triangles FH B, G K F, which have their angles equal to one another, each to each, because their sides are parallel (18.): therefore (5.), their other sides are also equal to one another, and F B is equal to G F. In like manner it may be shown, that each of the other five parts into which AB is divided, is equal to FB or GF. And it is evident that, by a similar process, A B may be divided into any other number of equal parts.

Therefore, &c.

It is convenient to draw the parallels DF, EG, &c. by making the angle A B a, equal to B A C (47.), taking Bd, de, &c. each equal to M, and joining corresponding points Dd, E e, &c. For the straight lines which join the extremities of equal and parallel straight lines, are also themselves equal and parallel (21.). A similar process may be recommended, whenever a number of parallels are to be drawn without the assistance of a parallel ruler.

PROP. 50. Prob. 9. (Euc. i. 22). To construct a triangle from three parts given, of which one, at least, is a side. Case 1. Let

If the given side is to be interjacent or

B

E

to lie between the given angles, make (47.) at the points C, D, the angles D CE, CDE, equal to A, B, respectively and the triangle ECD will evidently be the triangle required.

But if CD is to be opposite to one of the given angles, as B, make at the point C, the angles DCE, ECF, equal to A, B, respectively, and through D, draw DE parallel to C F (48.): then the angle CE Dis (15. Cor. 2.) equal to the alternate angle ECF, that is, to B, and ECD is the triangle required.

Case 2. Let the given parts be two sides and an angle; viz. the sides A, B, and the angle C.

If the given angle is to be included, take CD, CE, upon its sides, equal to A, B, respectively, and join DE; the triangle CDE will evidently be the triangle required.

A

B

But if C is to be opposite to one of the given sides as A, take CE equal to B as before: from the centre E, with a radius equal to A, describe a circle cutting CD in the points D, D; and join ED, ED. Then, if the points D, D fall both on the same side of C with the angle, each of the triangles E C D will satisfy the given conditions; but if they fall upon different sides, only one of them, as ECD, will satisfy those conditions, and therefore that one will be the triangle required.

If the points D, D, being upon the same side with the angle, coincide with one another, which will happen when EDC is a right angle (12. Cor. 3.), there is only one triangle ED C, which is the triangle required.

Case 3. Let the given parts be three sides, viz. A,

In each of these three cases we have supposed the data, or given parts, to be such that the problem is possible: and the same will be supposed in all future problems. Many of them, however, will be found to be of that description, that, if the relations of the data be not confined within certain limits, the required solutions will be impossible. In problems of a more complex nature than the present, the determination of these limits is facilitated by the consideration of loci, to be noticed hereafter (in Book III. Sect. 6), and which may here be explained as circumscribing the range, if it be limited, within which every particular datum confines the solution of the problem; for it is obvious that, if a second datum cannot be satisfied within that range, the two will be inconsistent, and the solution impossible.

In the mean time, it may be observed, the limits of the data are in many cases readily suggested by the known properties of the figure, or by the necessary construction of the problem. It is evident, for instance, that in every case of the problem before us, there are certain limits, beyond which the solution will be impossible. This will happen in the first case if the two given angles should be together equal to or

A B K, is equal to the figure ABCD EF, and it has the same side A B, and adjacent angle B with the figure. Therefore, &c.

PROP. 55. Prob. 14.

To bisect a given triangle or rectilineal figure by a straight line drawn from a given angle.

First, let it be required to bisect the triangle ABC by a straight line to be drawn from the angle A.

Bisect (43.) B C in D, and join AD. Then because B D is equal to D C, the triangle A B D is (27.) equal to ADC, and the triangle A B C is bisected by the straight line A D.

B

Next, let it be required to bisect the rectilineal figure ABCDEF by a straight line, to be drawn from the angle A.

With the same side A B and angle B, describe (54.) the triangle A BK, equal to the given rectilineal figure: bisect (43.) BK in L: through L draw (48.) LM parallel to A C, to meet C D produced in M: through M draw M N parallel to AD, to meet D E in N, and

given figure, and the latter is bisected by the straight line A N. Therefore, &c.

