A -B A B The radii must be taken greater than half AB, so that the arcs can cut each other. By the same construction a line may be bisected. 53. To describe a circle, passing through three given points A, B and C. Find the centre; it must be equidistant from A and B, therefore in a perpendicular on the middle of AB, likewise in a perpendicular on the middle of BC, and must therefore lie where these intersect; as there is only one point of intersection, there can only be described one circle passing through three given points. If the three points lie in a straight line, the two perpendiculars will be parallel and have no point of intersection, so that the problem is impossible; we usually say, that the parallel lines cut one another at an infinite distance, and that therefore, through three given points lying in a straight line, there can be described a circle, of which the centre is infinitely distant and the radius of which therefore is infinitely great; by this is only meant, that by taking the centre sufficiently far away, a circle can be got passing through the two points, and as near as we please, to the third point. If the centre of a given arc is to be found, we employ the same construction, by taking any three points in the arc. From this it follows, that two circles only can cut each other in two points, for if they had three points in common, they must coincide. The line joining the centres is called the line of centres; if it is produced, it divides both circles symmetrically, and therefore is perpendicular on the middle of the line joining the points of intersection of the circles. (51). If radii be drawn to one of the points of intersection, they will be sides of a triangle, of which the line of centres is the third side, and this must therefore be less than the sum of the radii and greater than their difference, when the circles cut one another (39), whilst it is greater than the sum of the radii, when the circles lie outside each other, and less than their difference, when one circle lies within the other. If the chord, which the circles A have in common, becomes infinitely small, so that the points of intersection coincide in one point, this is called a point of contact; it must lie in the line of centres, for, so long as there is one common point on the one side of the line of centres, there must also be one symmetrically with this on the other side. From this we see, that the line of centres is equal to the sum of the radii, when the circles touch one another externally, and equal to their difference, when they touch internally. Conversely, if the line of centres equals the sum or difference of the radii, the circles must touch one another. 54. To raise a perpendicular on a given line at a given point. a) From the given point O cut off the equal b) If the given point is the extremity of the B B नै 55. From a given point A to draw a perpendicular to a given line BC. Take any two points in the line as centres, and describe circles through the given point; the line joining the points of intersection of the circles will be the required line (53). 56. To bisect a given angle or arc. With the vertex of the angle C as centre, describe the arc AB; next find a point D equidistant from A and B. Then CD bisects both the angle and the arc, as it by construction is perpendicular on the middle point of the chord AB. Any angle cannot be divided by compasses and ruler into equal parts in any other way than by continued bisection, therefore only into 2, 4, 8, 16, &c. equal parts. B D 57. Through a given point in a line to draw a line, making an angle with the given line equal to a given angle. Let O be the given point and A the given angle, with O describe the arcs BC and DE with equal DE is made equal to BC, then 20 ~ 58. Through a given point to draw a line parallel to a given line. From the given point A draw any line AB, cutting the given line BC; then if we make ▲ A = LB, then DA BC. (31). 59. To construct a triangle, having given Mark off the one side AB; the vertex of the triangle will then be the point of intersection of two circles, with centres A and B, and with the other given sides as radii. The problem is possible, when the circles cut each other, therefore, A B A when the one side is less than the sum C B B A 60. To construct a triangle, having given two sides and the angle contained by them. Mark off the given angle, and mark off parts on its legs, equal to the given sides. As the problem only has one solution, two triangles are congruent, when they have two sides and the angle contained by them respectively equal. 61. To construct a triangle, having given an angle, an adjacent side, and the opposite side. Mark off the given angle A, and on one of its legs, the adjacent side AB. The point will be determined by a circle with centre B and radius BC; different cases can now occur: a) If BC is less than the perpendicular BD, the circle will not cut AD, and the problem is impossible. b) If BC = BD, the circle will touch AD at D, and there will be one solution, namely, the right-angled triangle ADB. c) If BC > BD but < BA, the circle will cut. AD in two points, so that the problem has two solutions. d) If BC BA, the one point of intersection falls in A, so that the one solution gives no triangle. e) If BC > BA, the one point of intersection falls on the opposite side of A, and the corresponding triangle does not contain the given angle A, but its adjacent angle; there is therefore in this case only one solution. From this it follows, that two triangles, having one angle, an adjacent side, and the opposite side respectively equal, are congruent, if the opposite side be greater than or equal to the adjacent side, but that, if this is not known, it is not certain that the triangles are congruent. 62. To construct a triangle, having given a side and the angles at the side. Mark off the given side and on it the given angles, and produce their legs till they intersect. The problem is always possible, when the sum of the given angles is less than 2R. As there only is one solution, two triangles are congruent, when they have one side and the angles at that side respectively equal. Y x A 63. To construct a triangle, having given a side, an angle at the side, and the opposite angle. Mark off the given side AB and on it the given A; thereupon make LC; then Zy equals ▲ B, as the sum of the angles is 2 R. draw BC parallel to AD. Lx Then The conditions of possibility for the construction of this is the same as for the preceding one. As there only is one solution, two triangles are congruent, when they have one side, an angle at the side, and the opposite angle equal respectively. An isosceles triangle can be constructed, having given the base and the vertical angle. Draw any isosceles triangle with the given vertical angle D, and on the base AB of this triangle mark off AC equal to the given base; thereupon through C draw a parallel to BD. 64. To describe a circle, touching three given straight lines. If O is the centre, Oc and Ob radii to the points of contact, then▲ AbO ▲ AcO (61, e; for we have Ob Oc; AO AO; Lb R. and AO > Ob), so that the centre must lie in the line bisecting A. Similarly the centre must lie = |