ID · IH= IP · IM ; but when the products of the sides about the equal angles of two triangles having a common angle are equal, those triangles are equal ; therefore A IHD= A IPM. Q. E. D. Comput. In the triangles ADI, ADG are given all the angles, and the side AD; whence at, AG, DI, and ic, DI-DC, become known. In the triangle ifc, all the angles and the side ic are known ; whence it becomes known, as well as Fh, since ah : HF : : min. Lastly, IP = V(1H . 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 it 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. I, 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 ptq; and then each of the quadrangles so formed into their respective ratios, of m to n, and p to g. 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 I prob. I. PROBLEM VL. To Divide any Polygon into Two Parts having a Given Ratio. Вес - 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. Reduce the part ABCD into an equivalent triangle nps, whose vertex shall be the given point P, and base ad WIDS produced : an operation which may be performed at once, if the portion ABCD be quadrangular; or by several opera F KME will . a tions (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 have ing the given ratio, by the line ph. In like manner, reduce ADEFGA into an equivalent triangle having a for its vertex, and pe produced for its base; and divide this triangle into the given ratio by a line from h, as HK The compound line divide the whole polygon into two parts PK, and tiirough n draw hu parallel to it ; join PM ; so will the right line pm divide the polygon as required, provided . fall beiween 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 problem, without subdividing the figure: but as it requires the computation of the area, it is not given here. 2dly. By the shortest line possible. FMME Construc. From any point pl 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 the points p and m, so that ip or im shall be a mean propora tional between iP', 1M' ; then will pu 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 isosa celes, or have AB = AC; then is BC shorter than EF. VOL. I. Dddd First, G First, the oblique base EF cannot pass through d, the middle point of BC, as in the annexed figure. For, draw.ng co parallel to AB, 10 meet EF produced in G Then the two triangles DBE, DCG are identical, or mutually equal in all B respects. Consequently the triangle DCF is less than DBE, and therefore ABC less than AEP. Ey must therefore cut bc in some point h between B and D, and cutting the perp. Ad in some point i above D, as in the 2d fig. Upon EF (produced if necessary) demit the perp. AK. Then in the right angled A AIK, the perp. Ak is less than >K the hypothenuse Al, and therefore much more less than the other perp. Ad. But, BA HD of equal triangles, that which has the greatest perpendicular, has !he least base. Therefore the base bc is less than the base Er Q. ED. 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. Er. Ex. 5. Two sides of a triangle, which include an angle of 70°, and 14 and 17 respectively. It is required to divide it into three equal parts, by lines drawn parallel to its longest side. Er. 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. Er. 7. The legs of a right-angled triangle are 28 and 45. Required the lengths of lines drayn from the middle of the hypothenuse, to divide it into four equal parts. Er. 8 The length and breadth of a rectangle are 15 and 9. It is proposed to cut off one-fifth of is 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 hall cut each other perpendicularly; * The student will find that some of these examples will admit of two answers. On the Construction of Geometrical Problems. Problems in Plane Geometry are solved either by means of the modern or algebraical analysis, or of the ancient or geometrical analysis. Of the former, some specimens are given in the Application of Algebra to Geometry, page 369, &c. of this volume. Of the latter, we bere present a few examples, premising a brief account of this kind of analysis. Geometrical analysis is the way by which we proceed from the thing demanded, granted for the moment, till we have connected it by a series of consequences with something anteriurly known, or placed it among the number of principles known to be true. Analysis Analysis may be distinguished into two kinds. In the one, which is named by Pappus, contemplative, it is proposed to ascertain the truth or the falsehood of a proposition advanced ; the other is referred to the solution of problems, or to the investigation of unknown truths. In the first we assume as true, or as previously existing, the subject of the proposition advanced, and proceed by the consequences of the hypothesis to something known; and if the result be thus found true, the proposition advanced is likewise true. The direct demonstration is afterwards formed, by taking up again, in an inverted order, the several parts of the analysis. If the consequence at which we arrive in the last place is found false, we thence conclude that the proposition analysed is also false. When a problem is under consideration, we first suppose it resolved, and then pursue the consequences thence derived till we come to something known. "If the ultimate result thus obtained be comprised in what the geometers call data, the question proposed may be resolveel : the demonstration (or rather the construction), is also constituted by taking the parts of the analysis in an inverted order. The impossibility of the last result of the analysis, will prove evidently, in this case as well as in the former, that of the thing required. In illustration of these remarks take the following examples. Ex. 1. It is required to draw, in a given segment of a circle, from the extremes of the base A and B, two lines AC, BC, meeting at a point c in the circumference, such that they shall have to each other a given ratio, viz. that of a to n. Anatysis. Suppose that the thing is af. fected, that is to say, that ac : CB ::M:N, B and let the base al of the segment be cut in the same ratio in the point E. Then Ec, being drawn, will bisect the angle ACB (by th. 83 Geom.); consequently, if the circle be completed, and ce be produced to meet it in r, the remaining circumference will also be bisected in F, or have FA = FB, because those arcs are the double measures of equal angles : therefore the point F, as well as E, being given, the point c is also given. Construction Let the given base of the segment AB be cut in the point E in the assigned ratio of m to N, and complete the circle ; bisect the remaining circumference in F; join FE, and produce it till it meet the circumference in c! then drawing CA, CB, the thing is done. Demonsiration. Since the arc FA = the arc FB, the angle angle BCF, by theor. 49 Geom. ; therefore AC : CB : : |