Proof: By Euclid
(related to Proposition: 3.24: Similar Segments on Equal Bases are Equal)
 For if the segment $AEB$ is applied to the segment $CFD$, and point $A$ is placed on (point) $C$, and the straight line $AB$ on $CD$, then point $B$ will also coincide with point $D$, on account of $AB$ being equal to $CD$.
 And if $AB$ coincides with $CD$ then the segment $AEB$ will also coincide with $CFD$.
 For if the straight line $AB$ coincides with $CD$, and the segment $AEB$ does not coincide with $CFD$, then it will surely either fall inside it, outside (it),^{1} or it will miss like $CGD$ (in the figure), and a circle (will) cut (another) circle at more than two points.
 The very thing is impossible [Prop. 3.10].
 Thus, if the straight line $AB$ is applied to $CD$, the segment $AEB$ cannot not also coincide with $CFD$.
 Thus, it will coincide, and will be equal to it [C.N. 4] .
 Thus, similar segments of circles on equal straight lines are equal to one another.
 (Which is) the very thing it was required to show.^{2}
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes