To circumscribe a circle about a given equilateral and equiangular pentagon.

Let *ABCDE* be the given pentagon, which is equilateral and equiangular.

It is required to circumscribe a circle about the pentagon *ABCDE.*

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

As in IV.13

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

Now, since the angle *BCD* equals 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* also equals the angle *CDF,* so that the side *FC* also equals the side *FD.*

Similarly it can be proved that each of the straight lines *FB, FA,* and *FE* also equals each of the straight lines *FC* and *FD.* Therefore the five straight lines *FA, FB, FC,* *FD,* and *FE* equal one another.

Therefore the circle described with center *F* and radius one of the straight lines *FA, FB, FC, FD,* or *FE* also passes through the remaining points, and is circumscribed.

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

Therefore a circle has been circumscribed about the given equilateral and equiangular pentagon.

Q.E.F.

This construction is used in propositions XIII.8 and XIII.18.