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.*

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.*

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.*

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.*

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.*

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.

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.*

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.*

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.

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.

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.