---
book: 3
number: 29
id: "III.29"
kind: "theorem"
uses: ["[[book-3/proposition-27]]", "[[book-1/proposition-4]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.29"
license: "CC-BY-SA-4.0"
---

# III.29

*In equal circles equal circumferences are subtended by equal straight lines*.

## Proof

Let *ABC*, *DEF* be equal circles, and in them let equal circumferences *BGC*, *EHF* be cut off; and let the straight lines *BC*, *EF* be joined; I say that *BC* is equal to *EF*.

For let the centres of the circles be taken, and let them be *K*, *L*; let *BK*, *KC*, *EL*, *LF* be joined.

Now, since the circumference *BGC* is equal to the circumference *EHF*, the angle *BKC* is also equal to the angle *ELF*. [[book-3/proposition-27|III. 27]]

And, since the circles *ABC*, *DEF* are equal, the radii are also equal; therefore the two sides *BK*, *KC* are equal to the two sides *EL*, *LF*; and they contain equal angles; therefore the base *BC* is equal to the base *EF*. [[book-1/proposition-4|I. 4]]

Therefore etc.