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If four Praight lines (AB, CD, E F, GH) le proportionals, the fimilar
re&ilineal figures & fimilarly described upon them (M, N, & P, fall also be proportionals. And if the fimilar re&ilineal figures ( M, N, & P,) similarly described upon four straight lines be proportionals, those straight lines shall be proportional.
Thefas. 1. AB : C D = EF: G H.
M: NEP: 11. The figures M&N described upon AB, CD. also the figures P& Q described upon EF,
GH. are fimilar, & fimilarly ftuated.
ECAUSE AB:CD= E F:GH. S(Hyp. 1).
CD:2 =GH:X. Hyp... Prep.&P.11.B.5).
AB: Z = EF: X.
P.20. B.6. & EF: X =P : Q.
Cor. 1. 3. Wherefore, M :N =P:Q. (Arg. 1).
II. Hypothesis. "1. M:N=P: Q. 11. Those figures are similar & fimilarly described
upon ibe firaight lines AB, CD & EF, GH.
Thesis. ABCD = EF: GH.
P 18. B.6.
ECAUSE AB: CD=EF:KL (Prep. 1), & upon those straight lines have been similarly described the figures M, N, & P, R, similar each to each (Hyp. 2. & Prep. 2).
M: N =P:R (IX. part of this proposition.) But
M:N = P: (Hyp. 1). 2. Consequently, P :R =P :0.
P.11. B.: 3. Wherefore, R=0
P. 9. B.5. Moreover, those figures being similar & fimilarly described upon the ftraight lines G HKL (Prep. 2).
SP.20. B.6. 4.
Q:R=of GH: 0 of K L
Š P.16. B.5. 4. The o of GĦ is = to the 0 of KL.
P.46. B.1. 5. Consequently, GH = KL.
P. 7. B.5-
PROPOSITION XXIII. THEOREM XVII. QUIANGULAR parallelograms (M & N) have to one another the
M ratio which is compounded of the ratios of their sides (AC, CD & EC, CG) about the equal angles. Hypothesis.
Thesis. The pgrs. M & N are equiangular, Pgr. M: Pgr. N=AC. CD:EC. CG so that VACD=VECG.
P.14. B.1. 2. Complete the pgr. P.
Pos.i.B.1. DEMONSTRATION. ECAUSE the pgrs. M, P, N form a series of three magnitudes M: MPEN : N.P.
D. 5. B 5. 2. And alternando M : N = M.P : N.P. 3. Consequently the ratio of the first M to the last N is compounded of the ratios M:P&P:N.
B.6. But since AC:CG= M : P & DC:CE = P : N.
P. i. B.6. 4. The ratio of the sides AC : CG is the same as that of the Pgrs.
M:P; & the ratio of the sides DC:CE, the same as that of the
& P:N (Arg. 1).
AC:CG & CD: EC, of the sides about the equal VACD, ECG. 6. Consequently, M:N = AC.CD: EC.CĠ.
D. 5. B.6.
Which was to be demonstrated. Cor. The same truth is applicable 10 the triangles (A C D, ECG) having an angle (AC D) equal to an angle (ECG): for the diagonals (AD, EG) divide the pgrs. into two equal parts (P. 34.
1. B D is a pgr.
PROPOSITION XXIV. THEOREM XVIII.
HE parallelograms (FH, IG) about the diagonal (AC) of any parallelogram (BD), are similar to the whole, and to one another. Hypothesis.
1. The pgrs. AFEH, EICG are II. FH,IG are pgrs about the
similar to the pgr. ABCD. diagonal AC.
II. And pimilar to one anotber.
P.29. B.1. 2. The AAHE is equiang. to the AADC, in the order of the letters. 3. Therefore the pgr. APEH is also equiangular to the pgr. ABCD, in
the order of the letters.
P. 4. B.6.
P. 4. B.6. 6. Therefore, ex æquo, HE: EF = DC : CB.
And because the V E AH, EFA are common to the
P. 4. B.6. 8. Therefore, ex æquo, HA:AF =DA: A B. 9. Wherefore the pgrs. AFEH, ABCD have their angles equal, each to
each in the order of the letters (Arg. 3) ; & the sides about the
equal angles, proportionals (Arg: 4. 6.8.). 10.Confequently, those pgrs. are similar.
D. 1. B.6. 11. It may be demonstrated after the same manner that the pgrs. EICG, ABC D are fimilar,
Which was to be demonstrated. 1. 12. Consequently, the pgrs. AFEH, EICG are also similar to one another. P.21. B.1 .
Which was to be demonstrated. 11.
PROPOSITION XXV. PROBLEM VII.
O describe a re&ilineal figure (N),which shall be similar to a given re&tilineal figure (L), and equal to another (M). Given.
Sought. I. The refilineal figure L.
The re&il. figure Nfimilar to the reáil. II. The rectilineal figure M.
figure L, &= 1o the redil. figure M
P-45. B.L. 3. Consequently, the sides AC, CI, & GH, H K will be in P.14. 29 a straight line.
& 34. B.1. 4. Between A C, & C I find a mean proportional D F.
Piz. B.6 5. Upon this straight line D F, describe the rectil. figure N, similarly & fimilar to the rectilineal figure L.
P.18. B.6 DEMONSTRATION. ECAUSE the pgrs. AH,CK have the same altitude (Ref.2.803). Pgr. AH : pgr. CK = AC:CI.
P. 1, 8.6 But the pgr. AH=rectil
. L, & the pgr. CK=rectil.M (Ref.1.&2). 2. Consequently, L: M = AC:CI.
AC:DF = DF :CI (Ref: 4.), & upon the straight lines AC, DF have been similarly described the similar
figures L & N, (Ref. 5). 3. Consequently, N = AC: CI.
P.20. B.6 4. Hence, L: NEL : M (Arg. 2).
Cor. . 5. Wherefore, N - M
SP.11, B.< 6. Therefore, there has been described a rectilineal figure N, fimilar P.14.B.
to the rectilineal figure L (Ref 5), & equal to the rectilineal figure M (Arg. 5).
Which was to be done.