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Thro' the point C draw CE parallel a to DA, and let BA pro- Book VI. duced meet CE in E. Because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle a. 31.1. CADb. but CAD, by the Hypothesis, is equal to the angle BAD; b. 29. 1. wherefore BAD is equal to the angle ACE. Again, because the straight line BAE meets the parallels AD, EC, the outward an

E gle BAD is equal to the inward

А
and opposite angle AEC. but the
angle ACE has been proved e-
qual to the angle BAD; there-
fore also ACE is equal to the
angle AEC, and consequently the
side AE is equal to the side AC.

B
D C

. 6. I. and because AD is drawn parallel to one of the sides of the triangle BCE, viz. to EC, BD is to DC, as BA to AE d; but AE is d. 2. 6. equal to AC; therefore as BD to DC, fo is BA to AC

Let now BD be to DC, as BA to AC, and joia AD; the angle BAC is divided into two equal angles by the straight line AD.

The same construction being made, because as BD to DC, so is BA to AC; and as BD to DC, so is BA to AE 4, because AD is parallel to EC; therefore BA is to AC, as BA to AE f. con- f. 11. s. sequently AC is equal to AE, and the angle AEC is therefore 8. 9. s. equal to the angle ACE h. but the angle AEC is equal to the out- h. sod, ward and opposite angle BAD; and the angle ACE is equal to the alternate angle CAD b. wherefore also the angle BAD is en qual to the angle CAD. therefore the angle BAC is cut into two equal angles by the straight line AD. Therefore if the angle, &c. Q. E. D.

e. 7. 5

K 4

Rock VI.

PROP. A. THEOR.
Ir the outward angle of a triangle made by producing

one of its sides, be divided into two equal angles, by a straight line which also cuts the base produced ; the segments between the dividing line and the extremities of the base have the same ratio which the other fides of the triangle have to one another, and if the segments of the base produced have the same ratio which the other sides of the triangle have, the straight line drawn from the vertex to the point of fection divides the cutward angle of the triangle into two equal angles.

*. 31. I.

b. 22. 1. 6. Hiyo.

Let tlie outward angle CAE of any triangle ABC be divided in:o two equal angles by the straight line AD which meets the Laie produced in D. BD is to DC, as BA to AC.

Through C draw CF parallel to AD“; and because the straight line AC meets the parallels AD, FC, the angle ACF is equal to the alternate angle CAD, but CAD is equal to the angle DAE“; therefore aito DAE is equal to the angle ACF. Again, because the straight line FAE meets the parallels AD, FC, the outward eli u DAE is equal to the in

E
ward and opposite angle CFA.
Lut the angle ACF has been
proved equal to the angle
DAF; therefore also the an-

T
E: ACH' is equal to the angle
CFA, and consequently the
1:23 AF is equal to the fide B

C

D AC4. and because AD is parallel to FC a side of the triangle BCF, !:D is to DC, as BA to Ale; but AF is equal to AC; as therefore BD is to DC, so is BA to AC.

Let now BD be to DC, as BA to AC, and join AD; the angle CAD is equal to the angle DAE.

The fine conítruction being made, because BD is to DC, as LA TI AC; and thit BD is also to DC, as BA to AF€; therefore BA is to AC, as BA to Alf, wherefore AC is equal to AF8, and the wyle AFC cqual to the angle ACF. but the angle AFC

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is equal to the outward angle EAD, and the angle ACF to the Book VI.
alternate angle CAD; therefore also EAD is equal to the angle
CAD. Wherefore if the outward, &c. Q. E. D.

PROP. IV. THEOR.
THE fides about the equal angles of equiangular tri-

angles are proportionals; and those which are op-
posite to the equal angles are homologous fides, that is,
are the antecedents or confequents of the ratios.

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Let ABC, DCE be equiangular triangles, having the angle ABC equal to the angle DCE, and the angle ACB to the angle DEC, and consequently a the angle BAC equal to the angle CDE. The 2. 32. 1. sides about the equal angles of the triangles ABC, DCE are proportionals; and thole are the homologous sides which are oppofite to the equal angles.

