*Book V.

PROP. E. THEOR.

If four magnitudes be proportionals, the sum of the first two is to their difference as the sum of the other two to their difference.

Let A: B:: C: D; then if AB,

A+B: A-B:: C+D: C-D; or if AB,

A+B: B-A:: C+D: D-C.

For, if AB, then because A: B:: C: D, by division, a 17. 5. A-B: B:: C-D: D, and by inversion,

B: A-B :: D: C-D. But, by composition,
-A+B: B:: C+D: D, therefore, ex æquali ",
A+B: A-B:: C+D: C-D.

In the same manner, if BA, it is proved, that
A+B: B—A::C+D: D-C. Therefore, &c.
Q. E. D.

b A. 5.

c 18. 5.

d 22..5.

PROP. F. THEOR.

Ratios which are compounded of equal ratios, are equal to one another.

Let the ratios of A to B, and of B to C, which compound the ratio of A to C, be equal, each to each, to the ratios of D to E, and E to F, which compound the ratio of D to F; A:C::D:F.

For, first, if the ratio of A to B be equal to that of D to E, and the ratio of B to C equal to that of E to F, ex æquali a, Ã : C: : D: F.

b

A, B, C,

D, E, F.

a 22. 5.

And next, if the ratio of A to B be equal to that of È to F, and the ratio of B to C equal to that of D to E, ex æquali inversely ", A: C:: D:F. In the same manner b 23. 5. may the proposition be demonstrated, whatever be the number of ratios. Therefore, &c. Q. E. D.

1

ELEMENTS

OF

GEOMETRY.

BOOK VI.

DEFINITIONS.

I.

SIMILAR rectilineal figures are those which have Book VI. their several angles

equal, each to each, and the sides about the equal angles proportionals.

II.

Two sides of one figure are said to be reciprocally proportional to two sides of another, when one of the sides of the first is to one of the sides of the second, as the remaining side of the second 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.

Book VI.

PROP. I. THEOR.

Triangles and parallelograms, of the same 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, 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, a 38. 1. AGB, ABC are all equal: 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 multiple is the triangle ALC of the triangle ADC. But if the base HC be equal to the

base CL, the tri-
angle AHC is al-

FA

F

so equal to the HGB

triangle ALCa,

and if the base HC be greater 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 equimultiples whatever have been taken, viz. the base HC, and the 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 since 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; Book VI. and if less, less: Therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD.

b def. 5. 5.

And because the parallelogram CE is double of the triangle ABC, and the parallelogram CF double of the c 41. 1triangle ACD, and because magnitudes have the same ratio which their equimultiples have d; as the triangle d 15.. ABC is 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 to the triangle ACD, so 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 paral- e 11. 5. lelogram CF. Wherefore triangles, &c. Q. E. D.

e

COR. From this it is plain, that triangles and 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 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 perpendiculars are both 133. 1. equal and 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 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.

Join BE, CD; then the triangle BDE is equal to the triangle CDE, because they are on the same base DE, a 37. 1.