* 11. 5. *33.1. † 28. 1. See N. 37.1. *7.5. * 1.6. * 11.5. 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. Q. E. D. * COR. From this it is plain, that triangles and parallelograms that 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 parallel † to one another. Then, if the same construction be made as in the proposition, the demonstration will be the same. PROP. II. 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. 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 triangle; and equal magnitudes have to the same* the same ratio; therefore, as the triangle BDE. is to the 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 triangle CDE to the triangle ADE, so is CE to EA: therefore, as BD to DA, so is CE to EA*. * 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 B E B E 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 *1. 6. EA, so is the triangle CDE to the triangle ADE; therefore the triangle BDE is to the triangle ADE, † 11.5. t as the triangle CDE to the triangle ADE; that is, the triangles BDE, CDE have the same ratio to the tri * angle ADE: therefore the triangle BDE is equal to *9. 5. same parallels therefore DE is parallel to BC. Where- *39. 1. Q. E. D. PROP. III. THEOR. If the angle of a triangle be divided into two equal angles, See N. Let ABC be a triangle, and let the angle BAC bedivided 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, * 31. 1. and let BA produced meet CE in E. Because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle CAD: but CAD, *29. 1. by the hypothesis, is equal to the angle BAD; where * E † 29. 1. fore BAD is equal + to the angle ACE. Again, be- +1 Ax. * B D +1 Ax. * 6.1. and consequently the side AE is equal to the side AC: Produce Bet to make A 2 = Ae then LA?C=LACE & because "Bete = 2 = LACEL 11 (29.1 2.6. 11.5. 9.5. 5.1. * 29.1. † 1 Ax. * 31. 1. 29.1. Hyp. † 1 Ax. † 29. 1. † 1 Ax. 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 to the angle CAD; that is, the angle BAC is cut intotwo equal angles by the straight line AD. Therefore, if the angle, &c. Q. E. D. 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 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 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 the angle ACF is equal to the angle CFA; and consequently E the side AF is equal to the side AC: and because * 6.1. AD is parallel to FC, a side of the triangle BCF, there 珧 fore BD is to DC, as BA to AF: but AF is equal 2. 6. 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. * * 11. 5. The same construction being made, because BD is to DC, as BA to AC; and that BD is also to DC*, as * 2.6. BA to AF; therefore* BA is to AC, as BA to AF; wherefore AC is equal to AF, and the angle AFC 9.5. equal to the angle ACF: but the angle AFC is equal ⚫ 5. 1. to the outward angle EAD †, and the angle ACF to the +29.1. alternate angle CAD; therefore also EAD is equal† † 1 Ax. to the angle CAD. Wherefore, if the outward, &c. * Q. E. D. 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* the angle BAC equal to the angle 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. * 32. 1. & 3 Ax. 17. 1. Let the triangle DCE be placed, so that its side 22.1. CE may be contiguous to BC, and in the same straight line with it: then, because the angle BCA is equal+ + Hyp. to the angle CED: add to each the angle ABC; there fore the two angles ABC, BCA are equal to the † 2 Ax. two angles ABC, CED: but the angles ABC, BCA are together less than two right angles; therefore the angles ABC, CED are also less than two right angles: wherefore BA, ED if produced will meet*: let * them be produced and meet 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 is equal to the angle DEC, AC is parallel to FE*: therefore FACD is a parallelogram; and consequently* * F B AF is equal to CD, and AC to FD: and because AC 12 Ax. 1. * 28. 1. * 28.1. *34. 1. *2.6. * 2.6. • 22.5. 23. 1. 3 Ax. * 4.6. + Hyp. EUCLID'S ELEMENTS. 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 +, 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: PROP. V. Q. E. D. 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. 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. * * 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 * 32. 1. & equal to the remaining angle EGF and the triangle ABC is therefore equiangular to the triangle GEF: consequently* they have their sides opposite to the equal angles proportionals: wherefore, as AB to BC, so is GE to EF: but as AB to BC+, so is DE to 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 DEF is equal* to the angle GEF, and the other angles to the other angles which are subtended by the equal sides*: therefore *9.5. 8.1. *4.1. 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