problem will be impossible, will have one solution, or two, according to the conditions laid down in Ex. 221. G HF: L' C B 2 M N E' P A D KF G 279. Let AB, CD intersect in O. Draw the bisector EE' of & AOC, BOD, and the bisector FF' of & AOD, BOC. EE', FF' constitute the locus of circles tangent to both AB and CD. Draw GG, HH', each || to CD at the given distance R, and KK', LL', Е. each || to AB at the distance R. GG, HH constitute the locus of circles with radius R tangent to CD, and KK', LL' the locus of circles with radius R tangent to AB. Hence, M, N, P, Q, the intersections of these loci, are the centers of the required circles, and there are four solutions. 280. On constructing the diagrams, it will be evident that four circles can be described as required, each having its center at the given distance from the intersection of the given lines, and in the bisector of the angle in which it lies, and having a radius equal to the perpendicular from that center to one of the lines to be touched. 281. Let✗ touch the given line AB in P and pass through Q. Join PQ. PQ e X in a perpendicular at the mid point of PQ; and AB being a tangent to X, the center will also lie in the perpendicular drawn to A C to PQ at its mid point C; at P draw PXI to AB and cutting CX in X; etc. 282. 1°. Let the given lines be parallel. point, draw PO || and APB I to the given line. Through P, the given PY, each = PA; then X, Y, are each equidistant from AB and the given lines, etc. 2°. Let the given lines meet in O. Join OP and draw APB 1 to OP. Draw AX, AY, bisecting the angles at A and meeting OP in X, Y, respectively. Xand Y are each equidistant from OA, OB, AB, etc. If the perpendicular at the mid point of DE is parallel to the bisector, there is no solution; if that perpendicular coincides with the bisector, there is an infinite number of solutions, since every circumference that touches AB, AC, will pass through D and E. 283. Let P be the required circumference, having its center at P and bisecting the circumference of O in X and Y. Since Pis equidistant from X and Y, the line joining PO will be perpendicular to the diameter XY. Hence join PO, and draw the diameter XY I to P0; then from P as center, with radius PX, etc. X 284. Let be the required circle touching and O', and having a radius equal to a given line M. Draw the central distance 00', and join OX, O'X. AX00' has for base the central distance 00', and for sides the sum of the radii of andX, and the sum of the Hence draw 00', and upon 00' as base radii of O' and X. construct a △ X00' with sides equal to rad. O+ M and rad. O' + M, respectively. In order that the problem may be possible, evidently we must have 00' not greater than rad. O + rad. O' + 2 М. or or 287. According to Avoirdupois weight, the ratios are: 16 16 1 9'33' 2 According to Troy weight, the ratios are: 288. 13 12 9 289. 1° F.: 1° C. = 100 : 180 = 5:9. 1° F.: 1o R. = 80: 180 = 4:9. 294. Let x denote the number of degrees in the angle. Then x° : 60° = 60° : 90° ; .. x° = 40°. EXERCISES, pp. 119-135. 295. Let ABCD be the parallelogram, and AC the diagonal bisecting LBCD. Since ∠ ACD = ∠ACB (Hyp.), and ∠ ACD = ∠ BAC (110), ∠BAC = ∠ACB; :. BA = BC = CD = AD. 296. Let ABCD be the par'm. Since AB + CD = AD + BC (Ex. 201), while AB = CD and AD = BC (136), AB=BC=CD=DA. 297. The two radii drawn to each point of contact are in the same straight line, being each I to the tangent at that point. Hence each coinciding with a pair of equal radii, the central distances are equal. 298. Join D and E with F, the mid point of BC. Then & ADE, DBF, DFE, EFC, are equal, and DBCE is equal to three of them. Hence ABC: DBCE = 4:3. 299. For ∠A: ∠B = 60°: 90° = 2:3. 300. Denoting the angles of the triangle by A, B, C, we have ∠A+ ∠ A + ∠ A = 180° ; .·. ∠ A = 70°, ∠ B = 30°, = ∠ C = 80°. Now ∠AOC = 180° − }(∠ A + ∠ C) = 180° — 75° ; ... ZAOC: ∠ABC = 105:30 = 7:2. 301. (1) ∠BOC = 180° − (∠ B + ∠ C') = 180° - 55° = 125° ; (2) ∠AOB = 180° − }(∠ A + ∠ B) = 180° − 50° = 130° ; 302. Since arc AC = arc BD, chd. AC = chd. BD. Hence ACBD is either a rectangle, or a trapezoid having its nonparallel sides equal; in either case ∠ BAC = ∠ ABD (Ex. 88). 303. Let AD, BC intersect in X. ACD = △ BDC (69); ... Z XCD = ∠ XDC; .