---
book: 6
number: 30
id: "VI.30"
kind: "construction"
uses: ["[[book-6/proposition-29]]", "[[book-6/proposition-14]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.30"
license: "CC-BY-SA-4.0"
---

# VI.30

*To cut a given finite straight line in extreme and mean ratio*.

## Proof

Let *AB* be the given finite straight line; thus it is required to cut *AB* in extreme and mean ratio.

On *AB* let the square *BC* be described; and let there be applied to *AC* the parallelogram *CD* equal to *BC* and exceeding by the figure *AD* similar to *BC*. [[book-6/proposition-29|VI. 29]]

Now *BC* is a square; therefore *AD* is also a square.

And, since *BC* is equal to *CD*, let *CE* be subtracted from each; therefore the remainder *BF* is equal to the remainder *AD*.

But it is also equiangular with it; therefore in *BF*, *AD* the sides about the equal angles are reciprocally proportional; [[book-6/proposition-14|VI. 14]] therefore, as *FE* is to *ED*, so is *AE* to *EB*.

But *FE* is equal to *AB*, and *ED* to *AE*.

Therefore, as *BA* is to *AE*, so is *AE* to *EB*.

And *AB* is greater than *AE*; therefore *AE* is also greater than *EB*.

Therefore the straight line *AB* has been cut in extreme and mean ratio at *E*, and the greater segment of it is *AE*. Q. E. F.