|Book III Proposition 31|
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]
the angle ABC is less than a right angle,
the angle ADC which remains is greater than a right angle;
it is the angle in the segment ADC less than the semicircle.
And again being what it was required to prove.
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