|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.||I.5|
|But the angle BEF equals the sum of the angles EAB and EBA, therefore the angle BEF, is also double the angle EAB.||I.32|
|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.|
Next proposition: III.21