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Boor Veach, 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
h, k, 1.
S,T, V, X.
Y, Z, a, b, c, d.
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 æquali, 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 tod; therefore, ex æquali, 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 æquali, h is to I, as T to X: In like manner, it may be demonstrated, that m is to
p, as a tod: And it has been shown, that T is to X, as a to * 11.5. d ; therefore a h is to I, as m top. Q.E.D.
The propositions G and K are usually, for the sake of brevity, expressed in the same terms with propositions Fand H: And therefore it was proper to show the true meaning of them when they are so expressed; especially since they are very frequently made use of by geometers,
SIMILAR rectilineal figures
Book VI. 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 See N.
“ such as have their sides about two of their angles pro“portionals 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.”.
. 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.
See N. TRIANGLES and parallelograms of the same alti
tude 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 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, AH, AK, AL: Then, because
CB, BG, GH are all equal, the triangles AHG, AGB, 8. 1. ABC, are all equala: Therefore, whatever multiple the
base HC is of the base BC, the same multiple is the tri-
triangle AHC is greater than the triangle ALC; and if equal, 5 Def. 5. equal; and if less, less: Thereforeb, 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 tri
Anthem because parallele triang ane Doo
angle ABC, and the parallelogram CF double of the tri- Book VI. angle ACD, and that magnitudes have the same ratio
• 41. 1. which their equimultiples haved, as the triangle ABC is à 15: 5: 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 is to the triangle ACD, so is the parallelogram EC to the parallelo- . gram CF; therefore, as the base BC is to the base CD, so ise the parallelogram EC to the parallelogram CF. Where- ' 11. 5. 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 their figures be placed so as to have their bases in the same straight line; and having draw'n 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 aref, because the perpendiculars are both equal and ' 53. 1. parallel to one another. Then, if the 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 See N. sides 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, . 37. 1. and between the same parallels DE, BC: ADE is another triangle, and equal magnitudes have to the same, the same ratiob; therefore, as the triangle BDE to the triangle ADE, 67.5. so is the triangle CDE to the triangle ADE, but as the triangle BDE to the triangle ADE, so is c BD to DA, be- ' l. 6. cause 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
triangle en the same use they are le BDE' is
Book VI. to the triangle ADE, so is CE to EA: Therefore, as BD w to DA, so is CE to EA. 11. 5.
Next, Let the sides AB, AC, of the triangle ABC, or
E B these 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 ? 1. 6. BDE to the triangle ADEe; 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 19. 5. same ratio to the triangle ADE; and thereforef 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 : 39. 1. between the same parallels8; therefore DE is parallel to BC.
Wherefore, if a straight line, &c. Q.E.D.
• PROP. III. THEOR.
See N. If the angle of a triangle be divided into two equal
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 into two equal angles by the straight line AD: BD is to DC, as BA to AC.