Cor. 1. By a similar construction, any part required, for example a fifth, may be cut off from a given triangle or rectilineal figure, by a straight line drawn from one of its angles. For in the case of the triangle, if the base BC be divided into five equal parts in the points D, &c. the triangles A BD, &c. will be equal to one another (27.); and therefore any one of them, as A B D, will be equal to one fifth of the given triangle. And hence the passage is easy to the division of the rectilineal figure; for it is evident (53.) that, whatever be the point L taken in AK, the figure A B CD N, constructed as above, will be equal to the triangle A B L, and will therefore be a fifth part of the given figure, if AB L be a fifth of the triangle ABK.

It is manifest in this case, that, if the given fractional part be such, that B L is less than BC, the problem will be solved at once by joining A L.

Cor. 2. And hence it appears in what manner a triangle or rectilineal figure may be divided into any number of equal parts by straight lines drawn from one of its angles.

PROP. 56. Prob. 15.

To bisect a given triangle or rectilineal figure by a straight line drawn from a given point in one of its sides.

First, let it be required to bisect the triangle ABC, by a straight line to be drawn from the point D in the side A B.

Join D C: bisect (43.)

B

E

D

AB in E: through E draw (48.) E F, parallel to DC, and join DF, EC. Then, because DC is parallel to E F, the triangle D E F is (27.) equal to CEF; therefore DEF, BEF together, are equal to CEF, BEF together, that is, the triangle DB F, is equal to the triangle C EB, or (27.) to half of the triangle ABC. Therefore, ABC is bisected by the straight line D F.

Next, let it be required to bisect the figure A B C D E F, by a straight line, to be drawn from the point G in the side AB.

E

M

Describe the triangle ABK (54.) equal to the given figure, and, as already shewn, bisect it by the straight line GL, drawn from the point G: join GC, GD, &c.; through L draw (48.) LM, parallel to G C, to meet CD produced in M: through M draw M N parallel to GD to meet DE in N, and join G N, GM, G L.* Then, as in the last proposition, it may be shown that the figure GB CDN is equal to the triangle G B L, that is to half of the given figure; therefore the latter is bisected by the straight line GN.

Therefore, &c.

Cor. 1. By a similar construction, any part required may be cut off from a given triangle or rectilineal figure, by a straight line drawn from a given point in one of its sides. For we have but to make CBE the same part of C B A in the first case, and G BL the same part of ABK in the case of the rectilineal figure, and proceed as before.

If the fractional part be such that BL is less than BC, the problem will be solved at once by joining G L.

Cor. 2. And hence a given triangle or rectilineal, figure may be divided into any number of equal parts, by straight lines drawn from a given point in one of its

sides.

PROP. 57. Prob. 16. (Euc. i. 44.) Upon a given base BD to describe a rectangle which shall be equal to a given triangle ABC.

H

M LA

F

From the point D draw D E, at right angles to BD (44.): through A draw A E parallel to BD, to meet DE in E (48.): join EB: bisect B C in F (43.): through F draw FG, parallel to DE, to meet BE in G: through G draw H K parallel to BD, and complete the rectangle DH: DH shall be the rectangle required.

Complete the rectangles DL, DM. Then, because GM, GD are complements of the rectangles F H, K L, which are about the diagonal of the rectangle DM, GD is (23.) equal to G M. Therefore, adding FH to each, the whole HD is equal to the whole M F, that is, (26.) to twice the triangle ABF, or to the triangle ABC; and HD is described upon the given base BD. Therefore, &c.

*GD and GL do not necessarily coincide, as in the figure.

Cor. 1. Hence, upon a given base, a rectangle may be described which shall be equal to a given rectilineal figure: for a triangle (54.) may be described equal to the figure, and a rectangle equal to the triangle.

Cor. 2. It is evident that, with a like construction, a parallelogram may be described upon a given base BD which shall be equal to a given triangle A B C, and shall have one of its angles equal to a given angle B D E.

