PROBLEM IX.

To find a triangle that shall be equivalent to a given polygon.

B

C

D

Let AEDCB be the given polygon. First. Draw the diagonal CE cutting off the triangle CDE; through the point D, draw DF parallel to CE, and meeting AE prolonged; draw CF: the polygon AEDCB is equivalent to the polygon AFCB, which has one side less than the given polygon.

Ꮐ A E F

For the triangles CDE, CFE, have the base CE common, they have also equal altitudes, since their vertices D and F, are situated in a line DF parallel to the base: these triangles are therefore equivalent (P. 2, C.) Add to each of them the figure AECB, and there will result the polygon AEDCB, equivalent to the polygon AFCB.

The angle B may in like manner be cut off, by substituting for the triangle ABC, the equivalent triangle AGC, and thus the pentagon AEDCB 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 (PROB. 6); 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 X.

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 other; take ED= A, and EG=B; and draw DG:

F

G

B

H E

D

this will be the required side of the square.

For the triangle DEG being right-angled, the square described upon the hypothenuse DG, is equivalent to the sum of the squares upon ED and EG (P. 11).

Secondly. If it is required to find a square equivalent to the difference of the given squares, form, as before, 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 equivalent to the difference of the squares described upon the lines A and B.

For, the triangle GEH is right-angled, the hypothenuse GH=A, and the side GE=B; hence, the square described upon EH, is equivalent to the difference of the squares A and B (P. 11, c. 1).

Scholium. A square may thus be found, equivalent to the sum of any number of squares; for a construction similar to that 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.

9

PROBLEM XI.

To find a square which shall be to a given square as a given line to a given line.

Let AC be the given square, and M and N the given

pendicular FH. 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 required square.

For, by reason of the parallels KI, GE, we have

hence,

HI : HK :: HE: HG;

2

:

HI ΗΚ

-2

:: HE2 :

HG2:

but in the right-angled triangle GHE, the square of HE is to the square of HG as the segment EF is to the seg ment FG (P. 11, c. 3), or as M is to N; hence,

But HK=AB; therefore, the square 'described upon HI is to the square described upon AB as M is to N.

PROBLEM XII.

Upon a given line, to describe a polygon similar to a given polygon.

Let FG be the given line, and AEDCB the given polygon.

GH will intersect each other in H, and the triangle FGH will be similar to ABC (P. 18). In the same manner upon FH, homologous to AC, describe the triangle FIH similar to ADC; and upon FI, homologous to AD, describe the triangle FIK similar to ADE. The polygon FGHIK will be similar to ABCDE, as required.

For, these two polygons are composed of the same number of similar triangles, similarly placed (p. 26,,s.)

PROBLEM XIII.

Two similar figures being given, to describe a similar figure which shall be equivalent to their sum or difference.

Let A and B be homologous sides of the given figures. Find a square equivalent

to the sum or difference of the squares described upon A and B; let X be the side of

sides A and B in the given figures. Let the figure itself, then, be constructed on the side X, as in the last problem. This figure will be equivalent to the sum or difference of the figures described on A and B (p. 27, c.)

PROBLEM XIV.

To describe a figure similar to a given figure, 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 required figure.

Find the value of X, such, that its square shall be to the square of A, as M to N (PROB. 11). Then upon X describe a figure similar to the given figure (PROB.12): this will be the figure required.

A

X

PROBLEM XV.

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 fig ure P, and N the side of a square equivalent to the figure Q (PROB. 9, s.) Let X be a fourth proportional to the

P

Q

A

B

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,

consequently, YQ; hence, the figure Y is similar to the figure P, and equivalent to the figure Q.

PROBLEM XVI.

To construct a rectangle equivalent to a given square, and having the sum of its adjacent sides equal to a given line.

Let be the square, and the line AB equal to the sum of the sides of the required rectangle.