because, as AB is to C, so is DE to F; and as C to BG, so F Booz V. to EH; by equality b, AB is to BG, as DE to EH: and, by compositions, as AG is to GB, fo is DH to HE ; but as GB b 22. 5. is to C, fo is HE to F; therefore, by equality, as AG is to c 18. 5. C, fo is DH to F. Wherefore, &, Q. E. D.

Cor. 1. If the same hypothesis be made as in the proposition, the excess of the first and fifth shall be to the second, as the excess of the third and fixth to the fourth. The demonstration of this is the faine with that of the proposition, if division be used instead of composition.

Cor. 2. The proposition holds trus, of two ranks of magnitudes, whatever be their number, of which each of the first rank has to the second magnitude tlie same ratio that the corresponding one of the second rank has to a fourth magnitude ; as is manifeft.

PROP. XXV. THEOR.

F four magnitudes of the same kind be propor

I ,

are greater than the other two together.

Let AB be to CD, as E is to F, and let AB be the greatest of them, and consequently F the leaft 2: AB, together with F, is a A. & 14, greater than CD, together with E.

5. Take AG equal to E, and CH equal to F: Then, because as AB is to CD, fo is E to F, and that AG is equal to E, and CH equal to F; AB is to CD, as AG

B to CH: wherefore the remainder GB is to

G the remainder DH, as the whole AB to

D the whole CD b: But AB is greater than E.

b 19: 5. CD; therefore GB is greater than HD , And because AG is equal to E, and CH to F; AG and F together are equal to CH

C E F and E together d. If therefore AG and F

à 2. Ax. 1. be added to the greater magnitude BG, and CHI and E to the lefs DH; the whole AB and F together are greater than the whole CD and Ee together. Wherefore, &c. e 4. Ax. 1. Q. E. D.

CA. 5.

PROP

BOOK V.

R

PROP. F. THEOR.
Atios which are compounded of the same ratios,

are the same with one another.

F.

Let the ratios of A to B, and B to C, be the same with the ratios of D to E and E to F; the ratio of A to A, B, C, C, which is compounded of the first ratios, is the same with the ratio of D to F, which is compounds D, E, ed of the other ratios.

First, Let A be to B, as D to E; and B to C, as E to F; a 22. 5. then, by equality ?, as A is to C, so is D to F.

Next, Let A be to B, as E to F; and B to C, as D to E; b 23. 5. therefore, by perturbate equality, A is to C, as D to F; that

is, the ratio compounded of the ratios of A to B, and B to C, is the same with the ratio compounded of the ratios of D to E, and E to F: and in like manner, the proposition may be demonstrated, whatever be the number of ratios in either cafe.

THE

II.

Two magnitudes are said to be reciprocally proportional to two See N.

others, when one of the first is to one of the other magni-
tudes, as the remaining one of the last two is to the remaining
one 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.

IV.
The altitude of any figure is the straight

line drawn from its vertex perpendicular
to the base.

PROP.

Book VI.

PROP. I. THEOR.

See N.

T

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, K, and take any number of ftraight lines BG, GH, each equal to the base BC, so that CH be greater than CD; and join AG, AH: and be

cause the bases CB, BG, GH are all equal, the triangles ABC, a 38. 1. AGB, AHG are all equal a; therefore the triangle AHC is the

same muliple of ABC, that the base HC is of BC. In like
manner, if CK be taken the least multiple of CD that is
greater than CH, and AK be joined, it may be proved, that the
triangle ACK is the same multiple

EA
of ACD: Wherefore the base DK
and the triangle ADK contain CD
and ACD equally. And if the bafe
HC be equal to DK, the triangle
AHC is equal a to the triangle ADK:
But, if not, the base HC is

greater
than DK, but less than CK, and the H GBC DK
triangle AHC is greater than ADK,

but less than ACK; therefore the base CH contains CD the fame b Def.A.5. number of times that DK contains it b; and the triangle AHC

contains ACD the same number of times that ADK does: But the base DK and the triangle ADK contain DC and ACD equally; therefore the base HC and triangle AHC contain them equally: and the base HC and tirangle AHC are any equimultiples of BC and ABO: as many times, therefore, as any multiple of the base BC contains the base CD, so many times does

the same multiple of the triangle ABC contain the triangle cs. Def.5. ACD; therefore C as the base BC is to the base CD, fo is the

triangle ABC to the triangle ACD.

And because the parallelogram CE is double of the triangle C41. 1. ABC , and the parallelogram CF double of the triangle ACD,

and that magnicudes have the same ratio which their equimuld 15. 5. tiples have d; as the triangle ABC is to ACD, so is the paralle

logram EC to CF: And as the base BC is to CD, so is the triangle ABC to ACD; therefore, as the base BC is to CD, fo is

the

e the parallelogram EC to the parallelogram CF. Wherefore Book VI. triangles, &c. Q. E. D.

Cor. From this it is plain, that triangles and parallelograms e 11. So 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 to the bases, the straight line which joins the vertices, is parallel to their bases f, because the perpendiculars are both equal f 33. 1, 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.

F a straight line be drawn parallel to one of the

sides of a triangle, it shall cut the other sides, or those produced, proportionally: And if the sides, or the fides 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 tri. angle ABC: BD is to DA, as CE to EA.

Join BE, CD; then the triangle BDE is equal to the tri.
angle CDE “, because they are on the same base DE, and be a 37. 1.
tween the same parallels DE, BC: ADE is another triangle,
and equal magnitudes have to the same, the same ratio b; there- 67.5.
fore, as the triangle BDE to the triangle ADE, so is the tri-
angle CDE to the triangle ADE; but as the triangle BDE to
the triangle ADE, fo is BD to DA, because having the same c 1. 6.
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 to the triangle ADE, fo is CE to
EA. Therefore, as BD to DA, fo is CE to EA d.

Next, Let the sides AB, AC of the tri 4
angle ABC, or these produced, be cut pro-
portionally 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 fame construction being made, be-
cause, as BD to DA, fo is CE to EA; and
as BD to DA, fo is the triangle BDE to
the triangle ADE ®; and as CE to EA, so B

C e 1. 6.
is the triangle CDE to the triangle ADE;
therefore the triangle BDE is to the triangle ADE, as the tri-
angle CDE to the triangle ADE; that is, the triangles BDE,

S

CDE

D