If there are two straight lines, and one of them is cut into any number of pieces whatsoever, then the rectangle contained by the two straight lines is equal to the (sum of the) rectangles contained by the uncut (straight line), and every one of the pieces (of the cut straight line). * Let $A$ and $BC$ be the two straight lines, and let $BC$ be cut, at random, at points $D$ and $E$. * I say that the rectangle contained by $A$ and $BC$ is equal to the rectangle(s) contained by $A$ and $BD$, by $A$ and $DE$, and, finally, by $A$ and $EC$.
Algebraically and more generally, with $a=A=BF,$ $b=BD,$ $c=DE,$ and $d=EC,$ this proposition is a geometric version of the algebraic identity: \[a \,(b+c+d+\cdots) = a\,b + a\,c + a\,d + \cdots.\]
See also distributivity law for real numbers.
Proofs: 1
Sections: 1
