consequently (V. 4.), the third and fourth terms of the analogy being equal, the first and second must also be equal. PROP. XXII. THEOR. Similar rectilineal figures may be divided into corresponding similar triangles. Let ABCDE and FGHIK be similar rectilineal figures, of which A and F are corresponding points; these figures may be resolved into a like number of triangles respectively similar. For, from the point A in the one figure, draw the straight lines AC, AD, and, from Fin the other, draw FH, FI; the triangles BAC, CAD, and DAE are respectively similar to GFH, HFI, and IFK. Because the polygon ABCDE is similar to FGHIK, the angle ABC is e A C H G EF K T Hence the angle BCA is equal to GHF; and the whole angle BCD being equal to GHI, the remaining angle ACD must be equal to FHI. But BC:AC:: GH: FH, and BC: CD :: GH: HI, consequently (V. 15.) AC: CD :: FH : HI, and the triangles CAD and HFI (VI. 13.) are similar. Whence, the angle CDA being equal to HIF and the angle CDE to HIK, the angle ADE is equal to FIK;and since CD: DA :: HI: IF, and CD: DE:: HI: IK, therefore (V. 15.) DA: DE :: IF : IK, and the triangles DAE and IFK are similar. -The same train of reasoning, it is obvious, would apply to polygons of any number of sides. PROP. XXIII. PROB. On a given straight line, to construct a rectilineal figure similar to a given rectilineal figure. Let FK be a straight line, on which it is required to construct a rectilineal figure similar to the figure ABCDE. Join AC and AD, dividing the given rectilineal figure into its component triangles. From the points F and K draw FI and KI, making the angles KFI and FKI equal to EAD and AED; from F and I draw FH and IH making the angles IFH and FIH equal to DAC and ADC; and lastly from F and H draw FG and HG making the angles HFG and FHG equal to CAB and ACB. The figure FGHIK is similar to ABCDE. For the several triangles KFI, IFH, and HFG, which compose the figure FGHIK, are, by the construction, evidently similar to the triangles EAD, DAC, and CAB, in to which the figure ABCDE was resolved. Whence FK:KI:: AE: ED; also KI: IF :: ED: DA, and 1 IF: IH :: DA: DC, and consequently (V. 16.) KI: IH ::. ED: DC. Again, IH: HF :: DC: CA, and HF: HG:. CA: CB;-and hence (V. 16.) IH : HG : :DC:CB. But HG:GF::CB: BA; and the ratio of GF to FK, being compounded of that of GF to FH, of FH to FI, and of FI to FK, is the same with the ratio of BA to AE, which is compounded of the like ratios of BA to AC, of AC to AD, and AD to AE. Wherefore all the sides about the figure FGHIK are proportional to those about ABCDE; but the several angles of the former, having a like composition, are respectively equal to those of the latter, Whence the figure FGHIK is similar to the given figure. The same reasoning, it is manifest, would extend to polygons of any number of sides. Scholium. The general solution of this problem is derived from the principle, that similar triangles, by their composition, form similar polygons. The mode of construction, however, admits of some variation. For instance, if the straight line FK be parallel to AE, or in the same extension with that homologous side, the several triangles FIK, FHI, and FGH may be more easily constituted in succession, by drawing the straight lines FI and KI, FH and IH, and FG and GH parallel to the corresponding sides in the original figure ABCDE; because (I. 29.) a corresponding equality of angles will be thus produced. But, if FK have no determinate position, the construc tion may be still farther simplified; For, having made AK equal to that base and joined AD and AC, draw KI, IН, and HG parallel to ED, DC, and CB. The figure AKIHG is evidently similar to AEDCB, D B C G KE since its component triangles have the same vertical angles as those of the original figure, and the angles at the bases equal (I. 22.). If the given base FK be parallel to the corresponding side AE of the original figure, a more general construction will result. Join AF, EK, and produce them to meet in O; join OB, OC, and OD, and draw FG, GH, HI, and therefore IK, parallel to AB, BC, CD, and DE: The figure FGHIK thus formed is similar to ABCDE. For the triangles KOF, FOG, GOH, HOI, and IOK are evidently similar to the triangles EOA, AOв, вос, COD, and DOE. But these triangles compose severally the two polygons, when the point O lies within the ori ginal figure; and when that point of concurrence lies without the figure ABCDE, the similar triangles IOK and DOE being taken away from the similar compound polygons FGHIOK and ABCDOE, there remains the figure FGHIK similar to the original one. It farther appears, from these investigations, that a rectilineal figure may have its sides_reduced or enlarged in a given ratio, by assuming any point O and cutting the diverging lines OЕ, ОА, ОВ, ОC, and OD in that ratio; the corresponding points of section being joined, will exhibit the figure required. Of similar figures, the perimeters are proportional to the corresponding sides, and the areas are in the duplicate ratio of those homologous terms. Let ABCDE and FGHIK be similar polygons, which have the corresponding sides AB and FG; the perimeter, or linear boundary, ABCDE is to the perimeter FGHIK, as AB to FG, BC to GH, CD to HI, DE to IK, or EA to KF; but the area of ABCDE, or the contained surface, is to the area of FGHIK, in the duplicate ratio of AB to FG, of BC to GH, of CD to HI, of DE to IK, or of EA to KF. For, by drawing the diagonals AC, AD in the one, and FH, FI in the other, AC: CD :: FH : HI, CD : AD :: HI : FI, and AD: DE:: FI: IK; wherefore, by equality and alternation, AB: FG:: BC:GH :: CD: HI:: DE: IK:: |