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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, in the first volume of this Course. Of the latter, we here 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 anteriorly, known, or placed it among the number of principles known to be true.

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 resolved : 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 those remarks take the following examples.

Er. 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 M to N.

Analysis. Suppose that the thing is af. fected, that is to say, that Ac : cB :: M : N, and let the base AB 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 F, 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, ch, the thing is done.

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Er. 2. From a given circle to cut off an arc, such that the sum of m times the sine, and n times the versed sine, may be equal to a given line.

Analysis. Suppose it done, and that AEE'B is the given circle, BE'E the required arc, ED its sine, BD its versed sine; in DA (produced if necessary) take BP and nth part of the given sum ; join PE, and produce it to meet BF 1 to Ab or ll to ED, in the point F. Then, since m . ED + n . BD = n . BP = n . PD + n . BD ; consequently m. Ed = n . PD ; hence PD : ED B. F. M. :: m : n. But Pd : Ed :: (by sim. tri.) Pb : BF ; therefore •PB : BF :: m : n. Now PB is given, therefore BF is given in magnitude, and, being at right angles to PB, is also given in position; therefore the point F is given and consequently pr given in position; and therefore the point P, its intersection


148 construction of GeoMETRicAL PRobleMs.

with the circumference of the circle AEE'B, or the arc BE is given. Hence the following

Construction. From B, the extremity of any diameter AB of the given circle, draw BM at right angles to AB; in AB (produced if necessary) take BP an nth part of the given sum ; an on BM take BF so that BF : BP :: n : m. Join PP, meeting the circumference of the circle in E and E', and BE or BE' is the arc required.

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Ea. 3. In the given triangle ABH, to inscribe another triangle abc, similar to a given one, having one of its sides parallel to a line men given by position, and the angular points a, b, c, situate in the sides AB, BH, AH, of the triangle ABB respectively.

Analysis. Suppose the thing done, and that abo is inscribed as required. Through any point c in Bh draw co parallel to man or to ab, and cutting AB in D ; draw ce parallel to be, and DE to ac, intersecting each other in E. The triangles DEc, acb, are similar, and Dc : o :: ce: be ; also bdc, Bab, are similar, and DC : ab :: Bc : Bb. Therefore Bc : ce:: Bb : be ; and they are about equal angles, consequently B, E, c, are in a right line.

Construction. From any point c in BH, draw cd parallel to nm ; on cd constitute a triangle cd E similar to the given one ; and through its angles I, draw BE, which produce till it cuts Ali in c : through c draw ca parallel to Ed and cb parallel to Ec; join ab, then abc is the triangle required, having its side ab parallel to mn, and being similar to the given triangle.

Demonstration. For, because of the parallel lines ac, DE, and cb, Ec, the quadrilaterals BDEc and Bacb, are similar; and therefore the proportional lines dc, ab, cutting off equal angles BDC, Bab ; BCD, Bba; must make the angles EDC, ECD, respectively equal to the angles cab, cba; while ab is parallel to DC, which is parallel to men, by construction.

Ex. 4. Given, in a plane triangle, the vertical angle, the perpendicular, and the rectangle of the segments of the base made by that perpendicular; to construct the triangle.


Analysis. Suppose ABC the triangle required, Bn the given perpendicular to the base Ac, produce it to meet the pe. riphery of the circumscribing circle ABch, whose centre is o, in H ; then, by th. 61 Geom. the rectangle BD. DH=AD . Dc, the given rectangle : hence, since BD is given, DH and BH are given; therefore bi = H1 is given : as also ID = or ; and the angle Eoc is = ABC the given one, because Eoc is measured by the arc Kc, and ABC by half the arc Akc or by Kc. Consequently Ec and Ac = 2Ec are given. Whence this

Construction. Find DH such, that DB. DH = the given

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rectangle, or find DH = ; then on any right line GF

take FE = the given perpendicular, and EG = DH ; bisect FG in o, and make Eoc = the given vertical angle; then will oc cut Ec, drawn perpendicular to oe, in c. W. centre o and radius oc, describe a circle, cutting ce produced in A : through F parallel to Ac draw FB, to cut the circle in B; join AB, ch, and ABC is the triangle required.

Remark. In a similar manner we may proceed, when it is required to divide a given angle into two parts, the rectangle of whose tangents may be of a given magnitude. See prob. 40, Simpson's Select Exercises.

Note. For other exercises, the student may construct all the problems except the 24th, in the Application of Algebra to Geometry, at page 371, vol. i. And that he may be the better able to trace the relative advantages of the ancient and the modern analysis, it will be advisable that he solve those problems both geometrically and algebraically.

Definitions and preliminary Notions.

1. Mechanics is the science of equilibrium and of motion.

2. Every cause which moves, or tends to move a body, is called a force.

3. When the forces that are applied simultaneously to a

body, destroy or annihilate each other's effects, then there is equilibrium.


4. Statics has for its object the equilibrium of forces applied to solid bodies.

5. By Dynamics we investigate the circumstances of the motion of solid bodies.

6. Hydrostatics is the science in which the equilibrium of

* fluids is considered.

7. Hydrodynamics is that in which the circumstances of their motion is investigated.

According to this division, Pneumatics, which relates to the properties of elastic fluids, is a branch of Hydrostatics.

For farther elucidation the following definitions, also, may advantageously find a place here, viz.

8. Body is the mass, or quantity of matter, in any material substance; and it is always proportional to its weight or gravity, whatever its figure may be.

Body is either Hard, Soft, or Elastic. A Hard Body is that whose parts do not yield to any stroke or percussion, but retains its figure unaltered. A Soft Body is that whose parts yield to any stroke or impression, without restoring themselves again; the figure of the body remaining altered. And an Elastic Body is that whose parts yield to any stroke, but which presently restore themselves again, and the body regains the same figure as before the stroke.

We know of no bodies that are absolutely, or perfectly, either hard, soft, or elastic; but all partaking these properties, more or less, in some intermediate degree.

9. Bodies are also either Solid or Fluid. A Solid Body is that whose parts are not easily moved among one another, and which retains any figure given to it. But a Fluid Body is that whose parts yield to the slightest impression, being easily moved among one another; and its surface, when left to itself, is always observed to settle in a smooth plane at the top.

10. Density is the proportional weight or quantity of matter in any body. So, in two spheres, or cubes, &c. of equal size or magnitude ; if the one weigh only one pound, but the other two pounds; then the density of the latter is double the density of the former; if it weigh three pounds, its density is triple; and so on.

11. Motion is a continual and successive change of place. —If the body move equally, or pass over equal spaces in equal times, it is called Equable or Uniform Motion. But if it increase or decrease, it is Variable Motion; and it is called Accelerated Motion in the former case, and Retarded Motion in the latter.—Also, when the moving body is considered with

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