duced; join CF, and the polygon ABCDE will be equivalent to the polygon ABCF, which has one side less.

For the triangles CDE, CFE, have the common base CE, they are also of the same altitude, for their vertices D, F, are in a line DF parallel to the base; therefore the triangles are equiv alent. Adding to each of these the figure ABCE and we shall have the polygon ABCDE equivalent to the polygon ABCF.

We can in like manner cut off the angle B by substituting for the triangle ABC the equivalent triangle AGC, and then the pentagon ABCDE will be transformed into an equivalent trian gle GCF.

The same process may be applied to any other figure; for, by making the number of sides one less at each step, we shall at length arrive at an equivalent triangle.

249. Scholium. As we have already seen, that a triangle may be transformed into an equivalent square (243), we may accordingly find a square equivalent to any given rectilineal figure; this is called squaring the rectilineal figure, or finding the quadrature of it.

The problem of the quadrature of the circle consists in finding a square equivalent to a circle whose diameter is given.

Fig. 147.

PROBLEM.

250. To make a square which shall be equal to the sum or the difference of two given squares.

Solution. Let A and B (fig. 147) be the sides of the given

squares.

1. If it is proposed to find a square equal to the sum of these squares, draw the two indefinite lines ED, EF, at right angles to each other; take EDA and EG = B; join DG, and DG will be the side of the square sought.

For the triangle DEG being right-angled, the square described upon DG will be equal to the sum of the squares described upon ED and EG.

2. If it is proposed to find a square equal to the difference of the given squares, form in like manner a right angle FEH, take GE equal to the smaller of the sides A and B; from the point G, as a centre, and with a radius GH equal to the other side, describe an arc cutting EH in H; the square described upon EH

will be equal to the difference of the squares described upon the lines A and B.

For in the right-angled triangle GEH the hypothenuse GHA, and the side GE B; therefore the square described = upon EH is equal to the difference of the squares described upon the given sides A and B.

251. Scholium. We can thus find a square equal to the sum of any number of squares; for the construction by which two are reduced to one, may be used to reduce three to two and these two to one, and so of a larger number. Also a similar method may be employed when certain given squares are to be subtracted from others.

PROBLEM.

252. To construct a square which shall be to a given square ABCD (fig. 150), as the line M is to the line N.

Solution. On the indefinite line EG take EFM, and FG = N ; on EG, as a diameter, describe a semicircle, and at the point F erect upon the diameter the perpendicular FH. From the point H draw the chords HG, HE, which produce indefinitely; on the first take HK equal to the side AB of the given square, and through the point K draw KI parallel to EG; HI will be the side of the square sought.

For, on account of the parallels KI, GE,

hence

-2

H:HK::HE:Hứ (v).

But, in the right-angled triangle EH,

the square upon HI: the square upon AB:: M: N.

Fig. 150

PROBLEM.

253. Upon a side FG (fig. 129), homologous to AB, to describe a Fig. 129. polygon similar to a given polygon ABCDE.

=

Solution. In the given polygon draw the diagonals AC, AD. At the point F make the angle GFH= BAC, and at the point G the angle FGH ABC; the lines FH, GH, will cut each other in H, and the triangle FGH will be similar to ABC. Likewise upon FH, homologous to AC, construct the triangle FIH similar to ADC, and upon FI, homologous to AD, construct the triangle FIK similar to ADE. The polygon FGHIK will be similar to ABCDE.

For these two polygons are composed of the same number of triangles, which are similar to each other and similarly disposed (219).

PROBLEM.

254. Two similar figures being given, to construct a similar figure which shall be equal to their sum or their difference.

Solution. Let A and B be two homologous sides of the given figures, find a square equal to the sum or the difference of the squares described upon A and B; let X be the side of this square, X will be, in the figure sought, the side homologous to A and B in the given figures. The figure may then be constructed by the preceding problem.

For similar figures are as the square of their homologous sides; but the square of the side X is equal to the sum or the difference of the squares described upon the homologous sides A and B; therefore the figure described upon the side X is equal to the sum or the difference of the similar figures described upon the sides A and B.

PROBLEM.

255. To construct a figure similar to a given figure, and which shall be to this figure in the given ratio of M to N.

Solution. Let A be a side of the given figure, and X the homologous side of the figure sought; the square of X must be to the square of A, as M is to N (221); X then may be found by art. 252; and, knowing X, we may finish the problem by art.

253.

PROBLEM.

ig. 151.

256. To construct a figure similar to the figure P (fig. 151) and equivalent to the figure Q.

Solution. Find the side M of a square equivalent to the figure P, and the side N of a square equivalent to the figure Q. Then let X be a fourth proportional to the three given lines M, N, AB; upon the side X, homologous to AB, describe a figure similar to the figure P; it will be equivalent to the figure Q. For, by calling Y the figure described upon the side X, we shall have

We have also, by construction, MP, and NQ;

hence Y=Q; therefore the figure Y is similar to the figure P and equivalent to the figure Q.

PROBLEM.

257. To construct a rectangle equivalent to a given square C (fig. 152), and whose adjacent sides shall make a given sum AB. Solution. On AB, as a diameter, describe a semicircle, and draw DE parallel to the diameter, and at a distance AD, equal to a side of the given square C. From the point E, in which the parallel cuts the circumference, let fall upon the diameter the perpendicular EF; AF and FB will be the sides of the rectangle sought.

For their sum is equal B, and their rectangle AF × FB is equal to the square of E15), or of AD; therefore the rectangle is equivalent to the given square C.

258. Scholium. It is necessary in order that the problem may be possible, that the distance AD should not exceed the radius, that is, that the side of the square should not exceed half of the line AB.

Fig. 152.

PROBLEM.

259. To construct a rectangle equivalent to a square C (fig. 153), Fig. 153. and whose adjacent sides shall differ by a given quantity AB.

Fig. 154.

Solution. On the given line AB, as a diameter, describe a circle; from the extremity of the diameter draw the tangent AD equal to the side of the square C. Through the point D and the centre O draw the secant DF; DE and DF will be the adjacent sides of the rectangle required.

For, 1. the difference of the sides is equal to the diameter EF or AB; 2. the rectangle DE × DF is equal to AD (228); therefore this rectangle will be equivalent to the given square C.

PROBLEM.

260. To find the common measure, if there be one, between the diagonal and side of a square.

Solution. Let ABCG (fig. 154) be any square, and AC its diagonal.

We are, in the first place, to apply CB to CA, as often as it can be done (157); and in order to this, let there be described, from the centre C, and with a radius CB, the semicircle DBE. It will be seen, that CB is contained once in AC with a remainder AD; the result of the first operation therefore is the quotient 1 with the remainder AD, which is to be compared with BC, or its equal AB.

We may take AF AD, and apply AF actually to AB; and we should find that it is contained twice with a remainder. But, as this remainder and the following ones go on diminishing and would soon become too small to be perceived, on account of the imperfection of the mechanical operation, we can conclude nothing with regard to the question, whether the lines AC, CB, have a common measure or not. Now there is a very simple method, by which we may avoid these decreasing lines, and which only requires an operation to be performed upon lines of the same magnitude.

The angle ABC being a right angle, AB is a tangent, and AE is a secant, drawn from the same point, so that

AD: AB::AB: AE (228).

Thus, in the second operation which has for its object to compare AD with AB, we may, instead of the ratio of AD to AB, take that of AB to AE. Now AB, or its equal CD, is contained twice in AE with a remainder AD; therefore the result of the second operation is the quotient 2 with the remainder AD, which is to be compared with AB.