BOOK FOURTH.

DEFINITIONS.

1. For the doctrine of ratios and proportions, we will refer the student to the method explained in the.Algebra. 2. There is this difference between geometrical ratios of magnitudes, and ratios of numbers: All numbers are commensurable; that is, their ratio can be accurately expressed but many magnitudes are incommensurable; that is, their ratio can be expressed only by approximation; which approximation may, however, be carried to any extent we desire. Such is the ratio of the circumference of a circle to its diameter, the diagonal of a square to its side, etc. Hence many have deemed the arithmetical method not sufficiently general to apply to geometry. This would be a safe inference, were it necessary in all cases to assign the specific ratio between the two terms compared. But this is not the case. Such ratios themselves may be unknown, indeterminate, or irrational, and still their equality or inequality may be as completely determined by the arithmetical method as by the more lengthy method of the Greek geometers.

(63.) PROBLEM. To find a common measure of two given lines, and consequently their numerical ratio.

Let AB and CD be the given lines.

From the greater AB cut off parts equal to the lesser line CD, as many times as possible; for example, twice, with the remainder FB.

From the line CD cut off parts equal to FB, as many times as possible; for example, once, with the remainder GD.

From the first remainder FB cut off parts equal to the remainder GD, as many times as possible; for example, once, with the remainder HB.

From the second remainder GD cut off parts equal to the third remainder HB, as many times as possible; for example, twice, without a remainder.

H

The last remainder HB will be a common measure of the given lines.

If we regard HB as a unit, GD will be 2, and

FB = FH+HB = GD+HB = 3;

CD

CG + GD = FB + GD =3+2 = 5;

AB

AF FB = 2 CD + FB = 10 +3 = = 13.

Therefore the line AB is to the line CD as 13 to 5.

If AB is taken for the unit, CD will be; but if CD be taken as the unit, AB will be 3.

If AB is of a yard, then CD will be of a yard, or of a yard.
Again, if CD is of a foot, then AB will be 3 of of a foot

of a foot; and so on for other comparisons.

(64.) As a case in which the magnitudes are incommensurable, we will take the following

PROBLEM. Find the ratio of the diagonal of a square to its side.

Let ABCD be the square, and AC D its diagonal.

Cutting off AF from the diagonal equal to AB a side of the square, we have the remainder CF which must be compared with CB.

If we join FB, and draw FG perpendicular to AC, the triangle BGF will be isosceles.

A

G

For, the angle ABG= AFG, each being a right angle; and since the triangle ABF is isosceles, the angle ABF = AFB. Therefore, subtracting the angle ABF from ABG, the remainder FBG will equal the angle BFG, found by subtracting the angle AFB from AFG. Consequently the triangle BGF is isosceles, and BG FG; but, since AC is the diagonal of a square, the angle FCG is half a right angle; but CFG is a right angle, and consequently FGC is also half a right angle, and CG is the diagonal of a square whose side is CF.

=

Hence, after CF has been taken once from CB, it remains to take CF from CG, that is, to compare the side of a square with its diagonal, which is the very question we set out with, and of course we shall find precisely the same difficulty in the next step of the process; so that, continue as far as we please, we shall never arrive at a term in which there will be no remainder. Therefore there is no common measure of the diagonal and side of a square.

If the side of a square be represented by 1, then arithmetically the diagonal will be √2, and this value can be found only approximately. The student may say, that in this case, we have two numbers which have not even a unit for their common measure. I reply, that √2 is not a number : it is simply an expression for a ratio, the arithmetical value of which can only be found approximately.

3. Similar figures are those which have the angles of the one equal to the angles of the other, each to each, and the sides about the equal angles proportional.

4. The perimeter, or contour of a figure, is the sum of all its sides, or the length of the bounding line.

5. Two magnitudes are said to be identical, when they: are equal in all their parts, and are also capable of coinciding in every part of their extent.

PROPOSITION I.

THEOREM. If a line be drawn parallel to one side of a triangle, cutting the other two sides, these sides will be divided into proportional parts.

Let DF be parallel to the side BC of the triangle ABC; then will AD : ᎠᏴ :: AF : FC.

For, draw BF and CD; then the triangles DBF, FCD are equal to each other, because they have the same base DF, and are

B

D

between the same parallels DF, BC (B. II, Prop. 1, Cor. 1). But the two triangles ADF, BDF, on the bases AD, DB, have the same altitude; and the two triangles ADF, CDF, on the bases AF, FC, have also the same altitude; and because triangles of the same altitude are to each other as their bases, therefore

But BDF=CDF; consequently, by equality of ratios, we AD DB:: AF: FC.

have

:

In a similar manner, the theorem is proved when the sides of the triangle are cut in prolongation beyond either the vertex or the base,

Cor. Hence, also, the whole lines AB, AC are proportional to their corresponding proportional segments. Thus, since AD : DB:: AF : FC, we have by composition AD+DB : AD :: AF+FC: AF, or

AB AD AC: AF; and

:

AD+DB DB:: AF+FC FC, or
AB: DB :: AC: FC.

PROPOSITION II.

THEOREM. If two sides of a triangle are cut proportionally by a straight line, this line will be parallel to the third side.

In the triangle ABC, let the line DF

be drawn, so that

AD: DB:: AF: FC;

then will DF be parallel to BC.

For if DF is not parallel to BC, suppose that from the point D the line DG B be drawn parallel to BC. Then we have (B. IV, Prop. 1)

AD

DB:: AG: GC. thesis, we have AD: DB :: AF : FC;

must have AG GC AG AF

AF: FC, or

GC FC, which is an impossible result, since the antecedent of the first couplet is less than its consequent, while the antecedent of the second couplet is greater than its consequent. Hence the line drawn from D parallel to BC cannot differ from the line DF; that is, DF is parallel to BC.