2o. Let the arcs AMD and ENG be equal: then will the angles ACD and EOG be equal. For, if the arcs AMD and ENG are equal, the chords AD and EG are equal (P. IV.); consequently, the triangles ACD and EOG have their sides equal, each to each; they are, therefore, M G equal in all their parts: hence, the angle ACD is equal to the angle EOG; which was to be proved. In equal circles, commensurable angles at the centre are proportional to their intercepted arcs. In the equal circles, whose centres are C and 0, let the angles ACB and DOE be commensurable; that is, let them have a common unit: then will they be proportional to the intercepted arcs AB and DE.. Let the angle M be a common unit; and suppose, for example, that this unit is contained 7 times in the angle ACB, and 4 times in the angle DOE ACB be divided into 7 angles, by the radii &c.; and DOE into 4 angles, by the radii Oz, each equal to the unit M. Then, suppose Ox, Oy, and 1 et From the last proposition, the arcs Am, mn, &c., Dx, xy, &c., are equal to each other; and because there are 7 of these arcs in AB, and 4 in DE, we shall have, arc AB : arc DE :: 7 : 4. But, by hypothesis, we have, angle ACB : angle DOE :: hence, from (B. II., P. IV.), we have, 7: 4; angle ACB : angle DOE :: arc AB : arc DE If any other numbers than 7 and 4 had been used, the same proportion would have been found; which was to be proved. Cor. If the intercepted arcs are commensurable, they will be proportional to the corresponding angles at the centre, as may be shown by changing the order of the couplets in the above proportion. 眼 e e 1 For, let the less angle FOH, be placed upon the greater angle ACB, SO that it shall take the position ACD. angle ACB : angle ACD :: arc AB : arc AO. Conceive the arc AB to be divided into equal parts; each less than DO: there will be at least one point of division between D and 0; let I be that point; and draw CI. Then the arcs AB, AI, will be commensurable, and we shall have (P. XVI.), angle ACB : angle ACI :: arc AB : arc AI Comparing the two proportions, we see that the antecedents are the same in both: hence, the consequents are proportional (B. II., P. IV., C.); hence, angle ACD : angle ACI :: arc 40 arc AI But, AO is greater than AI: hence, if this proportion is true, the angle ACD must be greater than the angle ACI. On the contrary, it is less: hence, the fourth term of the proportion cannot be greater than AD. In a similar manner, it may be shown that the fourth term cannot be less than AD: hence, it must be equal to AD; therefore, we have, angle ACB : angle ACD :: arc AB arc AD; which was to be proved. Cor. 1. The intercepted arcs are proportional to the cor responding angles at the centre, as may be shown by changing the order of the couplets in the preceding proportion. Cor. 2. In equal circles, angles at the centre are proportional to their intercepted arcs; and the reverse, whether they are commensurable or incommensurable. Cor 3. In equal circles, sectors are proportional to their angles, and also to their arcs. Scholium. Since the intercepted arcs are proportional to the corresponding angles at the centre, the arcs may be taken as the measures of the angles. That is, if a circumference be described from the vertex of any angle, as a centre, and with a fixed radius, the arc intercepted between the sides of the angle may be taken as the measure of the angle. In Geometry, the right angle which is measured by a quarter of a circumference, or a quadrant, is taken as a unit. If, therefore, any angle be measured by one-half or two-thirds of a quadrant, it will be equal to one-half or two-thirds of a right angle. An inscribed angle is measured by half of the arc included between its sides. There may be three cases: the centre of the circle may lie on one of the sides of the angle; it may lie within the angle; or, it may lie without the angle. 1o. Let EAD be an inscribed angle, one of whose sides AE passes through the centre: then will it be measured by half of the arc DE. E For, draw the radius CD. The external angle DCE, of the triangle DCA, is equal to the sum of the opposite interior angles CAD and CDA (B. I., P. XXV., C. 6). But, the triangle DCA being isosceles, the angles D and A are equal; therefore, the angle DCE is double the angle DAE. Because DCE is at the centre, it is measured by the arc DE (P. XVII., S.): hence, the, angle DAE is measured by half of the arc DE; which was to be proved. B E 2o. Let DAB be an inscribed angle, and let the centre lie within it: then will the angle be measured by half of the arc BED. For, draw the diameter AE. Then, from what has just been proved, the angle DAE is measured by half of DE, and the angle EAB by half of EB: hence, BAD, which is the sum of EAB and DAE, is measured by half of the sum of DE and EB, or by half of BED; which was to be proved. 3o. Let BAD be an inscribed angle, and let the centre lie without it: then will it be measured by half of the aro arc BD. For, draw the diameter AE. Then, from what precedes, the angle DAE is measured by half of DE, and the angle BAE by half of BE: hence, BAD, which is the difference of BAE A B and DAE, is measured by half of the D difference of BE and DE, or by half of the arc BD; which was to be proved. |