The chords AI, IB, BL, etc., are equal, since they are subtended by equal arcs.

Secondly. We obtain the corresponding circumscribed polygon, by drawing through the points A, B, two tangents which terminate at the points m, n, on the tangent MN, and repeating the operation at the points C, D, etc.

The line mn is the side of a circumscribed polygon of double the number of sides; and Am, nB are half sides.

In effect, the right-angled triangles OAm, OIm are equal, having the common hypotenuse Om and the side OA=OI; consequently Am Im, and the angle AOm angle IOm.

=

=

We see also that the angle mOn is half the angle at the centre of each of the given polygons, and it is therefore the angle at the centre of the polygon sought.

Thirdly. As to the inscribed and circumscribed polygons of half the number of sides, we may obtain them by drawing through the alternate points A, C, E, etc., chords AC, CE, etc. And through the same points drawing tangents, having no regard to the intermediate points B, D, F, etc.

Scholium. It is evidently necessary in the third case, in order that the problem may be possible, to admit that the given polygons have an even number of sides.

OF CONTACT

PROBLEM XII.

Through a given point without the circumference of a circle

to draw a tangent to this circumference.

Let E be the centre of the given circle, and A the given point. Draw AE, and upon it, as a diameter, describe a circumference which will cut the given circumference in two points M, M', AM, and AM', being drawn, will be the tangents required. For,

E

M

M

the angles EMA, EM'A being in a semicircle, are right (T., X. C. I.). Consequently, AM and AM' are respectively perpendicular to the radii EM and EM', and therefore tangent to the circumference (T. V.).

Scholium. The case when the given point is in the circum ference offers no difficulty, since we then draw a radius to the given point, and through its extremity draw a perpendicular.

OF COMMON MEASURE.

PROBLEM XIII.

To find the common measure of two given lines, provided they have one, and consequently their numerical ratio.

Let AB and CD be the given lines.

From the greater AB cut off parts equal to the less 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.

C

G

F

H

B

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.

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 of of a foot of a foot; and so on for other comparisons.

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Cor. Since arcs of the same circle, or of equal circles, are capable of superposition, this same method may be employed. when we wish to find a common measure of two arcs of the same circle, or of equal circles.

Scholium. Magnitudes frequently have no common measure; that is, they are incommensurable; in which case, we shall always, by the foregoing method, have a remainder, however far we carry our successive operations. But these successive remainders would finally become so small that they might practically be neglected. Consequently, we can always find two numbers whose ratio shall approximate as close as we please to the ratio of the given incommensurable magnitudes.

This being admitted, we say that two incommensurable magnitudes A and B are proportional to two other incommensurable magnitudes A' and B', or that the ratio of A to B is equal to the ratio of A' to B', when we obtain for each the same numerical ratio at each successive step of the approximation.

PROBLEM XIV.

Show, geometrically, that the side of a square and its diagonal are incommensurable.

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.

D

A

F

C

B

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 BGFG; 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 = FG = BG 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.

THIRD BOOK.

THE PROPORTIONS OF STRAIGHT LINES AND THE AREAS OF RECTILINEAL FIGURES.

DEFINITIONS.

I. Two figures are similar when they resemble each other in all their parts, differing only in their comparative magnitudes, so that every point of the one has a corresponding point in the other. If, in two similar figures, we conceive a line joining any two points of the one, and another line joining the corresponding points of the other, the numerical ratio of these lines, which is always the same, is called the ratio of similitude.

When the ratio of similitude is equal to unity, the two similar figures become equal in all respects.

II. Two triangles are similar when their corresponding sides are proportional.

III. Two polygons are similar when they are capable of being decomposed into the same number of similar triangles, each to each, having the same order of arrangement.

IV. In similar figures homologous points, lines, or angles, are those which have like positions, in regard to the two figures.

V. The area of a surface, or its superficial extent, is the nu merical ratio of this surface referred to its unit surface. We readily see that figures of very different forms may have the same superficial extent. When we wish to express the property that two surfaces have the same area, without, however, being equal, or capable of superposition, we say they are equivalent.

VI. In determining the area of surfaces, it is found convenient. to call the bases of a parallelogram the sides which are parallel. The altitude or height is the common perpendicular to these bases, of which one is called the inferior or lower base, and the other the superior or upper base.