L, L to F, &c.; and therefore the figure FGHKL is equiangular. It is also equilateral; for its sides are the doubles of the equal lines FB, GC, &c. Therefore, if any equilateral, &c. PROP. VI. PROB.-To inscribe a square in a given circle. Let ABCD be the given circle; it is required to inscribe a square in it. E Draw two diameters AC, BD at right angles to one another; and join AB, BC, CD, DA; ABCD is the square required. Because (I. def. 30) BE is equal to ED, and EA is common, and at right angles to BD; BA is equal (I. 4) to AD. Now, the angles CBA, BAD being angles in semicircles, are (III. 31) right angles; and therefore (I. 28, part 2) AD B is parallel to BC. For the same reason AB is parallel to DC; and therefore (I. def. 24) ABCD is a parallelogram. But it has been shown that BAD is a right angle, and that the sides BA, AD containing it are equal: ABCD is therefore (I. def. 27) a square, and (IV. def. 2) it is inscribed in the circle ABCD: which was to be done. C PROP. VII. PROB.-To describe a square about a given circle. Let ABCD be a given circle; it is required to describe a square about it. Draw two diameters, AC, BD, at right angles to one another, and (III. 17) through the points A, B, C, D draw FG, GH, HK, KF, touching the circle: FGHK is the square required. G A F B D E H C K For, by the preceding proposition, A, B, C, D are the angular points of the inscribed square; and therefore (IV. B) the figure FGHK formed by the tangents drawn through these points, is a square described about the circle: which was to be done. PROP. VIII. PROB. -To incribe a circle in a given square. Let ABCD be the given square; it is required to inscribe a circle in it. Find (IV. A, cor.) the centre E of the inscribed circle, and draw (I. 12) EF perpendicular to AD: from E as centre, with EF as radius, describe the circle FGHK: it is (IV. A, cor.) the circle required.* A F D B H C K * In practice, the construction is effected most easily, both in this proposition and the following, by drawing the diagonals, as their point of intersection is evidently the centre. PROP. IX. PHOB.-To describe a circle about a given square. Let ABCD be the given square; it is required to describe a circle about it. Find, as in the preceding proposition, the centre E; and from that centre, at the distance EA, describe the circle ABCD: it is (IV. A, cor.) the circle required. PROP. X. PROB.-To describe an isosceles triangle, having each of the angles at the base double of the third angle. Take any straight line AB, and divide it (II. 11) in C, so that 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. D B Join CD, and (IV. 5) about the triangle ADC describe the circle ACD. Then (1. 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 it, 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 given circle. Let ABG be a given circle; it is required to inscribe in it a regular pentagon. 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 two tenths, or one fifth of E H four right angles, that is, of all the angles (I. 13, cor. 2) made by any 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 the 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, also, 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 a 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 FGHKÍ is (IV. B) a regular pentagon described about the circle: which was to be done. B A F E D H K * 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. PROP. XIII. PROB. To inscribe a circle in a given regular pentagon. Let ABCDE be a given regular pentagon; it is required to inscribe a circle in it. 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 Fas 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. B H C K M E F L D PROP. XIV. PROB.-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 that centre, 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 a given circle; it is required to inscribe a regular hexagon in it. Find G the centre of a 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 B 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 D E 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. Schol. If, through the points A, B, C, E, F, there be drawn straight lines touching the circle, a regular hexagon will (IV. B) be described about it, and likewise a circle may be inscribed in a ་ given regular hexagon, and also one 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 a given circle; it is required to inscribe in it an equilateral and equiangular quindecagon. 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 E 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 each equal to one of them be placed around in the whole circle, a regular quindecagon will be inscribed in it: which was to be done. In the same manner, also as was done in case of 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 described about it: and likewise, as in the pentagon, a circle may be inscribed in a given regular quindecagon, and also one cir cumscribed 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 means of describing these, is too complex and difficult to be given here. |