332. The greater angle is measured by one half of the greater arc. 333. The line of centers, being perpendicular to the common chord (Ex. 195), bisects both pairs of arcs subtended by that chord (175). QUESTIONS, p. 136. 334. A = 90° × & = 60°; .'. ZA: ZB= 60° : 50° = 6 : 5. 335. Let A be the vertical angle. (1) Since ZA = 2 ZB = LB+ 2C, ZA+ZA = 180°; .. ZA = 90°, and is measured by a quadrant. (2) Since A=}<B={(ZB+ZC),ZA+4ZA=180°; .. ZA=36°, and is measured by 36 or 3 of a quadrant. (3) Since ▲ A (LB+ZC), ZA+2nZA = 180°; .. ZA = 180° 2n+1 -= 1 LB n 1 2 n and is measured 336. ≤ A = 360° × } = 72°, ≤ B = 360° × } = 45°; ... ≤ C = 63°. 337. Let A and B be the acute angles. Since 54°, ≤ B = 90° – 54° = 36° ; :. ZB : ZA = 36: 54 338. 360° 230° = 130°; .. < C = 65° = Z D = 339. LABE = 300 × 1 = 15°. LBAE = 340. If BAC = 65°, arc AC = 130°; also arc AD = 130°; .. arc CD = 100°. 341. Arc CA + arc AD + arc CD = 2 arc CA + 3 arc CA = 360°; .. arc CA 10290; .. BAC = 51° 51° 25' 42.85". -- 342. Arc AC + arc AB = 40° × 2 = 80°; ... arc AD + arc BC= 360° 80° 280°. 343. Let x denote the number of degrees in arc BC. Since хо − 1 x° = 25° × 2, x = 100°, and arc BC+ arc DE = 150°; .. arc BE +arc CD 210°. 344. Since DA = DB, LB = ZA = n° ; .. arc DE = 2 n°. But arc BC arc DE 2 n°; .. arc BC= 4 n° = 2 arc DE. 345. Arc BC – arc BD = 3 arc BD arc BD = 2 arc BD = 60°; .. arc BD = 30°; ... arc BC= 90°; .. arc CD 360° - 120° : the number of degrees in ZA; then / C = n° ; 2 n°;. arc BC: 347. If the angle formed by the tangents is 30°, the angle formed by the radii to the points of contact is 180° 30° 150°; ... the major arc 360° - 150° = 210°. 348. (180° 60°) = 60°. 349. 360° x 360° × 30°3 circumf. = 360 = 12 circumf. 350. The vertical angle being 54°, each base angle (180° – 54°) = 63°. Hence the ratio of the arcs is 54 x 2: 63 × 2 = 6 : 7. 351. The vertical angle being 37° 15' 32", each base angle=(180°— 37° 15' 32'') = 71° 22′ 14′′, and the ratio is 134132" : 256934"= 67066 128467. EXERCISES, pp. 139-154. 352. The bisector of the vertical angle divides the triangle into two equal right triangles with the right angles at the mid point of the base (77); hence the mid point of each arm is equidistant from the extremities of the arm and the mid point of the base (Ex. 144). 353. The altitude upon the third side divides the triangle into two right triangles having their common vertex in the third side, or the third side produced; and the mid point of each side is equidistant from its own extremities and that vertex (Ex. 144). 354. For the common chord, as it does not pass through the center, divides the intersected circumference unequally, the exterior segment being the greater, since it includes the center. Hence the angle in this segment is acute. 355. At P draw the common tangent PQ meeting AB in Q. Since QA = QP, and QB QP (Ex. 200), / PAQ=ZAPQ, and PBQ=ZBPQ; :. ¿PAB+ZPBA = LAPB; :. APB is a right angle (Ex. 100). A B 356. For the radius to the point of contact is perpendicular to the tangent, and also to the chord. 357. For BD': D'C = AB: AC = BD: DC (278, 280). 358. Since BD' : D'C = BD : DC (Ex. 357), (1) BC is divided internally in D' so that D'B: D'C = DB: DC; (2) BD is divided into three segments, BD', D'C, and CD, so that BD, the whole line, is to DC, one of the outer segments, as BD', the other outer segment, is to D'C, the inner segment. 359. Let MN be divided internally in D', and externally in D, so that D'M: D'N DM: DN; then, also, = = M D N D D'N: DN D'M: DM (244); that is, D'D is divided internally in N and externally in M, so as to be divided harmonically. 360. For the whole line DM is to one of the outer segments DN as the other outer segment D'M is to the inner segment D'N. F C E = 361. In any parallelogram AC, draw EF parallel to AB. Then AC and AF are mutually equiangular yet not similar, since the homologous sides are not proportional. Again, we may have two parallelograms, AC, AE, such that the determining sides DA, AB are respectively equal to FA AB; yet AC, AE are not similar, not being mutually equiangular. 16: 108: 5, p : p' = 8: 5. AG: AB=GH: BC (283). 364. Δ Α' ΑD is similar to Δ ΟΑΒ and Δ ΟΑΒ to Δ ΟΑ' Β'. 365. OB, OB' are homologous sides of similar ▲ OAB, OA'B' ; hence perimeter OAB: perimeter OA'B' = OB: OB' (296). 