10 Th. If from half the perimeter of any triangle the base be taken, and also each of the sides; the rectangle of the half and its excess above the base, is to the rectangle of the other two excesses, as the square of the radiu is to the square of the tangent of half the angle contained by the sides. Given the triangle ABC: bisect its angles at A and B, by the straight lines AG, BG; produce AG, AB, AC; bisect the exterior angles at B and C by the lines BK, CK; draw the perpendiculars GD, GE, GF, which are equal radii of the inscribed circle; draw also the perpendiculars KH, KM, KL; which are equal, because of the similar triangles. H K D A B E M G F C Because the triangles AHK, ALK, are equiangular, and have AK common, AH is equal to AL.; for the same reason BD is equal to BF, AD to AE, CE to CF, CM to CL, BH to BM. And, because AD+BF+CE=AE+CF+BD=AE+CE+CM, taking away the parts shown to be equal, the remainders BF=BD= CM, ax. 3. Therefore AH+AL = perimeter, and AH= p. BC therefore equals HD, and AC equals BD; and the excess of p. over BC=AD, over AB=HB, over AC=BD. Hence the hypothesis is AHXAD: BHXBD:: R2: (TABAC)2 Again, because the triangles BDG, KHB are similar, GD:BD:: BH: HK. Therefore GDXHK=BDXBH. But since AH: HK:: AD: DG. Therefore AHXAD: HKXDG::AD×AD:DGXDG (p. 22 of b. 6); or AHXAD : BDXBH :: AD AD: DGXDG. Now, if AD be made radius, DG will be the tangent of half the angle BAC. Wherefore, if from half, &c. Q. E. D. QUADRATURE OF THE CIRCLE. Definition. A convex line is any are of a circle, or any polygonal line, which has no re-entrant angles, or inward inflections; and which a straight line cannot cut in more than two points. Lemma. Any curve, or polygonal line, which passes round a convex line, from end to end, is longer than the convex line. Let AMB be a convex line, enveloped from A to B, by the polygonal line APMQB: the distance from A or B to M, through Por Q, is longer than the part of the convex line from A or B to M; because every point of the convex line, from A to M, or from B to M, would fall within the triangle APM, or BQM, if the points M, P, A-M, Q, B, were joined by straight lines (a). A P C MD E B Q In like manner, it may be proved that the distance from P to Q, through CDE, is longer than the straight line PQ. Therefore, any curve, or plolygonal line, which passes round, &c. Recite (a) p. 21 of b. 1. QE. D. Cor. 1. Hence the perimeter of any polygon inscribed in a circle is less than the circumference of the circle. Cor. 2. And the perimeter of any polygon described about a circle exceeds the circumference of the circle. 1 Th. If from the greater of two unequal magnitudes its half be taken; and from the remainder its half; and so on; there will remain at length, a magnitude less than the less of the two given magnitudes. Let AB and C be two unequal magnitudes; of which AB is the greater: take DE a multiple of C, which is greater than AB; divide DE into parts DF, FG, GE, each equal to C. From AB take its half BH; and from the remainder take its half K HK; until AB be divided into as many parts as DE. Then since DE is greater than AB; and EG is H not greater than the half of DE, but BH is the half of AB, the remainder DG is greater than the remainder AH. Again, since GF is not greater than the half of DG, but HK is the half of AH, the remainder DF is greater than the remainder AK. A B C ID E E 0 Now DF is equal to C, the less of the two given magnitudes. Therefore, AK, the part left of the greater, is less than C. Q. E. D. 2 Th. Equilateral polygons of the same number of sides, inscribed in circles, are similar, and are to one another as the squares of the diameters of the circles. From the centres N, O, draw NB, OH. And because the sides of each polygon are all equal to one another, the arcs which they subtend in each polygon are also equal (p. 28 of b. 3); and the number of them in one circle is equal to the number of them in the other; therefore, whatever part any arc, AB, is of the circumfer ence ABD, the same part is the arc GH, of the circumference GHK But the angle ANB is the same part of four right angles, that the arc AB is of the circumference ABD (p. 33 of b. 6); and the angle GOH is the same part of four right angles, that the arc GH is of the circumference GHK (p. 33 of b. 6): therefore, the angles ANB. GOH, are each of them the same part of four right angles, and are therefore equal to one another. The isosceles triangles, ANB, GOH. are therefore equiangular, and the angle ABN equals the angle GHO. In the same manner, by joining NC, OI, it may be proved, that the angles NBC, OHI, are equal to one another, and to the angle ABN. Therefore, the whole angle ABC is equal to the whole GHI; and the same may be proved of the angles BCD, HIK; and so of the rest. The polygons, ABCDEF and GHIKLM, are therefore equiangular to one another; and since they are equilateral, the sides about the equal angles are proportionals: these two polygons are therefore similar to one another (def. 1. of b. 6.) And because similar polygons are as the squares of their homologous sides, the polygon ABCDEF is to the polygon GHIKLM, as the square of AB to the square of GH: but because the triangles ANB, GOH are equiangular, the squares of AB, GH, are as the squares of AN, GO (p. 4 of b. 6); or as four times the square of AN to four times the square of GO (p. 15 of b. 5); that is, as the square of AD to the square of GK (cor. 2, p. 8, 2). Therefore also, the polygon ABCDEF is to the polygon GHIKLM, as the square of AD to the square of GK: and they have been shown to be similar. Cor. Every equilateral polygon inscribed in a circle is also equiangular: for the isosceles triangles which have their common vertex in the centre, are all equal and similar; therefore, the angles at their bases are all equal, and the angles of the polygons are therefore also equal. 