Kitap 13
10
But the rectangle AB, BN was also proved equal to the square on BF; therefore the rectangle AB, BN together with the rectangle BA, AN, that is, the square on BA [II. 2], is equal to the square on BF together with the square on AK.
10
And BA is a side of the pentagon, BF of the hexagon [IV. 15, Por.], and AK of the decagon.
10
Therefore etc. Q. E. D.
PROPOSITION 11.
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.
11
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.
11
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.
11
Now AF is rational; therefore FK is also rational.
11
But BF is also rational; therefore the whole BK is rational.
11
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.
11
And, if we join AD, we conclude that the angles at L are right, and CD is double of CL.
11
For the same reason the angles at M are also right, and AC is double of CM.
11
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.
11
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.
11
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.
