---
book: 3
number: 20
id: "III.20"
kind: "theorem"
uses: ["[[book-1/proposition-5]]", "[[book-1/proposition-32]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.20"
license: "CC-BY-SA-4.0"
---

# III.20

*In a circle the angle at the centre is double of the angle at the circumference*, *when the angles have the same circumference as base*.

## Proof

Let *ABC* be a circle, let the angle *BEC* be an angle at its centre, and the angle *BAC* an angle at the circumference, and let them have the same circumference *BC* as base; I say that the angle *BEC* is double of the angle *BAC*.

For let *AE* be joined and drawn through to *F*.

Then, since *EA* is equal to *EB*, the angle *EAB* is also equal to the angle *EBA*; [[book-1/proposition-5|I. 5]] therefore the angles *EAB*, *EBA* are double of the angle *EAB*.

But the angle *BEF* is equal to the angles *EAB*, *EBA*; [[book-1/proposition-32|I. 32]] therefore the angle *BEF* is also double of the angle *EAB*.

For the same reason the angle *FEC* is also double of the angle *EAC*.

Therefore the whole angle *BEC* is double of the whole angle *BAC*.

Again let another straight line be inflected, and let there be another angle *BDC*; let *DE* be joined and produced to *G*.

Similarly then we can prove that the angle *GEC* is double of the angle *EDC*, of which the angle *GEB* is double of the angle *EDB*; therefore the angle *BEC* which remains is double of the angle *BDC*.

Therefore etc. Q. E. D.