Sidebilder
PDF
ePub
[blocks in formation]

II.
Two sides of one figure are said to be reciprocally propor-

tional to two sides of another, when one of the sides of the
first is to one of the sides of the other, as the remaining
side of the other is to the remaining side of the first.

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 perpendicu-
lar to the base.

Book VI.

PROP. I. THEOR.

TRIANGLES of the same altitude are to one an. other as their bases; and parallelograms of the same altitude are to one another as their bases.

a 38. 1.

Let the triangles ABC, ACD and the parallelograms EC, CF, have the same altitude; as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and so is the parallelogram EC to the parallelogram CF.

Produce BD both ways to the points H, L; take any num
ber of straight lines BG, GH, each equal to the base BC, and
any number DK, KL, each equal to the base CD; and join
AG, AH, AK, AL. Because CB, BG, GH are all equal,
the triangles AHG, AGB, ABC are all equala; 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 multi-
ple is the triangle

E A
ALC of the triangle
ADC. But if the base
HC be equal to the
base CL, the triangle
AHC is equal to the
triangle ALCA; and if
the base HC be greater AGB C

D K

L than the base CL, the triangle AHC is greater than the triangle ALC; and if less, less. Now there are four magnitudes, 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 the triangle ACD, the second and fourth, any equimultiples whatever have been taken, viz. the base CL and the triangle ALC; and 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 as the base BC is to the base Book VI. CD, so is the triangle ABC to the triangle ACDb.

Because the parallelogram CE is double of the triangle b 5. Def. 5. ABC“, and the parallelogram. CF double of the triangle c 41. 1. ACD, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CFd. But it has been d 15. 5. shown that as the base BC is to the base CD, so is the triangle ABC to the triangle ACD; therefore as the base BC is to the base CD, so is the parallelogram EC to the parallelogram CF. Wherefore, triangles, &c. Q. E. D.

e 11. 5.

Cor. Triangles, and also 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 let perpendiculars be drawn 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, f 33. 1. because the perpendiculars are both equal and parallel to one 8 28.1., 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 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.

Book VI.

a 37.1. b 7. 5. c 1. 6.

Join BE, CD; then the triangle BDE is equal to the triangle CDEa. But ADE is another triangle, therefore as the triangle BDE to the triangle ADE, so is the triangle CDE to the triangle ADEb.. But, as the triangle BDE to the triangle ADE, so isc BD to DA; and as the triangle CDE to the triangle ADE, so is CE to EA. Therefore as BD to DA, so is CE to EA.

d 11. 5.

[blocks in formation]

Next, let the 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, as BD to DA, so is CE to EA, and as BD to DA, so is the triangle BDE to the triangle ADE; and as CE to EA, so is the triangle CDE to the triangle ADE; therefore the triangle BDE is to the triangle AĎE, as the triangle CDE to the triangle ADE*; therefore the triangle BDE is equal to the triangle CDEC. But they are on the same base DE, therefore DE is parallelf to BC. Wherefore, if a straight line, &c. Q. E. D.

e 9.5. f 39. 1.

* This proportion depends upon the principle, that ratios which are equal to equal ratios are equal to one another, which may be inferred from 11. 5.

[ocr errors]
« ForrigeFortsett »