For the triangles CDE, CFE have the base CE common; they have also the same altitude, since their vertices D, F, are situated in a line DF parallel to the base; these triangles are therefore equivalent. Add to each of them the figure ABCE, and there will result the polygon ABCDE equivalent to the polygon ABCF.

The angle B may in like manner be cut off, by substituting for the triangle ABC the equivalent triangle AGC, and thus the pentagon ABDE will be changed into an equivalent triangle GCF.

The same process may be applied to every other figure; for, by successively diminishing the number of its sides, one being retrenched at each step of the process, the equivalent triangle will at last be found.

Scholium. We have already seen that every triangle may be changed into an equivalent square; and thus a square may always be found equivalent to a given rectilineal figure, which operation 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.

PROBLEM XI.

To find the side of a square which shall be equivalent to the sum or the difference of two given squares.

Let A and B be the sides of the given squares.

First. If it is required to find a square equivalent to the sum of these squares, draw the two indefinite lines ED, EF at right angles to each

F

G

BH

E

other; take ED=A, and EC=B; join DG: this will be the side of the square required.

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

Secondly. If it is required to find a square equal to the difference of the given squares, form in the same manner the right angle FEH; take GE equal to the shorter of the sides A and B; from the point G as a centre, 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 the triangle GEH is right-angled, the hypotenuse GH-A, and the side GE=B; hence the square constructed upon EH, &c.

Scholium. A square may thus be found equal to the sum of any number of squares; for the construction which reduces two of them to one, will reduce three of them to two, and these two to one, and so of others. It would be the same, if any of the squares were to be subtracted from the sum of the others.

PROBLEM XII.

To construct a square which shall be to a given square ABCD as the line M is to the line N.

Upon the indefinite D line EG, take EF=M, and FG=N; upon EG as a diameter describe a semicircle, and

at the point F erect the perpendicular FH.

C M

N

F

From the point H, draw the chords HG, HE, which produce indefinitely upon 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 required.

:

For, by reason of the parallels KI, GE, we have HI: HK:: HE HĞ; hence HI: HK2:: HE2: HG2: but in the rightangled triangle EHG (23. III.) the square of HE is to the square of HG as the segment EF is to the segment FG, or as M is to N; hence HI2: HK2 :: M: N. But HK-AB; therefore the square described upon HI is to the square described upon AB as M is to N.

PROBLEM XIII.

Upon the side FG, homologous to AB, to describe a polygon similar to the given polygon ABCDE.

In the given polygon,

draw the diagonals AC,

AD; at the point F

make the angle GFH=

B

BAC, and at the point A

G the angle FGH=
ABC; the lines FG,

GH will cut each other

K

H

in H, and FGH will be a triangle similar to ABC. In the same manner 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, as required.

F

For, these two polygons are composed of the same number of triangles, which are similar and similarly situated (26. III.)

PROBLEM XIV.

Two similar figures being given, to construct a figure which shall be similar to one of them, and equal to their sum or their difference.

Let A and B be two homologous sides of the given figures. Find a square equal to the sum or to the difference of the squares described upon A and B; let X be the side of that square; then will X in the figure required, be the side which is homologous to the sides A and B in the given figures. The figure itself may then be constructed on X, by the last problem. For, the similar figures are as the squares of their homologous sides; now the square of the side X is equal to the sum, or to 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 to the difference, of the similar figures described upon the sides A and B.

PROBLEM XV.

To construct a figure similar to a given one, and bearing to it the given ratio of M to N.

Let A be a side of the given figure, X the homologous side of the figure required. The square of X must be to the square of A as M is to N; hence X will be found by Problem 12.; and knowing X, the rest will be accomplished by Problem 13.

PROBLEM XVI.

To construct a figure similar to the figure P and equivalent to the figure Q.

Find M the side of a square equivalent to the figure P, and N the side of a square equivalent to the figure Q. 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 also be equivalent to the figure Q.

For, calling Y the figure described upon the side X, we have P: Y:: AB: X2; but by construction, AB: X:: M: N, or AB2: X2: M2: N2; hence P: Y :: M2: N2. But by construction also, M2 P and N-Q; therefore P: Y:: P: Q; consequently Y=Q; hence the figure Y is similar to the figure P, and equivalent to the figure Q.

PROBLEM XVII.

To construct a rectangle equivalent to a given square C, and having its adjacent sides together equal to a given line AB.

D

Upon AB as a diameter, describe a semicircle; draw the line DE parallel to the diameter, at a distance AD equal to the side of the given square C; from the A

C

FB

point E, where the parallel cuts the circumference, draw EF perpendicular to the diameter; AF and FB will be the sides of the rectangle required.

For, their sum is equal to AB; and their rectangle AF.FB is equal to the square of EF, or to the square of AD; hence that rectangle is equivalent to the given square C.

Scholium. To render the problem possible, the distance AD must not exceed the radius; that is, the side of the square C must not exceed the half of the line AB.

PROBLEM XVIII.

To construct a rectangle that shall be equivalent to a given square C, and the difference of whose adjacent sides shall be equal to a given line AB.

Upon the given line AB as a diameter, describe a semicircle; at 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: then will DE and DF be the adjacent sides of the rectangle required.

For, first, the difference of their sides is equal to the diameter EF or AB; secondly, the rectangle DE, DF is equal to AD2 (30. III.); hence that rectangle is equivalent to the given square C.

D

PROBLEM XIX.

To find the common measure, if there is one, between the diago

nal and the side of a square.

Let ABCG be any square whatever, and AC its diagonal.

We must first (Prob. 17. II) apply CB upon CA, as often as it may be contained there. For this purpose, let the semicircle DBE be described, from the centre C, with the radius CB. It is evident that CB is contained once in AC, with the remainder AD; the re- A F sult of the first operation is therefore

D

B

the quotient 1 with the remainder AD, which latter must now be compared with BC, or its equal AB.

We might here take AF-AD, and actually apply it upon AB; we should find it to be contained twice with a remainder: but as that remainder, and those which succeed it, continue diminishing, and would soon elude our comparisons by their minuteness, this would be but an imperfect mechanical method, from which no conclusion could be obtained to determine whether the lines AC, CB have or have not a common measure. There is a very simple way, however, of avoiding these decreasing lines, and obtaining the result, by operating only upon lines which remain always of the same magnitude:

The angle ABC being right, AB is a tangent, and AE a secant drawn from the same point; so that, (30. III.) AD: AB:: AB: AE. Hence in the second operation, when AD is compared with AB, the ratio of AB to AE may be taken instead of that of AD to AB; now AB, or its equal CD, is contained twice in AE, with the remainder AD; the result of the second operation is therefore the quotient 2 with the remainder AD, which must be compared with AB.

Thus the third operation again consists in comparing AD with AB, and may be reduced in the same manner to the comparison of AB or its equal CD with AE; from which there will again be obtained 2 for the quotient, and AD for the remainder.

Hence, it is evident, the process will never terminate; and therefore there is no common measure between the diagonal and the side of a square: a truth which was already known by arithmetic (since these two lines are to each other:: 2 : 1), But which acquires a greater degree of clearness by the geometrical investigation.

Scholium. The impossibility of finding numbers to express