PROPOSITION IV. PROBLEM. 173. To find the center of a given circle. Join any two points A, B, in the circumf. Draw CD to AB at its mid point. (95) Bisect CD in O (80). o is the required center. For chord CD is to AB at its mid point, (Const.) (171) (Const.) .. the center of O ACB lies in CD; and since OC = OD, O is the center of O ACB, (it being the mid point of the diam. CD.) Q.E.F. SCHOLIUM. In order to avoid repetition, it will usually be assumed, henceforth, that the center o of any circle, if not given, has been found by this construction. EXERCISE 168. Show how to find the center of a circle, when an arc only of its circumference is given. 169. Apply Exercise 167 so as to find the center of a circle without the bisection of any chord. 170. Apply Exercise 168 to effect the same purpose. 171. Find the direction in which lies the longest, and also the shortest, distance from a point within a circle to its circumference. 172. Find the same for a point without the circle. PROPOSITION V. THEOREM. 174. In the same circle, or in equal circles, equal arcs are subtended by equal chords; and conversely. 1o. Given: In equal circles ADB, A'D'B', AB, A'B', chords of equal arcs ACB, A'C'B'; To Prove: Chord AB is equal to chord A'B'. o, o', being the centers, place O ADB upon O A'D'B', so that 00' and AA'. Then since arc ACB = arc A'C'B', B will coincide with B', (Hyp.) ... chd. AB will coincide with, and equal, chd. A'B'. Q.E.D. (14) 2°. Given: In equal circles ADB, A'D'B', equal chords AB, A'B'; To Prove: Arc ACB is equal to arc 'C'B'. .. ADB can be placed upon O A'D'B', so that ▲ OAB ▲ O'A'B'; then arc ACBarc A'C'B', .. arc ACB=arc A'C'B'. (69) (61) Q.E.D. (14) SCHOLIUM. If the arcs are in the same circle, the proof is similar to that above. PROPOSITION VI. THEOREM. 175. A radius bisecting a chord bisects also its subtended arc. 176. COR. 1. A radius bisecting an arc bisects its chord at right angles. (75) For the extremities of the radius are equally distant from those of the chord. 177. COR. 2. A radius bisecting a chord or its subtended arc is perpendicular to the chord. (75 and 176) EXERCISE 173. In the diagram for Prop. V., if the bisector of angle AOB be drawn, it will bisect arc ACB and also arc ADB. 174. In the same diagram, if the points where the bisector cuts the circumference be joined with A and B, the chords drawn from each point will be equal. 175. In the diagram for Prop. VI., if diameters from A and B meet the circumference in E and F, the line joining EF will be parallel to AB. 176. Two intersecting circles cannot have the same center. PROPOSITION VII. PROBLEM. 178. In a given circle, to place a chord equal to a given straight line not greater than a diameter. Given: A straight line MN not greater than the diameter of a circle ADB; Required: To place in ADB a chord equal to MN. From 4, any desired point in the circumference, as center, with radius equal to MN, describe an arc CB, cutting the circumference in B. Join AB. AB is the required chord. Since AB is a radius of arc BC, and AB = MN, (Const.) a chd. AB equal to MN has been placed in O ADB. Q.E.F. 179. SCHOLIUM. By this construction, any given arc may be laid off on a circumference having a radius equal to that of the arc. 180. DEFINITION. An arc is called a major or a minor arc according as it is greater or less than a semicircumference. EXERCISE 177. In the diagram for Prop. VII., if arc BC be produced to meet the circumference in D, and BD be drawn, the diameter drawn from A will bisect BD. 178. In the same diagram, place a chord that is equal and parallel to AB. 179. In the same diagram, place a chord that is equal and perpendicular to AB. 181. In the same circle, or in equal circles, the greater of two minor arcs is subtended by a greater chord; and conversely. F D E 1°. Given: In circle ADB, arc AFB greater than arc CED; To Prove: Chord AB is greater than chord CD. From 4 draw chd. AF = CD (178), and join 0, the center, arc AF = arc CED (174") and < arc AFB, (Const.) (Hyp.) .. F falls between A and B, and OF between OA and OB, .. ZAOB><AOF. Now OA, OB = OA, OF, respectively, .. AB > AF or CD. (Ax. 8) (162) Q.E.D. (90) 2o. Given: In circle ADB, chord AB greater than chord CD; To Prove: Arc AFB is greater than arc CED. With the same construction as in 1o, since OA, OBOA, OF, respectively, but AB > AF, ZAOBAOF, .. F falls between A and B, .. arc AFB > arc AF or arc CD. (162) (Hyp.) |