---
book: 5
number: 14
id: "V.14"
kind: "theorem"
uses: ["[[book-5/proposition-8]]", "[[book-5/proposition-13]]", "[[book-5/proposition-10]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:5.prop.14"
license: "CC-BY-SA-4.0"
---

# V.14

*If a first magnitude have to a second the same ratio as a third has to a fourth*, *and the first be greater than the third*, *the second will also be greater than the fourth*; *if equal*, *equal*; *and if less*, *less*.

## Proof

For let a first magnitude *A* have the same ratio to a second *B* as a third *C* has to a fourth *D*; and let *A* be greater than *C*; I say that *B* is also greater than *D*.

For, since *A* is greater than *C*, and *B* is another, chance, magnitude, therefore *A* has to *B* a greater ratio than *C* has to *B*. [[book-5/proposition-8|V. 8]]

But, as *A* is to *B*, so is *C* to *D*; therefore *C* has also to *D* a greater ratio than *C* has to *B*. [[book-5/proposition-13|V. 13]]

But that to which the same has a greater ratio is less; [[book-5/proposition-10|V. 10]] therefore *D* is less than *B*; so that *B* is greater than *D*.

Similarly we can prove that, if *A* be equal to *C*, *B* will also be equal to *D*; and, if *A* be less than *C*, *B* will also be less than *D*.

Therefore etc. Q. E. D.