If there be two straight lines, and one of them be cut into any number of segments whatever, the rectangle contained by the two straight lines is equal to the rectangles contained by the uncut straight line and each of the segments.
Proof
Let A and BC be the two straight lines, and let BC be cut at
random at the points D, E. Construct the rectangle contained by
A and BC as follows. From B draw BF at right angles to BC
(I.11) with BF equal to A. Through F draw FG parallel to BC
(I.31), and through C, D, E in turn draw CH, DK, EL
parallel to BF (I.31). Then the rectangle BH on the lines A,
BC is divided by the parallels DK, EL into the three rectangles
BK on A, BD; DL on A, DE; and EH on A, EC
(Definition II.1). By Common Notion 2, the whole rectangle equals
the sum of these parts: A⋅BC=A⋅BD+A⋅DE+A⋅EC. The argument generalises to any number of cuts on BC.