Proof: By Euclid

For let $BF$ have been drawn from point $B$, at right angles to $BC$ [Prop. 1.11], and let $BG$ be made equal to $A$ [Prop. 1.3], and let $GH$ have been drawn through (point) $G$, parallel to $BC$ [Prop. 1.31], and let $DK$, $EL$, and $CH$ have been drawn through (points) $D$, $E$, and $C$ (respectively), parallel to $BG$ [Prop. 1.31].
So the (rectangle) $BH$ is equal to the (rectangles) $BK$, $DL$, and $EH$.
And $BH$ is the (rectangle contained) by $A$ and $BC$.
For it is contained by $GB$ and $BC$, and $BG$ (is) equal to $A$.
And $BK$ (is) the (rectangle contained) by $A$ and $BD$.
For it is contained by $GB$ and $BD$, and $BG$ (is) equal to $A$.
And $DL$ (is) the (rectangle contained) by $A$ and $DE$.
For $DK$, that is to say $BG$ [Prop. 1.34], (is) equal to $A$.
Similarly, $EH$ (is) also the (rectangle contained) by $A$ and $EC$.
Thus, the (rectangle contained) by $A$ and $BC$ is equal to the (rectangles contained) by $A$ and $BD$, by $A$ and $DE$, and, finally, by $A$ and $EC$.
Thus, if there are two straight lines, and one of them is cut into any number of pieces whatsoever, then the rectangle contained by the two straight lines is equal to the (sum of the) rectangles contained by the uncut (straight line), and every one of the pieces (of the cut straight line).
(Which is) the very thing it was required to show.
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