PROPOSITION XXVIII. THEOREM. In equal circles, the arcs, which are subtended by equal chords, must be equal, the greater to the greater, and the less to the less. B K D H Let ABC, DEF be equal circles, and BC, EF equal chords, subtending the major arcs BAC, EDF, and the minor arcs BGC, EHF. Then must arc BAC-arc EDF, and arc BGC=arc EHF. Hence, if ABC be applied to © DEF, so that K coincides with L, and KB falls on LE, then and KC=LF, .. C will coincide with F. Then B coincides with E, and C with F, I. C. .. arc BAC will coincide with and be equal to arc EDF, and arc BGC...... EHF. Q.E.D. Ex. 1. If, in a circle ABCD, the arc AB be equal to arc DC, AD must be parallel to BC. Ex. 2. If a straight line, drawn from A the middle point of an arc BC, touch the circle, shew that it is parallel to the chord BC. Ex. 3. If two chords of a circle intersect at right angles, the portions of the circumference taken alternately are together equal to half the circumference. PROPOSITION XXIX. THEOREM. In equal circles, the chords, which subtend equal arcs, Let ABC, DEF be equal circles, and let BC, EF be chords, subtending the equal arcs BGC, EHF. Then must chord BC = chord EF. Then, if ABC be applied to © DEF, so that K coincides with Z, and B with E, and arc BGC falls on arc EHF, arc BGC=arc EHF, .. C will coincide with F. Then B coincides with E and C with F, .. chord BC must coincide with and be equal to chord EF. Q. E. D. Ex. 1. The two straight lines in a circle, which join the extremities of two parallel chords, are equal to one another. Ex. 2. If three equal chords of a circle cut one another in the same point, within the circle, that point is the centre. NOTE IV. On the Symmetrical properties of the Circle with regard to its diameter. The brief remarks on Symmetry in pp. 107, 108 of Part I. may now be extended in the following way: A figure is said to be symmetrical with regard to a line, when every perpendicular to the line meets the figure at points, which are equidistant from the line. Hence a Circle is Symmetrical with regard to its Diameter, because the diameter bisects every chord, to which it is perpendicular. B Further, suppose AB to be a diameter of the circle ACBD, of which O is the centre, and CD to be a chord perpendicular to AB. Then, if lines be drawn as in the diagram, we know that AB bisects (1) The chord CD, (2) The arcs CAD and CBD, (3) The angles CAD, COD, CBD, and the reflex angle DOC. These Symmetrical relations should be carefully observed, because they are often suggestive of methods for the solution of problems. :: AD=CD, and DB is common, and ▲ ADB = ▲ CDB, But, in the same circle, the arcs, which subtend equal chords, are equal, the greater to the greater and the less to the less; and ⚫. BD, if produced, is a diameter, III. 28. .. each of the arcs BA, BC, is less than a semicircle, and.. arc BA = arc BC. Thus the arc ABC is bisected in B. Q. E. F. Ex. If, from any point in the diameter of a semicircle, there be drawn two straight lines to the circumference, one to the bisection of the circumference, and the other at right angles to the diameter, the squares on these two lines are together double of the square on the radius. PROPOSITION XXXI. THEOREM. In a circle, the angle in a semicircle is a right angle; and the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle. B Let ABC be a O, O its centre, and BC a diameter. Draw AC, dividing the into the segments ABC, ADC. Join BA, AD, DC, AO. Then must the in the semicircle BAC be a rt. L, and in segment ABC, greater than a semicircle, less than a rt. 4, and 4 in segment ADC, less than a semicircle, greater than a rt. L. First, BO=AO, :. ▲ BAO= L ABO; .. 4 COA = twice BAO; and :: CO=AO, . L CAO= LACO ; .. L BOA=twice 4 CAO; 1. 32. I. 32. .. sum of 48 COA, BOA=twice sum of 48 BAO, CAO, that is, two right angles twice BAC. = .. BAC is a right angle. Next,': ▲ BAC is a rt. 4, .. ABC is less than a rt. 2. Lastly, sum of 48 ABC, ADC = two rt. 48, and ABC is less than a rt. 4, . ADC is greater than a rt. 4. NOTE. For a simpler proof see page 210. I. 17. III. 22. Q. E. D. |