Proof: By Euclid
(related to Proposition: 6.10: Construction of Similarly Cut Straight Line)
 Let $AB$ be the given uncut straight line, and $AC$ a (straight line) cut at points $D$ and $E$, and let ($AC$) be laid down so as to encompass a random angle (with $AB$).
 And let $CB$ have been joined.
 And let $DF$ and $EG$ have been drawn through (points) $D$ and $E$ (respectively), parallel to $BC$, and let $DHK$ have been drawn through (point) $D$, parallel to $AB$ [Prop. 1.31].
 Thus, $FH$ and $HB$ are each parallelograms.
 Thus, $DH$ (is) equal to $FG$, and $HK$ to $GB$ [Prop. 1.34].
 And since the straight line $HE$ has been drawn parallel to one of the sides, $KC$, of triangle $DKC$, thus, proportionally, as $CE$ is to $ED$, so $KH$ (is) to $HD$ [Prop. 6.2].
 And $KH$ (is) equal to $BG$, and $HD$ to $GF$.
 Thus, as $CE$ is to $ED$, so $BG$ (is) to $GF$.
 Again, since $FD$ has been drawn parallel to one of the sides, $GE$, of triangle $AGE$, thus, proportionally, as $ED$ is to $DA$, so $GF$ (is) to $FA$ [Prop. 6.2].
 And it was also shown that as $CE$ (is) to $ED$, so $BG$ (is) to $GF$.
 Thus, as $CE$ is to $ED$, so $BG$ (is) to $GF$, and as $ED$ (is) to $DA$, so $GF$ (is) to $FA$.
 Thus, the given uncut straight line, $AB$, has been cut similarly to the given cut straight line, $AC$.
 (Which is) the very thing it was required to do.
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"