the points A and B, with the distance AF, describe arcs intersecting in G: This point is the centre of the circle ADB. For the isosceles triangles BEC, BEF, being evidently equal, the angle FBC is equal to both the angles at the base; but FBC is (I. 32. El.) equal to the interior angles BAF and BFA of the iso qual to GBA; BG is, therefore, parallel to CD, and hence (I. 30. El.) the angle BDC, or BCD, is equal to GBD. The triangles BGA and BGD, having thus the side BA equal to BD, BG common, and equal contained angles GBA and GBD, are (I. 3. El.) equal, and therefore the side GA is equal to GD. The point G being thus equidistant from three points, A, D, and B in the circumference, is hence (III. 8. cor.) the centre of the circle. PROP. X. PROB. To divide the circumference of a given circle successively into four, eight, twelve, and twentyfour equal parts. 1. Insert the radius AB three times from A to D, E, and C; from the extremities of the diameter AC, and with a distance equal to the chord AE, describe arcs intersecting in the point F; and from A, with the distance BF, cut the circumference on opposite sides at G and H : AG, GC, CH, and HA are quadrants. For, as before, AF2=AE2=3AB2; and the triangle ABF being right-angled, 3AB'=AF2=AB +BF2, and therefore BF-AG=2AB; whence (II. 12.) ABG is a right angle, and AG a quadrant. 2. From the point F with the radius AB, cut the circle in I and K, and from A and C in sert the chord AI to L, H and M; For BF, being equivalent to 2AB2, is equivalent to the D H squares of BI and IF, and consequently BIF is a right angle; but the triangle BIF is also isosceles, and therefore the angle IBF at the base is half a right angle; whence the arc IG is an octant. 3. The arc DG, on being repeated, will form twelve equal sections of the circumference. For the arc AD is the sixth or two-twelfth parts of the circumference, and AG is the fourth or three-twelfths; consequently the difference DG is one-twelfth. 4. The arc ID is the twenty-fourth part of the circumference. For the octant AI is equal to three twenty-fourths, and the sextant AD is equal to four twenty-fourths; their difference ID is hence one twenty-fourth part of the circumference. PROP. XI. PROB. To divide the circumference of a given circle successively into five, ten, and twenty equal parts. Mark out the semicircumference ADEC, by the triple insertion of the radius, from A and C, with the double chord AE, describe arcs intersecting in F, from A, with the distance BF, cut the circle in G and K, insert the chords GH and GI equal to the radius AB, and, from the points H and I, with the dis tance BF or AG, describe arcs intersecting in L. It is evident from App. II. 7, that BL is the greater segment of the radius BHA divided by a medial section; wherefore (IV. 23. cor. 2. El.), H D C B XH AL is equal to the side of the inscribed pentagon, and M K BL to that of the decagon inscribed in the given circle. Hence AL may be inserted five times in the circumference, and BL ten times; and consequently the arc MK, or the excess of the fourth above the fifth, is equal to the twentieth part of the whole circumference. Scholium. This proposition, and the preceding, include the happiest application of the circle to the solution of such problems. PROP. XII. PROB. From a given side to trace out a square. Let the points A and B terminate the side of a square, which it is required to trace. From B as a centre describe the semicircle ADEC, from A and C, with the distance AE, H describe arcs intersecting in F, from A, with the distance BF,、 cut the circumference in G, and from A and G, with the radius AB, describe arcs intersecting in H: The points H and G are corners of the required square. For (App. II. 10.) the angle ABG is a right angle, and the distances AB, AH, HG, and GB, are, by construction, all equal. PROP. XIII. PROB. Given the side of a regular pentagon, to find the traces of the figure. From B describe through A the circle ADECF, in which the radius is inflicted four times, from A and C with the double chord AE describe arcs intersecting in G, from E and F, with the distance BG, describe arcs intersecting in H, from A, with the radius AB, describe a portion of a circle, inflect BH thrice from B to L and from 1 A to O, and lastly from L and O, with the radius AB, describe arcs intersecting in P: The points A, L, P, 0, B mark out the poly on. For, from App. II. 7, it is evident that BH is the greater segment of the distance each of them six-fifths of a right angle (IV. 4. cor.), and hence (I. 33. cor.) the points I and O are corners of the pentagon; but P is evidently the vertex of the pentagon since the sides LP and OP are each equal to AB. Scholium. The pentagon might also have been traced, as in Book IV. Prop. 5., by describing arcs from A and B with the distance HC, and again, from their intersection P, and with the radius AB, cutting those arcs in L and O. It is likewise evident, from Book IV. Prop. 8, that the same previous construction would serve for describing a decagon, P being made the centre of a circle in which AB is inserted ten times. |