« ZurückWeiter »
8. This is a particular case of Euc. I. 22. The triangle however may be described by means of Euc. 1. 1. Let AB be the given base, produce AB both ways to meet the circles in D, E (fig. Euc. I. 1.); with center A, and radius AE, describe a circle, and with center B and radius BD, describe another circle cutting the former in G. Join GA, GB.
9. Apply Euc. 1. 6, 8.
13. Let CAB be the triangle (fig. Euc. 1. 10.) CD the line bisecting the angle ACD and the base AB. Produce CD, and make DE equal to CD, and join AE. Then CB may be proved equal to AE, also AE to AC.
14. Let AB be the given line, and C, D the given points. From C draw CE perpendicular to AB, and produce it making EF equal to CE, join FD, and produce it to meet the given line in G, which will be the point required.
15. Make the construction as the enunciation directs, then by Euc. 1. 4, BH is proved equal to CK: and by Euc. 1. 13, 6, OB is shewn to be equal to OC.
16. This proposition requires for its proof the case of equal triangles omitted in Euclid:-namely, when two sides and one angle are given, but not the angle included by the given sides.
17. The angle BCD may be shewn to be equal to the sum of the angles ABC, ADC.
18. The angles ADE, AED may be each proved to be equal to the complements of the angles at the base of the triangle.
19. The angles CAB, CBA, being equal, the angles CAD, CBE are equal, Euc. 1. 13. Then, by Euc. I. 4, CD is proved to be equal to CE. And by Euc. I. 5, 32, the angle at the vertex is shewn to be four times either of the angles at the base.
20. Let AB, CD be two straight lines intersecting each other in E, and let P be the given point, within the angle AED. Draw EF bisecting the angle AED, and through P draw PGH parallel to EF, and cutting ED, EB in G, H. Then EG is equal to EH. And by bisecting the angle DEB and drawing through Pa line parallel to this line, another solution is obtained. It will be found that the two lines are at right angles to each other.
21. Let the two given straight lines meet in A, and let P be the given point. Let PQR be the line required, meeting the lines AQ, AR in Q and R, so that PQ is equal to QĒ. Through P draw PS parallel to AR and join RS. Then APSR is a parallelogram and AS, PR the diagonals. Hence the construction.
22. Let the two straight lines AB, AC meet in A. In AB take any point D, and from AC cut off AE equal to AD, and join DE. On DE, or DE produced, take DF equal to the given line, and through F draw FG parallel to AB meeting AC in G, and through G draw GH parallel to DE meeting AB in H. Then GH is the line required.
23. The two given points may be both on the same side, or one point may be on each side of the line. If the point required in the line be supposed to be found, and lines be drawn joining this point and the given points, an isosceles triangle is formed, and if a perpendicular be drawn on the base from the point in the line: the construction is obvious.
24. The problem is simply this—to find a point in one side of a triangle from which the perpendiculars drawn to the other two sides shall be equal. If all the positions of these lines be considered, it will readily be seen in what case the problem is impossible.
25. If the isosceles triangle be obtuse-angled, by Euc. 1. 5, 32, the truth will be made evident. If the triangle be acute-angled, the enunciation of the proposition requires some modification.
26. Construct the figure and apply Euc. 1. 5, 32, 15.
If the isosceles triangle have its vertical angle less than two-thirds of 3 right-angle, the line ED produced, meets AB produced towards the oase, and then 3. AEF = 4 right angles + AFE. If the vertical angle be greater than two-thirds of a right angle, ED produced meets AB produced towards the vertex, then 3. AEF right angles + AFE.
27. Let ABC be an isosceles triangle, and from any point D in the base BC, and the extremity B, let three lines DE, DF, BG be drawn to the sides and making equal angles with the base. Produce ED and make DH equal to DF and join BH.
28. In the isosceles triangle ABC, let the line DFE which meets the side AC in D and AB produced in E, be bisected by the base in the point E. Then DC may be shewn to be equal to BE.
29. If two equal straight lines be drawn terminated by two lines which meet in a point, they will cut off triangles of equal area. Hence the two triangles have a common vertical angle and their areas and bases equal. By Euc. I. 32 it is shewn that the angle contained by the bisecting lines is equal to the exterior angle at the base.
30. There is an omission in this question. After the words " making equal angles with the sides," add, “and be equal to each other respectively.” (1), (3) Apply Euc. 1. 26, 4. (2) The equal lines which bisect the sides may be shewn to make equal angles with the sides.
31. At C make the angle BCD equal to the angle ACB, and produce AB to meet CD in D.
32. By bisecting the hypotenuse, and drawing a line from the vertex to the point of bisection, it may be shewn that this line forms with the shorter side and half the hypotenuse an isosceles triangle.
33. Let ABC be a triangle, having the right angle at A, and the angle at C greater than the angle at B, also let AD be perpendicular to the base, and AE be the line drawn to E the bisection of the base. Then AE may be proved equal to BE or EC independently of Euc. III. 31.
