the equal angles of two triangles having a common angle are equal, those triangles are equal; therefore ▲ IHD = ▲ IPM. Q. E. D. Comput. In the triangles ADI, ADG, are given all the angles, and the side AD; whence AI, AG, DI, and IC, = DI DC, become known. In the triangle IFC, all the angles and the side Ic are known; whence IF becomes known, as well as FH, since AH : HF :: mn. Lastly, IP = √(IH. IG), and IG: ID: IP: IM. Cor. 1. When the line of division PM is to be perpendicu lar to a side, or parallel to a given side; we have only to draw DG accordingly so that those two cases are included in this. Cor. 2. When the line PM is to be the shortest possible, it must cut off an isosceles triangle towards the acutest angle; and in that case IG must evidently be equal to ID. 3dly. By a line drawn through a given point. The method will be the same as that to case 4th prob. 1, and therefore need not be repeated here. Scholium. If a quadrilateral were to be divided into four parts in a given proportion, m, n, p, q: we must first divide it into two parts having the ratio of m+n, to p + q; and then each of the quadrangles so formed into their respective ratios, of m to n, and p to q. PROBLEM V. To Divide a Pentagon into Two Parts having a Given Ratio, from a Given Point in one of the Sides. Reduce the pentagon to a triangle by prob. 37, Geometry, and divide this triangle in the given ratio by case 1 prob. 1. PROBLEM VI. To Divide any Polygon into Two Parts having a Given Ratio. 1st. From a given point in the perimeter of the polygon. Construc. Join any two opposite angles A, D, of the polygon, by the line AD. : HD Reduce the part ABCD into an equivalent triangle NPs, whose vertex shall be the given point P, and base AD produced an operation which may be performed at once, if the portion ABCD be quadrangular; or by several operations (as from 8 sides to 6, from 6 to 4, &c,) if the sides be more than four. Divide the triangle NPS into two parts having the given ratio, by the line PH. In like manner, reduce FKME ADEFGA ADEFGA into an equivalent triangle having H for its vertex, and FE produced for its base; and divide this triangle into the given ratio by a line from н, as HK. The compound line PHK will manifestly divide the whole polygon into two parts having the given ratio. To reduce this to a right line, join PK, and through H draw Hм parallel to it; join PM; so will the right line PM divide the polygon as required, provided м fall between F and E. If it do not, the reduction may be completed by the process described in cases 5th and 6th prob. 2d. All this is too evident to need demonstration. Remark. There is a direct method of solving this blem, without subdividing the figure; but as it requires the computation of the area, it is not given here. 2dly. By the shortest line possible. Construc. From any point P', in one of those two sides of the polygon which, when produced, meet in the most acute angle 1, draw a line P'M', to the other of those sides (EF), dividing the polygon in the given ratio. Find TPP C A EMME pro the points P and M, so that IP or IM shall be a mean proportional between IP', IM'; then will PM be the line of division required. The demonstration of this is the same as has been already given, at case 5 prob. 1. Those, however, who wish for a proof, independent of the arithmetic of sines, will not be displeased to have the additional demonstration below. The shortest line which, with two other lines given in position, includes a given area, will make equal angles with those two lines, or with the segments of them it cuts off from an . isosceles triangle. Let the two triangles ABC, AEF, having the common angle A, be equal in surface, and let the former triangle be isosceles, or have AB = AC; then is BC shorter than EF. First, the oblique base Er cannot pass through D, the middle point of BC, as in the annexed figure. For, drawing CG parallel to AB, to meet EF produced in. Then the two triangles DBE, DCG are identical, or mutually equal in all respects. Consequently the triangle DCF is less than DBE, and therefore ABC less than AEF. G. B EF must therefore cut BC in some point н between в and D, and cutting the perp. AD in some point I above D, as in the 2d fig. 2d fig. Upon EF (produced if necessary) HD Q. E. D. This series of problems might have been extended much further; but the preceding will furnish a sufficient variety, to suggest to the student the best method to be adopted in almost any other case that may occur. The following practical examples are subjoined by way of exercise. Ex. 1. A triangular field, whose sides are 20, 18, and 16 chains, is to have a piece of 4 acres in content