at a point A in the circumference, so that they shall bave a given ratio to each other; as, for instance, that of E to F.

Anal. Suppose the thing done, or that the point A is found: (that is, such as to render AC:AB::F:E) and let BD be drawn, making the angle CBD equal to CAB, and meeting CA produced in D.

D

B

E F

Then the angle C being common to the two triangles CBA, CBD, and the angles CAB, CBD, being equal, they are equiangular; and consequently AC AB:: CB: CD. Whence CB being given, BD is also given in position and magnitude.

Constr. Draw BB, making the angle CBD equal to that which is contained in the given segment CAD, and CB to BD in the given ratio of F to E. Join CD, cutting the segment in A; and draw BA, AC. These are the lines required.

Demon. The triangles CBA, CBD, being equiangular, AC: AB:: CB: BD :: F: E, which is the given ratio by construction. 6. From one angle C of a given rhombus ABCD, to draw a line, such that the part intercepted by the sides BA, AD, which contain the opposite angle A, shall be of a given length, as m.

Anal. Suppose the thing done; and make

the CEG CAF.

=

Then by sim. tria. BA: AF :: CE : EF, and CA AF:: CE: EG, whence CA. EG= AF.CE BA. EF, or CA: AB:: EF (=m) : EG. And since CAF = = ▲ BAC = L CAD, the whole GAE CAF or CEG. Hence the triangles AGE, CGE, being similar,

=

CG: EG EG: GA, or CG.GA EG2. But EG is already known, being a fourth, proportional to CA, BA, and m (as shown above) wherefore also the rectangle CG. GA is known. Whence arises the following

Const. Upon the diameter of the rhombus CA, describe a semicircle, and make AH perpendicular to AC, and to the fourth proportional to AC, AB, and m. From the centre of the circle K, draw KH, cutting the circumference in L, and make KG = KH; and join GL. Then, from the point G, apply GE = GL; and through C and E draw CEF, meeting BA produced in F. The line EF will be that required.

Demon. Since the sides KG, KL, of the triangle KGL, are equal to the sides KH, KA, of the triangle KHA, and the K is common, we have KGL = KAH, and the side GL (or GE) = AH. But the Z HAK being right (constr.), the KLG is also

=

Whence CG.GA GA; and therefore CEG EAG. CAD, the whole CAF. Hence the

right, and GL is a tangent to the circle at L. = · GL2 — AH2 — GE2, or CG:GE::GE: the triangles CGE, AGE, being similar, the Again, the GAF being equal to BAC or EAG = CAF; and consequently CEG triangles CGE, CFA, being similar, CA: AF :: CE: EG, or CA.EG=AF.CE. Also, since AE, BC, are parallel, BA ; AF CE: EF; or BA. EFAF.CE; whence, also, BA.EF BA.EF = CA.EG, or CA: BA :: EF: EG. But CA: BA :: m : AH (by constr.) whence EF: EG::m: AH; and therefore, EG, as before shown, being equal to AH, EF is also equal to m.

Analysis and Demonstration of Theorems.

THEOREM 1. The square of a line bisecting the vertical angle of any triangle, together with the rectangle of the segments of the base made by that line, is equal to the rectangle contained under the sides of the triangle.

B

Let ABC be any triangle, and BD drawn to bisect the vertical angle, and cut the base in D; then will BD2+ AD. DCAB. BC, Anal. Suppose the theorem to be true; that is, suppose BD2+ AD. DC = AB. BC. Then, by way of preparation, or construction, or to obtain something on which to found our analysis, let a circle be described about the triangle ABC, and let BD be produced to meet it in E; and join EC.

=

JB

But BD2

Then AD.DC BD.DE. Add BD2 to each of these, and BD2+ AD. DC BD2 + BD.DE. +BD.DE = EB.ED; and BD2 + AD. DC = AB.BC, by hypothesis. Hence, also, AB. BC=EB.BD.

Now this we shall find to be true by an elementary theorem ;* and hence the theorem which forms its foundation is also true. Demon. Describe a circle ABCE about the triangle, and produce BD to E, and join EC. Because the triangles ABD, EBC, are similar, having the angles at A and E equal, and likewise the angles at B, by hypothesis, equal, we shall have AB: BD :: BE: BC. Therefore, AB. BC = EB.BD. But EB. BD: BD2 + BD. DE, and BD. DE = AD. DC. Therefore, finally, AB.BC = BD2 + AD.DC.

=

*For the triangles ABD, EBC, are similar, having the angles at A and E equal, and the angles at B in each of them equal, by the hypothesis. Hence AB: BD :: BE: BC; that is, AB.BČ EB.BD.

=

2. If one of the angles C, of a right-angled triangle CAD, be bisected by a line CB meeting AD in B, then 2CA2 : CA2 — AB2 :: AD: AB.

D

B

A

Anal. Suppose it true, and through C draw CM perpendicular to CD, meeting DA produced in M; then, by hypothesis, 2CA2: CA AB :: AD: AB. But CA2: MA. AD, by similar triangles, DCA and ACM; therefore 2MA. AD: 2MA.AD AB AD AB:: 2MA. AD: 2MA. AB; whence 2MA. AD — AB2 = 2MA.AB. And adding AM2+ AB2 to each of these equals, we have DM. MA-AM2+ AB2+2AM. AB. Now DM. MA = MC2, and AM2+ AB2 + 2AM.AB = MB2. Hence MC MB, and the angle MBC= MCB. Therefore the angle BCA is equal to the angle BCD.

