---
book: 10
number: 35
id: "X.35"
kind: "construction"
uses: ["[[book-10/proposition-32]]", "[[book-10/proposition-18]]", "[[book-10/proposition-11]]", "[[book-3/proposition-31]]", "[[book-1/proposition-47]]", "[[book-10/proposition-13]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.35"
license: "CC-BY-SA-4.0"
---

# X.35

*To find two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them*.

## Proof

Let there be set out two medial straight lines *AB*, *BC* commensurable in square only, containing a medial rectangle, and such that the square on *AB* is greater than the square on *BC* by the square on a straight line incommensurable with *AB*; [[book-10/proposition-32|X. 32]] , *ad fin*. let the semicircle *ADB* be described on *AB*, and let the rest of the construction be as above.

Then, since *AF* is incommensurable in length with *FB*, [[book-10/proposition-18|X. 18]] *AD* is also incommensurable in square with *DB*. [[book-10/proposition-11|X. 11]]

And, since the square on *AB* is medial, therefore the sum of the squares on *AD*, *DB* is also medial. [[book-3/proposition-31|III. 31]] , [[book-1/proposition-47|I. 47]]

And, since the rectangle *AF*, *FB* is equal to the square on each of the straight lines *BE*, *DF*, therefore *BE* is equal to *DF*; therefore *BC* is double of *FD*, so that the rectangle *AB*, *BC* is also double of the rectangle *AB*, *FD*.

But the rectangle *AB*, *BC* is medial; therefore the rectangle *AB*, *FD* is also medial. [[book-10/proposition-32|X. 32, Por.]]

And it is equal to the rectangle *AD*, *DB*; [[book-10/proposition-32|Lemma after X. 32]] therefore the rectangle *AD*, *DB* is also medial.

And, since *AB* is incommensurable in length with *BC*, while *CB* is commensurable with *BE*, therefore *AB* is also incommensurable in length with *BE*, [[book-10/proposition-13|X. 13]] so that the square on *AB* is also incommensurable with the rectangle *AB*, *BE*. [[book-10/proposition-11|X. 11]]

But the squares on *AD*, *DB* are equal to the square on *AB*, [[book-1/proposition-47|I. 47]] and the rectangle *AB*, *FD*, that is, the rectangle *AD*, *DB*, is equal to the rectangle *AB*, *BE*; therefore the sum of the squares on *AD*, *DB* is incommensurable with the rectangle *AD*, *DB*.

Therefore two straight lines *AD*, *DB* incommensurable in square have been found which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them. Q. E. D.