PROBLEM IV. At a point in a given line, to make an angle equal to a given angle. Let A be the given point, AB the given line, and IKL, the given angle. K L I A B From the vertex K, as a centre, with any radius, describe the arc IL, terminating in the two sides of the angle. From the point A as a centre, with a distance AB, equal to KI, describe the indefinite arc BO; then take a radius equal to the chord LI, with which, from the point B as a centre, describe an arc cutting the indefinite arc BO, in D; draw AD; and the angle BAD will be equal to the given angle K. For, the two arcs BD, LI, have equal radii, and equal chords; hence, they are equal (P. 4); therefore, the angles BAD, IKL, measured by them, are also equal (P. 15). PROBLEM V. To bisect a given arc, or a given angle. First. Let it be required to divide the arc AEB into two equal parts. From the points A and B, as centres, with equal radii, describe two arcs cutting each other in D; through the point D and the centre C, draw CD: it will bisect the arc AB in the point E. For, the two points C and D are each equally distant from the extremities A and B of the chord AB; hence, the line CD bisects the chord at right angles (B. I., P. 16, c); and consequently, it bisects the arc AEB in the point E (p. 6). B E X D Secondly. Let it be required to divide the angle ACB into two equal parts. We begin by describing, from the vertex C, as a centre, the arc AEB; which is then bisect ed as above. It is plain that the line CD will divide the angle ACB into two equal parts (P. 17, s. 1). Scholium. By the same construction, each of the halves. AE, EB, may be divided into two equal parts; and thus, by successive subdivisions, a given angle, or a given arc, may be divided into four equal parts, into eight, into sixteen, and so on. PROBLEM VI. Through a given point, to draw a parallel to a given straight line. Let A be the given point, and BC the given line. F B E From the point A as a centre, with a radius greater than the shortest distance from A to BC, describe the indefinite arc EO; from the point E as a centre, with the same radius, describe the arc AF; lay off ED: AF, and draw AD: this will be the parallel required. A DI For, drawing AE, the angles AEF, EAD, are equal (P. 15); therefore, the lines AD, EF, are parallel (B. I., P. 19, c. 1). PROBLEM VII. Two angles of a triangle being given, to find the third. Let A and B be the given angles. D II E F at any point as E, make the angle DEC equal to the angle A, and the angle CEH equal to the other angle B: the remaining angle HEF will be the third angle required; because, these three angles are together equal to two right angles (B. I., P. 1, c. 3), and so are the three angles of a triangle (B. I., P. 25); consequently, HEF is equal to the third angle of the triangle. PROBLEM VIII. Two sides of a triangle, and the angle which they contain, being given, to describe the triangle. Let the lines B and C be equal to the given sides, and A the given angle. Having drawn the indefinite line DF, make at the point D, the angle FDE equal to the given angle A; then take DG=B, DH= C, and draw GH: DGH will be the triangle required (B. I., P. 5). PROBLEM IX. TB TC E Ꮐ A D H F ▲ side and two angles of a triangle being given, to describe the triangle. The two angles will either be both adjacent to the given side, or one will be adjacent, and the other opposite: in the latter case find the third angle (PROB. 7), D F H E and the two adjacent angles will be known. Then draw the straight line DE, and make it equal to the given side: at the point D, make an angle EDF, equal to one of the adjacent angles, and at E, an angle DEG equal to the other; the two lines DF, EG, will intersect each other in H; and DEH will be the triangle required (B. I., P. 6). PROBLEM X. The three sides of a triangle being given, to describe the triangle. Let A, B, and C, denote the three given sides. to the side A; from the point D D AH BH CH F E describe another arc intersecting the former in F; draw DF, EF; and DEF will be the triangle required (B. I., P. 10). Scholium. If one of the sides were greater than the sum of the other two, the arcs would not intersect each other, for no such triangle could exist (B. I., P. 7): but the solution will always be possible, when the sum of any two of the lines, is greater than the third. PROBLEM. XI. Two sides of a triangle, and the angle opposite one of them, being given, to describe the triangle. Let A and B be the given sides, and C the given angle. There are two cases. First. When the angle C is a right angle, or when it is obtuse. Draw DF and make the angle FDE=C; take DE=A: from the point E as a centre, with a radius equal to the given side B, describe an arc cutting DF in F; draw EF; then DEF will be the triangle required. In this case, the side B must E C A B D F be greater than A; for the angle C being a right angle, or an obtuse angle, is the greatest angle of the triangle (B. I., r. 25, c. 3), and the side opposite to it must, therefore, also be the greatest (B. I., P. 13). Secondly. If the angle C is acute, and B greater than A, the same construction will again apply, and DEF will be the triangle required. But if the angle C is acute, and the side B less than A, then the arc described from the centre E, with the radius EF-B, will Br E F D A C E cut the side DF in two points D F and G, lying on the same side of D: hence, there will be two G triangles DEF, DEG, either of which will satisfy all the conditions of the problem. Scholium. If the arc described with E as a centre, should be tangent to the line DG, the triangle would be right angled, and there would be but one solution. The problem will be impossible in all cases, when the side B is less than the perpendicular let fall from E on the line DF. PROBLEM XII. The adjacent sides of a parallelogram and their included angle being given, to describe the parallelogram. F Let A and B be the given sides, and C the given angle. Draw the line DH, and lay off DE equal to A: at the point D, make the angle EDF=C; take DF=B; describe two arcs, the one from Fas a centre, with a radius FG-DE, the other from E as a centre, with a radius EG D H E A+ B =DF; to the point G, where these arcs intersect each other, draw FG, EG; DEGF will be the parallelogram required. For, the opposite sides are equal, by construction; hence, the figure is a parallelogram (B. I., P. 29); and it is formed with the given sides and the given angle. Cor. If the given angle is a right angle, the figure will be a rectangle; if, in addition to this, the sides are equal. it will be a square. PROBLEM XIII. To find the centre of a given circle or arc. Take three points, A, B, C anywhere in the circumference, or in the arc; draw AB, BC, or suppose them to be drawn; bisect these two lines by the perpendiculars DE, FG (PROB. 1): the point O, where these perpendiculars meet, will be the centre sought (P. 6, S). E 犬 B A D |