triangles OTH, OHN, having the common hypotenuse OH, and the side OT-ON, must be equal, and consequently the angle TOH HON, wherefore the line OH passes through the middle point B of the arc TN. For a like reason, the point I is in the prolongation of OC; and so with the rest. But since BH is parallel to AB, and HI to BC, the angle GHI=ABC (26. Ï.); in like manner, HIK=BCD; and so with all the rest: hence the angles of the circumscribed polygon are equal to those of the inscribed one. And farther, by reason of these same parallels, we have GH: AB :: OH: OB, and HI BC:: OH: OB; therefore GH: AB :: HI: BC. But AB = BC, therefore GH HI. For the same reason, HI=IK, &c.; hence the sides of the circumscribed polygon are all equal; hence this polygon is regular, and similar to the inscribed one.

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Cor. 1. Reciprocally, if the circumscribed polygon GHIK &c. were given, and the inscribed one ABC &c. were required to be deduced from it, it would only be necessary to draw from the angles G, H, I, &c. of the given polygon, straight lines OG, OH, &c. meeting the circumference in the points A, B, C, &c.; then to join those points by the chords AB, BC, &c.; which would form the inscribed polygon. An easier solution of this problem would be simply to join the points of contact T, N, P, &c. by the chords TN, NP, &c. which likewise would form an inscribed polygon similar to the circumscribed one.

Cor. 2. Hence we may circumscribe about a circle any regular polygon, which can be inscribed within it; and conversely.

PROPOSITION VII. THEOREM.

The area of a regular polygon is equal to its perimeter multiplied by half the radius of the inscribed circle.

Let the regular polygon be GHIK &c. (see the last figure), the triangle GOH will be measured by GHXOT; the triangle OHI by HIXON: but ON=OT; hence the two triangles taken together will be measured by (GH+HI) × ¿OT. And, by continuing the same operation for the other triangles, it will appear that the sum of them all, or the whole polygon, is measured by the sum of the bases GH, HI, IK, &c. or the perimeter of the polygon, multiplied into OT, or half the radius of the inscribed circle.

Scholium. The radius OT of the inscribed circle is nothing else than the perpendicular let fall from the centre to one of the sides: it is sometimes named the apothem of the polygon.

PROPOSITION VIII. THEOREM.

The perimeters of two regular polygons, having the same number of sides, are to each other as the radii of the circumscribed circles, and also as the radii of the inscribed circles; their surfaces are to each other as the squares of those radii. Let AB be a side of the one polygon, O the centre, and consequently OA the a radius of the circumscribed circle, and OD, perpendicular to AB, the radius of the inscribed circle; let ab, in like manner, be a side of the other polygon, o its centre, oa and od the radii of the circumscribed and the inscribed circle. The perimeters of the two polygons are to

each other as the sides AB and ab: but the angles A and a are equal, being each half of the angle of the polygon; so also are the angles B and b; hence the triangles ABO, abo are similar, as are likewise the right-angled triangles ADO, ado; hence AB: ab:: AO: ao:: DO: do; hence the perimeters of the polygons are to each other as the radii AO, ao of the circumscribed circles, and also as the radii DO, do of the inscribed circles.

squares

The surfaces of those polygons are to each other as the of the homologous sides AB, ab; they are therefore likewise to each other as the squares of AO, ao the radii of the circumscribed circles, or as the squares of OD, od the radii of the inscribed circles.

PROPOSITION IX. LEMMA.

Any curve, or any polygonal line, which envelopes the convex line AMB from one end to the other, is longer than AMB the enveloped line.

We have already said that by the term convex line, we understand a line, polygonal or curve, or partly curve and partly polygonal, such that a straight line cannot cut it in more than two A points. If in the line AMB there

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B

were any sinuosities or re-entrant portions, it would cease to be convex, because a straight line might evidently cut it in more than two points. The arcs of a circle are essentially convex;

but the present proposition extends to any line which fulfils the required condition.

