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UPON

PROPOSITION XVIII. PROBLEM VI.

PON a given straight line (AD) to describe a re@ilineal figure (M) similar, and similarly situated to a given re&ilineal figure (N). Giyen,

Sought. 1. The straight line A D.

The rectilineal figure M fimilar 11. The rectilineal figure N,

to a redilineal figure N fimi

larly situated.

Resolution. 1. Join HF.

Pof.1. B... 2. At the points A & D in AD, make VA= VE & Vm=

V n, wherefore the remaining V ABD will be = to P. 23. 32.

the remaining VEFH.
3. At the points D & B in D B make vo=V&V

V r, consequently the remaining VC will be to the Lem. B.I.
remaining G.

DeMONSTRATION,
ECAUSE the A ABD is equiangular to the A EFH, & the
ADB C equiangular to the AHF Ğ (Ref. 2. & 3).
BD: FHBA: FE = AD: E H.

P. B.6. &

BD : FHDC: HG=CB : GF. 2. Consequently, BA :FE= AD: EH-DC: HG= CB: GF.

P.II.B.5. But Vm being = Vn (Ref. 2), & Vo = V p (Ref. 3). 3. The whole v m to is to the whole yn + p.

Ax.2. B.L. 4. Likewise V ABC= VEF G.

Moreover, VAVE (Ref. 2), & VCEVG (Ref. 3).
Wherefore, the rectilineal figure M is equiangular to the rectilineal

figure N, & their fides about the equal y are proportionals.
6. Therefore, the rectilineal figure M described upon the given line AD
is similar to the rectilineal figure EG, & is similarly situated. D. 1. B.6.

Which was to be donc.

Because

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SIMILAR triangles (ABC D E F) are to one another in the dupli

P.16. B.5.

PROPOSITION XIX. THEOREM XIII. IMIL

D cate ratio of their homologous fides (CB, F E or A C, DF, &c). Hypothesis.

Thesis. The triangles ABC, DEF are similar.

The A ABC is to the A DEF in So that' VC= VF, & the sides

the duplicate ratio of CB to FE AC, DF & CB, FE are homologous.

that is as CB2 : FEZ.* Preparation.

SP.11. B.6.
, .

Pos.1.B.1.
DeMoNSTRATION.
EC

FE .
1. Alternando AC: DF=CB: FE.
But

CB : FE=FE:CG (Prep.). 2. Consequently, AC:DF=FE:CG. 3. Therefore, the sides of the AGC, DEF about the equal VC & F

(Hyp.) are reciprocal (D. 2. B. 6).
4. Hence the A AGC is = to the ADEF.

P.15. B.6.
But the A ABC, AGC having the fame altitude.
A ABC:AAGC=CB:CG.

P. 1. B.6. 6. Consequently, the A ABC:ADEF=CB: CG.

P. 7.B.5. But since

CB: FE=FE: CG. (Prep.).
7. CB: CG in the duplicate ratio of CB to F E, or as CB2 : F E2® D.10. B.5.
8. Wherefore, the A ABC:ADEF in the duplicate ratio of C Beo
FE, or as CB2 : F E2*,

P.11.B.5.
Which was to be demonstrated.

COROLLA RY.
ROM this it is manifeft, that if three lines (CB, F E, CG) be proportionals,
as the first is to the third, fo is any upon tbe first to a similar, & fimilarly described
A upon the second.

• Su Cor. 2. of the following propofition.

P.11.B.5

5. The

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PROPOSITION XX. THEOREM XIV. IMILAR polygons (M & N) may be divided by the diagonals (AC, AD; FH,F) into the same number of fimilar triangles (ABC, ACD, ADE, & FGH, FHI, FIK) having the same ratio to one another, that the polygons (M& N) have; and the polygons (M& N) have to one another the duplicate ratio of that which their homologous sides (AB, FG; or BC, GH &c.) have. Hypothesis.

Thesis. The polyg.Mis similar to the polyg.N; 1. Those polygons may be divided into the so that the VA,B,C,&c. are = to the same number of fimilar A. VF,G,H, &c. each to each & the II. Whereof, each to each bas ibe fame ratis fides A B, FG; or BC, GH, &c. which the polygons have. are homologous.

