Proof: By Euclid

For let $FK$ be made equal to $BC$, and $FL$ to $D$.
And since $FK$ is equal to $BC$, of which $FH$ is equal to $BG$, the remainder $HK$ is thus equal to the remainder $GC$.
And since as $EF$ is to $D$, so $D$ (is) to $BC$, and $BC$ to $A$ [Prop. 7.13], and $D$ (is) equal to $FL$, and $BC$ to $FK$, and $A$ to $FH$, thus as $EF$ is to $FL$, so $LF$ (is) to $FK$, and $FK$ to $FH$.
By separation, as $EL$ (is) to $LF$, so $LK$ (is) to $FK$, and $KH$ to $FH$ [Prop. 7.11], [Prop. 7.13].
And thus as one of the leading (numbers) is to one of the following, so (the sum of) all of the leading (numbers is) to (the sum of) all of the following [Prop. 7.12].
Thus, as $KH$ is to $FH$, so $EL$, $LK$, $KH$ (are) to $LF$, $FK$, $HF$.
And $KH$ (is) equal to $CG$, and $FH$ to $A$, and $LF$, $FK$, $HF$ to $D$, $BC$, $A$.
Thus, as $CG$ is to $A$, so $EH$ (is) to $D$, $BC$, $A$.
Thus, as the excess of the second (number) is to the first, so the excess of the last (is) to (the sum of) all those (numbers) before it.
(Which is) the very thing it was required to show.
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