---
book: 10
number: 41
id: "X.41"
kind: "theorem"
uses: ["[[book-10/proposition-35]]", "[[book-2/proposition-4]]", "[[book-10/proposition-22]]", "[[book-6/proposition-1]]", "[[book-10/proposition-11]]", "[[book-10/proposition-36]]", "[[book-2/proposition-5]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.41"
license: "CC-BY-SA-4.0"
---

# X.41

*If two straight lines incommensurable in square which make the sum of the squares on them medial*, *and the rectangle contained by them medial and also incommensurable with the sum of the squares on them*, *be added together*, *the whole straight line is irrational*; *and let it be called the* *side of the sum of two medial areas*.

## Proof

For let two straight lines *AB*, *BC* incommensurable in square and satisfying the given conditions [[book-10/proposition-35|X. 35]] be added together; I say that *AC* is irrational.

Let a rational straight line *DE* be set out, and let there be applied to *DE* the rectangle *DF* equal to the squares on *AB*, *BC*, and the rectangle *GH* equal to twice the rectangle *AB*, *BC*; therefore the whole *DH* is equal to the square on *AC*. [[book-2/proposition-4|II. 4]]

Now, since the sum of the squares on *AB*, *BC* is medial, and is equal to *DF*, therefore *DF* is also medial.

And it is applied to the rational straight line *DE*; therefore *DG* is rational and incommensurable in length with *DE*. [[book-10/proposition-22|X. 22]]

For the same reason *GK* is also rational and incommensurable in length with *GF*, that is, *DE*.

And, since the squares on *AB*, *BC* are incommensurable with twice the rectangle *AB*, *BC*, *DF* is incommensurable with *GH*; so that *DG* is also incommensurable with *GK*. [[book-6/proposition-1|VI. 1]] , [[book-10/proposition-11|X. 11]]

And they are rational; therefore *DG*, *GK* are rational straight lines commensurable in square only; therefore *DK* is irrational and what is called binomial. [[book-10/proposition-36|X. 36]]

But *DE* is rational; therefore *DH* is irrational, and the side of the square which is equal to it is irrational. [[book-10/definitions#Definition 4|X. Def. 4]]

But *AC* is the side of the square equal to *HD*; therefore *AC* is irrational.

And let it be called the *side of the sum of two medial areas*. Q. E. D.

Lemma. And that the aforesaid irrational straight lines are divided only in one way into the straight lines of which they are the sum and which produce the types in question, we will now prove after premising the following lemma.

Let the straight line *AB* be set out, let the whole be cut into unequal parts at each of the points *C*, *D*, and let *AC* be supposed greater than *DB*; I say that the squares on *AC*, *CB* are greater than the squares on *AD*, *DB*.

For let *AB* be bisected at *E*.

Then, since *AC* is greater than *DB*, let *DC* be subtracted from each; therefore the remainder *AD* is greater than the remainder *CB*.

But *AE* is equal to *EB*; therefore *DE* is less than *EC*; therefore the points *C*, *D* are not equidistant from the point of bisection.

And, since the rectangle *AC*, *CB* together with the square on *EC* is equal to the square on *EB*, [[book-2/proposition-5|II. 5]] and, further, the rectangle *AD*, *DB* together with the square on *DE* is equal to the square on *EB*, [*id*.] therefore the rectangle *AC*, *CB* together with the square on *EC* is equal to the rectangle *AD*, *DB* together with the square on *DE*.

And of these the square on *DE* is less than the square on *EC*; therefore the remainder, the rectangle *AC*, *CB*, is also less than the rectangle *AD*, *DB*, so that twice the rectangle *AC*, *CB* is also less than twice the rectangle *AD*, *DB*.

Therefore also the remainder, the sum of the squares on *AC*, *CB*, is greater than the sum of the squares on *AD*, *DB*. Q. E. D.