---
book: 4
number: 7
id: "IV.7"
kind: "construction"
uses: ["[[book-3/proposition-16]]", "[[book-3/proposition-18]]", "[[book-1/proposition-28]]", "[[book-1/proposition-30]]", "[[book-1/proposition-34]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:4.prop.7"
license: "CC-BY-SA-4.0"
---

# IV.7

*About a given circle to circumscribe a square*.

## Proof

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

Let two diameters *AC*, *BD* of the circle *ABCD* be drawn at right angles to one another, and through the points *A*, *B*, *C*, *D* let *FG*, *GH*, *HK*, *KF* be drawn touching the circle *ABCD*. [[book-3/proposition-16|III. 16, Por.]]

Then, since *FG* touches the circle *ABCD*, and *EA* has been joined from the centre *E* to the point of contact at *A*, therefore the angles at *A* are right. [[book-3/proposition-18|III. 18]]

For the same reason the angles at the points *B*, *C*, *D* are also right.

Now, since the angle *AEB* is right, and the angle *EBG* is also right, therefore *GH* is parailel to *AC*. [[book-1/proposition-28|I. 28]]

For the same reason *AC* is also parallel to *FK*, so that *GH* is also parallel to *FK*. [[book-1/proposition-30|I. 30]]

Similarly we can prove that each of the straight lines *GF*, *HK* is parallel to *BED*.

Therefore *GK*, *GC*, *AK*, *FB*, *BK* are parallelograms; therefore *GF* is equal to *HK*, and *GH* to *FK*. [[book-1/proposition-34|I. 34]]

And, since *AC* is equal to *BD*, and *AC* is also equal to each of the straight lines *GH*, *FK*, while *BD* is equal to each of the straight lines *GF*, *HK*, [[book-1/proposition-34|I. 34]] therefore the quadrilateral *FGHK* is equilateral.

I say next that it is also right-angled.

For, since *GBEA* is a parallelogram, and the angle *AEB* is right, therefore the angle *AGB* is also right. [[book-1/proposition-34|I. 34]]

Similarly we can prove that the angles at *H*, *K*, *F* are also right.

Therefore *FGHK* is right-angled.

But it was also proved equilateral; therefore it is a square; and it has been circumscribed about the circle *ABCD*.

Therefore about the given circle a square has been circumscribed. Q. E. F.