Therefore the three straight lines DA,DB,DC are all equal;
and therefore D is the centre of the circle. [III. 9.
From the centre D, at the distance of any of the three DA, DB, DC, describe a circle; this will pass through the other points, and the circle of which ABC is a segment is described.
And because the centre D is in AC, the segment ABC is a semicircle.
Next, let the angles ABD, BAD be not equal to one
another.
At the point A, in the straight line AB, make the angle
BAE equal to the angle ABD ; [I, 23.
produce BD, if necessary, to meet AE at E, and join EC.
Then, because the angle BAE is equal to the angle
ABE, [Construction.
EA is equal to EB. [I. 6.
And because AD is equal to CD, [Construction.
and DE is common to the two triangles ADE, CDE ;
the two sides AD, DE are equal to the two sides CD, DE,
each to each ;
and the angle ADE is equal to the angle CDE, for each of
them is a right angle ; [Construction.
therefore the base EA is equal to the base EC. [I. 4.
But EA was shewn to be equal to EB ;
therefore EB is equal to EC. [Axiom 1.
Therefore the three straight lines EA, EB, EC are all equal ;
and therefore E is the centre of the circle. [III. 9.
From the centre E, at the distance of any of the three EA, EB, EC, describe a circle ; this will pass through the other points, and the circle of which ABC is a segment is described.
And it is evident, that if the angle ABD be greater than the angle BAD, the centre E falls without the seg- ment ABC, which is therefore less than a semicircle ; but if the angle ABD be less than the angle BAD, the centre E falls within the segment ABC, which is therefore greater than a semicircle.
Wherefore, a segment of a circle being given, the circle has been described of which it is a segment. q.e.f.