3 P. The side of any equilateral polygon inscribed in a circle being given, to find the side of a polygon of the same number of sides described about the circle. Let ABCDEF be an equilateral polygon inscribed in the circle ABD; it is required to find the side of an equilateral polygon of as many sides described about the circle. Find G the centre of the circle; join GA, GB; bisect the arc AB in H; and through H draw LK touching K the circle, and meeting GA, GB produced in K and L; KL is the side of the polygon required. Produce GF to N, so that GN may be equal to GL; join KN, and from G draw GM at right angles to KN, join D Because the arc AB is bisected in H, the angles AGH, BGH are equal (p. 27 of b. 3); and because KL touches the circle in H, the angles GHL, GHK are right angles (p. 18 of b. 3); therefore, the triangles HGK, HGL, have two angles in the one equal to two angles in the other, and the side GH is common to both; therefore they are equal (p. 26 of b. 1), and GL is equal to GK. Again, in the triangles KGL, KGN, because GN is equal to GL, and GK common, and also the angles KGL, KGN equal; therefore the bases KL and KN are equal (p. 4 of b. 1). But because the triangle KGN is isosceles, the angles GKN and GNK are equal; and the angles GMK, GMN were made right angles; wherefore the triangles GMK, GMN have two angles in the one equal to two in the other, and the side GM is common to both; therefore they are equal (p. 26 of b. 1), and KN is bisected in M. But KN is equal to KL, therefore their halves KM, KH are also equal. Wherefore, in the triangles GKH, GKM, the two sides GK and KH are equal to the two GK and KM, each to each; the angles GKH, GKM are also equal; therefore GM is equal to GH (p. 4 of b. 1). Wherefore the point M is in the circumference of the circle; and because KMG is a right angle, KM touches the circle. Therefore KL is the side of an equilateral polygon described about the circle, of as many sides as the inscribed polygon ABCDEF. Cor. 1. Because of the equal straight lines GL, GK, GN, if a circle be described from the centre G, through the points L, K, N, the polygon will be inscribed in that circle; and be similar to the polygon ABCDEF. Cor. 2. The sides of the inscribed and described polygons, have to each other the ratio of the perpendiculars let fall from G upon AB and LK. Therefore because magnitudes have the ratio of their equimultiples (p. 15 of b. 5), the perimeters which are equimultiples of the sides of the polygons, are to each other as the perpendicular from the centre upon a side of the inscribed polygon is to the radius of the circle. 4 Th. A circle being given, two similar polygons may be found, the one inscribed in the circle, the other described about it, whose difference shall be less than any given space. LF Let ABC be the given circle, and the square D any given space; a polygon may be inscribed in the circle K ABC, and a similar one described H/A about it, whose difference shall be less than the square of D. In the circle ABC, apply the straight line AE, equal to D, and let AB be a quadrant. From AB take its half, from the residue its half, and so on, until the arc AF, be less than the arc AE (p. 1 of this). Find the centre G, draw the diameter AC, as also the straight lines AF, FG: bisecting the arc AF in K, join KG; and draw HL touching the circle in K, and meeting GA, GF produced in Hand L: join CF. Because the isosceles triangles HGL, AGF have the common angle AGF, they are equiangular (p. 6 of b. 6); therefore the angles, GHK, GAF are equal; but the angles GKH, CFA are also equal as right angles; therefore the triangles HGK, ACF, are likewise equiangular (p. 32 of b. 1). And because the arc AF was found by taking from the arc AB its half, and from the residue its half, and so on; AF will be found a certain number of times exactly in AB, and therefore also in the whole circumference ABC. AF is therefore the side of an equilateral polygon inscribed in the circle ABC. Wherefore also, HL is the side of an equilateral polygon of the same number of sides described about the circle ABC (p. 3 of this). Let the polygon described be called M, and the polygon inscribed. be called N; and because they are similar, they are as the squares of the homologous sides HLand AF, (p. 20 of b. 6), that is (because the triangles HLG, AFG, are similar), as the square of HG to the square of AG, or its equal GK. But the triangles HGK, ACF were shown to be similar, and therefore the square of AC is to the square of CF, as the polygon M to the polygon N; and, by conversion, the square of AC is to its excess above the square of CF, viz. the square of AF (p. 47 of b. 1), as the polygon Mis to its excess above the polygon N. But the square of AC, which is about the circle ABC, is greater than the regular octagon which is about the same circle; because the square envelopes the octagon (Lemma); and for the same reason, the polygon of eight sides is greater than one of sixteen, and so on: therefore the square of AC is greater than any polygon described about the circle by the continual bisection of the arc ABC: it is therefore greater than the polygon M. Now, it has been demonstrated, that the square of AC is to the square of AF, as the polygon M to the difference of the polygons; therefore, since the square of AC is greater than M, the square of AF is greater than the difference of the polygons, (p. 14 of b. 5). The difference of the polygons is therefore less than the square of AF; but |