HINT. - Draw radii OA, OB, O'A', O'B', and prove the equality of A OAB and O'A'B'. Ex. 306. If chords AB, BC, CD, DE are equal, then chords AC, BD, CE are equal. B C E Ex. 307. If in the annexed diagram AB= CD, then BC AD. = Ex. 308. If in the same diagram two intersecting chords, AD and BC, are equal, then AB = CD. Ex. 309. The diagonals of an equilateral pentagon inscribed in a circle are equal. Ex. 310. The radii drawn to the vertices of an inscribed equilateral hexagon divide the figure into six equilateral triangles. Ex. 311. If two chords bisect each other, the arcs intercepted by the sides of a pair of vertical angles are equal. Ex. 312. If two chords bisect each other, they are diameters. PROPOSITION III. THEOREM 178. A radius perpendicular to a chord bisects the chord and the subtended arc. 179. COR. A perpendicular bisector of a chord passes through the center of the circle. Ex. 313. If two chords are equal, the perpendiculars from the center upon the chords are equal. Ex. 314. If the perpendiculars from the center upon the sides of an inscribed polygon are equal, the polygon is equilateral. Ex. 315. If from a point without a circle two equal lines are drawn to a circumference, the bisector of the angle they form passes through the center of the circle. Ex. 316. Two points, each equidistant from the ends of a chord, determine a line passing through the center of the circle, PROPOSITION IV. PROBLEM 180. To circumscribe a circle about a given triangle. Required. To circumscribe a circle about ▲ ABC. Construction. Erect perpendicular-bisectors upon BC and AC. They will intersect at some point E. (Why?) From E as a center, with a radius equal to EA, describe a O. It will pass through A, B, and C. .. ABC is the required O. (Why?) 181. COR. 1. Three points not in a straight line determine a circumference. 182. COR. 2. A circumference cannot be drawn through three points which lie in the same straight line. 183. COR. 3. A straight line cannot intersect a circumference in more than two points. 184. COR. 4. Two circumferences cannot meet in more than two points. Ex. 317. To find the center of a given circle. PROPOSITION V THEOREM 185. In the same circle, or in equal circles, equal chords are equally distant from the center; and, conversely, chords equally distant from the center are equal. chord AB = chord CD, OELAB, OHLCD. HINT. - What is the means of proving the equality of lines? CONVERSELY. Hyp. In ABCD: To prove OE OH, OE LAB, OH1 CD. AB = CD. The proof is similar to the above. 186. REMARK. - The equality of two chords is usually established by means of equal distances from the center or equal subtended arcs. Ex. 319. If from any point in the circumference two chords are drawn making equal angles with the radius to the point, these chords are equal. Ex. 320. If through any point in a radius two chords are drawn making equal angles with the radius, these chords are equal. Ex. 321. A line joining the point of intersection of two equal chords to the center bisects the angle formed by the chords. Ex. 322. In a given circle, to draw a chord equal and parallel to a given chord. PROPOSITION VI. THEOREM 187. In the same circle, or in equal circles, the greater of two minor arcs is subtended by a greater chord; and, conversely, the greater chord subtends the greater arc. CONVERSELY.-Hyp. In equal, O and O', AB > A'B'. To prove arc AB>arc A'B'. Proof. Draw radii OA, OB, O'A', O'B'. In A AOB and A'O'B', AOA'O', BO = B'0', AB > A'B'. B (175) (128) Q.E.D. (Hyp.) .LAOB>▲ A'O'B'. (129) ..arc AB > arc A'B'. (175) Q.E.D. G |