This being premised, if the line AMB is not shorter than any of those which envelope it, there will be found among the latter a line shorter than all the rest, which is shorter than AMB, or, at most, equal to it. Let ACDEB be this enveloping line: : any where between those two lines draw the straight line PQ, not meeting, or at least only touching, the line AMB. The straight line PQ is shorter than PCDEQ; hence if, instead of the part PCDEQ, we substitute the straight line PQ, the enveloping line APQB will be shorter than APDQB. But, by hypothesis, this latter was shorter than any other; hence that hypothesis was false; hence all of the enveloping lines are longer than AMB.

Scholium. In the very same way, it might be shewn that AMB, a convex line returning into itself, is shorter than any line enveloping it on all sides, whether the enveloping line FHG touch AMB in one or more points, or surround without touching it.

PROPOSITION X. LEMMA.

H

Two concentric circles being given, a regular polygon may always be inscribed within the greater, the sides of which shall not meet the circumference of the less; and likewise, a regular polygon may always be described about the less, the sides of which shall not meet the circumference of the greater.

subtended by its sides, and draw the chords of those half-arcs; a polygon will thus be found having twice as many sides. Continue the bisection, till an arc is obtained less than DBE. Let MBN be that arc, the middle point of it being supposed to lie at B: it is plain that the chord MN will be farther from the centre than DE; and that consequently the regular polygon, of which MN is a side, cannot meet the circumference, of which CA is the radius.

Now, the same construction remaining, join CM and CN, meeting the tangent DE in P and Q; PQ will be the side of a polygon described about the less circumference, similar to that polygon inscribed within the greater, of which the side is MN. And it is evident, that this circumscribed polygon having PQ for its side, can never meet the greater circumference, CP being less than CM.

Hence, by the same operation, a regular polygon may be inscribed within the greater circumference, and a similar one described about the less, both of which shall have their sides included between the two circumferences.

Scholium. If two concentric sectors FCG, ICH be given, a portion of a regular polygon may, in like manner, be inscribed in the greater, or circumscribed about the less, so that the perimeters of the two polygons shall be included between the two circumferences. For this purpose, it will be sufficient to divide the arc FBG successively into 2, 4, 8, 16, &c. equal parts, till a part smaller than DBĚ is obtained.

By the phrase, portion of a regular polygon, we here mean the figure terminated by a series of equal chords inscribed in the arc FG, from one of its extremities to the other. This portion has all the main properties of regular polygons; it has its angles equal, and its sides equal, it can be inscribed in a circle, or circumscribed about one: yet, properly speaking, it forms part of a regular polygon only in those cases where the arc subtended by one of its sides is an aliquot part of the circumference.

PROPOSITION XI. THEOREM.

The circumferences of circles are to each other as their radii, and their surfaces as the squares of those radii.

For the sake of brevity, let us designate the circumference whose radius is CA by circ. CA; we are to shew that circ. CA: cire. OB:: CA: OB.

If this proposition is not true, CA must be to OB as circ. CA is to a fourth term less or greater than circ. OB: suppose it less; and that, if possible, CA: OB:: circ. CA: circ. ÖD.

In the circle of which OB is the radius inscribe a regular polygon EFGKLE, such that the sides of it shall not meet the circumference of which OD is the radius (Prop. 10. IV.): inscribe a similar polygon, MNPFM, in the circle of which AC is the radius.

Then, since those polygons are similar, their perimeters MNPSM, EFGKE will be to each other (8. IV.) as CA, OB, the radii of the circumscribed circles, that is MNPSM: EFGKE :: CA : OB. But by hypothesis, CA: OB:: circ. CA: circ. OD; therefore MNPSM: EFGKE:: circ. CA: circ.OD; which proportion is false, because (9. IV.) the perimeter MNSPM is less than circ. CA, while on the contrary EFGKE is greater than circ. OD: therefore it is impossible that CA can be to OB as circ. CA is to a circumference less than circ. OB; or, in more general terms, it is impossible that one radius can be to another, as the circumference described with the former radius is to a circumference less than the one described with the latter radius.