Let the triangle DCE be placed fo that its fide CE may be contiguous to BC, and in the same straight line with it. and because the angles ABC, ACB are together less than two right angles b; b. 17.8. ABC and DEC, which is equal to ACB, are allo less than two right angles. wherefore BA, ED produced

0.12. Ax.s.
Shall meet " ; let them be produced A
and meet in the point F. and because

D
the angle ABC is equal to the angle
DCE, BF is parallel to CD. Again,

d. 28. 1.
because the angle ACB is equal to B
the angle DEC, AC is parallel to

CE FE d. therefore FACD is a parallelogram; and consequently AF is equal to CD, and AC to FDʻ. and because AC is parallel to FE e. 34. 1. one of the sides of the triangle FBE, BA is to AF, as BC to CEF. f. 2.6. but AF is equal to CD, therefore 8 as BA to CD, fo is BC to CE ; 6. 7.5. and alternately, as AB to BC, so DC to CE. Again, because CD is parallel to BF, as BC to CE, fo is FD to DE{; but FD is equal to AC; therefore as BC to CE, fo is AC to DE. and alternately, as BC to CA, fo CE to ED. therefore because it has been proved that AB is to BC, as DC to CE; and as BC to CA, fo CE to ED, ex acquali , BA is to AC, as CD to DE. Therefore the fides, &c. h. 22. 5. Q. E. D,

Book VI.

PRO P. V. THEOR.

IF F the sides of two triangles, about each of their an

gles, be proportionals, the triangles Mhall be equiangular, and have their equal angles opposite to the homologous sides.

R. 23. I.

b. 32. 1.

C. 4. 6.

Let the triangles ABC, DEF have their sides proportionals, so that AB is to BC, as DE to EF; and BC to CA, as EF to FD; and consequently, ex aequali, BA to AC, as ED to DF. the triangle ABC is equiangular to the triangle DEF, and their equal angles are apposite to the homologous sides, viz. the angle ABC equal to the angle DEF, and BCA to EFD, and also BAC to EDF.

At the points E, F in the straight line EF make the angle FEG
equal to the angle ABC, and the angle EFG equal to BCA; where-
fore the remaining angle BAC
is equal to the remaining angle

А
EGF b, and the triangle ABC

D
is therefore equiangular to the
triangle GEF; and consequently
they have their sides opposite to

E F the equal angles proportionalse. B

С wherefore as AB o BC, so is GE to EF; but as AB to BC, fo is DE to EF; therefore as DE to EF, so 4 GE to EF. therefore DE and GE have the same ratio to EF, and confequently are equal. for the same reason DF is equal to FG. and because, in the triangles DEF, GEF, DE is equal to EG, and EF common, the two sides DE, EF are equal to the two GE, EF, and the base DF is equal to the base GF; therefore the angle DEF is equal f to the angle GEF, and the other angles to the other angles which are subtended by the equal lides 8. Wherefore the angle DFE is equal to the angle GFE, and EDF to EGF. and because the angle DEF is equal to the angle GEF, and GEF to the angle ABC; therefore the angle ABC is equal to the angie DEF. for the fame reason, the angle ACB is equal to the angle DFE, and the angle at Ato the ang'e at D. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore if the sides, &c. Q. E: D,

d. 11. S.

e. 9. S.

f. 8. 1.

$. 4.

Book VI.

PROP. VI.

THEOR.

IF
F two triangles have one angle of the one equal to

one angle of the other, and the sides about the e-
qual angles proportionals, the triangles shall be equian-
gular, and shall have those angles equal which are op-
posite to the homologous sides.

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c. 4. 6.

Let the triangles ABC, DEF have the angle BAC in the one equal to the angle EDF in the other, and the sides about those angles proportionals; that is, В A to AC, as ED to DF. The triangles ABC, DEF are equiangular, and have the angle ABC equal to the angle DEF, and ACB to DFE.

At the points D, F, in the straight line DF, make the angle a. 23. I,
FDG equal to either of the angles BAC, EDF; and the angle DFG
equal to the angle ACB. where-
fore the remaining angle at B is

А.
equal to the reinaining one at
Gb, and consequently the tri-

b. 32. 1.
angle ABC is equiangular to
the triangle DGF; and there-
fore as BA to AC, fo is GD
to DF. but, by the Hypothesis,
as BA to AC, fo is ED to DF;.

CE as therefore ED to DF, fo is GD to DF; wherefore ED is equal to DG; and DF is common to the two triangles EDF,GDF. C. 9. 5. therefore the two sides ED, DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle GDF, wherefore the base EF is equal to the base FGf, and the triangle EDF to f. 4. I. the triangle GDF,and the remaining angles to the remaining angles, each to each, which are subtended by the equal sides, therefore the angle DFG is equal to the angle DFE, and the angle at G to the angle at E. but the angle DFG is equal to the angle ACB; therefore the angle ACB is equal to the angle DFE. and the angle BAC is equal to the angle EDF %; wherefore also the remaining g. Hyp. angle at B is equal to the remaining angle at E. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore if iwo triangles, &c. Q. E. D.

d.

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