therefore each of the straight lines AB, BC, CD, DE, EA touches the circle: PROPOSITION XIV. PROBLEM. To describe a circle about a given equilateral and equiangular pentagon. Let ABCDE be the given equilateral and equiangular pentagon. It is required to describe a circle about ABCDĒ. B С Bisect the angles BCD, CDE by the straight lines CF, FD, (1. 9.) and from the point F, in which they meet, draw the straight lines FB, FA, FE, to the points B, A, E. It may be demonstrated, in the same manner as in the preceding proposition, that the angles CBA, BAE, AED are bisected by the straight lines FB, FĂ, FE. and CDF the half of CDE; in like manner it may be demonstrated, one another; and the circle described from the centre F, at the distance of one of them, will pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Q.E.F. PROPOSITION XV. PROBLEM. To inscribe an equilateral and equiangular heragon in a given circle. Let ABCDEF be the given circle. It is required to inscribe an equilateral and equiangular hexagon in it. A B В G E H Find the centre G of the circle ABCDEF, and draw the diameter AGD; . (111. 1.) and from D, as a centre, at the distance DG, describe the circle EGCH, and join AB, BC, CD, DE, EF, FA: GE is equal to GD: DE is equal to DG: and the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG, are equal to one another: (1. 5. Cor.) but the three angles of a triangle are equal to two right angles ; (1. 32.) therefore the angle EGD is the third part of two right angles : in the same manner it may be demonstrated, that the angle DGC is also the third part of two right angles : and because the straight line GC makes with EB the adjacent angles EGC, CGB equal to two right angles; (1. 13.) the remaining angle CGB is the third part of two right angles: therefore the angles EGD, DGC, CGB are equal to one another: and to these are equal the vertical opposite angles BGA, AGF, FGE: (1.15.) therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE, are equal to one another: but equal angles stand upon equal circumferences; (111. 26.) therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one another: and equal circumferences are subtended by equal straight lines: (111. 29.) therefore the six straight lines are equal to one another, It is also equiangular: EDCBA: and the angle AFE upon EDCBA; in the same manner it may be demonstrated therefore the hexagon is equiangular; and it is equilateral, as was shewn; and it is inscribed in the given circle ABCDÉF. Q. E. F. CoR.-From this it is manifest, that the side of the hexagon is equal to the straight line from the centre, that is, to the semi-diameter of the circle. And if through the points A, B, C, D, E, F there be drawn straight lines touching the circle, an equilateral and equiangular 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 equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon. PROPOSITION XVI. PROBLEM. To inscribe an equilateral and equiangular quindecagon in a given circle. Let ABCD be the given circle. It is required to inscribe an equilateral and equiangular quindecagon in the circle ABCD. B E Let AC be the side of an equilateral triangle inscribed in the circle,(iv. 2.) and AB the side of an equilateral and equiangular pentagon inscribed in the same : (iv. 11.) therefore, of such equal parts as the whole circumference ABCDF contains fifteen, the circumference ABC, being the third part of the whole, contains five; and the circumference AB, which is the fifth part of the whole, contains three; bisect BC in E; (111. 30.) circumference ABCD: therefore if the straight lines BE, EC be drawn, and straight lines equal to them be placed round in the whole circle, (iv. 1.) an equilateral and equiangular quindecagon will be inscribed in it. Q. E. F. 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, an equilateral and equiangular quindecagon will be described about it: and likewise, as in the pentagon, a circle may be inscribed in a given equilateral and equiangular quindecagon, and circumscribed about it. NOTES TO BOOK IV. The fourth book of the Elements contains some particular cases of four general problems on the inscription, and the circumscription of triangles and regular figures in and about circles. Euclid has not given any instances of the inscription or circumscription of rectilinear figures in and about other rectilinear figures. Any rectilinear figure, of five sides and angles, is called a pentagon ; of seven sides and angles,' a heptagon ; of eight sides and angles, an octagon; of nine sides and angles, a nonagon; of ten sides and angles, a decagon ; of twelve sides and angles, a duodecagon ; of fifteen sides and angles, a quindecagon, &c. These figures are included under the general name of polygons; and are call equilateral, when their sides are equal; and equiangular, when their angles are equal ; also when both their sides and angles are equal, they are called regular polygons. Prop. III. An objection has been raised to the construction of this problem. "It is said that in this and other instances of a similar kind, the lines which touch the circle at A, B, and C, should be proved to meet one another. This may be done by joining AB, and then since the angles KAM, KBM are equal to two right angles (111. 