PROP. I. THEOR. TRIANGLES and parallelograms of the same altitude are one to another as their bases.* Let the triangles ABC, ACD, and the parallelograms EC, CF, have the same altitude, viz. the perpendicular drawn from the point A to BD: then, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF. Produce BD both ways to the points H, L, and take any number of straight lines BG, GH, each equal to the base BC; and DK, KL, any number of them, each equal to the base CD, and join AG, AH, AK, AL: then, because CB, BG, GH are all equal, the triangles AHG, AGB, ABC are all equal (38. 1.): therefore, whatever multiple the base HC is of the base BC, the same multiple is the triangle AHC of the triangle ABC; for the same reason, whatever multiple the base LC is of the base CD, the same multiple is the triangle ALC of the triangle ADC: and if the base HC be equal to the base CL, the triangle AHC is E A also equal to the triangle H G B C D ALC (38. 1.); and if the F K base HC be greater than the base CL, likewise the triangle AHC is greater than the triangle ALC; and if less, less: therefore, since there are four magnitudes, viz. the two bases BC, CD, and the two triangles ABC, ACD, and of the base BC and the triangle ABC the first and third, any equimultiples, whatever have been taken, viz. the base HC and triangle AHC; and of the base CD and triangle ACD, the second and fourth, have been taken any equimultiples whatever, viz. the base CL, and triangle ALC; and that it has been shown, that, if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ALC; and if equal, equal; and if less, less; therefore (5. def. 5.) as the base BC is to the base CD, so is the triangle ABC to the triangle ACD. And because the parallelogram CE is double of the triangle ABC (41. 1.) and the parallelogram CF double of the triangle * See Note. ACD, and that magnitudes have the same ratio which their equimultiples have (15. 5.); as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF; and because it has been shown, that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD; and as the triangle ABC 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 (11. 5.) 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 (33. 1), 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 those 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 is to DA, as CE to EA. Join BE, CD; then the triangle BDE is equal to the triangle CDE (37. 1.), 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 (7. 5.); therefore, as the triangle BDE to the triangle ADE, so is the triangle CDE to the triangle ADE; but as the triangle BDE to the triangle ADE, so is (1. 6.) 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 (11. 5.). * See Note. Next, let the sides AB, AC of the triangle ABC, or these produced, be cut proportionally in the points D, E, that is, so that BD be to DA, as CE to EA, and join DE; DE is 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 (1. 6.); and as-CE to EA, so is the triangle CDE to the triangle ADE; therefore the triangle BDE is to the triangle ADE, as the triangle CDE to the triangle ADE; that is, the triangles BDE, CDE have the same ratio to the triangle 'ADE; and therefore (9. 5.) the triangle BDE is equal to the triangle CDE; and they are on the same base DE; but equal triangles on the same base are between the same parallels (39. 1.) therefore DE is parallel to BC. Wherefore, if a straight line, &c. Q. E. D. PROP. III. THEOR. 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 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 the angle BAC of any triangle ABC be divided into two equal angles by the straight line AD: BD is to DC, as BA to AC. * See Note. E Through the point C draw CE parallel (31. 1.) to DA, 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 (29. 1.); but CAD, by the hypothesis is equal to the angle BAD; wherefore BAD is equal to the angle ACE. Again, because the straight line BAE meets the parallels AD, EC, the outward angle BAD is equal to the inward and opposite angle AEC but the angle ACE has been proved equal to the angle BAD; therefore also ACE is equal to the angle AEC, and consequently the side AE is equal B D to the side (6. 1.) AC; and because AD is drawn parallel to one of the sides of the triangle BCE, viz. to EC, BD is to DC, as BA to AE (2. 6.): but AE is equal to AC; therefore, as BD to DC, so is BA to AC (7. 5.). Let now BD be to DC, as BA to AC, and join AD; the angle BAC is divided into two equal angles by the straight line ÅD. 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 (2. 6.), because AD is parallel to EC; therefore BA is to AC, as BA to AE (11. 5.) consequently AC is equal to AE (9. 5.), and the angle AEC is therefore equal to the angle ACE (5. 1.): but the angle AEC is equal to the outward and opposite angle BAD: and the angle ACE is equal to the alternate angle CAD (29. 1.): wherefore also the angle BAD is equal to the angle CAD: therefore the angle BAC is cut into two 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 seg ments 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 the outward angle CAE of any triangle ABC be divided into two equal angles by the straight line AD which meets the base produced in D: BD is to DC, as BA to AC. E A Through C draw CF parallel to AD (31. 1.): and because the straight line AC meets the parallels AD, FC, the angle ACF is equal to the alternate angle CAD (29. 1.): but CAD is equal to the angle DAE (Hyp.); 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 angle ACF is equal to the angle CFA, and consequently the side AF is B C D equal to the side AC (6. 1.): and because AD is parallel to FC, a side of the triangle BCF, BD is to DC, as BA to AF (2. 6.): but AF is equal to AC; as therefore BD is to DC, so 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 same construction being made, because BD is to DC, as BA to AC and that BD is also to DC, as BA to AF (11. 5.): therefore BA is to AC, as BA to AF (9. 5.); wherefore AC is equal to AF (5. 1.), and the angle AFC equal (5. 1.) to the |