then GI, the difference between the radius and the perpendicular DI, is the half of EC, the difference between the radius and the perpendicular from the centre, on the side of that which has half the number of sides. For the As AEC, AIF are equiangular, since the LA is common to both, and the Ls at E and I are ri Ls, . AE:EC =AI:IF; but since AEC is a me , AC 7 AE, :. Al the half of AC (Prop. 46) is the half of AE; :. since the third term is half the first, the fourth is also half the second, that is, IF 7 the half of EC. Again, since DEF is a riL, DF 7 DE, but DC=DG, .. the remainder EC 7 FG; hence, from a quantity FG < EC there has been taken FI 7 the half of EC, the remainder IG < half EC. In the same manner, if AG were joined, and a perpendicular drawn from the centre D upon it, the difference between this perpendicular and the radius would be less than the half of IG, and still more less than the fourth of EC. Cor. By continually doubling the number of sides of the polygons, the difference between the radius and the perpendicular from the centre, on a side of the ultimate polygon, may be made less than any given line. For it has been shown in the proposition that this is equivalent to taking away from EC more than its half, and from the remainder more than its half, and so on continu. ally ; .. (Prop. 73, Cor.) the quantity ultimately left is less than any given quantity. PROPOSITION LXXV.-THEOREM. If there be two similar polygons, the one inscribed in a circle, and the other described about it, the perimeter of the inscribed is to that of the described as the perpendicular from the centre on a side of the inscribed is to the radius of the circle ; and the polygons are to each other as the squares of these lines. Let ABCDEF be a regular polygon inscribed in a circle, and GHIKLM a polygon described about the circle, having its sides parallel to the sides of the inscribed, then (Prop. 20) these polygons will be equiangular, and since their sides are equal, they are similar A also, let ON be drawn from the centre perpendicular to AB, and produced to P, OP will be perpendicular to GH, since AB is || GH. :. (Prop. 72, Cor. 3) the perimeter of the inscribed is to the perimeter of the described as ON is to OP; and (Prop. 72, Cor. 2) (the inscribed polygon): (the described polygon)=ON: OP2. Cor. í. If the number of sides of the polygons be continually doubled, their perimeters will become more nearly equal than by any given difference. For they are to one another as ON: OP, which is ultimately a ratio of equality. (Prop. 74, Cor.) Cor. 2. The polygons themselves, upon the same supposition, will become more nearly equal than by any given difference. Cor. 3. Since the circle is always greater than the inscribed, and less than the described, polygon, each of them will ultimately become more nearly equal to the circle than by any given difference. Cor. 4. The perimeters of each of the polygons will ultimately coincide with, and consequently be equal to, the circumference of the circle. Cor. 5. Circles are to one another as the squares of their radii. For (Prop. 72, Cor. 1) the polygons inscribed in them are to one another as the squares of the radii of the circles in which they are inscribed, and these polygons are ultimate equal to the circles ; .. the circles are one to another as the squares of their radii. B PROPOSITION LXXVI.-PROBLEM. Having given the perpendicular from the centre of a circle upon a side of a figure inscribed in it, to find the perpendicular upon the side of a figure having double the number of sides. Let CD be the given perpendicular, and CF the required one. Join AE; then because AB is a dia. meter, AEB is a semicircle ; :: the LAEB is a r' L, (Prop. 52); but since CF is perpendicular to EB, AE and CF are lli :: the 48 AEB, CFB, are similar; and since AB is double of CB, .. AE is double of CF; and since AD is perpendicular to EG, ADE is a ut L, and :: = the LAEB, and the angle at A is common to each of the As, BAE, G EAD, hence they are equiangular; . AB: AE=AE: AD, (Prop. 61), and AE’=AB AD; but AE was proved to be double of CF, and (Prop. 37, cor. 2), AE?