Page 196. DAB on arc BD = L ADC 1. Let AB, CD be par': chords. Then on arc AC [1. 29]. .. arc BD arc AC [III. 26]. 2. Let AC, BD be equal arcs. Then ADC on arc AC = L BAD on arc BD [III. 27]. .. AB is par1. to CD [1. 27]. = Let BGC EHF. Then the AS BGC, = arc ELF, and the .. remainder arc BAC = arc 3. See fig. III. 26. EHF are identically equal. Also arc BKC: whole ce KBAC whole ce LEDF, EDF. .. BKC = L ELF [III. 27]. similar and they are on equal chords. 24.] Also ABGC = ▲ EHF. .. sector BGC = sector EHF. = .. segments BKC, ELF are .. they are equal. [111. Then 4. Let chords AC, BD intersect at rt. s in X. LAXD: < " ABD, BAC together, .. the arcs AD, BC subtend at the circumferences together equal to a rt. 2. And AXB = ACB, CBD together, .. the arcs AB, CD subtend at the circumference Ls together equal to a rt. . .. arcs AD, BC together = arcs AB, the semicircumference. S CD = S 5. As in preceding ex., 4 AXD = sum of the subtended by AD, BC = subtended by an arc equal to sum of AD, BC. се 6. The AXB = difference of subtended at the Oce by AD, BC = difference of AD, BC. 7. Let bisector of APB cut the ▲ APQ = BPQ, .'. arc AQ= arc BQ. of the conjugate arc AQB. 8. Let PA, PB cut the other ABD, BAC = difference of ≤3 subtended by arc equal to Then conjugate arc in Q. (1) Let Q and R be in PA, PB produced. Then of BQA, BPA = sum of $ subtended at the stant. 8 in Q and R. (2) Let Q and R be in PA, PB. Then QBR 8 = LS BQA, BPA = difference of subtended at the ces of the two Os by AB = constant. (3) Let R be in PB and Q in PA produced. Then QBR = supplement of BQA, BPA supplement of subtended at the ces by AB = constant. 9. The AXY LABY = B. .. LZXY = (B+C) = complement of A. 11. PX and QY subtend at A opp. vertical 3. Hence arc PX arc QY. Hence chord PX 12. Each the common chord [Ex. 1]. 13. APB = Since the chord AB is common to the two equal ○3, the arc AB in one = the arc AB in the other [III. 28]; .. AQB [III. 27]. .. BP = BQ. 14. Each of the chords BX, XA, AY, YC subtends an ▲ equal to half the base 4. S Hence, if the base are each double of the vertical, the pentagon is equilateral. Page 199. 1. See fig. p. 199. Let tangent at D be parl. to AB. Then, if DC be perp. to tangent, the centre is in DC [III. 19]. But DC .. DC bisects AB [III. 3.] Hence arc ADB is is also perp. to AB. bisected at D [III. 30]. On AB describe 2. Let CB be the quadrant, A the centre. an equilateral ADB. Because AD = AB, .. D is on the circumference. Bisect DAB in E. Then the rt. BAC is trisected by AD, AE [Ex. 6, p. 60]. .. the arc BC is trisected at D and E. Page 201. 1. The opp. the diameter must be a rt. 4. Hence the vertex is on the Oce. [Converse of Prop. 21.] 2. The locus is the on hyp. as diameter. 3. The locus is a quadrant of the whose centre is the pt. of intersection of the rulers, and radius half the length of the rod [III. 31]. 4. Each of the 8 .. PB, QB are in a st. line. PBA, QBA in a semicircle is a rt. L. 5. The line joining the vertex of an isosceles to the middle pt. of the base is perp. to the base. Hence the O on a side as diameter passes through the middle pt. of the base [Ex. 1]. 8 6. Let A be pt. of contact, AB diameter of inner, and ABC of outer. Draw any chord ADE. Then each of the ADB, AEC in a semicircle is a rt. ... BD is par1. to CE. But B is middle pt. of AC. .. D is middle pt. of AE [Ex. 1, p. 96]. 7. Both the circles described on the sides of a ▲ as diameters must pass through the foot of the perp. from the vertex on the base or base produced. 8. The required locus is the whose diameter is the line joining the given pt. to the centre of the given ☺. If the pt. is without the ce, the locus is confined within the two tangents to the O. If the pt. is on the Oce, the locus is the on the radius through the point as diameter. [See solution of Ex. 40, p. 229.] 