For let AB be bisected at E.
Then, since AC is greater than DB, let DC be subtracted from each;
therefore the remainder AD is greater than the remainder CB.
But AE is equal to EB;
therefore DE is less than EC;
therefore the points C, D are not equidistant from the point of bisection.
And, since the rectangle AC, CB together with the square on EC is equal to the square on EB, [II. 5 ]
and, further, the rectangle AD, DB together with the square on DE is equal to the square on EB, [id.]
therefore the rectangle AC, CB together with the square on EC is equal to the rectangle AD, DB together with the square on DE.
And of these the square on DE is less than the square on EC;
therefore the remainder, the rectangle AC, CB, is also less than the rectangle AD, DB,
so that twice the rectangle AC, CB is also less than twice the rectangle AD, DB.
Therefore also the remainder, the sum of the squares on AC, CB, is greater than the sum of the squares on AD, DB.
Q. E. D.