Then, since AF is equal to FB, and FG is common,
the two sides AF, FG are equal to the two sides BF, FG;
and the angle AFG is equal to the angle BFG;
therefore the base AG is equal to the base BG. [I. 4]
Therefore the circle described with centre G and distance GA will pass through B also.
Let it be drawn, and let it be ABE; let EB be joined.
Now, since AD is drawn from A, the extremity of the diameter AE, at right angles to AE,
therefore AD touches the circle ABE. [III. 16, Por.]
Since then a straight line AD touches the circle ABE,
and from the point of contact at A a straight line AB is drawn across in the circle ABE,
the angle DAB is equal to the angle AEB in the alternate segment of the circle. [III. 32]
But the angle DAB is equal to the angle at C; therefore the angle at C is also equal to the angle AEB.