---
book: 3
number: 10
id: "III.10"
kind: "theorem"
uses: ["[[book-3/proposition-1]]", "[[book-3/proposition-5]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.10"
license: "CC-BY-SA-4.0"
---

# III.10

*A circle does not cut a circle at more points than two*.

## Proof

For, if possible, let the circle *ABC* cut the circle *DEF* at more points than two, namely *B*, *C*, *F*, *H*;

let *BH*, *BG* be joined and bisected at the points *K*, *L*, and from *K*, *L* let *KC*, *LM* be drawn at right angles to *BH*, *BG* and carried through to the points *A*, *E*.

Then, since in the circle *ABC* a straight line *AC* cuts a straight line *BH* into two equal parts and at right angles, the centre of the circle *ABC* is on *AC*. [[book-3/proposition-1|III. 1, Por.]]

Again, since in the same circle *ABC* a straight line *NO* cuts a straight line *BG* into two equal parts and at right angles, the centre of the circle *ABC* is on *NO*.

But it was also proved to be on *AC*, and the straight lines *AC*, *NO* meet at no point except at *P*; therefore the point *P* is the centre of the circle *ABC*.

Similarly we can prove that *P* is also the centre of the circle *DEF*; therefore the two circles *ABC*, *DEF* which cut one another have the same centre *P*: which is impossible. [[book-3/proposition-5|III. 5]]

Therefore etc. Q. E. D.