Proof: By Euclid
(related to Proposition: 6.24: Parallelograms About Diameter are Similar)
 For since $EF$ has been drawn parallel to one of the sides $BC$ of triangle $ABC$, proportionally, as $BE$ is to $EA$, so $CF$ (is) to $FA$ [Prop. 6.2].
 Again, since $FG$ has been drawn parallel to one (of the sides) $CD$ of triangle $ACD$, proportionally, as $CF$ is to $FA$, so $DG$ (is) to $GA$ [Prop. 6.2].
 But, as $CF$ (is) to $FA$, so it was also shown (is) $BE$ to $EA$.
 And thus as $BE$ (is) to $EA$, so $DG$ (is) to $GA$.
 And, thus, compounding, as $BA$ (is) to $AE$, so $DA$ (is) to $AG$ [Prop. 5.18].
 And, alternately, as $BA$ (is) to $AD$, so $EA$ (is) to $AG$ [Prop. 5.16].
 Thus, in parallelograms $ABCD$ and $EG$ the sides about the common angle $BAD$ are proportional.
 And since $GF$ is parallel to $DC$, angle $AFG$ is equal to $DCA$ [Prop. 1.29].
 And angle $DAC$ (is) common to the two triangles $ADC$ and $AGF$.
 Thus, triangle $ADC$ is equiangular to triangle $AGF$ [Prop. 1.32].
 So, for the same (reasons), triangle $ACB$ is equiangular to triangle $AFE$, and the whole parallelogram $ABCD$ is equiangular to parallelogram $EG$.
 Thus, proportionally, as $AD$ (is) to $DC$, so $AG$ (is) to $GF$, and as $DC$ (is) to $CA$, so $GF$ (is) to $FA$, and as $AC$ (is) to $CB$, so $AF$ (is) to $FE$, and, further, as $CB$ (is) to $BA$, so $FE$ (is) to $EA$ [Prop. 6.4].
 And since it was shown that as $DC$ is to $CA$, so $GF$ (is) to $FA$, and as $AC$ (is) to $CB$, so $AF$ (is) to $FE$, thus, via equality, as $DC$ is to $CB$, so $GF$ (is) to $FE$ [Prop. 5.22].
 Thus, in parallelograms $ABCD$ and $EG$ the sides about the equal angles are proportional.
 Thus, parallelogram $ABCD$ is similar to parallelogram $EG$ [Def. 6.1] .
 So, for the same (reasons), parallelogram $ABCD$ is also similar to parallelogram $KH$.
 Thus, parallelograms $EG$ and $HK$ are each similar to [parallelogram] $ABCD$.
 And (rectilinear figures) similar to the same rectilinear figure are also similar to one another [Prop. 6.21].
 Thus, parallelogram $EG$ is also similar to parallelogram $HK$.
 Thus, in any parallelogram the parallelograms about the diagonal are similar to the whole, and to one another.
 (Which is) the very thing it was required to show.
∎
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"