« ForrigeFortsett »
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 Q 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 lame with the ratios of G to HI, 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
, and k to 1, which are the fame 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 saine, 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 1 is the same with the ratio of m to p, or h is to l, as m to p.
h, k, 1.
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 Ed 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 kis tol, 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, as a to d. and it was shewn thar T is to X, as a to do therefore a h is to 1, as m to p.
m to p. 0. E. D. The Propositions G and K are usually, for the sake of brevity, expressed in the fame terms with Propofitions 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 uss of by Geometers.
are those which have their
II. Reciprocal figures, viz. triangles and parallelograms, are such as Sec 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
the whole is to the greater segment, as the greater segment is
drawn from its vertex perpendicular to the
PRO P. I. THEOR.
TRIANGLES and parallelograms of the same
altitude are one to another as their bates.
Let the triangles ABC, ACD, and the parallelograms EC, CF have the fame altitude, viz. the perpendicular drawn from the point A to BD. then as the base BC is to the base CD, fo 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, cach equal to the base CD; and join
AG, AH, AK, AL. then because CB, BG, GH are all equal, the a. 38. s. triangles AHG, AGB, ABC are all equal therefore whatever
multiple the bafe HIC is of the base DC, the fame multiple is the
E A F
greater than the base CL, the triangle AHC is greater than the b.s. Def.s. triangle ALC; and if equal, equal; and if lefs, lefs. Therefore b
as the bale 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 Book Vi. ABC , and the parallelogram CF double of the triangle ACD,
and that magnitudes have the same ratio which their equimultiples Ć have d; as the triangle ABC is to the triangle ACD, fo is the pa- d. 15. 5.
rallelogram EC to the parallelogran CF. and because it has been
shewn that as the bale BC is to the base CD, fo is the triangle ABC s
to the triangle ACD; and as the triangle ABC to the triangle ACD, so is the parallelogram FC to the parallelogram CF; therefore as the base BC is to the base CD, so is the parallelogram EC e. 11. 5. 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 f, because the perpen- f. 33. 5. diculars are both equal and parallel to one another. then, if the fame construction be made as in the Propofition, the Demonstration will be the fame.
I F a straight line be drawn parallel to one of the fidcs See N,
of a triangle, it shall cut the other sides, or these 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 “, because they are on the same base DE, and between the a. 37. 1, fame parallels DE, BC. ADE is another triangle, and cqual magnitudes have to the same, the same ratio "; 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 to 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
Bonk VI. for the same reason, as the triangle CDE to the triangle ADE, so
is CE to EA. Therefore as BD to DA; fo is CE to EA d. d. 11 Next, Let the sides AB, AC of the triangle ABC, or these proA А
C 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, so is CE to EA; and as BD to DA, fo is the triangle 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 e the triangle BDE is equal to the triangle CDE. and they are on the same bale DE ; but equal triangles on the same base are between the fame parallels f; therefore DE is parallel to BC. Wherefore if a straight line, &c. Qe 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 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 in two equal angles by the straight line AD, BD is to DC, as BA'to AC.