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a byy. and

lem. 54. 10. b hyp.

c fch. 12. 10.

d 20. 10.

e lem. 54.10.

f37.10.

a hyp. and lem. 54. 10. b byp.

cfch. 12. 10.

d 22. 10.

e lem. 54.10.

f hyp. 12.

10.

g 20. 10.

h 38. 10.

PROP. LV. Plate IV. Fig. 50.

If a space AD be contained under arational-line AB, and a first binomial-line AC (AE- EC) the right-line OP which containeth that space in power is irrational, and called a binomial-line.

All that being supposed which is described and demonstrated in the foregoing Lemma, it is manifest that the right-line OP containeth in power the space AD. (a) Likewife AG, GE, AE, are L. therefore feeing AE, (6) is LAB, (c) shall alfo AG and GE bef ΑΒ. (d) Therefore the rectangles AH, GI, that is, the squares LM, MN, are fa. therefore OM, MP are (e), (f) and conseqently OP is a binomial. Which was to be demonstrated.

In numbers, let there be AB5. AC4 -- 12. wherefore the rectangle AD = 20 + √ 300 = to the square LN. Therefore OP is 15 + ✓ 5. namely a fixth

binomial.

PROP. LVI. Fig. 50.

If a space AD be comprehended under a rational-line AB, and a fecond binomial AC (AE + EC) the right-line OP, which containeth that space AD in power, is irrational, and called a first medial-line.

The foresaid Lemma of the 54 of this Book being again supposed, then shall OP be = v AD. (a) alfo AE, AG, GE, are. therefore since AE (6) is AB, likewife AG, GE (c) shall be AB, therefore the rectangles AH, GI, i. e. OMq, MPq. (d) are μα, (e) Moreover OM MP. Lastly, EFTEC, and EC (f) AB. (8) Wherefore EK, i. e. SM, or OMP, is v, (b) Confequently OP is a first bimedial. Which was

to be dem.

In numbers, let there be AB 5, and AC√ 48: +6. then the rectangle AD = √ : 1200+30= OPq. therefore OP is v✓ 675+√ 75, viz. a first bimedial.

PROP. LVII. Fig. 50.

If a fpace AD be contained under a rational-line AB, and a third binomial-line AC (AE + EC) the right-line OP which containeth in power the space AD, is irrational, and called a Second bimedial-line.

As

As above, OPq=AD. Also the rectangles AH, GL, that is, OMq, MPq area (a) Likewise EK, or OMP is pr. (5) Therefore OP is a second bimedial.

In numbers, let there be AB 5. AC√ 32+√24. wherefore AD is ✓ 800 + √ 600 = OPq. and fo OP is υ√ 450+ v✓ 50, that is, a second bimedial.

PROP. LVIII. Plate IV. Fig. 50.

If a space AD be comprehended under a rational-line AB, and a fourth binomial AC (AE + EC) the right-line OP containing the space AD in power, is that irrational-line which is called a Major-line.

For again, OMq (a) MPq; and the rectangle AI, i. e. OMq+MPq (6) is pr. (c) also EK or OMP is μν. (d) therefore OP (✓ AD) is a Major line. Which was to be demonstrated.

In numbers, let there be AB 5. and AC 4+ √8. then the rectangle AD is 20 + √ 200. wherefore OP is √ : 20 + √ 200.

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If a space AD be contained under a rational-line AB, and a fifth binomial AC, the right-line OP which containeth the Space AD in power, is that irrational-line, which is a line containing a rational and a medial rectangle in power.

Again OMP MPq. and the rectangle AI or OMq + MPq is μν. (α) Likewise the rectangle EK or OMP is fv. (b) therefore OP (✓ AD) contains in power fv and μ'. Which was to be demonstrated.

In numbers, let there be AB 5, and AC 2 + √ 8; then the rectangle AD = 10 + √ 200 = OPq. Wherefore OP is √ 10 + √ 200.

