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. * For let the straight line $AB$ have been cut, at random, at (point) $C$. * I say that the rectangle contained by $AB$ and $BC$ is equal to the rectangle contained by $AC$ and $CB$, plus the square on $BC$.
With \(b=AC\) and \(a=CB\), we have \(b+a=AB\), and this proposition is a geometric version of the algebraic identity: \[(b+a)\,a = b\,a+a^2.\]
Proofs: 1
Proofs: 1