Proof: By Euclid

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 right-angled.
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 $CB$.

fig053ae

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 BY-SA 4.0!