Let ADB be a circle of which AB is a diameter and O the centre; and let C be any point on the semi-circumference ACB. It is required to prove that the LACB is a rt. angle. 1st Proof. The ACB at the O is half the straight angle AOB at the centre, standing on the same arc ADB; and a straight angle = two rt. angles: .. the ACB is a rt. angle. Join OC. Q.E.D. 2nd Proof. ... the whole ▲ ACB = the OAC + the OBC. = But the three angles of the ▲ ACB together two rt. angles; ... the ACB= one-half of two rt. angles COROLLARY. The angle in a segment greater than a semi-circle is acute; and the angle in a segment less than a semi-circle is obtuse. The ACB at the Oe is half the LAOB at the centre, on the same arc ADB. (i) If the segment ACB is greater than a semi-circle, then ADB is a minor arc; (ii) If the segment ACB is less than a semi-circle, then ADB is a major arc; .. the AOB is greater than two rt. angles; EXERCISES ON THEOREM 41. 1. A circle described on the hypotenuse of a right-angled triangle as diameter, passes through the opposite angular point. 2. Two circles intersect at A and B ; and through A two diameters AP, AQ are drawn, one in each circle: shew that the points P, B, Q are collinear. 3. A circle is described on one of the equal sides of an isosceles triangle as diameter. Shew that it passes through the middle point of the base. 4. Circles described on any two sides of a triangle as diameters intersect on the third side, or the third side produced. 5. A straight rod of given length slides between two straight rulers placed at right angles to one another; find the locus of its middle point. 6. Find the locus of the middle points of chords of a circle drawn through a fixed point. Distinguish between the cases when the given point is within, on, or without the circumference. DEFINITION. A sector of a circle is a figure bounded by two radii and the arc intercepted between them. о the same THEOREM 42. [Euclid III. 26.] In equal circles, arcs which subtend equal angles, either at the centres or at the circumferences, are equal. H B K Let ABC, DEF be equal circles, and let the BGC = the EHF at the centres; and consequently the BAC = the LEDF at the Oces. It is required to prove that the arc BKC = the arc ELF. Proof. Theor. 38. Apply the ABC to the O DEF, so that the centre G falls on the centre H, and GB falls along HE. Then because the BGC the EHF, = .. GC will fall along HF. And because the circles have equal radii, B will fall on E, and C on F, and the circumferences of the circles will coincide entirely. .. the arc BKC must coincide with the arc ELF; Q.E.D. COROLLARY. In equal circles sectors which have equal angles are equal. Obs. It is clear that any theorem relating to arcs, angles, and chords in equal circles must also be true in the same circle. THEOREM 43. [Euclid III. 27.] In equal circles, angles, either at the centres or at the circum ferences, which stand on equal arcs are equal. K Ε H Let ABC, DEF be equal circles; and let the arc BKC the arc ELF. It is required to prove that = the LBGC = the L EHF at the centres; also the L BAC = the LEDF at the ces Proof. Apply the ABC the ODEF, so that the centre G falls on the centre H, and GB falls along HE. Then because the circles have equal radii, ... B falls on E, and the two Oces coincide entirely. ... the BGC = the EHF. THEOREM 44. [EUCLID III. 28.] In equal circles, arcs which are cut off by equal chords are equal, the major arc equal to the major arc, and the minor to the minor. B K Let ABC, DEF be equal circles whose centres are G and H; and let the chord BC= the chord EF. EHF; Theor. 7. Theor. 42. .. the arc BKC = the arc ELF ; and these are the minor arcs. But the whole Oce ABKC = the whole Oce DELF; .. the remaining arc BAC = the remaining arc EDF: and these are the major arcs. Q.E.D. |