---
book: 5
number: 16
id: "V.16"
kind: "theorem"
uses: ["[[book-5/proposition-15]]", "[[book-5/proposition-11]]", "[[book-5/proposition-14]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:5.prop.16"
license: "CC-BY-SA-4.0"
---

# V.16

*If four magnitudes be proportional*, *they will also be proportional alternately*.

## Proof

Let *A*, *B*, *C*, *D* be four proportional magnitudes, so that, as *A* is to *B*, so is *C* to *D*; I say that they will also be so alternately, that is, as *A* is to *C*, so is *B* to *D*.

For of *A*, *B* let equimultiples *E*, *F* be taken, and of *C*, *D* other, chance, equimultiples *G*, *H*.

Then, since *E* is the same multiple of *A* that *F* is of *B*, and parts have the same ratio as the same multiples of them, [[book-5/proposition-15|V. 15]] therefore, as *A* is to *B*, so is *E* to *F*.

But as *A* is to *B*, so is *C* to *D*; therefore also, as *C* is to *D*, so is *E* to *F*. [[book-5/proposition-11|V. 11]]

Again, since *G*, *H* are equimultiples of *C*, *D*, therefore, as *C* is to *D*, so is *G* to *H*. [[book-5/proposition-15|V. 15]]

But, as *C* is to *D*, so is *E* to *F*; therefore also, as *E* is to *F*, so is *G* to *H*. [[book-5/proposition-11|V. 11]]

But, if four magnitudes be proportional, and the first be greater than the third, the second will also be greater than the fourth; if equal, equal; and if less, less. [[book-5/proposition-14|V. 14]]

Therefore, if *E* is in excess of *G*, *F* is also in excess of *H*, if equal, equal, and if less, less.

Now *E*, *F* are equimultiples of *A*, *B*, and *G*, *H* other, chance, equimultiples of *C*, *D*; therefore, as *A* is to *C*, so is *B* to *D*. [[book-5/definitions#Definition 5|V. Def. 5]]

Therefore etc. Q. E. D.