, so is
M to
L.
Therefore
K,
M,
L are continuously proportional in the ratio of
C to
F.
And since, as
C is to
D, so is
F to
G, alternately therefore, as
C is to
F, so is
D to
G. [
VII. 13]
For the same reason also,
as D is to G, so is E to H.
Therefore
K,
M,
L are continuously proportional in the ratio of
C to
F, in the ratio of
D to
G, and also in the ratio of
E to
H.
Next, let
E,
H by multiplying
M make
N,
O respectively.
Now, since
A is a solid number, and
C,
D,
E are its sides, therefore
E by multiplying the product of
C,
D has made
A.
But the product of
C,
D is
K; therefore
E by multiplying
K has made
A.
For the same reason also
H by multiplying L has made B.
Now, since
E by multiplying
K has made
A, and further also by multiplying
M has made
N, therefore, as
K is to
M, so is
A to
N. [
VII. 17]
But, as
K is to
M, so is
C to
F,
D to
G, and also
E to
H; therefore also, as
C is to
F,
D to
G, and
E to
H, so is
A to
N.
Again, since
E,
H by multiplying
M have made
N,
O respectively, therefore, as
E is to
H, so is
N to
O. [
VII. 18]
But, as
E is to
H, so is
C to
F and
D to
G; therefore also, as
C is to
F,
D to
G, and
E to
H, so is
A to
N and
N to
O.
Again, since
H by multiplying
M has made
O, and further also by multiplying
L has made
B, therefore, as
M is to
L, so is
O to
B. [
VII. 17]
But, as
M is to
L, so is
C to
F,
D to
G, and
E to
H.
Therefore also, as
C is to
F,
D to
G, and
E to
H, so not only is
O to
B, but also
A to
N and
N to
O.
Therefore
A,
N,
O,
B are continuously proportional in the aforesaid ratios of the sides.
I say that
A also has to
B the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of the ratio which the number
C has to
F, or
D to
G, and also
E to
H.
For, since
A,
N,
O,
B are four numbers in continued proportion, therefore
A has to
B the ratio triplicate of that which
A has to
N. [
V. Def. 10]
But, as
A is to
N, so it was proved that
C is to
F,
D to
G, and also
E to
H.
Therefore
A also has to
B the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of the ratio which the number
C has to
F,
D to
G, and also
E to
H. Q. E. D.
