---
book: 5
number: 17
id: "V.17"
kind: "theorem"
uses: ["[[book-5/proposition-1]]", "[[book-5/proposition-2]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:5.prop.17"
license: "CC-BY-SA-4.0"
---

# V.17

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

## Proof

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

For of *AE*, *EB*, *CF*, *FD* let equimultiples *GH*, *HK*, *LM*, *MN* be taken, and of *EB*, *FD* other, chance, equimultiples, *KO*, *NP*.

Then, since *GH* is the same multiple of *AE* that *HK* is of *EB*, therefore *GH* is the same multiple of *AE* that *GK* is of *AB*. [[book-5/proposition-1|V. 1]]

But *GH* is the same multiple of *AE* that *LM* is of *CF*; therefore *GK* is the same multiple of *AB* that *LM* is of *CF*.

Again, since *LM* is the same multiple of *CF* that *MN* is of *FD*, therefore *LM* is the same multiple of *CF* that *LN* is of *CD*. [[book-5/proposition-1|V. 1]]

But *LM* was the same multiple of *CF* that *GK* is of *AB*; therefore *GK* is the same multiple of *AB* that *LN* is of *CD*.

Therefore *GK*, *LN* are equimultiples of *AB*, *CD*.

Again, since *HK* is the same multiple of *EB* that *MN* is of *FD*, and *KO* is also the same multiple of *EB* that *NP* is of *FD*, therefore the sum *HO* is also the same multiple of *EB* that *MP* is of *FD*. [[book-5/proposition-2|V. 2]]

And, since, as *AB* is to *BE*, so is *CD* to *DF*, and of *AB*, *CD* equimultiples *GK*, *LN* have been taken, and of *EB*, *FD* equimultiples *HO*, *MP*, therefore, if *GK* is in excess of *HO*, *LN* is also in excess of *MP*, if equal, equal, and if less, less.

Let *GK* be in excess of *HO*; then, if *HK* be subtracted from each, *GH* is also in excess of *KO*.

But we saw that, if *GK* was in excess of *HO*, *LN* was also in excess of *MP*; therefore *LN* is also in excess of *MP*, and, if *MN* be subtracted from each, *LM* is also in excess of *NP*; so that, if *GH* is in excess of *KO*, *LM* is also in excess of *NP*.

Similarly we can prove that, if *GH* be equal to *KO*, *LM* will also be equal to *NP*, and if less, less.

And *GH*, *LM* are equimultiples of *AE*, *CF*, while *KO*, *NP* are other, chance, equimultiples of *EB*, *FD*; therefore, as *AE* is to *EB*, so is *CF* to *FD*.

Therefore etc. Q. E. D.