PROP. II. THEOR.

If a right line (DE, see all the fig. to this prop.) be drawn parallel to one of the sides (AB) of a triangle (ABC), it cuts the other sides (AC, BC), or these sides produced, proportionally.

And the right line (DE), which cuts two sides (AC, BC) a triangle (ABC), or these sides produced, proportionally, is parallel to the remaining side (AB).

Part 1.-Let DE be parallel to AB; AD is to DC, as BE is to EC.

Join AE, DB, and because the triangles DAE, DBE are on the same base DE, and between the same parallels DE and AB, they are equal [37. 1]; therefore ADE is to CDE, as BDE is to CDE [7. 5]; but ADE is to CDE, as AD is to DC [1. 6], and BDE is to CDE, as BE is to EC (by the same); therefore AD is to DC, as BE is to EC.

Part 2.-Let AD be to DC, as BE to EC; DE is parallel to AB.

For, the same construction remaining, because AD is to DC, as BE to EC [Hyp.]; and AD is to DC, as the triangle ADE to the triangle CDE [1. 6]; and BE to EC, as the triangle BDE to the triangle CDE [by the same]; the triangle ADE is to CDE, as BDE to the same CDE [11. 5]; therefore the triangles ADE, BDE are equal [9. 5]; and they are on the same base DE, and to the same part; therefore DE is. parallel to AB (39.. 1).

Schol.-In the second part it is understood, that the homologous segments are similarly situated on the sides, or sides produced, of the triangle.

PROP. III. THEOR.

A right line (CD), bisecting an angle (ACB) of a triangle (ABC), cuts the opposite side into segments (AD, DB), having to each other the same ratio, as the other sides (AC, CB) of the triangle have to each other.

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And a right line (CD), drawn from an angle (ACB) of a triangle (ABC), cutting the opposite side (AB) into segments (AD, DB), having to each other the same ratio, as the other sides (AC, CB) of the triangle have to each other, bisects that angle.

Part 1.-Through B, draw BE parallel to CD, meeting AC produced in E.

Because DC and BE are parallel, the angle CBE is equal to the alternate BCD (29. 1), or its equal (Hyp.) ACD, or, CE meeting the same parallels, to its equal, the internal remote an- A

B

gle CEB (29. 1]; therefore CB and CE are equal (6. 1); but, because DC is parallel to BE, AD is to DB, as AC is to CE (2. 6), whence, CB and CE being equal, AD is to DB, as AC is to CB.

Part 2.-The same construction remaining, AD is to DB, as AC to CB (Hyp.), and, because DC is parallel to BE, AD is to DB, as AC is to CE (2. 6), therefore AC is to CB, as AC is to CE (11. 5), of course CB and CE are equal (9. 5), and therefore the angle CBE is equal to the angle CEB (5. 1); whence, because of the parallels DC and BE, the angle ACD being equal to the internal remote CEB, and DCB to the alternate CBE (29. 1), the angles ACD and DCB are equal (Ax. 1. 1), and so the angle ACB is bisected by the right line CD.

Schol.-In the second part it is understood, that either segment of the divided side and the adjacent undivided side, are homologous terms.

Theorem.-If the external angle (BCF), of a triangle (ACB), made by producing one of its sides (AC), be bisected by a right line (CD) meeting the base produced; the segments (AD, BD) of the base produced, between its extremes (A, B), and the bisecting line, are to each other, as the other sides (AC, CB) of the triangle.

And if a right line (CD), be drawn from an angle (ACB) of a triangle (ABC), to a point (D) in the opposite side produced, the distances (AD, BD) of which point, from the extremes of that side, are to each other, as the other sides (AC, BC) of the triangle; the right so drawn bisects the external angle (BCF) formed at the first mentioned angle.

Part 1.-Through B draw BE parallel to CD.

Because BE and DC are parallel, the angle CBE is equal to the alternate angle BCD (29. 1), or, to its equal [Hyp.] DCF, or to the internal and opposite angle CEB [29. 1 and Ax. 1. 1], therefore CB is equal to CE [6. 1]; and, because in the triangle ACD, EB is parallel to CD, AE is to EC as AB is to BD [2. 6], and, by compounding, AC is to CE or its equal CB, as AD is to BD [18. 5].

