PROP. III. THEOR. If the chord of an arc of a circle less than a semicircle be drawn, and from the extremities of the arc tangents be drawn meeting each other, the arc shall be greater than the chord, and less than the sum of the tangents between the point of their meeting and the points of contact. Let ACB (fig. 3) be an arc of a circle, AB its chord, and AD, BD tangents to the circle at A and B, meeting each other in D: the arc ACB is greater than the chord AB, and less than the sum of the tangents AD, DB. Bisect the arc ACB in C, draw the chords AC, CB, and at C draw the tangent ECF, meeting AD, BD in E and F; bisect the arcs AC, CB in G, H, draw the chords AG, GC, CH, HB, and at G and H, draw the tangents IGK and LHM; and so on, bisecting the arcs, and drawing their chords and tangents. Then if the are ACB be not greater than the chord AB, it must either be equal to AB or less than it; that is, AB will either be equal to the arc ACB or greater than it; and in either case the sum of AC and CB, being greater than AB, will be greater than the arc ACB. Again, AG, GC being greater than AC, and CH, HB greater than CB, the sum of the chords AG, GC, CH, HG will exceed the arc ACB more than the sum of the chords AC, CB exceeds that arc. In like manner, the sum of the chords AN, NG, GO, OC, CP, PH, HQ, QB will exceed the arc ACB, more than the sum of the chords AG, GC, CH, HG exceeds that arc; and so on: that is, the more nearly the chords of the arcs into which the arc ACB is divided, approach to coincidence with that are, the more will their sum differ from it, which is manifestly absurd. Therefore the arc ACB is neither equal to the chord AB nor less than it: consequently the arc ACB is greater than the chord AB. If the arc ACB be not less than the sum of the tangents AD, DB, it must be either equal to that sum or greater than it; and in either case, EF being less than ED, DF, the arc ACB will be greater than the sum of AE, EF, FB. Again, IK being less than IE, EK, and LM less than LF, FM, the arc ACB will exceed the sum of the tangents AI, IK, KL, LM, MB more than it exceeds the sum of the tangents AE, EF, FB. In like manner, the arc ACB will exceed the sum of the tangents AR, RS, ST, TU, UV, VW, WX, XY, YB more than it exceeds the sum of the tangents AI, IK, KL, LM, MB; and so on: that is, the more nearly the tangents of the arcs into which the arc ACB is divided approach to coincidence with that arc, the more will their sum differ from it, which again is manifestly absurd. Therefore the arc ACB is neither equal to the sum of AD, DB nor greater than it consequently the arc ACB is less than the sum of the tangents AD, DB. Wherefore, if the chord of any arc of a circle, &c.: which was to be demonstrated. Cor. Hence the circumference of a circle is greater than the perimeter of any polygon inscribed in the circle, and less than the perimeter of any polygon described about the circle. PROP. IV. THEOR. The perimeters of similar polygons inscribed in circles are to one another as the diameters of the circles; and their areas are to one another as the squares of the diameters. Let ABCDEF, GHIKLM (fig. 4) be two similar polygons inscribed in the circles ABCDEF, GHIKLM, of which the diameters are AD, GK: the perimeter of the polygon ABCDEF is to the perimeter of the polygon GHIKLM as the diameter AD to the diameter GK. Join BD, AC, HK, GI. Then because the polygon ABCDEF is similar to the polygon GHIKLM, the angle ABC is equal to the angle GHI, and AB is to BC as GH to HI (VI. Def. 1): therefore the two triangles ABC, GHI, having one angle in the one equal to one angle in the other, and the sides about the equal angles proportionals, are equiangular (VI. 6); and therefore the angle BCA is equal to the angle HIG: but BCA is equal to BDA, because they are in the same segment (III. 21); and, for the same reason, the angle HIG is equal to the angle HKG. Therefore the angle BDA is equal to HKG: and the right angle ABD (III. 31) is equal to the right angle GHK; wherefore the remaining angles in the triangles ABD, GHK are equal, and the triangles are equiangular to one another; therefore AB is to GH as AD to GK (VI. 4). And because AB is to GH as BC to HI, as CD to IK, as DE to KL, as EF to LM, and as FA to MG, therefore as one antecedent is to its consequent, so are all the antecedents taken together to all the consequents taken together (V. 12), that is, AB is to