Page:The Elements of Euclid for the Use of Schools and Colleges - 1872.djvu/121

Rh 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.