Book 3
1
Then, since AD is equal to DB, and DG is common, the two sides AD, DG are equal to the two sides BD, DG respectively; and the base GA is equal to the base GB, for they are radii; therefore the angle ADG is equal to the angle GDB. [I. 8]
1
But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10] therefore the angle GDB is right.
1
But the angle FDB is also right; therefore the angle FDB is equal to the angle GDB, the greater to the less: which is impossible. Therefore G is not the centre of the circle ABC.
1
Similarly we can prove that neither is any other point except F. Therefore the point F is the centre of the circle ABC.
1
Porism. From this it is manifest that, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line. Q. E. F.
PROPOSITION 2.
2
If on the circumference of a circle two points be taken at random, the straight line joining the points will fall within the circle.
2
Let ABC be a circle, and let two points A, B be taken at random on its circumference; I say that the straight line joined from A to B will fall within the circle.
2
For suppose it does not, but, if possible, let it fall outside, as AEB; let the centre of the circle ABC be taken [III. 1], and let it be D; let DA, DB be joined, and let DFE be drawn through.
2
Then, since DA is equal to DB, the angle DAE is also equal to the angle DBE. [I. 5] And, since one side AEB of the triangle DAE is produced, the angle DEB is greater than the angle DAE. [I. 16]
2
But the angle DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE. And the greater angle is subtended by the greater side; [I. 19] therefore DB is greater than DE. But DB is equal to DF; therefore DF is greater than DE, the less than the greater : which is impossible.
2
Therefore the straight line joined from A to B will not fall outside the circle.
