Let the straight line AB be cut at random at C.
I say that the rectangle AB by BC equals the sum of the rectangle AC by CB and the square on BC.
Describe the square CDEB on CB. Draw ED through to F, and draw AF through A parallel to either CD or BE.
Then AE equals AD plus CE.
Now AE is the rectangle AB by BC, for it is contained by AB and BE, and BE equals BC; AD is the rectangle AC by CB, for DC equals CB; and DB is the square on CB.
Therefore the rectangle AB by BC equals the sum of the rectangle AC by CB and the square on BC.
Therefore if a straight line is cut at random, then the rectangle contained by the whole and one of the segments equals the sum of the rectangle contained by the segments and the square on the aforesaid segment.