---
book: 5
number: 15
id: "V.15"
kind: "theorem"
uses: ["[[book-5/proposition-7]]", "[[book-5/proposition-12]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:5.prop.15"
license: "CC-BY-SA-4.0"
---

# V.15

*Parts have the same ratio as the same multiples of them taken in corresponding order*.

## Proof

For let *AB* be the same multiple of *C* that *DE* is of *F*; I say that, as *C* is to *F*, so is *AB* to *DE*.

For, since *AB* is the same multiple of *C* that *DE* is of *F*, as many magnitudes as there are in *AB* equal to *C*, so many are there also in *DE* equal to *F*.

Let *AB* be divided into the magnitudes *AG*, *GH*, *HB* equal to *C*, and *DE* into the magnitudes *DK*, *KL*, *LE* equal to *F*; then the multitude of the magnitudes *AG*, *GH*, *HB* will be equal to the multitude of the magnitudes *DK*, *KL*, *LE*.

And, since *AG*. *GH*, *HB* are equal to one another, and *DK*, *KL*, *LE* are also equal to one another, therefore, as *AG* is to *DK*, so is *GH* to *KL*, and *HB* to *LE*. [[book-5/proposition-7|V. 7]]

Therefore, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents; [[book-5/proposition-12|V. 12]] therefore, as *AG* is to *DK*, so is *AB* to *DE*.

But *AG* is equal to *C* and *DK* to *F*; therefore, as *C* is to *F*, so is *AB* to *DE*.

Therefore etc. Q. E. D.