Proof: By Euclid
(related to Proposition: 3.28: Straight Lines Cut Off Equal Arcs in Equal Circles)
- For let the centers of the circles, $K$ and $L$, have been found [Prop. 3.1], and let $AK$, $KB$, $DL$, and $LE$ have been joined.
- And since ($ABC$ and $DEF$) are equal circles, their radii are also equal [Def. 3.1] .
- So the two (straight lines) $AK$, $KB$ are equal to the two (straight lines) $DL$, $LE$ (respectively).
- And the base $AB$ (is) equal to the base $DE$.
- Thus, angle $AKB$ is equal to angle $DLE$ [Prop. 1.8].
- And equal angles stand upon equal circumferences, when they are at the centers [Prop. 3.26].
- Thus, circumference $AGB$ (is) equal to $DHE$.
- And the whole circle $ABC$ is also equal to the whole circle $DEF$.
- Thus, the remaining circumference $ACB$ is also equal to the remaining circumference $DFE$.
- Thus, in equal circles, equal straight lines cut off equal circumferences, the greater (circumference being equal) to the greater, and the lesser to the lesser.
- (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"
Footnotes
