PROPOSITION IX.

PROBLEM I.

FROM a given ftraight line (A B) to cut off any part required.

(For example the third part).

Given.

The ftraight line A B.

Sought.

The abfcinded ftraight line' A D, which may be the third part of A B.

Refolution.

1. From the point A draw an indifinite ftraight line A C, mak-
ing with A B any VBA C.

Pof.1. B.1.

2. Take in AC three equal parts AE, EF, FC of any length.

P. 3. B.1.

3. Join C B.

Pof.1. B.1.

4. And thro' E, draw ED plle. to C B, which will cut the
straight line A B so that AD will be the third part.

P. 31. B.1.

DEMONSTRATION,

BECAUSE

ECAUSE ED is plle. to the fide CB of the A CAB (Prep. 4).

1.

But

CE: EABD: DA.

CE is double of E A (Ref. 2) ;

:

2. Confequently, B D is alfo double of D A.

3. Wherefore, A B is triple of A D.

4. And the abfcinded straight line A D is the third part of A B.

Which was to be demonstrated.

P. 2. B.6.

D. 8. B.5.

Dd

PROPOSITION X.

PROBLEM II.

To divide a given straight line (A B), fimilarly to a given straight line

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

1. Join the given ftraight lines A B, AC fo as to contain any
У ВАС.

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

BECAUSE

DEMONSTRATION.

ECAUSE DF is plle. to the fide E G of the AAGE (Ref. 2. & P. 30. B.1), and K E plle. to the fide HC of the ADHC (Ref. 2).

I.

And

AF: FG = AD: DE.

DK: KHDE: E C.

But the figures KF, HG being pgrms. (Ref. 2. & D. 35. B. 1.). 2. It follows, that F G is to DK & G D = KH.

3. Therefore, FG: GB =

DE EC.

Pof.1. B.1.

P.31. B.1.

P. 7. B.6.

P.34. B.1. P.7.& 11. B.5.

4. Confequently, the given ftraight line A B is divided in the points F & G ; fo that AF: FG=AD: DE & F G : GB = DE :EC.

Which was to be done.

B

Α

E

To

D

PROPOSITION XI. PROBLEM III.

O find a third proportional (CE) to two given straight lines (AB, AC),

Given.

The two ftraight lines

A B, A C.

1.

Join the two ftraight lines A B, AC fo as to contain any

VBA C.

2. Produce them, & make B DAC.

3. Join B C.

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

P. 3. B.1.
Pof.1. B.1.

P.31.B.I.

PROPOSITION XII. PROBLEM IV.

To find a fourth proportional (CE) to three given straight lines

(M, N, P).

1. Draw the two ftraight lines A D, A E, containing any

V DAE.

2. Make A BM; BD=N; ACP.

3. Join B C.

4. From the extremity D of the straight line AD, draw DE,
plle to B C.

I.

P. 2. B.6.

But

A B

2. Confequently, M

: N

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

P.7. & 11. B.5.

Which was to be done.

PROPOSITION XIII. PROBLEM V.

To find a mean proportional (BD); between two given straight lines

(A B, BC).

Given.

The two ftraight lines A B, BC,

Sought,

The ftraight line B D, a mean proportional between AB & B C, that is fuch that A B: BDBD: BC.

Refolution.

1. Place A B, B C in a straight line A C.

2. Defcribe upon AC the femi O AD C.

3. At the point B, in A C, erect the LBD meeting the
O in D.

Preparation,

Join A D, & CD.

DEMONSTRATION.

Pof.3. B.1.

P.11. B.I.

Pof.1. B.I.

BECAUS

ECAUSE the VA DC is in a femi O (Ref. 2. & Prep.).

1. It is a right angle.

2. Wherefore, the AADC is right angled in D, & B D is a L let fall from the vertex D of the right angle, on the base AC (Ref. 3).

3. Confequently, AB BD BD : BC.

P.31. B.3.

P. 8. B.6.

Cor.

Which was to be done.