18.), therefore the angles BAM, ABM are less than two right angles, and consequently (ax. 12.), AM and BM must meet one another, when produced far enough. Similarly, it may be shewn that AL and CL, as also CN and BN meet one another. Prop. v. The corollary to this position appears to have been already demona strated in Prop. 31, Book III. Prop. VI, VII. It is obvious that the square described about a circle is equal to double the square inscribed in the same circle. Also that the circumscribed square is equal to the square of the diameter, or four times the square of the radius of the circle. Prop. VII. It is manifest that a square is the only right-angled parallelogram which can be circumscribed about a circle, but that both a rectangle and a square may be inscribed in a circle. Prop. x. By means of this proposition, a right angle may be divided into five equal parts. Pro The arc subtending a side of the quindecagon, may be found by placing in the circle from the same point, two lines respectively equal to the sides of the regular hexagon and pentagon. The centres of the inscribed and circumscribed circles of any regular polygon are coincident. Besides the circumscription and inscription of triangles and regular polygons about and in circles, some very important problems are solved in the constructions respecting the division of the circumferences of circles into equal parts. By inscribing an equilateral triangle, a square, a pentagon, a hexagon, &c. in a circle, the circumference is divided into three, four, five, six, &c. equal parts. In Prop. XXVI, Book 111, it has been shewn that equal angles at the centres of equal circles, and therefore at the centre of the same circle, subtend equal arcs ; by bisecting the angles at the centre, the arcs which are subtended by them are also bisected, and hence, a sixth, eighth, tenth, twelfth, &c. part of the circumference of a circle may be found. XVI. By the aid of the first corollary to Prop. 32, Book 1, may be found the magnitude of an interior angle of any regular polygon whatever. Let o denote the magnitude of one of the interior angles of a regular polygon of n sides, then no is the sum of all the interior angles. But all the interior angles of any rectilinear figure together with four right angles, are equal to twice as many right angles as the figure has sides, that is, if we agree to assume a to designate two right angles, .. no + 2+ = nt, (n − 2) TT, n the magnitude of an interior angle of a regular polygon of n sides. By taking n= 3, 4, 5, &c. may be found the magnitude, in terms of two right angles, of an interior angle of any regular polygon whatever. Pythagoras was the first, as Proclus informs us in his commentary, who discovered that a multiple of the angles of three regular figures only, namely, the trigon, the square, and the hexagon, can fill up space round a point in a plane. It has been shewn that the interior angle of any regular polygon of n sides in terms of two right angles, is expressed by the equation 2 0 n Let 03 denote the magnitude of the interior angle of a regular figure of 3 sides. 3 2 Then 02 = one third of two right angles, 3 .. 303 = TT, and 603 = 27, that is, six angles each equal to the interior angle of an equilateral triangle are equal to four right angles, and therefore six equilateral triangles may be placed so as completely to fill up the space round the point at which they meet in a plane. In a similar way, it may be shewn that four squares and three hexagons may be placed so as completely to fill up the space round a point. Also it will appear from the results deduced, that no other regular figures besides these three, can be made to fill up the space round a point: for any multiple of the interior angles of any other regular polygon, will be found to be in excess above, or in defect from four right angles. The equilateral triangle or trigon, the square or tetragon, the pentagon, and the hexagon, were the only regular polygons known to the Greeks, capable of being inscribed in circles, besides those which may be derived from them. M. Gauss in his Disquisitiones Arithmeticæ, has extended the number by shewing that in general, a regular polygon of 2" + 1 sides is capable of being inscribed in a circle by means of straight lines and circles, in those cases in which 2" + 1 is a prime number. |