To inscribe a Square and a Hexagon.

211. Corollary. By bisecting the arcs AM, MB, BN, &c., a regular inscribed polygon is obtained. of 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. To inscribe a square in a given circle.

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

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

213. Corollary. Hence, by arts. 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.

Demonstration. 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 art. 69, equal to of two right angles..

To inscribe a Decagon.

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

216. Gorollary. 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.

Demonstration. Join AD, and we are to prove that the arc BD is of the circumference, or that the angle BAD is of 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 art. 192,

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

BC: BD = BD : AB.

The triangles BCD and ABD are therefore similar, by art. 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.

But, in the isosceles triangles BCD and ACD,

the angle BCD = the angle CBD

= the angle ADB

twice the angle A,

and the sum of the three angles ABD, ADB, and A of the triangle ABD, or, by art. 64, two right angles, is equal to five times the angle A. Hence, A is of two right angles.

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

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

Demonstration. For the arc BC is circumference.

To = o of the

To circumscribe a Circle about a Regular Polygon.

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

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

Solution. Find, by art. 148, the circumference of a circle which passes through the three vertices A, B, C; and this circle is circumscribed about the given polygon.

Demonstration. Suppose the circumference divided into the same number of equal arcs AB', B'C', &c. as the sides of the given polygon. The chords AB', B'C', &c. form, by art. 202, a regular polygon, which, by art. 206, is similar to ABCD, &c.

Hence,

the angle ABC= the angle AB'C' ;

and the arc, which remains after subtracting the arc ABC from the circumference, being the measure of the angle -ABC, is equal to the measure of the angle AB'C', or to the arc which remains after subtracting the arc AB'C' from the circumference; and, consequently, the arc ABC, which is twice the arc AB, is equal to the arc AB'C', which is twice the arc AB'. We have then,

the arc AB the arc AB',

and the chord AB is equal to the chord AB', and coincides with it. The polygons AB'C'D', &c., ABCD, &c., must, therefore, by art. 195, coincide; and the circle is circumscribed about the given polygon.

223. Corollary. There is a point O in every regular polygon equally distant from all its vertices, and which is called the centre of the polygon.

To inscribe in a Circle any Regular Polygon.

224. Corollary. If we join AO, BO, CO, &c., the angles AOB, BOC, CQD, &c. are all equal, and each has the same ratio to four right angles, which the arc AB has to the circumference.

225. Corollary. The isosceles triangles AOB, BOC, COD, &c. are all equal.

226. Corollary. The angles OAB, OBA, OBC, OCB, &c. are all equal, and each is half of the angle ABC.

227. Theorem. The sides of a regular polygon are all equally distant from its centre.

Demonstration. Let fall the perpendiculars OM, ON, OP, &c. (fig. 122), from the centre O, upon the sides AB, BC, &c. In the right triangles OAM, OBM, OBN, OCN, OCP, &c., the hypothenuses OA, OB, OC, &c. are all equal, by art. 223, and the legs AM, MB, BN, NC, CP, &c. are equal, since each is, by art. 115, half of AB, or of its equal BC, &c. The triangles OAM, OBM, OBN, &c. are, consequently, equal, by art. 63; and the perpendiculars OM, ON, OP, &c. are equal.

228. Problem. To inscribe in a given circle a regular polygon, similar to a given regular polygon ABCD, &c. (fig. 123).

Solution. From the centre of the given polygon draw the lines AO, BO; at the centre O' of the given circle make the angle A'O'B' equal to AOB, and the chord A'B', being applied round the circumference as many times as ABCD, &c. has sides, gives the required polygon A'B'C'D', &c., as is evident from art. 224.

229. Problem. To inscribe a circle in a given polygon ABCD, &c. (fig. 124).