---
book: 13
number: 9
id: "XIII.9"
kind: "theorem"
uses: ["[[book-6/proposition-33]]", "[[book-1/proposition-5]]", "[[book-1/proposition-32]]", "[[book-4/proposition-15]]", "[[book-6/proposition-4]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.9"
license: "CC-BY-SA-4.0"
---

# XIII.9

*If the side of the hexagon and that of the decagon inscribed in the same circle be added together*, *the whole straight line has been cut in extreme and mean ratio*, *and its greater segment is the side of the hexagon.*

## Proof

Let *ABC* be a circle; of the figures inscribed in the circle *ABC* let *BC* be the side of a decagon, *CD* that of a hexagon, and let them be in a straight line; I say that the whole straight line *BD* has been cut in extreme and mean ratio, and *CD* is its greater segment.

For let the centre of the circle, the point *E*, be taken, let *EB*, *EC*, *ED* be joined, and let *BE* be carried through to *A*.

Since *BC* is the side of an equilateral decagon, therefore the circumference *ACB* is five times the circumference *BC*; therefore the circumference *AC* is quadruple of *CB*.

But, as the circumference *AC* is to *CB*, so is the angle *AEC* to the angle *CEB*; [[book-6/proposition-33|VI. 33]] therefore the angle *AEC* is quadruple of the angle *CEB*.

And, since the angle *EBC* is equal to the angle *ECB*, [[book-1/proposition-5|I. 5]] therefore the angle *AEC* is double of the angle *ECB*. [[book-1/proposition-32|I. 32]]

And, since the straight line *EC* is equal to *CD*, for each of them is equal to the side of the hexagon inscribed in the circle *ABC*, [[book-4/proposition-15|IV. 15, Por.]] the angle *CED* is also equal to the angle *CDE*; [[book-1/proposition-5|I. 5]] therefore the angle *ECB* is double of the angle *EDC*. [[book-1/proposition-32|I. 32]]

But the angle *AEC* was proved double of the angle *ECB*; therefore the angle *AEC* is quadruple of the angle *EDC*.

But the angle *AEC* was also proved quadruple of the angle *BEC*; therefore the angle *EDC* is equal to the angle *BEC*.

But the angle *EBD* is common to the two triangles *BEC* and *BED*; therefore the remaining angle *BED* is also equal to the remaining angle *ECB*; [[book-1/proposition-32|I. 32]] therefore the triangle *EBD* is equiangular with the triangle *EBC*.

Therefore, proportionally, as *DB* is to *BE*, so is *EB* to *BC*. [[book-6/proposition-4|VI. 4]]

But *EB* is equal to *CD*.

Therefore, as *BD* is to *DC*, so is *DC* to *CB*.

And *BD* is greater than *DC*; therefore *DC* is also greater than *CB*.

Therefore the straight line *BD* has been cut in extreme and mean ratio, and *DC* is its greater segment. Q. E. D.