---
book: 5
number: 25
id: "V.25"
kind: "theorem"
uses: ["[[book-5/proposition-19]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:5.prop.25"
license: "CC-BY-SA-4.0"
---

# V.25

*If four magnitudes be proportional*, *the greatest and the least are greater than the remaining two*.

## Proof

Let the four magnitudes *AB*, *CD*, *E*, *F* be proportional so that, as *AB* is to *CD*, so is *E* to *F*, and let *AB* be the greatest of them and *F* the least; I say that *AB*, *F* are greater than *CD*, *E*.

For let *AG* be made equal to *E*, and *CH* equal to *F*.

Since, as *AB* is to *CD*, so is *E* to *F*, and *E* is equal to *AG*, and *F* to *CH*, therefore, as *AB* is to *CD*, so is *AG* to *CH*.

And since, as the whole *AB* is to the whole *CD*, so is the part *AG* subtracted to the part *CH* subtracted, the remainder *GB* will also be to the remainder *HD* as the whole *AB* is to the whole *CD*. [[book-5/proposition-19|V. 19]]

But *AB* is greater than *CD*; therefore *GB* is also greater than *HD*.

And, since *AG* is equal to *E*, and *CH* to *F*, therefore *AG*, *F* are equal to *CH*, *E*.

And if, *GB*, *HD* being unequal, and *GB* greater, *AG*, *F* be added to *GB* and *CH*, *E* be added to *HD*, it follows that *AB*, *F* are greater than *CD*, *E*.

Therefore etc. Q. E. D.