#### 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.

Then, since AB is a first bimedial divided at C, therefore AC, CB are medial straight lines commensurable in square only, and containing a rational rectangle, [X. 37] so that the squares on AC, CB are also medial. [X. 21]

Therefore DL is medial. [X. 15 and 23, Por.]

And it has been applied to the rational straight line DE; therefore MD is rational and incommensurable in length with DE. [X. 22]

Again, since twice the rectangle AC, CB is rational, MF is also rational.

And it is applied to the rational straight line ML; therefore MG is also rational and commensurable in length with ML, that is, DE; [X. 20] therefore DM is incommensurable in length with MG. [X. 13]

And they are rational; therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X. 36]

It is next to be proved that it is also a second binomial straight line.

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

And, since the square on AC is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [VI. 1, X. 11]

And the rectangle DK, KM is equal to the square on MN; therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. [X. 17]

And MG is commensurable is length with DE.

Therefore DG is a second binomial straight line. [X. Deff. II. 2]