Book 10
59
[LEMMA. If a straight line be cut into unequal parts, the squares on the unequal parts are greater than twice the rectangle contained by the unequal parts.
59
Let AB be a straight line, and let it be cut into unequal parts at C, and let AC be the greater; I say that the squares on AC, CB are greater than twice the rectangle AC, CB.
59
For let AB be bisected at D.
59
Since then a straight line has been cut into equal parts at D, and into unequal parts at C, therefore the rectangle AC, CB together with the square on CD is equal to the square on AD, [II. 5] so that the rectangle AC, CB is less than double of the square on AD.
59
But the squares on AC, CB are double of the squares on AD, DC; [II. 9] therefore the squares on AC, CB are greater than twice the rectangle AC, CB. Q. E. D.]
PROPOSITION 60.
60
The square on the binomial straight line applied to a rational straight line produces as breadth the first binomial.
60
Let AB be a binomial straight line divided into its terms at C, so that AC is the greater term; let a rational straight line DE be set out, and let DEFG equal to the square on AB be applied to DE producing DG as its breadth; I say that DG is a first binomial straight line.
60
For let there be applied to DE the rectangle DH equal to the square on AC, and KL equal to the square on BC; therefore the remainder, twice the rectangle AC, CB, is equal to MF.
60
Let MG be bisected at N, and let NO be drawn parallel [to ML or GF].
60
Therefore each of the rectangles MO, NF is equal to once the rectangle AC, CB.
60
Now, since AB is a binomial divided into its terms at C, therefore AC, CB are rational straight lines commensurable in square only; [X. 36] therefore the squares on AC, CB are rational and commensurable with one another, so that the sum of the squares on AC, CB is also rational. [X. 15]
60
And it is equal to DL; therefore DL is rational.
60
And it is applied to the rational straight line DE; therefore DM is rational and commensurable in length with DE. [X. 20]
