Geometrical Analysis. Introduc

tion.

Ancient

with the modern.

SECTION I.

(1.) ANALYSIS, or resolution, is a process by which, commencing with what is sought as if it were given, a chain of relations is pursued which terminates in what is given, (or may be obtained,) as if it were sought. SYNTHESIS, or composition, is a process the very reverse of this; being one in which the series of relations exhibited commences with what is given, and ends with what is sought. Consequently analysis is the instrument of invention, and synthesis that of instruction. The analysis of the ancients is distinguished from that of the moderns by being conducted without the aid of any calculus, or the use of any principles except those of Geometry, the latter being conducted entirely by the language and principles of Algebra. The ancient is, therefore, called the Geometrical Analysis. For its origin and history, the reader is referred to our HISTORY OF ANALYSIS.

The interest which the Geometrical Analysis derives analysis from its antiquity, and from having been the instrument compared by which the splendid results of the ancient Geometry were obtained, would alone be sufficient to render it an object of attention even after the discovery of the more powerful agency of Algebra. But this is not its only nor its principal claim upon our notice. Its inferiority, compared with the modern analysis, in power and facility, is balanced by its extreme purity and rigour; and though its value as an instrument of discovery be lost, yet it must ever be considered as a most useful exercise for the mind of a student; and it may be fairly questioned, whether it may not be more conducive to the improvement of the mental faculties than the modern analysis, unless the latter be pursued much farther than it usually is in the common course of academical education, in which the student acquires little more than a knowledge of its notation. Newton was fully aware of the advantages attending the cultivation of this branch of mathematical science, and in many parts of his works laments that the study of it has been so much abandoned. He considered, that, however inferior in power and despatch the ancient method might be, it had greatly the advantage in rigour and purity; and he feared, that by the premature and too frequent use of the modern analysis the mind would become debilitated and the taste vitiated. We must however confess, that the pretensions of the ancient method to superior rigour do not seem to us to be as well founded as they are sometimes considered. It would be no very difficult matter to expunge the algebraical symbols from a modern investigation, and substitute for them their meaning expressed in the language used in geometrical investigations; but would such a change confer upon them greater rigour, or would it give to the conclusions greater validity? And yet th recisely what Newton himself has done

in many parts of his great work, the Principia. His Section I theorems are, evidently, investigated algebraically; but in demonstrating them, the process is disguised by the substitution of lines and geometrical figures for the algebraical species and formulæ. It cannot but excite astonishment, that a man of his extraordinary sagacity could so far deceive himself, as to suppose that by such a proceeding his reasoning acquired greater rigour.

But, without reference to the modern analysis, we conceive that the ancient method has sufficient claims to our attention on the score of its own intrinsic beauty. It has this further advantage, that we can enter at once upon its most interesting discussions without the repelling task of learning any new language or system of notation.

Anair

In the application of the Geometrical Analysis to the No ge solution of problems, or the demonstration of theorems, rules no general rules nor invariable directions can be given Geome which will apply in all cases. The previous construction to be used, and the preparatory steps to be taken, depend on the particular circumstances of the question, and must be determined by the sagacity of the analyst; and his skill and taste will be evinced in the selection of the properties or affections of the given or sought quantities on which he founds his analysis; for the same question may frequently be investigated in many dif ferent ways.

Analys

In submitting a problem to analysis, its solution, in the first instance, is assumed; and from this assump- a procu tion a series of consequences are drawn, until at length something is found which may be done by established principles, and which if done will necessarily lead to the execution of what is required in the problem. Such is the analysis. In the synthesis, then, or the solution, we retrace our steps; beginning by the execu tion of the construction indicated by the final result of the analysis, and ending with the performance of what is required in the problem, and which constituted the first step of the analysis.

When a theorem is submitted to analysis, the thing of a to be determined is, whether the statement expressed by theor it be true or not. In the analysis this statement is, in the first instance, assumed to be true; and a series of consequences are deduced from it until some result is obtained, which either is an established or admitted truth, or contradicts an established or admitted truth. If the final result be an established truth, the theorem proposed may be proved by retracing the steps of the investigation, commencing with that final result, and concluding with the proposed theorem. But if the final result contradict an established truth, the proposed theorem must be false, since it leads to a false conclusion.

These general observations on the nature of the Geometrical Analysis, and the methods of proceeding in it, will be more clearly apprehended after the inves

Fig.

Fig. 1.

SECTION II.

Miscellaneous Problems.

(2.) Definition. A point is said to be given when its situation is either given or may be determined. (3.) Definition. 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. Let PM: PN:m: n. If any other line as PL be drawn intersecting A B and C D, and a parallel to CD be drawn from N, that parallel will divide PL similarly to P M, 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 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 CD from N meet PL in O, so that PL: PO:: m : n. But PL may be drawn, and the point O therefore may be determined; and since the direction of CD is given, the direction of ON 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 P L be cut at O, so that PL: PO :: m : n, and from O draw ON parallel to CD, meeting AB in N, and through N draw PNM. Then PM:PN:: PL: PO:: 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 O 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.