34. Produce EG, FG to meet the perpendiculars CE, BF, produced if necessary. The demonstration is obvious.
35. If the given triangle have both of the angles at the basé , acute angles; the difference of the angles at the base is at once obvious from Euc. 1. 32. If one of the angles at the base be obtuse, does the property hold good!
36. Let ABC be a triangle having the angle ACB double of the angle ABC, and let the perpendicular AD be drawn to the base BC. Take DE equal to DC and join AE. Then AE may be proved to be equal to EB.
If ACB be an obtuse angle, then AC is equal to the sum of the segments of the base, made by the perpendicular from the vertex A.
37. Let the sides AB, AC of any triangle ABC be produced, the exterior angles bisected by two lines which meet in D, and let AD be joined, then AD bisects the angle BAC. For draw DE perpendicular on BC, also DF, DG perpendiculars on AB, AC produced, if necessary. Then DF may be proved equal to DG, and the squares on DF, DA are equal to the squares on FG, GA, of which the squareon FD is equal to the square on DG; hence AF is equal to AG, and Euc. I. 8, the angle BAC is bisected by AD.
38. The line required will be found to be equal to half the sum of the two sides of the triangle.
39. Apply Euc. 1. 1, 9.
40. The angle to be trisected is one-fourth of a right angle. If an equilateral triangle be described on one of the sides of a triangle which contains the given angle, and a line be drawn to bisect that angle of the equilateral triangle which is at the given angle, the angle contained between this line and the other side of the triangle will be one-twelfth of a right angle, or equal to one-third of the given angle.
It may be remarked, generally, that any angle which is the half, fourth, eighth, &c. part of a right angle, may be trisected by Plane Geometry.
41. Apply Euc. 1. 20. 42. Let ABC, DBC be two equal triangles on the same base, of which ABC is isosceles, fig. Euc. I. 37. By producing AB and making AG equal to AB or AC, and joining GD, the perimeter of the triangle ABC may be shewn to be less than the perimeter of the triangle DBC,
43. Apply Euc. 1. 20.
44. For the first case, see Theo. 32, p. 76: for the other two cases, apply Euc. 1. 19.
45. This is obvious from Euc. 1. 26.
46. By Euc. 1. 29, 6, FC may be shewn equal to each of the lines EF, FG.
47. Join GA and AF, and prove GA and AF to be in the same straight line,
48. Let the straight line drawn through D parallel to BC meet the side AB in E, and AC in F. Then in the triangle EBD, EB is equal to ED, by Euc. 1. 29, 6. Also, in the triangle EAD, the angle EAD may be shewn equal to the angle EDA, whence EA is equal to ED, and therefore AB is bisected in E. In a similar way it may be shewn, by bisecting the angle C, that AC is bisected in F. Or the bisection of AC in F may be proved when AB is shewn to be bisected in E.
49. The triangle formed will be found to have its sides respectively parallel to the sides of the original triangle.
50. If a line equal to the given line be drawn from the point where the two lines meet, and parallel to the other given line; a parallelogram may be formed, and the construction effected.
51. Let ABC be the triangle; AD perpendicular to BC, AE drawn to the bisection of BC, and AF bisecting the angle BAC. Produce AD and make DA' equal to AD : join FA', EA'.
52. If the point in the base be supposed to be determined, and lines drawn from it parallel to the sides, it will be found to be in the line which bisects the vertical angle of the triangle.
53. Let ABC be the triangle, at Č draw CD perpendicular to CB and equal to the sum of the required lines, through D draw DE parallel to CB meeting AC in E, and draw EF parallel to DC, meeting BC in F. Then EF is equal to DC. Next produce CB, making CG equal to CE, and join EG cutting AB in H. From H draw HK perpendicular to EAC, and
HL perpendicular to BC. Then HK and HL.together are equal to DC, The proof depends on Theorem 27, p. 75.
54. Let Ĉ be the intersection of the circles on the other side of the base, and join AC, BC. Then the angles CBA, C'BA being equal, the angles CBP, C'BP are also equal, Euc. 1. 13: next by Euc. 1. 4, CP, PC are proved equal ; lastly prove CC to be equal to CR or PC'.
55. In the fig. Euc. 1. 1, produce AB both ways to meet the circles in D and E, join CD, CE, then CDE is an isosceles triangle, having each of the angles at the base one-fourth of the angle at the vertex, At E draw EG perpendicular to DB and meeting DC produced in G. Then CEG is an equilateral triangle.
56. Join CC', and shew that the angles CCF, CCʻG are equal to two right angles; also that the line FCG is equal to the diameter.
57. Construct the figure and by Euc. 1. 32. If the angle BAC be a right angle, then the angle BDC is half a right angle.
58. Let the lines which bisect the three exterior angles of the triangle ABC form a new triangle A'B'C'. Then each of the angles at A', B', C' may be shewn to be equal to half of the angles at A and B, B and C, C and A respectively. And it will be found that half the sums of every two of three unequal numbers whose sum is constant, have less differences than the three numbers themselves.
