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Proposition 11

If an equilateral pentagon is inscribed in a circle which has its diameter rational, then the side of the pentagon is the irrational straight line called minor.
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.
| Take the center F of the circle, join AF and FB and carry them through to the points G and H, join AC, and make FK a fourth part of AF. | III.1 VI.9
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 equals the circumference ADG, and in them ABC equals AED, therefore the remainder CG equals the remainder GD. And, if we join AD, then we conclude that the angles at L are right, and CD is double CL. For the same reason the angles at M are also right, and AC is double CM.
Since then the angle ALC equals the angle AMF, and the angle LAC is common to the two triangles ACL and AMF, therefore the remaining angle ACL equals the remaining angle MFA. | I.32
Therefore the triangle ACL is equiangular with the triangle AMF. Therefore, proportionally LC is to CA as MF is to FA. Taking the doubles of the antecedents, therefore double LC is to CA as double MF to FA. But double MF is to FA as MF is to the half of FA, therefore also double LC is to CA as MF is to the half of FA.
Taking the halves of the consequents, therefore double LC is to the half of CA as MF to the fourth of FA. And DC is double LC, CM is half of CA, and FK is a fourth part of FA, therefore DC is to CM as MF to FK.
Taken together, the sum of DC and CM is to CM as MK to KF. Therefore the square on the sum of DC and CM is to the square on CM as the square on MK is to the square on KF. | V.18
And since, when the straight line opposite two sides of the pentagon AC is cut in extreme and mean ratio, the greater segment equals the side of the pentagon, that is, DC, 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, and CM is half of the whole AC, therefore the square on DC and CM taken as one straight line is five times the square on CM. | XIII.8 XIII.1
But it was proved that the square on DC and CM taken as one straight line is to the square on CM as 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 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 and KM are rational straight lines commensurable in square only.
But, if from a rational straight line there is subtracted a rational straight line which is commensurable with the whole in square only, then 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, taken together, KB is commensurable with FB. But BF is commensurable with BH, therefore BK is also commensurable with BH. | X.15 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, in conversion, the square on BK has to the square on N the ratio which 5 has to 4, and this is not the ratio which a square number has to a square number. Therefore BK is incommensurable with N. Therefore the square on BK is greater than the square on KM by the square on a straight line incommensurable with BK. | V.19,Cor. X.9
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.Def.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 equals the rectangle HB by BM, because, when AH is joined, the triangle ABH is equiangular with the triangle ABM, and HB is to BA as AB is to BM. Therefore the side AB of the pentagon is the irrational straight line called minor.
Therefore, if an equilateral pentagon is inscribed in a circle which has its diameter rational, then the side of the pentagon is the irrational straight line called minor.
Q.E.D.

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#### Use of this proposition

This proposition is needed in XIII.16 after the construction of a dodecahedron to show the side of a pentagonal face is the irrational straight line called minor.

Book XIII Introduction - Proposition XIII.10 - Proposition XIII.12.

© 1996
D.E.Joyce
Clark University