PROB. 6. To draw a straight line through a given point parallel to a given straight line. Let A be the given point, BC the given straight line: it is required to draw through A a straight line parallel to BC. with centre A and radius AD draw a circle; Post. 3. Post. 3. from DC cut off DE equal to AD; with centre E and radius ED draw a circle cutting the former circle therefore the angle AFD is equal to the angle EDF, Constr. Constr. I. 18. and these are alternate angles made by DF with AF and BC, therefore AF is parallel to BC. I. 21. Q.E.F. Ex. 79. Through a given point draw a straight line so that the part intercepted between two given parallel straight lines may be equal to a given straight line. PROB. 7. To construct a triangle when three sides are given. Let A, B and C be the three given sides: it is required to construct a triangle having its sides equal to A, B and C. With centre D, and radius equal to B, draw a circle, Post. 3. with centre E, and radius equal to C, draw a circle cutting the Ex. 80. Point out how the construction fails if one side is greater than the sum of the other two. *Ex. 81. Construct an equilateral triangle on a given base. *Ex. 82. Trisect a right angle. PROB. 8. To construct a triangle when two sides and the angle between them are given. Let A, B be the two given sides, CED the given angle: it is required to construct a triangle having two of its sides equal to A and B, and the angle between them equal to CED. PROB. 9. To construct a triangle when two sides and an angle opposite to one of them are given. Let A, B be the two given sides, CED the given angle opposite to the side equal to B: it is required to construct a triangle having two of its sides equal to A and B, and the angle opposite to B equal to CED. A B From ED cut off EF equal to A, with centre F, and radius equal to B, draw a circle. If B is less than the perpendicular from F upon EC, Post. 3 Post. 3. the circle does not meet EC, and the solution is impossible. If B is equal to the perpendicular the circle meets EC in one point only, and there is one, and only one, solution. If B is greater than the perpendicular, and less than A, the circle cuts EC at two points on the same side of the vertex of the given angle, and there are two solutions, or the solution is ambiguous. If B is greater than A, the circle cuts EC at two points on opposite sides of the vertex of the given angle, and there is one, and only one, solution. Let G be a point in which the circle meets EC, Ex. 83. Construct a right-angled triangle having given the hypotenuse and one side. PROB. 10. To construct a triangle, when two angles and the side between their vertices are given. Let A, B be the two given angles, CD the given side : it is required to construct a triangle having two of its angles equal to A and B, and the side between their vertices equal to CD. At the point C, in the straight line CD, make the angle DCE equal to the angle A; Prob. 5. at the point D, in the straight line CD, make the angle CDE Ex. 84. Shew that the solution is possible only when the two given angles are together less than two right angles. Ex. 85. Construct an isosceles right-angled triangle on a given straight line as hypotenuse. Ex. 86. On a given straight line as hypotenuse construct a right G |