GH as the perimeter of the polygon ABCDEF to the perimeter of the polygon GHIKLM: but as AB is to GH so is AD to GK; consequently (V. 11) the perimeter of the polygon ABCDEF is to the perimeter of the polygon GHIKLM as the diameter AD to the diameter GK. Also the area of the polygon ABCDEF is to the area of the polygon GHIKLM as the square of AD is to the square of GK. Since AB is to GH as AD to GK, the duplicate ratio of AB to GH is the same as the duplicate ratio of AD to GK (V. Def. 10); but the ratio of the square of AD to the square of GK is the duplicate ratio of that which AD has to GK (VI. 20); and the ratio of the polygon ABCDEF to the polygon GHIKLM is the duplicate ratio of that which AB has to GH (VI. 20); therefore the polygon ABCDEF is to the polygon GHIKLM as the square of AD to the square of GK (V. 11). Wherefore the perimeters of similar polygons, &c.: which was to be proved. Cor. Since equilateral polygons inscribed in a circle are equiangular (IV. 11, 15, 16), equilateral polygons, of the same number of sides, inscribed in circles, are similar; and, therefore, their perimeters are to each other as the diameters of the circles in which they are inscribed; and their areas are to each other as the square of the diameters. PROP. V. THEOR. If, in a circle, an equilateral polygon be inscribed, and an equilateral polygon of the same number of sides be described about the circle, the perimeter of the inscribed polygon will be to the perimeter of the circumscribed, as the perpendicular from the centre on a side of the inscribed polygon is to the radius of the circle. Let ABCDEFGH (fig. 5) be an equilateral polygon inscribed in the circle ACEG, of which the centre is R, and radius is RA: if through the points A, B, C, D, E, F, G, H straight lines IK, KL, LM, MN, NO, OP, PQ, QI be drawn, IKLMNOPQ will be an equilateral polygon of the same number of sides described about the circle (IV. 12, 15. 16. Cor.): the perimeter of the inscribed polygon ABCDEFGH will be to the perimeter of the circumscribed polygon IKLMNOPQ as the perpendicular from R upon AB is to the radius RA. Join RB, RK, RL. Because KA is equal to KB (IV. 12), RA equal to RB and RK common to the two triangles KRA, KRB, the angle KRA is equal to the angle KRB; and the angle BRK is half the angle BRA. In like manner, the angle BRL is half the angle BRC and since the angle ARB is equal to BRC, the angle BRK is equal to BRL. Since the two triangles TRA, TRB have the two sides AR, RT equal to the two BR, RT, each to each, and the angle ART equal to the angle BRT, the angle ATR is equal to the angle BTR; and RT is at right angles to AB. And because the triangles BRK, BRL have the two angles BRK, KBR equal to the two BRL, LBR, each to each, and BR common, RK is equal to RL. In like manner, it may be shown that RM, RN, RO, &c. are all equal; and therefore a circle described with the radius RK will circumscribe the polygon IKLMNOPQ. And since the equilateral polygons ABCDEFGH, IKLMNOPQ are inscribed in circles, their perimeters are to each other as the diameters of the circles in which they are described (Prop. 4), and consequently (V. 15) as the radii of those circles; therefore the perimeter of the polygon ABCDEFGH is to the perimeter of the polygon IKLMNOPQ as AR to RK, that is as RT to RA, the triangles KRA and ART being similar. Wherefore, if in a circle, &c. which was to be proved. : Cor. 1. The area of the inscribed polygon ABCD &c. is equal to the rectangle contained by half its perimeter and the perpendicular RT drawn from the centre R upon one of its sides. For the area of the triangle ABR is equal to the rectangle contained by BT and RT, that is by the half of AB and RT; and the same is true of all the other equal triangles which have their vertices in R, and which together make up the polygon ABCD &c.; therefore the whole polygon is equal to the rectangle contained by half the perimeter and the perpendicular RT. In the same manner it appears that the area of the circumscribing polygon IKLMN &c. is equal to the rectangle contained by half its perimeter and the radius RC. Cor. 2. Since the equilateral polygons ABCD &c., IKLM &c., are similar polygons inscribed in circles, their areas are to each other as the squares of the diameters of their circumscribing circles (Prop. 4), and therefore as the squares of their radii; consequently the area of ABCD &c. is to the area of IKLM &c. as