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Lemma. Plate V. Fig. 4, 5.

Let AC be a rectangle contained under the right-lines AB, AD. Let AD be drawn forth to E, and DE equally divided in F; and let the rectangle AGE be = FEq, and the rectangles AI, DK, FH, finished. Then let the square LM AH be made, and the square NO = GI; and the lines NSR, OST, produced.

I say, 1. The rectangle AI = LM + NO = TOq+ SOq, which appears by the constr.

2. The rectangle DK LO. For because the rectangle AGE (a) = FEq (6) thence are AG, FE, GE(c) and fo AH, FIGI (a) that is, LM, FI, NO; but LM, LO, NO (d) are; therefore FI = (e) LO (f) = DK = (g) NM.

3.

LO. Hence, ACAI - DK – FI = LM + NO

-

4. It is manifest that DF, FE, DE, are I.

5. If AE. DE, and AE TL

then shall AG, GE, AE be .

AEq - DEQ, (k)

6. Also, because AE (1) DE, (m) thence shall AE, FE, be; (n) and fo AI, FI, that is, LM + NO and LO

are.

7. Becaufe AG*GE, (n) Shall AH, GI, that is, LM, NO be.

8. But beeaufe AE (1) DE, (0) therefore shall FE,GE be, (n) and so the rectangle FIGI, that is, LO NO; wherefore seeing LO: NO (p): : TS: SO; (q) therefore shall TS, SO be.

9. If AE be put

GE, AE be.

AEq - DEq, then shall AG,

IC. (S) Wherefore the rectangles AH, GI, that is, Toq, SOq shall be

PROP. XCII. Fig. 4, 5.

If a Space AC be contained under a rational-line AB, and a first residual-line AD (AE-DE) the right-line TS, which containeth the Space AC in power, is a refidual-line.

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Use the foregoing Lemma for a preparation to the demonstration of this prop. Therefore TS = √ AC. Also AG, GE, AF, are ; therefore since AEL (a) AB. (6) alfo AG and GE shall be LAB, (c) therefore the rectangles AH and GI, that is, TOq and SOq are pz. (d) Likewise TO, SO, are (e) and confequently TS is a refidual-line. Which was to be demonstrated.

PROP. XCIII. Plate V. Fig. 4, 5.

If a space AC be contained under a rational line AB, and a second refidual AD (AE-DE) the right-line TS, containing the space AC in power, is a first medial refidual-line.

Again, by the foregoing Lemma, AG, GE, AE are

; therefore (2) fince AE is AB, (6) alfo AG, GE, shall be AB, (c) therefore the rectangles AH, GI, that is, TOq, SOq are μα; (d) likewise TO SO. Lastly, because DE (e) AB, (f) the right-angle D', and the half thereof DK or LO, that is, TOS shall be ῥv; (g) from whence it follows that TS ( AC) is a firit medial refidual. Which was to be domonstrated.

PROP. XCIV. Fig. 4, 5.

If a space AC be contained under a rational-line AB ant a third refidual AD (AE_DE) the right-line TS containing in power the space AC is a second medial refidual-linc.

As in the former, TO and SO are μ. Therefore because DE (a) is AB, (b) the rectangle DI, (c) and so DK, or TOS, shall be ur; therefore TS = √ AC is a second medial refidual. Which was to be demonstrated,

PROP. XCV. Fig. 4, 5.

If a space AC be contained under a rational-line AB and a fourth refidual AD (AE-DE) the right-line TS containing the Space AC in power, is a Minor-line.

As before, TO (2) SO. Therefore because AE (b) is ὁ τι AB, c) shall AI (TOq+SOq) be fr, but, as before, the rectangle TOS is μν; (d) therefore TS = AC is a Minor-line. Which was to be demonstrated.

PROP.

PROP. XCVI. Plate V. Fig. 4 5.

If a space AChe contained under a rational-line AB and a fifth refidual AD (AE -DE) the right-line TS containing in power the space AC, is a line which maketh with a rational space the whole space medial.

For again TOSO. Therefore since AE (a) is ῥ p A B (6) alfo Al, that is, TOq + SOq shall be μ. But, as in 93. the rectangle TOS is pr; (c) whence TS = √ AC is a line which with fy makes a whole 19. Which was to be demonftrated.

PROP. XCVII. Fig. 4, 5.

If a space AB contained under a rational-line AB, and a fixth refidual AD (AE-DE) the right-line TS containing in power the space AC is a line waling with a medial rectangle, a whole space mediol.

As often above TO SO. Also, as in 96. TOq+ SOq is pr, but the rectangle TOS is pv, as in 94. (a) Laftly, TOq-+50qTOS (6) therefore, TS = AC is a line which with u. makes a whole uy. Which was to be dem.