(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 APC shall be equal to the angle BPD.

Produce the line B P beyond P, until PE is equal Section II. to PA, and join A E. The angles BPD and E PC are equal; but also (hyp.) B PD and APC are also equal, therefore the angle APC is equal 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 the 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 A F, and draw the right line E B meeting the line CD in P. Draw AP, and the lines AP and BP are those which are required. For since A F and FE are equal, and P F common to the triangles A FP and E FP, and the angles AFP and EFP are equal, the angles A P F and EPF are equal. But BPD and E PF are also equal, therefore the angles APF and BPD 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.

(8.) To inscribe a square in a triangle.

Let A B C be the triangle, and DFE the required Fig. 3. square. Draw the perpendicular B G, and draw A E to meet a parallel BH to AC at H. It is easy to see that DF:FE:: GB: BH; for the triangles AFD and A B G, AFE and ABH are respectively similar each to each. Hence, since DF is equal to FE, GB is also equal to B H. But GB is given in magnitude and position, and therefore BH 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 of the square.

(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 BG is given, and therefore BH is as before given in position and magnitude.

(10) Schol. If B H be drawn equal to B G and on Fig. 4. the same side of the vertex with A, then it will be necessary to produce A H and CB, 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, B GA to FDA, and H BA to EFA. Hence GB: BH: DF:FE.

(11.) Cor. 2. In the same manner the more general problem, "To inscribe a rectangle given in species," may be extended.

PROPOSITION.

(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 BD be a mean Fig. 5. proportional between A D and DC. Produce BD to E, so that D E shall be equal to BD, and join C E. Since

AD: BD ED: DC,

and the angles B D A and E D C are equal, the triangles BDA 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 FD be drawn, the angle FDB will be a right angle, and the point F will therefore be in a circle described on F B as diameter. But the point F is given, since it is the centre of a circle circumscribed about the given triangle, and the line F B 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 BDF in a semicircle is right, therefore BD DE; and therefore the square of B D is equal to the rectangle under AD and D C.

=

If the circle on B F 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 AC, the angle ABC will be right, and the point F itself is one of the points which solve the problem; for in that case AF, BF, and C F 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 hypothenuse is a mean proportional between the segments; and it is the only line which can be drawn from the right angle to the hypothenuse which is a mean, except the bisector of the hypo

thenuse.

Schol. It has been observed, 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 AC, 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 AD and CD, and yet the angle ABC 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 magnitude 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 FI and BH on A C, and from the centre of the lesser circle G draw the perpendicular G K. Since GF is equal to G B, G K is equal to half the sum of FI and B H. Hence it follows, that the solu

tion will only be possible when half the sum of FI Section II. and BH is not greater than B G, or when the sum of FI and B H is not greater than B F; 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.

PROPOSITION.

(14.) Right lines being drawn bisecting the internal and external angles of a triangle, and being produced to meet the base, and the production of the base to determine the conditions on which the rectangle under the sides of the triangle will be a geometric, arithmetic, or harmonic mean between the rectangle under the segments of the base by the internal bisector, and the rectangle under the segments of the base by the external bisector.

Let A B C be the triangle, B D the bisector of the Fig. 8. internal angle, and BE the bisector of the external angle. By the principles of Geometry we have

Hence it follows, that the three rectangles A E x CE, ABX BC, AD × DC are similar.

1. Let the rectangle under A B and B C be a geo If three similar metric mean between the other two. figures be in geometrical progression, their homologous sides must also be in geometrical progression; hence CE: CB: CD. But since the angle D B E is equal to A BD and E BF together, it is a right angle, and DC and CE, BCA must be a right angle, (12.) therefore since BC is a mean proportional between Hence the rectangle under the sides is a geometric mean, when either of the base angles is right.

2. Let the rectangle A B x B C be an arithmetic mean between the other two. In that case the rectangle A EEC should exceed A B x BC by as much the excess of AEXEC above AB x BC is the square as this last exceeds A D x DC. But by Geometry of B E, and the excess of AB x B C above A D X BC is the square of B D. Hence in the present instance the squares of BE and BD are equal, and therefore the lines themselves are equal. Hence the angles BDC and BE C are equal, and since DBE is a right angle, BDC must be half a right angle, and therefore the difference between BDC and BDA is a right BAD and BCD the equal halves of the vertical angles, angle. But since by adding to each of the base angles we obtain sums equal to the angles BDC and B DA, it follows that the difference between the base angles BCD and BA D is a right angle. Hence when the

difference of the base angles is right, the rectangle ABX BC is an arithmetic mean between the other two rectangles.

mean.