III. And the polyg. M: polyg. N in the dupli

çate ratio of the homologous fodes A B,

FG; or as Ą B? : F Ĝ2.

Preparation.
Draw AC, FH, likewise AD, FI.

pofir. B... DEMONSTRATION. EECAUSE VB=VG & AB : BC=FG:GH (Hyp. & D.1.B.6). 1. The AA B D is equiangular to the AFGH.

P. 6. B.6. 2. Wherefore those are similar, & Vm = Va.

SP. 4.B.6 But the whole Vm tnis = to the whole Vato (Hyp).

Cor. 3. Consequently, Vnis = to Vc,

Ax.3.B.I. Since then by the fimil. of the A ABC & FGH (Arg.2),

AC:BC=FH:GH, D. 1.B 6. & by the simil. of the polyg. M&N, BC:CD=GH:HI. 4. It follows, Ex Æquo, that AC:CD=FH:HI.

P.22. B.5. That is, the sides about the equal n&c are proportionals. s. Therefore the AACD is equiangular to the AFHI.

P. 6. B.6. And consequently is similar to it.

SP. 4. B.6 . 6. For the same reason, all the other A ADE, FIK, &c. are similar. Cor. 7. Therefore, similar polygons may be divided into the same number of . similar A.

Which was to be demonstrated: 1.
See Cor. 2. of this propofition.

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}P.19.B.6

.

Likewise, because the A ABC, F G H are similar (Arg. 2). 6. The

A ABC:AFGH=AC2 : F H2* And the AACD: AFHI = AC2: F H2.* 7. Therefore, the A ABC:AFGH=AACD: AFHI.

P.11. B.5 It

may be demonstrated after the same manner, that 8. The

AADE:AFIK =AACD: AFHI. 9. Wherefore, AABC: AFGH=AACD: AFHI=AADE:AFIK. P.11. B.5. 10. Therefore, coinparing the sum of the anteced. to that of the conseq.

AABC+AACD, &c. : AFGH+AFHI,&c.=QABC:AFGH,&c. P.12. B.5.
That is, the polyg. M : polyg. N= A A B-C : AFGH=
AACD:APHI, &c.

P. 7. B.5
Which was to be demonstrated. u.
Since then the ABC: AFGH=A B2: FG2* (P.19. B.6).
11. The polyg. M : polyg. N = A B2 : F G2*

P.U.B.5 Which was to be demonstrated. uni

I11.

COROLLA R r I. AS

S this Demonstration may be applied to quadrilateral figures, & the same truth has already been proved in triangles (P. 19), it is evident universally, ibat fimilar rectilineal figures are to one another in the duplicate ratio of their homologous fides. Wherefore, if to A B, F G two of ibe homologous fides a third proportional X be taken ; because A B is to X in the duplicate ratio of AB:FG; & that a rectilineal figure M is to another similar rečilineal figure N, in the duplicate ratio of the fame sides AB: FG; it follows, that if three straight lines be proportionals, as the first is to the third, fo is any rectilineal figure described upon the first to a similar & similarly deferibed rectilineal figure upon the second. (P.11. B.5).

* COROLL ART II. ALL

LL Squares being fimilar figures (D. 30. B. 1. & D. 1. B. 6), fimilar rettilineal figures ME N, are to one anorber as the squares of their homologous fides AB,'CD (expressed thus A B2: C D2.) for those figures are in the duplicate ratio of these fame hides.

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PROPOSITION XXI. THEOREM XV.
EC

, )
re&ilineal figure (B), are also similar to one another.
Hypothefis,

Thesis.
The rectilineal figures A & C.

The rectilineal figure A is fimilar
are fimilar to the figure B.

to the rectilineal figure C.

DeMoNSTRATION.
BECAUS

ECAUSE each of the figures A&C is similar to the figure B
(Hyp.).
1. Each of those figures will be also equiangular to the figure B, &

will have the sides about the equal , proportional to the sides of
the figure B.

D. 1. B.6.
2. Consequently, those figures A & C will be also equiangular to one s Ax.1. B.1.

another, and their sides about the equal , will be proportional. {P.11. B.5.
3. Consequently, the figures A & C are similar.

D. 1.B.6.
Which was to be demonstrated.

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