Then, because the angles DAF, FAE are equal to two right angles (I. 13.), and the angle Fae is equal to the angle DAG (I. 15.), the angles DAF, DAG are also equal to two right angles ; and consequently FAG is a right line. And since the parallelogram ab is equal to the parallelogram Ac (by Hyp.), and AK is another parallelogram, AB iş to AK as ac is to AK (V.9.) But AB is to AK as DA to AE (VI. 1.), and AC to AK as AG to AF; whence Da is to AE as AG is to AF (K11.) And, if Fe be joined, it may be shown, in like manner, that the triangle day is to the triangle Afe as the triangle GAE is to the triangle AFE; and DA to se as AG to AF. Again, let the angle day be equal to the angle GAE, and the lide da to the fide AE as the fide ag is to the side AF; then will the parallelogram AB be equal to the paral. lelogram Ac, and the triangle DFA to the triangle GAE. For fince da is tQ AE as AG to AF (by Hyp.), and DĄ to AE as AB to AK (VI. 1.), AG will be to AF as AB to AK (V. 11.) But AG is to AF as AC to AK (VI. 1.); whence AB is to AK as AC to AK (V. 11.); and consequently the parallelogram AB is equal to the parallelogram Aç (V. 10.) And since triangles are the halves of parallelogramas, which have the same base and altitude, the triangle DFA will be equal to the triangle GaĖ: Q. E. D. COROLL. The sides of equal rectangles are reciprocally proportional ; and if the sides are reciprocally proportional, the rectangles will be equal. PRO P. XV. PROBLEM. Upon a given right line to describe a rectilineal figure, similar, and similarly situated, to a given rectilineal figure. Let AB be the given right line, and CDEFG the given rectilineal figure; it is required to describe a rectilineal figure upon AB, which shall be fimilar, and funilarly fitu. ated to CDEFG. Join DG, DF; and at the points A, B, make the angles BAL, ABL equal to the angles DCG, CDG (I. 20.) In like manner, at the points B, l, make the angles BLK, Lek equal to the angles DGF, GDF; and BKH, KBH equal to DFE, FDE. Then, because two angles in one triangle ar equal to two angles in another, each to each, the remaining angles in each of the corresponding triangles, will also be equal (1. 28. Cor.) And since the angles ALB, BLK, are equal to the angles CGD, DGF, and LKB, BKH TO GFD, DFE, the angle ALK will be equal to the angle CGF, and the angle LKH to the angle GFE. And, in the same manner, it may be shewn that the angles KHB, HBA and BAL are equal to the angles FED, EDC and DCG. The figures ABHKL and CDEFG are, therefore, equiangular : and they have their sides about the equal angles, also, proportional. For, since the triangles ALB, CGD are equiangular, AL will be to LB as CG to GD (VI. 5.) ; or AL to cg as LB to ' GD (V. 15.) And, in like manner, 'LK will be to LB as GF to GD (VI. -5.); or LK to GF as LB to GD (V. 15.) But ratios which are the same to the same ratio are the same to each other (V. 11.); whence Al will be to co as'lk to GF;'or AL to LK as cg to GF (V. 15:) And, in the fame manner, it may be fhewn, that the fides about the angles K, H, B, A, are proportional to the sides about the angles F, E, D, C. The figure ABHKL is, therefore, similar, and similarly situated with the figure cbeFG (VI. Def. 1.); and it is described upon the right line AB, as was to be done. PRO P. XVI. THEOREM, Equiangular, or Ginilar triangles, are to each other as the squares of their homologous fides. Let ABC, DEF be two similar triangles, of which the fides AB, DE are homologous ; then will the triangle ĄBC be to the triangle der as the square of AB is to the square of DE. For, on AB, de describe the squares Al, DN (II. 1.), and let fall the perpendiculars CG, FH (1. 12.) Then, since the triangles ABC, DEF are similar (by Hyp.), Ac will be to AB as df to De (VI. Def. 1.); or AC to DF as AB to De (V. 15.) And, because the triangles AGC, DHF are equiangular, AC will be to cg as DF to FH (VI. 5.) ; or Aç to DF as CG to Fh (V. 15.) But ratios which are the fame to the same ratio, are the fame to each other (V. 11.); therefore cg is to Fh as AB to de; or cg to AB as FH to de (V. 15.) And since triangles which have the same base, are to each other as their altitudes (VI. 2.), the triangle ABC is to the triangle AKB as cg is to AK, or AB. In the same manner it may be shewn, that the triangle per is to the triangle DME as FH is to DM, or de. But cg has been thewn to be to AB as FH is to DE; therefore the triangle ABC is to the triangle AKB as the triangle der is to the triangle DME (V. 11.) And since the square Al is double the triangle AKB (I. 32.), and the square dn is double the triangle DME, the triangle ABC will be to the triangle der as the squarc Al is to the square DN (V. 13 and 15.) Q. E. D. Similar polygons are to each other as the squares of their homologous fides. Let ABCDE, FGHIK be similar polygons, of which AB, *'G are homologous sides, then will the polygon ABCDE be to the polygon FGHIK as the square of AB is to the fquare of FG. For join the points BE, BD, GK and GI : Then, since the angle a is equal to the angle F, and And if, from the equal angles AED, FKI, there be taken But ed is to ki as ea is to KF (VI. Def. 1. and V. 15.), and E A is to KF as EB to KG (VI. 5. and V. 15.); whence ED will be to ki as EB is to KG (V. 11.) Since, therefore, the angles Bed, GKI are equal to each other, and the sides about them are proportional, the triangles BED, GKI will, also, be equiangular, or similar (VI. 6.) And, in the same manner, it may be sħewn, that the triangles BĊD, GHI are equiangular, or fimilar. 1 |