NOTE. By this operation the arc A C is bisected; and in a similar manner may any given arc of a circle be bisected. PROBLEM III.. 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 and the same radius, cut the arc A C PROBLEM IV. B C To draw a line parallel to a given line AB. CASE I. When the parallel line is to be at a given distance C. From any two points m and n, in the line AB, with a rad D ius equal to G, describe the arcs r and o. Draw CD to A-T touch these arcs, without cut ting them, and it will be the parallel required.t B m n C m; then the sides of the triangle B A m are respectively equal to the sides of the triangle B C m. Therefore the angle AB m 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 For the points, where the right line CD touches the arcs 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 mĈ, describe the arc Cn. From the centre C, with the same radius, describe A the arc mr. Take the arc Cn D C 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 drawn with a Parallel Rule. E PROBLEM V. To erect a perpendicular from a given point A in a given line BC. CASE I. 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 m A. n 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 to AB. Suppose right lines drawn from r to my and rton; then the CASE 2. When the point is near the end of the line. n 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. m draw a right arc in r. Lastly, Through n and line cutting the draw A r, and it will be the perpendicular re quired.t m ייג 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, rto 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 & Ar. And the angle m An = the angle CAs. Therefore the angles at A are right angles, and Ar is perpendicular to BC. = For the angle BAr, being in a semicircle, is a right angle, or A r is perpendicular to BA. Or any other numbers in the same proportion, as 3, 4, 5, will answer the same purpose.* PROBLEM VI. From a given point A, out of a given line BC, to let fall a perpendicular. n. CASE I. When the point is nearly opposite the middle of the line. With the centre A, and any radius, describe an arc, cutting BC in m and With the centres m and n, and the same, or any other radius, describe arcs, intersecting in r. Draw ADr for the perpendicular requiredț CASE 2. When the point is nearly opposite the end of the line. From A draw any line A m tọ meet BC, in any point m. Bisect Am at n, and with the centre n, and radius A nor m n, describe an arc, cutting BC in D. Draw AD, the perpendicular required.§ B * If a right line be drawn from m to n; then is A m n a rightangled triangle. For the square of mn is equal to the square of m A, added to the square of An. Therefore An is perpendic ular to AB. + For if right lines be drawn from m to A, and A ton; then, the angle m An being bisected by AD, according to Problem 11, in the triangle m AD, the side m A and angle m AD are respectively equal to the side An and angle n AD in the triangle n AD, and AD is common to both. Therefore the angles at D are equal, and AD is perpendicular to BC. § BDA, being an angle in a semicircle, is a right angle, or AD is perpendicular to BD. ANOTHER METHOD. From B or any point in BC, as a centre, describe an arc through the point A. From any other centre m in B C, describe another arc through A, cutting the former arc again in n. Through A and n draw the line A D n; and AD will be the perpendicular required.* B C D NOTE. Perpendiculars may be more readily raised and 1et fall, in practice, by means of a square or other fit instru ment PROBLEM VII. To divide a given line A B in the same proportion, as anoth er line C D is divided. * Right lines being drawn from B to A, m to A, B to n, and m ton; then the sides of the triangle B A m being respectively equal to the sides of the triangle Bn m, the angles of the former are respectively equal to the corresponding angles of the latter; consequently in the triangle Am D the angle Am D is equal to the angle D m n in the triangle D m n; A m is equal to mn, and m D is common to the two triangles. Therefore the angles at D are right angles, and A D is perpendicular to B D. For A 1 1, A 2 2, A 3 3, &c. are similar triangles; there |