---
book: 10
number: 69
id: "X.69"
kind: "theorem"
uses: ["[[book-10/proposition-40]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.69"
license: "CC-BY-SA-4.0"
---

# X.69

*A straight line commensurable with the side of a rational plus a medial area is itself also the side of a rational plus a medial area*.

## Proof

Let *AB* be the side of a rational plus a medial area, and let *CD* be commensurable with *AB*; it is to be proved that *CD* is also the side of a rational plus a medial area.

Let *AB* be divided into its straight lines at *E*; therefore *AE*, *EB* are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [[book-10/proposition-40|X. 40]]

Let the same construction be made as before.

We can then prove similarly that *CF*, *FD* are incommensurable in square, and the sum of the squares on *AE*, *EB* is commensurable with the sum of the squares on *CF*, *FD*, and the rectangle *AE*, *EB* with the rectangle *CF*, *FD*; so that the sum of the squares on *CF*, *FD* is also medial, and the rectangle *CF*, *FD* rational.

Therefore *CD* is the side of a rational plus a medial area. Q. E. D.