Book V. c to d, which are the same, each to each, with the ratios of m G to H, K to L, M to N, O to P, and 0 to R: Therefore, by the hypothesis, S is to X, as Y to d: Also, let the ratio of A to B, that is, the ratio of S to T, which is one of the first ratios, be the same with the ratio of e to g, which is compounded of the ratios of e to f, and f to g, which, by the hypothesis, are the same with the ratios of G to H, and K to L, two of the other ratios ; and let the ratio of h to l be that which is compounded of the ratios of h to k, and k to 1, which are the same with the remaining first ratios, viz. of C to D, and E to F; also, let the ratio of m to p be that which is compounded of the ratios of m to n, n to o, and o to p, which are the same,, cach to each, with the remaining other ratios, viz of M to N, O to P, and Q to R: Then the ratio of h to l is the same with the ratio of m to p, or h is to i, as m to p. h, k, 1. S, T, V, X. e, f, g. m, n, o, p Because e is to f, as (G to H, that is, as) Y to Z; and fis to g, as (K to L, that is, as) Z to a ; therefore, ex aequali, e is to g, as Y to a: And, by the hypothefis, A is to B, that is, S to T, as e to g; wherefore S is to T, as Y to a; and, by inversion, T is to S, as a to Y ; and S is to X, as Y to di therefore, ex aequali, T is to X, as a tod: Also, because his to k, as (C to D, that is, as) T to V; and k is to !, as (E to F, that is, as) V to X; therefore, ex aequali, h is to !, as T 10 X: In like manner, it may be demonstrated, that m is to p, as a to d: And it has been shown, that T is to X, as a to d: Theree fore a h is to 1, as in to p. Q. E. D. The propoGtions G and K are usually, for the sake of brevity, expressed in the same terms with propositions F and H: And therefore it was proper to show the true meaning of them when they are so exprefied; especially since they are very frequently made use of by geometers. 1 Il. 3 THE I. are those which have their several angles equal, each to cach, and the Gdes about the equal angles proportionals. II. " Reciprocal figures, viz. triangles and parallelograms, are Sce N, « such as have their fides about two of their angles propor- III. when the whole is to the greater segment, as the greater seg. IV. drawn from its vertex perpendicular to the PROP. Book VL PRO P. I. THE O R. Sce N. TRIANGLES and parallelograms of the fame 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, lo 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 equ base CD; and join AG, AH, AK, AL: Tben, because CB, BG, GH are all equal, the triangles AHG, AGB, ABC a 38. 1. are all equal“ : Therefore, whatever multiple the base HC E A F D K L 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 bafes 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 bs. def. s. equal, equal ; and if lels, less : Therefore, as the bafe 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, dis. S. ABC“, and the parallelogram CF double of the triangle ACD, Book VI. and that magnitudes have the same ratio which their equimul. W tiples have d; as the triangle ABC is to the triangle ACD, fo 41. I. is the parallelogram EC to the parallelogram CF: And because it has been shown, that, as the base BC is to the base CD, fo is the triangle ABC to the triangle ACD ; and 'as the triangle ABC to the triangle ACD, fo is the parallelogram EC to the parallelogram CF; therefore, as the base BC is to the base CD, fo is the parallelogram EC to the parallelogram CF. Where. e II. S. fore 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 f, because the f 33. 1. perpendiculars are both equal and parallel to one another: Then, if the same construction be made as in the propofition, the demonstration will be the same. F a straight line be drawn parallel to one of the sides of See N. a triangle, it shall cut the other fides, 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 tri. angle ABC: BD is to DA, as CE to EA. Join BE, CD; then the triangle BDE is equal to the triangle CDE', because they are on the fame bale DE, and be * 37. 1. tween the same parallels DE, BC: ADE is another triangle, and equal magnitudes have to the same, the same ratiob; there. 6 7. 5. fore, as the triangle BDE to the triangle ADE, fo is the triangle CDE to the triangle ADE; but as the triangle BDE to the triangle ADE, fo is BD to DA, because having the same c 1. 6. altitude, viz. the perpendicular drawn from the point E to AB, they are to one another as their bases; and for the same reason, d II. 3• Book VI. reason, as the triangle CDE to the triangle ADE, so is CE ra Next, Let the fides AB, AC of the triangle ABC, or these E D А D E B • I. 6. B C 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 fame construction being made, because as BD to DA, so is CE to EA; and as BD to DA, so is the criangle BDE to the triangle ADE °; and as CE to EA, fo 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 the triangle BDE is equal to the triangle CDE : And they are on the fame base DE ; but e. qual triangles on the same base are between the same parallels 8 ; therefore DE is parallel to BC. Wherefore, if a straight line, &c. Q. E. D. f . s. 39. I. F the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; che segments of the base shall have the fame 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 Ades of the triangle have to one another, the straight line drawn from the vertex to the point of section, die vides 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 co AC. Through |