PROPOSITION 11.
If in a circle which has its diameter rational an equilateral pentagon be inscribed,
the side of the pentagon is the irrational straight line called minor.
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.
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.
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.
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.
Therefore the side
AB of the pentagon is the irrational straight line called minor. Q. E. D.