fenced off from it, by a right line drawn from the most obtuse angle to the opposite side. Required the length of the dividing line, and its distance from either extremity of the line on which it falls? Ex. 2. The three sides of a triangle are 5, 12, and 13. If two-thirds of this triangle be cut off by a line drawn parallel to the longest side, it is required to find the length of the. dividing line, and the distance of its two extremities from the extremities of the longest side. Ex. 3. It is required to find the length and position of the shortest possible line, which shall divide, into two equal parts, a triangle whose sides are 25, 24, and 7 respectively. Ex. 4. The sides of a triangle are 6, 8, and 10: it is required to cut off nine-sixteenths of it, by a line that shall pass through the centre of its inscribed circle. Ex. 5. Two sides of a triangle, which include an angle of 70°, are 14 and 17 respectively. It is required to divide it into three equal parts, by lines drawn parallel to its longest side. Ex. 6. The base of a triangle is 112.65, the vertical angle 57° 57′, and the difference of the sides about that angle is 8. It is to be divided into three equal parts, by lines drawn from the angles to meet in a point within the triangle. The lengths of those lines are required. Ex. 7. The legs of a right-angled triangle are 28 and 45. Required the lengths of lines drawn from the middle of the hypothenuse, to divide it into four equal parts. Ex. 8. The length and breadth of a rectangle are 15 and 9. It is proposed to cut off one-fifth of it, by a line which shall be drawn from a point on the longest side at the distance of 4 from a corner. Ex. 9. A regular hexagon, each of whose sides is 12, is to be divided into four equal parts, by two equal lines; both passing through the centre of the figure. What is the length of those lines when a minimum ? Ex. 10. The three sides of a triangle are 5, 6, and 7. How may it be divided into four equal parts, by two lines which shall cut each other perpendicularly? The student will find that some of these examples will admit of two answers. CHAPTER VIII. ON THE NATURE AND SOLUTION OF EQUATIONS IN GENERAL. 1. In order to investigate the general properties of the higher equations, let there be assumed between an unknown quantity, and given quantities a, b, c, d, an equation constituted of the continued product of uniform factors: thus (x-a) x (x-b) × (x−c) x (xd) = 0. This, by performing the multiplications, and arranging the final product according to the powers or dimensions of x, becomes 204 x= a) x3 + ab) x2 - abc) x + abcd = 0. (A) + bd +cd -abd -acd -bcd Now it is obvious that the assemblage of terms which compose the first side of this equation may become equal to nothing in four different ways; namely, by supposing either x = a, or b, or x = c, or rd; for in either case one or other of the factors x —a, x — b, x-c, x-d, will be equal to nothing, and nothing multiplied by any quantity whatever will give nothing for the product. If any other value e be put for x, then none of the factors e-a, e—b, e—c, e—d, being equal to nothing, their continued product cannot be equal to nothing. There are therefore, in the proposed equation, four roots 2d fig. Upon EF (produced if necessary) B This series of problems might have been exte further; but the preceding will furnish a suffici to suggest to the student the best method to be almost any other case that may occur. The foll tical examples are subjoined by way of exercise. Ex. 1. A triangular field, whose sides are 2 chains, is to have a piece of 4 acres in content fe it, by a right line drawn from the most obtus opposite side. Required the length of the div. its distance from either extremity of the lin falls? Ex. 2. The three sides of a triangle are 5, two-thirds of this triangle be cut off by a lin to the longest side, it is required to find t dividing line, and the distance of its two extr extremities of the longest side. Ex. 3. It is required to find the leng ngt the shortest possible line, which shall divid parts, a triangle whose sides are 25, 24, an Ex. 4. The sides of a triangle are 6, 8 quired to cut off nine-sixteenths of it, by a through the centre of its inscribed circle. Ex. 5. Two sides of a triangle, which 70°, are 14 and 17 respectively. It is r into three equal parts, by lines drawn pa side. Ex. 6. The base of a triangle is 112 57° 57', and the difference of the sides a It is to be divided into three equal parts, the angles to meet in a point within the t of those lines are required. Ex. 7. The legs of a right-angled Required the lengths of lines drawn fi hypothenuse, to divide it into four equ 1 HD |