---
book: 13
number: 10
id: "XIII.10"
kind: "theorem"
uses: ["[[book-1/proposition-5]]", "[[book-1/proposition-26]]", "[[book-3/proposition-26]]", "[[book-6/proposition-33]]", "[[book-1/proposition-32]]", "[[book-6/proposition-4]]", "[[book-6/proposition-17]]", "[[book-1/proposition-4]]", "[[book-2/proposition-2]]", "[[book-4/proposition-15]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.10"
license: "CC-BY-SA-4.0"
---

# XIII.10

*If an equilateral pentagon be inscribed in a circle*, *the square on the side of the pentagon is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the same circle.*

## Proof

Let *ABCDE* be a circle, and let the equilateral pentagon *ABCDE* be inscribed in the circle *ABCDE*.

I say that the square on the side of the pentagon *ABCDE* is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the circle *ABCDE*.

For let the centre of the circle, the point *F*, be taken, let *AF* be joined and carried through to the point *G*, let *FB* be joined, let *FH* be drawn from *F* perpendicular to *AB* and be carried through to *K*, let *AK*, *KB* be joined, let *FL* be again drawn from *F* perpendicular to *AK*, and be carried through to *M*, and let *KN* be joined.

Since the circumference *ABCG* is equal to the circumference *AEDG*, and in them *ABC* is equal to *AED*, therefore the remainder, the circumference *CG*, is equal to the remainder *GD*.

But *CD* belongs to a pentagon; therefore *CG* belongs to a decagon.

And, since *FA* is equal to *FB*, and *FH* is perpendicular, therefore the angle *AFK* is also equal to the angle *KFB*. [[book-1/proposition-5|I. 5]], [[book-1/proposition-26|I. 26]]

Hence the circumference *AK* is also equal to *KB*; [[book-3/proposition-26|III. 26]] therefore the circumference *AB* is double of the circumference *BK*; therefore the straight line *AK* is a side of a decagon.

For the same reason *AK* is also double of *KM*.

Now, since the circumference *AB* is double of the circumference *BK*, while the circumference *CD* is equal to the circumference *AB*, therefore the circumference *CD* is also double of the circumference *BK*.

But the circumference *CD* is also double of *CG*; therefore the circumference *CG* is equal to the circumference *BK*.

But *BK* is double of *KM*, since *KA* is so also; therefore *CG* is also double of *KM*.

But, further, the circumference *CB* is also double of the circumference *BK*, for the circumference *CB* is equal to *BA*.

Therefore the whole circumference *GB* is also double of *BM*; hence the angle *GFB* is also double of the angle *BFM*. [[book-6/proposition-33|VI. 33]]

But the angle *GFB* is also double of the angle *FAB*, for the angle *FAB* is equal to the angle *ABF*.

Therefore the angle *BFN* is also equal to the angle *FAB*.

But the angle *ABF* is common to the two triangles *ABF* and *BFN*; therefore the remaining angle *AFB* is equal to the remaining angle *BNF*; [[book-1/proposition-32|I. 32]] therefore the triangle *ABF* is equiangular with the triangle *BFN*.

Therefore, proportionally, as the straight line *AB* is to *BF*, so is *FB* to *BN*; [[book-6/proposition-4|VI. 4]] therefore the rectangle *AB*, *BN* is equal to the square on *BF*. [[book-6/proposition-17|VI. 17]]

Again, since *AL* is equal to *LK*, while *LN* is common and at right angles, therefore the base *KN* is equal to the base *AN*; [[book-1/proposition-4|I. 4]] therefore the angle *LKN* is also equal to the angle *LAN*.

But the angle *LAN* is equal to the angle *KBN*; therefore the angle *LKN* is also equal to the angle *KBN*.

And the angle at *A* is common to the two triangles *AKB* and *AKN*.

Therefore the remaining angle *AKB* is equal to the remaining angle *KNA*; [[book-1/proposition-32|I. 32]] therefore the triangle *KBA* is equiangular with the triangle *KNA*.

Therefore, proportionally, as the straight line *BA* is to *AK*, so is *KA* to *AN*; [[book-6/proposition-4|VI. 4]] therefore the rectangle *BA*, *AN* is equal to the square on *AK*. [[book-6/proposition-17|VI. 17]]

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* [[book-2/proposition-2|II. 2]], is equal to the square on *BF* together with the square on *AK*.

And *BA* is a side of the pentagon, *BF* of the hexagon [[book-4/proposition-15|IV. 15, Por.]], and *AK* of the decagon.

Therefore etc. Q. E. D.