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COR. If two of the nearest fides of an equilateral and equian- Book IV gular figure be bifected, and from the point where these lines cut each other, there be lines drawn to all the angles of the figure, thefe lines will bifect all the angles of the figure.

PRO P. XIV. PRO B.

To defcribe a circle about a given equilateral and equiangular

pentagon.

It is required to defcribe a circle about the equilateral and equiangular pentagon ABCDE.

Bifect the right lines AB, BC, in the points H and G; a 1o. I. from which draw the right lines HF, GF, interfecting each o

ther in the point F, and at right angles, to AB, BC; from the b 11. 1. point F draw the right lines BF, FĂ, FE, FD, FC, they will bifect the angles at A, B, C, D, E. Then, because AB is e- c cor. I 3. equal to AE, and AF common, and the angle BAF equal to EAF, the base BF will be equal to EFd. For the fame reason, d 4. 1. EF is equal to FC; therefore, if, with the center F, and distance B, E, or C, a circle is defcribed, it will pass through the points A, B, C, D, E, of the equilateral and equiangular pentagon : Which was required.

PROP. XV. PRO B.

To infcribe an equilateral and equiangular hexagon in a given

circle.

It is required to infcribe the equilateral and equiangular hexagon in the given circle ABCDEF.

Draw AD a diameter to the circle ABCDEF, whofe center is G; with the point D as a center, and distance DG, describe a circle EGCH; join EG, GC; which produce to the points B, F; join AB, BC, CD, DE, EF, FA; then ABCDEF is an equilateral and equiangular hexagon.

For, fince G is the center of the circle ABCDEF, GC is e

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qual to GD; and fince D is the center of the circle CGEH, a def 15.1. GD is equal to DC; therefore CGD is an equilateral triangle b;b 1. 1. but it is likewife equiangular. For the fame reafon, GDE is c cor. 5. x. an equilateral and equiangular triangle, and equal to the

triangle

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d 15. I.

C 13. I.

Book IV. triangle CGD; and, because BG, GA, are equal to DG, GE, and the angle BGA to DGE, the bafe AB is equal to DE. For the fame reason, AF is equal to CD; but CD is equal to DE; therefore AB is equal to AF. Again, becaufe CG falls upon BE, the angles CGB, CGE, are equal to two right angles; but CGD, DGE, are each one third of two right angles f; f cor. 32. 1. therefore CGB is likewife one third of two right angles; therefore BG, GC, are equal to CG, GF, and the angle BGC equal to the angle CGD; therefore the base BC is equal to CD; but likewife BG, GC, are equal to FG, GE, and the angle BGC to FGE 4; therefore the bafe BC is equal to FE; but BC is proved equal to CD, and CD to DE; therefore the fix fides BC, CD, DE, EF, FA, AB, are equal; therefore the figure is equilateral; it is likewise equiangular; for the two angles GDC, GDE, are equal to the two angles GCD, GCB; for each is one third of two right angles f; therefore the whole angle BCD, is equal to the whole CDE. For the fame reason, all the other angles are equal to one another; therefore the figure is likewife equiangular. Wherefore, &c.

COR. Hence the fide of a hexagon is equal to the femi-diameter of the circle. And, if through the points A, B, C, D, E, F, tangents to the circle, be drawn, an equilateral and equiangular hexagon will be defcribed about the circle, as may be proved in the fame manner as the pentagon: And fo likewise a circle may be infcribed and described about a given hexagon.

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PRO P. XVI. PRO B.

To
'O infcribe an equilateral and equiangular quindecagon in a
given circle.

It is required to infcribe an equilateral and equiangular quindecagon in the given circle ABCD.

Let AC be the fide of an equilateral and equiangular triangle infcribed in the circle, and AB the fide of an equilateral and equiangular pentagon, drawn from the point A; then, if the circle is divided into fifteen parts, the fide of the triangle AG will fubtend five of them, and the fide of the pentagon AB will fubtend three; therefore BC will be two of faid parts; therefore bifect BC in E; BE or CE will be one fifteenth part of the cir

cumference; if BC, CE, &c. be joined, the equilateral and Book IV. equiangular quindecagon will be infcribed: Which was requi

red.

COR. If, from what has been faid of the pentagon, right lines be drawn through the divifions of the circle, tangents to the fame, an equilateral and equiangular quindecagon will be defcribed about the circle; or a circle may be infcribed or de fcribed about the quindecagon.

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Book V.

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A

DEFINITION S.

1.

PART is a magnitude of a magnitude; the lefs of the greater, when the lefs measures the greater.

II.

A multiple is a magnitude of a magnitude; the greater of the lefs, when the lefs measures the greater.

III.

Ratio is a certain mutual habitude of magnitudes of the fame kind, according to quantity.

IV.

Magnitudes have proportion to each other; which, being multiplied, can exceed one another.

V.

Magnitudes have the fame ratio to each other, viz. the first to the fecond, and third to the fourth, when there are taken any equimultiples of the firft and third, and likewife any equimultiples of the fecond and fourth; if the multiple of the firft be equal to the multiple of the second, then the multiple of the third will be equal to the multiple of the fourth; if greater, greater; and, if lefs, lefs.

VI.

Magnitudes which have the fame proportion called Proportionals.

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