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the exterior angles will be equal to the sum of the angles occupying the whole space round B as a common vertex, that is, four right angles (30 Cor.).

COR. 3. Since the magnitude of each angle in a regular polygon depends only on the number, and not the length, of the sides, therefore all regular polygons of the same name have precisely the same angles, however much they may differ in their other dimensions. Hence all regular polygons of the same name are similar, in the sense in which certain triangles were defined to be similar, for besides equal angles, each to each, such polygons having equal sides throughout each, will, of course, have the sides about equal angles proportionals.

87. PROP. II. In every regular polygon if lines be drawn severally bisecting the angles, these lines will all meet in the same point within the polygon; and that point will be equidistant from all the angular points in the perimeter of the polygon.

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Let ABCD be a portion of the perimeter of a regular polygon. Bisect the angles at A and B by the straight lines OA, OB, meeting in O; and join OC. Then, since OAB half ▲ A, and OBA = half B, and the angles of the polygon at A and B are equal, OAB LOBA, and .. OA OB. Again, since AB = BC, and BO is common to the two triangles OAB, OBC, and OBC= ¿OBA, .. OC=OA, and OCB=OBA (24). But OBA = half < B = half <C,.. OC bisects C. Hence OA=OB=OC, and OA, OB, OC, bisect the angles at A, B, C. The same may be proved in the same manner for all the remaining angles of the polygon.

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COR. 1. Hence, if with centre O and radius OA a circle be described, its circumference will pass through all the angular points in the perimeter of the polygon.

In this case the circle is said to be described about' the polygon, or the polygon to be inscribed in' the

circle.

COR. 2. Since AB and BC are given chords of this

circle, it is obvious also (50), that the centre O may easily be determined by bisecting AB, and BC, and through the points of bisection drawing lines at right angles to AB, BC, and meeting, as they will do, in O.

COR. 3. If a circle be described with centre O and radius equal to the perpendicular from O upon AB, every side of the polygon will be a tangent to this circle. In this case the polygon is said to be described about' the circle, or the circle to be inscribed in' the polygon.

COR. 4. Hence every regular polygon may be inscribed in a given circle, or described about a given circle.

88. PROP. III. In every regular polygon of an even number of sides to each side there is another opposite side parallel to it; and to each angle there is an opposite angle such that the vertices of the two are in the same diameter of the circumscribing circle.

Let ABCDEF be a regular polygon of six sides, (the proof will be the same for eight sides, ten sides, &c.) Find O the centre of the circumscribing circle (87), and join OA, OB, OC, OD, OE, OF. F Then since one half of all the sides will be equal to the other half, and AB = BC = CD = &c., and equal chords in the same

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circle cut off equal arcs (58), the three arcs AB, BC, CD are together equal to the three arcs DE, EF, FA, that is, ABCD is a semicircle, and.. AOD is a straight line and a diameter. Similarly BOE is a diameter, and FOC a diameter, of the circumscribing circle.

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Again, since OA OBOE = OD, and AOB DOE (31), .. the triangles AOB, DOE are equal in all respects, and 40AB=ODE, .. AB is parallel to ED (34). Similarly BC is parallel to EF; and CD to AF.

COR. 1. Hence, in the case of a hexagon, AOB is an equilateral triangle. For, since AOD is a straight line, and <AOB = ¿BOC = ¿COD, :. ¿AOB = one-third of two right angles (30 Cor.), and .. ¿OAB+ ¿OBA =

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two-thirds of two right angles (37). But ▲ OAB = ¿OBA, .. each of them is one-third of two right angles; and .. the triangle AOB is equiangular; and because it is equiangular it is also equilateral (26 Cor.).

COR. 2. Hence, also, to construct a hexagon upon a given straight line AB, that is, having the given straight line for a side, it is only necessary to describe an equilateral triangle on the given line, as AOB; then produce AO, BO to D and E, making OD=OA=OE, which will determine the angular points D and E; then with centres B and D and radius OA describe two arcs intersecting in C, and with centres A and E and the same radius two arcs intersecting in F; join BC, CD, DE, EF, FA, and the required hexagon ABCDEF is constructed.

