Fig. 129. angles ACD, FHI, will be equal. Now, since the triangles ABC, FGH, are similar, AC: FH:: BC: GH; besides, on account of the polygons being similar (162), consequently BC: GH:: CD: HI; AC: FH:: CD: HI; but we have seen that the angle ACD = FHI; consequently the triangles ACD, FHI, have an angle of the one equal to an angle of the other and the sides about the equal angles proportional; they are therefore similar (208). We might proceed in the same manner to demonstrate, that the remaining triangles are similar, whatever be the number of the sides of the proposed polygons; therefore two similar polygons are composed of the same number of triangles, which are similar to each other and similarly disposed. 220. Scholium. The converse of this proposition is equally true; if two polygons are composed of the same number of triangles, which are similar to each other and similarly disposed, these two polygons will be similar. For, the triangles being similar, the angles ABC = FGH, BCA= GHF, ACD = FHI; consequently BCD = GHI, also CDE HIK, &c. Moreover, = AB: FG:: BC: GH:: AC: FH :: CD: HI, &c. ; consequently the two polygons have their angles respectively equal and their sides proportional; therefore they are similar. THEOREM. 221. The perimeters of similar polygons are as their homologous sides, and their surfaces are as the squares of these sides. Demonstration. 1. By the nature of similar figures we have AB: FG: BC: GH:: CD: HI, &c. (fig. 129), and from this series of equal ratios we may infer, that the sum of the antecedents AB + BC + CD, &c., the perimeter of the first figure is to the sum of the consequents FG+ GH+HI, &c., the perimeter of the second figure, as one antecedent, is to its consequent (iv), or as the side AB is to its homologous side FG. 2. The triangles ABC, FGH, being similar in like manner, ACD, FHI, being similar, ACD: FHI:: AC: FH 2 hence, on account of the common ratio AC: FH, By a similar process of reasoning it may be shown that and so on, if there should be a greater number of triangles. Hence, from this series of equal ratios, the sum of the antecedents ABC + ACD+ADE, or the polygon ABCDE, is to the sum of consequents FGH + FHI+FIK, or the polygon FGHIK, as one antecedent ABC is to its consequent FGH, or as AB is to FG (219). Therefore the surfaces of similar polygons are to each other, as the squares of their homologous sides. 2 222. Corollary. If three similar figures be constructed whose homologous sides are equal to the three sides of a right-angled triangle, the figure described upon the greatest side will be equal to the sum of the two others; for the three figures will be proportional to the squares of their homologous sides; now the square of the hypothenuse is equal to the sum of the squares of the two other sides; therefore, &c. THEOREM. 223. The parts of two chords which cut each other in a circle are reciprocally proportional; that is, AO : DO : : CO : OB (fig. 130). Fig. 130. Demonstration. Join AC and BD. In the triangles ACO, BOD, the angles at O are equal, being vertical angles, and the angle A is equal to the angle D, because they are inscribed in the same segment (127); for the same reason the angle C=B; therefore these triangles are similar, and the homologous sides give the proportion AO: DO:: CO: OB. 224. Corollary. Hence AO × OB == DO × CO; therefore the rectangle of the two parts of one of the chords is equal to the rectangle of the two parts of the other. THEOREM. 225. If from a point O (fig. 131), taken without a circle, secants OB, OC, be drawn terminating in the concave arc BC, the entire Fig. 131 Geom. 9 secants will be reciprocally proportional to the parts without the circle; that is, OB: OC:: OD: OA. Demonstration. Join AC and BD. The triangles OAC, OBD, have the angle O common; moreover the angle B = C (126); therefore the triangles are similar; and the homologous sides give the proposition OB: OC:: OD: OA. 226. Corollary. The rectangle OA × OB = OC × OD. 227. Scholium. It may be remarked, that this proposition has great analogy with the preceding; the only difference is, that the two chords AB, CD, instead of intersecting each other in the circle, meet without it. The following proposition may also be regarded as a particular case of this. Fig. 132. THEOREM. 