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53. ABC is a triangle, right-angled at A, and having the angle B double of the angle C: shew that the side CB is double of the side AB.

54. If the three angles of a triangle be bisected, and one of the bisecting lines be produced to meet the opposite side, the angle contained by this line produced and one of the others is equal to the angle contained by the third line and a perpendicular from their common point of intersection to the side aforesaid.

55. If the opposite sides or opposite angles of a quadrilateral be equal, the figure will be a parallelogram.

56. If in the sides of a square, at equal distances from the four angles, four points be taken, one in each side, the figure formed by joining them will also be a


57. Let AD, AE be squares upon the sides of the right-angled triangle ABC, and drop DF, EG, perpendiculars on the hypothenuse BC produced: then BC and the triangle ABC are respectively equal to the sum of DF and EG, and of the triangles DBF and ECG.

58. If two opposite sides of a parallelogram be bisected, the lines drawn from the points of bisection to the opposite angles will bisect the diagonal.

59. Given the perpendicular from the vertex on the base, and the difference between each side and the adjacent segment of the base: construct the triangle.

60. Let AD be drawn perpendicular to the base BC of a triangle, in which the angle B is double of the angle C; in AB, produced or not according as the angle B is greater or less than a right angle, take BE equal to BD, and draw EDF cutting AC in F: then FA, FC, FD are equal to one another, and the triangles ABC, AFE are equiangular.

61. In [60] shew that the less side AB is equal to the sum or difference of the segments of the base, according as the greater angle B is greater or less than a right angle.

62. The lines which bisect the angles of any parallelogram, form a rectangular parallelogram, whose diameters are parallel to the sides of the former.

63. AD, BC are two parallel lines cut obliquely by AB and perpendicularly by AC; BED is drawn cutting AC in E, so that ED is equal to twice AB: prove that the angle DBC is one-third of the angle ABC.

64. In a given triangle place a line which shall be terminated by the two sides, and be equal to one given line and parallel to another.

65. Any line drawn through the bisection of the diagonal of a parallelogram to meet the sides is bisected in that point, and also bisects the parallelogram.

66. Through a given point draw a line, so that the part of it intercepted between two given parallel lines may be equal to a given line.

67. In any right-angled triangle, the middle point of the hypothenuse is equally distant from the three angles.

68. In any triangle ABC, if BE, CF be perpendiculars on any line through A, and D be the bisection of BC, shew that DE DF.


69. If from the right angle of a triangle two lines be drawn, one bisecting the base and the other perpendicular to it, they will contain an angle equal to the difference of the two acute angles of the triangle.

70. Find the point in the base of a triangle, from which lines drawn parallel to the sides to meet them are equal.

71. The area of a trapezium is half that of a parallelogram, whose base is the sum of the two parallel sides, and altitude the perpendicular distance between them.

72. On the sides AB, AC of a triangle describe parallelograms ABDE, ACFG, and produce DE, FG to meet in H: then the area of these parallelograms together is equal to the area of the parallelogram on BC, whose side is equal and parallel to AH.

73. Upon a given base describe an isosceles triangle equal to a given triangle.

74. Shew that the perimeter of an isosceles triangle is less than that of any other equal triangle upon the same base.

75. Of all triangles having the same base and perimeter, the greatest is that which is isosceles.

76. From a given point, in one of the equal sides of an isosceles triangle, draw a line, meeting the other side produced, which shall make with these sides a triangle equal to the given triangle.

77. If one angle of a triangle be a right angle, and another be two-thirds of a right angle, shew that the equilateral triangle on the hypothenuse is equal in area to the sum of those on the sides.

78. Convert a trapezium into a triangle of equal area with one angle common; and hence shew how to transform any rectilineal figure into a triangle, whose vertex shall be in a given angle of the figure and base in one of the sides.

79. Given a triangle ABC and a point D in AB: construct another triangle ADE equal to the former, and having the common angle A.

80. Change a triangle into another equal one of given altitude.

81. In any given line, AB is taken half of AC: if through B, C, parallel lines be drawn, cutting any other line through A in D, E, then AD is half of AE, and BD of CE, and the triangle ABD a fourth of the triangle ACE; and, conversely, if BD be such that AD is half of AE, then BD is parallel to CE.

82. If the sides of any quadrilateral be bisected and the points of bisection joined, the included figure is

a parallelogram, and equal in area to half the original figure shew also that the lines joining the bisections of opposite sides bisect each other.

83. Through D, E, the bisections of the sides AB, AC of a triangle, draw DF, EF parallel to BE, AB; and shew that the sides of the triangle DCF are equal to the three lines drawn from the angles to bisect the sides.

84. Bisect a triangle by a line drawn from a given point in one of its sides.

85. If from any point in the diagonal of a parallelogram lines be drawn to the angles, the parallelogram will be divided into two pairs of equal triangles.

86. Through E, the bisection of the diagonal BD of a quadrilateral ABCD, draw FEG parallel to AC; and shew that AG will bisect the figure.

87. ABC is a given triangle; draw BD, CE perpendicular to BC and on the same side of it, each equal to twice the altitude of the triangle; bisect AB, AC in F, G; and shew that the triangle ABC is equal to the sum or difference of the triangles BDF, CEG, according as the angles at the base of ABC are both or only one acute.

88. If of the four triangles, into which the diagonals divide a quadrilateral, two opposite ones are equal, the quadrilateral has two opposite sides parallel.

89. Upon stretching two chains AC, BD across a field, ABCD, I find that AC, BD make equal angles with CD, and that AC makes with AD the same angle that BC does with BD: hence prove that AB is parallel to CD.

90. The three lines, joining the angular points of a triangle with the middle points of the opposite sides, intersect in one point, and trisect the triangle.

91. Shew that any one of the lines in [90] is divided in the point of intersection, so that one of the parts is double of the other.

92. If two sides of a triangle be given, its area will be greatest when they contain a right angle.

93. The two triangles, formed by drawing lines from any point within a parallelogram to the extremities of two opposite sides, are together half the parallelogram.

94. If from the ends of one of the oblique sides of a trapezium two lines be drawn to the bisection of the opposite side, the triangle thus formed with the first side is half the trapezium.

95. If from the extremities of the base of an isosceles triangle lines be drawn perpendicular to the sides, the line which joins the vertex with their point of intersection will bisect the base at right angles.

96. If two exterior angles of a triangle be bisected, the line drawn from the point of intersection of the bisecting lines, to the opposite angle of the triangle, will bisect it.

97. In the figure, Euc. 1. 47, shew that, if BG and CH be joined, these lines will be parallel.

98. In ditto, if DB, EC be produced to meet FG and KH in M, N, the triangles BFM, CKN are equiangular and equal to the triangle ABC.

99. In ditto, if GH, KE, FD be joined, each of the triangles so formed is equal to the given triangle ABC.

100. In ditto, produce FG, KH to meet in M, join MB, MC, and produce MA to cut BC in L: then shew that ML is perpendicular to BC, and thence, assuming [6. 20], that the three lines AL, BK, CF intersect in one point.

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