These general observations on the nature of the Geometrical Analysis, and the methods of proceeding in it, will be more clearly apprehended after the investigations contained in the subjoined treatise have been examined. SECTION II. Problems respecting right lines. (2) Dec.--A point is said to be given when its position is either given or may be determined. (3) Dep.-A right line is said to be given in position when it is either actually exhibited and drawn, or may be exhibited and drawn by previously established principles. PROPOSITION. (4) To draw from a given point a right line intersecting two right lines given in position, so that the segments between the point and the right lines shall have a given ratio. Let the given point be P, A B and C D the right lines given in position, and m : n the given ratio. А с Let PM:PN=m:n. If any other line as PL be drawn intersecting A B and CD, and a parallel to CD be drawn from N, that parallel will divide PL similarly to PM, and therefore in the required ratio. This parallel may, or may not, coincide with the line N K. First, let us suppose that it does. In that B! case the two lines given in position will be parallel, and the line PL, or any other line, drawn intersecting them, will be cut similarly to PM, and therefore all such lines will be cut in the required ratio. Hence it appears, that in this case the problem is indeterminate, since every line which can be drawn intersecting the given lines will equally solve it. Secondly, if the given lines A B, CD be not parallel, let the parallel to C D from N meet PL in G, so that PL:PG=m:n. But PL may be drawn, and the point G therefore may be determined; and since the direction of C D is given, the direction of GN is determined, and therefore the point N may be found. Hence, the solution is as follows: let any line PL be drawn. If PL:PK= m:n, the problem is solved. If not, let PL be cut at G, so that PL:PG = m:n, and from G draw G N parallel to C D, meeting AB in N, and through N draw PNM. Then PM:PN=PL:PG m :n. (5) Cor. 1.-The same solution will apply if the line A B be a curve of any kind. (6) Cor. 2.-If the parallel to CD through G do not meet the line A B, the solution is impossible. If A B be a right line, this happens when it is parallel to C D. And therefore we conclude in general, that when the two right lines A B and C D are parallel, the problem is either indeterminate or impossible. PROPOSITION. A R (7) From two given points to draw to the same point in a right line given in position, two lines equally inclined to it. Let the given points be A and B, and let C D be the line given in position. Let P be the sought point, so that the angle A PC shall be equal to the angle BP D. Produce the line B P beyond P, until PE is equal to PA, and join AE. The angles BPD and EPC are equal; but also (hyp.) BPD and APC are also equal, therefore the angle A PC is qual to the angle EPC. But also the sides PA and PE are equal, and the side PF is common to the triangles A PF and EPF. Therefore the angles AFP and EFP are equal, and therefore are right angles, and also AF is equal to EF. But since A and C D are given, the perpendicular AF is given, and hence the solution of the problem may be derived. From either of the given points A draw a perpendicular AF to the given right line CD, and produce it through F, until FE is equal to AF, and draw the right line EB meeting the line CD in P. Draw A P, and the lines A P and B P are those which are required. For since A F and FE are equal, and PF common to the triangles AFP and EFP, and the angles AFP and EFP are equal, the angles APF and EPF are equal. But B P D and EPF are also equal, therefore the angles APF and B PD are equal. SCHOLIUM.—If the given points lie at different sides of the given right line, the problem is solved by merely joining the points. PROPOSITION. B (8) To inscribe a square in a triangle. Let A B C be the triangle, and D F E the required square. Draw the perpendicular BG, and draw A E to meet a parallel B H to AC at H. It is easy to see that DF: FE GB:BH; for the triangles AFD and A B G, A FE and A BH are respectively similar each to each. Hence, since D F is equal to FE, Ā D G GB is also equal to B H. But G B is given in magnitude and position, and therefore B H is given in magnitude and position. To solve the problem therefore it is only necessary to draw B H and join A H, and the point E where A H meets B C will be the vertex of the angle (9) Cor. 1.--It is evident that the same analysis will solve the more general problem,“ To inscribe in a triangle a rectangle given in species." For in this case the ratio BH:B G is given, and therefore BH is as before given in position and magnitude. of the square. (10) SCHOL.-If B H be drawn equal to B G and on the same side of the vertex with A, then it will be necessary to produce A H and C B, in order to obtain their point of intersection E. In this case, however, DFE will still be a square, for the corresponding triangles will be similar, BGA to FDA, and HBA to EFA. Hence GB:BH=DF:FE. (11) Cor. 2.