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. Cor. 4.—The line joining the centres of two intersecting circles bisects their common chord perpendicularly. Exercises. 1. If a chord of a circle subtend a right angle at a given point, the locus of its middle point is a circle. 2. Every circle passing through a given point, and having its centre on a given line, passes through another given point. 3. Draw a chord in a given circle which shall subtend a right angle at a given point, and be parallel to a given line. PROP. IV.—Theorem. Two chords of a circle (AB, CD) which are not both diameters cannot bisect each other, though either may bisect the other. Dem.—Let O be the centre. Let AB, CD intersect in E; then since AB, CD are not both diameters, join OE. If possible let AE be equal to EB, and CE equal to ED. Now, since OE passing through the centre bisects AB, which does not pass through the centre, it is at right angles to it; therefore the angle AEO is right. In like manner the angle CEO is right. Hence AEO is equal to CEO—that is, a part equal to the whole—which is absurd. Therefore AB and CD do not bisect each other. Cor.—If two chords of a circle bisect each other, they are both diameters. PROP. V.—Theorem. If two circles (ABC, ABD) cut one another in any point (A), they are not concentric. Dem.—If possible let them have a common centre at O. Join OA, and draw any other line OD, cutting the circles in C and D respectively. Then because O is the centre of the circle ABC, OA is equal to OC. Again, because O is the centre of the circle ABD, OA is equal to OD. Hence OC is equal to OD—a part equal to the whole—which is absurd. Therefore the circles are not concentric. Exercises. 1. If two non-concentric circles intersect in one point, they must intersect in another point. For, let O, O′ be the centres, A the point of intersection; from A let fall the ⊥ AC on the line OO′. Produce AC to B, making BC = CA: then B is another point of intersection. 2. Two circles cannot have three points in common without wholly coinciding. PROP. VI.—Theorem. If one circle (ABC) touch another circle (ADE) internally in any point (A), it is not concentric with it.