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. 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. 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 = 2ACD, and .. 4bcd = ¿BCD. Similarly cde CDE, def = DEF, and zefA = EFA. Hence Abcdef and ABCDEF are equiangular. 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; WV W A D B 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). = one of the polygons, od radius of inscribed circle to the other polygon (84). Again, since each side of a regular polygon subtends the same angle at the centre of the inscribed and circumscribing circle, AOB = aob, being angles which are the same part of 4 right angles. L Also, since AO BO, and ao = bo, OAB = ▲ OBA, and oab = < oba; but ‹ OAB+ ‹ OBA + ‹ AOB = two right angles = oab + 2 oba + ≤ aob, .. ¿OAB = 2≤ oab, and OBA = oba,.. OAB and oab are similar triangles. Hence AB: ab :: OA: oa, or :: OD: od; and every pair of sides is in the same ratio; therefore (80) sum of the sides of one polygon: sum of the sides of the other :: OA: oa, or :: OD: od, that is, the perimeters of the polygons are as the radii of the inscribed or circumscribing circles. (2) Again, since the polygons are made up of the same number of similar triangles, as AOB, aob; and since AOB: aob :: square of AO: square of ao, or :: square of OD: square of od, sum of these triangles in one polygon: sum of them in the other :: square of AO: square of ao, or :: square of OD: square of od; that is, the areas of the polygons are as the squares the radii of the inscribed or circumscribing circles. of 92. PROP. VII. The areas of similar polygons are to one another as the squares of any homologous sides, or corresponding lines within the polygons. Let ABCDEF, abcdef be two similar polygons, of which AB, ab are any two corresponding sides; then area ABCDEF: abcdef :: square of AB : square of ab. From A, a, draw the diagonals AC, AD, AE, ac, ad, These will divide the polygons into the same number of triangles, similar and similarly situated, each to each, see fig. (89). ae. .. by (76), triangle ABC: triangle abc :: square of AB : square of ab, ACD: ..... acd: square of CD: square of cd, ade: square of DE : square ... ADE: AEF: ...... ...... of de, aef: square of EF: square of ef. 3ut 1B :: EC. 10:a: DE: de: EF. 7(71), square or ca:quare or 5: square of ab, square of mare of E. square or de:: quare of 35 : square of 27 .. 30, LEC —¿ € I → at E-LEF· 20 jav— and — par — 301, : square of B square of ab, Tarea LECD EF docder square 4 square of ab. :: 13: ae, Again, since area ABCDEF Biol. 20 :: 0 ocder square or C, or 4). x 45 • square of ic, ora, re, respectively. OP. II. Me ircumferences of ircies are to one mother is their "aator umeters; inu heir treas We Tonortionat to the auares raose "ault r Hameters. incnose my two similar regular polygons to have heir cumsericing circies irawn out them: these arries will represent in "WO circies. Bisect each of the arcs subtended by each or he sides of the two polygona, ma join the points or section with the adjacent angular joints of the polygons; then two polygons of tonble the number of sides will be formed, while the creamscribing circies remain the same: and the perimeters and areas of these latter polygons will obviously anomach nearer to the perimeters and areas of the circles than those of the former polygons. Again the ares subtended by the sides of these polygons may be bisected, and other polygons described with double the number of idea, while the circles remain the same; and so on without Limit, until the polygons are made to approach as near as me please to the circles. Now the perimeters of similar regular polygons are *the radii of their circumscribing circles, and the areas * the aquares of those radii, whatever be the number of des, and therefore when that number, as above, is supposed to be indefinitely increased. But, by thus inCreasing the number of sides the polygons may be made to differ from the circles by less than any assignable magnitude, both as to perimeter and area. Hence the perimeters, that is, the circumferences of the circles will as their radii, and the areas as the squares of those Also, since the diameters will obviously have the same ratio to each other as the radii, the circumferences of circles will be as their diameters, and the areas as the squares of those diameters. COR. Since circumf. of one circle circumf. of another diameter of the former: diameter of the latter, .. alternately, circumf. of one : its diameter :: circumf. of the other its diameter; that is, the ratio of the circumference of every circle to its diameter is the same. EXERCISES D. (1) Define 'hexagon,' and 'diagonal' of a polygon. How many different diagonals has the hexagon? (2) Define 'angle of a polygon'; and shew that in every polygon the sum of all the angles is a multiple of a right angle. (3) Shew that the angle of a regular polygon is always greater than a right angle; and that it increases as the number of sides increases. (4) Shew that the angle of a regular octagon is equal to one right angle and a half. Hence construct a regular octagon upon a given straight line. (5) Shew that the side of a regular hexagon is equal to the radius of the circumscribing circle. (6) What is the number of diagonals which may be drawn in a polygon of ten sides? (7) Dividing a polygon by means of certain diagonals into the triangles of which it may be supposed to be made up, shew that the number of these triangles will always be less by 2 than the number of sides of the polygon. (8) Shew that in a regular pentagon each diagonal is parallel to a side; and that, if all the diagonals be drawn another regular pentagon will be formed by their intersections within the former one. |