---
book: 3
number: 24
id: "III.24"
kind: "theorem"
uses: ["[[book-3/proposition-10]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.24"
license: "CC-BY-SA-4.0"
---

# III.24

*Similar segments of circles on equal straight lines are equal to one another*.

## Proof

For let *AEB*, *CFD* be similar segments of circles on equal straight lines *AB*, *CD*; I say that the segment *AEB* is equal to the segment *CFD*.

For, if the segment *AEB* be applied to *CFD*, and if the point *A* be placed on *C* and the straight line *AB* on *CD*, the point *B* will also coincide with the point *D*, because *AB* is equal to *CD*; and, *AB* coinciding with *CD*, the segment *AEB* will also coincide with *CFD*.

For, if the straight line *AB* coincide with *CD* but the segment *AEB* do not coincide with *CFD*, it will either fall with it, or outside it; or it will fall awry, as *CGD*, and a circle cuts a circle at more points than two : which is impossible. [[book-3/proposition-10|III. 10]]

Therefore, if the straight line *AB* be applied to *CD*, the segment *AEB* will not fail to coincide with *CFD* also; therefore it will coincide with it and will be equal to it.

Therefore etc. Q. E. D.