·. △ XCD is isosceles, and a perpendicular from X will pass through the mid point of CD (77) and coincide with MN; i.e., Xlies in MN. 304. For that perpendicular will pass through O (171) and coincide with the diameter MN. 305. Let AB, CD, each tangent to O at B, D respectively, make equal angles OAB, OCD with an intercept AOC. Join OB, OD. Since & AOB, COD are equal (44), and OB = OD, △ AOB = ∆ COD; etc, 306. Proof same as in Ex, 305, 307. Let AC, BD be the diagonals of parallelogram ABCD inscribed in circle O. Since AB = CD (136), arc AB = arc CD; and since AD = BC, arc AD = arc BC; ... arc BAD = arc BCD; ... BD is a diameter. Similarly AC is a diameter; ... AC, BD intersect in O. 308. Since OA = OB = OC (Ex. 307), ∠ OAB = ∠ OBA and ∠OAD = ∠ODA;... ∠ BAD = ∠ OAB + ∠ ODA = a right angle (Ex. 100); ... ABCD is a rectangle. 309. This follows from Ex. 308, but may be proved independently from the consideration that OA = OB = OC = OD. 310. Draw a perpendicular at the mid point of the line joining the given points. Its intersection with the given line will be the required center; etc. If the line joining the given points is perpendicular to the given line, there will be no solution. P AQ B 311. Let P be the given point, and AB the given line. From P draw PQ perpendicular to AB, and produce to P', so that QP' = QP. Since P and P' are equally distant from every point in AB (213), any circumference that has its center in AB, and passes through P, must also pass through P' (212). P' 312. Join P, the point, with O, the center of the given circle, cutting the circumference in X and again in Y. From Pas center, with radii PX, PY, describe circles. Both will satisfy the given conditions. 313. For if a radius be drawn cutting off from the greater angle an angle equal to the less, the arc subtended by this part will be less than the arc subtended by the whole angle. 314. For two diameters intersecting at right angles divide the circumference into four equal arcs (257), or quadrants; i.e., a right angle has a quadrant for arc; an acute angle, an arc less than a quadrant; etc. 315. For the vertical angles formed by the intersecting diameters being equal, the intercepted arcs are equal. 316. Let ACBD be the figure formed by joining the extremities of the diameters AB and CD, O being the center. It is easily seen that ZOCA = ∠OAC = ∠ OBD = ∠ ODB, and ∠OAD=∠ODA=∠OCB = ∠OBC. Hence ∠ACB = ∠ CBD; etc. By making use of Art. 266, the proof is, of course, easier. 317. of a right angle = 90° × 1 = 108° ; ... arc ACB: arc ADB = 108°: (360° - 108°) = 3 : 7. 318. Since ∠AOE=180°, and ∠AOB=108°, ∠ BOE=72° ; ... arc BE = = a circumference. 319. Find another point O' equidistant from A and B (78); then 00' bisects ∠AOB and also arc АСВ. 320. Since B is an angle of 32°, arc AC is an arc of 64° (264). 321. Arc CD = 360° (arc AC + arc AD) = 360° - 2 arc AC; ... arc AC = 1 (360° − 102°) = 129° ; .. ∠ BAC = 64° 30′. 322. Let x denote the number of degrees in arc BD. Then since (30° + x) = 25°, x = 20°. B A F E C D 323. Let A, B, C, be three points known to be points of a circumference. Join AB, BC, CA. At B draw BD making any angle ABD with AB, and cutting AC in F. At C draw CE making ∠ACE=∠ABE, and meeting BD in E. In & AFB, CFE, since ∠ABE = ∠ACE, and ∠AFB = ZCFE, ∠A = ∠E (121); ... E is a point on the same circumference as A, B, and C, for otherwise C being joined with the point where BD does cut the circumference, there would be an exterior equal to an interior remote angle (266). 324. In O let OPR, OP'R' be radii drawn to AB, CD, opposite sides of an inscribed par R' D C A B R 325. OA and O'B will be parallel, being both perpendicular to AB (191). 326. Join PO cutting AB in C. It is easily shown that rt. A PAO, ACO are equiangular, so that ∠ CAO = ∠ APO = + ∠ P (Ex. 200). 327. Let x denote the number of degrees in arc BC. Then :. ∠ A + ∠ C + ∠E is measured by a circumference (AB + BC + CD + ED + EF + AF); .:. ∠ A + ∠ C+∠E is measured by a circumference; etc. 329. The problem is the same as to find the center of a circumference passing through three given points. It is impossible if the points are in the same straight line. 330. Join the points so as to form a quadrilateral. If the opposite angles of this figure are supplementary, the point required can be found (see Ex. 398). 331. The sides of the triangle being chords of the circumference that can be passed through its vertices, the perpendiculars at the mid points of these chords intersect at the center (171). |