# Proof: By Euclid

• For let the equal multiples $GH$, $HK$, $LM$, and $MN$ have been taken of $AE$, $EB$, $CF$, and $FD$ (respectively), and the other random equal multiples $KO$ and $NP$ of $EB$ and $FD$ (respectively).
• And since $GH$ and $HK$ are equal multiples of $AE$ and $EB$ (respectively), $GH$ and $GK$ are thus equal multiples of $AE$ and $AB$ (respectively) [Prop. 5.1].
• But $GH$ and $LM$ are equal multiples of $AE$ and $CF$ (respectively).
• Thus, $GK$ and $LM$ are equal multiples of $AB$ and $CF$ (respectively).
• Again, since $LM$ and $MN$ are equal multiples of $CF$ and $FD$ (respectively), $LM$ and $LN$ are thus equal multiples of $CF$ and $CD$ (respectively) [Prop. 5.1].
• And $LM$ and $GK$ were equal multiples of $CF$ and $AB$ (respectively).
• Thus, $GK$ and $LN$ are equal multiples of $AB$ and $CD$ (respectively).
• Thus, $GK$, $LN$ are equal multiples of $AB$, $CD$.
• Again, since $HK$ and $MN$ are equal multiples of $EB$ and $FD$ (respectively), and $KO$ and $NP$ are also equal multiples of $EB$ and $FD$ (respectively), then, added together, $HO$ and $MP$ are also equal multiples of $EB$ and $FD$ (respectively) [Prop. 5.2].
• And since as $AB$ (is) to $BE$, so $CD$ (is) to $DF$, and the equal multiples $GK$, $LN$ have been taken of $AB$, $CD$, and the equal multiples $HO$, $MP$ of $EB$, $FD$, thus if $GK$ exceeds $HO$ then $LN$ also exceeds $MP$, and if ($GK$ is) equal (to $HO$ then $LN$ is also) equal (to $MP$), and if ($GK$ is) less (than $HO$ then $LN$ is also) less (than $MP$) [Def. 5.5] .
• So let $GK$ exceed $HO$, and thus, $HK$ being taken away from both, $GH$ exceeds $KO$.
• But (we saw that) if $GK$ was exceeding $HO$ then $LN$ was also exceeding $MP$.
• Thus, $LN$ also exceeds $MP$, and, $MN$ being taken away from both, $LM$ also exceeds $NP$.
• Hence, if $GH$ exceeds $KO$ then $LM$ also exceeds $NP$.
• So, similarly, we can show that even if $GH$ is equal to $KO$ then $LM$ will also be equal to $NP$, and even if ($GH$ is) less (than $KO$ then $LM$ will also be) less (than $NP$).
• And $GH$, $LM$ are equal multiples of $AE$, $CF$, and $KO$, $NP$ other random equal multiples of $EB$, $FD$.
• Thus, as $AE$ is to $EB$, so $CF$ (is) to $FD$ [Def. 5.5] .
• Thus, if composed magnitudes are proportional then they will also be proportional (when) separated.
• (Which is) the very thing it was required to show.

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