Scholium 4. 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 heir 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. Hence, it is evident that the arcs of equal circles, which serve as a measure of corresponding angles, are proportional to their sectors. PROPOSITION XVIII. THEOREM. Any inscribed angle is measured by half the arc included between its sides. Let BAD be an inscribed angle, and let us first sup pose the centre of the circle to lie within the angle BAD. Draw the diameter ACE, and the radii CB, CD. B A The angle BCE, being exterior to the triangle ABC, is equal to the sum of the two interior angles CAB, ABC (B. I., P. 25, c. 6): but the triangle BAC being isosceles, the angle CAB is equal to ABC; hence, the angle BCE is double BAC Since BCE is at the centre, it is measured by the arc BE (P. 17, s. 1); 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 half of BED. Secondly. Suppose the centre C to lie without the angle BAD. Then, draw- B E DE Cor. 2. Every angle BAD, inscribed in a semicircle is a right angle; because it is measured by half the semicircumference BOD, that is, by the fourth part of the whole circumference (P. 17, s. 2). Cor. 3. Every angle BAC, inscribed in a segment greater than a semicircle, is an acute angle; for it is measured by half the arc BOC, less than a semicircumference (P. 17, s. 2). And every angle BOC, inscribed in a segment less than a semicircle, is A B B B A an obtuse angle; for it is measured by half the arc 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 angle BAD is measured by half the arc BCD, the angle BCD is D measured by half the arc BAD; hence, B the two angles BAD, BCD, taken together, are measured by half the circumference; hence, their sum is equal to two right angles (P. 17, s. 2). 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. Let AB, CD, be two chords intersecting each other at E: then will the angle AEC, or DEB, be measured by half of AC+DB. Draw AF parallel to DC: the arc DF will be equal to AC (P. 10), and the angle FAB equal to the angle DEB (B. I., P. 20, c. 3). But the angle FAB is measured by half the arc FDB (P. 18); therefore, DEB is measured by half of FDB; that is, by half of DB+DF, or half of DB+AC. E B To prove the same for the angle DEA, or its equas BEC. Draw the chord AC. Then, the angle DCA will be measured by half the arc DFA; and the angle BAC by half the arc CB (P. 18). But the outward angle AED, of the triangle EAC, is equal to the sum of the angles A and C (B. I., P. 25, c. 6); hence, this angle is measured by one-half of BC plus one-half of AFD; that is, by half the sum of the intercepted arcs. By drawing a chord BC, similar reasoning would apply to the angle AEC or DEB. PROPOSITION XX. THEOREM. The angle formed by two secants, is measured by half the difference 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: the arc EC will be equal to DF (P. 10), and the angle BDE equal to the angle BAC (B. I., P. 20, c. 3). But BDE is measured by half the arc BE (P. 18); hence, BAC is also measured by half the arc BE; that is, by half the difference of BEC and EC, and consequently, by half the difference of BEC and DF B A D E C Any angle formed by a tangent and a chord passing through the point of contact, is measured by half the arc included between its sides. M Let BE be a tangent, and AC a chord. From A, the point of contact, draw the diameter AD. The angle BAD is a right angle (P. 9), and is measured by half the semicircumference AMD (P. 17, s. 2); the angle DAC is measured by the half of DC; hence, BAD+DAC, or BAC, is measured by the half of AMD plus B the half of DC, or by half the whole arc AMDC. It may be shown, by taking the difference of 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 bisect a given straight line. 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, B and the point C, where this line meets AB, will be equally distant from A and B. For, the two points D and E, being each equally dislant from the extremities A and B, must both lie in the perpendicular raised at the middle point of AB (B. I., P. 16, C). But only one straight line can be drawn through two given points (A. 11); hence, the line DE must itself be that perpendicular, which divides AB into two equal parts. PROBLEM II. At a given point, in a given straight line, to erect a perpendic ular to that line. Let BC be the given line, and A the given point. 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 at D; draw AD and it will be the perpendicular required. B A For, the point D, being equally distant from B and C, must be in the perpendicular raised at the middle of BC (B. I., P. 16); 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 that line. Let A be the point, and BD the given straight line. From the point A as a centre, and with a radius sufficiently great, describe an arc cutting the line BD in 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. +A C E 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 (B. I., P. 16, c). |