But if the angle C be not a right angle, at the point A, in the straight line AB, make the angle BAD equal to the angle C; [I. 23.
from the point A, draw AE at right angles to AD;[I.ll.
bisect AB at F; [I. 10. from the point F, draw FG at right angles to AB [1. 11.
and join GB.
Then, because AF is equal to BF, [Const.
and FG is common to the two triangles AFG,BFG;
the two sides AF, FG are equal to the two sides BF, FG, each to each;
and the angle AFG is equal to the angle BFG; [I. Definition 10.
therefore the base AG is equal to the base BG; [I. 4.
and therefore the circle described from the centre G, at the distance GA, will pass through the point B.
Let this circle be described; and let it be AHB.
The segment AHB shall contain an angle equal to the given rectilineal angle G.
Because from the point A, the extremity of the diameter AE, AD is drawn at right angles to AE, [Construction.
therefore AD touches the circle. [III. 16. Corollary.
And because AB is drawn from the point of contact A, the angle DAB is equal to the angle in the alternate segment AHB. [III. 32.
But the angle DAB is equal to the angle O. [Constr.
Therefore the angle in the segment AHB is equal to the angle C. [Axiom 1.
Wherefore, on the given straight line AB, the segment AHB of a circle has been described, containing an angle equal to the given angle C. q.e.f.
