If a straight line is cut at random then the sum of the squares on the whole (straight line), and one of the pieces (of the straight line), is equal to twice the rectangle contained by the whole, and the said piece, and the square on the remaining piece. * For let any straight line $AB$ have been cut, at random, at point $C$. * I say that the (sum of the) squares on $AB$ and $BC$ is equal to twice the rectangle contained by $AB$ and $BC$, and the square on $CA$.
With \(a:=AB\) and \(b:=CB\), this proposition is a geometric version of the algebraic binomial formula: \[(a-b)^2=a^2-2ab+b^2.\]
Proofs: 1
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Sections: 19
