For let the straight line AB be cut at random at the point C; I say that the rectangle contained by AB, BC together with the rectangle contained by BA, AC is equal to the square on AB.

For let the square ADEB be described on AB [I. 46], and let CF be drawn through C parallel to either AD or BE. [I. 31]

Then AE is equal to AF, CE.

Now AE is the square on AB;

AF is the rectangle contained by BA, AC, for it is contained by DA, AC, and AD is equal to AB; and CE is the rectangle AB, BC, for BE is equal to AB.

Therefore the rectangle BA, AC together with the rectangle AB, BC is equal to the square on AB.

Therefore etc. Q. E. D.