angles. Thus, (Fig. 9,) EFB and DGH are equal to two right angles, as are also AFE and CGH. So also the pairs of interior angles AFG and FGC, BFG and FGD, are each equal to two right angles. (29.1.) 74. The angles at the base of an isosceles triangle are equal to each other. (5.1.) Fig. 10. A. 75. If one side of a triangle be produced, the exterior angle so formed will be equal to the two angles adjacent to the opposite side, and the three interior angles are equal to two right angles. Thus, (Fig. 10,) ACD = ABC + BAC, and ABC + BAC + ACB = two right angles. (32.1.) B C 76. The interior angles of any rectilineal figure are equal to twice as many right angles as the figure has sides, diminished by four right angles. The interior angles of a quadrilateral are therefore equal to four right angles. (Cor. 1, 32.1.) 77. The opposite sides and angles of a parallelogram are equal to each other. (34.1.) 78. Conversely, any quadrilateral of which the opposite sides or the opposite angles are equal is a parallelogram. 79. Parallelograms having equal bases and altitudes, and also triangles having equal bases and altitudes, are equal to each other. (35–38.1.) 80. A parallelogram is double a triangle having the same base and altitude. (41.1.) 81. The square on the hypothenuse of a right-angled triangle is equal to the sum of the squares of the legs. (47.1.) 82. Any figure described on the hypothenuse of a rightangled triangle is equal to the sum of the similar figures similarly described on the sides. (31.6.) 83. The angle at the centre of a circle is double the angle at the circumference on the same base. Thus, the angle at C (Fig. 11) is double either D or E. (20.3.) D Fig. 11. B E 84. Angles in the same segment of a circle are equal. Thus, D and E (Fig. 11) are equal. 85. The angle in a semicircle is a right angle; the angle in a segment greater than a semicircle is acute; and that in a segment less than a semicircle is obtuse. 86. The sides about the equal angles of equiangular triangles are proportional. (4.6.) B.-GEOMETRICAL PROBLEMS. Under this head are given those methods of construction which are applicable to paper drawings. The methods to be used in field operations will be given in a subsequent chapter. 87. Problem 1.-To bisect a given straight line. Let AB (Fig. 12) be the given line. With the centres A and B, and radius greater than half AB, describe arcs cutting in C and D. Join CD cutting AB in E, and the thing is done. (10.1.) Fig. 12. ¡E C B Problem 2. To draw a perpendicular to a straight line from a given point in it. a. When the point is not near the end. 88. Let AB (Fig. 13) be the line and C the given point. Lay off CD = CE, and with D and E as centres, and any radius. greater than DC, describe arcs cutting in F. Draw CF, and the thing is done. (11.1) b. When the point is near the end of the line. 89. First Method.-Take any point D (Fig. 14) not in the line, and with the centre D and radius DC describe the circle ECF, cutting AB in E. Join ED and produce it to F. Then will CF be the perpendicular. For ECF, being an angle in a semicircle, is a right angle. (85.) 90. Second Method.-With C (Fig. 15) and any radius describe DEF; with D and the same radius cross the circle in E; and with E as a centre, and the same radius, cross it in F. With E and F as centres, and any radius, describe arcs cutting in G. Then will CG be the perpendicular. A Problem 3.-To let fall a perpendicular to a line from a point without it. a. When the point is not nearly opposite the end of the line. Fig. 16. с 91. Let AB (Fig. 16) be the line and C the given point. With the centre C describe an arc cutting AB in D and E. With the centres D and E and any radius describe arcs cutting in F. Join CF, and the thing is done. (12.1.) G b. When the point is nearly opposite the end of the line. 93. Second Method.-Let F (Fig. 14) be the point. From F to any point E in the line AB draw FE. On it describe a semicircle cutting AB in C. Join F and C, and FC will be the perpendicular (85.) Problem 4.-At a given point in a given straight line to make an angle equal to a given angle. 94. Let BCD (Fig. 18) be the given angle, and A the given point in AE. With the centre C and any radius describe BD, cutting the sides of the angle in B and D. With A as a centre and the same radius describe EF; make EF = = DB; draw AF, and the thing is done. Fig. 18. B D E Problem 5.-To bisect a given angle. 95. Let BAC (Fig. 19) be the given angle. With the centre A and any radius describe an arc cutting the sides in B and C. With the centres B and C, and the same or any other radius, describe arcs cutting in D. Join AD, and the thing is done. (9.1.) Problem 6.-To draw a straight line touching a circle from a given point without it. 96. Let ABC be the given circle, and D the given point. Join D and the centre E. On DE describe a semicircle cutting the circumference in B. Join DB, and it will be the tangent required. Fig. 20. E A For DBE, being an angle in a semicircle, is a right angle, (31.3;) therefore, DB touches the circle, (16.3.) If the point were in the circumference at B. Join EB, and draw BD perpendicular to it. BD will be the tangent. Problem 7.-Through a given point to draw a line parallel to a given straight line. F Fig. 21. A 97. First Method.-Let A (Fig. 21) be the given point, and BC the given line. From A to BC let fall a perpendicular AD; and at any other point E in BC erect a perpendicular EF equal to AD. Through A and F draw AF, which will be the parallel required. 98. Second Method.-From A (Fig. 22) to D, any point in BC, draw AD. Make DAE = parallel to BC. ADC, and AE will be B E D Fig. 22. E A |