---
book: 3
number: 26
id: "III.26"
kind: "theorem"
uses: ["[[book-1/proposition-4]]", "[[book-3/proposition-24]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.26"
license: "CC-BY-SA-4.0"
---

# III.26

*In equal circles equal angles stand on equal circumferences*, *whether they stand at the centres or at the circumferences*.

## Proof

Let *ABC*, *DEF* be equal circles, and in them let there be equal angles, namely at the centres the angles *BGC*, *EHF*, and at the circumferences the angles *BAC*, *EDF*; I say that the circumference *BKC* is equal to the circumference *ELF*.

For let *BC*, *EF* be joined.

Now, since the circles *ABC*, *DEF* are equal, the radii are equal.

Thus the two straight lines *BG*, *GC* are equal to the two straight lines *EH*, *HF*; and the angle at *G* is equal to the angle at *H*; therefore the base *BC* is equal to the base *EF*. [[book-1/proposition-4|I. 4]]

And, since the angle at *A* is equal to the angle at *D*, the segment *BAC* is similar to the segment *EDF*; [[book-3/definitions#Definition 11|III. Def. 11]] and they are upon equal straight lines.

But similar segments of circles on equal straight lines are equal to one another; [[book-3/proposition-24|III. 24]] therefore the segment *BAC* is equal to *EDF*. But the whole circle *ABC* is also equal to the whole circle *DEF*; therefore the circumference *BKC* which remains is equal to the circumference *ELF*.

Therefore etc. Q. E. D.