As the radicals may be taken either positive or negative, their various combinations, rightly disposed, will produce eight distinct results. Let denote the circumference; then 2. Pythagoras was the first who remarked the simple property, that only three regular figures, the square, the equilateral triangle, and the hexagon,-can be constituted about a point. Here the mystic philosopher might again admire the union of the monad with the triad.-It may not be superfluous perhaps to observe, that on this property is founded the adaptation of patchwork, and the construction of tessellated pavement. 3. Several interesting propositions may be annexed to this Book. PROP. I. THEOR. The square of the side of a regular octagon inscribed in a circle, is equivalent to the rectangle contained by the radius and the difference between the diameter and the side of the inscribed square. Let ABCD be square inscribed in a circle, and AEBFCGDH an octagon, which is formed evidently by the bisection of the quadrants AB, BC, CD, and DA: The square of AE is equivalent to the rectangle under AO and the difference between AB and AC. A E B C For draw the diameter EG. It is manifest, that the triangles AIO and BIO are right-angled and isosceles; and because AO is equal to EO, and AI perpendicular to it, the square of AE (II. 23. cor. El.) is equivalent to twice the rectangle under EO and EI, or the rectangle under AO and twice EI. But EI is the difference of EO and IO, and twice EI is, therefore, (equal to the difference of twice EO or AC and twice IO or AB. Whence the square of AE, the side of the octagon, is equivalent to the rectangle under the radius and the difference of the diameter and AB the side of the inscribed square. PROP. II. THEOR. D In and about a given circle, to inscribe and circumscribe an equilateral triangle. Let AEB be a circle, in which it is required to nscribe an isosceles triangle. Draw the diameter AB, describe (I. 1.) the equilateral triangle ADB, join CD meeting the circumferénce in E, draw (1. 23.) EF, EG parallel to AD, BD, and join FG: The triangle EFG is equilateral. For the triangles ADC, BDC having the two sides DA, AC equal to DB, BC, and the third side DC common to both, are (I. 2.) equal, and the angle DCA is equal to DCB; whence the arc AE is (III. 12.) equal to BE. And the triangle ADB (I. 10. cor.) being likewise equiangular, the angle DBA is equal to DAB, and the arc AEM equal to BEL, and the re maining arc ME equal to LE. But EF and EG being parallel to LA and MB, the arcs AF and BG are equal to LE and ME, and to each other; hence FG is parallel to AB, and the inscribed triangle FEG is (I. 29.) equiangular, and consequently equilateral. Again, let it be required to describe an equilateral triangle about the circle AEB. H F D K M The same construction remaining; at the points F, E, and G, apply the tangents HI, HK, and KI, to form the circumscribing triangle IHK: This triangle is equilateral. For because IH is a tangent and FG is inflected from the point of contact, the angle IFG is equal to the angle FEG in the alternate segment (III. 21.), and therefore IH is parallel to EG (I. 22. cor.). In like manner it is proved, that HK, KI are parallel to GF, FE, and consequently (I. 29.) the angles of the triangle IHK are equal to those of FEG, and therefore equal to each other. Cor. Hence the circumscribing equilateral triangle contains four times that which is inscribed; for the figures EFIG, EHFG, and EFGK are evidently equal rhombuses, and contain equilateral triangles which are all equal. Hence also the side of the circumscribing, is double of that of the inscribed, equilateral triangle. PROP. III. THEOR. To inscribe and circumscribe a circle in and about a given regular pentagon. Let ABCDE be a regular pentagon, in which it is required to inscribe a circle. Draw AO and EO to bisect the angles at A and E, let fall the perpendicular OF, and from O as a centre, with the distance OF, describe a circle FGHIK: This circle will touch the pentagon internally. QGB and the common side OG, are (I. 20.) equal. Again, the triangles BOG and BOH have now the angles OBG and OGB equal to OBH and OHB, with the side BO common to both, and are therefore equal. In like manner, all the triangles about the centre O are proved to be equal; conse quently the perpendiculars OF, OG, OH, OI, and OK are equal, and the circle touches the pentagon in the points F, G, H, I, and K. Next, let it be required to describe a circle about the pentagon. From the same centre O, with the distance OA, describe a circle: It will pass through the points B, C, D, E; for the triangles about O being all equal, the straight lines OA, OB, OC, OD, and OE must be likewise equal. PROP. IV: THEOR. In and about a regular hexagon to inscribe and circumscribe a circle. Let ABCDEF be a regular hexagon, in which it is required to inscribe a circle. Draw AO and FO, bisecting the angles BAF and AFE (1.5.); and from the point of intersection O, with its distance from the side AF, describe a circle: This circle will touch the hexagon internally. For let fall perpendiculars from O upon the sides of the figure. It may be demonstrated, as in the last proposition, that the triangles AOB, BOC, COD, DOE, and EOF are all equal to AOF; and, in like manner, it will appear that the intermediate bisected triangles are equal. Hence the perpendiculars OG, OH, B H M E OI, OK, OL, and OM, are all equal, and a circle must touch these at the points, G, H, I, K, L, and M. Again, let it be required to describe a circle about the hexagon. From the same point O, as a centre, with the distance OA, describe a circle, which must pass through the points B, C, D, E, and F; for the straight lines OA, OB, OC, OD, OE, and OF were proved to be equal. Cor. Hence, in any regular polygon, the centre of the inscribing and circumscribing circle is the same, and may be determined in general, by drawing lines to bisect the adjacent angles of the figure. BOOK V. DEFINITIONS. 1. The words λoyos in Greek and ratio in Latin, signifying reason or manner of thought, indicate vaguely a philosophical conception. The compound term vaλoya comes nearer to |