---
book: 1
number: 46
id: "I.46"
kind: "construction"
uses: ["[[book-1/proposition-11]]", "[[book-1/proposition-31]]", "[[book-1/proposition-34]]", "[[book-1/proposition-29]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.46"
license: "CC-BY-SA-4.0"
---

# I.46

On a given straight line to describe a square.

![I.46](figures/I-46.svg)

## Proof

Let *AB* be the given straight line; thus it is required to describe a square on the straight line *AB*.

Let *AC* be drawn at right angles to the straight line *AB* from the point *A* on it [[book-1/proposition-11|I. 11]], and let *AD* be made equal to *AB*; through the point *D* let *DE* be drawn parallel to *AB*, and through the point *B* let *BE* be drawn parallel to *AD*. [[book-1/proposition-31|I. 31]]

Therefore *ADEB* is a parallelogram; therefore *AB* is equal to *DE*, and *AD* to *BE*. [[book-1/proposition-34|I. 34]]

But *AB* is equal to *AD*; therefore the four straight lines *BA*, *AD*, *DE*, *EB* are equal to one another; therefore the parallelogram *ADEB* is equilateral.

I say next that it is also right-angled.

For, since the straight line *AD* falls upon the parallels *AB*, *DE*, the angles *BAD*, *ADE* are equal to two right angles. [[book-1/proposition-29|I. 29]]

But the angle *BAD* is right; therefore the angle *ADE* is also right.

And in parallelogrammic areas the opposite sides and angles are equal to one another; [[book-1/proposition-34|I. 34]] therefore each of the opposite angles *ABE*, *BED* is also right. Therefore *ADEB* is right-angled.

And it was also proved equilateral.

Therefore it is a square; and it is described on the straight line *AB*.

Q. E. F.