---
book: 10
number: 44
id: "X.44"
kind: "theorem"
uses: ["[[book-10/proposition-38]]", "[[book-2/proposition-4]]", "[[book-10/proposition-41]]", "[[book-10/proposition-22]]", "[[book-10/proposition-11]]", "[[book-10/proposition-15]]", "[[book-10/proposition-6]]", "[[book-10/proposition-13]]", "[[book-6/proposition-1]]", "[[book-10/proposition-36]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.44"
license: "CC-BY-SA-4.0"
---

# X.44

*A second bimedial straight line is divided at one point only*.

## Proof

Let *AB* be a second bimedial straight line divided at *C*, so that *AC*, *CB* are medial straight lines commensurable in square only and containing a medial rectangle; [[book-10/proposition-38|X. 38]] it is then manifest that *C* is not at the point of bisection, because the segments are not commensurable in length.

I say that *AB* is not so divided at another point.

For, if possible, let it be divided at *D* also, so that *AC* is not the same with *DB*, but *AC* is supposed greater; it is then clear that the squares on *AD*, *DB* are also, as we proved above [Lemma], less than the squares on *AC*, *CB*; and suppose that *AD*, *DB* are medial straight lines commensurable in square only and containing a medial rectangle.

Now let a rational straight line *EF* be set out, let there be applied to *EF* the rectangular parallelogram *EK* equal to the square on *AB*, and let *EG* equal to the squares on *AC*, *CB* be subtracted; therefore the remainder *HK* is equal to twice the rectangle *AC*, *CB*. [[book-2/proposition-4|II. 4]]

Again, let there be subtracted *EL*, equal to the squares on *AD*, *DB*, which were proved less than the squares on *AC*, *CB* [[book-10/proposition-41|Lemma]] ; therefore the remainder *MK* is also equal to twice the rectangle *AD*, *DB*.

Now, since the squares on *AC*, *CB* are medial, therefore *EG* is medial.

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

For the same reason *HN* is also rational and incommensurable in length with *EF*.

And, since *AC*, *CB* are medial straight lines commensurable in square only, therefore *AC* is incommensurable in length with *CB*.

But, as *AC* is to *CB*, so is the square on *AC* to the rectangle *AC*, *CB*; therefore the square on *AC* is incommensurable with the rectangle *AC*, *CB*. [[book-10/proposition-11|X. 11]]

But the squares on *AC*, *CB* are commensurable with the square on *AC*; for *AC*, *CB* are commensurable in square. [[book-10/proposition-15|x. 15]]

And twice the rectangle *AC*, *CB* is commensurable with the rectangle *AC*, *CB*. [[book-10/proposition-6|X. 6]]

Therefore the squares on *AC*, *CB* are also incommensurable with twice the rectangle *AC*, *CB*. [[book-10/proposition-13|X. 13]]

But *EG* is equal to the squares on *AC*, *CB*, and *HK* is equal to twice the rectangle *AC*, *CB*; therefore *EG* is incommensurable with *HK*, so that *EH* is also incommensurable in length with *HN*. [[book-6/proposition-1|VI. 1]] , [[book-10/proposition-11|X. 11]]

And they are rational; therefore *EH*, *HN* are rational straight lines commensurable in square only.

But, if two rational straight lines commensurable in square only be added together, the whole is the irrational which is called binomial. [[book-10/proposition-36|X. 36]]

Therefore *EN* is a binomial straight line divided at *H*.

In the same way *EM*, *MN* will also be proved to be rational straight lines commensurable in square only; and *EN* will be a binomial straight line divided at different points, *H* and *M*.

And *EH* is not the same with *MN*.

For the squares on *AC*, *CB* are greater than the squares on *AD*, *DB*.

But the squares on *AD*, *DB* are greater than twice the rectangle *AD*, *DB*; therefore also the squares on *AC*, *CB*, that is, *EG*, are much greater than twice the rectangle *AD*, *DB*, that is, *MK*, so that *EH* is also greater than *MN*.

Therefore *EH* is not the same with *MN*. Q. E. D.