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• 31. 1.

Through the point C draw CE parallel a to DA, and let Boox VI. BA produced meet CE in E. Because the straight line AC meets the parallels AD, EC the angle ACE is equal to the to alternate angle CADb: But CAD, by the hypothesis, is b 29. 1. equal to the angle BAD; wherefore BAD is equal to the angle ACE. Again, because the straight line BAE meets the parallels AD, EC, the outward angle BAD is equal to the inward and opposite angle AEC: but the angle AČE has been proved equal to the angle BAD; therefore } also ACE is equal to the angle AEC, and consequently the side AE is equal to the side o AC: And because AD is 6. 1. drawn parallel to one of the sides of the triangle BCE, viz. to EC, BD is to DC, as BA to AE4, but AE is equal to 2. 6. AC; therefore, as BD to DC, so is BA to ACe.

7.5. Let now BD be to DC, as BA to AC, and join 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 AEJ, because AD is parallel to EC; therefore BA is to AC, as,

$f 11.5. BA to AEf; Consequently AC is equal to AEs, and the : 9.5. angle AEC is therefore equal to the angle ACE h: But the h 5. 1. angle AEC is equal to the outward and opposite angle BAD; and the angle ACE is equal to the alternate angle CADb: Wherefore also the angle BAD is equal 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.

Book VI.

PROP. A. THEOR. If 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 sides 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 section divides the outward angle of the triangle into two equal angles.

Let the outward angle CAE of any triangle ABC be di

vided into two equal angles by the straight line AD which : meets the base produced in D: BD is to DC, as BA to AC. a 31. 1. Through C draw CF parallel to ADa; and because the

straight line AC meets the parallels AD, FC, the angle b29. 1. ACF is equal to the alternate angle CADb; But CAD is • Hyp. equal to the angle DAE'; therefore also DAE is equal to

the angle ACF. Again, because the straight line FAE
meets the parallels AD, FC, the outward angle DAE is
equal to the inward and
opposite angle CFA: But
the angle ACF has been
proved equal to the angle
DAE; therefore also the
angle ACF is equal to the
angle CFA, and conse-

quently the side AF is e- me » 6. 1. qual to the side ACd: And

because AD is parallel to FC, a side of the triangle BCF, * %. 6. BD is to DC, as BA to AFe, 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 same construction being made, because BD is to DC,

as BA to AC; and that BD is also to DC, as BA to Afe; ' 11. 5. therefore BA is to AC, as BA to Aff; wherefore AC is la 5. 1. 9. equal to AF5, and the angle AFC equalb to the angle ACF:

But the angle AFC is equal to the outward angle EAD,

A

89..

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

• 17. 1.

PROP. IV. THEOR. The sides about the equal angles of equiangular triangles are proportionals; and those which are opposite to the equal angles are homologous sides, that is, are the antecedents or consequents of the ratios.

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 59. 1. angle CDE. The sides about the equal angles of the triangles ABC, DCE are proportionals; and those are the homologous sides which are opposite to the equal angles.

Let the triangle DCE be placed, so that its side 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, ABC, and in DEC, which is equal to ACB, are also less than two right angles; wherefore BA, ED pro- A duced shall meet"; let them be

'12 Ax. 1. produced and meet in the point F; and because the angle ABC I is equal to the angle DCE, BF is BL paralleld to CD. Again, because

C E - 28. 1. the angle ACB is equal to the angle DEC, AC is parallel to FE«: Therefore FACD is a parallelogram, and conse-, quently AF is equal to CD, and AC to FDe: And because e 34. 1. AC is parallel to FE, one of the sides of the triangle FBE, BA is to AF, as BC to CEf: But AF is equal to CD;f 2. 6. therefore 8, as BA to CD, so is BC to CE; and alternately, 8 7. 5. “as AB to BC, so is DC to CE: Again, because CD is parallel to BF, as BC to CE, so is FD to DEf; but FD is equal to AC; therefore, as BC to CE, so is AC to DE: And alternately, as BC to CA, so CE to DE: Therefore, because it has been proved that AB is to BC, as DC to CE, and as BC to CA, so CE to ED, ex æqualib, BA is to AC, 5 22. 5. as CD to DE. Therefore the sides, &c. R.E.D.,

Book VI.

PROP. V. THEOR.

If the sides of two triangles, about each of their angles, be proportionals, the triangles shall be equiangular, and have their equal angles opposite to the homologous sides.

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 æquali, BA to AC, as ED to DF; the triangle ABC is equiangular to the tri. angle DEF, and their equal angles are opposite to the homologous sides, viz. the angle ABC equal to the angle

DEF, and BCA to EFD, and also BAC to EDF. 223. 1. At the points E, F, in the straight line EF, makea the

angle FEG equal to the angle ABC, and the angle EFG equal to BCA ; wherefore the

remaining angle BAC is equal • 32. 1. to the remaining angle EGFb,

and the triangle ABC is there-
fore equiangular to the triangle
GEF; and consequently they 3

have their sides opposite to the € 4. 6. equal angles proportionals €.

Wherefore, as AB to BC, so is

GE to EF; but as AB to BC, so is DE to EF; therefore 11. 5. as DE to EF, sod GE to EF: Therefore DE and GE have e 9. 5. the same ratio to EF, and consequently are equale: 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 f 8. 1. DEF is equal to the angle GEF, and the other angles to 84. 1. the other angles which are subtended by the equal sides 5.

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 angle DEF: For the same reason, the angle ACB is equal to the angle DFE, and the argle at A, to the angle at D. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if the sides, &c. Q. E. D. : .

Boon VI.

PROP. VI. THEOR.

If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous sides..

DK

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, BA 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 a the• 23, 1. angle FDG equal to either of the angles BAC, EDF; and the angle DFG equal to the angle ACB : Wherefore the remaining angle at B is equal to the remaining one at Gb, and

632. 1. consequently the triangle ABC is equiangular to the triangle DGF; and therefore as BA to AC, B so ise GD to DF; but by the hypothesis, as BA to AC, so e 4. 6. is ED to DF; as therefore ED to DF, so isd GD to DF; d 11. 5. wherefore ED is equale to DG; and DF is common to the c 9. 5. two triangles EDF, GDF: 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 the tri- f 4. 1. angle 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 EDFs; wherefore also the remaining angle at : Hyp. B is equal to the remaining angle · at E. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. Q. E. D.

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