PROBLEM.

157. To find the numerical ratio of two given straight lines AB, CD (fig. 90), provided, however, these two lines have a common Fig. 90,

measure.

Solution. Apply the smaller CD to the greater AB, as many times as it will admit of, for example, twice with a remainder BE.

Apply the remainder BE to the line CD, as many times as it will admit of, for example, once with a remainder DF.

Apply the second remainder DF to the first BE, as many times as it will admit of, once, for example, with a remainder BG. Apply the third remainder BG to the second DF, as many

times as it will admit of.

Proceed thus, till a remainder arises, which is exactly contained a certain number of times in the preceding.

This last remainder will be the common measure of the two proposed lines; and, by regarding it as unity, the values of the preceding remainders are easily found, and, at length, those of the proposed lines from which their ratio in numbers is deduced.

If, for example, we find that GB is contained exactly twice in FD, GB will be the common measure of the two proposed lines. Let GB = 1, we have FD= 2; but EB contains FD once plus GB; therefore EB=3; CD contains EB once plus FD; therefore CD = 5; AB contains CD twice plus EB; therefore AB = 13; consequently the ratio of the two lines AB, CD, is as 13 to 5. If the line CD be considered as unity, the line AB would be ; and, if the line AB be considered as unity, the line CD would be 5

13

13°

158. Scholium. The method, now explained, is the same as that given in arithmetic for finding the common divisor of two numbers (Arith. 61), and does not require another demonstration.

It is possible, that, however far we continue the operation, we may never arrive at a remainder, which shall be exactly contained a certain number of times in the preceding. In this case the two lines have no common measure, and they are said to be incommensurable. We shall see, hereafter, an example of this in the ratio of the diagonal to the side of a square. But, although the exact ratio cannot be found in numbers, by neglecting the last remainder we may find an approximate ratio to a greater Geom.

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or less degree of exactness, according as the operation is more or less extended.

PROBLEM.

15

Fig, 91.

159. Two angles A and B (fig. 91) being given, to find their common measure, if they have one, and from this their ratio in

numbers.

Solution. Describe, with equal radii, the arcs CD, EF, which may be regarded as the measure of these angles; in order then to compare the arc CD, EF, proceed as in the preceding problem; for an arc may be applied to an arc of the same radius, as a straight line is applied to a straight line. We shall thus obtain a common measure of the arcs CD, EF, if they have one, and their ratio in numbers. This ratio will be the same as that of the given angles (122); if DO is the common measure of the arcs, DAO will be the common measure of the angles.

160. Scholium. We may thus find the absolute value of an angle by comparing the arc, which serves as its measure, with the whole circumference. If, for example, the arc CD is to the circumference as 3 to 25, the angle A will be of four right angles, or of one right angle.

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It may happen, as we have seen with respect to straight lines, that arcs also, which are compared, have not a common meas. ure; we can then obtain, for the angles, only an approximate ratio in numbers, more or less exact, according to the degree to which the operation is extended.

SECTION THIRD.

Of the Proportions of Figures.

DEFINITIONS.

161. I SHALL call those figures equivalent whose surfaces are equal.

Two figures may be equivalent, however dissimilar; thus a circle may be equivalent to a square, a triangle to a rectangle, &c. The denomination of equal figures will be restricted to those, which being applied, the one to the other, coincide entirely; thus two circles having the same radius are equal; and two triangles

having the three sides of the one equal to the three sides of the other, each to each, are also equal.

162. Two figures are similar, which have the angles of the one equal to the angles of the other, each to each, and the homologous sides proportional. By homologous sides are to be understood those, which have the same position in the two figures, or which are adjacent to equal angles. The angles, which are equal in the two figures, are called homologous angles. Equal figures are always similar, but similar figures may be very unequal.

163. In two different circles, similar arcs, similar sectors, similar segments, are such as correspond to equal angles at the centre. Thus, the angle A (fig. 92) being equal to the angle O, the Fig. 92. arc BC is similar to the arc DE, the sector ABC to the sector ODE, &c.

164. The altitude of a parallelogram is the perpendicular which measures the distance between the opposite sides AB, CD (fig. 93), considered as bases.

Fig. 93. The altitude of a triangle is the perpendicular AD (fig. 94), Fig. 94. let fall from the vertex of an angle A to the opposite side taken for a base.

The altitude of a trapezoid is the perpendicular EF (fig. 95) Fig. 95. drawn between its two parallel sides AB, CD.

