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 AΑ and B; let X be the side of that square; then will X be that side in the figure required, which is homologous to the A B X 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 w 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 Р 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, P: Y: AB: x2; but by construction, :: AB : X :: M: N, or, AB2 X2 M2 : N2; hence, : P: Y :: M : N But, by construction also, therefore, MP, and N3~Q• P: Y: P Q; :: 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 hav ing 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. then draw the line DE parallel to the diameter AB; from the point E, where the parallel cuts the circumference, draw EF perpendicular to the diameter; AF and FB will be the sides of the required rectangle. For, their sum is equal to AB; and their rectangle AFXFB is equivalent to the square of EF, or to the square of AD; hence, this rectangle is equivalent to the given square C. Scholium. The problem is impossible, if the distance AD exceeds the radius; that is, the side of the square C must not exceed half the line AB. PROBLEM XVII. To construct a rectangle that shall be equivalent to a given square, and the difference of whose adjacent sides shall be equal to a given line. Let C denote the given square, and AB the difference of the sides of the rectangle. Upon the given line AB, as a diameter, describe a circumference. At the extremity of the diameter, draw the tangent AD, and make it equal to the side of the square C; through the point D and the centre O draw the secant DOF, intersecting the circumference in E and F; then will DE ånd DF be the adjacent sides of the required rectangle. D E B For, the difference of these lines is equal to the diame ter EF or AB; and the rectangle DE, DF is equivalent t AD2 (P. 30); hence, the rectangle DFX DE, is equivalent to the given square C PROBLEM XVIII. To find the commm measure, between the side and diagonal of a square. Let ABCG be any square, and AC its diagonal. the centre with the radius CB, G A F B E This remainder must now be compared with BC, or its equal AB. Since the angle ABC is a right angle, AB is a tangent, and since AE is a secant drawn from the same point, we have († 30), AD : AB :: AB : AE. tience, in the second operation, where AD is compared with AB, the equal ratio of AB to AE may be taken instead: but AB, or its equal CD, is contained twice in AE, with the remainder AD; the result of the second operation is therefore a quotient 2 with the remainder AD, and this must be again compared with AB. Thus, the third operation consists in comparing again 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 a quotient 2, and the remainder AD. Hence, it is evident that the process will never termi nate, and consequently that no remainder is contained in its divisor an exact number of times; therefore, there is no common measure between the side and the diagonal of a square. This property has already been shown, since (P. 11, c. 5), AB : AC:: 1 : √2, but it acquires a greater degree of clearness by the geometrical investigation. BOOK V. REGULAR POLYGONS-MEASUREMENT OF THE CIRCLE DEFINITION. A REGULAR POLYGON is one which is both equilateral and equiangular. A regular polygon may have any number of sides. The equilateral triangle is one of three sides; the square, is one of four. ·PROPOSITION I. THEOREM. Regular polygons of the same number of sides are similar figures. Let ABCDEF, abcedf, be two such polygons. Then, either angle, as A, of the polygon ABCDEF, is equal to twice as many right angles less four, as the fig ure has sides, divided by F E A D B the number of sides; and the same is true of either angle of the other polygon (B. I., P. 26, c. 4); hence (A. 1), the angles of the polygons are equal. Again, since the polygons are regular, the sides AB, BC, CD, &c., are equal, and so likewise the sides ab, bc, cd (D.), &c.; hence AB ab: BC: bc :: CD: cd, &c.; therefore, the two polygons have their angles equal, and their sides taken in the same order proportional; conse quently, they are similar (B. IV., D. 1). Cor. 1. The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their surfaces are to each other as the squares of those sides (B. IV., p. 27). |