59. An angle in a segment is that, which is contained by two lints, 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 circle, or the circle is inscribed in it, when all the sides of the figure touch the circumference of the circle. 62. One right-lined figure is inscribed in another, 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 >r 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. PROBLEM. I. To divide a given line A B into two equal parts. From the centres A and B. with any m n PROBLEM II.、 To divide a given angle A B C into two equal parts. From the centre B, with any radius, de^ Scribe the arc A C. From A and C, with one and the same radius, dtscribe arcs, intersecting in m. Draw the line B m, and it will bisect the angle, as required.t B m B * Suppose right lines to be drawn from A to m, m to B, B to M, and « to A; then Am B n is a parrallelogram, and its diago nals A B, mn, mutually bisect each other. bisected in C. Therefore AB is ↑ Suppose right lines to be drawn from A to m and from G to NOTE. By this operation the arc A C is bisected; and in a similar manner may any given arc of a circle be bi sected. PROBLEM lit. To divide a right angle A B C into three equal parts. From the centre B, with any radius, describe the arc A C. From the cen tre A, with the same radius, cross the A arc A C in n; and with the centre C, in m. and the same radius, cut the arc A C Then through the points m and a draw B m and B n, and they will trisect the angle, as required.* B C PROBLEM IV. To draw a line parallel to a given line AB. Case r. When the parallel line is to be at a given distance C* m; then the sides of the triangle B A m are respectively equal to th« sides of the triangle B C m. Therefore the angle A B = the angle CB m. * The number of degrees in a right angle is 90; and the radius of a circle being equal to the chord of 60 degrees, the arc Cm An 60 degrees. Therefore mn = 30 degrees ➡ = Am=nC = of a right angle. For the points, where the right line GD touches the a^cs CASE 2. When the parallel line is to pass through a given point C. From any point m, in the line AB, with the radius mC, describe the arc Cn. From the centre C, with the same radius, describe A the arc mr. Take the arc Cn D n 322 in the compasses, and apply it from m to r. Through C and r draw DE, the parallel required.* Note. In practice, parallel lines are more easily drawa with a Parallel Rule. E PROBLEM v. To erect a perpendicular from a given point A in a given line BC. Case t. 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 and n, 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.t B A rando, are equally distant from the line AB. Therefore all the other points, through which CD passes, are equally distant from AB, that is, CD is parallel to AB. * For if Cm be joined by a right line, it is evident, that the angle nmC the angle m Cr. Therefore DE is parallel t» AB. + Suppose right lines drawn from r to m, and rton; then the CASE 2. WAen *Ae poinf is near fAe 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. Again, with the centres n and s, describe arcs, intersecting in r. Then draw Ar, and it will be the perpendicular required.* B 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 in r. Lastly, draw A r, and it will be the perpendicular required.f ANOTHER METHOD. From any plane scale of equal parts, set off Am equal to 4 parts. With centre A, and radius of three parts, describe an arc. And with centre m, and radius of 5 parts, cross it at n. Draw An for the perpendicular required. B m C A A sides of the triangle mrA are respectively equal to the sides of the triangle nr A. Therefore the angles at A are equal to each other, and r A is perpendicular to BC. *Right lines being drawn from A to n, n to r, r to s, and s to A, the sides An, nr, of the triangle Anr, are respectively equal to the sides As, sr, of the triangle Asr, and Ar is common to both triangles. Consequently the angle n Ar the angle s Ar. And the angle m An the angle CAS. Therefore the angles at A are ri rht angles, and Ar is perpendicular to BC, 8 † For the angle BAr, being in a semicircle, is a right angle, or A r is perpendicular to BA. |