# 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

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