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. |
Observation.—The three theorems, namely, Cor. 1., Prop. i., and Parts 1, 2, of Prop. iii., are so related, that any one being proved directly, the other two follow by the Rule of Identity.
Cor. 1.—The line which bisects perpendicularly one of two parallel chords of a circle bisects the other perpendicularly.
Cor. 2.—The locus of the middle points of a system of parallel chords of a circle is the diameter of the circle perpendicular to them all.
Cor. 3.—If a line intersect two concentric circles, its intercepts between the circles are equal.