Proof: By Euclid

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

∎

Thank you to the contributors under CC BY-SA 4.0!

Both, this possibility and the previous one, are precluded by [Prop. 3.23] (translator's note). ↩
It has been shown that similar segments are congruent if their bases are equal in length (editor's note). ↩