In a circle the angle at the center is double the angle at the circumference when the angles have the same circumference as base.

Let *ABC* be a circle, let the angle *BEC* be an angle at its center, 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 the angle *BAC.*

Join *AE,* and draw it through to *F.*

Then, since *EA* equals *EB,* the angle *EAB* also equals the angle *EBA.* Therefore the sum of the angles the angles *EAB* and *EBA* is double the angle *EAB.*

But the angle *BEF* equals the sum of the angles *EAB* and *EBA,* therefore the angle *BEF,* is also double the angle *EAB.*

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

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

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

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

Therefore *in a circle the angle at the center is double the angle at the circumference when the angles have the same circumference as base.*

Q.E.D.