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a

Thro' the point C draw CE parallel to DA, and let BA produ- Book VI. ced meet CE in E. Because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle CAD b. but a 31. 1. CAD, by the Hypothefis, is equal to the angle BAD; wherefore b. 29. 1. BAD is equal to the angle ACE. Again, because the straight line

E

A

BAE meets the parallels AD, EC, the outward angle BAD is equal to the inward and oppofite angle AEC. but the angle ACE has been proved equal to the angle BAD; therefore alfo ACE is equal to the angle AEC, and consequently the fide AE is equal to the fide AC. and becaufe AD is drawn parallel to one of the fides of the triangle BCE, viz. to EC, BD is to DC, as BA to AE; but AE is equal d. 2. 6. to AC; therefore as BD to DC, fo is BA to AC.

C

B

D C

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

c. 6. I.

e. 7.5.

The fame construction being made, because as BD to DC, fo is BA to AC; and as BD to DC, fo is BA to AE, because AD is parallel to EC; therefore BA is to AC, as BA to AEf. confequently f. 11. 5. AC is equal to AE, and the angle AEC is therefore equal to the g. 9. 5. angle ACE. but the angle AEC is equal to the outward and op- h. 5. 1. pofite angle BAD; and the angle ACE is equal to the alternate angle CAD. wherefore alfo the angle BAD is equal to the angle CAD. therefore the angle BAC is cut into two equal angles by the ftraight line AD. Therefore if the angle, &a. Q. E. D.

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Book VI.

2. 31. 1.

b. 29. 1.

c. Hyp.

d. 6. I.

c. 2. 6.

f. 11. 5.

2. 9. 5.

PROP. A. THEOR.

IF the outward angle of a triangle made by producing one of its fides, be divided into two equal angles, by a straight line which alfo cuts the base produced; the fegments between the dividing line and the extremities of the bafe have the fame ratio which the other fides of the triangle have to one another. and if the fegments of the base produced have the fame ratio which the other fides of the triangle have, the straight line drawn from the vertex to the point of fection divides the outward angle of the triangle into two equal angles.

Let the outward angle CAE of any triangle ABC be divided into two equal angles by the ftraight line AD which meets the bafe 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 CADb. but CAD is equal to the angle DAE; therefore alfo DAE is equal to the angle ACF. Again, because the ftraight line FAE meets the parallels AD, FC, the outward angle DAE is equal to the inward

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E

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becaufe AD is parallel to FC a fide of the triangle BCF, BD is to DC, as BA to AF; but AF is equal to AC; as therefore BD is to DC, fo 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 fame conftruction being made, because BD is to DC, as BA to AC; and that BD is also to DC, as BA to AF; therefore BA is to AC, as BA to AF. wherefore AC is equal to AF, and the angle AFC equal to the angle ACF. but the angle AFC is equal

to

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

PRO P. IV. THEOR.

THE fides about the equal angles of equiangular trịangles are proportionals; and thofe which are oppofite to the equal angles are homologous fides, that is, are the antecedents or confequents of the ratios,

a

Let ABC, DCE be equiangular triangles, having the angle ABC equal to the angle DCE, and the angle ACB to the angle DEC, and confequently the angle BAC equal to the angle CDE. The fides a. 38. 1. about the equal angles of the triangles ABC, DCE are proportionals; and those are the homologous fides which are opposite to the equal angles.

F

A

B

c. 12. Ax. 1.

d. 28. 1.

C

E

Let the triangle DCE be placed fo that its fide CE may be contiguous to BC, and in the fame straight line with it. and because the angles ABC, ACB are together lefs than two right anglesb; ABC b. 17. 1. and DEC, which is equal to ACB, are alfo lefs than two right angles. wherefore BA, ED produced shall meet; let them be produced and meet in the point F. and because the angle ABC is equal to the angle DCE, BF is parallel to CD. Again, because the angle ACB is equal to the angle DEC, AC is parallel to FE. therefore FACD is a parallelogram; and confequently AF is equal to CD, and AC to FD. and because AC is parallel to FE one of the fides of the triangle FBE, BA is to AF, as BC to CE f. but AF is equal to CD, therefore & as BA to CD, fo is BC to CE; 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 f; to AC; therefore as BC to CE, fo is AC to DE. 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,

g

but FD is equal
and alternately,

c. 34. 1.

f. 2. 6.

g. 7. 5.

ex aequali h, BA is to AC, as CD to DE. Therefore the fides, &c. h. 22. 5. Q. E. D.

PROP.

154 Book VI.

a. 23. I.

b. 32. I.

IF

PROP. V. THEOR.

F the fides of two triangles, about each of their angles, be proportionals, the triangles fhall be equiangular, and have their equal angles oppofite to the homologous fides.

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

a

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; wherefore the remaining angle BAC

is equal to the remaining angle EGF, and the triangle ABC is therefore equiangular to the triangle GEF; and confequently they have their fides oppofite to the equal c. 4. 6. angles proportionals. wherefore as AB to BC, fo is GE

d. 11. 5. c. 9. 5.

B

A

D

E

F

G

to EF; but as AB to BC, fa is DE to EF; therefore as DE to EF, fod GE to EF. therefore DE and GE have the fame ratio to EF, and confequently are equal. for the fame reason DF is equal to FG. and becaufe, in the triangles DEF, GEF, DE is equal to EG, and EF common, the two fides DE, EF are equal to the two GE, EF, and the bafe DF is equal to the bafe GF; therefore the f. 8. 1. angle DEF is equal to the angle GEF, and the other angles to the 8.4.1. other angles which are fubtended by the equal fides. 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 fame reason, the angle ACB is equal to the angle DFE, and the angle at A to the angle at D. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore if the fides, &c. Q. E. D.

PROP.

IF

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two triangles have one angle of the one equal to one angle of the other, and the fides about the equal angles proportionals, the triangles fhall be equiangular, and shall have those angles equal which are oppofite to the homologous fides.

Let the triangles ABC, DEF have the angle BAC in the one equal to the angle EDF in the other, and the fides 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.

a

Book VI.

At the points D, F, in the straight line DF, make the angle a. 23. z. FDG equal to either of the angles BAC, EDF; and the angle DFG equal to the angle ACB.

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AC, fo is ED to DF; as therefore ED to DF, fo is 4 GD to DF; d. 11.5. wherefore ED is equal to DG; and DF is common to the two e. 9. 5. triangles EDF, GDF. therefore the two fides ED, DF are equal to the two fides GD, DF; and the angle EDF is equal to the angle GDF, wherefore the base EF is equal to the base FGf, and the tri- f. 4. 1. angle EDF to the triangle GDF, and the remaining angles to the remaining angles, each to each, which are fubtended by the equal fides. 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 alfo the g. Hyp. remaining angle at B is equal to the remaining angle at E. fore the triangle ABC is equiangular to the triangle DEF. fore if two triangles, &c. Q. E. D.

There

Where

PROP.

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