Page 247, Ex. 7. Read, “P and Q" instead of "A and B.” Page 258, Ex. 20. Read, "Three circles" instead of “Two circles.” Page 268. Ex. 14 is removed ; Ex. 15 becomes Ex. 14. Page 277. The letters E, E, are interchanged with F, F, in the figure and in Ex. 1. Page 280. In place of Ex. 27 read, “Given a vertex, the centre of the circumscribed circle, and the centre of the inscribed circle, construct the triangle.” Page 283. Exx. 38 and 39 are removed to page 382, where they take the place of Exx. 59 and 60. For Ex. 38 read, “Given the base and vertical angle of a triangle, shew that one angle and one side of the pedal triangle are constant." For Ex. 39 read, “Given the base and vertical angle of a triangle, find the locus of the centre of the circle which passes through the three escribed centres." Page 284. In place of Ex. 13 read, "Given the orthocentre, the centre of the nine-points-circle, and the middle point of the base, construct the triangle.” H. S. HALL. October, 1892. INTRODUCTION. SOLUTIONS TO EXERCISES ON PAGES 17-17 B. Page 17. 1. From centre C with rad. L describe a O cutting AB in E, F. Then CE = CF [Def. 11.] Thus E and F are the required pts. The pts. can only be found provided the given length L is such that the circle meets AB. 2. Join CA; from centre c, with rad. equal to CA, describe a o cutting PQ in B. Then CA = CB, [Def. 11.] .. ACAB is isosceles. The o will generally cut PQ in another pt. D, so that CAD is a second triangle satisfying the given conditions. 3. Join Ac, and let L be the given length of each side. From centre c, with rad. L, draw the O EBD. From centre A, with rad. , draw the OFBD, cutting the former o in B, D. Then ABCD shall be the required rhombus. For by constr. and Def. 11, each of the sides AB, BC, CD, DA is equal to L. 4. From centre A, with rad. AN draw ONCL. From centre B, with rad. BM draw o MCK, cutting the former o in c. Join AC, BC. Then AC= AN, [Def. 11.] and BC BM. [Def, 11.] :. ACB is the required triangle. Join AC; 5. and on AC describe an equil. A DAC. From centre c, with rad. CB, describe O BGH, cutting cd in G. From centre D, with rad. DG, describe OGFK, cutting AD at F. .. CB=CG. And because D is the centre of O GFK, .. DF = DG; and DA, DC are equal; [Def. 19.] .. the remainder AF = the remainder CG. And it has been shewn that CG = CB. .. AF = CB. Page 17 A. 1. (i) Because o is the centre of the larger 0, .. OD = OE. Because o is the centre of the smaller 0, .. OA= OB. (ii) In the AS ODB, OEA, (Def. 11.) and the contd. L at A is common to the two AS; .. DB = AE [1. 4.] and the As are equal in all respects. (iii) Because OAB is an isosceles A, ..LDAB = LEBA. [1. 5.] (iv) The A S ODB, OEA are equal in all respects, (proved in (ii)] .. LODB = LOEA. 2. (i) In the A8 BLM, CMN, LB = MC, and BM= CN, [Def. 28, Ax. 7.1 and LBM = Ź MON, being rt. _ $; [1. 4.] (ii) In the A $ ABM, DCM, AB = DC, and BM = · MC, and LABM=L DCM; [1. 4.] The other two cases follow in a similar manner by 1. 4. (iii) from the equal AS AND, AMB. Page 17 B. 3. In the As OAM, OBM, OA = OB, being radii of a 0, OM is common to the two As, and L AOM = _ BOM; .. AM BM. [Hyp.] [1. 4.] The L ABC = the 2 ACB, and the _ DBC = the _ DCB; [1. 5.] 5. In the A8 ABD, ACD, AB = AC, and BD = DC, and LABD = L ACD L BAD=6 CAD, and LBDA=LCDA. [1. 5.] [Ex. 4.] [1. 4.] so that [1. 5.] 6. Because A PQR is isosceles, ... PQR=PRQ; and because ASQR is isosceles, ... SQR = L SRQ. [Ax. 3.] |