Part 2. The same construction remaining, AD is to BD, as AC to CB (Hyp.), and, because EB and CD are parallel, AD is to BD, as AC is to EC [2. 6], therefore AC is to CB, as AC to EC (11. 5), and so CB and CE are equal [9. 5], and the angle CBE is equal to the angle CEB [5. 1; whence, because of the parallels EB and CD, the angle BCD being equal to the alternate CBE, and the angle FCD to the internal remote CEB (29. 1), the angles BCD and FCD are equal (Ax. 1. 1), and so the right line CD bisects the external angle BCF.

PROP. IV. THEOR.

The sides about the equal angles of equiangular triangles are proportional, the sides opposite equal angles, being homologous.

Let ABC and DEF be equiangular triangles, having the angle A equal to D, the angle ABC to E, and therefore [32. 1] the angle ACB to F. AC is to AB, as DF to DE, AB to BC, as DE to EF, and AC to CB, as DF to FE.

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On AB produced take BG equal to DE, at the point B, with the right line BG, make the angle GBH equal to D or A, take BH equal to DF, and join GH and since, in the triangles GBH, EDF, the sides GB, BH and the angle GBH, are severally equal to ED, DF and the angle EDF [Constr.], the triangles BGH and DEF are equal in all respects, and the angle G is equal to E [4. 1], or its equal [Hyp.] ABC; whence, the angles A and ABC being less than two right angles [17. 1], the angles A and G are less than two right angles, therefore AC and GH may be so produced as to meet [Theor. at 29. 1], let them be produced to meet, as in K; and, because the angle ABC is equal to G, BC and GK are parallel [28. 1], and, because the angle GBH is equal to A, BH and AK are parallel [by the same], therefore BHKC is a parallelogram, and of Course, CK is equal to BH or DF, and HK to BC [34. 1].

And because, in the triangle AGK, BC is parallel to ̊ GK, AC is to CK, or its equal DF, as AB to BG or DE [2. 6], and by alternating, AC is to AB, as DF is to DE [16. 5]; and, because BH is parallel to AK, AB is to BG, or its equal DE, as KH, or its equal BC, is to HG or its equal EF [2. 6], and, by alternating, AB is to BC, as DE to EF [16. 5]; and since AC is to AB,eas DF to DE, and AB to BC, as DE to EF, by ordinate equality, AC is to CB, as DF to FE [22.5]: therefore the sides about the equal angles of the triangles ABC, DEF are proportional, the sides opposite equal angles being homologous [see def. 15. 5].

Schol.-Equiangular triangles are similar: see def. 1. 6.

PROP. V. THEOR.

If the sides of two triangles (ABC, DEF), about each of their angles, be proportional (AB to BC, as DE to EF; BC to JC, as EF to DF; and therefore, by ordinate equality, AB to AC, as DE to DF), the triangles are equiangular, having their equal angles opposite to the homologous sides.

At the extremes of any side DE,of either triangle, as DEF, make angles EDG and DEG equal to the angles A and B at the extremes of the side AB, which is homologous to DE; the remaining angle G of the tri

A A

B

angle DEG, is equal to the remaining angle C of the triangle ABC (32. 1).

And, because the triangles ABC, DEG are equiangular, BA is to AC, as ED to DG (4. 6), and BA is to AC, as ED to DF (Hyp.), therefore ED is to DG, as ED to DF (11. 5), and so DG and DF are equal (9. 5); in like manner it may be proved, that EG and EF are equal, therefore the triangle DEG is equilateral to the triangle DEF, and of course equiangular to it [8. 1], and DEG is equiangular to ABC [Constr.], therefore the triangle ABC is equiangular to DEF, having the angle A equal to EDF, B to DEF, and C to F [Ax. 1. 1], namely, having those angles equal, which are opposite to the homologous sides.

PROP. VI. THEOR.

If two triangles (ABC, DEF, see fig. to preced. prop.) have an angle (A) of one, equal to an angle (EDF) of the other, and the sides about the equal angles proportional (BA to AC, as ED to DF); the triangles are equiangular, having those angles equal, which are opposite to homologous sides.

With either leg DE, of either of the equal angles A and EDF, and at either extreme of it D, make the angle EDG equal to A, and at E, the angle DEG equal to B; the remaining angle G of the triangle DEG, is equal to the remaining angle C of the triangle ABC (32. 1).