9. On the side of the greater of the two given squares as diameter describe a semicircle: from its extremity draw a chord equal to the side of the other given square. The chord completing the is the side of the required square [III. 31, 1. 47]. 10. Let A be a pt. of intersection of two ; B the centre of one of them. Let the other cut the described on AB as diameter in C. The chord AC produced will be bisected at C [III. 31, 3]. 11. 8 Since the diagonals of a rhombus are at rt. 3 to one another [Ex. 11, p. 27], . the required locus is the C described on the given st. line as diameter. Page 204. 1. If, from one extremity of a chord of a circle, a straight line be drawn making an angle with the chord equal to the angle in the alternate segment, this straight line shall touch the circle. Take fig. p. 203. Let BA be the diameter through B. Then, DBF is acute, the alternate segment must be > a semi.. the diameter BA must fall in this segment. if the circle. .. L DBF: = BAD: add ▲ ABD. 8 .. LABF = BAD, ABD = a rt. ≤ [1. 32, 111. 31]. .. BF is a tangent. Similarly, if EBD is an obtuse angle, EB must be a tangent. 2. Each of the their pts. of contact .. these are equal. 3. made by the tangents with the line joining equal to the angle in the alt. segment. ..the tangents are equal [1. 6]. is S Let A be pt. of contact, AB, AC the diameters of the given Draw ADE to cut the " in D and E. Then SADB, AEC in semicircles are rt. 9. .. BD, CE are par1. and DAB, EAC are in (i) coincident and in (ii) opp. vertical. Hence the remaining ABD, ACE are equal [i. 32]: i. e. the segments DBA, ECA are similar. S 4. Draw T'AT the common tangent to the two at A. [Ex. 3, p. 181.] Let AX be between AP and AT. 5. Tangent at A to first makes with AO an equal to OBA in alternate segment. But because O is centre of the other O, OA = OB. .. LOBAL OAB. .. AO bisects between AB and the tangent at A. L PBA = 6. The tangent at P makes with PAC an 4 equal to = LABD (or its supplement) = ACD (or its supplement). .. tangent at P is par1. to CD. 7. Let A be pt. of contact, AB chord through A, C the middle pt. of arc cut off by AB; CM, CN perps. on the tangent at A and the chord AB. Then CAM ABC [III. 32] and ABC = CAB .. CAM = L CAB. .. the $ CAM, CAN are identi = [III. 30]. L= Page 206. 1. On the given base describe a segment containing an to the given 2. The pt. or pts. in which the arc of the segment cuts the given st. line give the required vertex. 2. The required vertex is the intersection of the arc of the segment described on the base and containing an equal to the given ▲, and (i) The circle, whose centre is an extremity of the base and radius is equal to the given side; (ii) The st. line parallel to the base at a distance from it equal to the given altitude; (iii) The circle whose centre is the middle pt. of the base and radius equal to the given median ; (iv) The perp. to the base drawn through the given point. 3. Because arc AP: =arc BP; .. LACP = BCP. .. CB = CX [1. 6]; .. AC + CB = AX = required length. 5. On AB, the given base, describe a segment containing an equal to the given K; also another segment containing an angle greater by a right than K. From centre A, with radius equal to the given difference of the sides, describe a cutting the last drawn segment in X. Join AX and produce it to cut the first segment in C. Then ABC shall be the required triangle. Let the bisector of ACB cut BX in D. greater by the XDC than XCD, i.e. than Then ext. K... rt. ... the ▲ XCD, BCD are identically equal [1. 26]. fore CX CB. .. AX the difference of AC and CB. = Page 207. ▲ AXD is XDC is a There 1. Let AB be the base of given segment, produced to C. From A draw AP to meet the arc of the segment at P; join PB, and through C draw CQ par1. to PB to meet AP produced at Q. Then a segment described on the base AC to pass through Q [Ex. 4, p. 156] is that required. For APB AQC [1. 29]. = |