Cor. 2.—The locus of the centres of the circles which pass through two fixed points is the line bisecting at right angles that connecting the two points. Cor. 3.—If A, B, C be three points in the circumference of a circle, the lines bisecting perpendicularly the chords AB, BC intersect in the centre. PROP. II.—Theorem. 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. xxxiii.). 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. If a line (AB) passing through the centre of a circle bisect a chord (CD), which does not pass through the centre, it cuts it at right angles. 2. If it cuts it at right angles, it bisects it. Dem.—1. Let O be the centre of the circle. Join OC, OD. Then the triangles CEO, DEO have CE equal to ED (hyp.), EO common, and OC equal to OD, because they are radii of the circle; hence [I. viii.] the angle CEO is equal to DEO, and they are adjacent angles. Therefore [I. Def. xiii.] each is a right angle. Hence AB cuts CD at right angles. 2. The same construction being made: because OC is equal to OD, the angle OCD is equal to ODC [I. v.], and CEO is equal to DEO (hyp.), because each is right. Therefore the triangles CEO, DEO have two angles in one respectively equal to two angles in the other, and the side EO common. Hence [I. xxvi.] the side CE is equal to ED. Therefore CD is bisected in E. 2. May be proved as follows:— OC2 = OE2+ EC2 [I. xlvii.], and OD2 = OE2 + ED2; but OC2 = OD2; ∴ OE2 + EC2 = OE2 + ED2. Hence EC2 = ED2, and EC = ED.