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Biseet AF and BG, in H and I, and draw HC and IC perpendicular to AF and BG, intersecting in C; which, is the Center of a circumscribing Circle, as before.

It is evident, that the external Angle DAF (equal EBG) is equal to the Angle ACB at the Center.

For, ACB, BCG, and ACF, are congruous Isosceles Triangles; whose Angles at the Base, AB and BG, are 70 Degrees each ; two, of which, form an internal Angle of the Nonagon; as ABC + CBG= ABG; and, the external Angle, EBG, is the Compliment of two Right Angles; consequently, it is equal to the remaining Angle ACB, of the Triangle ACB, or BCG.

Hence it is also evident; that, if any Right Line, as DE, cuts a Circle, and if from either Point, A or B, of the part AB, within the Circle, another Chord, AF or BG, be drawn, equal AB ; the external Angle, DAF or EBG, made by that Chord and the Line DE, is equal to the Angle at the Center of the Circle, subtended by the Chord, AF or BG (equal AB).

From what I have advanced, is deduced the Table, after Prob. 47, Practical Geometry, for constructing Poligons ; by dividing the Ark of a Quadrant, or Right Angle, into as many equal parts as the Poligon has Sides.

For, the Ark of a Right Angle being divided into five equal parts, each will be 18 Deg. wherefore, the Angle of a Pentagon being fix of those parts, or one added to a Right Angle, , i. e. 18 +90 = 108; which being subtracted from two Right Angles, or 180 Degrees, the Difference is 72, equal four times 18; which is the Compliment of two Right Angles; consequently, the external Angle of a Pentagon is equal to an Angle at the Center, subtended by a Side.

For a Nonagon, the Ark of a Right Angle is divided into 9 equal parts, each equal 10 Degrees.

Now, the external Angle being 40 Degrees, equal to the Angle at the Center, the Angle of the Nonagon is 50+go-or

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hve ninth parts added to a Right Angle; for, an internal and external Angle are, together, equal to two Right Angles - 1. I

Therefore a Right Angle is to the Angle of a Nonagon as 9 to 14; difference 5, as in the Table.

The reason of all this is so very obvious, it is almost needless to say more about it; seeing it is manifeft, that all the external Angles of any Right-lined Figure, whatever, are, together, equal to all the Angles about a Point, fubtended by the Sides. i. e. equal to four Right Angles. And, all the internal Angles of any Right-lined Figure are equal to twice as many Right Angles as the Figure has Sides, wanting four, (Th. 1. 10. 1.) consequently, the external Angles being equal to those four (Th. 2. of the fame) are equal to all the Angles at the Center; and, being equal, in number, they are equal in quantity,

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Hence, the Angle of any regular Poligon, whatever, may be readily obtained.

For, in a Pentagon, all the internal Angles, together, are equal to fix Right Angles; consequently, each Angle is 6 fifths of a Right Angle, i.e. it is equal to one Right Angle and one fifth part of a Right Angle; seeing, there are five Angles in the Figure, and they are, altogether, equal to fix Right Angles.

The Angles of a Hexagon are, altogether, equal to eight Right Angles ; confequently, each Angle is equal to eight fixths of a Right Angle, or four thirds; i.e. equal to one Right Angle and one third part of a Right Angle.

A Heptagon has all its Angles, together, equal to ten Right Angles ; wherefore, each Angle is equal to ten sevenths of a Right Angle, i. e. equal to a Right Angle and three seventh parts of a Right Angle.

All the Angles of an Octagon being equal to twelve Right Angles, it is evident that each Angle is equal to one Right Angle and a half; i. e. to 12 eights, i. e. to fix fourths, or three seconds, i. e. one and a half. 4

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To particularize more would be unnecessary; as it is easy, from what I have said, to calculate the quantity of the Angle of any Poligon whatever, by the Proportion it bears to a Right Angle, or to two Right Angles; which must ever be more than the Angle of any Polygon whatever.

