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• 11. 5.

Q. E. D.

CD, so is the triangle ABC to the triangle ACD; and as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF; therefore, as the base BC is to the base CD, so is * the parallelogram EC to the parallelogram CF. Wherefore, triangles, &c.

COROLLARY. From this it is plain, that triangles and parallelograms which have equal altitudes, are one to another as their bases.

Let the figures be placed so as to have their bases in the same straight line; and having drawn perpendiculars from the vertices of the triangles to the bases, the straight line which joins the vertices is parallel to that in which their bases are *, because the perpendiculars are both equal and parallelt to one another. Then, if the same construction be made as in the proposition, the demonstration will be the same.

• 33. 1. † 28. 1.

PROPOSITION II.

See N.

THEOR.-If a straight line be drawn parallel to one of the

sides of a triangle, it shall cut the other sides, or these produced, proportionally : and if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section, shall be parallel to the remaining side of the triangle.

* 37. 1.

7.5.

* the

Let DE be drawn parallel to BC, one of the sides of the triangle ABC: BD shall be to DA, as CE to EA.

Join BE, CD; then the triangle BDE is equal * to the triangle CDE, because they are on the same base DE, and between the same parallels DE, BC: ADE is another

D triangle; and equal magnitudes have to the same same ratio ; therefore, as

E B the triangle BDE is to the

E B triangle ADE, so is the triangle CDE to the triangle ADE: but as the triangle BDE to the triangle ADE, so is * BD to DA, because, having the same altitude, viz. the perpendicular drawn from the point E to AB, they are to one another as their bases; and for the same reason, as the angle CDE to the triangle ADE, so is CE to EA: therefore as BD to DA, so is CE to EA *.

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B

• 1. 6.

•11, 5.

Next, let the sides AB, AC, of the triangle ABC, or these sides produced, be cut proportionally in the points D, E, that is, so that BD may be to DA as CE to EA; and join DE: DE shall be parallel to BC.

The same construction being made; because as BD to DA, so is CE to EA; and as BD to DA, so is the triangle BDE to the triangle ADE*: and as CE to EA, so is the triangle CDE * 1.6. to the triangle ADE; therefore t the triangle BDE is to the † 11. 5. triangle ADE, as the triangle CDE to the triangle ADE; that is, the triangles BDE, CDE have the same ratio to the triangle ADE: therefore * the triangle BDE is equal to the triangle • 9.5. CDE: and they are on the same base DE: but equal triangles on the same base are between the same parallels *, therefore 39. 1. DE is parallel to BC. Wherefore, if a straight line, &c. Q. E. D.

PROPOSITION III.

THEOR.-If the angle of a triangle be divided into two equal See N. 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 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: BD shall be to DC, as BA to AC.

Through the point C, draw CE parallel * to DA, and let BA • 31. 1. produced, meet CE in E. Because the straight line AC meets the parallels AD, EC, the angle ACE is

E

• 29. 1. equal * to the alternate angle CAD: but CAD, by the hypothesis, is equal to the angle BAD: wherefore BAD is equal + to the angle ACE. Again, because the

D straight line BAE meets the parallels AD, EC, the outward angle BAD is equal + to the inward and op- † 29. 1. posite angle AEC: but the angle ACE has been proved equal to the angle BAD; therefore also ACE is equal † to the angle † 1 Ax. AEC, and consequently the side AE is equal * to the side AC: * 6. 1. and because AD is drawn parallel to one of the sides of the

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• 2. 6.

7.5.

• 2. 6. • 11. 5. • 9.5. • 5. I.

triangle BCE, viz. to EC, therefore * BD is to DC, as BA to AE: but AE is equal to AC; therefore *, as BD to DC, so is BA to AC.

Next, let BD be to DC, as BA to AC, and join AD; the angle BAC shall be 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*, because AD is parallel to EC; therefore * BA is to AC, as BA to AE: consequently AC is equal * to AE, and therefore the angle AEC is equal * to the angle ACE; but the angle AEC is equal to the outward and opposite angle BAD; and the angle ACE is equal to the alternate angle CAD*; wherefore also the angle BAD is equal t to the angle CAD; that is, the angle BAC is cut into two equal angles by the straight line AD. Therefore, if the angle, &c.

• 29. 1.

† 1 Ax

Q. E, D.

PROPOSITION A.

THEOR.-If the outward angle of a triangle made by produc

ing 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 hare, the straight line drawn from the verter 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: BD shall be 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 also DAE is equal + to the angle ACF. Again, because the straight line FAE meets

B 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

• 31. 1.

• 29. I.

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+ 29. 1.

• 6. 1.

Iso the angle ACF is equal † to the angle CFA ; and conse- † 1 Ax.
quently the side AF is equal to the side AC: and because
AD is parallel to FC, a side of the triangle BCF, therefore * * 2. 6.
BD is to DC, as BA to AF: but AF is equal to AC; therefore
as BD is to DC , so is BA to AC.

† 7. 5.
Next, let BD be to DC, as BA to AC, and join AD: the
angle CAD shall be 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 AF; 2.6. 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 the outward angle EAD+, and the angle † 29. 1. ACF to the alternate angle CAD; therefore also EAD is equal + † 1 Ax. to the angle CAD. Wherefore, if the outward, &c. Q. E. D.

11. 5. 9. 5. 5. 1.

PROPOSITION 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.

3 Ax.

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 • 32. 1. and CDE: 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.

Let the triangle DCE be placed * so that its side CE may be • 22. 1. contiguous to BC, and in the same straight line with it: then, because the angle BCA is equal + to the angle

+ Hyp. CED, add to each the angle ABC; therefore

F the two angles ABC, BCA are equal t to the

+ 2 As. two angles ABC, CED: but the angles ABC,

D BCA are together less * than two right an

• 17. 1. gles; therefore the angles ABC, CED are also B less than two right angles : wherefore BA, ED if produced will meet * : let them be produced and meet

12 Ax. 1. in the point F: then because the angle ABC is equal to the angle DCE, BF is parallel * to CD; and because the angle ACB • 28. 1. is equal to the angle DEC, AC is parallel to FE*: therefore FACD is a parallelogram; and consequently, AF is equal to 34, 1.

A

28. 1.

* 2. 6. * 7.5.

*

+ 16. 5.

• 2. 6.

CD, and AC to FD: and because AC is parallel to FE, one of the sides of the triangle FBE, BA is to AF*, as BC to CE: but AF is equal to CD; therefore as BA to CD, so is BC to CE; and alternately +, as AB to BC, so is DC to CE: again, because CD is parallel to BF, as BC to CE, so is FD to DE: but FD is equal to AC; therefore t, as BC to CE, so is AC to DE ; and alternately +, as BC to CA, so CE to ED: 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 æquali *, BA is to AC, as CD to DE. Therefore, the sides, &c. Q. E. D.

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PROPOSITION V.

Theor.-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.

3 Ax.

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 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. • 23. 1.

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 • 32. 1. and BAC is equal to the remaining angle

D EGF, and the triangle ABC is therefore

EA equiangular to the triangle GEF: conse- B 4.6.

quently * they have their sides opposite to

the equal angles proportionals: wherefore, + Hyp: as AB to BC, so is GE to EF: but as AB to BC +, so is DE to * 11.5.

EF; therefore, as DE to EF, so GE to EF; that is, DE and GE have the same ratio to EF, and consequently * 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, each to each ;

and the base DF is equal to the base GF ; therefore the angle • 6, 1.

DEF is equal to the angle GEF, and the other angles to the • 4. 1. other angles which are subtended by the equal sides * : there

* 9.5.

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