Abbildungen der Seite
PDF
EPUB

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

[merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small]

to which the figure ABCDE was resolved. Whence

FK:KI:: AE: ED; also KI : IF :: ED : DA, and

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 construction may be still farther simplified; For, having made AK equal to that base and joined AD and AC, draw KI, IH, and HG parallel to ED, DC, and CB. The figure AKIHG is evidently similar to AEDCB,

B

G

A

K E

D

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, AOB, BOC, COD, and DOE. But these triangles compose severally

[merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][ocr errors]

the two polygons, when the point O lies within the original 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 OE, OA, OB, OC, and OD in that ratio; the corresponding points of section being joined, will exhibit the figure required.

PROP. XXIV. THEOR.

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,

these polygons will be resolved into similar triangles. Whence the several analogies AB: BC:: FG: GH,

C

B.

H

BC: AC:: GH: FH,

AC CD

[blocks in formation]

FH: HI, CD: AD: HI: FI, and

AD: DE:: FI: IK; wherefore, by equality and alternation, AB FG:: BC: GH:: CD: HI:: DE: IK ::

AE: FK, and consequently (V. 19.) as one of the antecedents AB, BC, CD, DE or AE, is to its consequent FG, GH, HI, IK or FK, so is the amount of all those antecedents, or the perimeter ABCDE, to the amount of all the consequents, or the perimeter FGHIK.

Again, the triangle CAB is to the triangle HFG (VI. 21. cor. 1.) in the duplicate ratio of AB to FG,—the triangle DAC is to the triangle IFH in the duplicate ratio of AC to FH, or of AB to FG,—and the triangle EAD is to KFI in the duplicate ratio of AD to FI or of AB to FG; wherefore (V. 19.) the aggregate of the triangles CAB, DAC, and EAD, or the area of the polygon ABCDE, is to the aggregate of the triangles HFG, IFH, and KFI, or the area of the polygon FGHIK, in the duplicate ratio of AB to FG, of BC to GH, of CD to HI, or of DE to IK.

Cor. Hence also the perimeter ABCDE is to the perimeter FGHIK, as any diagonal AD to the corresponding diagonal FI, and the area ABCDE is to the area FGHIK in the duplicate ratio of AD to FI.

PROP. XXV. PROB.

To construct a rectilineal figure that shall be similar to one, and equivalent to another, given rectilineal figure.

Let it be required to describe a rectilineal figure similar to A, and equivalent to B.

On CD, a side of A, describe (II. 8.) the rectangle CDFE, equivalent to that figure, and on DF describe the

« ZurückWeiter »