59. The first case may be shewn by Euc. 1. 4; and the second by Euc. I. 32, 6, 15.
60. At D any point in a line EF, draw DC perpendicular to EF and equal to the given perpendicular on the hypotenuse. With centre C and radius equal to the given base describe a circle cutting EF in B. At C draw CA perpendicular to CB and meeting EF in A. Then ABC is the triangle required.
61. Let ABC be the required triangle having the angle ACB a right angle. In BC produced, take CE equal to AC, and with center B and radius BA describe a circular arc cutting CE in D, and join AD. Then DE is the difference between the sum of the two sides AC, CB and the hypotenuse AB; also one side AC the perpendicular is given. Hence the construction. On any line EB take ÈC equal to the given side, ED equal to the given difference. At C, draw CA perpendicular to CB, and equal to EC, join AD, at A in ĄD make the angle DAB equal to ADB, and let AB meet EB in B. Then ABC is the triangle required.
62. (1) Let ABC be the triangle required, having ACB the right angle. Produce AB to B making AD equal to AC or CB : then BD is the sum of the sides. Join DC: then the angle ADC is one-fourth of a right angle, and DBC is one-half of a right angle. Hence to construct: at B in BD make the angle DBM equal to half a right angle, and at D the angle BDC equal to one-fourth of a right angle, and let DC meet BM in c. At C draw CA at right angles to BC meeting BD in A : and ABC is the triangle required.
(2) Let ABC be the triangle, C the right angle: from AB cut off AD equal to AC; then BD is the difference of the hypotenuse and one side. Join CD; then the angles ACD, ADC are equal, and each is half the supplement of DAC, which is half a right angle. Hence the construction.
63. Take any straight line terminated at A. Make AB equal to the difference of the sides, and AC equal to the hypotenuse. At B make the angle CBD equal to half a right angle, and with center A and radius AC describe a circle cutting BD in D: join AD, and draw DE perpendicular to AC. Then ADE is the requiredtriangle.
64. Let BC the given base be bisected in D. Åt D draw DE at right angles to BC and equal to the sum of one side of the triangle and the perpendicular from the vertex on the base : join DB, and at B in BE make the angle EBA equal to the angle BED, and let BA meet DE in A: join AC, and ABC is the isosceles triangle.
65. This construction may be effected by means of Prob. 4, p. 71.
66. The perpendicular from the vertex on the base of an equilateral triangle bisects the angle at the vertex which is two-thirds of one right angle.
67. Let ABC be the equilateral triangle of which a side is required to be found, having given BD, CD the lines bisecting the angles at B, C. Since the angles DBC, DCB are equal, each being one-third of a right angle, the sides BD, DC are equal, and BDC is an isosceles triangle having the angle at the vertex the supplement of a third of two right angles. Hence the side BC may be found.
68. Let the given angle be taken, (1) as the included angle between the given sides; and (2) as the opposite angle to one of the given sides. In the latter case, an ambiguity will arise if the angle be an acute angle, and opposite to the less of the two given sides.
69. Let ABC be the required triangle, BC the given base, CD the given difference of the sides AB, AC: join BD, then DBC by Euc. 1. 18, can be shewn to be half the difference of the angles at the base, and AB is equal to AD. Hence at B in the given base BC, make the angle CBD equal to half the difference of the angles at the base. On CB take CE equal to the difference of the sides, and with center C and radius CE, describe a circle cutting BD in D: join CD and produee it to A, making DA equal to DB. Then ABC is the triangle required.
70. On the line which is equal to the perimeter of the required triangle describe a triangle having its angles equal to the given angles. Then bisect the angles at the base ; and from the point where these lines meet, draw lines parallel to the sides and meeting the base.
71. Let ABC be the required triangle, BC the given base, and the side AB greater than AC. Make AD equal to AC, and draw CD. Then the angle BCD may be shewn to be equal to half the difference, and the angle DCA equal to half the sum of the angles at the base. Hence ABC, ACB the angles at the base of the triangle are known.
72. Let the two given lines meet in A, and let B be the given point.
If BC, BD be supposed to be drawn making equal angles with AC, and if AD and DC be joined, BCD is the triangle required, and the figure ACBD may be shewn to be a parallelogram. Whence the construction.
73. It can be shewn that lines drawn from the angles of a triangle to bisect the opposite sides, intersect each other at a point which is twothirds of their lengths from the angular points from which they are drawn. Let ABC be the triangle required, AD, BE, CF the given lines from the angles drawn to the bisections of the opposite sides and intersecting in G. Produce GD, making DH equal to DG, and join BH, CH: the figure GBHC is a parallelogram. Hence the construction.
74. Let ABC (fig. to Euc. 1. 20.) be the required triangle, having the base BC equal to the given base, the angle ABC equal to the given angle, and the two sides BA, AC together equal to the given line BD. Join DC, then since AD is equal to AC, the triangle ACD is isosceles, and therefore the angle ADC is equal to the angle ACD. Hence the construction.
75. Let ABC be the required triangle (fig. to Euc. I. 18), having the angle ACB equal to the given angle, and the base BC equal to the given