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

# III.1

*To find the centre of a given circle*.

## Proof

Let *ABC* be the given circle; thus it is required to find the centre of the circle *ABC*.

Let a straight line *AB* be drawn through it at random, and let it be bisected at the point *D*; from *D* let *DC* be drawn at right angles to *AB* and let it be drawn through to *E*; let *CE* be bisected at *F*; I say that *F* is the centre of the circle *ABC*.

For suppose it is not, but, if possible, let *G* be the centre, and let *GA*, *GD*, *GB* be joined.

Then, since *AD* is equal to *DB*, and *DG* is common, the two sides *AD*, *DG* are equal to the two sides *BD*, *DG* respectively; and the base *GA* is equal to the base *GB*, for they are radii; therefore the angle *ADG* is equal to the angle *GDB*. [[book-1/proposition-8|I. 8]]

But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [[book-1/definitions#Definition 10|I. Def. 10]] therefore the angle *GDB* is right.

But the angle *FDB* is also right; therefore the angle *FDB* is equal to the angle *GDB*, the greater to the less: which is impossible. Therefore *G* is not the centre of the circle *ABC*.

Similarly we can prove that neither is any other point except *F*. Therefore the point *F* is the centre of the circle *ABC*.

Porism. From this it is manifest that, 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. Q. E. F.