---
book: 3
number: 3
id: "III.3"
kind: "theorem"
uses: ["[[book-1/proposition-8]]", "[[book-1/proposition-5]]", "[[book-1/proposition-26]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.3"
license: "CC-BY-SA-4.0"
---

# III.3

*If in a circle a straight line through the centre bisect a straight line not through the centre*, *it also cuts it at right angles; and if it cut it at right angles*, *it also bisects it*.

## Proof

Let *ABC* be a circle, and in it let a straight line *CD* through the centre bisect a straight line *AB* not through the centre at the point *F*; I say that it also cuts it at right angles.

For let the centre of the circle *ABC* be taken, and let it be *E*; let *EA*, *EB* be joined.

Then, since *AF* is equal to *FB*, and *FE* is common, two sides are equal to two sides; and the base *EA* is equal to the base *EB*; therefore the angle *AFE* is equal to the angle *BFE*. [[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 each of the angles *AFE*, *BFE* is right.

Therefore *CD*, which is through the centre, and bisects *AB* which is not through the centre, also cuts it at right angles.

Again, let *CD* cut *AB* at right angles; I say that it also bisects it. that is, that *AF* is equal to *FB*.

For, with the same construction, since *EA* is equal to *EB*, the angle *EAF* is also equal to the angle *EBF*. [[book-1/proposition-5|I. 5]]

But the right angle *AFE* is equal to the right angle *BFE*, therefore *EAF*, *EBF* are two triangles having two angles equal to two angles and one side equal to one side, namely *EF*, which is common to them, and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [[book-1/proposition-26|I. 26]] therefore *AF* is equal to *FB*.

Therefore etc. Q. E. D.