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terior 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 DG, GA, of which the square on FD is equal to the square on DG; hence AF is equal to AG, and Euc. 1. 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. I. 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. 1. 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. I. 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 II. 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 C' 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 CP 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 CC'F, CC'G are equal to two right angles; also that the line FC'G 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. 1. 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 EC 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 AD 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 D 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 required triangle.
64. Let BC the given base be bisected in D. At 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.
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 CED 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 produce 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
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
line, also CD equal to the difference of the two sides AB, AC. If BD be joined, then ABD is an isosceles triangle. Hence the synthesis. Does this construction hold good in all cases?
76. Let ABC be the required triangle, (fig. Euc. 1. 18), of which the side BC is given and the angle BAC, also CD the difference between the sides AB, AC. Join BD; then AB is equal to AD, because CD is their difference, and the triangle ABD is isosceles, whence the angle ABD is equal to the angle ADB; and since BAD and twice the angle ABD are equal to two right angles, it follows that ABD is half the supplement of the given angle BAC. Hence the construction of the triangle.
77. Let AB be the given base: at A draw the line AD to which the line bisecting the vertical angle is to be parallel. At B draw BE parallel to AD; from A draw AE equal to the given sum of the two sides to meet BE in E. At B make the angle EBC equal to the angle BEA, and draw CF parallel to AD. Then ACB is the triangle required.
78. Take any point in the given line, and apply Euc. 1. 23, 31.
79. On one of the parallel lines take EF equal to the given line, and with center E and radius EF describe a circle cutting the other in G. Join EG, and through A draw ABC parallel to EG.
80. This will appear from Euc. 1. 29, 15, 26.
81. Let AB, AC, AD, be the three lines. Take any point E in AC, and on EC make EF equal to EA. through F draw FG parallel to AB, join GE and produce it to meet AB in H. Then GE is equal to GH.
82. Apply Euc. 1. 32, 29.
83. From E draw EG perpendicular on the base of the triangle, then ED and EF may each be proved equal to EG, and the figure shewn to be equilateral. Three of the angles of the figure are right angles.
84. The greatest parallelogram which can be constructed with given sides can be proved to be rectangular.
85. Let AB be one of the diagonals: at A in AB make the angle BAC less than the required angle, and at A in AC make the angle CAD equal to the required angle. Bisect AB in E and with center E and radius equal to half the other diagonal describe a circle cutting AC, AD in F, G. Join FB, BG: then AFBG is the parallelogram required.
86. This problem is the same as the following; having given the base of a triangle, the vertical angle and the sum of the sides, to construct the triangle. This triangle is one half of the required parallelogram.
87. Draw a line AB equal to the given diagonal, and at the point A make an angle BAC equal to the given angle. Bisect AB in D, and through D draw a line parallel to the given line and meeting AC in C. This will be the position of the other diagonal. Through B draw BE parallel to CA, meeting CD produced in E; join AE, and BC. Then ACBE is the parallelogram required.
88. Construct the figures and by Euc. 1. 24.
89. By Euc. 1. 4, the opposite sides may be proved to be equal.
90. Let ABCD be the given parallelogram; construct the other parallelogram A'B'C'D' by drawing the lines required, also the diagonals AC, A'C', and shew that the triangles ABC, A'B'C' are equiangular.
91. A'D' and B'C' may be proved to be parallel. 92. Apply Euc. 1. 29, 32.
93. The points D, D', are the intersections of the diagonals of two rectangles: if the rectangles be completed, and the lines OD, OD' be produced, they will be the other two diagonals.
94. Let the line drawn from A fall without the parallelogram, and
let CC', BB', DD', be the perpendiculars from C, B, D, on the line drawn from A; from B draw BE parallel to AC', and the truth is manifest. Next, let the line from A be drawn so as to fall within the parallelogram. 95. Let the diagonals intersect in E. In the triangles DCB, ČDA, two angles in each are respectively equal and one side DE: wherefore the diagonals DB, AC are equal: also since DE, EC are equal, it follows that EA, EB are equal. Hence DEC, AEB are two isosceles triangles having their vertical angles equal, wherefore the angles at their bases are equal respectively, and therefore the angle CDB is equal to DBA.
96. (1) By supposing the point P found in the side AB of the parallelogram ABCD, such that the angle contained by AP, PC may be bisected by the line PD; CP may be proved equal to CD; hence the solution.
(2) By supposing the point P found in the side AB produced, so that PD may bisect the angle contained by ABP and PC; it may be shewn that the side AB must be produced, so that BP is equal to BD.
97. This may be shewn by Euc. 1. 35.
98. Let D, E, F be the bisections of the sides AB, BC, CA of the triangle ABC: draw DE, EF, FD; the triangle DEF is one-fourth of the triangle ABC. The triangles DBE, FBE are equal, each being one-fourth of the triangle ABC: DF is therefore parallel to BE, and DBEF is a parallelogram of which DE is a diagonal.
99. This may be proved by applying Euc 1. 38. 100. Apply Euc. 1. 37, 38.
101. On any side BC of the given triangle ABC, take BD equal to the given base; join AD, through C draw CE parallel to AD, meeting BA produced if necessary in E, join ED; then BDE is the triangle required. By a process somewhat similar the triangle may be formed when the altitude is given.
102. Apply the preceding problem (101) to make a triangle equal to one of the given triangles and of the same altitude as the other given triangle. Then the sum or difference can be readily found.
103. First construct a triangle on the given base equal to the given triangle; next form an isosceles triangle on the same base equal to this triangle.
104. Through A draw AD parallel to BC the base of the triangle; from B draw BD at right angles to BC to meet AD and join DC.
105. Make a triangle equal to the given parallelogram upon the given line, and then a triangle equal to this triangle, having an angle equal to the given angle.
106. If the figure ABCD be one of four sides; join the opposite angles A, C of the figure, through D draw DE parallel to AC meeting BC produced in E, join AE:-the triangle ABE is equal to the foursided figure ABCD.
If the figure ABCDE be one of five sides, produce the base both ways, and the figure may be transformed into a triangle, by two constructions similar to that employed for a figure of four sides. If the figure consists of six, seven, or any number of sides, the same process must be repeated.
107. Draw two lines from the bisection of the base parallel to the two sides of the triangle.
108. This may be shewn ex absurdo.
109. On the same base AB, and on the same side of it, let two triangles ABC, ABD be constructed, having the side BD equal to BC, the angle ABC a right angle, but the angle ABD not a right angle; then the triangle ABC is greater than ABD, whether the angle ABD be acute or obtuse.
110. Let ABC be a triangle whose vertical angle is A, and whose