For, since the straight line CD has been cut at random at the point A, the square on DC is equal to the squares on CA, AD and twice the rectangle contained by CA, AD. [II. 4]
Let the square on DB be added to each;
therefore the squares on CD, DB are equal to the squares on CA, AD, DB and twice the rectangle CA, AD.
But the square on CB is equal to the squares on CD, DB, for the angle at D is right; [I. 47]
and the square on AB is equal to the squares on AD, DB; [I. 47]
therefore the square on CB is equal to the squares on CA, AB and twice the rectangle contained by CA, AD;
so that the square on CB is greater than the squares on CA, AB by twice the rectangle contained by CA, AD.