69. If the base of an isosceles triangle be produced to any point, the difference of the distances of the point from the two sides is constant. 70. Shew that the sum of the distances of any point within an equilateral triangle from the three sides is constant. 71. ABCD is a parallelogram, AD is bisected at E, and BC at F: shew that BE and DF trisect the diagonal AC. *72. In the triangle ABC, D and E are the middle points of BC and CA shew that BE cuts off a third part from DA. *73. Shew that the three straight lines drawn from the vertices of a triangle to the middle points of the opposite sides meet in a point. SECTION IV PROBLEMS A Geometrical Problem is a proposition, of which the object is to effect some Geometrical construction. The solution of a Problem depends on the instruments, the use of which is allowed; and it will be readily understood that the ïmore restricted the choice of instruments, the more limited will be the Problems which can be solved by their use, and the more difficult will the solution of many that are possible be found. Owing to the existence of this arbitrary element in the treatment of Problems, they are grouped together in a separate section. Though important as applications of Geometrical truths, it should be clearly understood that Problems form no part of the chain of connected truths embodied in the Theorems of Geometry, so that, though they may advantageously be studied in connection with the Theorems on which they directly depend, they are not a necessary part of the pure Science of Geometry. It is the recognised convention of Elementary Geometry that the only instruments to be employed are the ruler, for drawing and producing straight lines, and the compasses for describing circles and for the transference * of distances. This Convention is embodied in the following POSTULATES OF CONSTRUCTION Let it be granted that 1. A straight line may be drawn from any one point to any other point. 2. A terminated straight line may be produced to any length in a straight line. 3. A circle may be drawn with any centre, with a radius equal to any finite straight line. DEF. 41. A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another. This point is called the centre of the circle. DEF. 42. A radius of a circle is a straight line drawn from the centre to the circumference. DEF. 43. A diameter of a circle is a straight line drawn through the centre and terminated both ways by the circumference. It follows from the definition of a circle that a point is within * NOTE.-Euclid restricts the use of the compasses to describing circles, and shews in his Props. 2 and 3 how with this restriction to draw a line of given length and to cut off a given length from a given line. a circle, when its distance from the centre is less than the radius of the circle. It is evident that (1) If a point in a straight line is within a closed figure the straight line if produced in either sense from the point will meet the boundary of the figure, and thus intersect it in two points at least ; (2) If a point in the boundary of one closed figure lie within another closed figure, and also a point in the boundary of the latter lie within the former, the two boundaries intersect in two points at least ; For they cannot lie wholly outside each of the other and if one were wholly inside the other, no point in the boundary of the second would lie within the first. Hence they must lie partly inside and partly outside each of the other, and their boundaries (circumferences) must meet in two points at least. By the help of the above, it may be shewn that the straight lines and circles drawn in the Problems of this section intersect; or the conditions that must be satisfied in order that they may intersect may be determined. PROB. 1. To bisect a given angle. Let BAC be the given angle: it is required to bisect it. With centre A, and with any radius, draw a circle cutting AB at D, and AC at E. Post. 3. With centres D and E, and with any radius greater than half the straight line DE, draw circles, Post. 3. let F be a point of intersection of these circles which lies within the angle BAC; join AF: then shall AF bisect the angle BAC. Post. I, Join DF, EF. Then in the triangles DAF, EAF, the side AD is equal to the side AE, the side AF is common to both, and the side DF is equal to the side EF, therefore the angle DAF is equal to the angle EAF, therefore AF bisects the angle BAC. Constr. Constr. I. 18. Q.E.F. Ex. 74. Divide a given angle into four equal parts. Ex. 75. Prove that FA produced will bisect the major conjugate angle BAC. PROB. 2. To draw a perpendicular to a given straight line from a given point in it. Let BAC be the given straight line, A the given point in it: it is required to draw from A a perpendicular to BAC. B With centre A, and with any radius, draw a circle cutting AB at D, and AC at E, Post. 3. With centres D and E, and with any radius greater than AD or AE, draw circles, Post. 3. let F be a point of intersection of these circles; and the side DF is equal to the side EF, therefore the angle DAF is equal to the angle EAF, therefore AF is perpendicular to BAC. Constr. Constr. I. 18. Defs. 10, 11. Q.E.F. Ex. 76. Shew that Prob. 2 is a particular case of Prob. 1. PROB. 3. To draw a perpendicular to a given straight line from a given point outside it. Let AB be the given straight line, C the given point outside it: it is required to draw a perpendicular to AB from C. F |