=arc KF - arc LF; that is, the intercepted arcs GH and KL are equal. Secondly. When one of the parallels is a secant and the other a tangent. Draw the radius EF to the point of contact, and it will be perpendicular to the tangent (T. V., C. I.), and consequently perpendicular to its parallel CD. Hence the arc GF = are KF (T. IV.). C F A B .D K E Thirdly. When both parallels are tangents. The line FG, joining the points of contact, will be a diameter (T. V., C. IV.), consequently the intercepted arcs will be semi-circumferences, and therefore equal. G A C F B E D G Cor. Conversely, if two secants intercept equal arcs of the same circumference they will be parallel, provided they do not intersect, within the circumference. If a secant and a tangent intercept equal arcs of the same circumference, they will be parallel. If two tangents intercept equal arcs of the same circumference they will be parallel. THEOREM VII. In the same circle, or in equal circles, if two arcs are equal, they will have equal chords, which will be equally distant from the centre. If the arcs are unequal and each less than the semi-circumference, the greater arc will have the greater chord, which will be nearer the centre. Two circles having equal radii, by superposition, may be made exactly to coincide. Hence we may confine our demonstration to the case of chords in the same circle. AH C F G B D K First. Suppose the arc ABCD. From the centre E draw EF and EG perpendicular to the chords AB and CD. Also through H, the middle point of the arc AC, draw the diameter HK. If we revolve the portion of the figure on the left of HK over upon the portion on the right, the arcs HA and HAB will coincide with their equal arcs HC and HCD, consequently the chords AB and CD will be equal, and they will have the same perpendicular; that is, EF will equal EG, and the chords will be equally distant from the centre. B F A H C N M E G L D K Secondly. Suppose the arc AB less than CD. Applying the portion of the figure on the left of HK upon the portion on the right, as before, we see that the point B will fall at some point L, between C and D, since the arc AB is less than CD. Hence, the middle of the arc CL, which is in the prolongation of the perpendicular EM, will be nearer C than the middle of the arc CD which is in the prolongation of the perpendicular EG; consequently the point G is between D and N, the point where the perpendicular EM intersects CD. Therefore CG, the half of CD, is greater than CN, but CN being an oblique line in reference to EM is greater than the perpendicular CM; consequently, CD > CL, or CD > AB. That is, the greater arc has the greater chord. Again, we have EG <EN, and EN <EM, consequently EG <EM, or EG < EF; that is, the chord corresponding to the greater arc is nearer the centre. Cor. Conversely, in the same circle, or in equal circles: 1. If the chords are equal, the arcs will be equal. And these chords will be equally distant from the centre. 2. The greater chord will correspond to the greater arc. And the greater chord will be nearer the centre. We may also add: 3. Chords equally distant from the centre are equal, and subtend equal arcs. 4. Of two chords unequally distant from the centre, the one nearer the centre is the greater, and it subtends the greater arc. Scholium. We must keep in mind that the arcs considered in this Theorem are in each case less than a semi-circum ference. OF THE MEASURE OF ANGLES. THEOREM VIII. In the same circle, or equal circles, equal angles at the centre correspond to equal arcs. Conversely, equal arcs correspond to equal angles. A E B C F D First. Suppose the angle at E to equal the angle at F. Apply the angle AEB upon the angle CFD; EA and EB will take, respectively, the directions of FC and FD; and since the radii are equal, the points A and B will coincide respectively with the points C and D. AB arc CD. Hence the arc Then, as be Secondly. Suppose the arcs AB and CD equal. fore, applying the figure ABE upon CDF, so that the radius AE may coincide with its equal radius CF, then will the arcs AB and CD coincide, since they are equal, and the point B will coincide with the point D; consequently the angle AEB is equal to the angle CFD. Cor. In the same circle, or equal circles, the greater angle at the centre corresponds to the greater arc, and, conversely, the greater arc corresponds to the greater angle. THEOREM IX. In the same circle, or in equal circles, angles at the centre are to each other as their included arcs. A E F B D C First. When the two arcs AB and CD are commensurablethat is, when they are to each other as two whole numbers, for instance, as 7 to 4. If we divide the arc AB into seven equal portions, and the arc CD into four, and draw radii to these points of division, we shall divide the angle AEB into seven angles and CFD into four, and these angles will be equal, since their arcs are (T. VIII.). Hence the angle AEB will be to the angle CFD as 7 to 4; that is, as the arc AB to the arc CD. position AEG. If now, these angles are not to each other as their arcs, suppose we have angle AEB: angle AEG:: arc AB : arc AH greater than arc AG. If we conceive the arc AB to be divided into equal portions, each of which shall be less than arc GH, there will be at least one point of division between G and H, as at K. Draw the radius EK, and then since the arcs AB and AK are commensurable, we have by the first part of this Theorem, angle AEB: angle AEK :: arc AB : arc AK. From these two proportions, since their antecedents are the same, we deduce angle AEG: angle AEK:: arc AH: arc AK, which cannot be, for the antecedent of the first couplet is less. than its consequent, while in the second couplet the antecedent is greater than its consequent. Hence the angle AEB cannot be to the angle AEG as the arc AB is to an arc greater than AG. And by a similar process we may show that the angle AEB cannot be to the angle AEG as the arc AB is to an arc less than AG. Consequently we must have in the same circle, or in equal circles, the angles at the centre to each other as their corresponding arcs. Scholium. Since, in the same circle, or in equal circles, the angles at the centre are to each other as their corresponding arcs, we may use the arcs as the measure of their corresponding angles. In the case of a right angle, its measuring arc is a quadrant; in the case of an angle equal to two right angles, the measuring arc is a semi-circumference. We sometimes regard the right angle as the unit angle, so that acute angles would all be less than 1, and obtuse angles would exceed 1 and be less than 2. THEOREM X. An inscribed angle is measured by half the arc included between its sides. First. When the centre of the circle is within the angle. A D B E. Draw the diameter CE and the radii DA and DB. Now, since the triangle DAC is isosceles, the exterior angle ADE is double the angle DCA. The angle ADE is measured by the arc AE (T. IX., S.); hence the angle ACE is measured by one half of the arc AE. In the same way we may show that the angle ECB is measured by one half of the arc EB; consequently the angle ACB is measured by one half its included arc AEB. Secondly. When the centre of the circle is without the angle. A D As before, draw the diameter CE, and we shall have the angle ACE, measured by one half of the arc ABE, also the angle BCE measured by half the arc BE; consequently, their difference, the angle ACB, is measured by half the difference of these arcs; that is, by half the included arc AB. B E Cor. I. Each angle inscribed in a semicircle is a right angle, since it is measured by half the semi-circumference, or by a quadrant. Each angle is a segment greater than a semicircle is acute, since half the included arc, which measures it, is less than a quadrant. Each angle in a segment less than a semicircle is obtuse, since half the included arc, which measures it, is greater than a quadrant. |