-In the same manner the more general problem, inscribe a rectangle given in species," may be extended. A G с D 66 To PROPOSITION. B В C E (12) To draw a line from the vertex of a given triangle to the base, so that it will be a mean proportional between the segments. Let A B C be the triangle, and let B D be a mean proportional between A D and DC. Produce B D to E, so that D E shall be equal to BD, and join CE. Since AD:BD=ED:DC, and the angles B D A and EDC are equal, the triangles B D A and CDE are similar. Therefore the angles E and A are equal, and are in the same segment of a circle described on C B. If from the centre of this circle F D be drawn, the angle FDB will be a right angle, and the point D will therefore be in a circle described on FB as diameter. But the point F is given, since it is the centre of a circle circumscribed about the given triangle, and the line FB is therefore given, and the circle on it is as diameter is given, and therefore the point D is given. The solution of the problem is therefore effected by circumscribing a circle about the given triangle, and drawing from its centre to the angle B a radius. On that radius, as diameter, describe a circle; and to a point D, where this circle meets the base, draw the line B D, and it will be a mean proportional between the segments. For the angle B DF in a semicircle is right, therefore BD = DE; and therefore the square of B D is equal to the rectangle under A D and D C. If the circle on BF intersect A C, there will be two points in the base to which a line may be drawn, which will be a mean proportional between the segments. If this circle touch the base there will be but one such line, and it may happen that the circle may not meet the base at all, in which case the solution is impossible. If the centre F be upon the base A C, the angle A B C will be -right, and the point F itself is one of the points which solve the problem ; for in that case AF, BF, and CF are equal. The other point D is the foot of a perpendicular B D from the vertex on the base. (13) CoR.-Hence, in a right angled triangle, the perpendicular on the hypotenuse is a mean proportional between the segments; and it is the only line which can be drawn from the right angle to the hypotenuse which is a mean, except the bisecior of the hypotenuse. SCHOL.-It has been observed by some elementary writers, that the solution of the problem to draw a line to the base which shall be a mean proportional between the segments is impossible when the vertical angle is acute. That this is erroneous, must be evident from the preceding analysis. For let one circle be described upon the radius of another as diameter. Let any line, as A C, be drawn not passing through F, but intersecting the inner circle ; and so that the point of contact B and the centre F shall lie at the same side of it. Draw A B and C B, and also B D. It is evident that B D is a mean proportional between A D and C D, and yet the angle A B C is acute, being in a segment greater than a semicircle. The possibility of the solution of this problem does not at all depend on the maguitude of the vertical angle. It may be obtuse, right, or acute, and may be equal in fact to any given angle, and yet the solution be possible. Let it be required to determine the conditions on which the solution is possible. If the circle on BF meet the base, the perpendicular distance of its centre from the base must be less than its radius; that is, less than half the radius of the circle which circumscribes the given triangle. From F and B draw perpendiculars F I and B H on A C, and from the centre of the lesser circle G draw the perpendicular GK. Since G F is equal to G B, GK is equal to half the sum of FI and В н. Hence it follows, that the solution will only be possible when half the sum of FI and B H is not greater than BG, or when the sum of F I and B H is not greater than BF; that is, when the sum of the perpendiculars on the base from the vertex and the centre of the circumscribed circle is not greater than the radius of that circle. HKI SECTION III. Propositions respecting circles. (14) PROBLEMS of contact of right lines and circles furnished the 1. Passing through a given point. 5. Touching two given right lines. 13. Having a given centre. Every combination of three which can be formed from these data, may be taken as the limiting circumstances in problems for the determination of a circle. In the invention of such problems it should however be observed that 2, 5, 8, and 13 are each to be counted as two data, and 3, 6, 9 are each to be counted as three data. Each of the latter is, therefore, itself sufficient to determine the circle, but each of the former ought to be combined with some one of the data 1, 4, 7, 10, 11, 12 We cannot here enter at large on this class of problems, we shall therefore confine ourselves to a few examples. PROPOSITION. (15) To describe a circle passing through two given points, and touching a right line given in position. If the given points be at different sides of the given line, the solution is manifestly impossible. Let them then be A, B at the same side of the given right line C D. Let the required circle be ABC, and let A B be produced to meet the right line at D. The square of CD is equal to the rectangle A D * D B. But this rectangle is given, therefore the square of C D is given, and therefore C D itself is given in magnitude and position, and hence the point C is given. But also the points A, B being given, the circle through these points A, B, C is given. The solution, therefore, is effected by producing A B to D, and taking DC equal to a mean proportional between AD and D B, and then describing a circle through A, B, C. But it may happen, that the line A B is parallel to CD, and will not meet it when produced. In this case draw A C and BC. The angle BCD is equal to the angle A in the alternate segment, and also equal to the alternate angle B. Hence the angles A and B are equal, and therefore the sides A C and B C are equal. Draw CE perpendicular to A B, and AE and B É are equal. The point E is therefore given, and the perpendicular EC is given in position, and therefore the point C is given. To solve the problem in this case therefore, bisect A B at E, and draw the perpendicular through E, intersecting C D in C. A circle passing through A, B, C will be that which is required. E B Б D |