228. If from the same point O (fig. 132), taken without the circle, a tangent OA be drawn and a secant OC, the tangent will be a mean proportional between the secant and the part without the circle; that is, OC: OA :: OA : OD, or, OA = OC × OD. --2 Demonstration. By joining AD and AC, the triangles OAD, OAC, have the angle O common; moreover, the angle OAD formed by a tangent and a chord (131) has for its measure the half of the arc AD, and the angle C has the same measure; consequently the angle OAD C; therefore the two triangles are = --2 similar, and OC: OA :: OA: OD, which gives OA = OC × OD. Fig. 133. THEOREM. 229. In any triangle ABC (fig. 133), if the angle A be bisected by the line AD, the rectangle of the sides AB, AC, will be equal to the rectangle of the segments BD, DC, plus the square of the bisecting line AD. Demonstration. Describe a circle the circumference of which shall pass through the points A, B, C; produce AD till it meet the circumference, and join CE. The triangle BAD is similar to the triangle EAC; for, by hypothesis, the angle BAD = EAC; moreover the angle B = E, since they have each for their measure the half of the arc AC; consequently the triangles are similar; and the homologous sides give the proportion BA: AE:: AD: AC; whence BAX AC = AE × AD; but AE AD + DE, and, by -2 multiplying each by AD, we have AE × AD=AD+AD × DE ; besides, AD × DE = BD × DC (224); therefore --2 BA × AC = AD + BD × DC. THEOREM. 230. In every triangle ABC (fig. 134) the rectangle of two of Fig. 134. the sides AB, AC, is equal to the rectangle contained by the diameter CE of the circumscribed circle and the perpendicular AD, let fall upon the third side BC. Demonstration. Join AE, and the triangles ABD, AEC, are right-angled, the one at D and the other at A; moreover the angle BE; consequently the triangles are similar; and they give the proportion, AB: CE :: AD: AC; whence ABX ACCE× AD. 231. Corollary. If these equal quantities be multiplied by BC, we shall have AB × AC × BC = CE × AD × BC. Now ADX BC is double the surface of the triangle (176); therefore the product of the three sides of a triangle is equal to the surface multiplied by double the diameter of the circumscribed circle. The product of three lines is sometimes called a solid, for a reason that will be given hereafter. The value of this product is easily conceived by supposing the three lines reduced to numbers and these numbers multiplied together. 232. Scholium. It may be demonstrated also, that the surface of a triangle is equal to its perimeter multiplied by half of the radius of the inscribed circle. For the triangles AOB, BOC, AOC (fig. 87), which have their Fig 87. common vertex in O, have for their common altitude the radius of the inscribed circle; consequently the sum of these triangles will be equal to the sum of the bases AB, BC, AC, multiplied by half of the radius OD; therefore the surface of the triangle ABC is equal to the product of its perimeter by half of the radius of the inscribed circle. Fig. 135. THEOREM. 233. In every inscribed quadrilateral figure ABCD (fig. 135), the rectangle of the two diagonals AC, BD, is equal to the sum of the rectangles of the opposite sides; that is, AC × BD = AB × CD + AD × BC. Demonstration. Take the arc CO= AD, and draw BO meeting the diagonal AC in I. = The angle ABD CBI, since one has for its measure half of the arc AD, and the other half of CO equal to AD. The angle ADB = BCI, because they are inscribed in the same segment AOB; consequently the triangle ABD is similar to the triangle IBC, and AD: CI :: BD : BC; whence ADX BCCI × BD. Again, the triangle ABI is similar to the triangle BDC; for the arc AD being equal to CO, if we add to each of these OD we shall have the arc AO= DC; consequently the angle ABI is equal to DBC; moreover the angle BAI BDC, because they are inscribed in the same segment; therefore the triangles ABI, BDC, are similar, and the homologous sides give the proportion AB: BD :: AI: CD; whence, AB × CD=AI × BD. Adding the two results above found, and observing that AIX BD + CIX BD=(Al+CI) × BD = AC × BD, we have AD × BC+AB × CD = AC × BD. 234. Scholium. We may demonstrate, in a similar manner, another theorem with respect to an inscribed quadrilateral figure. The triangle ABD being similar to BIC, BD: BC : : AB : B1, whence BIX BD=BC x AB. join CO, the triangle ICO, similar ABI, is similar to BDC, and gives the proportion BD: CO:: DC: OI, whence we have OI × BD=CO x DC, or CO being equal to AD, OIX BD=AD × DC. Adding these two results, and observing that BI × BD +OI × BQ reduces itself to BOX BD, we obtain BO × BD=AB × BC+AD× DC. |