PROP. XXIII. THEOR. If there be any number of magnitudes, and as many others, which, taken two and two, in a cross order, have the same ratio; the first will have to the last of the first magnitudes the same ratio which the first of the others has to the last.* First, let there be three magnitudes, A. B, C, and other three, D, E, and F, which, taken two and two in a cross order, have the same ratio, viz. A: B::E: F, and B:C::D: E, then A:C::D:F. Take of A, B. and D, any equimultiples mA, mB, mD; and of C, E, F any equimultiples nC, nE, nF. A, B, C, Because A: B:: E: F, and because also A : B::mA:mB (15.5.), and E:F::nE: F; therefore, MA: MB::nE: nF (11. 5.). Again, because B: C:: D: E, mB:nC::mD: nE (4.5.); and it has been just shewn that mA : MB::nE:nF; therefore if mA7nC, mD 7nF (21.5.); if mA=nC, mD=nF; and if mAnC, mDnF. Now, mA and mD are any equimulti D, E, F, mA, MB, nC, mD, nE, nF. ples of A and D, and nC, nF any equimultiples of C and F; therefore, A: C:: D: F (def. 5. 5.). * Next, Let there be four magnitudes, A, B, C, and D, and other four, E, F, G, and H, which taken two and two, in a cross order, have the same ratio, viz. A: B::G:H;B:C:: F: G, and C:D:: E: F, then A: D::E: H. For, since A, B, C, are three magnitudes, and F, G, H other three, which taken two and A, BC, D, E, F, G, H. two, in a cross order, have the same ratio, by the first case, A: C:: F: H. But C:D::E:F, therefore, again, by the first case, A: D::E:H. In the same manner, may the demonstration be extend. ed to any number of magnitudes. Therefore, &c. Q. E. D. PROP. XXIV. THEOR. If the first has to the second the same ratio which the third has to the fourth; and the fifth to the second, the same ratio which the sixth has to the fourth; the first and fifth, together, shall have to the second, the same ratio which the third and sixth together, have to the fourth. Let A: B:: C: D, and also E: B:: F: D, then A+E: B:: C+F: D. Because E: B:: F: D, by inversion, B: E:: D: F. But by hypothesis, A: B:: C: D, therefore, ex æquali (22, 5.), A: E:: C:F: and by composition (18. b.), A+E:E::C+F: F. And again by hypothesis, E: B:: F: D, therefore, ex æquali (22. 5.), A+E: B::C+F:D Therefore &c. Q. E. D. * N. B. This proposition is usually cited by the words "ex æquali in proportione pertur "bata;" or, " ex æquo inversely." PROP. E. THEOR. If four magnitudes be proportionals, the sum of the first two is to their difference as the sum of the other two to their difference. Let A: B::C: D; then if A7B, AB:A-B::C+D:CD; or if A∠B A+B:B-A::C+D:D-C. For, if A7B, then because A: B:: C: D, by division (17.5.), A-B:B::C-D: D, and by inversion (A. 5.), B:A-B::D:C-D. But, by composition (18. 5.), A+B:B::C+D: D, therefore, ex æquali (22. 5.), A+B:A-B::C+D:C-D. In the same manner, if B7A, it is proved, that A+B:B-A::C+D:D-C. Therefore, &c. Q. E. D. PROP. F. THEOR. Ratios which are compounded of equal ratios, are equal to one another. Let the ratios of A to B, and of B to C, which compound the ratio of A to C, be equal, each to each, to the ratios of D to E, and E to F, which compound the ratio of D to F, A: C::D:F. For, first, if the ratio of A to B be equal to that of D to E, and the ratio of B to C equal to that of E to F, ex æquali (22. 5.), A: C:: D: F. A, B, C, And next, if the ratio of A to B be equal to that of E to F, and the ratio of B to C equal to that of D to E, ex æquali inversely (23. 5.), A: C:: D: F. In the same manner may the proposition be demonstrated, whatever be the number of ratios. Therefore, &c. Q. E. D. Two sides of one figure are said to be reciprocally proportional to two sides of another, when one of the sides of the first is to one of the sides of the second, as the remaining side of the second is to the remaining side of the first. III. 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. IV. The altitude of any figure is the straight line drawn from its vertex perpendicular to the base, 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 parallellogram 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, AН, АК, 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. But if the base HC be equal to the base CL, the triangle AHC is also equal to the triangle ALC (38. 1.): and if the base HC be greater than the base CL, likewise the triangle AHC is EAF HGBCDKL 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 the 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 since 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 (def. 5. 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 ACD, and because 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 to one 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 J 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 will cut the other sides, or the other sides produced, proportionally: And if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section will 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: but 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 ΕΑ (11. 5.). N A. he sides AB, AC of the triangle ABC, or these sides 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 |