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

# V.7

*Equal magnitudes have to the same the same ratio*, *as also has the same to equal magnitudes*.

## Proof

Let *A*, *B* be equal magnitudes and *C* any other, chance, magnitude; I say that each of the magnitudes *A*, *B* has the same ratio to *C*, and *C* has the same ratio to each of the magnitudes *A*, *B*.

For let equimultiples *D*, *E* of *A*, *B* be taken, and of *C* another, chance, multiple *F*.

Then, since *D* is the same multiple of *A* that *E* is of *B*, while *A* is equal to *B*, therefore *D* is equal to *E*.

But *F* is another, chance, magnitude.

If therefore *D* is in excess of *F*, *E* is also in excess of *F*, if equal to it, equal; and, if less, less.

And *D*, *E* are equimultiples of *A*, *B*, while *F* is another, chance, multiple of *C*; therefore, as *A* is to *C*, so is *B* to *C*. [[book-5/definitions#Definition 5|V. Def. 5]]

I say next that *C* also has the same ratio to each of the magnitudes *A*, *B*.

For, with the same construction, we can prove similarly that *D* is equal to *E*; and *F* is some other magnitude.

If therefore *F* is in excess of *D*, it is also in excess of *E*, if equal, equal; and, if less, less.

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

Therefore etc.

Porism. From this it is manifest that, if any magnitudes are proportional, they will also be proportional inversely. Q. E. D.