If a straight line is cut into equal and unequal (pieces) then the rectangle contained by the unequal pieces of the whole (straight line), plus the square on the (difference) between the (equal and unequal) pieces, is equal to the square on half (of the straight line). * For let any straight line $AB$ have been cut - equally at $C$, and unequally at $D$. * I say that the rectangle contained by $AD$ and $DB$, plus the square on $CD$, is equal to the square on $CB$.
With \(a:=AC=CB\) and \(b:=CD\), this propositions states that \[(a+b)(a-b)=a^2-b^2\]
which is one of the binomial formulae.