For, if we join DF, it will be parallel to AC (B. IV, Prop. 11), and the angles KFD and KAC will be equal, and also the angles KDF and KCA will be equal (B. I, Prop. xvii). Moreover the angle DKF will be equal to AKC (B. I, Prop. 11); and therefore the two triangles DKF and AKC, being equiangular, are similar (B. IV, Prop. vi). These triangles being similar, we have DF: AC :: DK : KC; DF AC:: FK : KA. But since DF is drawn bisecting the two sides BA, BC, it is equal to half of AC; therefore we have DK | KC, FKKA, which proves that AF and CD trisect each other. In the same way it may be shown that AF and BG, as well as CD and BG, trisect each other, and consequently the three lines mutually trisect each other. Cor. This proposition shows that the three lines drawn from the vertices of any triangle, so as to bisect the opposite sides, will all pass through the same point. (73.) THEOREM. If lines be drawn from the vertices of a triangle, bisecting the opposite sides, these lines will bisect the triangle itself. Since CD, drawn from the vertex C, bisects the base AB, it will of necessity D B The point K in which all these lines meet, is obviously the centre of gravity of the triangle. Hence the centre of gravity of a triangle may be found by drawing a line from any vertex to the middle of the opposite side, and then taking two-thirds of this line from the vertex. PROPOSITION XV. THEOREM. Two similar polygons are composed of the same number of similar triangles, similarly situated. Let ABCDF, GHKLM be two similar polygons. From any angle A in the polygon ABCDF, draw the diagonals AC, AD; and from the angle G in the other polygon, homologous with A, draw the diagonals GK, GL. These polygons being similar, the angles ABC, GHK, which are homologous, are equal, and the sides AB, BC must also be proportional to the sides GH, HK (B. IV, Def. 3). Therefore the two triangles ABC, GHK have an angle of the one equal to an angle of the other, and the sides about those angles proportional, and consequently these triangles are similar; and being similar, we have the angle BCA equal to the angle HKG. But since the polygons are similar, the angle BCD is equal to the angle HKL; therefore ACD, which is the difference between the angles BCD and BCA, is equal to the angle GKL, which is the difference between the angles HKL and HKG (Ax. III). Since the triangles ABC and GHK are similar, we have BC HK: AC: GK; and since the polygons are similar, we have Therefore, by equality of ratios, we have Hence the two triangles ACD and GKL have an angle of the one equal to an angle of the other, and the sides about those angles proportional, and consequently the triangles are similar. In the same manner it might be shown that all the remaining triangles are similar, whatever be the number of sides of the proposed polygons. Therefore two similar polygons are composed of the same number of similar triangles, similarly situated. Scholium. The converse of this proposition is also true : If two polygons are composed of the same number of similar triangles similarly situated, those polygons will be similar. For, the similarity of the respective triangles will give the angles ABC=GHK, BCA=HKG, ACD=GKL, &c. Hence BCD=HKL; likewise CDF KLM, &c. Moreover we shall have AB: GH:: BC: HK :: AC : GK :: CD : KL, &c. Hence the two polygons have their angles equal and their sides proportional, and consequently they are similar. PROPOSITION XVI. THEOREM. The perimeters of similar polygons are to each other as their homologous sides; and their areas are to each other as the squares of those sides. First. By the nature of similar polygons, we have Now the sum of these antecedents AB+BC+CD, &c., which makes the perimeter of the first polygon, is to the sum of their consequents GH + HK + KL, &c., which makes the perimeter of the second polygon, as any one antecedent is to its corresponding consequent, and therefore as AB is to GH. Secondly. Since the triangles ABC, GHK are similar, we have the triangle ABC GHK :: AC: GK (B. IV, Prop. xii); also since the triangles ACD and GKL are similar, we have triangle ACD : GKL :: AC2 : GK2; therefore, by reason of the common ratio AC2 : GK2, we have ABC GHK :: ACD: GKL. : By the same mode of reasoning, we should find ACD: GKL :: ADF : GLM; and so on, if there were more triangles. From this series of equal ratios, we conclude that the sum of the antecedents ABC + ACD + ADF, or the polygon ABCDF, is to the sum of the consequents GHK + GKL + GLM, or the polygon GHKLM, as any one antecedent ABC, is to its corresponding consequent GHK, or as AB is to GH'. Hence the areas of similar polygons are to each other as the squares of their homologous sides. Cor. If three similar rectilineal figures are constructed on the three sides of a right-angled triangle, the figure on the hypothenuse will be equivalent to the sum of the other two; for the three figures are to each other as the squares of their homologous sides, and the square of the hypothenuse is equivalent to the sum of the squares of the other two sides. (74.) PROBLEM. Through a given point P, to draw a straight line PED to cut two straight lines AB, AC given in position, so that the triangle ADE thus formed may be of a given magnitude. |