3. Let the rectangle A B x B C be an harmonic In that case, by the nature of harmonic proportion, we have

A EX EC: ADX DC: AEXEC-AB X BC ABX BC AD x DC;

:

we have the rectangle A E x EC to the rectangle AD x DC as the square of B E is to the square of Analysis. BD. Since then the similar rectangles of which CE and C D are homologous sides, are proportional to the squares of BE and BD, these lines themselves are proportional. Therefore

Fig. 9.

Fig 10.

BE BD:CE: CD.

Hence the line BC bisects the angle D BE; but since D BE is right, CBD is half a right angle, and therefore A B C is a right angle. Hence if the bisected angle be right, the rectangle A B x B C is an harmonic mean between the other two rectangles.

PROPOSITION.

(15.) To draw a right line from the vertex of a triangle to the base, or to the base produced, so that its square shall be equal to the difference between the rectangle under the sides, and the rectangle under the segments into which it divides the base.

Let the triangle be ABC, and let the required line be BD, and let a circle be circumscribed about the triangle.

1. Let the line be drawn to the base itself, and let it be produced to meet the opposite circumference at E, and draw CE. By hypothesis, the square of BD, together with the rectangle A D x DC, is equal to the rectangle A B x BC. But the rectangle A D x DC is equal to the rectangle B D x DE. Add to both the square of B D; and the rectangle A D x DC, together with the square of BD, is equal to the rectangle BDX DE, together with the square of B D. But the former rectangle and square are together equal to the rectangle ABX BC, and the latter rectangle and square are together equal to the rectangle BE× B D. Hence the rectangle A B × BC is equal to the rectangle BE x B D. Hence we have

AB: BD:: EB: BC;

and the angles A and E are equal. Therefore in the triangles A B D and E B C the sides A B, B D are proportional to E B, B C, and the angles opposite to one pair of homologous sides BD and B C are equal, and therefore the angles opposite the other pair of homologous sides must be either equal or supplemental. If they be equal, the triangles ABD and EBC are similar, and therefore the line BD bisects the angle ABC. If the angles BDA and BCE be supplemental, the sum of the arcs which they subtend must be equal to the whole circumference. Hence the arcs BA, A E, BA, and CE are together equal to the

circumference. But BA, A E, B C, and CE are also together equal to the circumference. Take away from both the arcs BA, A E, and C E, and the remaining arcs BC and BA are equal; and therefore their chords are equal, and therefore the triangle is isosceles.

Hence we infer, that" if a line be drawn from the vertex of a triangle to the base, so that its square, together with the rectangle under the segments, shall be equal to the rectangle under the sides, that line will bisect the vertical angle, except when the triangle is isosceles, in which case any line drawn from the vertex to the base will have the required property."

By

2. Let the line B D meet the base produced. hypothesis, the rectangle A D x DC is equal to the rectangle A B B C, together with the square of B D.

Draw C E, and the angles E and A are equal. Hence in the triangles EBC and ABD there are two sides EB and BC proportional to two A B, BD, and the angles opposite one pair of homologous sides equal, and therefore the angles opposite to the other homologous sides must be either equal or supplemental. If they be equal, take A B C from both, and the remainders EBA and CBD are equal; but EBA and FBD are also equal, and therefore BD bisects the external angle CBF of the given triangle.

If the angles ABD and EBC be supplemental. Since the angles A B D and FBD are also supplemental, we should have the angles FBD and EBC equal; and therefore E B A and E B C equal; and therefore the point B cannot in this case lie between E and D. It must therefore be placed as in fig. 11. Here the Fig. 11. square of B D is manifestly greater than the rectangle CD × D A, and therefore the proposed condition must be that the rectangle CD x DA, together with the rectangle AB x B C, is equal to the square of B D. But the rectangle C D x DA is equal to the rectangle BDX DE; and taking these equals from the former, the remainders, viz. the rectangles A B x BC and BDX BE are equal. Hence

EB: BC::AB: BD.

Draw C E, and in the triangles EBC and A B D the two sides E B, B C are proportional to two A B, BD, and the angles BEC and BAD opposite to one pair of homologous sides are supplemental, (for BAC and BEC are equal,) and therefore the angles BCE and D opposite the other pair of homologous sides are equal. Hence the difference of the arcs subtended by D is equal to the arc subtended by BC E, that is, the difference between the arcs BC and A E is equal to the arc BE; or the arcs B E and A E together, that is, the arc A E B is equal to the arc A B, and therefore their chords are equal, but their chords are the sides A B, BC of the triangle, which is therefore isosceles.

Hence it follows, that "if a line be drawn from the vertex of a triangle to the produced base, so that its square, together with the rectangle under the sides, shall equal the rectangle under the segments of the base, that line will bisect the vertical angle, except when the given triangle is isosceles, in which case there is no line which has the required property. In this case, however, the square of every line drawn from the vertex to the produced base is equal to the sum of the rectangles under the sides and segments."

SECTION III.

Of the Contact of Right Lines and Circles.

(16.) PROBLEMS of contact of right lines and circles furnished the ancients with an extensive subject for the exercise of the Geometrical Analysis. In general