The Elements of Euclid, containing the first six books, with a selection of geometrical problems. To which is added the parts of the eleventh and twelfth books which are usually read at the universities. By J. Martin

PROPOSITION 3.-Theorem.

If the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another; and conversely, if the segments of the base have the same ratio which the other sides of the triangle have to one another; the straight line drawn from the vertex to the point of section, divides the vertical angle into two equal angles.

Let ABC be a triangle, and let the angle BAC be divided into two equal angles by the straight line AD.

Then BD shall be to DC, as BA to AC.

Construction. Through the point C draw CE parallel to DA (I. 31), and let BA produced meet CE in E.

Demonstration. Because the straight line AC meets the parallels AD, EC,

1. The angle ACE is equal to the alternate angle CAD (I. 29); but CAD, by the hypothesis, is equal to the angle BAD; wherefore 2. BAD is equal to the angle ACE (Ax. 1).

Again, because the straight line BAE meets the parallels AD, EC, 3. The outward angle BAD is equal to the inward and opposite angle AEC (I. 29);

but the angle ACE has been proved equal to the angle BAD; therefore also

4. ACE is equal to the angle AEC (Ax. 1);

and consequently,

5. The side AE is equal to the side AC (I. 6);

and because AD is drawn parallel to EC, one of the sides of the triangle BCE, therefore

6. BD is to DC, as BA to AE (VI. 2) ;

but AE is equal to AC; therefore

7. As BD to DC, so is BA to AC (V. 7).

Next. Let BD be to DC, as BA to AC, and join AD.

Then the angle BAC shall be divided into two equal angles by the straight line AD.

Demonstration. 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, because AD is parallel to EC (VI. 2); therefore

1. BA is to AC, as BA to AE (V. 11);

consequently AC is equal to AE (V. 9), and therefore

2. The angle AEC is equal to the angle ACE (I. 5) ;

but the angle AEC is equal to the outward and opposite angle BAD; and the angle ACE is equal to the alternate angle CAD (I. 29) ; wherefore also the angle BAD is equal to the angle CAD (Àx. 1); that is,

3. The angle BAC is cut into two equal angles by the straight line AD.

Therefore, if the angle, &c. Q.E.D.

PROPOSITION A.-Theorem.

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 ex'remities of the base, have the same ratio which the other sides of the triangle have to one another; and conversely, 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 ABC be a triangle, and let one of its sides BA be produced to E; and let the outward angle CAE be divided into two equal angles by the straight line AD which meets the base produced in D.

Then BD shall be to DC, as BA to AC.

Construction. Through C draw CF parallel to AD (I. 31). Demonstration. Because the straight line AC meets the parallels AD, FC,

1. The angle ACF is equal to the alternate angle CAD (I. 29); but CAD is equal to the angle DAE (hyp.); therefore also 2. The angle DAE is equal to the angle ACF (Ax. 1). Again, because the straight line FAE meets the parallels AD, FC, 3. The outward angle DAE is equal to the inward and opposite angle CFA (I. 29) ;

but the angle ACF has been proved equal to the angle DAE; therefore also

4. The angle ACF is equal to the angle CFA (Ax. 1);

and consequently,

5. The side AF is equal to the side AC (I. 6);

and because AD is parallel to FC, a side of the triangle BCF, therefore

6. BD is to DC, as BA to AF (VI. 2) ;

but AF is equal to AC; therefore

7. As BD is to DC, so is BA to AC (V. 7) ;

Next. Let BD be to DC, as BA to AC, and join AD.

Then the angle CAD shall be equal to the angle DAE.

Demonstration. The same construction being made, because BD is to DC, as BA to AC; and that BD is also to DC, as BA to AF (VI. 2); therefore

1. BA is to AC, as BA to AF (V. 11); wherefore AC is equal to AF (V. 9); and therefore

2. The angle AFC is equal to the angle ACF (I. 5);

but the angle AFC is equal to the outward angle EAD (I. 29), and the angle ACF to the alternate angle CAD; therefore also

3. EAD is equal to the angle CAD (Ax. I).

Wherefore, if the outward, &c. Q.E.D.

PROPOSITION 4.-Theorem.

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 the angle BAC equal to the angle CDE (I. 32).

Then the sides about the equal angles of the triangles ABC, DCE shall be proportionals; and those shall be the homologous sides which are opposite to the equal angles.

Construction. Let the triangle DCE be placed, so that its side CE may be contiguous to BC, and in the same straight line with it (I. 22). Then, because the angle BCA is equal to the angle CED (hyp.), add to each the angle ABC; therefore

1. The two angles ABC, BCA are equal to the two angles ABC, CED (Ax. 2) ;

but the angles ABC, BCA are together less than two right angles (I. 17); therefore also

wherefore

2. The angles ABC, CED are less than two right angles;

3. BA, ED if produced will meet (I. Ax. 12); let them be produced, and meet in the point F.

Demonstration. Then, because the angle ABC is equal to the angle DCE (hyp.),

4. BF is parallel to CD (I: 28) ;

and because the angle ACB is equal to the angle DEC, 5. AC is parallel to FE (I. 28);

therefore FACD is a parallelogram; and consequently

6. AF is equal to CD and AC to FD (I. 34);

and because AC is parallel to FE, one of the sides of the triangle FBE,

7. BA is to AF, as BC to CE (VI. 2) ; but AF is equal to CD; therefore

8. As BA to CD, so is BC to CE (V. 7) ;

and alternately,

9. As AB to BC, so is DC to CE (V. 16). Again, because CD is parallel to BF,

10. As BC to CE, so is FD to DE (VI. 2) ;

but FD is equal to AC; therefore

11. As BC to CE, so is AC to DE (V. 7);

and alternately,

12. As BC to CA, so is CE to ED (V. 16);

therefore, because it has been proved that AB is to BC, as DC to CE, and as BC to CA, so CE to ED, therefore ex æquali,

13. BA is to AC, as CD to DE (V. 22).

Therefore, the sides, &c. Q.E.D.

PROPOSITION 5.-Theorem.

If the sides of two triangles, about each of their angles, be proportionals, the triangles shall be equiangular; and the equal angles shall be those which are 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.

Then the triangle ABC shall be equiangular to the triangle DEF, and the angles which are opposite to the homologous sides shall be equal, viz., the angle ABC equal to the angle DEF, and BCA to EFD, and also BAC to EDF.

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The Elements of Euclid, containing the first six books, with a selection of ...