Therefore the angle ABF is equal to the angles BAD, ABD.
From each of these equals take away the common angle ABD;
therefore the remaining angle DBF is equal to the remaining angle BAD, [Axiom 3.
which is in the alternate segment of the circle.
And because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are together equal to two right angles. [III. 22.
But the angles DBF, DBE are together equal to two right angles. [I. 13.
Therefore the angles DBF, DBE are together equal to the angles BAD, BCD.
And the angle DBF has been shewn equal to the angle BAD;
therefore the remaining angle DBE is equal to the remaining angle BCD, [Axiom 3.
which is in the alternate segment of the circle.
Wherefore, if a straight line &c. q.e.d.
PROPOSITION 33. PROBLEM.
On a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle.
Let AB be the given straight line, and C the given rectilineal angle: it is required to describe, on the given straight line AB, a segment of a circle containing an angle equal to the angle C.
First, let the angle G be a right angle.
Bisect AB at F, [1. 10.
and from the centre F, at the distance FB, describe the semicircle AHB.
Then the angle AHB in a semicircle is equal to the right angle C. [III. 31.
