---
book: 3
number: 28
id: "III.28"
kind: "theorem"
uses: ["[[book-1/proposition-8]]", "[[book-3/proposition-26]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.28"
license: "CC-BY-SA-4.0"
---

# III.28

*In equal circles equal straight lines cut off equal circumferences*, *the greater equal to the greater and the less to the less*.

## Proof

Let *ABC*, *DEF* be equal circles, and in the circles let *AB*, *DE* be equal straight lines cutting off *ACB*, *DFE* as greater circumferences and *AGB*, *DHE* as lesser; I say that the greater circumference *ACB* is equal to the greater circumference *DFE*, and the less circumference *AGB* to *DHE*.

For let the centres *K*, *L* of the circles be taken, and let *AK*, *KB*, *DL*, *LE* be joined.

Now, since the circles are equal, the radii are also equal; therefore the two sides *AK*, *KB* are equal to the two sides *DL*, *LE*; and the base *AB* is equal to the base *DE*; therefore the angle *AKB* is equal to the angle *DLE*. [[book-1/proposition-8|I. 8]]

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

And the whole circle *ABC* is also equal to the whole circle *DEF*; therefore the circumference *ACB* which remains is also equal to the circumference *DFE* which remains.

Therefore etc. Q. E. D.