PROP. 58. Prob. 17. (Euc. ii. 14).

To describe a square which shall be equal to a given rectangle ABCD. Produce A B to E, so that BE may be equal to B C: bisect the centre F, with AE in F (43.): from the radius FA or

A

D

to meet the circumference in G: the FE, describe a circle, and produce C B square of BG shall be equal to the rectangle ABCD.

to FE, AB is equal to the sum, and Join FG, then, because FA is equal BE to the difference, of FE, FB: therefore (34.) the rectangle AB, BE, that is, the rectangle A B C D, is equal to the difference of the squares of FE, F B, that is to the difference of the squares of F G, F B, or (36. Cor. 1.) to the square of B G.

Therefore, &c.

Cor. Hence a square may be described which shall be equal to a given rectilineal figure (57. Cor. 1.).

equal to the sum of two, three, or any number of given squares, viz. the squares of A, B, C, &c.

Take D E equal to A; from the point D, draw DF (44.) at right angles to DE, and equal to B: join EF: from F, draw F G at right angles to EF, and equal to C; and so proceed for the rest of the given squares. Join E G. Then, because E D F is a right angled triangle, the square of EF is equal to the squares of E D, D F (36.), that is, to the squares of A, B: again, because E F G is a right angled triangle, the square of EG is equal to the squares of E F, FG, that is, to the squares of A, B, C; and

$1. Ratios of Commensurable Magnitudes-§ 2. Proportion of Commensurable Magnitudes — § 3. General Theory of Proportion-§ 4. Propor tion of the sides of Triangles- 5. Proportion of the surfaces of Rectilineal Figures-§ 6. Properties of Lines divided Harmonically-§ 7. Problems.

SECTION 1. Ratios of Commensurable Magnitudes.

IN the language of Mathematics, the Latin word ratio has been adopted to express what is more generally understood by the term proportion: thus, instead of "the proportion which" one thing bears to another, we say "the ratio which" one bears to the other, meaning its comparative magnitudeinstead of saying that A is to B "in the proportion of 5 to 6," we say "in the ratio of 5 to 6."

The word proportion has on the other hand been appropriated to express the equality of ratios, as hereafter defined; or, as it may be here less minutely explained, the case in which one magnitude is as many times greater or less than another, as a third

magnitude is greater or less than a fourth.

The ratio of one magnitude to another is independent of the kind of magnitudes compared; for it is obvious that one may contain the other, or the sixth, or twelfth, or hundredth part of the other the same number of times, whether they be lines, or surfaces, or solids, or again, weights, or parts of duration.

It is required only for the comparison we speak of, that the magnitudes be of the same kind, containing the same magnitude, each of them, a certain number of times, or a certain number of times nearly. Upon these numbers, and upon these only, the ratio depends.

Hence it appears that this theory pertains in truth to Arithmetic. The use of Proportion is, however, so indispensable in Geometry, that it has been usual, either to introduce its theorems into the body of the science, after the example of Euclid, or to premise a few of the most important of them as a manual for reference. In the present, and two following sections, the subject will be discussed at a length commensurate with its importance.

Def. 1. When one magnitude is compared with another of the same kind, the first is called the antecedent, and the second the consequent.

2. One magnitude is said to be a multiple of another, when it contains that other a certain number of times exactly and the other magnitude, which is contained in the first a certain number of times exactly, is said to be a submultiple, or measure, or part of the first.

Hence, also, one magnitude is said to measure another when it is contained in the other a certain number of times exactly.

3. Two magnitudes are said to be equimultiples of two others, when they contain those others the same number of times exactly: and the other magnitudes which are contained in the first the same number of times exactly, are said to be like parts of the two first.

Thus, 7 A, 7 B, are equimultiples of A, B; and A, B, are like parts of 7 A, 7 B.

4. Two magnitudes are said to be commensurable with one another, when a common measure of the two may be found, i. e. a magnitude which is contained in each of them a certain number of times exactly.

In like manner, any number of mag