---
book: 10
number: 72
id: "X.72"
kind: "theorem"
uses: ["[[book-10/proposition-22]]", "[[book-6/proposition-1]]", "[[book-10/proposition-11]]", "[[book-10/proposition-36]]", "[[book-10/proposition-56]]", "[[book-10/proposition-59]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.72"
license: "CC-BY-SA-4.0"
---

# X.72

*If two medial areas incommensurable with one another be added together, the remaining two irrational straight lines arise, namely either a second bimedial or a side of the sum of two medial areas*.

## Proof

For let two medial areas *AB*, *CD* incommensurable with one another be added together; I say that the side of the area *AD* is either a second bimedial or a side of the sum of two medial areas.

For *AB* is either greater or less than *CD*.

First, if it so chance, let *AB* be greater than *CD*.

Let the rational straight line *EF* be set out, and to *EF* let there be applied the rectangle *EG* equal to *AB* and producing *EH* as breadth, and the rectangle *HI* equal to *CD* and producing *HK* as breadth.

Now, since each of the areas *AB*, *CD* is medial, therefore each of the areas *EG*, *HI* is also medial.

And they are applied to the rational straight line *FE*, producing *EH*, *HK* as breadth; therefore each of the straight lines *EH*, *HK* is rational and incommensurable in length with *EF*. [[book-10/proposition-22|X. 22]]

And, since *AB* is incommensurable with *CD*, and *AB* is equal to *EG*, and *CD* to *HI*, therefore *EG* is also incommensurable with *HI*.

But, as *EG* is to *HI*, so is *EH* to *HK*; [[book-6/proposition-1|VI. 1]] therefore *EH* is incommensurable in length with *HK*. [[book-10/proposition-11|X. 11]]

Therefore *EH*, *HK* are rational straight lines commensurable in square only; therefore *EK* is binomial. [[book-10/proposition-36|X. 36]]

But the square on *EH* is greater than the square on *HK* either by the square on a straight line commensurable with *EH* or by the square on a straight line incommensurable with it.

First, let the square on it be greater by the square on a straight line commensurable in length with itself.

Now neither of the straight lines *EH*, *HK* is commensurable in length with the rational straight line *EF* set out; therefore *EK* is a third binomial. [[book-10/definitions#Definition 3 (part 2)|X. Deff. II. 3]]

But *EF* is rational; and, if an area be contained by a rational straight line and the third binomial, the side of the area is a second bimedial; [[book-10/proposition-56|X. 56]] therefore the side of *EI*, that is, of *AD*, is a second bimedial.

Next, let the square on *EH* be greater than the square on *HK* by the square on a straight line incommensurable in length with *EH*.

Now each of the straight lines *EH*, *HK* is incommensurable in length with *EF*; therefore *EK* is a sixth binomial. [[book-10/definitions#Definition 6 (part 2)|X. Deff. II. 6]]

But, if an area be contained by a rational straight line and the sixth binomial, the side of the area is the side of the sum of two medial areas; [[book-10/proposition-59|X. 59]] so that the side of the area *AD* is also the side of the sum of two medial areas.

Therefore etc. Q. E. D.