For in the circle ABCDE which has its diameter rational let the
equilateral pentagon ABCDE be inscribed; I say that the side of
the pentagon is the irrational straight line called minor.
For let the centre of the circle, the point F, be taken, let AF,
FB be joined and carried through to the points, G, H, let AC be
joined, and let FK be made a fourth part of AF.
Now AF is rational; therefore FK is also rational.
But BF is also rational; therefore the whole BK is rational.
And, since the circumference ACG is equal to the circumference ADG,
and in them ABC is equal to AED, therefore the remainder CG is equal
to the remainder GD. [p. 462]
And, if we join AD, we conclude that the angles at L are right,
and CD is double of CL.
For the same reason the angles at M are also right, and AC is double
of CM.
Since then the angle ALC is equal to the angle AMF, and the angle
LAC is common to the two triangles ACL and AMF, therefore the remaining
angle ACL is equal to the remaining angle MFA; [I. 32] therefore
the triangle ACL is equiangular with the triangle AMF; therefore,
proportionally, as LC is to CA, so is MF to FA.
And the doubles of the antecedents may be taken; therefore, as the
double of LC is to CA, so is the double of MF to FA.
But, as the double of MF is to FA, so is MF to the half of FA; therefore
also, as the double of LC is to CA, so is MF to the half of FA.
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.
[p. 463]
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.
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.
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.
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.
But the square on KF is rational, for the diameter is rational;
therefore the square on MK is also rational; therefore MK is rational.
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.
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]
And each of them is rational.
Therefore BK, KM are rational straight lines commensurable in square
only. [p. 464]
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]
I say next that MB is also a fourth apotome.
Let the square on N be equal to that by which the square on BK is
greater than the square on KM; therefore the square on BK is greater
than the square on KM by the square on N.
And, since KF is commensurable with FB, componendo also, KB is commensurable
with FB. [X. 15]
But BF is commensurable with BH; therefore BK is also commensurable
with BH. [X. 12]
And, since the square on BK is five times the square on KM, therefore
the square on BK has to the square on KM the ratio which 5 has to
1.
Therefore, convertendo, the square on BK has to the square on N
the ratio which 5 has to 4 [V. 19, Por.], and this is not the ratio
which a square number has to a square number; therefore BK is incommensurable
with N; [X. 9] therefore the square on BK is greater than the square
on KM by the square on a straight line incommensurable with BK.
Since then the square on the whole BK is greater than the square
on the annex KM by the square on a straight line incommensurable
with BK, and the whole BK is commensurable with the rational straight
line, BH, set out, therefore MB is a fourth apotome. [X. Deff. III.
4]
But the rectangle contained by a rational straight line and a fourth
apotome is irrational, and its square root is irrational, and is
called minor. [X. 94]
But the square on AB is equal to the rectangle HB, BM, because,
when AH is joined, the triangle ABH is equiangular with the triangle
ABM, and, as HB is to BA, so is AB to BM. [p. 465]
Therefore the side AB of the pentagon is the irrational straight
line called minor.
Q. E. D.
