Proof: By Euclid
(related to Proposition: 6.25: Construction of Figure Similar to One and Equal to Another)
 For let the parallelogram $BE$, equal to triangle $ABC$, have been applied to (the straight line) $BC$ [Prop. 1.44], and the parallelogram $CM$, equal to $D$, (have been applied) to (the straight line) $CE$, in the angle $FCE$, which is equal to $CBL$ [Prop. 1.45].
 Thus, $BC$ is straighton to $CF$, and $LE$ to $EM$ [Prop. 1.14].
 And let the mean proportion3 $GH$ have been taken of $BC$ and $CF$ [Prop. 6.13].
 And let $KGH$, similar, and similarly laid out, to $ABC$ have been described on $GH$ [Prop. 6.18].
 And since as $BC$ is to $GH$, so $GH$ (is) to $CF$, and if three straight lines are proportional then as the first is to the third, so the figure (described) on the first (is) to the similar, and similarly described, (figure) on the second [Prop. 6.19 corr.] 4, thus as $BC$ is to $CF$, so triangle $ABC$ (is) to triangle $KGH$.
 But, also, as $BC$ (is) to $CF$, so parallelogram $BE$ (is) to parallelogram $EF$ [Prop. 6.1].
 And, thus, as triangle $ABC$ (is) to triangle $KGH$, so parallelogram $BE$ (is) to parallelogram $EF$.
 Thus, alternately, as triangle $ABC$ (is) to parallelogram $BE$, so triangle $KGH$ (is) to parallelogram $EF$ [Prop. 5.16].
 And triangle $ABC$ (is) equal to parallelogram $BE$.
 Thus, triangle $KGH$ (is) also equal to parallelogram $EF$.
 But, parallelogram $EF$ is equal to $D$.
 Thus, $KGH$ is also equal to $D$.
 And $KGH$ is also similar to $ABC$.
 Thus, a single (rectilinear figure) $KGH$ has been constructed (which is) similar to the given rectilinear figure $ABC$, and equal to a different given (rectilinear figure) $D$.
 (Which is) the very thing it was required to do.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

 nonGithub:
 @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"