Limitation of axiom of Art. 99.

the arcs coincide with the chords, this polygon is the circle itself.

205. Scholium. The two preceding demonstrations contain the following obvious and necessary limitation of the axiom of § 99.

The infinitely small quantities, which are neglected by the axiom of § 99, must be infinitely small in comparison with those which are retained.

In the present case, indeed, the difference between the infinitely small arc and its chord is infinitely small, and yet it could not be neglected if it were not infinitely small in comparison with the arc. For, as the sum of all these differences corresponding to all the arcs of the circle has the same ratio to the sum of all the arcs, that is, to the entire circumference, which each difference has to its arc; the sum of the differences, that is, the difference between the circumference of the circle and the perimeter of the polygon of an infinite number of sides, would not be infinitely small, and, therefore, capable of being neglected, unless each difference were infinitely small in comparison with its arc.

206. Theorem. Two regular polygons ABCD, &c. A'B'C'D', &c. (fig. 115), of the same number of sides, are similar.

Proof. For, they are equiangular with respect to each other, since the sum of their angles is the same, by § 72, and each angle of each polygon is found by dividing this common sum by the number of sides.

Their homologous sides are, moreover, proportional;

To inscribe a Regular Polygon of twice the number of Sides, &c.

we have

AB: AB': =

BC: B'C' CD: C'D', &c.

=

207. Corollary. Hence, and by § 197, the perimeters of regular polygons are to each other as their homologous sides.

208. Theorem. Two circles are similar regular polygons.

Proof. The number of sides of each circle is any infinite number whatever, and, if we choose, the same infinite number for all circles.

209. Theorem. A regular polygon of any number of sides may be inscribed in a given circle.

Proof. Suppose the circumference ABCD &c. (fig. 116) to be divided into any number of equal arcs AB, BC, CD, &c. Their chords AB, BC, CD, &c. are also equal, by § 112; and the polygon ABCD, &c. formed by these chords is, by § 202, a regular polygon of a number of sides equal to that of the arcs AB, BC, CD, &c.

210. Problem. To inscribe in a given circle a regular polygon, which has double the number of sides of a given inscribed regular polygon ABCD &c. (fig. 116).

Solution. Bisect the arcs AB, BC, CD, &c. at the points M, N, O, P, &c. Join AM, MB, BN, NC, &c., and AMBNCO, &c. is the required polygon.

Proof. For the sides AM, MB, BN, NC, &c. being the sides of equal arcs, are equal, and, by § 202, the polygon is regular.

211. Corollary. By bisecting the arcs AM, MB, BN, &c., a regular inscribed polygon is obtained of

To inscribe a Square and a Hexagon.

4 times the number of sides of the given polygon; and, by continuing the process, regular inscribed polygons are obtained of 8, 16, 32, &c. times the number of sides of the given polygon.

212. Problem.

circle.

To inscribe a square in a given

Solution. Draw the two diameters AB and CD (fig. 117) perpendicular to each other; join AD, DB, BC, CA; and ADBC is the required square.

Proof. The arcs AD, BD, BC, and AC are equal, being quadrants; and therefore their chords AD, DB, BC, and CA are equal, and, by §§ 201 and 202, ADBC is a square.

213. Corollary. Hence, by §§ 210 and 211, a polygon may be inscribed in a circle of 8, 16, 32, 64, &c. sides.

214. Problem. To inscribe in a given circle a regular hexagon.

Solution. Take the side BC (fig. 118) of the hexagon equal to the radius AC of the circle, and, by applying it six times round the circumference, the required hexagon BCDEFG is obtained.

Proof. Join AC, and we are to prove that the arc BC is one sixth of the circumference, or that the angle BAC is of four right angles, or of two right angles.

Now, in the equilateral triangle ABC, each angle, as BAC, is, by § 70, equal to of two right angles. 13

215. Corollary. Hence regular polygons of 12, 24, 48, &c. sides may, by §§ 210 and 211, be inscribed in a given circle.

To inscribe a Decagon.

216. Corollary. An equilateral triangle BDF is inscribed by joining the alternate vertices, B, D, F.

217. Problem. To inscribe in a given circle a regular decagon.

Solution. Divide the radius AB (fig. 119) in extreme and mean ratio at the point C. Take BD for the side of the decagon equal to the larger part AC, and, by applying it ten times round the circumference, the required decagon BDEF &c. is obtained.

Proof. Join AD, and we are to prove that the arc BD is of the circumference, or that the angle BAD is of to four right angles, or of two right angles.

Join DC. The triangles BCD and ABD have the angle B common; and the sides BC and BD, which include this angle in the one triangle, are proportional to the sides BD and AB, which include the same angle in the other triangle. For, by § 192,

but, by construction, BD is equal to AC, and, being substituted for it in this proportion, gives

The triangles BCD and ABD are therefore similar, by § 179.

Now the triangle ABD is isosceles, and therefore BCD must also be isosceles; and the side DC is equal to BD, which is equal to AC; so that the triangle ACD is also isosceles.

We have, therefore,

and, by § 71,

the angle Athe angle ADC;

the angle BCD=the angle A+ the angle ADC

To inscribe a Pentagon.

twice the angle A.

But, in the isosceles triangles BCD and ACD,

the angle BCD=the angle CBD

218. Corollary. Hence, regular polygons of 20, 40, 80, &c. sides may, §§ 210 and 211, be inscribed in a given circle.

219. Corollary. A regular pentagon BEGIL is inscribed by joining the alternate vertices B, E, G, I, L.

220. Problem.

To inscribe in a given circle a regular polygon of 15 sides.

Solution. Find, by § 217, the arc AB (fig. 120) equal to of the circumference, and, by § 214, the arc AC equal to of the circumference, and the chord BC, being applied 15 times round the circumference, gives the required polygon.

៖

=

Proof. For the arc BC is of the circum

ference.

221. Corollary. Hence, regular polygons of 30, 60, 120, &c. sides may, by §§ 210 and 211, be inscribed in a given circle.

222. Problem. To circumscribe a circle about a given regular polygon ABCD &c. (fig. 121).