165. The area and the surface of a figure are terms nearly synonymous. Area, however, is more particularly used to denote the superficial extent of the figure considered as measured, or compared with other surfaces.

THEOREM.

166. Parallelograms, which have equal bases and equal altitudes, are equivalent.

Demonstration. Let AB (fig. 96) be the common base of the Fig. 96. two parallelograms ABCD, ABEF; since they are supposed to have the same altitude, the sides DC, FE, opposite to the bases, will be situated in a line parallel to AB (69). Now, by the nature of a parallelogram, AD= BC (84), and AF BE; for the same reason, DC AB, and FE AB; therefore DC FE. If DC be taken from DE, there will remain CE; and if FE, equal to DC, be taken also from DE, there will remain DF; consequently CE = DF.

=

=

Fig. 97.

Fig. 98.

Hence the triangles DAF, CBE, have the three sides of the one equal to the sides of the other, each to each; they are therefore equal (43).

But, if from the quadrilateral ABED the triangle ADF be taken, there will remain the parallelogram ABEF; and, if from the same quadrilateral ABED the triangle CBE, equal to the former, be taken, there will remain the parallelogram ABCD; therefore the two parallelograms ABCD, ABEF, which have the same base and the same altitude, are equivalent.

167. Corollary. Every parallelogram ABCD (fig. 97) is equivalent to a rectangle of the same base and altitude.

THEOREM.

168. Every triangle ABC (fig. 98) is half of a parallelogram ABCD of the same base and altitude.

Demonstration. The triangles ABC, ACD, are equal (87), therefore each is half of the parallelogram ABCD.

169. Corollary 1. A triangle ABC is half of a rectangle BCEF of the same base BC and the same altitude AO; for the rectangle BCEF is equivalent to the parallelogram ABCD (167). 170. Corollary 11. All triangles, which have equal bases and equal altitudes, are equivalent.

Fig. 99.

THEOREM.

171. Two rectangles which have the same altitude, are to each other as their bases.

Demonstration. Let ABCD, AEFD (fig. 99), be two rectangles, which have a common altitude AD; they are to each other as their bases AB, AE:

Let us suppose, in the first place, that the bases AB, AE, are commensurable, and that they are to each other, as the numbers 7 and 4, for example; if we divide AB into 7 equal parts, AE will contain four of these parts; erect at each point of division, a perpendicular to the base, we shall thus form seven partial rectangles which will be equal to each other, since they will have the same base and the same altitude (166). The rectangle ABCD will contain seven partial rectangles, while AEFD will contain four; therefore the rectangle ABCD is to the rectangle AEFD, as 7 is to 4, or as AB is to AE. The same rea

soning may be applied to any other ratio beside that of 7 to 4; nence, whatever be the ratio, provided it is commensurable, we have

ABCD: AEFD::AB: AE.

Let us suppose, in the second place, that the bases AB, AE (fig. 100), are incommensurable; we shall have notwithstanding Fig. 100. ABCD: AEFD::AB: AE.

For, if this proportion be not true, the three first terms remaining the same, the fourth will be greater or less than AE. Let us suppose that it is greater, and that we have

ABCD: AEFD::AB: AO.

Divide the line AB into equal parts smaller than EO, and there will be at least one point of division I between E and 0: at this point erect the perpendicular IK; the bases AB, AI, will be commensurable, and we shall have, according to what has just been demonstrated,

ABCD: AIKD::AB: AI.

But we have, by hypothesis,

ABCD: AEFD::AB: AO.

In these two proportions the antecedents are equal, therefore the consequents are proportional (I); that is

AIKD: AEFD::AI:AO.

Now AO is greater than AI; it is necessary then, in order that this proportion may take place, that the rectangle AEFD should be greater than AIKD; but it is less; therefore the proportion is impossible, and ABCD cannot be to AEFD, as AB is to a line greater than AE.

By a process entirely similar it may be shown, that the fourth term of the proportion cannot be smaller than AE; consequently it is equal to AE.

Whatever therefore be the ratio of the bases, two rectangles ABCD, AEFD, of the same altitude, are to each other as their bases AB, AE.

THEOREM.

172. Any two rectangles ABCD, AEGF (fig. 101), are to each Fig. 101. other, as the products of their bases by their altitudes, that is,

ABCD: AEGF :: AB × AD: AE x AF.