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PROPOSITION IX. PROBLEM I. ROM a given straight line (A B) to cut off any part required. (For example the third part). Given.

Sought. The straight line A B.

The abscinded sraight line' A D,

which may be the third part of A B.

Resolution.
1. From the point A draw an indifinite straight line A C, mak-
ing with A B any V B A C.

Pos. 1. B.1. 2. Take in AC three equal parts AE, EF, FC of any length. P. 3. B.1.

Pos.1.B.1. 4. And thro' E, draw ED plle. to C B, which will cut the P. 31. B.1.

straight line A B so that A D will be the third part.

3. Join CB

DEMONSTRATION.

Because

1.

ECAUSE ED is plle. to the fide CB of the ACAB (Prep. 4).
CE:E A = BD:D A.

P. 2. B.6. But

CE: is double of E A (Ref. 2); 2. Consequently, B D is also double of D A. 3. Wherefore, A B triple of A D. 4. And the abscinded straight line A D is the third part of A B.

Which was to be demonstrated.

D. 8. B.5.

Dd

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PROPOSITION X. PROBLEM II.
O divide a given Araight line (A B), fimilarly to a given straight line

A (AC) divided in the points (D, E &c) Given.

Sought. 1. The Araigbt line A B.

To divide A B fimilarly to AC II. The ftraight line AC divided

in the points F & G, so that in the points D, E EC.

AF:FG=AD:DE E tbat

F G:GB=DE:EC.

Resolution.
1. Join the given straight lines A B, AC fo as to contain any

Pos.1. B...
2. Draw C B, & from the points D & E, the straight lines
DF, EG plle. to CB, allo D H plle. to A B.

P.31. B...

V BAC.

DEMONSTRATION

Because of is plte. to the fide e G of the AAGE (Res. 2.

ECA
& P.30. B.1), and KE plle, to the fide HC of the ADHC (Ref. 2).

AF : FG = AD: DE.
And DK : KH - DE: EC.

P. 7. B.6. But the figures KF, HG being pgrms. (Ref. 2. & D. 35. B. 1.). 2. It follows, that F G is = to D K & GDEKH.

P.34. B.1. 3. Therefore, F G:GB = DE: EC.

P.7. & II. B.5. 4. Consequently, the given straight line AB is divided in the points

F&G; fo that AF:FG=AD:DE&FG:GB=DE :EC.

Which was to be done.

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PROPOSITION XI.

PROBLEM III.

To fred a third proportional (CE) to two given Straight lines (AB, AC).

Given.
The two straight lines
AB, AC.

Sought. The straight line CE, a third proportional to the two straight lines A B, A C that is such that AB: AC = AC:CE,

Resolution.
1. Join the two straight lines A B, AC so as to contain any

VBAC.
2. Produce them, & make B D=AC.

P.
3.

B.I.

Pol.1. B.1. 4. And from the extremity D of the straight line A D draw DE plle. to B C.

P.31. B.:

3. Join B C.

DEMONSTRATION.

1.

ECAUSE BC is plle. to D E (Ref. 4).
AB: BD = AC:CE.

P. 2. B.6: But

BD is = to AC (Ref. 2); 2. Consequently, AB : AC=AC: CE.

P.7.& 11. B.SE Which was to be done.

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T.

PROPOSITION XII. PROBLEM IV.

O find a fourth proportional (CE) to three given straight lines (M, N, P). Given.

Sought. The straight lines M, N, P.

The straight line C E, a fourth proportional to M, N, P that is such, that M:N=P:CE.

Resolution.
1. Draw the two straight lines A D, A E, containing any
VDAE.

P. 3. B.1. 2. Make A B = M; B D=N; AC = P.

Po/.1. B.T. 3. Join B C. 4. From the extremity D of the straight line AD, draw DE, plle to B C.

P.zi. B.1. DEMONSTRATION.

BECAUSE B C is pile. to DE (RES: 4).

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4. AB: B D = AC:CE.

P. 2. B.6. But

A B = M, BD = N, & AC= P (Ref. 2); 2. Consequently, M = P : CE.

P.7. & 11. B.5. Which was to be done.

= P

:N

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PROPOSITION XIII. PROBLEM V. To

O find a mean proportional (BD); between two given straight lines (A B, BC). Given.

Sought. The two straight lines A B, BC,

The Araight line B D, a mean proportional between AB & B C, that is

such that A B:BD=BD: BC.

Resolution.
1. Place A B, B C in a straight line A C.
2. Describe upon A C the semi O ADC.

Pof:3.B.1.
3. At the point B, in A C, erect the IBD meeting the
O in D.

P.11. B.1,
Preparation,
Join A D, & CD.

Pos.1. B.1.
DEMONSTRATION.
ECAUSE the V ADC is in a semi o (Ref. 2. & Prep.).
1. It is a right angle.

P.31. B.3. 2. Wherefore, the AADC is right angled in D, & B D is a I let fall

from the vertex D of the right angle, on the base AC (Ref. 3). 3. Consequently, A B : BD = BD : BC.

SP. 8. B.6.

Cor.
Which was to be done.

BE CAUS

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