---
book: 4
number: 11
id: "IV.11"
kind: "construction"
uses: ["[[book-4/proposition-10]]", "[[book-4/proposition-2]]", "[[book-1/proposition-9]]", "[[book-3/proposition-26]]", "[[book-3/proposition-29]]", "[[book-3/proposition-27]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:4.prop.11"
license: "CC-BY-SA-4.0"
---

# IV.11

*In a given circle to inscribe an equilateral and equiangular pentagon*.

## Proof

Let *ABCDE* be the given circle; thus it is required to inscribe in the circle *ABCDE* an equilateral and equiangular pentagon.

Let the isosceles triangle *FGH* be set out having each of the angles at *G*, *H* double of the angle at *F*; [[book-4/proposition-10|IV. 10]] let there be inscribed in the circle *ABCDE* the triangle *ACD* equiangular with the triangle *FGH*, so that the angle *CAD* is equal to the angle at *F* and the angles at *G*, *H* respectively equal to the angles *ACD*, *CDA*; [[book-4/proposition-2|IV. 2]] therefore each of the angles *ACD*, *CDA* is also double of the angle *CAD*.

Now let the angles *ACD*, *CDA* be bisected respectively by the straight lines *CE*, *DB* [[book-1/proposition-9|I. 9]], and let *AB*, *BC*, *DE*, *EA* be joined.

Then, since each of the angles *ACD*, *CDA* is double of the angle *CAD*, and they have been bisected by the straight lines *CE*, *DB*, therefore the five angles *DAC*, *ACE*, *ECD*, *CDB*, *BDA* are equal to one another.

But equal angles stand on equal circumferences; [[book-3/proposition-26|III. 26]] therefore the five circumferences *AB*, *BC*, *CD*, *DE*, *EA* are equal to one another.

But equal circumferences are subtended by equal straight lines; [[book-3/proposition-29|III. 29]] therefore the five straight lines *AB*, *BC*, *CD*, *DE*, *EA* are equal to one another; therefore the pentagon *ABCDE* is equilateral.

I say next that it is also equiangular.

For, since the circumference *AB* is equal to the circumference *DE*, let *BCD* be added to each; therefore the whole circumference *ABCD* is equal to the whole circumference *EDCB*.

And the angle *AED* stands on the circumference *ABCD*, and the angle *BAE* on the circumference *EDCB*; therefore the angle *BAE* is also equal to the angle *AED*. [[book-3/proposition-27|III. 27]]

For the same reason each of the angles *ABC*, *BCD*, *CDE* is also equal to each of the angles *BAE*, *AED*; therefore the pentagon *ABCDE* is equiangular.

But it was also proved equilateral; therefore in the given circle an equilateral and equiangular pentagon has been inscribed. Q. E. F.