(related to Lemma: Lem. 13.18: Angle of the Pentagon)

- For let $ABCDE$ be an equilateral and equiangular pentagon, and let the circle $ABCDE$ have been circumscribed about it [Prop. 4.14].
- And let its center, $F$, have been found [Prop. 3.1].
- And let $FA$, $FB$, $FC$, $FD$, and $FE$ have been joined.

- Thus, they cut the angles of the pentagon in half at (points) $A$, $B$, $C$, $D$, and $E$ [Prop. 1.4].
- And since the five angles at $F$ are equal (in sum) to four right angles, and are also equal (to one another), (any) one of them, like $AFB$, is thus one less a fifth of a right angle.
- Thus, the (sum of the) remaining (angles in triangle $ABF$), $FAB$ and $ABF$, is one plus a fifth of a right angle [Prop. 1.32].
- And $FAB$ (is) equal to $FBC$.
- Thus, the whole angle, $ABC$, of the pentagon is also one and one-fifth of a right angle.
- (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"