G.321 similar solid numbers

VIII.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.

similar_solid_numbers(A, B, C, D)

given A, B be two similar solid numbers, let C, D, E be the sides of A, F, G, H of B

prove 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; 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

given A, B be two similar solid numbers, let C, D, E be the sides of A, F, G, H of B

∵ similar_plane_solid_defassert similar solid numbers are those which have their sides proportional

assert as C is to D, so is F to G, and, as D is to E, so is G to H

assert Let C by multiplying D make K, and let F by multiplying G make L

∵ similar_plane_solid_defassert 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

∵ similar_plane_numbersassert between K, L there is one mean proportional number

assert Let it be M

∵ similar_plane_numbersassert M is the product of D, F, as was proved in the theorem preceding this

assert D by multiplying C has made K, and by multiplying F has made M,

∵ number_multiplying_numbersassert as C is to F, so is K to M

assert But, as K is to M, so is M to L

assert K, M, L are continuously proportional in the ratio of C to F

assert Since, as C is to D, so is F to G,

∵ numbers_proportional_proportionalassert Alternately therefore, as C is to F, so is D to G

assert For the same reason also,

assert As D is to G, so is E to H

assert 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

assert Next, let E, H by multiplying M make N, O

assert A is a solid number, and C, D, E are its sides,

assert E by multiplying the product of C, D has made A

assert The product of C, D is K

assert E by multiplying K has made A

assert For the same reason also

assert H by multiplying L has made B

assert E by multiplying K has made A, and further also by multiplying M has made N,

∵ number_multiplying_numbersassert as K is to M, so is A to N

assert But, as K is to M, so is C to F, D to G, and also E to H

assert also, as C is to F, D to G, and E to H, so is A to N

assert E, H by multiplying M have made N, O,

∵ numbers_multiplying_numberassert as E is to H, so is N to O

assert But, as E is to H, so is C to F and D to G

assert also, as C is to F, D to G, and E to H, so is A to N and N to O

assert H by multiplying M has made O, and further also by multiplying L has made B,

∵ number_multiplying_numbersassert as M is to L, so is O to B

assert But, as M is to L, so is C to F, D to G, and E to H

assert 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

assert A, N, O, B are continuously proportional in the aforesaid ratios of the sides

assert A, N, O, B are four numbers in continued proportion,

∵ magnitudes_proportional_said_def_v_def_10assert A has to B the ratio triplicate of that which A has to N

assert But, as A is to N, so it was proved that C is to F, D to G, and also E to H

∴ 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

Dependency Graph

Required for

G.327, G.329, G.333, G.334, G.335, G.339