366. ZB+≤ C = 2 DAC + 2 DAB = & 2 DAB + 2 DAB = 90° ; :. ZDAB = 56° 15', Z DAC = 33° 45'. 367. BD: AD = AD : DC; .. 10 – 2 : AD = AD : 2 ; .. AD2 = 16, AD = 4. 368. BC: AB : = AB: BD; :. AB2 = 10 × 8 ; .. AB=4√5. Similarly AC210 × 2 ; .. AC = 2√5. D 369. Prove as in Art. 301. 370. For A= LD and ZB = ZC (266). A A B B 371. Let AB, CD be equal chords intersecting in E. Join AC. Since AB = CD, arc CAD; ... arc BC = arc AD (Ax. 3); .'. ACB = arc LA= 20; = ... EC = EA; .. EB = ED (Ax. 3). = 374. If OA, OD are equally distant from the center, then AB CD; then OA OD and OC= OB (Ex. 372); .'. OA: OB = OD: OC. 375. Let PT, PT be tangents drawn from any point P in the production of AB, the common chord of the intersecting circles. Then PT PT', each being a mean proportional between PA and PB (303). QUESTIONS, p. 155. 376. AB: AD = AC: AE; or 3 + 7 : 7 = 55: AE;.. AE= 38.5. 377. √11+7:7 = 55: AE;.. AE = 37.32. 378. AB: AC = BD: DC; .'. AB + AC : AB = BD + DC : BD; ... 9 + 7 : 9 = 12 : BD; ... BD = 6.75, DC = 5.25. 379. AB AC = BD: DC; .. 9:720: DC; ... DC = 155, BC= 44. 380. If ABC is isosceles, the bisector of the exterior vertical angle is parallel to the base (Ex. 146), and cannot meet it. The proposition may then be said to hold in this way, that, the external segments being infinitely great lines, their ratio is indefinitely near to a ratio of equality, which is that of AB to AC. 381. They are similar if their third angles are both either greater or less than a right angle, or if one of them is a right angle. bc α 382. AB: AC= A'B' : A'C', or a: b = c : A'C' ; ... A'C' 383. AD: DE = A'D': D'E', or, 4: :: 33.2: D'E'; .. D'E' = 2.4. AD: AC A'D': A'C', or, 4:5 = 3.2: A'C' ; :. A'C' = 4. AB2 ÷ BC= 3.2. = 384. AB2 = BC · BD; ... BD = AC2 ÷ BC = 1.8. AC2 = BC · DC; ... DC = AD2 BD DC, = 1 × † ; ... AD = 12 = 2.4. 385. OA: OD = OC : OB; i.e., & OD : OD = OC:8; .'. OC = 55%. 386. CG: AG=BF: AF = BD : AD + 2 BD=BD:mBD + 2 BD ; ... CG: AG=1: m + 2; .. CG = 1 = = 387. AD: EC = AB: BE = AB : AB + AC 10: 10+7 10:17. 388. If ▲ ABC is isosceles, then ZBCE is a right angle (Ex. 140); i.e., ZBCE = 90°. 389. BD: BC = AD: AE = 3:2; .. BD: BD– BC = 3: 3 — 2, that is, BD: DC=3:1= AB: AC. 390. Since CE would be parallel to AD (Ex. 146), E would coincide with B. 391. AB+BC: BC=AD+EC: EC; ... AD+EC: AC-EC: BC. Now EC: BC= EC' or BB': OB = OB – OB': OB = 85:8; .'. AD EC3 AC÷8 = 4.5. 392. OA': OC = OA: OC AB: BC10:8 = = : 5:4. 393. AB AC = A'B' : A'C' ; .'. AB – AC : AB = A'B' — A'C' : ... AB AC: A'B' A'C' AB: A'B' = BC: B'C' = m : n. A'B' ; 394. AC: A'C' : = AB: A'B' = 18:12 = 3 : 2. 395. ABDA + BD: AC + DC + DA = BD: DA = m : n. measured by B is measured by straight angle. THEOREMS, p. 160. 396. Let AD, BC join the extremities of the equal chords AB, CD. Since AB= CD, arc AB=arc CD; .. arc ABC =arc BCD; :. ZD=ZA. Similarly 2 C = LB; :. ZA + ZB = 2C+LD; .. ZA+ LB= = a straight angle; ... AD is || to CB (113). 397. Let ABCD be the quadrilateral. A is arc BCD, and 4C by arc BAD; .. ZA+ZC circumference ABCD; ... ZA+ 2 C is equal to a 398. Let ABCD be a quadrilateral such that A+≤ C= a straight angle. Describe a circumference through D, A, and B (185); it will. also pass through the fourth vertex C. For if C should fall without the circumference, so that the latter cut CD in C', join BC'. Then as ZA is supplementary to BC'D (Ex. 397), and is also supplementary to C (Hyp.), int. ≤ C' ext. 4C (44), which is impossible (83). Hence C cannot lie without circumference DAB; and, in the same way, it may be D A B shown that C cannot lie within circumference DAB. cides with C", and is concyclic with A, B, and D. Hence C coin 399. Join AC, CB, BD, DA. Since OA: OC = OD: OB, ▲ AOC, BOD are similar (290); .. LOBD = LOCA and OAC LODB. In the same way we show that OBC = LODA and OAD = LOCB; .. ZOBD + ≤ OBC+ ZOAC+ZOAD = ZOCA+ZOCB+2ODB+ ZODA; i.e., 4 CAD and CBD are supplementary; .. A, B, C, D, are concyclic (Ex. 398). REMARK.-In preparing the diagram for this exercise, describe an erasible circumference, and draw in it any chords AB, CD. Proceed in a similar manner in the preparation of the next diagram. C 402. Let AB be the tangent, AC the chord, D the mid point of |