For Poligons which have an even number of Sides, it may be observed, that the division of a Right Angle, may be reduced to a lower Denomination ; as for a Duodecagon, for instance; all its Angles, together, being equal to 20 twelfths, i. e. 10 fixths, or 5 third parts, i. e. to one Right Angle and two thirds; the Right Angle need be divided into three equal Parts, only, instead of twelve.

Whereas, those which have an odd number of Sides, cannot be reduced lower; but, in forming them, a Right Angle must necessarily be divided into as many parts as the Figure has Sides,

The method of inscribing a Quindecagon in a Circle, according to Euclid, is a matter of mere curiofity, first, to inscribe an equilateral Triangle, and afterward a Pentagon, in the fame Circle, in order to get a fifteenth part of the Circumference; which, notwithstanding it admits of perfe& Demonstration, is liable to great error in Practice.

GO

PROPOSITION XIII.
To find the Side of a Quindecagon inscribed in a Circle.

Let ABC be a given Circle, in which it is required to
inscribe a Quindecagon.

Draw, DG, at pleasure, touching the Circle ABC in A. With any Radius, AD, on A, the Point of contact, describe a Seme-circle, DEFG.

Inscribe the equilateral Triangle ABC; by making the Arks DE, EF and FG, each equal to the Radius AD, and drawing AE, AF to B and C, and joining the Points BandC.

DEM.

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Dem. For, the Arks, DE, EF, & FG, are

equal by Construction; wherefore, the
Angies DAE, EAF, and FAG, are equal. C
But, the Angle BCA is equal to BAD,
i.e. EAD; and, CBA is eq.toCAG -13.3.
cons. BCA, CBA, BAC, being equal,
respectively, to DAE, EAF, and FAG;
the Triangle ABC is equiangular; and
therefore it is equilateral C. 3.9. 1.

Now, if the Ark of the Semicircle, DEFG, be divided into five equal parts, for inscribing a Pentagon, as, before into three (in E and F) and, if A1, A2 be drawn, cuting the Circumference, at H and I; AH and H I are Sides of a Pentagon, inscribed.

In respect of the Operation, I leave it to the discretion of the Practitioner, to take what method he

most approves.

Granting AB to be the side of an equilateral Triangle, and AH, HI, to be the sides of a Pentagon inscribed; BI is the side of a Quindecagon, or Poligon of 15 Sides.

For, because AB is the side of an equilateral Triangle, the Ark AHB is one third part of the whole Circumference.

Also, AH and HI being Sides of a Pentagon, the Arks AH and HBI, are each equal to one fifth part of the Circumference, i. e. the Ark AHI is equal to two fifths, equal fix fifteenth parts; for HI, one fifth, isequal to three fifteenths.

Now, fince AHB is one third part of the Circumference, it contains five fifteenths; for three times five is fifteen.

Wherefore, BI is equal to one fifteenth part, and is, therefore, the Side of a Quindecagon, inscribed in the Circle ABC.

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THE

HE fifth Book of the Elements of Euclid contains the

fublime Doctrine of Proportion; which is, psrhaps, the most subtle manner of reasoning, the most brief, solid and convincing, that the Art of Man could devise. It is essentially necessary in demonstrating all the Propofitions of the fixth Book, which alone is sufficient to recommend it. But, where we consider its general and extenfive use, throughout the Mathematics, it is impossible to be dispensed with. The Doctrine it contains is not only useful in, but, it is the Criterion of almost every mathematical Science; infomuch, that, without the knowledge it teaches, we should not be able to advance one step further.

I must ever be of opinion that the readieft and easiest way of acquiring knowledge, is the best. The Doctrine of Proportion is, in some measure, born with us; it is a portion of Reason, implanted in us by Nature; which does not require a formal Demonstration, so much as barely to be illustrated, by some familiar Examples; and which, indeed, may pass for Axioms, for most of them are really such. 4

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