the square of RA to the square of RK, or as the square of RT to the square ᎡᎪ. PROP. VI. THEOR. of In a given circle an equilateral polygon may be inscribed, and a similar polygon described about the circle, such that the perimeters of the polygons shall differ from one another by less than any given line, however small. Let ABC (fig. 6) be a given circle, and DE a given line; an equilateral polygon may be inscribed in the circle ABC, and a similar polygon described about it, such that the difference of their perimeters shall be less than DE. From DE cut off DF equal to the eighth part of DE (VI. 10), and in the diameter AC take CG equal to DF; through G draw IGH at right angles to AC; and from A apply the straight line AK equal to IH (IV. 1). Bisect the circumference ABC in B (III. 30), so that AB is a fourth part of the whole circumference of the circle; and from the circumference AB take away its half, and from the remainder its half, and so on, until the remaining circumference AL is found less than the circumference AK (Prop. 1). Bisect the circumference AL in M, and take the circumference AN equal to the circumference AM: join AL, and at A draw PQ touching_the circle; find the centre O, and join OL, ON and OM cutting AL in R; and produce OM, ON to meet the tangent at A, in P and Q. And because the arc AL was found by continued bisection, of the semi-circumference ABC, its half AB, and the half of that half, and so on, AL will be contained a certain number of times exactly in the whole circumference: the straight line AL is therefore the side of an equilateral polygon inscribed in the circle ABC; and consequently (Prop. 5) PQ is the side of an equilateral polygon of the same number of sides described about ABC: the difference of the perimeters of the equilateral circumscribed and inscribed polygons of which PQ and AL are the sides, is less than the given line DE. Let the perimeter of the equilateral circumscribed polygon of which PQ is the side be designated by C, and that of the similar inscribed polygon by I, then C is to I, as OM to OR (Prop. 5); and therefore by conversion (V. E), C is to its excess above I, as OM to MR, and therefore as eight times OM to eight times MR. And because the perimeter of a square described about the circle ABC is equal to four times the diameter AC, that is to eight times the radius OM, and the perimeter of the circumscribed equilateral polygon of eight sides is less than the perimeter of the circumscribed square (Prop. 2. Cor. 2); and again, the perimeter of the equilateral circumscribed polygon of sixteen sides is less than that of the polygon of eight sides; and so on; the perimeter of the equilateral polygon of which PQ is the side, that is C, is less than eight times OM: consequently (V. 16, A) the excess of C above I is less than eight times MR. And because AL is less than AK or HI, OR is greater than OG, and therefore MR is less than GC, that is than DF; and consequently eight times MR is less than eight times DF, that is than DE. And since it has been shown that the excess of Ć above I is less than eight times MR, with much greater reason is the excess of C above I less than DE. Wherefore in a given circle, &c.: which was to be proved. Cor. 1. Because the circumference of the circle is less than the perimeter of the circumscribing polygon, and greater than that of the inscribed polygon (Prop. 3. Cor.), it differs from either of these perimeters by less than they differ from one another, and therefore the difference between each of them and the circumference of the circle is less than DE. Consequently, however small a straight line may be, a polygon may be inscribed in the circle, and another described about it, the perimeter of each of which shall differ from the circumference of the circle by less than the given line. Cor. 2. The straight line S, which is greater than the perimeter of any polygon that can be inscribed in the circle ABC, and less than the perimeter of any polygon that can be described about the circle, is equal to the circumference of the circle. For if not, let them be unequal; and first let S exceed the circumference, by the line Z. Then, because the perimeters of the circumscribing polygons are all greater than S, by hypothesis, and because S is greater than the circumference by Z, no polygon can be described about the circle but that its perimeter will exceed the circumference by a |