---
book: 13
number: 11
id: "XIII.11"
kind: "theorem"
uses: ["[[book-1/proposition-32]]", "[[book-5/proposition-18]]", "[[book-13/proposition-8]]", "[[book-13/proposition-1]]", "[[book-10/proposition-9]]", "[[book-10/proposition-73]]", "[[book-10/proposition-15]]", "[[book-10/proposition-12]]", "[[book-5/proposition-19]]", "[[book-10/proposition-94]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.11"
license: "CC-BY-SA-4.0"
---

# XIII.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.*

## Proof

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*; [[book-1/proposition-32|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*; [[book-5/proposition-18|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*, [[book-13/proposition-8|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, [[book-13/proposition-1|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*. [[book-10/proposition-9|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. [[book-10/proposition-73|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*. [[book-10/proposition-15|X. 15]]

But *BF* is commensurable with *BH*; therefore *BK* is also commensurable with *BH*. [[book-10/proposition-12|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 [[book-5/proposition-19|V. 19, Por.]], and this is not the ratio which a square number has to a square number; therefore *BK* is incommensurable with *N*; [[book-10/proposition-9|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. [[book-10/definitions#Definition 4 (part 3)|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. [[book-10/proposition-94|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.