---
book: 13
number: 3
id: "XIII.3"
kind: "theorem"
uses: []
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.3"
license: "CC-BY-SA-4.0"
---

# XIII.3

*If a straight line be cut in extreme and mean ratio*, *the square on the lesser segment added to the half of the greater segment is five times the square on the half of the greater segment.*

## Proof

For let any straight line *AB* be cut in extreme and mean ratio at the point *C*, let *AC* be the greater segment, and let *AC* be bisected at *D*; I say that the square on *BD* is five times the square on *DC*.

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

Since *AC* is double of *DC*, therefore the square on *AC* is quadruple of the square on *DC*, that is, *RS* is quadruple of *FG*.

And, since the rectangle *AB*, *BC* is equal to the square on *AC*, and *CE* is the rectangle *AB*, *BC*, therefore *CE* is equal to *RS*.

But *RS* is quadruple of *FG*; therefore *CE* is also quadruple of *FG*.

Again, since *AD* is equal to *DC*, *HK* is also equal to *KF*.

Hence the square *GF* is also equal to the square *HL*.

Therefore *GK* is equal to *KL*, that is, *MN* to *NE*; hence *MF* is also equal to *FE*.

But *MF* is equal to *CG*; therefore *CG* is also equal to *FE*.

Let *CN* be added to each; therefore the gnomon *OPQ* is equal to *CE*.

But *CE* was proved quadruple of *GF*; therefore the gnomon *OPQ* is also quadruple of the square *FG*.

Therefore the gnomon *OPQ* and the square *FG* are five times *FG*.

But the gnomon *OPQ* and the square *FG* are the square *DN*.

And *DN* is the square on *DB*, and *GF* the square on *DC*.

Therefore the square on *DB* is five times the square on *DC*. Q. E. D.