If a straight line is cut in half, and any straight line added to it straight-on, then the rectangle contained by the whole (straight line) with the (straight line) having being added, and the (straight line) having being added, plus the square on half (of the original straight line), is equal to the square on the sum of half (of the original straight line) and the (straight line) having been added. * For let any straight line $AB$ have been cut in half at point $C$, and let any straight line $BD$ have been added to it straight-on. * I say that the rectangle contained by $AD$ and $DB$, plus the square on $CB$, is equal to the square on $CD$.
Algebraically, with \(a:=AB\) and \(b:=BD\), the proposition states that \[\left(\frac a2\right)^2+(a+b)b=\left(\frac a2+b\right)^2.\]
Proofs: 1
