Again, in the As PQS, PRS, PQ = PR, and QS = SR, and the contd. – PQS = the contd. PRS; .. APQS = A PRS in all respects so that L QPS= = L RPS. [1. 4.] 7. In the AS BAE, CAE, BA= AC and AE is common to both; also - BAE = L CAE, [Ex. 5.] [1. 4.] 8. Because DA = DC, :: LDACOL DCA. [Def. 29.] Similarly, L BAC = L BCA, ... DAB=L DCB. 9. In the AS BCD, ADC, BC = AD, and DC is common to both, and LBCD= L ADC; 10. In the AS BLM, CNM, BL = CN, and BM = MC, and LLBM = L NCM, Join LM. ...ALN= L ANL, and 2 MLN=MNL, LALM = L ANM [1. 5.] KEY TO EXERCISES. BOOK I. Page 13. 1. Let AB be the given line, x the line to which the sides are to be equal. From centre A with rad. X draw O FCD. From centre B with rad. X draw OGCE cutting the former o in C. Join CA, CB. Then CAB is the A, for by constr, and def. 11 each of the sides CA, CB=X. 2. Let AB be the given line; produce AB to D making BD equal to AB, and produce BA to E making AE equal to AB. From centre A with rad. AD draw O DCF; from centre B with rad. BE draw O ECG. Join AC, BC. Then ACB is the A. 3. Since DBA is equilat., BD=BA. If BA = BC, then BD = BC, which is the rad. of OCGH. Thus D lies on its o co. [For solutions to the Exercises on Pages 17—17B see Introduction.] Page 23. 1. The point F would fall in one of the following ways: (1) above De and below A, in which case the proof would still hold, (2) on the pt. A, in which case the constr. fails, (3) above the pt. A, in which case the constr. fails because the line AF does hot fall within the given angle. 2. The A$ DAF, EAF are equal in all respects (1. 8, Cor.]. H. K. E. 1 91 Page 24. 1. With fig. of Prop. 10, CA = CB, CD is common, and LACD= į BCD. :: ASACD, BCD are equal in all respects (1. 4]. 2. Bisect the given st. line, and with the half line for the equal sides describe an isosc. A as in Ex. 1, page 13. Page 25. 3. Take a pt. X on CF, or CF produced. Then DC = CE, CX is common, and < DCX = _ ECX, being rt. _ $. ... DX = EX. Page 27. 1. Let A be the vertex and BC the base bisected at D. Then BD=DC, AD is common, and AB=AC. .. LBDA= LCDA (1. 8]. 2. Let X, Y be the middle pts. of the equal sides AC, AB. Then in As BXC, CYB the sides XC, CB are equal to YB, BC respectively, and the contained < $ are equal; .. BX = CY. Then 3. Let BD=CE in BC the base of isosc. A ABC. AS ABD, ACE are equal in all respects (1. 4]. 4. Let BD be a diagonal of a quadril. which has AB = DC, BC=DA. Then AS ABD, BCD have their sides respectively equal; .. DAB= _ DCB. Similarly for the other pair of angles. 5. Here L YAB = _ YBA; and _ XAB LXBA; .. XAY: L XBY. Join XY, then it easily follows that A XAY = AXBY in all respects. 6. ABCD is a rhombus and BD a diagonal. Then A* ABD, CBD are equal in all respects [1. 8]. 7. Let BX, CY bisect _ S ABC, ACB of isosc. A ABC, and let them meet in o. Then _ S OBC, OCB being halves of equal angles are themselves equal; .. OBC is isosceles (1. 6]. 8. Here BA, AO = CA, AO respectively, and BO=00. [Ex. 7.] .. 2 BAO = 2 CAO. 9. Let D, E, F be middle pts. of BC, CA, AB respectively; then in AS AFE, BFD, AF = BF, AE = BD, and _FAE= ZFBD. :. FE=FD. Similarly DE=FE= FD. 10. In A8 CBF, BCE, CF = BE by constr., BC is common, and - FCB= · EBC, :. BF = CE. 11. ABCD the rhombus has diags. BD, CA meeting at X. Then AB=BC, BX is common and L ABX = L CBX [Ex. 6]. :: A$ABX, CBX are equal in all respects [1. 4]. 12. In AS BAY, CAX, at A is common and BA, AY = CA, AX. As are equal in all respects; .. _ ABY = = LACY. But ABC = L ACB; .. OBC = LOCB; that is, BOC is isosceles. Again AB = AC and AO is common, and BO = OC; .. 4 BAO = Z CAO. Let AO meet BC in Z. Then BA, AZ = CA, AZ, and Ż BAZE = L CAZ; .. A$ BAZ, CAZ are equal in all respects. 13. AB the base, p the length of perp. Bisect AB in C. Draw cx perp. to AB and equal to P. Join AX, BX. Then AS ACX, BCX are equal in all respects; .. AX = BX. 14. A, B the given pts., XY the given line. Join AB, and bisect it in C. Draw CP perp. to AB meeting XY in P. Then ASACP, BCP are equal in all respects; .. AP = BP. The. construction fails when CP is par!. to XY; this will be the case when AB is perp. to XY, as will be seen later. Page 29. 1. In AABC let BC be produced both ways to X and Y. Then _ s ABC, ACB are supplementary to equal angles, and are therefore equal. 2. In the fig. the _ * XOB, YOB together make up half the _ * AOB, BOC; that is, half of 2 rt. _ s. 3. Since XoY is a rt. 1, and AOB, BOC together = 2 rt. 28; .. AOX and COY = a rt. L. 4. The cox is supplementary to Z AOX, and LAOX = LBOX. The second case is similar. Page 30. By Ex. 6, p. 27, · BAO DAO. :: AS BAO, DAO are equal in all respects [1. 4]; .. - DOA = L AOB. Again, the As AOB, COB are equal in all respects (1. 8]; .; AOB = L COB = a rt. L. :: < $ DOA, AOB are together equal to 2 rt. _ $. Page 33. 1. If any two st. lines would meet at a pt. A if produced, and are cut by another st. line BC, the interior angles on the same side, viz. < ABC, ACB are together less than 2 rt. 28. 2. In the fig. to the Prop. let CB be produced to E. Then the 28 DCA, ACB, CBA, ABE together = 4 rt. _$ [1. 13]. Of these, _ S ABC, ACB are less than 2 rt. _ *; .. LS ACD, ABE are together greater than 2 rt. 29. 3. Join A to X in BC; then AXC is greater than - ABC, and L AXB is greater than · ACB.::$ ABC, ACB are together less than < 8 AXC, AXB; that is, less than 2 rt. 48 [1. 13]. Page 38. 1. A A must have two acute _ $ [1. 17]. .. the rt. L is the greatest L, and has the greatest side opposite to it. 2. Let ABC be a A having LABC= L ACB. Then AB cannot be > AC, for then the < ACB would be > the 2 ABC. Similarly AB cannot be < AC. 3. Here the - ACB > the ADC [1. 16]; .. the - ABC > the LADC; :. in A ABD, AD > AB. 4. Let ABCD be the quadril. having AB the least and CD the greatest side. Join BD. Then the < ABD > the L ADB because AD > AB; and the L CBD > the LCDB, because DC > BC. That is, the whole 2 ABC > the whole 2 ADC. The other case can be proved similarly by joining AC. 5. Let X be the pt. in base BC; then the L AXB> the L ACX, and L ACX is not less than ABC; .. the L AXB > the L ABX; that is, AB > AX. |