---
book: 4
number: 14
id: "IV.14"
kind: "construction"
uses: ["[[book-1/proposition-6]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:4.prop.14"
license: "CC-BY-SA-4.0"
---

# IV.14

*About a given pentagon*, *which is equilateral and equiangular*, *to circumscribe a circle*.

## Proof

Let *ABCDE* be the given pentagon, which is equilateral and equiangular; thus it is required to circumscribe a circle about the pentagon *ABCDE*.

Let the angles *BCD*, *CDE* be bisected by the straight lines *CF*, *DF* respectively, and from the point *F*, at which the straight lines meet, let the straight lines *FB*, *FA*, *FE* be joined to the points *B*, *A*, *E*.

Then in manner similar to the preceding it can be proved that the angles *CBA*, *BAE*, *AED* have also been bisected by the straight lines *FB*, *FA*, *FE* respectively.

Now, since the angle *BCD* is equal to the angle *CDE*, and the angle *FCD* is half of the angle *BCD*, and the angle *CDF* half of the angle *CDE*, therefore the angle *FCD* is also equal to the angle *CDF*, so that the side *FC* is also equal to the side *FD*. [[book-1/proposition-6|I. 6]]

Similarly it can be proved that each of the straight lines *FB*, *FA*, *FE* is also equal to each of the straight lines *FC*, *FD*; therefore the five straight lines *FA*, *FB*, *FC*, *FD*, *FE* are equal to one another.

Therefore the circle described with centre *F* and distance one of the straight lines *FA*, *FB*, *FC*, *FD*, *FE* will pass also through the remaining points, and will have been circumscribed.

Let it be circumscribed, and let it be *ABCDE*.

Therefore about the given pentagon, which is equilateral and equiangular, a circle has been circumscribed. Q. E. F.