the rectangle AB.BC may be equal to the square of AC; describe (I. 22.) the triangle ABD, having AD equal to AB, and BD to AC: ABD is a triangle, such as is required; that is, each of the angles ABD, ADB is double of A. B Join CD, and (IV. 5.) about the triangle ADC describe the circle ACD. Then (I. 5.) the angles ABD, ADB are equal. But (const.) the rectangle AB.BC is equal to the square of AC, or (const.) to the square of BD: therefore since the rectangle under AB, the whole line which cuts the circle, and BC, the part of it without the circle, is equal to the square of BD, which meets the circle, BD (III. 37.) touches the circle in D; and therefore (III. 32.) the angle BDC contained by BD which touches the circle, and CD which cuts the circle, is equal to A, the angle in the alternate segment. To each of these equals add the angle ADC; then (I. ax. 2.) the whole angle ADB, or its equal B, is equal to the two angles A and ADC: but the exterior angle BCD is also equal (I. 32.) to the two angles A and ADC: therefore (I. ax. 1.) the angles B and BCD are equal, and therefore (I. 6.) the sides BD, CD are equal. Hence also the sides AC, DC are equal, each being equal to BD; and therefore (I. 5.) the angles A and ADC are equal. But B has been proved to be equal to A and ADC; and therefore since these are equal to one another, B and its equal ADB are each double of A; wherefore a triangle ABD has been described, which has each of the angles at its base BD double of the third angle A: which was to be done. Cor. Since (I. 32.) the three angles of a triangle are together equal to two right angles, it is plain, that the angle A is one-fifth, and each of the angles at the base BD, two-fifths of two right angles. Hence also the angle A is one-tenth of four right angles. PROP. XI. PROB. To inscribe a regular pentagon, that is, an equilateral and equiangular pentagon, in a given circle. Let ABC be a given circle: it is required to inscribe in it a regular pentagon. H Draw any radius DC, and (II. 11.) divide it in E, so that the rectangle DC.CE may be equal to the square of DE: inscribe (IV. 1.) the chords CF, CG each equal to DE, and join FG FG is a side of the required pentagon. For (IV. 10. cor.) each of the angles FDC, GDC is one-tenth of four right angles; and therefore FDG is twotenths, or one-fifth of four right angles, that is, of all the angles (I. 13. cor. 2.) made by any E number of lines meeting in D. Make therefore (I. 23.) the angles GDH, HDA, ADB, each equal to FDG, and BDF will also be equal to FDG. Join GH, HA, AB, BF. Now (III. 26.) equal angles at the centre of a circle stand on equal arcs, and (III. 29.) equal arcs have equal chords; therefore the chords FG, GH, HA, AB, BF are all equal, and the pentagon ABFGH is equilateral. It is also equiangular: for each of its angles stands on an arc which is treble of the arc FG; and (III. 27.) the angles which stand on equal arcs are equal to one another, whether they be at the centre or circumference. In the given circle, therefore, a regular pentagon has been described; which was to be done.* Cor. 1. It is evident from this demonstration, that any equilateral polygon described in a circle is also equiangular. Cor. 2. It is evident that FC or CG is a side of a regular decagon inscribed in the circle; and thus we have the means of describing that figure in a given circle. PROP. XII. PROB. To describe a regular pentagon about a given circle. Let ABCDE be the given circle; it is required to describe an equilateral and equiangular pentagon about the circle ABCDE. Let the angular points of a pentagon, inscribed in the circle, by the last proposition, be A, B, C, D, E; and through these points draw (III. 17.) the straight lines FG, GH, HK, KI, IF, touching the circle: the figure FGHKI is (IV. B.) a regular pentagon described about the circle which was to be done. : PROP. XIII. PROB. B F E H K To inscribe a circle in a given regular pentagon. D Let ABCDE be the given regular pentagon; it is required to inscribe a circle in it. * In the construction given by Euclid, a triangle is described as in the last proposition; then a triangle equiangular to it is described (IV. 2.) in the circle: the angular points of this triangle are three of the angular points of the pentagon, and the remaining points are found by bisecting the angles at the base of the inscribed triangle. The construction here given is considerably easier in practice. It may be remarked, that after the chord FG is drawn, the construction of the pentagon is completed simply by inscribing the chords, GH, HA, &c., each equal to FG. H B G M E F L Find (III. 1. schol.) F the centre of the circle whose circumference would pass through the three points A, B, C; this (IV. A. cor.) is the centre of the circle inscribed in the pentagon. Draw FG perpendicular to AB, and from F as centre, with FG as radius, describe the circle GHKLM: this (by the same corollary) is the circle described in the given pentagon: which was to be done. PROP. XIV. PROB. H C K D To describe a circle about a given regular pentagon. Let ABCDE be the given regular pentagon; it is required to describe a circle about it. Find the centre F, as in the preceding proposition; and from the centre F, at the distance FA, describe the circle ABCDE: this (IV. A. cor.) is a circle described about the given pentagon: which was to be done. B E • F PROP. XV. PROB. To inscribe a regular hexagon in a given circle. Let ABCDEF be the given circle; it is required to inscribe a regular hexagon in it. B E Find G the centre of the given circle, and draw any radius AG: inscribe (IV. 1.) the chord AB equal to AG, and join GB. Then, (I. 32. and I. 5. cor.) since the triangle AGB is equilateral, the angle AGB is a third of two right angles, or a sixth of four. But (I. 13. cor. 2.) the angles that can be made about the point G, are equal to four right angles: therefore, if the angles BGC, CGD, DGE, EGF be each made equal to AGB, AGF will also be equal to AGB. The angles at G, therefore, are all equal; and (III. 26.) equal