PROP. LX.

If a space AD be contained under a rational-line AB and a fixth binomial AC (AE + EC) the line OP containing the Space AD in power is irrational, which containeth in power two medial-rectangles.

As often before, OMqMPq and OMq + MPq is ur. and also the rectangle (EK) OMP is ur. (a) therefore OP = AD contains in power 2 μα. Which was to be demonftrated.

In numbers, let there be AB 5, AC

12 +18,

300+208

therefore the rectangle AD or OPq is 300 + 200

and fo OP is ✓ : √300+√ 200,

Lemma

a hyp. and

22.10.

b 39. 10.

a lem. 54.

10.

b hyp. and

20.10.

c hyp. ard

22.10.

d 40. 10.

a as in the

prec.

b 41. 10.

a 42. 10.

Lemma. Plate IV. Fig. 51.

Let a right-line AB be unequally divided in C, and let AC be the greater segment, and upon fome line DE apply the rectangles DF = ABq, and DH=ACq, and IK == CBq, and let LG, be divided equally in M, and also MN drawn parallel to GF.

I say, 1. The rectangle ACB is LN or MF. (a)

a 4. 2. and

3. ax. 1.

For 2 ACB = LF.

b 7. 2.

2. DL

C 1.6.

d 16. 10.

e lem, 26.

jo.

f 10. 10.

g 1. 6.

h17.6.

k byp.

110.10.

m 18.10.

n 19. 10.

a byp. blem. 60.

LG. for DK (ACq+CBq) (6)

LF (2

ACB) therefore fince DK, LF are of equal altitude, (c)
DL shall be

LG.

3. If ACCB, (d) then shall the rectangle DK be ACq and CBq.

4. Alo DLLG. For ACq+CBq (e) 2 ACB, i. e. DKLF, but DK: LF (e) :: DL: LG, (f) therefore DLLG.

5. Moreover DL/DLq-LGq. For ACq: ACB (g):: ACB: CBq. that is DH: LN :: IK, (c) LN : wherefore DL:LM::LM: IL; (b) therefore DI x IL =LMq. therefore feeing ACq (4) CBq, that is, DHL IK, and (1) fo DIII, (m) shall DL be TL / DLqLGq. Whith was to be demonftrated.

6. But if ACq be put CBq, (n) then shall DL be
DLq-LGq.

This Lemma is preparatory to the fix following Propafitions.

PROP. LXI.

The Square of a binomial-line (AC+CB) applied unto a rational-line DE, makes the latitude DG a first binomiailine.

Those things being supposed, which are described and demonftrated in the preceding Lemma, because AC, CB (a) are, (b) the rectangle DK shall be L cfch. 12. 10. ACq; (c) and to DK is pr; (d) therefore DLL DE d 21.10.

10.

But the rectangle ACB, and fo 2 ACB (LF) (e) is μν. (f) therefore the latitude LG is DE, (g) therefore alfo DLLG also DLL DLq - LGq. From whence (k) it follows that DG is a first binomial. Which was to be demonstrated.

e 22. and

24.10.

f 23. 10.

g 13, 10. h lem. 60.

10.

k I def. 48.

10.

PROP.

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The square of a first bimedial-line (AC+CB) being applied to a rational-line DE, makes the latitude DG a fecond binomial-line.

The aforesaid Lemma being again supposed; The rectangle DK TL ACq; (a) therefore DK is μν; (6) therefore the latitude DL is DE. But because the rectangle ACB, and so LF (2 ACB) (c) is pv, (d, shall LG beL DE; (e) therefore DL, LG are 1; (f) alfo DL DLq - LGq; (g) from whence it is clear that DG is a second binomial. Which was to be demonstrated,

PROP. LXIII.

The Square of a second bimedial-line (AC+CB) applied to a rational-line DE makes the breadth LG a third binomial-line.

