Proof: By Euclid
(related to Proposition: 7.08: Division with Quotient and Remainder Obeys Distributivity Law (Difference))
 For let $GH$ be laid down equal to $AB$.
 Thus, which(ever) parts $GH$ is of $CD$, $AE$ is also the same parts of $CF$.
 Let $GH$ have been divided into the parts of $CD$, $GK$ and $KH$, and $AE$ into the part of $CF$, $AL$ and $LE$.
 So the multitude of (divisions) $GK$, $KH$ will be equal to the multitude of (divisions) $AL$, $LE$.
 And since which(ever) part $GK$ is of $CD$, $AL$ is also the same part of $CF$, and $CD$ (is) greater than $CF$, $GK$ (is) thus also greater than $AL$.
 Let $GM$ be made equal to $AL$.
 Thus, which(ever) part $GK$ is of $CD$, $GM$ is also the same part of $CF$.
 Thus, the remainder $MK$ is also the same part of the remainder $FD$ that the whole $GK$ (is) of the whole $CD$ [Prop. 7.5].
 Again, since which(ever) part $KH$ is of $CD$, $EL$ is also the same part of $CF$, and $CD$ (is) greater than $CF$, $HK$ (is) thus also greater than $EL$.
 Let $KN$ be made equal to $EL$.
 Thus, which(ever) part $KH$ (is) of $CD$, $KN$ is also the same part of $CF$.
 Thus, the remainder $NH$ is also the same part of the remainder $FD$ that the whole $KH$ (is) of the whole $CD$ [Prop. 7.5].
 And the remainder $MK$ was also shown to be the same part of the remainder $FD$ that the whole $GK$ (is) of the whole $CD$.
 Thus, the sum $MK$, $NH$ is the same parts of $DF$ that the whole $HG$ (is) of the whole $CD$.
 And the sum $MK$, $NH$ (is) equal to $EB$, and $HG$ to $BA$.
 Thus, the remainder $EB$ is also the same parts of the remainder $FD$ that the whole $AB$ (is) of the whole $CD$.
 (Which is) the very thing it was required to show.
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"