---
book: 3
number: 2
id: "III.2"
kind: "theorem"
uses: ["[[book-3/proposition-1]]", "[[book-1/proposition-5]]", "[[book-1/proposition-16]]", "[[book-1/proposition-19]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.2"
license: "CC-BY-SA-4.0"
---

# III.2

*If on the circumference of a circle two points be taken at random*, *the straight line joining the points will fall within the circle*.

## Proof

Let *ABC* be a circle, and let two points *A*, *B* be taken at random on its circumference; I say that the straight line joined from *A* to *B* will fall within the circle.

For suppose it does not, but, if possible, let it fall outside, as *AEB*; let the centre of the circle *ABC* be taken [[book-3/proposition-1|III. 1]], and let it be *D*; let *DA*, *DB* be joined, and let *DFE* be drawn through.

Then, since *DA* is equal to *DB*, the angle *DAE* is also equal to the angle *DBE*. [[book-1/proposition-5|I. 5]] And, since one side *AEB* of the triangle *DAE* is produced, the angle *DEB* is greater than the angle *DAE*. [[book-1/proposition-16|I. 16]]

But the angle *DAE* is equal to the angle *DBE*; therefore the angle *DEB* is greater than the angle *DBE*. And the greater angle is subtended by the greater side; [[book-1/proposition-19|I. 19]] therefore *DB* is greater than *DE*. But *DB* is equal to *DF*; therefore *DF* is greater than *DE*, the less than the greater : which is impossible.

Therefore the straight line joined from *A* to *B* will not fall outside the circle.

Similarly we can prove that neither will it fall on the circumference itself; therefore it will fall within.

Therefore etc. Q. E. D.