¶ The 8. Theoreme. The 10. Proposition. If betwene two numbers and vnitie fall numbers in continuall proportion: how many numbers in continuall proportion fal betwene either of them & vnitie so many also shall there fall in continuall proportion betwene them.
SVppose that betwene the two numbers A, B, and vnitie C•• do fall these numbers in continuall proportion D, E, and F, G. Then I say that how many numbers in con∣tinuall proportion there are betwene either of these A, B, and vnitie C, so many
numbers also in continuall proportion shall there fall betwene
A and
B. Let
D multiply∣ing
F produce
H, and let
D multiplying
H produce
K, and like wise let
F multiplying
H produce
L. And for that by supposition as vnitie
C is to the number
D, so is
D to
E, there∣fore how many times vnitie
C measureth the number
D, so many times doth
D measure
E. But vnitie
C measureth
D by those vnities which are in
D•• wherefore
D measureth
E by those vnities which are in
D. Wherefore
D multiplying himselfe produceth
E. Againe for that as vnitie
C is to the number
D, so is
E to
A, therefore how many times vnitie
C mea∣sureth the number
D, so many times
E measureth
A. But vnitie
C measureth
D, by those vnitie
•• which are in
D, therefore
E measureth
A by those vnities which are in
D. Where∣fore
D multilying
E produced
A. And by the same reason
F multiplying himselfe produced
G, and multiplying
G produced
B. And forasmuch as
D multiplying himselfe produced
E, and multiplying
F produced
H, therefore (by the
17. of the seuenth) as
D is to
F, so is
E to
H. And by the same reason as
D is to
F, so is
H to
G. Wherefore as
E is to
H, so is
H to
G. Agayne forasmuch as
D multiplying
E produced
A, and multiplying
H produced
K, there∣fore (by the
17. of the seuēth) as
E is to
H, so is
A to
K. But as
E is to
H, so is
D to
F, there∣fore as
D is to
F, so is
A to
K. Againe forasmuch as
D multiplying
H produced
K, and
F multiplying
H produced
L, therefore (by the
17. of the seuenth) as
D is to
F, so is
K to
L. But as
D is to
F, so is
A to
K, wherfore as
A is to
K, so is
K to
L. Againe forasmuch as
F multiplying
H produced
L and multiplying
G produced
B, therefore (by the
17. of the seuenth) as
H is to
G, so is
L to
B. But as
H is to
G so is
D to
F, wherefore as
D is to
F so i
•• L to
B. And it is proued that as
D is to
F, so is
A to
K, and
K to
L, and
L to
B. Wherfore the numbers
A, K, L, B, are continuall proportion. Wherefore how many numbers in conti∣nuall proportion fall betwene either of these numbers
A, B, & vnitie
C, so many also in con∣tinuall proportion fall there betwene the numbers
A and
B: which was required to be proued.