Proof: By Euclid
(related to Lemma: Lem. 10.053: Construction of Rectangle with Area in Mean Proportion to two Square Areas)
 For since $DB$ is equal to $BF$, and $BE$ to $BG$, the whole of $DE$ is thus equal to the whole of $FG$.
 But $DE$ is equal to each of $AH$ and $KC$, and $FG$ is equal to each of $AK$ and $HC$ [Prop. 1.34].
 Thus, $AH$ and $KC$ are also equal to $AK$ and $HC$, respectively.
 Thus, the parallelogram $AC$ is equilateral.
 And (it is) also rightangled.
 Thus, $AC$ is a square.
 And since as $FB$ is to $BG$, so $DB$ (is) to $BE$, but as $FB$ (is) to $BG$, so $AB$ (is) to $DG$, and as $DB$ (is) to $BE$, so $DG$ (is) to $BC$ [Prop. 6.1], thus also as $AB$ (is) to $DG$, so $DG$ (is) to $BC$ [Prop. 5.11].
 Thus, $DG$ is in mean proportion to $AB$ and $BC$.
 So I also say that $DC$ [is] [in mean proportion]bookofproofs$6453 to $AC$ and $CB$.
 For since as $AD$ is to $DK$, so $KG$ (is) to $GC$.
 For [they are] respectively equal.
 And, via composition, as $AK$ (is) to $KD$, so $KC$ (is) to $CG$ [Prop. 5.18].
 But as $AK$ (is) to $KD$, so $AC$ (is) to $CD$, and as $KC$ (is) to $CG$, so $DC$ (is) to $CB$ [Prop. 6.1].
 Thus, also, as $AC$ (is) to $DC$, so $DC$ (is) to $BC$ [Prop. 5.11].
 Thus, $DC$ is in mean proportion to $AC$ and $CB$.
 Which (is the very thing) it was prescribed 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"