angles at the centre stand on equal arcs, and (III. 29.) equal arcs have equal chords; therefore the figure ABCDEF is equilateral. It is also inscribed in the circle; and therefore (IV. 11. cor. 1.) it is equiangular: wherefore in the given circle a regular hexagon has been described: which was to be done. Cor. From this it is manifest, that each side of the hexagon is equal to the radius of the circle. And if, through the points A, B, C, D, E, F, there be drawn straight lines touching the circle, a regular hexagon will be described about it, which may be demonstrated from what has been said of the pentagon: and likewise a circle may be inscribed in a given regular hexagon, and circumscribed about it by a method similar to that used for the pentagon. PROP. XVI. PROB. To inscribe a regular quindecagon in a given circle. Let ABCD be the given circle; it is required to inscribe in it an equilateral and equiangular quindecagon. E B Let AC be a side of an equilateral triangle inscribed (IV. 2.) in the circle, and AB a side of a regular pentagon inscribed (IV. 11.) in the same; therefore, of such equal parts as the whole circumference ABCDF contains fifteen, the arc ABC, being the third part of the whole, contains five; and the arc AB, which is the fifth part of the whole, contains three; therefore BC, their difference, contains two of the same parts. Bisect therefore (III. 30.) BC in E; then BE, EC are, each of them, the fifteenth part of the whole circumference ABCD: and therefore, if the straight lines BE, EC be drawn, and chords equal to them be placed around in the whole circle a regular quindecagon will be inscribed in it: which was to be done. D And, in the same manner as was done in the pentagon, if, through the points of division made by inscribing the quindecagon, straight lines be drawn touching the circle, a regular quindecagon will be deser bed about it. And likewise, as in the pentagon, a circle may be inscribed in a given regular quindecagon, and circumscribed about it. Schol. Any regular polygon being inscribed in a circle, if the arcs cut off by its sides be bisected, and the points of bisection be joined with the nearest angular points, a polygon of double the number of sides is obtained: and, by repetitions of the process, as many regular polygons as we please, may be inscribed, each having twice as many sides as the one immediately preceding it. * The polygons discussed in this book, and those which may be derived from them by the process pointed out in the preceding scholium, are the only ones which geometers were able, till lately, to describe by elementary geometry, that is, by means of the straight line and circle. In 1801, however, M. Gauss of Gottingen, in a work entitled Disquisitiones Arithmetica, showed that in a circle, by elementary geometry, every regular polygon may be inscribed, the number of whose sides is a power of 2 increased by unity, and is a prime number, that is, a number which is not produced by the multiplication of any two whole numbers; such as 17, which is the fourth power of 2 increased by one. Such also are polygons of 257, and 65537 sides. The investigation of the methods of describing these, is too complex and difficult to be given here. BOOK V. DEFINITIONS. 1. A LESS number or magnitude is said to measure a greater, or to be a measure, a part, or a submultiple of the greater, when the less is contained a certain number of times, exactly, in the greater and 2. The greater is said to be a multiple of the less.* 3. Magnitudes which can be compared in respect of quantity, that is, which are either equal to one another, or unequal, are said to be of the same kind.† 4. If there be two magnitudes of the same kind, the relation which one of them bears to the other in respect of quantity, is called its ratio to the other. The first term or magnitude is called the antecedent of the ratio, and the second the consequent.‡ * Thus, a line of 15 inches is a multiple of a line of 5 inches, containing ít exactly three times; and a line of 5 inches is a measure, a part, or a submultiple of a line of 15 inches. On the contrary a line of 14 inches is not a multiple of a line of 5 inches, nor a line of 5 inches a submultiple of one of 14 inches. It may be remarked, that the word part in this restricted use of it, is improperly applied, and should be laid aside, 5 being, in the true sense of the word, a part of 14 as well as of 15. To obviate this, the expression aliquot part is sometimes employed. The term submultiple, however, seems to be preferable to any other, from its relation to the corresponding term multiple. A number or magnitude which is a multiple of two or more others, is said to be a common multiple of them. Thus, 12 is a common multiple of 2, 3, 4, and 6. In like manner, a number or magnitude which is a measure of two or more others, is said to be a common measure of them. Thus, 4 is a common measure of 16, 20, and 36. Numbers or magnitudes which have a common measure, are said to be commensurable to one another; while those which have no common measure, are incommensurable. †Thus, lines, whether straight or curved, are magnitudes of the same kind, or are homogeneous, since they may be equal or unequal. In like manner, surfaces, solids, and angles form three other classes of homogeneous magnitudes. On the contrary, lines and surfaces, lines and angles, surfaces and solids, &c., are heterogeneous. Thus, it is obviously improper to say, that the side and the area of a square are equal to one another, or are unequal; and the same is the case with regard to the circumference of a circle and its area. So likewise we cannot say that the area and one of the angles of a triangle are equal to each other, or are unequal; and they are therefore heterogeneous. Thus, a line of 10 inches has a certain relation in respect of magnitude to a line of 5 inches, being double of it. This relation is called the ratio of the first to the second; and the line of 10 inches is called the antecedent of the ratio, and the other its consequent. |