BOOK VI. DEFINITIONS. 1. SIMILAR rectilineal figures are those which have their several angles equal, each to each, and the sides about the equal angles proportionals. 2. Reciprocal figures, namely, triangles and parallelograms, are such as have their sides about two of their angles proportionals in such a manner, that a side of the first figure is to a side of the other, as the remaining side of this other is to the remaining side of the first. 3. A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less. 4. The altitude of any figure is the straight line drawn from its vertex perpendicular to the basc. PROPOSITION 1. THEOREM. Triangles and parallelograms of the same altitude are to one another as their bases. Let the triangles ABC, ACD, and the parallelograms EC, CF have the same altitude, namely, the perpendicular drawn from the point A to BD: as the base BC is to the base CD, so shall the triangle ABC be to the triangle ACD, and the parallelogram EC to the parallelogram CF. Produce BD both ways; take any number of straight lines BG, GH, each equal to BC, and any number of straight lines DK, KL, each equal to CD; [I. 3. and join AG, AH, AK, AL. HG B E F Then, because CB, BG, GH are all equal, [Construction. the triangles ABC, AGB, AHG are all equal. [I. 38. 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 CL is of the base CD, the same multiple is the triangle ACL of the triangle ACD. And if the base HC be equal to the base CL, the triangle AHC is equal to the triangle ACL; and if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ACL; and if less, less. [I. 38. Therefore, since there are four magnitudes, namely, the two bases BC, CD, and the two triangles ABC, ACD; and of the base BC, and the triangle ABC, the first and the third, any equimultiples whatever have been taken, namely, the base HC and the triangle AHC; and of the base CD and the triangle ACD, the second and the fourth, any equimultiples whatever have been taken, namely, the base CL and the triangle ACL; and since it has been shewn that if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ACL; and if equal, equal; and if less, less; therefore as the base BC is to the base triangle ABC to the triangle ACD. CD, so is the [V. Definition 5. And, because the parallelogram CE is triangle ABC, and the parallelogram CF is triangle ACD; double of the double of the [I. 41. and that magnitudes have the same ratio which their equimultiples have; [V. 15. therefore the parallelogram EC is to the parallelogram CF as the triangle ABC is to the triangle ACD. But it has been shewn that the triangle ABC is to the triangle ACD as the base BC is to the base CD; therefore the parallelogram EC is to the parallelogram CF as the base BC is to the base CD. [V. 11. Wherefore, triangles &c. Q.E.D. COROLLARY. From this it is plain that triangles and parallelograms which have equal altitudes, are to one another as their bases. For, let the figures be placed so as to have their bases in the same straight line, and to be on the same side of it; 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; [I. 33. because the perpendiculars are both equal and parallel to one another. [I. 28. Then, if the same construction be made as in the proposition, the demonstration will be the same. PROPOSITION 2. THEOREM. If a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those sides 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 is to EA. A A A 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. [I. 37. And ADE is another triangle; and equal magnitudes have the same ratio to the same magnitude; [V. 7. therefore the triangle BDE is to the triangle ADE as the triangle CDE is to the triangle ADE. But the triangle BDE is to the triangle ADE as BD is to DA; because the triangles have the same altitude, namely, the perpendicular drawn from E to AB, and therefore they are to one another as their bases. [VI. 1. For the same reason the triangle CDE is to the triangle ADE as CE is to EA. Therefore BD is to DA as CE is to EA. [V. 11. Next, let BD be to DA as CE is to EA, and join DE: DE shall be parallel to BC. For, the same construction being made, because BD is to DA as CE is to EA, [Hypothesis. and as BD is to DA, so is the triangle BDE to the triangle ADE, [VI. 1. and as CE is to EA so is the triangle CDE to the triangle ADE; [VI. 1. therefore the triangle BDE is to the triangle ADE as the triangle CDE is to the triangle ADE; [V. 11. that is, the triangles BDE and CDE have the same ratio to the triangle ADE. Therefore the triangle BDE is equal to the triangle CDE. [V. 9. And these triangles are on the same base DE and on the same side of it; but equal triangles on the same base, and on the same side of it, are between the same parallels; [I. 39. therefore DE is parallel to BC. Wherefore, if a straight line &c. Q.E.D. PROPOSITION 3. THEOREM. If the vertical angle of a triangle be bisected 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 shall bisect the vertical angle. Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD, which meets the base at D: BD shall be to DC as BA is to AC. Through C draw CE parallel to DA, [I. 31. and let BA produced meet CE at E. Then, because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle CAD; [I. 29. D C but the angle CAD is, by hypothesis, equal to the angle BAD; therefore the angle BAD is equal to the angle ACE. [Ax. 1. |