(i) If PQ is without the given O, then the derived from AD has external contact, that derived from AB internal contact (the given being within the other). (ii) If PQ touches the given O, then the derived from AD has external contact, that from AB is impossible. (iii) If PQ cuts the given, then both touch externally, or both internally, according as the point A is without or within the given O. the Draw 37. Let PQ be the given st. line, and E the given point on of which C is the centre. Draw the diameter BD perp. to Join DE (or BE), and produce it to meet PQ at A. AF perp. to PQ; and join CE, producing it to cut AF at F. F shall be the centre of the required. [Proof as in Ex. 36.] PQ. 38. Let BD be given st. line, and Let F be the centre of the given circle. To the given draw a tangent AP point of contact. Then D the given point in it. [See fig. p. 221.] се perp. to BD, A being the Join AD, meeting the Oce at E. Join FE and produce it to meet BD in C. Then C shall be the centre of the required circle. [Proof as in Ex. 36.] Two solutions, since two tangents may be drawn to the given perp. to BD. 39. Let A and B be the centres of the two Os, and C, c' their intersections. Then ACB = AC'B [1. 8]. And the angles between the tangents at C, and the tangents at C ́are respectively supplementary to the ACB, AC'B. S 40. This follows immediately from III. 19. 41. This follows from Ex. 40, by the aid of 1. 47. 42. It follows from Ex. 40 that the required locus is the tangent to the given circle at the given point. 43. Let A be the centre of the given O, P the point on its Oce, and Q another point. Draw PR the tangent at P. Then the centre of the required must lie on this tangent [Ex. 40]. Again, the centre of the required must lie on the line which bisects PQ at rt. ▲3. Hence the centre is determined. III. ON ANGLES IN SEGMENTS, AND ANGLES AT THE CENTRES AND CIRCUMFERENCES OF CIRCLES. Page 222. 2. Let the chords AB, CD intersect without the at E. Join AD. by the arcs AC, BD; or the at the centre subtended by half the difference of the arcs AC, BD. 8 3. Let AB, CD two chords of a intersect at rt. at E. AED is equal to the sum of the Then by Ex. 1, the tended at the Oce by AC, BD. subThat is the sum of the arcs AC, BD subtend a rt. angle at the ce; or, the sum of the arcs is equal to a semi-circumference [III. 31. Converse]. [111. But by hyp. the arcs AQ, PB = the arcs QC, AP respectively. .. LAXY = LAYX ; .. AX =AY. 5. Let ABCD be a quad'. inscribed in a O, having one side DA produced to E. DAB, DCB together two rt. angles [111. 22], L BAE = 4 DCB. 8 6. Let the two intersect at A, B, and let PAQ, XBY be the two st. lines terminated at the O Join AB. [In the ces. figure taken A lies between P and Q, and B between X and Y.] XPA, XBA together two rt. angles [111. 22], and the ext. XBA = the AQY. [Ex. 5.] .. the XPA, AQY together two rt. angles. .. PX and QY are par1. [1. 28]. 7. Join PR, QR. Then PR, QR shall be in one st. line. 8. on AB Let ABC be the ▲, rt.-angled at B, and let the as diameter meet AC at D. Then the tangent at D shall bisect BC at E. Join BD. Since ABC is a rt. 4, BC is the tangent at B [III. 16] .. BE = DE. [III. 17, Cor.] And since BDC is a rt. angle [11. 31], it follows that 9. Let A, B, C be the three points. st. line BX, in which take any point P on A. At P in BP make BPQ equal to CD par1. to PQ. For BDC = = Through B draw any the same side of BC as BAC. Through C draw Then D is a point on the . BPQ [1. 29]= BAC [constr.]. Hence the points B, A, D, C are concyclic [111. 21, Cor.]. 10. Let A, B, C be the given points. On the side of CB remote from A make CBD equal to ▲ BAC. Then BD is the tangent at B [III. 32. Converse]. 11. and EC. Let E be the centre of the second . Join AB, EB, DE [In the fig. taken ACD lies between E and B.] S Hence DCE, BCE are identically equal [1. 4]. 14. Let AB be the chord, C any point on the exterior seg ment. Let AC, BQ meet interior segment at P and Q. Then shall PQ be constant. Join AQ. Then AQB = sum of 4 ACQ, CAQ [1. 32]. .. ▲ CAQ = diff. of AQB, ACB, both of which are of constant magnitude [111. 21]. .. CAQ, i.e. the PAQ, is constant. Hence the arc PQ is constant [III. 26]. S 15. If all the given ▲ stand on a fixed base BC, and have a given vertical angle, they also have the same circumscribed circle [III. 21. Converse]. Take BAC, any one of these As, and let the bisector of the A meet the circum-circle at X. ▲ CAX (hyp.), ..arc BX = CX [III. 26]. Then since BAX: .. X, being the middle point of the arc BC, is same for all triangles of the series. 16. [III. 27]. Draw CF perp. to AE. 8 [Ex. 7, p. 101]. Now DE, EA are respectively perp. to BC, AE. .. < DEA = ▲ BCF [Ex. 3, p. 59] = half the diff. of the S at B and C. 17. Let BC be the chord of the ext. O, and D its point of contact with the int. O. Then shall AD bisect BAC. At A draw the common tangent AT. Then 18. DAC = L DAT - CAT =ADC ABD [111. 17, Cor., III. 32] = BAD [1. 32]. Let BC, the chord of the ext. O, cut the int. O at P, Q. Let A be the point of contact of the two Then shall ▲ BAP = 4 CAQ. At A, draw the common tangent AT. 3. = QAC [1. 32]. ON THE ORTHOCENTRE OF A TRIANGLE. Page 226. In an acute-angled ▲ the orthocentre is within the ▲. 22. For, in the fig. of p. 225, produce ED to X. It has been shown that ▲ EDC = ▲ FDB [Ex. 20, p. 221]. That is, the ext. FDX is bisected by BD: and so on for the others of the pedal ▲. The latter part of the proposition may be solved in a similar manner. 8 23. For, with the fig. of p. 227, since the AFO, AEO are rt. angles (hyp.), .. the four points A, F, O, E are concyclic. ..the FAE, FOE together two rt. angles [III. 22]. That is, the BAC, BOC together two rt. angles [1. 15]. S |