It happens frequently, as in this instance, that by deducing consequences from one or more propositions, we are led back to some proposition already proved. In fact, the chief characteristic of geometrical theorems, and one indubitable proof of their certainty is, that, however we combine them together, provided only our reasoning be correct, the results we obtain always agree with each other. The case would be different, if any proposition were false or only approximately true: it would frequently happen that on combining the propositions together, the error would increase and become perceptible. Examples in which the conclusions do not agree with each other, are to be seen in all the demonstrations, in which the reductio ad absurdum is employed. In such demonstrations, if the hypothesis is untrue, a train of accurate reasoning leads to a manifest absurdity that is, to a conclusion in contradiction to a principle previously established: and from this we conclude that the hypothesis is false. B D Cor. If from the point A, in the circumference of a circle, two chords. BA, AC, be drawn to the extremities of a diameter BC, the triangle BAC will be right-angled at A (B. III., P. 18, c. 2); hence, first, the perpendicular AD is a mean proportional between the two segments BD, DC, of the diameter, hence, ADBDX DC. Furthermore, by the proposition, the chord BA is a mean proportional between the diameter BC, and the adjacent segment BD, that is, BABCXBD, and ACBCXCD. BC× Two triangles having an angle in each equal, are to each other as the rectangles of the adjacent sides. Let ABC, ADE, be two triangles having the equal angles A, placed, the one. on the other; then the triangle ABC ADE :: ABXAC : ADX AE. Draw BE. Then, the triangles ABE, ADE, having the common vertex E, are to each other as their bases (P. 6, c.) that is, A D E B In like manner, since B is a common vertex, the triangle BAC BAE: AC: Multiply together the corresponding terms of these proportions, omitting the common term ABE; and we have (B. II., P. 13), BAC : DAE BAXAC : Cor. If the two triangles are equiva lent, we have, BAX AC=DAXAE: DA : AE : AC: hence (B. II., P. 2), BA D A consequently, DC and BE are parallel B (P. 16). E PROPOSITION XXV. THEOREM. Similar triangles are to each other as the squares described on their homologous sides. Let ABC, DEF, be two similar triangles, having the angle A equal to D, and the angle B=E: then will the triangle BAC be to the triangle EDF, as a square described on any side of BAC to a square described on the homologous side of EDF First, by reason of the equal an gles A and D, we have (P. 24), BAC: DEF :: BAXAC : DEXDF. Also, because the triangles are similar (P. 18), AB : DE :: AC: DF, And multiplying the terms of this proportion by the corresponding terms of the identical proportion, AC: DF: AC: DF, there will result :: Therefore, the similar triangles BAC, EDF, are to each other as the squares described on their homologous sides AC, DF, or as the squares described on any other two homologous sides. PROPOSITION XXVI. THEOREM. Two similar polygons may be divided into the same number of triangles, similar each to each, and similarly placed. B Let AEDCB, FKIHG, be two similar polygons. From any angle A, in the polygon AEDOB, draw diagonals, AD, AC. From the homologous angle F, in the other polygon, draw the diagonals FI, FH, to the other angles. A E D K H The polygons being similar, the homologous angles, ABC, FGH, are equal, and the sides AB, BC, proportional to FG, GH, that is, AB : FG :: BC : GH (D. 1). Wherefore, the triangles ABC, FGH, have an angle in each equal, and the adjacent sides proportional: hence, they are similar (P. 20); consequently, the angle BCA is equal to GHF. Taking away these equal angles from the equal angles BCD, GHI, and there remains ACD=FHI But since the triangles ABC, FGH, are similar, we have AC : FH :: BC: GH; and since the polygons are similar, BC : : GH :: CD: HI; FH :: CD : HI. hence, The angle ACD, we already know, is equal to FHI; hence, the triangles ACD, FHI, are similar (P.20). In the same manner, it may be shown that all the remaining triangles are similar, whatever be the number of sides in the polygons proposed: therefore, two similar polygons may be divided into the same number of triangles, similar, and similarly placed. Scholium. The converse of the proposition is equally true: If two polygons are composed of the same number of triangles similar and similarly situated, the two polygons are similar. For, the similarity of the respective triangles will give the angles, ABC=FGH, BOA=GHF, ACD=FHI: hence, BCD=GHI, likewise, CDE=HIK, &c. Moreover, we have, AB: FG :: BC: GH :: CD : HI :: DE: IK, &c.; hence, the two polygons have their angles equal and their sides proportional; consequently, they are similar. PROPOSITION XXVII. THEOREM. The perimeters of similar polygons are to each other as their homologous sides: and the polygons are to each other as the squares described on these sides. Let AEDCB and FKIHG, be two similar polygons: then will per. AEDCB : per. FKIHG: AE FK. : First. Since the fig CD, &c., which makes up the perimenter of the first polygon, is to the sum of the consequents FG+GH+HI, &c., which makes up the perimeter of the second polygon, as any one antecedent is to its consequent (B. II., P. 10); that is, as AB to FG, or as any other two homologous sides. Secondly. Since the triangles ABC, FGH, are similar, we have (P. 25), ABC : FGH :: AC2 FH2; and from the similar triangles ACD, FHI, ᎪᏛ ACD : FHI :: AC2 : FH2; therefore, by reason of the common ratio, AC2 to FH2, we have (B. II., P. 4, c.) ABC FGH :: ACD FHI. By the same reasoning, we should find ACD : FHI :: ADE : FIK; and so on, if there were more triangles. And from this series of equal ratios, we conclude that the sum of the antecedents ABC+ACD+ADE, which make up the polygon AEDCB, is to the sum of the consequents FGH+ FHI+FIK, which make up the polygon FKIHG, as one antecedent ABC, is to its consequent FGH (B. II., P. 10), or as AB2 is to FG3 (P. 25); hence, similar polygons are to each other as the squares described on their homologous sides. : Cor. If three similar figures are described on the three sides of a right-angled triangle, the figure on the hypothenuse is equivalent to the sum of the other two. Let A, B, C, denote three similar figures described on the hypothenuse and sides of a right-angled triangle, and a, b, c, the corresponding squares; then, and, A : B+C :: a : b+c (B. II., P. 9) : but, hence, a is equivalent to b+c (P. 11); A is equivalent B+C. |