PROPOSITION 31.
In a circle the angle in the semicircle is right,
that in a greater segment less than a right angle,
and that in a less segment greater than a right angle; and further the angle of the greater segment is greater than a right angle,
and the angle of the less segment less than a right angle.
Let
ABCD be a circle, let
BC be its diameter, and
E its centre, and let
BA,
AC,
AD,
DC be joined; I say that the angle
BAC in the semicircle
BAC is right, the angle
ABC in the segment
ABC greater than the semicircle is less than a right angle, and the angle
ADC in the segment
ADC less than the semicircle is greater than a right angle.
Let
AE be joined, and let
BA be carried through to
F.
Then, since
BE is equal to
EA,
the angle ABE is also equal to the angle BAE. [I. 5]
Again, since
CE is equal to
EA,
the angle ACE is also equal to the angle CAE. [I. 5]
Therefore the whole angle
BAC is equal to the two angles
ABC,
ACB.
But the angle
FAC exterior to the triangle
ABC is also equal to the two angles
ABC,
ACB; [
I. 32]
therefore the angle BAC is also equal to the angle FAC; therefore each is right; [I. Def. 10] therefore the angle BAC in the semicircle BAC is right.
Next, since in the triangle
ABC the two angles
ABC,
BAC are less than two right angles, [
I. 17] and the angle
BAC is a right angle,
the angle ABC is less than a right angle; and it is the angle in the segment
ABC greater than the semicircle.
Next, since
ABCD is a quadrilateral in a circle, and the opposite angles of quadrilaterals in circles are equal to two right angles, [
III. 22] while the angle
ABC is less than a right angle, therefore the angle
ADC which remains is greater than a right angle; and it is the angle in the segment
ADC less than the semicircle.
I say further that the angle of the greater segment, namely that contained by the circumference
ABC and the straight line
AC, is greater than a right angle; and the angle of the less segment, namely that contained by the circumference
ADC and the straight line
AC, is less than a right angle.
This is at once manifest. For, since the angle contained by the straight lines
BA,
AC is right,
the angle contained by the circumference ABC and the straight line AC is greater than a right angle.
Again, since the angle contained by the straight lines
AC,
AF is right,
the angle contained by the straight line CA and the circumference ADC is less than a right angle.
Therefore etc. Q. E. D.
