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. E 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 6 draw bc parallel to BC meeting AC in c; through c draw cd a = 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, i. - Abc = -ABC; -Acb = LACB; 2 Acd = ZACD, and .. <bcd = <BCD. Similarly zcde = -CDE, -def = _DEF, and Zef A = 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: Af; .. 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; : : 99 0, 0, the centres of their inscribed and circumscribing circles *. Join 0A, OB, oa, ob; and draw OD perpendicular to AB, and od perpendicular to ab. Then OA = 0B = 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, 2 AOB = 2 aob, being angles which are the same part of 4 right angles. Also, since AO= BO, and ao = bo, - OAB = 2 OBA, and coab = = z oba; but _OAB + <OBA + ZAOB = two right angles = 2 oab + z oba + 2 aob, .. 20AB = 2 oab, and LOBA = oba, .. OAB and oab are similar triangles. = Hence AB : ab :: 0A : 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 A0 : square of ao, or :: square of OD : square of od; that is, the areas of the polygons are as the squares of the radii of the inscribed or circumscribing circles. 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). .. by (76), triangle ABC : triangle abc :: square of AB : square of ab, ... ACD: of CD: square of cd, . ADE: of DE: square of de, AEF: aef :: square of EF : square of ef. ae. acd :: square But iB : E.cer :: Eler FTI), ., are o quare 102 cunare or 15: sciare I LO, Maze ot' LE cagare orie.. Graze of Icquare Of ?-30: _E1, ---£F 100+20---me-lef ** square 3: square or ab, rarea 30LEX cest auare-13: quare of ih. izan since 13: 26.0:33:ae, i ara ECJEE Locaer square of or .r *E square tie, rr se, espectively. 2. JP. I. De ITCUMTerences rzreies tre A ne innther is zeer za r zumetery, Inu metr tres ITP Troportional a ne anares T 205C L r iameters. arnase mw osmiar regrar Toyons to have herr Inser:bcr rcies LEIWII DOU: Den: sese Tries 1: "ppresent My Worces. Zisect each of he ucs surtendea iv zachoi me siues or she wo goiymon. Ha join the ICS I visection with the mujacent aga Joints I te JOIVOIS; ben wo Joir qons of Lonnie tie jumpe? I sides wi je urmedi. wmie the umscribing arries semain the same : inu the perimetry n erebese after poirzons wil viously aims leared the perimeters iu reas or the circies than hoge ot he ormez yaygons Igun the res subtedest jy hie ses or nese Jiy qons nav je bisected, md sther Joiy qons iescabei v.11 jousie ne number of vinos, while the urcies semain the same; uni yon withant imit. intii che poiy tons rre naue to approach is rer ** mo, piense to the circles. Vor the permeters of similar reqılar poivgods are 19 Pne radii of their circumscribing circies, and the areas 14 he aquares of those mcii, urutever be the number of Ador, ud therefore wien that number, as above, is supposed to be indefinitely increased. But, by thus inmnogoing the number of sides the polygons may be made toy difepe from the circles by less than any assignable magnade, both as to perimeter and aree. Hence the pepithetery, that is the circumferences of the circles will byte na their radii, and the areas as the squares of those same Also, since the diameters will obviously have the 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. a (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. |