But the angle BEF is equal to the angles EAB, EBA; [1. 32. therefore the angle BEF is double of the angle EAB.
For the same reason the angle FEC is double of the angle BAC.
Therefore the whole angle BEC is double of the whole angle BAC.
Next, let the centre of the circle be without the angle BAC.
Then it maybe shewn, as inthe first case, that the angle FEC is double of the angle FAC, and that the angle FEB, a part of the first, is double of the angle FAB, a part of the other;
therefore the remaining angle BEC is double of the remaining angle BAC.
Wherefore, the angle at the centre &c. q.e.d.
PROPOSITION 21. THEOREM.
The angles in the same segment of a circle are equal to one another.
Let ABGD be a circle, and BAD, BED angles in the same segment BAED: the angles BAD, BED shall be equal to one another.
Take F the centre of the circle ABGD. [III. 1.
First let the segment BAED be greater than a semicircle.
Join BF, DF.
Then, because the angle BFD is at the centre, and the angle BAD is at the circumference, and that they have the same arc for their base, namely, BCD;
therefore the angle BFD is double of the angle BAD.[III.20.
For the same reason the angle BFD is double of the angle BED.
Therefore the angle BAD is equal to the angle BED. [Ax. 7.