As in the prec. DL is DE. Furthermore because the rectangle ACB, and fo LF (2 ACB) (a) is μν; (6) therefore shall LG bef DE. (c) Moreover DL LG. and alfo DL DLq-LGq; (d) therefore DG

is a third binomial, Which was to be demonstrated.

PROP. LXIV.

The Square of a Major-line (AC+CB) applied to a rational-line DE, makes the breadth DG a fourth binomialline.

a 24. 10. b 23.10. c byp. and Sch. 22. 10. d 21. 10.

e 13. 10. f lem. 60.

10.

g2. def.48.

10.

a byp. and
24.10.
b 23. 10.
c lem. 60.

10.

d 3. def. 48.

10.

Again ACq + CBq. i. e. DK (a) is pv, (b) therefore a hyp. and

fo

DL is DE, also ACB, and so LF (2 ACB) (c) is

μν; (d) therefore LG is
DE, (e) and confequently
DLLG. Lastly, because ACBC, (f) fhall DL
be DLq- LGq, (g) whence DG is a fourth bino-
mial. Which was to be demonstrated.

PROP. LXV.

The Square of a line containing in power a rational and a medial rectangle (AC + CB) applied to a rational-line DE makes the latitude DG a fifth binomial.

DE; (c) therefore

fch. 12. 10.
b 21. 10.

c hyp. and
24. 10.
d 23. 10.

e 13. 10.
f lem. 60.

10.

g4.def.48.

10.

Again, DK is μν; (a) therefore DL is

DE; also

LF is ov; (6) therefore LG is

a 23. 10. b 21.10. C13.10. d lem. 60.

10.

e 5. def.48.

PROP.

10.

DLG; (d) likewise DLDLq - LGq; (e) and fo by consequence DG is a fifth binomial. Which was to be demonstrated.

a hyp.
b 14. 10,
CI. 6.

d 10. 10.

e lem. 60.

PROP. LXVI.

The square of a line containing in power tawo mediul rectangles (AB+CB) applied to a rational line DE, makes the latitude DG a fixth binomial-line.

As before, DL and LG are DE. But because ACq+CBq (1)K) (a) TL ACB, (6) and fo DKLF (2 ACB) and alfo DK: LF (c) :: DL: LG. (d) therefore shall DL be LG. (e) Lastiy DL/DLq - LGq; (f) by which it appears that DG is a fixth binomial.

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a 19.5.

b 1.6.

c before

d 10. 10.

e zz. 6.

f 10. 10.

g 10. 10.

Let AB, DE be . and make AB: DE::AC: DF.
I fay 1. AC DF, as appears by 10. 10. alfo CB

FE; (a) because AB: DE:: CB: FE.

2. AC:CB:: DF: FE. For AC: DF: : AB: DE ::CB: FE; therefore, by permutation, AC:CB:: DF: FE.

3. The Rectangle ACS DFE. For ACq: ACB (6) : : AC: CB(c): : DF: EF: : DFq: DFE, wherefore by permutation ACq: DFq:: ACB: DFE, therefore fince ACqDFq, (d) shall ACB be DFE.

4. ACq + CBqDFq + FEq, For because ACq : CBq. (e):: DFq: FEq; therefore by compounding ACq--CBq: CBq:: DFq+FEq: FEq, therefore since CBqFEq, (f) shall alfo ACq + CBq be DFq +FEq.

5. Hence, If AC or CB, (g) then likewise shall DE be or EF.

PROP. LXVII.

A line DE, commenfu

rable in length to a binomial
line (AC+CB) is it felf a

ACB

D FE

a lem. 66. binomial-line, and of the same order.

10.

b byp.

c lem. 66.

10. and

fch. 12. 10. d 15. 10,

Make AB: DE:: AC: DF; (a) then are AC, DF L (a) and CB, FEL; whence fince AC and CB, (6) are, thence, DF, FE; therefore DF is a binomial. But because AC: CB (a):: DE: FE. If ACL or ACq - BCq, (d) then in like manner

DE

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