Proof: By Euclid

For let the square $ADEB$ have been described on $AB$ [Prop. 1.46], and let $CF$ have been drawn through $C$, parallel to either of $AD$ or $BE$ [Prop. 1.31].
So the (square) $AE$ is equal to the (rectangles) $AF$ and $CE$.
And $AE$ is the square on $AB$.
And $AF$ (is) the rectangle contained by the (straight lines) $BA$ and $AC$.
For it is contained by $DA$ and $AC$, and $AD$ (is) equal to $AB$.
And $CE$ (is) the (rectangle contained) by $AB$ and $BC$.
For $BE$ (is) equal to $AB$.
Thus, the (rectangle contained) by $BA$ and $AC$, plus the (rectangle contained) by $AB$ and $BC$, is equal to the square on $AB$.
Thus, if a straight line is cut at random then the (sum of the) rectangle(s) contained by the whole (straight line), and each of the pieces (of the straight line), is equal to the square on the whole.
(Which is) the very thing it was required to show.
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