Proof: By Euclid

For let the square $CDEB$ have been described on $CB$ [Prop. 1.46], and let $ED$ have been drawn through to $F$, and let $AF$ have been drawn through $A$, parallel to either of $CD$ or $BE$ [Prop. 1.31].
So the (rectangle) $AE$ is equal to the (rectangle) $AD$ and the (square) $CE$.
And $AE$ is the rectangle contained by $AB$ and $BC$.
For it is contained by $AB$ and $BE$, and $BE$ (is) equal to $BC$.
And $AD$ (is) the (rectangle contained) by $AC$ and $CB$.
For $DC$ (is) equal to $CB$.
And $DB$ (is) the square on $CB$.
Thus, the rectangle contained by $AB$ and $BC$ is equal to the rectangle contained by $AC$ and $CB$, plus the square on $BC$.
Thus, if a straight line is cut at random then the rectangle contained by the whole (straight line), and one of the pieces (of the straight line), is equal to the rectangle contained by (both of) the pieces, and the square on the aforementioned piece.
(Which is) the very thing it was required to show.
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