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DEF. Í straight line, irawn from the vertex of any angle o che erter tiny other angle not adiacent to be tormer. .. called a lingonat.

Thus, in the innexed : AB - IR BODE - EL 3 ne verte meter, and each ot the straignt ines IC, ID, BE. BD, IC. 9 . tiagonal, of the holygon IBCDE.

V. 3. Plıroughout this seetion wil those puiugms He excludent which iave wnat are called "re-entrand angles, such as the polygons annexed :

where A, B, C are re-entrant ugies. They are called 79_ontraint mgles, because if the ines forming them te prastucent trongh the vertex, these ines enter med the polyzon, which is not the case with ordinary polygone.

88. Paop. I. All the mgles of a polugon are tageThor sual to twice as many right angies as the polygon hoc sides, timinishest by four right angies.

For every polyzon, as 4BCDEF, may be divided inta triangles by taking any point

mithin the polygon, and joining OA, OB, OC, OD, OE, OF ; and the number of triangles will ob- r vionely be the same as the number of the sides of the polygoo. Bert the three angles of each triHugle are together equal to two

B right angles ; i. the angles of all the triangles are together equal to twice as many right angles as the polygon has sides; that is, all the angles of the polygon, together with the angles having the comtheri vertex 0, are equal to twice as many right angles

the polygon has sides. But the angles at Oare

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equal to four right angles (30 Cor.); .. all the angles of the polygon are equal to twice as many right angles as the polygon has sides, diminished by four right angles. Cor. 1. Hence, all the angles of a pentagon = 6 right angles;

hexagon = 8
octagon = 12

; and so on, whatever be the number of sides of the polygon*

Hence, also, since all the angles are equal to one another in a regular polygon,

6 each angle of a regular pentagon = of a right angle;

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hexagon =

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CoR. 2. If ABCDEF be a portion of the perimeter of any polygon; and if the sides

id AB, BC, CD, &c., be produced to b, c, d, &c., since each inte

E rior angle, as 2 ABC, + its exte

T rior angle, as 26BC, = two right angles, .. all the interior angles + all the exterior angles = twice as many right angles as the polygon has sides ; and .., by what has been proved, all the exterior angles of a polygon are together equal to four right angles.

The same result may also be made to appear from a very simple consideration. From B draw B: parallel to CD, By parallel to DE, Bx parallel to EF, &c., taking every side of the polygon in succession. Then < DCc -CB2, -EDd=2yBz, 2 FEe = 2xBy, &c.; and the last of the lines Bz, By, Bx, &c., will be Bb; •, the sum of

• The triangle, and quadrilateral, as we might expect, both follow the same rule. Thus all the angles of a triangle are equal to 6 right angles diminished by 4 right angles, that is, are equal to 2 right angles, And all the angles of a quadrilateral are equal to 8 right angles dimi. nished by 4 right angles, that is, are equal to 4 right angles.

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the exterior angles will be equal to the sum of the angles occupying the whole

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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.

Let ABCD be a portion of the perimeter of a regular polygon. Bisect the angles

D at A and B by the straight lines OA, OB, meeting in 0; and join OC. Then, since 20AB= half LA, and _OBA= half _B, and the angles of the polygon at A and B are equal, :: COAB= 2OBA, and .. 0A = OB. Again, since AB = BC,

=, and BO is common to the two tri- А. angles OAB, OBC, and _OBC= LOBA, .. OC = 0A, and _OCB= ZOBĂ (24). But _OBA = half _B = half <C,.. OC bisects <C. Hence OA= OB=OC, and 0A, 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.

Cor. 1. Hence, if with centre 0 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

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circle, it is obvious also (50), that the centre_ () 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 0.

COR. 3. If a circle be described with centre 0 and radius equal to the perpendicular from 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. I

C 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 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.

Again, since OA = OB = 0E = OD, and 2AOB = DOE (31), .. the triangles AOB, DOE are equal in all respects, and _OABE LODE, .. 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 ZAOB = 2 BOC = COD, .. ZAOB = one-third of two right angles (30 Cor.), and .. ZOAB+ 2OBA=

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two-thirds of two right angles (37). But _OAB= LOBA, .. cach 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=0A=0E, 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.

[Der. 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.] E

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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 2ABC= zabc, and also AB : ab :: BC : bc, by Definition, .. the triangles ABC, abc, are similar (71 Cor. 1), .. also ACB= zacb, and AC : ac :: BC : bc. But BCD=<bcd, ..LACD = cacd; and BC:bc:: CD:cd,

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