Fig. 127. 215. Corollary. If from the point A (fig. 127) of the circumference of a circle two chords AB, AC, be drawn to the extremities of the diameter BC, the triangle ABC will be right-angled at A (128); whence, 1. the perpendicular AD is a mean proportional between the segments BD, DC, of the diameter, or, which amounts to the same thing, --2 AD BD x DC. 2. The chord AB is a mean proportional between the diameter BC and the adjacent segment BD; --2 Also AC = DC × BC; therefore AB: AC:: BD: DC. If we These ratios of the squares of the sides to each other and to the square of the hypothenuse have already been given in articles. 189, 190. THEOREM. 216. Two triangles, which have an angle in the one equal to an angle in the other, are to each other as the rectangles of the sides Eig. 128. which contain the equal angles; thus, the triangle ABC (fig. 128) is to the triangle ADE, as the rectangle AB × AC is to the rectangle AD X AE. Demonstration. Draw BE; the two triangles ABE, ADE, whose common vertex is E, have the same altitude, and are to each other as their bases AB, AD (177); hence In like manner, ABE: ADE :: AB : AD. ABC: ABE:: AC: AE; multiplying the two proportions in order and omitting the common term ABE, we have, ABC: ADE :: AB × AC : AD × AE. 217. Corollary. The two triangles would be equivalent, if the rectangle AB × AC were equal to the rectangle AD × AE, ΟΙ if AB: AD:: AE: AC, which is the case when the line DC is parallel to BE. THEOREM. 218. Two similar triangles are to each other as the squares of their homologous sides. Demonstration. Let the angle A=D (fig. 122), and the an- Fig. 192. gle B= E, then, by the preceding proposition, ABC: DEF:: AB × AC : DE × DF; and, because the triangles are similar, AB: DE:: AC: DF. This proportion being multiplied in order by the identical preportion. Therefore two similar triangles ABC, DEF, are to each other as the squares of the homologous sides AC, DF, or as the squares of any other two homologous sides. THEOREM. 219. Two similar polygons are composed of the same number of triangles, which are similar to each other and similarly disposed. Demonstration. In the polygon ABCDE (fig. 129) draw from Fig. 129. an angle A the diagonals AC, AD, to the other angles. In the other polygon FGHIK draw, in like manner, from the angle F, homologous to A, the diagonals FH, FI, to the other angles. Since the polygons are similar, the angle ABC is equal to the homologous angle FGH (162), moreover the sides AB, BC, are proportional to the sides FG, GH, so that AB: FG:: BC: GH. It follows from this, that the triangles ABC, FGH, having an angle of the one equal to an angle of the other and the sides. about the equal angles proportional, are similar (208), cọnsequently the angle BCA GHF. These equal angles being subtracted from the equal angles BCD, GHI, the remaining = 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), 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 --2 --2 ABC: FGH :: AC: FH (218); in like manner, ACD, FHI, being similar, -2 ACD: FHI:: AC: FH; 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, -2 as one antecedent ABC is to its consequent FGH, or as AB is to FG 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 Fig. 131. OB, OC, be drawn terminating in the concave arc BC, the entire secants will be reciprocally proportional to the parts without the cir cle; 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 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 langent 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 similar, and OC: OA:: OA: OD, which gives OA = OC × OD. --2 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; |