Lemma. Fig. 6.

Upon a right-line DE * apply the rectangles DF = = ABq, and DH=ACq, and IK=BCq. and let GL be bifelted in M, and the line Me drawn parallel to GF.

Then 1. The rectangle DK is = ACq-- BCq. as the construction manifests.

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2. The rectangle ACE=GN or MK. For DK (a) = ACq+BCq (b) = 2 ACB + ABq; but ABq (a) = DF. b 7.2. therefore GK (c) = 2 ACB; and confequently GN or

MK = ACB.

3. The rectangle DIL = MLq. For because ACq:

C 3. ax. 1. d 7. ax. 1.

e

ACB (e):: ACB: BCq, that is, DH: MK:: MK: £ 17.6.

IK. (e thence is DI: ML:: ML: IL (f) therefore DIL =MLq.

4- If AC be taken BC, then DK shall be For ACq | BCq (DK) (5) ACq.

ACq.

5. Likewise DL DLq - GLq. For because DH (ACq) IK (BCq) (16) thence shall Di beIL; (k) therefore DLq - GLqDL.

6. Alo DL GL. For ACq+BCq(1) 2 ACB. That is DGK; (m) therefore DLGL.

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PROP. XCVIII. Plate V. Fig. 6.

The Square of a refidual-line AB (AC-BC) applyed to a rational-line DE, makes the breadth DG a first residual-line. Do as is enjoined in the Lemma next preceding. Then because AC, BC, (a) are 7; (6) alfo DK (ACq+ BCq) shall be ACq. (c) Therefore DK is ῥν; (d) wherefore DL is DE. (e) Likewise the rectangle GK (2 ACB) is pr; (f) therefore GL is DE, (g) and consequently DL TL GL. (6) But DLqTL GLq, (k therefore DG is a refidual, (1) and that of the first order (because (m) AC BC, and therefore DLI √ DLq-GLq.) W'hich was to be demonftrated.

PROP. XCIX. Fig. 6.

The Square of a first medial refidual-line AB (AC-BC) applied to a rational-line DE, makes the breadth DG a fecond refidual-line.

Supposing the foregoing Lemma; because AC and BC
(a) are μ.τ, (b) thence shall DK (ACq+BCq) be
ACq; (c) wherefore DK is μν; (d) therefore DL is
DE; (e) alfo GK (2 ACB) is ; (f) therefore GL is f

E; (g) wherefore DL GL. (6) But DLq
GLq; (4) therefore DG is a refidual-line. And because
DL is DLq - GLq, (m) therefore shall DG be
a second refidual. Which was to be demonstrated.

PROP. C. Fig. 6.

The square of a fecond medial ref.dual-line AB (AC-BC) applied to a rational-line DE, makes the breadth DG a third refidual-line.

Again DK is μν. (a) wherefore DL is ; DE; also GK is pr, (a) whence GL is DE; (6) likewife DK GK; (c) wherefore DLGL; (d) but DLq TGLq. (e) therefore DG is a refidual-line, and that of (f) the third order, (g) because DL

GLq. Which was to be demonstrated.

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PROP. CI. Plate V. Fig. 6.

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The Square of a Minor-line AB (AC-BC) applied to a rational-line DE, makes the breadth DG a fourth refidual. As before, ACq+BCq, that is DK, is pv. (a) therefore DL is DE. but the rectangle ACB, and fo GK (2 ACB) * is μν. (6) wherefore GL is TL DE. (c) b 23.10. therefore DL GL. (d) but DLqGLq. and be- C 13.10. cause * ACqBCq, (e) thence shall DL bedsch. 12. DLq-GLq. (f) therefore DG has the conditions required to a fourth residual, Which was to be demonPrated.

PROP. CII. Fig. 6.

The Square of a line AB (AC-BC) which makes with à rational space the whole space medial, applied to a rationalline DE, makes the breadth DG a fifth refidual-line.

For, as above, DK is ur. (a) wherefore DL is p DE also GK is pv. (6) whence GL is DE. (e) therefore DLGL. (d) but DLqGLq. Moreover DL (e) DLq-GLq. wherefore DG (f) is a fifth refidual. Which was to be demonstrated,

PROP. CIII. Fig, 6.

The square of a line AB (AC - BC) making with a medial space the whole space medial, applied to a rationalline DE, makes the breadth DG a fixth refiduai-line.

As above, DK and GK are μα; (a) wherefore DL and GL are . DE. alfo DK (1) GK. (c) whence DGL. (d) therefore DG is a refidual. (6) And whereas ACqBCq. and fo DL (e) therefore DG shall be a fixth refidual. Which was (e) to be demonstrated,

PROP. CIV.

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