For, since the straight line CB has been cut at random at D,
the squares on CB, BD are equal to twice the rectangle contained by CB, BD and the square on DC. [II. 7]
Let the square on DA be added to each;
therefore the squares on CB, BD, DA are equal to twice the rectangle contained by CB, BD and the squares on AD, DC.
But the square on AB is equal to the squares on BD, DA,
for the angle at D is right; [I. 47]
and the square on AC is equal to the squares on AD, DC;
therefore the squares on CB, BA are equal to the square on AC and twice the rectangle CB, BD,
so that the square on AC alone is less than the squares on CB, BA by twice the rectangle contained by CB, BD.