---
book: 13
number: 5
id: "XIII.5"
kind: "theorem"
uses: ["[[book-6/proposition-17]]", "[[book-5/proposition-14]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.5"
license: "CC-BY-SA-4.0"
---

# XIII.5

*If a straight line be cut in extreme and mean ratio*, *and there be added to it a straight line equal to the greater segment*, *the whole straight line has been cut in extreme and mean ratio*, *and the original straight line is the greater segment.*

## Proof

For let the straight line *AB* be cut in extreme and mean ratio at the point *C*, let *AC* be the greater segment, and let *AD* be equal to *AC*.

I say that the straight line *DB* has been cut in extreme and mean ratio at *A*, and the original straight line *AB* is the greater segment.

For let the square *AE* be described on *AB*, and let the figure be drawn.

Since *AB* has been cut in extreme and mean ratio at *C*, therefore the rectangle *AB*, *BC* is equal to the square on *AC*. [[book-6/definitions#Definition 3|VI. Def. 3]], VI. 17

And *CE* is the rectangle *AB*, *BC*, and *CH* the square on *AC*; therefore *CE* is equal to *HC*.

But *HE* is equal to *CE*, and *DH* is equal to *HC*; therefore *DH* is also equal to *HE*.

Therefore the whole *DK* is equal to the whole *AE*.

And *DK* is the rectangle *BD*, *DA*, for *AD* is equal to *DL*; and *AE* is the square on *AB*; therefore the rectangle *BD*, *DA* is equal to the square on *AB*.

Therefore, as *DB* is to *BA*, so is *BA* to *AD*. [[book-6/proposition-17|VI. 17]]

And *DB* is greater than *BA*; therefore *BA* is also greater than *AD*. [[book-5/proposition-14|V. 14]]

Therefore *DB* has been cut in extreme and mean ratio at *A*, and *AB* is the greater segment. Q. E. D.