---
book: 5
number: 18
id: "V.18"
kind: "theorem"
uses: ["[[book-5/proposition-17]]", "[[book-5/proposition-11]]", "[[book-5/proposition-14]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:5.prop.18"
license: "CC-BY-SA-4.0"
---

# V.18

*If magnitudes be proportional* separando, *they will also be proportional* componendo.

## Proof

Let *AE*, *EB*, *CF*, *FD* be magnitudes proportional separando, so that, as *AE* is to *EB*, so is *CF* to *FD*; I say that they will also be proportional componendo, that is, as *AB* is to *BE*, so is *CD* to *FD*.

For, if *CD* be not to *DF* as *AB* to *BE*, then, as *AB* is to *BE*, so will *CD* be either to some magnitude less than *DF* or to a greater.

First, let it be in that ratio to a less magnitude *DG*.

Then, since, as *AB* is to *BE*, so is *CD* to *DG*, they are magnitudes proportional componendo; so that they will also be proportional separando. [[book-5/proposition-17|V. 17]]

Therefore, as *AE* is to *EB*, so is *CG* to *GD*.

But also, by hypothesis, as *AE* is to *EB*, so is *CF* to *FD*.

Therefore also, as *CG* is to *GD*, so is *CF* to *FD*. [[book-5/proposition-11|V. 11]]

But the first *CG* is greater than the third *CF*; therefore the second *GD* is also greater than the fourth *FD*. [[book-5/proposition-14|V. 14]]

But it is also less: which is impossible.

Therefore, as *AB* is to *BE*, so is not *CD* to a less magnitude than *FD*.

Similarly we can prove that neither is it in that ratio to a greater; it is therefore in that ratio to *FD* itself.

Therefore etc. Q. E. D.