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And, since the squares on AC, CB are greater than twice the rectangle AC, CB, [Lemma] therefore DL is also greater than MF, so that DM is also greater than MG. [VI. 1]

And the rectangle DK, KM is equal to the square on MN, that is, to the fourth part of the square on MG, and DK is commensurable with KM.

But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into commensurable parts, the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater; [X. 17] therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM.

And DM, MG are rational, and DM, which is the greater term, is commensurable in length with the rational straight line DE set out.

Therefore DG is a first binomial straight line. [X. Deff. II. 1] Q. E. D.


PROPOSITION 61.

The square on the first bimedial straight line applied to a rational straight line produces as breadth the second binomial.

Let AB be a first bimedial straight line divided into its medials at C, of which medials AC is the greater; let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth; I say that DG is a second binominal straight line.

For let the same construction as before be made.

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