89. PROP. IV. Two similar polygons may be divided into the same number of similar triangles, each to each, and similarly situated.

[DEF. Two polygons are similar, when they have the same number of sides, and all the angles of the one are separately equal to all the angles of the other, each to each, and the sides also about equal angles proportionals.]

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Let ABCDEF, abcdef, be two similar polygons, the angles at A, B, C, D, E, F, being equal to the angles at a, b, c, d, e, f, each to each. From A draw the diagonals AC, AD, AE; and from a draw the diagonals ac, ad, ae. Then since ABC= 4abc, and also AB: ab :: BC: bc, by Definition, .. the triangles ABC, abc, are similar (71 Cor. 1), .. also 4 ACB = Lacb, and AC : ac :: BC : bc. But BCD=2bcd,.. ACD = Lacd; and BC: bc :: CD: cd,

by supposition, .. AC: ac :: CD: cd. Hence again the triangle ACD is similar to the triangle acd. And in the same way it may be shewn that the triangles ADE, ade, are similar and also that the remaining triangles AEF, aef are similar.

N. B. It is not enough in polygons, as in triangles, to make them similar, that the angles of the one are respectively equal to those of the other, because two triangles cannot have their angles respectively equal without having the sides about equal angles proportional; whereas this does not hold for polygons, seeing that we can alter the sides in an almost endless number of ways, without altering any angle. For instance, suppose we cut off a large part of the polygon ABCDEF by a line parallel to BC and near to AD, the angles of the new polygon will be the same as those of ABCDEF, but it is obvious that the new polygon is not similar to abcdef, not having its sides in the same proportion.

COR. The converse will easily follow, viz. that, if two polygons are composed of the same number of similar triangles, arranged in the same order in each polygon, the polygons shall be similar.

90. PROP. V. Upon a given straight line to construct a polygon similar to a given polygon.

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Let ABCDEF be the given polygon, and G the given straight line; it is required to construct upon G, that is, upon a base equal to G, a polygon similar to ABCDEF.

(1) Suppose G less than AB; with centre A and radius equal to G describe a circle cutting AB in b, making Ab equal to G; join AC, AD, AE; through b draw be parallel to BC meeting AC in c; through c draw cd

parallel to CD meeting AD in d; through d draw de parallel to DE meeting AE in e; and through e draw ef parallel to EF meeting AF in f. Then Abcdef shall be similar to ABCDEF, and it stands upon the base Ab equal to G.

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For, since bc is parallel to BC, the triangles Abc, ABC are similar. So also Acd is similar to ACD; Ade to ADE; and Aef to AEF, .. ▲ Abc=▲ABC; ▲Acb= LACB; Acd = LACD, and .. 4bcd: LBCD. Similarly cde CDE, def = DEF, and zefA = EFA. Hence Abcdef and ABCDEF are equiangular.

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Again, by similarity of triangles, AB : Ab :: BC : bc; AC: Ac:: CD cd, and AC: Ac :: BC: bc,..BC : bc :: CD : cd. Similarly CD: cd :: DE: de; and DE: de :: EF : ef ; and EF: ef :: AF : Aƒ; .. the sides about the equal angles are proportionals.

Hence ABCDEF and Abcdef are similar polygons.

(2) If G be greater than AB, produce AB, AC, AD, AE, AF indefinitely, and in AB produced take Ab equal to G, and proceed as before.

91. PROP. VI. The perimeters of regular polygons of the same number of sides are proportional to the radii of their inscribed or circumscribing circles; and their areas are proportional to the squares of those radii.

(1) Let AB, ab be sides of two regular polygons of the same name, that is, of the same number of sides;

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O, o, the centres of their inscribed and circumscribing circles. Join OA, OB, oa, ob; and draw OD perpendicular to AB, and od perpendicular to ab. Then OA = OB = radius of circumscribing circle to one of the polygons, and oa = ob = radius of circumscribing circle to the other polygon; OD = radius of inscribed circle to

That the inscribed and circumscribing circles in the same regular polygon have the same centre appears from (80).

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