M

Demon. The triangle MBC is isosceles, having the defects from a right angle of the angles at the base BC equal. Hence MB (MC2)=CA2+ AM2. And taking MA+ AB2 from each of their equals, there remains 2MA. AB MA.AD CA2, we have 2CA2: CA2 2MA.AB:: AD: AB.

=

CA2

AB2; and as · AB2 :: 2MA.AD:

It will be perceived, that the steps of the analysis and demonstration are, all through, the reverse of each other.

Application of Algebra to Geometry.

When it is proposed to resolve a geometrical problem algebraically, or by algebra, it is proper, in the first place, to draw a figure that shall represent the several parts or conditions of the problem, and to suppose that figure to be the true one. Then, having considered attentively the nature of the problem, the figure is next to be prepared for a solution, if necessary, by producing or drawing such lines in it as appear most conducive to that end. This done, the usual symbols or letters, for known and unknown quantities, are employed to denote the several parts of the figure, both the known and unknown parts, or as many of them as necessary, as also such unknown line or lines as may be easiest found, whether required or not. Then proceed to the operation, by observing the relations that the several parts of the figure have to each other; from which, and the proper theorems in Nulty's Elements of Geometry, make out as many equations independent of each other, as there are unknown quantities employed in them: the resolution of which equations, in the same manner as in arithmetical problems, will determine the unknown quantities, and resolve the problem proposed.

As no general rule can be given for drawing the lines and selecting the fittest quantities to substitute for, so as always to bring out the most simple conclusions, because different problems require different modes of solution; the best way to gain experience, is to try the solution of the same problem in different ways, and then apply that which succeeds best, to other cases of the same kind, when they afterwards occur. The following particular directions, however, may be of some use.

1st. In preparing the figure, by drawing lines, let them be either parallel or perpendicular to other lines in the figure, or so as to form similar triangles. And if an angle be given, it will be proper to let the perpendicular be opposite to that angle, and to fall from one end of a given line, if possible.

2d. In selecting the quantities proper to substitute for, those are to be chosen, whether required or not, which lie nearest the known or given parts of the figure, and by means of which the next adja-, cent parts may be expressed by addition and subtraction only, without using surds.

3d. When two lines or quantities are alike related to other parts of the figure or problem, the best way is, not to make use of either of them separately, but to substitute for their sum, or difference, or rectangle, or the sum of their alternate quotients, or for some line or lines, in the figure, to which they have both the same relation.

4th. When the area, or perimeter, of a figure, is given, or such parts of it as have only a remote relation to the parts required, it is sometimes of use to assume another figure similar to the proposed one, having one side equal to unity, or some other unknown quantity. For, hence the other parts of the figure may be found, by the known proportions of the like sides, or parts, and so an equation be obtained. For examples, take the following problems; and for the principles of their geometrical construction, pp. 577-582.

Problem 1. In a right-angled triangle, having given the base (3), and the sum of the hypotenuse and perpendicular (9), to find both these two sides.

Let ABC represent the proposed triangle right-angled at B. Put the base AB-3=b, and the sum AC+ BC of the hypotenuse and perpendicular=9=s; also, let z denote the hypotenuse AC, and y the perpendicular BC.

Then by the question,

And by theorem 34

x+y=s,
x2= y2+b2,

By transposing y in the 1st. equation gives xs-y,

C

This value of a substituted in the 2d, gives s-2sy+y2=y2+b2, Taking away y' on both sides leaves

Hence xs-y=5—AC.

N. B. In this solution and some of the following ones, the notation is made by using as many unknown letters, x and y, as there are unknown sides of the triangle, a separate letter for each: in preference to using only one unknown letter for one side, and expressing the other unknown side in terms of that letter and the given sum or difference of the sides; though this latter way would render the solution shorter and sooner; because the former way gives occasion for more diversified practice in reducing equations; which is the very end and reason for which these problems are given at all.

2. In a right-angled triangle, having given the hypotenuse (5), and the sum of the base and perpendicular (7), to find both these two sides.

Let ABC represent the proposed triangle right-angled at B. Put the given hypotenuse AC 5a, and the sum AB+BC of the base and perpendicular =7 = s; also let x denote the base AB, and y the perpendicular Then by the question And by theorem 34

BC.

By transposing y in the 1st, gives
By substitu. this value for x, gives
By transposing s2, gives
By dividing by 2, gives

By completing the square, gives
By extracting the root, gives
By transposing the s, gives

x+y=s,
x2 + y2 = a2,

x=s y,

· 2sy + 2y2 =

s2-2sy

2sy: a2

=4 and 3, the values of x and y. 3. In a rectangle, having given the diagonal (10), and the perimeter, or sum of all the four sides (28), to find each of the sides sev erally.

Let ABCD be the proposed rectangle; and put the diagonal AC: 10d, and half the perimeter AB+ BC or AD + DC=14=a; also put one side AB: =x, and the other side BC=y.

Hence, by right-angled triangles
And by the question

D

C