therefore it is exactly AD, and we have the proportion angle ACB angle ACD:: arc AB arc AD. 123. Corollary. Since the angle at the centre of a circle and the arc intercepted between its sides have such a connexion, that when one increases or diminishes in any ratio whatever, the other increases or diminishes in the same ratio, we are authorized to establish one of these magnitudes as the measure of the other; thus we shall, in future, take the arc AB as the measure of the angle ACB. The only thing to be observed in the comparison of angles with each other is, that the arcs, which are used to measure them, must be described with equal radii. This is to be understood in the preceding propositions. 124. Scholium. It may seem more natural to measure a quantity by another quantity of the same kind, and upon this principle it would be convenient to refer all angles to the right angle; and thus, the right angle being the unit of measure, the acute angle would be expressed, by a number comprehended between O and 1, and an obtuse angle by a number between 1 and 2. But this manner of expressing angles would not be the most convenient in practice. It has been found much more simple to measure them by arcs of a circle on account of the facility of making arcs equal to given arcs and for many other reasons. Besides, if the measure of angles by the arcs of a circle be in some degree indirect, it is not the less easy to obtain, by means of them, the direct and absolute measure; for, if we compare the arc, which is used as the measure of an angle, with the fourth part of the circumference, we have the ratio of the given angle to a right angle, which is the absolute measure. 125. Scholium 11. All that has been demonstrated in the three preceding propositions, for the comparison of angles with arcs, is equally applicable to the purpose of comparing sectors with arcs; for sectors are equal, when their arcs are equal, and in general they are proportional to the angles; hence two sectors ACB, ACD, taken in the same circle or in equal circles, are to each other, as the arcs AB, AD, the bases of these sectors. It will be perceived therefore, that the arcs of a circle, which are used as a measure of angles, will also serve as the measure of different sectors of the same circle or of equal circles. THEOREM. 65. 126. The inscribed angle BAD (fig. 64, 65), has for its measure Fig. 64, the half of the arc BD comprehended between its sides. Demonstration. Let us suppose, in the first place, that the centre of the circle is situated in the angle BAD (fig. 64); we Fig. 64. draw the diameter AE, and the radii CB, CD. The angle BCE, being the exterior angle of the triangle ABC, is equal to the sum of the two opposite interior angles, CAB, ABC. But, the triangle BAC being isosceles, the angle CAB = ABC; hence the angle BCE is double of BAC. The angle BCE, having its vertex at the centre, has for its measure the arc BE; therefore the angle BAC has for its measure the half of BE. For a similar reason the angle CAD has for its measure the half of ED; therefore BAC + CAD, or BAD, has for its measure the half BE + ED, or the half of BD. Let us suppose, in the second place, that the centre C (fig. 65), Fig. 65. is situated without the angle BAD; then, the diameter AE being drawn, the angle BAE will have for its measure the half of BE, and the angle DAE the half of DE; hence their difference BAD will have for its measure the half of BE minus the half of ED, or the half of BD. Therefore every inscribed angle has for its measure the half of the arc comprehended between its sides. 127. Corollary 1. All the angles BAC, BDC (fig. 66), &c., Fig. 66. inscribed in the same segment, are equal; for they have each for their measure the half of the same arc BOC. 128. Corollary 11. Every angle BAD (fig. 67), inscribed in Fig. 67. a semicircle, is a right angle; for it has for its measure the half of the semicircumference BOD, or the fourth of the circumfer ence. To demonstrate the same thing in another way, draw the radius AC; the triangle BAC is isosceles, and the angle BAC=ABC; the triangle CAD is also isosceles, and the angle CAD=ADC; hence BAC+ CAD, or BAD = ABD+ADB. But, if the two angles B and D of the triangle ABD are together equal to the third BAD, the three angles of the triangle will be equal to twice the angle BAD; they are also equal to two right angles; there fore the angle BAD is a right angle. Fig. 66. Fig. 68. Fig. 69. 129. Corollary 1. Every angle BAC (fig. 66), inscribed in a segment greater than a semicircle, is an acute angle; for it has for its measure the half of the arc BOC less than a semicircumference. And every angle BOC, inscribed in a segment less than a semicircle, is an obtuse angle; for it has for its measure the half of the arc BAC greater than a semicircumference. 