---
book: 10
number: 33
id: "X.33"
kind: "construction"
uses: ["[[book-10/proposition-30]]", "[[book-6/proposition-28]]", "[[book-10/proposition-18]]", "[[book-1/proposition-47]]", "[[book-10/proposition-21]]", "[[book-10/proposition-23]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.33"
license: "CC-BY-SA-4.0"
---

# X.33

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

## Proof

Let there be set out two rational straight lines *AB*, *BC* commensurable in square only and such that the square on the greater *AB* is greater than the square on the less *BC* by the square on a straight line incommensurable with *AB*, [[book-10/proposition-30|X. 30]] let *BC* be bisected at *D*, let there be applied to *AB* a parallelogram equal to the square on either of the straight lines *BD*, *DC* and deficient by a square figure, and let it be the rectangle *AE*, *EB*; [[book-6/proposition-28|VI. 28]] let the semicircle *AFB* be described on *AB*, let *EF* be drawn at right angles to *AB*, and let *AF*, *FB* be joined.

Then, since *AB*, *BC* are unequal straight lines, and the square on *AB* is greater than the square on *BC* by the square on a straight line incommensurable with *AB*, while there has been applied to *AB* a parallelogram equal to the fourth part of the square on *BC*, that is, to the square on half of it, and deficient by a square figure, making the rectangle *AE*, *EB*, therefore *AE* is incommensurable with *EB*. [[book-10/proposition-18|X. 18]]

And, as *AE* is to *EB*, so is the rectangle *BA*, *AE* to the rectangle *AB*, *BE*, while the rectangle *BA*, *AE* is equal to the square on *AF*, and the rectangle *AB*, *BE* to the square on *BF*; therefore the square on *AF* is incommensurable with the square on *FB*; therefore *AF*, *FB* are incommensurable in square.

And, since *AB* is rational, therefore the square on *AB* is also rational; so that the sum of the squares on *AF*, *FB* is also rational. [[book-1/proposition-47|I. 47]]

And since, again, the rectangle *AE*, *EB* is equal to the square on *EF*, and, by hypothesis, the rectangle *AE*, *EB* is also equal to the square on *BD*, therefore *FE* is equal to *BD*; therefore *BC* is double of *FE*, so that the rectangle *AB*, *BC* is also commensurable with the rectangle *AB*, *EF*.

But the rectangle *AB*, *BC* is medial; [[book-10/proposition-21|X. 21]] therefore the rectangle *AB*, *EF* is also medial. [[book-10/proposition-23|X. 23, Por.]]

But the rectangle *AB*, *EF* is equal to the rectangle *AF*, *FB*; [Lemma] therefore the rectangle *AF*, *FB* is also medial.

But it was also proved that the sum of the squares on these straight lines is rational.

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