PROBLEM II. At a given point, in a given straight line, to erect a perpendicular to this line. Let A be the given point, and BC the given line. A Take the points B and C at equal distances from A; then from the points B and C as centres, with a radius greater than B BA, describe two arcs intersecting each other in D; draw AD: it will be the perpendicular required. For, the point D, being equally distant from B and C, must be in the perpendicular raised from the middle of BC (Book I. Prop. XVI.); and since two points determine a line, AD is that perpendicular. Scholium. The same construction serves for making a right angle BAD, at a given point A, on a given straight line BC. PROBLEM III. From a given point, without a straight line, to let fall a perpendicular on this line. Let A be the point, and BD the straight line. From the point A as a centre, and with a radius sufficiently great, describe an arc cutting the line BD in the two points B and D; then mark a point E, equally distant from the points B and D, and draw AE: it will be the perpendicular required. C For, the two points A and E are each equally distant from the points B and D; hence the line AE is a perpendicular passing through the middle of BD (Book I. Prop. XVI. Cor.). 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. From the vertex K, as a cen tre, with any radius, describe the tance 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 DAB will be equal to the given angle K. For, the two arcs BD, LI, have equal radii, and equal chords; hence they are equal (Prop. IV.); therefore the angles BAD, IKL, measured by them, are equal. PROBLEM V. To divide a given arc, or a given angle, into two equal parts. First. Let it be required to divide the arc AEB into two equal parts. From the points A and B, as centres, with the same radius, 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 bi sects the chord at right angles (Book I. Prop. XVI. Cor.); hence, it bisects the arc AB in the point E (Prop. VI.). 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 bisected as above. It is plain that the line CD will divide the angle ACB into two equal parts. 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. B A E DI 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; make ED AF, and draw AD: this will be the parallel required. For, drawing AE, the alternate angles AEF, EAD, are evidently equal; therefore, the lines AD, EF, are parallel (Book I. Prop. XIX. Cor. 1.). PROBLEM VII. Two angles of a triangle being given, to find the third. Draw the indefinite line DEF; at any point as E, make the angle DEC equal to one of the given angles, and the angle CEH equal to the other: the remaining angle HEF will be the third angle required; be- D cause those three angles are XXV). together equal to two right angles (Book I. Prop. I. and PROBLEM VII1. 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 TB TC the given sides, and A the given an gle. E G Having drawn the indefinite line DE, at the point D, make the angle EDF equal to the given angle A; then take DG=B, DH=C, and draw GH; DGH will be the triangle required (Book I. Prop. V.). PROBLEM IX. A 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 the one adjacent, and the other opposite: in the latter case, find the third angle (Prob. VII.); and the two adjacent angles will thus be known: draw the straight line D G I H DE 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 cut each other in H; and DEH will be the triangle required (Book I. Prop. VI.). PROBLEM X. The three sides of a triangle being given, to describe the triangle. Scholium. If one of the sides were greater than the sum of the other two, the arcs would not intersect each other: but the solution will always be possible, when the sum of two sides, any how taken, is greater than the third. F 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, make the angle EDF=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 first case, the side B must be greater than A; for the angle C, being a right angle, or an obtuse an- D gle, is the greatest angle of the tri B A C angle, and the side opposite to it must, therefore, also be the greatest (Book I. Prop. XIII.). 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 would be impossible in all cases, if the side B were less than the perpendicular let fall from E on the line DF. |