Let ABC, DEF be equal circles, and let arc BC = arc EF: it is required to prove that BGC ▲ EHF, and ▲ A = If BGC be not = EHF, one of them must be the greater. Let BGC be the greater, and make ▲ BGK Because the circles are equal, and ▲ BGK ... arc BK = arc EF But arc BC = .. arc BK = arc BC, which is impossible. = L EHF. COR.-In equal circles, or in the same circle, those sectors are equal which have equal arcs. 1. If AC and BD be two equal arcs in a circle ACDB, prove chord AB || chord CD. 2. In equal circles, or in the same circle, if two arcs be unequal, that angle, whether at the centre or at the Oce, is the greater which stands on the greater arc. 3. The angle in a semicircle is a right angle. 4. 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. 5. If the sum of two arcs of a circle be equal to the whole Oce, the angles at the Oce which stand on them are supplementary. 6. Prove the proposition by superposition. 7. Two circles touch each other internally, and a chord of the greater circle is a tangent to the less. Prove that the chord is divided at its point of contact into segments which subtend equal angles at the point of contact of the circles. In equal circles, or in the same circle, if two chords be equal, the arcs they cut off are equal, the major arc equal to the major arc, and the minor equal to the minor. Let ABC, DEF be equal circles, and let chord BC chord EF: = it is required to prove that major arc BAC= major arc EDF, and minor arc BGC = minor arc EHF. Find K and L the centres of the circles, III. 1 = III. Def. 1, Cor. 4 But Oce ABC Oce DEF; .. remaining arc BAC remaining arc EDF = 1. If AC and BD be two equal chords in a circle ACDB, prove chord AB || chord CD. 2. Hence devise a method of drawing through a given point a straight line parallel to a given straight line. 3. If two equal circles cut one another, any straight line drawn through one of the points of intersection will meet the circles again in two points which are equidistant from the other point of intersection. In equal circles, or in the same circle, if two arcs be equal, the chords which cut them off are equal. Let ABC, DEF be equal circles, and let arc BGC = arc EHF: it is required to prove that chord BC Find K and L the centres of the circles, and join BK, KC, EL, LF. = chord EF. III. 1 Because the circles are equal, Hyp. .. their radii are equal. III. Def. 1, Cor. 4 And because the circles are equal, and arc BGC = .. L K = L L. = arc EHF, III. 27 1. If AC and BD be two equal arcs in a circle ACDB, prove chord AD = chord BC. 2. Prove the proposition by superposition. Draw the chord AB, and bisect it at C; from C draw CD 1 AB, and meeting the arc at D. D is the point of bisection. Join AD, BD. In As ACD, BCD, ... AD = BD. I. 10 I. 11 ། I. 4 But in the same circle equal chords cut off equal arcs, the and AD and BD are both minor arcs, since DC if produced would be a diameter; ... arc AD = arc BD. III. 1, Cor. 1 1. If two circles cut one another, the straight line joining their centres, being produced, bisects all the four arcs. 2. A diameter of a circle bisects the arcs cut off by all the chords to which it is perpendicular. 3. Bisect the arc ADB without joining AB. 4. Prove ▲ DAB greater than any other triangle on the same base AB, and having its vertex on the arc ADB. PROPOSITION 31. THEOREM. An angle in a semicircle is a right angle; an angle in a segment greater than a semicircle is less than a right angle; and an angle in a segment less than a semicircle is greater than a right angle. B Let ABC be a circle, of which E is the centre and BC a diameter; and let any chord AC be drawn dividing the circle into the segment ABC which is greater than a semicircle, and the segment ADC which is less than a semicircle : it is required to prove (1) ▲ in semicircle BAC = a rt. L ; (2) ▲ in segment ABC less than a rt. L; Join AB; take any point D in arc ADC, and join AD, CD. ce (1) Because an angle at the Oe of a circle is half of the angle at the centre which stands on the same arc; III. 20 .. BAC = half of the straight ▲ BEC, (2) Because and .. III. Def. 21 BAC + ▲ B is less than two rt. ▲s, I. 17 BAC = a rt. ; B is less than a rt. L. |