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PROPOSITION XXIX. PROBLEM IX.

O a given straight line (A B), to apply a parallelogram (A G), equal to a given re&ilineal figure (V), exceeding by a parallelogram (MI), similar to another given (T). Given

Sought. 1. The Braight line AB, & the pgr. T. The constrution of a pgr. AG, applied to II. The rectilineal figure V.

A B, equal to the reailineal figure V, &

baving for excess a pgr. MI, similar to T.

Resolution.
1. Divide A B into two equal parts in E.

P.10. B.I.
2. Upon E B, describe a pgr. È D, fimilar to the pgr. T, &
fimilarly situated

P.18. B.I. 3. Describe a pgr. X (or PS) =V+ED, fimilar & fimi

larly situated to the pgr. T; & consequently fimilar to the
pgr. ED (Ref 2.P 21.B.6); & let the sides RS, FD; RP, FE

be homologous.
4. Since X, (as =V+E D), is > E D; the fide R Sis > FD,

& the fide RPFE, wherefore having produced FD
& Fe, make FNERS & FK=RP; & complete the
pgr.
FKG N, which will be equal & fimilar to the pgr. X. P.31. B.1,

DEMONSTRATION.
HE pgr. K N being equal and fimilar to the pgr. X, which is
itself similar to the pgr. E D (Ref. 3).
1. The

pgr.
KN is similar to the

Pgr.
ED.

P.21. B.6. 2. Wherefore those two pgrs KN, E D are about the same diagonal. P.26. B.6.

Draw this diagonal F B G, & complete the description of the figure.
Since X is = io V +ED; & X = pgr. KN (Ref. 3. & 4).

=.
3.
The
pgr.
KN = V + ED.

Ax.I.B.1, Therefore taking away from both sides the common pgr. E D. 4. The remaining gnomon a b c is = to the rectilineal figure V.

Ax.3.B.1. But because À È = EB (Ref. 1). 5 The pgr. AK = the pgr. E I.

P.36. B.1. 6. Consequently, this pgr. A K is = to the pgr. NB.

P.43. B.1.

THE

Therefore adding to both sides the common pgr. MK 7. There will retult the pgr. AG = to the gnomon a be.

2.2. B.I.
But the gnomon abc is I to thu rectilineal figure V (Arg. 4).
8. Consequently, the pgr. AG is 3 to the reetilineal figure V. Axi.B.I.

Since then this pgr. A G has for excess the pgr. Mi, fimilar to the
pgr. ED (P. 24. B.6.) ; & coalquently similar to the pgr. T

(Ref. 2. P. 21. B. 6).
9. There has been applied to A B, a pgt, AG = to the rectilineal
figure V, having for excess a pgr. M 1, similar to the pgr. T.

Which was to be done.

R E M A R K.

Ir as in the foregoing case A B be made = a, the given square V (reduced to a

pgr. equiangular to the pgr. T)=nl; the ratio of the fides QP, PR of the pgr. X (which is the same as that of the sides of the pgr. T)'m:n; &AM=x, consequently, MB =* a. there will result an equation of the same kind.

For fince the defe& M I jould be fimilar to the pgr. Tor X, we will have as before the following proportion.

QP:PR = MB : MG (D. 1. B. 6).
m : n

f* -a).
And because the pgr. A GEAM. MG) Jould be equal to the given space V
lanl), there results the following equation,

(* a) x=V (P. 23. B. 6).

m

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V = 0.

which is reduced to
And substituting for V its value n h rben multiplying by m. & dividing by n.

ml = 0.
From whence it appears that the XXIXıb Prop. corresponds to the Cafe, in which
be last term of tbe equation is negative.

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Ax.3. B.I.

PROPOSITION XXX. PROBLEM X. O cut a given straight line (A B) in extreme and mean ratio (in E). Given.

Sought. Tbe straight line A B.

The point E, such that

BA: AEE AE:BE.

Resolution.
1. Upon the straight line A B describe a square B C. P.46. B.1.
2. Apply to the side CA, a pgr. C D= to the square B C, P.29. B.6.

whose excess A D is similar to BC, which will conse-
quently be a square.

