Proof: By Euclid
(related to Lemma: Lem. 13.18: Angle of the Pentagon)
![fig18ae](https://github.com/bookofproofs/bookofproofs.github.io/blob/main/_sources/_assets/images/euclid/Book13/fig18ae.png?raw=true)
- 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.
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"