---
book: 3
number: 27
id: "III.27"
kind: "theorem"
uses: ["[[book-1/proposition-23]]", "[[book-3/proposition-26]]", "[[book-3/proposition-20]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.27"
license: "CC-BY-SA-4.0"
---

# III.27

*In equal circles angles standing on equal circumferences are equal to one another*, *whether they stand at the centres or at the circumferences*.

## Proof

For in equal circles *ABC*, *DEF*, on equal circumferences *BC*, *EF*, let the angles *BGC*, *EHF* stand at the centres *G*, *H*, and the angles *BAC*, *EDF* at the circumferences; I say that the angle *BGC* is equal to the angle *EHF*, and the angle *BAC* is equal to the angle *EDF*.

For, if the angle *BGC* is unequal to the angle *EHF*, one of them is greater. Let the angle *BGC* be greater : and on the straight line *BG*, and at the point *G* on it, let the angle *BGK* be constructed equal to the angle *EHF*. [[book-1/proposition-23|I. 23]]

Now equal angles stand on equal circumferences, when they are at the centres; [[book-3/proposition-26|III. 26]] therefore the circumference *BK* is equal to the circumference *EF*.

But *EF* is equal to *BC*; therefore *BK* is also equal to *BC*, the less to the greater : which is impossible.

Therefore the angle *BGC* is not unequal to the angle *EHF*; therefore it is equal to it.

And the angle at *A* is half of the angle *BGC*, and the angle at *D* half of the angle *EHF*; [[book-3/proposition-20|III. 20]] therefore the angle at *A* is also equal to the angle at *D*.

Therefore etc. Q. E. D.