semicircle, shall be less than a right angle; and the angle in the segment ADC, which is less than a semicircle, shall be greater than a right angle.
Join AE, and produce BA to F.
Then, because EA is equal to EB, [I. Definition 15.
the angle EAB is equal to the angle EBA; [1.5.
and, because EA is equal to EC, the angle EAC is equal to the angle ECA; therefore the whole angle BAC is. equal to the two angles, ABC, ACB. [Axiom 2.
But FAC, the exterior angle of the triangle ABC, is equal to the two angles ABC, ACB; [I. 32.
therefore the angle BAC is equal to the angle FAC, [Ax. I.
and therefore each of them is a right angle. [I. Def. 10.
Therefore the angle in a semicircle BAC is a right angle.
And because the two angles ABC, BAC, of the triangle ABG, are together less than two right angles, [I. 17.
and that BAC has been shewn to be a right angle, therefore the angle ABC less than a right angle.
Therefore the angle in a segment ABC, greater than a semicircle, is less than a right angle.
And because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are together equal to two right angles; [III. 22.
therefore the angles ABC, ADC are together equal to two right angles.
But the angle ABC has been shewn to be less than a right angle;
therefore the angle ADC is greater than a right angle.
Therefore the angle in a segment ADC, less than a semicircle, is greater than a right angle.
Wherefore, the angle &c. q.e.d.
