---
book: 4
number: 6
id: "IV.6"
kind: "construction"
uses: ["[[book-1/proposition-4]]", "[[book-3/proposition-31]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:4.prop.6"
license: "CC-BY-SA-4.0"
---

# IV.6

*In a given circle to inscribe a square*.

## Proof

Let *ABCD* be the given circle; thus it is required to inscribe a square in the circle *ABCD*.

Let two diameters *AC*, *BD* of the circle *ABCD* be drawn at right angles to one another, and let *AB*, *BC*, *CD*, *DA* be joined.

Then, since *BE* is equal to *ED*, for *E* is the centre, and *EA* is common and at right angles, therefore the base *AB* is equal to the base *AD*. [[book-1/proposition-4|I. 4]]

For the same reason each of the straight lines *BC*, *CD* is also equal to each of the straight lines *AB*, *AD*; therefore the quadrilateral *ABCD* is equilateral.

I say next that it is also right-angled.

For, since the straight line *BD* is a diameter of the circle *ABCD*, therefore *BAD* is a semicircle; therefore the angle *BAD* is right. [[book-3/proposition-31|III. 31]]

For the same reason each of the angles *ABC*, *BCD*, *CDA* is also right; therefore the quadrilateral *ABCD* is right-angled.

But it was also proved equilateral; therefore it is a square; [[book-1/definitions#Definition 22|I. Def. 22]] and it has been inscribed in the circle *ABCD*.

Therefore in the given circle the square *ABCD* has been inscribed. Q. E. F.