Kitap 13
11
And the halves of the consequents may be taken; therefore, as the double of LC is to the half of CA, so is MF to the fourth of FA.
11
And DC is double of LC, CM is half of CA, and FK a fourth part of FA; therefore, as DC is to CM, so is MF to FK.
11
Componendo also, as the sum of DC, CM is to CM, so is MK to KF; [V. 18] therefore also, as the square on the sum of DC, CM is to the square on CM, so is the square on MK to the square on KF.
11
And since, when the straight line subtending two sides of the pentagon, as AC, is cut in extreme and mean ratio, the greater segment is equal to the side of the pentagon, that is, to DC, [XIII. 8] while the square on the greater segment added to the half of the whole is five times the square on the half of the whole, [XIII. 1] and CM is half of the whole AC, therefore the square on DC, CM taken as one straight line is five times the square on CM.
11
But it was proved that, as the square on DC, CM taken as one straight line is to the square on CM, so is the square on MK to the square on KF; therefore the square on MK is five times the square on KF.
11
But the square on KF is rational, for the diameter is rational; therefore the square on MK is also rational; therefore MK is rational
11
And, since BF is quadruple of FK, therefore BK is five times KF; therefore the square on BK is twenty-five times the square on KF.
11
But the square on MK is five times the square on KF; therefore the square on BK is five times the square on KM; therefore the square on BK has not to the square on KM the ratio which a square number has to a square number; therefore BK is incommensurable in length with KM. [X. 9]
11
And each of them is rational.
11
Therefore BK, KM are rational straight lines commensurable in square only.
11
But, if from a rational straight line there be subtracted a rational straight line which is commensurable with the whole in square only, the remainder is irrational, namely an apotome; therefore MB is an apotome and MK the annex to it. [X. 73]