130. Corollary IV. The opposite angles A and C (fig. 68) of an inscribed quadrilateral ABCD are together equal to two right angles; for the angle BAD has for its measure the half of the arc BCD, and the angle BCD has for its measure the half of the arc BAD; hence the two angles BAD, BCD, taken together, have for their measure the half of the circumference; therefore their sum is equal to two right angles. THEOREM. 131. The angle BAC (fig. 69), formed by a tangent and a chord, has for its measure the half of the arc AMDC, comprehended between its sides. Demonstration. At the point of contact A draw the diameter AD; the angle BAD is a right angle (110), and has for its measure the half of the semicircumference AMD; the angle DAC has for its measure the half of DC; therefore BAD + DAC, or BAC, has for its measure the half of AMD plus the half of DC, or the half of the whole arc AMDC. It may be demonstrated, in like manner, that CAE has for its measure the half of the AC, comprehended between its sides. Fig. 70. Problems relating to the two first Sections. PROBLEM. 132. To divide a given straight line AB (fig. 70) into two equal parts. Solution. From the points A and B, as centres, and 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 the points A and B; find in like manner, either above or below 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; this line will divide the line AB into two equal parts in the point C. For, the two points D and E being each equally distant, from the extremities A and B, they must both be in the perpendicular which passes through the middle of AB. But through two given points only one straight line can be drawn; therefore the line DE will be this perpendicular, which divides the line AB into two equal parts in the point C. PROBLEM. 133. From a given point A (fig. 71), in the line BC, to erect a Fig. 71. perpendicular to this line. Solution. Take the points B and C, at equal distances from A; and from B and C, as centres, with a radius greater than BA, describe two arcs cutting each other in D; draw AD, which will be the perpendicular required. For the point D, being equally distant from B and C, must be in a perpendicular to the middle of BC (55); therefore AD is this perpendicular. 134. Scholium. The same construction will serve to make a right angle BAD at a given point A in a given line BC. PROBLEM. 135. From a given point A (fig. 72) without the straight line BD, Fig. 72. to let fall a perpendicular upon this line, Solution. From A, as a centre, with a radius sufficiently great, describe an arc cutting the line BD in two points B and D ; then find a point E, equally distant from the points B and D (132), and draw AE, which will be the perpendicular required. For the two points A and E are each equally distant from the points B and D; therefore the line AE is perpendicular to the middle of BD. PROBLEM. 136. At a given point A (fig. 73) in the line AB, to make an angle Fig. 73. equal to a given angle K. Solution. From the vertex K, as a centre, with any radius, describe an arc IL meeting the sides of the angle, and from the point A, as a centre, with the same radius, describe an indefinite Fig. 74. Fig. 75. Fig. 76. arc BO; from B, as a centrs, with a radius equal to the chord LI, describe an arc cutting the arc BO in D; draw AD, and the angle DAB will be equal to the given angle K. For the arcs BD, LI, have equal radii and equal chords; they are therefore equal (102), and the angle BAD = IKL. PROBLEM. 137. To bisect a given arc or angle. Solution 1. If it is proposed to bisect the arc AB (fig. 74); from the points A and B, as centres, with the same radius, describe two arcs intersecting each other in D; through the point D and the centre C draw CD, which will divide the arc AB into two equal parts in the point E. For, since the points C and D are each equally distant from the extremities A and B of the chord AB, the line CD is perpendicular to the middle of this chord; therefore it bisects the arc ᎯᏴ (105). 2. If it is proposed to bisect the angle ACB; from the vertex C, as a centre, describe the arc AB, and complete the construction, as above described. It is evident that the line CD will bisect the angle ACB. 138. Scholium. By the same construction, we may bisect each of the halves AE, EB, and thus, by successive subdivisions, we may divide an angle or arc into four, eight, sixteen, &c., equal parts. PROBLEM. 139. Through a given point A (fig. 75), to draw a straight line parallel to a given straight line BC. Solution. From the point A, as a centre, with a radius sufficiently great, describe the indefinite arc EO; from the point E, as a centre, with the same radius, describe the arc AF; take ED=AF, and draw AD, which will be the parallel required. For, AE being joined, the alternate angles AEF, EAD, are equal; therefore AD, EF, are parallel (67). PROBLEM. 140. Two angles A and B (fig. 76) of a triangle being given, to find the third. |