Scholium 2. All that has been demonstrated in the last three propositions, concerning the comparison of angles with ares, holds true equally, if applied to the comparison of sectors with arcs; for sectors are not only equal when their angles are so, but are in all respects proportional to their 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 those sectors. It is hence evident that the arcs of the circle, which serve as a measure of the different angles, are proportional to the different sectors, in the same circle, or in equal circles. PROPOSITION XVIII. THEOREM. An ribed angle is measured by half the arc included between its sides. Let BAD be an inscribed angle, and let us first suppose that the centre of the circle lies within the angle BAD. Draw the diameter AE, and the radii CB, CD. The angle BCE, being exterior to the triangle ABC, is equal to the sum of the two interior angles CAB, ABC (Book I. B Prop. XXV. Cor. 6.): but the triangle BAC D being isosceles, the angle CAB is equal to E ABC ; hence the angle BCE is double of BAC. Since BCE lies ; at the centre, it is measured by the arc BE; hence BAC will be measured by the half of BE. For a like reason, the angle CAD will be measured by the half of ED; hence BAC+CAD, or BAD will be measured by half of BE+ED, or of BED. Suppose, in the second place, that the centre C lies without the angle BAD. Then drawing the diameter AE, the angle BAE will be measured by the half of BE; the angle DAE by the half of DE: hence their C difference BAD will be measured by the half of BE minus the half of ED, or by the B half of BD. Hence every inscribed angle is measured by half of the arc included between its sides. Cor. 2. Every angle BAD, inscribed in a semicircle is a right angle ; because it is measured by half the semicircumference BOD, ВК that is, by the fourth part of the whole circumference. D Cor. 3. Every angle BAC, inscribed in a А segment greater than a semicircle, is an acute angle ; for it is measured by 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 is measured by half of the arc B BAC, greater than a semicircumference. Cor. 4. The opposite angles A and C, of an inscribed quadrilateral ABCD, are together equal to two right angles : for the an A gle BAD is measured by half the arc BCD, the angle BCD is measured by half the arc BAD; hence the two angles BAD, BCD, ta- D ken together, are measured by the half of the circumference ; hence their sum is equal to two right angles. PROPOSITION XIX. THEOREM. The angle formed by two chords, which intersect each other, is measured by half the sum of the arcs included between its sides. KE Let AB, CD, betwochords intersecting A C each other at E: then will the angle AEC, or DEB, be measured by half of AC+DB. Draw AF parallel to DC: then will the arc DF be equal to AC (Prop. X.); and the angle FAB equal to the angle DEB (Book I. Prop. XX. Cor. 3.). But I the angle FAB is measured by half the D B arc FDB (Prop. XVIII.); therefore, DEB is measured by half of FDB; that is, by half of DB+DF, or half of DB+ AC. In the same manner it might be proved that the angle AED is measured by half of AFD+BC. PROPOSITION XX. THEOREM. The angle formed by two secants, is measured by half the diffe rence of the arcs included between its sides. Let AB, AC, be two secants : then will the angle BAC be measured by half the difference of the arcs BEC and DF. Draw DE parallel to AC: then will the arc EC be equal to DF, and the angle BDE equal to the angle BAC. But BDE is measured by half the arc B BE; hence, BAC is also measured by half the arc BE ; that is, by half the difference of BEC and EC, or half the difference of BEC and DF. PROPOSITION XXI. THEOREM. The angle formed by a tangent and a chord, is measured by half of the arc included between its sides. 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 M 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 ] А E AMDC. 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 distan, from A and B. Find, in like manner, above or beneath the line AB, a second point E, equally distant from the C Bt 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. 8 PROBLEM II. At a given point, in a given straight line, to erect a perpendicum lar to this line, Let A be the given point, and BC the given line. 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 A ct 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 HA line. From the point A as a centre, and with a radius sufficiently great, describe an с B arc cutting the line BD in the two points B and D; then mark a point E, equally distant from the points B and I), and XE draw AE: it will be the perpendicular required. 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. Ai a point in a given line, to make an angle equal to a given angle. |