---
book: 8
number: 19
id: "VIII.19"
kind: "theorem"
uses: ["[[book-8/proposition-18]]", "[[book-7/proposition-17]]", "[[book-7/proposition-13]]", "[[book-7/proposition-18]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:8.prop.19"
license: "CC-BY-SA-4.0"
---

# 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.*

## Proof

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, [[book-7/definitions#Definition 21|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; [[book-7/definitions#Definition 21|VII. Def. 21]] therefore between *K*, *L* there is one mean proportional number. [[book-8/proposition-18|VIII. 18]]

Let it be *M*

Therefore *M* is the product of *D*, *F*, as was proved in the theorem preceding this. [[book-8/proposition-18|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*. [[book-7/proposition-17|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*. [[book-7/proposition-13|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*. [[book-7/proposition-17|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*. [[book-7/proposition-18|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*. [[book-7/proposition-17|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*. [[book-5/definitions#Definition 10|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.