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Book V. c, and c to d, which are the same, each to each, with the ratios of
G to H, K to L, M to N, O to P, and 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 l, 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 fame, each to each, with the remaining other ratios, viz. of M to N, O to P, and Q to R. then the ratio of h tol is the same with the ratio of m to p, or h is to l, as m to p.
h, k, 1.
e, f, g. m, n, o, p.
S, T, V, X.
Because e is to f, as (G to H, that is as) Y to Z; and f is 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 Hypothesis, 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 d; therefore, ex aequali, T is to X, as a to d.
also because h is to k, as (C to D, that is as) T to V; and k is to 1, as (E to F, that is as) V to X; therefore, ex aequali, h is to
1, as T to X. in like manner it may be demonstrated that m is to P, di 11. si as a to d. and it was shewn that T is to X, as a to d. therefore a h
is to 1, as m to p. 0. E. D.
The Propositions G and K are usually, for the sake of brevity, expressed in the fame terms with Propositions F and H. and therefore it was proper to shew the true meaning of them when they are so expressed; especially since they are very frequently made use of by Geometers.
SIMILAR rectilineal figures
are those which have their several angles equal, each to each, and the sides about the equal angles proportionals.
II. Reciprocal figures, viz. triangles and parallelograms, are such as See N. “ have their fides about two of their angles proportionals in such
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.”
the whole is to the greater segment, as the greater segment is to
line drawn from its vertex perpendicular
PROP. I. THEOR.
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 K, 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 tri2. 38. 1. angles AHG, AGB, ABC are all equal'. therefore whatever mul
tiple the base HC is of the base BC, the same multiple is the tri-
D K L 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 cquimultiples 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 equiinultiples whatever, viz. the base CL and triangle ALC; and that it has been shewn that if the base
HC be greater than the base CL, the triangle AHC is greater than t. 5. Def.5. the triangle ALC; and if equal, equal; and if less, less. Therefore b
as the base BC is to the base CD, fo is the triangle ABC to the tri-
6. 41. 1.
and the parallelogram CF double of the triangle ACD, and that Book VI. magnitudes have the same ratio which their equimultiples have d; as the triangle ABC is to the triangle ACD, so is the parallelogram d. 35. ó. EC to the parallelogram CF. and because it has been shewn 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, fo 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 pa- e. 11. 5. rallelogram 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 perpendi- f. 33. 1. culars are both equal and parallel to one another. then, if the same construction be made as in the Proposition, the Demonstration will be the fame.
PROP. II. THEOR.
IF a straight line be drawn parallel to one of the sides of See N.
a triangle, it shall cut the other fides, or these prodaced, proportionally, and if the sides, or the fides produtced 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 tiimgie CDE", because they are on the same base DE, and between the a. 37.1. fame parallels DE, BC. ADE is another triangle, and equal magnitudes have to the same, the same ratio b; therefore as the triangle b. 7. 5. BDE to the triangle ADE, so is the triangle CDE to the triangle ADE; but as the triangle BDE to the triangle ADE, fo is BD 19 C. 1.6. 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 K 3
Book VI. for the same reason, as the triangle CDE to the triangle ADE, fo
įs CE to EA. Therefore as BD to DA; so is CE to EA d. d. ii. Next, Let the sides AB, AC of the triangle ABC, or these pro
duced, 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, fo is
CE to EA; and as BD to DA, so is the triangle BDE to the trit. 1.6. angle ADE"; and as CE to EA, fo is the triangle CDE to the tri
angle ADE; therefore the triangle BDE is to the triangle ADE, as
the triangle CDE to the triangle ADE, that is, the triangles BDE, c. 9. §. 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 equal triangles on the same base are between F. 39. 1. the same parallels f; therefore DE is parallel to BC. Wherefore if
a straight line, &c. Q. E. D.
PROP. III. THEOR.
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 fame 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 aogles by the straight line AD. BD is to DC, as BA to AC.