COROLLARY. Hence it is evident that every equiangular figure inscribed in a circle is equilateral, and that every equilateral figure is equiangular. SCHOLIUM. In the above proposition the pentagon is inscribed in the circle by the aid of an isosceles triangle, the angles at whose base are each double its vertical angle; and in like manner any other equilateral figure of any number of sides may be inscribed in a circle, by the aid of an isosceles triangle, in which each of the angles at its base is to its vertical angle as half the number of its sides minus half, is to unity; thus a square may be inscribed by the aid of an isosceles triangle having the ratio between each of the angles at its base and its vertical angle as ( − 1 =) 1 : 1; a pentagon, as (31⁄2 — § =) 2; 1; a hexagon, as =) 2:1; and so on. PROPOSITION XII. PROBLEM. To circumscribe an equilateral and equiangular pentagon about a given circle (ABCDE). SCLUTION. Inscribe within the given circle the equilateral and equiangular pentagon ABCDE (a), then through the angular points of the same, A, B, C, &c., draw tangents KF, FG, GH, &c., to the given circle (b), and they G will form an equilateral and equiangular pentagon, FGHIK, circumscribing the given circle (c). B F A K E I II PROBLEM. To inscribe a circle in a given equilateral and equiangular pentagon (ABCDE). SOLUTION. Bisect any two adjacent angles A and B by the straight lines AF and BF (a), and from their point of intersection F draw FG perpendicular to AB (b); from the B center F with the radius FG describe a circle, and it will be inscribed in the given pentagon. Draw FC, FD, FE, and from Flet fall the perpendiculars FH, FK, FL, and FM (b). DEMONSTRATION. In the triangles ABF and AEF the sides AB and AE are equal (c), AF common to both, and FAB and FAE are equal (d), therefore the angles ABF and AEF are equal (e); but the angles ABC and AED are also equal (c), therefore, since H G A M E D K (a) I. 9. L B H M L E ABF is half of ABC (d), AEF is half of AED; and in the same manner it may be shown that the other angles BCD and CDE are bisected by the lines FC and FD. Therefore in the triangles FBH and FBG the angles FBH and FBG are equal, the angles BHF and BGF are right angles (d), and the side FB common, therefore the sides FH and FG are equal (ƒ); and in the same manner it may be shown that all the perpendiculars FH, FK, FL, &c., are equal, therefore the circle described from F as a center with the radius FG will pass through the points H, K, L, and M, and the sides of the given pentagon are tangents to it because the angles at those points are right angles (g). K D (d) Solution. SCHOLIUM. This problem is only a particular case of the more general proposition given at p. 6. PROPOSITION XIV. PROBLEM.-To circumscribe a circle about a given equila teral and equiangular pentagon (ABCDE). SOLUTION. Bisect the angles A and E by the straight lines AF and EF (a); from the point of intersection F as a center with the radius AF, describe a circle ABCDE which shall circumscribe the given pentagon. Draw the straight lines BF, CF, and DF. B (a) I. 9. E DEMONSTRATION. It may be shown in the same manner as in the preceding proposition that the angles of the pentagon are bisected by the straight lines drawn from F. Therefore in the triangle AFE the angle EAF is equal to AEF, and therefore the side AF is equal to FE (6); and in the same manner it may be shown that all the lines AF, BF, CF, DF, and EF are equal, and therefore the circle described from F as a center with the radius AF will pass through the points B, C, D, E, and circumscribe the pentagon. PROPOSITION XV. PROBLEM. To inscribe an equilateral and equiangular hexagon in a given circle F (ABCDEF). SOLUTION. Find the center G of the given circle (a), and through it draw the diameter AD. From D as a center with the radius DG describe a circle GEHC, join EG and GC, and produce them to the points B and F. Join AF, FE, ED, DC, CB, BA, with straight lines, and they will form an equilateral and equiangular hexagon. E D H I. Def. 15. (a) III. 1. I. 32 B, cor. 5. (d) I. 13. e) I. 15. (ƒ) III. 26. III. 29. IV. 11. cor. B DEMONSTRATION. The straight lines GD and DC, being radii of the same circle, are equal (6), and for the same reason DG and GC are equal, therefore the triangle DGC is equilateral, and the angle CGD is the third part of two right