Section 1. · - Central Angles. Theorem 1. In the same circle or in equal circles, if two central angles are equal, the arcs on which they stand are equal also, and of two unequal central angles the greater stands on the greater arc. 2. Then A' coincides with A, and B' with B. its points are equidistant from O. AOC> < AOB, ·.· ZAOC>ZA'O'B'. .. C is not in ▲ AOB, and AC > AB. 4. Also, 5. Def. O Def. O Given Ax. 8 Step 3 The proof is essentially the same for one circle, and so in general when equal circles are involved. COROLLARIES. 1. Sectors of the same circle or of equal circles, which have equal angles, are equal. For, by steps 1, 2, 3, they coincide. 2. Sectors of the same circle, or of equal circles, which have unequal angles, are unequal, the greater having the greater angle. Proved by superposition in steps 4, 5, 6. 3. The two arcs into which the circumference is divided by a diameter are equal. (Why?) 4. The two figures into which a circle is divided by a diameter are equal. (Why?) This corollary is attributed to Thales. DEFINITION. The figure formed by a semicircumference and the diameter joining its extremities is called a semicircle. It is proved (cor. 4) that all semicircles, cut from the same circle, are equal. Hence the name, semi- meaning half. NOTE. Since the 360 equal angles, into which the perigon at the center of a circle is imagined to be divided, stand on equal arcs, by th. 1, the common mensuration of angles by degrees is also used for arcs. Similarly for minutes, seconds, and other measurements. Theorem 2. In the same circle or in equal circles, if two arcs are equal, the central angles which they subtend are equal also, and of two unequal arcs the greater subtends the greater central angle. Proof. If O and O' are two central angles, and A, A' are the arcs on which they stand, it has been proved in th. 1 that Hence the converses are true, by the Law of Converse. COROLLARIES. 1. In the same circle, or in equal circles, equal sectors have equal angles; and of two unequal sectors, the greater has the greater angle. Law of Converse, from th. 1, cors. 1, 2. 2. A central angle is greater than, equal to, or less than a right angle, according as the arc on which it stands is greater than, equal to, or less than a quadrant. (Why?) Theorem 3. In the same circle or in equal circles, if two arcs are equal they are subtended by equal chords, and of two unequal minor arcs the greater is subtended by the greater chord. Given two equal circles, M, M'; two equal arcs, K, K'; and two unequal minor arcs, K> K". Proof. 1. Draw the radii OA, OB, O'A', O'B', O'C. To prove that, as lettered in the figure, chords AB A'B', AB CA'. Then 3. Why? ... Th. 2 5. .. AB CA'. I, th. 10 2. But OA OB=O'A' O'B' ..A OABA O'A'B', and AB A'B', 4. Also, KK", ... ZAOB><CO'A'. COROLLARY. In the same circle, or in equal circles, of two unequal major arcs, the greater is subtended by the less chord. Theorem 4. In the same circle, or in equal circles, if two chords are equal they subtend equal major and equal minor arcs; and of two unequal chords the greater subtends the greater minor and the less major arc. Proof. Let C, C' be two chords of the same circle or of equal circles; N, N' their corresponding minor arcs; Hence the converses are true, by the Law of Converse. Theorem 5. A diameter which is perpendicular to a chord bisects the chord and its subtended arcs. Given the diameter BD perpendicular to chord AC at E. To prove that (1) AE EC, (2) AB=BC, (3) DA = CD. Proof. 1. Drawing radii OA, OC, then 2. 3. And 4. 5. 6. And 7. OE=OE. ZOEA CEO, = ... BD LAC. ...A AEO A CEO, AOEEOC, AE= EC. ... AB BC. D E B Given DOA = COD. I, th. 19, cor. 5 ... DA = CD. Why? COROLLARIES. 1. Conversely, a diameter which bisects a 2. The perpendicular bisector of a chord passes through the center of the circle and bisects the subtended arcs. For the center is equidistant from the ends of the chord, by definition of a circle; .. it lies on the perpendicular bisector of the chord, by I, th. 29. Theorem 6. All points in a chord lie within the circle; and all points in the same line, but not in the chord, lie without the circle. Given the points P1 in a chord 1 OA, OB, OP1, OP2 drawn, and OM AB. 2. Then M is between A and B. Th. 5 COROLLARY. A straight line cannot meet a circumference in more than two points. For every other point on that line must be either between or not between those two points, and hence must lie either within or without the circle. Theorem 7. In the same circle or in equal circles, equal chords are equidistant from the center; and of two unequal chords the greater is nearer the center. Given two equal M, M', with chords ABA'B', AE> A'B', and OC, OD, O'C' Is from center O to AB, AE, and from center O' to A'B'. To prove = that (1) OC O'C', (2) OD < O'C'. |
