PROPOSITION 64.
The square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial.
Let
AB be the side of a rational plus a medial area, divided into its straight lines at
C, so that
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 fifth binomial straight line.
Let the same construction as before be made.
Since then
AB is the side of a rational plus a medial area, divided at
C, therefore
AC,
CB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [
X. 40]
Since then the sum of the squares on
AC,
CB is medial, therefore
DL is medial, so that
DM is rational and incommensurable in length with
DE. [
X. 22]
Again, since twice the rectangle
AC,
CB, that is
MF, is rational, therefore
MG is rational and commensurable with
DE. [
X. 20]
Therefore
DM is incommensurable with
MG; [
X. 13] therefore
DM,
MG are rational straight lines commensurable in square only; therefore
DG is binomial. [
X. 36]
I say next that it is also a fifth binomial straight line.
For it can be proved similarly that the rectangle
DK,
KM is equal to the square on
MN, and that
DK is incommensurable in length with
KM; therefore the square on
DM is greater than the square on
MG by the square on a straight line incommensurable with
DM. [
X. 18]