If a straight line be cut at random, the rectangle contained by the whole and one of the segments is equal to the rectangle contained by the segments and the square on the aforesaid segment.
Proof
Let AB be cut at C; consider the rectangle on AB, AC. By
Proposition II.1, taking AB as the uncut line and the two segments
AC, CB of AB itself as the cut line, the rectangle on AB and
AC equals the rectangle on AC and AC together with the
rectangle on AB and CB — but the rectangle on AC and AC is
the square on AC by Definition II.1. Re-expressing the rectangle
on AB and CB via II.1 again as the rectangle on AC, CB plus
the rectangle on CB, CB (which is the square on CB, again by
Definition II.1) and combining, we obtain:
the rectangle on AB and AC equals the rectangle on AC and
CB plus the square on AC. By Common Notion 2 the equality is
preserved when rearranged.