---
book: 10
number: 70
id: "X.70"
kind: "theorem"
uses: ["[[book-10/proposition-41]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.70"
license: "CC-BY-SA-4.0"
---

# X.70

*A straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas*.

## Proof

Let *AB* be the side of the sum of two medial areas, and *CD* commensurable with *AB*; it is to be proved that *CD* is also the side of the sum of two medial areas.

For, since *AB* is the side of the sum of two medial areas, let it 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, the rectangle contained by them medial, and furthermore the sum of the squares on *AE*, *EB* incommensurable with the rectangle *AE*, *EB*. [[book-10/proposition-41|X. 41]]

Let the same construction be made as before.

We can then prove similarly that *CF*, *FD* are also incommensurable in square, 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, the rectangle *CF*, *FD* is medial, and moreover the sum of the squares on *CF*, *FD* is incommensurable with the rectangle *CF*, *FD*.

Therefore *CD* is the side of the sum of two medial areas. Q. E. D.