89. Theorem. In the same or equal circles, of two unequal arcs, the greater subtends the greater angle at the center. 3 2 1 Proof. If the first arc is greater than the second, it equals the second plus a third arc, and so the angle which the first subtends is greater than the angle which the second subtends by the angle which the third arc subtends at the FIG. 37. center. 90. Inversely: Of two unequal angles at the center, the greater intercepts the greater arc. 91. Two arcs which together equal the whole circle are called explemental. Thus the explemental angles at the center of a circle, whose arms are the same radii, are said to stand upon the explemental arcs which would be described simultaneously with the angles, the greater angle upon the greater arc. 92. Explemental arcs equal in magnitude are called semi circles. 93. A semicircle subtends a straight angle. For two subtend a perigon, and are equal. 94. Any straight through the center cuts the circle into two semicircles. For it makes at the center straight angles which together are subtended by the whole circle. 95. If we fold over about a straight through the center of a circle, the semicircles it makes are brought into coincidence. For every point on the turned semicircle must fall on some point of the other, as its sect from the center is a radius. 96. Two arcs which together equal a semicircle are called supplemental. 97. Half a straight angle is called a right A angle. 98. All right angles are equal in magnitude. 99. The arc subtending a right angle is called a quadrant. It is one quarter of a circle. 100. Two straights which make a right c angle are said to be perpendicular to one another. 101. Two angles whose sum is a right angle are called complementai. FIG. 39. B FIG. 40. 102. An angle less than a right angle is called acute. 103. An angle greater than a right angle, but less than a straight angle, is called obtuse. 104. An angle greater than a straight angle, but less than a perigon, is called reflex. 105..An angle which is either acute, right, or obtuse, is called a minor angle. B B O FIG. 41. FIG. 42. A 106. An arc less than a semicircle is called a minor arc. 107. An arc less than a circle, but greater than a semicircle, is called a major arc. 108. Theorem. If two circles have one common point not on the straight through their centers, they have also another such point. Proof. Let C and O O have the the point A in common. Fold the figure over along the straight through their centers, CO. Then the semicircles which have A in common are A C FIG. 43. O brought into coincidence with the other semicircles. There fore these also have a common point, A'. A M 109. Theorem. If two circles have a common point not on the straight through their centers, and therefore another such point, then the center-straight bisects the angles made at the centers by the radii to these two common points, and is the perpendicular bisector of the common chord. FIG. 44. Proof. For by folding over along CO we bring A into coin =sect A'M. * OMA AOM MOA'. = CHAPTER III. THE FUNDAMENTAL PROBLEMS. 110. Problem. To bisect any given sect. Construction. With its end points, A and A', as centers, and itself as radius, describe two circles. They will have one common point not on their center straight, and therefore a second such. Join these two common points, C and O. Then CO bisects the given sect AA'. Proof. For AA' is to CO, and if from C and O as centers, with radii FIG. 45. equal to AA', two circles were described, then AA' would be a common chord, bisected by the center-straight CO. III. Theorem. The straight through the mid point of a chord, and the center of the circle, is perpendicular to the chord, and bisects the explemental arcs, and their angles at the center. B FIG. 46. A Proof. A and B are any points on 0. Turn the whole figure over through half a revolution around BA. Then BA is the common chord of the new circle and its trace the old circle. Therefore their center-straight contains the points O and the bisection point C, and so is identical with the straight OC, through center and mid point. 112. Problem. To bisect any given angle. M A FIG. 47. Then MC bisects Construction. With its vertex, C, as center, and any sect, r, as radius, describe a circle cutting the arms of the angle at A and A'. Bisect the chord AA', and join its mid point, M, to the center C. ACA'. 113. Problem. At a given point on a given straight, to draw a perpendicular to that straight. Construction. Bisect the straight angle at the point. M C FIG. 48. D Construction. In the hemiplane not containing the given point, C, take any point D. Call A and A' points where OC[CD] cuts the given straight. Bisect the chord AA' at M. Then is CML to AA'. Determination. Through a given point only one perpendicular can be drawn to a given straight. For, if the plane were folded over along the given straight, the given point would fall on the production of any perpendicular from it to the straight. 115. Since now the perpendicular from the centre to a chord of a circle is identical with the st' through the center bisecting that chord, and also the explemental arcs and the explemental angles. pertaining to that chord, therefore the r't bi' of any chord passes through the center; and the straight which possesses any two of these seven properties possesses also the other five. 116. Problem. To bisect any given FIG. 49. FIG. 50. A |