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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.
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
ECAUSE ED is plle. to the fide CB of the ACAB (Prep. 4).
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.
PROPOSITION X. PROBLEM II.
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
Because of is plte. to the fide e G of the AAGE (Res. 2.
AF : FG = AD: DE.
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.
To fred a third proportional (CE) to two given 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,
Pol.1. B.1. 4. And from the extremity D of the straight line A D draw DE plle. to B C.
3. Join B C.
ECAUSE BC is plle. to D E (Ref. 4).
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.
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.
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).
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.
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.
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.