280. Theorem. The perpendicular bisector of a chord passes through the center of the circle. CD is the perpendicular bisector of chord AB. .. O is in the line CD produced, and CD must pass through O. § 263 DA 1. The sides of the polygon ABCDEF are chords of the circle. Show that the perpendicular bisectors of the sides pass through the same point. 2. In a circle draw five parallel chords. Are the centers of these chords in the same straight line? Why? Where are the centers of all parallel chords in a circle? A 3. Locate a point within a given circle. Through this point draw a chord which shall be bisected at that point. B C D 281. Theorem. In the same circle or in equal circles, equal chords are equidistant from the center. Conversely: Chords equidistant from the center H D E are equal. Given chord CD equal F to chord EF in the equal circles A and B. To prove that CD and EF are equidistant from A and B. Proof. Draw the radii AC and BF. Also draw AH 1 CD, and BK EF. Then AH bisects CD and BK bisects EF. .. CD and EF are equidistant from the centers. Given AH=BK in the equal circles A and B. § 277 Why? § 189 Why? Why? Why? EXERCISES 1. Where are the middle points of all equal chords of a circle? 2. The diameter AC is the perpendicular bisector of the chord BD. Prove that ≤B=LD. 3. Chord AB = chord ED. Prove that ACBD is isosceles. 4. Two equal chords intersect within a circle. Prove that the corresponding segments are equal. 5. If two intersecting chords make equal angles with the diameter drawn through the point of intersection, the chords are equal. 6. Two equal chords intersect within a circle. The center is joined to the point of intersection by the line OF. Prove that OF makes equal angles with each of the chords. 7. If two chords intersect so that a segment of one is equal to a segment of the other, the chords are equal. 8. Through a given point within a circle, draw two equal chords. 9. If two chords are bisected by a diameter of the circle, prove that the chords are parallel. 282. Since all points on a circle are equidistant from the center, the distance from the center of a circle to any point without the circle is greater than the radius, and the distance from the center to any point within the circle is less than the radius. BA>radius BE and BC <radius BD. § 110 E A 283. A point is without a circle when its distance from the center is greater than the radius. Also, a point is within a circle when its distance from the center is less than the radius. These must be true from the definition of a circle. § 263 TANGENTS AND SECANTS 284. Theorem. A perpendicular to a radius at its outer extremity can touch the circle at only one point. Given BD perpendicular to the radius OA at its outer extremity A. To prove BD can touch the circle at only one point. B A C Proof. Draw OC, any other line, from the center O to the line BD. Then OC>OA (§ 185), and the point C must lie without the circle. § 283 Hence every point in the line BD, except A, lies without the circle. .. BD can touch the circle at only one point. 285. Tangent. A straight line which touches a circle at only one point is a tangent. Thus CE is a tangent to the circle O. The circle is also said to be tangent to the line CE. The point D is called the point of contact or point of tangency. In speaking of a tangent to a circle from A Secant D Tangent an external point, only that part of the tangent is C considered which lies between the external point and the point of contact. CD is the tangent to the circle O from the point C. 286. Secant. A straight line which intersects a circle in two points is a secant. In the figure, AB is a secant. 287. Theorem. A straight line perpendicular to a radius at the point where the radius meets the circle is tangent to the circle. § 284 and § 285 288. Theorem. The radius drawn to the point of tangency is perpendicular to the tangent. Since only one radius can be drawn to a tangent, this must be true from § 287. EXERCISE Tangents to a circle at the extremities of a diameter are parallel. 289. Theorem. Parallel lines intercept equal arcs on a circle. CASE I. When a tangent and a secant are parallel. (Fig. 1.) Given the secant EF parallel to the tangent CD. To prove AH-AK. Proof. Draw the radius OA to the point of contact. CASE II. When two secants are parallel. (Fig. 2.) Given the parallel secants DE and HK. LN To prove ĹN=MO. Proof. Draw the tangent AB || DE. Then AB HK. CN=CO and CL=CM. .. LN=MO. CASE III. When two tangents are parallel. (Fig. 3.) § 288 § 123 § 277 $ 124 Case I § 106 EXERCISES A 1. The lines BC and AD are parallel chords. Prove that x = Lz. 4. AB | DC. Prove that ABCD is an isosceles trapezoid. 290. Theorem. Tangents to a circle from an external point, are equal, and make equal angles with the line joining that point with the center. Given AB and AC, tangents to the circle O, from the external point A, and AO, the line joining A with the center 0. To prove AB=AC and Zx= Lz. OB LAB and OC LAC. B § 288 Why? 1. Prove that HK=MH+NK in the first figure. 2. The sides of a quadrilateral ABCD are tangent to the circle O. Prove that AD+BC=AB+DC. 3. The lines AB, BC, and CD are tangent to the circle O. AB || DC. Prove that 20 is a right angle. 4. The lines AC and AD are tangent to the circle K. A is joined to K and the line is produced to meet the circle at B. CD is tangent to the circle at B. Prove that ACAD is isosceles. 291. Line of Centers. The line joining the centers of two circles is the line of centers. Thus AB is the line of centers of the two circles A and B. |