If any two points (A, B) be taken in the circumference of a circle—1. The segment (AB) of the indefinite line through these points which lies between them falls within the circle. 2. The remaining parts of the line are without the circle.
Dem.—1. Let C be the centre. Take any point D in AB. Join CA,CD,CB. Now the angle ADC is [I. xvi.] greater than ABC; but the angle ABC is equal to CAB [I. v.], because the triangle CAB is isosceles; therefore the angle ADC is greater than CAD. Hence AC is greater than CD [I. xix.]; therefore CD is less than the radius of the circle, consequently the point D must be within the circle (note on I. Def. xxiii.).
In the same manner every other point between A and B lies within the circle.
2. Take any point E in AB produced either way. Join CE. Then the angle ABC is greater than AEC [I. xvi.]; therefore CAB is greater than AEC. Hence CE is greater than CA, and the point E is without the circle.
We have added the second part of this Proposition. The indirect proof given of the first part in several editions of Euclid is very inelegant; it is one of those absurd things which give many students a dislike to Geometry.
Cor. 1.—Three collinear points cannot be concyclic.
Cor. 2.—A line cannot meet a circle in more than two points.
Cor. 3.—The circumference of a circle is everywhere concave towards the centre.
PROP. III.–Theorem.