SECTION IV. PROBLEMS. In the Science of Geometry there are not only theorems to be proved, but constructions to be effected, which are called problems. Geometers have always imposed certain limitations on themselves with respect to the instruments which might be used in these constructions. There is no reason why any convenient instrument used in the Art of Geometry, such as the square, parallel ruler, sector, protractor, should not be supposed to be used also in the Science; but the ruler and compasses suffice for nearly all the simpler constructions, and those which cannot be effected by their means are considered as not forming a part of Elementary Geometry. These instruments are therefore postulated or requested (vid. p. 4). There are some problems, that seem at first sight not very difficult, that cannot be solved by the use of these instruments. We can, for example, bisect an angle; but we cannot, in general, trisect it, that is, divide it into three equal parts, by any combination of ruler and compasses. It may be observed that the ruler is simply a straight edge, not graduated, and the compasses are supposed to be transferable from one part of the figure to another, the distance between the points being unaltered. The solution of a problem in Elementary Geometry as above defined consists (1) in indicating how the ruler and compasses are to be used in effecting the construction required; (2) in proving that the construction so given is correct; (3) in discussing the limitations, which sometimes exist, within which alone the solution is possible. We shall give several examples of such problems, and then discuss the principles of the methods we have used. PROBLEM I. : . To bisect a given angle. Construction. Let ABC be the given angle. Take any equal lengths BA, BC, along its arms, and join AC. With centre A, and any radius greater than half AC, describe a circle, and with centre C, and the same radius, describe another circle intersecting the former circle on the side of AC remote from B in D. Join AD, CD, and BD; BD bisects the angle ABC. Proof. In the triangles ABD, CBD, because AB= BC, (Constr.) and BD is common, and the base AD = the base DC, (Constr.) therefore the angle ABD= the angle CBD, that is, BD bisects the angle ABC*. . * Eucl. 1. 9, PROBLEM 2. To draw a perpendicular to a given straight line from a given point given in it. Construction. With centre C and any radius describe a circle to cut the straight line in two points D, E, so that CD= CE. With centre D, and any radius ADO B greater than DC, describe a circle, and with centre E and the same radius describe a circle, cutting the former in F. Join FC; it is required to prove that FC is perpendicular to AB. Proof. In the triangles DCF, ECF, because DC=CE, (Constr.) CF is common, and the base DF= the base EF, (Constr.) therefore the angle DCF=the angle ECF, (Th. 15.) and therefore DCF and ECF are right angles*. (Def. 14.) NOTE.—This construction is usually effected in practice by means of the square. It may be observed that this problem is only a special case of Prob. I, the given angle being a straight angle. PROBLEM 3. To draw a perpendicular to a given straight line from a given point outside it. Let BC be the given straight line, A the given point. * Euclid, 1. 11. Construction. With centre À describe a circle with any sufficient radius to cut BC in two points D, E. Bisect the angle DAE by the line AF. (Prob. 1.) Proof. In the triangles AFD, AFE, (Constr.) and AF is common, and the contained angle DAF= the contained angle EAF; (Constr.) therefore the angle AFD= the angle AFE; (Th. 5.) therefore AF is perpendicular to DE*. NotE.—This construction also is usually effected in practice by means of the square. PROBLEM 4. To bisect a given straight linet. Construction. With centre A and any * Euclid, 1. 12. + Euclid, 1. 10. Then 0 will be the 'point of bisection. Join AD, DB. Proof. Because AC = CB; and CD is common to the two triangles ACB, DCB; and the base AD is equal to the base DB; therefore the angle ACD= the angle BCD; therefore in the two triangles ACO, BCO, we have AC = BC, (Constr.) CO common, and the included angles ACO, BCO equal; therefore the base AO= the base BO, (Th. 5.) or the line AB is bisected in 0. ABC PROBLEM 5. To construct a triangle, having given the lengths of the three sides. Let the three given lengths be the lines A, B, C. Construction. Draw a line PQ equal to one of them A. With centre Pand radius equal to B describe a circle; and with centre Q and radius equal to C describe a circle. Let these circles intersect in R. Join RP, RQ. RPQ is the triangle required, Proof. For RPQ has its three sides respectively equal to A, B and C*. * Euclid, 1. 22. |