---
book: 3
number: 9
id: "III.9"
kind: "theorem"
uses: ["[[book-1/proposition-8]]", "[[book-3/proposition-1]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.9"
license: "CC-BY-SA-4.0"
---

# III.9

*If a point be taken within a circle*, *and more than two equal straight lines fall from the point on the circle*, *the point taken is the centre of the circle*.

## Proof

Let *ABC* be a circle and *D* a point within it, and from *D* let more than two equal straight lines, namely *DA*, *DB*, *DC*, fall on the circle *ABC*; I say that the point *D* is the centre of the circle *ABC*.

For let *AB*, *BC* be joined and bisected at the points *E*, *F*, and let *ED*, *FD* be joined and drawn through to the points *G*, *K*, *H*, *L*.

Then, since *AE* is equal to *EB*, and *ED* is common, the two sides *AE*, *ED* are equal to the two sides *BE*, *ED*; and the base *DA* is equal to the base *DB*; therefore the angle *AED* is equal to the angle *BED*. [[book-1/proposition-8|I. 8]]

Therefore each of the angles *AED*, *BED* is right; [[book-1/definitions#Definition 10|I. Def. 10]] therefore *GK* cuts *AB* into two equal parts and at right angles.

And since, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line, [[book-3/proposition-1|III. 1, Por.]] the centre of the circle is on *GK*.

For the same reason the centre of the circle *ABC* is also on *HL*.

And the straight lines *GK*, *HL* have no other point common but the point *D*; therefore the point *D* is the centre of the circle *ABC*.

Therefore etc. Q. E. D.