Proof: By Euclid
(related to Proposition: 6.18: Construction of Similar Polygon)
 Let $DF$ have been joined, and let $GAB$, equal to the angle at $C$, and $ABG$, equal to (angle) $CDF$, have been constructed on the straight line $AB$ at the points $A$ and $B$ on it (respectively) [Prop. 1.23].
 Thus, the remaining (angle) $CFD$ is equal to $AGB$ [Prop. 1.32].
 Thus, triangle $FCD$ is equiangular to triangle $GAB$.
 Thus, proportionally, as $FD$ is to $GB$, so $FC$ (is) to $GA$, and $CD$ to $AB$ [Prop. 6.4].
 Again, let $BGH$, equal to angle $DFE$, and $GBH$ equal to (angle) $FDE$, have been constructed on the straight line $BG$ at the points $G$ and $B$ on it (respectively) [Prop. 1.23].
 Thus, the remaining (angle) at $E$ is equal to the remaining (angle) at $H$ [Prop. 1.32].
 Thus, triangle $FDE$ is equiangular to triangle $GHB$.
 Thus, proportionally, as $FD$ is to $GB$, so $FE$ (is) to $GH$, and $ED$ to $HB$ [Prop. 6.4].
 And it was also shown (that) as $FD$ (is) to $GB$, so $FC$ (is) to $GA$, and $CD$ to $AB$.
 Thus, also, as $FC$ (is) to $AG$, so $CD$ (is) to $AB$, and $FE$ to $GH$, and, further, $ED$ to $HB$.
 And since angle $CFD$ is equal to $AGB$, and $DFE$ to $BGH$, thus the whole (angle) $CFE$ is equal to the whole (angle) $AGH$.
 So, for the same (reasons), (angle) $CDE$ is also equal to $ABH$.
 And the (angle) at $C$ is also equal to the (angle) at $A$, and the (angle) at $E$ to the (angle) at $H$.
 Thus, (figure) $AH$ is equiangular to $CE$.
 And (the two figures) have the sides about their equal angles proportional.
 Thus, the rectilinear figure $AH$ is similar to the rectilinear figure $CE$ [Def. 6.1] .
 Thus, the rectilinear figure $AH$, similar, and similarly laid down, to the given rectilinear figure $CE$ has been constructed on the given straight line $AB$.
 (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"