# 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$. • For since as$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.

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