---
book: 3
number: 14
id: "III.14"
kind: "theorem"
uses: ["[[book-3/proposition-1]]", "[[book-3/proposition-3]]", "[[book-1/proposition-47]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.14"
license: "CC-BY-SA-4.0"
---

# III.14

*In a circle equal straight lines are equally distant from the centre*, *and those which are equally distant from the centre are equal to one another*.

## Proof

Let *ABDC* be a circle, and let *AB*, *CD* be equal straight lines in it; I say that *AB*, *CD* are equally distant from the centre.

For let the centre of the circle *ABDC* be taken [[book-3/proposition-1|III. 1]], and let it be *E*; from *E* let *EF*, *EG* be drawn perpendicular to *AB*, *CD*, and let *AE*, *EC* be joined.

Then, since a straight line *EF* through the centre cuts a straight line *AB* not through the centre at right angles, it also bisects it. [[book-3/proposition-3|III. 3]] Therefore *AF* is equal to *FB*; therefore *AB* is double of *AF*.

For the same reason *CD* is also double of *CG*; and *AB* is equal to *CD*; therefore *AF* is also equal to *CG*.

And, since *AE* is equal to *EC*, the square on *AE* is also equal to the square on *EC*. But the squares on *AF*, *EF* are equal to the square on *AE*, for the angle at *F* is right; and the squares on *EG*, *GC* are equal to the square on *EC*, for the angle at *G* is right; [[book-1/proposition-47|I. 47]] therefore the squares on *AF*, *FE* are equal to the squares on *CG*, *GE*, of which the square on *AF* is equal to the square on *CG*, for *AF* is equal to *CG*; therefore the square on *FE* which remains is equal to the square on *EG*, therefore *EF* is equal to *EG*.

But in a circle straight lines are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal; [[book-3/definitions#Definition 4|III. Def. 4]] therefore *AB*, *CD* are equally distant from the centre.

Next, let the straight lines *AB*, *CD* be equally distant from the centre; that is, let *EF* be equal to *EG*.

I say that *AB* is also equal to *CD*.

For, with the same construction, we can prove, similarly, that *AB* is double of *AF*, and *CD* of *CG*.

And, since *AE* is equal to *CE*, the square on *AE* is equal to the square on *CE*. But the squares on *EF*, *FA* are equal to the square on *AE*, and the squares on *EG*, *GC* equal to the square on *CE*. [[book-1/proposition-47|I. 47]]

Therefore the squares on *EF*, *FA* are equal to the squares on *EG*, *GC*, of which the square on *EF* is equal to the square on *EG*, for *EF* is equal to *EG*; therefore the square on *AF* which remains is equal to the square on *CG*; therefore *AF* is equal to *CG*. And *AB* is double of *AF*, and *CD* double of *CG*; therefore *AB* is equal to *CD*.

Therefore etc. Q. E. D.