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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 beas completely determined by the arithmetical method as by the more lengthy method of the Greek geometers.

(71.) 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 one or more parts equal to FB, as many times as possible; for example, once, with the remainder GD.

From the first remainder FB cut off one or more 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.

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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 13.

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 3 of a foot=2 of a foot; and so on for other comparisons.

(72.) 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 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.

For, the angle ABG=AFG, each

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B

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 rightangle, 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.

THEOREM.

PROPOSITION I.

Two rectangles of the same altitude are to each other as their bases.

Let ABCD, AFGD be two rectangles having the common altitude AD; then will they be to each other as their bases AB, AF.

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A

F

G

- C

We will first suppose the bases AB, AF to be commensurable; as, for example, suppose they are to each other as 7 to 4. If we divide AB into 7 equal parts, AF will contain 4 of these parts; at each point of division drawn lines perpendicular to the base, thus forming 7 partial rectangles, which will be equal, since they have equal bases and the same altitude, (B. II, Prop. 1, Cor. 2.)

Now, as the rectangle ABCD contains 7 of these equal rectangles, while AFGD contains only 4, it follows ABCD : AFGD :: 7 : 4;

that

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therefore ABCD: AFGD :: AB: AF.

We will suppose, in the second place, that the bases AB, AF are incommensurable, still we shall

have

ABCD : AFGD :: AB : AF.

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For, if this proposition is not true, the first three terms remaining the same, the fourth term will be either greater or less than AF. Suppose it to be greater, and that we have

ABCD

Divide the line AB

AFGD: AB : AL.

into equal parts, each of which shall be less than FL. There will be at least one point of division, as at H, between F and L. Through this point H draw the perpendicular HK, then will the bases AB and AH be commensurable; and we shall have by the first part of this proposition

ABCD AHKD :: AB : AH.

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But by supposition,

ABCD AFGD :: AB : AL.

Since, in these two proportionals, the antecedents are equal, the consequents are proportional, and we have AHKD AFGD :: AH: AL.

But AL is greater than AH; hence, if this proportion is correct, we must have AFGD greater than AHKD; on the contrary, it is less, hence the above proportion is impossible. Therefore ABCD cannot be to AFGD as AB is to a line greater than AF.

By similar reasoning, we can show that the fourth term of the proportion cannot be smaller than AF.

Hence, whatever may be the ratio of the bases, the two rectangles ABCD, AFGD, of the same altitude, will be to each other as their bases AB, AF.

PROPOSITION II.

THEOREM. Any two rectangles are to each other as the product of their bases multiplied by their altitudes.

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