angles (c); and in the same manner it may be shown that the angle EGD 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 (d), the remaining angle CGB is the third part of two right angles, and the three angles EGD, DGC, and CGB are equal to one another; and to these the vertical opposite angles BGA, AGF, and FGE are also equal (e); therefore the six angles at the center G are equal, and the arcs on which they stand are equal (f), and also the lines subtending those arcs are equal (g), and therefore the hexagon ABCDEF is equilateral, and also, since it is inscribed in a circle, equiangular (h). COROLLARY. It is evident that the side of the hexagon is equal to the radius of the circumscribing circle. PROPOSITION XVI. PROBLEM. To inscribe an equilateral and equiangular quindecagon in a given circle (ABCD). SOLUTION. Let AC be the side of an equilateral triangle inscribed in the circle (a), and AB the side of an equilateral pentagon inscribed in the same (b); bisect the arc BC in E (c), join BE and EC, and in the given circle place chords equal to BE, and they will form an equilateral and equiangular quindecagon inscribed in it. B E IV. 2. (b) IV. 11. III. 30. DEMONSTRATION. For if the whole circumference of the given circle be divided into fifteen equal parts, the arc AC, because it is the third part of the whole circumference, contains five of these parts; in like manner the arc AB contains three of them, therefore the arc BC contains two, and therefore the arc BE is the fifteenth part of the whole circumference, and BE is the side of the required equilateral and equiangular quindecagon. SCHOLIUM. The only regular polygons which the Greek Geometers could incribe geometrically in the circle were the trigon, or equilateral triangle, the tetragon, or square, the pentagon, the hexagon, and any others, such as the quindecagon, derived from them. M. Gause, however, in his Disquisitiones Arithmetica, has shown that a regular polygon of 2+1 sides is always capable of being inscribed geometrically in a circle, when 2" + 1 is a prime number. THE ELEMENTS OF EUCLID. BOOK V. DEFINITIONS. 1. A LESS magnitude is said to be a part of a greater magnitude, when the less measures the greater, that is, when the less is contained a certain number of times exactly in the greater. SCHOLIUM. In ordinary use the word "part" means "any portion whatever," but its geometrical sense in the above definition, and wherever subsequently employed, is that of an aliquot part or submultiple. It has already been explained in the scholium to the first proposition of the second book, that one magnitude is said to measure another when it is exactly contained in it any number of times without any remainder. The lesser magnitude is then said to be a part or submultiple of the greater, while the greater is said to be a multiple of the less. In the four preceding books magnitudes have been compared simply as to their equality or inequality, but in the latter case no attempt has been made to determine how great or how small that inequality might be. The object, however, of the fifth book is to compare unequal magnitudes, and to determine with greater exactness their relative value. Now there are two ways in which two unequal magnitudes or quantities might be compared, namely,-1o, by subtracting the lesser from the greater, and so ascertaining how much one exceeded the other; thus if one line were represented by 50 and the other by 40, their difference thus estimated would be 10; this method, however, would fail to convey any idea of their relative values ;— 2o, by ascertaining how often the greater contained the less, or, in other words, what multiple the greater was of the less; this latter method is the one employed by Euclid in the fifth book, and by it we are enabled to ascertain their relative value. 2. A greater magnitude is said to be a multiple of a less, when the greater is measured by the less, that is, when the greater contains the less a certain number of times exactly. 66 SCHOLIUM. It is necessary to observe the distinction between the expressions measures" and "is contained in;" for example, 3 measures 15, being contained in it exactly 5 times without any remainder, but 3 does not measure 13, although it is contained in it 4 times, because there is a remainder of 1 over. It has already been explained, in the scholium to |