NOTE 8. On Loci. Suppose we have to determine the position of a point, which is equidistant from the extremities of a given straight line BC. There is an infinite number of points satisfying this condition, for the vertex of any isosceles triangle, described on BC as its base, is equidistant from B and C. Let ABC be one of the isosceles triangles described on BC. If BC be bisected in D, MN, a perpendicular to BC drawn through D, will pass through A. It is easy to shew that any point in MN, or MN produced in either direction, is equidistant from B and C. It may also be proved that no point out of MN is equidistant from B and C. The line MN is called the Locus of all the points, infinite in number, which are equidistant from B and C. DEF. In plane Geometry Locus is the name given to a line, straight or curved, all of whose points satisfy a certain geometrical condition (or have a common property), to the exclusion of all other points. Next, suppose we have to determine the position of a point, which is equidistant from three given points A, B, C, not in the same straight line. E A B If we join A and B, we know that all points equidistant from A and B lie in the line PD, which bisects AB at right angles. If we join B and C, we know that all points equidistant from B and C lie in the line QE, which bisects BC at right angles. Hence O, the point of intersection of PD and QE, is the only point equidistant from A, B and C. PD is the Locus of points equidistant from A and B, QE..... B and C, and the Intersection of these Loci determines the point, which is equidistant from A, B and C. Find the loci of Examples of Loci. (1) Points at a given distance from a given point. (2) Points at a given distance from a given straight line. (3) The middle points of straight lines drawn from a given point to a given straight line. (4) Points equidistant from the arms of an angle. (5) Points equidistant from a given circle. (6) Points equally distant from two straight lines which intersect. NOTE 9. On the Methods employed in the solution of In the solution of Geometrical Exercises, certain methods may be applied with success to particular classes of questions. We propose to make a few remarks on these methods, so far as they are applicable to the first two books of Euclid's Elements. The Method of Synthesis. In the Exercises, attached to the Propositions in the preceding pages, the construction of the diagram, necessary for the solution of each question, has usually been fully described, or sufficiently suggested. The student has in most cases been required simply to apply the geometrical fact, proved in the Proposition preceding the exercise, in order to arrive at the conclusion demanded in the question. This way of proceeding is called Synthesis (ovveσis=composition), because in it we proceed by a regular chain of reasoning from what is given to what is sought. This being the method employed by Euclid throughout the Elements, we have no need to exemplify it here. The Method of Analysis. The solution of many Problems is rendered more easy by supposing the problem solved and the diagram constructed. It is then often possible to observe relations between lines, angles and figures in the diagram, which are suggestive of the steps by which the necessary construction might have been effected. This is called the Method of Analysis (áváλvois=resolution). It is a method of discovering truth by reasoning concerning things unknown or propositions merely supposed, as if the one were given or the other were really true. The process can best be explained by the following examples. Our first example of the Analytical process shall be the 31st Proposition of Euclid's First Book. Ex. 1. To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC be the given straight line. Suppose the problem to be effected, and EF to be the straight line required. Now we know that any straight line AD drawn from A to meet BC makes equal angles with EF and BC. (1. 29.) This is a fact from which we can work backward, and arrive at the steps necessary for the solution of the problem; thus: Take any point D in BC, join AD, make ▲ EAD= 2 ADC, and produce EA to F: then EF must be parallel to BC. Ex. 2. To inscribe in a triangle a rhombus, having one of its angles coincident with an angle of the triangle. Let ABC be the given triangle. Suppose the problem to be effected, and DBFE to be the rhombus. D B Then if EB be joined, ▲ DBE ▲ FBE. This is a fact from which we can work backward, and deduce the necessary construction; thus: Bisect ABC by the straight line BE, meeting AC in E. Draw ED and EF parallel to BC and AB respectively. Then DBFE is the rhombus required. (See Ex. 4, p. 59.) Ex. 3. To determine the point in a given straight line, at which straight lines, drawn from two given points, on the same side of the given line, make equal angles with it. Let CD be the given line, and A and B the given points. Suppose the problem to be effected, and P to be the point required. D We then reason thus: If BP were produced to some point A', ▲ CPA', being= ▲ BPD, will be= ▲ APC. Again, if PA' be made equal to PA, AA' will be bisected by CP at right angles. This is a fact from which we can work backward, and find the steps necessary for the solution of the problem; thus: From A draw AO 1 to CD. Produce AO to A', making OA'=0A. Join BA', cutting CD in P. Then P is the point required. NOTE 10. On Symmetry. The problem, which we have just been considering, suggests the following remarks: If two points, A and A', be so situated with respect to a straight line CD, that CD bisects at right angles the straight line joining A and A', then A and A' are said to be symmetrical with regard to CD. The importance of symmetrical relations, as suggestive of methods for the solution of problems, cannot be fully shewn |