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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, inflect the chords GH and GI equal to the radius AB, and, from the points H and I, with the distance

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BL to that of the decagon inscribed in the given circle. Hence AL may be inflected 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.

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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, 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.

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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 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 L and 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 segment of the distance

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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.

PART II.

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PROP. XIV. PROB.

The side of a regular octagon being given, to mark out the figure.

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, IM equal to BH, and the sides equal to AB; and, on MH and NI,

describe the rhombus

M

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D

F

L

E

A

B

For (by App. II. 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 AB, BK, KO, 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 B as 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.) AG2 +CG2 = 2AB2 + 2BG2; but, (by App. II. Prop. 10.) CG2=

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B

BF2=2AB2; whence AG2=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.

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PROP. XVI. PROB.

Two distances being given, to find a third

portional.

pro

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