DEMONSTRATION.
ECAUSE BC=CD (Ref. 2); by taking away the common

rgle. C E from each.
1. The remainder BF = A D.

But B F is also equiangular with AD (P. 15. B. 1). 2. Therefore their fides FoE, E B, ED, A E about the equal angles,

are reciprocally proportional, that is FE:ED=AE: É B.

But FÈ is =CA (P. 34. B. 1), or = to BA, & ED= A E. D.30. B.1. 3. Wherefore, BA: AE = AE : EB.

P.7.811.B.5. But because B A is > AE (Ax. 8. B. 1). 4. The straight line A Eis > EB.

P.14. B.5. 5. Consequently, the straight line A B is cut in extreme & mean ratio in E.

Which was to be done.

Otherwise.
Divide B A in E, so that the rect. A B. BE be = to the of A E P.11. B2.

DEMONSTRATION.
BA. BE is = to the o of A E (Ref.).

BA: AE = AE : BE. And because B A is > AE (Ax. 8 B. 1). 2. The straight line AE is BE.

P.14. B.5: 3. Consequently, the straight line AB is cut in extreme & mean ratio in E. D. 3. B.6.

Which was to be done.

P.14. B.6.

Because

P.17. B.6

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IN

PROPOSITION XXXI. THEOREM XXI. N every right angled triangle (A B C), the re&ilineal figure (E) described upon the hypothenuse (A C) is equal to the sum of the similar and similarly described figures (G & H), upon the sides (A B, BC) containing the right angle. Hypothesis.

Thesis. 1. A B C is a rgle. A in B.

fig. Erfg. G+H II. The fig. E is described upon the hypoth. A C of this A. III. And the figures G&H are similar to E, & fimilarly

described upon the two otber fides AB, BC.

DeMoNSTRATION.

BECAUS

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1.

Cor. 2.

ECAUSE the figures E, G, H are fimilar, & similarly described
upon the homologous fides A C, AB, BC (Hyp. 3).
G : E = 0 of A B : of AC.?

P 20. B.6
And

H:E= of BC:n of A C.)
2. Consequently, G+H: E= of AB+ of BC: of A C. P.24. B.5.

But because the AA B C is rgle. in B (Hyp. 1).
3. The o of AB + of B C is = to the o of A C.

P.47. B.1. 4. Therefore, the figure E is = to the figures G + H.

SP.16. B.5.

Cor.
Which was to be demonstrated.

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PROPOSITION XXXII. Τ Η Ε Ο R Ε Μ ΧΧΙΙ. . F two triangles (A B C, C D E), which have two sides (A B, B C) of the one, proportional to two sides (C D, D E) of the other, be joined at one angle (C), so as to have their homologous fides (A B, CD, BC, D. E) parallel to one another, the remaining fides (A C, CE) shall be in a straight line. Hypothefis.

Thesis. 1. AB :BC=CD:DE.

The remaining fides AC, CE of bofe A II. The A ABC,CDE, are joined in C. are in a straight line A E. III. So that AB is plle.co CD, & BC plle. 10 DE

DemonSTRATION. BECAUSE

ECAUSE the plles. A B C D are cut by the straight line BC,

& the plles. B C, D E by the straight line DC (Hyp. 2). 1. The Ÿ B is = 10 VBCD & VĎ is = to VBCD.

P.29. B.1. 2. Consequently, V B is = to VD.

Ax. 1.B.1. And besides A B : BC=CD:DE (Hyp. 1). 3. The AABC, C D E are equiangular.

P. 6. B.6 4. Therefore, V Ais = to Voice, being opposite to the

homologous fides B C D E.
Adding then to both sides VB, or its = VBCD (Arg.1), together

with the common VBC A. 5. The VA+B+B C A will be = to the VDCE+BCD+BCA. Ax.2. B.1. 6. Consequently the y'DCE + BCD + BC A are also = to 2 L. Ax.1. B.1. 7. Wherefore the straight lines AC, C E are in the same straight line A E.

P.14. B.I. Which was to be demonstrated.

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