---
book: 6
number: 33
id: "VI.33"
kind: "theorem"
uses: ["[[book-3/proposition-27]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.33"
license: "CC-BY-SA-4.0"
---

# VI.33

*In equal circles angles have the same ratio as the circumferences on which they stand*, *whether they stand at the centres or at the circumferences*.

## Proof

Let *ABC*, *DEF* be equal circles, and let the angles *BGC*, *EHF* be angles at their centres *G*, *H*, and the angles *BAC*, *EDF* angles at the circumferences; I say that, as the circumference *BC* is to the circumference *EF*, so is the angle *BGC* to the angle *EHF*, and the angle *BAC* to the angle *EDF*.

For let any number of consecutive circumferences *CK*, *KL* be made equal to the circumference *BC*, and any number of consecutive circumferences *FM*, *MN* equal to the circumference *EF*; and let *GK*, *GL*, *HM*, *HN* be joined.

Then, since the circumferences *BC*, *CK*, *KL* are equal to one another, the angles *BGC*, *CGK*, *KGL* are also equal to one another; [[book-3/proposition-27|III. 27]] therefore, whatever multiple the circumference *BL* is of *BC*, that multiple also is the angle *BGL* of the angle *BGC*.

For the same reason also, whatever multiple the circumference *NE* is of *EF*, that multiple also is the angle *NHE* of the angle *EHF*.

If then the circumference *BL* is equal to the circumference *EN*, the angle *BGL* is also equal to the angle *EHN*; [[book-3/proposition-27|III. 27]] if the circumference *BL* is greater than the circumference *EN*, the angle *BGL* is also greater than the angle *EHN*; and, if less, less.

There being then four magnitudes, two circumferences *BC*, *EF*, and two angles *BGC*, *EHF*, there have been taken, of the circumference *BC* and the angle *BGC* equimultiples, namely the circumference *BL* and the angle *BGL*, and of the circumference *EF* and the angle *EHF* equimultiples, namely the circumference *EN* and the angle *EHN*.

And it has been proved that, if the circumference *BL* is in excess of the circumference *EN*, the angle *BGL* is also in excess of the angle *EHN*; if equal, equal; and if less, less.

Therefore, as the circumference *BC* is to *EF*, so is the angle *BGC* to the angle *EHF*. [[book-5/definitions#Definition 5|V. Def. 5]]

But, as the angle *BGC* is to the angle *EHF*, so is the angle *BAC* to the angle *EDF*; for they are doubles respectively.

Therefore also, as the circumference *BC* is to the circumference *EF*, so is the angle *BGC* to the angle *EHF*, and the angle *BAC* to the angle *EDF*.

Therefore etc. Q. E. D.