---
book: 4
number: 12
id: "IV.12"
kind: "construction"
uses: ["[[book-4/proposition-11]]", "[[book-3/proposition-16]]", "[[book-3/proposition-1]]", "[[book-3/proposition-18]]", "[[book-1/proposition-47]]", "[[book-1/proposition-8]]", "[[book-3/proposition-27]]", "[[book-1/proposition-26]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:4.prop.12"
license: "CC-BY-SA-4.0"
---

# IV.12

*About a given circle to circumscribe an equilateral and equiangular pentagon*.

## Proof

Let *ABCDE* be the given circle; thus it is required to circumscribe an equilateral and equiangular pentagon about the circle *ABCDE*.

Let *A*, *B*, *C*, *D*, *E* be conceived to be the angular points of the inscribed pentagon, so that the circumferences *AB*, *BC*, *CD*, *DE*, *EA* are equal; [[book-4/proposition-11|IV. 11]] through *A*, *B*, *C*, *D*, *E* let *GH*, *HK*, *KL*, *LM*, *MG* be drawn touching the circle; [[book-3/proposition-16|III. 16, Por.]] let the centre *F* of the circle *ABCDE* be taken [[book-3/proposition-1|III. 1]], and let *FB*, *FK*, *FC*, *FL*, *FD* be joined.

Then, since the straight line *KL* touches the circle *ABCDE* at *C*, and *FC* has been joined from the centre *F* to the point of contact at *C*, therefore *FC* is perpendicular to *KL*; [[book-3/proposition-18|III. 18]] therefore each of the angles at *C* is right.

For the same reason the angles at the points *B*, *D* are also right.

And, since the angle *FCK* is right, therefore the square on *FK* is equal to the squares on *FC*, *CK*.

For the same reason [[book-1/proposition-47|I. 47]] the square on *FK* is also equal to the squares on *FB*, *BK*; so that the squares on *FC*, *CK* are equal to the squares on *FB*, *BK*, of which the square on *FC* is equal to the square on *FB*; therefore the square on *CK* which remains is equal to the square on *BK*.

Therefore *BK* is equal to *CK*.

And, since *FB* is equal to *FC*, and *FK* common, the two sides *BF*, *FK* are equal to the two sides *CF*, *FK*; and the base *BK* equal to the base *CK*; therefore the angle *BFK* is equal to the angle *KFC*, [[book-1/proposition-8|I. 8]] and the angle *BKF* to the angle *FKC*. Therefore the angle *BFC* is double of the angle *KFC*, and the angle *BKC* of the angle *FKC*.

For the same reason the angle *CFD* is also double of the angle *CFL*, and the angle *DLC* of the angle *FLC*.

Now, since the circumference *BC* is equal to *CD*, the angle *BFC* is also equal to the angle *CFD*. [[book-3/proposition-27|III. 27]]

And the angle *BFC* is double of the angle *KFC*, and the angle *DFC* of the angle *LFC*; therefore the angle *KFC* is also equal to the angle *LFC*.

But the angle *FCK* is also equal to the angle *FCL*; therefore *FKC*, *FLC* are two triangles having two angles equal to two angles and one side equal to one side, namely *FC* which is common to them; therefore they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle; [[book-1/proposition-26|I. 26]] therefore the straight line *KC* is equal to *CL*, and the angle *FKC* to the angle *FLC*.

And, since *KC* is equal to *CL*, therefore *KL* is double of *KC*.

For the same reason it can be proved that *HK* is also double of *BK*.

And *BK* is equal to *KC*; therefore *HK* is also equal to *KL*.

Similarly each of the straight lines *HG*, *GM*, *ML* can also be proved equal to each of the straight lines *HK*, *KL*; therefore the pentagon *GHKLM* is equilateral.

I say next that it is also equiangular.

For, since the angle *FKC* is equal to the angle *FLC*, and the angle *HKL* was proved double of the angle *FKC*, and the angle *KLM* double of the angle *FLC*, therefore the angle *HKL* is also equal to the angle *KLM*.

Similarly each of the angles *KHG*, *HGM*, *GML* can also be proved equal to each of the angles *HKL*, *KLM*; therefore the five angles *GHK*, *HKL*, *KLM*, *LMG*, *MGH* are equal to one another.

Therefore the pentagon *GHKLM* is equiangular.

And it was also proved equilateral; and it has been circumscribed about the circle *ABCDE*. Q. E. F.