56. The circumference of every circle is supposed to be divided into 360 equal parts, called degrees; and each degree into 60 minutes, each minute into 60 seconds, and so Hence a semicircle contains 180 degrees, and a quadrant 90 degrees. on. 57. The measure of a rightlined angle is an arc of any circle, contained between the two lines, which form that angle, the angular point being the centre; and it is estimated by the num ber of degrees, contained in that arc. Hence a right angle is an angle of 90 degrees. 58. Identical figures are such, as have all the sides and all the angles of one respectively equal to all the sides and all the angles of the other, each to each; so that, if one figure were applied to, or laid upon, the other, all the sides of it would exactly fall upon and cover all the sides of the other; the two becoming coincident. 59. An angle in a segment is that, which is contained by two lines, drawn from any point in the arc of the segment to the extremities of the arc. 60. A right-lined figure is inscribed in a circle, or the circle circumscribes it, when all the angular points of the figure are in the circumference of the circle. 61. A right-lined figure circumscribes a cirele, or the circle is inscribed in it, when all the sides of the figure touch the circumference of the circle. 62. One 62. One right-lined figure is inscribed in anether, or the latter circumscribes the former, when all the angular points of the former are placed in the sides of the latter. 63. Similar figures are those, that have all the angles of one equal to all the angles of the other, each to each, and the sides about the equal angles proportional. 64. The perimeter of a figure is the sum of all its sides, taken together. 65. A proposition is something, which is either proposed to be done, or to be demonstrated, and is either a problem or a theorem. 66. A problem is something proposed to be done. 67. A theorem is something proposed to be demonstrated. 68. A lemma is something, which is premised, or previously demonstrated, in order to render what follows more easy. 69. A corollary is a consequent truth, gained immediately from some preceding truth, or demonstration. 70. A scholium is a remark, or observation, made upon something preceding it. PROBLEMS. : PROBLEMS. PROBLEM I. To divide a given line A B înto two equal parts. From the centres A and B, with any radius greater than half A B, describe arcs, cutting each other in m and n. Draw the line men, A and it will cut the given line into two equal parts in the middle point C. B C PROBLEM II. To divide a given Angle ABC into two equal parts. From the centre B, with any radius, describe the arc AC. From A and C, with one and the same radius, describe arcs, in- A tersecting in m. Draw the line Bm, and / it will bisect the angle, as required. B A C m NOTE. By this operation the arc AC is bisected; and in a similar manner may any given arc of a circle be bisected. PROBLEM PROBLEM 111. To divide a right angle ABC into three equal parts. From the centre B, with any radius, describe the arc AC. From the centre A, with the same radius, A cross the arc AC in n; and with Then the centre C, and the same radius, PROBLEM IV. To draw a Line parallel to a given Line AB. C CASE 1.-When the parallel Line is to be at a given Dis tance C. CASE 2.-When the parallel Line is to pass through a given Take the arc Cn in the compasses, and apply it from m to r. Through C and r draw DE, the parallel required. Beb : NOTE, NOTE. In practice, parallel lines are more easily drawn with a Parallel Rule. PROBLEM V. To erect a Perpendicular from a given Point A in a given Line BC. CASE 1.-When the Point is near the middle of the Line. On each side of the point A, take any two equal distances Am, An. From the centres m anden, with any radius. greater than Am or An, describe two arcs intersecting in r. Through A and r draw the line Ar, and it will be the perpendicular required. B 711 1 CASE 2.-When the Point is near the end of the Line. With the centre A, and any radius, describe the arc mns. From the point m, with the same radius, turn the compasses twice over on the arc, at n and s. A n S C 11 gain, with the centres n and s, describe arcs intersecting in r. C B m A Then draw Ar, and it will be the perpendicular required. Another Method. From any point m, as a centre, with the radius or distance m A, describe an arc cutting the given line in n and A. Through n and m draw a right line cutting the arc inr: Lastly, draw Ar, and it will be the perpendicular required. B |