---
book: 10
number: 34
id: "X.34"
kind: "construction"
uses: ["[[book-10/proposition-31]]", "[[book-6/proposition-28]]", "[[book-10/proposition-18]]", "[[book-10/proposition-11]]", "[[book-3/proposition-31]]", "[[book-1/proposition-47]]", "[[book-10/proposition-6]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.34"
license: "CC-BY-SA-4.0"
---

# X.34

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

## Proof

Let there be set out two medial straight lines *AB*, *BC*, commensurable in square only, such that the rectangle which they contain is rational, and the square on *AB* is greater than the square on *BC* by the square on a straight line incommensurable with *AB*; [[book-10/proposition-31|X. 31]], *ad fin*. let the semicircle *ADB* be described on *AB*, let *BC* be bisected at *E*, let there be applied to *AB* a parallelogram equal to the square on *BE* and deficient by a square figure, namely the rectangle *AF*, *FB*; [[book-6/proposition-28|VI. 28]] therefore *AF* is incommensurable in length with *FB*. [[book-10/proposition-18|X. 18]]

Let *FD* be drawn from *F* at right angles to *AB*, and let *AD*, *DB* be joined.

Since *AF* is incommensurable in length with *FB*, therefore the rectangle *BA*, *AF* is also incommensurable with the rectangle *AB*, *BF*. [[book-10/proposition-11|X. 11]]

But the rectangle *BA*, *AF* is equal to the square on *AD*, and the rectangle *AB*, *BF* to the square on *DB*; therefore the square on *AD* is also incommensurable with the square on *DB*.

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 *BC* is double of *DF*, therefore the rectangle *AB*, *BC* is also double of the rectangle *AB*, *FD*.

But the rectangle *AB*, *BC* is rational; therefore the rectangle *AB*, *FD* is also rational. [[book-10/proposition-6|X. 6]]

But the rectangle *AB*, *FD* is equal to the rectangle *AD*, *DB*; [Lemma] so that the rectangle *AD*, *DB* is also rational.

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