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PROBLEM.

239. To find a mean proportional between two given lines A and B (fig. 140).

Solution. On the indefinite line DF take DE=A, and EF-B; on the whole line DF, as a diameter, describe the semicircumference DGF; at the point E erect upon the diameter the perpendicular EG meeting the circumference in G; EG will be the mean proportional sought.

For the perpendicular GE, let fall from the point in the circumference upon the diameter, is a mean proportional between the two segments of the diameter DE, EF (215), and these two segments are equal to the two given lines A and B.

Fig. 140.

PROBLEM.

240. To divide a given line AB (fig. 141) into two parts in Fig. 141. such a manner, that the greater shall be a mean proportional between the whole line and the other part.

Solution. At the extremity B of the line AB erect the perpendicular BC equal to half of AB; from the point C as a centre, and with the radius CB, describe a circle; draw AC cutting the circumference in D, and take AF AD; the line AB will be divided at the point Fin the manner required; that is,

=

AB: AF::AF: FB.

For AB, being a perpendicular to the radius CB at its extremity CB, is a tangent; and, if AC be produced till it meet the circumference in E, we shall have

AE : AB :: AB : AD (228),

and hence AE—AB: AB :: AB-AD: AD (Iv).

But, since the radius BC is half of AB, the diameter DE is equal to AB, and consequently AE-ABAD = AF; also, since AF AD, AB-ADFB; therefore,

=

and, by inversion,

AF:AB:: FB : AD or AF,

AB: AF::AF: FB.

241. Scholium. When a line is divided in this manner, it is said to be divided in extreme and mean ratio. Its application will be seen hereafter.

It

may be remarked, that the secant AE is divided in extreme and mean ratio at the point D; for, since AB = DE,

GEOM.

AE: DE::DE: AD.

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Fig. 142.

PROBLEM.

242. Through a given point A (fig. 142), in a given angle BCD, to draw a line BD in such a manner that the parts AB, AD, comprehended between the point A and the two sides of the angle, shall be equal.

Solution. Through the point A draw AE parallel to CD; take BE CE, and through the points B and A draw BAD, which will be the line required.

=

For AE being parallel to CD, BE: EC:: BA: AD; but BE EC; therefore BA=AD.

=

PROBLEM.

Fig. 143.

Fig 144.

Fig.145.

243. To make a square equivalent to a given parallelogram, or to a given triangle.

Solution. 1. Let ABCD (fig. 143) be the given parallelogram, AB its base, and DE its altitude; between AB and DE find a mean proportional XY (239); the square described upon XY will be equivalent to the parallelogram ABCD. For, by construction, AB : XY:: XY: DE; hence

2
XY=AB × DE;

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but AB × DE is the measure of the parallelogram, and XY is that of the square; therefore they are equivalent.

2. Let ABC (fig. 144) be the given triangle, BC its base, and AD its altitude; find a mean proportional between BC and half of AD, and let XY be this mean proportional; the square described upon XY will be equivalent to the triangle ABC.

-2

For, since BC: XY:: XY: ¦ AD, XY = BC × AD; therefore the square described upon XY is equivalent to the triangle ABC.

PROBLEM.

244. Upon a given line AD (fig. 145) to construct a rectangle ADEX equivalent to a given rectangle ABFC.

Solution. Find a fourth proportional to the three lines AD, AB, AC (137), and let AX be this fourth proportional; the rectangle contained by AD and AX will be equivalent to the rectangle ABFC.

For, since AD: AB:: AC: AX, AD × AX=AB × AC therefore the rectangle ADEX is equivalent to the rectangle ABFC.

PROBLEM.

245. To find in lines the ratio of the rectangle of two given lines A and B (fig. 148) to the rectangle of two given lines C and D. Fig. 148.

Solution. Let X be a fourth proportional to the three given lines B, C, D; the ratio of the two lines A and X will be equal to that of the two rectangles A × B, C × D.

For, since B: C:: D: X, C × D=Bx X; therefore

A × B: C × D :: A × B: B × X :: A : X.

246. Corollary. In order to obtain the ratio of the squares described upon two lines A and C, find a third proportional X to the lines A and C, so that we may have the proportion

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247. To find in lines the ratio of the product of three given lines A, B, C (fig. 149), to the product of three given lines P, Q, R. Solution. Find a fourth proportional X to the three given lines P, A, B; and a fourth proportional Y to the three given The two lines X and Y will be to each other as

lines C, Q, R.

the products

AX BX C, P× Q× R.

For, since P:A:: B: X, AX B=PX X; and, by multiplying each of these by C, we shall have

In like manner,

AX BX CC X PX X. since

C:QR: Y, Q × R = C × Y;

and, by multiplying each of these by P, we shall have

PXQXR=P × C× Y;

therefore the product

AXBX C: the product P× QXR::C×P×X:P× C× Y:: X: F.

Fig. 149.

PROBLEM.

248. To make a triangle equivalent to a given polygon.

Solution. Let ABCDE (fig. 146) be the given polygon. Fig. 146. Draw the diagonal CE, which cuts off the triangle CDE; through the point D draw DF parallel to CE to meet AE pro

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 equivalent. 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 triangle 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.

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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 lines A and B.

the

For in the right-angled triangle GEH the hypothenuse GH=A, 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.

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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 FGN; 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, III will be the side of the square sought.

For, on account of the parallels KI, GE,

hence

HI: HK :: HE : HG;

-2

HỈ: HK:: HẺ: HG (V).

HE

But, in the right-angled triangle EHG,

HE: HG :: segment EF: the segment FG (215),

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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 Fig. 129. a polygon similar to a given polygon ABCDE.

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