Proof: By Euclid
(related to Proposition: 3.27: Angles on Equal Arcs are Equal)
- For if $BGC$ is unequal to $EHF$, one of them is greater.
- Let $BGC$ be greater, and let the (angle) $BGK$, equal to angle $EHF$, have been constructed on the straight line $BG$, at the point $G$ on it [Prop. 1.23].
- But equal angles (in equal circles) stand upon equal circumferences, when they are at the centers [Prop. 3.26].
- Thus, circumference $BK$ (is) equal to circumference $EF$.
- But, $EF$ is equal to $BC$.
- Thus, $BK$ is also equal to $BC$, the lesser to the greater.
- The very thing is impossible.
- Thus, angle $BGC$ is not unequal to $EHF$.
- Thus, (it is) equal.
- And the (angle) at $A$ is half $BGC$, and the (angle) at $D$ half $EHF$ [Prop. 3.20].
- Thus, the angle at $A$ (is) also equal to the (angle) at $D$.
- Thus, in equal circles, angles standing upon equal circumferences are equal to one another, whether they are standing at the center or at the circumference.
- (Which is) the very thing it was required to show.
Thank you to the contributors under CC BY-SA 4.0!
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"