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. radius AB, describe arcs intersecting in H: The points H and Gare 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. PROВ. 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 inflected 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 Land from A to O, and lastly from L and O, with the radius AB, describe arcs intersecting in P: The points A, L, P, O, B mark out the polygon. For, from App. II. 7, it is evident that BH is the greater each of them six-fifths of a right angle (IV. 4. cor.), and hence (I. 33. cor.) the points L 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 inflected ten times. PROP. XIV. PROB. The side of a regular octagon being given, to mark out the figure. 1 Let the side of an octagon terminate in the points A and B; to find the remaining corners of the figure. From the centres A and B, with the radius AB, describe the two semicircles AEFC and, BEGD; with the double chord AF, and from A; C and B, D describe arcs intersecting in H, I; from these points, with the radius AB, cut the semicircles in K, L: on HI describe the square HMNI, by making the diagonals HN, Prop. 10.) BH, AI are both of them perpendicular to BA, and BKH, ALI are right angled isosceles triangles; HI is therefore parallel to BA, and HMNI, consisting of triangles equal to BKH, is a square; whence all the sides АВ, ВК, КО, OM, MN, NP, PL, and LA of the octagon are equal: But they likewise contain equal angles; for ABK, composed of ABH and HBK, is equal to three half right angles, and BKO, by reason of the parallels BH and KO, being the supplement of HBK, is also equal to three half right angles. In the same manner, the other angles of the figure may be proved to be equal. PROP. XV. PROB. On a given diagonal to describe a square. Let the points A and B be the opposite corners of a square which it is required to trace. From Bas a centre describe the semicircle ADEC, from A and C with the double chord AE describe arcs intersecting in F, from C with the distance BF describe an arc and cut this from A with the radius AD in G, and lastly from B and A with the distance BG describe arcs intersecting in H and I: ABHI is the required square. For, in the triangle AGC, the straight line GB bisects the base, and consequently (II. 22.) AG +CG2 = 2AB2 + 2BG2 ; but, (by App. II. Prop. 10.) CG2 = D HG F E A B C BF2=2AB2; whence AG=AB2=2BG2, and (II. 11.) AHB is a right angle; and the sides AH, HB, BI, and IA being all equal, the figure is therefore a square. PROP. XVI. PROB. Two distances being given, to find a third proportional. 1 Let it be required to find a third proportional to the dis them double the angle GFH or IFH at the circumference (III. 17. El.); whence the triangles GEH and IGH must also have the angles at the base equal, and are consequently similar: Wherefore (VI. 12. El.) EG : GH :: GH: HI. If the first term AB be less than half the second term CD, this construction, without some help, would evidently not succeed. But AB may be previously doubled, or assumed 4, 8, or 16 times greater, so that the circle FGH shall always cut FHI; and in that case, HI, being likewise doubled, or taken 4, 8, or 16 times greater, will give the true result. PROP. XVII. PROВ. To find a fourth proportional to three given distances. |