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 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, 今回 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 :: AB2 : 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 N2=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 XVI. To construct a rectangle equivalent to a given square, and having the sum of its adjacent sides equal to a given line. Let C be the square, and AB equal to the sum of the sides of the required rectangle. Upon AB as a diameter, describe a semicircle; draw the line DE parallel to the diameter, at a distance AD from it, equal to the side of the given square C; 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 rectangle required. A TB 1 For their sum is equal to AB; and their rectangle AF.FB is equivalent to the square of EF, or to the square of AD; hence that rectangle is equivalent to the given square C. D с 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 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. Suppose C equal to the given square, and AB the difference of the sides. 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. F For, first, the difference of these sides is equal to the diameter EF or AB; secondly, the rectangle DE, DF, is equal to AD2 (Prop. XXX.): hence that rectangle is equivalent to the given square C. PROBLEM XVIII. D G To find the common measure, if there is one, between the diagonal and the side of a square. B Let ABCG be any square whatever, and AC its diagonal. We must first 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 result of the first operation is therefore the quotient 1, with the remainder AD, which latter must now be compared with BC, or its equal AB. A F B 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, con E tinue diminishing, and would soon elude our comparisons by their minuteness, this would be but an imperfect mechanical method, from which G 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, A F and obtaining the result, by operating only upon lines which remain always of the same magnitude. The angle ABC being a right angle, AB is a tangent, and AE a secant drawn from the same point; so that AD: AB :: AB AE (Prop. XXX.). 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. B E 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 that 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 (Prop. XI. Cor. 4.), but which acquires a greater degree of clearness by the geometrical investigation. BOOK V. REGULAR POLYGONS, AND THE MEASUREMENT OF THE CIRCLE. Definition. A POLYGON, which is at once equilateral and equiangular, is called a regular polygon. Regular polygons may have any number of sides: the equilateral triangle is one of three sides; the square is one of four. PROPOSITION I. THEOREM. Two regular polygons of the same number of sides are similar figures. Suppose, for example, that ABCDEF, abcdef, are two regular hexagons. The sum of all the angles is the same in both figures,being in each equal to eight right angles (Book I. Prop. XXVI. Cor. 3.). The angle A is the sixth part of that sum; so is the angle a: hence the angles A and a are equal; and for the same reason, the angles B and b, the angles C and c, &c. are equal. B Again, since the polygons are regular, the sides AB, BC, CD, &c. are equal, and likewise the sides ab, bc, cd, &c. (Def.); it is plain that AB ab: BC: bc:: CD: cd, &c.; hence the two figures in question have their angles equal, and their homologous sides proportional; consequently they are similar (Book IV. Def. 1.). E D 0 f K d Cor. 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 (Book IV. Prop. XXVII.). Scholium. The angle of a regular polygon, like the angle of an equiangular polygon, is determined by the number of its sides (Book I. Prop. XXVI.). PROPOSITION II. THEOREM. Any regular polygon may be inscribed in a circle, and circumscribed about one. Let ABCDE &c. be a regular polygon describe a circle through the three points A, B, C, the centre being O, and OP the perpendicular let fall from it, to the middle point of BC: draw AO and H OD. B F If the quadrilateral OPCD be placed upon the quadrilateral OPBA, they will coincide; for the side OP is common; the angle OPC=OPB, each being a right angle; hence the side PC will apply to its equal PB, and the point C will fall on B: besides, from the nature of the polygon, the angle PCD= PBA; hence CD will take the direction BA; and since CD= BA, the point D will fall on A, and the two quadrilaterals will entirely coincide. The distance OD is therefore equal to AO; and consequently the circle which passes through the three points A, B, C, will also pass through the point D. By the same mode of reasoning, it might be shown, that the circle which passes through the three points B, C, D, will also pass through the point E; and so of all the rest: hence the circle which passes through the points A, B, C, passes also through the vertices of all the angles in the polygon, which is therefore inscribed in this circle, Again, in reference to this circle, all the sides AB, BC, CD, &c. are equal chords; they are therefore equally distant from the centre (Book III. Prop. VIII.): hence, if from the point O with the distance OP, a circle be described, it will touch the side BC, and all the other sides of the polygon, each in its middle point, and the circle will be inscribed in the polygon, or the polygon described about the circle. Scholium 1. The point Othe common centre of the inscribed and circumscribed circles, may also be regarded as the centre of the polygon; and upon this principle the angle AOB is called the angle at the centre, being formed by two radii drawn to the extremities of the same side AB. Since all the chords AB, BC, CD, &c. are equal, all the angles at the centre must evidently be equal likewise; and therefore the value of each will be found by dividing four right angles by the number of sides of the polygon. |