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2.6.

a 16. 5.

b22. 5.

a

BC: CE. and fo by permutation a AB : BC :: CD: CE. alfo " BC: CE :: FD (AC): DE. whence by Permutation BC: AC:: CE : DE. and therefore by Equality AB: AC:: CD: DE. Therefore, &c.

COROL.

Hence AB: DC:: BC: CE:; AC: DE.

SCHOL.

Hence if a Line AC be drawn in a Triangle (FBE) parallel to one Side FE; the Triangle ABC fhall be fimilar to the whole one FBE.

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If the Sides of two Triangles ABC, DEF, are proportional, viz. AB: BC :: DE: EF. and AC: BC: DF: EF. alfo AB AC:: DE: DF. the faid Triangles are Equiangular, and their Angles under which the Homologous Sides are fubtended, are equal.

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At the Points E and F, with the Side EF make the Ang. FEG B, and the Ang. EFG =C, and fo is the Ang. GA. therefore GE : EF :: AB: BC:: DE: EF. whence GE DE. alfo GF: FE:: AC: CB:: DF FE. therefore & GF=DF. Therefore the Triangles DEF, GEF, are mutually equilateral

g

f

=

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to each other, and confequently the Ang, D=

G=A. and the Ang. FED FEG= B. a 8. 1. whence the Ang. DFE C. Therefore, &c. 32. I, PROP. VI.

If two Triangles ABC, DEF, have one Angle B of the one equal to one Angle DEF of the other, and if the Sides about the equal Angles B, DEF, be propormonal, viz. AB: BC:: DE: EF; the Triangles ABC, DEF fhall be equiangular, and have thofe Angles equal, under which are fubtended the Homologous Sides,

AA

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F

therefore

At the Points E, F, with the Line EF make the Ang. FEGB, and the Ang. EFG = C. whence alfo is the Ang. G=A. therefore GE: EF:: AB: BC:: DE: EF; f DE=GE. But the Ang. DEF & GEF. Whence the Ang. Di G= fo the Ang. EFD* = C. Q. E.D.

PROP. VII.

B

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32. 1. d 4. 6. byp. 9. 5.

A. and 8 11. 5. n conft.

If there are two Triangles ABC, DEF, having one Angle A of the one equal to one Angle D of the other, and the

AA

F Sides about the o

ther

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4. I.

32. I.

a hyp.

b

32. I.

C

4. 6.

d

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9. 5.

ther Angles ABC, D, proportional, viz. AB: BC :: DE: EF. and if the remaining third Angles C, F, are either both lefs or both greater than right Angles; the Triangles ABC, DEF fhall be equiangular, and have thofe Angles equal about which are the proportional Sides.

For if poffible, fet the Ang. ABC be E. then make the Ang. ABGE. therefore fince the Ang. "A=D; the Ang. AGB fhall be F. Therefore AB : BG:: DE: EF:: AB: BC. whenced BG BC; therefore the Ang. ABC 5. I. BCG. And fo the Ang. f BGC, or C, is lefs cor. 13.1. than a right one; whence the Ang. AGB or F is greater than a right one, and fo the Angles C and F are not of the fame kind; which is contrary to the Hypothefis.

Cor. 17.1.=

n hyp.

k

1

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gle into two others ADB, ADC, which will be fimilar to each other, as likewife to the whole Triangle ABC.

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For because the Angles BAC, ADB are 12. ax. right Angles; and fo equal, and B is common; 32. 1& the Triangles BAC, ADB are fimilar. After the fame manner are the Triangles BAC, ADC fi1 See 21.6. milar, and confequently ADB, ADC fhall be fimilar. QE. D.

4. 6.

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From the Point A draw any indefinite right Line AC any

how, in which take

any how the three parts AD, DE, EF, equal a 3. 1. to each other; join the Points F, B, by the Line FB, to which draw DG parallel, and the thing " 31. 1. is done.

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d

2. 6.

For GB: AG:: FD: AD. therefore by compounding AB: AG: AF: AD. Whence 18. 5. fince AD AF, the part AD fhall be =

AB. QE. F.

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e

Join the Extremities of the Lines by the Line BC, and from the

Points E, D, draw the right Lines EG, DF, 31. 1. parallel to BC, meeting the undivided right Line AB in the Points G and F; I fay the thing required is done.

e

For draw DH parallel to AB. then is AD: 2. 6. DE: AF: FG, and DE : EC:: DI: IH :: 8 34. I. FG: GB. QE. D.

SCHOL.

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Then draw the right Lines

33. I.

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€ 2.6.

LR, TS, XV, ZN, which will divide the given Line AB into five equal Parts.

C

For RL, ST, VX, NZ, are parallel. Therefore fince AR, RS, SV, VN are equal; AM, MO, OP, PQ, are alfo equal. In like manner because BZ-ZX, BQ fhall be = QP. therefore AB is divided into five equal parts. Q. E. F.

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in AB continued out take BCAD; thro' C draw CE parallel to BD, and let AD continued out meet it in E; then fhall DE be the Line fought.

For

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