=ACF; substituting this value, we have 4CF-ABAD ; -, CF2 ={AB AD, and consequently CF="JABAD. SCHOLIUM. If a numerical result be required, let EG be the side of an equilateral triangle, and let the radius of the circle be 1, then CD is equal to ]; for if EC were joined, ECB would evidently be an equilateral triangle, and the perpendicular on CB would bisect it," (Prop. 7, cor. 2.) :: AD would be (1+1)= , and AB would be two; putting these values in the expression above, it becomes CF= įv/2(1+1)= (1+1). Generally, if p be the perpendicular from the centre, upon a side of a figure inscribed in a circle, pi the perpendicular upon a side of one having double the number of sides, P, the perpendicular on a side of one having four times the number of sides, and pm the perpendicular on a side of one having 2m number of sides as the first; then PIEN /ltp m-1 P2= 2 Also, (Prop. 68), EB=VAB:BD=1/2(1-p), radius being one ; in the same manner, if s be a side of the original polygon 81, a side of that of double the number of sides, and generally sm a side of one having 2 the number of sides as the first, we will have sm+1=2(14Pm), and if n be the number of sides of the original polygon with which we commence, the whole perimeter of the polygon will be =n2m+1/2(1-Pm). Hence, (Prop. 72, cor. 3), if this operation be repeated often enough, the result will be the circumference of the circle to radius one. 1+p 1+P1, and pm 1+P 2 PROPOSITION LXXVII.- PROBLEM. To divide a given line AB into two parts, such that the greater part, AF, shall be a mean proportional between the whole line, AB, and the remaining part, BF. Let BC be drawn to to AB, from the extremity B, of the straight line AB, and = the half of AB; join AC, and produce it to E, and from the centre C, with the radius A CB, describe the semicircle DBE, and from AB cut off AF=AD; then since AB is perpendicular to CB, it is a tangent to the circle at B; :. (Prop. 67), EA : AB=AB: AD; but since DE and AB are each double of CB, they are equal, and AF=AD; :. (Alg. 107), EA-AB: AB= AB-AD: AD, which, from the above mentioned equalities, gives AF: AB=BF: AF; hence, (Alg. 105), AB: AF =AF: BF. Cor. Since AB=DE, the first proportion gives AE: ED =ED: AD; .. the line AE is also divided in the same manner. SCHOLIUM. A line divided as in the proposition, is said to be divided in extreme and mean ratio, and also to be cut in medial section, or to be divided medially. GEOMETRY OF PLANES. DEFINITIONS.—I. A solid is that which hath length, breadth, and thickness. II. The boundaries of a solid are superficies. III. A straight line is perpendicular, or at right angles to a plane, when it makes right angles with every line which meets it in that plane. IV. A plane is perpendicular to a plane, when the straight lines drawn in one of the planes perpendicular to the common secti of the two planes, are perpendicular to the other plane. V. The inclination of a straight line to a plane is the acute angle contained by that straight line, and another drawn from the point in which the first line meets the plane, to the point in which a perpendicular to the plane drawn from any point in the first line, meets the same plane. VI. The inclination of a plane to a plane is the acute angle contained by two straight lines drawn from any the same point of their common section, at right angles to it, one upon one plane, and the other upon the other plane. VII. Two planes are said to have the same, or like inclinations to one another, which two other planes have, when their angles of inclination are equal to one another. VIII. Parallel planes are such as do not meet one another though produced. IX. A straight line and a plane are parallel, if they do not meet when produced. X. The angle formed by two intersecting planes is called a dihedral angle, and is measured as in Def. 6. XI. Any two angles are said to be of the same affection, when they are either both greater or both not greater than a right angle. The same term is applied to arcs of the same or equal circles, when they are either both greater or both not greater than a quadrant. B D PROPOSITION LXXVIII.-THEOREM. Let the straight lines AB, BC, Let any plane pass through the straight line AB, and let the plane, produced if necessary, be turned about AB, till it pass through the point C. Then, because the points B and C are in the plane, the straight line BC is in the plane, (Def. 7); and because the points C and E are in the plane, the | CE or CD is in the plane, and by hypothesis AB is in the plane; .. the three straight lines AB, BC, CD are all in one plane. Cor. 1. Any two straight lines that cut one another are in one plane. Cor. 2. Only one plane can pass through three points, or through a straight line and a point. Cor. 3. Any three points are in one plane. PROPOSITION LXXIX.-THEOREM. B If two planes cut one another, their common section is a straight line. Let two planes AB, BC cut one another, and let B and D be two points in the line of their common section. From B to D draw the straight line BD, then since B and D are points in the plane AB, the line BD is in that plane, (Def. 7); for the same reason it is in the plane CB; .. being in each of the planes, it is their common section ; hence the common section of the two planes is a straight line. |