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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 AC, 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.
Being what it was required to prove.
