Let BE be the tangent, and AC the chord. From A, the point of contact, draw the diameter AD. The angle BAD is a right angle (Prop. IX.), and is measured by half the semicircumference AMD; the angle DAC is measured by the half of DC: hence, BAD+DAC, or BAC, is measured by the half of AMD plus the half of DC, or by half the whole arc AMDC. M B It might be shown, by taking the difference between the angles DAE, DAC, that the angle CAE is measured by half the arc AC, included between its sides. PROBLEMS RELATING TO THE FIRST AND THIRD BOOKS. PROBLEM I. To divide a given straight line into two equal parts. Let AB be the given straight line. From the points A and B as centres, with a radius greater than the half of AB, describe two arcs cutting each other in D; the point D will be equally distant from A and B. Find, in like manner, above or beneath the line AB, a second point E, equally distant from the points A and B; through the two points D and E, draw the line DE: it will bisect the line AB in C. For, the two points D and E, being each equally distant from the extremities A and B, must both lie in the perpendicular raised from the middle of AB (Book I. Prop. XVI. Cor.). But only one straight line can pass through two given points; hence the line DE must itself be that perpendicular, which divides AB into two equal parts at the point C. PROBLEM II. At a given point, in a given straight line, to erect a perpendicu lar to this line. Let A be the given point, and BC the given line. B 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 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 perpen dicular 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 centre, with any radius, describe the arc IL, terminating in the two sides of the angle. From the K point A as a centre, with a dis 44 ΙΑ B 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. 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 E 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 E H F together equal to two right angles (Book I. Prop. I and cause those three angles are XXV). 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 DE, at the point D, make the angle TB TC E G 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 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. Let A, B, and C, be the sides. Draw DE equal to the side A; from the point Ê as a centre, with a radius equal to the second side B, describe an arc; from D as a centre, with a radius equal to the third side C, describe another arc intersecting the former in F; draw DF, EF; and DEF will be the triangle required (Book I. Prop. X.). D AH B C T Scholium. If one of the sides were greater han 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. |