If a straight line is cut into equal and unequal (pieces) then the (sum of the) squares on the unequal pieces of the whole (straight line) is double the (sum of the) square on half (the straight line) and (the square) on the (difference) between the (equal and unequal) pieces. * For let any straight line $AB$ have been cut - equally at $C$, and unequally at $D$. * I say that the (sum of the) squares on $AD$ and $DB$ is double the (sum of the squares) on $AC$ and $CD$.
Algebraically, with \(a=AC\) and \(b=CD\), this proposition states that
\[(a+b)^{2}+(a-b)^{2}=2(a^2+b^2).\]
Proofs: 1
Proofs: 1
